While preparing for Class XII Board Exams, many students often burn the midnight oil by the sidewise preparation of JEE
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English Pages 465 Year 2019
Table of contents :
Cover
Title page
Copyright page
Preface
Contents
Syllabus
Measurement
Kinematics
Laws of Motion
Work, Energy and Power
Rotational Motion
Gravitation
Properties of Solids and Liquids
Thermodynamics
Kinetic Theory of Gases
Oscillations and Waves
Electrostatics
Current Electricity
Magnetic Effect of Current
Electromagnetic Induction and AC
Electromagnetic Waves
Optics
Dual Nature of Radiation
Atoms and Nuclei
Electronic Devices
Communication Systems
Practical Physics
Practice Set 1
Practice Set 2
Practice Set 3
Practice Set 4
Practice Set 5
JEE Main CHAPTERWISE
SOLUTIONS 20192002
Physics All the 16 Question Papers of JEE Main Online 2019 (Jan & Apr Attempt)
ARIHANT PRAKASHAN (Series), MEERUT
Arihant Prakashan (Series), Meerut All Rights Reserved
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PREFACE JEE Main is a gateway examination for candidates expecting to seek admission in Bachelor in Engineering (BE), Bachelor of Technology (B.Tech) and Bachelor of Architecture (B.Arch) at Indian Institutes of Information Technology (IIITs), National Institutes of Technology (NITs), Delhi Technological University and other Centrally Funded Technical Institutes (CFTIs). JEE Main is also an examination which is like screening examination for JEE Advanced (The gateway examination to India's most reputed Technical Institutes, Indian Institutes of Technology— IITs). To make the students wellversed with the pattern as well as the level of the questions asked in the exam, this book contains Chapterwise Solutions of Questions asked in Last 18 Years’ Examinations of JEE Main (formerly known as AIEEE). Solutions to all the questions have been kept very detailed and accurate. Along with the indication of level of the exam, this book will also teach you to solve the questions objectively in the examination. To give the students a complete practice, along with Chapterwise Solutions, this book contains 5 Practice Sets, based exactly on JEE Main Syllabus and Pattern. By practicing these sets, students can attain efficiency in Time Management during the examination. We hope this book would be highly beneficial for the students. We would be grateful if any discrepancy or mistake in the questions or answers is brought to our notice so that these could be rectified in subsequent editions.
Publisher
CONTENTS 1. Measurement 2. Kinematics
18 930
3. Laws of Motion
3146
4. Work, Energy and Power
4761
5. Rotational Motion
6292
6. Gravitation
93105
7. Properties of Solids and Liquids
106130
8. Thermodynamics
131147
9. Kinetic Theory of Gases
148159
10. Oscillations and Waves
160192
11. Electrostatics
193231
12. Current Electricity
232264
13. Magnetic Effect of Current
265298
14. Electromagnetic Induction and AC
299317
15. Electromagnetic Waves
318326
16. Optics
327358
17. Dual Nature of Radiation
359371
18. Atoms and Nuclei
372390
19. Electronic Devices
391406
20. Communication Systems
407413
21. Practical Physics
414415
Practice Sets for JEE Main Practice Set 1 Practice Set 2 Practice Set 3 Practice Set 4 Practice Set 5
416423 424431 432439 440448 449456
SYLLABUS NOTE The syllabus contains two Sections  A & B. Section A pertains to the Theory Part, having 80% weightage, while Section B contains Practical Component (Experimental Skills) having 20% weightage.
SECTION A UNIT 1 Physics and Measurement Physics, technology and society, SI units, Fundamental and derived units. Least count, accuracy and precision of measuring instruments, Errors in measurement, Significant figures. Dimensions of Physical quantities, dimensional analysis and its applications.
UNIT 2 Kinematics Frame of reference. Motion in a straight line: Positiontime graph, speed and velocity. Uniform and nonuniform motion, average speed and instantaneous velocity. Uniformly accelerated motion, velocitytime, position time graphs, relations for uniformly accelerated motion. Scalars and Vectors, Vector addition and Subtraction, Zero Vector, Scalar and Vector products, Unit Vector, Resolution of a Vector. Relative Velocity, Motion in a plane, Projectile Motion, Uniform Circular Motion.
UNIT 3 Laws of Motion Force and Inertia, Newton's First Law of motion; Momentum, Newton's Second Law of motion; Impulse; Newton's Third Law of motion. Law of conservation of linear momentum and its
applications, Equilibrium of concurrent forces. Static and Kinetic friction, laws of friction, rolling friction. Dynamics of uniform circular motion: Centripetal force and its applications.
UNIT 4 Work, Energy and Power Work done by a constant force and a variable force; kinetic and potential energies, workenergy theorem, power. Potential energy of a spring, conservation of mechanical energy, conservative and nonconservative forces; Elastic and inelastic collisions in one and two dimensions.
UNIT 5 Rotational Motion Centre of mass of a twoparticle system, Centre of mass of a rigid body; Basic concepts of rotational motion; moment of a force, torque, angular momentum, conservation of angular momentum and its applications; moment of inertia, radius of gyration. Values of moments of inertia for simple geometrical objects, parallel and perpendicular axes theorems and their applications. Rigid body rotation, equations of rotational motion.
UNIT 6 Gravitation The universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Kepler's laws of planetary motion. Gravitational potential energy; gravitational potential. Escape velocity. Orbital velocity of a satellite. Geostationary satellites.
UNIT 7 Properties of Solids & Liquids Elastic behaviour, Stressstrain relationship, Hooke's. Law, Young's modulus, bulk modulus, modulus of rigidity. Pressure due to a fluid column; Pascal's law and its applications. Viscosity, Stokes' law, terminal velocity, streamline and turbulent flow, Reynolds number. Bernoulli's principle and its applications. Surface energy and surface tension, angle of contact, application of surface tension  drops, bubbles and capillary rise. Heat, temperature, thermal expansion; specific heat capacity, calorimetry; change of state, latent heat. Heat transferconduction, convection and radiation, Newton's law of cooling.
UNIT 8 Thermodynamics Thermal equilibrium, zeroth law of thermodynamics, concept of temperature. Heat, work and internal energy. First law of thermodynamics. Second law of thermodynamics: reversible and irreversible processes. Camot engine and its efficiency.
S.H.M.  kinetic and potential energies; Simple pendulum  derivation of expression for its time period; Free, forced and damped oscillations, resonance. Wave motion Longitudinal and transverse waves, speed of a wave. Displacement relation for a progressive wave. Principle of superposition of waves, reflection of waves, Standing waves in strings and organ pipes, fundamental mode and harmonics, Beats, Doppler effect in sound.
UNIT 11 Electrostatics Electric charges Conservation of charge, Coulomb's lawforces between two point charges, forces between multiple charges; superposition principle and continuous charge distribution. Electric field Electric field due to a point charge, Electric field lines, Electric dipole, Electric field due to a dipole, Torque on a dipole in a uniform electric field. Electric flux, Gauss's law and its applications to find field due to infinitely long, uniformly charged straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Electric potential and its calculation for a point charge, electric dipole and system of charges; Equipotential surfaces, Electrical potential energy of a system of two point charges in an electrostatic field. Conductors and insulators, Dielectrics and electric polarization, capacitor, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, Energy stored in a capacitor.
UNIT 9 Kinetic Theory of Gases
UNIT 12 Current Electricity
Equation of state of a perfect gas, work done on compressing a gas. Kinetic theory of gases  assumptions, concept of pressure. Kinetic energy and temperature: rms speed of gas molecules; Degrees of freedom, Law of equipartition of energy, applications to specific heat capacities of gases; Mean free path, Avogadro's number.
Electric current, Drift velocity, Ohm's law, Electrical resistance, Resistances of different materials, VI characteristics of Ohmic and nonohmic conductors, Electrical energy and power, Electrical resistivity, Colour code for resistors; Series and parallel combinations of resistors; Temperature dependence of resistance. Electric Cell and its Internal resistance, potential difference and emf of a cell, combination of cells in series and in parallel. Kirchhoff's laws and their applications. Wheatstone bridge, Metre bridge.
UNIT 10 Oscillations And Waves Periodic motion  period, frequency, displacement as a function of time. Periodic functions. Simple harmonic motion (S.H.M.) and its equation; phase; oscillations of a spring  restoring force and force constant; energy in
Potentiometer  principle and its applications.
UNIT 13 Magnetic Effects of Current and Magnetism BiotSavart law and its application to current carrying circular loop. Ampere's law and its applications to infinitely long current carrying straight wire and solenoid. Force on a moving charge in uniform magnetic and electric fields Cyclotron. Force on a currentcarrying conductor in a uniform magnetic field. Force between two parallel currentcarrying conductorsdefinition of ampere. Torque experienced by a current loop in uniform magnetic field, Moving coil galvanometer, its current sensitivity and conversion to ammeter and voltmeter. Current loop as a magnetic dipole and its magnetic dipole moment. Bar magnet as an equivalent solenoid, magnetic field lines; Earth's magnetic field and magnetic elements. Para, dia and ferromagnetic substances Magnetic susceptibility and permeability, Hysteresis, Electromagnets and permanent magnets.
UNIT 14 Electromagnetic Induction and Alternating Currents Electromagnetic induction; Faraday's law, induced emf and current; Lenz's Law, Eddy currents. Self and mutual inductance. Alternating currents, peak and rms value of alternating current/ voltage; reactance and impedance; LCR series circuit, resonance; Quality factor, power in AC circuits, wattless current. AC generator and transformer.
UNIT 15 Electromagnetic Waves Electromagnetic waves and their characteristics. Transverse nature of electromagnetic waves. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, Xrays, gamma rays). Applications of e.m. waves.
UNIT 16 Optics Reflection and refraction of light at plane and spherical surfaces, mirror formula, Total internal reflection and its applications, Deviation and Dispersion of light by a prism, Lens Formula, Magnification, Power of a Lens, Combination of thin lenses in contact, Microscope and Astronomical Telescope (reflecting and refracting) and their magnifying powers. Wave optics wave front and Huygens' principle, Laws of reflection and refraction using Huygen's principle.
Interference, Young's double slit experiment and expression for fringe width, coherent sources and sustained interference of light. Diffraction due to a single slit, width of central maximum. Resolving power of microscopes and astronomical telescopes, Polarisation, plane polarized light; Brewster's law, uses of plane polarized light and Polaroids.
UNIT 17 Dual Nature of Matter and Radiation Dual nature of radiation. Photoelectric effect, Hertz and Lenard's observations; Einstein's photoelectric equation; particle nature of light. Matter waveswave nature of particle, de Broglie relation. DavissonGermer experiment.
UNIT 18 Atoms and Nuclei Alphaparticle scattering experiment; Rutherford's model of atom; Bohr model, energy levels, hydrogen spectrum. Composition and size of nucleus, atomic masses, isotopes, isobars; isotones. Radioactivityalpha, beta and gamma particles/rays and their properties; radioactive decay law. Massenergy relation, mass defect; binding energy per nucleon and its variation with mass number, nuclear fission and fusion.
UNIT 19 Electronic Devices Semiconductors; semiconductor diode: IV characteristics in forward and reverse bias; diode as a rectifier; IV characteristics of LED, photodiode, solar cell, and Zener diode; Zener diode as a voltage regulator. Junction transistor, transistor action, characteristics of a transistor transistor as an amplifier (common emitter configuration) and oscillator. Logic gates (OR, AND, NOT, NAND & NOR). Transistor as a switch.
UNIT 20 Communication Systems Propagation of electromagnetic waves in the atmosphere; Sky and space wave propagation, Need for modulation, Amplitude and Frequency Modulation, Bandwidth of signals, Bandwidth of Transmission medium, Basic Elements of a Communication System (Block Diagram only)
SECTION B UNIT 21 Experimental Skills Familiarity with the basic approach and observations of the experiments and activities 1. Vernier callipers  its use to measure internal and external diameter and depth of a vessel. 2. Screw gauge  its use to determine thickness/ diameter of thin sheet/wire. 3. Simple Pendulum  dissipation of energy by plotting a graph between square of amplitude and time. 4. Metre Scale  mass of a given object by principle of moments. 5. Young's modulus of elasticity of the material of a metallic wire. 6. Surface tension of water by capillary rise and effect of detergents. 7. Coefficient of Viscosity of a given viscous liquid by measuring terminal velocity of a given spherical body. 8. Plotting a cooling curve for the relationship between the temperature of a hot body and time. 9. Speed of sound in air at room temperature using a resonance tube. 10. Specific heat capacity of a given (i) solid and (ii) liquid by method of mixtures. 11. Resistivity of the material of a given wire using metre bridge. 12. Resistance of a given wire using Ohm's law. 13. Potentiometer (i) Comparison of emf of two primary cells. (ii) Internal resistance of a cell.
14. Resistance and figure of merit of a galvanometer by half deflection method. 15. Focal length of (i) Convex mirror (ii) Concave mirror (iii) Convex lens 16. Using parallax method. Plot of angle of deviation vs angle of incidence for a triangular prism. 17. Refractive index of a glass slab using a travelling microscope. 18. Characteristic curves of a pn junction diode in forward and reverse bias. 19. Characteristic curves of a Zener diode and finding reverse break down voltage. 20. Characteristic curves of a transistor and finding current gain and voltage gain. 21. Identification of Diode, LED, Transistor, IC, Resistor, Capacitor from mixed collection of such items. 22. Using multimeter to (i) Identify base of a transistor. (ii) Distinguish between npn and pnp type transistor. (iii) See the unidirectional flow of current in case of a diode and an LED. (iv) Check the correctness or otherwise of a given electronic component (diode, transistor or IC).
1
Measurement
1 Measurement 1. In SI units, the dimensions of −1
3
2
(a) A TML (c) AT −3 ML3/ 2
ε0 is µ0
−1 −1
(b) AT M L (d) A2T3 M−1 L−2 [JEE Main 2019, 8 April ShiftI]
Exp. (d) Dimensions of ε0 (permittivity of free space) are [ε0 ] = M−1L−3 T4 A 2 As, c = speed of light. ∴Dimension of [c ] = [LT−1 ] ε0 are µ0 ε20 = [ε0c ] ε0µ 0
2 1 Qc = µ 0 ε0
= [M−1L−3 T4 A 2 ][LT−1 ] = [M−1L−2 T3 A 2 ]
2. In a simple pendulum, experiment for determination of acceleration due to gravity ( g ), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained 55.0 cm. The percentage error in the determination of g is close to [JEE Main 2019, 8 April ShiftII] (a) 0.7%
(b) 6.8%
(c) 3.5%
…(i)
Given, ∆l = 01 . cm, l = 55 cm, ∆T = 1s and T for 20 oscillations = 30 s Substituting above values in Eq. (i), we get . 1 ∆g 01 = +2× 30 g 55 Hence, percentage error in g is 10 20 ∆g = × 100 = + = 6.8% 55 3 g
3. If surface tension (S ), moment of inertia (I )
So, dimensions of ε0 = µ 0
So, fractional error in value of g is ∆g ∆l 2 ∆T = + g l T
(d) 0.2%
Exp. (b) Relation used for finding acceleration due to gravity by using a pendulum is 4π2l g= 2 T
and Planck’s constant (h ), were to be taken as the fundamental units, the dimensional formula for linear momentum would be [JEE Main 2019, 8 April ShiftII] 1/ 2 1/ 2 −1
(a) S I h (c) S 1/ 2 I 1/ 2h0
(b) S 3/ 2 I 1/ 2h0 (d) S 1/ 2 I 3/ 2h−1
Exp. (c) Suppose, linear momentum ( p) depends upon the Planck’s constant (h) raised to the power (a), surface tension(S) raised to the power (b) and moment of inertia (I) raised to the power (c). Then, p ∝ (h)a (S )b(I)c or p = khaS b Ic where, k is a dimensionless proportionality constant. Thus, [p] = [h]a [S]b[I]c
…(i)
Then, the respective dimensions of the given physical quantities, i.e. [p] = [mass × velocity] = [MLT−1 ] [I] = [mass × distance 2 ] = [ ML2 T0 ] [S] = [force × length] = [ML0 T−2 ] [h] = [ML2 T−1 ]
2
JEE Main Chapterwise Physics Then, substituting these dimensions in Eq. (i), we get [MLT−1 ] = [ML2 T−1 ]a [MT−2 ]b[ML+2 ]c For dimensional balance, the dimensions on both sides should be same. Thus, equating dimensions, we have a+ b+ c =1 1 2(a + c ) = 1 or a + c = 2 − a − 2 b = − 1 or a + 2 b = 1 Solving these three equations, we get 1 1 a = 0, b = , c = 2 2 1 1
∴
p = h0S 2 I 2 or
1 1
p = S 2 I 2 h0
4. In the density measurement of a cube, the mass and edge length are measured as (10.00 ± 0.10) kg and (0.10 ± 0.01) m, respectively. The error in the measurement of density is [JEE Main 2019, 9 April ShiftI] (a) 0.01 kg/m3 (c) 0.07 kg/m3
(b) 0.10 kg/m3 (d) 0.31 kg/m3
∴ Permissible error in density is ∆ρ ∆M ∆l = ± 3 ρ M l
7 such squares taking into account the significant figures is [JEE Main 2019, 9 April ShiftII]
(a) 37.030 cm 2 (c) 37.03 cm 2
(b) 37.0 cm 2 (d) 37 cm 2
Exp. (c) Area of 1 square = 529 . cm2 Area of seven such squares = 7 times addition of area of 1 square = 529 . + 529 . + 529 . K 7 times = 37.03 cm2 As we know that, if in the measured values to be added/subtracted the least number of significant digits after the decimal is n. Then, in the sum or difference also, the number of significant digits after the decimal should be n. Here, number of digits after decimal in 5.29 is 2, so our answer also contains only two digits after decimal point. ∴Area required = 37.03 cm2
6. In the formula X = 5YZ 2, X and Z have
Exp. (*) Given, mass = (10.00 ± 010 . ) kg Edge length = (010 . ± 0.01) m Error in mass, ∆M 01 . = M 10 and error in length, ∆l 0.01 = 01 . l Density of the cube is given by M Mass = r= Volume l 3
5. The area of a square is 5.29 cm 2. The area of
dimensions of capacitance and magnetic field, respectively. What are the dimensions ofY in SI units? [JEE Main 2019, 10 April ShiftII] …(i)
(a) [M− 1 L− 2 T 4 A 2 ] (c) [M− 3 L− 2 T 8 A 4 ]
(b) [M− 2 L0 T − 4 A − 2 ] (d) [M− 2 L− 2 T 6 A 3 ]
Exp. (c) …(ii)
…(iii)
Substituting the value from Eqs. (i) and (ii) in Eq. (iii), we get . 0.01 ∆ρ 01 1 3 31 = + 3× = + = 01 . 10 ρ 100 10 100 31 ∆ρ = = 0.31 ⇒ 100 ρ ∆ρ Since, should be unitless quantity. ρ But there is no option with unitless error. Hence, no option is correct.
To find dimensions of capacitance in the given relation, we can use formula for energy. 1 Capacitors energy is U = CV 2 2 So, dimensionally, U [C ] = 2 ⇒ V potential energy As, V = potential = charge We have, [U ] [q 2 ] A 2 T2 [C ] = = = U 2 [U ] ML2 T−2 2 q ∴
X = [M−1L−2 A 2 T4 ]
To get dimensions of magnetic field, we use force on a current carrying conductor in magnetic field, [F] MLT−2 F = BIl ⇒ [B] = = [I][l] AL
3
Measurement Z = [M1L0 T−2 A −1 ]
∴
8. The pitch and the number of divisions, on
Now, using given relation, X = 5YZ 2 [Y ] =
M−1L−2 A 2 T4 M−1L−2 A 2 T4 = 1 0 −2 −1 2 = 2 −4 −2 [Z ] (M L T A ) M T A [X] 2
∴ [Y ] = [M−3L−2 A 4 T8 ]
7. Which of the following combinations has the dimension of electrical resistance (ε 0 is the permittivity of vacuum and µ 0 is the permeability of vacuum)? [JEE Main 2019, 12 April ShiftI]
(a)
µ0 ε0
(b)
µ0 ε0
(c)
ε0 µ0
(d)
ε0 µ0
Exp. (a) Key Idea A formula is valid only, if the dimensions of LHS and RHS are same. So, we need to balance dimensions of given options with the dimension of electrical resistance.
Let dimensions of resistance R, permittivity ε0 and permeability µ 0 are [R], [ε0 ] and [µ 0 ], respectively. …(i) So, [R ] = [ε0 ]α [µ 0 ]β [R ] = [M1 L2 T−3 A −2 ] , [ε0 ] = [M−1 L−3 T4 A 2 ], [µ 0 ] = [M1 L1 T−2 A −2 ]
the circular scale for a given screw gauge are 0.5 mm and 100, respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet are 5.5 mm and 48 respectively, the thickness of this sheet is [JEE Main 2019, 9 Jan ShiftII]
(a) 5.950 mm (c) 5.755 mm
(b) 5.725 mm (d) 5.740 mm
Exp. (c) For a measuring device, the least count is the smallest value that can be measured by measuring instrument. Minimum reading on main  scale Least count = Total divisions on the scale Here, screw gauge is used for measurement therefore, Pitch LC = number of division 0.5 mm LC = 100 LC = 5 × 10−3 mm …(i) Anvil Spindle
Sleeve
Circular scale
Now, from Eq. (i), we get [M1 L2 T −3 A −2 ] = [M−1 L−3 T 4 A 2 ]a [M1 L1 T −2 A −2 ]b
[M1 L2 T−3 A −2 ] = M− α + β L−3 α + β T4 α − 2 β A 2 α − 2 β
On comparing both sides, we get …(ii) −α + β = 1 …(iii) −3 α + β = 2 …(iv) 4 α − 2 β = −3 …(v) 2 α − 2 β = −2 Value of α and β can be found using any two Eqs. from (ii) to (v), On subtracting Eq. (iii) from Eq. (ii), we get (− α + β ) − (−3α + β ) = 1 − 2 −1 ⇒ 2α = −1 or α = 2 Put the value of α in Eq. (ii), we get 1 β=+ 2 µ ∴ [R ] = [ε0 ]−1/ 2 [µ 0 ]1/ 2 = 0 ε0
Ratchet Main scale
According to question, the zero line of its circular scale lies 3 division below the mean line and the readings of main scale = 5.5 mm The reading of circular scale = 48 then the actual value is given by actual value of thickness (t ) = (main scale reading) + (circular scale reading + number of division below mean line) × LC ⇒t = 5.5 mm + (48 + 3) × 5 × 10−3 mm mm ⇒t = 5755 .
9. The density of a material in SI units is
128 kg m −3. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is [JEE Main 2019, 10 Jan ShiftI] (a) 40
(b) 16
(c) 640
(d) 410
4
JEE Main Chapterwise Physics Exp. (a) To convert a measured value from one system to another system, we use N1u1 = N2u 2 where, N is numeric value and u is unit. We get 50 g kg Q density = mass 128 ⋅ 3 = N2 volume (25 cm)3 m N2 × 50 g 128 × 1000 g = ⇒ 100 × 100 × 100 cm3 25 × 25 × 25 cm3 128 × 1000 × 25 × 25 × 25 = 40 ⇒ N2 = 50 × 100 × 100 × 100
10. The diameter and height of a cylinder are
measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ? [JEE Main 2019, 10 Jan ShiftII]
(a) 4300 ± 80 cm 3
(b) 4260 ± 80 cm 3
(c) 4264.4 ± 81.0 cm 3 (d) 4264 ± 81 cm 3
Exp. (b) Volume of a cylinder of radius ‘r’ and height h is given by V = πr 2 h 1 or V = πD2 h, where D is the diameter of circular 4 surface. Here, D = 12.6 cm and h = 342 . cm π 2 . ) ⇒ V = × (12.6) × (342 4 V = 4262.22 cm3 V = 4260 (in three significant numbers) Now, error calculation can be done as . ∆V ∆D ∆h 2 × 01 0.1 = 2 = + + D V h 12.6 34.2 ∆V = 0.0158 + 0.0029 ⇒ V ⇒ ∆V = (0.01879) × (4262.22 ) ⇒ ∆V = 797 . ~ − 80 cm3 ∴For proper significant numbers, volume reading will be V = 4260 ± 80 cm3
11. The force of interaction between two atoms
x2 is given by F = αβ exp − ; where x is the αkT distance, k is the Boltzmann constant and T
T is temperature and α and β are two constants. The dimension of β is [JEE Main 2019, 11 Jan ShiftI] −2
(b) [M0 L2T −4 ] (d) [M2L2T −2]
(a) [MLT ] (c) [M2LT −4 ]
Exp. (c) Force of interaction between two atoms is given as F = αβ exp (− x2 / αkT ) As we know, exponential terms are always dimensionless, so − x2 = [M0L0 T0 ] dimensions of α kT ⇒ Dimensions of α = Dimension of ( x2 / kT ) Now, substituting the dimensions of individual term in the given equation, we get [M0L2 T0 ] = 1 2 − 2 {Q Dimensions of kT equivalent to [M L T ] the dimensions of energy = [M1L2 T− 2 ] } …(i) = [M− 1L0 T2 ] Now from given equation, we have dimensions of F = dimensions of α × dimensions of β F ⇒ Dimensions of β = Dimensions of α [M1L1T− 2 ] [Qusing Eq. (i)] = −1 0 2 [M L T ] = [M2L1T− 4 ]
12. If speed (v ), acceleration ( A ) and force ( F ) are considered as fundamental units, the dimension of Young’s modulus will be [JEE Main 2019, 11 Jan ShiftII]
(a) [v −4 A − 2 F] −2
2
(c) [v A F
−2
(b) [v −2 A 2 F 2 ] ]
(d) [v −4 A 2 F]
Exp. (d) Dimensions of speed are , [v ] = [LT−1 ] Dimensions of acceleration are, [A ] = [LT−2 ] Dimensions of force are, [F] = [MLT−2 ] Dimension of Young modulus is,, [Y] = [ML−1T−2 ] Let dimensions of Young’s modulus is expressed in terms of speed, acceleration and force as; …(i) [Y ] = [v ] α [A ] β [F ] γ Then substituting dimensions in terms of M, L and T
5
Measurement we get, [ML−1T−2 ] = [LT−1 ]α [LT−2 ]β [MLT−2 ]γ
l are rcv 2 −2 −2 l = [M L T A ] = [A −1 ] rcv [M1 L2 T−2 A −1 ]
So, dimensions of
= [MγLα + β + γ T− α − 2β − 2γ ] Now comparing powers of basic quantities on both sides we get, γ = 1 α + β + γ = −1 and −α − 2β − 2 γ = −2 Solving these, we get α = −4, β = 2, γ = 1 Substituting α,β, and γ in Eq. (i) we get; [Y] = [V −4 A 2F1 ]
13. The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 µm diameter of a wire is [JEE Main 2019, 12 Jan ShiftI]
(a) 50 (c) 500
(b) 200 (d) 100
Exp. (b) In a screw gauge, Least count Measure of 1 main scale division (MSD) = Number of division on circular scale Here, minimum value to be measured/least count is 5 µm. = 5 × 10−6 m ∴ According to the given values, 1 × 10−3 5 × 10−6 = N 10−3 1000 or N= = = 200 divisions 5 5 × 10−6
14. Let l , r , c , andv represent inductance, resistance, capacitance and voltage, l respectively. The dimension of in SI rcv units will be [JEE Main 2019, 12 Jan ShiftII] (a) [LT 2 ]
(b) [LTA]
(c) [A−1 ]
Exp. (c) Dimensions of given quantities are l = inductance = [M1 L2 T−2 A −2 ]
(d) [LA− 2 ]
15. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1% , the maximum error in determining the density is [JEE Main 2018]
(a) 2.5%
(c) 4.5%
(d) 6%
Exp. (c) Mass M M or ρ = 3 = Volume L3 L ∆ρ ∆M 3∆L = + ⇒ Error in density ρ M L
Q Density, ρ =
So, maximum % error in measurement of ρ is ∆M 3∆L ∆ρ × 100 = × 100 + × 100 M L ρ or % error in density = 15 . + 3×1 % error = 4.5%
16. The following observations were taken for determining surface tension T of water by capillary method. Diameter of capillary, m rise of water, d = 1.25 × 10−2 h = 1.45 × 10−2m. Using g = 9.80 m/s 2 and the rhg simplified relation T = × 103N/ m, the 2 possible error in surface tension is closest to [JEE Main 2017 (Offline)]
(a) 1.5%
(b) 2.4%
(c) 10%
(d) 0.15%
Exp. (a) By ascent formula, we have surface tension, N rhg N dhg Q r = d T= × 103 = × 103 m 2 m 4 2 ∆T ∆d ∆h [given, g is constant] = + ⇒ T d h So, percentage =
r = resistance = [M1 L2 T−3 A −2 ] c = capacitance = [M− 1 L− 2 T4 A 2 ] v = voltage = [M1 L2 T−3 A −1 ]
(b) 3.5%
∴
∆T × 100 = T
∆d + ∆h × 100 d h 0.01 × 10−2 × 100 = 15 . % + . × 10−2 145
0.01 × 10−2 = −2 1.25 × 10 ∆T × 100 = 15 . % T
6
JEE Main Chapterwise Physics
17. A student measures the time period of 100
19. The current voltage relation of diode is given
oscillations of a simple pendulum four times. The data set is 90s, 91s, 92s and 95s. If the minimum division in the measuring clock is 1s, then the reported mean time should be
by I = (e1000V /T − 1) mA, where the applied voltage V is in volt and the temperature T is in kelvin. If a student makes an error measuring ± 0.01V while measuring the current of 5 mA at 300K, what will be the error in the value of current in mA?
[JEE Main 2016 (Offline)]
(a) ( 92 ± 2 ) s (c) ( 92 ± 18 . )s
(b) ( 92 ± 5 ) s (d) ( 92 ± 3) s
Exp. (a) Arithmetic mean time of a oscillating simple pendulum Σ xi = N 90 + 91 + 92 + 95 = = 92 s 4 Mean deviation of a simple pendulum Σ  x − xi = N 2 + 1+ 3 + 0 = = 1.5 4 Given, minimum division in the measuring clock, i.e. simple pendulum = 1s. Thus, the reported mean time of a oscillating simple pendulum = (92 ± 2 ) s
18. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? (a) A meter scale (b) A vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in 1 cm (c) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm (d) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm [JEE Main 2014]
Exp. (b) If student measure 3.50 cm, it means that there in an uncertainly of order 0.01 cm. For vernier scale with 1 MSD = 1mm and 9 MSD =10 VSD LC of VC = 1 MSD − 1 VSD ∴ 9 1 = 1 − 10 10 1 cm = 100
[JEE Main 2013]
(a) 0.2 mA (c) 0.5 mA
(b) 0.02 mA (d) 0.05 mA
Exp. (a) Given,
I = (e1000 V / T − 1) mA dV = ± 0.01 V T = 300 K I = 5 mA I = e1000V /T − 1 I + 1 = e1000 V / T
Taking log on both sides, we get 1000V log (I + 1) = T d (I + 1) 1000 ⇒ = dV I+1 T dI 1000 = dV I+1 T 1000 dI = × (I +1) dV ⇒ T 1000 dI = × (5 + 1) × 0.01 300 = 0.2 mA So, error in the value of current is 0.2 mA.
20. Let [ ε0] denotes the dimensional formula of the permittivity of vacuum. If M = mass, L = length,T = time and A = electric current, then [JEE Main 2013] (a) [ε0 ] = [ M −1 L−3 T 2 A]
(b) [ε0 ] = [ M −1 L−3 T 4 A 2 ]
(c) [ε0 ] = [ M −2 L2 T −1 A −2 ] (d) [ε0 ] = [ M −1 L2 T −1 A 2 ]
Exp. (b) Electrostatic force between two charges, q q 1 q1 q 2 F= ⇒ ε0 = 1 2 2 4 πε0 R 2 4 πFR Substituting the units.
7
Measurement Hence, ε0 =
C2 2
N m
=
[AT ]2
the smallest division of the main scale is half a degree (= 0.5°), then the least count of the instrument is [AIEEE 2009]
[MLT−2 ] [L2 ]
= [M −1L−3 T 4 A 2 ]
21. A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading : 58.5 degree Vernier scale reading : 09 divisions Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data is [AIEEE 2012] (a) 58.59 degree (c) 58.65 degree
(b) 58.77 degree (d) 59 degree
29 main scale division 30 29 29 ° 1 VSD = × 0.5° = 60 30
1 Vernier scale division =
Thus, least count = 1 MSD − 1 VSD 1 ° 29 ° 1 ° = − = 60 60 2
22. The respective number of significant figures 23.023,
(a) 5, 1, 2
(c) 5, 5, 2
(b) 5, 1, 5
0.0003
Exp. (a) Least count Value of main scale division Number of divisions on vernier scale 1 1° 1° 1 MSD = = × = = 1 min 30 2 60 30
=
24. The dimensions of magnetic field in M, L, T and C (coulomb ) is given as
[AIEEE 2008]
(b) [MT 2 C −2 ] (d) [MT 2 C −1 ]
Exp. (c) From the relation F = qvB ⇒ [MLT−2 ] = [C][LT−1 ][B] ⇒
[B] = [MC −1T−1 ]
25. A body of mass m = 3.513 kg is moving along
Reading = Main scale reading + Vernier scale reading = MSR + n × LC 1 ° = 58.5° + 9 × = 58.65° 60
for the numbers 2.1 × 10−3 are
(b) half minute (d) half degree
(a) [MLT −1 C −1 ] (c) [MT −1 C −1 ]
Exp. (c)
∴
(a) one minute (c) one degree
and
[AIEEE 2010]
(d) 4, 4, 2
Exp. (a) The reliable digit plus the first uncertain digit is known as significant figures. For the number 23.023, all the nonzero digits are significant, hence 5. For the number 0.0003, number is less than 1, the zero(s) on the right of decimal point but to the left of the first nonzero digit are not significant, hence 1. For the number 2.1 × 10−3 , significant figures are 2.
23. In an experiment, the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If
the xaxis with a speed of 5.00 ms −1. The magnitude of its momentum is recorded as [AIEEE 2008]
(a) 17.6 kg ms−1 (c) 17.56 kg ms−1
(b) 17.565 kg ms−1 (d) 17.57 kg ms−1
Exp. (a) So, momentum, p = mv = 17.565 kg ms −1 where m = 3.513 kg and v = 5.00 ms −1 As the number of significant digits in mis 4 and in v is 3, so, p must have 3 (minimum) significant digits. Hence, p = 17.6 kg ms −1
26. Which of the following units denotes the
dimensions [ML2 /Q2], where Q denotes the electric charge? [AIEEE 2006] (a) Wb/m 2 (c) H/m 2
(b) henry (H) (d) weber (Wb)
Exp. (b) Magnetic energy =
1 2 Lq 2 LI = 2 2 2t
where L = inductance, I = current Energy has the dimensions = [ML2 T–2 ]
as I = q t
8
JEE Main Chapterwise Physics Equate the dimensions, we have [Q 2 ] [ML2 T–2 ] = [henry] × 2 [T ] [ML2 ] [henry] = ⇒ [Q 2 ]
30. Dimensions of 1 /µ 0ε0, where symbols have their usual meaning, are –1
(a) [L T] (c) [L2 T –2 ]
27. The ‘rad’ is the correct unit used to report the measurement of
[AIEEE 2006]
(a) the ability of a beam of gamma ray photons to produce ions in a target (b) the energy delivered by radiation to a target (c) the biological effect of radiation (d) the rate of decay of a radioactive source
‘Rad’ is used to measure biological effect of radiation.
28. Out of the following pairs, which one does not have identical dimensions? [AIEEE 2005] (a) (b) (c) (d)
Angular momentum and Planck’s constant Impulse and momentum Moment of inertia and moment of a force Work and torque
Moment of inertia, I = mr 2 [I] = [ML2 ]
and τ = Moment of force = r × F ∴ [τ ] = [L] [MLT–2 ] = [ML2 T–2 ]
29. Which one of the following represents the correct dimensions of the coefficient of viscosity? [AIEEE 2004] (a) [ML–1 T –2 ] (c) [ML–1 T –1 ]
(b) [L T ] (d) [LT –1 ]
As we know that formula of velocity is 1 v= µ 0 ε0 ⇒
v2 =
1 = [LT–1 ]2 µ 0 ε0
1 = [L2 T–2 ] µ 0 ε0
31. The physical quantities not having same dimensions are (a) (b) (c) (d)
[AIEEE 2003]
torque and work momentum and Planck’s constant stress and Young’s modulus speed and ( µ 0 ε0 )−1/ 2
Exp. (b) Planck’s constant (in terms of unit) h = Js = [ML2 T–2 ][T] = [ML2 T–1 ]
Exp. (c) ∴
2
Exp. (c)
∴
Exp. (c)
[AIEEE 2003] 2
(b) [MLT –1 ] (d) [ML–2 T –2 ]
Momentum ( p) = kg  ms –1 = [M][L][T–1 ] = [MLT–1 ]
32. Identify the pair whose dimensions are equal. (a) (b) (c) (d)
[AIEEE 2002]
Torque and work Stress and energy Force and stress Force and work
Exp. (a) Exp. (c) By Newton’s formula, η =
F A (∆v x / ∆z)
∴ Dimensions of η Dimensions of force = Dimensions of area × Dimensions of velocity gradient [MLT–2 ] –1 –1 = 2 –1 = [ML T ] [L ][T ]
The dimensions of torque is given by Angular momentum Torque (τ ) = Time [ML2 T−1 ] Torque (τ ) = = [ML2 T−2 ] [T] ∴ Work = Force × Distance = [MLT−2 ] [L] = [ML2 T–2 ] Hence, both have same dimensions.
2 Kinematics 1. Ship A is sailing towards northeast with velocity v = 30$i + 50$j km/h, where $i points east and $j north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/h. A will be at minimum distance from B in [JEE Main 2019, 8 April ShiftI]
(a) 4.2 h
(b) 2.6 h
(c) 3.2 h
(d) 2.2 h
Exp. (b) Considering the initial position of ship A as origin, so the velocity and position of ship will be v A = (30 $i + 50 $j ) and rA = (0 $i + 0 $j ) Now, as given in the question, velocity and position of ship B will be, v B = −10 $i and rB = (80 $i + 150 $j ) Hence, the given situation can be represented graphically as B vB=–10 ˆi
(80,150)
y vA=30^ i + 50^ j
A
d = (80 − 10 t − 30 t )2 + (150 − 50 t )2 ⇒
d 2 = (80 − 40 t )2 + (150 − 50 t )2 d Distance is minimum when (d 2 ) = 0 dt After differentiating, we get d [(80 − 40 t )2 + (150 − 50 t )2 ] = 0 ⇒ dt ⇒ 2(80 − 40 t )(−40) + 2(150 − 50 t ) (−50) = 0 ⇒ −3200 + 1600t − 7500 + 2500t = 0 10700 ⇒ 4100 t = 10700 ⇒t = = 2.6 h 4100 Alternate Solution Time after which the distance is minimum between A and B can be calculated as r . v  t = BA BA v BA2 where, rBA = rB − rA = 80$i + 150$j and v BA = − 10$i − (30$i + 50$j ) = −40$i − 50$j (80$i + 150$j ) ⋅ (−40$i − 50$i ) t = ⇒  − 40$i − 50$j 2 3200 + 7500 10700 = = = 2 .6 h 4100 4100
2. Let  A 1 = 3,  A2 = 5 and  A1 + A2 = 5. The value of (2 A1 + 3 A 2 ) ⋅ ( 3 A1 − 2 A 2 ) is
x (0,0)
[JEE Main 2019, 8 April ShiftII]
(a) −106.5 (b) −112 . 5 (c) −99.5
(d) −118. 5
Exp. (d) After time t, coordinates of ships A and B are (80 − 10 t , 150) and (30 t , 50 t ). So, distance between A and B after time t is d = ( x2 − x1 ) + ( y2 − y1 ) 2
2
For vector A1 + A 2 , we have A1 + A 22 = (A1 + A 2 ) ⋅ (A1 + A 2 ) [Qx ⋅ x =x2 ] ⇒ A1 + A 22 = A12 + A 22 + 2 A1 ⋅ A 2 Given, A1 = 3,A 2 = 5 andA + A 2 = 5
10
JEE Main Chapterwise Physics So, we have (5)2 = 9 + 25 + 2 A1 ⋅ A 2 9 A1 ⋅ A 2 = − ⇒ 2 Now, (2 A1 + 3A 2 ) ⋅ (3A1 − 2 A 2 ) = 6A12 − 4A1 ⋅ A 2 + 9A1 ⋅ A 2 − 6A 22 = 6A12 − 6A 22 + 5A1 ⋅ A 2 Substituting values, we have (2 A1 + 3A 2 ) ⋅ (3A1 − 2 A 2 ) 9 = 6 (9) − 6(25) + 5 − = − 118.5 2 Alternate Solution As we know,A1 + A 2 can also be written as A + A 2 = A12 + A 22 + 2A1A 2cos θ Substituting the given values, we get 5 = (3)2 + (5)2 + 2 × 3 × 5cosθ 9 3 or cosθ = − =− 2 ×3×5 10
4. The stream of a river is flowing with a speed
(2 A1 + 3A 2 ) ⋅ (3A1 − 2 A 2 ) = 6A12 − 6A 22 + 5A1 ⋅ A 2
So,
Since, the particle starts from rest, this means, initial velocity, u = 0 Also, it moves with uniform acceleration along positive Xaxis. This means, its acceleration (a) is constant. ∴Given, a  t graph in (A) is correct. As we know, for velocitytime graph, slope = acceleration. Since, the given vt graph in (B) represents that its slope is constant and nonzero. ∴Graph in (B) is also correct. Also, the displacement of such a particle w.r.t. time is given by 1 1 x = ut + at 2 = 0 + at 2 ⇒ x ∝ t 2 2 2 So, x versus t graph would be a parabola with starting from origin. This is correctly represented in displacementtime graph given in (D).
= 6A12 − 6A 22 + 5A1A 2cos θ −3 = 6 × 9 − 6 × 25 + 5 × 2 × 3 × = − 118.5 10
3. A particle starts from originO from rest and moves with a uniform acceleration along the positive Xaxis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration,v = velocity, x = displacement,t = time) [JEE Main 2019, 8 April ShiftII]
of 2 km/h. A swimmer can swim at a speed of 4 km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight ? [JEE Main 2019, 9 April ShiftI]
(a) 60°
(b)120°
(c) 90°
Exp. (b) Let the velocity of the swimmer is v s = 4 km/h and velocity of river is v r = 2 km/h Also, angle of swimmer with the flow of the river (down stream) is α as shown in the figure below α=90°+θ vr
(B) v
(A) a O
vsr
O
t
(C) x
t
(D) x
(d) 150°
vr θ vs
From diagram, angle θ is 2 km / h 1 v sinθ = r = = ⇒θ = 30° v sr 4 km / h 2 Clearly,
α = 90° + 30° = 120°
5. The position vector of particle changes with O
(a) (A) (c) (B), (C)
Exp. (d)
t
O
t
(b) (A), (B), (C) (d) (A), (B), (D)
time according to the relation What is the r(t ) = 15t 2$i + ( 4 − 20t 2 )$j. magnitude of the acceleration (in ms−2 ) att = 1? [JEE Main 2019, 9 April ShiftII]
11
Kinematics (a) 50
(b) 100
(c) 25
(d) 40
2γ v 0 g γ tan −1 v 0 g γ sin −1 v 0 g γ ln 1 + v 0 g
(a)
1 tan −1 2 γg
Position vector of particle is given as r = 15 t 2 $i + (4 − 20 t 2 )$j
(b)
1 γg
Velocity of particle is dr d v= = [15 t 2 $i + (4 − 20t 2 )$j ] dt dt = 30 t $i − 40 t $j Acceleration of particle is d d a = (v ) = (30 t $i − 40 t$j ) = 30$i − 40$j dt dt So, magnitude of acceleration at t = 1s is
(c)
1 γg
(d)
1 γg
Exp. (a)
 a t
= 1s
=
a2x + ay2 =
302 + 402
= 50 ms − 2
6. The position of a particle as a function of timet , is given by x (t ) = at + bt 2 − ct 3 where a ,b and c are constants. When the particle attains zero acceleration, then its velocity will be [JEE Main 2019, 9 April ShiftII] (a) a +
b2 2c
(b) a +
b2 b2 (c) a + 4c 3c
(d) a +
b2 c
Exp. (b) Given, drag force, F = mγv 2
… (i)
As we know, general equation of force = ma Comparing Eqs. (i) and (ii), we get a = γ v2 ∴ Net retardation of the ball when thrown vertically upward is dv anet = − (g + γv 2 ) = dt dv ⇒ = − dt (g + γv 2 )
… (ii)
… (iii)
Position of particle is, x(t ) = at + bt 2 − ct 3 dx So, its velocity is, v = = a + 2 bt − 3 ct 2 dt dv and acceleration is, a = = 2 b − 6 ct dt Acceleration is zero, then 2 b − 6 ct = 0 2b b = t = ⇒ 6c 3c
By integrating both sides of Eq. (iii) in known limits, i.e. When the ball thrown upward with velocity v 0 and then reaches to its zenith, i.e. for maximum height at time t = t , v = 0 0 t dv ⇒ ∫v0 (γ v 2 + g ) = ∫0 − dt t 1 0 1 or dv = − ∫ dt ∫ 2 0 v 0 g γ 2 + v γ
Substituting this ‘t’ in expression of velocity, we get
⇒
Exp. (c)
2
b b v = a + 2 b − 3c 3c 3c 2 b2 b2 b2 =a+ − =a+ 3c 3c 3c
7. A ball is thrown upward with an initial velocityv 0 from the surface of the earth. The motion of the ball is affected by a drag force equal to mγv 2 (where, m is mass of the ball,v is its instantaneous velocity and γ is a constant). Time taken by the ball to rise to its zenith is [JEE Main 2019, 10 April ShiftI]
⇒
1 ⋅ γ
1 . tan−1 g / γ
0
v = − t g / γ v 0
1 1 x dx = tan−1 Q ∫ 2 2 a a x + a γ v0 1 ⋅ tan−1 = +t γg g
8. A plane is inclined at an angle α = 30º with respect to the horizontal. A particle is projected with a speed u = 2 ms− 1, from the base of the plane, making an angle θ = 15º with respect to the plane as shown in the
12
JEE Main Chapterwise Physics figure. The distance from the base, at which the particle hits the plane is close to [Take, g = 10 ms− 2] [JEE Main 2019, 10 April ShiftII]
Here, u = 2 ms −1, g = 10 ms −2 , θ = 15°, α = 30° 2 × 2 sin15° 2u sinθ So, T = = g cos α 10 × cos 30° 2 × 2 × 0258 . ×2 = 01191 . = 10 × 1732 . Now, . R = 2 × cos 15° × 01191 −
° 15 θ= α=30°
u
(a) 26 cm
5 (01191 . )2 2 m ≈ 020 = 0229 . − 0.0354 = 01936 . . m = 20 cm
. = 2 × 0.965 × 01191 −
(b) 20 cm (c) 18 cm
(d) 14 cm
Exp. (b) When a projectile is projected at an angle θ with an inclined plane making angle α with the horizontal, then y
u sin θ = uy
u cos
θ
θ = ux
x
9. A bullet of mass 20 g has an initial speed of 1 ms− 1, just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5 × 10− 2N, the speed of the bullet after emerging from the other side of the wall is close to [JEE Main 2019, 10 April ShiftII]
θ α
Components of u along and perpendicular to plane are u x = u cos θ and u y = u sinθ We can also resolve acceleration due to gravity into its components along and perpendicular to plane as shown below y
g sin α
1 . )2 × 10sin 30° (01191 2
x
= ax α
(b) 0.4 ms− 1 (d) 0.7 ms− 1
Exp. (d) Given, resistance offered by the wall = F = − 2.5 × 10−2 N So, deacceleration of bullet, F −2.5 × 10−2 5 a= = = − ms −2 m 4 20 × 10−3 (Qm = 20 g = 20 × 10−3 kg)
g cos α = ay
g
So, we can now apply formula for range, i.e. net horizontal displacement of the particle as 1 …(i) R = u xT + a xT 2 2 where, T = time of flight. Using formula for time of flight, we have 2u 2u sinθ …(ii) T= y = ay g cos α From Eqs. (i) and (ii), we have Range up the inclined plane is 1 R = u xT + a xT 2 2 2u sinθ 1 2u sinθ = u cos θ − g sinα g cos α 2 g cos α
(a) 0.3 ms− 1 (c) 01 . ms− 1
Now, using the equation of motion, v 2 − u 2 = 2 as We have, 5 v 2 = 1 + 2 − (20 × 10−2 ) 4 ⇒ ∴
(Qu = 1 ms −1 and s = 20 cm = 20 × 10−2 m) 1 v2 = 2 1 . ms −1 v= ≈ 07 2
10. A shell is fired from a fixed artillery gun with
2
an initial speedu such that it hits the target on the ground at a distanceR from it. Ift 1 and t 2 are the values of the time taken by it to hit the target in two possible ways, the product [JEE Main 2019, 12 April ShiftI] t 1t 2 is
13
Kinematics R 4g
(a)
(b)
R g
(c)
R 2g
(d)
2R g
11. The trajectory of a projectile near the surface of the earth is given as y = 2 x − 9x 2.
Exp. (d)
If it were launched at an angle θ 0 with speed v 0, then (Take, g = 10 ms −2)
Key Idea Range of a projectile motion is given by
R=
u 2 sin2 θ g
[JEE Main 2019, 12 April ShiftI] −1
(a) θ0 = sin (b) θ0 = cos−1 −1 (c) θ0 = cos −1 (d) θ0 = sin
2u sinθ Time of flight is given by t = g
t2 θ2
t1
θ1
Exp. (c)
R
Given, range of the fired shell, R=R and time of flights are t 1 and t 2 . Let θ1 and θ2 are the two angles at which shell is fired. As, range in both cases is same, i.e. R1 = R 2 = R u 2 sin2 θ1 u 2 sin2 θ2 Here, R1 = and R 2 = g g ⇒
R=
u sin2 θ1 u sin2 θ2 = g g 2
2
…(i)
⇒ sin2 θ1 = sin2 θ2 ⇒ sin2 θ1 = sin(180 − 2 θ2 ) [Qsin(180 − θ) = sinθ] ⇒ 2(θ1 + θ2 ) = 180 or θ1 + θ2 = 90 …(ii) ⇒ θ2 = 90 − θ1 So, time of flight in first case, 2u sinθ1 …(iii) t1 = g and time of flight in second case, 2u sinθ2 2u sin(90 − θ1 ) 2u cos θ1 …(iv) = = t2 = g g g
4u 2 sin θ1 cos θ1 g
2
From Eq. (i), we get
2u 2 sin2 θ1
g2 (Qsin2 θ = 2 sinθcos θ) …(v)
u 2 sin2 θ1 g 2R t1 t 2 = g R=
∴
⇒t 1t 2 =
Given, g = 10 m/s 2 Equation of trajectory of the projectile, y = 2 x − 9 x2
…(i)
In projectile motion, equation of trajectory is given by g x2 …(ii) y = x tanθ0 − 2 2 v 0 cos 2 θ0 By comparison of Eqs. (i) and (ii), we get tanθ0 = 2 g and =9 2 v 02 cos 2 θ0 g or v 02 = 9 × 2 cos 2 θ0
…(iii)
…(iv)
From Eq. (iii), we can get value of cosθ and sinθ 1 2 and sinθ0 = …(v) cos θ0 = 5 5 √5
2
θ0 1
From Eqs. (iii) and (iv), we get 2u sin θ1 2u cos θ1 × t 1t 2 = g g ⇒t 1t 2 =
1 5 −1 and v 0 = ms 5 3 2 3 −1 and v = ms 0 5 5 1 5 −1 and v 0 = ms 5 3 2 3 −1 and v 0 = ms 5 5
Using value of cosθ0 from Eq. (v) to Eq. (iv), we get 10 × ( 5 )2 10 × 5 = v 02 = 2 × (1)2 × 9 2 × 9 25 5 or v 0 = m/s …(vi) ⇒ v 02 = 9 3 From Eq. (v), we get 1 θ0 = cos −1 5
14
JEE Main Chapterwise Physics
12. Two particles are projected from the same point with the same speed u such that they have the same rangeR, but different maximum heights h1 and h2. Which of the following is correct? [JEE Main 2019, 12 April ShiftII] (a) R 2 = 4h1h2
(b) R 2 = 16h1h2
(c) R = 2 h1h2
(d) R 2 = h1h2
2
Exp. (b) Key Idea Same range for two particles thrown with some initial speed may occurs when they are projected at complementary angle. ∴ θ 1 + θ 2 = 90 ° where, θ 1 and θ 2 are angles of projection.
h1
4g 2
Using 2sin A ⋅ sin B = cos( A − B) − cos( A + B), we have 1 sin(45° + θ) sin(45° − θ) = (cos 2 θ − cos 90° ) 2 cos 2 θ ⇒sin(45° + θ) sin(45° − θ) = [Qcos 90° = 0] 2 So, we have h1h2 =
u 4
2
cos 2 θ 4 2 2 ⇒ h h = u cos 2 θ …(ii) 1 2 2 4g 16g 2
From Eqs. (i) and (ii), we get R2 ⇒ h1h2 = ⇒ R 2 = 16 h1h2 16
[JEE Main 2019, 12 April ShiftII]
As maximum range occurs at θ = 45° for a given initial projection speed, we take angles of projection of two particles as 45°+θ
θ 45°
θ1 = 45° + θ, θ2 = 45° − θ where, θ is angle of projectiles with 45° line. So, range of projectiles will be u 2 sin2 (θ1 ) R = R1 = R 2 = g ⇒R =
u 2 sin2(45° + θ) u 2 sin(90° + 2 θ) ⇒R = g g
⇒R =
u 2 cos 2 θ g
u 4 cos 2 2 θ g2
Maximum heights achieved in two cases are u 2 sin2 (45° + θ) h1 = 2g u 2 sin2 (45° − θ) 2g
b2τ (b) 2
(c) b 2 τ
(d)
b2τ 2
Given, speed,
θ
⇒ R2 =
b2τ (a) 4
Exp. (b)
45° 45°–θ
h2 =
u 4 sin2 (45° + θ)sin2 (45° − θ)
along positive Xaxis. Calculate the speed of the particle at time t = τ (assume that the particle is at origin att = 0).
u
and
h1h2 =
13. A particle is moving with speed v = b x
h2
u 45+θ 45–θ
So,
…(i)
v=b x Now, differentiating it with respect to time, we get dv d = b x dt dt Now, acceleration b dx Q dv = a ⇒ ⋅ a= 2 x dt dt b a= ⋅v ⇒ 2 x b b2 = ⋅b x= 2 x 2 As acceleration is constant, we use …(i) v = u + at Now, it is given that x = 0 at t = 0. So, initial speed of particle is u=b x = b × 0= 0 x=0
Hence, when time t = τ, speed of the particle using Eq. (i) is b2 b2 ⋅τ = ⋅τ v = u + at = 0 + 2 2
15
Kinematics 14. A particle is moving with a velocity v = k ( y$i + x$j), where k is a constant.
Let car B takes time (t 0 + t ) and car A takes time t 0 to finish the race. x
The general equation for its path is [ JEE Main 2019, 9 Jan ShiftI]
(c) xy = constant
(d) y 2 = x 2 + constant
Exp. (d) Given, velocity of a particle is v = k( y $i + x $j ) Suppose, it’s position is given as r = x $i + y $j dr d $ ( x i + y $j ) v= = ∴ dt dt dx dy $ j = $i + dt dt Comparing Eqs. (i) and (ii), we get dx =y dt dy and = x dt Dividing Eq. (iii) and Eq. (iv), we get dx dt = y ⇒ x dx = y dy dy x dt dt dt or xdx = ydy Integrating both sides, we get
...(i)
...(ii)
...(iii)
or
Substituting the value of t 0 from Eq. (ii) into Eq. (i), we get a2 t v = (a1 − a2 ) − a2t a1 − a2 = ( a1 −
15. In a car race on a straight path, car A takes a timet less than car B at the finish and passes finishing point with a speed ‘v’ more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a 2 respectively. Then ‘v ’ is equal to [ JEE Main 2019, 9 Jan ShiftII]
2a1a 2 (a) t a1 +a 2
(b) 2a1a 2 t
(c) a1a 2 t
(d)
a2 ) ( a1 +
a2 ) ⋅
a2 t ( a1 −
a2 )
− a2t
or v = ( a1 + a2 ) ⋅ a2 t − a2t = a1a2 ⋅ t + a2t − a2t or v = a1 ⋅ a2 t
16. In three dimensional system, the position coordinates of a particle (in motion) are given below [JEE Main 2019, 9 Jan (ShiftII)] x = a cosωt y = a sinωt z = aωt
or
y2 = x2 + constant
Exp. (c)
Then, Given, v A − v B = v = (a1 − a2 )t 0 − a2 t …(i) 1 2 1 2 sB = sA = a1t 0 = a2 (t 0 + t ) 2 2 or a1 t 0 = a2 (t 0 + t ) or a1t 0 = a2 t 0 + a2 t or ( a1 − a2 ) t 0 = a2 t a2 . t …(ii) or t0 = ( a1 − a2 )
...(iv)
x2 c1 y2 c 2 + = + 2 2 2 2 where, c1 and c 2 are the constants of integration. [here, c (constant) = c1 − c 2 ] ⇒ x 2 + c = y2
∫ xdx = ∫ ydy
vA=a1.t0 vB=a2(t0+t)
u=0
(a) y = x 2 + constant (b) y 2 = x + constant
The velocity of particle will be (a) 2 aω (c) aω
(b) 2 aω (d) 3 aω
Exp. (a) Given that the position coordinates of a particle x = acos ωt …(i) y = asinωt z = aωt So, the position vector of the particle is r$ = x $i + y$j + zk$
a1 + a 2 t 2 ⇒
r$ = acos ωt $i + asinωt $j + a ωt k$
16
JEE Main Chapterwise Physics z
r$ = a[cos ωt $i + sinωt $j + ωt k$ ] therefore, the velocity of the particle is dr d [a] [cos ωt $i + sinωt$j + ωt k$ ] = Q v$ = dt dt ⇒ v$ = − aω sinωt $i + aω co sωt $j + aωk$ ) The magnitude of velocity is v = or
v 2x
+
v y2
+
B A
= ωa (− sinωt )2 + (cos ωt )2 + (1)2 = 2 ωa
17. Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s, respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets on the ground fired by the two guns is
F
O
x
a
D
1 $) (a) a ( i$ − k 2 1 $) (c) a ( $j − k 2
1 $ $ a ( j − i) 2 1 $ $ (d) a ( k − i) 2 (b)
Exp. (b) In the given cube, coordinates of point G(centre of face ABOD) are a a x1 = , y1 = 0, z1 = 2 2 where, a = side of cube Z
(b) 1 : 16 (d) 1 : 2
Exp. (b)
Maximum range for gun u2 B = R2 = 2 g So, ratio of covered areas π(R1 )2 π ⋅ u14 / g 2 u14 = = = 2 π(R 2 ) π ⋅ u 24 / g 2 u 24 Here, u1 = 1 km / s and u 2 = 2 km / s 14 1 So, ratio of areas = 4 = 16 2 = 1 : 16
18. In the cube of side ‘a’ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be [JEE Main 2019, 10 Jan (ShiftI)]
y
a
[JEE Main 2019, 10 Jan (ShiftI)]
Bullets from guns can reach upto a distance of maximum range which occurs when projection is made at angle of 45º. Maximum range for gun u2 A = R1 = 1 g
a
H
G
v z2
v = (− aω sinωt )2 + (aω cos ωt )2 + (aω)2
(a) 1 : 4 (c) 1 : 8
E
B A
H
E
G a ,0 , a 2 2
O
F
D
Y
X
and coordinates of point H are a a x2 = 0, y2 = , z2 = 2 2 So, vector GH is GH = ( x2 − x1 )$i + ( y2 − y1 )$j + ( z2 − z1 ) k$ a a a = − $i + $j = ($j − $i ) 2 2 2
19. Two vectors A and B
have equal magnitudes. The magnitude of ( A + B) is ‘n’ times the magnitude of ( A − B). The angle between A and B is [JEE Main 2019, 10 Jan (ShiftII)]
n2 −1 (a) sin −1 2 n + 1 n2 −1 (c) cos−1 2 n + 1
n −1 (b) sin −1 n + 1 n −1 (d) cos−1 n + 1
17
Kinematics Exp. (c) Given,  A  =  B or …(i) A=B Let magnitude of ( A + B ) is R and for (A − B) is R′. Now, R = A + B and R 2 = A 2 + B2 + 2 ABcos θ R 2 = 2 A 2 + 2 A 2 cos θ
…(ii) [Qusing Eq. (i)]
Again, R′ = A − B ⇒ R ′2 = A 2 + B2 − 2 ABcosθ
…(iii) [Qusing Eq. (i)]
Given, R = nR ′ 2
R = n2 R′
n2 − 1 n2 + 1
=
2 cos θ = cos θ 2
3 2 1
(a) 6 m (c) 10 m
4
5 6
7
∴Displacement of particle = Area of OPA + Area of PABSP + Area of QBCRQ 1 = × 2 × 2 + (2 × 2 ) + (3 × 1) 2
8
9 10
(b) 2.8 m (d) 2.5 m
Exp. (b)
[JEE Main 2019, 10 Jan (ShiftII)]
3
1 2 3 4 5 6 7 8 Time (s) A B C
[JEE Main 2019, 11 Jan (ShiftI)]
v (m/s) 4
2
1
(a) 10.3 m (c) 5.1 m
and moves along the positive X axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at timet = 5 s ?
1
Q R S
ms −1 at an angle of 60° with the horizontal. The radius of curvature of its trajectory at t = 1 s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 ms −2, the value of R is
20. A particle starts from the origin at time t = 0
0
2
P
21. A body is projected att = 0 with a velocity 10
n2 − 1 θ = cos − 1 2 n + 1
or
3
= 2 + 4 + 3= 9m
Dividing Eq. (ii) with Eq. (iii), we get n2 1 + cos θ = 1 1 − cos θ n2 − 1 (1 + cos θ) − (1 − cos θ) or = n2 + 1 (1 + cos θ) + (1 − cos θ) ⇒
v (m/s)
0
R ′2 = 2 A 2 − 2 A 2 cosθ
or
To get exact position at t = 5 s, we need to calculate area of the shaded part in the curve as shown below
t (s)
(b) 3 m (d) 9 m
Exp. (d) Key Idea Area under the velocitytime curve represents displacement.
Components of velocity at an instant of time t of a body projected at an angle θ is v x = u cosθ + g xt and v y = u sinθ + g y t Here, components of velocity at t = 1s, is [as g x = 0] v x = u cos 60º + 0 1 = 10 × = 5 m / s 2 and v y = u sin 60º + (−10) × (1) 3 = 10 × + (− 10) × (1) 2 = 5 3 − 10 ⇒ v y = 10 − 5 3  m / s Now, angle made by the velocity vector at time of t = 1s v 10 − 5 3  tanα  = y = vx 5 tanα = 2 − 3 or ⇒ α = 15º ∴ Radius of curvature of the trajectory of the projected body R = v 2 / g cosα
18
JEE Main Chapterwise Physics =
⇒
(5)2 + (10 − 5 3 )2 10 × 0.97 [Qv 2 = v 2x + v y2 and cos 15º = 0.97]
R = 2.77 m ≈ 2.8 m
22. A particle is moving along a circular path
with a constant speed of 10 ms −1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60° around the centre of the circle? [JEE Main 2019, 11 Jan (ShiftI)]
(a) 10 2 m/s (c) 10 3 m/s
(b) 10 m/s (d) Zero
Exp. (b) Let v1 be the velocity of the particle moving along the circular path initially, v1 and v 2 be the velocity when it moves through an angle of 60º as shown below. v2 v1 θ O
23. A
particle moves from the point (2.0 $i + 4.0 $j ) m at t = 0 with an initial velocity (5.0 i$ + 4.0 $j) ms− 1. It is acted upon by a constant force which produces a constant acceleration ( 4.0 i$ + 4.0 $j) ms− 2. What is the distance of the particle from the origin at time 2 s? [JEE Main 2019, 11 Jan (ShiftII)] (a) 5 m
(b) 20 2 m (c) 10 2 m (d) 15 m
Exp. (b) Given, initial position of particle, r0 = (2 $i + 4 $j )m, Initial velocity of particle, u = (5 $i + 4 $j )m / s Aceleration of particle, a = (4 $i + 5 $j )m / s 2 According to second equation of motion, position of particle at time t is, 1 r = r0 + ut + at 2 2 At t = 2s, position of particle is, 1 r = (2 $i + 4 $j ) + (5 $i + 4 $j ) × 2 + (4 $i + 4 $j ) × 4 2 or r = (2 + 10 + 8) $i + (4 + 8 + 8) $j ⇒ r = 20 $i + 20 $j ∴ Distance of particle from origin is,r  = 20 2 m
24. A passenger train of length 60 m travels at a
v2 120° –v1
∆v
30° 30°
O
v1
From the figure, ∆v = v 2 − v1 θ  ∆ v = 2 v sin = 2 v sin 30º ⇒ 2 [Q v1  =  v 2 ] 1 = 2 v × = v (Given, v = 10 m / s) 2 ⇒  ∆v  = 10 m / s Alternate Solution Q ∆v = v 2 − v1 = v 2 + (− v1 ) ∴  ∆v 2 = v12 + v 22 + 2 v1v 2 cos 120º 1 = v 2 + v 2 + 2 v × v × − 2 ⇒  ∆ v  = v = 10 m / s.
speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when : (i) they are moving in the same direction and (ii) in the opposite direction is [JEE Main 2019, 12 Jan (ShiftI)] (a)
3 2
(b)
25 11
(c)
11 5
(d)
5 2
Exp. (c) When trains are moving in same direction relative speed = v1 − v 2 and in opposite direction relative speed = v1 + v 2 Hence, ratio of time when trains move in same direction with time when trains move in opposite direction is l + 1 v − t1 1 = t2 l1 + v + 1
v + v2 = 1 v1 − v 2 l2 v 2
l2 v2
19
Kinematics where, l1 + l2 = sum of lengths of trains which is same as distance covered by trains to cross each other t 1 80 + 30 110 11 So, = = = t2 80 − 30 50 5
25. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up. [JEE Main 2018]
one of the following graphs correctly represent the velocity vs time? [JEE Main 2017 (Offline)] v
v
(a)
(b)
t
t
Distance
Velocity (a)
27. A body is thrown vertically upwards. Which
Position
Time
(b)
(c)
v
(d)
t
v
t
Position
Velocity
Exp. (b) (c)
Time
(d)
Time
Exp. (b) If velocity versus time graph is a straight line with negative slope, then acceleration is constant and negative. With a negative slope distancetime graph will be 1 parabolic s = ut − at 2 . 2 So, option (b) will be incorrect.
26. C p and C V are specific heats at constant
pressure and constant volume, respectively. It is observed that C p − C V = a for hydrogen gas C p − C V = b for nitrogen gas. The correct relation between a andb is
Initially velocity keeps on decreasing at a constant rate, then it increases in negative direction with same rate.
28. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10m /s2 ) [JEE Main 2015] 240
(y2 – y1)m
(a)
[JEE Main 2017 (Offline)]
(a) a = b (c) a = 28 b
(b) a =14 b 1 (d) a = b 14
t 240
12
t(s)
(y2 – y1)m
(b)
Exp. (b) By Mayor’s relation, for 1 g mole of a gas, Cp − CV = R So, when n gram moles are given, R Cp − CV = n As per given question, R a = C p − C V = ; for H2 2 R b = C p − C V = ; for N2 28 From Eqs. (i) and (ii), we get a = 14b
8
8 12 240
t(s)
(y2 – y1)m
(c)
K (ii)
t(s)
12
K (i) 240
(y2 – y1)m
(d) 8
12
t(s)
20
JEE Main Chapterwise Physics Exp. (c)
Exp. (b) / Concept of relative motion can be applied to predict the nature of motion of one particle with respect to the other. Consider the stones 40 m/s thrown up simultaneously as shown in the diagram 10 m/s below. Considering motion of the 240 m second particle with respect to the first we have relative acceleration a 21 = a 2 − a 1 = g − g = 0 Thus, motion of first particle is straight line with respect to second particle till the first particle strikes ground at a time given by 1 − 240 = 10 t − × 10 × t 2 2 or t 2 − 2t − 48 = 0 or
Time taken to reach the maximum height u t1 = g If t 2 is the time taken to hit ground 1 − H = ut 2 − gt 2 2
i.e.,
t1 u H t2
t 2 = nt 1 nu 1 n2u 2 − H=u − g 2 g g2
But So,
−H=
t = 8, − 6
(not possible)
Thus, distance covered by second particle with respect to first particle in 8 s is s12 = (v 21 ) t = (40 − 10) (8s) = 30 × 8 = 240 m Similarly, time taken by second particle to strike the ground is given by 1 − 240 = 40t − × 10 × t 2 2 or − 240 = 40t − 5t 2 5t 2 − 40t − 240 = 0
or
t 2 − 8t − 48 = 0
or
nu 2 1 n2u 2 − 2 g g
2 gH = nu 2 (n − 2 )
t 2 − 8t + 6t − 48 = 0
or
[given]
t − 12 t + 4t − 48 = 0
30. A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical support. About the point of suspension, [JEE Main 2014] (a) angular momentum is conserved (b) angular momentum changes in magnitude but not in direction (c) angular momentum changes in direction but not in magnitude (d) angular momentum changes both in direction and magnitude
2
or t (t − 12 ) + 4 (t − 12 ) = 0 or (not possible) t = 12, − 4 Thus, after 8 s, magnitude of relative velocity will increase upto 12 s when second particle strikes the ground.
29. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is [JEE Main 2014] (a) 2 gH = n 2 u 2
Exp. (c) Angular momentum of the pendulum about the suspension point O is O
O r m
L Lrad axial
Vrad
Laxial O
v Vaxial
L
L
(b) gH = (n − 2 )2 u 2
(c) 2 gH = nu 2 (n − 2 ) (d) gH = (n − 2 )2 u 2
L = m(r×v)
21
Kinematics Then, v can be resolved into two components, radial component vrad and axial component vaxial . Due to vrad , will be axial and due to vaxial , L will be radially outwards as shown.
F0 mb
(a)
( i + 2 j ) m/s, where i is along the ground and j is along the vertical. If g = 10 m/s 2, the equation of its trajectory is [JEE Main 2013] (a) y = x − 5x 2 (c) 4y = 2 x − 5x 2
(b) y = 2 x − 5x 2 (d) 4y = 2 x − 25x 2
Exp. (b) As, initial velocity v = (i + 2 j) m/s For projectile Motion along horizontal direction, …(i) x=t Motion along vertical direction, 1 y = ut − gt 2 2 where, u is a vertical component of velocity. 1 …(ii) y = 2 t − 10t 2 2 From Eqs. (i) and (ii), we get y = 2 x − 5 x2
32. A boy can throw a stone upto a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone upto will be [AIEEE 2012] (a) 20 2 m (b) 10 m (c) 10 2 m
(d) 20 m
Exp. (d) Maximum speed with which the boy can throw stone is u = 2 gh = 2 × 10 × 10 = 10 2 m/s Range is maximum when projectile is thrown at an angle of 45°. Thus, u 2 (10 2 )2 Rmax = = = 20 m g 10
33. A particle of mass m is at rest at the origin at
time t = 0. It is subjected to a force F (t ) = F0e − bt in the xdirection. Its speedv (t ) is depicted by which of the following curves? [AIEEE 2012]
(b) v(t)
So, net angular momentum will be as shown in figure whose value will be constant ( L = mv ). But its direction will change as shown in the figure. L = m (r × v ) where, r = position vector of particle point O.
31. A projectile is given an initial velocity of
F0 mb v(t) t
(c)
t
F0 mb
F0 mb
(d) v(t)
v(t) t
t
Exp. (c) As the force is exponentially decreasing, so its acceleration, i.e., rate of increase of velocity will decrease with time. Thus, the graph of velocity will be an increasing curl with decreasing slope with time. F F dv ∴ = 0 e − bt = a= M m dt v t F − bt 0 ⇒ ∫0 dv = ∫0 m e dt t
⇒
F0 1 − bt F − bt 0 e = 0 [e ]t m − b 0 mb F F = 0 (e 0 − e − bt ) = 0 (1 − e − bt ) mb mb F = 0 mb
v=
vmax
34. A particle of mass m is projected with a velocity v making an angle of 30° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is [AIEEE 2011] 3 mv 2 2 g mv 3 (c) 2g
(a)
(b) zero (d)
3 mv 3 16 g
Exp. (d) Angular momentum of the projectile as given by L = mv h r⊥ = m(v cos θ) h v 30°
h
22
JEE Main Chapterwise Physics ⇒
where, h is the maximum height. v 2 sin2 θ = m (v cos θ) 2g L=
mv 3 sin2 θcos θ = 2g
⇒
wrapped around a pulley on a frictionless bearing. The pulley has massm and radiusR. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is [AIEEE 2011]
(b)
2 g 3
(c)
g 3
(d)
3 g 2
For the motion of the block,
(a) π
g
2
π v4 (b) 2 g2
(c) π
Exp. (a) Maximum range of water coming out of the fountain, Rm = v 2 / g
(c) xy = constant
(c) 8 s
(d) 1 s
…(i)
Exp. (d) Given,
…(ii)
(d) y 2 = x 2 + constant
⇒
…(iii)
v = kyi + kx j dy dx = ky, = kx dt dt dy dy dt kx = × = dx dt dx ky ydy = xdx
Integrating and using ∫ xn dx =
dv = − 2 .5 v dt dv = − 2 .5d t v −1/ 2
t
dv = − 2 .5 ∫ d t 0
xn + 1 , we have n+1
y2 = x 2 + c
39. For a particle in uniform circular motion the
acceleration at a point P (R , θ ) on the circle of radius R is (here, θ is measured from the [AIEEE 2010] xaxis) v2 v2 cos θ + sin θ R R v2 v2 (b) − sin θ + cos θ R R
(a) −
Exp. (a)
0
v2 g
(a) y = x 2 + constant (b) y 2 = x + constant
decelerated at a rate given by dv = − 2 . 5 v , wherev is the instantaneous dt speed. The time taken by the object to come to rest, would be [AIEEE 2011]
∫ 6. 25 v
(d) π
T
36. An object moving with a speed of 6.25 m/s, is
⇒
g
2
v = k ( y i + x j ), where k is a constant. The general equation for its path is [AIEEE 2010]
mg − T = ma For the rotation of the pulley, 1 τ = TR = Iα ⇒ T = mR 2 α 2 As string does not slip on the pulley a = Rα On solving Eqs. (i), (ii) and (iii),we get a = 2g /3
Given,
v2
T m
⇒
[AIEEE 2011]
v4
38. A particle is moving with velocity
R m
(b) 4 s
water all around it. If the speed of water coming out of the fountain isv, the total area around the fountain that gets wet is
∴ Total area around fountain, v4 2 = π 2 A = πR m g
Exp. (b)
(a) 2 s
t =2 s
37. A water fountain on the ground sprinkles
3 mv 3 16 g
35. A mass m hangs with the help of a string
(a) g
− 2 .5 [t ] 0t = [2 v1/ 2 ] 06. 25
v2 v2 cos θ − sin θ R R v2 v2 (d) + R R (c) −
23
Kinematics Exp. (c)
41. A point P moves in
For a particle in uniform circular motion, y ax ac
P (R, θ) ay
O
a =
40.
x
v2 towards centre of circle R [centripetal acceleration] 2
v (− cos θi − sinθ j) R
∴
a =
or
a =−
v2 v2 cos θ i − sinθ j R R
A small particle of mass m is projected at an angle θ with the xaxis with an intial velocityv 0 in the xy plane as shown in the figure. At a time v sin θ t< 0 , the angular momentum of the g [AIEEE 2010] particle is
y
counterclockwise B direction on a circular P (x, y) path as shown in the m figure. The movement 20 of P is such that it x sweeps out a length O A s = t 3 + 5, where s is in metre and t is in second. The radius of the path is 20 m. The acceleration of P when [AIEEE 2010] t = 2 s is nearly (a) 13 ms −2 (c) 7.2 ms −2
(b) 12 ms −2 (d) 14 ms −2
Exp. (d) s = t3 + 5
Given,
Speed, v =
∴
ds = 3t2 dt
and rate of change of speed, a t =
dv = 6t dt
∴ Tangential acceleration at t = 2 s, a t = 6 × 2 = 12 ms −2 and at t = 2 s, v = 3(2 )2 = 12 ms −1
y
∴ Centripetal acceleration, v 2 144 ms −2 ac = = R 20
v0
∴ θ
x
(a) − mg v 0 t cosθ (b) mg v 0 t cosθ 1 1 2 (c) − mg v 0 t cosθ (d) mg v 0 t 2 cosθ 2 2 2
Exp. (c) Angular momentum of the particle is, L = m (r × v ) 1 ⇒ L = m v 0 cos θt i + (v 0 sinθt − gt 2 ) j 2 × [v 0 cos θ i + (v 0 sinθ − gt ) j] 1 = mv 0 cosθt − gt k 2 1 = − mgv 0 t 2 cosθ k 2
Net acceleration =
a2t + a2c ≈ 14 ms −2
42. A particle has an initial velocity 3 + 4 and an acceleration of 0.4 + 0.3 . Its speed after 10 s is [AIEEE 2009] (b) 7 2 units (d) 8.5 units
(a) 10 units (c) 7 units
Exp. (b) Given,
u = 3i + 4 j; a = 0.4i + 0.3 j v = u + at = 3i + 4 j + (0.4i + 0.3 j) 10 = 3i + 4j + 4i + 3 j = 7i + 7 j
So, speed is equal to magnitude of velocity = 72 + 72
= 7 2 units
24
JEE Main Chapterwise Physics
43. Consider a rubber ball freely falling from a
height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then, the velocity as a function of time the height as function of time will be [AIEEE 2009] v v1
(a)
44. A body is at rest at x = 0. At t = 0, it starts moving in the positive xdirection with a constant acceleration. At the same instant, another body passes through x = 0 moving in the positive xdirection with a constant speed. The position of the first body is given by x1(t ) after time t and that of the second body by x 2(t ) after the same time interval. Which of the following graphs correctly describes ( x1 − x 2 ) as a function of time?
t
O
[AIEEE 2008] (x1 – x2)
(x1 – x2)
v
(b)
(a)
+v1
(b)
t
O
t
O
–v1
t
O
(x1 – x2)
(x1 – x2)
v
(c)
(c)
+v1 O t 1 2t1 3t1 4t1 –v1
t
(d)
Here, +v1 O
t
O
Exp. (b)
v
(d)
t
O
3t1 t1
∴
t
4t1
2t1
–v1
and
x1 =
From the above expression, it is clear that at t = 0, x1 − x2 = 0 further for increasing values, the graph is as follows.
Exp. (c) As we know that for vertical motion, 1 [parabolic] h = gt 2 2 v = − gt and after the collision, v = gt (straight line). Collision is perfectly elastic, then ball reaches to same height again and again with same velocity. y
v
3t1
t
–v1
Hence, option (c) is true.
(x1 – x2)
t
O
Hence, option (c) is true.
45. The velocity of a particle isv = v 0 + gt + ft 2. If
h
+v1 t1 2t1
at 2 2 at 2 x1 − x2 = − v t − 2
x2 = vt
its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is
[AIEEE 2007]
t
(a) v 0 + 2 g + 3 f (c) v 0 + g + f
g f + 2 3 g (d) v 0 + + f 2 (b) v 0 +
25
Kinematics Exp. (b)
48. A projectile can have the same range R for
v = v 0 + gt + ft 2
Given,
After differentiating with respect to time, we get dx = v 0 + gt + ft 2 dt ⇒
dx = (v 0 + gt + ft 2 ) dt 1
x
∫ 0 dx = ∫ 0
So,
(a) R 2
(v 0 + gt + ft ) dt
(b)
2
1
(c)
R2
Since, vertical component of velocity is zero.
y
horizontal with a kinetic energy K . The kinetic energy at the highest point is
u u
[AIEEE 2007]
(b) zero
(c)
K 4
(d)
K 2
Exp. (c)
θ
= K cos 2 θ = K(cos 60° )2 = K / 4
47. A particle located at x = 0 at time t = 0, starts moving along the positive xdirection with a velocity v that varies as v = α x . The displacement of the particle varies with time as [AIEEE 2006] (b) t
(c) t 1/ 2
(d) t 3
v =α x dx =α x dt dx = α dt x
or or
Q v = dx dt
On integrating, we get x dx t ∫ 0 x = ∫ 0 α dt [Qat t = 0, x = 0 and let at any time t, particle be at x] x
x1/ 2 α 1/ 2 = α t or x = t 2 1 / 2 0 x=
α2 × t 2 or x ∝ t 2 4
…(i)
…(ii)
From Eqs. (i) and (ii), we get 4u 2 sin θ cos θ , t1t 2 = g2 t1t 2 =
Given,
or
x
The time of flights are 2 u sin θ t1 = g 2 u sin (90° − θ) and t2 = g 2u cos θ = g
Exp. (a)
⇒
90° – θ
O
Kinetic energy at highest point, 1 (KE)H = mv 2 cos 2 θ 2
(a) t 2
(d) R
A projectile can have same range, if angles of projection are complementary of each other i.e., θ and (90° − θ). Thus, in both cases,
46. A particle is projected at 60° to the
(a) K
1 R
Exp. (d)
g f x = v0 + + 2 3
⇒
two angles of projection. If t 1 and t 2 are the times of flights in the two cases, then the product of the two times of flights is proportional to [AIEEE 2005, 04]
2 u 2 sin 2 θ g2 2R g
∴
t1t 2 =
or
t1t 2 ∝ R
=
2 u 2 sin 2 θ g g
u 2 sin 2 θ Q R = g
49. The relation between timet and distance x is
t = ax 2 + bx ,wherea and b are constants. The acceleration is [AIEEE 2005] (a) −2 abv 2 (b) 2 bv 3
(c) −2 av 3
Exp. (c) Given, t = ax2 + bx Differentiating it w.r.t. t, we get dx dx dt = 2 ax + b dt dt dt
(d) 2 av 2
26
JEE Main Chapterwise Physics v=
dx 1 = dt (2 ax + b )
Again, differentiating w.r.t. t, we get −2 a d2x dx ⋅ = dt 2 (2 ax + b )2 dt −1
d2x
2a f= 2 = ⋅ dt (2 ax + b )2 (2 ax + b ) −2 a f= (2 ax + b )3
∴ or ⇒
f = − 2 av 3
50. A car starting from rest, accelerates at the rate f through a distance s, then continues at constant speed for time t and then decelerates at the rate f /2 to come to rest. If the total distance travelled is 15 s, then [AIEEE 2005]
1 (b) s = f t 2 6
(a) s = f t (c) s =
1 ft2 2
(d) None of these
The velocitytime graph for the given situation can be drawn as below. Magnitudes of slope of OA = f y
A
s + (ft 1 )t + ft 12 = 15 s
or
s + ( ft 1 )t + 2 s = 15 s
Q s = 1 ft 2 1 2 …(ii)
or ( ft 1 )t = 12 s From Eqs. (i) and (ii), we have ( ft 1 ) t 12 s = 1 s (ft 1 ) t 1 2 t or t1 = 6 From Eqs. (i) and (ii), we get 1 s = f(t 1 )2 2 2
1 t f 2 6 1 2 ft = 72 Hence, none of the given options is correct. s=
⇒
51. A particle is moving Eastwards with a
Exp. (d)
v (m/s)
s1 + s2 + s3 = 15 s
Thus, ⇒
B
velocity of 5 ms −1. In 10 s, the velocity changes to 5 ms −1 Northwards. The average acceleration in this time is [AIEEE 2005] 1 ms−2 towards NorthEast 2 1 (b) ms−2 towards North 2 (c) zero 1 (d) ms−2 towards NorthWest 2 (a)
Exp. (d) C O
t1
t D t (s)
E
t2
x
f f and slope of BC = ⇒ v = f t 1 = t 2 2 2 ∴ t 2 = 2 t1 In graph, area of ∆OAD gives distance, 1 …(i) s = ft 12 2 Area of rectangle ABED gives distance travelled in time t. s2 = (ft 1 )t Distance travelled in time t 2 , 1 f s3 = (2 t 1 )2 2 2
As, velocity of eastward v1 = + 5i y
N ∆v
W
x
v2
v1
–v1
E
S
and velocity of northward v 2 = + 5 j ∆v = v − v1 = 5 j − 5i 2
∆v =
52 + 52 = 5 2
27
Kinematics a=
∴
∆v 5 2 1 ms −2 = = 10 2 t
s = ut +
1 gT 2 2 t=0
For direction, 5 = −1 5 1 Average acceleration is ms −2 towards 2 NorthWest. tan α = −
u=0 s
t=
T 3
h
52. A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 ms −2. He reaches the ground with a speed of 3 ms −1. At what height, did he bail out? [AIEEE 2005] (a) 91 m
(b) 182 m (c) 293 m
0 t=T Ground
or
(d) 111 m ⇒
Exp. (c) Parachute bails out at height H from ground. Velocity at A v = 2 gh = 2 × 9.8 × 50 =
980 ms −1
Velocity at ground, v1 = 3 ms −1 Acceleration = − 2 ms
[given] −2
[given]
h = 50 m H
1 h = 0 + gT 2 [Q u = 0] 2
A v = √2gh
T=
2 h g 2
1 T T s, s = 0 + g 2 3 3 1 T2 or s= g⋅ 2 9 2h g 2h ⇒ × s= QT = g 18 g h or s= m 9 Hence, the position of ball from the ground h 8h m =h− = 9 9 At t =
54. If A × B = B × A , then the angle between A and B is
Ground
H−h=
∴ ⇒
v 2 − v12 980 − 9 971 = = = 242 .75 2 ×2 4 4
H = 242 .75 + h = 242 .75 + 50 ≈ 293 m
53. A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball T in second? 3 (a) (b) (c) (d)
h / 9 m from the ground 7h / 9 m from the ground 8h / 9 m from the ground 17h /18 m from the ground
Exp. (c) Second law of motion gives
[AIEEE 2004]
(a) π (c) π/2
[AIEEE 2004]
(b) π/ 3 (d) π/4
Exp. (a) [given] (A × B) = (B × A) (A × B) − (B × A) = 0 (A × B) + (A × B) = 0 [Q( B × A ) = −(A × B)] 2(A × B) = 0 2 AB sin θ = 0 or sin θ = 0 [QA = A ≠ 0,B = B ≠ 0] ⇒ θ = 0 or π
As ⇒ or or ⇒
55. Which of the following statements is false for a particle moving in a circle with a constant angular speed? [AIEEE 2004] (a) The velocity vector is tangent to the circle (b) The acceleration vector is tangent to the circle
28
JEE Main Chapterwise Physics (c) The acceleration vector points to the centre of the circle (d) The velocity and acceleration vectors are perpendicular to each other
Exp. (b) For a particle moving in a circle with constant angular speed, velocity vector is always along the tangent to the circle and the acceleration vector always points towards the centre of circle or is always along radius of the circle. Since, tangential vector is perpendicular to radial vector, therefore velocity vector will be perpendicular to the acceleration vector. But in no case, acceleration vector is tangent to the circle.
56. An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e, 120 km/h, the stopping distance will be [AIEEE 2004]
(a) 20 m
(b) 40 m
(c) 60 m
(d) 80 m
Exp. (d) The braking retardation will remain same and assumed to be constant, let it be a. From third equation of motion, v 2 = u 2 + 2 as 2
Case I
5 0 = 60 × − 2 a × s1 18 (60 × 5/18) 2a
2
⇒
s1 =
Case II
5 0 = 120 × − 2 a × s2 18
Exp. (c) When a force of constant magnitude acts on velocity of particle perpendicularly, then there is no change in the kinetic energy of particle. Hence, kinetic energy remains constant.
58. A ball is thrown from a point with a speedv 0
at an angle of projection θ. From the same point and at the same instant, a person v starts running with a constant speed 0 to 2 catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection? [AIEEE 2004] (a) Yes, 60° (c) No
(b) Yes, 30° (d) Yes, 45°
Exp. (a) Man will catch the ball, if the horizontal component of velocity becomes equal to the constant speed of man, i.e, v v 0 cos θ = 0 2 1 or cos θ = 2 ⇒ cos θ = cos 60° or θ = 60°
59. A car moving with a speed of 50 km/h, can be stopped by brakes after atleast 6 m. If the same car is moving at a speed of 100 km/h, the minimum stopping distance is [AIEEE 2003]
2
⇒ ∴ ⇒
(120 × 5/18)2 2a s1 1 = s2 4 s2 =
s2 = 4s1 = 4 × 20 = 80 m
57. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane, it follows that [AIEEE 2004] (a) (b) (c) (d)
its velocity is constant its acceleration is constant its kinetic energy is constant it moves in a straight line
(a) 12 m (c) 24 m
(b) 18 m (d) 6 m
Exp. (c) As the first equation of motion v 2 = u 2 + 2 as 2
⇒
5 0 = 50 × + 2 a × 6 18
or
a = − 16 m/s 2
Again
[Q a is retardation]
v = u + 2 as 2
2
2
⇒
5 0 = 100 × − 16 × 2 × s 18
or
s=
or
s = 241 . ≈ 24 m
(100 × 5)2 18 × 18 × 32
29
Kinematics =
60. A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of 30° with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? 1 [g = 10 m/s 2, sin 30° = , cos 30° = 3/2] 2 [AIEEE 2003]
(a) 5.20 m (c) 2.60 m
(b) 4.33 m (d) 8.66 m
The ball will be at point P when it is at a height of 10 m from the ground. So, we have to find the distance OP, which can be calculated directly by considering it as a projectile on a levelled plane (OX ). 10 m/s
30°
P
62. Speeds of two identical cars areu and 4u at a specific instant. The ratio of the respective distances at which the two cars are stopped from that instant is [AIEEE 2002] (a) 1 :1
10 m
2
61. The coordinates of a moving particle at any
time t are given by x = αt 3 and y = βt 3. The speed of the particle at timet is given by (a) 3t α + β
(c) t
2
α +β 2
2
(b) 3t (d)
2
α +β 2
2
α + β2 2
Exp. (b) The given coordinate x = αt 3 , y = βt 3 dx dy Then, v x = = 3α t 2 and v y = = 3β t 2 dt dt ∴ Resultant velocity, v =
(d) 1 :16
⇒
0 = u 2 − 2 µg × s1 ⇒ s1 =
and
0 = (4u )2 − 2 µ g × s2
u2 2µ g
16u 2 = 16 s1 2µ g
⇒
s2 =
∴
s1 1 = s2 16
at an angle of 45° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be [AIEEE 2002] (a) E
v 2x + v y2
(b)
E 2
(c)
E 2
(d) zero
Exp. (c) At the highest point of its flight, vertical component of velocity is zero and only horizontal component is left which is Given,
u x = u cos θ θ = 45°
∴
u x = u cos 45° =
[AIEEE 2003] 2
(c) 1 : 8
63. A ball whose kinetic energy is E , is projected
10 × sin ( 2 × 30° ) 10 10 3 = = 5 3 = 8.66 m 2
2
(b) 1 : 4
In this question, the cars are identical means coefficient of friction between the tyre and the ground is same for both the cars, as a result retardation is same for both the cars equal to µg. Let first car travel distance s1, before stopping while second car travel distance s2 , then from v 2 = u 2 − 2 as
x
Ground u 2 sin 2 θ OP = R = g =
= 3t 2 α 2 + β 2
Exp. (d)
Exp. (d)
O
9 α 2 t 4 + 9β 2 t 4
u 2
Hence, at the highest point, kinetic energy is 1 E ′ = mu 2x 2 2 u 1 1 u2 = m = m 2 2 2 2 =
E 2
Q 1 mu 2 = E 2
30
JEE Main Chapterwise Physics
64. From a building, two balls A and B are
65. If a body losses half of its velocity on
thrown such that A is thrown upwards andB downwards (both vertically). Ifv A and v B are their respective velocities on reaching the ground, then [AIEEE 2002]
penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest? [AIEEE 2002]
(a) (b) (c) (d)
vB > vA vA = vB vA > vB their velocities depend on their masses
(a) 1 cm
Note In this question, it is not mentioned that magnitude of thrown velocity of both balls are same which is assumed in solution.
(c) 3 cm
(d) 4 cm
Exp. (a) Let initial velocity of body at point A be v, AB is 3 cm. v A
Exp. (b) From conservation of energy, Potential energy at height h = Kinetic energy at ground Therefore, at height h, PE of ball A = mA gh 1 KE at ground = mA v A2 2 1 So, mA gh = mA v A2 2 or v A = 2 gh Similarly, v B = 2 gh Therefore, vA = vB
(b) 2 cm
B 3 cm
v/2
C x
v 2 = u 2 − 2 as
From 2
2 v = v 2 − 2 a × 3 or a = v 2 8 Let on penetrating 3 cm in a wooden block, the body moves x distance from B to C. v So, for B to C, u = , v = 0, 2 v2 [deceleration] s = x, a = 8 2 v v2 ∴ (0)2 = − 2 ⋅ ⋅x 2 8 or x = 1cm
⇒
Note Here, it is assumed that retardation is uniform.
3 Laws of Motion 1. A ball is thrown vertically up (taken as + Zaxis) from the ground. The correct momentumheight (ph) diagram is p
p (a)
h
O
(b)
h
O
So, for the flight when the ball is thrown till is reaches the maximum height (h). v → changes from u to 0 ⇒ p → changes from mu to 0 Similarly, when it reacher it’s initial point, then h → changes from hmax to 0 Also, p → changes from 0 to some values. Thus, these conditions are only satisfied in the plot given in option (d).
2. Two blocks A and B of masses m A = 1 kg and p
p (c)
h
O
(d)
O
h
[JEE Main 2019, 9 April ShiftI]
m B = 3 kg are kept on the table as shown in figure. The coefficient of friction between A andB is 0.2 and betweenB and the surface of the table is also 0.2. The maximum force F that can be applied onB horizontally, so that the block A does not slide over the blockB is [Take, g = 10 m / s2]
Exp. (d)
[JEE Main 2019, 10 April ShiftII]
When a ball is thrown vertically upward, then the acceleration of the ball, a = acceleration due to gravity (g ) (acting in the downward direction). Now, using the equation of motion, v 2 = u 2 − 2 gh or
−v 2 + u 2 h= 2g
…(i)
As we know, momentum, p = mv or v = p / m So, substituting the value of v in Eq. (i), we get u 2 − ( p / m)
2
h=
2g
As we know that, at the maximum height, velocity of the ball thrown would be zero.
A B
(a) 12 N (c) 8 N
F
(b) 16 N (d) 40 N
Exp. (b) Acceleration a of system of blocks A and B is F − f1 Net force = a= Total mass mA + mB where, f1 = friction between B and the surface = µ(mA + mB )g
32
JEE Main Chapterwise Physics So, a=
Free body diagram of block is
F − µ(mA + mB )g (mA + mB )
N
…(i)
Here, µ = 02 . , mA = 1kg, mB = 3 kg, g = 10 ms −2 Substituting the above values in Eq. (i), we have F − 02 . (1 + 3) × 10 a= 1+ 3 F−8 …(ii) a= 4 Due to acceleration of block B, a pseudo force F′ acts on A. This force F′ is given by F ′ = mA a where, a is acceleration of A and B caused by net force acting on B. For A to slide over B; pseudo force on A, i.e. F′ must be greater than friction between A and B. ⇒ mA a ≥ f2 We consider limiting case, mA a = f2 ⇒ mA a = µ(mA )g …(iii) ⇒ a = µg = 02 . × 10 = 2 ms−2 Putting the value of a from Eq. (iii) into Eq. (ii), we get F−8 =2 4 ∴ F = 16 N
F cos 30° 30°
f=µN
F mg F sin 30°
In this case, normal reaction, 1 = 60 N 2 [Given, m = 5 kg, F = 20 N] Force of function, f = µN = 02 . × 60 [Qµ = 0.2 ] = 12 N So, net force causing acceleration (a1 ) is Fnet = ma1 = F cos 30° − f 3 ⇒ ma1 = 20 × − 12 2 10 3 − 12 ≈ 1 ms −2 a1 = ∴ 5 Case II Block is pulled over the surface N = mg + F sin30° = 5 × 10 + 20 ×
F
3. A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20 N, making an angle of 30° with the horizontal, as shown in the figures. The coefficient of friction between the block, the floor is µ = 0.2. The difference between the accelerations of the block, in case (B) and case (A) will be (Take, g = 10 ms−2)
30°
Free body diagram of block is, F sin 30° N F
[JEE Main 2019, 12 April ShiftII] 30°
F=20 N
30° 30°
F=20 N (A) −2
(a) 0.4 ms
f=µN
(B) −2
(b) 3.2 ms
(c) 0.8 ms−2 (d) 0 ms−2 mg
Exp. (c) Case I Block is pushed over surface 30° F
Net force causing acceleration is Fnet = F cos 30° − f = F cos 30° − µN ⇒ Fnet = F cos 30° − µ (mg − F sin 30° ) If acceleration is now a2 , then F a2 = net m
F cos 30°
33
Laws of Motion F cos 30° − µ (mg − F sin 30° ) m 1 3 . 5 × 10 − 20 × 20 × − 02 2 2 = 5 10 3 − 8 = 5 ⇒ a 2 ≈ 1.8 ms −2 =
So, difference = a2 − a1 = 1.8 − 1 = 0.8 ms −2
4. A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force F , such that the block does not move downward ? (Take, g = 10ms−2) [JEE Main 2019, 9 Jan ShiftI] F
Substituting the value of ‘R’ from Eq. (ii) to Eq. (i), we get 3 + Mg sinθ = F + µ( Mg cos θ) Here, M = 10 kg, θ = 45°, g = 10 m/s 2
...(iii)
and µ = 0.6 Substituting these values is Eq. (iii), we get 3 + (10 × 10 sin 45° ) − (0.6 × 10 × 10 cos 45° ) = F 100 60 ⇒ F= 3 + − 2 2 40 =3+ 2 = 3 + 20 2 = 31.8 N or F ~ − 32 N
5. A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the mass, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (Take, g = 10 ms −2) [JEE Main 2019, 9 Jan ShiftII]
10 3N
(a) 32 N
(a) 70 N (c) 100 N
kg
(b) 200 N (d) 140 N
Exp. (c) 45°
FBD of the given system is follow
(b) 25 N
(c) 23 N
(d) 18 N 45°
Exp. (a)
y
T
T cos 45° x
Free body diagram, for the given figure is as follows, F
R Mg sin θ
θ
f
Mg cos θ
3N θ=45º
Mg
For the block to be in equilibrium i.e., so that it does not move downward, then Σfx = 0 ∴ 3 + Mg sinθ − F − f = 0 or 3 + Mg sinθ = F + f As, frictional force, f = µR ...(i) ∴ 3 + Mg sin θ = F + µR Similarly, Σfy = 0 − Mg cosθ + R = 0 or ...(ii) Mg cos θ = R
F T sin 45° Mg = 10 × 10 N
Let T = tension in the rope. For equilibrium condition of the mass, Σ Fx = 0 (force in xdirection) Σ Fy = 0 ( force in ydirection) When Σ Fx = 0, then Q F = T sin45° When Σ Fy = 0, then Mg = T cos 45° Using Eqs. (i) and (ii), we get F T sin 45° ⇒ = Mg T cos 45° 1 F = 2 =1 ⇒ 1 Mg 2 ⇒ F = Mg = 10 × 10 = 100 N
…(i) …(ii)
34
JEE Main Chapterwise Physics
6. To mopclean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is µ, the torque applied by the machine on the mop in (N m) is [JEE Main 2019, 10 Jan ShiftI] 2 (a) µFR 3
µ FR (c) 3
µFR (b) 6
µ FR (d) 2
Exp. (a) Let a small strip of mop has width dx and radius x, as shown below,
F x
Mop
Torque applied to move this strip is dτ = Force on strip × Perpendicular distance from the axis ⇒dτ = Force per unit area × Area of strip × Perpendicular distance from the axis. µF = ⋅ 2 πxdx ⋅ x πR 2 2 µFx2 ⇒ ⋅ dx dτ = R2 So, total torque to be applied on the mop is x=R R 2 µFx2 τ=∫ dτ = ∫ ⋅ dx x=0 0 R2 2 µF R 3 2 = 2 × = µFR (N  m) 3 3 R
7. Two forces P andQ of magnitude 2F and 3F , respectively,are at an angle θ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle [JEE Main 2019, 10 Jan ShiftII] θ is
Exp. (b)
Fr
F2=Q θ
F2=Q α F1=P
Fr2
=
F12
+ F22 + 2 F1F2 cos θ
… (i)
In first case F1 = 2 F and F2 = 3 F ⇒ Fr2 = 4F 2 + 9F 2 + 2 × 2 × 3F 2 cos θ Fr2 = 13F 2 + 12 F 2 cos θ
⇒
… (ii)
In second case F1 = 2 F and F2 = 6 F (QForce Q gets doubled) and Fr′ = 2 Fr (Given) By putting these values in Eq. (i), we get (2 Fr )2 = (2 F )2 + (6F )2 + 2 × 2 × 6 F 2 cos θ 4Fr2 = 40F 2 + 24F 2 cos θ
… (iii)
From Eq. (ii) and Eq. (iii), we get; 52 F 2 + 48F 2 cos θ = 40F 2 + 24F 2 cos θ
R
(b)120°
F1=P
⇒
dx
(a) 60°
Resultant force Fr of any two forces F1 (i.e. P) and F2 (i.e. Q) with an angle θ between them can be given by vector addition as
(c) 30°
(d) 90°
⇒ or
12 + 24 cos θ = 0 or cos θ = − 1 / 2 (Qcos 120º = − 1 / 2 ) θ = 120º
8. A particle of mass m is moving in a straight line with momentump. Starting at timet = 0, a force F = kt acts in the same direction on the moving particle during time interval T , so that its momentum changes from p to 3p. Here, k is a constant. The value of T is [JEE Main 2019, 11 Jan ShiftII]
(a) (c)
2p k 2k p
p k k (d) 2 p (b) 2
Exp. (b) Here, F = kt When t = 0, linear momentum = p When t = T, linear momentum = 3p According to Newton’s second law of motion, dp applied force, F = dt or dp = F ⋅ dt or dp = kt ⋅ dt
35
Laws of Motion Now, integrate both side with proper limit 3p
T
t 2 3p ∫ dp = k∫ t dt or [ p]p = k 2 0 p 0 T
1 (3 p − p) = k [T 2 − 0] 2 4p T2 = k p T =2 k
or or or
9. A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is (Take, g = 10 m / s2) [JEE Main 2019, 12 Jan ShiftII] 10
N
Now, equating forces, we have mg sinθ + f = 10 N or mg sinθ + µmg cos θ = 10 Now, solving Eqs. (i) and (ii), we get mg sinθ = 4 and µ mg cos θ = 6 Dividing, Eqs. (iii) and (iv) we get 3 µ cot θ = 2 3 3 tanθ 3 tan 30° = ⇒µ= ⇒ µ= 2 2 2
connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m 2 to stop the motion is [JEE Main 2018] m
T
2N 30°
2 3
T
3 2
(b)
...(iii) ...(iv)
10. Two masses m1 = 5 kg and m 2 = 10 kg
m2
(a)
…(ii)
3 4
(c)
(d)
1 2
m1 m 1g
Exp. (b) Block does not move upto a maximum applied force of 2N down the inclined plane. f
N
2+
mg
si
nθ
θ mg
θ
mg cos θ
So, equating forces, we have 2 + mg sinθ = f or …(i) 2 + mg sinθ = µ mg cos θ Similarly, block also does not move upto a maximum applied force of 10 N up the plane. 10 N
N
(a) 18.3 kg (c) 43.3 kg
(b) 27.3 kg (d) 10.3 kg
Exp. (b) Motion stops when pull due to m1 ≤ force of friction between m and m 2 and surface. ⇒ m1g ≤ µ(m2 + m)g ⇒ 5 × 10 ≤ 015 . (10 + m) × 10 ⇒ m ≥ 23.33 kg Here, nearest value is 27.3 kg So, mmin = 27.3 kg
11. A particle is moving in a circular path of radius a under the action of an attractive k potentialU = − 2 . Its total energy is 2r [JEE Main 2018]
mg
sin
θ
θ
f
(a) − θ mg
mg cos θ
k 4a 2
Exp. (c)
(b)
k 2a 2
(c) zero
(d) −
3 k 2 a2
36
JEE Main Chapterwise Physics dU dr d −k k F = − 2 = − 3 ⇒ dr 2 r r As particle is on circular path, this force must be centripetal force. mv 2 F = ⇒ r k k mv 2 1 So, mv 2 = 2 = ⇒ r 2 r3 2r ∴ Total energy of particle = KE + PE k k = 2 − 2 =0 2r 2r Total energy = 0 ∴ Force = −
12. A body of mass m = 10−2kg is moving in a medium and experiences a frictional force F = − kv 2. Its initial speed is v 0 = 10 ms −1. If, after 10 s, its energy is 1 mv 02, the value of k 8 will be [JEE Main 2017 (Offline)] (a)10−3 kgs−1
(b)10−4 kgm −1
(c)10−1 kgm −1 s−1
(d)10−3 kgm −1
Exp. (b) Given, force, F = − kv 2 −k 2 v ∴ Acceleration, a = m dv − k 2 dv k or = v ⇒ 2 = − . dt dt m m v Now, with limits, we have v dv k ∫10 v 2 = − m
t
∫0 dt
v
kt − 1 = − k t ⇒ 1 = 01 . + v 10 m v m 1 1 = v= ⇒ kt 01 . + 1000k . + 01 m v 1 1 × m × v 2 = m v 20 ⇒ v = 0 = 5 ⇒ 2 2 8 1 =5 ⇒ 01 . + 1000 k ⇒
⇒ ⇒ ⇒
1 = 0.5 + 5000 k 0.5 k= 5000 k = 10−4 kg/m
13. A point particle of
P
mass m, moves along the h=2m uniformly rough R 30º track as PQR O Q shown in the Horizontal surface figure. The coefficient of friction between the particle and the rough track equals µ . The particle is released, from rest , from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ andQR , of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(= QR), are respectively close to [JEE Main 2016 (Offline)] (a) 0.2 and 6.5 m (c) 0.29 and 3.5 m
(b) 0.2 and 3.5 m (d) 0.29 and 6.5 m
Exp. (c) Energy lost over path PQ = µ mg cos θ × 4 P h=2m
O
4m 30º 2 3 mQ
x
R
Energy lost over path QR = µ mg x i.e. µ mg cos 30° × 4 = µ mg x (Qθ = 30° ) x = 2 3 = 3.45 m From Q to R energy loss is half of the total energy loss. 1 i.e. µ mg x = × mgh ⇒ µ = 0.29 2 The values of the coefficient of friction µ and the distance x(= QR ) are 0.29 and 3.5.
14. Given in the figure are two blocks A and B of F A B weight 20 N and 100 N respectively. These are 20N 100N being pressed against a wall by a force F as shown in figure. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15,
37
Laws of Motion the frictional force applied by the wall in
⇒
block B is
Now, putting the value of x in y = x3 / 6, we get
(a) 100 N
[JEE Main 2015]
(b) 80 N
(c) 120 N
(d) 150 N
Exp. (c) / In vertical direction, weights are balanced by frictional forces. Consider FBD of block A and B as shown in diagram below. fB
fA
20 N
fA 100 N
As the blocks are in equilibrium, balance forces are in horizontal and vertical direction. For the system of blocks ( A + B). F=N
15. A block of mass m is placed on a surface with
a vertical crosssection given by y = x 3 / 6. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is 2 m 3
(c)
1 m 3
(d)
pushing a body up a rough (frictional coefficient µ) inclined plane is F1 while the minimum force needed to prevent it from sliding down is F2. If the inclined plane makes an angle θ from the horizontal such F that tanθ = 2 µ , then the ratio 1 is F2 [AIEEE 2011]
(b) 1
(c) 2
(d) 3
Exp. (d)
fB = fA + 100 = 120N
(b)
So, the maximum height above the ground at which the block can be placed without slipping is 1 m. 6
(a) 4
For block A, fA = 20 N and for block B,
1 (a) m 6
When x = −1 (−1)3 −1 y= = 6 6
When x = 1 (1)3 1 y= = ∴ 6 6
16. The minimum force required to start N
F
x2 = 1 ⇒ x = ± 1
F1 = mg (sinθ + µ cos θ) [as body just in position to move up, friction force downward] F2 = mg (sinθ − µ cos θ) [as body just in position to slide down, friction upward]
m
1 m 2
F1
[JEE Main 2014] s mg
Exp. (a) A block of mass m is placed on a surface with a vertical crosssection, then
in
θ+
µm
gc
o
sθ
θ
F1 sinθ + µ cos θ tanθ + µ 2 µ + µ = =3 = = F2 sinθ − µ cos θ tanθ − µ 2 µ − µ
∴
y
17. The figure shows the positiontime (xt)
m
graph of onedimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is [AIEEE 2010]
y θ
x
x3 d 6 x2 dy = tan = = dx dx 2 At limiting equilibrium, we get µ = tanθ , 0.5 = x2 / 2
2 x (m) 0
2
4
6
8 t (s)
10 12 14 16
38
JEE Main Chapterwise Physics (a) 0.4 N s (c) 1.6 Ns
(b) 0.8 Ns (d) 0.2 Ns
Exp. (b) From the graph, it is a straight line, so uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph. 2 Initial velocity, v1 = = 1ms −1 2 2 Final velocity, v 2 = − = − 1ms −1 2 pi = mv1 = 0.4 Ns pf = mv 2 = − 0.4 Ns J = pf − pi = − 0.4 − 0.4 = − 0.8 Ns [J = impulse] ∴ J = 0.8 Ns
19. A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then, a constant force F starts acting on the block of mass M to pull it. Find the force on the block of mass m. [AIEEE 2007] (b)
Exp. ( c ) Acceleration of system, F a= m+ M k
18. Two fixed frictionless inclined plane making the angles 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B? [AIEEE 2010]
( M + m) F m MF (d) (m + M )
mF M mF (c) (m + M ) (a)
m
M
F
So, force acting on mass, mF F = ma = m+ M
20. A mass of M kg is suspended by a weightless
A
string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is
B
[AIEEE 2006]
60°
(a) (b) (c) (d)
30°
4.9 ms−2 in horizontal direction 9.8 ms−2 in vertical direction zero 4.9 ms−2 in vertical direction
Exp. (d) Force applying on the block F = mg sinθ or mg sinθ = ma ∴ a = g sinθ where, a is along the inclined plane. ∴ Vertical component of acceleration is g sin2 θ. ∴ Relative vertical acceleration of A with respect to B is g g (sin2 60° − sin2 30° ] = = 4.9 ms −2 2 [in vertical direction]
(a) Mg ( 2 + 1 ) Mg (c) 2
(b) Mg 2 (d) Mg ( 2 − 1 )
Exp. (d) Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculated by using workenergy theorem. Let us assume that body be taken slowly, so that its speed does not change, then
45°
l
2 1 M Mg
F
39
Laws of Motion ∆ K = 0 = WF + WMg + Wtension [symbols have their usual meanings] Fl WF = F × l sin 45° = 2 WMg = Mg (l − l cos 45° ), Wtension = 0 F = Mg ( 2 − 1)
∴
21. A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s 2. [AIEEE 2006] (a) 4 N (c) 20 N
(b) 16 N (d) 22 N
Exp. (d)
Then, for PA, v = 0 + 2 a × 0.2 2
2
This is the question based on impulsemomentum theorem. F ⋅ ∆t =  Change in momentum ⇒ F × 01 . =  pf − pi As the ball will stop after catching pi = mv i = 015 . × 20 = 3, pf = 0 ⇒ F × 01 . =3 ⇒ F = 30 N
2m
surface with angle of inclination α. The incline is given an acceleration a to keep the block stationary. Then, a is equal to
a
P
(a)
g tan α
(b) g cosec (d) g tan α
(c) g
F
From above equations,
Exp. (d)
a = 10g = 100 ms −2
⇒
Exp. (c)
A
0= v −2 × g ×2
Then, for PA,FBD of ball is F − mg = ma [F is the force exerted by hand on ball]
(b) 3 N (d) 300 N
[AIEEE 2005]
2
v2 = 2 g × 2
(a) 150 N (c) 30 N
23. A block is kept on a frictionless inclined
B
For AB, ⇒
moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to [AIEEE 2006]
0.2 m
The situation is shown in figure. At initial time, the ball is at P, then under the action of a force (exerted by hand) from P to A and then from A to B let acceleration of ball during PA be a ms −2 (assumed to be constant) in upward direction and velocity of ball at be is v m/s.
22. A player caught a cricket ball of mass 150 g
a = 10g
mg
F = m(g + a) = 0.2 (11 g) = 22 N
Alternate Solution Using workenergy theorem Wmg + WF = 0 ⇒ − mg × 2 .2 + F × 0.2 = 0 or F = 22 N
In the frame of wedge, the force diagram of block is shown in figure. From free body diagram of wedge, N
ma cos α ma
α α
mg cos α + ma sin α
mg sin α mg
α
a
40
JEE Main Chapterwise Physics situated on the inner and outer parts of the ring, F1 / F2 is [AIEEE 2005]
For block to remain stationary, ma cos α = mg sin α or a = g tan α
(a)
24. Consider a car moving on a straight road with a speed of 100 ms −1. The distance at which car can be stopped, is[ µ k = 0.5]
(a) 800 m (c) 100 m
(b) 1000 m (d) 400 m
[AIEEE 2005]
Exp. (b)
R (b) 1 R2
R2 R1
2
(c) 1
Since, ω is constant, so no net force or torque is acting on ring. The force experienced by any particle is only along radial direction or we can say the centripetal force.
F2
Given, v = 0 [car is stopped] As friction provide the retardation a = µg , v = 100 ms −1 ⇒
R2
R1
F1 v
ω
(100)2 = 2 µ gs 100 × 100 s= 2 × 0.5 × 10 =
100 × 100 = 1000 m 5×2
25. A particle of mass 0.3 kg is subjected to a
force F = − kx with k = 15 Nm −1. What will be its initial acceleration, if it is released from a point 20 cm away from the origin? [AIEEE 2005]
(a) 3 ms −2 (c) 5 ms −2
(b) 15 ms −2 (d) 10 ms −2
The force experienced by inner part, F1 = mω2 R1 and the force experienced by outer part, F R F2 = mω2 R 2 ⇒ 1 = 1 F2 R 2
27. A smooth block is released at rest on a 45° incline and then slides a distanced.The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient of friction is [AIEEE 2005] (a) µ k = 1 −
Exp. (d) Given,
R1 R2
Exp. (d)
From Newton’s equations, we have v 2 = u 2 − 2 as
∴
(d)
m = 0.3 kg,
x = 20 cm
k = 15 N / m …(i) F = − kx …(ii) F = ma ma = − kx 15 a=− × 20 × 10−2 ⇒ 0.3 15 a=− × 2 = − 10 ms −2 ⇒ 3 Negative sign indicates that acceleration is always towards the mean position. ∴ Initial acceleration, a = 10 ms −2 and given and ∴
26. An annular ring with inner and outer radii R1 and R 2 is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles
(c) µ s = 1 −
1
(b) µ k = 1 −
n2 1
(d) µ s = 1 −
n2
1 n2 1 n2
Exp. (a) When friction is absent ma1 = mg sinθ a = g sin θ 1 ∴ s1 = a1 t 12 2 m
mg cos θ θ = 45°
θ mg sin θ
…(i)
41
Laws of Motion When friction is present, friction is in opposite to the direction of motion a2 = g sin θ − µ k g cos θ 1 s2 = a 2 t 22 2
∴
From Eqs. (i) and (ii), we get 1 1 a1t 12 = a2 t 22 2 2 ⇒ a1 t 12 = a2 (nt 1 )2
…(ii)
Force exerted by machine gun on man’s hand in firing a bullet = Change in momentum per second on a bullet or rate of change of momentum 40 = × 1200 = 48 N 1000 Force exerted by man on machine gun = Force exerted on man by machine gun = 144 N
[Q t 2 = nt 1]
or
a1 = n a2
⇒
a2 g sin θ − µ k g cos θ 1 = = 2 a1 g sin θ n
or
g sin 45° − µ k g cos 45° 1 = 2 g sin 45° n
Hence, number of bullets fired =
2
1 − µk =
or
30. Two masses m1 = 5 kg and m 2 = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift is free to move ? (g = ms −2) [AIEEE 2004]
1 n2
µk = 1 −
or
144 = 3. 48
1 n2
28. The upper half of an inclined plane with
inclination φ is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by [AIEEE 2005] (a) 2 sin φ (b) 2 cos φ (c) 2 tan φ (d) tan φ
(a) 0.2 ms −2 (c) 5 ms −2
Exp. (c)
Exp. (d)
According to workenergy theorem, Work done = Change in kinetic energy W = ∆K = 0 ⇒ Work done by friction + Work done by gravity =0 l ⇒ −( µ mg cos φ) + mgl sin φ = 0 2 µ or cos φ = sin φ 2 or µ = 2 tan φ
29. A machine gun fires a bullet of mass 40 g −1
with a velocity 1200 ms . The man holding it, can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most? [AIEEE 2004] (a) 1
Exp. (d)
(b) 4
(c) 2
(b) 9.8 ms −2 (d) 4.8 ms −2
On releasing, the motion of the system will be according to figure. …(i) m1g − T = m1a and T − m2 g = m2 a m − m2 On solving, a = 1 g m1 + m2
a
T
T
m1 g
(d) 3 m2 g
a
…(ii) …(iii)
42
JEE Main Chapterwise Physics Here, m1 = 5 kg,
m2 = 4.8 kg, g = 9.8 ms −2
5 − 4.8 a= × 9.8 5 + 4.8
∴
0.2 = × 9.8 = 0.2 ms −2 9.8
31. A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is ( g = 10 m/s 2) [AIEEE 2004] (a) 2.0 (c) 1.6
(b) 4.0 (d) 2.5
49 = 5 kg 9.8 When lift moves downward, mg − T = ma Reading of balance, T = mg − ma = 5(9.8 − 5) = 5 × 4.8 = 24.0 N m=
or
T a
mg
33. Three forces start acting simultaneously on a particle moving with velocity v. These forces are represented in magnitude and direction by the three sides of a ∆ABC (as shown). The particle will now move with velocity
Exp. (a)
[AIEEE 2003] C
Let mass of the block be m. R F
mg sin 30° 30°
mg
Frictional force in rest position F = mg sin 30° [this is static frictional force and may be ∴ or
less than the limiting frictional force] 1 10 = m × 10 × 2 2 × 10 m= = 2 kg 10
32. A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 ms −2, the reading of the spring balance will be [AIEEE 2003] (a) (b) (c) (d)
B
A
mg cos 30°
(a) (b) (c) (d)
less than v greater than v  v  in the direction of largest force BC v, remaining unchanged
Exp. (d) Resultant force is zero, as three forces acting on the particle can be represented in magnitude and direction by three sides of a triangle in same order. dv Hence, by Newton’s 2nd law F = m , the dt velocity (v ) of particle will be same.
34. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is
[AIEEE 2003]
24 N 74 N 15 N 49 N
F 10 N
Exp. (a) In stationary position, spring balance reading = mg = 49
(a) 20 N
(b) 50 N
(c) 100 N
(d) 2 N
43
Laws of Motion Exp. (d)
37. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is [AIEEE 2003]
f 10 N
R
(a) w
Let R be the normal contact force by wall on the block. R = 10 N f L = w and fL = µ R µR = w or w = 0.2 × 10 = 2 N
∴
Pm M+m
Pm M −m PM (d) M+m (b)
(c) P
Exp. (d) Let acceleration of system (rope + block) be a along the direction of applied force. Then, P a= M+ m
35. A marble block of mass 2 kg lying on ice
when given a velocity of 6 ms −1 is stopped by friction in 10 s. Then, the coefficient of friction is [AIEEE 2003] (a) 0.02
(b) 0.03
(c) 0.06
M
P Frictionless surface
(d) 0.01
Exp. (c) Let coefficient of friction be µ, then retardation will be µ g. From equation of motion, v = u + at ⇒ 0 = 6 − µ g × 10 6 or µ= = 0.06 100
36. Consider the following two statements. I. Linear momentum of a system of particles is zero. II. Kinetic energy of a system of particles is zero. Then, [AIEEE 2003] (a) (b) (c) (d)
a m
I does not imply II and II does not imply I I implies II but II does not imply I I does not imply II but II implies I I implies II and II implies I
Exp. (c) Here, II is implying I but I is not implying II as kinetic energy of a system of particles is zero means speed of each and every particle is zero which says that momentum of every particle is zero. But statement I means linear momentum of a system of particles is zero, which may be true even, if particles have equal and opposite momentums and hence having nonzero kinetic energy.
Draw the FBD of block and rope as shown in figure. a
a m
M
T T
P
where, T is the required parameter. For block, T = Ma MP T= ⇒ M+ m
38. A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then, the true statement about the scale reading is [AIEEE 2003] (a) Both the scales read M kg each (b) The scale of the lower one reads M kg and of the upper one zero (c) The reading of the two scales can be anything but the sum of the readings will be M kg (d) Both the scales read M/2 kg
Exp. (a) The arrangement is shown in figure. Now, draw the free body diagram of the spring balances and block.
44
JEE Main Chapterwise Physics For equilibrium of block, T1 = Mg where, T1 = Reading of S 2 For equilibrium of S 2 , T2 = T1
40. A lift is moving down with acceleration a . A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively [AIEEE 2002]
(b) g − a , g − a (d) a , g
(a) g , g (c) g − a , g
Exp. (c) S1
Light spring balance
S2
Light spring balance
Apparent weight of ball, w ′ = w − R [where, R=normal reaction] R = ma [acting upward] w ′ = mg − ma = m(g − a) Hence, apparent acceleration in the lift is g − a. Now, if the man is standing stationary on the ground, then the apparent acceleration of the falling ball is g.
41. Two forces are such that the sum of their
M
magnitudes is 18 N and their resultant which has magnitude 12 N, is perpendicular to the smaller force. Then, the magnitudes of the forces are [AIEEE 2002]
where, T2 = Reading of S1 For equilibrium of S1, T2 = T3 T3
T2 T1
S1
S2
(b) 13 N, 5 N (d) 16 N, 2 N
Exp. (b)
Block Mg
T2
(a) 12 N, 6 N (c) 10 N, 8 N
T1
The sum of the two forces A + B = 18 12 =
Hence, T1 = T2 = Mg So, both scales read M kg.
39. A rocket with a liftoff mass 3.5 × 104 kg is blasted upwards with an initial acceleration of 10 ms −2. Then, the initial thrust of the blast is [AIEEE 2003] (a) 3.5 × 105 N (b) 7.0 ×105 N
⇒ ⇒
(d) 1.75 ×105 N
Exp. (a) Here, thrust force is responsible to accelerate the rocket, so initial thrust of the blast = ma = 3.5 × 104 × 10 = 3.5 × 105 N
…(i) …(ii)
…(iii)
Solving Eqs. (i), (ii) and (iii), we get A = 5 N, B = 13 N A B Squaring both sides in Eq. (i), we get 144 = A 2 + B2 + 2 ABcosθ −A On putting cosθ = in above equation, we get B A 144 = A 2 + B2 + 2 AB − B
Since,
(c) 14.0 ×105 N
A 2 + B2 + 2 AB cos θ B sin θ tan α = A + B cos θ B sin θ tan 90° = A + B cos θ −A cos θ = B
A + B = 18 and cosθ = −
= A 2 + B2 − 2 A 2
45
Laws of Motion ⇒
144 = B2 − A 2
44. A light string passing over a smooth light
⇒ 144 = (B − A) ⋅ (B + A) On putting B + A = 18, we have 144 B− A= =8 18 Solving B− A= 8 B + A = 18 We have, B = 13 N, A = 5 N
pulley connects two blocks of masses m 1 and m 2 (vertically). If the acceleration of the system is g/8, then the ratio of the masses is [AIEEE 2002] (a) 8 :1 (c) 4 : 3
Exp. ( b )
42. When forces F1 , F2 , F3 are acting on a particle of mass m such that F2 and F3 are mutually perpendicular, then the particle remains stationary. If the force F1 is now removed, then the acceleration of the particle is [AIEEE 2002]
F2 F3 mF1 F2 (d) m
F1 m ( F2 − F3 ) (c) m (a)
(b)
Exp. ( a ) The particle remains stationary under the acting of three forces F1, F2 and F3 , it means resultant force is zero, F1 + F2 + F3 = 0 Since, in second cases F1 is removed (in terms of magnitude we are talking now), the forces acting are F2 and F3 the resultant of which has the F magnitude as F1, so acceleration of particle is 1 in m the direction opposite to that of F1.
43. The minimum velocity (in ms −1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is [AIEEE 2002] (a) 60
(b) 30
(c) 15
(d) 25
Exp. (b)
mv 2 = µ R, r
⇒ or
As the string is inextensible, both masses have the same acceleration a. Also, the pulley is massless and frictionless, hence the tension at both ends of the string is the same. Suppose, the mass m2 is a greater than mass m1, so the heavier mass is m2 accelerating downward and the lighter mass m1 is accelerating upwards. Therefore, by Newton’s 2nd law T − m1g = m1a and
m1 g
m 2g
…(i)
m2 g − T = m2 a
…(ii)
After solving Eqs. (i) and (ii), we get (m − m1 ) g ⋅g = a= 2 8 (m1 + m2 ) So, Let
a
T
T
[given]
g m2 (1 − m1/m2 ) = ⋅g 8 m2 (1 + m1/m2 )
…(iii)
m1 = x m2
Thus, Eq. (iii) becomes m2 9 1− x 1 7 or = = or x = 9 m1 7 1+ x 8 So, the ratio of the masses is 9 : 7.
As centrifugal force is balanced by the centripetal force i.e., frictional force. Using the relation
⇒
(b) 9 : 7 (d) 5 : 3
R = mg
mv 2 = µ mg r
or
45. Three identical blocks of masses m = 2 kg are drawn by a force F = 10.2 N with an acceleration of 0.6 ms −2 on a frictionless surface, then what is the tension (in N) in the string between the blocks B and C ? [AIEEE 2002]
v 2 = µ rg
F
v 2 = 0.6 × 150 × 10 v = 30 ms −1
(a) 9.2
(b) 7.8
(c) 4
(d) 9.8
46
JEE Main Chapterwise Physics The free body diagram of the person can be drawn as
Exp. (b) The system of masses is shown below. C
T2
B
T1
T A
a
F
Person
From the figure, and Eq. (i) gives,
F − T1 = ma
…(i)
T1 − T2 = ma
…(ii)
10.2 − T1 = 2 × 0.6 ⇒ T1 = 10.2 − 1.2 = 9 N Again, from Eq. (ii), we get 9 − T2 = 2 × 0.6 ⇒ T2 = 9 − 1.2 = 7.8 N
46. One end of massless rope, which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that the rope can bear is 360 N. With what value of maximum safe acceleration (in ms −2) can a man of 60 kg climb on the rope? [AIEEE 2002]
60 g
Let the person move up with an acceleration a, then T − 60g = 60a T − 60g a max = max ⇒ 60 360 − 60g or a max = → − ve value 60 That means, it is not possible to climb up on the rope. Even in this problem, it is not possible to remain at rest on rope. Hence, no option is correct. But, if they will ask for the acceleration of climbing down, then T a
P
Person
C 60 g
(a) 16
Exp. (c)
(b) 6
(c) 4
(d) 80
⇒ or
60g − T = 60 a 60g − Tmax = 60 a min a min =
60g − 360 = 4 ms −2 60
4 Work, Energy and Power 1. A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled 3 m is [JEE Main 2019, 8 April ShiftI] 3
Force (in N)
2
2. A body of mass m1 moving with an unknown
1
1
(a) 4 J
∴ Work done on the particle = Area under the curve ABC W = Area of square ABFO + Area of ∆BCD + Area of rectangle BDEF 1 = 2 × 2 + × 1 × 1 + 2 × 1 = 6.5 J 2 Now, from workenergy theorem, ∆W = K f − K i [QK i = 0] ⇒ K f = ∆W = 6.5 J
(b) 2.5 J
2 Distance (in m)
3
(c) 6.5 J
(d) 5 J
velocity of v1$i, undergoes a collinear collision with a body of mass m 2 moving with a velocityv 2$i. After collision, m1 and m 2 move with velocities of v 3$i and v 4 $i , respectively. If m2 = 0. 5m1 and v3 = 0. 5 v1, then v1 is [JEE Main 2019, 8 April ShiftII]
Exp. (c) Key Idea Area under forcedisplacement graph gives the value of work done.
(a) v 4 + v 2 (c) v 4 −
F(N)
2
B D 2
0 O
E
F 2
3
(d) v 4 − v 2
Key Idea Total linear momentum is conserved in all collisions, i.e. the initial momentum of the system is equal to final momentum of the system.
1 A
v2 4
Exp. (d)
C
3
v2 2
(b) v 4 −
Given, x(m)
m2 = 0.5m1 ⇒ m1 = 2 m2
48
JEE Main Chapterwise Physics Let m2 = m, then, m1 = 2 m Also, v 3 = 0.5v1 Given situation of collinear collision is as shown below Before collision, 2m
m
v1
v2
After collision, 2m
m v3
v4
∴According to the conservation of linear momentum, Initial momentum = Final momentum m1v1$i + m2 v 2 $i = m1v 3 $i + m2 v 4 $i ⇒ 2 mv1$i + mv 2 $i = 2 m(0.5v1 )$i + mv 4 $i ⇒ v 4 = v1 + v 2 ⇒ v1 = v 4 − v 2
Using value from Eq. (ii) into Eq. (i), we get 5v mv mv = + M 4 4 5v v m v − = M ⇒ 4 4 3 5 mv = Mv ⇒ 4 4 3 3 M = m = × 2 = 12 . kg 5 5
4. A uniform cable of mass M and length L is placed on a horizontal surface such that its 1 th part is hanging below the edge of the n surface. To lift the hanging part of the cable upto the surface, the work done should be [JEE Main 2019, 9 April ShiftI]
3. A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with onefourth of its original speed. What is the mass of the second body ? [JEE Main 2019, 9 April ShiftI]
(a) 1.5 kg
(b) 1.2 kg
(c) 1.8 kg
(a)
2MgL n2
(b) nMgL
(c)
MgL n2
(d)
MgL 2n 2
Exp. (d) (L – L /n)
(d) 1.0 kg
Exp. (b) L /n
Key Idea For an elastic collision, coefficient of restitution (e), i.e. the ratio of relative velocity of separation after collision to the relative velocity of approach before collision is 1.
Given, mass of small body, m =2 kg Given situation is as shown At rest v m M 14243 Before collision
v/4 v′ M m 14243 After collision
Using momentum conservation law for the given system, (Total momentum)before collision = (Total momentum)after collision v …(i) ⇒ m(v ) + M (0) = m + M(v ′) 4 Q
⇒ 1= − or
e = 1and we know that, v − v1 e=− 2 u 2 − u1
v′ − v / 4 ⇒ v = v′ − v / 4 0−v v ′ = 5v / 4
…(ii)
Given, mass of the cable is M. 1 So, mass of th part of the cable, i.e. hanged n part of the cable is = M / n …(i) Now, centre of mass of the hanged part will be its middle point. L /2n So, its distance from the top of the table will be L / 2 n. L /n ∴Initial potential energy of the hanged part of cable, L M U i = (− g ) 2n n MgL …(ii) ⇒ Ui = − 2 n2 When whole cable is on the table, its potential energy will be zero. …(iii) ∴ Uf = 0
49
Work, Energy and Power M
Now, using workenergy theorem, Wnet = ∆U = U f − U i MgL Wnet = 0 − − ⇒ 2 n2 [using Eqs. (ii) and (iii)] MgL Wnet = ⇒ 2 n2
The speed of each of the moving particle will be [JEE Main 2019, 9 April ShiftII] (a) (c)
(b)
2v v
v 2
30°
30°
M
6.5 m/s and 3.2 m/s 3.2 m/s and 6.3 m/s 3.2 m/s and 12.6 m/s 6.5 m/s and 6.3 m/s
Exp. (d) The given condition can be drawn as shown below v1 sin 30° 10 sin 30°
30° 45°
5 m/s 2M m
v′ 45°
m v f =0
m
45° v′
As we know that, in collision, linear momentum is conserved in both x and y directions separately. So, ( px )initial = ( px )final m(2 v ) + 2 m(v ) = 0 + mv ′cos 45º + mv ′cos 45º 2m v′ 4mv = ⇒ 2 ⇒ v′ = 2 2v
6. Two particles of masses M and 2 M , moving as shown, with speeds of 10 m/s and 5 m/s, collide elastically at the origin. After the collision, they move along the indicated directions with speedv1 andv 2 are nearly [JEE Main 2019, 10 April ShiftI]
2M
v1
10 m/s
5 cos 45° 10 cos 30°
Final condition,
v2
2M
(a) (b) (c) (d)
Exp. (d) According to the questions, Initial condition,
45°
5m/s
M
(d) 2 2 v
(2 2 )
v1
45°
5. A particle of massm is moving with speed 2v and collides with a mass 2m moving with speedv in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass m , which move at angle 45º with respect to the original direction.
2M 10 m/s
30° 45° v1 cos 30° v2 cos 45°
5 sin 45° v2 sin 45°
v2 M
Applying linear momentum conservation law in xdirection, we get Initial momentum = Final momentum (M × 10 cos 30° ) + (2 M × 5 cos 45° ) = (M × v 2 cos 45° ) + (2 M × v1 cos 30° ) 1 3 ⇒ + 2 M × 5 × M × 10 × 2 2 3 1 = M × v 2 × + 2 M × v1 × 2 2 v … (i) 5 3 + 5 2 = 2 + v1 3 ⇒ 2 Similarly, applying linear momentum conservation law in ydirection, we get (M × 10 sin 30° ) − (2 M × 5 sin 45° ) = (M × v 2 sin 45° ) − (2 M × v1 sin 30° ) M × 10 × 1 − 2 M × 5 × 1 ⇒ 2 2 1 1 = M × v 2 × − 2 M × v1 × 2 2
50
JEE Main Chapterwise Physics v2 − v1 2 Subtracting Eq. (ii) from Eq. (i), we get (5 3 + 5 2 ) − (5 − 5 2 ) v v = 2 + v1 3 − 2 − v1 2 2 5−5 2 =
⇒
… (ii)
⇒
5 3 +10 2 − 5 = v1 3 + v1 5 3 + 10 2 − 5 v1 = ⇒ 1+ 3 8.66 + 14142 . −5 = 1 + 1732 . 17.802 = ⇒ v1 = 6.516 m/s ≈ 6.5 m/s 2.732 … (iii) Substituting the value from Eq. (iii) in Eq. (i), we get v 5 3 + 5 2 = 2 + 6.51 × 3 2 ⇒ v 2 = (5 3 + 5 2 − 6.51 × 3 ) 2 v 2 = (8.66 + 7.071 − 11215 . ) 1414 . ⇒ v 2 = 4.456 × 1414 . ⇒ v 2 ≈ 6.3 m/s
7. A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son, so that he starts moving at a speed of 0.70 ms −1 with respect to the man. The speed of the man with respect to the surface is
Using momentum conservation law, (Total momentum)before collision = (Total momentum)after collision (m1 × 0) + (m2 × 0) = m1v1 + m2 v 2 0 = m1(− v1 )$i + m2 v 2 $i ⇒ m1v1 = m2 v 2 ⇒ 50v1 = 20v 2 …(i) ⇒ v 2 = 2 . 5v1 Again, relative velocity = 070 . m/s But from figure, relative velocity = v1 + v 2 …(ii) ∴ v1 + v 2 = 07 . From Eqs. (i) and (ii), we get v1 + 2 . 5v1 = 07 . ⇒ v1(3.5) = 07 . 07 . v1 = = 0.20 m/s 3.5
8. Three blocks A , B and C are lying on a smooth horizontal surface as shown in the figure. A and B have equal masses m whileC has mass M . Block A is given an initial speed v towards B due to which it collides with B perfectly inelastically. The combined mass collides with C , also perfectly inelastically 5 th of the initial kinetic energy is lost in 6 M whole process. What is value of ? m
[JEE Main 2019, 12 April ShiftI]
(a) 0.28 ms −1 (c) 0.47 ms −1
(b) 0.20 ms −1 (d) 0.14 ms −1
(a) 4
A
[JEE Main 2019, 9 Jan ShiftI] B C
m
m
(b) 2
M
(c) 3
(d) 5
Exp. (a)
Exp. (b) The given situation can be shown as below m1=50 u1=0
m2=20 u2=0 µ=0
Initially, block A is moving with velocity v as shown in the figure below,
Man Son Before collision m1=50 v1 m2=20 v2
A m
µ=0 Man Son After collision
Key Idea For a perfectly inelastic collision, the momentum of the system remains conserved but there is some of loss of kinetic energy. Also, after collision the objects of the system are stuck to each other and move as a combined system.
C M
B m
v
Now, A collides with B such that they collide inelastically. Thus, the combined mass (say) move with the velocity ‘v’ as shown below, m
m
v′
C M
51
Work, Energy and Power Then, if this combined system is collided inelastically again with the block C. So, now the velocity of system be v ′′ as shown below. m
m
M
v′′
(a)
Thus, according to the principle of conservation of momentum, initial momentum of the system = final momentum of the system ⇒ mv = (2 m + M ) v ′ ′ mv or … (i) v′′= 2 m+ M Initial kinetic energy of the system, 1 (KE)i = mv 2 2 Final kinetic energy of the system, (KE)f =
m
mv 1 1 (2 m + M )(v ′ ′ )2 = (2 m + M ) 2 2 2m + M
…(ii)
2
[Qusing Eq. (i)] 1 v 2 m2 = ⋅ 2 (2 m + M )
…(iii)
Dividing Eq. (iii) and Eq. (ii), we get 1 2 2 m v (KE)f 2 m … (iv) = = (KE)ii (2 m + M ) 2 m + M 1 mv 2 2 5 It is given that th of (KE)i is lost in this process. 6 1 (KE)f = (KE)i ⇒ 6 (KE)f 1 … (v) = ⇒ (KE)i 6 Comparing Eq. (iv) and Eq. (v), we get m 1 = ⇒ 6m = 2 m + M 2m + M 6 M 4m = M ⇒ =4 m
9. A block of mass m lying on a smooth
πF mk
(b)
F mk
(c)
F
2F mk
(d)
F π mk
Exp. (b) In a springblock system, when a block is pulled with a constant force F, then its speed is maximum at the mean position. Also, it’s acceleration will be zero. In that case, force on the system is given as, …(i) F = kx where, x is the extension produced in the spring. F or x= k Now we know that, for a system vibrating at its mean position, its maximum velocity is given as, vmax = Aω where, A is the amplitude and ω is the angular velocity. Since, the block is at its mean position. F So, A= x= k k F k v max = Qω = m k m F = km Alternate Solution According to the workenergy theorem, net work done = change in the kinetic energy Here, net work done = work done due to external force (Wext ) + work done due to the spring (Wspr ). As, Wext = F ⋅ x −1 2 and Wspr = kx 2 1 ∆KE = F ⋅ x + − kx2 ⇒ 2 1 2 (∆KE)f − (∆KE)i = F ⋅ x − kx 2 ⇒
horizontal surface is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force F , the maximum speed of the block is
⇒
[JEE Main 2019, 9 Jan ShiftI]
⇒
or
F 1 F 1 1 2 − m(0)2 = F ⋅ − k mvmax k 2 k 2 2
2
[ using Eq. (i)] F2 F2 F2 1 2 − = mvmax = k 2 2k 2k F2 2 = vmax km vmax = F / km
52
JEE Main Chapterwise Physics
10. A force acts on a 2 kg object, so that its position is given as a function of time as x = 3t 2 + 5. What is the work done by this force in first 5 seconds? [JEE Main 2019, 9 Jan ShiftII]
(a) 850 J
(b) 900 J
(c) 950 J
(d) 875 J
m
(a)
a=
3mg 2t 2 8 mg 2t 2 (d) − 8
mg 2t 2 8
(b)
(c) 0
Exp. (b) Here, the displacement of an object is given by x = (3 t 2 + 5) m Therefore, velocity (v ) = or
Exp. (b) Normal reaction force on the block is
dx d (3 t + 5) = dt dt 2
v = 6t m/s
N g/2
...(i) m
The work done in moving the object from t = 0 to t = 5s x W =
g 2
5
∫ F ⋅ dx
…(ii)
x0
The force acting on this object is given by dv F = ma = m × dt d (6t ) = m× [using Eq. (i)] dt F = m × 6 = 6 m = 12 N Also,
x0 = 3 t 2 + 5 = 3 × (0)2 + 5 = 5 m
and at t = 5 s, x5 = 3 × (5)2 + 5 = 80 m Put the values in Eq. (ii), we get W = 12 ×
x5
∫
dx = 12 [80 − 5]
x0
W = 12 × 75 = 900 J Alternate Solution To using work – kinetic energy theorem is, 1 W = ∆K ⋅ E = m(v f2 – v i2 ) 2 1 = m × (302 – 02 ) 2 1 = × 2 × 900 = 900 J 2
11. A block of mass m is kept on a platform which starts from rest with constant g acceleration upwards as shown in figure. 2 Work done by normal reaction on block in timet is [JEE Main 2019, 10 Jan ShiftI]
mg
N = manet where, anet = net acceleration of block. =g+ a g 3g =g+ = 2 2 g 3mg N = m g + = ⇒ 2 2 Now, in time ‘t’ block moves by a displacement s given by 1 1 g s = 0 + at 2 = t 2 (Qu = 0) 2 22 g Here, a = (given) 2 ∴Work done = Force × Displacement 3mg gt 2 3mg 2t 2 W = × = ⇒ 2 4 8
12. A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward with a velocity 100 ms −1 from the ground. The bullet gets embedded in the wood. Then, the maximum height to which the combined system reaches above the top of the building before falling below is (Take, g = 10 ms −2) [JEE Main 2019, 10 Jan ShiftI]
(a) 20 m
(b) 30 m
(c) 10 m
(d) 40 m
53
Work, Energy and Power Exp. (b)
Exp. (d) Key Idea As bullet gets embedded in the block of wood so, it represents a collision which is perfectly inelastic and hence only momentum of the system is conserved.
Velocity of bullet is very high compared to velocity of wooden block so, in order to calculate time for collision, we take relative velocity nearly equal to velocity of bullet. So, time taken for particles to collide is d 100 t = = = 1s vrel 100 Speed of block just before collision is; v1 = gt = 10 × 1 = 10 ms − 1 Speed of bullet just before collision is v 2 = u − gt = 100 − 10 × 1= 90 ms − 1 Let v = velocity of bullet + block system, then by conservation of linear momentum, we get − (0.03 × 10) + (0.02 × 90) = (0.05) v ⇒ v = 30 ms − 1 Now, maximum height reached by bullet and block is 30 × 30 v2 h= ⇒ h= 2 × 10 2g ⇒ h = 45 m ∴ Height covered by the system from point of collision = 45 m Now, distance covered by bullet before collision in 1 s. 1 = 100 × 1 − × 10 × 12 = 95 m 2 Distance of point of collision from the top of the building = 100 − 95 = 5 m ∴ Maximum height to which the combined system reaches above the top of the building before falling below = 45 − 5 = 40 m
13. A particle which is experiencing a force, is
given by F = 3$i − 12$j, undergoes a displacement of d = 4$i. If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ? [JEE Main 2019, 10 Jan ShiftII]
(a) 9 J
(b) 15 J
(c) 12 J
(d) 10 J
We know that, work done in displacing a particle at displacement d under force F is given by ∆W = F ⋅ d By substituting given values, we get ⇒ ∆W = (3$i − 12 $j ) ⋅ (4$i ) … (i) ⇒ ∆W = 12 J Now, using workenergy theorem, we get work done (∆W ) = change in kinetic energy (∆K ) or … (ii) ∆W = K 2 − K1 Comparing Eqs. (i) and (ii), we get K 2 − K1 = 12 J or K 2 = K1 + 12 J Given, initial kinetic energy, K1 = 3 J ∴ Final kinetic energy, K 2 = 3 J + 12 J = 15 J
14. A body of mass 1 kg falls freely from a height of 100 m on a platform of mass 3 kg which is mounted on a spring having spring constant k = 1.25 × 106 N/m. The body sticks to the platform and the spring’s maximum compression is found to be x. Given that g = 10 ms −2, the value of x will be close to [JEE Main 2019, 11 Jan ShiftI]
(a) 8 cm
(b) 4 cm
(c) 40 cm
(d) 80 cm
Exp. (*) Initial compression of the spring, x (x0 in cm) mg = k 0 100 3 × 10 × 100 3 x0 = = ⇒ 1250 125 . × 106 Which is very small and can be neglected. Applying conservation of momentum before and after the collision i.e., momentum before collision = momentum after collision. m × 2 gh = (m + M ) v (Qvelocity of the block just before the collision is v 2 − 02 = 2 gh or v = 2 gh) After substituting the given values, we get 1 × 2 × 10 × 100 = 4v or 4v = 20 5 So, v = 5 5 m/s Let this be the maximum velocity, then for the given system, using 1 1 mv 2 = kx2 2 2
54
JEE Main Chapterwise Physics ∴
x 1 1 . × 106 × × 4 × 125 = × 125 100 2 2 4 = 104 ×
⇒
2
x2
or x = 2 cm 104 ∴ No option given is correct.
Then if pendulum is deflected back upto angle θ1, then …(ii) v1 = 2 gl (1 − cos θ1 ) Using definition of coefficient of restitution to get velocity of separation e= velocity of approach
15. A simple pendulum is made of a string of length l and a bob of mass m, is released from a small angle θ 0 . It strikes a block of mass M , kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle θ1. Then, M is given by [JEE Main 2019, 12 Jan ShiftI]
m θ0 − θ1 2 θ0 + θ1 m θ0 + θ1 (d) 2 θ0 − θ1
θ + θ1 (a) m 0 θ0 − θ1
(b)
θ −θ (c) m 0 1 θ0 + θ1
1=
v 2 − (− v1 ) ⇒ v = v 2 + v1 v−0
…(iii)
From Eqs. (i), (ii) and (iii), we get ⇒
2 gl(1 − cos θ0 ) = v 2 +
2 gl(1 − cos θ1 )
⇒ v 2 = 2 gl ( 1 − cos θ0 − 1 − cos θ1 ) …(iv) According to the momentum conservation, initial momentum of the system = final momentum of the system ⇒ mv = Mv 2 − mv1 ⇒
Mv 2 = m(v + v1 )
Mv 2 = m 2 gl ( 1 − cos θ0 +
1 − cos θ1 )
Dividing Eq. (v) and Eq. (iv), we get
Exp. (a) Pendulum’s velocity at lowest point just before striking mass m is found by equating it’s initial potential energy (PE) with final kinetic energy (KE). Initially, when pendulum is released from angle θ0 as shown in the figure below, θ0
We have, 1 mgh = mv 2 2 Here, h = l − l cosθ0 …(i) So, v = 2 gl (1 − cos θ0 ) With velocity v, bob of pendulum collides with block. After collision, let v1 and v 2 are final velocities of masses m and M respectively as shown m v
Before collision
M m
M
v2
After collision
θ sin2 0 + 2
θ sin2 1 2
θ θ sin2 0 − sin2 1 2 2
θ θ sin 0 + sin 1 2 2 M = m sin θ0 − sin θ1 2 2
l –h
v
1 − cos θ0 + 1 − cos θ1 M = m 1 − cos θ0 − 1 − cos θ1 =
cos θ = l – h l
h
v1
⇒
For small θ0 , we have θ0 θ1 + M 2 or M = m θ0 + θ1 = 2 m θ0 − θ1 θ0 − θ1 2 2
16. An αparticle of mass m suffers onedimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64% of its initial kinetic energy. The mass of the nucleus is [JEE Main 2019, 12 Jan ShiftII] (a) 1.5 m
(b) 4 m
(c) 3.5 m
(d) 2 m
Exp. (b) We have following collision, where mass of α particle = m and mass of nucleus = M
55
Work, Energy and Power v
m
,
m
M
m v0
α α m
Before collision
M
Let α particle rebounds with velocity v1, then Given; final energy of α = 36% of initial energy 1 1 mv12 = 0.36 × mv 2 ⇒ 2 2 …(i) ⇒ v1 = 0.6 v As unknown nucleus gained 64% of energy of α, we have 1 1 2 Mv 2 = 0.64 × mv 2 2 2 m …(ii) ⇒ × 0.8 v v2 = M From momentum conservation, we have mv = Mv 2 − mv1 Substituting values of v1 and v 2 from Eqs. (i) and (ii), we have m × 0.8 v − m × 0.6 v mv = M M 16 . mv = mM × 0.8 v ⇒ ⇒ 2m = mM ⇒ 4m2 = mM ⇒
M = 4m
17. In a collinear collision, a particle with an initial speedv 0 strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles after collision, is [JEE Main 2018] v0 4 v0 (c) 2
(a)
(b) 2 v 0 (d)
v0 2
Exp. (c) / Momentum is conserved in all type of collisions, Final kinetic energy is 50% more than initial kinetic energy 150 1 1 1 …(i) × mv 02 mv 22 + mv12 = ⇒ 100 2 2 2
m
m
v2
v1
v2
v1 After collision
Conservation of momentum gives, mv 0 = mv1 + mv 2 v 0 = v 2 + v1 From Eqs. (i) and (ii), we have
…(ii)
− v 02 2 (v1 − v 2 )2 = (v1 + v 2 )2 − 4 v1v 2 = 2 v 02
v12 + v 22 + 2 v1v 2 = v 02 ⇒ 2 v1v 2 = ∴
vrel = 2 v 0
or
18. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 s will be [JEE Main 2017 (Offline)] (a) 22 J
(b) 9 J
(c) 18 J
(d) 4.5 J
Exp. (d) ∆p = F ⇒ ∆p = F∆t ∆t F dt
From Newton’s second law, ∴
p = ∫ dp =
⇒
p=
1
1
∫0
m
∫0 6t dt = 3 kg s
Also, change in kinetic energy 32 ∆p2 = = 4.5 ∆k = 2m 2 × 1 From workenergy theorem, work done = change in kinetic energy. So, work done = ∆k = 4.5 J
19. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m, 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 10 7 J of energy per kg which is converted into mechanical energy with a 20% of efficiency rate. (Take, g = 9.8 ms −2) [JEE Main 2016 (Offline)] (a) 2.45 × 10−3 kg
(b) 6.45 × 10−3 kg
(c) 989 . × 10−3 kg
(d)12.89 × 10−3 kg
56
JEE Main Chapterwise Physics Exp. (d) Given, potential energy burnt by lifting weight = mgh = 10 × 9.8 × 1 × 1000 = 9.8 × 104 If mass lost by a person be m, then energy dissipated 2 = m× × 3.8 × 107 J 10 1 9.8 × 104 = m × × 3.8 × 107 ⇒ 5 5 −3 m= × 10 × 9.8 = 12 .89 × 10−3 kg ⇒ 3.8
20. A particle of mass m moving in the xdirection with speed 2v is hit by another particle of mass 2m moving in the ydirection with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to [JEE Main 2015] (a) 44%
(b) 50%
(c) 56%
(d) 62%
Loss in the energy ∆E = Ei − Ef 4 5 = mv 2 3 − = mv 2 3 3 Percentage loss in the energy during the collision (5/ 3) mv 2 5 ∆E − 56% × 100 = × 100 = × 100 ~ 9 Ei 3mv 2
21. When a rubber band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx 2, where a and b are constants. The work done in stretching the unstretched rubber band by L is [JEE Main 2014] (a) aL2 + bL3 aL2 bL3 (c ) + 2 3
1 (aL2 + bL3 ) 2 1 aL2 bL3 (d) + 2 2 3
(b)
/ As, we know that change in potential energy of a system corresponding to a conservative t
Exp. (c) / Conservation of linear momentum can be applied but energy is not conserved. Consider the movement of two particles as shown below.
Conserving linear momentum in xdirection ( pi )x = ( pf )x or 2 mv = (2 m + m) v x 2 or vx = v 3 Conserving linear momentum in ydirection ( pi )y = ( pf )y or 2 mv = (2 m + m) v y 2 or vy = v 3 Initial kinetic energy of the two particles system is 1 1 Ei = m (2 v )2 + (2 m) (v )2 2 2 1 1 = × 4mv 2 + × 2 mv 2 2 2 = 2 mv 2 + mv 2 = 3mv 2 Final energy of the combined two particles system is 1 Ef = (3m) (v 2x + v y2 ) 2 4v 2 4v 2 3m 8 v 2 4mv 2 1 = (3m) + = = 9 2 2 9 3 9
internal force as U f − Ui = − W = − ∫ F. d r i
Exp. (c) Given, F = ax + bx2 According to workenergy theorem, we know that work done in stretching the rubber band by t is dW = Fdx L bx3 L ax2 L W = ∫ (ax + bx2 ) dx = + 3 0 2 0 0 aL2 a × (0)2 b × L3 b × (0)3 = − − + 2 3 2 3 ∴
W =
aL2 bL3 + 2 3
22. This question has Statement I and Statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement I A point particle of mass m moving with speed v collides with stationary point particle of mass M . If the maximum 1 energy loss possible is given as f mv 2 , 2 m then f = . M + m
57
Work, Energy and Power Statement II Maximum energy loss occurs when the particles get stuck together as a result of the collision. [JEE Main 2013]
Alternate Solution
(a) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true
As,
Exp. (d) 2
p , where p is momentum, m is the 2m mass moving of the particle. Maximum energy loss occurs when the particles get stuck together as a result of the collision. p2 p2 Maximum energy loss (∆E) = − 2 m 2 (m + M ) where, (m + M ) is the resultant mass when the particles get stuck. p2 m p2 M = 1− ∆E = m + M 2 m m + M 2m Also, p = mv m2 v 2 M mv 2 M ∆E = = ∴ 2 m m + M 2 m + M Comparing the expression with 1 M ∆E = f mv 2 , f = 2 m+ M Energy E =
(Q. Nos. 2425) contain Statement I and Statement II. Of the four choices given after the statements, choose the one that best describes the two statements. (a) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I (c) Statement I is true, Statement II is true (d) Statement I is true, Statement II is false
24. Statement I Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement II Principle of conservation of momentum holds true for all kinds of collisions. [AIEEE 2010]
Exp. (a) If it is a completely inelastic collision, then m1v1 + m2 v 2 = m1v + m2 v m1
Exp. (d) Given, k ∝ t ⇒ ⇒
dk = constant dt K ∝t
1 mv 2 ∝ t ⇒ v ∝ t 2 dK Also, P = Fv = = constant dt 1 F∝ ⇒ v 1 F∝ ⇒ t
m2 v1
v2
m v + m2 v 2 , v= 11 m1 + m2
xaxis. If its kinetic energy increases uniformly with timet , the net force acting on it must be proportional to [AIEEE 2011] (b) constant (d)1/ t
1 dv = k t −1/ 2 2 dt 1 1 a∝ , so F ∝ t t
a=
æ Directions
23. At timet = 0 , particle starts moving along the
(a) t (c) t
dv dt
K ∝ t v ∝ t , F = ma = m
KE =
p12 p2 + 2 2 m1 2 m2
As p1 and p 2 both simultaneously cannot be zero. Therefore, total KE cannot be lost.
25. An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range [AIEEE 2008]
(a) 200 J − 500 J
(b) 2 × 105 J − 3 × 105 J
(c) 20000 J − 50000 J
(d) 2000 J − 5000 J
Exp. (d) The given question is somewhat based on approximations. Let mass of athlete be 65 kg. Approx velocity from the given data is 10 m/s.
58
JEE Main Chapterwise Physics 65 × 100 = 3250 J 2 So, option (d) is the most probable answer. KE =
So,
28. The potential energy of a 1 kg particle free to move along the xaxis is given by x4 x2 V (x ) = − J 2 4
26. A block of mass 0.50 kg is moving with a
speed of 2 .00 ms −1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is [AIEEE 2008] (a) 0.16 J (c) 0.67 J
(b) 1.00 J (d) 0.34 J
Exp. (c)
[assumed that 2nd body is at rest]
∴
2 v= 3 ∆K = K f − K i 2
2 1.5 × 3 22 2 − (0.5) × =− J = 2 3 2 = −0.67 J So, energy lost is 0.67 J.
27. A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10000 N/m. The spring compresses by [AIEEE 2007] (a) 5.5 cm (c) 11.0 cm
(b) 2.5 cm (d) 8.5 cm
Exp. (a) According to workenergy theorem Loss in kinetic energy = Work done against friction + Potential energy of spring 1 1 mv 2 = f x + kx2 2 2 1 1 × 2 (4 )4 = 15 x + × 10000 x2 ⇒ 2 2 ⇒ 5000 x2 + 15 x − 16 = 0 ∴
[AIEEE 2006]
3 (a) 2
(b)
2
1 (c) 2
(d) 2
Exp. (a)
From law of conservation of momentum, we have m1v1 + m2 v 2 = (m1 + m2 ) v Given, m1 = 0.50 kg, v1 = 2 ms–1, [at rest] m2 = 1 kg, v 2 = 0 0.5 × 2 + 1 × 0 = 1.5 × v ⇒
The total mechanical energy of the particle is 2 J. Then, the maximum speed (in ms −1) is
x = 0.055 m = 5.5 cm
x4 x2 V ( x) = − 2 4 dV For minimum value of V, =0 dx 4 x3 2 x − =0 ⇒ 4 2 ⇒ x = 0, x = ± 1 1 1 −1 J So, Vmin ( x = ± 1) = − = 4 2 4 Given,
Now, Kmax + Vmin = Total mechanical energy 9 1 ⇒ Kmax = + 2 or Kmax = 4 4 mv 2 9 = 2 4
or
or
v=
3 ms −1 2
29. A bomb of mass 16 kg at rest explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is 4ms −1. The kinetic energy of the other mass is [AIEEE 2006]
(a) 144 J
(b) 288 J
(c) 192 J
(d) 96 J
Exp. (b) Here, momentum of the system is remaining conserved as no external force is acting on the bomb (system). 16 kg
4 ms–1
v 4 kg
12 kg
Initial momentum (before explosion) = Final momentum (after explosion)
59
Work, Energy and Power Let velocity of 4 kg mass be v ms −1. From momentum conservation, we can say that its direction is opposite to velocity of 12 kg mass. From p i = p f ⇒ 0 = 12 × 4 − 4 × v or v = 12 m/s 4 × (12 )2 ∴ KE of 4 kg mass = = 288 J 2
30. A mass m moves with a velocity v and collides inelastically with another identical mass. After collision, the 1st mass moves v with velocity in a direction perpendicular 3 to the initial direction of motion. Find the speed of the second mass after collision. [AIEEE 2005]
(a) v
(b)
3v
2 (c) v 3
(d)
v 3
–F × 3=
Case I
2
1 v0 1 m – mv 02 2 2 2
where, F is resistive force and v 0 is initial speed. Case II Let the further distance travelled by the bullet before coming to rest be s. 1 – F(3 + s ) = K f − K i = – mv 02 ∴ 2 1 1 − mv 02 (3 + s ) = − mv 02 ⇒ 2 8 1 or (3 + s ) = 1 4 3 s or + =1 4 4 or s = 1 cm
32. A body of mass m is accelerated uniformly
Exp. (c) In xdirection, Apply conservation of momentum, we get mu1 + 0 = mv x ⇒
mv = mv x ⇒ v x = v 1
2
1
2
4 2 v 3
mv 2 T
2
t
1 mv 2 t 2 T2
(b) (d)
mv 2 T2
t2
1 mv 2 2 t 2 T2
A body of mass m with uniform acceleration, then force mv ∴ a = v − 0 F = ma = T T
In ydirection, apply conservation of momentum, we get v v 0 + 0 = m − mv y ⇒ v y = 3 3 Velocity of second mass after collision 2
(a)
Exp. (a) vx
vy After collision
v + v2 = 3
from rest to a speed v in a time T . The instantaneous power delivered to the body as a function of time, is given by [AIEEE 2005, 04]
(c)
Before collision v/ 3 m
v′=
According to workenergy theorem, Total work done = Change in kinetic energy W = ∆K
or v ′ =
2 v 3
31. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest, assuming that it faces constant resistance to motion? [AIEEE 2005]
Instantaneous power = Fv = mav mv = ⋅ at T mv v mv 2 = ⋅ ⋅t = 2 t T T T
33. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is [AIEEE 2005]
(a) 3.0 cm (b) 2.0 cm (c) 1.5 cm (d) 1.0 cm
(a) 40 m/s (c) 10 m/s
Exp. (d)
Exp. ( a )
(b) 20 m/s (d) 10 30 m/s
60
JEE Main Chapterwise Physics According to conservation of energy, potential energy at height H is sum of kinetic energy and potential energy at h2 .
H = 100 m
h1 = 3 0 m
h2 = 20 m
1 mv 2 + mgh2 2 1 mg (H − h2 ) = mv 2 2 v = 2 g (100 − 20) v = 2 × 10 × 80 = 40 m/s
∴
mgH =
⇒ or or
From given information a = − kx, where a is acceleration, x is displacement and k is a proportionality constant, vdv = − kx dx ⇒ v dv = − kx dx Let for any displacement from 0 to x, the velocity changes from v 0 to v. v
∫v
⇒
v dv = −
0
x
∫0
k x dx
v
34. The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collision is [AIEEE 2005]
x
⇒
v2 x2 =−k 2 v0 2 0
⇒
v 2 − v 02 kx2 =− 2 2 v 2 − v 02 mkx2 =− m 2 2
⇒
⇒
1 1 mkx2 mv 2 − mv 02 = − 2 2 2 [∆K is loss in KE] ∆ K ∝ x2
36. A uniform chain of length 2 m is kept on a M
(a)
Mk L (b)
kL2 2M
(c) zero
(d)
ML2 k
Exp. (a) Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted into KE. According to conservation of energy, 1 2 1 kL = Mv 2 2 2 (Mv )2 kL2 = ⇒ M or MkL2 = p2 [Q p = Mv ]
table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table? [AIEEE 2004] (a) 7.2 J (c) 120 J
(b) 3.6 J (d) 1200 J
Exp. (b) Mass per unit length =
M 4 = = 2 kg/m L 2
1.4 m 0.6 m
retardation proportional to its displacement. Its loss of kinetic energy for any displacement [AIEEE 2004] x is proportional to
The mass of 0.6 m of chain = 0.6 × 2 = 1.2 kg The height of centre of mass of hanging part 0.6 + 0 h= = 0.3 m 2 Hence, work done in pulling the chain on the table = Work done against gravity force
(a) x 2
i .e,
⇒
p = L Mk
35. A particle moves in a straight line with
Exp. ( a )
(b) e x
(c) x
(d) loge x
W = mgh = 1.2 × 10 × 0.3 = 3.6 J
61
Work, Energy and Power 37. A force F = (5 i + 3 j + 2 k ) N is applied over a particle which displaces it from its origin to the point r = (2 i − j) m. The work done on the particle in joule is [AIEEE 2004] (a) –7 (c) +10
(b) +7 (d) +13
Exp. (b) Work done in displacing the particle W = F⋅ r = (5i + 3 j + 2 k ) ⋅ (2 i − j) = 5 × 2 + 3 × (−1) + 2 × 0 = 10 − 3 = 7 J
Delivering power of a machine P = constant s w P = F ⋅ u QPower = = F ⋅ = F ⋅ v t t dv v = mav = m dt dv P = mv dt P vdv = dt m Integrating on both sides, we get v
∫ v dv = 0
38. A spring of spring constant 5 × 103 N/m is
v=
(b) 18.75 Nm (d) 6.25 Nm
= 25 − 6 ⋅ 25 = 18.75 J = 18.75 Nm
39. A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to [AIEEE 2003] (b) t 3/ 2 (d) t 1/ 2
s
∫0 ds = s= ∴
1/ 2
ds 2 Pt = dt m
2 Pt ds = m
Work done to stretch the spring by 5 cm from 1 mean position = W1 = k x12 2 1 = × 5 × 103 × (5 × 10−2 )2 2 = 6.25 J Work done to stretch the spring by 10 cm from 1 mean position, W2 = k( x1 + x2 )2 2 1 = × 5 × 103 (5 × 10−2 + 5 × 10−2 )2 2 = 25 J Net work done to stretch the spring from 5 cm to 10 cm = W2 − W1
Exp. (b)
0
2 Pt v = m
Exp. (b)
(a) t 3/ 4 (c) t 1/ 4
P
∫ m dt
v 2 Pt = 2 m
stretched initially by 5 cm from the unstretched position. Then, the work required to stretch it further by another 5 cm is [AIEEE 2003] (a) 12.50 Nm (c) 25.00 Nm
t
2P m
1/ 2
1/ 2
dt t
∫0 t
1/ 2
dt
2P t 3/ 2 m 3/2
s ∝ t 3/ 2
40. A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is [AIEEE 2002]
(a) 16 J
(b) 8 J
(c) 32 J
(d) 24 J
Exp. (b) The work done on the spring is stored as the PE of the body and is given by U=
x2
∫x
1
or
U=
x2
∫x
Fext dx kx dx
1
1 k( x22 − x12 ) 2 800 [(015 . )2 − (0.05)2 ] = 2 = 400(0.2 × 0.1) = 8 J =
5 Rotational Motion 1. Four particles A , B ,C and D with masses m A = m , m B = 2m , mC = 3m and m D = 4m are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles (in ms−2 ) is [JEE Main 2019, 8 April ShiftI] a
where, mA = m, mB = 2 m, mC = 3m and mD = 4m, a A = a B = a C = a D= a (according to the question) − ma $i + 2 ma $i + 3ma $i − 4ma$j aCM = m + 2 m + 3m + 4m 2 a $i − 2 a$j a $ $ = = ⋅ ( i − j ) ms −2 10 5
2. A thin circular plate of mass M and radius R
Y
B
has its density varying as ρ(r ) = ρ0r with ρ0 as constant and r is the distance from its centre. The moment of inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I = aMR 2. The value of the coefficient a is
a
C
X D
a A
[JEE Main 2019, 8 April ShiftI] a
a (a) ( $i − $j) (b) a( $i + $j) (c) zero 5
a (d) ( $i + $j) 5
Exp. (a)
mB=2m
a^ j
For a system of discrete masses, acceleration of centre of mass (CM) is given by m a + mBa B + mC a C + mDa D a CM = A A mA + mB + mC + mD
A mA=m
mC=3m C
3 (b) 5
(c)
8 5
(d)
3 2
Exp. (*) Consider an elementary ring of thickness dx and radius x. dx
x
a^ i R
D
mD=4m
–ai
B
1 (a) 2
– a^ j
Moment of inertia of this ring about a perpendicular axes through centre is dIc = dm ⋅ x2 = ρ0 x(2 πx)dx ⋅ x2 = 2 πρ0 x4dx Moment of inertia of this elementary ring about a perpendicular axes at a point through edge, (by parallel axes theorem)
63
Rotational Motion dI = dmx2 + dmR 2 = 2 πρ0 x4dx + 2 πρ0 R 2 x2dx Moment of inertia of complete disc is R
R
R
0
0 5
0
I = ∫ dI = ∫ 2 πρ0 x4dx + ∫ 2 πρ0 R 2 x2dx 2 πρ0 R 2 πρ0 R 5 16 πρ0 R 5 + = 5 3 15 16 a= 15 =
∴
3. A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph and hcyl on the hsph is given by incline. The ratio hcyl [JEE Main 2019, 8 April ShiftII]
(a)
2 5
(b)
14 15
(c) 1
(d)
4 5
Exp. (b) When a spherical/circular body of radius r rolls without slipping, its total kinetic energy is K total = K translation + Krotation 1 1 = mv 2 + Iω2 2 2 1 1 v2 Qω = v = mv 2 + I ⋅ 2 r 2 2 r Let v be the linear velocity and R be the radius for both solid sphere and solid cylinder. ∴Kinetic energy of the given solid sphere will be 1 1 v2 Ksph = mv 2 + Isph 2 2 2 R v2 1 1 2 2 = mv + × mR 2 × 2 2 2 5 R 7 2 …(i) mv = 10 Similarly, kinetic energy of the given solid 1 1 v2 cylinder will be Kcyl = mv 2 + Icyl 2 2 R 2 2 mR v 1 1 3 = mv 2 + × × 2 = mv 2 …(ii) 2 2 2 4 R
Now, from the conservation of mechanical energy, mgh = K total ∴For solid sphere, 7 mghsph = mv 2 …(iii) [using Eq. (i)] 10 Similarly, for solid cylinder, 3 …(iv) [using Eq. (ii)] mghcyl = mv 2 4 Taking the ratio of Eqs. (iii) and (iv), we get 7 2 mghsph 10 mv h 7 4 14 = ⇒ sph = × = 3 2 mghcyl 10 3 15 h cyl mv 4 Alternate Solution Total kinetic energy for a rolling body without slipping can also be given as 1 K2 K total = mv 2 1 + 2 2 R where, K is the radius of gyration. ∴From law of conservation, 1 K2 mgh = mv 2 1 + 2 2 R 2 K or h ∝ 1 + 2 R As we know that, for solid sphere, 2 K= R 5 K2
2 = R2 5 Similarly, for solid cylinder, K2 1 R K= ⇒ 2 = 2 2 R 2 7 hsph 1 + 5 5 7 2 14 So, = = = × = 1 3 5 3 15 hcyl 1+ 2 2
⇒
4. A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time τ = 0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to [JEE Main 2019, 8 April ShiftII]
64
JEE Main Chapterwise Physics l
Y H
(0, b) A
h
(3—4a,3—4b(
E
(a) 0.02
(b) 0.3
(c) 0.5
(d) 0.28
Exp. (c) Key Idea The rectangular box rotates due to torque of weight about its centre of mass.
Now, angular impulse of weight = Change in angular momentum. 3g × τ l ml 2 ∴ ω⇒ ω = mg × τ = 2×l 2 3 Substituting the given values, we get 3 × 10 × 0.01 = 0.5 rad s −1 = 2 × 0.3
So, angle turned by box in reaching ground is θ = ωt = 0.5 × 1 = 0.5 rad
5. A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cutoff, the coordinates of the centre of mass of the remaining portion will be [JEE Main 2019, 8 April ShiftII] (a , b ) B
H a b 2 2
E
D (0, 0)
2a 2 b (a) , 3 3 3a 3b (c) , 4 4
G
O
C (a, 0)
F
5a (b) 12 5a (d) 3
( —2a , —b2 (
D (0, 0)
F
G
C (a, 0)
X
Here, Area of complete lamina, A1 = ab a b ab Area of shaded part of lamina = × = 2 2 4 ( x1, y1 ) = coordinates of centre of mass of a a complete lamina = , 2 2 ( x2 , y2 ) = coordinates of centre of mass of 3a 3a shaded part of lamina = , 4 4 ∴Using formula for centre of mass, we have A x − A2 x2 XCM = 1 1 A1 − A2
Time of fall of box, 2×5 2h ≈ 1s = t = 10 g
(0, b) A
B(a, b)
5b 12 5b , 3
,
a ab 3a 8a2 b − 3a2 b ab − 2 5a 4 4 = 16 = = ab 3ab 12 ab − 4 4 A1 y1 − A2 y2 Similarly, YCM = A1 − A2 b ab 3b ab − 2 4 4 = 5b = ab 12 ab − 4 Alternate Solution Let m be the mass of entire rectangular lamina. So, the mass of the shaded portion of lamina m = 4 Using the relation, m x − m2 x2 , we get XCM = 1 1 m1 − m2 a m 3a a 3a m − − 4 2 4 2 16 = 5a XCM = = m 3 12 m− 4 4 m y − m2 y2 , we get Similarly, YCM = 1 1 m1 − m2
Exp. (b) The given rectangular thin sheet ABCD can be drawn as shown in the figure below,
YCM
a m 3b m − b − 3b 2 4 4 = 2 16 = 5b = 3 m 12 m− 4 4
65
Rotational Motion ∴The coordinates of the centre of mass of 5a 5b remaining portion will be , . 12 12
6. A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function ofθ, whereθ is the angle by which it has rotated, is given as kθ 2. If its moment of inertia is I, then the angular acceleration of the disc is [JEE Main 2019, 9 April ShiftI]
k (a) θ 2I k (c) θ 4I
k (b) θ I 2k (d) θ I
Exp. (d) Given, kinetic energy = kθ2 We know that, kinetic energy of a rotating body 1 about its axis = Iω2 2 where, I is moment of inertia and ω is angular velocity. 1 2 So, Iω = kθ2 2 2 kθ2 or ω2 = I 2k … (i) ω= θ ⇒ I Differentiating the above equation w.r.t. time on both sides, we get dω 2 k dθ 2k Qω = dθ = ⋅ = ⋅ω dt I dt I dt ∴ Angular acceleration, dω 2k 2k 2k α= = ⋅ω = ⋅ θ [using Eq. (i)] dt I I I 2k or α = θ I Alternate Solution 2 kθ2 As, ω2 = I dω 2 k dθ = ⋅ 2θ 2ω ⇒ dt I dt 2k or ω dω = θdθ I 2k dω (= α ) = ⋅θ ω dθ I 2k or α= θ I
7. The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane : (i) a ring of radius R, (ii) a solid cylinder of radius R/2 and (iii) a solid sphere of radiusR/4.If in each case, the speed of the centre of mass at the bottom of the incline is same, the ratio of the maximum height they climb is [JEE Main 2019, 9 April ShiftI]
(a) 10 :15 : 7 (c) 14 :15 : 20
(b) 4 : 3 : 2 (d) 2 : 3 : 4
Exp. (c) From question, let height attained by ring = h1 Height attained by cylinder = h2 Height attained by sphere = h3 As we know that for a body which is rolling up an inclined plane (without slipping), follows the law of conservation of energy. ∴ For ring, using energy conservation law at its height h1. (KE)linear + (KE)rotational = (PE) 1 1 ⇒ m1v 02 + I1ω2 = m1gh1 2 2 1 1 m1v 02 + m1R 2ω2 = m1gh1 ⇒ 2 2 (Q I = mR 2 for ring) ⇒ ⇒
v 02 v2 + 0 2 2 h1 = v 02 / g
gh1 =
(Q v 0 =ωR ) …(i)
Similarly, for solid cylinder, applying the law of conservation of energy, 1 1 m2 v 02 + I2ω2 = m2 gh2 2 2 2 1 11 R m2 v 02 + × m2 × ω2 = m2 gh2 ⇒ 2 2 2 2 Q I = 1 mR 2 for cylinder 2 R and R = 2 ⇒ v 02 1 1 1 m2 v 02 + × m2 R 2 × = m2 gh2 2 2 8 (R / 2 )2 ⇒
1 2 1 2 v 0 + v 0 = gh2 2 4
66
JEE Main Chapterwise Physics 3 2 v0 4 3 v2 h2 = 0 4g
gh2 =
⇒ ⇒
⇒ …(ii)
Similarly, for solid sphere, 1 1 m3 v 02 + I3ω2 = m3 gh3 2 2 2 1 1 2 R ⇒ m3 v 02 + m3 ω2 = m3 gh3 4 2 2 5 Q I = 2 mR 2 for solid sphere 5 R and R = 4 v 02 1 1 2 R2 × = m3 gh3 ⇒ m3 v 02 + × × m3 2 2 5 16 (R / 4)2 1 2 1 2 v 0 + v 0 = gh3 ⇒ 2 5 7 2 ⇒ gh3 = v0 10 7 v 02 or …(iii) h3 = 10 g ∴Taking the ratio of h1, h2 and h3 by using Eqs. (i), (ii) and (iii), we get v 2 3 v 02 7 v 02 3 7 h1 : h2 : h3 = 0 : : = 1: : g 4 g 10 g 4 10 ⇒ h1 : h2 : h3 = 40 : 30 : 28 = 20 : 15 : 14 ∴ The most appropriate option is (c). Although, it is still not in the correct sequence. Alternate Solution Total kinetic energy of a rolling body is also given as 1 K2 Etotal = mv 2 1 + 2 2 R where, K is the radius of gyration. Using conservation law of energy, v2 K2 K2 1 mv 2 1 + 2 = mgh or h = 1 + 2 2g 2 R R For ring, ⇒
K2 R
2
=1
h1 =
K2
2 1 + 1 = 3v 2 4g
2 5 v2 2 7v2 h3 = = 1+ 2 g 5 10g
For solid sphere, ⇒
R2
=
So,the ratio of h1, h2 and h3 is v 2 3v 2 7 v 2 : : h1 : h2 : h3 = g 4g 10 g 3 7 = 1: : = 20 : 15 : 14 4 10
8. A wedge of mass M = 4 m lies on a frictionless plane. A particle of mass m aproaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by [JEE Main 2019, 9 April ShiftII] 2
(a)
2
2v 7g
(b)
v g
(c)
2 v2 5g
K
2
R
2
2
=
(R / 2 2 )
=
4 1 R2 × 2 = 8 2 R
(R / 2 )2
(d)
v2 2g
Exp. (c) Key Idea Since, the ground is frictionless, so when the particle will collide and climb over the wedge, then the wedge will also move. Thus, by using conservation laws for momentum and energy, maximum height climbed by the particle can be calculated.
Initial condition can be shown in the figure below
m v
4m
As mass m collides with wedge, let both wedge and mass move with speed v′. Then, m v′ h 4m
v2 2v2 v2 = [1 + 1] = g 2g 2g
For solid cylinder,
v2 2g
h2 =
v′
By applying linear momentum conservation, we have Initial momentum of the system = Final momentum of the system mv + 0 = (m + 4m) v ′ v … (i) v′ = ⇒ 5
67
Rotational Motion Now, if m rises upto height h over wedge, then by applying conservation of mechanical energy, we have Initial energy of the system = Final energy of the system 2 1 1 1 mv 2 + 0 = mv ′2 + mgh + (4m)v ′ 2 2 2
So, final angular speed of system is ML2 ⋅ ω0 Mω0 12 = ω= ⇒ ML2 + 6mL2 M + 6m 12
2
mv 2 = (m + 4m)v ′ + 2 mgh ⇒
v 2 = 5v ′2 + 2 gh 1 v 2 = v 2 + 2 gh 5 4 2 2v2 v = 2 gh ⇒ h = 5 5g
⇒ ⇒
10. Moment of inertia of a body about a given [using Eq. (i)]
9. A thin smooth rod of length L and mass M is
rotating freely with angular speed ω 0 about an axis perpendicular to the rod and passing through its centre. Two beads of mass m and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be [JEE Main 2019, 9 April ShiftII] M ω0 M + 3m M ω0 (c) M + 2m
M ω0 M+m M ω0 (d) M + 6m
(a)
(b)
Exp. (d) As there is no external torque on system. ∴Angular momentum of system is conserved. ⇒ Iiωi = Ifωf Initially, ω M
axis is 1.5 kg m 2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s 2 must be applied about the axis for a duration of [JEE Main 2019, 9 April ShiftII] (a) 5 s (c) 3 s
(b) 2 s (d) 2.5 s
Exp. (b) Rotation kinetic energy of a body is given by 1 KErotational = Iω2 2 where, ω = ω0 + αt 1 So, … (i) KErotational = I(ω0 + αt )2 2 Here, I = 1.5 kgm2 , KE = 1200 J and α = 20rad / s 2 and ω0 = 0 Substituting these values in Eq. (i), we get 1 . ) (20 × t )2 1200 = (15 2 2 × 1200 t2 = =4 ⇒ 15 . × 400 ∴
t =2s
11. Two coaxial discs, having moments of I1 are rotating with respective 2 ω angular velocities ω1 and 1 , about their 2 common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If E f and E i are the final and initial total energies, then ( E f − E i ) is
inertia I 1 and m
m
L
Finally, ω m
m
[JEE Main 2019, 10 April ShiftI] M, L
⇒
ML2 ⋅ ω0 + 0 = 12
2 ML2 L + 2(m) ω 12 2
I ω2 (a) − 1 1 24 3 (c) I 1ω12 8
I 1ω12 12 I 1ω12 (d) 6 (b) −
68
JEE Main Chapterwise Physics dr
Exp. (a) Initial kinetic energy of the given system, 1 2 1 I1 ω1 I1ω1 + 2 2 2 2 1 1 2 9 2 I1ω1 = + Iω = 2 16 1 1 16
KE i =
2
O
… (i)
Now, using angular momentum conservation law (assuming angular velocity after contact is ω) Initial angular momentum = Final angular momentum I ω I I1ω1 + 1 1 = I1ω′ + 1 ω′ 2 2 2 5 3 ω1 = ω′ 4 2 5 or ω′ = ω1 6 Now, final kinetic energy (after contact) is 1 1 I KEf = I1ω′2 + 1 ω′2 2 2 2 ⇒
2
1 5 1 5 I1 ω + I1 ω1 2 6 1 4 6 25 25 25 2 2 Iω = + Iω = 72 144 1 1 48 1 1 Hence, change in KE, ∆KE = KE f − KE i 25 2 9 2 I1ω1 − I1ω1 = 48 16 1 2 I1ω1 ∆KE = − 24 =
R
∫ dm = M = ∫0 (kr
2
[using Eq. (ii)] …(iii)
[using Eqs. (i)
per unit area σ(r ) = kr , where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is [JEE Main 2019, 10 April ShiftI] MR 2 6 2 MR 2 (d) 3
(b)
Exp. (d) Given, surface mass density, σ = kr 2 So, mass of the disc can be calculated by considering small element of area 2 πrdr on it and then integrating it for complete disc, i.e.
2
) 2 πrdr
R4 1 … (i) = πkR 4 4 2 Moment of inertia about the axis of the disc, I = ∫ dI = ∫ dmr 2 = ∫ σdAr 2 M = 2 πk
⇒
=
R
∫0 kr
2
(2 πrdr ) r 2
πkR 6 2 πkR 6 … (ii) = 6 3 From Eqs. (i) and (ii), we get 2 I = MR 2 3 I = 2 πk
⇒
2
MR 2 2 MR 2 (c) 3
dm = σ dA = σ × 2 πrdr
… (ii)
12. A thin disc of mass M and radius R has mass
(a)
r
R 5
∫0 r
dr =
13. A particle of mass m is moving along a trajectory given by and x = x 0 + a cosω1t y = y 0 + b sinω 2t. The torque acting on the particle about the origin at t = 0 is [JEE Main 2019, 10 April ShiftI] (a) zero $ (b) m ( − x 0 b + y 0a ) ω12 k $ (c) − m ( x 0 bω22 − y 0aω12 ) k 2 $ (d) + my a ω k 0
1
Exp. (d) Given equations of trajectory are x = x0 + a cos ω1t y = y0 + b sinω2t Now, v x = dx / dt = − aω1 sin ω1t and dy = bω2 cos ω2t vy = dt Similarly, dv a x = x = − aω12 cos (ω1t ) ⇒ dt dv y Similarly, ay = = − bω22 sin (ω2t ) dt Now, at t = 0 ; x = x0 + a ; y = y0 and a x = − aω12 ; ay = 0
… (i)
69
Rotational Motion Torque acting on the particle is given as τ = r × F = m(r × a ) $ Here, r = ( x0 + a)i$ + y0 $j + 0k $ and a = − aω2 i$ + 0$j + 0k 1
So, torque at t = 0, $) τ = m (− aω12 ) × y0 (− k [using Eq. (i)] $ ⇒ τ = + my aω2 k 0
$i × i$ = $j × $j = 0 $ + 6t 2 k $) L = m(−12t 2 k $i × $j = k $ $ and $j × i$ = − k
As, ⇒ As, ⇒
$ L = − m(6t 2 )k
So, angular momentum of particle of mass 2 kg at time t = 2 s is $ = − 48 k $ L = (−2 × 6 × 2 2 )k
1
Alternate Solution We can calculate total force from Eq. (i), F = − m (aω12 cos ω1t $i + bω22 sinω2t $j) and r = x$i + y$j = ( x + acos ω t )$i 0
1
+ ( y0 + b sinω2t )$j
Now, torque, τ = r × F = [( x0 + acos ω1t ) i$ + ( y0 + b sinω2t )$j] × [− m(aω12 cos ω1 ti$ + bω22 sinω2t $j)] $i $j $ k = − m x0 + acos ω1t y0 + b sinω2t 0 aω12 cos ω1t bω22 sinω2t 0 = − m [ x0 bω22 sinω2t + abω2 sin ω2t cos ω1t
15. A solid sphere of mass M and radius R is divided into two unequal parts. The first part 7M and is converted into a has a mass of 8 uniform disc of radius2R. The second part is converted into a uniform solid sphere. Let I 1 be the moment of inertia of the disc about its axis and I 2 be the moment of inertia of the new sphere about its axis. The ratio I 1 / I 2 is given by [JEE Main 2019, 10 April ShiftII] (a) 285
(b) 185
(c) 65
(d) 140
Exp. (d) The given situation is shown in the figure given below
− y0 a ω12 cos ω1t $ − ab ω12 sin ω2t cos ω1t ] k On solving it at t = 0 , we get $ τ = ma ω12 y0 k
R
7 M 8
M
1 M 8
Solid sphere
14. The time dependence of the position of a
particle of mass m = 2 is given by r(t ) = 2ti$ − 3t 2$j. Its angular momentum, with
2R r
respect to the origin, at timet = 2 is [JEE Main 2019, 10 April ShiftII]
$ (a) 36 k $ (c) − 48 k
$ − i$ ) (b) − 34( k $ (d) 48( i + $j)
Exp. (c) Position of particle is, r = 2ti$ − 3t 2 $j where, t is instantaneous time. Velocity of particle is dr v= = 2 $i − 6t$j dt Now, angular momentum of particle with respect to origin is given by L = m(r × v) = m{(2t$i − 3t 2 $j) × (2 i$ − 6t$j)} = m(−12t 2 ($i × $j) − 6t 2 ($j × i$ ))
Disc (radius= 2R) Solid sphere (radius = r)
Density of given sphere of radius R is Mass M ρ= = Volume 4 πR 3 3 Let radius of sphere formed from second part is r, then mass of second part = volume × density 1 4 M M = πr 3 × 4 8 3 πR 3 3 R3 R r3 = ⇒ r= ∴ 2 8
70
JEE Main Chapterwise Physics Now, I1 = moment of inertia of disc (radius 2R 7 and mass M) about its axis 8 7 M × (2 R )2 Mass × (Radius)2 8 7 = = = MR 2 2 2 4 and I2 = moment of inertia of sphere R 1 (radius and mass M) about its axis 2 8 2 2 = × Mass × (Radius) 5 2 R 2 1 MR 2 = × M × = 5 8 2 80 7 MR 2 I = 140 ∴ Ratio 1 = 4 1 I2 MR 2 80
Moment of inertia of disc about it’s axis, MR 2 I1 = 2
R
From perpendicular axes theorem, moment of inertia of disc about an axis along it’s diameter is I x + Iy = Iz ⇒ 2 I2 = I1 I MR 2 I2 = 1 = ⇒ 2 4 Iy=I2
16. A metal coin of mass 5g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second in 5s, is close to [JEE Main 2019, 10 April ShiftII] A
1 cm
B
(a) 40 . × 10− 6 N m
(b) 2.0 × 10− 5 N m
(c) 16 . × 10− 5 N m
(d) 7.9 × 10− 6 N m
Exp. (b) Moment of inertia (MI) of a disc about a tangential axis in the plane of disc can be obtained as below. A M=5 g = 5×10–3 kg
R =1 cm
Ix=I2
I1=Iz
So, moment of inertia about a tangential axis from parallel axes theorem is MR 2 I = I2 + MR 2 = + MR 2 4 5 = MR 2 4 Now, using torque, τ = Iα, we have ω − ωi 5 τ = Iα = MR 2 f ∆t 4
I2
I
R
Here, M = 5 × 10−3 kg, R = 1 × 10−2 m rad rad , = 50π ωf = 25 rps = 25 × 2 π s s ωi = 0, ∆t = 5 s 5 × 5 × 10−3 × (10−2 )2 × 50 π So, τ = 4 5 ≈ 2 × 10−5 Nm
17. A circular disc of radiusb has a hole of radius B
a at its centre (see figure). If the mass per σ unit area of the disc varies as 0 , then the r
71
Rotational Motion radius of gyration of the disc about its axis passing through the centre is [JEE Main 2019, 12 April ShiftI] b
M
b
0
a
∫ dm = ∫ σ 0 × 2 π × dr
⇒ M = σ 0 2 π(b − a) Calculation of Moment of Inertia M
b
I = ∫ r 2dm = ∫ r 2 ⋅ 0
a
a
a 2 + b 2 + ab 2
(b)
a+b 2
(c)
a 2 + b 2 + ab 3
(d)
a+b 3
b
r3 = σ 0 2 π ∫ r 2dr = σ 0 2 π 3 a a 1 I = σ 0 2 π [b 3 − a3 ] ⇒ 3 Now, radius of gyration, 2 πσ 0 3 ( b − a3 ) I 3 K= = 2 πσ 0 (b − a) M K=
⇒
Exp. (c) Key Idea Radius of gyration K of any structure is given by 2 I I = MK or K = M To find K, we need to find both moment of inertia I and mass M of the given structure.
Given, variation in mass per unit area (surface mass density), σ …(i) σ= 0 r Calculation of Mass of Disc
…(iv)
1 ( b 3 − a3 ) 3 b−a
As we know, b 3 − a3 = (b − a) (b 2 + a2 + ab ) ∴K =
1 2 (b + a2 + ab ) or K = 3
a2 + b 2 + ab 3
18. A uniform rod of lengthl is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T ( x ) at a distance x from the axis, then which of the following graphs depicts it most closely? [JEE Main 2019, 12 April ShiftI]
dm dr b
σ0 × 2 πrdr [from Eq. (i)] r
b
(a)
…(iii)
r
T(x)
T(x)
(a)
(b)
a l
x
l
T(x)
T(x)
(c)
Let us divide whole disc in small area elements, one of them shown at r distance from the centre of the disc with its width as dr. Mass of this element is dm = σ ⋅ dA σ [from Eq. (i)] dm = 0 × 2 πrdr ⇒ r …(ii) Mass of the disc can be calculated by integrating it over the given limits of r,
x
(d) l
x l
Key Idea In a uniform rod, mass per unit length remains constant. If it is denoted by λ, then m λ = = constant for all segments of rod. l l dm (mass)
x dx
x
72
JEE Main Chapterwise Physics To find tension at x distance from fixed end, let us assume an element of dx length and dm mass. Tension on this part due to rotation is …(i) dT = Kx As, K = mω2 For this element, K = (dm)ω2 …(ii) ∴
dT = (dm)ω2 x
…(iii)
To find complete tension in the rod, we need to integrate Eq. (iii), T
m
0
0
2 ∫ dT = ∫ (dm) ω x
…(iv)
Using linear mass density, m dm λ= = l dx m …(v) ⇒ dm = ⋅ dx l Putting the value of Eq. (v) in Eq. (iv), we get
⇒
l
T=
m 2 x2 m 2 ∫x l ⋅ ω x ⋅ dx = l ⋅ ω 2 x
T=
mω2 2 [l − x2 ] or T ∝ − x2 2l
l
19. A person of mass M is sitting on a swing to length L and swinging with an angular amplitude θ 0. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance l (l 0)]. The terminal speed of the ball is [AIEEE 2008] Vg (ρ1 − ρ2 ) k Vgρ1 (c) k (a)
Vgρ1 k Vg (ρ1 − ρ2 ) (d) k
B
A
B
A
B
A
B
(a)
(b)
(b)
(c)
Exp. (a)
Vρ2g
The forces acting on the ball are gravity force, buoyancy force and viscous force. When ball acquires terminal speed, it is in dynamic equilibrium, let terminal speed of ball be vT . So, kvT2 2 Vρ2 g + kvT = Vρ1g or
A
vT =
(d)
Exp. (c)
V(ρ1 − ρ2 ) g k
Vρ1g
55. A jar is filled with two nonmixing liquids 1 and 2 having densities ρ1 and ρ2 , respectively. A solid ball, made of a material of density ρ3, is dropped in the jar. It comes to equilibrium in the position shown in the figure.
Liquid 1
ρ1
ρ3 Liquid 2 ρ2
Soap solution has lower surface tension, T as compared to pure water and capillary rise 2 T cos θ , so h is less for soap solution. h= ρrg
57. One end of a thermally insulated rod is kept at a temperature T1 and the other at T2. The rod is composed of two sections of lengths l1 and l 2 and thermal conductivitiesK 1 and K 2 respectively. The temperature at the interface of the two sections is [AIEEE 2007] l1
T1
Which of the following is true for ρ1 , ρ2 and [AIEEE 2008] ρ3 ? (a) ρ3 < ρ1 < ρ2 (c) ρ1 > ρ2 > ρ3
(b) ρ1 > ρ3 > ρ2 (d) ρ1 < ρ3 < ρ2
K1
(a)
Exp. (d) ρ1 < ρ2 as denser liquid acquires lowest position of vessel. ρ3 > ρ1 as ball sinks in liquid 1 and ρ3 < ρ2 as ball does not sink in liquid 2, so ρ1 < ρ3 < ρ2
56. A capillary tube (A) is dipped in water. Another identical tube (B ) is dipped in a soapwater solution. Which of the following shows the relative nature of the liquid columns in the two tubes? [AIEEE 2008]
l2
(c)
T2
K2
( K 2 l 2 T1 + K 1 l1 T2 ) ( K 1 l1 + K 2 l 2 ) ( K 1 l 2T1 + K 2 l1T2 ) ( K 1 l 2 + K 2 l1 )
(b) (d)
( K 2 l1T1 + K 1 l 2T2 ) ( K 2l1 + K 1 l 2 ) ( K 1 l1 T1 + K 2 l 2T2 ) ( K 1 l1 + K 2 l 2 )
Exp. (c) Let temperature at the interface be T. For part AB, T1
A
l1
T
K1
B
l2
K2
T2
C
126
JEE Main Chapterwise Physics As rate of flow of heat, Q1 (T1 − T )K1 ∝ t l1 Q2 (T − T2 )K 2 For part BC, ∝ l2 t Q1 Q2 At equilibrium, = t t (T1 − T ) K1 (T − T2 ) K 2 ⇒ = l1 l2
πr02 × σR 2T 4 r
2
=
πr02 R 2 σT 4 r2
59. If the terminal speed of a sphere of gold
(density = 19. 5 kgm −3) is 0.2 ms −1 in a viscous liquid (density = 1.5 kgm −3), find the terminal speed of a sphere of silver (density = 10. 5 kg/m 3) of the same size in the same liquid. [AIEEE 2006]
T1K1l2 + T2 K 2 l1 K1l2 + K 2 l1
T=
or
=
(a) 0.4 ms −1 (c) 0.1 ms −1
(b) 0.133 ms −1 (d) 0.2 ms −1
Exp. (c)
58. Assuming the sun to be a spherical body of radiusR at a temperature ofT K, evaluate the total radiant power, incident on earth, at a distance r from the sun. [AIEEE 2006] (a)
4 πr02 R 2 σT 4
(c)
r02 R 2 σT 4 2
2
r
(b)
πr02 R 2 σT 4 r2 R σT 4
Terminal speed of spherical body in a viscous liquid is given by vT =
where, ρ = density of substance of a body σ = density of liquid.
2
(d)
4 πr
r
2 r 2 (ρ − σ ) g 9η
From given data,
2
ρ − σl vT (Ag) = Ag vt (Gold) ρGold − σ l
where, r0 is the radius of the earth and σ is Stefan’s constant. ⇒
Exp. (b) From Stefan’s law, the rate at which energy is radiated by sun at its surface is
vT (Ag) = =
P = σ × 4 πR 2 × T 4 [sun is a perfectly black body as it emits radiations of all wavelengths and so for it, e = 1] The intensity of this power at earth’s surface (under the assumption r > > r0 ) is TK
r
r0 Earth
R
9 . ms −1 × 0.2 = 01 18
60. A wire elongates by l mm when a load w is hanged from it. If the wire goes over a pulley and two weightsw each are hung at the two ends, the elongation of the wire will be (in mm) [AIEEE 2006] (a) l
(b) 2l
(c) zero
(d)
l 2
Exp. (a) Let us consider the length of wire as L and crosssectional area A, the material of wire has Young’s modulus as Y. Case 1
Sun
σ × 4 πR T P σR T = = 4 πr 2 4 πr 2 r2 The area of earth which receives this energy is only one half of total surface area of earth, whose projection would be πr02 . I=
10.5 − 1.5 × 0.2 19.5 − 1.5
2 4
Case 2
2 4
∴ The total radiant power as received by earth = πr02 × I
L 2
L 2
127
Properties of Solids and Liquids w/ A l /L w/ A l Y = ⇒ l′ = 2 2 l ′/L
the surface through which heat will flow is not constant.
Then, for 1st case, Y = For 2nd case,
x
dx
So, total elongation of both sides = 2 l ′ = l
61. A 20 cm long capillary tube is dipped in water. The water rises upto 8 cm. If the entire arrangement is put in a freely falling elevator, the length of water column in the capillary tube will be [AIEEE 2005] (a) 8 cm
(b) 10 cm (c) 4 cm
(d) 20 cm
Exp. (d) Water fills the tube entirely in gravity less condition. Hence, length of water column in the capillary tube is 20 cm.
62. If S is stress and Y is Young’s modulus of material of a wire, the energy stored in the wire per unit volume is [AIEEE 2005] (a) 2 S 2Y
(b)
S2 2Y
(c)
2Y S2
(d)
S 2Y
Exp. ( b ) Energy stored in wire 1 = × Stress × Strain × Volume 2 Stress S and Young’s modulus = ⇒ Strain = Strain Y Energy stored in wire 1 = × Stress × Strain ∴ 2 Volume S S2 1 = ×S × = Y 2Y 2
63. The figure shows a system of two concentric spheres r1 of radii r1 and r2 and kept at T1 temperatures T1 and T2 , r2 T2 respectively. The radial rate of flow of heat in a substance between the two concentric spheres, is proportional to [AIEEE 2005] (r2 − r1 ) (r1 r2 ) r1 r2 (c) (r2 − r1 ) (a)
r (b) ln 2 r1 (d) (r2 − r1 )
Exp. (c) To measure the radial rate of heat flow, we have to go for integration technique as here, the area of
T1 T2
r1 r2
Let us consider an element (spherical shell) of thickness dx and radius x as shown in figure. Let us first find the equivalent thermal resistance between inner and outer spheres. dx Thermal resistance of shell = dR = K × 4 πx 2 l from R = KA where, K is thermal conductivity r2 dx ⇒ ∫ dR = R = ∫ r1 4 πKx2 r2 − r1 1 1 1 = − = 4 πK r1 r2 4πK(r1r2 ) T1 − T2 R T1 − T2 = × 4πK(r1r2 ) r2 − r1 rr ∝ 12 r2 − r1
Rate of heat flow = H =
64. A wire fixed at the upper end stretches by lengthl by applying a force F .The work done in stretching is [AIEEE 2004] (a)
F 2l
(b) F l
(c) 2 F l
(d)
Fl 2
Exp. (d) As work done in stretching the wire = Potential energy stored 1 = × Stress × Strain × Volume 2 l 1 F 1 = × × × AL = Fl 2 A L 2
65. Spherical balls of radius R are falling in a
viscous fluid of viscosity η with a velocity v. The retarding viscous force acting on the spherical ball is [AIEEE 2004]
128
JEE Main Chapterwise Physics (a) directly proportional to R but inversely proportional to v (b) directly proportional to both radius R and velocity v (c) inversely proportional to both radius R and velocity v (d) inversely proportional to R but directly proportional to velocity v
Exp. (b)
In 2nd case, P2 = σe × 4 π(2 R )2 × (2 T )4 = σe × 4 πR 2 × T 4 × 64 = 64P1 The rate at which energy received at earth is P × AE E= 2 4 πRSE where, AE = area of earth and RSE = distance between sun and earth So, in 1st case,
Retarding force acting on a ball falling into a viscous fluid where, and
E1 =
F = 6πηRv R = radius of ball, v = velocity of ball η = coefficient of viscosity.
∴ F ∝ R and F ∝ v Or in other words, retarding force is directly proportional to both R and v.
66. If two soap bubbles of different radii are connected by a tube
[AIEEE 2004]
(a) air flows from the bigger bubble to the smaller bubble till the sizes become equal (b) air flows from bigger bubble to the smaller bubble till the sizes are interchanged (c) air flows from the smaller bubble to the bigger (d) there is no flow of air
Exp. (c) The excess pressure inside the soap bubble is inversely proportional to radius of soap bubble i.e., p ∝ 1/r, r being the radius of bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble. Thus, if these two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the bigger bubble grows at the expense of the smaller one.
67. If the temperature of the sun were to increase from T to 2 T and its radius from R to 2 R , then the ratio of the radiant energy received on earth to what it was previously, will be [AIEEE 2004] (a) 4 (c) 32
(b) 16 (d) 64
Exp. (d) From Stefan’s law, the energy radiated by sun is given by P = σeAT 4 . In 1st case,
P1 = σe × 4 πR 2 × T 4
E2 =
P1 2 4 πRSE
P2 2 4 πRSE
× AE × AE = 64E1
Hence, ratio is 64.
68. The temperature
x
4x
of the two outer surfaces of a T K 2K T1 2 composite slab, consisting of two materials having coefficients of thermal conductivity K and 2 K and thickness x and 4x respectively are T2 and T1(T2 > T1 ). The rate of heat transfer through the slab, in a steady state is A (T2 − T1 )K f , then f is equal to x [AIEEE 2004]
(a) 1
(b) 1/2
(c) 2/3
(d) 1/3
Exp. (d) Let the temperature of common interface be T°C. Rate of heat flow Q KA ∆T H= = t l 2 KA(T − T1 ) Q H1 = = ∴ t 1 4x ( KA T Q 2 − T) and H2 = = t 2 x In steady state, the rate of heat flow should be same in whole system i.e., H1 = H2 2 KA(T − T1 ) KA(T2 − T ) = ⇒ 4x x T − T1 or = T2 − T 2 or T − T1 = 2 T2 − 2 T
129
Properties of Solids and Liquids 2 T2 + T1 …(i) 3 Hence, heat flow from composite slab is 2 T2 + T1 KA(T2 − T ) KA H= = T2 − x 3 x KA …(ii) (T2 − T1 ) = 3x A(T − T1 )K …(iii) Accordingly, H = 2 f x or
T=
Exp. (a) Energy radiated per second by a body which has surface area A at temperature T is given by Stefen’s law, E = σAT 4 [as A = π r 2 ] 2
Therefore,
[since, bodies are of same material, so e1 = e 2 ] E1 16 1 = = = 1: 1 ⇒ E2 16 1
By comparing Eqs. (ii) and (iii), we get 1 f= 3
69. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then, the elastic energy stored in the wire is [AIEEE 2003] (a) 0.2 J
(b) 10 J
(c) 20 J
(d) 0.1 J
Exp. (d) Elastic energy stored in the wire is 1 U = × Stress × Strain × Volume 2 1 F ∆l 1 = × × × AL = F ∆l 2 A L 2 1 = × 200 × 1 × 10−3 = 01 . J 2
70. According to Newton’s law of cooling, the rate of cooling of a body is proportional to ( ∆θ )n , where ∆θ is the difference of the temperature of the body and the surroundings, then n is equal to [AIEEE 2003] (a) 2
(b) 3
(c) 4
(d) 1
Exp. (d)
72. A cylinder of height 20 m is completely filled with water. The velocity of efflux of water (in ms −1) through a small hole on the side wall of the cylinder near its bottom, is (a) 10 (c) 25.5
(b) 20 (d) 5
[AIEEE 2004]
Exp. (d) Applying the Bernoulli’s theorem just inside and outside the hole. 20 m Take reference line for gravitational potential energy at the bottom of the vessel. Let p0 be the atmospheric pressure, ρ be the density of liquid and v be the velocity at which water is coming out. ρv 2 pinside + ρgh + 0 = poutside + 2 ρv 2 p0 + ρgh = p0 + ⇒ 2 or v = 2 gh = 2 × 10 × 20 = 20 ms −1
According to Newton’s law of cooling, dQ Rate of loss of heat, ∝ ∆θ dt dQ But ∝ (∆θ)n dt ∴ n=1
73. If massenergy equivalence is taken into [given]
71. Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K, respectively. The ratio of the energy radiated per second by the first sphere to that by the second is [AIEEE 2002] (a) 1 : 1
4
2 4 E1 r1 T1 1 4000 = = E2 r2 T2 4 2000
(b) 16 : 1
(c) 4 : 1
(d) 1 : 9
account, when water is cooled to form ice, the mass of water should [AIEEE 2002] (a) (b) (c) (d)
increase remain unchanged decrease first increase then decrease
Exp. ( a ) According to the massenergy equivalence, mass and energy remain conserved. So, when water is cooled to form ice, water loses its energy, so change in energy increases the mass of water.
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JEE Main Chapterwise Physics
74. Which of the following is more close to a black body ?
[AIEEE 2002]
(a) Black board paint (b) Green leaves (c) Black holes (d) Red roses
Exp. (c) Black body is one which absorbs all radiations and emits none, out of the four examples given black holes are more close to a black body.
75. Infrared radiations are detected by
76. Heat given to a body which raises its temperature by 1°C is (a) (b) (c) (d)
[AIEEE 2002]
water equivalent thermal capacity specific heat temperature gradient
Exp. (b) The thermal capacity of a substance is defined as the amount of heat required to raise its temperature by 1°C.
[AIEEE 2002]
(a) spectrometer (c) nanometre
(b) pyrometer (d) photometer
Exp. (b) Spectrometer is an instrument which is used to obtain a pure spectrum of white light. It is used to determine the wavelength of different colours of white light (or wavelength of monochromatic light source), the refractive index of the material of the prism and the dispersive power of the material of the prism. Pyrometer is infrared sensitive device, so it is used to detect infrared radiations. Nanometre is the small unit of distance and is not a device. Photometer is used to measure luminous intensity, illuminance and other photometric quantities.
77. Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will [AIEEE 2002]
(a) (b) (c) (d)
increase decrease remain same decrease for some, while increase for others
Exp. (c) Temperature of a gas is determined by the total transactional kinetic energy measured with respect to the centre of mass of the gas. Therefore, the motion of centre of mass of the gas does not affect the temperature. Hence, the temperature of gas will remain same.
8 Thermodynamics a
1. A thermally insulated vessel contains
p
150 g of water at 0°C. Then, the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to (Latent heat of vaporisation of water = 2.10 × 106 Jkg−1 and latent heat of fusion of water = 3.36 × 105 Jkg−1) [JEE Main 2019, 8 April ShiftI]
(a) 150 g (c) 130 g
(b) 20 g (d) 35g
b c
d
V
(a) d a b c (b) a d b c (c) d a c b (d) a d c b
Exp. (a) Given processes are p a
Exp. (b) Let x grams of water is evaporated. According to the principle of calorimetry, Heat lost by freezing water (that turns into ice) = Heat gained by evaporated water Given, mass of water = 150 g ⇒ (150 − x) × 10−3 × 3.36 × 105 = x × 10−3 × 2 .10 × 106 ⇒ ⇒ ∴
(150 − x) × 3.36 = 21x 150 x= 7.25 = 20.6 x ≈ 20 g
b c d
V
For process a, pressure is constant. ∴a is isobaric. For process d, volume is constant. ∴d is isochoric. Also, as we know that, slope of adiabatic curve in pV diagram is more than that of isothermal curve. ∴b is isothermal and c is adiabatic.
3. Following figure shows two processes A and
isochoric, isobaric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by
B for a gas. If ∆Q A and ∆Q B are the amount of heat absorbed by the system in two cases, and ∆U A and ∆U B are changes in internal energies respectively, then
[JEE Main 2019, 8 April ShiftII]
[JEE Main 2019, 9 April ShiftI]
2. The given diagram shows four processes, i.e.
132
JEE Main Chapterwise Physics θ2 + θ1 2 θ1 5θ2 (c) + 6 6
p
A B
2θ2 3 9θ2 10
Let interface temperature in steady state conduction is θ, then assuming no heat loss through sides; Rate of heat Rate of heat flow through = flow through second slab first slab
V
∆Q A > ∆Q B , ∆U A > ∆U B ∆Q A < ∆Q B , ∆U A < ∆U B ∆Q A > ∆Q B , ∆U A = ∆U B ∆Q A = ∆Q B ; ∆U A = ∆U B
Exp. (c) According to the first law of thermodynamics, Heat supplied (∆Q ) = Work done (W ) + Change in internal energy of the system (∆U ) ∆ QA = ∆ U A + W A Similarly, for process B, ∆ QB = ∆ U B + W B Now, we know that, work done for a process = area under it’s pV curve p
θ2
⇒ ⇒ ⇒
f
q
d 3K
3d K
θ1
(3K ) A(θ2 − θ) KA(θ − θ1 ) = 3d d 9(θ2 − θ) = θ − θ1 ⇒ 9θ2 + θ1 = 10 θ 9 1 θ = θ2 + θ1 10 10
5. A cylinder with fixed capacity of 67.2 L contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20°C is [Take, R = 8.31 J mol−1K −1]
p A
(b)
Exp. (d)
i
(a) (b) (c) (d)
θ1 + 3 θ (d) 1 + 10
(a)
f
f
[JEE Main 2019, 10 April ShiftI] B
(a) 700 J
i O
i Vi
Vf
V
O
Vi
Vf
V
Thus, it is clear from the above graphs, …(i) WA > WB Also, since the initial and final state are same in both process, so …(ii) ∆U A = ∆U B So, from Eqs. (i) and (ii), we can conclude that ∆ QA > ∆ QB
4. Two materials having coefficients of thermal conductivity ‘3K ’ and ‘K ’ and thickness ‘d’ and ‘3d’ respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ‘θ 2’ and ‘θ1’ respectively, (θ 2 > θ1 ). The temperature at the interface is [JEE Main 2019, 9 April ShiftII] d 3d θ2 3K K
θ1
(b) 748 J
(c) 374 J
(d) 350 J
Exp. (b) Given, capacity of cylinder is = 67.2 L and ∆T = 20° C At STP volume = 22.4 L 67.2 ∴Number of moles = =3 22.4 Now, change in heat is given as ∆Q = nC V ∆T Substituting the given values, we get 3R 3 ∆Q = 3 × × 20 (Qfor He gas, C V = R ) 2 2 = 90 R J Given, R = 8.31 J mol −1K −1 ∴
∆Q = 90 × 8.31 = 747.9 J = 748 J
6. When heatQ is supplied to a diatomic gas of rigid molecules, at constant volume, its temperature increases by ∆T . The heat required to produce the same change in temperature, at a constant pressure is [JEE Main 2019, 10 April ShiftII]
133
Thermodynamics 2 Q 3 3 (c) Q 2
5 Q 3 7 (d) Q 5
(b)
(a)
Exp. (d) Given, heat supplied at constant volume is Q. ∴ Q = nC V ∆T For same change in temperature, if heat supplied at constant pressure is Q, then Q ′ = nC p ∆T So, we have Q ′ nC p ∆T Q′ C p = = ⇒ Q nC V ∆T Q CV ⇒
Cp Q γ = CV
Q ′ = Qγ
As given gas is diatomic, so γ = Q′ =
∴
7 5
7 Q 5
7. A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is −180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the pathbc. The work done by the gas along the path abc is [JEE Main 2019, 12 April ShiftI] c
p
a
…(ii)
…(iii)
…(iv) 0…(v)
…(vi)
…(vii)
8. A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62°C, its efficiency is doubled. The temperatures of the source and the sink are respectively, [JEE Main 2019, 12 April ShiftII] (a) 62°C, 124°C (c) 124°C, 62°C
(b) 99°C, 37°C (d) 37°C, 99°C
Exp. (b)
b V
(a) 120 J (c) 100 J
In Process a → b Given, ∆Qab = 250 J ∴ 250 J = ∆U ab + dWab In Process b → c Given, ∆Qbc = 60 J Also, V is constant, so dV = 0 ⇒ dWbc = p(dV )bc = 0 ∴ 60 J = ∆U bc + 0 ⇒ ∆U bc = 60 J In Process c → a Given, ∆Uca = −180 J Now, for complete cycle, ∆U abca = ∆U ab + ∆U bc + ∆Uca = From Eqs. (iii), (iv) and (v), we get ∆U ab = − ∆U bc − ∆Uca ∆U ab = −60 + 180 = 120 J From Eq. (ii), we get 250 J = 120 J + dWab ⇒ dWab = 130 J From Eqs. (i) and (vii), we get Work done by the gas along the path abc, dWabc = dWab + dWbc = 130 J + 0 K ⇒ dWabc = 130 J
(b) 130 J (d) 140 J
Exp. (b) Key Idea In pV curve, work done dW , change in internal energy ∆U and heat absorbed ∆Q are connected with first law of thermodynamics, i.e. …(i) ∆Q = ∆U + dW and total change in internal energy in complete cycle is always zero. Using this equation in different part of the curve, we can solve the given problem.
Efficiency of a Carnot engine working between source of temperature T1 and sink of temperature T2 is given by T η = 1− 2 T1 Here, T2 and T1 are absolute temperatures. Initially, 1 η= 6 T T 1 5 ∴ = 1− 2 ⇒ 2 = T1 6 T1 6 Finally, efficiency is doubled on reducing sink temperature by 62°C.
134
JEE Main Chapterwise Physics ∴ So, ⇒
2 , Tsink = T2′ = T2 − 62 6 T′ η = 1− 2 T ′1 T2 − 62 T2 − 62 4 2 = 1− = ⇒ T1 6 6 T1 η=
T2 62 4 5 62 4 − = ⇒ − = T1 T1 6 6 T1 6 T2 5 Q = T1 6 ⇒T1 = 6 × 62 = 372 K = 372 − 273 = 99° C 5 – 310 K T2 = × T1 ~ ⇒ 6 = 310 − 273 = 37 ° C ⇒
When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If path ADB is used work done by the system is 10 J the heat flow into the system in path ADB is [ JEE Main 2019, 9 Jan ShiftI]
(a) 80 J
(b) 40 J
(c) 100 J
(d) 20 J
Exp. (b) For the ABC as shown in the figure below, p
C
B
A
9. A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process? [JEE Main 2019, 12 April ShiftII]
(a) 25 J (c) 30 J
(b) 35 J (d) 40 J
Exp. (b) Work done by a gas is (given) …(i) W = p ∆V = nR∆T = 10J As the gas is diatomic, C p = specific heat of gas at constant pressure 7 = R 2 So, heat absorbed by gas at constant pressure is 7 ∆Q = nC p ∆T = n R ∆T 2 7 = (nR ∆T ) 2 7 [From Eq. (i)] = × 10 2 = 35J
10. A gas can be taken from A to B via two different processes ACB and ADB. p C
B
A
D
V
According to the first law of thermodynamics, heat supplied, ∆Q = work done (∆W ) + internal energy (∆U) ⇒ ∆QCB = ∆WACB + (U B − U A ) [where, ∆U = U B − U A ] Substituting the given values, ...(i) U B − U A = 60 − 30 = 30 J Similarly for the ADB as shown in the figure below,
B
p
A
D V
⇒
∆QADB = ∆WADB + (U B − U A ) [using Eq. (i)] ∆QADB = 10 + 30 = 40 J
11. Temperature
difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of 3L is same crosssection as AB and length 2 connected across AB (see figure). In steady state, temperature difference between P andQ will be close to [ JEE Main 2019, 9 Jan ShiftI]
V
135
Thermodynamics
Calculate the temperature T2 if the work outputs of the two engines are equal.
L 4 A
L
P
(a) 45° (c) 75°C
[ JEE Main 2019, 9 Jan ShiftII]
B
Q
(a) 600 K (c) 400 K
(b) 35°C (d) 60°C
Exp. (b)
Exp. (a) According to the given question, the given figure with its length for each section is given as below L L/4
Key idea In a Carnot engine the heat flow from higher temperature source (at T1) to lower temperature sink (at T2) and give the work done equal to the W = Q 1 − Q 2.
L/4
A
T1
B L/2
P
L
Q
L/2
R R/4 R/2
R
W=Q1 –Q2 (Work output) Q2 T2
Sink
S
For the given condition, Carnot engine A and B are operated in series as shown below
R/4
R P
Source Q1
The above figure considering that every section has the same thermal conductivity, then in terms of thermal resistance is shown in the figure below,
A
(b) 500 K (d) 300 K
Q
R/2
Net resistance of the section PQRS is 3R R× 2 = 3R = 5R 5 2 Total resistance of the net network, Rnet R R 3R 8R = + + = 2 2 5 5 ∆T ∴Thermal current, I = AB Rnet 120 − 0 120 × 5 I= = 8R 8R 5
B
W1
T1=600K
...(i)
Thus, the net temperature difference between point P and Q is 3R [using Eq. (i)] TP − TQ = I × 5 120 × 5 3R = 45°C = × 8R 5
12. Two Carnot engines A and B are operated in series. The first one, A receives heat at T1( = 600 K ) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and in turn rejects to a heat reservoir at T3 ( = 400 K ).
A
Q1
Q2
W2
T2
B
Q2 T3=400K Q3
Q3
C
where, Q1 = heat rejected by engine A at T1K, Q2 = heat received by engine B at T2K and Q3 = heat rejected by engine B to source C at T3K. According to Carnot engine principle, W1 = Q1 − Q2 (workoutput from source A and B) W2 = Q2 − Q3 (work output from source B and C) As per the given condition, if the work outputs of the two engines are equal, then Q1 − Q2 = Q2 − Q3 ⇒ Q1 + Q3 = 2 Q2 Q1 Q3 …(i) + =2 Q2 Q2 T1 T3 Therefore + =2 T2 T2 T1 + T3 So, =2 T2 T + T3 600 + 400 ⇒ T2 = 1 = 2 2 T2 = 500 K
136
JEE Main Chapterwise Physics
13. A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 WK −1m −1, the energy flux through it in the steady state is [ JEE Main 2019, 10 Jan ShiftI] −2
−2
(a) 90 Wm (c) 120 Wm −2
(a) T2 = (T13T4 )1/ 4 ; T3 = (T1T43 )1/ 4 (b) T2 = (T12T4 )1/ 3 ; T3 = (T1T42 )1/ 3 (c) T2 = (T1 T4 )1/ 2 ; T3 = (T12T4 )1/ 3 (d) T2 = (T1 T42 )1/ 3 ; T3 = (T12T4 )1/ 3
Exp. (b) Given, Carnot engines operates as,
(b) 65 Wm (d) 200 Wm −2
T1
Exp. (a)
ε1
Energy flux is the rate of heat flow per unit area through the rod. Also, rate of flow of heat per unit time through a material of area of crosssection A and thermal conductivity k between the temperatures T1 and T2 (T1 > T2 ) is given as, ∆Q kA(T1 − T2 ) = ∆t l Energy flux using Eq. (i), we get 1 ∆Q k(T1 − T2 ) = ⋅ = A ∆t l Here, k = 01 . K K − 1m− 1, l = 1m Energy flux =
011000 .( − 100) 1 −2
= 90 Wm .
14. Three Carnot engines operate in series between a heat source at a temperature T1 and a heat sink at temperature T4 (see figure). There are two other reservoirs at temperatures T2 and T3, as shown with T1 > T2 > T3 > T4 . The three engines are equally efficient if [ JEE Main 2019, 10 Jan ShiftI] T1
ε2 T3
… (i)
T1 = 1000 K and T2 = 100 K ∴
T2
ε3 T4
As, efficiency of a Carnot’s engine is given by T η = 1 − sink Tsource We have, T2 T1 T η2 = efficiency of engine ε2 = 1 − 3 T2 T η3 = efficiency of engine ε3 = 1 − 4 T3 η1 = efficiency of engine ε1 = 1 −
For equal efficiencies, η1 = η2 = η3 T T T 1− 2 = 1− 3 = 1− 4 ⇒ T1 T2 T3 T2 T3 T4 = = ⇒ T1 T2 T3 ⇒
ε1 T2
⇒
2 T22 = TT 1 3 and T3 = T2T4
T24 = T12T32 T24 = T12T2T4
ε2
T23 = T4T12 T3
ε3 T4
1
or
T2 = (T12 T4 )3
Also,
2 T34` = T22 T42 = TT 1 3T4 2 T33 = TT 1 4
1
2 3 or T3 = (TT 1 4 )
137
Thermodynamics 15. An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4°C. Calculate the specific heat of the unknown metal, if water temperature stabilises at 21.5°C. (Take, specific heat of brass is 394 J kg −1 −1 [ JEE Main 2019, 10 Jan ShiftII] K ) (a) 916 J kg −1 K −1 (c) 1232 J kg −1 K −1
(b) 654 J kg −1 K −1 (d) 458 J kg −1 K −1
Exp. (a) Key Idea The principle of calorimetry states that total heat lost by the hotter body equals to the total heat gained by colder body, provided that there is no exchange of heat with the surroundings.
Let specific heat of unknown metal is s and heat lost by this metal is ∆Q. Heat lost and specific heat of a certain material/substance are related as … (i) ∆Q = ms∆T For unknown metal, m = 192 g and ∆T = (100 − 21.5) º C ...(ii) ∴ ∆Q ′ = 192(100 − 21.5) × s Now, this heat is gained by the calorimeter and water inside it. As, heat gained by calorimeter can be calculated by Eq. (i). So, for brass specific heat, (given) s = 394 J kg − 1 K − 1 = 0.394 J g − 1 K − 1 Mass of calorimeter, m = 128 g Change in temperature, ∆T = (215 . − 8.4)º C So, using Eq. (i) for calorimeter, heat gained by brass …(iii) ∆Q1 = 128 × 0.394 × (21.5 − 8.4) Heat gained by water can be calculated as follows mass of water, m = 240 g, specific heat of water, s = 4.18 J g − 1K − 1, change in temperature, ∆T = (21.5 − 8.4)º C Using Eq. (i) for water also, we get heat gained by water, …(iv) ∆Q2 = 240 × 418 . × (215 . − 8.4) Now, according to the principle of calorimeter, the total heat gained by the calorimeter and water must be equal to heat lost by unknown metal ∆Q ′ = ∆Q1 + ∆Q2
Using Eqs. (ii), (iii) and (iv), we get ⇒ = 192(100 − 21.5) × s = 128 × 0.394 × (21.5 − 8.4) + 240 × 418 . × (21.5 − 8.4) ⇒ 15072 s = 660.65 + 13142 ⇒ s = 0.916 J g − 1 K − 1 or
s = 916 J Kg − 1K − 1.
16. Ice at − 20°C is added to 50 g of water at 40°C. When the temperature of the mixture reaches 0°C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Take, specific heat of water = 4.2 J/g/°C specific heat of ice = 2.1 J/g/°C and heat of fusion of water at 0°C = 334 J/g) [ JEE Main 2019, 11 Jan ShiftI]
(a) 40 g
(b) 50 g
(c) 60 g
(d) 100 g
Exp. (a) Let amount of ice be ‘x’ gm. According to the principle of calorimeter, heat lost by water = heat gained by ice Here, heat lost by water, ∆Q = mswater ∆ T Substituting the given values, we get ∆Q = 50 × 42 . × 40 Heat gained by ice, ∆Q = x sice ∆T + ( x − 20) L = x × 2.1 × 20 + ( x − 20) × 334 = 20 x × 2.1 + 334 x − 6680 ∴ 20 x × 2.1 + 334 x − 6680 = 50 × 42 . × 40 42 x + 334 x − 6680 = 8400 ⇒ 376 x = 15080 or x = 4010 . g x~ − 40 g
17. Two rods A and B of identical dimensions are at temperature 30ºC. If A is heated upto 180ºC and B upto T ºC, then new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is [ JEE Main 2019, 11 Jan ShiftII] (a) 230ºC
(b) 270ºC
(c) 200ºC
(d) 250ºC
Exp. (a) Let initial length of identical rods is l0 Thermal expansion in length of rod due to heating is given by the relation ∆l = l0 α (∆T ) = l0 α (T2 − T1 )
138
JEE Main Chapterwise Physics Here, α is coefficient of linear expansion. So, change in length of rods are ∆l1 = l0 α1(180 − 30) ∆l2 = l0α 2 (T − 30) Because new lengths are same, so change in lengths of both rods are equal. i.e. ∆l1 = ∆l2 ⇒ l0 α1 (180 − 30) = l0 α 2 (T − 30) or
α1 (T − 30) = α2 150
Given, α1 : α 2 = 4 : 3 T − 30 4 4 ∴ = ⇒ T − 30 = × 150 = 200 150 3 3 or T = 200 + 30 = 230°C
18. A metal ball of mass 0.1 kg is heated upto 500ºC and dropped into a vessel of heat capacity 800 JK − 1 and containing 0.5 kg water. The initial temperature of water and vessel is 30ºC. What is the approximate percentage increment in the temperature of the water? [Take, specific heat capacities of water and metal are respectively 4200 Jkg− 1K − 1 and 400 Jkg− 1K − 1] [ JEE Main 2019, 11 Jan ShiftII] (a) 25%
(b) 15%
(c) 30%
(d) 20%
Exp. (d) Using heat lost or gained without change in state is ∆Q = ms∆T, where s is specific heat capacity and T = change in temperature Let final temperature of ball be T. Then heat lost by ball is, ...(i) ∆Q = 01 . × 400(500 − T ) This lost heat by ball is gained by water and vessel and given as Heat gained by water, ...(ii) ∆Q1 = 0.5 × 4200(T − 30) and heat gained by vessel is ∆Q2 = heat capacity ×∆T ...(iii) = 800 × (T − 30) According to principle of calorimetry, total heat lost = total heat gained ⇒ 01 . × 400(500 −T ) = 0.5 × 4200(T − 30) + 800(T − 30) (2100 + 800)(T − 30) ⇒ (500 − T ) = 40 ⇒ 500 − T = 72 . 5(T − 30)
⇒ 500 + 217.5 = 72.5T or T = 36.39 K So, percentage increment in temperature of water 36.39 − 30 = × 100 ≈ 20% 30
19. A thermometer graduated according to a linear scale reads a value x 0, when in contact with boiling water and x 0 / 3, when in contact with ice. What is the temperature of an object in ºC, if this thermometer in the contact with the object reads x 0 / 2 ? [ JEE Main 2019, 11 Jan ShiftII]
(a) 35
(b) 60
(c) 40
(d) 25
Exp. (d) By principle of thermometry for any liner temperature scale, T − TLFP = a (constant) TUFP − TLFP where, T = temperature measured TLFP = temperature of melting ice or lower fixed point. TUFP = temperature of boling water or upper fixed point. If, T = temperature of given object, then we have, x0 xo − T − 0° C 3 or T = 1 or = 2 100° C − 0° C x − x0 100 4 0 3 T = 25° C
20. For the given cyclic process CAB as shown for a gas, the work done is [ JEE Main 2019, 12 Jan ShiftI] 6.0
C
A
5 p (Pa)
4 3 2 1
B 1
(a) 5 J (c) 1 J
Exp. (b)
2
3
(b) 10 J (d) 30 J
4
5 3 V (m )
139
Thermodynamics Key Idea In a cyclic thermodynamic process work done = area under p – V diagram. Also in clockwise cycle, work done is positive. In the given cyclic process, work done = ∫ pdV = area enclosed by the cycle 1 × base × height of triangle (CAB) made by cycle 2 1 = (V2 − V1 ) ( p2 − p1 ) 2 From graph, given V2 = 5 m 3 , V1 = 1 m 3 , =
p2 = 6 Pa, p1 = 1 Pa 1 = (5 − 1) (6 − 1) 2 1 = × 4 × 5 = 10 J 2
cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K 1 and that of the outer cylinder is K 2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is [ JEE Main 2019, 12 Jan ShiftI] K1 + K 2 2 2 K 1 + 3K 2 (c) 5
(b)
K 1 + 3K 2 4
Exp. (c) For adiabatic process relation of temperature and volume is, T2 V2γ − 1 = T1V1γ − 1 T2 (2 V )2 / 3 = 300(V )2 / 3 5 [γ = for monoatomic gases] 3 300 T2 = 2 / 3 ≈ 189 K ⇒ 2 Also, in adiabatic process, ∆Q = 0, ∆U = − ∆W − nR(∆T ) 3 25 or ∆U = = − 2 × × (300 − 189) γ −1 2 3 ≈ − 2.7 kJ T2 ≈ 189 K, ∆U ≈ 2.7kJ
pressure and constant volume, respectively. It is observed that C p − C V = a for hydrogen gas C p − C V = b for nitrogen gas. The correct relation between a andb is
(d) K 1 + K 2
[JEE Main 2017 (Offline)]
Both the given cylinders are in parallel as heat flow is given along length. In parallel, equivalent thermal conductivity of system is K A + K 2 A2 Keq = 1 1 A1 + A2 So, in given system
Heat
or
[JEE Main 2018]
(a) (i) 189 K (ii) 2.7 kJ (b) (i) 195 K (ii) −2.7 kJ (c) (i) 189 K (ii) −2.7 kJ (d) (i) 195 K (ii) 2.7 kJ
23. C p and C V are specific heats at constant
Exp. (b)
Keq =
occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V . Calculate (i) the final temperature of the gas and (ii) change in its internal energy.
⇒
21. A cylinder of radius R is surrounded by a
(a)
22. Two moles of an ideal monoatomic gas
K1( πR 2 ) + K 2 [ π(2 R )2 − πR 2 ] ( πR 2 ) + ( 4 πR 2 − πR 2 ) K + 3K 2 Keq = 1 4
(a) a = b
(b) a =14 b (c) a = 28 b (d) a =
1 b 14
Exp. (b) By Mayor’s relation, for 1 g mole of a gas, Cp − CV = R So, when n gram moles are given, R Cp − CV = n As per given question, R a = C p − C V = ; for H2 2 R b = C p − C V = ; for N2 28 From Eqs. (i) and (ii), we get a = 14b
K (i) K (ii)
140
JEE Main Chapterwise Physics
24. A copper ball of mass 100 g is at a temperature T . It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is (Given, room temperature = 30°C, specific heat of copper = 0.1 cal/g°C) [JEE Main 2017 (Offline)]
(a) 885°C (c) 825°C
(b) 1250°C (d) 800°C
T = 885° C
25. An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure p and volumeV is given by pV n = constant, then n is given by (Here C p and C V are molar specific heat at constant pressure and constant volume, respectively) [JEE Main 2016 (Offline)]
Cp CV Cp − C C − CV
(b) n =
C − CV
A
p0
B
mw sw ∆T + mc sc ∆T = mB sB ∆T
∴
C − CV C − Cp
p 2 p0
= 100 × 01 . × (T − 75)
C − Cp
A and B as shown in the figure. The maximum temperature of the gas during the process will be [JEE Main 2016 (Offline)]
⇒ 170 × 1 × (75 − 30) + 100 × 01 . × (75 − 30)
(c) n =
C − CV
⇒ n=
26. n moles of an ideal gas undergoes a process
Heat gained (water + calorimeter) = Heat lost by copper ball
(a) n =
Cp − CV
Thus, number of moles n is given by n =
Exp. (a) ⇒
n = 1−
⇒
2V0
V0
(a)
9 p 0V 0 4 nR
(b)
3 p 0V 0 2 nR
(c)
9 p 0V 0 2 nR
V
(d)
9p 0V 0 nR
Exp. (a) As, T will be maximum temperature where product of pV is maximum p A
2p0 p0
B
C − Cp
C − CV C − CV (d) n = C − Cp
Exp. (b) For polytropic process, specific heat for an ideal gas, R + CV C= 1− n R ∴ + CV = C 1− n R = C − CV ⇒ 1− n R = 1 − n (where, R = C p − C V ) ⇒ C − CV Cp − CV = 1− n ⇒ C − CV
V V0
2V0
Equation of line AB, we have y − y1 y − y1 = 2 ( x − x1 ) x2 − x1 2 p0 − p0 (V − 2 V0 ) p − p0 = ⇒ V0 − 2 V0 − p0 p − p0 = (V − 2 V0 ) ⇒ V0 − p0 ⇒ V + 3 p0 p= V0 − p0 2 pV = V + 3 p0 V V0 −p nRT = 0 V 2 + 3 p0 V V0 T=
1 − p0 2 V + 3 p0 V nR V0
141
Thermodynamics For maximum temperature, ∂T =0 ∂V − p0 (2 V ) + 3 p0 = 0 V0 − p0 3 (2 V ) = − 3 p0 ⇒ V = V0 V0 2 (condition for maximum temperature) Thus, the maximum temperature of the gas during the process will be 1 − p0 9 2 3 Tmax = × V0 + 3 p0 × V0 nR V0 4 2 =
1 nR
− 9 p V + 9 p V = 9 p0 V0 4 0 0 2 0 0 4 nR
Alternate Solution Since, initial and final temperature are equal, hence maximum temperature is at the middle of line.
T0 Tmax
pV = nRT ⇒
V
3 p 3V0 0 2 2 = Tmax nR
9 p0 V0 = Tmax 4 nR
27. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit 1 U U volume u = ∝ T 4 and pressure p = . 3 V V If the shell now undergoes an adiabatic expansion, the relation between T and R is [JEE Main 2015] −R
(a) T ∝ e 1 (c) T ∝ R
(b) T ∝ e
− 3R
(d) T ∝
1
Exp. (c) According to question, 1 U p = 3 V
R3
or ⇒
VT 3 = constant
4 π R 3T 3 = constant or TR = constant 3 1 T∝ R
28. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways [JEE Main 2015] (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.
(a) ln 2, 4 ln 2 (c) ln 2, 2 ln 2
T0 V0 3V0 2V0 2
i.e.
or
[Q pV = nRT ]
In both the cases, body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively, is
p 2p0 3p0 2 p0
nRT 1 U = 3 V V nRT 1 4 ∝ T or 3 V
⇒
(b) ln 2, ln 2 (d) 2 ln 2, 8 ln 2
Exp. (b) Since, entropy is a state function, therefore change in entropy in both the processes must be same. Therefore, correct option should be (b).
29. One
B 800 K mole of P diatomic ideal gas undergoes a cyclic process ABC as shown in figure. 600 K The process BC is C A 400 K adiabatic. The V temperatures at A, B and C are 400 K, 800 K and 600 K, respectively.
Choose the correct statement. [JEE Main 2014] (a) The change in internal energy in whole cyclic process is 250 R (b) The change in internal energy in the process CA is 700 R (c) The change in internal energy in the process AB is − 350 R (d) The change in internal energy in the process BC is − 500R
142
JEE Main Chapterwise Physics
Exp. (d)
31. Helium gas goes through a cycle ABCDA
According to first law of thermodynamics, we get (i) Change in internal energy from A to B i.e., ∆U AB ∆ U AB = nC V (TB − TA ) 5R (800 − 400) = 1000 R = 1× 2 (ii) Change in internal energy from B to C ∆U BC = nC V (TC − TB ) 5R (600 − 800) = − 500 R = 1× 2 (iii) ∆ U total = 0 (iv) Change in internal energy from C to A i.e., ∆UCA ∆ UCA = nC v (TA − TC ) 5R (400 − 600) = − 500 R = 1× 2
30. The given pV diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is [JEE Main 2013]
(consisting of two isochoric and isobaric lines) as shown in figure. Efficiency of this cycle is nearly (Assume the gas to be close to ideal gas) [AIEEE 2012] 2p0 p0
(a) 15.4%
2V0
(a) p 0V 0 11 (c) p 0V 0 2
(d) 4p 0V 0
Exp. (b) Heat supplied = nC V (2T0 − T0 ) + nC p + (2T0 − T0 )
2p0
2T0
D
(b) 9.1%
2V0
(c) 10.5%
(d) 12.5%
Exp. (a) Efficiency of a process is defined as the ratio of work done to energy supplied. Here, ∆W Area under p  V diagram η= = ∆Q ∆QAB + ∆QBC p0 V0 ∴ η= nC V ∆T1 + nC p ∆T2 p0 V0 3 5 nR(TB − TA ) + nR(TC − TD ) 2 2 p0 V0 = 5 3 (2 p0 V0 − p0 V0 ) + (4 p0 V0 − 2 p0 V0 ) 4 2 [as pV = nRT] p0 V0 1 = = = 15.4% 3 5 p0 V0 + ⋅ 2 p0 V0 6.5 2 2 =
p0
13 (b) p 0V 0 2
C
A V0
2p0
V0
2T0 B
32. A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be [AIEEE 2012] (a) efficiency of Carnot engine cannot be made larger than 50% (b) 1200 K (c) 750 K (d) 600 K
Exp. (c) p0
T0 V0
2T0 2V0
Now, =
n3RT0 n5R 13 13 2T0 = nRT0 = p0 V0 + 2 2 2 2
Tsink Tsource T 0.4 = 1 − sink 500 K
Efficiency, η = 1 −
⇒
Tsink = 0.6 × 500 K = 300 K
143
Thermodynamics
⇒
300 K T ′source 300 K = = 750 K 0.4
other part is completely evacuated. If the valve is suddenly opened, the pressure and temperature of the gas will be [AIEEE 2011]
η = 1−
Thus,
T ′source
(a)
33. The specific heat capacity of a metal at low temperature
is given as (T ) 3 T C p kJ K − kg − 1 ) = 32 . A 100 g vessel of 400 this metal is to be cooled from 20 K to 4 K by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is [AIEEE2011] (a) equal to 0.002 kJ (b) greater than 0.148 kJ (c) between 0.148 kJ and 0.028 kJ (d) less than 0.028 kJ
Heat required to change the temperature of vessel by a small amount dT, − dQ = mC p dT Total heat required 3 4 T dT − Q = m ∫ 32 20 400 =
4
× 32 T (400) 4 20
100 × 10
(c) p , T
p T , 2 2 T (d) p , 2
(b)
Exp. (a) Internal energy of the gas remains constant, hence T2 = T Using at constant temperature, p1 V1 = p2 V2 p p2 = 2
35. An aluminium sphere of 20 cm diameter is
Exp. (c)
−3
p ,T 2
4
3
⇒ Q = 0.001996 kJ Work done required to maintain the temperature of sink to T2 , T Q − Q2 Q2 = 1 − 1 Q2 W = Q1 − Q2 = 1 T Q2 2 T − T2 W = 1 Q2 T2 For T2 = 20 K, 300 − 20 × 0.001996 = 0.028 kJ W1 = 20 For T2 = 4 K, 300 − 4 W2 = × 0.001996 = 0.148 kJ 4 As temperature is changing from 20 K to 4 K, work done required will be more than W1 but less than W2 . ⇒
34. A container with insulating walls is divided into two equal parts by a partition fitted with a valve. One part is filled with an ideal gas at a pressurep and temperatureT , whereas the
heated from 0°C to 100°C. Its volume changes by (given that coefficient of linear expansion for aluminiumα A = 23 × 10−6 /° C) [AIEEE 2011]
(a) 28.9 cc (c) 9.28 cc
(b) 2.89 cc (d) 49.8 cc
Exp. (a) Cubical expansion ∆ V = γ V∆ T = 3 α V∆ T [as γ = 3α ] 4 3 −6 = 3 × 23 × 10 × π (10) × 100 3 Q r = d = 10 cm 2 = 28. 9 cc
36. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats γ. It is moving with speedv and its suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by [AIEEE 2011] ( γ −1) Mv2 K 2γ R ( γ −1) (c) Mv 2 K 2R (a)
γ Mv 2 K 2R ( γ − 1) (d) Mv2 K 2( γ + 1)R
(b)
Exp. (c) As no heat is lost, Loss of kinetic energy = Gain of internal energy of gas
144
JEE Main Chapterwise Physics 1 mv 2 = nC V ∆T 2 m R 1 mv 2 = ⋅ ∆T M γ −1 2
⇒ ⇒
∆T =
θ
(a)
Mv (γ − 1) K 2R
(b) 330 K and 268 K (d) 372 K and 310 K
η1 = 1 −
⇒
T2 5 = T1 6
T2 T1
⇒
θ
θ
(c)
(d)
Exp. (b) Rate of flow of heat, dθ dQ = kA dx dt In steady state, flow of heat,
T 1 = 1− 2 6 T1
dθ =
…(i)
T2 − 62 T1 T2 − 62 1 = 1− T1 3
x
x
We know that,
Exp. (d) As efficiency,
x
x
1 temperatures T1 and T2 has efficiency . 6 When T2 is lowered by 62 K, its efficiency increases to 1/3. Then, T1 and T2 are respectively [AIEEE 2011]
dQ 1 ⋅ dx dt kA
⇒ θH − θ = k ′ x ⇒ θ = θH − k ′ x Equation θ = θH − k ′ x represents a straight line.
η2 = 1 − ⇒
(b)
2
37. A Carnot engine operating between
(a) 372 K and 330 K (c) 310 K and 248 K
θ
…(ii)
æ Direction (Q.Nos. 40 to 42) are based on the
following figure.
On solving Eqs. (i) and (ii), we get T1 = 372 K
and T2 = 310 K
38. 100g of water is heated from 30°C to 50°C. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K) [AIEEE 2011] (a) 8.4 kJ
(b) 84 kJ
(c) 2.1 kJ
(d) 4.2 kJ
Exp. (a) As work done = 0 [as W = p∆V and ∆V = 0, so W = 0] ∆U = mC∆T = 100 × 10−3 × 4184 × (50 − 30)≈ 8.4 kJ
39. A long metallic bar is carrying heat from one of its ends to the other end under steadystate. The variation of temperature θ along the length x of the bar from its hot end is best described by which of the following figure. [AIEEE 2009]
A
2×105
B
n = 2, γ = 1.67
p (Pa)
1×105
C
D
T 300 K
500 K
40. Assume the gas to be ideal, the work done on the gas in taking it from A to B is [AIEEE 2009] (a) 200 R
(b) 300 R
(c) 400 R
(d) 500 R
Exp. (c) WAB = ∆Q − ∆U = nC p dT − nC V ∆T [at constant pressure] = n(C p − C V ) dT = nR dT = 2 × R × (500 − 300) = 400 R
41. The work done on the gas in taking it from D to A is (see above figure) (a) − 414 R (c) − 690 R
(b) + 414 R (d) + 690 R
[AIEEE 2009]
145
Thermodynamics Exp. (b)
For path iaf,
At constant temperature (isothermal process), p p WDA = nRT ln 1 = 2.303 nRT log10 1 p2 p2 10 = 2.303 × 2 R × 300 log 5 2 × 10 1 = 2.303 × 600 R log 2
50 = ∆U + 20 ∴
∆U = U f − U i = 30 cal
For path ibf, ∆Q = ∆U + W
5
= 0.693 × 600R = − 414 R So, work done on the gas is + 414 R.
42. The net work on the gas in the cycle ABCDA is (see above figure) (a) zero
(b) 276 R
[AIEEE 2009]
(c) 1076 R
(d) 1904 R
Exp. (c)
2 × 105 = 2.303 × R × 500 log 5 10 = 2 × 346 R = 692 R WCD = nR∆T = 2 R × 200 = 400 R 105 WDA = 2.303 × R × 300 log 5 2 × 10 = − 415.963 R ∴ Net work done in a cycle = WAB + WBC + WCD + WDA = 400 R + 692 R + 400 R − 415.963 R = 1076 R
43. When a system is taken from statei to state f along the path iaf , it is found thatQ = 50 cal and W = 20 cal. Along the pathibf ,Q = 36 cal. [AIEEE 2007] W along the path ibf is a
f
i
b
(b) 16 cal
(c) 66 cal
Exp. (a) From first law of thermodynamics, ∆ Q = ∆U + W
= 36 − 30 = 6 cal
44. A Carnot engine, having an efficiency of 1 as heat engine is used as a refrigerator. 10 If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is [AIEEE 2007]
η=
(a) 99 J
(b) 90 J
(c) 1 J
(d) 100 J
Exp. (b)
WAB = nR∆T = 2 R × 200 = 400 R p WBC = nRT log 1 p2
(a) 6 cal
W = Q − ∆U
or
(d) 14 cal
For Carnot engine using as refrigerator, T W = Q2 1 − 1 T2 η=
and
W1 T1 = − 1 Q2 T2
It is given that, η =
1 10 T2 T1
⇒
η = 1−
or
T2 9 = T1 10
So,
Q2 = 90 J
[as W = 10 J]
45. The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically and in this process, the temperature of the gas increases by 7°C. The gas is (R = 8.3 J mol −1 K −1 ) [AIEEE 2006] (a) (b) (c) (d)
diatomic triatomic a mixture of monatomic and diatomic monatomic
Exp. (a) For adiabatic process, no exchange of heat. ∴ So, ⇒
dQ = 0 dU = − ∆W nC V dT = + 146 × 103 J
146
JEE Main Chapterwise Physics nfR × 7 = 146 × 103 2 [where, f is degree of freedom]
⇒
⇒
For process 1, Q1 = T0 S 0 +
103 × f × 8.3 × 7 = 146 × 103 2
∴
For process 2, Q2 = T0 (2S 0 − S 0 ) = T0 S 0
f = 5.02 ≈ 5
For process 3, Q3 = 0 W Q1 − Q2 η= = ∴ Q1 Q1
So, it is a diatomic gas.
46. A system goes from A to B via two processes I and II as shown in figure. If ∆U 1 and ∆U 2 are the changes in internal energies in the processes I and II respectively, then
[AIEEE 2005] (a) ∆U 1 = ∆U 2 (b) relation between ∆U 1 and ∆U 2 cannot be determined (c) ∆U 2 > ∆U 1 (d) ∆U 2 < ∆U 1
Exp. (a) The change in internal energy does not depend upon path followed by the process. It only depends on initial and final states.
47. The temperatureentropy diagram of a reversible engine cycle is given in the figure. Its efficiency is [AIEEE 2005] T
S0
(b)
48. Which of the following is incorrect regarding the first law of thermodynamics? [AIEEE 2005] (a) It is not applicable to any cyclic process (b) It is a restatement of the principle of conservation of energy (c) It introduces the concept of the internal energy (d) Both (a) and (c)
Statements (a) and (d) are wrong. Concept of entropy is associated with second law of thermodynamics.
49. Which of the following statements is correct (a) The internal energy changes in all processes (b) Internal energy and entropy are state functions (c) The change in entropy can never be zero (d) The work done in an adiabatic process is always zero
T0
1 2
Q2 2 1 = 1− = Q1 3 3
for any thermodynamic system? [AIEEE 2004]
2T0
(a)
= 1−
Exp. (d)
∆ U1 = ∆ U 2
Hence,
1 3 T0 S 0 = T0 S 0 2 2
1 4
2S0
(c)
1 3
S
(d)
Exp. (c)
Exp. (b) 2 3
Internal energy does not change in isothermal process. ∆S can be zero for adiabatic process. Work done in adiabatic process may be nonzero.
50. “Heat cannot be itself flow from a body at lower temperature to a body at higher temperature” is a statement or consequence of [AIEEE 2003]
According to the figure, T
(a) (b) (c) (d)
2T0 1
3 T0
2 S0
2S0
S
second law of thermodynamics conservation of momentum conservation of mass first law of thermodynamics
Exp. (a)
147
Thermodynamics Heat cannot flow itself from a body at lower temperature to a body at higher temperature. This corresponds to second law of thermodynamics.
51. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratioC p /C V for the gas is [AIEEE 2003] (a) 4/3
(b) 2
(c) 5/3
T2 × Q1 T1 300 × 3 × 106 = 1 × 106 cal Q2 = ⇒ 900 ∴ Work done = Q1 − Q2 = 3 × 106 − 1 × 106 = 2 × 106 cal or
Q2 =
= 2 × 4.2 × 106 J = 8.4 × 106 J
(d) 3/2
54. Even Carnot engine cannot give 100%
Exp. (d) p ∝ T3
Given,
…(i)
In an adiabatic process, T γ p1 − γ = constant T∝
1
Cp as γ = CV
…(ii)
Comparing Eqs. (i) and (ii), we get γ =3 γ −1 C 3 or 3γ − 3 = γ or 2 γ = 3 or p = γ = 2 CV
52. Which of the following parameters does not characterise the thermodynamic state of matter? [AIEEE 2003] (a) Temperature (c) Work
(b) Pressure (d) Volume
Exp. (c) Work does not characterise the thermodynamic state of matter, it is a path function giving only relationship between two quantities.
53. A Carnot engine takes 3 × 106 cal of heat from a reservoir at 627°C and gives it to a sink at 27°C. The work done by the engine is (a) 4.2 ×106 J
(b) 8.4 ×106 J [AIEEE 2003]
(c) 16.8 ×10 J
(d) zero
6
Exp. (b) T1 = 627 + 273 = 900 K, Q1 = 3 × 106 cal T2 = 27 + 273 = 300 K Now, from Carnot theorem, Q1 Q2 = T1 T2
(a) (b) (c) (d)
[AIEEE 2002]
prevent radiation find ideal sources reach absolute zero temperature eliminate friction
Exp. (c)
p( 1 − γ )/ γ
T( γ / γ − 1) ∝ p
efficiency, because we cannot
The efficiency of Carnot engine is η = 1 − (T2 / T1 ) where, T1 is the temperature of the source and T2 that of sink. T2 Q2 Since, = T1 Q1 Q So, η = 1− 2 Q1 To obtain 100% efficiency (i.e., η = 1), Q2 must be zero i.e., if a sink at absolute zero would be available, all the heat taken from the source would have been converted into work. The temperature of sink means a negative temperature on the absolute scale at which the efficiency of engine is greater than unity. This would be a violation of the 2nd law of thermodynamics. Hence, a negative temperature on the absolute scale is impossible. Hence, we cannot reach absolute zero temperature.
55. Which statement is incorrect?
[AIEEE 2002]
(a) All reversible cycles have same efficiency (b) Reversible cycle has more efficiency than an irreversible one (c) Carnot cycle is a reversible one (d) Carnot cycle has the maximum efficiency in all cycles
Exp. (a) Efficiency of all reversible cycles depends upon temperature of source and sink which will be different. Rest of the statements are correct.
9 Kinetic Theory of Gases 1. If 1022 gas molecules each of mass
kg collide with a surface 10−26 (perpendicular to it) elastically per second over an area 1 m 2 with a speed 104 m/s, the pressure exerted by the gas molecules will be of the order of [JEE Main 2019, 8 April ShiftI] (a) 104 N / m2 (b) 108 N / m2 (c) 103 N / m2 (d) 1016 N / m2
2. The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to : [Boltzmann constant k B = 1.38 × 10−23 J/K, Avogadro number Radius of earth N A = 6.02 × 1026 /kg, = 6.4 × 106 m, Gravitational acceleration on earth = 10 ms −2] [JEE Main 2019, 8 April ShiftII] (a) 104 K (c) 3 × 105 K
Exp. (*) Momentum imparted to the surface in one collision, …(i) ∆p = ( pi − pf ) = mv – (− mv ) = 2 mv Force on the surface due to n collision per n second, F = (∆p) = n∆p (Qt = 1s) t [from Eq. (i)] = 2 mnv So, pressure on the surface, F 2 mnv p= = A A Here, m = 10−26 kg, n = 1022 s −1, v = 104 ms −1, A = 1 m2 2 × 10−26 × 1022 × 104 1 = 2 N / m2
∴Pressure, p =
So, pressure exerted is of order of 100 .
(b) 650 K (d) 800 K
Exp. (a) Root mean square velocity of hydrogen molecule is given as 3kBT vrms = m Escape velocity of hydrogen molecule from the earth is given as ve = 2 gRe Given, vrms = ve or ⇒
3kBT = 2 gRe m 2 gRe m T= 3 × kB
Substituting the given values, we get 2 × 10 × 6.4 × 106 × 2 T= ≈ 104 K 3 × 138 . × 10−23 × 6.02 × 1026
149
Kinetic Theory of Gases Alternate Solution At rms speed, average thermal kinetic energy of 3 a hydrogen gas molecule is = kBT 2 and if ve = escape velocity of the gas molecule from the earth, then its kinetic energy is 1 KE = mve2 , where m is the mass of the gas 2 molecule. Equating the above thermal and kinetic energies, we have 3 1 mve2 …(i) kBT = mve2 or T = 2 2 3kB Here,
kB = 138 . × 10−23 JK −1, ve = 112 . × 103 ms −1, 2 m= = 0.332 × 10−26 kg 6.02 × 1026
Substituting these value in Eq. (i), we get . × 103 )2 0.332 × 10−26 × (112 ≈ 104 K T= . × 10−23 3 × 138
3. For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127°C. At 2 atm pressure and at 227°C, the rms speed of the molecules will be [JEE Main 2019, 9 April ShiftI]
(a) 100 5 m/s (c) 100 m/s
(b) 80 m/s (d) 80 5 m/s
Exp. (a)
has rotational, HCl molecule translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase isv , m is its mass and k B is Boltzmann constant, then its temperature will be [JEE Main 2019, 9 April ShiftI]
mv 2 (a) 3k B
mv 2 (b) 7k B
(c)
mv 2 5k B
(d)
mv 2 6k B
Exp. (b) Key Idea According to the law of equipartition n 1 2 of energy, mvrms = kBT 2 2 where, n is the degree of freedom.
Since, HCl is a diatomic molecule that has rotational, translational and vibrational motion. So, n=7 1 7 2 mvrms = kBT ⇒ 2 2 2 Here, =v vrms ⇒
T=
mv 2 7 kB
5. The specific heats, C p and C V of a gas of
diatomic molecules, A are given (in units of J mol −1 K −1) by 29 and 22, respectively. Another gas of diatomic molecules B, has the corresponding values 30 and 21. If they are treated as ideal gases, then [JEE Main 2019, 9 April ShiftII]
Key Idea For a gas molecule, 3RT vrms = M
∴
4. An
vrms ∝ T
Let unknown rms speed be vrms, 2 at (or 500 K) T2 = 227 °C and at (or 400 K) T1 = 127 °C vrms,1 = 200 m/s ∴Using the relation vrms ∝ T , we can write vrms ,2 T … (i) = 2 vrms ,1 T1 Substituting these given values in Eq. (i), we get 500 vrms, 2 = × 200 m/s ∴ 400 1 5 × 200 m/s = 100 5 m/s = 2
(a)A has a vibrational mode but B has none (b)Both A and B have a vibrational mode each (c)A has one vibrational mode and B has two (d)A is rigid but B has a vibrational mode
Exp. (a) Key Idea A diatomic gas molecule has 5 degrees of freedom, i.e. 3 translational and 2 rotational, at low temperature ranges (~ 250 K to 750 K). At temperatures above 750 K, molecular vibrations occurs and this causes two extra degrees of freedom.
Now, in given case, For gas A, C p = 29, C V = 22 For gas B, C p = 30, C V = 21 By using C 2 γ = p = 1+ CV f
150
JEE Main Chapterwise Physics We have, For gas A, 2 29 = ≈ 13 . ⇒ f = 6.67 ≈ 7 f 22 So, gas A has vibrational mode of degree of freedom. For gas B, 2 30 ≈ 14 . ⇒f = 5 1+ = f 21 Hence, gas B does not have any vibrational mode of degree of freedom. 1+
6. n moles of an ideal gas with constant volume heat capacity C V undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is [JEE Main 2019, 10 April ShiftI] 4nR CV + nR nR (c) CV − nR
4nR CV − nR nR (d) CV + nR
(a)
∴ Average collision per second, 8 × vrms 1 vav 3 π = = τ λ λ 1 8 = × τ 3π
V 2 Nπr 2
0.09 × 10−18 4
1 = 4.4 × 108 per second ≈ 108 τ ∴No option given is correct. ⇒
1 mole of O2 gas at room temperature (300 K). The molecular diameter ofO2 and its root mean square speed are found to be 0.3 nm and 200 m/s, respectively. What is the average collision rate (per second) for an O2 molecule? [JEE Main 2019, 10 April ShiftI] (d) ~1013
Exp. (*) Given, Volume, V = 25 × 10−3 m3 N = 1mole of O 2 = 6.023 × 1023 atoms of O 2 , T = 300 K
⇒λ=
25 × 10−3
7. A 25 × 10−3 m 3 volume cylinder is filled with
(c) ~1011
RT V
200 × 2 × 6.023 × 023 × π ×
For isobaric process, work done is given as W = nR∆T Heat supplied, ∆Q = nC p ∆T and C Cp − V = R ⇒ Cp = R + CV / n n W nR∆T R R = = = ∴ C ∆Q nC p ∆T C p R+ V n W nR ⇒ = ∆Q nR + C V
(b) ~1012
p=
As,
(b)
Exp. (d)
(a) ~1010
Root mean square velocity of a gas molecule of O 2 , vrms = 200 m / s 0.3 0.3 Radius, r = nm = × 10−9 m 2 2 1 v Now, average time, = av τ λ RT where, λ= 2 Nπr 2 p
8. One mole of an ideal gas passes through a process, where pressure and volume obey 1 V 2 the relation p = p 0 1 − 0 . Here, p 0 and 2 V V 0 are constants. Calculate the change in the temperature of the gas, if its volume changes fromV 0 to 2V 0. [JEE Main 2019, 10 April ShiftII] (a)
1 p 0V 0 1 p 0V 0 3 p 0V 0 5 p 0V 0 (b) (c) (d) 2 R 4 R 4 R 4 R
Exp. (d) Given process equation for 1 mole of an ideal gas is 2 1 V …(i) p = p0 1 − 0 2 V Also, for 1 mole of ideal gas, pV = RT RT ∴ p= V So, from Eqs. (i) and (ii), we have 2 1 V RT = p0 1 − 0 2 V V
…(ii)
151
Kinetic Theory of Gases T=
∴
p0 V 1 V 1 − 0 2 V R
2
…(iii)
When volume of gas is V0 , then by substituting V = V0 in Eq. (iii), we get Temperature of gas is T1 =
p0 V0 R
2 1 − 1 V0 = p0 V0 2R 2 V0
Similarly, at volume, V = 2 V0 Temperature of gas is T2 =
2 p0 (2 V0 ) 7 p0 V0 1 V 1− 0 = 2 2 V0 R 4 R
So, change in temperature as volume changes from V0 to 2 V0 is ∆T = T2 − T1 7 1 pV 5 p0 V0 = − 0 0 = 4 2 R 4 R
9. Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? [Take, R = 8.3 J/molK]
21R 2 21R CV = 10 21 × 8.3 174 . 3 = = 10 10 C V = 17.4 J/molK
5C V =
⇒ or
or
10. The number density of molecules of a gas depends on their distance r from the origin 4 as, n(r ) = n 0e −ar .Then, the total number of molecules is proportional to [JEE Main 2019, 12 April ShiftII]
(a) n 0α −3/ 4
(b) n 0 α1/ 2
(c) n 0α
(d) n 0α −3
1/ 4
Exp. (a) Number density of gas molecules, Number of molecules n= Volume of gas ⇒ Number of molecules, N = n × volume of gas
[JEE Main 2019, 12 April ShiftI]
(a) 19.7 J/molK (c) 17.4 J/molK
dr
(b) 15.7 J/molK (d) 21.6 J/molK
r
Exp. (c) Let molar specific heat of the mixture is C V . Total number of molecules in the mixture =3+2 =5 ∴C V can be determined using n C V dT = n1 C V1 dT + n2 C V2 dT or (n1 + n2 ) (C V )mix = n1C V + n2 C V2 1 [here, n = n1 + n2 ] 3R Here, C V1 = (for helium); n1 = 2 2 5R (for hydrogen); n2 = 3 C V2 = 2 3 [For monoatomic gases, C V = R and for 2 5 diatomic gases, C V = R] 2 5R 3R ∴5 × C V = 2 × + 3× 2 2
Now, consider a shell of radius r and thickness dr with centre at r = 0. Volume of shell of differentiable thickness (dr ), dV = surface area × thickness = 4 πr 2 dr Now, number of molecules in this shell is 4
dN = n(r ) ⋅ dV = n0e − αr ⋅ 4 πr 2 dr So, total number of molecules present in given volume (extending from r = 0 to r = ∞) is N= =
∞
∞
∫0 n(r )⋅ dV = ∫0 ∞
∫0
4
4
n0 e − αr ⋅ 4 πr 2 dr
4 πn0 e − α r ⋅ r 2 dr
…(i)
152
JEE Main Chapterwise Physics Here, we take αr = t ⇒ r = t 4
⇒ ⇒
1 4
12. A 15 g mass of nitrogen gas is enclosed in a
−1 ⋅α 4
vessel at a temperature 27°C. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled is about (Take, R = 8.3 J/ K  mole)
4αr dr = dt dt dt dt r dr = = = 1 1 1 4αr − 4α t 4 ⋅ α 4 4 α 3 / 4 t 4 3
2
[ JEE Main 2019, 9 Jan ShiftII]
(a) 10 kJ
Also, when r = 0, t = 0 and when r = ∞, t = ∞ substituting in Eq. (i), we get ∞ dt N = ∫ 4 π n0 e −t ⋅ 3 1 0
−
3 4
∞
. n0 . ∫ e −t ⋅ t
−
1 4
0
1 − ∞ −t e ⋅t 4 0
As value of definite integral ∫
dt dt is a
constant (= k let), we have N = π kn0α
−
3 4
⇒ N ∝ n0α
−
3 4
11. A mixture of 2 moles of helium gas (atomic mass = 4u) and 1 mole of argon gas (atomic mass = 40u) is kept at 300 K in a container. The ratio of their rms speeds v rms (helium) is close to v rms (argon) [ JEE Main 2019, 9 Jan ShiftI]
(a) 0.32
(b) 2.24
(c) 3.16
(d) 0.45
Exp. (c) Root mean square (rms) velocity of the molecules of a gas is given as 3RT vrms = M where, M is the atomic mass of the gas. 1 vrms ∝ ⇒ M Margon vrms (helium) = ∴ vrms (argon) Mhelium Given, Margon = 40 u and Mhelium = 4u Substituting these values in Eq. (i), we get vrms (helium) 40 = 4 vrms (argon) = 10 = 316 .
(c) 14 kJ
(d) 6 kJ
Exp. (a)
4 α4 t 4
N = πα
(b) 0.9 kJ
We know that from kinetic theory of gases, 3RT or Vrms ∝ T Vrms = M where, R is gas constant, T is temperature and M is molecular mass of a gas. Here, to double the v rms, temperature must be 4 times of the initial temperature. ∴ New temperature, T2 = 4 × T1 As, T1 = 27 ° C = 300 K ∴ T2 = 4 × 300 = 1200 K As, gas is kept inside a closed vessel. ∴ Q = nC V ⋅ dT 15 5R (1200 − 300) = × 28 2 As, n = m = 15 for nitrogen M 28 15 5 = × × 8.3 × 900 28 2 [Given, R = 8.3 J/Kmol] or Q = 10004.4 J ~ − 10 kJ
13. Halfmole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20°C to 90°C . Work done by gas is close to (Take, gas constant, R = 8.31 J/molK) [ JEE Main 2019, 10 Jan ShiftII]
(a) 291 J (c) 146 J
(b) 581 J (d) 73 J
Exp. (a) …(i)
Work done by gas during heat process at constant pressure is given by ∆W = p∆V Using ideal gas equation, pV = nRT ⇒ p∆V = nR∆T So, ∆W = nR∆T
… (i)
153
Kinetic Theory of Gases 1 and 2 ∆T = 90° C − 20° C = 363 K − 293 K = 70 K and R (gas constant) = 8.31J/molK Substituting these values in Eq. (i), we get 1 ∆W = × 8.31 × 70 = 290.85 J 2 ∆W ~ − 291 J
Now, it is given that, n =
14. 2 kg of a monoatomic gas is at a pressure of
4 × 104 N/m 2. The density of the gas is 8 kg/m 3. What is the order of energy of the gas due to its thermal motion ? [ JEE Main 2019, 10 Jan ShiftII]
(a)106 J (c)104 J
Exp. (c) Given, mass of gas, m = 2 kg Pressure on gas, p = 4 × 104 N / m2
⇒
nRT = 10
4
… (i)
1 4
… (ii)
15. A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TV x = constant, then x is [ JEE Main 2019, 11 Jan ShiftI]
Exp. (a)
2 (b) 3
(c)
5 3
16. A gas mixture consists of 3 moles of oxygen
(b) 15 RT
(c) 20 RT
(d) 4 RT
Exp. (b)
Internal energy of n moles of a monoatomic gas is given by, 3 U = nRT 2 3 U = × 104 J = 1.5 × 104 J ⇒ 2 i.e. in order of 104 J.
2 (a) 5
…(ii)
[ JEE Main 2019, 11 Jan ShiftI]
ρ = 8 kg / m
nRT = 4 × 104 ×
and it is given that, here TV x = constant Comparing Eqs. (i) and (ii), we get 7 γ − 1= x ⇒ − 1= x 5 or x=2/5
(a) 12 RT
3
Mass ⇒ Volume of gas = Density 2 1 V = = m3 ⇒ 8 4 Using ideal gas equation, pV = nRT
For a diatomic gas, degree of freedom, f = 5 2 7 ∴ γ = 1+ 2 / f = 1+ = 5 5 As for adiabatic process, TV γ − 1 = constant …(i)
and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is
(b)103 J (d)105 J
Density of gas,
Key Idea For an ideal gas undergoing an adiabatic process at room temperature, pV γ = constant or TV γ − 1 = constant
(d)
3 5
Internal energy of a gas with f degree of freedom is nf RT , where n is the number of moles. U= 2 Internal energy due to O 2 gas which is a n f RT 5 diatomic gas is U1 = 1 1 = 3 × RT 2 2 (Qn1 = 3 moles, degree of freedom for a diatomic gas f1 = 5) Internal energy due to Ar gas which is a monoatomic gas is n f RT 3 U2 = 2 2 = 5 × RT 2 2 (Qn2 = 5 moles, degree of freedom for a monoatomic gas f2 = 3) ∴Total internal energy = U1 + U 2 ⇒ U = 15 RT
17. In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = k , where k is a constant. In this process, the temperature of the gas is increased by ∆T . The amount of heat absorbed by gas is (where, R is gas constant) [ JEE Main 2019, 11 Jan ShiftII]
154
JEE Main Chapterwise Physics (a)
1 2k 1 kR∆T (b) ∆T (c) R∆T 2 3 2
(d)
3 R∆T 2
Given, VT = k, (k is constant) 1 or …(i) T∝ V Using ideal gas equation, pV = nRT pV ∝ T 1 pV ∝ or pV 2 = constant …(ii) ⇒ V i.e. a polytropic process with x = 2. (Polytropic process means, pV x = constant) We know that, work done in a polytropic process is given by p V − p1V1 ( for x ≠ 1) ...(iii) ∆W = 2 2 1− x V ∆W = pV ln 2 (for x = 1) V1
and, Here, x =2 ,
⇒
p V − p1V1 nR(T2 − T1 ) ∆W = 2 2 = 1− x 1− x nR∆T ...(iv) = − nR∆T ∆W = 1 −`2
Now, for monoatomic gas change in internal energy is given by 3 ...(v) ∆U = R∆T 2 Using first law of thermodynamics, heat absorbed by one mole gas is 3 ∆Q = ∆W + ∆U = R∆T − R∆T 2 1 ∆Q = R∆T ⇒ 2
18. An ideal gas occupies a volume of 2 m3 at a pressure of 3 × 106 Pa. The energy of the gas is [ JEE Main 2019, 12 Jan ShiftI] (a) 6 × 104 J (c) 9 × 106 J
3 nRT 2
3 (Q from pV = nRT) p⋅ V 2 Substituting the given values, we get 3 = × 3 × 106 × 2 2 = 9 × 106 J =
Exp. (c)
∴
For an ideal gas, internal energy, E =
(b) 108 J (d) 3 × 102 J
Exp. (c) Key Idea Internal energy of ‘n’ moles of a gas with degree of freedom f ( = 3 for an ideal gas), at temperature T is 3 f E = ⋅ nRT = nRT 2 2
19. An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is 6 × 10− 8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to [ JEE Main 2019, 12 Jan ShiftII]
(a) 4 × 10− 8 s
(b) 3 × 10− 6 s
(c) 2 × 10− 7 s
(d) 0.5 × 10− 8 s
Exp. (a) Mean time elapsed between two successive, λ collisions is t = v where, λ = mean free path length and v = mean speed of gas molecule kBt 2 2 π d p C T ∴ t = ⇒t = p 8 kBT . π M 1 kB M where, C = = a constant for a gas. π 4d 2 T p t2 So, … (i) = 2 . 1 t1 T1 p2 Here given, and
p1 1 T2 = , = p2 2 T1
500 = 300
5 3
t 1 = 6 × 10−8 s
Substituting there values in Eq. (i), we get 5 1 t 2 = 6 × 10−8 × × 3 2 = 3.86 × 10−8 s ≈ 4 × 10−8 s
20. A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston
155
Kinetic Theory of Gases is l1 and that below the piston is l 2, such that l1 > l 2. Each part of the cylinder contains n moles of an ideal gas at equal temperatureT . If the piston is stationary, its mass m, will be given by (where, R is universal gas constant and g is the acceleration due to gravity)
N RT NA pVNA pVNA We have, nf − ni = − RTf RTi
From pV = nRT =
105 × 30 1 1 − × 6.02 × 1023 . 300 290 8.3 = − 2 . 5 × 1025
⇒ nf − ni =
[ JEE Main 2019, 12 Jan ShiftII]
1 1 + l 2 l1 RT l1 − 3l 2 (d) ng l1 l 2
nRT l1 − l 2 (a) g l1l 2 RT 2l1 + l 2 (c) g l1 l 2
(b)
nRT g
Exp. (a) Key Idea As piston is in equilibrium, so net force on piston is zero.
When the piston is stationary, i.e. on equilibrium as shown in the figure below, mg
p1A
∴
∆n = − 2 . 5 × 1025
22. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as V q , where V is the volume of the gas. The Cp value of q is γ = CV 3γ + 5 (a) 6
3γ − 5 (b) 6
γ +1 (c) 2
[JEE Main 2015]
(d)
γ −1 2
Exp. ( c ) / For an adiabatic process TV γ − 1 = constant.
p 2A
then ⇒ or
p1 A + mg = p2 A mg = p2 A − p1 A nRTA nRTA − mg = V1 V2
[Q pV = nRT (ideal gas equation)] l − l A A = nRT − = nRT 1 2 Al1 l1l2 Al2 or
m=
nRT l1 − l2 g l1l2
21. The temperature of an open room of volume
30 m 3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and n f are the number of molecules in the room before and after heating, then n f − ni will be [JEE Main 2017 (Offline)]
(a)1.38 × 1023
(b) 2.5 × 1025
(c) −2.5 × 1025
(d) −161 . × 1023
Exp. ( c )
We know that average time of collision between molecules 1 τ= nπ 2 vrms d 2 where, n = number of molecules per unit volume vrms = rms velocity of molecules 1 As n ∝ and vrms ∝ T V V τ∝ T Thus, we can write n = K1V −1 and vrms = K 2 T1/ 2 where, K1 and K 2 are constants. For adiabatic process, TV γ − 1 = constant. Thus, we can write τ ∝ VT −1/ 2 ∝ V (V1 − Y )−1/ 2 γ+1
or
τ ∝V
2
23. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional
156
JEE Main Chapterwise Physics 46 cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg) [JEE Main 2014]
(a) 16 cm
(b) 22 cm
(c) 38 cm
(d) 6 cm
/ In this question, the system is accelerating horizontally, hence the pressure in the vertical direction will remain unaffected. i.e., p1 = p0 + ρgh Again, we have to use the concept that the pressure in the same level will be same.
Exp. (a) 8 cm
x
p2 54 cm
54–x
Exp. ( a ) F F F n1kT1 + n2 kT2 + n3 kT3 2 2 2 F = (n1 + n2 + n3 ) kT 2 [QF is degree of freedom] Temperature of mixture, n T + n2T2 + n3T3 T = 11 n1 + n2 + n3
25. The potential energy function for the force between two atoms in a diatomic molecule a b is approximately given by U ( x ) = 12 − 6 , x x where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is D = [U ( x = ∞ ) − U at equilibrium ], D is [AIEEE 2010]
For air trapped in tube, p1V1 = p2 V2 p1 = patm = ρg 76 V1 = A × 8 [QA = area of crosssection] p2 = patm − ρg (54 − x) = ρg (22 + x) V2 = A ⋅ x At constant temprature, p1 V1 = p2 V2 ρg 76⋅ A8 = ρg (22 + x) Ax x2 + 22 x − 78 × 8 = 0 ⇒
x = 16 cm
24. Three perfect gases at absolute temperatures T1 , T2 and T3 are mixed. The masses of molecules arem1 , m 2 andm 3 and the number of molecules are n1 , n 2 and n 3 respectively. Assuming no loss of energy, the final temperature of the mixture is [AIEEE 2011] (a)
n1T1 + n 2T2 + n 3T3 n1 + n 2 + n 3
(b)
n1T12 + n 2T22 + n 3T32 n1T1 + n 2T2 + n 3T3
n 2T 2 + n 22T22 + n 32T32 (c) 1 1 n1T1 + n 2T2 + n 3T3 (T1 + T2 + T3 ) (d) 3
b2 (a) 2a
b2 (b) 12 a
b2 (c) 4a
(d)
b2 6a
Exp. ( c ) a b − 6 x12 x U( x = ∞ ) = 0 dU 12 a 6 b As, F=− = − 13 + 7 x dx x At equilibrium, F = 0 2a x6 = ∴ b − b2 b a ∴ − Uat equilibrium = = 2 2a 4a 2a b b U( x ) =
∴
D = [U ( x = ∞ ) − Uat equilibrium ] =
b2 4a
26. One kg of a diatomic gas is at a pressure of 8 × 104 Nm −2. The density of the gas is 4 kgm −3. What is the energy of the gas due to its thermal motion? [AIEEE 2009] (a) 3 × 104 J (c) 6 × 104 J
(b) 5 × 104 J (d) 7 × 104 J
Exp. ( b ) Thermal energy corresponds to internal energy Mass = 1 kg Density = 4 kg m−3
157
Kinetic Theory of Gases ⇒
Volume =
1 Mass = m3 Density 4
Pressure = 8 × 104 Nm−2 f f As internal energy = nRT = pV 2 2 Degree of freedom for diatomic = 5 5 ∴ Internal energy = p × V 2 Putting the values of p = 8 × 104 Nm −2 and 1 V = m3 , we get 4 1 5 Internal energy = × 8 × 104 × = 5 × 104 J 4 2
27. An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume V1 and contains ideal gas at pressure p1 and temperature T1. The other chamber has volumeV 2 and contains ideal gas at pressure p 2 and temperature T2. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be [AIEEE 2008, 04] (a)
T1T2 (p1V1 + p 2V 2 ) p1V1T2 + p 2V 2T1
(b)
p1V1T1 + p 2V 2T2 p1V1 + p 2V 2
(c)
p1V1T2 + p 2V 2T1 p1V1 + p 2V 2
(d)
T1T2 (p1V1 + p 2V 2 ) p1V1T1 + p 2V 2T2
nitrogen per unit mass at constant pressure and constant volume respectively, then [AIEEE 2007]
R 28 (c) C p − CV = R
(a) C p − CV =
R 14 (d) C p − CV = 28 R
(b) C p − CV =
Exp. ( a ) According to Mayer’s relation, R Cp − CV = m For nitrogen, m = 28 R Cp − CV = ∴ 28
1 2 gases are placed on a table. Box A contains one mole of nitrogen at temperature T0 , while box B contains one mole of helium at temperature (7/3) T0.The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases,T f ,in terms ofT0 is
29. Two rigid boxes containing different ideal
[AIEEE 2006]
3 (a) T f = T0 7 3 (c) T f = T0 2
7 (b) T f = T0 3 5 (d) T f = T0 2
Exp. ( c )
Exp. ( a ) As no work is done and system is thermally insulated from surrounding, it means sum of internal energy of gas in two partitions is constant i .e., U = U1 + U 2 . Assuming both gases have same degree of freedom, then f(n + n2 ) RT Internal energy, U = 1 2 fn1RT1 fn RT and , U2 = 2 2 U1 = 2 2 Solving, we get Common temperature, T =
28. If C p andC V denote the specific heats of
( p1V1 + p2 V2 ) TT 1 2 p1VT 1 2 + p2 V2T1
Here, change in internal energy of the system is zero, i.e., increase in internal energy of one is equal to decrease in internal energy of other.
B
Thermal contact a mole N2
a mole He
Change in internal energy in box A, 5 ∆U A = 1 × R (Tf − T0 ) 2 Change in internal energy in box B 3R 7 ∆U B = 1 × Tf − T0 2 3
158
JEE Main Chapterwise Physics Now, ∆U A + ∆U B = 0 7T0 5R 3R (Tf − T0 ) + ⇒ Tf − =0 2 2 3 or or or
5Tf − 5T0 + 3Tf − 7 T0 = 0 8Tf = 12 T0 12 3 Tf = T0 = T0 8 2
⇒
30. A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio
Cp CV
of the
mixture is (a) 1.59 (c) 1.4
(b) 1.62 (d) 1.54
[AIEEE 2005]
Exp. ( b ) For mixture, C V =
n1C V1 + n2C V2 n1 + n2
For helium,
5 31. One mole of ideal monoatomic gas γ = is
3 7 mixed with one mole of diatomic gas γ = . 5 What is γ for the mixture ? γ denotes the ratio of specific heat at constant pressure, to that at constant volume. [AIEEE 2004]
(a)
3 2
(b)
23 15
(c)
35 23
(d)
4 3
Exp. ( a ) Mayer’s formula is
16 =4 4 5 γ1 = 3
Number of moles, n1 = and For oxygen,
16 1 = 32 2 7 γ2 = 5 3 R R = = R = γ1 − 1 5 − 1 2 3 5 R R = = = R γ2 − 1 7 − 1 2 5
Number of moles, n2 = and Then,
C V1
and
C V2
∴ For mixture,
Now,
R R + 1= +1 29 CV R 18 C p 18 = +1 C V 29 18 + 29 = = 1.62 29 γ=
or
3 1 5 4× R+ × R 2 2 2 CV = 1 4+ 2 5 6R + R 4 = 9 2 29 R × 2 29 R = = 9×4 18 R R CV = ⇒ γ − 1= CV γ −1
C p − C V = R and
γ=
Cp CV
Therefore, using above two relations, we find R CV = γ −1 For a mole of monatomic gas, 5 R 3 γ= = R ∴ CV = 5 − 1 2 3 3 7 For a mole of diatomic gas, γ = 5 R 5 CV = = R ∴ 7 − 1 2 5 When these two moles are mixed, then heat required to raise the temperature to 1°C is 3 5 C V = R + R = 4R 2 2 4R Hence, for one mole, heat required = = 2R 2 3 R ∴ CV = 2 R ⇒ = 2 R or γ = 2 γ −1 Alternate Solution n1 + n2 n1 n2 = + γ −1 γ1 − 1 γ 2 − 1 5 7 Here, n1 = 1, n2 = 1, γ 1 = , γ 2 = 3 5
159
Kinetic Theory of Gases ∴
⇒ or Hence,
1+ 1 1 1 = + γ − 1 5 − 1 7 − 1 3 5 2 3 5 2 8 or = + = γ −1 2 2 γ −1 2 2 2 =4 ⇒ γ= +1 γ −1 4 3 γ= 2
33. At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C ? [AIEEE 2002] (a) 80 K (c) 3 K
Exp. ( d )
7 5
32. 1 mol of a gas with γ = is mixed with 1 mol 5 of a gas with γ = , then the value of γ for the 3 resulting mixture is [AIEEE 2002] (a)
7 5
(b)
2 5
(c)
24 16
(d)
12 7
The rms velocity of the molecule of a gas of molecular weight M at temperature T is given by 3RT crms = M Let MO and MH be molecular weights of oxygen and hydrogen and TO and TH be the corresponding kelvin temperatures at which crms is same for both gases. i .e.,
Exp. ( c ) Using the relation, n1 + n2 n1 n2 = + γ −1 γ1 − 1 γ 2 − 1 ⇒
1+ 1 1 1 = + γ − 1 (5/3 − 1) (7/5 − 1)
or
2 3 5 = + =4 γ −1 2 2
⇒
3 24 γ= = 2 16
(b) –73 K (d) 20 K
crms (O) = crms (H) 3RTO = MO
TO T = H MO MH
Hence, Given, and ∴
3RTH MH
TO = 273 + 47 = 320 K M O = 32 , M H = 2 TH =
2 × 320 = 20 K 32
10 Oscillations and Waves A
B
L
L
1. A wire of length 2L, is made by joining two wires A and B of same length but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q , then the ratio p : q is [JEE Main 2019, 8 April ShiftI]
(a) 3 : 5 (c) 1 : 2
(b) 4 : 9 (d) 1 : 4
Exp. (c) Wire B
Wire A
Node L
L
Let mass per unit length of wires are µ A and µ B , respectively. Q For same material, density is also same. ρπr 2 L ρ4 πr 2 L So, µ A = = µ and µ B = = 4µ L L Tension (T ) in both connected wires are same. So, speed of wave in wires are T T vA = = µA µ [Qµ A = µ and µ B = 4 µ] and
vB =
T = µB
T 4µ
So, nth harmonic in such wires system is pv fnth = 2L pv A p T (for p antinodes) ⇒ = fA = 2L 2L µ Similarly,
fB =
qv B 1 q T q T = = 2L 2L 4µ 2 2L µ
(for q antinodes) As frequencies fA and fB are given equal. p T q 1 T So, fA = fB ⇒ = 2 L µ 2 2 L µ p 1 = ⇒ p : q = 1: 2 q 2
2. A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1 of the original amplitude is close to 1000 [JEE Main 2019, 8 April ShiftII]
(a) 20 s (c) 100 s
(b) 50 s (d) 10 s
Exp. (a) Given, frequency of oscillations is f = 5 osc s −1 ⇒ Time period of oscillations is T = So, time for 10 oscillations is =
1 1 = s f 5
10 = 2s 5
161
Oscillations and Waves Now, if A0 = initial amplitude at t = 0 and γ = damping factor, then for damped oscillations, amplitude after t second is given as A = A0e − γ t ∴ After 2 s, A0 = A0e −γ( 2 ) ⇒ 2 = e 2 γ 2 log2 …(i) γ= ⇒ 2 1 Now, when amplitude is of initial amplitude, 1000 i.e. A0 = A0e −γt 1000 ⇒ log(1000) = γt ⇒ log(103 ) = γt 3log 10 = γt 2 × 3log 10 t = log 2
⇒ ⇒
t = 19.93 s
⇒
16 × 2π 15 4 = ×T 15
T′ =
[using Eq. (i)]
or t ≈ 20 s
3. A simple pendulum oscillating in air has period T . The bob of the pendulum is completely immersed in a nonviscous 1 liquid. The density of the liquid is th of the 16 material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is [JEE Main 2019, 9 April ShiftI] (a) 2T
(where, V = volume of the bob = volume of displaced liquid by the bob when immersed in it). If effective value of gravitational acceleration on the bob in this liquid is geff , then net force on the bob can also be written as …(iv) Fnet = Vρgeff Equating Eqs. (iii) and (iv), we have Vρg eff = Vρg − Vρg / 16 15 …(v) geff = g − g / 16 = g ⇒ 16 Substituting the value of g eff from Eq. (v) in Eq. (i), the new time period of the bob will be 16 L L = 2π T′ = 2 π 15 g geff
1 1 1 1 (b) 2T (c) 4T (d) 4T 10 14 14 15
Exp. (d) We know that, Time period of a pendulum is given by …(i) T = 2 π L / geff Here, L is the length of the pendulum and g eff is the effective acceleration due to gravity in the respective medium in which bob is oscillating. Initially, when bob is oscillating in air, g eff = g . L So, initial time period, T = 2 π …(ii) g Let ρbob be the density of the bob. When this bob is dipped into a liquid whose density is given as ρ ρ (given) ρliquid = bob = 16 16 ∴ Net force on the bob is ρ …(iii) Fnet = Vρg − V ⋅ ⋅g 16
L g [using Eq. (ii)]
4. A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is Y = 0.3 sin(0.157x )cos(200πt ). The length of the string is (All quantities are in SI units) [JEE Main 2019, 9 April ShiftI]
(a) 60 m
(b) 40 m
(c) 80 m
(d) 20 m
Exp. (c) Given equation of stationary wave is Y = 0.3 sin(0157 . x) cos(200 πt ) Comparing it with general equation of stationary wave, i.e. Y = asin kxcosωt , we get 2π . k = = 0157 λ 2π Q 1 ≈ 0157 …(i) . = 4π2 ⇒ λ= 2π 0157 . 2π 1 and ω = 200 π = s ⇒T = T 20 As the possible wavelength associated with nth harmonic of a vibrating string, i.e. fixed at both ends is given as λ 2l λ = or l = n 2 n Now, according to question, string is fixed from both ends and oscillates in 4th harmonic, so λ 4 = l ⇒2λ = l 2 or Now,
l = 2 × 4π2 = 8π2 π ≈ 10 ⇒ l ≈ 80 m 2
[using Eq. (i)]
162
JEE Main Chapterwise Physics
5. The pressure wave p = 0.01 sin[1000 t − 3x ]
Nm −2, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0°C. On some other day when temperature is T , the speed of sound produced by the same blade and at the same frequency is found to be 336 ms −1. Approximate value of T is
Exp. (d) The given situation is shown in the figure given below
k=800N/m m
[JEE Main 2019, 9 April ShiftI]
(a) 15° C
(b) 11° C
(c) 12° C
(d) 4° C
Exp. (d) Given, p = 0.01sin (1000 t − 3 x) N/m 2 Comparing with the general equation of pressure wave of sound, i.e. p0 sin(ωt − kx), we get ω = 1000 and k = 3 ω Also, k= ⇒ v =ω/k v 1000 ∴Velocity of sound isv1 = 3 Or Speed of sound wave can also be calculated as (coefficient of t ) 1000 1000 m/s v=− =− = (coefficient of x) (− 3) 3 Now, relation between velocity of sound and temperature is γRT v= ⇒v ∝ T m or
v2 v2 T = 2 ⇒ T2 = 22 ⋅ T1 v1 T1 v1
Here, v 2 = 336 m/s, v1 = 1000 / 3 m/s, T1 = 0°C = 273 K (336)2 ∴ T2 = × 273 = 277.38 K (1000 / 3)2 − 4°C ∴ T2 = 4.38°C ~
6. A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The spring is stretched by 2 cm and released, so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K) [JEE Main 2019, 9 April ShiftII] (a) 10−4 K
(b) 10−3 K
(c) 10−1 K
(d) 10−5 K
When vibrations of mass are suddenly stopped, oscillation energy (or stored energy of spring) is dissipated as heat, causing rise of temperature. So, conversation of energy gives 1 2 kxm = (m1s1 + m2 s2 )∆T 2 where, xm = amplitude of oscillation, s1 = specific heat of mass, s2 = specific heat of water and ∆T = rise in temperature. Substituting values given in question, we have 1 × 800 × (2 × 10− 2 )2 2 500 = × 400 + 1 × 4184 ∆T 1000 16 × 10− 2 = 3.65 × 10− 5 K 4384 So, the order of magnitude of change in temperature is approx 10− 5 . ⇒
∆T =
7. Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of 20 ms −1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B ? (speed of sound in air = 340 ms −1) [JEE Main 2019, 9 April ShiftII]
(a) 2060 Hz (c) 2300 Hz
(b) 2250 Hz (d) 2150 Hz
Exp. (b) The given condition can be shown below as, f
20 ms–1
20 ms–1
f0 A
B
163
Oscillations and Waves Here, source and observer both are moving away from each other. So, by Doppler’s effect, observed frequency is given by v + vo … (i) f = f0 v − vs where, v = speed of sound = 340 ms − 1, vo = speed of observer = − 20 ms − 1, v s = speed of source = − 20 ms − 1, f0 = true frequency and f = apparent frequency = 2000 Hz Substituting the given values in Eq. (i), we get 340 − 20 2000 = × f0 340 + 20 2000 × 360 ⇒ = 2250 Hz f0 = 320
8. A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is [JEE Main 2019, 9 April ShiftII] (a) 180 m/s, 80 Hz (c) 320 m/s, 120 Hz
(b) 320 m/s, 80 Hz (d) 180 m/s, 120 Hz
Exp. (b) Frequency of vibration of a string in nth harmonic is given by v … (i) fn = n ⋅ 2l where, v = speed of sound and l = length of string. Here, f3 = 240 Hz, l = 2 m and n = 3 Substituting these values in Eq. (i), we get 4 × 240 v ∴ 240 = 3 × = 320 ms − 1 ⇒ v= 2 ×2 3 Also, fundamental frequency is f f 240 = 80 Hz f= n = 3 = n 3 3
9. The displacement of a damped harmonic
oscillator is given by x (t ) = e − 0.1t cos (10 πt + φ ). Here,t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to [JEE Main 2019, 10 April ShiftI]
(a) 27 s (c) 4 s
(b) 13 s (d) 7 s
Exp. (d) Given, displacement is x (t ) = e − 0.1t cos (10πt + φ) Here, amplitude of the oscillator is A = e − 0.1 t
… (i)
Let it takes t seconds for amplitude to be dropped by half. At t = 0⇒ A =1 [from Eq. (i)] A 1 At t = t ⇒ A′ = = 2 2 So, Eq. (i) can be written as 1 e − 0.1t = 2 or e 0.1t = 2 01 . t = ln (2) 1 or ln (2) = 10 ln (2) t = 01 . Now, ln (2) = 0.693 ∴ t = 10 × 0.693 = 6.93 s or t ≈7s or
10. A stationary source emits sound waves of frequency 500 Hz. Two observers moving along a line passing through the source detect sound to be of frequencies 480 Hz and 530 Hz. Their respective speeds are in ms−1, (Take, speed of sound = 300 m/s) [JEE Main 2019, 10 April ShiftI]
(a) 12, 16
(b) 12, 18
(c) 16, 14
(d) 8, 18
Exp. (b) Given, Frequency of sound source (f0 ) = 500 Hz Apparent frequency heard by observer 1, f1 = 480 Hz and apparent frequency heard by observer 2, f2 = 530 Hz. Let v 0 be the speed of sound. When observer moves away from the source, v − v′ 0 Apparent frequency, f1 = f0 … (i) v When observer moves towards the source, v + v ′′ 0 … (ii) Apparent frequency, f2 = f0 v Substituting values in Eq. (i), we get 300 − v ′ 0 480 = 500 300
164
JEE Main Chapterwise Physics ⇒
96 × 3 = 300 − v 0′
⇒
v 0′ = 12 m/s
Here, f1 = 11 Hz and f2 = 9 Hz ⇒ Beat frequency is, fbeat = 11 − 9 = 2 Hz Hence, time period of beats or time interval 1 1 between beats, T = ⇒ T = = 0.5 s 2 fbeat
Substituting values in Eq. (ii), we get 330 + v ′ ′ 530 = 500 300 ⇒ ⇒
So, resultant wave has a time period of 0.5 s which is correctly depicted in option (a) only.
106 × 3 = 300 + v ′ ′ v 0′ ′ = 18 m/s
12. A source of soundS is moving with a velocity
Thus, their respective speeds (in m/s) is 12 and 18.
11. The correct figure that shows schematically, the wave pattern produced by superposition of two waves of frequencies 9Hz and 11 Hz, is [JEE Main 2019, 10 April ShiftII]
of 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take, velocity of sound in air is 350 m/s) [JEE Main 2019, 10 April ShiftII]
(a) 807 Hz (c) 750 Hz
y
(b) 1143 Hz (d) 857 Hz
Exp. (c) Initially,
(a) 0 y
t(s) 1
vs=50 ms–1
2 Sound source
fobserved = 1000Hz
Stationary observer
After sometime,
(b) 0 y
1
Stationary observer
2
(c) 0
y
vs = 50 ms–1
t(s)
t(s) 1
2
(d) 0
t(s) 1
2
Exp. (a) When waves of nearby frequencies overlaps, beats are produced. Beat frequency is given by, fbeat = f1 − f2
When source is moving towards stationary observer, frequency observed is more than source frequency due to Doppler’s effect, it is given by v fobserved = f v − vs where, f = source frequency, fo = observed frequency = 1000 Hz, v = speed of sound in air = 350 ms −1 and v s = speed of source = 50 ms −1 f (v − v s ) 1000(350 − 50) 6000 Hz So, f = obs = = v 7 350 When source moves away from stationary observer, observed frequency will be lower due to Doppler’s effect and it is given by v 6000 × 350 f0 = f = v + v s 7 × (350 + 50) 6000 × 350 = 750 Hz = 7 × 400
165
Oscillations and Waves 13. A progressive wave travelling along the positive xdirection is represented by y ( x ,t ) = A sin (kx − ωt + φ ). Its snapshot at t = 0 is given in the figure. [JEE Main 2019, 12 April ShiftI] y
and velocity of submarine (B), 27000 m/s v B = 27 km/h = 3600 or …(ii) v B = 7.5 m/s Signal sent by submarine (B) is detected by submarine (A) can be shown as 7.5 m/s
5 m/s
B
A
A x
Frequency of the signal, fo = 500 Hz So, in this relative motion, frequency received by submarine (A) is v − vA 1500 − 5 f1 = S fo = 500 Hz 1500 − 7.5 vS − v B 1495 f1 = × 500 Hz ⇒ 1492 . 5
For this wave, the phase φ is (a) −
π 2
(c) 0
(b) π (d)
π 2
Exp. (b) From the given snapshot at t = 0, y = 0 at x = 0 and y = − ve when x increases from zero. Standard expression of any progressive wave is given by y = A sin(kx − ωt + φ) Here, φ is the phase difference, we need to get at t = 0 y = A sin (kx + φ) Clearly φ = π, so that y = A sin (kx + π ) y = − A sin (kx) ⇒ y = 0 at x = 0 and y = −ve at x > 0 Which satisfies the given snapshot.
14. A submarine A travelling at 18 km/h is being chased along the line of its velocity by another submarine B travelling at 27 km/h. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency ν. The value of ν is close to (Speed of sound in water = 1500 ms −1) [JEE Main 2019, 12 April ShiftI]
(a) 504 Hz (c) 499 Hz
(b) 507 Hz (d) 502 Hz
The reflected frequency f1 is now received back by submarine (B). So, frequency received at submarine (B) is v + vB 1500 + 7.5 1495 f2 = S f1= 500 1500 + 5 1492 . 5 vS + v A Hz 1507.5 1495 f2 = 500 Hz 1505 1492 . 5
⇒ ⇒ ⇒
f2 = 1. 00166 × 1. 00167 × 500 f2 = 501. 67 Hz ≈ 502 Hz
15. Two sources of sound S1 and S 2 produce sound waves of same frequency 660 Hz. A listener is moving from source S1 towards S 2 with a constant speedu m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then,u equal to [JEE Main 2019, 12 April ShiftII] (a) 5.5 m/s (c) 2.5 m/s
Exp. (c) When observer moves away from S1 and towards S2, fs′
Exp. (d) Given, velocity of submarine (A), 18000 m/s v A = 18 km/h = 3600 or v A = 5 m/s
(b) 15.0 m/s (d) 10.0 m/s
S1 fs
1
…(i)
fs′
1
2
u
S2 fs
2
166
JEE Main Chapterwise Physics then due to Doppler’s effect observed frequencies of sources by observer are v − vo ⋅ fS1 f ′S1 = v (observer moving away from source) v + vo and f ′S 2 = ⋅f v S2 (observer moving towards source) (where, v = speed to sound, vo = speed of observer) So, beat frequency heard by observer is fb = f ′S 2 − f ′S1 Here, vo = u, v = 330 ms −1 fb = 10 Hz, fS1 = fS 2 = 660 Hz On putting the values, we get fb = f ′S 2 − f ′S1 v + vo v − vo f = ⋅f − v S 2 v S1 v + vo v − vo 2 vo = fS1 − = fS1 ⋅ v v v 660 × 2u [Q vo = u ] 10 = ⇒ 330 330 × 10 ⇒ u= ⇒u = 2 . 5 ms −1 2 × 660
16. A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched lengths l1 and l 2 where, l1 = nl 2 and n is an integer. The ratio k1 / k 2 of the corresponding force constants k1 and k 2 will be [JEE Main 2019, 12 April ShiftII]
(a) n
(b)
1 n2
(c)
1 n
17. A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound (v ) in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column l1 = 30 cm and l 2 = 70 cm. Then,v is equal to [JEE Main 2019, 12 April ShiftII]
(a) 332 ms −1 (c) 338 ms −1
Exp. (b) In a resonance tube apparatus, first and second resonance occur as shown λ l 1= 4 l2= 3λ 4
Ist resonance
If parameters like material, number of loops per unit length, area of crosssection, etc., are kept same, then force constant of spring is inversely proportional to its length. In given case, all other parameters are same for both parts of spring. 1 1 So, k1 ∝ and k2 ∝ l1 l2 k1 l2 = ∴ k2 l1 l 1 = 2 = [Q l1 = nl2 ] nl2 n
IInd resonance
As in a stationary wave, distance between two λ successive nodes is and distance of a node 2 λ and an antinode is . 4 3λ λ λ − = l2 − l1 = 4 4 2 So, speed of sound, v = fλ = f × 2(l2 − l1 ) = 480 × 2 × (70 − 30) × 10−2 = 384 ms −1
(d) n 2
Exp. (c)
(b) 384 ms −1 (d) 379 ms −1
18. A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? [Take, reference intensity of sound as 10−12 W/m 2] [JEE Main 2019, 12 April ShiftII]
(a) 40 cm (c) 10 cm
(b) 20 cm (d) 30 cm
Exp. (a) Loudness of sound in decible is given by I β = 10 log10 I0 where, I = intensity of sound in W/m 2 ,
167
Oscillations and Waves I0 = reference intensity (= 10−12 W/m 2 ), chosen because it is near the lower limit of the human hearing range. Here, β =120 dB I So, we have 120 = 10 log10 −12 10
θ τ = kθ = I ′θ′ 2l/3
I 12 = log10 −12 10
⇒
⇒
1012 =
I
10−12 I = 1W/m 2
⇒
This is the intensity of sound reaching the observer. P Now, intensity, I = 4π r2 where, r = distance from source, P = power of output source. Here, P = 2 W, we have 2 1= 4 πr 2 1 ⇒ r2 = 2π 1 m = 0.398 m ≈ 40 cm r= ⇒ 2π
m are connected at the 2 two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rodmass system (see figure). Because of torsional constant k, the restoring torque is τ = kθ for angular displacementθ. If the rod is rotated by θ 0 and released, the tension in it when it passes through its mean position
19. Two masses m and
will be
m
l
Exp. (b)
m/2 l/3 2l/3
(b)
O
θ0 T
l θ 3 0 A (m)
At some angular displacement ‘θ0 ’, at point ‘A ’ the maximum velocity will be l l k ...(i) vmax = θ0ω = θ0 3 I 3 Then, tension in the rod when it passes through mean position will be 2 ml 2 θ20 k × 3 m × vmax [using Eq. (i)] = T= l 9 × l ×I 3 ml θ20 k = 3I The moment of inertia I at point O, 2
2
2 2 m 2l l 2 l m + ml + m = 3 2 3 9 9 3ml 2 ml 2 = = 9 3 ml θ20 k × 3 θ20 k kθ20 T= = = l l 3 × ml 2
=
⇒
m/2
m
Thus, when it will be released, the system will I execute SHM with a time period, T = 2 π k (Where I is moment of inertia and k is torsional constant) k and the angular frequency is given as, ω = . I If we know look at the top view of the above figure, we have
[ JEE Main 2019, 9Jan ShiftI]
l
2kθ20
l/3
m/2
Taking antilog, we have
(a)
Since in the given question, rotational torque, τ ∝ angular displacement.
20. A heavy ball of mass M is suspended from kθ20 l
(c)
3kθ20 l
(d)
kθ20 2l
the ceiling of a car by a light string of mass m (m (2 n + 1) f0 ⇒ 20000 > (2 n + 1) 1500 ⇒ 2 n + 1 < 13.33 ⇒ 2 n < 13.33 − 1 ⇒ 2 n < 12.33 or n < 616 . or n = 6 (integer number) Hence, total six overtones will be heard.
28. A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time (in seconds) is (a)
4π 3
Exp. (b)
(b)
8π 3
[JEE Main 2019, 10 Jan ShiftII]
(c)
7 π 3
v = ω A 2 − x2 or v = Aωcos ωt and
a = − ω2 x
or a = − ω2 A sinωt
Given, amplitude A = 5 cm and displacement x = 4 cm. At this time (when x = 4 cm), velocity and acceleration have same magnitude. ⇒  v x = 4  =  a x = 4  or  ω 52 − 42  =  − 4ω2
27. A closed organ pipe has a fundamental
is 20,000 Hz)
In simple harmonic motion, position ( x), velocity (v ) and acceleration (a) of the particle are given by x = A sinω t
(d)
3 π 8
⇒
3ω = + 4ω2 ⇒ ω = (3 / 4) rad/s 8π 2π 2π So, time period, T = s. ×4= ⇒T = 3 ω 3
29. A particle undergoing simple harmonic motion has time dependent displacement πt given by x (t ) = A sin . The ratio of kinetic 90 to potential energy of this particle att = 210 s will be [JEE Main 2019, 11 Jan ShiftI] (a) 2
(b) 1
(c)
1 9
(d) 3
Exp. (*) πt 90 where A is amplitude of SHM, t is time taken by particle to reach a point where its potential energy 1 1 U = kx2 and kinetic energy = k( A 2 − x2 )here k 2 2 is force constant and x is position of the particle. Here given, displacement, x(t ) = A sin
Potential energy (U ) at t = 210 s is 1 210 1 π U = kx2 = kA 2 sin2 90 2 2 1 3 = kA 2 sin2 2 π + π 2 9 1 2 2 π = kA sin 3 2 Kinetic energy at t = 210 s , is 1 K = k( A 2 − x2 ) 2 210 π 1 = kA 2 1 − sin2 90 2 1 = kA 2 cos 2 (210 π / 90) 2 1 K = kA 2 cos 2 ( π / 3) ⇒ 2
172
JEE Main Chapterwise Physics So, ratio of kinetic energy to potential energy is 1 2 kA cos 2 ( π / 3) 1 K 2 = = cot 2 ( π / 3) = 1 2 2 3 U kA sin ( π / 3) 2 ∴ No option given is correct.
30. Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin( 450t − 9x ), where distance and time are measured in SI units. The tension in the string is [JEE Main 2019, 11 Jan ShiftI] (a) 5 N (c) 7.5 N
(b) 12.5 N (d) 10 N
Exp. (b) Given, equation can be rewritten as, 9x y = 0.03sin 450 t − 450
(Qgiven, µ = 5 g / m = 5 × 10− 3 kg / m)
31. The mass and the diameter of a planet are three times the respective values for the earth. The period of oscillation of a simple pendulum on the earth is 2 s. The period of oscillation of the same pendulum on the planet would be [JEE Main 2019, 11 Jan ShiftII]
2 s 3
(c) 2 3 s
On the another planet, let period of motion is Tp and gravitational acceleration is g p l …(iii) Tp = 2 π ∴ gp (QPendulum is same, so l will be same) From Eqs. (ii) and (iii), we get l 2π gp ge Te = = ge l Tp 2π gp Now,
ge =
and
gp =
…(i)
We know that the general equation of a travelling wave is given as, …(ii) y = A sinω (t − x / v ) Comparing Eqs. (i) and (ii), we get 450 velocity, v = = 50 m / s 9 and angular velocity, ω = 450 rad / s As, the velocity of wave on stretched string with tension (T) is given as v = T /µ where, µ is linear density ∴ T = µv 2 = 5 × 10− 3 × 50 × 50 = 12.5 N
(a)
On the surface of earth, let period of motion is Te and acceleration due to gravity is ge l …(ii) Te = 2 π ∴ ge
3 s 2 3 (d) s 2
(b)
Exp. (c) Period of motion of a pendulum is given by l …(i) T = 2π g
…(iv)
GMe Re2 GM p R p2
M p = 3Me R p = 3Re G × 3Me 1 GMe 1 = ⋅ 2 = ge gp = 3 Re 3 9Re2
Given, and ∴
gp
⇒
ge
=
1 or 3
gp ge
=
1 3
From Eqs. (iv) and (v), Tp = 3 Te or Tp = 2 3 s
…(v)
(QTe = 2 s)
32. A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10− 2 m. The relative change in the angular frequency of the pendulum is best given by [JEE Main 2019, 11 Jan ShiftII]
(a) 1 rad/s −3
(c) 10
rad / s
(b) 10− 5 rad / s (d) 10− 1 rad / s
Exp. (c) We know that time period of a pendulum is given by l T = 2π g
173
Oscillations and Waves 2π g = T l Now, differentiate both side w.r.t g dω 1 = ∴ dg 2 g l dg dω = 2 g l So, angular frequency ω =
..(i)
...(ii)
By dividing Eq. (ii) by Eq. (i), we get dω dg = ω 2g Or we can write
∆ω ∆g = ω 2g
...(iii)
As ∆g is due to oscillation of support. ∴ ∆g = 2 ω 2 A (ω1 → 1 rad/s, support) Putting value of ∆g in Eq. (iii) we get ∆ω 1 2 ω12 A ω12 A ; (A = 10−2 m2 ) = ⋅ = ω 2 g g ⇒
∆ω 1 × 10−2 = = 10−3 rad/s ω 10
33. A pendulum is executing simple harmonic motion and its maximum kinetic energy is K 1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K 2. Then [JEE Main 2019, 11 Jan ShiftII]
K1 2
(a) K 2 = 2 K 1
(b) K 2 =
K (c) K 2 = 1 4
(d) K 2 = K 1
As amplitude is same in both cases so; Kmax ∝ ω2 1 or [Qg is constant] Kmax ∝ l 1 According to given data, K1 ∝ l 1 and K2 ∝ 2l K1 1 / l = ∴ =2 K2 1/ 2l K or K1 = 2 K 2 ⇒ K 2 = 1 2
34. A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60° with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, then speed of the plane is [JEE Main 2019, 12 Jan ShiftI]
3 (a) v 2
2v (c) 3
(b) v
v (d) 2
Exp. (d) Let P1 be the position of plane at t = 0, when sound waves started towards person A and P2 is the position of plane observed at time instant t as shown in the figure below. vp
P1 North
P2
60°
South
v A
Exp. (b) Kinetic energy of a pendulum is maximum at its mean position. Also, maximum kinetic energy of pendulum 1 Kmax = mω2 a2 2 where, angular frequency 2π 2π ω= = T l 2π g or
ω=
and a = amplitude.
g g or ω2 = l l
In triangle P1P2 A, P1P2 = speed of plane × time = v P × t P1 A = speed of sound × time = v × t Now, from ∆P1P2 A base cosθ = hypotenues PP v ×t cos 60° = 1 2 = P P1 A v ×t 1 vP = 2 v v vP = 2
174
JEE Main Chapterwise Physics A
35. Two light identical
springs of spring y constant k are attached horizontally at the O x two ends of an uniform horizontal rod AB of length l and mass m. The B rod is pivoted at its centre ‘O’ and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is [JEE Main 2019, 12 Jan ShiftI]
1 2k 1 (a) (b) 2π m 2π
3k 1 6k 1 k (c) (d) m 2π m 2π m
kl 2 θ For small deflection, τ = 2 Q For small deflections, sinθ =
…(iv)
x ≈θ (l / 2 )
lθ 2 Also, cosθ ≈ 1 comparing Eqs. (iv) and (i), we get kl 2 C= 2 6k (kl 2 / 2 ) α= ⋅θ ⋅θ ⇒ α = ⇒ 1 ml 2 m 12 ⇒
x=
Hence, time period of oscillation is T = 2 π
m 6k
Frequency of oscillation is given by 1 1 6k f= = T 2π m
36. A travelling harmonic wave is represented
Exp. (c) When a system oscillates, the magnitude of restoring torque of system is given by …(i) τ = Cθ where, C = constant that depends on system. Also, …(ii) τ = Iα where, I = moment of inertia and α = angular acceleration From Eqs. (i) and (ii), C ...(iii) α = ⋅θ I and time period of oscillation of system will be I C In given case, magnitude of torque is τ = Force × perpendicular distance T = 2π
x θ l/2
kx
τ = 2 kx ×
l cos θ 2
(a) The wave is propagating along the negative Xaxis with speed 25 ms−1 . (b) The wave is propagating along the positive Xaxis with speed 25 ms−1 . (c) The wave is propagating along the positive Xaxis with speed100 ms−1 . (d) The wave is propagating along the negative Xaxis with speed100 ms−1
Exp. (a) Wave equation is given by, y = 10−3 sin(50t + 2 x) kx
l cos θ 2
by the equation y ( x ,t ) = 10−3 sin (50t + 2 x ), where x and y are in metre andt is in second. Which of the following is a correct statement about the wave? [JEE Main 2019, 12 Jan ShiftI]
Speed of wave is obtained by differentiating phase of wave. Now, phase of wave from given equation is φ = 50t + 2 x = constant Differentiating ‘φ’ w.r.t. ‘t’, we get d d (constant) (50 t + 2 x) = dx dt dx 50 + 2 = 0 ⇒ dt dx −50 = = − 25 ms −1 ⇒ 2 dt
175
Oscillations and Waves So, wave is propagating in negative x−direction with a speed of 25 ms −1. Alternate Solution The general equation of a wave travelling in negative x direction is given as …(i) y = asin(ωt + kx) Given equation of wave is …(ii) y = 10−3 sin(50 + 2 x) Comparing Eqs. (i) and (ii), we get ω = 50 and k = 2 ω 50 Velocity of the wave, v = = = 25 m/s k 2
37. A simple harmonic motion is represented by
The y = 5(sin 3 πt + 3 cos 3πt ) cm. amplitude and time period of the motion are [JEE Main 2019, 12 Jan ShiftII]
3 (a) 10 cm, s 2 3 (c) 5 cm, s 2
2 s 3 2 (d) 10 cm, s 3
(b) 5 cm,
Exp. (d) Equation for SHM is given as y = 5 (sin 3 πt + 3 cos 3 πt ) 1 3 cos 3 πt = 5 × 2 × sin 3 πt + 2 2 π π = 5 × 2 cos . sin 3 πt + sin . 3 πt 3 3 π = 5 × 2 sin 3 πt + 3 or
[using, sin(a + b ) = sin acos b + cos asin b] π y = 10sin 3 πt + 3
Comparing this equation with the general equation of SHM, i.e. 2 πt + φ , y = A sin T We get, amplitude, A = 10 cm 2π 2 and or Time period, T = s 3π = 3 T
38. A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark. near the
open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment is close to [JEE Main 2019, 12 Jan ShiftII]
(a) 328 ms− 1 (c) 322 ms− 1
(b) 341 ms− 1 (d) 335 ms− 1
Exp. (a) Key idea To overcome the error occured in measurement of resonant length.We introduce end correction factor e in length
In first resonance, length of air λ coloumn = . 4 λ So, l1 + e = l +e 4 or 11 × 4 + 4e = λ So, speed of sound is ⇒ v = f1λ = 512 (44 + 4e ) …(i) And in second case, λ′ l1 ′ + e = 4 or 27 × 4 + 4 e = λ ′ ⇒ v = f2 λ ′ = 256 (108 + 4 e ) Dividing both Eqs. (i) and (ii), we get 512 (44 + 4e ) 1= 256 (108 + 4e )
…(ii)
⇒ e = 5 cm Substituting value of e in Eq. (i), we get Speed of sound v = 512 (44 + 4e ) = 512 (44 + 4 × 5) = 512 × 64cm s −1 = 327.68 ms −1 ≈ 328 ms −1
39. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012 per second. What is the force constant of the bonds connecting one atom with the other? (Take, molecular weight of silver = 108 and Avogadro number = 6.02 × 1023 g mol −1) [JEE Main 2018] (a) 6.4 N/m (c) 2.2 N/m
(b) 7.1 N/m (d) 5.5 N/m
176
JEE Main Chapterwise Physics (a) 12.1 GHz (b) 17.3 GHz (c) 15.3 GHz (d) 10.1 GHz
Exp. (b) For a harmonic oscillator, m T = 2π k where, k = force constant and T = ∴
1 f
k = 4π2f 2m 108 × 10−3 22 = 4 × × (1012 )2 × 7 6.02 × 1023 2
⇒
k = 7.1 N/m
40. A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 × 103 kg/m 3 and its Young’s modulus is 9.27 × 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations? [JEE Main 2018]
(a) 5 kHz
(b) 2.5 kHz (c) 10 kHz (d) 7.5 kHz λ 4
Exp. (a)
Exp. (b) As the observer is moving towards the source, so frequency of waves emitted by the source will be given by the formula 1 + v /c fobserved = factual ⋅ 1− v /c v 1 Here, frequency = c 2 So,
3/2 fobserved = factual 1/ 2
∴
fobserved = 10 ×
1/ 2
1/ 2
3 = 17.3 GHz
42. A particle is executing simple harmonic
motion with a time period T . At timet = 0, it is at its position of equilibrium. The kinetic energytime graph of the particle will look, like [JEE Main 2017 (Offline)] KE
(a)
T
O
t
L KE
From vibration mode, λ = L ⇒ λ = 2L 2 Y ∴ Wave speed, v = ρ v So, frequency f = λ 1 Y f= ⇒ 2L ρ =
(b)
O
T/2
T
t
KE
(c)
O
T
T/4 T/2 t
1
9.27 × 1010
2 × 60 × 10−2
2.7 × 103
≈ 5000 Hz f = 5 kHz
KE
(d)
O
T/2
T
T t
41. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 108 ms −1) [JEE Main 2017 (Offline)]
Exp. (c) KE is maximum at mean position and minimum T at extreme position at t = . 4
177
Oscillations and Waves 43. A particle performs simple harmonic motion with amplitude A . Its speed is trebled at the instant that it is at a distance 2 A from equilibrium position. The new 3 amplitude of the motion is [JEE Main 2016 (Offline)]
A (a) 3
41
(b) 3A
(c) A 3
So, velocity at point P =
i.e.
⇒
7 (d) A 3
20
∫0
mgx l m/ l
v = gx dx = gx dt t dx g dt =∫ 0 x
(l, m) P
dx x
[2 x ]20 0 = 10 t ⇒ 2 20 = 10 t t =2 2 s
Exp. (d) / The velocity of a particle executing SHM at any instant, is defined as the time rate of change of its displacement at that instant. v = ω A 2 − x2 where, ω is angular frequency, A is amplitude and x is displacement of a particle. Suppose that the new amplitude of the motion be A′. Initial velocity of a particle performs SHM, 2 2A ... (i) v 2 = ω2 A 2 − 3 where, A is initial amplitude and ω is angular frequency. Final velocity, 2 2 A ...(ii) (3v )2 = ω2 A′2 − 3 From Eqs. (i) and (ii), we get 4 A2 1 9 ⇒ A′ = 7 A = 3 9 4 A2 2 A′ − 9 A2 −
44. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is (Take, g = 10 ms− 2) [JEE Main 2016 (Offline)]
(a) 2 π 2 s (b) 2 s
(c) 2 2 s
(d) 2 s
Exp. (c) A uniform string of length 20 m is suspended from a rigid support. Such that the time taken to reach the support, mgx T= l
45. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water, so that half of it is in water. The fundamental frequency of the air column is now [JEE Main 2016 (Offline)] (a)
f 2
(b)
3f 4
(c) 2 f
(d) f
Exp. (d) For open ends, fundamental frequency f in air, we have l λ =l 2 ⇒ λ = 2l v= fλ v v …(i) ⇒ f= = λ 2l When a pipe is dipped vertically in l/2 water, so that half of it is in water, we have l l λ = 4 2 ⇒ λ = 2l ⇒ v = f ′ λ v v …(ii) ⇒ f′ = = =f λ 2l Thus, the fundamental frequency of the air column is now, f = f′
46. The period of oscillation of a simple pendulum isT = 2 π
L . Measured value of L g
is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. The accuracy in the determination of g is [JEE Main 2015] (a) 2%
(b) 3%
(c) 1%
(d) 5%
178
JEE Main Chapterwise Physics
Exp. (b)
Exp. (b) During oscillation, motion of a simple pendulum KE is maximum of mean position where PE is minimum. At extreme position, KE is minimum and PE is maximum. Thus, correct graph is depicted in option (b).
/ Given time period T = 2π L/ g Thus, changes can be expressed as ±
2∆T ∆L ∆g =± ± T L g
According to the question, we can write 0.1cm 1 ∆L = = 20.0 cm 200 L
48. A train is moving on a straight track with speed 20ms − 1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is close to (speed of sound = 320 ms − 1)
Again time period 90 1 s and ∆T = s 100 100 ∆T 1 = T 90 T=
⇒
[JEE Main 2015]
(a) 6%
Now, Q
T = 2π
Q
g = 4π
c 320 f1 = f0 = 1000 c − vs 320 − 20
1 1 × 100% = × 100 % + 2 × 200 90 ~ − 2.72% ~ − 3%
Similarly, apparent frequency heard, after crossing the trains c 320 f2 = f0 = 1000 + c v 320 + 20 s [c = speed of sound] 2 cvs f ∆f = f1 − f2 = 2 2 0 c − vs
Thus, accuracy in the determination of g is approx 3%.
47. For a simple pendulum, a graph is plotted
or
∆f × 100 = f0
between its Kinetic Energy (KE) and Potential Energy (PE) against its displacement d. Which one of the following represents these correctly? (graphs are schematic and not drawn to scale) [JEE Main 2015] E PE
KE
(a)
PE
(b) KE
d E
E KE
d
PE
(d) KE
(c)
d PE
(d) 24%
Apparent frequency heard by the person before crossing the train.
2 ∆T ∆g ∆L or × 100% = × 100% + × 100% T L g
E
(c) 18%
Exp. (b)
L g 2 L
T2 ∆g ∆L 2 ∆T = + g L T
∴
(b) 12%
2cvs 2 × 100 2 c − vs
=
2 × 320 × 20 × 100 300 × 340
=
2 × 32 × 20 3 × 34
= 12.54% = 12%
49. A particle moves with simple harmonic motion in a straight line. In first t sec, after starting from rest, it travels a distance a and in next t sec, it travels 2a in same direction, then [JEE Main 2014] (a) (b) (c) (d)
amplitude of motion is 3a time period of oscillations is 5π amplitude of motion is 4a time period of oscillations is 6π
Exp. (d)
179
Oscillations and Waves In SHM, a particle starts from rest, we have x = A cos ωt , at t = 0, x = A When t = t , then x = A − a When t = 2t , then x = A − 3a So, A − a = A cos ωt A − 3a = A cos 2ωt As, cos 2ωτ = 2 cos 2 ωt − 1 ⇒ ⇒
A − 3a 2 A 2 + 2 a2 − 4 Aa − A 2 = A A2 2 2 2 A − 3aA = A + 2 a − 4 Aa A − a = A cos ωt cos ωt = 1/2 2π π = T 3 T = 6π
(c) 6
(a) 0.7 (c) 0.729
Exp. (c) Amplitude of damped oscillator is given by A = A0 e After 5 s,
For closed organ pipe =
(2 n + 1) v (n = 0, 1, 2, ...) 4l
(2 n + 1) v < 1250 4l
(2 n + 1) < 1250 × (2 n + 1) < 12.5 2n < 11.50 n < 5.25
(b) 0.81 [JEE Main 2013] (d) 0.6
(d) 4
Exp. (c)
⇒ ⇒
∴Possible frequencies = 10 Hz, 300 Hz, 500 Hz, 700 Hz, 900 Hz, 1100 Hz below 1250 Hz.
decreases to 0.9 times its original magnitude in 5s. In another 10s, it will decrease to α times its original magnitude, whereα equals
end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s. [JEE Main 2014] (b) 8
λ = 0.85 ⇒ λ = 4 × 0.85 4 c 340 As we know, ν = = = 100 Hz λ 4 × 0.85
i.e.,
51. The amplitude of a damped oscillator
50. A pipe of length 85 cm is closed from one
(a) 12
l = 0.85 = 4l
2
A − 3a A − a = 2 −1 A A
a2 = 2 aA ⇒ A = 2 a Now, ⇒
Alternate Method In closed organ pipe, fundamental node
4 × 0.85 340
⇒
0.9 A0 = A0 e 0.9 = e
After, 10 s,
A=
⇒
A=
−
−
bt 2m
−
b( 5 ) 2m
b( 5 ) 2m
( 15 ) −b A0e 2 m 5b − A0 (e 2 m )3
…(i)
…(ii)
From Eqs. (i) and (ii), we get A = 0.729 A0 Hence, α = 0.729
85 cm
52. An ideal gas enclosed in a vertical cylindrical
So, n = 0, 1, 2, 3, ...,5 So, we have 6 possibilities.
container supports a freely moving piston of mass M . The piston and the cylinder have equal crosssectional area A. When the piston is in equilibrium, the volume of the gas isV 0 and its pressure is p 0. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency [JEE Main 2013]
180
JEE Main Chapterwise Physics (a) (c)
1 A γ p0 2 π V0M 1 2π
(b)
A γ p0 M V0 2
(d)
Solution of above equation is
1 V 0 Mp 0 2 π A2γ 1 2π
−
bt 2
k b2 − m 4 where, amplitude drops exponentially with time. x = A0 e
M V0 A γ p0
sin ωt ; with ω2 =
A τ = A0 e
i.e.,
Exp. (c) Pressure applied by piston,
−
bτ 2
Average time τ is that duration when amplitude drops by 63%, i.e., becomes A0 /e.
Mg = p0 A
bτ
− A0 = A0 e 2 e 2 bτ = 1 or τ = b 2
Aτ =
Thus,
x
or
x0
54. A cylindrical tube open at both ends, has a …(i) Mg = p0 A As no exchange of heat, so process is adiabatic. p0 V0γ = pV γ , p0 Ax0γ = pA( x0 − x)γ p=
p0 x0γ
(a) f
( x0 − x)γ
Let piston be displaced by x. p0 xγ 0 A = f Mg − restoring ( x − x)γ 0
f=
1 2π
γp0 A 1 = x0 M 2 π
γp0 A 2 MV0
53. If a simple pendulum has significant amplitude (upto a factor of 1/e of original) only in the period between t = 0 s to t = τ s, then τ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity withb as the constant of proportionality, the average life time of the pendulum (assuming damping is small) in seconds is [AIEEE 2012]
0.693 (a) b
(b) b
1 (c) b
Exp. (d) For damped harmonic motion, ma = − kx − mbv or ma + mbv + kx = 0
(b)
f 2
(c)
3f 4
(d) 2 f
Exp. (d)
x0γ = Frestoring [Qx0 − x ≈ x0 ] p0 A 1 − γ ( x0 − x) γ p0 Ax F=− x0 ∴
fundamental frequency f, in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the aircolumn is now [AIEEE 2012]
(d)
2 b
Initially for open organ pipe, fundamental frequency, v [given] =f νo = 2l But when it is half dipped in water, then it becomes l closed organ pipe of length . In this case, 2 fundamental frequency, v v v = = =f νc = 4l ′ 4 l 2 l 2
55. Statement I Two longitudinal waves given
by equations—y1 ( x ,t ) = 2a sin (ωt − kx ) and y 2 ( x ,t ) = a sin(2ωt − 2kx ) will have equal intensity. Statement II Intensity of waves of given frequency in same medium is proportional to the square of amplitude only. [AIEEE 2011] (a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is false (c) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I (d) Statement I is true, Statement II is true; Statement II is not correct explanation of Statement I
Exp. (d)
181
Oscillations and Waves 1 ρω2 A 2 v 2 Here, ρ = density of medium, A = amplitude, ω = angular frequency and v = velocity of wave. ∴ Intensity depend upon amplitude, frequency as well as velocity of wave. Also, I1 = I2 As Intensity, I =
56. A
travelling wave represented y = A sin(ωt − kx ) is superimposed another wave represented y = A sin (ωt + kx ). The resultant is
by on by
[AIEEE 2011]
(a) a standing wave having nodes at 1 λ x = n + , n = 0,1, 2 2 2 (b) a wave travelling along + x direction (c) a wave travelling along −x direction
nλ (d) a standing wave having nodes at x = ; 2 n = 0,1, 2
∴
By superposition principle, y = y1 + y2 = A sin (ωt − kx) + A sin(ωt + kx) y = 2 A sinωt cos kx Clearly, it is equation of standing wave for position of nodes y = 0. λ 1 λ i .e., x = (2 n + 1) ⇒ n + , n = 0, 1, 2 , 3 4 2 2
57. A wooden cube (density of wood d ) of side l
floats in a liquid of density ρ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period, T . Then, T is equal to [AIEEE 2011] (a) 2 π
lρ (ρ − d )g
(b) 2 π
ld ρg
(c) 2 π
lρ dg
(d) 2 π
ld (ρ − d )g
Exp. (b) Let at any instant, cube be at a depth x from the equilibrium position, then Net force acting on the cube = Upthrust on the portion of length x
...(i)
F ∝ − x (SHM) Negative sign shows that force is opposite to x . Hence, equation of SHM ...(ii) F = − kx Comparing Eqs. (i) and (ii), we get k = ρl 2 g ∴
T = 2π
l 3d m ld = 2π = 2π k ρg ρl 2 g
58. If a spring of stiffnessk is cut into two parts A
andB of lengthl A : l B = 2 : 3, then the stiffness of spring A is given by [AIEEE 2011] (a)
5 k 2
(b)
3k 5
(c)
2k 5
(d) k
Exp. (a) As
F = − kx ⇒ k ∝
For spring,
k∝
∴
Exp. (a)
F = − ρl 2 xg = − ρl 2 g x
1 1 ∝ l x
1 l
kA l = lA k
⇒ kA =
lA + lB 5 k= k lA 2
59. The transverse displacement y ( x ,t ) of a wave on a string is 2 2 y ( x ,t ) = e −( ax + bt + 2 ab xt ). This represents a
given
b
[AIEEE 2011]
(a) wave moving in −x direction with speed
b a
(b) standing wave of frequency b 1 (c) standing wave of frequency b a (d) wave moving in + x direction with speed b
Exp. (a) y ( x, t ) = e − ( ax
2
+ bt 2 + 2 ab xt )
= e −(
a x+
b t )2
It is a function of type y = f(ωt + kx) So, y ( x, t ) represents wave travelling along − x direction. ω b b = ∴ Speed of wave = = k a a
60. Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the xaxis. Their mean position is separated by distance X 0 ( X 0 > A ). If the maximum separation between them is
182
JEE Main Chapterwise Physics
( X 0 + A ), the phase difference between their motion is [AIEEE 2011] (a)
π 3
(b)
π 4
(c)
π 6
(d)
π 2
Exp. (a) Let
x1 = A sin(ωt + φ 1) and x2 = A sin (ωt + φ 2 )
x2 − x1 = A[sin(ωt + φ 2 ) − sin (ωt − φ 1)] 2 ωt + φ1 + φ2 φ2 − φ1 = 2 A cos sin 2 2 The resultant motion can be treated as a simple φ − φ1 harmonic motion with amplitude2 Asin 2 . 2 Given, maximum distance between the particles = X0 + A ∴ Amplitude of resultant SHM = X 0 + A − X 0 = A φ − φ1 2 A sin 2 ∴ = A 2 ⇒ Phase difference, φ2 − φ1 = π/ 3
61. A mass M , attached to a horizontal spring, executes SHM with amplitude A1. When the mass M passes through its mean position, then a smaller mass m is placed over it and both of them move together with amplitude A A 2. The ratio of 1 is [AIEEE 2011] A2 (a)
M (b) M + m
M+m M
(c)
M + m M
1/ 2
(d)
1/ 2
M M+m
between two atoms in a diatomic molecule a b is approximately given by U ( x ) = 12 − 6 , x x where a and b are constants and x is the distance between the atoms. If the dissociation energy of the molecule is D = [U ( x = ∞ ) − U at equilibrium ], then D is b2 (a) 2a
b2 (b) 12a
b2 (c) 4a
[AIEEE 2010]
(d)
b2 6a
Exp. (c) a b − 6 ⇒U( x = ∞ ) = 0 x12 x dU 12 a 6b As, = − 13 + 7 F=− x dx x At equilibrium, F = 0 2a ∴ x6 = b − b2 b a = − ∴ Uat equilibrium = 2 2 a 4a 2a b b U( x ) =
∴
D = [U ( x = ∞ ) − Uat equilibrium ] =
b2 4a
63. The equation of a wave on a string of linear mass density 0.04 kgm −1 is given by t x y = 0.02 (m) sin 2 π − . 0.04 (s) 0.50 (m)
The tension in the string is
[AIEEE 2010]
(a) 4.0 N
(d) 6.25 N
(b) 12.5 N (c) 0.5 N
Exp. (d)
Exp. (c) At mean position, Fnet = 0 ∴ By conservation of linear momentum, Mv1 = (M + m)v 2 Mω1 A1 = (M + m) ω2 A2 But angular velocity, k ω1 = M k and ω2 = M+ m On solving,
62. The potential energy function for the force
A1 = A2
m+ M M
As
v=
T µ
So, T = µv 2 = µ
ω2 k
2
= 0.04
(2 π / 0.004)2 (2 π / 0.50)2
= 6.25 N
64. If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then which of the following does not change with time? (a) a 2T 2 + 4 π 2v 2 (c) aT + 2 πv
aT x aT (d) v
(b)
[AIEEE 2009]
183
Oscillations and Waves Exp. (b) 4π aT ω xT 4 π = = 2 ×T = = constant x x T T Hence, aT/x does not change with time. 2
2
65. A motor cycle starts from rest and
accelerates along a straight path at 2 ms −2. At the starting point of the motor cycle, there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (Speed of sound = 330 ms −1 ) [AIEEE 2009] (a) 49 m
(b) 98 m
(c) 147 m
(d) 196 m
Exp. (b) For motor cycle, u = 0, a = 2 ms −2 Observer is in motion and source is at rest, then apparent frequency, v − vo 330 − vo 94 ⇒ n′ = n n=n v + vs 100 330 330 × 94 330 − vo = ⇒ 100 94 × 33 33 × 6 ms −1 ⇒ vo = 330 − = 10 10 v 2 − u 2 9 × 33 × 33 9 × 1089 = s= = ≈ 98 m 100 2a 100
66. Three sound waves of equal amplitudes
have frequencies ( ν − 1), ν, ( ν + 1). They superpose to give beat. The number of beats produced per second will be [AIEEE 2009] (a) 4
(b) 3
(c) 2
(d) 1
vO
∴
2
vHe
=
7 RT 5 32
vHe =
5 RT 3 4
7 × 3× 4 5 × 32 × 5
vHe = 460 ×
or
and
5 × 32 × 5 ≈ 1420 ms −1 7 ×3×4
68. While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, she measures the column length to be x cm for the second resonance. Then, [AIEEE 2008] (a)18> x (c) 54 > x > 36
(b) x > 54 (d) 36 > x > 18
Exp. (b) l1 = 18 cm v f= 1 4l1 3 v2 f= 4 l2
[for first resonances] [for second resonances]
where, l2 = x according to given situation and also v1 < v 2 as during summer temperature would be higher. v 3 v2 v = 1 or l2 = 3l1 × 2 ⇒ 4l2 4l1 v1 ⇒ x = 54 × (A quantity greater than 1) So, x > 54
69. A wave travelling along the xaxis is
Exp. (c) Maximum number of beats = ν + 1 − (ν − 1) = 2
67. The speed of sound in oxygen (O2 ) at a
certain temperature is 460 ms −1. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal) [AIEEE 2008]
(a) 460 ms−1 (c) 650 ms−1
vO 2 =
2
(b) 500 ms−1 (d) 1420 ms−1
Exp. (d) Speed of sound is given by γRT v= M
described by the equation y ( x ,t ) = 0.005cos(αx − βt ). If the wavelength and the time period of the wave are 0.08 m and 2.0 s respectively, then α and β in appropriate units are [AIEEE 2008] (a) α = 25.00 π , β = π (c) α =
0.04 1.0 , β= π π
(b) α =
008 . 2.0 ,β = π π
(d) α = 12.50 π , β =
π 2.0
Exp. (a) As general equation is y ( x, t ) = a cos (kx − ωt ) 2π 2π and ω = where, k = λ T ∴ y( x, t ) = 0.005cos(αx − βt )
184
JEE Main Chapterwise Physics ⇒ So,
2π 2π = α and =β λ T 2π 2π α= = 25 π m−1 and β = = π s −1 0.08 2
70. A point mass oscillates along the xaxis
according to the law x = x 0 cos (ωt − π/4). If the acceleration of the particle is written as [AIEEE 2007] a = A cos (ωt + δ ), then (a) A = x 0 ,δ = −
π 4
π 4 3π (d) A = x 0ω2 , δ = 4 (b) A = x 0ω2 , δ =
π (c) A = x 0ω , δ = − 4 2
Exp. (d) π x = x0 cos ωt − 4 Acceleration,
So,
π = − ω2 x0 cos ωt − 4 3 π = ω2 x0 cos ωt + 4
a=
d2x dt 2
A = ω2 x0
and δ =
3π 4
71. A sound absorber attenuates the sound level by 20 db. The intensity decreases by a factor of [AIEEE 2007] (a) 1000
(b) 10000
(c) 10
(d) 100
Exp. (d) Let intensity of sound be I and I′. Loudness of sound initially, l β1 = 10 log l0 Later, Given,
l′ β 2 = 10 log l0 β1 − β 2 = 20
l l 20 = 10 log or l ′ = ∴ l′ 100 Therefore, intensity decreases by a factor of 100.
72. Two springs of force constants k1 and k 2 , are connected to a mass m as shown. The frequency of oscillation of the mass is f . If both k1 and k 2 are made four times their original values, the frequency of oscillation becomes [AIEEE 2007]
(a)
f 2
(b)
k1
m
f 4
(c) 4f
k2
(d) 2 f
Exp. (d) 1 k1 + k2 1 and f ′ = 2π m 2π k1 + k2 1 f′ = ⋅2 = 2f 2π m f=
⇒
4k1 + 4k2 m
73. A particle of mass m executes simple harmonic motion with amplitude a and frequency ν. The average kinetic energy during its motion from the position of equilibrium to the end is [AIEEE 2007] (a) π 2ma 2 ν2
1 ma 2 ν2 4 (d) 2 π 2ma 2 ν2 (b)
(c) 4 π 2ma 2 ν2
Exp. (a) Average kinetic energy of particle 1 [as ω = 2 πν] = ma2ω2 4 1 = ma2 (2 πν)2 = π 2 ν2 ma2 4
74. The displacement of an object attached to a spring and executing simple harmonic motion is given by x = 2 × 10−2 cos πt metre. The time at which the maximum speed first occurs is [AIEEE 2007] (a) 0.5 s
(b) 0.75 s
(c) 0.125 s (d) 0.25 s
Exp. (a) x = (2 × 10−2 ) cos πt Here,
a = 2 × 10−2 m = 2 cm
At t = 0, x = 2 cm, i.e., the object is at positive extreme, so to acquire maximum speed (i.e., to 1 reach mean position), it takes th of time period. 4 T ∴ Required time = 4 As x = a cos ωt 2π where, ω= = π ⇒ T =2 s T T 2 So, required time = = = 0.5 s 4 4
185
Oscillations and Waves 75. A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is [AIEEE 2006]
(a) 105 Hz (b) 1.05 Hz (c) 1050 Hz (d) 10.5 Hz
Exp. (a) For string fixed at both the ends, resonant nv frequency are given by f = , where symbols 2L have their usual meanings. It is given that 315 Hz and 420 Hz are two consecutive resonant frequencies, let these be nth and (n + 1th ) harmonics. nv Then, …(i) 315 = 2L (n + 1)v …(ii) and 420 = 2L Dividing Eq. (i) by Eq. (ii), we get 315 n or n = 3 = 420 n + 1 From Eq. (i), lowest resonant frequency v 315 = = 105 Hz f0 = 2L 3
76. A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms −1. The velocity of sound in air is 300 ms −1. If the person can hear frequencies upto a maximum of 10000 Hz, the maximum value ofv upto which he can hear the whistle is [AIEEE 2006] (a) 15 2 ms −1 (c) 15 ms −1
(b) 15/ 2 ms −1 (d) 30 ms −1
Exp. (c) Given, velocity of sound in air = 300 m/s If a source of sound is moving towards a stationary listener, the frequency heard by the listener would be different from the actual frequency of the source, this apparent frequency vsound in air , where is given by fapp = vsound in air ± vsource symbols have their usual meanings.
In the denominator, +ve sign would be taken when source is receding away from the listener, while –ve sign would be taken when source is approaching the listener. Let v be the maximum value of source velocity for which the person is able to hear the sound, then 300 10000 = fapp = × 9500 300 − v or
v = 15 ms −1
77. The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 ms −1. The period of oscillation is [AIEEE 2006] (a) 0.01 s
(b) 10 s
(c) 0.1 s
(d) 100 s
Exp. (a) The maximum velocity of a particle performing SHM is given by v = Aω, where A is the amplitude and ω is the angular frequency of oscillation. ∴ 4.4 = (7 × 10−3 ) × 2 π/ T or
T=
7 × 10−3 2 × 22 × = 0.01 s 4.4 7
78. Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy ? [AIEEE 2006] 1 s 6 1 (c) s 3 (a)
1 s 4 1 (d) s 12 (b)
Exp. (a) KE of a body undergoing SHM is given by 1 mω2 A 2 KE = mω2 A 2 cos 2 ωt and KEmax = 2 2 [symbols represent standard quantities] 75 From given information, KE = (KEmax ) × 100 mω2 A 2 mω2 A 2 3 cos 2 ωt = × ⇒ 2 2 4 3 π or or ωt = cos ωt = ± 2 6 2π π T 1 or t = or ×t = = s 6 T 12 6
79. The function sin 2 (ωt ) represents (a) a periodic but not simple harmonic motion with a period 2 π/ω [AIEEE 2005]
186
JEE Main Chapterwise Physics −π 6 −π (c) 3
(b) a periodic but not simple harmonic motion with a period π/ω (c) a simple harmonic motion with a period 2 π/ω (d) a simple harmonic motion with a period π/ω
(a)
Exp. (b)
Exp. (a) y = sin ωt 2
Here,
Given,
y
O
π ω
2π ω
3π ω
4π ω
dy = 2 ω sin ωt cos ωt = ω sin2 ωt dt d2y dt 2 For SHM,
∝−y
dt 2 Hence, function is not SHM but periodic. From the 2π = 2ω yt graph, time period is T = π /ω. Q T
80. The intensity of gamma radiation from a given source is I .On passing through 36 mm of lead, it is reduced to I/8. The thickness of lead, which will reduce the intensity to I/2 will be [AIEEE 2005] (a) 6 mm
(b) 9 mm
(c) 18 mm (d) 12 mm
Exp. (d)
As intesity, I′ = Ie −µx Taking log on both sides, we get I′ I …(i) or − µ ⋅ 36 = log − µx = log I 8I I …(ii) and − µx′ = log 2I From Eqs. (i) and (ii), we get 1 3 log 2 36 or = 1 x′ log 2
π . sin 100 πt + y1 = 01 3
⇒ Velocity of particle 1, dy1 π . × 100 π cos 100 πt + = v1 = 01 dt 3 π π or v1 = 10 π sin 100 πt + + 3 2 5π or v1 = 10 π sin 100 πt + 6 and y2 = 01 . cos πt
= 2 ω2 cos 2 ωt d2y
π 3 π (d) 6 (b)
⇒ Velocity of particle 2, dy2 = v 2 = − 01 . sin πt dt or
v 2 = 01 . sin ( πt + π )
Hence, phase difference 5π ∆φ = φ1 − φ 2 = 100 πt + − ( πt + π ) 6 5π π [at t = 0 ] = − π=− 6 6
82. When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what will be the original frequency of fork 2? [AIEEE 2005]
(a) 200 Hz (b) 202 Hz (c) 196 Hz (d) 204 Hz x′ = 12 mm
81. Two simple harmonic motions are represented π by the equations y1 = 0.1 sin 100πt + and 3 y 2 = 0.1 cos πt . The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is [AIEEE 2005]
Exp. (b) The frequency of fork 2 = 200 ± 4 = 196 or 204 Hz Since, on attaching the tape on the prong of fork 2, its frequency decreases but now the number of beats per second is 6 i.e., the frequency difference now increases. It is possible only when before attaching the tape, the frequency of fork 2 is less than the frequency of tuning fork 1. Hence, the frequency of fork 2 is 196 Hz.
187
Oscillations and Waves 83. If a simple harmonic motion is represented 2
by
d x dt 2
+ αx = 0, its time period is [AIEEE 2005]
2π α
(a)
(b)
2π α
(c) 2 πα
(d) 2 π α
Exp. (b) d2x
= −αx dt 2 We know that, for SHM d2x a = 2 = − ω2 x dt From Eqs. (i) and (ii), we have ω2 = α or ω = α 2π = α ⇒ T 2π or T= α
…(i)
So, time period also increases but upto half empty. As water is coming out after half empty, CG goes up. Effective length decreases and for complete hollow sphere, CG is also at the initial position, so effective length as well as time period decreases and same as initial at last. T2 = 2 π
…(ii)
84. The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would [AIEEE 2005] (a) first increase and then decrease to the original value (b) first decrease and then increase to the original value (c) remain unchanged (d) increase towards a saturation value
l
G Spherical hollow ball
l − ∆l g
and T1 < T2 Hence, time period first increases and then decreases to the original value.
85. An observer moves towards a stationary source of sound, with a velocity onefifth of the velocity of sound. What is the percentage increase in the apparent frequency? [AIEEE 2005] (a) Zero
(b) 0.5%
(c) 5%
(d) 20%
Exp. (d) v 320 = 64 ms −1 ⇒ vo = 5 5 When observer moves towards the stationary source, then v + vo 320 + 64 n n′ = n ⇒ n′ = v 320 n′ 384 384 or n′ = n or = 320 n 320 vo =
Exp. (a) Hence, percentage increase n′ − n = 384 − 320 × 100 n 320
l + ∆l
l
64 = × 100 = 20% 320
G G
G'
Spherical hollow ball filled with water
Spherical hollow ball half filled with water
l g As water is coming out, CG of the system goes down, effective length increase. l + ∆l ,T1 > T2 T1 = 2 π g T =2π
86. The bob of a simple pendulum executes simple harmonic motion in water with a periodt ,while the period of oscillation of the bob ist 0 in air. Neglecting frictional force of water and given that the density of the bob is (4 / 3) × 1000 kgm −3. What relationship [AIEEE 2004] betweent andt 0 is true? (a) t = t 0
Exp. (c)
(b) t =
t0 2
(c) t = 2 t 0 (d) t = 4t 0
188
JEE Main Chapterwise Physics ∴
The time period of simple pendulum in air, l g
T = t0 = 2π
…(i)
So, the effective force constant in their series combination is kk k= 1 2 k1 + k2
Here, l being the length of simple pendulum, In water, effective weight of bob, w′ = Weight of bob in air − Upthrust ⇒ ρVgeff = mg − m′ g = ρVg − ρ′ Vg = (ρ − ρ′ )Vg where, ρ′ = density of water, ρ = density of bob. ρ − ρ′ ρ′ ∴ geff = g = 1 − g ρ ρ
∴
t = 2π
t Thus, = t0
=
l ρ′ − g 1 ρ
1 = ρ′ 1 − ρ 4 4−
Time period of combination, T =2π
…(ii)
= 2 or t = 2 t 0 3
[AIEEE 2004]
(b)
(c) T −1 = t 1−1 + t 2−1
(d)
t 22 + t 2−2
Exp. (b) Time period of spring, T =2π
m k
m k1
…(i)
In series, f is same.
From Eqs. (i) and (ii), we obtain m m t 12 + t 22 = 4 π 2 + k1 k2 or
1 1 t 12 + t 22 = 4 π 2 m + k1 k2
or
t 12 + t 22 =
⇒
4 π 2 m(k1 + k2 ) k1k2
t 12 + t 22 = T 2
[from Eq. (iii)]
(a) ∝ x
[AIEEE 2004]
(b) ∝ x 2
(c) independent of x (d) ∝ x1/ 2 where, x is the displacement from the mean position.
Exp. (c) In simple harmonic motion, when a particle is displaced to a position from its mean position, then its kinetic energy gets converted into potential energy. Hence, total energy of a particle remains constant or the total energy in simple harmonic motion does not depend on displacement x.
m k2
…(ii)
medium can be expressed as π y = 10−6 sin 100t + 20x + m 4 where, t is in second and x in metre. The speed of the wave is [AIEEE 2004]
For second spring, t2 = 2 π
…(iii)
89. The displacement y of a particle in a
Here, k being the force constant of spring. For first spring, t1 = 2 π
4 π 2 m(k1 + k2 ) k1k2
simple harmonic motion is
simple harmonic motion with a period t 1 , while the corresponding period for another spring is t 2. If the period of oscillation with the two springs in series is T , then (a) T = t 1 + t 2
m(k1 + k2 ) k1k2
88. The total energy of a particle, executing
87. A particle at the end of a spring executes
T = t 12 + T −2 = t 1−2
T2 =
⇒
1 1000 1 − (4/3) × 1000
2
x = x1 + x2 k k −F −F F ⇒k = 1 2 = − k k1 k2 k1 + k2
(a) 2000 ms −1 (c) 20 ms −1
Exp. (b)
(b) 5 ms −1 (d) 5π ms −1
189
Oscillations and Waves π As given y = 10−6 sin 100t + 20 x + 4 Comparing it with y = a sin (ωt + kx + φ), we obtain ω = 100 rad s −1, k = 20 m −1 v=
∴
…(i)
…(ii)
Exp. (a)
ω 100 = = 5 ms −1 k 20
90. A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω 0. An external force F (t ) proportional to cos ωt (ω ≠ ω 0 ) is applied to the oscillator. The time displacement of the oscillator will be proportional to [AIEEE 2004] (a) (c)
m
(b)
ω20 − ω2 1
(d)
m(ω20 + ω2 )
1 m(ω20 − ω2 )
92. A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T . If the mass is increased by m , the time period becomes 5T /3, then the ratio of m / M is [AIEEE 2003]
m
3 5
ω20 + ω2
Exp. (c)
(b)
25 9
(c)
16 9
5 3
(d)
M k M+ m T′ = 2 π k M+ m 5T ⇒ =2π k 3 Dividing Eq. (i) by Eq. (ii), we have 3 M 9 M or = = 5 M+ m 25 M + m As time period, T = 2 π
Initial angular velocity of particle = ω0 and at any instant t, angular velocity = ω Therefore, for a displacement x, the resultant acceleration …(i) f = (ω20 − ω2 )x External force,
F = m(ω20 − ω2 )x
Since,
F ∝ cos ωt
…(ii) [given]
From Eq. (ii), we get m(ω20 − ω2 )x ∝ cos ωt
…(iii)
Now, from equation of simple harmonic motion, x = A sin (ωt + φ) At t = 0; x = A
…(iv) π 2
∴
A = A sin (0 + φ) ⇒ φ =
∴
π x = A sin ωt + = A cos ωt 2
…(v)
Hence, from Eqs. (iii) and (v), we get
⇒
For amplitude of oscillation and energy to be maximum, frequency of force must be equal to the initial frequency and this is only possible in case of resonance. In resonance state, ω1 = ω2 .
(a)
Exp. (b)
m(ω20
(b) ω1 > ω2 (a) ω1 = ω2 (c) ω1 < ω2 , when damping is small and ω1 > ω2 , when damping is large (d) ω1 < ω2
− ω )A cos ωt ∝ cos ωt 2
A∝
1 m(ω20 − ω2 )
…(ii)
or 9 M + 9 m = 25 M or 16 M = 9 m m 16 or = 9 M
93. Two particles A and B of equal masses are suspended from two massless springs of spring constants k1 and k 2 , respectively. If the maximum velocities, during oscillations are equal, the ratio of amplitudes of A and B is [AIEEE 2003] (a)
k1 k2
(b)
k1 k2
(c)
k2 k1
(d)
k2 k1
Exp. (c) vmax = aω = a
91. In forced oscillation of a particle, the
amplitude is maximum for a frequency ω1 of the force, while the energy is maximum for a frequency ω 2 of the force, then [AIEEE 2004]
…(i)
Hence,
vmax 1 vmax 2
2π = T
2 πa =a m 2π k a1 k1 = a2 k2
k m
190
JEE Main Chapterwise Physics Qvmax 1 = vmax 2 ⇒
[given] a1 = a2
Now, comparing Eqs. (i) and (ii), we get ω = 600 rad s −1 and
k2 k1
∴ Velocity of wave =
94. The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased length is [AIEEE 2003] (a) 11%
(b) 21% (c) 42%
(d) 10%
Exp. (d)
T = 10 × 9.8 N = 98 N, m = 9.8 × 10−3 kg m−1
% Change = x + y +
the xdirection is given by π y = 10−4 sin 600t − 2 x + metre 3
Frequency of wire, f = =
(b) 600 (d) 200
Exp. (a) …(i)
…(ii)
1 2 ×1
T m
1 2L
98 = 50 Hz −3 9.8 × 10
97. A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was [AIEEE 2003] (a) ( 256 + 2 ) Hz (c) ( 256 − 5) Hz
Exp. (c)
where, x is expressed in metres and t in seconds. The speed of the wavemotion, in ms−1 is [AIEEE 2003]
The given equation of wave, π y = 10−4 sin 600t − 2 x + 3 Standard equation of wave y = a sin (ωt − kx + φ)
(b) 100 Hz (c) 200 Hz (d) 25 Hz
The wire will vibrate with the same frequency as that of source. This can be considered as an example of forced vibration.
95. The displacement y of a wave travelling in
(a) 300 (c) 1200
9.8 g /m is stretched with a tension of 10 kgwt between two rigid supports 1 m apart. The wire passes at its middle point between the poles of a permanent magnet and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is [AIEEE 2003]
Exp. (a)
xy 100 Vaild only for two variables in terms of percentage. x → % change in first variable x → % change in second variable x2 % Increase in length = x + x + 100 x2 21 = 2 x + 100 On solving, x = 10% [by cross check method]
∴
ω 600 = = 300 ms −1 k 2
96. A metal wire of linear mass density of
(a) 50 Hz l g l T2 = 4π2 g g l = 2 T 2 4π T = 2π
k = 2 m−1
(b) ( 256 − 2 ) Hz (d) ( 256 + 5) Hz f1 = 256 Hz
For tuning fork, where f2 − f1 = ± 5 f2 = frequency of piano f2 = (256 + 5) Hz or (256 − 5) Hz When tension is increased, the beat frequency decreases to 2 beats per second. If we assume that the frequency of piano string is 261 Hz, then on increasing tension, frequency, more than 261 Hz. But it is given that beat frequency decreases to 2, therefore 261 Hz is not possible. Hence, 251 Hz i.e., (256 − 5) was the frequency of piano string before increasing tension.
191
Oscillations and Waves 98. A body executes simple harmonic motion. The potential energy (PE), the kinetic energy (KE) and total energy (TE) are measured as function of displacement x. Which of the following statements is true? [AIEEE 2003] (a) (b) (c) (d)
KE is maximum when x = 0 TE is zero when x = 0 KE is maximum when x is maximum PE is maximum when x = 0
Exp. (a) At x = 0, kinetic energy is maximum and potential energy is minimum. 1 Q Kinetic energy, K = mω2 (a2 − x2 ) 2 Kmax when x = 0 1 Kmax = mω2 a2 2 and potential energy, 1 U = mω2 x2 , 2 at x = 0 , U = 0
99. The displacement of a particle varies according to the relation x = 4(cos πt + sin πt ). The amplitude of the particle is [AIEEE 2003] (a) – 4
(b) 4
(c) 4 2
(d) 8
Exp. (c) x = 4 (cos πt + sin πt ) 4 = × 2 [cos πt + sin πt ] 2 1 1 x = 4 2 sin πt ⋅ + cos πt 2 2 π π x = 4 2 sin π t + cos + cos πt sin 4 4 π x = 4 2 sin πt + 4 So, amplitude = 4 2
100. A child swinging on a swing in sitting position, stands up, then the time period of the swing will [AIEEE 2002] (a) increase
(b) decrease
(c) remain same (d) increase, if the child is long and decrease, if the child is short
Exp. (b) As the child stands up, the effective length of pendulum decreases due to the reason that the centre of gravity rises up. Hence, according to l , T will decrease. T = 2π g
101. In a simple harmonic oscillator, at the mean position
[AIEEE 2002]
(a) kinetic energy is minimum, potential energy is maximum (b) both kinetic and potential energies are maximum (c) kinetic energy is maximum, potential energy is minimum (d) both kinetic and potential energies are minimum
Exp. (c) Kinetic energy of particle of mass m in SHM at any 1 = mω2 (a2 − x2 ) 2 1 and potential energy = mω2 x2 2
point
where, a is amplitude of particle and x is the distance from mean position. So, at mean position, x = 0 1 [maximum] ∴ KE = mω2 a2 2 PE = 0
[minimum]
102. When
temperature increases, the frequency of a tuning fork [AIEEE 2002] (a) increases (b) decreases (c) remains same (d) increases or decreases depending on the material
Exp. (b) 1 T 2l m When temperature As, f =
increases, then length 1 increases and frequency f ∝ decreases. l
103. Length of a string tied to two rigid supports is 40 cm. Maximum length (wavelength in cm) of a stationary wave produced on it, is [AIEEE 2002]
192
JEE Main Chapterwise Physics (a) 20
(b) 80
(c) 40
If Eq. (i), propagates with Eq. (iii), then we get …(v) y = − 2 a sin kx cos ωt After putting x = 0 in Eqs. (iv) and (v) respectively, we get y = 2 a sin ωt and y = 0 Hence, Eq. (iii) is an equation of unknown wave.
(d) 120
Exp. (b) When the string is plucked in the middle, it vibrates in one loop with nodes at fixed ends and an antinode in the middle. N
So that, or
N
A
106. Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tubes [AIEEE 2002] A and B is
λ1 =l 2 λ1 = 2 × 40 = 80 cm
(a) 1 : 2
(d) 4 : 1
Let the tubes A and B have equal length called as l. Since, tube A is opened at both the ends, therefore, its fundamental frequency, v …(i) nA = 2l
equal parts, then the time period of each part will be [AIEEE 2002]
(c) nT
(c) 2 : 1
Exp. (c)
104. If a spring has time period T and is cut into n (a) T n
(b) 1 : 4
T n (d) T
(b)
Since, tube B is closed at one end, therefore its v fundamental frequency, nB = . …(ii) 4l From Eqs. (i) and (ii), we get nA v/ 2 l 4 = = = 2 :1 nB v/4l 2
Exp. (b) As we know that spring constant of spring is inversely proportional to length of spring, so new spring constant for each part is given by k ′ = nk, where nk is the spring constant of whole spring. From the theory of spring pendulum, we know that time period of spring pendulum is inversely proportional to square root of spring constant i.e., 1 1 and T ′ ∝ T∝ k nk T So, T′ = n
105. A wave y = a sin (ωt − kx ) on a string meets with another wave producing a node at x = 0. Then, the equation of the unknown wave is [AIEEE 2002]
(a) y = a sin (ωt + kx ) (b) y = − a sin (ωt + kx ) (c) y = a sin (ωt − kx ) (d) y = − a sin (ωt − kx )
Exp. (b) Equation of a wave, …(i) y = a sin (ωt − kx) Let equations of another wave may be …(ii) y = a sin (ωt + kx) and …(iii) y = − a sin (ωt + kx) If Eq. (i) propagates with Eq. (ii), then we get …(iv) y = 2 a cos kx sin ωt
107.
A tuning fork arrangement (pair) produces 4 beats per second with one fork of frequency 288 cps. A little wax is placed on the unknown fork and then it produces 2 beats per second. The frequency of the unknown fork is [AIEEE 2002] (a) 286 cps (c) 294 cps
(b) 292 cps (d) 288 cps
Exp. (b) The tuning fork of frequency 288 Hz is producing 4 beats per second with the unknown tuning fork i.e., the frequency difference between them is 4. Therefore, the frequency of unknown tuning fork. = 288 ± 4 = 292 or 284 On placing a little wax on unknown tuning fork, its frequency decreases but now the number of beats produced per second is 2 i.e., the frequency difference now decreases. It is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of given tuning fork. Hence, the frequency of unknown tuning fork = 292 Hz.
11 Electrostatics 1. The bob of a simple pendulum has mass 2g
and a charge of 5.0 µC. It is at rest in a uniform horizontal electric field of intensity 2000 V/m. At equilibrium, the angle that the pendulum makes with the vertical is (take g = 10m/ s2)
[JEE Main 2019, 8 April ShiftI]
(a) tan −1 ( 2.0) (c) tan −1 ( 50 . )
(b) tan −1 (0.2 ) (d) tan −1 (0.5)
2. A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V . If the shell is now given a charge of −4 Q, the new potential difference between the same two surfaces is [JEE Main 2019, 8 April ShiftI]
(a) −2 V
Exp. (d)
E
θ
+Q –Q
qE T sinθ
Q VB
mg
For equilibrium, T cosθ = mg and T sinθ = qE Dividing Eq. (ii) by Eq. (i), we get qE tanθ = mg Here,
m = 2 g = 2 × 10−3 kg, g = 10 ms −2 5 × 10
× 2000
1 = 0.5 2 2 × 10 × 10 So, the angle made by the string of the pendulum with the vertical is θ = tan−1 (0.5)
∴
…(i) …(ii)
q = 5µC = 5 × 10−6 C, E = 2000 V / m, −6
tanθ =
−3
(d) V
Initially when uncharged shell encloses charge Q, charge distribution due to induction will be as shown,
T cosθ T
(c) 4 V
Exp. (d)
Forces on the bob are as shown θ
(b) 2 V
=
VA
a
b
The potential on surface of inner shell is kQ k(−Q ) kQ VA = + + a b b where, k = proportionality constant. Potential on surface of outer shell is kQ k(−Q ) kQ + + VB = b b b Then, potential difference is 1 1 ∆VAB = VA − VB = kQ − a b Given, ∆VAB = V 1 1 So, kQ − = V a b
…(i)
…(ii)
…(iii)
194
JEE Main Chapterwise Physics Finally after giving charge d− 4Q to outer shell, potential difference will be ∆VAB = VA − VB kQ k(−4Q ) kQ k(−4Q ) = + + − a b b b 1 1 [from Eq. (iii)] = kQ − = V a b Hence, we obtain that potential difference does not depend on the charge of outer sphere, hence potential difference remains same.
3. Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 106 V/m. The plate area is 10−4 m 2. What is the dielectric constant, if the capacitance is 15 pF? (Take, ε 0 = 8.86 × 10−12C2 / Nm 2) [JEE Main 2019, 8 April ShiftI]
(a) 3.8
(b) 8.5
(c) 4.5
4. The electric field in a region is given by
E = ( Ax + B )$i, where E is in NC−1 and x is in metres. The values of constants are A = 20 SI unit andB = 10 SI unit. If the potential at x = 1 isV1 and that at x = − 5 isV 2, thenV1 − V 2 is [JEE Main 2019, 8 April ShiftII]
(a) − 48 V
(b) − 520 V (c) 180 V
Given,
E = ( Ax + B)$i NC −1
The relation between electric field and potential is given as dV = − E ⋅ dx Integrating both sides within the specified limits, we get 2
x2
∴∫ dV = V2 − V1 = − ∫ E ⋅ dx 1
⇒
x1
V1 − V2 =
(d) 6.2
x2
∫ E ⋅ dx
x1
=
Exp. (b) As we know, capacitance of a capacitor filled with dielectric medium, ε KA …(i) C= 0 d
(d) 320 V
Exp. (c)
x2
x2
x1
x1
∫ (Ax + B)$i ⋅(dx $i ) = ∫ (Ax + B)⋅ dx
Here, A = 20 SI unit, B = 10 SI unit, x1 = 1 and x2 = − 5 ⇒ V1 − V2 =
d
−5
−5
∫ (20 x + 10)⋅ dx 1
20 x = + 10 x = 10 [ x2 + x]1−5 2 1 = 10 [(−5)2 + (−5) − (1)2 − (1)] 2
C
= 10 (25 − 5 − 2 ) = 180 V
A=area
5. An electric dipole is formed by two equal V
and potential difference between plates is V V …(ii) ⇒ d= E= d E So, by combining both Eqs. (i) and (ii), we get CV …(iii) K= ε0 AE Given, C = 15pF = 15 × 10−12 F, V = 500 V, E = 106 Vm −1, A = 10−4 m2 and
ε0 = 8.85 × 10−12 C 2 N −1m −2
Substituting values in Eq. (iii), we get 15 × 10−12 × 500 K= = 8.47 ≈ 8.5 8.85 × 10−12 × 10−4 × 106
and opposite charges q with separation d. The charges have same massm. It is kept in a uniform electric field E . If it is slightly rotated from its equilibrium orientation, then its angular frequency ω is [JEE Main 2019, 8 April ShiftII]
(a)
2qE md
qE (b) 2 (c) md
qE md
(d)
qE 2md
Exp. (a) Key Idea When an electric dipole is placed in an electric field E at some angle θ, then two forces equal in magnitude but opposite in direction acts on the + ve and −ve charges, respectively. These forces forms a couple which exert a torque, which is given as τ =p×E where, p is dipole moment.
195
Electrostatics ∴Torque on the dipole can also be given as τ = Iα = − pEsinθ where, I is the moment of inertia and α is angular acceleration. For small angles, sinθ ≈ θ pE …(i) ∴ α = − θ I Moment of inertia of the given system is +q,m
E
F=qE d/2
d
θ d/2
F=qE
– q,m 2
2
d d I = m + m 2 2 2 md 2 md 2 = = 4 2 Substituting the value of I in Eq. (i), we get 2 pE ...(ii) α = − ⇒ ⋅θ md 2 The above equation is similar to the equation for a system executing angular SHM. Comparing Eq. (ii), with the general equation of angular SHM, i.e. α = − ω2θ where, ω is the angular frequency, we get 2 pE 2 pE or ω = ω2 = 2 md md 2 As, p = qd 2qdE 2qE ω= = ∴ md md 2
6. A positive point charge is released from rest at a distance r0 from a positive line charge with uniform density. The speed (v ) of the point charge, as a function of instantaneous distance r from line charge, is proportional to [JEE Main 2019, 8 April ShiftII]
r0
+
r r0
r (a) v ∝ r0
(b) v ∝ e
r (c) v ∝ ln r0
r (d) v ∝ ln r0
Exp. (d) For a positive line charge or charged wire with uniform density λ, electric field at distance x is 2 kλ λ …(i) = E= 2 πε0 x x So, force on charge q which is at a distance r0 2 kqλ …(ii) due to this line charge is F = qE = x [using Eq. (i)] Now, work done when charge is pushed by field by a small displacement dx is 2 kqλ dW = F ⋅ dx = ⋅ dx [using Eq. (ii)] x ∴Total work done by field of wire in taking charge q from distance r0 to distance r will be r r 2 kqλ W = ∫ dW = ∫ ⋅ dx r0 r0 x = 2 kqλ[log x]rr 0 = 2 kqλ(log r − log r0 ) r = 2 kqλlog r0
…(iii)
As we know, from workkinetic energy theorem, K final − Kinitial = W r 1 [using Eq. (iii)] mv 2 − 0 = 2 kq × log ⇒ r0 2 1
⇒
4kqλ r 2 log v = m r0
∴
r 2 v ∝ log r0
1
7. A parallel plate capacitor has 1µF capacitance. One of its two plates is given + 2 µC charge and the other plate + 4µC charge. The potential difference developed across the capacitor is [JEE Main 2019, 8 April ShiftII]
(a) 1 V (c) 2 V
Exp. (a)
(b) 5 V (d) 3 V
196
JEE Main Chapterwise Physics Net value of charge on plates of capacitor after steady state is reached is q − q1 qnet = 2 2 where, q 2 and q1 are the charges given to plates. (Note that this formula is valid for any polarity of charge.) Here, q 2 = 4 µC, q1 = 2 µC ∴Charge of capacitor is 4−2 q = ∆qnet = = 1µC 2 Potential difference between capacitor plates is Q 1 µC = 1V V= = C 1 µF
8. A system of three charges are placed as shown in the figure D +q
–q
Q
d
If D >> d, the potential energy of the system is best given by [JEE Main 2019, 9 April ShiftI] 1 q 2 2qQd 1 q 2 qQd (b) (a) + + − + 2 4 πε0 d 4 πε0 d D D2 1 q 2 qQd 1 q 2 qQd (c) (d) − 2 − − − 2 4 πε0 d 4 πε0 d 2D D
⇒
(PE)total = −
9. A capacitor with capacitance 5µF is charged to 5 µC. If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done? [JEE Main 2019, 9 April ShiftI] (a) 6.25 × 10−6 J −6
(c) 2.55 × 10
–q
+q d
(2) Potential energy of charge Q and dipole system KQq …(ii) (PE)2 = − ⋅d D2 Hence, total potential energy of the system Kq 2 KQq − ⋅d (PE)total = (PE)1 + (PE)2 = − d D2
(b) 216 . × 10−6 J (d) 3.75 × 10−6 J
J
Exp. (d) Potential energy stored in a capacitor is 1 1 Q2 U = QV = 2 2 C 1 So, initial energy of the capacitor, U i = Q 2 / C1 2 1 Final energy of the capacitor, U f = Q2 / C2 2 As we know, work done, W = ∆U = U f − U i 1 1 1 = Q2 − 2 C C 2 1 Here,
Q = 5 µC = 5 × 10−6 C, C1 = 5 µF = 5 × 10−6 F,
⇒
Exp. (d) The system of two charges, i.e. + q and − q that are separated by distance d can be considered as a dipole. Thus, the charge Q would be at D distance from the centre of an electric dipole on its axial line. So, the total potential energy of the system will be due to two components. (1) Potential energy of dipole’s own system Kq 1q 2 Kq 2 …(i) (PE)1 = =− d d
1 q 2 Qqd + 4 πε0 d D2
C 2 = 2 µF = 2 × 10−6 F 1 ∆U = × (5 × 10−6 )2 2 1 1 − –6 5 × 10−6 2 × 10 1 5 × 5 × 10−12 3 × × −6 2 10 10 25 × 3 = × 10−6 J 20 ∆U = 375 . × 10−6 J =
⇒
∴ Work done in reducing the capacitance from 5 µF to 2 µF by pulling plates of capacitor apart is 375 . × 10−6 J.
10. The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V . When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is [JEE Main 2019, 9 April ShiftII] (a)
(n + 1)V (K + n)
(b)
nV K +n
(c) V
(d)
V K +n
197
Electrostatics (a) 60 µF and 40 µF (c) 20 µF and 30 µF
Exp. (a) Key Idea When a dielectric slab is inserted to fill the space between the plates, then the charge on the capacitor plates remains same. However, the capacitance increase, i.e. C = KC 0 , where C 0 is the capacitance of the capacitor without slab and K is the dielectric constant of slab.
When parallel combination is fully charged, charge on the combination is C
nC +
– V
Q = Ceq V = C(1 + n) V When battery is removed and a dielectric slab is placed between two plates of first capacitor, then charge on the system remains same. Now, equivalent capacitance after insertion of dielectric is K
KC
nC
Ceq = KC + nC = (n + K ) C If potential value after insertion of dielectric is V′, then charge on system is Q ′ = Ceq V ′ = (n + K ) CV ′ As Q = Q ′, we have C(1 + n)V = (n + K ) CV ′ (1 + n) V ∴ V′ = (n + K )
11. Figure shows charge (q ) versus voltage (V ) graph for series and parallel combination of two given capacitors. The capacitances are [JEE Main 2019, 10 April ShiftI] q(µC)
A
Exp. (d) Key Idea For a capacitor, Q = CV ⇒ C = Q / V Hence, slope of the QV curve should represent the capacitance.
In the given figure, Slope of OA > Slope of OB Since, we know that, net capacitance of parallel combination > net capacitance of series combination ∴ Parallel combination’s capacitance, 500 µC ... (i) C P = C1 + C 2 = = 50µF 10 V Series combination’s capacitance, C1C 2 80µC = = 8 µF CS = 10 V C1 + C 2
B
C1C 2 = 8 × (C1 + C 2 ) = 8 × 50 µF [using Eq. (i)] = 400 µF From Eqs. (i) and (iii), we get C1 = 50 − C 2 and C1C 2 = 400 ⇒ C 2 (50 − C 2 ) = 400 ⇒ 50 C 2 − C 22 = 400 or
10 V
V(volt)
…(iii)
C 22 − 50C 2 + 400 = 0 + 50 ±
2500 − 1600
+ 50 ± 30 = 2 2 ⇒C 2 = + 40 µF or + 10 µF Also, C1 = 50 − C 2 ⇒C1 = + 10 µF or + 40 µF Hence, capacitance of two given capacitors is 10 µF and 40 µF. ⇒ C2 =
12. A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10−8 Ωm) of radius of crosssection 5 mm. Find the mobility of the charges, if their drift velocity is 1.1 × 10−3 m/s. [JEE Main 2019, 10 April ShiftI] (b) 1.3 m2 / Vs (d) 1.8 m2 / Vs
Exp. (c) Given,
80
… (ii)
or
(a) 1.5 m2 / Vs (c) 1.0 m2 / Vs 500
(b) 50 µF and 30 µF (d) 40 µF and 10µF
I = 5A, ρ = 17 . × 10−8 Ωm,
198
JEE Main Chapterwise Physics r = 5 mm = 5 × 10−3 m, vd = 11 . × 10−3 m/s Mobility of charges in a conductor is given by v … (i) µ= d E and resistivity is given by E E 1 (QJ = σ E = × E ) ρ= = J I/ A ρ EA ρ= ⇒ I ρI … (ii) or E= A From Eqs. (i) and (ii), we get v A µ= d ρI Substituting the given values, we get . × 10−3 × π × (5 × 10−3 )2 11 = . × 10−8 × 5 17 =
86.35 × 10−9 8.5 × 10−8
⇒
. × 10−1 = 101
µ ≈ 1 m2 / V s
13. In free space, a particle A of charge 1 µC is held fixed at a point P. Another particle B of the same charge and mass 4 µg is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm fromP 1 is Take, = 9 × 109 Nm 2C− 2 4 πε 0 [JEE Main 2019, 10 April ShiftII]
(a) 1.5 × 102 m / s
(b) 30 . × 104 m / s
(c) 1.0 m/s
(d) 2.0 × 103 m / s
Now, potential energy of system of charges at separation of 1 mm is Kq1q 2 U1 = r Here, q1 = q 2 = 1 × 10−6 C r = 1mm = 1 × 10−3 m ∴ U1 =
By energy conservation, Change in potential energy of system of A and B = Kinetic energy of charged particle B 1 U1 − U 2 = mB v B2 ⇒ 2 where, mB = mass of particle B = 4 µg = 4 × 10−6 × 10−3 kg = 4 × 10−9 kg and v B = velocity of particle B at separation of 9 mm 1 ⇒ 9 − 1 = × 4 × 10−9 × v B2 2 ⇒ v B2 = 4 × 109 ⇒ v B = 2 × 103 ms −1
14. A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E , as shown in figure. Its bob has mass m and charge q. The time period of the pendulum is given by [JEE Main 2019, 10 April ShiftII] + + + + + + + + + + +
Given situation is shown in the figure below, 1 mm B
+1µC
+1 µ C
Fixed charge P
m = 4µg
When charged particle B is released due to mutual repulsion, it moves away from A. In this process, potential energy of system of charges reduces and this change of potential energy appears as kinetic energy of B.
= 9J
1 × 10−3
Potential energy of given system of charges at separation of 9 mm is Kq1q 2 9 × 109 × (1 × 10−6 )2 = 1J = U2 = r 9 × 10−3
Exp. (d)
A
9 × 10 × 1 × 10−6 × 1 × 10−6 9
– – – – – m – – q – – – – L
E
(a) 2 π
L qE g 2 + m
(c) 2 π
L g + qE m
2
(b) 2 π
L g2 −
(d) 2 π
q 2E 2 m2
L g − qE m
199
Electrostatics Exp. (b)
Exp. (a) When pendulum is oscillating between capacitor plates, it is subjected to two forces; (i) Weight downwards = mg (ii) Electrostatic force acting horizontally = qE So, net acceleration of pendulum bob is resultant of accelerations produced by these two perpendicular forces.
Electric charge distribution at inner and outer surface of spherical shell due to the electric dipole can be shown as below
Q P
–q +q
a b
qE mg
Net acceleration is, anet = =
anet
a12 + a22 qE g 2 + m
2
So, time period of oscillations of pendulum is l L T = 2π = 2π 2 anet qE g 2 + m
15. Shown in the figure is a shell made of a conductor. It has inner radius a and outer radiusb and carries chargeQ. At its centre is a dipole p as shown. In this case,
[JEE Main 2019, 12 April ShiftI]
p
(a) surface charge density on the inner surface is Q uniform and equal to 2 2 4 πa (b) electric field outside the shell is the same as that of a point charge at the centre of the shell (c) surface charge density on the outer surface depends on p (d) surface charge density on the inner surface of the shell is zero everywhere
Here, we need to consider two different factors (i) charge on the spherical shell is +Q which will be distributed on its outer surface as shown in figure. (ii) Electric dipole will create nonuniform electric field inside the shell which will distribute the charges on inner surface as shown in figure. But its net contribution to the outer side of the shell will be zero as net charge of a dipole is zero. ∴Net charge on outer surface of shell will be +Q. Hence, using (ii), option (a) is incorrect as field inside shell is not uniform. Option (b) is correct, as net charge on outer surface is +Q even in the presence of dipole. Option (c) is incorrect, as surface charge density Q Q at outer surface is uniform = = . A 4 πb 2 Option (d) is incorrect, as surface charge density at inner surface is nonzero. So, option (b) is correct. Alternate Solution Using Gauss’ law at outer surface, let charge on dipole is q , Σq 1 Σq φ= = E ⋅ A or E = ε0 A ε0 (+ Q + q − q ) Q σ = = = = constant Aε0 Aε0 ε0
16. A point dipole p = − p 0 x$ is kept at the origin. The potential and electric field due to this dipole on the Yaxis at a distance d are, respectively [Take,V = 0 at infinity] [JEE Main 2019, 12 April ShiftI]
200
JEE Main Chapterwise Physics (a)
p
p , 4 πε0d 2 4 πε0d 3
(c) 0,
(b) 0,
−p
4 πε0d 3 p −p (d) , 2 4 πε0d 4 πε0d 3
p 4 πε0d 3
Exp. (b) The given problem can be shown as clearly potential difference at point P due to dipole is
filled with three dielectrics of equal thickness and dielectric constants K 1 , K 2 and K 3. The first capacitor is filled as shown in Fig. I, and the second one is filled as shown in Fig. II. If these two modified capacitors are charged by the same potentialV , the ratio of the energy stored in the two, would be (E 1 refers to capacitor (I) and E 2 to capacitor (II)) : [JEE Main 2019, 12 April ShiftI]
P K1 K2
r A –q
p a
⇒
B +q
…(i)
a2 + r 2 kq kq …(ii) V=− + =0 ∴ a2 + r 2 a2 + r 2 Now, electric field at any point on Yaxis, i.e. equatorial line of the dipole can be given by kp (standard expression) E=− 3 r 1 p ⇒ E=− 4 πε0 r 3 Given,
r =d E=−
1 p 4 πε0 d 3
K3
(I) (II) E1 K 1K 2K 3 (a) = E 2 ( K 1 + K 2 + K 3 )( K 2K 3 + K 3K 1 + K 1K 2 )
(b)
E 1 ( K 1 + K 2 + K 3 )( K 2K 3 + K 3K 1 + K 1K 2 ) = E2 K 1K 2K 3
(c)
9 K 1K 2K 3 E1 = E 2 ( K 1 + K 2 + K 3 )( K 2K 3 + K 3K 1 + K 1K 2 )
(d)
E 1 ( K 1 + K 2 + K 3 )( K 2K 3 + K 3K 1 + K 1K 2 ) = E2 9 K 1K 2K 3
Here, AP = BP =
∴
K2
K3
a
V = VAP + VBP (scalar addition) k(−q ) k(q ) + V= AP BP
K1
Exp. (d) Key Idea A capacitor filled with dielectrics can be treated/compared as series/parallel combinations of capacitor having individual dielectric. e.g.
…(iii)
From Eqs. (ii) and (iii), correct option is (b). Alternate Solution Electric field at any point at θ angle from axial line of dipole is given by kp E = − 3 3cos 2 θ + 1 r Here, θ = 90º ⇒cos θ = cos 90° = 0 and r = d kp p E=− 3 =− ∴ d 4πε0d 3
17. Two identical parallel plate capacitors of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is
A/3
KK11
Area A
KK22
A/3
K3
A/3
d ⇓
d K1
A/3
d K2
A/3
d K3
A/3
201
Electrostatics and
d/3
d/3
Case II
d/3
d/3 Area A
K1
K2
C1
d ⇓ A
d/3
K3
A
d/3
Case I K1 K2 K3
A/3 A/3 A/3
d C1
≈
K1
K3
K2
C2
C3
⇒
d A
K2 d/3
A
d/3
K3
A
K1
d/3
C2
Capacitance of equivalent circuit are 3ε A ε A C1 = 0 ⋅ K1 = 0 K1 d d 3 3ε A ε A C2 = 0 K 2 = 0 K 2 d d 3 3ε A ε A and C 3 = 0 K 3 = 0 K 3 d d 3 So, equivalent capacitance, 1 1 1 1 = + + C II C1 C 2 C 3 d d d = + + 3ε0 AK1 3ε0 AK 2 3ε0 AK 3 ⇒
C3
Capacitance in the equivalent circuit are A ε0 ε A 3 C1 = K1 = 0 K1 3d d A ε0 ε A 3 C2 = K2 = 0 K2 3d d A ε0 ε A 3 and C3 = K3 = 0 K3 d 3d So, equivalent capacitance, C I = C1 + C 2 + C 3 ε A ε A ε A = 0 K1 + 0 K 2 + 0 K 3 3d 3d 3d ε A …(i) C I = 0 (K1 + K 2 + K 3 ) 3d
1 1 1 + + K K K 1 2 3 d K 2 K 3 + K1K 3 + K1K 2 = 3ε0 A K1K 2 K 3 3ε0 A K1 K 2 K 3 C II = d K1 K 2 + K 2 K 3 + K 3 K1 1 d = C II 3ε0 A
…(ii)
From Eqs. (i) and (ii), we get ε A CI = 0 (K1 + K 2 + K 3 ) 3d CII d (K1K 2 + K 2 K 3 + K 3 K1 ) × 3ε0 A(K1K 2 K 3 ) (K1 + K 2 + K 3 )(K1K 2 + K 2 K 3 + K 3 K1 ) = 9K1K 2 K 3 1 Now, energy stored in capacitor, E = CV 2 2 ⇒ E ∝C EI C ∴ = I EII CII
202
JEE Main Chapterwise Physics
18. Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ(r ) = kr , where r is the distance from the centre. Two charges A and B, of −Q each, are placed on diametrically opposite points, at equal distance a, from the centre. If A andB do not experience any force, then [JEE Main 2019, 12 April ShiftII] −1/ 4
(a) a = 8
3R
(b) a =
R
1/ 4
2 (d) a = R / 3
(c) a = 2 −1/ 4 R
Exp. (a) Key Idea Force on A is zero only when repulsion of A and B = attraction of positive charge distribution of radius a and charge A. + + + + + + + + + + + + + + + + + + + + + + + ++ + + + + + + + + A B C + a + + a + + + + + + + + + ++ + + + + ++ + + + + + + + + + +
2Q, R ρ(r)=kr
⇒
k=
2Q
πR 4 Now, using Gauss’ law, electric field on the surface of sphere of radius a is a
E ∫ dA = 0
⇒
E ⋅ 4 πa2 =
⇒
E=
a
1 ⋅ (kr 4 πr 2 dr ) ε0 ∫0
a4 1 ⋅ 4 πk ε0 4
2Q a2 ka2 = 4ε0 4 πε0 R 4
Force of attraction on charge A (or B) due to this field is 2 Q 2 a2 F1 = QE = 4 πε0 R 4 Force of repulsion on charge A due to B is 1 Q2 Q2 1 F2 = = ⋅ 4 πε0 (2 a)2 4 πε0 4a2 If charge A(or B) does not feel any force, then F1 = F2 Q2 2Q 2 a2 1 = ⋅ 2 ⇒ 8a4 = R 4 ⇒ 4 4 πε0 4a 4 πε0 R a = 8−1/ 4 R
⇒
19. In the given circuit, the charge on 4 µF capacitor will be [JEE Main 2019, 12 April ShiftII] 1 mF 4 mF 5 mF
In given charge distribution, let r is radius of a shell of thickness dr. dr
3 mF R
10 V
r
(a) 5.4 µC
ρ(r)=kr
(b) 9.6 µC
(c) 13.4 µC (d) 24 µC
Exp. (d) 2Q
Given circuit is 1µF
Charge dQ present in shell of thickness dr = Volume of shell × Volumetric charge density ⇒ dQ = (4 πr 2 × dr ) × (kr ) = 4 πkr 3 dr R
0
0
5µF 3µF
Total charge in sphere is R
4µ F
2Q = ∫ dQ = ∫ 4 πkr 3 dr R
⇒
r4 2Q = 4 πk 4 0
+ – 10V
203
Electrostatics Ceq = 5 + 1 = 6 µF 6× 4 = 2.4 µF 6+ 4 This is equivalent to In parallel,
+Q , q , + Q are placed d respectively at distance 0, and d from the 2 origin on the X axis. If the net force experienced by +Q placed at x = 0 is zero, then value of q is [JEE Main 2019, 9 Jan ShiftI]
20. Three
′ = and in series, Ceq
2.4µF
+Q 2 −Q (c) 2
Exp. (d)
+ –
The given condition is shown in the figure given below,
10V
So, potential difference across upper branch =10 V Using Q = C × V, charge delivered to upper branch is ′ ⋅V Q = C eq = 2.4 µF × 10V = 24 µC As we know, in series connection, same charge is shared by capacitors, so charge on 4 µF capacitor and 6 µF capacitor would be same, i.e. Q4′ µF = 24µ C Alternate Solution The circuit obtained, 2.4µF
x=0 q
+Q
+Q
d/2 d
Then, according to the Coulomb’s law, the electrostatic force between two charges q1 and q 2 such that the distance between them is (r ) given as, 1⋅ q1q 2 F= 4 πε0 ⋅ r 2 ∴Net force on charge ‘Q’ placed at origin i.e. at x = 0 in accordance with the principle of superposition can be given as Q×q Q ×Q 1 1 ⋅ Fnet = + ⋅ 4 πε0 (d )2 4 πε0 d 2 2
B
3µ F C
+Q 4 −Q (d) 4 (b)
(a)
3µ F
A
charges
D 10V + –
This can be further simplified as, 2.4 µF and 3 µF are in parallel. So, net capacitance, Cnet = 2.4 + 3 = 5.4 µF Net charge flow through circuit, Q = Cnet V = 5.4 × 10 = 54µC ∴This charge will be distributed in the ratio of capacitance in the two branches AB and CD as Q1 2.4 4 = = 3 5 Q2 ⇒ 9 x = 54µC or x = 6 µC ∴Charge on 4 µF capacitor is = 4 × 6 µC = 24 µ C
Since, it has been given that, Fnet = 0. 1 Q ×q 1 Q ×Q + ⋅ =0 ⋅ ⇒ 4 πε0 d 2 4 πε0 (d )2 2 Q×q 1 Q×Q 1 ⋅ =− ⋅ ⇒ 4 πε0 (d )2 4 πε0 d 2 2 Q or q = − 4
21. A parallel plate capacitor is made of two square plates of side ‘a’ separated by a distance d (d 2a 3 [JEE Main 2019, 10 Jan ShiftII]
(a)
F 3
(b) 3F
Here,
y > > a2
(c) 9F
(d) 27F
Exp. (d) Electric field on the equatorial line of a dipole at any point, which is at distance r from the centre is given by 2 kP … (i) E= 2 (r + a2 )3 / 2 where, P is the dipole moment of the charges.
( y2 + a2 )3 / 2
2
y2 + a 2 ≈ y2 2 kP E1 = 3 y
… (ii)
So, force on the charge in its position at P will be 2 kPQ … (iii) F = QE1 = y3 In second case r = y / 3 From Eq. (i), electric field at point P′ will be 2 kP E2 = 3/ 2 y 2 2 + a 3 y Again, >> a 3 ⇒
O –q
E1 =
or
respectively, constitute an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A chargeQ is placed at P, where OP = y and y > > 2a. The charge Q experiences an electrostatic force F .
r= y 2 kP
⇒
⇒
2R R = 2 +1 2 +1
29. Charges –q and +q located at A and B,
A
B +q
O
2qa
2x=R− x R x= 2 +1
⇒
y/3
( x)2
2 x2 = (R − x)2 ⇒
A –q
⇒
2
2
y + a2 ≈ y 3 3 2 kP 2 kP ⇒ E2 = 27 × 3 E2 = ( y / 3)3 y
Force on charge in this position, 2 kPQ F ′ = QE2 = 27 × y3
… (iv)
From Eqs. (iii) and (iv), we get F ′ = 27 F
30. Four equal point charges Q each are placed in the xyplane at (0, 2), (4, 2), (4, −2) and (0, − 2 ). The work required to put a fifth charge Q at the origin of the coordinate system (in joule) will be [JEE Main 2019, 10 Jan ShiftII]
Q2 (a) 4πε0 (c)
Q2 2 2 πε0
(b)
Q2 1 1 + 3 4 πε0
(d)
Q2 1 1 + 5 4 πε0
209
Electrostatics ⇒
Exp. (d) The four charges are shown in the figure below
(0, 2)
√ r 3=
r1=2 (0,0) r2=2 (0, – 2)
(4, 2) Q
Q
20
r4 =
√ 20 Q
Q
(4,–2)
Electric potential at origin (0, 0) due to these charges can be found by scalar addition of electric potentials due to each charge. KQ KQ KQ KQ … (i) ∴ V= + + + r1 r2 r3 r4 1 1 1 1 1 + = KQ 1 + ⇒ V = KQ + + 2 2 20 20 5 ( 5 + 1) volt …(ii) ⇒ V = KQ 5 Now, if another charge Q is placed at origin, then work done to get the charge at origin …(iii) W = QV By putting value of V from Eq. (ii) in Eq. (iii), we get ( 5 + 1) W = KQ 2 joule 5 or
W=
Q2 1 joule 1 + 5 4 πε0
31. A
parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is [JEE Main 2019, 10 Jan ShiftII]
(a) 560 pJ (c) 692 pJ
(b) 508 pJ (d) 600 pJ
1 × 12 × 10− 12 × 100 2 U = 6 × 10− 10 J
U=
After insertion of slab, capacitance will be C ′ = KC and final energy, 1 Q2 1 Q2 U′ = ⋅ = 2 C ′ 2 KC 1 1 … (iv) U′ = U = × 6 × 10− 10 J ⇒ 6.5 K (Qgiven, K = 6.5) So, energy dissipated in the process will be equal to work done on the slab, i.e. 1 − 10 ∆U = U − U ′ = 1 − × 6 × 10 J 6.5 5.5 ⇒ ∆U = × 6 × 10− 10 J 6.5 ≅ 5.08 × 10− 10 J or 508 pJ
32. Three chargesQ, +q and +q are placed at the vertices of a right angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value ofQ is [JEE Main 2019, 11 Jan ShiftI] Q
+q
(a) −2q
−q (b) 1+ 2
+q
(c) +q
(d)
− 2q 2 +1
Exp. (d) Electrostatic energy between two charges q1 and q 2 such that the distance between them is r is given as K q1q 2 U= r In accordance to the principle of superposition, total energy of the charge system as shown in the figure below is Q
Exp. (b) Energy stored in a charged capacitor is given by 1 1 Q2 … (i) U = CV 2 = ⋅ 2 2 C Here, C = 12 × 10− 12 F and V = 10 V.
… (ii)
√2 a
a 90° +q
a
+q
210
JEE Main Chapterwise Physics Kq 2 KQq KQq + + a a 2a It is given that, U=0 Kq Q q+Q+ ∴ =0 a 2 − 2 ×q Q= ⇒ ( 2 + 1) U=
ro
ro
33. In the figure shown below, the charge on the left plate of the10 µF capacitor is − 30 µC. The charge on the right plate of the 6 µF capacitor is [JEE Main 2019, 11 Jan ShiftI] 6 µF 10 µF
Exp. (b) 2 µF
4 µF
(a) + 12 µC (c) −12 µC
(b) + 18µC (d) −18µC
Exp. (b) Applying the concept of charge conservation on isolated plates of10µF, 6µF and 4µF. Since, 6µF and 4µF are in parallel, so total charge on this combination will be 30 µC
– + 10µF
– + C1= 6µF – + C2= 4µF
– + 2 µF
∴Charge on 6 µF, capacitor C1 6 = × 30 = 18 µC q= 6+ 4 C1 + C 2 Since, the charge has been asked on the right plate of the capacitor. Thus, it would be + 18 µC. Alternate Solution Let charge on 6 µF capacitor is q µC. Now, V at 6 µF = V at 4 µF 30 − q q ∴ = (Q V = q / C ) 6 µF 4 µF ⇒ ⇒
r
(a) electric field of a uniformly charged spherical shell (b) potential of a uniformly charged spherical shell (c) electric field of a uniformly charged sphere (d) potential of a uniformly charged sphere
4q = − 6q + 180 q = 18 µC
34. The given graph shows variation (with distance r from centre) of [JEE Main 2019, 11 Jan ShiftI]
For a uniformly charged spherical shell, electric potential inside it is given by Vinside = Vsurface = kq / r0 = constant, (where r0 = radius of the shell). and electric potential outside the shell at a distance r is kq Voutside = r ⇒ V ∝ 1/ r ∴ The given graph represents the variation of r and potential of a uniformly charged spherical shell.
35. An electric field of 1000 V/m is applied to an electric dipole at angle of 45º. The value of electric dipole moment is 10− 29 Cm. What is the potential energy of the electric dipole? [JEE Main 2019, 11 Jan ShiftII]
(a) − 9 × 10− 20 J − 18
(c) − 20 × 10
J
(b) − 10 × 10− 29 J (d) − 7 × 10− 27 J
Exp. (d) Given, and
E = 1000 V/m θ = 45° p = 10−29 Cm
We know that, electric potential energy stored in an electric dipole kept in uniform electric field is given by the relation U = − p ⋅ E = − pEcosθ = − 10−29 × 1000 × cos 45° ⇒
U ≈ − 7 × 10−27 J
211
Electrostatics A
36. Seven capacitors, each of capacitance 2 µF are to be connected in a configuration to 6 obtain an effective capacitance of µF. 13 Which of the combinations shown in figures below will achieve the desired value? [JEE Main 2019, 11 Jan ShiftII]
(a)
(b)
B
S ε 3C
C
3 Q2 ⋅ 4 C 1 Q2 (c) ⋅ 8 C
5 Q2 . 8 C 3 Q2 (d) ⋅ 8 C (b)
(a)
Exp. (d) (c)
(d)
Exp. (c) 6 Here, effective capacitance = µF 13 In the given options some capacitors are joined in parallel and some are in series. For parallel combination, C P = C1 + C 2 + C 3 + .... 1 1 1 1 For series combination, = + + ... CS C1 C 2 C 3 Here, capacity of each capacitor is 2 µF. ∴Three capacitors must be in parallel to get 6 µF. Now, we consider the combinations with three capacitors in parallel (as shown in the figure below)
In position ‘A’ of switch, we have a capacitor joined with battery.
ε
C
1 2 Cε 2 When switch is turned to position B, we have a charged capacitor joined to an uncharged capacitor. So, energy stored in position, U1 =
++ ++ C –– ––
3C
2 µF 2 µF
2 µF 2µ F 2µF 2µF
2 µF
∴
1 1 1 1 1 1 = + + + + Ceq 6 2 2 2 2 13 1 6 or Ceq = µF = +2= 6 6 13
37. In the figure shown, after the switch ‘S ’ is turned from position ‘A’ to position ‘B ’, the energy dissipated in the circuit in terms of capacitance ‘C ’ and total charge ‘Q ’ is [JEE Main 2019, 12 Jan ShiftI]
Common potential in steady state will be total charge Cε ε V= = = total capacity 4C 4 Now, energy stored will be 1 U 2 = (Ceq ) (Vcommon )2 2 =
2
ε 1 1 4C × = Cε2 4 2 8
So, energy dissipated is ∆U = U1 − U 2 1 1 = Cε2 − Cε2 2 8 =
2
3 2 3 Q 3 Q2 Cε = C = ⋅ 8 8 C 8 C
212
JEE Main Chapterwise Physics
38. There is uniform spherically symmetric surface charge density at a distance R 0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speedv[R (t )] of the distribution as a function of its instantaneous radius R (t ) is [JEE Main 2019, 12 Jan ShiftI] v [R(t)]
v [R(t)]
(a)
–2q
y l
l +q
+q
x
l
$j − $i (a) 3 ql 2
(b) 2ql $j
(c) − 3 ql $j
(d) (ql )
(b)
$i + $j 2
Exp. (c) Ro
R(t)
Ro
v [R(t)] vo
R(t) v [R(t)]
Given system is equivalent to two dipoles inclined at 60° to each other as shown in the figure below y
(d)
(c) Ro
R(t)
Ro
p
R(t)
Exp. (c) Key Idea As, electrostatic force is conserved in nature so, total energy of charge distribution remains constant in absence of any external interaction
39. Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle as shown in the figure. [JEE Main 2019, 12 Jan ShiftI]
p
60°
+ O
+ B
x
Now, magnitude of resultant of these dipole moments is pnet =
p2 + p2 + 2 p ⋅ pcos 60° =
y
Let radius of distribution at some instant t is R. At t = 0, radius is given R 0 Now by conservation of energy, we have kQ 2 kQ 2 1 = mv 2 + 0+ 2 R0 2 2R (QThe distribution starts from rest, so, initial kinetic energy is zero.) Differentiating this equation with respect to R, we get dv kQ 2 1 dv kQ 2 − = 0 or m2 v = 2 dR 2 R 2 dR 2 mvR 2 dv Here, = slope of v versus R graph. It dR decreases with increasing v and R. Also, slope → 0 as R → ∞. From above conclusions, we can see that the best suited graph is given in option (c).
A – –
3p =
p
3ql
p
x
pnet
As, resultant is directed along negative ydirection pnet = − 3 p$j = − 3ql $j
40. The charge on a capacitor plate in a circuit as a function of time is shown in the figure. [JEE Main 2019, 12 Jan ShiftII] 6 5 q(µC)
4 3 2 0
2
4 t(s)
6
8
213
Electrostatics What is the value of current att = 4 s?
(a) 2 µA (c) Zero
Key Idea Two capacitors C 1 and C 2, if connected in series, then their equivalent capacitance is C × C2 C eq = 1 C1 + C2
(b) 1.5 µA (d) 3µA
Exp. (c) dq dt = Slope of q versus t graph = Zero at t = 4s; (as graph is a line parallel to time axis at t = 4s)
If they are connected in parallel, then their equivalent capacitance is, C eq = C 1 + C 2.
As we know, current, I =
41. A parallel plate capacitor with plates of area 1 m 2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is C2 − 12 Take, ε 0 = 8.85 × 10 N − m2
We simplify given circuit as Ceq = 2×2 =1 2+2
2
C A 2
2
2
1 Series combination
2
2
Parallel combination
B
[JEE Main 2019, 12 Jan ShiftII]
(a) 985 . × 10− 10 C (c) 785 . × 10− 10 C
Ceq =2+2=4
(b) 885 . × 10− 10 C (d) 685 . × 10− 10 C
Exp. (b)
ε0 = 8.85 × 10−12
C
1
Ceq = 2×4 = 4 2+4 3
N  m2
= 8.85 × 10
2 A 4/3
[JEE Main 2019, 12 Jan ShiftII] C 2
2
1
6 µF 5
Exp. (d)
(b) 4µF
(c)
7 µF 10
Ceq=
So, ⇒ ⇒
2
(d)
2×2 + 4 7 = 2+2 3 3
C ×7/3 C + 7/3 14 − 7C C 7 = + ⇒ 6 3 2 6 7 µF C= 11 C AB =
=
1 2
(given)
3 7 C= 6
43. Three concentric metal shells A , B and C of
B
(a)
B
C
capacitance of the whole circuit is to be 0.5 µF. All values in the circuit are in µF.
2
2
C
42. In the circuit shown, find C if the effective
2
B
2
−10
2
1
4
So, by substituting given values, we get Q = 8.85 × 10−12 × 1 × 100
A
Ceq = 1+1=2 2
If Q = charge on each plate, then ε A Q = CV = 0 ⋅ Ed d = ε0 AE Here, A = 1 m2 , E = 100 N/C and
2
C A
7 µF 11
respective radii a ,b and c (a < b < c ) have surface charge densities + σ , − σ and +σ, respectively. The potential of shell B is [JEE Main 2018]
214
JEE Main Chapterwise Physics (a)
σ a 2 − b 2 + c ε0 a
(b)
σ a 2 − b 2 + c ε0 b
(c)
σ b2 − c 2 + a ε0 b
(d)
σ b2 − c 2 + a ε0 c
withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is [JEE Main 2017 (Offline)]
(a) 16
/ Potential of B = Potential due to charge on A + Potential due to charge on B + Potential due to charge on C. C B
A
+σ –σ +σ
k(QA + QB ) kQC + b c 1 σ 4 πa2 σ 4 πb 2 σ 4 πc 2 = − + 4 πε0 b b c σε a2 − b 2 c 2 σ a2 − b 2 = = + + b b c ε0 ε0
VB =
VB =
(c) 32
(d) 2
Exp. (c)
Exp. (b)
∴
(b) 24
σ ε0
c
a2 − b 2 + c b
44. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant 5 K = is inserted between the plates, the 3 magnitude of the induced charge will be [JEE Main 2018]
(a) 1.2 nC (c) 2.4 nC
(b) 0.3 nC (d) 0.9 nC
Exp. (a)
As each capacitors cannot withstand more than 300 V, so there should be four capacitors in each row became in this condition 1 kV i.e. 1000 V will be divided by 4 (i.e. 250 not more than 300 V). Now, equivalent capacitance of one row 1 = × 1µF = 0.25µF 4 [Qin series combination, Ceq = c / n] Now, we need equivalent of 2µF, so let we need n such rows ∴ n × 0.25 = 2 µF [Qin parallel combination Ceq = nc ] 2 =8 n= 0.25 ∴ Total number of capacitors = number of rows × number of capacitors in each row = 8 × 4 = 32
46. An electric dipole has a fixed dipole moment
p, which makes angle θ with respect to Xaxis. When subjected to an electric field E1 = E$i, it experiences a torque T1 = τk$ . When subjected to another electric field E2= 3E 1$j, it experiences a torque T2 = − T1. The angle θ is [JEE Main 2017 (Offline)] (a) 45°
Exp. (b)
(b) 60°
(c) 90°
Y
p 90° – q q
Magnitude of induced charge is given by 5 Q ′ = (K − 1) CV0 = − 1 90 × 10−12 × 20 3 −9
= 1.2 × 10 C ⇒Q′ = 1.2nC
45. A capacitance of 2 µF is required in an electrical circuit across a potential difference of 1kV. A large number of 1 µF capacitors are available which can
(d) 30°
X
Torque applied on a dipole τ = pE sin θ where, θ = angle between axis of dipole and electric field. For electric field E1 = E $i it means field is directed along positive X direction, so angle between dipole and field will remain θ, therefore torque in this direction E1 = pE1 sin θ
215
Electrostatics In electric field E2 = 3 E$j, it means field is directed along positive Yaxis, so angle between dipole and field will be 90° − θ. Torque in this direction τ 2 = pE sin (90° − θ) = p 3 E1 cos θ According to question τ 2 = − τ1 ⇒τ 2 =  τ1 pE1 sin θ = p 3 E1 cos θ ∴ tanθ = 3 ⇒ tan θ = tan 60° ∴ θ = 60°
47. The region between two concentric spheres of radiia andb, respectively a Q (see the figure), has volume charge density b A ρ = , where A is a r constant and r is the distance from the centre. At the centre of the spheres is a point chargeQ. The value of A such that the electric field in the region between the spheres will be constant is [JEE Main 2016 (Offline)] (a) (c)
Q
(b)
2 πa 2 2Q
(d)
π(a 2 − b 2 )
∫ a
A 4 πr 2dr r ε0
= E 4 πr
a
b
4 π r 2 − a2 r 2 2
r2 1 Q E= 2 + A 2 π 4 πε0 r 1 E= 4 πε0
Exp. (c) Resultant circuit, 3 µF
4 µF
9 µF
4µ F
12µF
As, charge on 3µF = 3µF × 8V = 24µC ∴ Charge on 4µ F = Charge on 12 µ F = 24µC Charge on 3µ F = 3µ F × 2 V = 6µ C Charge on 9µ F = 9µ F × 2 V = 18 µ C Charge on 4 µ F + Charge on 9µF
E 4 πε0 r 2 =Q+ A
(b) 360 N/C (d) 480 N/C
3µ F
πa 2
Q 2
[JEE Main 2016 (Offline)]
(a) 240 N/C (c) 420 N/C
2 π( b 2 − a 2 ) 2Q
As, Gaussian surface at distance r from centre, r
4 µF capacitors is setup 9 µF as shown in the figure. The magnitude 2 µF of the electric field, due to a point charge + – Q (having a charge equal to the sum of 8V the charges on the 4 µF and 9µF capacitors), at a point distance 30 m from it, would equal to
Q
Exp. (a)
Q+
3 µF
48. A combination of
− a2 r 2 Q A 2 π a2 2 + A 2 π − r 2 r
1 × A × 2π 4 πε0 At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is As, Q = 2 πAa2 Q i.e. A= 2 πa2
= (24 + 18)µC = 42 µC ∴ Electric field at a point distant 30 m 9 × 103 × 42 × 10− 6 = 420N/C = 30 × 30
49. A long cylindrical shell carries positive surface
charge σ in the upper half and negative surface charge − σ in the lower half. The electric field lines around the cylinder will look like figure given in (figures are schematic and not drawn to scale) [JEE Main 2015]
E=
(a)
+++++ + –– –– –– ––
(b)
(c)
++ +++ + –– –– –– ––
(d)
+++ ++ + –– – – –– ––
++++ + – –– –+ ––––
216
JEE Main Chapterwise Physics
Exp. (a)
Similarly,
Field lines should originate from positive charge and terminate to negative charge. Thus, (b) and (c) are not possible. Electric field lines cannot form corners as shown in (d). Thus, correct option is (a).
or
50. A uniformly charged solid sphere of radius R
or
has potential V 0 (measured with respect to ∞) on its surface. For this sphere, the equipotential surfaces with potentials 3V 0 5V 0 3V 0 V and 0 have radius R1, R 2 , R 3 , , , 4 2 4 4 [JEE Main 2015] and R 4 respectively. Then,
(a) R1 = 0 and R 2 > ( R 4 − R 3 ) (b) R1 ≠ 0 and ( R 2 − R 1 ) > ( R 4 − R 3 ) (c) R1 = 0 and R 2 < ( R 4 − R 3 ) (d) 2 R < R 4
Exp. (c, d) Potential at the surface of Q the charged sphere KQ V0 = R KQ R ,r ≥ R V= r KQ = 3 (3R 2 − r 2 ); Charged sphere 2R r≤ R KQ Vcentre = Vc = × 3R 2 2R3 3KQ 3V0 = = 2R 2 ⇒ R1 = 0 As potential decreases for outside points. Thus, according to the question, we can write 5V KQ VR 2 = 0 = (3R 2 − R 22 ) 4 2R3 V 5V0 = 02 (3R 2 − R 22 ) 4 2R or
R 5 = 3 − 2 R 2
VR 3 =
or
⇒
4 R 3 KQ V0 = = 4 R4
KQ 3 KQ = × R3 4 R
R3 = VR 4
KQ 1 KQ = × R4 4 R
⇒
R 4 = 4R
51. In the given circuit,
1µF charge Q 2 on the C capacitor 2 µF 2µF changes as C is varied from 1 µF to 3µF.Q 2 as a function E of C is given properly by (figures are drawn schematically and are not to scale) [JEE Main 2015]
Charge
(a)
Charge
Q2
(c) Q2 1µF
3µF
C
1µ F
(b)
Q2
C
3µ F
C
(d) Q2 1µF
3µF
C
1µF
Exp. (b) / Assume negative terminal of the battery as grounded (0 V). Suppose, potential of point x is V. 1 µF
x C 2 µF E
2
R2 = 3 − 5 = 1 R 2 2 R R2 = 2
3µ F
Charge
Charge
From the circuit diagram, we can write
2
or
3V0 4
QC = Q1 + Q2 or
C (E − V ) = 1 × V + 2 × V
or
V[C + 3] = CE
217
Electrostatics V=
or ∴
CE 3+C
Q2 = C 2 ( V ) 2 CE 2E = = 3 + C 1 + 3 /C
Exp. (b)
C1 + – 120 V
As C1 varied from 1 µF to 3 µF, charge increases with decreasing slope.
C2
– +
Note As C → ∞ , Q 2 → 2 E = constant
200 V
52. Assume that an electric field E = 30x 2 i exists in space. Then, the potential difference V A − VO , where VO is the potential at the origin andV A is the potential at x = 2 m, is [JEE Main 2014]
(a)120 J
(b) −120 J
(c) −80J
(d) 80J
Exp. (c) As we know, potential difference VA − VO is dV = − Edx VA
∫V
O
55. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at distance L from the end A is [JEE Main 2013]
dV = − ∫ 30 x2dx
A
O
B
L
0
2
x3 VA − VO = −30 × 3 0
3Q 4 π ε0 L Q ln 2 (d) 4 π ε0 L
Q 8 π ε0 L Q (c) 4 π ε0 L ln 2
= −10 × 8 = −80 J
53. A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to [JEE Main 2014]
(b) 3 × 10−7 C/m 2 (d) 6 × 104 C/m 2
Exp. (a) Electric field inside dielectric,
L
(a)
= −10 × [2 3 − (0)3 ]
⇒
For potential to be made zero after connection, the charge of both capacitors are equal. ∴ q1 = q 2 ⇒ C1V1 = C 2 V2 ⇒ 120 C1 = 200 C 2 ⇒ 3C1 = 5C 2
2
(a) 6 × 10−7 C/m 2 (c) 3 × 104 C/m 2
(b) 3C1 = 5C 2 (d) 9C1 = 4C 2
(a) 5C1 = 3C 2 (c) 3C1 + 5C 2 = 0
σ = 3 × 104 Kε0
σ = 2.2 × 8.58 × 10−12 × 3 × 104 = 6.6 × 8.85 × 10−8 = 5.841 × 10−7 = 6 × 10−7 C/m 2
54. Two capacitors C 1 and C 2 are charged to
120 V and 200 V, respectively. It is found that by connecting them together, the potential on each one can be made zero. Then, [JEE Main 2013]
(b)
Exp. (d) L O
L x
A
dx
B
Let an element of length dx, charge dq, at distance x from point O. dq dV = k x Q where, dq = dx L Q dx 2L L 2 L kdq =∫ ∴ V=∫ L L x x 2 L 1 Q = ∫L x dx 4 πε Q0 L [loge x]2L L = 4 πε0 L Q [loge 2 L − loge L] = 4 πε0 L Q 2 2L loge 4 πε0 L L Q = ln(2 ) 4 πε0 L =
218
JEE Main Chapterwise Physics
56. Two charges each equal to q, are kept at
x = − a and x = a on the xaxis. A particle of mass m and charge q 0 = q / 2 is placed at the origin. If charge q 0 is given, a small displacement ( y V1, so electric field will point from plate 2 to plate 1.
226
JEE Main Chapterwise Physics The electron will experience an electric force, opposite to the direction of electric field and hence move towards the plate 2. Use workenergy theorem to find speed of electron when it strikes the plate 2. me v 2 − 0 = e ( V2 − V1 ) 2 where, v is the required speed.
0.1 m P
E
(a) cosθ 1
2
911 . × 10−31 2 v = 1.6 × 10−19 × 20 2 v=
or
−19
. × 10 16
(b) cot θ
θ
T cos θ θ T sin θ
= 2 .65 × 106 ms −1
to a nonuniform electric field. The dipole will experience [AIEEE 2006] (a) a translational force only in the direction of the field (b) a translational force only in a direction normal to the direction of the field (c) a torque as well as a translational force (d) a torque only
Exp. (c)
F1
F2
F = QE
B mg
where, ε0 is the permittivity in vacuum and εr is the relative permittivity of medium. Here, electrostatic force on B, Qσ QE = ε0 εr FBD of B is shown in figure. In equilibrium, T cos θ = mg Qσ and T sin θ = ε0 εr tan θ =
Thus,
In a nonuniform electric field, the dipole may experience both nonzero torque as well as translational force. For example, as shown in figure
(d) tan θ
Electric field due to a charged conducting sheet of σ surface charge density σ is given by E = . ε0 εr
× 40
84. An electric dipole is placed at an angle of 30°
(c) sin θ
Exp. (d)
. × 10−31 911
+q
S B
∴
⇒
θ
Qσ ⇒ tan θ ∝ σ ε0 εr mg
86. Two point charges +8 q and –2 q are located at x = 0 and x = L , respectively. The location of a point on the xaxis at which the net electric field due to these two point charges is zero, is [AIEEE 2005] (a) 2 L
(b)
–q
L 4
(c) 8 L
(d) 4L
Exp. (a) Suppose that a point B, where net electric field is zero due to charges 8q and − 2 q .
F1 ≠ F2 as E is nonuniform. Torque would also be nonzero.
85. A charged ball B hangs from a silk thread S , which makes an angle θ with a large charged conducting sheet P , as shown in the figure. The surface charge density σ of the sheet is proportional to [AIEEE 2005]
+8q B x=a
O x=0
E BO =
–2q x=L
−1 8q ⋅ i 4 πε0 a2
A
227
Electrostatics E BA =
1 +2q ⋅ i 4 πε0 (a + L)2
=
1 4 πε0
According to condition, ⇒
E BO + E BA = 0 1 2q 1 8q = 4 πε0 (a + L)2 4 πε0 a2
or
2 1 = a a+ L
q − R
− 1 R + d 4 πε0 q
2
2
=
1 q 1 q 1 + − 4 πε0 R 4 πε0 R 4 πε0
=
q 2 πε0
2a + 2L = a
or
2 2 R +d
−q + R
or 2L = − a Thus, at distance 2 L from origin, net electric field will be zero.
1 − R
q
q R2 + d 2 1 q − 4 πε0 R 2 + d 2
R + d 1
2
2
87. Two thin wire rings each having a radius R
88. A parallel plate capacitor is made by
are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and −q.The potential difference between the centres of the two rings is
stacking n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C , then the resultant capacitance is [AIEEE 2005]
[AIEEE 2005]
(a)
q 1 1 (b) − 2 πε0 R R 2 + d 2 q 1 1 (d) − 4 πε0 R R 2 + d 2
qR 4 πε0d
2
(c) zero
(a) (n −1)C
(b) (n + 1)C
(c) C
(d) nC
Exp. (a) Each plate is taking part in the formation of two capacitors except the plates at the ends. A
Exp. (b) VA = Potential due to charge + q on ring A + Potential due to charge −q on ring B +q
–q
d1
d2
R
R d
A
B
1 q q = − 4 πε0 R d1 =
1 4 πε0
q − R
R 2 + d 2 q
[Qd1 = Similarly, VB =
1 4 πε0
q − + R
Potential difference, VA − VB
R 2 + d 2 ] …(i)
R 2 + d 2 q
B
These capacitors are in parallel and n plates form (n − 1) capacitors. Thus, equivalent capacitance between A and B = (n − 1)C
89. A fully charged capacitor has a capacitance C . It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. If the temperature of the block is raised by ∆T , the potential difference V across the capacitance is [AIEEE 2005] (a) (c)
2mC∆T s ms∆T C
(b) (d)
mC∆T s 2ms∆T C
228
JEE Main Chapterwise Physics q
Exp. (d) 1 …(i) E = CV 2 2 The energy stored in capacitor is lost in form of heat energy. …(ii) ∴ H = ms ∆T From Eqs. (i) and (ii), we have 2 ms ∆T 1 ms ∆T = C V 2 or V = 2 C
90. Two spherical conductors B andC having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B andC is [AIEEE 2004]
F (a) 4
3F (b) 4
F (c) 8
(d)
3F 8
Exp. (d) Let the spherical conductors B and C have same charge as q. The electric force between them is 1 q2 4 πε0 r 2 Here, r being the distance between them. When third uncharged conductor A is brought in contact with B, then charge on each conductor q + qB 0 + q q qA = qB = A = = 2 2 2 When this conductor A is now brought in contact with C, then charge on each conductor q + qC (q /2 ) + q 3q q A = qC = A = = 2 2 4 Hence, electric force acting between B and C is 1 (q /2 )(3q /4) 1 q B qC F′ = = 4 πε0 4 πε0 r 2 r2 F=
=
3 8
1 q 2 3F = 2 8 4 πε0 r
91. A charged particle q is shot towards another charged particle Q which is fixed with a speed v. It approaches Q upto a closest distance r and then returns. If q was given a speed 2v , the closest distance of approach would be
v
Q r
(a) r r (c) 2
(b) 2r r 4
[AIEEE 2004]
(d)
Exp. (d) Let a particle of charge q having velocity v approaches Q upto a closest distance r and if the velocity becomes 2 v, the closest distance will be r ′. The law of conservation of energy yields, kinetic energy of particle = electric potential energy between them at closest distance of approach. 1 1 Qq or mv 2 = 2 4 πε0 r kQq 1 …(i) or mv 2 = r 2 1 Q k = constant = 4 πε0 kQq 1 m(2 v )2 = r′ 2 Dividing Eq. (i) by Eq. (ii), we get kQq 1 mv 2 2 = r kQq 1 m(2 v )2 r′ 2 1 r′ ⇒ = 4 r r ⇒ r′ = 4 and
…(ii)
92. Four charges equal to −Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium, the value of q is [AIEEE 2004] Q (1 + 2 2 ) 4 Q (c) − (1 + 2 2 ) 2
(a) −
Q (1 + 2 2 ) 4 Q (d) (1 + 2 2 ) 2 (b)
Exp. (b) The system is in equilibrium means the force experienced by each charge is zero. It is clear that charge placed at centre would be in equilibrium for any value of q, so we are considering the equilibrium of charge placed at any corner.
229
Electrostatics FCB FCA a
D –Q
C
√2a A –Q
–Q B
FCD + FCA cos 45° + FCO cos 45° = 0 (− Q )(− Q ) 1 1 (− Q )(− Q ) ⇒ ⋅ + × 4 πε0 4 πε0 ( 2 a)2 a2 (− Q )q 1 + × 4 πε0 ( 2 a/2 )2
1 2 1 =0 2
1 Q 1 Q 1 + ⋅ 4 πε0 a2 4 πε0 2 a2 2 1 2 Qq 1 − × =0 4 πε0 a2 2 Q Q+ − 2q = 0 2 2 2 2 Q + Q − 4q = 0 4q = (2 2 + 1)Q Q q = (2 2 + 1) 4
or or or or
93. If the electric flux entering and leaving an enclosed surface respectively is φ1 and φ2 ,the
electric charge inside the surface will be [AIEEE 2003]
(a) ( φ 2 − φ1 ) ε0 (c)
(b)
( φ 2 − φ1 )
( φ1 + φ 2 ) ε0
(d) ( φ1 + φ 2 ) ε0
ε0
Exp. (a) From Gauss’ law, ⇒
K
Alternate Solution From the formula,
2
2
or
t
Aluminium is a metal, so when we d insert an aluminium foil, equal and opposite charges appear on its two surfaces. Since, it is of negligible thickness, it will not affect the capacitance.
a
O
a
45° FCD –Q
decreases remains unchanged becomes infinite increases
Exp. ( b )
q
a
(a) (b) (c) (d)
FCO
Charge enclosed = Net flux ε0
q = φ 2 − φ1 or q = (φ 2 − φ1 ) ε0 ε0
94. A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor
[AIEEE 2003]
C=
Here, So,
ε0 A d −t +
t K
K = ∞ and t → 0 ε A ε A C = 0 = 0 = C0 d+ 0 d
95. A thin spherical conducting shell of radius R has a charge q. Another chargeQ is placed at the centre of the shell. The electrostatic potential at a point P at a distance R/2 from the centre of the shell is [AIEEE 2003] 2Q 4 πε0 R 2Q q (c) + 4 πε0 R 4 πε0 R
2Q 2q − 4 πε0 R 4 πε0 R (q + Q ) 2 (d) 4 πε0 R
(a)
(b)
Exp. (c) At P due to shell, potential q V1 = 4 πε0 R At P due to Q, potential 2Q Q V2 = = R 4 πε0 R 4 πε0 2
P R
R/2 Q
q
∴ Net potential at P, V = V1 + V2 =
2Q q + 4 πε0 R 4 πε0 R
96. The work done in placing a charge of 8 × 10−18 C on a condenser of capacity 100 µF is [AIEEE 2003] (a) 16 × 10−32 J
(b) 3.1 × 10−26 J
(c) 4 × 10−10 J
(d) 32 × 10−32 J
230
JEE Main Chapterwise Physics
Exp. ( d )
E
1 (8 × 10 ) 1q W = = × 2 2 C 100 × 10−6 =
D
1 64 × 10−36 × = 32 × 10−32 J 2 100 × 10−6
C
O q
q
H
97. Three charges −q1 , + q 2 and − q 3 are placed as shown in the figure. The xcomponent of the force on −q 1 is proportional to [AIEEE 2003] –q3
F
−18 2
2
y
G
A
B L
q 4 πε0 L q (c) 2 πε0 L
(b)
(a)
q 2 πε0 L
(d) None of these
Exp. (d) a
Electric flux for any surface is defined as φ = ∫ E ⋅ ds
θ b –q1
(a) (c)
q2 b2 q2 b2
− +
q3 a2 q3 a2
cosθ cosθ
(b) (d)
q2 b2 q2 b2
+ −
E
x
+q2
D
q3 a2 q3 a2
sin θ sin θ
So, flux on the face ABCD =
1 q 6 ε0
The options (a), (c) and (d) are dimensionally incorrect, so they cannot be answers.
θ +q2
F=
B
q ε0 1 q Flux on each face = 6 ε0
y
–q1
G
q P
As flux on the cube =
Force on −q1,
a
C
q O H
A
Exp. (b) –q3
F
b
99. If there are n capacitors in parallel x
connected toV volt source, then the energy stored is equal to [AIEEE 2002]
1 q1q 2 1 q1q 3 i+ [sin θi − cos θ j] 4 πε0 b 2 4 πε0 a2
From above, x′ component of force –q1 is q1 q 2 q 3 Fx = + 2 sin θ θ 4πε0 b 2 a q q Fx ∝ 22 + 23 sin θ a b
(a) CV (c) CV 2
F12
1 nCV 2 2 1 (d) CV 2 2n (b)
Exp. (b) F13
98. A charged particle q is placed at the centreO of cube of length L ( ABCDEFGH ). Another same charge q is placed at a distance L from O. Then, the electric flux through ABCD is [AIEEE 2002]
Energy stored by any system of capacitors 1 = Cnet V 2 2 where, V is source voltage. Thus, n capacitors are connected in parallel. So, Cnet = nC 1 ∴ Enet = nC V 2 2
231
Electrostatics 100. If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium, then the value of q is [AIEEE 2002] Q (a) 2 Q (c) 4
Q (b) − 2 Q (d) − 4
Let charge q be placed at midpoint of line AB as shown below. Q
q C x/2
Also, ∴
having radius 1 m, is
[AIEEE 2002]
(a) 1.1 × 10−10
(b) 10−6
(c) 9 × 10−9
(d) 10−3
Exp. ( a )
Exp. (d)
A
101. Capacitance (in F) of a spherical conductor
Q B
Capacitance of spherical conductor = 4 πε0 a where, a is radius of conductor. 1 Therefore, C = ×1 9 × 109 1 = × 10−9 = 011 . × 10−9 F = 1.1 × 10−10 F 9
102. On moving a charge of 20 C by 2 cm, 2 J of
x/2
AB = x x AC = , 2 x BC = 2
[say]
work is done, then the potential difference between the points is [AIEEE 2002] (a) 0.1 V
(b) 8 V
(c) 2 V
(d) 0.5 V
Exp. (a) Potential difference between two points in an W electric field is VA − VB = q0
For the system to be in equilibrium, FQq + FQQ = 0 ⇒
1 Qq 1 QQ + =0 4 πε0 ( x /2 )2 4 πε0 x2
or
q=−
Q 4
where, W is work done by moving charge q 0 from point A to B. Here, W = 2 J, q 0 = 20 C 2 So, VA − VB = = 0.1V 20
12 Current Electricity 1. A 200 Ω resistor has a certain colour code. If
Given circuit is 1Ω
one replaces the red colour by green in the code, the new resistance will be (a) 100 Ω
(b) 400 Ω
(c) 300 Ω
(d) 500 Ω
2Ω
2V
[JEE Main 2019, 8 April ShiftI]
1Ω
a
4V
4V
Exp. (d) Given, resistance is 200 Ω.= 20 × 101 Ω R
B
1Ω
b
1Ω
Above circuit can be viewed as
Br
2V
2Ω
a
So, colour scheme will be red, black and brown. Significant figure of red band is 2 and for green is 5. When red (2) is replaced with green (5), new resistance will be 200 ohm → 500 ohm.
2. For the circuit shown with R1 = 1.0 Ω,
R 2 = 2.0Ω, E 1 = 2 V and E 2 = E 3 = 4 V, the potential difference between the points a andb is approximately (in volt) R1
[JEE Main 2019, 8 April ShiftI] R1
a
E3 R2
E1
E2 R1
(a) 2.7 (c) 3.7
Exp. (d)
b
(b) 2.3 (d) 3.3
R1
b 4V 4V
2Ω 2Ω
This is a parallel combination of three cells or in other words, a parallel grouping of three cells with internal resistances. E1 E2 E + + 3 Ieq r1 r2 r3 So, Vab = Eeq = = 1 1 1 req + + r1 r2 r3 2 4 4 + + 10 V ≈ 3.3 V =2 2 2 = 1 1 1 3 + + 2 2 2
3. In the figure shown, what is the current (in ampere) drawn from the battery? You are given : R1 = 15 Ω, R 2 = 10 Ω, R 3 = 20 Ω, R 4 = 5 Ω, R 5 = 25 Ω, R 6 = 30 Ω, E = 15 V [JEE Main 2019, 8 April ShiftII]
233
Current Electricity R3
(a) R = 2r (c) R = 0001 . r
R1 E
+ –
R2
R4
(b) R = r (d) R =1000 r
Exp. (b) Given circuit is shown in the figure below E,r
R5
R6
(a) 13/24 (c) 20/3
(b) 7/18 (d) 9/32
I
Exp. (d) Given circuit is redrawn and can be simplified as R1=15Ω
R
Net current,
I=
R3=20Ω
E R+ r
…(i)
+
R4=5Ω
E=15 V
Power across R is given as –
R2=10Ω
R6=30Ω
In series, Req = R3+R4+R5 = 20+5+25 = 50Ω
R5=25Ω
⇒ Req=50Ω
In parallel,
R2=10Ω
+
–
E=15 V
R1=15Ω
50 R′eq = — Ω 6
–
E=15 V
+
R6=30Ω
= E2
In series, R′′eq =R1+R′eq+R6 50 = 30 + — + 15 6 320 160 = —— = ——Ω 3 6
So, current drawn through cell is Voltage i= Net resistance of the circuit 15 9 V A = = = (160 / 3) 32 Req ′′
4. A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when [JEE Main 2019, 8 April ShiftII]
d R dR (R + r )2
(R + r )2 × 1 − 2 R × (R + r ) = E2 =0 (R + r )4 ⇒
R1=15Ω
[using Eq. (i)]
For the maximum power, dP =0 dR 2 dP d E = ⇒ ⋅ R dR dR R + r
50×10 = ——— 50+10 50 =— Ω 6
⇒
R6=30Ω
Req×R2 R′eq = ———— Req+R2
2
E P = I2 R = ⋅R R + r
(R + r )2 = 2 R(R + r )
or R + r = 2R ⇒ r=R ∴The power delivered by the cell to the external resistance is maximum when R = r. Alternate Solution From maximum power theorem, power dissipated will be maximum when internal resistance of source will be equals to external load resistance, i.e. r = R.
5. In
the circuit shown, a fourwire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is r = 0.01 Ω/cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be [JEE Main 2019, 8 April ShiftII]
234
JEE Main Chapterwise Physics 1.5 V, 1.5 V, 0.5 Ω 0.5 Ω
A
V
B
A 50 cm D
R=1Ω B
3 R 4 1 (d) R 16 (b)
100 cm
(a) 0.20 V (c) 0.25 V
(b) 0.75 V (d) 0.50 V
Exp. (a) Let the length of each side of square ABCD is a. R 4a
∴Resistance per unit length of each side =
Exp. (c) In given potentiometer, resistance per unit length is x = 0.01 Ωcm−1. 1.5 V, 1.5 V, 0.5 Ω 0.5 Ω
C
E
7 (a) R 64 (c) R
A
B R2
V A 50 cm
R=1Ω
D
B 100 cm
Length of potentiometer wire is L = 400 cm Net resistance of the wire AB is R AB = resistance per unit length × length of AB = 0.01 × 400 ⇒ R AB = 4 Ω Net internal resistance of the cells connected in series, r = 0.5 + 0.5 = 1Ω ∴Current in given potentiometer circuit is Net emf Net emf I= = Total resistance r + R + R AB 3 = = 0.5 A 1+ 1+ 4
square ABCD as shown in the figure. The effective resistance between E and C is [E is midpoint of arm CD] [JEE Main 2019, 9 April ShiftI]
C
R1
R1 E
C R2
R EC
Reading of voltmeter when the jockey is at 50 cm (l′ ) from one end A, V = IR = I( xl ′ ) = 0.5 × 0.01 × 50 = 0.25 V
6. A wire of resistance R is bent to form a
E
R a R R Now, [EC ] = × = R1 = 4a 4a 2 8 R R 7a 7R Similarly, R 2 = [EDABC ] = × = 4a 4a 2 8 Now, effective resistance between E and C is the equivalent resistance of R1 and R 2 that are connected in parallel as shown below.
R1R 2 = R1 + R 2 (R / 8) × (7 R / 8) = (R / 8) + (7 R / 8) =
7R2 8 7 R × = 64 8R 64
7. Determine the charge on the capacitor in the following circuit [JEE Main 2019, 9 April ShiftI]
(a) 2 µC (c) 10 µC
(b) 200µC (d) 60 µC
235
Current Electricity Exp. (b)
8A
6Ω
Given circuit is 6Ω
10µF
10Ω
To find charge on capacitor, we need to determine voltage across it. In steady state, capacitor will acts as open circuit and circuit can be reduced as 6Ω
2Ω
4Ω
72V
10Ω
In series, Req = 2 Ω + 10 Ω = 12 Ω
⇒ 6Ω
In parallel, Req =
G V=0
G V=0
2Ω
Q
12Ω
G V=0
Q = 2 × 10−4 C = 200 µC
8. In a conductor, if the number of conduction
[JEE Main 2019, 9 April ShiftII]
6Ω
3Ω
72V
G V=0
10µF
electrons per unit volume is 8.5 × 1028 m −3 and mean free time is 25 fs (femto second), it’s approximate resistivity is (Take, me = 9.1 × 10−31 kg)
4 × 12 = 3Ω 4 + 12 I
10Ω
Current in 4 Ω branch is, &I = VP − 0 2 4 24 − 0 = = 6Ω 4 So, current in 2 Ω resistance is I1 = 8 − I2 [Q I = I1 + I2 ] = 8 − 6 = 2A ∴Potential difference across 10Ω resistor is VQG = 2 A × 10 Ω = 20 V Same potential difference will be applicable over the capacitor (parallel combination). So, charge stored in the capacitor will be Q = CV = 10 × 10−6 × 20 ⇒
4Ω
72V
4Ω
72V
2Ω 4Ω
72V
P I1 I2
(a) 10−7 Ωm (c) 10−6 Ωm
(b) 10−5 Ωm (d) 10−8 Ωm
Exp. (d) In series, Req = 6 Ω + 3 Ω = 9 Ω 9Ω
Resistivity of a conductor is m … (i) ρ = e2 ne τ where,me = mass of electron = 91 . × 10− 31 kg, n = free charge density = 8.5 × 1028 m− 3 ,
72V
τ = mean free time = 25 fs = 25 × 10− 15 s V 72 = = 8A 9 R Now, by using current division, at point P, current in 6 Ω branch is 72 − VP = 8A 6Ω So, current in steady state, I =
⇒
VP = 72 − 48 = 24 V
e = charge of electron = 16 . × 10− 19 C
and
Substituting values in Eq. (i), we get . × 10− 31 91 ρ= 28 . × 10− 19 )2 × 25 × 10− 15 8.5 × 10 × (16 =
. × 10− 6 91 = 0.016 × 10− 6 8.5 × 2.56 × 25
= 1. 6 × 10− 8 Ωm ~ − 10− 8 Ωm
236
JEE Main Chapterwise Physics
9. A metal wire of resistance 3 Ω is elongated to
10. In an experiment, the resistance of a
make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60º at the centre, the equivalent resistance between these two points will be [JEE Main 2019, 9 April ShiftII]
material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line.
7 Ω 2 12 (c) Ω 5
[JEE Main 2019, 10 April ShiftI]
InR(T)
5 Ω 2 5 (d) Ω 3
(b)
(a)
1/T2
One may conclude that
Exp. (d) Initial resistance of wire is 3 Ω. Let its length is l and area is A. l Then, … (i) Rinitial = ρ = 3 Ω A When wire is stretched twice its length, then its area becomes A′, on equating volume, we have A Al = A′2 l ⇒ A′ = 2 So, after stretching, resistance of wire will be l′ l R ′ = R final = ρ = 4 ρ = 12 Ω [using Eq. (i)] A′ A Now, this wire is made into a circle and connected across two points A and B (making 60º angle at centre) as R2
60°
R1
(c) R(T ) = R0e
− T 2/T 2 0
− T 2/T 2 0
(b) R(T ) = R0e R (d) R(T ) = 02 T
B
Now, above arrangement is a combination of two resistances in parallel, 60 × R ′ R1 = 360 1 = × 12 = 2 Ω 6 300 and R2 = × R′ 360 5 = × 12 = 10 Ω 6 Since, R1 and R 2 are connected in parallel. R1R 2 10 × 2 5 So, R AB = = = Ω R1 + R 2 12 3
T 2/T 2 0
Exp. (c) From the given graph, ln R(T0)
ln R(T) 1/T02 1/T2
We can say that, lnR (T ) ∝ −
300°
A
(a) R(T ) = R0e
1
T2 Negative sign implies that the slope of the graph is negative. 1 or ln R(T) = constant − 2 T exp (const.) R(T ) = ⇒ 1 exp 2 T T2 R(T ) = R 0 exp − 02 ⇒ T Alternate Solution 1 From graph,
⇒ or
T 2 + ln R (T ) = 1 1 ln R(T0 ) T0 2
T 2 ln R (T ) = [ln R(T0 )]⋅ 1 − 02 T 2 −T R(T ) = R 0 exp 20 T
237
Current Electricity 11. In a meter bridge experiment, the circuit
12. In the given circuit, an ideal voltmeter
diagram and the corresponding observation table are shown in figure
connected across the 10 Ω resistance reads 2 V. The internal resistance r, of each cell is
[JEE Main 2019, 10 April ShiftI] X
[JEE Main 2019, 10 April ShiftI] 15 Ω
R Resistance box
2Ω G
Unknown resistance
10 Ω
l
1.5 V, 1.5 V, rΩ rΩ
K
E
S. No.
R (Ω)
l (cm)
1.
1000
60
2.
100
3.
13 10
4.
(a) 1.5 Ω
(c) 1 Ω
(d) 0 Ω
Exp. (b) For the given circuit I
1.5
1
(b) 0.5 Ω
1.0
15Ω
A I2
B
2Ω
I
I1 10Ω
Which of the readings is inconsistent? (a) 3
(b) 2
(c) 1
(d) 4 rΩ
Exp. (d) Unknown resistance ‘X’ in meter bridge experiment is given by 100 − l X = R l Case (1) When R = 1000 Ω and l = 60 cm, then (100 − 60) 40 × 1000 X= × 1000 = 60 60 2000 Ω ≈ 667 Ω X= ⇒ 3 Case (2) When R = 100 Ω and l = 13 cm, then 100 − 13 100 × 87 X = × 100 = 13 13 8700 Ω ≈ 669 Ω = 13 Case (3) When R = 10 Ω and l = 15 . cm, then . 100 − 15 X= × 10 15 . 98.5 9850 Ω ≈ 656 Ω = × 10 = . 15 15 Case (4) When R = 1 Ω and l = 10 . cm, then 100 − 1 X= ×1 1 ∴ X = 99 Ω Thus, from the above cases, it can be concluded that, value calculated in case (4) is inconsistent.
1.5 V
rΩ 1.5 V
Given, VAB = 2V ∴Current in circuit, I = I1 + I2 2 2 [Q V = IR or I = V / R ] = + 15 10 4+ 6 1 … (i) = = A 30 3 Also, voltage drop across (r + r ) resistors is = voltage of the cell − voltage drop across AB = 3 − 2 = 1V Using V = IR over the entire circuit ⇒ 1 = I (2 + 2 r ) 1 [using Eq. (i)] = (2 + 2 r ) 3 ⇒ 3 = 2 + 2 r or 2 r = 1 Ω 1 or r = Ω = 0. 5 Ω 2 Alternate Solution Equivalent resistance between A and B is 1 1 1 1 = = + R AB 10 15 6 ⇒
R AB = 6 Ω
238
JEE Main Chapterwise Physics A 6Ω B
2Ω
Resistance, dR = ρ
2V
dR = ρ
⇒
dx
4 πx 2 So, resistance of complete arrangement is b b dx ρ b −2 R = ∫ dR = ∫ ρ = x dx a a 4 πx 2 4 π ∫a
2r
3V
l A
∴Equivalent resistance of the entire circuit is, Req = 6 Ω + 2 Ω + 2 r = 8 + 2r Now, current passing through the circuit is given Enet E as, = net I= R + req Req where, R is external resistance, req is net internal resistance and Enet is the emf of the cells. Here, Enet = 15 . + 15 . = 3V 3 req = r + r = 2 r ⇒ I = 8 + 2r Also, reading of the voltmeter, V = 2V = I ⋅ R AB 3 1 2= × 6 ⇒ 8 + 2 r = 9 or r = = 0.5 Ω 2 8 + 2r
b
⇒
R=
b ρ 1 ρ x −1 = − 4 π −1 a 4 π x a
=
ρ 1 1 ρ 1 1 − ohm − − − = 4π a b 4π b a
14. To verify Ohm’s law, a student connects the voltmeter across the battery as shown in the figure. The measured voltage is plotted as a function of the current and the following graph is obtained [JEE Main 2019, 12 April ShiftI] V
13. Space between two concentric conducting
spheres of radii a andb (b > a ) is filled with a medium of resistivity ρ. The resistance between the two spheres (in ohm) will be
Internal resistance
Ammeter
[JEE Main 2019, 10 April ShiftII]
ρ 1 1 (a) + 2 π a b ρ 1 1 (c) − 2 π a b
ρ 4π ρ (d) 4π
(b)
1 − 1 a b 1 + 1 a b
R V 1.5V
Exp. (b) Key Idea Resistance between surface of inner shell and a circumferential point of outer shell can be formed by finding resistance of a thin (differentially thin) shell in between these two shells. Then, this result can be integrated (summed up) to get resistance of the complete arrangement.
For a elemental shell of radius x and thickness dx,
ρ
dx
x a
b
V0
I
1000 mA
If V 0 is almost zero, then identify the correct statement. (a) The emf of the battery is 1.5 V and its internal resistance is 1.5 Ω (b) The value of the resistance R is 1.5 Ω (c) The potential difference across the battery is 1.5 V when it sends a current of 1000 mA (d) The emf of the battery is 1.5 V and the value of R is 1.5 Ω
Exp. (a) Given circuit in a series combination of internal resistance of cell (r ) and external resistance R.
239
Current Electricity ∴Effective resistance in the circuit, Reff = r + R ∴Current in the circuit, E or E = IR + Ir I= R+ r Voltage difference across resistance R is V, so …(i) E = V + Ir Now, from graph at I = 0 , V = 15 . V From Eq. (i) at I = 0, E = V = 15 …(ii) . V At I = 1000 mA (or 1A), V = 0 From Eq. (i) at I = 1A and V = 0 …(iii) ⇒ E = I×r = r From Eqs. (ii) and (iii), we can get r = E = V = 15 . V ∴ r = 15 . Ω
∴Equivalent circuit is 2R
4R
6R R
R
4R
E = 16 V
8R
E=16 V
According to question, power consumed by the network, P = 4 W From Eq. (i), we get 16 × 16 16 × 16 ∴ = 8Ω = 4 ⇒R = 8× 4 8R
16. Drift speed of electrons, when 1.5 A of
12 R
current flows in a copper wire of crosssection 5 mm 2 is v. If the electron density in copper is 9 × 1028 / m 3, the value of v (in mm/s) is close to (Take, charge of
E = 16 V
(a) 6 Ω (c) 1 Ω
(b) 8 Ω (d) 16 Ω
electron to be = 1.6 × 10−19 C)
Exp. (b)
[JEE Main 2019, 9 Jan ShiftI]
(a) 0.02
Given circuit is 4R 4R
R
∴Total resistance of the given network is R s = 2 R + R + 4R + R = 8R As we know, power of the circuit, E 2 (16)2 16 × 16 …(i) = = P= 8R 8R Rs
15. The resistive network shown below is connected to a DC source of 16 V. The power consumed by the network is 4 W. The value of R is [JEE Main 2019, 12 April ShiftI]
4R
R
6R A
R 12R
(b) 0.2
(c) 2
(d) 3
Exp. (a) B
E=16 V
Equivalent resistance of part A, 4R × 4R = 2R RA = 4R + 4R Equivalent resistance of part B, 6R × 12 R RB = 6R + 12 R 72 = R = 4R 18
R
Relation between current (I) flowing through a conducting wire and drift velocity of electrons (vd ) is given as I = neAvd where, n is the electron density and A is the area of crosssection of wire. I vd = ⇒ neA Substituting the given values, we get 1.5 v= 9 × 1028 × 1.6 × 10− 19 × 5 × 10− 6 15 . × 103 m/s = 02 . × 10− 4 m/s 72 v = 0.02 mm/s v=
or
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JEE Main Chapterwise Physics
17. A copper wire is stretched to make it 0.5%
The table for color code of carbon resistor is given below,
longer. The percentage change in its electrical resistance, if its volume remains unchanged is [JEE Main 2019, 9 Jan ShiftI]
Color Codes
(a) 2.0%
Black
0
10 0
Brown
1
101
Red
2
10 2
Orange
3
10 3
Yellow
4
10 4
Green
5
10 5
Blue
6
10 6
Violet
7
10 7
Grey
8
10 8
White
9
10 9
(b) 1.0%
(c) 0.5%
(d) 2.5%
Exp. (b) Electrical resistance of wire of length ‘l’, area of crosssection ‘A’ and resistivity ‘ρ’ is given as l ...(i) R =ρ A Since we know, volume of the wire is ...(ii) V = A×l ∴From Eqs. (i) and (ii), we get l2 ...(iii) R =ρ V As, the length has been increased to 0.5%. ∴ New length of the wire, l ′ = l + 0.5% of l = l + 0.005 l = 1.005 l But V and ρ remains unchanged. ρ[(1.005) l]2 So, new resistance, R ′ = ...(iv) V Dividing Eq. (iv) and Eq. (iii), we get R′ = (1.005)2 R R′ ⇒ % change in the resistance = − 1 × 100 R
Multiplier (C)
Values (A, B)
Gold
± 5%
Silver
± 10%
No colour
± 20%
∴Comparing the colors given in the question and the table and writing in the manner of Eq. (i), we get R = 27 × 103 Ω , ± 10% = 27 k Ω , ± 10%
19. When the switch S in the circuit shown is closed, then the value of current i will be [JEE Main 2019, 9 Jan ShiftI]
= [(1.005)2 − 1] × 100 = 1.0025% ~ − 1%
20 V i1 A
18. A resistance is shown in the figure. Its value
i2 10 V
C 2Ω
4Ω
i
and tolerance are given respectively by
B
2Ω
[ JEE Main 2019, 9 Jan ShiftI] Red
Tolerance (D) (%)
Orange
S V=0 Violet
(a) 270 Ω, 5% (c) 27 k Ω, 10%
Silver
(b) 27 k Ω, 20% (d) 270 k Ω, 10%
Exp. (c) The value of a carbon resistor is given as, … (i) R = AB × C ± D% where, color band A and B (first two colors) indicate the first two significant figures of resistance in ohms, C (third color) indicates the decimal multiplier and D (forth color) indicates the tolerance in % as per the indicated value.
(a) 4A
(b) 3A
(c) 2A
(d) 5A
Exp. (d) When the switch ‘S ’ is closed the circuit, hence formed is given in the figure below, VA=20 V A
2Ω
4Ω
VC i1 C
VB=10 V i2
i 2Ω V=0
B
241
Current Electricity Then, according to Kirchhoff’s current law, which states that the sum of all the currents directed towards a point in a circuit is equal to the sum of all the currents directed away from that point. Since, in the above circuit, that point is ‘C’ ∴ i1 + i 2 = i VA − VC V − VC V −V ⇒ + B = C 2 4 2 (Qusing Ohm‘s law, V = iR) 20 − VC 10 − VC V −0 or + = C 2 4 2 ⇒ 20 − VC + (10 − VC )2 = VC 40 = VC + 3VC 40 = 4 VC or VC = 10 V V 10 The current, i = C = = 5A ∴ 2 2
20. In the given circuit, the internal resistance of
the 18 V cell is negligible. If R1 = 400 Ω, R 3 = 100 Ω and R 4 = 500 Ω and the reading of an ideal voltmeter across R 4 is 5 V, then the value of R 2 will be [JEE Main 2019, 9 Jan ShiftII] R3
R1
R4
(b) 230 Ω
(c) 300 Ω
(d) 450 Ω
Exp. (c) According to question, the voltage across R 4 is 5 volt, then the current across it R3 =100Ω I1 R1 =400Ω I
V1
So current through R 2 is, 2 1 3 A A = – I2 = I – I1 = 100 100 100 Now, from V = IR, we have, V 6 = 300Ω. R2 = 2 = I2 (2 / 100)
21. A carbon resistance has a following color code. What is the value of the resistance? [JEE Main 2019, 9 Jan ShiftII]
G O Y
(a) 5.3 MΩ ± 5% (c) 6.4 MΩ ± 5%
Golden
(b) 64 kΩ ± 10% (d) 530 kΩ ± 5%
Exp. (d)
R2
18 V
(a) 550 Ω
The potential difference across series combination of R 3 and R 4 1 ⇒ = 6V V2 = (R 3 + R 4 )I = 600 × 100 So, potential difference (across R1 ) V1 = 18 – 6 = 12 V Current through R1 is, V 12 3 I= 1 = = A R1 400 100
I2
The value of a carbon resistor is given as …(i) R = AB × C ± D% where, colour band A and B (first two colour) indicates the first two significant figures of resistance in Ohms. C (third colour) indicate the decimal multiplies and D (fourth colour) indicates the tolerance in % as per the indicated value. The table for colour code,
R4 =500Ω
Colour code Values (AB)
5V
Black
0
10 0
Brown
1
101
Red
2
10 2
Orange
3
10 3
Yellow
4
10 4
Green
5
10 5
Blue
6
10 6
Voilet
7
10 7
Grey
8
10 8
R2 V2 18 V
According to Ohm’s law, ⇒ V = IR ⇒ 5 = I1 × R 4 ⇒ 5 = I1 × 500 5 1 A I1 = = 500 100
Multiplier (C) Tolerance
242
JEE Main Chapterwise Physics White
10 9
9
Gold


± 5%
Silver


± 10%
No colour


± 20%
QComparing the colours given in the question and table and writing in the manner of Eq. (i), we get R = 53 × 104 ± 5% Ω or
R = 530 × 103 Ω ± 5%
or
R = 530 k Ω ± 5%
22. A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of EMF ε and internal resistance r. A cell C ε having emf and internal resistance 3r is 2 connected. The length AJ at which the galvanometer as shown in figure shows no deflection is [JEE Main 2019, 10 Jan ShiftI] +
D (ε,r) –
J
A
B
Potential drop across the balance length AJ of potentiometer wire is VAJ = i × R AJ ⇒ VAJ = i (Resistance per unit length of potentiometer wire × length AJ) 12 r VAJ = i × x ⇒ L where, x = balance length AJ. As null point occurs at J so potential drop across balance length AJ = EMF of the cell ‘C ’. ε VAJ = ⇒ 2 ε 12 r × x = i ⇒ L 2 ε 12 r ε 13 × × x= ⇒ x= L ⇒ 13r L 2 24
23. A 2 W carbon resistor is color coded with
+ – G C ε, 2 3r
5 L 12 13 (c) L 24
Internal resistance of cell ‘D’ = r ε 2 Internal resistance of cell ‘C ’ = 3r Current in potentiometer wire EMF of cell of potentiometer i = Total resistance of potentiometer circuit ε ε ⇒i = = r + 12 r 13r EMF of cell ‘C ’ =
green, black, red and brown respectively. The maximum current which can be passed through this resistor is
11 L 12 11 (d) L 24
(b)
(a)
[JEE Main 2019, 10 Jan ShiftI]
(a) 0.4 mA (c) 20 mA
(b) 63 mA (d) 100 mA
Exp. (c)
Exp. (c) Given, length of potentiometer wire ( AB) = L Resistance of potentiometer wire ( AB) = 12 r EMF of cell D of potentiometer = ε
Colour code of carbon resistance is shown in the figure below Green Black Red Brown
D (ε,r) + –
J A
B
+
–
ε C , 3r 2
G
So, resistance value of resistor using colour code is R = 502 × 10 = 502 . × 102 Ω Here, we must know that for given carbon resistor first three colours give value of resistance and fourth colour gives multiplier value. Now using power, P = i 2 R
243
Current Electricity P = R
i =
we get,
(a) 12 Ω (c) 2 Ω
2 502 . × 102
(b) 8 Ω (d) 4 Ω
Exp. (d)
~ – 20 × 10− 3 A = 20 mA.
24. In the given circuit, the cells have zero internal resistance. The currents (in Ampere) passing through resistances R1 and R 2 respectively are
Resistance of each arm of equilateral triangle will be 18 R= =6Ω 3 So we have following combination will be
[JEE Main 2019, 10 Jan ShiftI] 6Ω R2
20 Ω
R1
20 Ω
A –
+
+
10 V
(a) 0.5, 0
B
6Ω
Equivalent resistance is 12 × 6 12 × 6 ∴ = =4Ω R AB = 12 + 6 18
–
10 V
(b) 1, 2
6Ω
(c) 2, 2
(d) 0, 1
26. A current of 2 mA was passed through an
Exp. (a) By Kirchhoff’s loop rule in the given loop ABEFA, we get I1
A
[JEE Main 2019, 10 Jan ShiftII]
C
B I1+I2
I2
R1=20Ω
unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is
R2=20Ω
(a) 11 × 10−4 W
(b) 11 × 10−5 W
(c) 11 × 105 W
(d) 11 × 10−3 W
Exp. (b) I1
– F
+
10 V
+ E
–
I2
Power dissipated by any resistor R, when I current flows through it is, P = I2 R
D
10 V
10 − (I1 + I2 )R1 = 0 ⇒ 10 − (I1 + I2 ) 20 = 0 1 or I1 + I2 = 2 and from loop BCDEB, we get 10 − (I1 + I2 )R1 − I2 R 2 = 0 ⇒ 10 − (I1 + I2 + I2 ) 20 = 0 1 ⇒ I1 + 2 I2 = 2 From Eqs. (i) and (ii), we get I2 = 0 and I1 = 0.5 A
−3
Given I = 2 mA = 2 × 10 …(i)
…(ii)
25. A uniform metallic wire has a resistance of
18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is [JEE Main 2019, 10 Jan ShiftI]
… (i)
A and P = 4.4 W
Using Eq. (i), we get 4.4 = (2 × 10− 3 )2 × R .4 or R= ⋅Ω 4 × 10− 6
…(ii)
When this resistance R is connected with 11 V supply then power dissipated is V2 P= R (11)2 or P= × 4 × 10− 6 4.4 [Q Using Eq. (ii)] 11 × 11 × 4 × 10− 6 ⇒ P= W 44 × 10− 1 or
P = 11 × 10− 5 W
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JEE Main Chapterwise Physics
27. The actual value of resistance R , shown in the figure is 30 Ω. This is measured in an experiment as shown using the standard V formula R = , where V and I are the I readings of the voltmeter and ammeter, respectively. If the measured value ofR is 5% less, then the internal resistance of the voltmeter is [JEE Main 2019, 10 Jan ShiftII] V A
28. The Wheatstone bridge shown
R1
in figure here, gets balanced when the carbon resistor is used as R1 has the color code R 3 (orange, red, brown). The resistorsR 2 and R 4 are 80 Ω and + 40 Ω, respectively.
R2 G R4 –
Assuming that the color code for the carbon resistors gives their accurate values, the color code for the carbon resistor is used as [JEE Main 2019, 10 Jan ShiftII] R 3 would be (a) brown, blue, black (b) brown, blue, brown (c) grey, black, brown (d) red, green, brown
R
Exp. (b) (a) 600 Ω
(b) 570 Ω
(c) 350 Ω
(d) 35 Ω
Exp. (b) Measured value of R = 5% less than actual value of R. Actual values of R = 30 Ω So, measured value of R is 5 R′ = 30 − (5% of 30) = 30 − × 30 100 … (i) ⇒ R′ = 28.5 Ω Now, let us assume that internal resistance of voltmeter R V . Replacing voltmeter with its internal resistance, we get following circuit.
The value of R1 (orange, red, brown) ⇒ = 32 × 10 = 320 Ω Given, R 2 = 80 Ω and R 4 = 40 Ω In balanced Wheatstone bridge condition, R R1 R 3 = ⇒ R3 = R4 × 1 R2 R2 R4 40 × 320 ⇒ R3 = 80 or R 3 = 160 Ω = 16 × 101 Comparing the value of R 3 with the colours assigned for the carbon resistor, we get R3=16 × 10
RV A
R
V
It is clear that the measured value, R′ should be equal to parallel combination of R and R V . Mathematically, RR V = 28.5 Ω R′ = R + RV Given, ⇒ ⇒ ⇒ ⇒ or
R = 30 Ω 30R V = 28.5 30 + R V 30R V = (28.5 × 30) + 28.5 R V 1.5R V = 28.5 × 30 28.5 × 30 RV = = 19 × 30 1.5 R V = 570 Ω
Brown
1
Blue Brown
29. The resistance of the meter bridge AB in
given figure is 4 Ω. With a cell of emf ε = 0.5 V and rheostat resistanceR h = 2 Ω .The null point is obtained at some point J. When the cell is replaced by another one of emf ε = ε 2, the same null point J is found for R h = 6 Ω. The emf ε 2 is ε
A
J
6V
(a) 0.6 V (c) 0.5 V
Exp. (b)
[JEE Main 2019, 11 Jan ShiftI]
Rh
(b) 0.3 V (d) 0.4 V
B
245
Current Electricity Let length of null point ‘J’ be ‘x’ and length of the potentiometer wire be ‘L’. In first case, current in the circuit 6 = 1A I1 = 4+2 1× 4 ∴ Potential gradient = I × R = L ⇒ Potential difference in part ‘AJ’ 1× 4 = × x = ε1 L 4x x 1 or = …(i) Given, ε1 = 0.5 = L L 8 In second case, current in the circuit 6 = 0.6 A I2 = 4+ 6 0.6 × 4 ∴ Potential gradient = L ⇒ Potential difference in part ‘AJ’ 0.6 × 4 = × x = ε2 L 0.6 × 4 L [using Eq. (i)] × ε2 = ⇒ L 8 ⇒ ε2 = 0.3 V
P=
V2 = 60 W 2R
V2 = 120 W R
⇒
New resistance in parallel = P′ =
V2 =2 R /2
[JEE Main 2019, 11 Jan ShiftI]
(a) 60 W
(b) 30 W
(c) 240 W
(d) 120 W
Exp. (c) Let P1 and P2 be the individual electric powers of the two resistances, respectively. In series combination, power is PP P0 = 1 2 = 60W P1 + P2 Since, the resistances are equal and the current through each resistor in series combination is also same. Then, P1 = P2 = 120 W In parallel combination, power is P = P1 + P2 = 120 + 120 = 240 W Alternate Solution Let R be the resistance. ∴ Net resistance in series = R + R = 2 R
V2 = 240 W R
31. In a Wheatstone bridge (see figure), resistances P and Q are approximately equal. When R = 400 Ω, the bridge is balanced. On interchanging P and Q , the value ofR for balance is 405 Ω. The value of X is close to [JEE Main 2019, 11 Jan ShiftI] B P
Q G
A
C R
K2
X
D
30. Two equal resistances when connected in series to a battery consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be
R×R = R /2 R+ R
K1
(a) 404.5 Ω (c) 402.5 Ω
(b) 401.5 Ω (d) 403.5 Ω
Exp. (c) For a balanced Wheatstone bridge, P Q = R X In first case when R = 400 Ω, the balancing equation will be P Q = R X P Q = ⇒ 400 Ω X 400 × Q …(i) X In second case, P and Q are interchanged and R = 405 Ω
⇒
P=
∴
Q P = R X
⇒
Q P = 405 X
…(ii)
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JEE Main Chapterwise Physics R1
Substituting the value of P from Eq. (i) in Eq. (ii), we get Q × 400 Q = 405 X2 ⇒
G A
X 2 = 400 × 405
X = 400 × 405 = 402.5 ⇒ The value of X is close to 402.5 Ω.
32. A galvanometer having a resistance of 20 Ω and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt is [JEE Main 2019, 11 Jan ShiftII]
(a) 100 Ω (c) 120 Ω
(b) 125 Ω (d) 80 Ω
Exp. (d) Resistance of galvanometer, (G ) = 20 Ω Number of divisions on both side = 30 Figure of merit = 0.005 ampere/division
G
R
Voltmeter
∴ Full scale deflection current (Ig ) = Number of divisions × figure of merit ⇒ Ig = 30 × 0.005 = 015 . A Now, for measuring 15V by this galvanometer, let we use a resistance of R ohm in series with the galvanometer. ∴Effective resistance of voltmeter, Reff = R + 20 As,maximum potential measured by voltmeter. V = Ig ⋅ Reff or 15 = 015 . (R + 20) or R + 20 = 100 or R = 80 Ω
33. In the experimental set up of meter bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10 Ω resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10) Ω such that the null point shifts back to its initial position is [JEE Main 2019, 11 Jan ShiftII]
B ()
(a) 60 Ω
(b) 20 Ω
(c) 30 Ω
(d) 40 Ω
Exp. (a) For meter bridge, if balancing length is R l l cm, then in first case, 1 = R 2 (100 − l ) It is given that, l = 40 cm; R1 R2 So; = 40 100 − 40 R1 2 or = R2 3
…(i)
In second case, R1′ = R1 + 10, and balancing length is now 50 cm then R1 + 10 R2 = (100 − 50) 50 R1 + 10 = R 2
or I
R2
…(ii)
Substituting value of R 2 from (ii) to (i) we get, R1 2 or = 10 + R1 3 ⇒ 3R1 = 20 + 2 R1 or R1 = 20Ω ⇒ R 2 = 30Ω Let us assume the parallel connected resistance is x. x (R1 + 10) Then equivalent resistance is x + R1 + 100 So, this combination should be again equal to R1. (R1 + 10)x 30 x = R1 ⇒ = 20 R1 + 10 + x 30 + x 30 x = 600 + 20 x x = 60Ω
or or
34. In the circuit shown, the potential difference between A and B is [JEE Main 2019, 11 Jan ShiftII] M 5Ω A
1Ω
1V
1Ω
2V
D N
1Ω
3V
10Ω C
B
247
Current Electricity (a) 3 V (c) 6 V
(b) 1 V (d) 2 V
R2
K2 R1=220Ω
Exp. (d) G
In the given circuit, let’s assume currents in the arms are i 1, i 2 and i 3 , respectively. M 5Ω A
i1
1Ω
1V
i2
1Ω
2V
D
N
C
i3
1Ω
K1
10Ω B
3V
V 1 Now, i1 = 1 = = 1 A R1 1 2 Similarly, i 2 = =2 A 1 3 and i3 = = 3 A 1 Total current in the arm DA is i = i1 + i 2 + i 3 = 6 A As all three resistors between D and C are in parallel. ∴ Equivalent resistance between terminals D and C is 1 1 1 1 = + + R DC 1 1 1 1 R DC = Ω ∴ 3 So, potential difference across D and C is 1 VDC = iR DC = 6 × 3 ⇒ VDC = 2 V Now, VAD and VCB = 0 (In case of open circuits, I = 0) So, VAB = VAD + VDC + VCB = VDC So, VAB = 2V
35. The galvanometer deflection, when keyK 1 is
closed but K 2 is open equals θ 0 (see figure). On closing K 2 also and adjusting R 2 to 5 Ω, θ the deflection in galvanometer becomes 0 . 5 The resistance of the galvanometer is given by (neglect the internal resistance of battery) : [JEE Main 2019, 12 Jan ShiftI]
(a) 22 Ω
(b) 5Ω
(c) 25Ω
(d)12 Ω
Exp. (a) For a galvanometer, i g ∝ θ or i g = C θ where, Ig = current through coil of galvanometer, θ = angle of deflection of coil and C is the constant of proportionality. Now, K1 is closed and K 2 is opened, circuit resistance is R eq = R1 + R g where, R g = galvanometer resistance. Hence, galvanometer current is V i.e. ig = R1 + R g where, V = supply voltage. As deflection is given θ0 , we have V = Cθ0 ig = R1 + R g
…(i)
When both keys K1 and K 2 are closed, circuit resistance. R × Rg R eq = R1 + 2 R + R 2 g Current through galvanometer will be R2 V × ig2 = R2 Rg (R2 + Rg ) R1 + R 2 + R g VR 2 θ = = C⋅ 0 R1R 2 + R1R g + R 2 R g 5
…(ii)
Now, dividing Eq. (i) by Eq. (ii), we get R1R 2 + R1R g + R 2 R g =5 R 2 (R1 + R g ) Substituting R1 = 220 Ω and R 2 = 5 Ω, we get 1100 + 220 R g + 5 R g =5 5 (220 + R g ) 1100 + 225 R g = 5500 + 25R g
248
JEE Main Chapterwise Physics ⇒
200 R g = 4400 4400 Rg = = 22 Ω 200 R g = 22 Ω
⇒ ∴
36. Two electric bulbs rated at 25 W, 220 V and 100 W, 220 V are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then [JEE Main 2019, 12 Jan ShiftI]
(a) P1 = 16 W , P2 = 4W (b) P1 = 4 W , P2 = 16W (c) P1 = 9 W , P2 = 16W (d) P1 = 16 W , P2 = 9W
Exp. (a) Resistance of a bulb of power P and with a voltage source V is given by V2 R= P Resistance of the given two bulbs are V 2 (220)2 V 2 (220)2 and R 2 = = = R1 = P1 25 P2 100 Since, bulbs are connected in series. This means same amount of current flows through them. ∴ Current in circuit is V 220 1 i = = = A 2 2 R total (220) 11 (220) + 25 100 Power drawn by bulbs are respectively, 2
220 × 220 1 P1 = i 2 R1 = × = 16 W. 11 25
Hence, potential drop across 1 m length of potentiometer wire is V VAB = 5 × 10−2 × 1 m = 5 × 10−2 V Now, potential drop that must occurs across resistance R is 395 V VR = 4 − 5 × 10−2 = 100 V 4 Now, circuit current is i = = R total R+ 5 Hence, for resistance R, using VR = iR, we get 395 4 = ×R 100 R + 5 395 (R + 5) = 400R 395 × 5 = (400 − 395)R ⇒ R = 395 Ω
38. In a meter bridge, the wire of length 1m has a nonuniform crosssection such that the dR of its resistance R with length l variation dl 1 dR is ∝ . Two equal resistance are dl l connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP ? [JEE Main 2019, 12 Jan ShiftI]
2
220 × 220 1 and P2 = i 2 R 2 = × = 4 W. 11 100
R′
37. An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 Ω. The value of R to give a potential difference of 5 mV across 10 cm of potentiometer wire is
R′
G P A
B l
(a) 0.3 m
1–l
(b) 0.25 m (c) 0.2 m
(d) 0.35 m
[JEE Main 2019, 12 Jan ShiftI]
(a) 395 Ω (c) 490 Ω
(b) 495 Ω (d) 480 Ω
Exp. (a) Given, potential difference of 5 mV is across 10 m length of potentiometer wire. So potential drop per unit length is −3 5 × 10 V = 5 × 10−2 = m 10 × 10−2
Exp. (b) As, galvanometer shows zero deflection. This means, the meter bridge is balanced. R′ R′ …(i) = ⇒ R AP = R PB R AP R PB dR k Now, for meter bridge wire, = dl l where, ‘k’ is the constant of proportionality.
249
Current Electricity
At junction P, I5 = I6 ⇒ I6 = 0.4 A At junction Q, I2 = I1 + I3 + I6 = 0.3 + 0.4 + 0.4 = 1.1 A
k dl l Integrating both sides, we get k R=∫ dl ⇒ l dR =
⇒
l
k ∫0 l dl = k(2 l ) 0 = 2k l 1 1 k and dl = 2 k( l ) R PB = ∫ l l l = 2k ( 1 − l ) = 2 k (1 − l ) Substituting values of R AP and R PB in Eq. (i), we get R AP = R PB 2 k l = 2 k(1 − l ) ⇒ 1 1 or l = = 025 . m l = ⇒ 2 4 R AP =
So,
l
39. In the given circuit diagram, the currents I 1 = − 0.3 A, I 4 = 0.8 A and I 5 = 0.4 A, are flowing as shown. The currents I 2, I 3 and I 6 respectively, are
[JEE Main 2019, 12 Jan ShiftII] I6 Q
P
40. A galvanometer whose resistance is 50 Ω, has 25 divisions in it. When a current of 4 × 10− 4 A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of [JEE Main 2019, 12 Jan ShiftII]
(a) 250 Ω
(b) 6200 Ω (c) 200 Ω
Exp. (c) Current for deflection of pointer by 1 division = 4 × 10−4 A So, current for fullscale deflection = Ig = Number of divisions × Current for 1 division ⇒ Ig = 25 × 4 × 10−4 = 1 × 10−2 A Now, let a resistance of R is put in series with galvanometer to make it a voltmeter of range 2.5 V. R
A I2 S I 4
I1
R
(a) 1.1 A, 0.4 A, 0.4 A (b) 1.1 A, − 0.4 A, 0.4 A (c) 0.4 A, 1.1 A, 0.4 A (d) − 0.4 A, 0.4 A, 1.1 A
Given circuit with currents as shown in the figure below, [In the question I1 = 0.3 A is given, due to it we change the direction of I1, in this figure] I6
I1
Q I3
B
Then, Ig (R + G ) = VAB 1 × 10−2 (R + 50) = 2.5 [QGiven, G = 50 Ω, VAB = 2.5 V] ∴ R = 250 − 50 = 200 Ω
connected in parallel across a load resistor of 10 Ω. The internal resistances of the two batteries are 1 Ω and 2 Ω, respectively. The voltage across the load lies between [JEE Main 2018]
(a) 11.6 V and 11.7 V (c) 11.4 V and 11.5 V
(b) 11.5 V and 11.6 V (d) 11.7 V and 11.8 V
Exp. (b)
I5=0.4 A
10 Ω
I2 S
VAB=2.5 V
41. Two batteries with emf 12 V and 13 V are
Exp. (a)
P
G
Ig
I3
I5
(d) 6250 Ω
I4=0.8 A
R
From Kirchoff’s junction rule, ΣI = 0 At junction S, I4 = I5 + I3 ⇒ 0.8 = 0.4 + I3 ⇒ I3 = 0.4 A
I1=0.3 A
+ 12V –
1Ω
– + 2Ω 13V
250
JEE Main Chapterwise Physics For parallel combination of cells, E1 E2 + r r2 Eeq = 1 1 1 + r1 r2
Exp. (b) With only the cell, E′ 52 cm
12 13 + 2 = 37 V ∴ Eeq = 1 1 1 3 + 1 2 Potential drop across 10 Ω resistance, E V= × 10 R total =
∴
G E, r
On balancing, …(i) E = 52 × x where, x is the potential gradient of the wire. When the cell is shunted, E′
37 / 3 × 10 = 11.56 V 10 + 2 3
40 cm
V = 1156 . V
G
Alternate Solution I2
E F
I1
E, r 13 V, 2 Ω R=5 Ω
D
12V, 1Ω
A
Similarly, on balancing,
C 10 Ω
V = E−
B
Applying KVL, in loop ABCFA, −12 + 10 (I1 + I2 ) + 1 × I1 = 0 ⇒ 12 = 11I1 + 10 I2 Similarly, In loop ABDEA, −13 + 10 (I1 + I2 ) + 2 × I2 = 0 ⇒ 13 = 10 I1 + 12 I2 Solving Eqs. (i) and (ii), we get 7 23 A, I2 = A I1 = 16 32 ∴ Voltage drop across 10 Ω resistance is 7 23 . V V = 10 + = 1156 16 32
…(i)
…(ii)
42. In a potentiometer experiment, it is found
43. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ. How much was the resistance on the left slot before interchanging the resistances? [JEE Main 2018] (a) 990 Ω
Exp. (c)
[JEE Main 2018]
Initially,
(b)1.5 Ω
(c) 2 Ω
(d) 2.5 Ω
…(ii)
Solving Eqs. (i) and (ii), we get E 1 52 = = V 1− r 40 R+ r 5 + r 52 E R + r 52 = = = ⇒ ⇒ 5 40 V R 40 3 ⇒ r = Ω r = 1.5 Ω 2
that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell. (a)1 Ω
Er = 40 × x (R + r )
(b) 505 Ω
(c) 550 Ω
(d) 910 Ω
We have, X + Y = 1000 Ω Y=1000 – X
X G l
X 1000 − X = l 100 − l
100 – l
…(i)
251
Current Electricity When X and Y are interchanged, then Y=1000 – X
Exp. (b) In steady state, no current flows through the capacitor. So, resistance rp becomes ineffective. So, the current in circuit, E I= r + r2 (Total Resistance)
X G
(l – 10)
(110 – l)
1000 − X X = l − 10 100 − (l − 10) 1000 − X X = l − 10 110 − l
or
QPotential drop across capacitor = Potential drop …(ii)
From Eqs. (i) and (ii), we get 100 − l l − 10 = l 110 − l (100 − l ) (110 − l ) = (l − 10) l 11000 − 100 l − 110 l + l 2 = l 2 − 10 l
46. In the below circuit, the current in each resistance is 2V
(a) In a balanced Wheatstone bridge, if the cell and the galvanometer are exchanged, the null point is disturbed (b) A rheostat can be used as a potential divider (c) Kirchhoff’s second law represents energy conservation (d) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude
Exp. (a) In a balanced Wheatstone bridge, there is no effect on position of null point, if we exchange the battery and galvanometer. So, option (a) is incorrect. E r the given circuit diagram, when the r1 current reaches steady C r2 state in the circuit, the charge on the capacitor of capacitanceC will be [JEE Main 2017 (Offline)]
r1 (r2 + r ) r1 (c) CE (r1 + r )
(a) CE
(b) CE (d) CE
r2 (r + r2 )
[JEE Main 2017 (Offline)] 2V
1Ω
2V
44. Which of the following statements is false? [JEE Main 2017 (Offline)]
Er2 r + r2
∴ Stored charge of capacitor, r Q = CV = CE 2 r + r2
⇒ 11000 = 200 l ∴ l = 55 cm Substituting the value of l in Eq. (i), we get X 1000 − 55 = ⇒ 20 X = 11000 55 100 − 55 ∴ X = 550 Ω
45. In
= Ir2 =
across r2
(a) 0.25 A (c) 0 A
2V
2V
1Ω
1Ω
2V
(b) 0.5 A (d) 1 A
Exp. (c) Each resistance is converted with two cells combined in opposite direction, so potential drop across each resistor is zero. Hence, the current through each of resistor is zero.
47. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300400 K, is best described by [JEE Main 2016 (Offline)]
(a) linear increase for Cu, linear increase for Si (b) linear increase for Cu, exponential increase for Si (c) linear increase for Cu, exponential decrease for Si (d) linear decrease for Cu, linear decrease for Si
Exp. (c) As, we know Cu is a conductor, so increase in temperature, resistance will increase. Then, Si is semiconductor, so with increase in temperature, resistance will decrease.
252
JEE Main Chapterwise Physics
48. When 5V potential difference is applied across a wire of length 0.1m, the drift speed of electrons is 2.5 × 10− 4 ms −1 . If the electron density in the wire is 8 × 1028 m − 3 the resistivity of the material is close to
Exp. (c) / Connect point Q to ground and apply KCL. Consider the grounded circuit as shown below. 6V
P(V )
[JEE Main 2015]
(a) 1.6 × 10− 8 Ωm
(b) 1.6 × 10− 7 Ωm
(c) 1.6 × 10− 6 Ωm
(d) 1.6 × 10− 5 Ωm
1Ω
Exp. (d)
Q 3Ω
According to the question, 0.1m
5V
or
vd = 2.5 × 10− 4 m/s ⇒
n = 8 × 1028 /m 3
or
We know that J = nevd or I = nevd A where, symbols have their usual meaning. V = nevd A ⇒ R V V or = nevd = nevd A or ρL ρL A V or ρ= nevd L
or
5Ω
Applying KCL of point Q we can write Incoming current at Q = outgoing current from Q ⇒
=
9V
5 8 × 1028 × 16 . × 10− 19 × 2.5 × 10− 4 × 01 .
ρ = 16 . × 10− 5 Ωm
V+ 6 V 9−V + = 3 1 5 9 1 1 V + + 1 = − 2 5 3 5 5 + 3 + 15 9 − 10 = V 15 5
or
−1 23 V = 5 15
−3 . V = − 013 23 Thus, current in the 1 Ω resistance is 0.13 A, from Q to P. V=
or
50. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. Then, the minimum current should be [JEE Main 2014] (a) 8 A
(b) 10 A
(c) 12 A
(d) 14 A
Exp. (c)
49. In the circuit shown below, the current in the 1Ω resistor is 6V
(a) 1.3 A, from P to Q (c) 0.13 A, from Q to P
= (15 × 40) + (5 × 100 )
P 2Ω 1Ω
3Ω
Total power (P) consumed
240 W
[JEE Main 2015]
9V
Q 3Ω
(b) 0 A (d) 0.13 A, from P to Q
60 W
+ (5 × 80) + (1 × 1000) = 2500 120 V W As, we know that P = VI 2500 125 I= = = 11.3 A ⇒ 200 11 Hence, minimum capacity should be 12 A.
6Ω
253
Current Electricity 51. The supply voltage to room is 120 V. The
resistance of the lead wires is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb? [JEE Main 2013] (a) Zero
(b) 2.9 V
(c) 13.3 V
(d) 10.04 V
(a) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I (c) Statement I is true, Statement II is false (d) Statement I is false, Statement II is true
Exp. (d)
Exp. (d) P=
2
V R
Resistance of the bulb, 120 × 120 R= = 240 Ω 60 Req = 240 + 6 = 246 Ω V 120 ⇒ I1 = = Req 246 V1 = I1R1 =
240 W
6Ω
53. Two electric bulbs marked 25 W220 V and 100 W220 V are connected in series to a 440 V supply. Which of the bulbs will fuse? [AIEEE 2012] 120 V
120 × 240 246
= 117.073 V Resistance of the heater V 2 120 × 120 = = = 60 Ω 240 P As bulb and heater are connected in parallel. 240 × 60 Net resistance = = 48 Ω 300 Total resistance, R 2 = 48 + 6 = 54 Ω Total current, I2 = V / R 2 = 120 / 54 Potential across heater = Potential across bulb 48 W
To increase the range of ammeter, resistance should be decreased (so additional shunt is connected in parallel).
6Ω
(a) Both (c) 25 W
Exp. (c) As the rated power of 25 W is less than 100 W, it implies that 25 W bulb has higher resistance. As in series connection, current through both the bulbs is same but heating in 25 W bulb is more than that of 100 W bulb. So, 25 W bulb will get fused.
54. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is (a) 6%
120 V
120 × 48 = 106.66 V 54 V1 − V2 = 117.073 − 106.66 = 10.04 V V2 =
52. This question has Statement I and Statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement I Higher the range, greater is the resistance of ammeter. Statement II To increase the range of ammeter, additional shunt needs to be used across it. [JEE Main 2013]
(b) 100 W (d) None of these
(b) zero
(c) 1%
(d) 3%
Exp. (a) From Ohm’s law, R = V / I ∆R ∆V ∆I By error method, = + R V I = 3% + 3% = 6%
55. If 400 Ω of resistance is made by adding four 100 Ω resistance of tolerance 5%, then the tolerance of the combination is [AIEEE 2011] (a) 20%
(b) 5%
(c) 10%
Exp. (b) Resistance of combination, Re = 4R ∆Re ∆R 5 × 100 = ⇒ = = 5% 100 Re R
(d) 15%
254
JEE Main Chapterwise Physics
56. The current in the primary circuit of a potentiometer is 0.2 A. The specific resistance and crosssection of the potentiometer wire are 4 × 10−7 Ωm and 8 × 10−7 m 2, respectively. The potential gradient will be equal to [AIEEE 2011] (a) 0.2 V/m (b) 1 V/m
Exp. ( b ) A moving conductor is equivalent to a battery of emf = vBl [motion emf]
R
R
R
(c) 0.3 V/m (d) 0.1 V/m I
Exp. ( d ) Potential gradient of a potentiometer, K =
V l
Iρ 0.2 × 4 × 10−7 K= = A 8 × 10−7
I1
Equivalent circuit, I = I1 + I2 Applying Kirchhoff’s law,
= 0.1 V/m
57. If a wire is stretched to make it 0.1% longer, its resistance will
⇒ ⇒
Exp. (d)
⇒
∴
58. A rectangular loop has a sliding connector PQ of length l and resistance R Ω and it is moving with a speedv as shown. The setup is placed in a uniform magnetic field going into the plane of the paper. The three currents I 1 , I 2 and I are [AIEEE 2010] P
RΩ
R
v
RΩ I I1
Q
2 Blv Blv ,I = R R 2 Blv Blv (b) I 1 = I 2 = ,I = 3R 3R Blv (c) I 1 = I 2 = I = R Blv Blv (d) I 1 = I 2 = ,I = 6R 3R (a) I 1 = − I 2 =
RΩ I2
I1R + IR − vBl = 0
...(ii)
I2 R + IR − vBl = 0
...(iii)
I1R + IR − VBl + I2 R + IR − vBl = 0
(a) increase by 0.2% (b) decrease by 0.2% (c) decrease by 0.05% (d) increase by 0.05%
ρl ρl 2 [Q V = volume] = A V ∆R ∆l = 2 = + 0.2% R l
...(i)
Adding Eqs. (ii) and (iii), we get
[AIEEE 2011]
R=
I2
⇒
I1R + I2 R + 2 IR − 2 vBl = 0 (I1 + I2 )R + 2 IR − 2 vBl = 0 2 IR + IR = 2 vBl 2 vBl I= 3R
[From Eq. (i)]
Subtracting Eq.(iii) from Eq.(ii), we get I1R + IR − VBl = 0 − I2 R
+ − IR − + VBl = − 0 I1R − I2 R = 0
⇒ R(I1 − I2 ) = 0 ⇒ I1 − I2 = 0 ⇒ I1 = I2 From Eq. (i), we get I = 2 I1 ⇒ 2 I1 = I I 2 vBl vBl ⇒ = I1 = = 2 2 × 3R 3R vBl I1 = I2 = ∴ 3R
[Q R ≠ 0] ...(iv) [Q I1 = I2 ]
59. In the circuit shown below, the key K is
closed at t = 0. The current through the battery is [AIEEE 2010]
255
Current Electricity (a)
VR1 R2 R12
+
R22
at t = 0 and
V at t = ∞ R2
V ( R1 + R2 ) V at t = ∞ at t = 0 and R1 R2 R2 VR1 R2 V at t = ∞ at t = 0 and (c) R2 R12 + R22 V ( R1 + R2 ) V (d) at t = 0 and at t = ∞ R1 R2 R2 (b)
∴
As α1 and α 2 are small quantities. ∴ α1 α 2 is negligible. α1 + α 2 or αp = 2 + (α1 + α 2 )θ =
At t = 0, inductor behaves like an infinite resistance. V So, at t = 0, I = R2 and at t = ∞, inductor behaves like a conducting wire. V (R1 + R 2 ) V ∴ = I= R eq R1R 2
60. Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are α1 and α 2. The respective temperature coefficients of their series and parallel combinations are nearly [AIEEE 2010]
α1 + α 2 , α1 + α 2 2 αα (c) α1 + α 2 , 1 2 α1 + α 2
α1 + α 2 2 α1 + α 2 α1 + α 2 (d) , 2 2
(b) α1 + α 2 ,
Exp. (d) Let R 0 be the initial resistance of both conductors. ∴ At temperature θ, their resistances will be R1 = R 0 (1 + α1θ) and R 2 = R 0 (1 + α 2θ) For series combination, R s = R1 + R 2 R s0 (1 + α sθ) = R 0 (1 + α1θ) + R 0 (1 + α 2 θ) where, R s 0 = R0 + R0 = 2 R0 ∴ 2 R 0 (1 + α sθ) = 2 R 0 + R 0θ(α1 + α 2 ) α + α2 or αs = 1 2 For parallel combination, R1R 2 Rp = R1 + R 2 R 0 (1 + α1θ)R 0 (1 + α 2θ) R p 0 (1 + α p θ) = R 0 (1 + α1θ) + R 0 (1 + α 2θ)
R0 R0 R = 0 2 R0 + R0
R0 R 2 (1 + α1θ + α 2θ + α1α 2θ2 ) (1 + α p θ) = 0 R 0 (2 + α1θ + α 2θ) 2
Exp. (b)
(a)
Rp 0 =
where,
α1 + α 2 2
1 −
α1 + α 2 2
θ
As (α1 + α 2 )2 is negligible. α + α2 αp = 1 ∴ 2
61. This question contains Statement I and Statement II. Of the four choices given after the statements, choose the one that best describes the two statements. [AIEEE 2009] Statement I The temperature dependence of resistance is usually given as R = R 0(1 + α∆t ). The resistance of a wire changes from 100 Ω to 150 Ω when its temperature is increased from 27°C to 227°C. This implies that α = 2.5 × 10−3 /° C. Statement II R = Ri (1 + α∆T ) is valid only when the change in the temperature ∆T is small and ∆R = (R − R 0 ) R1 ). If the potential difference across the source having internal resistance [AIEEE 2005] R 2 is zero, then R2 × ( R1 + R2 ) (b) R = R2 − R1 ( R2 − R1 ) R1 R2 R1 R2 (c) R = (d) R = ( R1 + R2 ) ( R2 − R1 ) (a) R =
Exp. ( b ) As R1, R 2 and R in series. Req = R1 + R 2 + R ∴ Net current, I =
(a) 200 Ω
(b) 100 Ω
(c) 500 Ω
12 ×R 500 + R
1000 + 2 R = 12 R
or
zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be [AIEEE 2005]
12V B
2=
⇒
73. In the circuit, the galvanometer G shows
G
12 500 + R
(d) 1000 Ω
2E R1 + R 2 + R
According to the question,
Exp. (b)
− (VA − VB ) = E − IR 2
The galvanometer shows zero deflection i.e., current through XY is zero 500Ω
X
G
∴
0 = E − IR 2 R1
Y
E
R2 A
E
B
2V R
12V
R
As a result potential drop across R is 2 V. Circuit can be redrawn as 500Ω I 2V 12V
R
or
E = IR 2
or
E=
or or
2E R2 R1 + R 2 + R
R1 + R 2 + R = 2 R 2 R = R 2 − R1
75. A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be [AIEEE 2005] (a) doubled (c) onefourth
(b) four times (d) halved
260
JEE Main Chapterwise Physics
Exp. (a) Is heat generated V2 V2 H1 = t and H2 = t R ( R /2 ) H2 = 2 or H2 = 2 H1 ∴ H1
79. The total current supplied to the circuit by the battery is
[AIEEE 2004] 2Ω
6V
6Ω
3Ω
1.5 Ω
76. An energy source will supply a constant current into the load, if its internal resistance is [AIEEE 2005] (a) equal to the resistance of the load (b) very large as compared to the load resistance (c) zero (d) nonzero but less than the resistance of the load
Exp. (c) E R+ r E I = = constant R where, R = external resistance r = internal resistance = 0 I=
(a) 1 A (c) 4 A
(b) 2 A (d) 6 A
Exp. (c) The equivalent circuit can be drawn as 6 Ω and 2 Ω are in parallel. 2×6 . Ω R′ = = 15 8 1.5Ω
2Ω 3Ω
–
77. In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2 Ω, the balancing length becomes 120 cm. The internal resistance of the cell is [AIEEE 2005] (a) 1 Ω
(b) 0.5 Ω
(c) 4 Ω
+ 6V
As 1. 5 Ω and 1. 5 Ω are in series. R′ ′ = 1. 5 + 1. 5 = 3 Ω 1.5Ω
1.5Ω
(d) 2 Ω 3Ω
Exp. (d) The internal resistance of the cell, l − l 240 − 120 r = 1 2 R = ×2 =2 Ω 120 l 2
78. The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100 W and 200 V lamp, when not in use? [AIEEE 2005] (a) 40 Ω
6Ω
(b) 20 Ω
(c) 400 Ω
–
+ 6V
As 3 Ω and 3 Ω are in parallel. 3× 3 Req = = 1. 5 Ω 6 3Ω
(d) 200 Ω
1.5Ω
3Ω
Exp. (a) P = V2 /R ∴
Rhot Rcold
V 2 200 × 200 = 400 Ω = = 100 P 400 = = 40 Ω 10
–
+ 6V
–
+ 6V
Hence, current supplied by the battery is 6 V I= = =4A . Req 15
261
Current Electricity 80. The resistance of the series combination of two resistances is S. When they are joined in parallel, the total resistance is P. If S = nP , then the minimum possible value of n is [AIEEE 2004]
(a) 4 (c) 2
(b) 3 (d) 1
Exp. (a) Let resistances be R1 and R 2 . Then, S = R1 + R 2 R1R 2 and P= R1 + R 2 n × R1R 2 (R1 + R 2 ) = ∴ R1 + R 2 or
[Q S = nP ]
(R1 + R 2 )2 = nR1R 2
⇒
R 2 + R 22 + 2 R1R 2 n= 1 R1R 2 R1 R2 = + + 2 R1 R2
We know, Arithmetic Mean ≥ Geometric Mean R1 R + 2 R2 R1 R1 R 2 ≥ × 2 R2 R1 ⇒
R1 R + 2 ≥2 R2 R1
So, n (minimum value) = 2 + 2 = 4
81. In a meter bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y . If X < Y , then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X againstY ? [AIEEE 2004] (a) 50 cm (c) 40 cm
(b) 80 cm (d) 70 cm
Exp. (a) Meter bridge is an arrangement which works on Wheatstone’s principle, so the balancing condition is R l1 = S l2 where,
l 2 = 100 − l 1
Case I
R = X, S = Y , l1 = 20 cm,
l 2 = 100 − 20 = 80 cm X 20 …(i) = Y 80 Case II Let the position of null point be obtained at a distance l from same end. ∴ R = 4 X, S = Y , l 1 = l, l 2 = 100 − l So, from Eq. (i), we get 4X l = Y 100 − l X l …(ii) = Y 4 (100 − l )
∴
Therefore, from Eqs. (i) and (ii), we get l 20 l 1 = ⇒ = 4 (100 − l ) 80 4(100 − l ) 4 or or Hence,
l = 100 − l 2 l = 100 l = 50 cm
82. The thermistors are usually made of [AIEEE 2004]
(a) metals with low temperature coefficient of resistivity (b) metals with high temperature coefficient of resistivity (c) metal oxides with high temperature coefficient of resistivity (d) semiconducting materials having low temperature coefficient of resistivity
Exp. (c) Thermistors are the resistors made up of semiconductors whose resistance decreases with the increase in temperature. This implies that they have negative and high temperature coefficient of resistivity. They are usually made of metal oxides with high temperature coefficient of resistivity.
83. Time taken by a 836 W heater to heat 1 L of water from 10°C to 40°C is
[AIEEE 2004]
(a) 50 s
(d) 200 s
(b) 100 s
(c) 150 s
Exp. (c) Let time taken in boiling the water by the heater be t second. Then, Q = ms ∆T
262
JEE Main Chapterwise Physics ⇒ ∴ or or
Pt = ms ∆T J 836 t = 1 × 1000(40 − 10) 4.2 836 t = 1000 × 30 4.2 . 1000 × 30 × 42 t = = 150 s 836
∴
I = current flowing = 3 A t =2 s m = 3.3 × 10−7 × 3 × 2 = 19.8 × 10−7 kg
86. The length of a wire of a potentiometer is
84. The thermoemf of a thermocouple varies
with the temperature θ of the hot junction as E = a θ + b θ 2 in volts, where the ratio a/b is 700°C. If the cold junction is kept at 0°C, then the neutral temperature is [AIEEE 2004] (a) 700°C (b) 350°C (c) 1400°C (d) no neutral temperature is possible for this thermocouple
Exp. (d)
100 cm and the emf of its stand and cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5 Ω. If the balance point is obtained at l = 30 cm from the positive end, the emf of the battery is [AIEEE 2003] (a)
30E 100.5
(b)
30E 100 − 0.5
30( E − 0.5 I ) , where I is the current in the 100 potentiometer wire 30E (d) 100
(c)
Exp. (d) E = aθ + bθ2
[given]
For neutral temperature (θn ), dE =0 dθ ⇒ a + 2 b θn = 0 a or θn = − 2b 700 θn = − ∴ 2
Q a = 700° C b
= − 350° C < 0° C But neutral temperature can never be negative (less than zero) i.e., θn > l1 µ 2 M(d / 2 ) µ 0 2 M = ⇒ BX = 0 ⋅ 4 π (d / 2 )4 4 π (d / 2 )3 r=
Here,
Z
Let the current I is flowing in anticlockwise direction, then the magnetic moment of the coil is m = NIA where, N = number of turns in coil and A = area of each coil = πr 2 . Its direction is perpendicular to the area of coil and is along Yaxis. Then, torque on the current coil is τ = m × B = mBsin90° = NIAB = NIπr 2 B(N  m)
3. Two magnetic dipoles X andY are placed at a separationd, with their axes perpendicular to each other. The dipole moment of Y is twice that of X . A particle of charge q is
µ0 2 Mr , along OP ⋅ 4 π (r 2 − l12 )2
Similarly, for dipole Y, point P lies on its equatorial line. So, magnetic field strength at P due to Y is BY BN
S O′
BS P
2M
N d/2
BY =
µ0 2M , ⋅ 4 π (r 2 + l22 )3 / 2
Here,
(along a line perpendicular to O ′ P) d r= 2
267
Magnetic Effect of Current d >> l2 µ 0 2M 4 π (d / 2 )3
Also,
Magnetic field due to an infinitely long straight wire at point P is given as
BY =
⇒
Thus, the resultant magnetic field due to X and Y at P is
M
P
I
BY
r
45° P
2M
BX
d/2
d/2
Bnet = BX + BY Since, BY = BX Thus, the resultant magnetic field (Bnet ) at P will be at 45° with the horizontal. This means, direction of Bnet and velocity of the charged particle is same. ∴Force on the charged particle moving with velocity v in the presence of magnetic field which is B = q(v × B) = q vBsinθ where, θ is the angle between B and v. According to the above analysis, we get θ=0 ∴ F=0 Thus, magnitude of force on the particle at that instant is zero.
4. Two very long, straight and insulated wires are kept at 90° angle from each other in xyplane as shown in the figure. y I
d
P
x
B =
µ0 I
2πr Thus, in the given situation, magnetic field due to wire 1 at point P is
I
d
P d
1
(0,0)
2
µ0 I ,⊗ 2π d Similarly, magnetic field due to wire 2 at point P is µ I B2 = 0 , u 2π d Resultant field at point P is Bnet = B1 + B2 Since,B1 = B2, but they are opposite in direction. Thus, Bnet = 0 ∴Net magnetic field at point P will be zero. B1 =
d
5. A moving coil galvanometer has resistance
I
These wires carry currents of equal magnitude I , whose directions are shown in the figure. The net magnetic field at point P will be [JEE Main 2019, 8 April ShiftII] (a) zero (c) −
µ 0I $ $ (x + y) 2 πd
Exp. (a)
+µ 0 I $ ( z) πd µ I $ $ (d) 0 ( x + y) 2 πd
(b)
50 Ω and it indicates full deflection at 4 mA current. A voltmeter is made using this galvanometer and a 5 kΩ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to [JEE Main 2019, 9 April ShiftI]
(a) 40 V
(b) 10 V
(c) 15 V
(d) 20 V
Exp. (d) Given, resistance of galvanometer, R g = 50Ω Current, Ig = 4 mA = 4 × 10−3 A Resistance used in converting a galvanometer in voltmeter, R = 5 kΩ = 5 × 103 Ω
268
JEE Main Chapterwise Physics I1
Ig
Rg
I2
R a
Voltmeter
∴Maximum current in galvanometer is E Ig = R + Rg ∴
E = Ig (R + R g ) = 4 × 10−3 × (5 × 103 + 50) = 5050 × 4 × 10−3 − 20 V = 20.2 V ~
6. A rectangular coil (dimension 5 cm × 2.5 cm) with 100 turns, carrying a current of 3A in the clockwise direction, is kept centred at the origin and in the XZ plane. A magnetic field of 1 T is applied along Xaxis. If the coil is tilted through 45° about Zaxis, then the torque on the coil is [JEE Main 2019, 9 April ShiftI] (a) 0.27 Nm (c) 0.42 Nm
(b) 0.38 Nm (d) 0.55 Nm
Exp. (a) Given, Area of the rectangular coil, A = 5 cm × 2.5 cm ⇒ A = 12.5 cm2 = 12.5 × 10−4 m2 Number of turns, N = 100 turns Current through the coil, I = 3 A Magnetic field applied, B = 1 T Angle between the magnetic field and area vector of the coil, θ = 45° As we know that, when a coil is tilted by an angle θ in the presence of some external magnetic field, then the net torque experienced by the coil is, τ = M × B = NI(A × B) = NIAB sinθ Substituting the given values, we get τ = 100 × 3 × 12.5 × 10−4 × 1 × sin 45° τ = 0707 . × 100 × 3 × 12.5 × 10−4 Nm = 2.651 × 10−1 Nm ≈ 0.27 Nm
7. A rigid square loop of side a and carrying current I 2 is lying on a horizontal surface near a long current I 1 carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be [JEE Main 2019, 9 April ShiftI]
a
µ 0 I1 I 2 2π µ I I (b) attractive and equal to 0 1 2 3π (c) zero µ I I (d) repulsive and equal to 0 1 2 4π (a) repulsive and equal to
Exp. (d) Key Idea Net force experienced by two wires separated by same distance is attractive, if current flow in them in same direction. However, this force is repulsive in nature, if current in them flows in opposite direction.
Force on a wire 1 in which current I1 is flowing due to another wire 2 which are separated by a distance r is given as F = I1(l × B2 ) µ II or F = 0 1 2 ⋅ l sinθ 2 πr
QB = µ 0 I2 2 2 πr
Thus, the given square loop can be drawn as shown below P I1
I2
B
C
FBC a
FCD FAB Q
a
A
FAD D
µ 0 I1 ⋅ I2 a 2 πa (away from wire PQ) FBC = FAD = 0 [Qθ = 0° ] µ 0 I1 FCD = ⋅ I2 a 2 π(2 a) µ 0 I1 (towards the wire PQ) = ⋅ I2 a 4 πa ∴ Fnet = FAB − FCD µ II µ II (away from wire) = 012 − 012 2π 4π µ II (repulsive in nature) = 012 4π FAB =
269
Magnetic Effect of Current 8. The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 00.5 A? [JEE Main 2019, 9 April ShiftII] (a) 0.2 ohm (c) 0.002 ohm
(b) 0.5 ohm (d) 0.02 ohm
Torque of torsion band, T = kθ where, k = torsion constant of torsion band and θ = deflection of coil in radians or angle of twist of restoring torque. kθ … (i) ∴ BINA = kθ or B = INA Here, k = 10− 6 N  m / rad, I = 1 × 10− 3 A,
Exp. (a) Key Idea An ammeter is a type of galvanometer with a shunt connected in parallel to the galvanometer.
Ammeter circuit is shown in the figure below S
N = 175, A = 1 cm2 = 1 × 10− 4 m2 π rad θ = 1º = 180 Substituting values in Eq. (i), we get 10− 6 × 22 B= −3 1 × 10 × 175 × 7 × 180 × 10− 4 = 0.998 × 10− 3
I – Ig I
Ig
~ − 10− 3 T
G
10. A uniformly charged ring of radius 3a and
Ig G = (I − Ig ) S Ig = 0.002 A, I = 0.5 A, G = 50 Ω So, shunt resistance required is I G 0.002 × 50 ≈ 02 . Ω S= g = I − Ig (0.5 − 0.002 ) So, Here,
total charge q is placed in xyplane centred at origin. A point chargeq is moving towards the ring along the Z axis and has speedv at z = 4a . The minimum value of v such that it crosses the origin is [JEE Main 2019, 10 April ShiftI]
9. A moving coil galvanometer has a coil with 175 turns and area 1 cm 2. It uses a torsion band of torsion constant 10−6 Nm/rad. The coil is placed in a magnetic fieldB parallel to its plane. The coil deflects by1º for a current of 1 mA. The value of B (in tesla) is approximately [JEE Main 2019, 9 April ShiftII]
(a) 10−3
(b) 10−4
(c) 10−1
(a)
2 m
1 q2 5 4 πε0a
(c)
2 m
1 q2 15 4 πε0a
2 m
4 q2 15 4 πε0a
1/ 2
(b)
2 m
2 q2 15 4 πε0a
1/ 2
(d)
1/ 2
Exp. (d) Potential at any point at distance x from the centre of the ring is given by
(d) 10−2
+q
Exp. (a) In a moving coil galvanometer in equilibrium, torque on coil due to current is balanced by torque of torsion band. As, torque on coil, τ = M × B = NIAB sinα where, B = magnetic field strength, I = current, N = number of turns of coil Since, plane of the coil is parallel to the field. ∴ α = 90º ⇒ τ = NIBA
1/ 2
R C
R = 3a P x = 4a
Vp = Given, ∴
Kq R + x2 2
R = 3a and x = 4a Kq Kq = VP = 5a 9a2 + 16a2
… (i)
270
JEE Main Chapterwise Physics At centre, x=0 So, potential at centre is Kq Kq … (ii) = VC = R 3a Now, energy required to get this charge from x = 4a to the centre is Kq Kq − ∆U = q ∆V = q [VC − VP ] = q 3a 5a =
Net force on the third wire, carrying current I in the following first case is A
B d
I1
I2
Kq 2 1 1 − a 3 5
x
d–x
2
2 Kq … (iii) 15 a This energy must be equal to (or less than) the kinetic energy of the charge, i.e. 1 2 Kq 2 mv 2 ≥ 2 15 a So, minimum energy required is q2 1 2 1 (put K = 1 / 4 πε0 ) × × mv 2 = a 2 15 4 πε0 ∆U =
∴Minimum velocity, v2 =
Exp. (d)
2 2 q2 or v = × × m 4 πε0 15a
2 × m
2q 2 4 πε0 a × 15
1
3
2
F13 + F23 = 0 Using thumb rule, direction of B at inside region of wires A and B will be same. µ I I µ 0 I2 I ∴ 01 + =0 2 πx 2 π (d − x) I1 I I I + 2 = 0⇒ 1 = 2 ⇒ x d− x x x−d or ⇒
( x − d ) I1 = x I2 ⇒ x (I1 − I2 ) = dI1 I1 x= ⋅d (I1 − I2 )
Second case of balanced force can be as shown A
andI 2 as shown in the figure. The separation between them is d. A third wireC carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are [JEE Main 2019, 10 April ShiftI] C
I1
B
I2 x d
I2 I2 (a) x = d and x = d I1 + I 2 I1 − I 2 I2 I1 (b) x = d d and x = I1 + I 2 I1 − I 2 I2 I1 (c) x = d d and x = I + I I1 − I 2 1 2 I 1d (d) x = ± ( I1 − I 2 )
B
d
11. Two wires A and B are carrying currents I 1
A
… (i)
I1
I2
x d–x 3
1
2
Using thumb rule, directions of B at any point on wires A and B will be opposite, so net force, µ 0 I1I µ 0 I2 I − = 0 2 πx 2 π (d + x) I1 I2 or − =0 x (d + x) I1 I = 2 ⇒ x d+ x ⇒ ⇒ ⇒
(d + x) I1 = x I2 (I2 − I1 ) x = dI1 x=−
I1 ⋅d (I1 − I2 )
From Eqs. (i) and (ii), it is clear that I1 x=± d (I1 − I2 )
… (ii)
271
Magnetic Effect of Current 12. A proton, an electron and a helium nucleus,
Exp. (*)
have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane.
Given data,
Let rp , re and rHe be their respective radii, then [JEE Main 2019, 10 April ShiftI]
Vmax = 5 V Let internal resistance of galvanometer is RG .
(b) re > rp = rHe (d) re > rp > rHe
(a) re < rp = rHe (c) re < rp < rHe
I = 10−4 A, RS = 2 MΩ = 2 × 106 Ω,
I
RG
Exp. (c) When a moving charged particle is placed in a magnetic field B. Then, the net magnetic force acting on it is Fm = q (v × B) Fm = q vBsinθ θ = 90°
Here,
Fm = q vB Also, due to this net force, the particle transverses a circular path, whose necessary centripetal force is being provided by Fm , i.e. mv 2 mv 2 mv = q vB ⇒ r = = ⇒ r∝m r qvB qB So, for electron, re =
me v or re ∝ me eB
rp =
mp v
For proton, eB
Clearly,
rHe
4mp v
2eB > rp
=
I × RS + I × RG = Vmax 2 × 106 × 10−4 + 10−4 × RG = 5
⇒
10−4 RG = 5 − 200 = − 195
or
RG = − 195 × 104 Ω
Resistance cannot be negative. ∴ No option is correct.
14. A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop (in Am) will be [JEE Main 2019, 10 April ShiftII] 4m π
(b)
3m π
(c)
2m π
(d)
m π
Exp. (a) 2 mp v eB (Q rHe = 2 rp ) Q mp ≈ 10−27kg, me ≈ 10−31 kg
rp > re ⇒ rHe > rp > re
13. A moving coil galvanometer allows a full −4
scale current of 10 A. A series resistance of 2 MΩ is required to convert the above galvanometer into a voltmeter of range 05 V. Therefore, the value of shunt resistance required to convert the above galvanometer into an ammeter of range 0.10 mA is [JEE Main 2019, 10 April ShiftI]
(a) 100 Ω (c) 200 Ω
⇒
or rp ∝ mp
and we know that, mp > me ∴
Voltmeter
Then,
(a)
For Heparticle, rHe =
Rs
(b) 500 Ω (d) 10 Ω
Key Idea Magnetic dipole moment of a current carrying loop is m = IA (A  m2) where, I = current in loop and A = area of loop.
Let the given square loop has side a, then its magnetic dipole moment will be m = Ia2 When square is converted into a circular loop of radius r,
a r
Then, wire length will be same in both area, 4a 2 a ⇒ 4a = 2 πr ⇒ r = = 2π π
272
JEE Main Chapterwise Physics Hence, area of circular loop formed is, A′ = πr 2 2
2
2a 4a = π = π π
∴ sinθ1 = sinθ2 =
µ 3 , i = 10A, 0 = 10−7NA −2 2 4π
1 × altitude 3 1 3 1 1 m = × × sides length = × 1= 3 2 2 3 2 3
and r =
Magnitude of magnetic dipole moment of circular loop will be 4a2 m′ = IA′ = I π Ratio of magnetic dipole moments of both shapes is, 4a2 I⋅ m′ π = 4 ⇒ m′ = 4m (A  m) = m π π Ia2
So,
Bnet
3 3 × 10−7 × 10 × 2 2 = 18 × 10−6 T = 1 2 3
⇒ Bnet = 18 µT
15. The magnitude of the magnetic field at the
16. A thin ring of 10 cm radius carries a
centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is
uniformly distributed charge. The ring rotates at a constant angular speed of 40π rad s −1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10−9 T, then the charge carried by the ring is close to (µ 0 = 4π × 10−7 N/A 2).
[Take, µ 0 = 4π × 10− 7 NA − 2] [JEE Main 2019, 10 April ShiftII]
(a) 9µT
(b) 1µT
(c) 3µT
(d) 18µT
Exp. (d)
[JEE Main 2019, 12 April ShiftI]
For a current carrying wire, from result obtained by BiotSavart’s law, magnetic field at a distance r is given by B θ1
θ2
(a) 2 × 10−6 C −5
(c) 4 × 10
C
µ 0 = 4 π × 10−7 N/A 2 ,
Given,
ω = 40 π rad/s = 3.8 × 10−9 T
Bat centre l
µ 0i (sinθ1 + sinθ2 ) 4 πr Now, in given case, B=
i r
θ2 θ1
Due to symmetry of arrangement, net field at centre of triangle is Bnet = Sum of fields of all wires (sides) µ i = 3 × 0 (sinθ1 + sinθ2 ) 4 πr Here, θ1 = θ2 = 60°
(d) 7 × 10−6 C
Exp. (b)
r i
(b) 3 × 10−5 C
and R = 10 cm = 01 . m Now, we know that, magnetic field at the centre of a current carrying ring is given by µ I …(i) B= 0 2r Here, I can be determined by flow of charge per rotation, i.e. Q …(ii) I= T 2π Here, T= ω Qω …(iii) I= ⇒ 2π By putting value of I from Eq. (iii) to Eq. (i), we get µ Qω 2 Br × 2 π or Q = B= 0 2r × 2 π µ 0ω =
. × 2π 2 × 3.8 × 10−9 × 01 4 π × 10−7 × 40 π
273
Magnetic Effect of Current . 2 × 3.8 × 01 × 10−2 2 × 40 π = 0.003022 × 10−2 C
=
or
⇒
= 3.022 × 10−5 C Q = 3 × 10−5 C
18. A galvanometer of resistance 100 Ω has 50
17. A magnetic compass needle oscillates 30 times per minute at a place, where the dip is 45° and 40 times per minute, where the dip is 30°. If B1 and B 2 are respectively, the total magnetic field due to the earth at the two B places, then the ratio 1 is best given by B2 [JEE Main 2019, 12 April ShiftI]
(a) 1.8 T (c) 3.6 T
B1 . 9 × 1732 15.588 = = . B2 16 × 1414 22.624 B1 = 0.689 ≈ 07 . T B2
⇒
(b) 0.7 T (d) 2.2 T
divisions on its scale and has sensitivity of 20 µA/division. It is to be converted to a voltmeter with three ranges of 02 V, 010 V and 020 V. The appropriate circuit to do so is [JEE Main 2019, 12 April ShiftI] G
R1
R2
R3 R1 = 2000 Ω
(a)
R2 = 8000 Ω 2V
10 V
R3 = 10000 Ω 20 V
Exp. (b) Given, at first place, angle of dip, θ1 = 45º 60 = 2s 30 At second place, angle of dip, θ2 = 30º 60 3 Time period, T2 = = s 40 2 Now, at first place, B …(i) BH1 = B1 cos θ1 = B1 cos 45º = 1 2 and at second place, 3 BH2 = B2 cos θ2 = B2 cos 30º = B2 …(ii) 2 Also, time period of a magnetic needle is given by I …(iii) T = 2π MBH Time period, T1 =
∴
T∝
T 1 or 1 = T2 BH
BH2 BH1
B1 9 3 = B2 16 2
R2
R1
R3
(b) 2V
G
R1
10 V R2
R1 = 1900 Ω R2 = 9900 Ω R3 = 19900 Ω 20 V
R3 R1 = 1900 Ω R2 = 8000 Ω
(c) 2V
G
R1
10 V
R2
R3
(d) 20 V
10 V
R3 = 10000 Ω 20 V
R1 = 19900 Ω R2 = 9900 Ω R3 = 1900 Ω 2V
…(iv)
By putting the values from Eqs. (i) and (ii) into Eq. (iv), we get B 3 2 2 2 2 or 4 = 3 × 2 B2 = B1 3 3 2 B1 2 2 B1 3× 2 9 = × ⇒ B2 2 16 ⇒
G
Exp. (c) Given, divisions in scale of galvanometer, n = 50 Sensitivity of galvanometer, Ig = 20 µA / division n ∴Current in galvanometer, I Ig = g × n = 20 µA × 50 n ⇒ Ig = 1000 µA = 1 mA
274
JEE Main Chapterwise Physics Now, for R, it should be converted into 2V voltmeter. ∴ V1 = Ig (R1 + G ) 2 = 10−3 (R1 + 100) ⇒ 2000 = R1 + 100 ∴ R1 = 1900 Ω For R 2 , it should be converted into 10V voltmeter. ∴ V2 = Ig [(R1 + R 2 ) + G ] ⇒ 10 = 10−3 [(R1 + R 2 ) + 100]
Exp. (*)
θ
…(i)
⇒ 10000 = R1 + R 2 + 100 = 2000 + R 2 …(ii) ∴ R 2 = 8000 Ω For R 3 , it should be converted into 20V voltmeter. ∴ V3 = Ig [(R1 + R 2 + R 3 + G ] ⇒ 20 = 10−3 [1900 + 8000 + R 3 + 100] ⇒ 20000 = R 3 + 10000 …(iii) ∴ R 3 = 10000 Ω From Eqs. (i), (ii) and (iii), it is clear that option (c) is correct.
19. An electron moving along the Xaxis with an initial energy of 100 eV, enters a region of $ at S (see magnetic field B = (1.5 × 10−3 T ) k figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P andQ (on the screen) is (Take, electron’s charge = 1.6 × 10−19 C, mass of electron = 9.1 × 10−31 kg) [JEE Main 2019, 12 April ShiftII] Q
R
d
θ
q, m v L D
When electron enters the region of magnetic field, it experiences a Lorentz force which rotates electron in a circular path of radius R. So, Lorentz force acts like a centripetal force and we have mv 2 = Bqv R where, m = mass of electron, q = charge of electron, v = speed of electron, R = radius of path, and B = magnetic field intensity. Radius of path of electron, mv R= Bq Now, from geometry of given arrangement, comparing values of tanθ, we have L d LD Bq LD tanθ = = ⇒d = = R D R mv BqLD [Qmv = 2 mk ] d= ⇒ 2 mk where, k = kinetic energy of electron Here, B = 15 . × 10−3 T, q = 1. 6 × 10−19 C, L = 2 × 10−2 m, D = 6 × 10−2 m,
d
m = 91 . × 10−31 kg, k = 100 × 16 . × 10−19 J So,
P
S
=
2 cm 8 cm
(a) 11.65 cm (c) 1.22 cm
d=
(b) 12.87 cm (d) 2.25 cm
(15 . × 10−3 × 1. 6 × 10−19 × 2 × 10−2 × 6 × 10−2 ) (2 × 91 . × 10−31 × 100 × 16 . × 10−19 ) 28.8 × 10−26 −48
2912 . × 10
=
28.8 × 10−26 5.39 × 10−24
= 5.34 cm No option is matching.
= 5.34 × 10−2 m
275
Magnetic Effect of Current 20. A moving coil galvanometer, having a resistance G , produces full scale deflection when a current I g flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I 0(I 0 > I g ) by connecting a shunt resistance R A to it and (ii) into a voltmeter of range 0 toV (V = GI 0 ) by connecting a series resistance RV to it. Then, [JEE Main 2019, 12 April ShiftII] (a) RA RV = G
2 I0
Ig − Ig and RA = RV ( I 0 − I g ) I g RA I g = RV I 0 − I g
(b) RA RV = G 2 and
Ig (c) RA RV = G 2 I −I 0 g
2
and
I − Ig and RA = 0 RV I g
(given) …(ii)
From Eqs. (i) and (ii), we have Ig G I − I Ig2 RA 0 g = = R V (I0 − Ig )G (I0 − Ig )2 Ig
2
RA × RV =
Ig G (I0 − Ig )
(I0 − Ig ) G
×
Ig
= G2
2
Ig RA = RV ( I 0 − I g )
(d) RA RV = G 2 and
Equating potential across point AB, VAB = (G + R V ) Ig But VAB = I0G So, I0G = (G + R V ) Ig (I − Ig )G RV = 0 ⇒ Ig
21. Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A (See figure). (Take, µ 0 = 4π × 10−7 NA −2 ) [JEE Main 2019, 12 April ShiftII]
Exp. (b) P
5c Ammeter
I0–Ig A Ig
G
A
B
−5
In ammeter, if Ig = full scale deflection current, then equating potential drops across points marked AB, we have VAB = Ig G = (I0 − Ig )R A I G …(i) ⇒ RA = g I0 − Ig
(c) 30 . × 10
RV
Voltmeter
Ig A I0 – Ig
R
B
T
(b)1.5 × 10−5 T (d) 2.5 × 10−5 T
Exp. (b) The given figure can be drawn as shown below P θ1 θ2
Here, G = resistance of galvanometer coil. When a galvanometer is used as a voltmeter, a high resistance (R V ) in series is used.
G
B 6 cm
(a) 2.0 × 10−5 T
I0
m
RA
I0
5c
m
To use galvanometer as an ammeter, a low resistance in parallel is used.
5 cm
5 cm
I=5A A
3 cm
3 cm
B
By BiotSavart’s law, magnetic field due to a wire segment at point P is µ I B = 0 (sinθ1 + sinθ2 ) 4 πd
276
JEE Main Chapterwise Physics θ1 = θ2 = θ µ I B = 0 × 2sinθ 4 πd From given data, I = 5A, µ = 4 π × 10−7 NA −2 Here,
(say)
Then,
…(i)
23. An infinitely long currentcarrying wire and a small currentcarrying loop are in the plane of the paper as shown. The radius of the loop is a and distance of its centre from the wire is d (d >> a ). If the loop applies a force F on the wire, then
d=
52 − 32 = 16 = 4 cm 3 sinθ = 5 On substituting these values in Eq. (i), we get µ I B = 0 × 2sinθ 4 πd
= 1. 5 × 10−5 T
22. A bar magnet is demagnetised by inserting it inside a solenoid of length 0.2 m, 100 turns and carrying a current of 5.2 A. The coercivity of the bar magnet is [ JEE Main 2019, 9 Jan ShiftI]
(a) 1200 A/m (c) 2600A/m
d
−7
4 π × 10 × 5 3 × 2 × 5 4 π × 4 × 10−2 5×2 × 3 = × 10−5 4×5 =
[ JEE Main 2019, 9 Jan ShiftI]
(b) 285 A/m (d) 520A/m
Exp. (c) Coercivity of a bar magnet is the value of magnetic field intensity (H) that is needed to reduce magnetisation to zero. Since, for a solenoid magnetic induction is given as, ...(i) B = µ 0 nI where, n is the number of turns (N) per unit, length (l ) and I is the current. Also, ...(ii) B = µ 0H ∴From Eqs. (i) and (ii), we get µ 0 nI = µ 0 H N or H = nI = I l Substituting the given values, we get 100 . H= × 52 02 . = 2600 A/m Thus, the value of coercivity of the bar magnet is 2600 A/m.
a2 (a) F ∝ 3 d
(b) F = 0
a (c) F ∝ d
a (d) F ∝ d
2
Exp. (d) In the given condition, the currentcarrying loop is at a very large distance from the long currentcarrying conducting wire. Thus, it can be considered as a dipole (a magnet with north pole facing in upward direction and south in the downward direction). Suppose the effective length of this dipole be ‘l ’. Thus, the top view of the condition can be shown in the figure given below. F
θ θ Magnetic field of the wire
l
d
θ F
Now, the net force on the loop (i.e. at the two poles) due to the wire is given as, Fnet = 2 F cosθ = 2mBcos θ where, m is the pole strength. From the figure, we have l 2 cos θ = l2 d2 + 4
277
Magnetic Effect of Current ⇒
2 mB l
Fnet =
…(i) l2 2 d + 4 Since, the magnetic moment of a loop of radius r is ...(ii) M = IA = Iπr 2 = ml 2
and magnetic field due to a straight infinitely long currentcarrying conductor at a distance µ I′ ...(iii) x is B= 0 2 πx ∴Using Eqs. (ii) and (iii), rewriting Eq. (i), we get Iπa2 µ 0 I′ Fnet = 2 l2 2π d2 + 4 ⇒
Fnet ∝
a 2 l2 d + 4
a Fnet ∝ d
2
joined by two radial lines as shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to [JEE Main 2019, 9 Jan ShiftI]
m
m 3c
3c
θ=45
°
O
R
m
m
2c
2c
P
S i=10 A
B
I
⇒
R1 A,B
L
⇒
From the given figure as shown below. O 3 cm
θ = 45°
3 cm
Q
R
2 cm
2 cm
P
S
The magnetic field at point ‘O’ due to wires PQ and RS will be zero. Magnetic field due to arc QR at point ‘O’ will be θ µ 0i B1= 2 π 2a π Here, θ = 45° = rad, i = 10 A 4 and a = 3 cm = 3 × 10− 2 m ⇒
B1 =
π 2π × 4
=
(b) 1.0 × 10−7 T
(c) 1. 5 × 10−7 T
(d) 1. 5 × 10−5 T
µ 0 × 10 −2 2 × 3 × 10
µ0 × 5
=
2 × 12 × 10− 2
5 × µ 0 × 102 24
Direction of field B1 will be coming out of the plane of figure. Similarly, field at point ‘O’ due to arc SP will be µ 0 × 10 π 1 B2 = 4 2 π 2 × (2 + 3) × 10− 2 = =
(a) 1.0 × 10−5 T
Exp. (a)
I
i=10 A
24. A current loop, having two circular arcs
Q
A
2
l can be neglected as the loop is kept at very large distance. or
Key Idea When a point ‘P’ lies on the axial position of currentcarrying conductor, then magnetic field at P is always zero.
µ0 × 5
2 × 20 × 10− 2 µ0
−2
2 × 4 × 10
=
µ 0 × 102 8
Direction of B2 is going into the plane of the figure. ∴The resultant field at O is µ0 1 5 ×µ0 B = B1 − B2 = − 2 12 × 10− 2 4 × 10− 2
278
JEE Main Chapterwise Physics 1 5 µ 0 − 3µ 0 1 = = 2 12 × 10− 2 2 =
4 π × 10− 7 12 × 10− 2
2µ0 −2 12 × 10
≅ 1 × 10− 5 T
25. A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Take, charge of electron = 1.6 × 10−19 C) [JEE Main 2019, 9 Jan ShiftII]
(a) 1.6 × 10−19 kg −31
(c) 9.1 × 10
(b) 1.6 × 10−27 kg (d) 2.0 × 10−24 kg
kg
In velocity selector, Fm = Fe …(iii) ⇒ qvB = qE Initially particle moves under the magnetic field, So the radius of circular path taken by the particle is mv …(iv) R= qB From Eqs. (iii) and (iv), we get qB2 R m= E . × 0.5 × 10−2 1.6 × 10−19 × 025 m= 102 −24 m = 2 × 10 kg
26. One of the two identical conducting wires of
Exp. (d) According to given condition, when a particle having charge same as electron move in a magnetic field on circular path, then the force always acts towards the centre and perpendicular to the velocity.
length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the centre of the loop(B L ) to that B at the centre of the coil (BC ), i.e. L will be BC
v
[JEE Main 2019, 9 Jan ShiftI]
1 (a) N
F v
F
(b) N
(c)
v
B = 0.5 T R = radius of circular path = 5 cm Now, the magnetic force is Fm = q (v × B) = q vB sin90° …(i) Fm = qvB When the electric field applied, then the particle moves in a straight path, then this is the case of velocity selector.
Let consider the length of first wire is L, then according to question, if radius of loop formed is R1, then, For wire 1,
Here,
× –
+
+
+ × q ×
–
–
–
(d) N 2
Exp. (c)
F
+ ×
1 N2
+ × × –
Here, the electric force on charge, Fe = q E
+Q
A
B
R1 A,B
⇒
L 2π The magnetic field due to this loop at its centre is µ I µ I …(i) BL = 0 = 0 × 2 π 2 R1 2 L Now, several wire is made into a coil of N turns. L = 2 πR1 ⇒ R1 =
I L
…(ii)
I
⇒
L
A –Q
I
B
⇒
⇒ I
R2
(R2=radius of coil having N loops)
279
Magnetic Effect of Current L 2 πN The magnetic field due to this circular coil of Nturns is µ I µ I ⋅ (2 πN) …(ii) BC = 0 N = N ⋅ 0 2L 2 R2 Thin,
L = N(2 πR 2 ) ⇒ R 2 =
Using Eqs. (i) and (ii), the ratio of
BL is : BC
µ 0I BL 2 R1 = µ I BC N 0 2 R2 µ 0I ⋅ (2 π ) 1 = 2 = 2L µ 0I 2 N ⋅ (2 π )N 2L
⇒
27. An insulating thin rod of length l has a linear x on it. The rod is l rotated about an axis passing through the origin ( x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is [JEE Main 2019, 10 Jan ShiftI]
charge density ρ( x ) = ρ0
(a) n ρl 3 π (c) n ρ l 3
(b) π n ρ l 3 π (d) n ρ l 3 4
3
Key Idea A rotating charge constitutes a current. Hence, a rotating charged rod behaves like a current carrying coil. If charge q rotates with a frequency n, then equivalent current is I = qn and magnetic moment associated with this current is M = IA where, A = area of coil or area swept by rotating rod.
Let dq be the charge on dx length of rod at a distance x from origin as shown in the figure below. y
O
l
=
π n ρ0 x4 π n ρ0 l 3 π n ρ0 l 3 ⋅ ∫ x dx = = 0 l 4 l 4 0
at ∴
x=l ρ = ρ0 π M = nρl 3 4
28. A solid metal cube of edge length 2 cm is
Exp. (d)
x
The magnetic moment dm of this portion dx is given as dm = (dI) A [QI = qn, ∴dI = n ⋅ dq ] dm = ndqA = n ρ dx A x where, ρ = charge density of rod = ρ0 . l nρ0 x dx πx2 So, dm = l π n ρ0 3 = ⋅ x ⋅ dx l Total magnetic moment associated with rotating rod is sum of all the magnetic moments of such differentiable elements of rod. So, magnetic moment associated with complete rod is x=l l π nρ 0 ⋅ x3 dx M=∫ dm = ∫ x=0 0 l
dx x
moving in a positive Y direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive Z direction. The potential difference between the two faces of the cube perpendicular to the X axis is [JEE Main 2019, 10 Jan ShiftI]
(a) 2 mV (c) 6 mV
(b) 12 mV (d) 1 mV
Exp. (b) Potential difference between opposite faces of cube is V = induced emf = B l v where, B = magnetic field = 0.1 T, l = distance between opposite faces of cube = 2 cm = 2 × 10− 2 m and v = speed of cube = 6 ms − 1. Hence,
V = 01 . × 2 × 10− 2 × 6 = 12 mV
280
JEE Main Chapterwise Physics
29. A hoop and a solid cylinder of same mass
30. In an experiment, electrons are accelerated,
and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then
from rest by applying a voltage of 500 V. Calculate the radius of the path, if a magnetic field 100 mT is then applied. (Take, charge of the electron = 1.6 × 10−19 C and mass of the electron = 9.1 × 10−31 kg)
[JEE Main 2019, 10 Jan ShiftII]
(a) Th = 0.5 Tc (c) Th = 2 Tc
(b) Th = Tc (d) Th =1.5 Tc
The timeperiod of oscillations made by a magnet of magnetic moment M, moment of inertia I, placed in a magnetic field is given by I … (i) MB For the hoop, let us assume its moment of inertia Ih and magnetic moment M h , then its time period will be Ih … (ii) Th = 2 π Mh B T = 2π
Similarly, for solid cylinder, time period is,
(d) 7.5 m
During the circular motion of accelerated electron in the presence of magnetic field, its radius is given by mv 2 meV r= = Be eB where, v is velocity and V is voltage. After substituting the given values, we get =
. × 10− 31 × 16 . × 10− 19 × 500 2 × 91 . × 10− 19 × 100 × 10− 3 16 = 10
… (iii)
31. In the circuit shown,
Ih Mc M h Ic
R
S2
and we know that, moment of inertia of hoop Ih = mR 2 and moment of inertia of solid 1 mR 2 2 Substituting these values in Eq. (iv), we get cylinder Ic =
⇒
Th = Tc
mR 2 × Mc 1 mR 2 × 2 Mc 2
L
… (iv)
Now, it is given that, M h = 2 Mc
Th = Tc
1/ 2
2 × 91 . × 500 × 10− 12 1.6
r = 7.5 × 10− 4 m
Dividing Eq. (ii) by Eq. (iii), we get Th = Tc
(b) 7.5 × 10−4 m (c) 7.5 × 10−3 m
Exp. (b)
Exp. (b)
Ic Tc = 2 π Mc B
[JEE Main 2019, 11 Jan ShiftI]
(a) 7.5 × 10−2 m
=1
S1 ε
The switch S1 is closed at time t = 0 and the switch S 2 is kept open. At some later time (t 0 ), the switch S1 is opened and S 2 is closed. The behaviour of the current I as a function of time ‘t ’ is given by
281
Magnetic Effect of Current I
Given, side of cube = 1cm = 10−2 m
I
(a)
∴Volume, V = 10−6m3 Dipole moment, M = 20 × 10−6 J/T
(b) to
t
I
Applied magnetic intensity, H = 60 × 103 A/m
to
t
to
t
I
(c)
(d) to
t
[JEE Main 2019, 11 Jan ShiftI]
Exp. (b) Initially in the given RL circuit with a source, when S1 is closed and S 2 is open at t ≤ t 0 . − R V 1 − exp t I1 = L R In this case, inductor L is charging. When switch S 2 is closed and S1 is open (after t > t 0 ), the inductor will be discharged through resistor. In this case (t > t 0 ), V R I2 = exp − (t − t 0 ) L R Thus, the variation of I with t approximately is shown below
t0
t
32. A paramagnetic substance in the form of a cube with sides 1 cm has a magnetic dipole moment of 20 × 10− 6 J/ T when a magnetic intensity of 60 × 103 A / m is applied. Its magnetic susceptibility is [JEE Main 2019, 11 Jan ShiftII]
(a) 3.3 × 10− 4
(b) 3.3 × 10− 2
−2
(d) 2.3 × 10− 2
Exp. (a)
33. A particle of mass m and charge q is in an electric and magnetic field is given by $. E = 2 i$ + 3$j, B = 4$j + 6k The charged particle is shifted from the origin to the point P ( x = 1; y = 1) along a straight path. The magnitude of the total work done is [JEE Main 2019, 11 Jan ShiftII] (a) (0.35) q (c) ( 2.5) q
(b) (015 . )q (d) 5q
Exp. (d) Here,
I
(c) 4.3 × 10
Intensity of magnetisation M 20 × 10−6 I= = = 20 A/m V 10–6 Now, magnetic susceptibility χ is I Intensity of magnetisation = χ= Applied magnetic intensity H 20 = 60 × 103 1 χ = × 10−3 = 3.3 × 10−4 ⇒ 3
$ E = 2 $i + 3$j, B = 4$j + 6k, q = charge on a particle. Initial position, r1 = (0,0) Final position, r2 =( 1, 1) Net force experienced by charge particle inside electromagnetic field is Fnet = qE + q(v × B) = q(2 $i + 3$j ) [here, v × B = 0] = (2q $i + 3q $j ) ∴ dW = Fnet .dr ⇒
r2
∫ dW = ∫ (2q $i + 3q $j )⋅(dx $i + dy $j ) r1
[here dr = dx $i + dy $j] 1
1
0
0
⇒
W = 2q ∫ dx + 3q ∫ dy
or
W = 2 q + 3q or W = 5q
282
JEE Main Chapterwise Physics
34. The region between y = 0 and y = d contains $ . A particle of mass m a magnetic field B = Bk and charge q enters the region with a mv velocity v =v$i. If d = then the 2qB acceleration of the charged particle at the point of its emergence at the other side is [JEE Main 2019, 11 Jan ShiftII]
qvB 3 $ 1 $ i + j m 2 2 $ $ qvB − j + i (c) m 2
(a)
1 $ 3 i− 2 2 qvB $i + $j (d) m 2
(b)
qvB m
$j
Exp. (*) Situation given in question is shown below;
y=d
Bk
90–θ θ
r–d C
vxf
P
O θ vyf
θ
y=0 xaxis
vf 90–θ
Centre of circular path
So, acceleration of charged particle at the point its emergence is; 1 3 $ j − v $i + 2 2 ∆v Acceleration a = = 3 m ∆t 2 Bq − Bqv $i = + $j ms −2 m 3
identical wires are bent by 90° and placed in such a way that the segments LP andQM are along the X axis, while segments PS andQN are parallel to the Y axis. If OP = OQ = 4 cm and the magnitude of the magnetic field atO is 10−4 T and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at will be O (Take, µ 0 = 4π × 10−7 NA −2 ) [JEE Main 2019, 12 Jan ShiftI]
Path taken by particle of charge ‘q’ and mass ‘m’ is a circle of radius r where, mv r= Bq L
= vcos 60 i$ − v sin 60° $j 1 3 $ = v i$ − j 2 2 So, change of velocity of charged particle is ∆v = vf − vi 1 3 $ j − v$i = v $i − 2 2
y
S
Here final velocity vf = v x $i + vyf (− $j)
1 3 $ j = − v $i + 2 2
3 mv × 3 2 = 3m 2 = Bq v 2 Bq v
35. As shown in the figure, two infinitely long,
Path of charged particle
d
=
r×
None of the option given matches with answer.
yaxis v1
It t = time taken by charged particle to cross region of magnetic field then, distance OP t = speed in direction OP
P
O
Q M
x
N
(a) 40 A, perpendicular out of the page (b) 20 A, perpendicular into the page (c) 20 A, perpendicular out of the page (d) 40 A, perpendicular into the page
Exp. (b) There is no magnetic field along axis of a currentcarrying wire. Also, magnetic field near one of end of an µ I infinitely long wire is 0 tesla. 4 πr
283
Magnetic Effect of Current Hence, magnetic field due to segments LP and MQ at ‘O’ is zero. Using right hand rule, we can check that magnetic field due to segments PS and QN at ‘O’ is in same direction perpendicularly into the plane of paper. Hence, BO = BPS + BQN µ i µ i µ i = 0 + 0 = 0 4 πr 4 πr 2 πr 2 πrB0 So, i = µ0 Here, r = OP = OQ = 4 cm and B0 = 10−4 T. Substituting values, we get 2 π × 4 × 10−2 × 10−4 i = ⇒ 4 π × 10−7 ⇒ i = 20A, Also, magnetic field points perpendicular into the plane of paper.
36. A proton and an αparticle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform magnetic field B is set up perpendicular to their velocities, the ratio of the radiirp : rα of the circular paths described by them will be [JEE Main 2019, 12 Jan ShiftI]
(a)1 : 2
(b)1 : 3
(c)1 : 3
(d)1 : 2
Exp. (a) Radius of path of charged particle q in a uniform magnetic field B of mass ‘m’ moving with velocity v is mv m (2qV / m) r= = Bq Bq m so, required ratio is ⇒ r∝ q ⇒
rp rα
=
mp
=
1 2 2 1 × = = 4 1 2 2
qα × mα qp
Exp. (d) From Curie’s law for paramagnetic substance, we have 1 Magnetic susceptiblity χ ∝ T χ .T χ 2 T1 = ⇒ χ2 = 1 1 ∴ χ1 T2 T2 ⇒ χ2 =
2.8 × 10−4 × 350 = 3.267 × 10−4 300
38. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re , rp , rα respectively, in a uniform magnetic field B. The relation between re , rp , rα is [JEE Main 2018]
(b) re < rp = rα (d) re < rα < rp
(a) re > rp = rα (c) re < rp < rα
Exp. (b) mv 2 , we have r 2 mK mv = r= Bq Bq where, K is the kinetic energy. As, kinetic energies of particles are same; m r∝ q From Bqv =
⇒
re : rp : rα =
me
:
mp
e e Clearly, rp = rα and re is least So, rp = rα > re
:
(a) 3.726 × 10− 4 (c) 2.672 × 10− 4
(b) 3672 . × 10− 4 (d) 3.267 × 10− 4
2e [Q me < mp ]
39. The dipole moment of a circular loop carrying a current I , is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the B centre of the loop is B 2. The ratio 1 is B2
37. A paramagnetic material has1028 atoms/m3 . Its magnetic susceptibility at temperature 350 K is 2.8 × 10− 4 . Its susceptibility at 300 K is [JEE Main 2019, 12 Jan ShiftII]
4mp
[JEE Main 2018]
(a) 2
Exp. (c)
(b) 3
(c) 2
(d)
1 2
284
JEE Main Chapterwise Physics
/ As m = IA , so to change dipole moment (current is kept constant), we have to change radius of loop. Initially,
m = IπR 2 and B1 =
Finally,
m′ = 2 m = IπR 22
µ 0I 2 R1
⇒ So,
=
IπR 22
µ 0I 2 R1
B Hence, ratio 1 = = 2 B2 µ 0 I 2 2 R1 ∴
Ratio
Time period of oscillation is I T = 2π MB ⇒
or R 2 = 2 R1 µ 0I µ 0I = B2 = 2(R 2 ) 2 2 R1
2 IπR12
Exp. (d)
B1 = 2 B2
7.5 × 10−6
T = 2π
. × 10−2 × 0.01 67
Hence, time for 10 oscillations is t = 6.65 s.
42. Two identical wires A and B, each of length l, carry the same current I . Wire A is bent into a circle of radius R and wire B is bent to form a square of side a. If B A and B B are the values of magnetic field at the centres of the circle B and square respectively, then the ratio A is BB [JEE Main 2016 (Offline)]
40. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 010 V is [JEE Main 2017 (Offline)] (a) 2.045 × 103 Ω
(b) 2.535 × 103 Ω
(c) 4005 . × 103 Ω
(d)1.985 × 103 Ω
π2 (a) 8
π2 (b) 16 2
(c)
O I R
⇒ R=
µ0 I 2R As, 2 πR = l, (where l is length of a wire) l R= 2π µ0 I µ Iπ BA = = 0 ⇒ l l 2× 2π BA =
V −G Ig
R = 1985 = 1985 . kΩ R = 1985 . × 103 Ω
41. A magnetic needle of magnetic moment
6.7 × 10−2 Am 2 and moment of inertia 7.5 × 10−6 kg m 2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is [JEE Main 2017 (Offline)] (a) 8.89 s (c) 8.76 s
π2 8 2
Rs G
V
⇒ or
(d)
Magnetic field in case of circle of radius R, we have
Suppose a resistance R s is connected in series with galvanometer to convert it into voltmeter.
Ig (G + R s ) = V
π2 16
Exp. (d)
BA
Exp. (d)
Ig
= 0.665 s
(b) 6.98 s (d) 6.65 s
...(i)
BB a 45º 45º
Magnetic field in case of square of side a, we get µ I 1 1 BB = 4 × 0 × + a 2 2 4π 2
285
Magnetic Effect of Current µ 2 2I 4Iµ 0 = 0 aπ πa 2 8 2 µ0 I l As, ... (ii) 4a = l, a = ⇒ BB = πl 4 Dividing Eq. (i) by Eq. (ii), we get BA π2 = BB 8 2 ⇒
BB =
Exp. (a) Maximum voltage that can be applied across the galvanometer coil = 100 Ω × 10−3 A = 01 . V. 1mA
Ammeter
100 Ω
Rs
43. Hysteresis loops for two magnetic materials A and B are as given below B
If R s is the shunt resistance, then R s × 10 A = 01 . V ⇒ R s = 0.01 Ω
B
H
H
(A)
(B)
These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then, it is proper to use [JEE Main 2016 (Offline)]
(a) A for electric generators and transformers (b) A for electromagnets and B for electric generators (c) A for transformers and B for electric generators (d) B for electromagnets and transformers
Exp. (d) / Area of hysteresis loop is proportional to the net energy absorbed per unit volume by the material, as it is taken over a complete cycle of magnetisation. For electromagnets and transformers, energy loss should be low. i.e. thin hysteresis curves. Also,B → 0 when H = 0 andH should be small when B → 0.
44. A galvanometer having a coil resistance of
100 Ω gives a full scale deflection when a current of 1 mA is passed through it. The value of the resistance which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is [JEE Main 2016 (Offline)]
(a) 0.01 Ω
(b) 2 Ω
(c) 0.1 Ω
(d) 3 Ω
45. Two coaxial solenoids of different radii carry current I in the same direction. Let F1 be the magnetic force on the inner solenoid due to the outer one and F2 be the magnetic force on the outer solenoid due to the inner one. Then, [JEE Main 2015] (a) F1 = F2 = 0 (b) F1 is radially inwards and F2 is radially outwards (c) F1 is radially inwards and F2 = 0 (d) F1 is radially outwards and F2 = 0
Exp. (a) Consider the two coaxial solenoids. Due to one of the solenoids magnetic field at the centre of the other can be r2 assumed to be constant. Due to symmetry, forces on upper and lower part of a solenoid will be equal and opposite and hence resultant is zero. Therefore, F1 = F2 = 0
I I r1
46. Two long current carrying thin wires, both with current I, are held by L insulating threads of θ length L and are in equilibrium as shown in the figure, with threads making an angleθ with the I I vertical. If wires have mass λ per unit length then, the value of I is (g = gravitational acceleration) [JEE Main 2015]
286
JEE Main Chapterwise Physics πλgL µ 0 cos θ
(a) sin θ
πgL tan θ µ0
(c) 2
πλgL µ 0 cos θ
(b) 2 sin θ
If there is a uniform magnetic field of 0.3 T in the positive zdirection in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?
πλgL tan θ µ0
(d)
[JEE Main 2015]
Exp. (b) Consider free body diagram of the wire. As the wires are in equilibrium, they must carry current in opposite direction.
(a) (a) and (b) respectively (b) (a) and (c) respectively (c) (b) and (d) respectively (d) (b) and (c) respectively
Exp. (c) θ
θ
T cos θ
Since B is uniform only torque acts on a current carrying loop. As, τ = M × B ⇒τ = MB sin θ
T θ
FB
T sin θ
d
For orientation shown in (b) θ = 00 , τ = 0 (stable equilibrium) and for (d) θ = π, τ = 0 (unstable equilibrium)
mg= λlg
µ 0I l , where l is length of each wire 2 πd and d is separation between wires. From figure, d = 2 L sinθ T = cosθ = mg = λlg (in vertical direction)…(i) µ 0 I2 l T sinθ = FB = 4 πL sinθ (in horizontal direction)…(ii) From Eqs. (i) and (ii), µ 0 I2 l T sinθ = T cos θ 4 πL sinθ × λ lg 2
Here, FB =
I=
∴
4 πλLg sin2 θ πλLg = 2 sinθ µ 0 cos θ µ 0 cos θ
z I
B
I
B
(b)
I
I
x
B I
(c) I x
y
B
(d)
I
I
y I
I I z
x
z I
I
I
y
x
I
I
required to move the conductor at constant speed to x = 2.0 m, y = − 0 in 5 × 10−3 s. Assume parallel motion along the xaxis. [JEE Main 2014] z I B 1.5
2.0
carrying a current I of 12 A is placed in different orientations as shown in the figures below. [JEE Main 2015]
(a)
− 1.5 ≤ z < 1.5 m and carries a fixed current of 10.0 A in − a z direction (see figure). For a field B = 3.0 × 10−4 e − 0.2 x a y T, find the power
y
x
47. A rectangular loop of sides 10 cm and 5 cm
z
48. A conductor lies along the zaxis at
y
(a) 1.57 W
(b) 2.97 W (c) 14.85 W (d) 29.7 W
Exp. (b) Force exerted on a current carrying conductor, Fext = B( x) IL Work done Average power = Time taken 1 2 1 2 P = ∫ Fext ⋅ dx = ∫ B( x) ILdx t 0 t 0 2 1 3 × 10−4 e −0. 2 x × 10 × 3dx = 5 × 10−3 ∫ 0 = 9 [1 − e −0. 4 ] 1 = 9 1 − 0. 4 = 2.967 ≈ 2.97 W e
287
Magnetic Effect of Current 49. The coercivity of a small magnet where the
ferromagnet gets demagnetised is 3 × 10 Am −1. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetised when inside the solenoid is [JEE Main 2014]
(a) 30 mA (b) 60 mA (c) 3 A
(d) 6 A
Exp. (c) For solenoid, the magnetic field needed to be magnetised, the magnet is given by B = µ 0 nI where, n is number of turns per unit length = N/ l 10 m = 0.1m and N = 100, l = 10 cm = 100 100 × I ⇒ I = 3A 3 × 103 = ⇒ 0.1
50. Two short bar magnets of length 1 cm
[JEE Main 2013]
(a) 3.6 × 10−5 Wb/m 2 (b) 2.56 × 10−4 Wb/m 2 (c) 3.50 × 10−4 Wb/m 2 (d) 5.80 × 10−4 Wb/m 2
Exp. (b) Net magnetic field, Bnet = B1 + B2 + BH N
N
O
= 2.56 × 10−4 Wb/m 2
51. A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is [JEE Main 2013] (a) 9.2 × 10−11 Wb (c) 3.3 × 10−11 Wb
(b) 6 × 10−11 Wb (d) 6.6 × 10−9 Wb
Exp. (a) The magnetic field due to the bigger circular loop is µ 0 IR 2 B= 2 2 ( x + R 2 )3 / 2
3 2A 15 cm
Given,
I = 2 A, R = 20 cm = 20 × 10−2 m, x = 15 cm = 15 × 10−2 m B=
µ 0 × 2 × (20 × 10−2 )2
[(0.2 )2 + (015 . )2 ]3 / 2 Flux linked, φ = BA where, A is area of small circular loop. A = πr 2 = π × (0.3 × 10−2 )2 ∴
φ=
µ 0 × 2 × (20 × 10−2 )2 2 [(0.2 )2 + (015 . )2 ]3 / 2
× π × (0.3 × 10−2 )2
52. Proton, deuteron and alpha particles of S N
S
Bnet
+ 3.6 × 10−5
φ = 9.2 × 10−11 Wb BH B1 B2
S
(0.1)3
20 cm
each have magnetic moments 1.20 Am 2 and 1.00 Am 2, respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the midpointO of the line joining their centres is close to (Horizontal component of the earth’s magnetic induction is 3.6 × 10−5 Wb/m 2)
10−7 (1.2 + 1)
=
3
µ (M1 + M 2 ) = 0 + BH 4π r3
same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively rp , rd and rα . Which one of the following relation is correct? [AIEEE 2012] (a) rα = rp = rd (c) rα > rd > rp
Exp. (b)
(b) rα = rp < rd (d) rα = rd > rp
288
JEE Main Chapterwise Physics For charged particle in magnetic field, radius 2 Km mv = r= qB qB [as mv = p ⇒ K =
r∝
⇒
and ⇒
1 m v mv 2 = = p2 /2 m 2 2m p = 2 Km ]
m q
rp : rd : rα =
mp qp
:
2 mp qp
:
4mp 2q p
= 1 : 2 :1 rα = rp < rd
⇒
r
O
Q ⋅ 2 πr dr µ0 µ dq µ 0 πR 2 = ⋅ dI = 0 ⋅ 2π 2r 2r T 2r ω µ 0Qω = dr 2 πR 2 So, net magnetic field at the centre of disk, R µ ωQ R µ ωQ B = ∫ dB = 0 2 ∫ dr = 0 0 2 πR 2 πR 0 1 i.e., B∝ R dB =
md = 2 mp qd = q p ; mα = 4mp q α = 2q p
or and
dr
R
2 2
53. A charge Q is uniformly distributed over the
54. A horizontal straight wire 20 m long
surface of nonconducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity ω. As a result of this rotation, a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc, then the variation of the magnetic induction at the centre of the disc will be represented by the figure [AIEEE 2012]
extending from East to West is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth’s magnetic field 0.30 × 10−4 Wb/m 2. The instantaneous value of the emf induced in the wire will be [AIEEE 2011] (a) 6.0 mV (b) 3 mV
(c) 4.5 mV (d) 1.5 mV
Exp. (b) Induced emf, e = BH l v = 0.30 × 10−4 × 20 × 5.0 = 3mV
55. A thin circular disk of radius R is uniformly (a)
B
(b)
charged with density σ > 0 per unit area. The disk rotates about its axis with a uniform angular speed ω. The magnetic moment of the disk is [AIEEE 2011]
B
R
R
(a) 2 πR 4 σω (b) πR 4 σω (c)
πR 4 πR 4 σω (d) σω 2 4
Exp. (d) (c)
B
(d)
R
B
Let us consider the disc to be made up of large number of concentric elementary rings. R
x
Exp. (a) Taking an elemental ring of radius r and thickness dr, we obtain magnetic field at the centre of the ring,
dx
289
Magnetic Effect of Current Consider one such ring of radius x and thickness dx. Charge on this elementary ring, dq = σ × 2 πxdx = 2 πσxdx Current associated with this elementary ring, dq = dq × f = σωxdx dI = dt [Qf is frequency and ω = 2 πf ] Magnetic moment of this elementary ring, dM = dI πx2 = πσω x3dx ∴ Magnetic moment of the entire disc, R R 1 M = ∫ d M = πσω∫ x3 dx = πR 4 σω 0 0 4
dB =
µ 0 2dI µ 0I ⋅ = dθ 4π R 2 π2R
Net magnetic field at the centre, π/2 µ 0I B= ∫ dBcos θ = −π / 2 2 π2R π/2 µ I ∫ − π / 2 cosθ d θ = π 20R
58. Two long parallel wires are at a distance 2d apart. They carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field along the line xx ′ is given by [AIEEE 2010]
56. A boat is moving due East in a region where
the earth’s magnetic field is 5.0 × 10−5 NA −1 m −1 due North and horizontal. The boat carries a vertical aerial 2m long. If the speed of the boat is 1.50 ms −1, the magnitude of the induced emf in the wire of aerial is [AIEEE 2011]
B
(a) x
x′ d
d
B
(a) 0.75 mV (b) 0.50 mV (c) 0.15 mV (d) 1 mV
Exp. (c) Induced emf, e = BH l v = 5.0 × 10−5 × 2 × 150 .
(b) x
x′
= 015 . × 10−3 V = 015 . mV d
57. A current I flows in an infinitely long wire with crosssection in the form of a semicircular ring of radius R. The magnitude of the magnetic induction along its axis is [AIEEE 2011] (a)
µ 0I
2π R 2
(b)
µ 0I 2 πR
(c)
µ 0I 4 πR
(d)
d B
(c) x
x′
µ 0I
d
π2R
d
B
Exp. (d)
θ
θ
dθ
Consider the wire to be made up of large number of thin wires of infinite length. Consider such wire of thickness d l subtending an angle dθ at centre. Current through this wire, dθ dI = l, π ∴ Magnetic field at centre due to this portion,
θ θ
dB
(d) x
x′ d
Exp. (a) The magnetic field in between because of each will be in opposite direction. µ i µ 0i ( − j) ∴ Bin between = 0 j − 2 πx 2 π (2d − x) =
dB
d
µ 0i 2π
1 1 x − 2d −
x
( j)
290
JEE Main Chapterwise Physics At x = d , Bin between = 0 For x < d , Bin between = (j ) For x > d , Bin between = (− j) Towards x, net magnetic field will add up and direction will be (−j ). Towards x′, net magnetic field will add up and direction will be (j ).
59. The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is [AIEEE 2009] B a
A
Exp. (b) The forces on AD and BC are zero because magnetic field due to a straight wire on AD and BC is parallel to elementary length of the loop.
61. Relative permittivity and permeability of a
material are εr andµ r , respectively. Which of the following values of these quantities are allowed for a diamagnetic material? [AIEEE 2008]
I1 O
(a) εr = 0.5, µ r = 1.5 (c) εr = 0.5, µ r = 0.5
I
30°
b
Exp. (b)
µ I(b − a ) (b) 0 24ab µ 0I π (d) 2( b − a ) + (a + b ) 4 π 3
(a) zero µ 0I b − a 4 π ab
(b) εr = 1.5, µ r = 0.5 (d) εr = 1.5, µ r = 1.5
D
C
(c)
(d) the magnitude of the net force on the loop is µ II given by 0 1 ( b − a ) 24ab
Exp. (b) Net magnetic field due to loop ABCD at O is B = BAB + BBC + BCD + BOA µ I π µ I π = 0+ 0 × + 0− 0 × πb 6 4 πa 6 µ 0I µ 0I µ 0I = − = ( b − a) 24a 24b 24ab
For diamagnetic material, 0 < µ r < 1 and for any material, εr > 1.
62. A horizontal overhead powerline is at a height of 4 m from the ground and carries a current of 100 A from East to West. The magnetic field directly below it on the ground is (take µ 0 = 4π × 10−7 T mA −1 ) [AIEEE 2008] (a) 2.5 × 10−7 T, Southward (b) 5 × 10−6 T, Northward (c) 5 × 10−6 T, Southward (d) 2.5 × 10−7 T, Northward
Exp. (c) B=
60. Due to the presence of the current I 1 at the origin in the figure,
[AIEEE 2009]
W
µ 0I 2 πR E
B a I1 O
A I
30°
b
Direction is given by right hand palm rule No. 1. B=
D C
2 × 10−7 × 100 T, Southward 4
= 5 × 10−6 T, Southward
(a) the forces on AB and DC are zero
63. Two identical conducting wires AOB and
(b) the forces on AD and BC are zero
COD are placed at right angles to each other. The wire AOB carries an electric current I 1 and COD carries a currentI 2.The magnetic field on a point lying at a distance d
(c) the magnitude of the net force on the loop is µ II π given by 0 1 2 ( b − a ) + (a + b ) 4 π 3
291
Magnetic Effect of Current from O , in a direction perpendicular to the plane of the wires AOB and COD , will be given by [AIEEE 2007] 1/ 2
I + I2 µ (a) 0 1 2π d µ0 (c) ( I1 + I 2 ) 2 πd
µ0 (b) ( I 12 + I 22 )1/ 2 2 πd µ0 (d) ( I 12 + I 22 ) 2 πd
65. A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields E and B with a velocity v perpendicular to both E and B and comes out without any change in magnitude or direction of v . Then, [AIEEE 2007]
Exp. (b) y x
For r > a, B × 2 πr = µ 0 I µ I B= 0 2 πr µ I At r = 2 a, B2 = 0 4 πa B1 =1 ∴ B2 ⇒
dx x
O
(a) v = E × (b) v = B ×
The magnetic field inductions at a point P, at a distance d from O in a direction perpendicular to the plane of the wires due to currents through AOB and COD are perpendicular to each other, is 1/ 2
B=
B12
+
B22
steady current I . The current is uniformly distributed across its crosssection. The a ratio of the magnetic field at and 2 a is 2 [AIEEE 2007]
(b) 4
(c) 1
(d) 1/2
I
From Ampere’s circuital law,
∫ B ⋅ dl = µ 0 ⋅ Ienclosed
r
For r < a, ⇒ At r =
a , 2
E2
As v of charged particle is remaining constant, it means force acting on charged particle is zero. f = q E + q ( v × B) = 0 q E − q ( v × B) = 0 So, q ( v × B) = q E ⇒ v ×B = E E ×B v= ⇒ B2
66. A current I flows along the length of an infinitely long, straight, thin walled pipe. Then, [AIEEE 2007]
(b) the magnetic field is different at different points inside the pipe
πa2
B × 2 πr = µ 0 × J × πr µ I r B = 02 × 2 πa µ I B1 = 0 4 πa
E2 E
(a) the magnetic field is zero only on the axis of the pipe
Exp. (c) Current density, J =
(d) v = B ×
B2 B
Exp. (a)
2 µ 2I 2 µ 2I = 0 1 + 0 2 4π d 4 π d µ0 2 2 (I1 + I2 ) = 2 πd
64. A long straight wire of radius a carries a
(a) 1/4
(c) v = E ×
B B2 E
2
(c) the magnetic field at any point inside the pipe is zero (d) the magnetic field at all points inside the pipe is the same but not zero
Exp. (c) a
By using Ampere’s circuital law, the magnetic field at any point inside the pipe is zero.
292
JEE Main Chapterwise Physics y
67. A charged particle moves through a magnetic field direction. Then,
perpendicular
to
its
[AIEEE 2007]
(a) the momentum changes but the kinetic energy is constant (b) both momentum and kinetic energy of the particle are not constant (c) both momentum and kinetic energy of the particle are constant (d) kinetic energy changes but the momentum is constant
B
x
E z
(a) helix (c) ellipse
(b) straight line (d) circle
Exp. (b)
and by workenergy theorem, W = ∆KE, the kinetic energy and hence speed v remains constant. But v changes, so momentum changes.
Let E and B be along xaxis. When a charged particle is released from rest, it will experience an electric force along the direction of electric field or opposite to the direction of electric field depending on the nature of charge. Due to this force, it acquires some velocity along xaxis. Due to this motion of charge, magnetic force cannot have nonzero value because angle between v and B would be either 0° or 180°.
68. Needles N 1 , N 2 and N 3 are made of a
So, only electric force is acting on particle and hence it will move along a straight line.
Exp. (a) In case of motion of a charged particle perpendicular to the motion, i.e., displacement, the work done [as θ = 90°] W = ∫ F ⋅ ds = ∫ Fds cos θ = 0
ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will [AIEEE 2006]
70. A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 V, the resistance in Ohms needed to be connected in series with the coil will be [AIEEE 2005]
(a) attract N 1 and N 2 strongly but repel N 3 (b) attract N 1 strongly, N 2 weakly and repel N 3 weakly (c) attract N 1 strongly, but repel N 2 and N 3 weakly (d) attract all three of them
(a) 103
Exp. (b)
Exp. (d)
Ferromagnetic substances have strong tendency to get magnetised (induced magnetic moment) in the same direction as that of applied magnetic field, so magnet attracts N1 strongly. Paramagnetic substances get weakly magnetised (magnetic moment induced is small) in the same direction as that of applied magnetic field, so magnet attracts N2 weakly. Diamagnetic substances also get weakly magnetised when placed in an external magnetic field but in opposite direction and hence N3 is weakly repelled by magnet.
69. In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other. A charged particle is released from rest in this region. The path of the particle will be a [AIEEE 2006]
(b) 105
(c) 99995
(d) 9995
150 mA = 15 mA 10 150 Full scale deflection voltage = mV = 75 mV 2 75 mV Galvanometer resistance, G = =5Ω 15 mA
Full scale deflection current =
Required full scale deflection voltage, V = 1 × 150 = 150 V Ig
R
G V
Let resistance to be connected in series be R. ⇒
V = Ig (R + G )
∴
150 = 15 × 10−3 (R + 5)
or
104 = R + 5
or
R = 10000 − 5 = 9995
293
Magnetic Effect of Current = (4 × 10−5 )2 + (3 × 10−5 )2
71. Two thin, long, parallel wires, separated by a distance d carry a current of I ampere in the same direction. They will [AIEEE 2005] (a) attract each other with a force of (b) repel each other with a force of
µ0 I 2 ( 2 πd )
(c) attract each other with a force of (d) repel each other with a force of
µ0 I 2 ( 2 πd )
µ0 I 2 ( 2 πd 2 ) µ0 I 2
( 2 πd 2 )
Exp. (a) The force per unit length between the two wires is F µ 0 2 I2 µ 0 I2 = = ⋅ 2 πd 4π d l The force will be attractive as current directions in both are same.
72. Two concentric coils each of radius equal to 2 π cm are placed at right angles to each other. 3 A and 4A are the currents flowing in each coil respectively. The magnetic induction in Wb/m 2 at the centre of the coils will be (µ 0 = 4π × 10−7 Wb/Am) [AIEEE 2005] (a) 12 × 10−5
(b) 10−5
(c) 5 × 10−5
(d) 7 × 10−5
= 5 × 10−5 Wb /m2
73. A magnetic needle is kept in a nonuniform magnetic field. It experiences (a) (b) (c) (d)
[AIEEE 2005]
a torque but not a force neither a force nor a torque a force and a torque a force but not a torque
Exp. (c) Magnetic needle is placed in nonuniform magnetic field. It experiences force and torque both due to unequal forces acting on poles.
74. A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B. The time taken by the particle to complete one revolution is [AIEEE 2005] (a)
2 πmq 2 πq 2 B 2 πqB (b) (c) B m m
(d)
2 πm qB
Exp. (d) Magnetic force, F = qvB
[Qθ = 90°] …(i) v
Exp. (c) Magnetic field or magnetic induction at P, µ I 4 π × 10−7 × 4 BP = 0 2 = = 4 × 10−5 Wb/m 2 2R 2 × 0.02 π BP B
I1 = 3 A BQ
P I2 = 4 A
µ I 4 π × 10−7 × 3 BQ = 0 1 = 2 × 0.02 π 2R = 3 × 10−5 Wb/m 2
∴
B=
BP2
…(ii)
The time taken by the particle to complete one revolution, 2 πr 2 πmv 2 πm T= = = v vqB qB
75. A current I ampere flows along an infinitely
Q
and
mv 2 r From Eqs. (i) and (ii), we have mv mv 2 = qvB or r = qB r
Centripetal force, F =
+
BQ2
long straight thin walled tube, then the magnetic induction at any point inside the tube is [AIEEE 2004] (a) infinite µ 2I (c) 0 ⋅ T 4π r
(b) zero 2I (d) T r
294
JEE Main Chapterwise Physics
Exp. (b)
77. The magnetic field due to a current carrying
Let R be the radius of a long thin cylindrical shell. To calculate the magnetic induction at a distance r, (r < R ) from the axis of cylinder, a circular shell of radius r is shown.
circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 µT. What will be its value at the centre of the loop? [AIEEE 2004] (a) 250 µT (c) 125 µT
Exp. (a)
r R
Since, no current is enclosed in the circle, so from Ampere’s circuital law, magnetic induction is zero at every point of circle. Hence, the magnetic induction at very point inside the infinitely long straight thin walled tube (cylindrical) is zero.
76. A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be [AIEEE 2004]
(a) nB (c) 2nB
(b) 150 µT (d) 75 µT
(b) n 2 B (d) 2 n 2 B
Exp. (b) The magnetic field at the centre of circular coil is µ I B= 0 2r l [Ql = 2 πr] where, r = radius of circle = 2π µ I 2 π µ 0 Iπ …(i) B= 0 × = ∴ l 2 l When wire of length l bents into a circular loops of n turns, then l = n × 2 πr ′ l ⇒ r′ = n × 2π Thus, new magnetic field, µ nI B′ = 0 2r′ µ 0 nI n × 2 π = × 2 l µ 0 Iπ 2 = × n = n2 B l [from Eq. (i)]
The magnetic field at a point on the axis of a circular loop at a distance x from the centre is µ 0 IR 2 …(i) B= 2(R 2 + x2 )3 / 2 Given, B = 54 µT, x = 4 cm, R = 3 cm Putting the given values in Eq. (i), we get µ 0 I × (3)2 54 = 2 (32 + 42 )3 / 2 9µ 0 I 9µ 0 I 54 = = ⇒ 2 (25)3 / 2 2 × (5)3 54 × 2 × 125 µ0 I = ∴ 9 …(ii) ⇒ µ 0 I = 1500 µ Tcm Now, putting x = 0 in Eq. (i), magnetic field at the centre of loop is µ IR 2 B= 0 3 2R µ 0 I 1500 [from Eq. (ii)] = = 2R 2 × 3 = 250 µT
78. Two long conductors, separated by a distance d carry currents I 1 and I 2 in the same direction. They exert a force F on each other. Now, the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d.The new value of the force between them is [AIEEE 2004] (a) −2 F (c) −
2F 3
(b)
F 3
(d) −
F 3
Exp. (c) Force acting between two current carrying conductors µ II …(i) F = 0 12 l 2π d
295
Magnetic Effect of Current where d = distance between the conductors l = length of small conductor µ (−2 I1 )(I2 ) Again, F′ = 0 ⋅l 2 π (3d ) µ 2 I1I2 …(ii) =− 0 ⋅l 2 π 3d Thus, from Eqs. (i) and (ii), we have 2 F′ 2 =− ⇒ F′ = − F 3 F 3
79. The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. The time period of this combination will be [AIEEE 2004]
(a) 2 s
(b)
2 s 3
(c) 2 3 s
(d)
2 s 3
Exp. (b) The time period of oscillations of magnet, I T = 2 π MH
…(i)
where, I = moment of inertia of magnet mL2 [m, being the mass of magnet] = 12 M = Pole strength × L and H = horizontal component of the earth’s magnetic field. When the three equal parts of magnet are placed on one another with their like poles together, then
80. The materials suitable for making electromagnets should have (a) (b) (c) (d)
Electromagnets are made of soft iron. The soft iron has high retentivity and low coercivity.
81. A thin rectangular magnet suspended freely has a period of oscillation equal to T .Now, it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation isT ′, the ratioT ′/T is [AIEEE 2003]
(a)
1 2 2
(b)
1 2
(c) 2
[AIEEE 2004]
high retentivity and high coercivity low retentivity and low coercivity high retentivity and low coercivity low retentivity and high coercivity
(d)
1 4
Exp. (b) When magnet is divided into two equal parts, the magnetic dipole moment l M M′ = Pole strength × = 2 2 [Qpole strength remains same] Also, the mass of magnet becomes half i .e., m m′ = 2 ml 2 Moment of inertia of magnet, I = 12 New moment of inertia, 1 m 12 2 I I′ = 8
I′ = ∴ Now,
T =2π
2
1 m L I 1 mL2 = × ×3 = 12 3 3 12 9 9 L and M′ = Pole strength × × 3 = M 3 / 9 I or T ′ = 1 × T Hence, T ′ = 2 π MH 3 2 or T′ = s 3 I′ =
Exp. (c)
T′ = 2 π ∴
T′ =
2
2 l = ml 2 12 × 8
I MB I′ = 2 π M ′ B
I/ 8 MB/2
T T′ 1 ⇒ = 2 T 2
82. A magnetic needle lying parallel to a magnetic field requires W unit of work to turn it through 60°. The torque needed to maintain the needle in this position will be [AIEEE 2003]
(a)
3W 3 (c) W 2
Exp. (a)
(b) W (d) 2W
296
JEE Main Chapterwise Physics ⇒ or Torque,
W = MB(1 − cos θ) W = MB(1 − cos 60° ) [Q θ = 60° ] MB or MB = 2 W W = 2 τ = MB sin 60° MB 3 2 W 3 = = =W 3 2 2
83. The magnetic lines of force inside a bar magnet (a) are from Northpole to Southpole of the magnet (b) do not exist (c) depend upon the area of crosssection of the bar magnet (d) are from Southpole to Northpole of the magnet
Exp. (d)
θ = 90° P=0 W But ⇒ W = P⋅t P= t Hence, work done, W = 0 [everywhere]
Here, ∴
86. A particle of charge − 16 × 10−18 C moving with velocity 10 ms −1 along the xaxis enters a region where a magnetic field of induction B is along the yaxis and an electric field of magnitude 104 Vm −1 is along the negative zaxis. If the charged particle continues moving along the xaxis, the magnitude of B is [AIEEE 2003] (a) 103 Wb/m 2 (c) 1016 Wb/m 2
(b) 105 Wb/m 2 (d) 10−3 Wb/m 2
Exp. (a)
Inside bar magnet, lines of force are from South to North.
The force on a particle is y
84. Curie temperature is the temperature above which
[AIEEE 2003]
(a) a ferromagnetic material becomes paramagnetic (b) a paramagnetic material becomes diamagnetic (c) a ferromagnetic material becomes diamagnetic (d) a paramagnetic material becomes ferromagnetic
Exp. (a)
B
85. A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when subjected to a uniform transverse magnetic field of inductionB.The work done by the field when the particle completes one full circle is [AIEEE 2003] Mv 2 2 πR (a) R
(b) zero
(c) BQ 2 πR
(d) BQv 2 πR
Exp. (b) When particle describes circular path in a magnetic field, its velocity is always perpendicular to the magnetic force. Power, P = F ⋅ v = Fv cos θ
x
E z
So, ∴
F = q (E + v × B) or F = Fe + Fm Fe = q E = − 16 × 10−18 × 104 (− j) = 16 × 10−14 k
and
At Curie temperature, a ferromagnetic substance transits into paramagnetic substance.
v
Fm = −16 × 10−18 (10i × B j) = − 16 × 10−17 × B(+ k )
= − 16 × 10−17 Bk Since, particle will continue to move along + xaxis, so resultant force is equal to zero. Fe + Fm = 0 ⇒ 16 × 10−14 = 16 × 10−17 B 16 × 10−14
⇒
B=
or
B = 103 Wb/m 2
16 × 10−17
= 103
87. An ammeter reads upto 1 A. Its internal
resistance is 0.81 Ω. To increase the range to 10 A, the value of the required shunt is [AIEEE 2003]
(a) 0.03 Ω (b) 0.3 Ω
Exp. (d)
(c) 0.9 Ω
(d) 0.09 Ω
297
Magnetic Effect of Current To increase the range of ammeter, we have to connect a small resistance in parallel (shunt), let its value be R. 1A
A 0.81Ω
Apply KCL at junction to divide the current.
9A
A 0.81Ω
Voltage across R = Voltage across ammeter i.e. , 9R = 0.81 × 1 0.81 or R= = 0.09 Ω 9
88. If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a [AIEEE 2002] (a) low resistance in parallel (b) high resistance in parallel (c) high resistance in series (d) low resistance in series
89. If in a circular coil A of radius R , current I is flowing and in another coil B of radius 2 R a current 2 I is flowing, then the ratio of the magnetic fields, B A and B B produced by them will be [AIEEE 2002] (c)
1 2
Exp. (a) Magnetic field in circular coil A is µ NI BA = 0 2R
Exp. (a) Since, momenta are same and masses of electron and proton are different, so they will attain different velocities and hence experience different forces. But radius of circular path is depending on momenta, so both will be moving on same trajectory (curved path).
undergoing a circular motion in a uniform magnetic field is independent of its
A voltmeter is a high resistance device and is always connected in parallel with the circuit. While an ammeter is a low resistance device and is always connected in series with the circuit. So, to use ammeter in place of voltmeter, a high resistance must be connected in series with the ammeter to make its resistance high.
(b) 2
(a) curved path of electron and proton will be same (ignoring the sense of revolution) (b) they will move undeflected (c) curved path of electron is more curved than that of proton (d) path of proton is more curved
91. The time period of a charged particle
Exp. (c)
(a) 1
90. If an electron and a proton having same momenta enter perpendicularly to a magnetic field, then [AIEEE 2002]
R
10 A 1 A
where, R is radius and I is current flowing in coil. µ N(2 I) µ 0 NI Similarly, BB = 0 = 2 ⋅ (2 R ) 2R BA ∴ =1 BB
(d) 4
(a) speed (b) mass (c) charge (d) magnetic induction
[AIEEE 2002]
Exp. (a) In a circular motion of a uniform magnetic field, the necessary centripetal force to the charged particle is provided by the magnetic force. mv 2 i .e., = qvB r mv or r= qB Thus, the time period T is 2 πr T= v 2 π mv 2 πm = = v qB qB So, T is independent of its speed.
298
JEE Main Chapterwise Physics 1
92. At a specific instant, emission of radioactive compound is deflected in a magnetic field. The compound can emit [AIEEE 2002] (i) electrons (ii) protons (iii) He2 + (iv) neutrons The emission at the instant can be (a) (i), (ii), (iii) (b) (i), (ii), (iii), (iv) (c) (iv) (d) (ii), (iii)
Exp. (a) Electrons, protons and helium atoms are deflected in magnetic field, so the compound can emit electrons, protons and He 2+ .
93. Wires 1 and 2 carrying currents I 1 and I 2 respectively are inclined at an angle θ to each other. What is the force on a small element dl of wire 2 at a distance r from wire 1 (as shown in figure) due to the magnetic field of wire 1? [AIEEE 2002]
2
I1
I2 r
dl
θ
µ0 I 1 I 2 dl tan θ 2 πr µ0 (c) I 1 I 2 dl cos θ 2 πr
(a)
µ0 I 1 I 2 dl sin θ 2 πr µ0 (d) I 1 I 2 dl sin θ 4 πr (b)
Exp. (c) The component dl cos θ of elementdl is parallel to the length of the wire 1. Hence, force on this elemental component µ 2I I F = 0 ⋅ 1 2 (dl cos θ) r 4π µ 0 I1I2 dl cos θ = 2 πr
14 Electromagnetic Induction and AC 1. A 20 H inductor coil is
i
10 W
connected to a 10 ohm resistance in 20 H E series as shown in figure. The time at which rate of dissipation of energy (Joule’s heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is [JEE Main 2019, 8 April ShiftI] (a)
2 ln 2
(b)
1 ln 2 2
(c) 2 ln 2
(d) ln2
Exp. (c) Given circuit is a series LR circuit R=10Ω
Given, rate of energy stored in inductor is equal to the rate of energy dissipation in resistor. So, after differentiating, we get di di R iL = i 2 R ⇒ = i dt dt L R R − t − t E R L R E ⇒ = ⋅ 1 − e L ⋅ e R L L R ⇒
2e
R − t L
=1 ⇒ e
R − t L
=
1 2
Taking log on both sides, we have 1 −R R t = ln ⇒ t = ln2 ⇒ 2 L L 20 L ⇒ t = ln2 = ln2 ⇒t = 2 ln2 10 R
2. An alternating voltage V (t ) = 220 sin 100πt
i L= 20H
E
volt is applied to a purely resistive load of 50 Ω. The time taken for the current to rise from half of the peak value to the peak value is [JEE Main 2019, 8 April ShiftI]
In an L  R circuit, current increases is R
− ⋅t E (1 − e L ) R Now, energy stored in inductor is 1 U L = Li 2 2 where, L = self inductance of the coil and energy dissipated by resistor is UR = i 2R
i =
(a) 5 ms
(b) 2.2 ms (c) 7.2 ms (d) 3.3 ms
Exp. (d) In an AC resistive circuit, current and voltage are in phase. V 220 So, …(i) ⇒ I= sin(100 πt ) I= R 50 ∴Time period of one complete cycle of current is 2π 2π 1 s T= = = ω 100 π 50
300
JEE Main Chapterwise Physics I
Imax
Imax —— 2
Comparing the above equation with general equation of emf, i.e. e = e 0 sinωt , we get ω = 100 rad/s = 102 rad/s
3T/4
T/4 T/2
1 T= — s 50
t
So, current reaches its maximum value at T 1 s t1 = = 4 200 When current is half of its maximum value, then from Eq. (i), we have I I = max = Imax sin(100 πt 2 ) 2 1 5π ⇒sin(100 πt 2 ) = ⇒100 πt 2 = 2 6 So, instantaneous time at which current is half of 1 maximum value is t 2 = s 120 Hence, time duration in which current reaches half of its maximum value after reaching maximum value is 1 1 1 s = 3.3 ms − = ∆t = t 2 − t 1 = 120 200 300
3. A circuit connected to an AC source of emf
e = e 0 sin(100t ) with t in seconds, gives a π phase difference of between the emf e and 4 current i. Which of the following circuits will exhibit this? [JEE Main 2019, 8 April ShiftII] (a) (b) (c) (d)
RC circuit with R =1 kΩ and C =1µF RL circuit with R =1 kΩ and L =1 mH RC circuit with R =1 kΩ and C =10µF RL circuit with R =1 kΩ and L =10 mH
Exp. (c) π 4 As we know, for RL or RC circuit, Capacitive reactance ( XC ) or inductive
Given, phase difference, φ =
tan φ =
Resistance (R ) π XC or X L tan = R 4 XC or X L 1= R ⇒ R = XC or X L Also, given e = e 0 sin(100 t )
reactance ( X L )
Now, checking option wise, For RC circuit, with R = 1kΩ = 103 Ω and C = 1µF = 10−6 F 1 1 So, XC = = = 104 Ω ⇒ R ≠ XC ωC 102 × 10−6 For R  L circuit, with R = 1kΩ = 103 Ω L = 1mH = 10−3 H
and
So, X L = ωL = 102 × 10−3 = 10−1 Ω ⇒
R ≠ XL
For R  C circuit, with R = 1kΩ = 103 Ω and
C = 10µF = 10 × 10−6 F = 10−5 F
So, XC =
1 10 × 10−5 2
= 103 Ω ⇒ R = C
For R  L circuit, with R = 1kΩ = 103 Ω and L = 10 mH = 10 × 10−3 H = 10−2 H X L = 102 × 10−2 = 1 Ω ⇒ R ≠ X L Alternate Solution X or X L π Since, tan = 1 = C R 4 ∴For RC circuit, we have 1 1 …(i) or ω = 1= CR CωR Similarly, for RL circuit, we have ωL R ...(ii) ⇒ω = 1= R L It is given in the question that, ω = 100 rad/s Thus, again by substituting the given values of R, C or L option wise in the respective Eqs. (i) and (ii), we get that only for option (c), 1 1 ω= = CR 10 × 10−6 × 103 or
ω = 100 rad/s
4. The total number of turns and crosssection area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to [JEE Main 2019, 9 April ShiftI]
(a) 1/ L (c) L
(b) L2 (d) 1 / L2
301
Electromagnetic Induction and AC Exp. (a) Self inductance Lsol of a solenoid is given by Lsol = µ 0 n2 πr 2 L (Here, n = N / L and L = length of solenoid) µ N2 πr 2 or Lsol = 0 L 1 Clearly, Lsol ∝ L (QAll other parameters are fixed) Note We can determine expression of L as follows φ = NBA = Lsol I
At t = 0, I = − ve ⇒ ∴The correct graph representing this condition is given in option (d). Alternate Solution Given solenoid is shown below as, 2R
R I=kte
But for a solenoid, B = µ 0 nI, A = πr 2 ∴
Lsol I = µ 0 nIπr 2 N
or
Lsol
2
Graph of e αt and kt versus time can be shown as,
5. A very long solenoid of radius R is carrying
I
(a) t=0
t
(b) t=0
I (c) t=0
t I
t
(d) t=0
t
Exp. (d) Magnetic flux associated with the outer coil is φouter = µ 0 πNR ⋅ I = µ 0 NπR(kte − αt ) = Cte − αt where, C = µ 0 NπRk = constant Induced emf, − dφouter e= = Ce − αt + (− α C t e − αt ) dt = Ce − αt (1 − αt ) e ∴Induced current, I = Resistance
y
eαt
kt
t
kt ; e αt Initially, kt > e αt So, current is increasing in magnitude. Finally, after a short time kt < e αt . So, current is decreasing in magnitude. But in both cases, it remains positive or counter clockwise. So, current induced is at first anticlockwise (following Lenz’s law) and then it becomes clockwise and finally reduces to zero as t → ∞. So, correct graph of induced current is As,
I=
Current in solenoid reaches maximum
Current of solenoid reducing and tends to zero
Induced current
current I (t ) = kte − αt (k > 0), as a function of time (t ≥ 0). Counter clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by [JEE Main 2019, 9 April ShiftII] I
t = 0, current in solenoid = I(t = 0) = k(0) e − α ⋅0 = 0
At
N2 = µ 0 n πr L = µ 0 πr 2 L 2
–αt
t
Current of solenoid grows in counter clockwise direction
302
JEE Main Chapterwise Physics
6. Two coils P and Q are separated by some distance. When a current of 3 A flows through coil P, a magnetic flux of 10−3 Wb passes through Q. No current is passed through Q. When no current passes through P and a current of 2 A passes throughQ, the flux through P is [JEE Main 2019, 9 April ShiftII] −3
(a) 667 . × 10
(b) 667 . × 10−4 Wb
Wb
(c) 367 . × 10−3 Wb
(d) 367 . × 10−4 Wb
Output power, P2 = 2.2 kW = 2.2 × 103 W Current in secondary coil, I2 = 10 A Output power, P2 = I2 V2 P 2.2 × 103 ⇒ = 220V V2 = 2 = 10 I2
We know that, N N1 V Input voltage = = 1 ⇒ V1 = 1 V2 N2 Output voltage V2 N2 300 ⇒ V1 = × (220 V ) [using Eq. (i)] 150 V1 = 440 V V1 I2 = V2 I1
Exp. (b) As, coefficient of mutual induction is same for both coils Q P
… (i)
Again,
… (ii)
V 220 × 10 [using Eqs. (i) and (ii)] ⇒ I1 = 2 I2 = 440 V1 ⇒
I1 = 5A
8. A coil of self inductance 10 mH and
∴
M PQ = MQP NP φPQ NQ φQP ⇒ = IQ IP
Here,
… (i)
NP = NQ = 1, φPQ = ?, φQP = 10− 3 Wb
IQ = 2A, IP = 3A Substituting values in Eq (i), we get I ⋅φ 2 φPQ = Q QP = × 10− 3 3 IP = 0.667 × 10− 3 = 6.67 × 10− 4 Wb
7. A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coil is 10 A, then the input voltage and current in the primary coil are [JEE Main 2019, 10 April ShiftI]
(a) 440 V and 5 A (c) 220 V and 10 A
resistance 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω. After the switch is closed, the time taken for the current to attain 80% of the saturation value is [Take, ln 5 = 1.6] [JEE Main 2019, 10 April ShiftII]
(a) 0.002 s (c) 0.103 s
(b) 0.324 s (d) 0.016 s
Exp. (d) Key Idea In an LR circuit, current during charging of inductor is given by i = i 0 (1 − e
where, i 0 = saturation current.
In given circuit, Inductance of circuit is L = 10 mH = 10 × 10−3 H Resistance of circuit is R = (R s + r ) = 01 . + 0.9 = 1 Ω
(b) 220 V and 20 A (d) 440 V and 20 A
RS=0.1Ω
L=10 mH
Exp. (a) Given, Number of turns in primary, N1 = 300 Number of turns in secondary, N2 = 150
R − ⋅t L )
+ – r=0.9Ω
303
Electromagnetic Induction and AC Now, from i = i 0 (1 −
R − ⋅t e L
...(i)
)
So, equivalent resistance of the Wheatstone bridge is
i = 80% of i 0
Given,
80 i 0 = 0.8 i 0 100 Substituting the value of i in Eq. (i), we get ⇒
i =
0.8 = 1 − ⇒
e
R t eL
⇒ ⇒ ln ⇒
R − t L
R t (e ) L
1Ω
R − ⋅t e L
2Ω 2Ω
⇒
= 02 .
1Ω
=5 = ln 5 ⇒
= 1.6 × 10−2 s = 0.016 s
9. The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of 1 cm s −1 . At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 Ω, the current in the loop at that instant will be close to [JEE Main 2019, 12 April ShiftI] v=1 cm/s L 1Ω A 1Ω
B
2Ω
3Ω D
2Ω
2×4 8 4 = = Ω 2+4 6 3 Again, resistance of conductor is 1.7 Ω. So, effective resistance will be 4 17 4 , R eq = + 1. 7 = + 3 10 3 40 + 51 91 ~ R eq = = − 3Ω 30 30 By putting given values of Req , B and v in Eq. (ii), we have (1) (5 × 10−2 ) × 10−2 i = 3 [here, L = 5 × 10−2 m, v = 1cm/s = 10−2 m/s] 5 × 10−4 i = = 1. 67 × 10−4 A 3 i = 167 µA ≈ 170 µA
10. Consider the LR circuit shown in the figure. If the switch S is closed at t = 0, then the amount of charge that passes through the L battery betweent = 0 andt = is R
C
[JEE Main 2019, 12 April ShiftII]
2Ω
L
R
5 cm
(a) 60µ A (c)150µ A
4Ω
RW =
R t = ln5 L 10 × 10−3 L t = ⋅ ln(5) = × ln(5) R 1 = 10 × 10−3 × 1.6
B
Now, given network is a balanced Wheatstone P R bridge = . Q S
(b)170µ A (d)115µ A
i
Exp. (b) Induced emf in the conductor of length L moving with velocity of 1 cm/s in the magnetic field of 1T is given by …(i) V = BLv If equivalent resistance of the circuit is Req , then current in the loop will be V BLv …(ii) = i = Req Req
S
E
(a) (c)
2.7 EL R2 7.3 EL R2
Exp. (b)
(b) (d)
EL 2.7 R 2 EL 7.3 R 2
304
JEE Main Chapterwise Physics In an LR circuit, current during charging is given by R − t I = I0 1 − e L E where, I0 = = saturation current. R R − t dq So, we have = I = I0 1 − e L dt R − t dq = I0 1 − e L dt
⇒
= 1.4 × 10−4 C = 14 mC
So, charge q that passes through battery from L time t = 0 to t = is obtained by integrating the R above equation within the specified limits, i.e. R L − t Q t= q = ∫ dq = ∫ R I0 1 − e L dt 0 t =0 L R
E R
⇒
Q=
(a) 3.39 × 103 J
(b) 565 . × 102 J
(c) 2.26 × 10 J
(d) 517 . × 102 J
The given series RLC circuit is shown in the figure below. R=60Ω L=20 mH C=120 µF
E L EL × = R Re R 2e EL
VR
2.7 R 2
VL +
[Qe ≈ 2.72]
11. A conducting circular loop is made of a thin
wire has area 3.5 × 10−3 m 2 and resistance 10 Ω. It is placed perpendicular to a time dependent magnetic field B (t ) = (0.4T )sin( 0.5 πt ). The field is uniform in space. Then the net charge flowing through the loop duringt =0 s and t = 10 ms is close to [JEE Main 2019, 9 Jan ShiftI] (a) 6 mC (c) 7 mC
(20 mH), a capacitor (120 µF) and a resistor (60 Ω ) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is [JEE Main 2019, 9 Jan ShiftII]
Exp. (d)
L L L + − 0 + R Re ′ R =
12. A series AC circuit containing an inductor
3
R − t 1 ⋅ e L = I0 t − − R L 0 =
As, ∆φB = B A = BA cos θ where, A is the surface area of the loop and ‘θ’ is an angle between B and A. Here, θ = 0 ⇒ ∆φB = BA ∴For the time interval, t = 0 ms to t = 10 ms, ∆φ A Q = B = (Bf at 0. 01 s − Bi at 0 s ) R R Substituting the given values, we get 3.5 × 10−3 = [0.4sin(0.5 π ) − 0.4sin 0] 10 = 3.5 × 10−4 (0.4sin π / 2 )
I
Since, the magnetic field is dependent on time, so the net charge flowing through the loop will be given as change in magnetic flux, ∆φB Q= resistance, R
–
24 V, 50 Hz
Here, VR = potential across resistance (R) VL = potential across inductor (L) and VC = potential across capacitor (C ). Impedance of this series circuit is, Z=
R 2 + ( X L − XC )2
…(i)
X L = ωL = (2 πf )(L) = 2 π × 50 × 20 × 10−3 Ω
Q
(b) 21 mC (d) 14 mC
Exp. (d)
VC
and
X L = 628 . Ω 1 1 XC = = ωC 2 πfC 1
250 Ω = 3π 2 π × 50 × 120 × 10−6 250 and X L − XC = 628 . – . Ω = – 2023 3π =
…(ii)
...(iii) …(iv)
305
Electromagnetic Induction and AC RMS value of current in circuit is, 24 V Irms = rms = 2 Z R + ( X L − XC )2 Irms =
24 . ) 60 + (–2023 2
2
=
14. A magnet of total magnetic moment
10−2 $i Am 2 is placed in a time varying magnetic field, B$i(cos ωt ), where B = 1 T and ω = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at [JEE Main 2019, 10 Jan ShiftI] t = 1 s is
24 6318 .
Irms = 0.379 A Therefore, energy dissipated is 2 = Irms × R ×t
(a) 0.01 J
Exp. (c)
E = (0.379)2 × 60 × 60 = 517.10 = 517 . × 102 J
or
13. A power transmission line feeds input power at 2300 V to a stepdown transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be [JEE Main 2019, 9 Jan ShiftII]
(a) 45 A
(b) 50 A
(c) 25 A
(d) 35 A
Exp. (a)
Work done in reversing dipole is W = 2 MB where, M = magnetic dipole moment = 10− 2 A m2 B = external field = B cos ωt = 1 × cos (0.125 × 1) = cos(7º ) = 0.992 Substituting these values, we get W = 2 × 10− 2 × 0.992 and
= 0.0198 J which is nearest to 0.014 J.
15. At some location on earth, the horizontal
For a transformer, there are two circuits which have Np and N s (number of coil turns), Ip and IS (currents) respectively as shown below. +
NP
VP –
IS
VS RV
IP Primary
NS Secondary
Here, input voltage, Vp = 2300 V Number of turns in primary coil, NP = 4000 Output voltage, VS = 230 volt Output power, PS = VS ⋅ IS Input power, PP = VP IP ∴ The efficiency of the transformer is Output (secondary) power η= Input (primary) power VS ⋅ IS ⇒ × 100 η= VP ⋅ IP (230) (IS ) ⇒ η= × 100 (2300) (5) 230 IS × 100 90 = (2300) × 5 ⇒
(b) 0.007 J (c) 0.014 J (d) 0.028 J
component of earth’s magnetic field is 18 × 10−6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 Am is suspended from its mid point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is [JEE Main 2019, 10 Jan ShiftII]
(a) 6. 5 × 10−5 N
(b) 3.6 × 10−5 N
(c) 1. 3 × 10−5 N
(d) 1.8 × 10−5 N
Exp. (a) Without applied forces, (in equilibrium position) the needle will stay in the resultant magnetic field of earth. Hence, the dip ‘θ’ at this place is 45° (given). BV Bearth
IS = 45 A
θ O
BH
306
JEE Main Chapterwise Physics We know that, horizontal and vertical components of earth’s magnetic field (BH and BV ) are related as BV = tanθ BH Here, θ = 45º and BH = 18 × 10− 6 T ⇒ ⇒
BV = BH tan 45º BV = BH = 18 × 10− 6 T
(Qtan 45° = 1)
Now, when the external force F is applied, so as to keep the needle stays in horizontal position is shown below, Magnetic needle mBH
mBV
S
P mBV
N
mBH
F
Taking torque at point P, we get mBV × 2 l = Fl ∴ F = 2 × mBV Substituting the given values, we get = 2 × 18 . × 18 × 10−6 = 6.48 × 10−5 = 6.5 × 10−5 N
16. The selfinduced emf of a coil is 25 V. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is [JEE Main 2019, 10 Jan ShiftII]
(b) 740 J
(c) 637.5 J
(d) 540 J
25 = L ×
17. There are two long coaxial solenoids of same lengthl. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n 2, respectively. The ratio of mutual inductance to the selfinductance of the inner coil is [JEE Main 2019, 11 Jan ShiftI] (a)
0.12 m
(a) 437.5 J
15 25 or L = = 5/ 3H 1 15 Putting values of L, I1 and I2 in Eq. (i), we get 1 5 1 5 ∆E = × × [252 − 102 ] = × × 525 2 3 2 3 ∆E = 437.5 J ⇒
n 2 r1 ⋅ n1 r2
(b)
n 2 r22 ⋅ n1 r12
(c)
n2 n1
(d)
n1 n2
Exp. (c) Mutual inductance for a coaxial solenoid of radius r1 and r2 and number of turns n1 and n2 , respectively is given as, M = µ 0 n1n2 π r12 l (for internal coil of radius r1) Self inductance for the internal coil, L = µ 0 n12 π r12 l M n1n2 n2 ∴ = 2 = L n1 n1
18. A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the selfinductance of the coil [JEE Main 2019, 11 Jan ShiftII]
Exp. (a) Energy stored in an inductor of inductance L and current I is given by 1 E = LI 2 2 When current is being changed from I1 to I2 , change in energy will be 1 1 …(i) ∆E = E2 − E1 = LI22 − LI12 2 2 As I1 and I2 are given, we need to find value of L. Now, induced emf in a coil is dI ε=L dt Here, ε = 25 V, dI = I2 − I1 = (25 − 10) = 15 A and dt = 1 s
(a) increases by a factor of 3 (b) decreases by a factor of 9 3 (c) increases by a factor of 27 (d) decreases by a factor of 9
Exp. (c) Selfinductance of a coil is given by the relation L = µ 0 n2 A ⋅ l where, n is number of turns per unit length. Shape of the wooden frame is equilateral triangle. ∴ Area of equilateral triangle, 3 2 A= a 4 (where, a is side of equilateral triangle)
307
Electromagnetic Induction and AC
a
a
+ 15 V
5Ω
–
5Ω
a
∴ Selfinductance, 3 2 L = µ 0 n2 a . l 4 Here, l = 3a × N (where, N is total turns) 3 2 L = µ 0 n2 a × 3aN or L ∝ a3 ∴ 4 When each side of frame is increased by a factor 3 keeping the number of turns per unit length of the frame constant. Then, a′ = 3a ∴ L′ ∝ (a′ )3 or
L′ ∝ (3a)3
or
L′ ∝ 27 a3
∴ Current in circuit is 15 × 2 15 E = I= = =6A 5 × 5 Req 5 5 + 5
20. A 10 m long horizontal wire extends from NorthEast to SouthWest. It is falling with a speed of 5.0 ms− 1 at right angles to the horizontal component of the earth’s magnetic field of 0.3 × 10− 4 Wb/ m 2. The value of the induced emf in wire is [JEE Main 2019, 12 Jan ShiftII]
(a) 1.5 × 10− 3 V
or L′ =27 L So, selfinductance will becomes 27 times.
−3
(c) 0.3 × 10
(d) 2.5 × 10− 3 V
V
Exp. (a)
19. In the figure shown, a circuit contains two
Wire falls perpendicularly to horizontal component of earth’s magnetic field, so induced electromotive force (ε) = Blv Substituting the given values, we get ε = 0.3 × 10−4 × 10 × 5 = 15 . × 10−3 V
identical resistors with resistance R = 5 Ω and an inductance with L = 2 mH. An ideal battery of 15 V is connected in the circuit. [JEE Main 2019, 12 Jan ShiftI]
I2
21.
S
L
C L
R 15 V
(b) 11 . × 10− 3 V
R2 R1 I1
R
What will be the current through the battery long after the switch is closed?
In the above circuit, C =
(a) 6 A (b) 3 A (c) 5.5 A (d) 7.5 A
L=
Exp. (a) After a sufficiently long time, in steady state, resistance offered by inductor is zero. So, circuit is reduced to
3 µF, R 2 = 20 Ω, 2
3 H and R1 = 10 Ω. Current in L  R1 10 path is I 1 and in C  R 2 path is I 2.The voltage of AC source is given by V = 200 2 sin(100t ) volts. The phase difference betweenI 1 andI 2 is [JEE Main 2019, 12 Jan ShiftII] (a) 30º (c) 0º
(b) 60º (d) 90º
308
JEE Main Chapterwise Physics ∴
Exp. (a) Phase difference between I2 and V, i.e. C − R 2 circuit is given by X 1 tanφ = C ⇒ tanφ = R2 CωR 2 Substituting the given values, we get 1 103 tanφ = = 3 3 × 10−6 × 100 × 20 2 ∴φ1 , is nearly 90°. Phase difference between I1 and V, i.e. in L − R1 circuit is given by X Lω tanφ2 = − L = − R1 R Substituting the given values, we get 3 × 100 =− 3 tanφ2 = − 10 10 As, tanφ2 = − 3 ∴ φ2 = 120° Now, phase difference between I1 and I2 is ∆φ = φ2 − φ1 = 120° − 90° = 30°
22. In an AC circuit, the instantaneous emf and current are given by π e = 100 sin 30t , i = 20 sin 30t − 4 In one cycle of AC, the average power consumed by the circuit and the wattless current are, respectively [JEE Main 2018] (a) 50 , 10
(b)
1000 50 , 10 (c) , 0 (d) 50 , 0 2 2
Exp. (b) Given, e = 100sin 30t and
and
Iwattless
23. For an RLC circuit driven with voltage of 1 , the LC current exhibits resonance. The quality factor,Q is given by [JEE Main 2018] amplitudevm and frequency ω 0 =
(a)
ω0 L R
(b)
ω0 R L
(c)
R ω0C
(d)
CR ω0
Exp. (a) Sharpness of resonance of a resonant LCR circuit is determined by the ratio of resonant frequency with the selectivity of circuit. This ratio is also called ‘‘Quality Factor’’ or Qfactor. ω L ω 1 Qfactor = 0 = 0 = R ω0CR 2 ∆ω
24. In
a coil of 10 resistance 100 Ω, a Current current is induced by changing the (A) magnetic flux through it as shown in the Time (s) 0.5 figure. The magnitude of change in flux through the coil is [JEE Main 2017 (Offline)] (a) 225 Wb (b) 250 Wb (c) 275 Wb (d) 200 Wb
Exp. (b) Induced constant, I =
e R
dφ dt l dφ 1 I = = ⋅ R dt R
Here, e = induced emf =
π i = 20 sin 30 t − 4
∴ Average power , Pav = Vrms Irms cos φ π 100 20 = × × cos 2 2 4 1000 watt = 2 Wattless current is, I = Irms sin φ π 20 20 = × sin = = 10 A 2 4 2
1000 watt 2 = 10 A
Pav =
dφ = IRdt φ = ∫ IRdt ∴ ∴
Here, R is constant φ = R ∫ Idt
∫ I ⋅ dt = Area under I − t graph 1 × 10 × 0.5 = 2.5 2 φ = R × 2.5 = 100 × 2.5 = 250 Wb =
∴
309
Electromagnetic Induction and AC 25. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to [JEE Main 2016 (Offline)]
(a) 80 H (c) 0.044 H
(b) 67 mA (d) 0.67mA
Exp. (d) / After long time inductor behaves as shortcircuit.
(b) 0.08 H (d) 0.065 H
At t = 0, the inductor behaves as shortcircuited. The current E 15 V = 100 mA I0 = 0 = . kΩ 015 R
Exp. (d) Given, I = 10 A, V = 80 V, V 80 = 8 Ω and ω = 50 Hz R= = I 10 For AC circuit, we have R=8 Ω
(a) 100 mA (c) 6.7 mA
As K 2 is closed, current through the inductor starts decay, which is given at any time t as − t × 15000
− tR
I = I0 e
L
At
= (100 mA) e
L
t = 1ms
10A
I = (100 mA) e
−
3
1×10 −3 ×15 ×10 3 3
I = (100 mA ) e − 5 = 0.6737 mA or
220 V
I= ⇒
V 8 + 2
X L2
⇒ 10 =
27. An LCR circuit is equivalent to a damped
220 64 +
I = 0.67 mA
pendulum. In an LCR circuit, the capacitor is charged toQ 0 and then connected to the L and R as shown below.
X L2
64 + X L2 = 22
Squaring on both sides, we get 64 + X L2 = 484 ⇒
L
R
X L2 = 484 − 64 = 420
X L = 420 ⇒ 2 π × ωL = Series inductor on an arc lamp, 420 L= = 0.065 H (2 π × 50)
420
26. An inductor ( L = 0.03 H ) and a resistor (R = 0.15 kΩ ) are connected in series to a battery of 15V EMF in a circuit shown below. The key K 1 has been kept closed for a long time. Then att = 0,K 1 is opened and keyK 2 is closed simultaneously. At t = 1ms, the current in the circuit will be (e 5 ~ = 150) [JEE Main 2015] 0.03H
C
If a student plots graphs of the square of 2 maximum charge (Q Max ) on the capacitor with time (t) for two different values L 1 and L 2 ( L 1 > L 2) of L, then which of the following represents this graph correctly? (plots are schematic and not drawn to scale) [JEE Main 2015] 2 (a) QMax
L2
(c)
L1
Max
L2 t
t
0.15 kΩ 2 (b) QMax
K2 K1 15V
Q2
L1
Exp. (a)
L2 L1
Q2
(d) t
Max
Q0 (For both L1and L2) t
310
JEE Main Chapterwise Physics Considert the LCR circuit at any time t i
R
+
(a)
–
e 1−e
(c) −1
(b) 1
(d)
1−e e
Exp. (c) After connecting C to B hanging the switch, the circuit will act like LR discharging circuit.
+ – q C
R
Now, applying KVL Ldi q We have − iR − =0 dt C As current is decreasing with time we can write dq i =− dt Ld 2q q dq + =0 R+ ⇒ C dt dt 2 2 d q R dq q or + + =0 2 L dt LC dt This equation is equivalent to that of a damped oscillator Thus, we can write the solution as Rt 2 Qmax (t ) = Q0 ⋅ e − Rt / 2 L or Qmax = Q02 e − L As L1 > L2 damping is faster for L2 Aliter Inductance is inertia of circuit. It means inductance opposes the flow of charge, more inductance means decay of charge is slow. In option (a), in a given time to, Q12 > Q22 .
L
Applying Kirchhoff’s loop equation, VR + VL = 0 ⇒ VR = − VL VR = −1 ∴ VL
29. A metallic rod of length l is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the emf induced across the ends of the rod is [JEE Main 2013]
L1 2 Q1 2
Q2
L2
(a)
2 B ωl 3 3 B ωl 3 4 Bωl 2 5 Bωl 2 (b) (c) (d) 2 2 2 2
t0
So, L1 > L2 . Hence, option (a) is correct.
28. In the circuit shown here, the pointC is kept connected to point A till the current flowing through the circuit becomes constant. Afterward, suddenly pointC is disconnected from point A and connected to point B at time t = 0. Ratio of the voltage across resistance and the inductor at t = L / R will be equal to [JEE Main 2014] A
Exp. (d) QInduced emf is rate of change of magnetic flux. ω
dx
x
e = ∫ Bv dx ⇒ Using
R
C
l
2l
e=
[Q v = ω x ]
3l
∫2 l (ωx)⋅ B ⋅ dx xn +1
∫ x dx = n + 1 n
3l
⇒ L B
x2 e = Bω 2 2l e = Bω
[(3l )2 − (2 l )2 ] 5Bl 2ω = 2 2
311
Electromagnetic Induction and AC 30. In an LCR circuit as
V
shown below, both S1 R switches are open initially. Now, switch C S1 is closed and S 2 S2 kept open (q is L charge on the capacitor and τ = RC is capacitance time constant). Which of the following statement is correct? [JEE Main 2013] (a) Work done by the battery is half of the energy dissipated in the resistor (b) At t = τ, q = CV /2 (c) At t = 2τ,q = CV (1 − e −2 ) τ (d) At t = , q = CV (1 − e −1 ) 2
For charging capacitor, q is given as q = q 0 (1 − e − t / τ ) = CV(1 − e − t / τ ) At t = 2τ, q = CV (1 − e
−2 τ τ
) ⇒q = CV (1 − e −2 )
31. The figure shows an experimental plot discharging of a capacitor in an RC circuit. The time constant τ of this circuit lies between Potential difference (in volt)
(a) development of air current when the plate is placed (b) induction of electrical charge on the plate (c) shielding of magnetic lines of force as aluminium is a paramagnetic material (d) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping
Exp. (d) According to Lenz’s law, electromagnetic induction takes place in the aluminium plate due to which eddy current is developed which oppose the motion or vibrations of coil. This causes loss in energy which results in damping of oscillatory motion of the coil.
33. In a series LCR circuit R = 200 Ω and the
Exp. (c)
voltage and the frequency of the main supply is 220 V and 50 Hz, respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the LCR circuit is [AIEEE 2010]
(a) 305 W
(b) 210 W (c) zero
(d) 242 W
Exp. (d) The given circuit is under resonance as X L = XC . [Qsame phase change in LR and CR circuits] Hence, power dissipated in the circuit is V2 = 242 W P= R
25 20 15 10 5 0
aluminium plate is placed near to the coil, it stops. This is due to [AIEEE 2012]
50 100 150 200 250 300 Time (in second)
(a) 150 s and 200 s (c) 50 s and 100 s
(b) 0 and 50 s (d) 100 s and 150 s
Exp. (d) Time constant τ is the duration when the value of potential drops by 63% of its initial maximum value (i.e., V0 /e ). Here, 37% of 25 V = 9.25 V which lies between 100 s to 150 s in the graph.
34. An inductor of inductance L = 400 mH and resistors of resistances R1 = 4 Ω and R 2 = 2 Ω are connected to battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed att = 0. The potential drop across L as a function of time is [AIEEE 2009] A
12 V
R1
L
32. A coil is suspended in a uniform magnetic field with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil, it starts oscillating; it is very difficult to stop. But, if an
R2 S
312
JEE Main Chapterwise Physics (a) 6e −5t V (c) 6(1 − e
12 −3t V e t (d) 12 e −5t V (b)
−t / 0. 2
)V
36. In an AC circuit, the voltage applied is
E = E 0 sin ωt . The resulting current in the π circuit is I = I 0 sin ωt − . The power 2 consumption in the circuit is given by
Exp. (d) I1 =
E 12 = = 3A R1 4
E0 I0 2 E0 I0 (c) P = 2
(a) P =
I I1
A
R1
I2 L
(b) P = zero
[AIEEE 2007]
(d) P = 2 E 0 I 0
Exp. (b) For given circuit, current is lagging the voltage by π , so circuit is purely inductive and there is no 2 power consumption in the circuit. The work done by battery is stored as magnetic energy in the inductor.
R2 S
Q Potential drop = E − I2 R 2 I2 = I0 (1 − e −T / t c ) ⇒
[current as a function of time] E 12 I0 = = =6A R2 2
and
tc =
L 400 × 10−3 = = 0.2 R2 2
I2 = 6(1 − e −t ) Potential drop across L = E − R 2 I2 = 12 − 2 × 6(1 − e −5 t )
37. An ideal coil of 10 H is connected in series with a resistance of 5 Ω and a battery of 5 V. After 2 s, the connection is made, the current flowing (in ampere) in the circuit is (a) (1 − e ) (c) e −1
(b) e (d) (1 − e −1 )
[AIEEE 2007]
Exp. (d) Rise of current in LR circuit is given by 10H
= 12 e −5 t
5Ω
35. Two coaxial solenoids are made by winding thin insulated wire over a pipe of crosssectional area A = 10 cm 2 and length = 20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is ( µ 0 = 4π × 10−7 TmA −1 ) [AIEEE 2008]
(a) 2.4 π × 10−5 H
(b) 4.8 π × 10−4 H
(c) 4.8 π × 10−5 H
(d) 2.4 π × 10−4 H
Exp. (d)
µ 0 N1 × N2 × A l where, N1 = 300 turns, N2 = 400turns, A = 10 cm2 = 10 × 10−4 m2 and I = 20 cm = 20 × 10−2 m Substituting the values in the given formula, we get M = 2 .4 π × 10−4 H. M=
5V
I = I0 (1 − e − t / τ ) E 5 where, I0 = = = 1 A R 5 L 10 Now, τ= = =2 s R 5 After 2s i.e., at t = 2s, Rise of current, I = (1 − e −1 ) A
38. An
L inductor a ( L = 100 mH ), resistor (R = 100 Ω ) R and a battery B A are ( E = 100 V) E initially connected in series as shown in the figure. After a long time, the battery is disconnected after short
313
Electromagnetic Induction and AC circuiting the points A and B . The current in the circuit 1 millisecond after the short circuit is [AIEEE 2006] (a) 1/e A
(b) e A
(c) 0.1 A
(d) 1 A
Exp. (a) This is a combined example of growth and decay of current in an LR circuit. L = 100 mH
R = 100Ω A
B E = 100 V
The current through circuit just before shorting the battery, E 100 I0 = = = 1A R 100 [as inductor would be shorted in steady state] After this decay of current starts in the circuit, according to the equation, I = I0e − t / τ τ = L/R
where,
40. In a series resonant
LCR circuit, the voltage across R is 100 V N S and with R = 1 kΩ C = 2 µF. The resonant ω frequency is ω 200 rad/s. At resonance, the voltage across L is [AIEEE 2006] (a) 2.5 × 10−2 V
(b) 40 V
(c) 250 V
(d) 4 × 10−3 V
Exp. (c) At resonance, ω L = 1/ ωC Current flowing through the circuit, V 100 . A I= R = = 01 R 1000 So, voltage across L is given by VL = I X L = Iω L. But ω L = 1/ ωC . I 01 ∴ VL = = VC = = 250 V ωC 200 × 2 × 10−6
41. The flux linked with a coil at any instant t is
given by φ = 10t 2 − 50t + 250. The induced emf att = 3 s is [AIEEE 2006] (a) –190 V (b) –10 V
L
θ = 0°
(c) 10 V
(d) 190 V
Exp. (b) R
I = 1 × e − ( 1 × 10
−3
)/( 100 × 10 −3 / 100 )
1 = A e
[Q t = 1millisecond = 1 × 10−3 s and L = 100 × 10−3 H]
39. In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is [AIEEE 2006] (a) NABR ω (b) NAB
(c) NABR
(d) NAB ω
Exp. (d) The emf generated would be maximum when flux (cutting) would be maximum i.e., angle between area vector of coil and magnetic field is 0°. The emf generated is given by (as a function of time) e = NBAω cos ωt ⇒ emax = NABω [Qcos ωt = cos θ = 1 ⇒ θ = 0° ]
φ = 10t 2 − 50t + 250 From Faraday’s law of electromagnetic induction, e = − dφ/dt ∴ e = − [10 × 2 t − 50] ∴ et = 3 s = − [10 × 6 − 50] = − 10 V
42. A long solenoid has 200 turns per cm and carries a current I . The magnetic field at its centre is 6.28 × 10−2 Wb/m 2. Another long solenoid has 100 turns per cm and it carries a current I/3. The value of the magnetic field at its centre is [AIEEE 2006] (a) 1.05 × 10−2 Wb/m 2 (b) 1.05 × 10−5 Wb/m 2 (c) 1.05 × 10−3 Wb/m 2 (d) 1.05 × 10−4 Wb/m 2
Exp. (a) Magnetic field due to a long solenoid is given by B = µ 0 nI From given data, …(i) 6.28 × 10−2 = µ 0 × 200 × 102 × I and
I B = µ 0 × 100 × 102 × 3
…(ii)
314
JEE Main Chapterwise Physics Solving Eqs. (i) and (ii), we get B ≈ 1.05 × 10−2 Wb / m2
43. A coil of inductance 300 mH and resistance
2 Ω is connected to a source of voltage 2 V. The current reaches half of its steady state value in [AIEEE 2005] (a) 0.05 s
(b) 0.1 s
(c) 0.15 s
(d) 0.3 s
Exp. (b) The current at any instant is given by I = I0 (1 − e − Rt / L ) L, R I0 − Rt / L = I0 (1 − e ) ⇒ 2 1 or = (1 − e − Rt / L ) 2 1 or e − Rt / L = 2 2V Rt or = ln 2 L 300 × 10−3 L t = ln 2 = ∴ × 0.693 R 2 −3 = 150 × 0.693 × 10 = 01039 . 5 s = 0.1 s
44. The selfinductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of [AIEEE 2005] (a) 4 µF
(b) 8 µF
(c) 1 µF
(d) 2 µF
Exp. (c) Given, L = 10 H, f = 50 Hz For maximum power, [Qresonance condition] X C = XL 1 1 or = ω L or C = 2 ωC ω L 1 C= ∴ 4 π 2 × 50 × 50 × 10 or
impedance of 15 Ω. The power factor of the circuit will be [AIEEE 2005] (b) 0.4
(c) 1.25
(d) 0.125
Exp. (a) Power factor = cos φ =
phase difference between the alternating current and emf is π/2. Which of the following cannot be the constituent of the circuit ? [AIEEE 2005] (a) C alone (b) R , L
(c) L , C
R 12 4 = = = 0.8 Z 15 5
(d) L alone
Exp. (c) (a) In a circuit having C alone, the voltage lags the π current by . 2 (b) In circuit containing R and L, the voltage leads π the current by . 2 (c) In LC circuit, the phase difference between current and voltage can have any value π depending on the values of between 0 to 2 L and C. (d) In a circuit containing L alone, the voltage leads the current by π / 2.
47. One
conducting B A Utube can slide v inside another as v shown in figure, C D maintaining electrical contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed v , then the emf induced in the circuit in terms ofB , l and v ,wherel is the width of each tube, will be [AIEEE 2005] (a) Blv
(b) −Blv
(c) zero
(d) 2 Blv
Exp. (d) dl dt Bl dl dφ e= ⇒ e= dt dt
Relative velocity = v − (− v ) = 2 v = Now,
Induced emf, e = 2 B l v
C = 01 . × 10−5 F = 1 µF
45. A circuit has a resistance of 12 Ω and an (a) 0.8
46. The
[Q φ = BA] Q dl = 2 v dt
48. Alternating current cannot be measured by DC ammeter because
[AIEEE 2004]
(a) AC cannot pass through DC ammeter (b) AC changes direction (c) average value of current for complete cycle is zero (d) DC ammeter will get damaged
Exp. (c)
315
Electromagnetic Induction and AC The full cycle of alternating current consists of two half cycles. For one half, current is positive and for second half, current is negative. Therefore, for an AC cycle, the net value of current average out to zero. While the DC ammeter, read the average value. Hence, the alternating current cannot be measured by DC ammeter.
51. In a uniform magnetic field of induction B ,a wire in the form of semicircle of radius r rotates about the diameter of the circle with angular frequencyω.If the total resistance of the circuit is R , the mean power generated per period of rotation is [AIEEE 2004] (a)
B πr 2ω 2R
(b)
( B πr 2ω)2 8R
across each of the components L ,C and R is 50 V. The voltage across the LC combination will be [AIEEE 2004]
(c)
( B π rω )2 2R
(d)
( B π rω 2 )2 8R
(a) 50 V
Exp. (b)
49. In an LCR series AC circuit, the voltage
(b) 50 2 V (c) 100 V
(d) zero
Exp. (d) In an LCR series AC circuit, the voltage across inductor L leads the current by 90° and the voltage across capacitor C lags behind the current by 90°. [QV = VL ~ VC = 50 − 50 = 0] VL VC
90° 90°
VR
I
Hence, the voltage across LC combination will be zero.
50. A coil having n turns and resistance R Ω is connected with a galvanometer of resistance 4R Ω. This combination is moved for time t seconds from a magnetic field W1 weber to W 2 weber. The induced current in the circuit is [AIEEE 2004] W − W1 (a) 2 5 Rnt (W 2 − W1 ) (c) − Rnt
n(W 2 − W1 ) (b) − 5 Rt n(W 2 − W1 ) (d) − Rt
Exp. (b) The rate of change of flux or emf induced in the dφ coil is e = − n . dt e n dφ …(i) ∴ Induced current, I = =− R′ R ′ dt Given, R ′ = R + 4R = 5R,dφ = W2 − W1,dt = t [here, W1 and W2 are flux associated with one turn] Putting the given values in Eq. (i), we get n (W2 − W1 ) I=− t 5R
The flux associated with coil of area A and magnetic induction B is 1 Q A = 1 πr 2 φ = BA cos θ = Bπr 2 cos ωt 2 2 d 1 dφ 2 ∴ einduced = − =− Bπr cos ωt dt 2 dt 1 = Bπr 2ω sin ωt 2 e2 [Q P = V 2 / R ] ∴ Power, P = induced R B2 π 2 r 4ω2 sin2 ωt = 4R Hence, Pmean = < P > =
B2 π 2 r 4ω2 1 ⋅ 4R 2
=
(Bπr 2ω)2 8R
Q< sin2 ωt > = 1 2
52. In an LCR circuit, capacitance is changed from C to 2C . For the resonant frequency to remains unchanged, the inductance should be changed from L to [AIEEE 2004]
(a) 4L
(b) 2 L
(c) L/2
(d) L/4
Exp. (c) In the condition of resonance, X L = XC or ω L =
1 ωC
…(i)
Since, resonance frequency remains unchanged, so LC = constant or LC = constant ∴ L1C1 = L2C 2 or L × C = L2 × 2 C or L2 = L /2
316
JEE Main Chapterwise Physics
53. A metal conductor of length 1 m rotates
56. In an oscillating LC circuit, the maximum
vertically about one of its ends at angular velocity 5 rad/s. If the horizontal component of earth’s magnetic field is 0.2 × 10−4 T, then the emf developed between the two ends of the conductor is [AIEEE 2004]
charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is [AIEEE 2003]
(a) 5 µV
(b) 50 µV
(c) 5 mV
(d) 50 mV
Q 2
(b)
Q 3
(c)
Q 2
(d) Q
Exp. (c)
Exp. (b) The emf induced between ends of conductor
1 1 . × 10−4 × 5 × (1)2 e = BωL2 = × 02 2 2 = 0.5 × 10−4 V = 5 × 10−5 V = 50 µV
54. Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon [AIEEE 2003] (a) the rates at which currents are changing in the two coils (b) relative position and orientation of the two coils (c) the materials of the wires of the coils (d) the currents in the two coils
Exp. (b) Mutual inductance M between two coils is given two coils is given by M = µ 0 n1n2 πr12 L where, n1, n2 are number of turns, r1 is the radius of, coil and L is the length. From the above formula, it is clear that mutual inductance depends on distance between the coils and geometry ( πr 2 = area) of two coils.
55. When the current changes from +2 A to –2 A in 0.05 s, an emf of 8 V is induced in a coil. The coefficient of selfinduction of the coil is [AIEEE 2003]
(a) 0.2 H
(a)
(b) 0.4 H
(c) 0.8 H
(d) 0.1 H
In an LC circuit, the energy oscillates between + UE inductor (in the magnetic UB − field) and capacitor (in the electric field). U Emax [Maximum energy stored in capacitor]
Q2 2C U Bmax [Maximum energy stored in inductor] Li2 = 2 where, I is the current at this time. For the given instant, U E = U B =
q2 LI 2 = 2C 2
i.e.,
…(i)
From energy conservation, U E + U B = U Emax = U Bmax
q 2 1 2 Q2 + LI = 2C 2 2C Q or q= 2
⇒
2 q 2 Q2 [from Eq. (i)] = 2C 2C
57. The core of any transformer is laminated so as to
[AIEEE 2003]
(a) reduce the energy loss due to eddy currents (b) make it light weight (c) make it robust and strong (d) increase the secondary voltage
Exp. (a) The core of transformer is laminated to reduce energy loss due to eddy currents because induction is reduced by laminating.
58. The inductance between A and D is
Exp. (d) e = −L ∴
( −2 − 2 ) (4) dI = −L ⇒ 8=L dt 0.05 0.05 8 × 0.05 L= = 0.1 H 4
A
D 3H
3H
3H
317
Electromagnetic Induction and AC (a) 3.66 H (c) 0.66 H
(b) 9 H (d) 1 H
[AIEEE 2002]
E B
F
3H
C
3H
3H
(a) 4 A
Here, inductors are in parallel. 1 1 1 1 or L = 1 H = + + L 3 3 3
62.
the spring will
[AIEEE 2002]
(b) compress (d) None of these
Exp. (b) Due to flow of current in same direction in two adjacent sides, an attractive magnetic force will be produced due to which spring will get compressed.
⊗
60. The power factor of an AC circuit having resistanceR and inductance L (connected in series) and an angular velocity ω is [AIEEE 2002]
(b) (d)
Is = 2 A
So,
59. If a current is passed through a spring, then
R ( R + ω2 L2 )1/ 2 R 2
A conducting square loop of side L and resistanceR moves in v L its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced emf is [AIEEE 2002] (a) zero vBL (c) R
(b) RvB (d) vBL
Exp. (d) As the side BC is outside the field, no emf is induced across BC. Since, AB and CD are not cutting any flux, the emf induced across these two sides will also be zero. B A
(R − ω L ) 2
2 2 1/ 2
B v
L
Exp. (b) From the relation, tan φ = Power factor, cos φ = A
ωL B
(d) 10 A
Ip = 4 A, Np = 140 and N s = 280 I N 4 280 From the formula, p = s or = I s Np I s 140
∴
R ωL ωL (c) R
(c) 6 A
Given,
A B
(a)
(b) 2 A
Exp. (b)
C
3Ω
(a) expand (c) remain same
2 2
primary are 140 and that in the secondary are 280. If current in primary is 4 A, then that in the secondary is [AIEEE 2002]
3Ω 3Ω
2
61. In a transformer, number of turns in the
D
F D
E
1
=
R +ω L 1 + (ωL / R )2 1 1 R = = = 1 + tan2 φ 1 + (ωL / R )2 R 2 + ω2 L2
Exp. (d) A
R
⇒cos φ =
ωL R 1
D
1 + tan2 φ √(ωL)2 + (R)2 φ R
C
C
The side AD is cutting the flux and emf induced across this side is BvL with corner A at higher potential. [according to Lorentz force, force on the charges will be towards A and on − ve charges will be towards D]
15 Electromagnetic Waves 1. A plane electromagnetic wave travels in free space along the xdirection. The electric field component of the wave at a particular point of space and time is E = 6Vm −1 along ydirection. Its corresponding magnetic field component, B would be (a) (b) (c) (d)
2. The magnetic field of an electromagnetic wave is given by B = 1.6 × 10−6 cos(2 × 10 7 z + 6 × 1015t ) (2 $i + $j) Wbm −2 The associated electric field will be
[JEE Main 2019, 8 April ShiftII]
−8
2 × 10 T along z  direction 6 × 10−8 T along x  direction 6 × 10−8 T along z  direction 2 × 10−8 T along y  direction
(a) E = 48 . × 102 cos( 2 × 107 z − 6 × 1015 t ) $ ( −2 j + i$ ) Vm−1
[JEE Main 2019, 8 April ShiftI]
Exp. (a) Key Idea For an electromagnetic wave, ratio of magnitudes of electric and magnetic field is E =c B where, c is the speed of electromagnetic wave in vacuum. 8
Given,
E = 6 V/m, c = 3 × 10 ms
So,
B=
−1
6 E = = 2 × 10−8 T c 3 × 108
Also, direction of propagation of electromagnetic wave is given by n$ = E × B Here, n$ = $i and E = Unit vector of electric field ($j ) B = unit vector of magnetic field. $i = $j × B ⇒ ∴ B = k$ Hence, magnetic field components B = 2 × 10−8 k$ T = 2 × 10−8 T (along zdirection)
(b) E = 48 . × 102 cos( 2 × 107 z − 6 × 1015 t ) ( 2 $j + $i ) Vm−1 (c) E = 48 . × 102 cos( 2 × 107 z + 6 × 1015 t ) ( i$ − 2 $j) Vm−1 (d) E = 48 . × 102 cos( 2 × 107 z + 6 × 1015 t ) $ ( − i + 2 $j)Vm−1
Exp. (c) (c) Given, B = 16 . × 10−6 cos(2 × 107 z + 6 × 1015 t ) (2 $i + $j ) Wbm −2 From the given equation, it can be said that the electromagnetic wave is propagating negative $ zdirection, i.e. − k. Equation of associated electric field will be E = (Bc )cos(kz + ωt ) ⋅ n$ where, n$ = a vector perpendicular to B. So, E = B⋅c = 16 . × 10−6 × 3 × 108 = 4.8 × 102 V/m Since, we know that for an electromagnetic wave, E and B are mutually perpendicular to each other. So, E ⋅B = 0
319
Electromagnetic Waves From the given options, when n$ = $i − 2 $j E ⋅ B = (2 $i + $j ) ⋅ ($i − 2 $j ) = 0 $ = − $i + 2 $j Also, when n E ⋅ B = (2 $i + $j ) ⋅ (− $i + 2 $j ) = 0 But, we also know that the direction of propagation of electromagnetic wave is perpendicular to both E and B, i.e. it is in the direction of E × B. Again, when n$ = $i − 2 $j E × B = (2 $i + $j ) × ($i − 2 $j ) = − k$ and when n$ = − $i + 2 $j E × B = (2 $i + $j ) × (− $i + 2 $j ) = k$ But, it is been given in the question that the $ direction of propagation of wave is in −k. Thus, associated electric field will be E = 4.8 × 102 cos(2 × 107 z + 6 × 1015 t )($i − 2 $j ) Vm −1
3. The magnetic field of a plane electromagnetic wave is given by B = B [cos(kz − ωt )]i$ + B cos(kz + ωt )$j 0
1
where,B 0 = 3 × 10−5 T andB1 = 2 × 10−6 T. The rms value of the force experienced by a stationary chargeQ = 10−4 C at z = 0 is closest to [JEE Main 2019, 9 April ShiftI] (a) 0.1 N
(b) 3 × 10−2 N
(c) 0.6 N
(d) 0.9 N
Exp. (c) Given, magnetic field of an electromagnetic wave is B = B0 [cos(kz − ωt ] $i + B1 [cos(kz + ωt ] $j Here, B0 = 3 × 10−5 T and B1 = 2 × 10−6 T Also, stationary charge, Q = 10−4 C at z = 0 As charge is released from the rest at z = 0, in this condition. Maximum electric field, E0 = cB0 and E1 = cB1 So, E0 = c × 3 × 10−5 and
E1 = c × 2 × 10−6
Now,the direction of electric field of an electromagnetic wave is perpendicular to B and to the direction of propagation of wave (E × B) $ which is k. So, for E0 , ⇒
E 0 × B0 = k$ E 0 × $i = k$
⇒ E 0 = − $j Similarly,for E1 E1 × B1 = k ⇒ E1 × $j = k$ ⇒ E1 = $i −5 $ ∴E 0 = c × 3 × 10 (− j )NC −1 E = c × 2 × 10−6 (+ $i )NC −1 ∴ Maximum force experienced by stationary charge is Fmax = QE = Q(E 0 + E1 ) = Q × c [−3 × 10−5 $j + 2 × 10−6 $i ] ⇒ Fmax= 10–4 × 3 × 108 × (3 × 10−5 )2 + (2 × 10−6 )2 = 3 × 104 × 10−6 900 + 4 = 3 × 10−2 × 904 ≈ 0. 9 N ∴ rms value of experienced force is F 0.9 Frms = max = = 0707 . × 0.9 2 2 = 0. 6363 N ≈ 0. 6 N
4. The electric field of a plane electromagnetic wave is given by E = E 0i$ cos (kz ) cos (ωt )
The corresponding magnetic field B is then given by [JEE Main 2019, 10 April ShiftI] E0 $ j sin (kz ) sin (ωt ) C E (b) B = 0 $j sin (kz )cos(ωt ) C E0 $ (c) B = k sin (kz )cos(ωt ) C E (d) B = 0 $j cos(kz ) sin (ωt ) C (a) B =
Exp. (a) Key Idea For an electromagnetic wave, its electric field vector (E) and magnetic field vector (B) is mutually perpendicular to each other and also to its direction of propagation.
We know that, E × B represents direction of propagation of an electromagnetic wave ⇒ (E × B) v ∴From the given electric field, we can state that direction of propagation is along Zaxis and direction of E is along Xaxis. Thus, from the above discussion, direction of B must be Yaxis. From Maxwell’s equation,
320
JEE Main Chapterwise Physics ∂B ∂t ∂E ∂B Here, … (i) =− ∂Z ∂t and … (ii) B0 = E0 / C Given, E = E0 $i cos kz cos ωt − ∂E ⇒ = kE0 sin kz cos ωt ∂Z ∴Using Eq. (i), we get ∂B = kE0 sin kz cos ωt ∂t Integrating both sides of the above equation w.r.t. t, we get k B = E0sin kz sin ωt ⇒ ω E = 0 sin kz sin ωt C E0 ⇒ B= sin(kz) sin (ωt ) $j C ∇×E = −
5. An electromagnetic wave is represented by the electric field $ sin[ωt + (6y − 8z )]. E= E0 n
Taking
unit $ $ vectors in x , y and z directions to be i, $j, k, the direction of propagation s$ , is [JEE Main 2019, 12 April ShiftI]
3$i − 4$j (a) s$ = 5 $ − 3$j + 4k (c) s$ = 5
$ + 3$j − 4k (b) s$ = 5 $j − 3k $ 4 (d) s$ = 5
Exp. (c) Key Idea Standard expression of electromagnetic wave is given by …(i) E = E 0n$ [sin (ωt − k ⋅ r$)] Here, k is the propagation vector. Direction of propagation in this case is k$ .
Given expression of electromagnetic wave, E = E0n$ sin [ωt + (6 y − 8 z) ] …(ii) ⇒ E = E0n$ sin [ωt − (8 z − 6 y) ] Comparing Eq. (ii) with Eq. (i), we get k ⋅ r$ = 8 z − 6 y Here, r$ = x $i + y$j + zk$ and k = k $i + k $j + k k$ x
∴
y
=
−6$j + 8k$ −3$j + 4k$ = 10 5
6. A plane electromagnetic wave having a
frequency ν = 23.9 GHz propagates along the positive z direction in free space. The peak value of the electric field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave? [JEE Main 2019, 12 April ShiftII]
(a) B = 2 × 107 sin(0.5 × 103 z + 1.5 × 1011 t ) i$ (b) B = 2 × 10–7 sin(0.5 × 103 z − 1.5 × 1011 t ) $i $ (c) B = 60 sin(0.5 × 103 x + 1.5 × 1011 t )k (d) B = 2 × 10–7 sin(1.5 × 102 x + 0.5 × 1011 t ) $j
Exp. (b) In an electromagnetic wave, magnetic field and electric field are perpendicular to each other and both are also perpendicular to the direction of propagation of wave. Now, given direction of propagation is along zdirection. So, magnetic field is in either x or y direction. Also, angular wave number for wave is 2 π 2 πν k= = λ c 2 π × 23.9 × 109 = 3 × 108 ≈ 0.5 × 103 m −1 and angular frequency ω for wave is ω = 2 πν = 2 π × 23.9 × 109 Hz = 1. 5 × 1011 Hz
…(iii)
Magnitude of magnetic field is E 60 B0 = 0 = = 2 × 10−7 T c 3 × 108
…(iv)
As the general equation of magnetic field of an electromagnetic wave propagating in + z direction is given as,
z
k ⋅ r$ = xk x + yky + zkz
From Eqs. (iii) and (iv), we get xk x = zero ⇒ k x = 0 yky = − 6 y ⇒ ky = −6 zkz = 8 z ⇒ kz = 8 Hence, k = − 6$j + 8k$ So, direction of propagation, −6$j + 8k$ k s$ = k$ = = k 62 + 82
321
Electromagnetic Waves B = B0 sin(kz − ωt )$i or $j Thus, substituting the values of B0 , k and ω, we get ⇒ B = 2 × 10−7 sin(0. 5 × 103 z − 1. 5 × 1011t )$i or $j
7. A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x  direction. At a particular point in space and time, E = 6.3 $j V/m. The corresponding magnetic field B, at that point will be [ JEE Main 2019, 9 Jan ShiftI] $T (a) 18.9 × 108 k −8 $ (c) 18.9 × 10 k T
$T (b) 6.3 × 10−8 k −8 $ (d) 21 . × 10 k T
Exp. (d) Since, the magnitude of electric field vector and magnetic field vector of electromagnetic wave is related as, E  =c B where, c is the speed of light in vacuum. Here, E = 6. 3 and c = 3 × 108 m/s.  E 6.3  B = = ⇒ c 3 × 108 = 2.1 × 10− 8 Now, as we know for a EM wave, E and B are perpendicular to each other as well as to direction of propagation of wave (E × B). As here, direction of E is + ydirection i.e. $j and direction of propagation of wave is in + xdirection i.e., $i ∴ If B has to be perpendicular to E and direction of propagation, then it must in + zdirection $ i.e., k. $ T. Thus, magnetic field, B = 2 .1× 10− 8 k
8. In free space, the energy of electromagnetic wave in electric field is U E and in magnetic field is U B . Then [ JEE Main 2019, 9 Jan ShiftII] (a)U E = U B (c)U E < U B
(b)U E > U B U (d)U E = B 2
Exp. (a) Energy density of an electomagnetic wave in electric field, 1 …(i) U E = ε0 ⋅ E 2 2
Energy density of an electromagnetic wave in magnetic field, B2 …(ii) UB = 2µ 0 where, E = electric field, B = magnetic field, ε0 = permittivity of medium and µ 0 = magnetic permeability of medium. From the theory of electromagnetic waves, the relation between µ 0 and ε0 is 1 …(iii) c= µ 0 ε0 where, c = velocity of light = 3 × 108 m/s E and =c B Dividing Eq. (i) by Eq. (ii), we get 1 ε0 E 2 µ ε E2 UE = 2 = 0 02 1 2 1 UB B B × 2 µ0
…(iv)
…(v)
Using Eqs. (iii), (iv) and (v), we get UE c 2 = =1 UB c 2 Therefore,
UE = UB
9. If the magnetic field of a plane electromagnetic wave is given by x B = 100 × 10−6 sin 2 π × 2 × 1015 t − , c then the maximum electric field associated with it is (Take, the speed of light = 3 × 108 m/s) [ JEE Main 2019, 10 Jan ShiftI]
(a) (b) (c) (d)
6 × 104 N/C 4 × 104 N/C 3 × 104 N/C 4.5 × 104 N/C
Exp. (c) Given, instantaneous value of magnetic field x B = 100 × 10− 6 sin 2 π × 2 × 1015 t − c and speed of light, c = 3 × 108 ms − 1 For an electromagnetic wave, Emax = Bmax × c where, Emax = maximum value of the electric field. We get, Emax = 100 × 10− 6 × 3 × 108
322
JEE Main Chapterwise Physics = 3 × 104
N C
10. The electric field of a plane polarised electromagnetic wave in free space at time t = 0 is given by an expression. E ( x , y ) = 10$j cos[(6x + 8z )] The magnetic field B( x , z ,t ) is given by (where, c is the velocity of light) [ JEE Main 2019, 10 Jan ShiftII]
1 $ (6k − 8$i )cos[(6x + 8z + 10ct )] c 1 $ (b) (6k − 8i$ )cos[(6x + 8z − 10ct )] c 1 $ (c) (6k + 8$i )cos[(6x − 8z + 10ct )] c 1 $ (d) (6k + 8i$ )cos[(6x + 8z − 10ct )] c (a)
… (i)
Let magnetic field vector, $ , then direction of wave B = a$i + b$j + dk propagation is given by $) 10$j × (ai$ + b$j + dk E× B =  E′ B 10 × (a2 + b 2 + d 2 )1/ 2 $ + 10di$ − 10 ak … (iii) = 2 10(a + b 2 + d 2 )1/ 2 E 10 As,  B = = c c a2 + b 2 + d 2 = 10 / c
By putting this value in Eqs. (iii) and (ii), we get direction of propagation $ ) 6 i$ × 8 k $ c10(d i$ − ak = ⇒ 10 × 10 10 ⇒
d = 6 / c and a = − 8 / c
$ ) = 6 $i + 8 k $ E × B = $j × ( x $i + z k $ $ $ + zi = 6i + 8k $ or − x k then
where, phase angle is independent of time, i.e., phase angle at t = 0 is φ = 6 x + 8 z. Phase angle for B will also be 6 x + 8 z because for an electromagnetic wave E and B oscillate in same phase. Thus, direction of wave propagation $ $ 6$i + 8k 6$i + 8k … (ii) = = 2 2 10 6 + 8
We get,  B =
B=
Since, the magnetic field given is B( x, z, t ), this means B is in xzplane. ∴ Propagation of wave is in ydirection. [Qfor an electromagnetic wave, E ⊥ B ⊥ propagation direction] As Poynting vector suggests that E × B is $ ). parallel to (6 i$ + 8 k $ ), Let B = ( x$i + zk
Exp. (b) We are given electric field as E = 10 $jcos(6 x + 8 z)
6 $ 8$ 1 $ k − i = (6 k − 8 $i ) c c c As the general equation of magnetic field of an EM wave propagating in positive ydirection is given as, B = B0 cos (Ry − ωt )) 1 $ B = (6 k − 8 $i )cos(6 x + 8 z − 10ct ) ∴ c Alternate Solution Given, electric field is E( x, y), i.e. electric field is in xyplane which is given as QE = 10$jcos(6 x + 8 z) Hence,
x = − 8 and z = 6 1 $ ∴ B = (6 k − 8i$ )cos(6 x + 8 z − 10ct ) c  E Q B = c or
11. An electromagnetic wave of intensity
50 Wm −2 enters in a medium of refractive index ‘n’ without any loss. The ratio of the magnitudes of electric fields and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by [ JEE Main 2019, 11 Jan ShiftI]
(a) (c)
1 , n n 1 1 , n n
(b) ( n , n ) 1 (d) n , n
Exp. (d) In the free space, the speed of electromagnetic wave is given as, E 1 …(i) c= = 0 µ 0 ε0 B0 where, E0 and B0 are the amplitudes of varying electric and magnetic fields, respectively.
323
Electromagnetic Waves Now, when it enters in a medium of refractive index ‘n’, its speed is given as, 1 1 E …(ii) v= = = µε Kε0µ B where, K is dielectric strength of the medium. Using Eqs. (i) and (ii), we get v 1 …(iii) = c K (QFor a transparent medium, µ 0 ≈ µ ) Also, refractive index of medium is ‘n’ and is given as c v 1 …(iv) = n or = v c n ∴ From Eqs. (iii) and (iv), we get ...(v) n = K or K = n2 The intensity of a EM wave is given as, 1 I = ε0 E02c 2 1 and in the medium, it is given as I′ = Kε0 E 2 v 2 It is given that, I = I′ 1 1 ε0 E02c = Kε0 E 2 v ⇒ 2 2 2
or
E0 = Kv E c
…(vi)
From Eqs. (iv), (v) and (vi), we get 2
2 E0 = n = n E n
or E0 / E = n
1 B02 1 B2 ⋅ c= ⋅ v 2 µ0 2 µ0 B0 1 ⇒ = B n Alternate Solution We know that, E0 E =v = c and B B medium 0 Similarly,
air / vacuum
Also,
n=
c v
E0 / B0 c = =n E/B v E0 / E ⇒ =n B0 / B E B 1 . This is possible only if 0 = n and 0 = E B n
12. A 27 mW laser beam has a crosssectional area of 10 mm 2. The magnitude of the maximum electric field in this electromagnetic wave is given by [Take, permittivity of space, ε 0 = 9 × 10− 12 SI units and speed of light, c = 3 × 108 m / s] [ JEE Main 2019, 11 Jan ShiftII]
(a) 1 kV/m (c) 2 kV/m
(b) 0.7 kV/m (d) 1.4 kV/m
Exp. (d) Given, Power of laser beam (P) = 27mW = 27 × 10−3 W Area of crosssection (A) = 10 m m2 = 10 × 10−6 m2 Permittivity of free space (ε0 ) = 9 × 10−12 SI unit Speed of light (c) = 3 × 108 m/s Intensity of electromagnetic wave is given by the relation 1 I = ncε0 E 2 2 where, n is refractive index, for air n = 1. 1 ...(i) I = c ⋅ ε0 E 2 ∴ 2 P ...(ii) Also, I= A From Eqs. (i) and (ii), we get P 1 c ε0 E 2 = A 2 2P 2 or E = Ac ε0 or
E=
2 × 27 × 10−3 −6
10 × 10 × 3 × 108 × 9 × 10−12 ~ −1. 4 × 103 V/m = 1. 4 k V/m
13. A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propogating in the glass medium will be [ JEE Main 2019, 12 Jan ShiftI]
324
JEE Main Chapterwise Physics (a) 30 V/m (c) 10 V/m
(b) 6 V/m (d) 24 V/m
=
108 8 3 −12 × (3 × 10 ) 8.85 × 10
=
108 12 8.85 × 27 × 10
Exp. (d) Energy of a light wave ∝ Intensity of the light wave Since, intensity = εvA 2 where, ε is the permittivity of the medium in which light is travelling with velocity v and A is its amplitude. Since, only 4% of the energy of the light gets reflected. ∴96% of the energy of the light is transmitted. Etransmitted (Et ) = 96% of Eincident (Ei ) 96 ε0 εr vAt2 = × ε0 × c × A10 100 96 ε0 c 2 . . Ai At2 = 100 εr v ε0 v 96 v 2 c 2 = . 2 . Ai At2 = Q c εr 100 c v 96 v 2 2 . . Ai At = 100 c 96 1 Q µ = c = 15 . = × × 302 v 100 3 / 2 64 At = × 302 100 At = 24 V/m
14. The mean intensity of radiation on the surface of the sun is about 108 W / m 2. The rms value of the corresponding magnetic field is closest to [ JEE Main 2019, 12 Jan ShiftII] (a) 1 T
(c) 10− 4 T
(b) 102 T
(d) 10− 2 T
Exp. (c) Mean radiation intensity is 2 I = ε0c Erms = ε0c (cBrms )2
2 = ε0c 3 Brms
⇒
Brms =
I ε0c 3
Substituting the given values, we get
Erms = c Q B rms
≈ (10−8 ) ≈ 10−4 T
15. An EM wave from air enters a medium. The
z electric fields are E1 = E 01x$ cos 2 πν − t c in air and E2 = E 02x$ cos[k (2 z − ct )] in medium, where the wave number k and frequency ν refer to their values in air. The medium is nonmagnetic. If εr1 and εr2 refer to relative permittivities of air and medium respectively, which of the following options is correct? [JEE Main 2018] (a) (c)
εr 1
εr 2 εr 1 εr 2
=4
(b)
1 4
(d)
=
εr 1 εr 2 εr 1 εr 2
=2 =
1 2
Exp. (c) / Speed of progressive wave is given by, v = As electric field in air is, 2 πνz − 2 πνt E1 = E01x cos c 2 πν =c ∴ Speed in air = 2 πν c 1 Also, c= µ 0 εr1 ε0 In medium, E2 = E02 xcos (2 kz − kct ) kc c = ∴ Speed in medium = 2k 2 c 1 Also, = 2 µ 0 εr 2 ε0 As medium is nonmagnetic medium, µ medium = µ air On dividing Eq. (i) by Eq. (ii), we have εr 2 2= εr1
ω k
…(i)
…(ii)
325
Electromagnetic Waves εr1
⇒
εr 2
=
1 4
(d) Ultraviolet rays are absorbed by ozone.
18. During the propagation of electromagnetic
16. Arrange the following electromagnetic radiations per quantum in the order of increasing energy. A. Blue light C. Xray
B. Yellow light D. Radio wave [JEE Main 2016 (Offline)]
(a) D, B, A, C (c) C, A, B, D
(b) A, B, D, C (d) B, A, D, C
Exp. (a) As, we know energy liberated, E =
hc λ
1 λ So, lesser the wavelength, than greater will be energy liberated by electromagnetic radiations per quantum. As, order of wavelength is given by Xray, VIBGYOR, Radio waves (C) (A) (B) (D) ∴Order of electromagnetic radiations per quantum. D sin −1 −1
µ2 µ2 µ 22 −1 1 − 12 = µ1 µ2 µ 12
(d) θ > sin −1
µ 22 µ 12
−1
µ1 µ2
[JEE Main 2019, 12 April ShiftII]
(a) 5.33 (c) 10.67
Exp. (c) Key Idea The critical angle is defined as the angle of incidence that provides an angle of refraction of 90°. µ So, θ c = sin−1 2 µ1
(b) 16.00 (d) 1.80
Exp. (c) P1 I0 = I1 2
I0
P2
I0 I 2= 2
For total internal reflection, angle of incidence(i) at medium interface must be greater than critical angle (C). E µ1 µ2 D θ
30
°
P3
30°
90°– C
sinC =
2
s
I=I2cos260°
C
where,
co
60°
µ1 µ2
…(i)
When unpolarised light pass through polaroid P1, intensity obtained is I I1 = 0 2 where, I0 = intensity of incident light.
336
JEE Main Chapterwise Physics Now, this transmitted light is polarised and it pass through polariser P2 . So, intensity I2 transmitted is obtained by Malus law. ⇒ I2 = I1 cos 2 θ As angle of pass axis of P1 and P3 is 90° and angle of pass axis of P2 and P3 is 60°, so angle between pass axis of P1 and P2 is (90°− 60° ) = 30°. So, 2
I2 =
3 I I0 3 cos 2 30° = 0 × = I0 2 2 8 2
When this light pass through third polariser P3 , intensity I transmitted is again obtained by Malus law. 3 So, I = I2 cos 2 60° = I0 cos 2 60° 8
21. Consider a tank made of glass (refractive index is 1.5) with a thick bottom. It is filled with a liquid of refractive index µ. A student finds that, irrespective of what the incident angle i (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarised. For this to happen, the minimum value of µ is [ JEE Main 2019, 9 Jan ShiftI] i
2
3 1 3 I0 × = I0 2 8 32 I 32 So, ratio 0 = = 10.67 3 I =
n =1.5
different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16.
3 5 4 (c) 3
The intensity of the waves are in the ratio
Exp. (a)
20. Two coherent sources produce waves of
(a)
5 3 5 (d) 3
(b)
[ JEE Main 2019, 9 Jan ShiftI]
(a) 16 : 9
(b) 5 : 3
(c) 25 : 9
(d) 4 : 1
Exp. (c) Let the intensity of two coherent sources be I1 and I2 , respectively. It is given that, maximum intensity Imax 16 = = Imin 1 minimum intensity Since, we know, Imax = ( I1 + I2 )2 and Imin = ( I1 − Imax ( I1 + = Imin ( I1 −
∴We can write, ⇒
I1 +
I2
I1 −
I2
=
I2 )2 I2 )2
I2 )2 = 16
I2 = 4 I1 − 4 I2 I 5 5 I2 = 3 I1 , 1 = I2 3
Squaring both the sides, we get
In the given condition, the light reflected irrespective of an angle of incidence is never completely polarised. So, iC > i B where, iC is the critical angle. ...(i) ⇒ sin iC < sin i B From Brewster’s law, we know that µ 1.5 ...(ii) tan i B = wµ g = glass = µ µ water From Eqs. (i) and (ii), we get 1.5 1 < µ (1.5)2 + (µ)2
16 4 = 1 1
I1 + ⇒
Key Idea When a beam of unpolarised light is reflected from a transparent medium of refractive index µ, then the reflected light is completely plane polarised at a certain angle of incidence i B , which is known as Brewster’s angle.
⇒ I1 25 = 9 I2
(1.5)2 + µ 2 < 1.5 µ µ 2 + (1.5)2 < (1.5 µ)2 or µ
sin − 1 µ sin A − sin − 1 µ 1 (b) θ < sin − 1 µ sin A − sin − 1 µ 1 (c) θ > cos− 1 µ sin A + sin − 1 µ 1 (d) θ < cos− 1 µ sin A + sin − 1 µ
347
Optics Exp. (a)
∴
/ The ray will get transmitted through face AC if iAC < iC Consider the ray diagram as shown below.
r1 r2 µ
B
(a) 1 µm (c) 100 µm
C
A ray of light incident on face AB at an angle θ. r1 = Angle of refraction on face AB r2 = Angle of incidence at face AC For transmission of light through face AC i AC < iC
or
As µ increases, θ decreases. Hence, beam will bend upward.
0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is [JEE Main 2015]
N2
A θ
A − r1 < iC
or sin ( A − r1 ) < sin iC or sin ( A − r1 )
sin A − sin− 1 µ
1 sin θ > sin A − sin− 1 µ µ 1 θ > sin− 1 µ sin A − sin− 1 µ
45. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens principle leads us to conclude that as it travels, the light beam [JEE Main 2015]
(a) becomes narrower (b) goes horizontally without any deflection (c) bends downwards (d) bends upwards
Exp. (d) According to Snell's law, µ sinθ = constant
Y
25 cm
Now, applying Snell’s law at the face AB sin θ 1 × sin θ = µ sin r1 or sin r1 = µ ⇒
(b) 30 µm (d) 300 µm
Exp. (b)
1 A − r1 < sin−1 µ or
1 µ
46. Assuming human pupil to have a radius of
A N1
sinθ ∝
λ D where, D is the diameter of eye lens. θ = 122 .
Y
or
−2
25 × 10
=
. × 500 × 10− 9 122 . × 2 × 10− 2 025
Y = 30 × 10− 6 m = 30 µm
47. A thin convex lens made from crown glass
3 µ = has focal length f. When it is 2 measured in two different liquids having 4 5 refractive indices and . It has the focal 3 3 lengths f1 and f 2, respectively. The correct relation between the focal lengths is (a) (b) (c) (d)
[JEE Main 2014] f1 = f 2 < f f 1 > f and f 2 becomes negative f 2 > f and f 1 becomes negative f 1 and f 2 both become negative
/ It is based on lens maker’s formula i.e.,
1 1 1 = ( µ − 1) − R1 R2 f
Exp. (b) According to lens maker’s formula, when the lens in the air
348
JEE Main Chapterwise Physics 1 1 3 1 − = − 1 R1 R 2 f 2 ⇒ Here,
(ii) Angle of incidence (θ) must be greater than or equal to critical angle (C) i.e., µ C = sin −1 rarer µ denser
1 1 = ⇒f = 2x f 2x 1 1 1 = − x R1 R 2
Exp. (b)
In case of liquid,where refractive indices are 4/3 and 5/3, we get Focal lenght in first liquid, 1 1 µ s 1 = − 1 − R f1 µ l1 R 2 1 1 1 3/2 = − 1 f1 4/ 3 x
1 nwater b and nwater = a + 2 λ If frequency is less, then λ is greater and hence RI n( water) is less and therefore critical angle increases. So, they do not suffer reflection and come out at angle less than 90° of the air medium. sinC =
Here,
Air
⇒ F1 is positive. Nature of lens is not change. 1 1 1 1 = = = ⇒ f1 = 4 f f1 8 x 4(2 x) 4 f Focal length in second liquid, 1 1 µ s 1 = − 1 − f2 µ l 2 R1 R 2 3 /2 1 1 = − 1 f2 5/ 3 x ⇒ f2 is negative. Nature of lens change i.e., convex behave as concave.
48. A green light is incident from the water to the airwater interface at the critical angle (θ). Select the correct statement. [JEE Main 2014]
(a) The entire spectrum of visible light will come out of the water at an angle of 90° to the normal (b) The spectrum of visible light whose frequency is less than that of green light will come out of the air medium (c) The spectrum of visible light whose frequency is more than that of green light will come out to the air medium (d) The entire spectrum of visible light will come out of the water at various angles to the normal
/ For total internal reflection of light take place, then following conditions must be obeyed. (i) The ray must travel from denser to rarer medium.
Water
C Green
49. Two beams, A and B, of plane polarised light with mutually perpendicular planes of polarisation are seen through a polaroid. From the position, when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are I A and I B respectively, then I A /I B equals [JEE Main 2014]
(a) 3 (c) 1
(b) 3/2 (d) 1/3
Exp. (d) By law of Malus i.e., I = I0 cos 2 θ Finally
Initially IA
IA IB
Transmission axis
Now,
IB
Polaroid
Polaroid
Transmission axis
IA ′ = IA cos 2 30° IB ′ = IB cos 2 60°
As, ⇒
IA′ = IB′ IA cos 2 30° = IB cos 2 60° I 1 3 1 ⇒ A = IA = IB 4 4 IB 3
349
Optics 50. A beam of unpolarised light of intensity I 0 is
Exp. (c)
passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is [JEE Main 2013] I0 2 I (d) 0 8 (b)
(a) I 0 (c)
3 mm
I0 4
n=
3 2
Q n = c v
32 + (R − 3 mm)2 = R 2 ⇒ 32 + R 2 − 2 R(3 mm) + (3 mm)2 = R 2
Exp. ( c )
⇒
Relation between intensities is
∴
B 45°
(I0/2)
I0
IR
(Unpolarised)
A
⇒
I I 1 I IR = 0 cos 2 (45° ) = 0 × = 0 2 2 2 4
R ≈ 15 cm 1 3 1 = − 1 15 f 2 1 1 1 Q = (n − 1) − f R R 1 2 and R = ∞, R = −15 1 2 f = 30 cm
53. The graph between angle of deviation (δ )
51. Two coherent point sources S1 and S 2 are separated by a small distance d as shown. The fringes obtained on the screen will be [JEE Main 2013]
and angle of incidence (i ) for a triangular prism is represented by [JEE Main 2013] δ
δ
(a)
(b)
Screen O
d S1
δ
S2 D
(a) points (c) semicircle
O
I
(c)
(b) straight lines (d) concentric circles
For bright fringe, S1P − S 2 P = nλ So, fringes are concentric S1 S2 circles (centre of origin). d D [Q due to point source, wave fronts will be spherical]
(d) O
Exp. (d)
O
i
thickness at the centre is 3 mm. If speed of light in material of lens is 2 × 108 m /s, the focal length of the lens is [JEE Main 2013] (b) 20 cm (d) 10 cm
i
Exp. ( c ) P
We know that the angle of deviation depends upon the angle of incidence. If we determine experimentally, the angles of deviation corresponding to different angles of incidence and then plot i (on xaxis) and δ (on yaxis), we get a curve as shown in figure. y
52. Diameter of a planoconvex lens is 6 cm and
(a) 15 cm (c) 30 cm
i
δ
δ
δ δm i1
i
i
i2
x
350
JEE Main Chapterwise Physics Clearly, if angle of incidence is gradually increased from a small value, the angle of deviation first decreases, becomes minimum for a particular angle of incidence and then begins to increase.
54. In Young’s double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If I m is the maximum intensity, the resultant intensity I when they interfere at phase difference φ, is given by [AIEEE 2012]
Im 9 Im (b) 3 Im (c) 5 Im (d) 9 (a)
( 4 + 5cos φ) 1 + 2 cos2 φ 2 1 + 4 cos2 φ 2 1 + 8 cos2 φ 2
Exp. ( d ) Given,
a1 = 2 a2
⇒
I1 = 4I2 = 4I0
∴
Imax = ( I1 +
[Q I ∝ a2 ] I2 )2
= (3 I2 )2 = 9I2 = 9I0 Now,
I = I1 + I2 + 2 I1I2 cos φ = 4I0 + I0 + 2 4I0 ⋅ I0 cos φ = 5I0 + 4I0 cos φ I = m (5 + 4 cos φ) 9
Q I = Im 0 9
Im [1 + 4(1 + cos φ)] 9 I = m (1 + 8 cos 2 φ /2 ) 9 Qcos φ = 2 cos 2 φ − 1 2 =
55. An object 2.4 m infront of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus on film? [AIEEE 2012] (a) 7.2 m
(b) 2.4 m
(c) 3.2 m
(d) 5.6 m
Exp. ( d ) Shift in image position due to glass plate, 1 1 1 S = 1 − t = 1 − × 1 cm = cm µ 1.5 3 For focal length of the lens, 1 1 1 1 1 = − = − f v u 12 −240 1 20 + 1 or = f 240 240 cm ⇒ f= 21 Now, to get back image on the film, lens has to 1 35 form image at 12 − cm = cm such that the 3 3 glass plate will shift the image on the film. 1 1 1 As = − f v u 1 1 1 3 21 = − = − ⇒ u v f 35 240 48 × 3 − 7 × 21 = 1680 1 =− 560 ⇒ u = −5.6 m
56. At two points P and Q on screen in Young’s double slit experiment, waves from slits S1 λ and S 2 have a path difference of 0 and , 4 respectively. The ratio of intensities at P and [AIEEE 2011] Q will be (a) 3 : 2 (c) 2 :1
(b) 2 :1 (d) 4 :1
Exp. ( b ) Let I0 be intensity of light emitted from the source, then Resultant intensity I = 4I0 cos 2 φ / 2 ∴ Now, ∴ ⇒ and ∴
I1 = 4I0 cos 2 0 / 2 = 4I0
[Q φ = 0]
∆x = λ/ 4 2π 2π λ φ= × ∆x = × λ λ 4 π φ= 2 2π π ∆x I2 = 4 I0 cos 2 = 2 I0 Q φ = λ 4 I1 : I2 = 2 : 1
351
Optics 57. A beaker contains water upto a heighth1 and kerosene of heighth2 above water so that the total height of (water + kerosene) is (h1 + h2 ). Refractive index of water is µ 1 and that of kerosene is µ 2. The apparent shift in the position of the bottom of the beaker when viewed from above is [AIEEE 2011] 1 1 (a) 1 − h2 + 1 − h1 µ µ 1 2 1 1 (b) 1 + h2 h1 + 1 + µ1 µ2 1 1 (c) 1 − h1 + 1 − h2 µ1 µ2 1 1 (d) 1 + h1 h2 − 1 + µ1 µ2
1 We know, apparent shift, ∆h = 1 − h µ ∴ Apparent shift produced by water, 1 ∆h1 = 1 − h1 µ1 and apparent shift produced by kerosene, 1 ∆h2 = 1 − h2 µ2 1 1 ∴ ∆h = ∆h1 + ∆h2 = 1 − h1 + 1 − h2 µ1 µ2
58. When monochromatic red light is used instead of blue light in a convex lens, its focal length will [AIEEE 2011] (a) not depend on colour of light (b) increase (c) decrease (d) remain same
Exp. ( b ) 1 1 1 = (µ − 1) − f R1 R 2 Also, by Cauchy’s formula, B C 1 µ = A + 2 + 4 + ... ⇒ µ ∝ λ λ λ Hence,
portion of the sky through a Calcite crystal, the intensity of transmitted light varies as the crystal is rotated. Statement II The light coming from the sky is polarised due to scattering of sun light by particles in the atmosphere. The scattering is largest for blue light. [AIEEE 2011] (a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is false (c) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I (d) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I
Exp. ( c )
Exp. ( c )
As
59. Statement I On viewing the clear, blue
λblue < λred ⇒ µ blue > µ red fred > f blue
Scattering for blue light is largest and it is polarised by scattering. Also, for polarised light, I = I0 cos 2 θ
60. In a Young’s double slit experiment, the two slits act as coherent sources of waves of equal amplitude A and wavelength λ. In another experiment with the same arrangement, the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is I 1 and in the second case is I 2 , then I the ratio 1 is [AIEEE 2011] I2 (a) 4
(b) 2
(c) 1
(d) 0.5
Exp. ( b ) φ = 4I0 2 For incoherent sources, I2 = I0 + I0 = 2 I0 For coherent sources, I1 = 4I0 cos 2
∴
I1 =2 I2
61. Let the zxplane be the boundary between two transparent media. Medium 1 in z ≥ 0 has a refractive index of 2 and medium 2 with z < 0 has a refractive index of 3. A ray of light in medium 1 given by the vector
352
JEE Main Chapterwise Physics A = 6 3 i + 8 3 j − 10 k is incident on the plane of separation. The angle of refraction in medium 2 is [AIEEE 2011] (a) 45°
(b) 60°
(c) 75°
(d) 30°
Exp. ( a ) As refractive index for z > 0 and z ≤ 0 is different, xyplane should be boundary between two media. Angle of incidence, Az 1 = cos i = 2 2 2 2 A x + Ay + Az ∴
i = 60° sin i µ 2 3 From Snell’s law, = = sin r µ 1 2
sin r = ⇒
2 3 1 2 × sin 60° = × = = 45° 3 2 2 3 r = 45°
63. A car is fitted with a convex sideview mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a relative speed of 15m/s. The speed of the image of the second car as seen in the mirror of the first one is [AIEEE 2011] (a)
1 m/s 15
(b)10 m/s
(c) 15 m/s (d)
1 m/s 10
Exp. ( a ) 1 1 1 + = u v f Differentiate this equation w.r.t. t, we get 1 du 1 dv − 2 − 2 =0 u dt v dt dv v 2 du = − 2 ⇒ dt u dt v f But = u u −f For the mirror,
2
62. This question has a paragraph followed by two statements, Statement I and Statement II. Of the given four alternatives after the statements, choose the one that describes the statements. A thin air film is formed by putting the convex surface of a planeconvex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film. [AIEEE 2011] Statement I When light reflects from the airglass plate interface, the reflected wave suffers a phase change of π. Statement II The centre of the interference pattern is dark. (a) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I (c) Statement I is false, Statement II is true (d) Statement I is true, Statement II is false
Exp. ( b ) Both statements are correct but Statement II does not explain Statement I.
∴
f du dv = − dt u − f dt 2
1 0.2 ms −1 = × 15 = 15 − . − . 2 8 02
64. As the beam enters the medium, it will (a) diverge [AIEEE 2010] (b) converge (c) diverge near the axis and converge near the periphery (d) travel as a cylindrical beam
Exp. ( b ) As intensity is maximum at axis. Therefore, µ will be maximum and speed will be minimum on the axis of the beam. Hence, beam will converge.
65. The initial shape of the wavefront of the beam is
[AIEEE 2010]
(a) convex (b) concave (c) convex near the axis and concave near the periphery (d) planar
Exp. ( d ) For a parallel cylindrical beam, wavefront will be planar.
353
Optics 66. The speed of light in the medium is [AIEEE 2010]
(a) (b) (c) (d)
minimum on the axis of the beam the same everywhere in the beam directly proportional to the intensity I maximum on the axis of the beam
Exp. (c)
(a) 393.4 nm (c) 442.5 nm
(b) 885.0 nm (d) 776.8 nm
Exp. (c) 3 3 λ1 = × 590 4 4 1770 = = 442.5 nm 4
λ2 =
⇒
The speed of light in the medium is directly proportional to the intensity I.
67. In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45° with the xaxis meets the experimental curve at P. The coordinates of P will be f f (a) ( 2 f , 2 f ) (b) , (c) ( f , f ) 2 2
[AIEEE 2009]
(d) ( 4 f , 4 f )
Exp. (a)
v
(v ) u
For
69. A transparent solid cylinder rod has a refractive index of
2
. It is surrounded by 3 air. A light ray is incident at the midpoint of one end of the rod as shown in the figure. θ
The incident angle θ for which the light ray grazes along the wall of the rod is [AIEEE 2009]
(a) (b)
It is possible when object kept at centre of curvature.
Q 3Dλ1 = 4 Dλ 2 d d
3λ1 = 4λ 2
(c) (d)
1 sin 2 3 sin −1 2 2 sin −1 3 1 sin −1 3 −1
Exp. ( d ) u (u > f )
u=v u = 2 f, v = 2 f
68. A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young’s double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is [AIEEE 2009]
sin C =
3 2
sin r = sin (90° − C ) = cos C =
…(i) 1 2
sin θ µ 2 = sin r µ 1 sin θ =
1 2 1 × ⇒ θ = sin−1 3 3 2
70. A student measures the focal length of a convex lens by putting an object pin at a distanceu from the lens and measuring the distance v of the image pin. The graph between u and v plotted by the student should look like [AIEEE 2008]
354
JEE Main Chapterwise Physics v (cm)
v (cm)
Power of a lens is reciprocal of its focal length. Power of combined lens is
(b)
(a) O
P = P1 + P2 = − 15 + 5 O
u (cm)
v (cm)
u (cm)
v (cm)
(d)
(c) O
u (cm)
Exp. ( c ) Experimental observation at v = ∞, u = − f At u = ∞, v = f 1 1 1 − = = constant v u f So, (c) is the correct graph.
71. In a Young’s double slit experiment, the intensity at a point where the path λ (λ being the wavelength of difference is 6 the light used) is I . If I 0 denotes the maximum intensity, then I /I 0 is equal to [AIEEE 2007]
(a)
1 2
(b)
3 2
(c)
1 2
(d)
∴
f=
1 100 cm or f = − 10 cm = P −10
light and 1.525 for blue light. Let D1 and D2 be angles of minimum deviations for red and blue light respectively in a prism of this glass. Then, [AIEEE 2006] (b) D1 = D 2 (a) D1 < D 2 (c) can be less than or greater than D 2 depending upon the angle of prism (d)D1 > D 2
Exp. ( a ) D = ( µ − 1) A For blue light, µ is greater than that for red light, so D2 > D1. Aliter D1 = (1.520 − 1) A ⇒
3 4
D2 = (1.525 − 1) A D1 D1
74. A Young’s double slit experiment uses a
Exp. ( d ) 2π Phase difference = × Path difference λ 2π λ π i.e., φ= × = λ 6 3 φ As, I = Imax cos 2 2 φ I or = cos 2 2 Imax or
= − 10 D
73. The refractive index of glass is 1.520 for red O
u (cm)
Exp. ( b )
3 π I = cos 2 = 6 4 I0
[Q Imax = I0 ]
72. Two lenses of power –15D and +5D are in contact with each other. The focal length of the combination is [AIEEE 2007] (a) –20 cm (b) –10 cm (c) +20 cm (d) +10 cm
monochromatic source. The shape of the interference fringes formed on a screen is (a) hyperbola (c) straight line
(b) circle [AIEEE 2005] (d) parabola
Exp. (d) The shape of the interference fringes formed on a screen is parabola.
75. A fish looking up through the water sees the outside world, contained in a circular horizon. If the refractive index of water is 4/ 3 and the fish is 12 cm below the water surface, the radius of this circle (in cm) is [AIEEE 2005]
(a) 36 7
36 (b) 7
(c) 36 5
(d) 4 5
355
Optics Exp. ( b )
and
The situation is shown in figure. R
A
B
Thus,
θc θc
12 cm
O
or
1 AB ⇒ tan θC = µ OA sinθc sinθc = × OA ∴ AB = OA tan θC = OA cos θc 1 − sin2 θc sin θC =
1/ µ
=
1− 1/ µ 2 AB =
or
× OA =
OA µ2 − 1
=
OA 12 2
=
36 7
[AIEEE 2005]
black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [Take wavelength of light = 500 nm] [AIEEE 2005] (b) 1 m
(c) 6 m
(d) 3 m
∴
D=
1.22 × 5 × 10−7
(c) zero
Exp. ( a ) I = I0 cos 2 θ Intensity of polarised light = I0 / 2 ∴ Intensity of untransmitted light = I0 −
= 5 m ⇒ Dmax = 5 m A
C
77. A thin glass (refractive index 1.5) lens has optical power of –5 D in air. Its optical power in a liquid medium with refractive index 1.6 will be [AIEEE 2005] (a) 1 D
(b) –1 D
(d) I 0
(c) 25 D
45° 45°
(d) –25 D
B
Exp. ( a ) 1 1 1 = (µ − 1) − fa R1 R 2 1 1 = (1.5 − 1) − R1 R 2
I0 I0 = 2 2
face of a 90° prism and is totally internally reflected at the glassair interface. If the angle of reflection is 45°, we conclude that for the refractive index n as [AIEEE 2004]
yd λ y ≥ 1.22 ⇒ D≤ 1.22 λ D d
10−3 × 3 × 10−3
1 (b) I 0 4
79. A light ray is incident perpendicular to one
Exp. ( a ) We know
1 1 1.5 1 − = − 1 R1 R 2 fm 1.6 (1.5 − 1) fm = = −8 1.5 − 1 fa 1.6 −1 = 1.6 m fm = − 8 × fa = −8 × 5 Q f = 1 = − 1 m a 5 P µ 1.6 = = 1D Pm = fm 1.6
78. When an unpolarised light of intensity I 0 is
1 (a) I 0 2
76. Two point white dots are 1 mm apart on a
(a) 5 m
…(ii)
incident on a polarising sheet, the intensity of the light which does not get transmitted is
µ2 −1 4 − 1 3
∴
1 µg − µm 1 1 − = fm µ m R1 R 2
1 2 1 (c) n > 2 (a) n
2 (d) n < 2
356
JEE Main Chapterwise Physics
Exp. ( b ) For total internal reflection from glassair interface, critical angle C must be less than angle of incidence. i.e., C or n > 2 n> sin 45° (1/ 2 ) 1 sin C
80. A planoconvex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now, this lens has been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object [AIEEE 2004] (a) 20 cm (c) 60 cm
(b) 30 cm (d) 80 cm
Exp. ( a ) A planoconvex lens behaves as a concave mirror, if its one surface (curved) is silvered. The rays refracted from plane surface are reflected from curved surface and again refract from plane surface. Therefore, in this lens two refractions and one reflection occur. Let the focal length of silvered lens be F. 1 1 1 1 = + + F f f fm =
2 1 + f fm
where, f = focal length of lens before silvering, fm = focal length of spherical mirror. 1 2 2 [Q R = 2 fm ]…(i) = + F f R
Again given that, Size of object = Size of image i.e.,
O=I
∴
m= −
⇒
v = −1 u
v I = u O
or v = −u Thus, from lens formula, 1 1 1 = − F v u 1 1 1 = − ⇒ 10 − u u 1 2 =− 10 u ∴ u = − 20 cm Hence, to get a real image, object must be placed at a distance 20 cm on the left side of lens.
81. The angle of incidence at which reflected light is totally polarised for reflection from air to glass (refractive index n), is [AIEEE 2004]
(a) sin −1 (n ) 1 (b) sin −1 n 1 (c) tan −1 n (d) tan −1 (n )
Exp. ( d ) The particular angle of incidence for which reflected light is totally polarised for reflection from air to glass, is called the angle of polarisation (i p ) or Brewster’s law.
357
Optics Accordingly, ⇒
n = tan i p
85. To get three images of a single object, one
i p = tan−1 (n)
should have two plane mirrors at an angle of (a) 60° (c) 120°
where, n is refractive index of glass.
82. The
maximum number of possible interference maxima for slitseparation equal to twice the wavelength in Young’s doubleslit experiment, is [AIEEE 2004] (a) infinite (c) three
(b) five (d) zero
Exp. ( b ) For possible interference maxima on the screen, the condition is d sin θ = nλ
…(i)
Given, d = slitwidth = 2λ ∴
2λ sin θ = nλ
or 2sin θ = n The maximum value of sin θ is 1, hence n = 2 × 1= 2 Thus, Eq. (i) must be satisfied by 5 integer values i.e., –2, –1, 0, 1, 2. Hence, the maximum number of possible interference maxima is 5.
83. The image formed by an objective of a compound microscope is (a) (b) (c) (d)
[AIEEE 2003]
virtual and diminished real and diminished real and enlarged virtual and enlarged
Exp. ( c ) Objective of compound microscope is a convex lens. Convex lens form real and enlarged image when an object is placed between its focus and lens.
84. The earth radiates in the infrared region of the spectrum. The spectrum is correctly given by [AIEEE 2003] (a) (b) (c) (d)
Rayleigh Jeans law Planck’s law of radiation Stefan’s law of radiation Wien’s law
Exp. (a) The spectrum is correctly given by Rayleigh Jeans law
(b) 90° (d) 30°
[AIEEE 2003]
Exp. ( b ) 360° −1 θ where, θ is angle between mirrors. 360° ∴ 3= − 1 or θ = 90° θ Number of images, n =
86. To demonstrate the phenomenon of interference, we require two sources which emit radiations of [AIEEE 2003] (a) nearly the same frequency (b) the same frequency (c) different wavelength (d) the same frequency and having a definite phase relationship
Exp. ( d ) Sustained interference is possible with coherent sources only i.e., sources having same frequency and constant or zero phase difference.
87. If two mirrors are kept at 60° to each other, then the number of images formed by them is [AIEEE 2002] (a) 5 (c) 7
(b) 6 (d) 8
Exp. ( a ) 360° −1 θ where, θ is angle between mirrors. Thus, θ = 60° 360° So, number of images, n = − 1= 5 60°
Number of images, n =
[given]
88. Wavelength of light used in an optical instrument are λ 1 = 4000 Å and λ 2 = 5000 Å, then ratio of their respective resolving powers (corresponding to λ 1 and λ 2) is
[AIEEE 2002]
(a) 16 : 25 (c) 4 : 5
(b) 9 : 1 (d) 5 : 4
358
JEE Main Chapterwise Physics
Exp. ( d )
Exp. (b)
Resolving power of an optical instrument is inversely proportional to λ i.e., 1 RP ∝ λ Resolving power at λ1 λ 2 5000 = = 5: 4 = ∴ Resolving power at λ 2 λ1 4000
89. An astronomical telescope has a large aperture to (a) (b) (c) (d)
reduce spherical aberration have high resolution increase span of observation have low dispersion
[AIEEE 2002]
Resolving power of telescope, RP =
d 1.22 λ
90. Which of the following is used in optical fibres ? (a) (b) (c) (d)
[AIEEE 2002]
Total internal reflection Scattering Diffraction Refraction
Exp. ( a ) Optical fibres work on the principle of total internal reflection.
17 Dual Nature of Radiation 1. Two particles move at right angle to each
other. Their deBroglie wavelengths are λ 1 and λ 2, respectively. The particles suffer perfectly inelastic collision. The deBroglie wavelength λ of the final particle, is given by
So, by conservation of momentum and vector addition law, net momentum after collision, p12 + p22 + 2 p1 p2 cos 90° =
pnet = Since, p1 =
h h and p2 = λ1 λ2
[JEE Main 2019, 8 April ShiftI]
(a)
1 λ2
=
1
+
1
λ12 λ 22 λ + λ2 (c) λ = 1 2
(b) λ = λ1 λ 2 (d)
2 1 1 = + λ λ1 λ 2
Exp. (a) Given, deBroglie wavelengths for particles are λ1 and λ 2 . h h So, and λ 2 = λ1 = p1 p2 and momentum of particles are h h and p2 = p1 = λ1 λ2 Given that, particles are moving perpendicular to each other and collide inelastically. So, they move as a single particle.
h2
pnet =
So,
λ21
p2 pnet
+
h2
…(i)
λ22
Let the deBroglie wavelength after the collision is λ net , then h …(ii) pnet = λnet From Eqs. (i) and (ii), we get h λnet
=
h2 λ21
+
h2 λ22
⇒
1 λ2net
=
1 λ21
+
1 λ22
2. The electric field of light wave is given as 2 πx E = 10 −3 cos − 2 π × 6 × 1014 t x$ NC −1 . 5 × 10 −7
This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is [JEE Main 2019, 9 April ShiftI] Given, E (in eV) =
p1
p12 + p22
12375 λ( inA° )
(a) 0.48 V
(b) 0.72 V
(c) 2.0 V
(d) 2.48 V
Exp. (a)
360
JEE Main Chapterwise Physics Given, −3
E = 10
2 πx − 2 π × 6 × 1014 t x$ NC −1 cos −7 5 × 10
By comparing it with the general equation of electric field of light, i.e. $ we get E = E0 cos (kx − ωt ) x, 2π k= = 2π/λ 5 × 10−7 (from definition, k = 2 π / λ) …(i) ⇒ λ = 5 × 10−7 m = 5000 Å Or The value of λ can also be calculated as, after comparing the given equation of E with standard equation, we get ω = 6 × 1014 × 2 π ⇒
ν = 6 × 1014
As,
c = νλ 3 × 108 c = 5 × 10−7 m = 5000 Å λ= = ν 6 × 1014
⇒
[Q2 πν = ω]
According to Einstein’s equation for photoelectric effect, i.e. & hc …(ii) − φ = (KE )max = eV0 λ For photon, substituting the given values, hc 12375 eV [given] E= = λ λ hc 12375 or eV [using Eq. (i)] …(iii) = 5000 λ Now, substituting the values from Eq. (iii) in Eq. (ii), we get 12375 eV − 2eV = eV0 5000 ⇒ 2 .475 eV − 2 eV = eV0 or V0 = 2.475 V − 2 V = 0.475 V ⇒ V0 ≈ 0.48 V
3. A particle P is formed due to a completely inelastic collision of particles x and y having deBroglie wavelengths λ x and λ y , respectively. If x and y were moving in opposite directions, then the deBroglie wavelength of P is [JEE Main 2019, 9 April ShiftII]
(a) λ x − λ y (c)
λx λ y λx + λ y
(b)
λx λ y
λx − λ y
(d) λ x + λ y
Exp. (b) Initially, x
py
px
y
We have, deBroglie wavelengths associated with particles are h h and λ y = λx = px py ⇒
px =
h h and py = λx λy
Finally, particles collided to form a single particle. x y p
As we know that, linear momentum is conserved in collision. So, pp =p x − py ⇒pp =
h h − λ x λy
So, deBroglie wavelength of combined particle is h h h λp = = =  pp  hλ y − hλ x h h − λ x λy λ xλy =
λ xλy  λ x − λy 
4. 50 W/m 2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m 2 surface area will be close to (c = 3 × 108 m/s) [JEE Main 2019, 9 April ShiftII]
(a) 20 × 10−8 N
(b) 35 × 10−8 N
(c) 15 × 10−8 N
(d) 10 × 10−8 N
Exp. (a) Radiation pressure or momentum imparted per second per unit area when light falls is 2 I ; for reflection of radiation p = c I ; for absorption of radiation c where, I is the intensity of the light. In given case, there is 25% reflection and 75% absorption, so radiation pressure = force per unit area
361
Dual Nature of Radiation 25 2 I 75 I × + × 100 c 100 c 1 I 3 I 5 I 5 50 = × + × = × = × 2 c 4 c 4 c 4 3 × 108 =
= 20.83 × 10− 8 N ≈ 20 × 10− 8 N
5. In a photoelectric effect experiment, the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be Given, E (in eV) =
1237 λ( in nm)
[JEE Main 2019, 10 April ShiftI]
(a) 15.1 eV (b) 3.0 eV
(c) 1.5 eV
(d) 4.5 eV
Exp. (c) Given, threshold wavelength, λ 0 = 380 nm Wavelength of incident light, λ = 260 nm Using Einstein’s relation of photoelectric effect, …(i) (KE)max = eV0 = hν − hν0 1237 But ev (Given) hν = E = λ (nm) 1237 …(ii) E0 = eV ∴ λ 0 (nm) From Eqs. (i) and (ii), we get 1237 1237 (KE) max = E − E0 − eV λ0 λ 1 1 = 1237 − eV (λ in nm) λ λ0
Exp. (d) Radiation pressure over an absorbing surface is, I p= c where, I = intensity or energy flux and c = speed of light. If A = area of surface, then force due to radiation on the surface is IA F = p× A = c If force F acts for a duration of ∆t seconds, then momentum transferred to the surface is IA × ∆t ∆p = F × ∆t = c Here, I = 25 W cm −2 , A = 25 cm 2 , c = 3 × 108 ms −1, ∆t = 40 min = 2400 s So, momentum transferred to the surface, 25 × 25 × 2400 ∆p = 3 × 108 = 5 × 10−3 Ns
7. A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is [Given, Planck’s constant h = 6.6 × 10− 34 Js, speed of light c = 3.0 × 108 m / s] [JEE Main 2019, 10 April Shift2]
… (iii)
By putting values of λ and λ 0 in Eq. (iii), we get 1 1 (KE)max = 1237 − eV 260 380
(a) 1 × 1016 (b) 5 × 1015 (c) 1.5 × 1016 (d) 2 × 1016
Exp. (b)
380 − 260 = 1237 × eV 380 × 260 ⇒
(KE)max = 15 . eV
⇒
6. Light is incident normally on a completely absorbing surface with an energy flux of 25 W cm − 2. If the surface has an area of 25 cm 2, the momentum transferred to the surface in 40 min time duration will be [JEE Main 2019, 10 April ShiftII]
(a) 3.5 × 10− 6 N s
(b) 6.3 × 10− 4 N s
(c) 1.4 × 10− 6 N s
(d) 50 . × 10− 3 N s
Energy Time Number of photons emitted × Energy of one photon = Time NE N P= = ⋅E t t
Power of laser is given as, P =
So, number of photons emitted per second N P = = t E P Pλ Q E = hν = hc = = λ hc / λ hc Here, h = 6.6 × 10−34 Js, λ = 500 nm = 500 × 10−9 m c = 3 × 108 ms −1
362
JEE Main Chapterwise Physics P = 2 mW = 2 × 10−3 W ∴
−3
−9
N 2 × 10 × 500 × 10 = 5.56 × 1015 = t 6.6 × 10−34 × 3 × 108 ≈ 5 × 1015 photons per second
8. The stopping potential V 0 (in volt) as a
function of frequency ( ν ) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be (Take, Planck’s constant( h ) = 6.63 × 10−34 Js, electron charge, e = 1.6 × 10−19 C] [JEE Main 2019, 12 April ShiftI] 3.0
∴Work function, W = hν0 ⇒ W = 6.63 × 10−34 × 4 × 1014 J 6.63 × 4 × 10−20 eV ⇒ W = . × 10−19 16 eV ≈ 166 = 1657 . . eV
9. In an electron microscope, the resolution
2.0 V0
W 6.63 × 10−34 × 4 × 1014 J = e e or W = 6.63 × 4 × 10−20 J 6.63 × 4 × 10−20 eV or eV = 1657 . W = . × 10−19 16 ∴ W = 166 . eV Alternate Solution From graph, threshold frequency, ν0 = 4 × 1014 Hz (where, V0 = 0) ⇒
that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10−12 m, the minimum electron energy required is close to
1.0
2
4
6 8 ν(1014Hz)
(a) 1.82 eV (b) 1.66 eV (c) 1.95 eV (d) 2.12 eV
Exp. (b) Given, Planck’s constant, h = 6.63 × 10−34 Js, e = 16 . × 10−19 C and there is a graph between stopping potential and frequency. We need to determine work function W . Using Einstein’s relation of photoelectric effect, (KE)max = eV0 = hν − hν0 = hν − W [QW = hν0 ] h W or V0 = ν − e e From graph at V0 = 0 and ν = 4 × 1014 Hz (V0)
[JEE Main 2019, 10 Jan ShiftI]
10
(a) 500 keV (c) 100 keV
Exp. (d) Given, resolution achieved in electron microscope is of the order of wavelength. So, to resolve 7.5 × 10− 12 m seperation wavelength associated with electrons is λ = 7.5 × 10− 12 m ∴Momentum of electrons required is h p= λ or kinetic energy of electron must be p2 (h / λ )2 KE = = 2m 2m Substituting the given values, we get
Stopping potential
= = (ν0)
ν
Threshold frequency
∴
0=
W 6.63 × 10−34 × 4 × 1014 − e e
(b) 1 keV (d) 25 keV
6.6 × 10− 34 − 12 7.5 × 10
2
. × 10− 31 2 × 91
J (6.6 × 10− 34 )2
− 31
2 × 91 . × 10
eV × (7.5 × 10− 12 )2 × (16 . × 10− 19 ) (Q1 eV = 1.6 × 10− 19 J)
= 26593.4 ~ − 26.6 × 103 eV ~ − 26 keV which is nearest to 25 keV.
363
Dual Nature of Radiation 10. A metal plate of area 1 × 10−4
m 2 is illuminated by a radiation of intensity 16 m W/m 2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photoelectrons. The number of emitted photoelectrons per second and their maximum energy, respectively will be (Take, 1 eV = 1.6 × 10−19 J) [JEE Main 2019, 10 Jan ShiftII]
(a)1011 and 5 eV (c)1010 and 5 eV
(b)1012 and 5 eV (d)1014 and 10 eV
Exp. (a) We know that, intensity of a radiation I with energy ‘E’ incident on a plate per second per unit area is given as dE dE = IdA or IA ⇒ ⇒ I= dA × dt dt i.e., energy incident per unit time = IA Substituting the given values, we get dE = 16 × 10− 3 × 1 × 10− 4 dt dE …(i) = 16 × 10− 7 W dt Using Einstein’s photoelectric equation, we can find kinetic energy of the incident radiation as 1 E = mv 2 + φ 2 (Here, φ is work function of metal) or E = KE + φ KE = E − φ = 10 eV − 5 eV …(ii) ⇒ KE = 5 eV Now, energy per unit time for incident photons will be Q E = Nhν • dN dE or hν N …(iii) = hν ∴ dt dt From Eqs. (i) and (iii), we get •
•
E = 10 eV, so
equal to 10−3 times, the wavelength of a photon of frequency 6 × 1014 Hz, then the speed of electron is equal to (Take, speed of light = 3 × 108 m/s, Planck’s constant = 6.63 × 10−34 Js and mass of electron = 9.1 × 10−31 kg) [JEE Main 2019, 11 Jan ShiftI]
(a) 1.45 × 106 m/s
(b) 1.8 × 106 m/s
(c) 11 . × 106 m/s
(d) 1.7 × 106 m/s
Exp. (a) Wavelength of the given photon is given as, 3 × 108 c = m λp = νp 6 × 1014 = 5 × 10− 7m
•
N(10 × 1.6 × 10− 19 ) = 16 × 10− 7 ⇒ N = 1012 QOnly 10% of incident photons emit electrons. So, emitted electrons per second are 10 × 1012 = 1011 100
…(i)
As, it is given that, deBroglie wavelength of the electron is [Qusing Eq. (i)] λe = 10− 3 × λ p = 5 × 10− 10 m Also, the deBroglie wavelength of an electron is given as, h h λe = = p mve ⇒
ve =
h λe me
Substituting the given values, we get 6.63 × 10− 34 = m/s . × 10− 31 5 × 10− 10 × 91 = 145 . × 106 m / s
12. In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close hc to = 1240 nmV e
•
hν N = 16 × 10− 7 or E N = 16 × 10− 7 But
11. If the deBroglie wavelength of an electron is
[JEE Main 2019, 11 Jan ShiftII]
(a) 0.5 V
(b) 2.0 V
Exp. (d) Given, λ1 = 300 nm λ 2 = 400 nm hc = 1240 nm e
(c) 1.5 V
(d) 1.0 V
364
JEE Main Chapterwise Physics Using Einstein equation for photoelectric effect, …(i) E = hν = φ + eV0 (here, φ is work function of the metal and V0 is stopping potential) For λ1 wavelength’s wave, E1 = hν1 = φ + eV01 hc ...(ii) or = φ + eV01 λ1 hc …(iii) Similarly, = φ + eV02 λ2 From Eqs. (ii) and (iii), we get 1 1 hc − = e(V01 − V02 ) λ λ 1 2 1 hc 1 or − = ∆V e λ1 λ 2 By using given values, 1 1 nm V ∆V = 1240 − 300 400 nm 1 V = 1240 × 1200 ⇒ ∆V = 1.03. V ≈ 1V
13. A particle A of mass ‘m’ and charge ‘q’ is accelerated by a potential difference of 50 V. Another particle B of mass ‘4m’ and charge ‘q’ is accelerated by a potential difference of 2500 V. The ratio of deBroglie λ wavelengths A is close to λB [JEE Main 2019, 12 Jan ShiftI]
(a) 4.47 (c) 0.07
(b) 10.00 (d) 14.14
Exp. (d) deBroglie wavelength associated with a moving charged particle of charge q is h h λ= = p 2 mqV where, V = accelerating potential. Ratio of deBroglie wavelength for particle A and B is, mBq B VB λA = mA q A VA λB =
mB q V . B . B mA q A VA
Substituting the given values, we get 4m q 2500 . . = m q 50 = 2 × 1 × 5 × 1414 . = 14.14
14. In a FrankHertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to [JEE Main 2019, 12 Jan ShiftII]
(a) 250 nm (c) 1700 nm
(b) 2020 nm (d) 220 nm
Exp. (a) Minimum wavelength occurs when mercury atom deexcites from highest energy level. ∴ Maximum possible energy absorbed by mercury atom = ∆E = 5.6 − 07 . = 4.9 eV Wavelength of photon emitted in deexcitation is hc 1240 eVnm λ= ≈ ≈ 250 nm E 4.9 eV Note FrankHertz experiment was the first electrical measurement to show quantum nature of atoms. In a vacuum tube energatic electrons are passed through thin mercury vapour film. It was discovered that when an electron collided with a mercury atom, it loses only a specific quantity (4.9 eV) of it’s kinetic energy. This experiment shows existence of quantum energy levels.
15. When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photocurrent is −V 0 / 2. When the surface is illuminated by monochromatic light of frequency ν /2, the stopping potential is −V 0. The threshold frequency for photoelectric emission is [JEE Main 2019, 12 Jan ShiftII]
4 ν 3 3ν (c) 2 (a)
(b) 2 ν (d)
5ν 3
Exp. (c) Relation between stopping potential and incident light’s frequency is eV0 = hf − φ0 where, V0 is the stopping potential and φ0 is the the work function of the photosensitive surface.
365
Dual Nature of Radiation
…(i) …(ii)
hydrogen atom emit radiation to come to the ground state. Let λ n , λ g be the deBroglie wavelength of the electron in the nth state and the ground state, respectively. Let Λn be the wavelength of the emitted photon in the transition from thenth state to the ground state. For large n, (A , B are constants) [JEE Main 2018]
(c) Λ2n ≈ A + Bλ2n
λ2 1 − g 2 mλ2g λ2n h2
⇒ ⇒
Λn =
2 mλ2g c λ2 1 + g h λ2n
⇒
Λn ~ − A+
2 mcλ2g where, A = h constants.
B λ2n
2 mcλ4g and B = are h
17. An electron beam is accelerated by a potential differenceV to hit a metallic target to produce Xrays. It produces continuous as well as characteristic Xrays. If λ min is the smallest possible wavelength of Xrays in the spectrum, the variation of log λ min with log V is correctly represented in [JEE Main 2017 (Offline)]
(b) log V
(c)
logλmin
n
−1
log V
(d)
v 3
i.e.
1/ 2
20. Match List I (fundamental experiment) with
Here, u B = 0 (Particle at rest) and for elastic collision e = 1 v − vA 1= B ∴ ⇒ v = vB − vA K (ii) v From Eqs. (i) and (ii), we get v 4v and v B = vA = 3 3 h v λA 4/ 3 mv A Hence, = = B = =2 h v 2 2 /3 λB A m . v B 2
4 (a) > v 3
According to Einstein’s photoelectric emission of light, E = (KE)max + φ hc As, = (KE)max + φ λ 3λ If the wavelength of radiation is changed to , 4 then 4 hc 4 φ = (KE)max + + φ ⇒ 3 3 λ 3
1/ 2
1/ 2
Exp. ( a ) / According to the law of conservation of energy, i.e. Energy of a photon (hν) = Work function (φ) + Kinetic energy of the 1 photoelectron mv2 max 2
List II (its conclusion) and select the correct option from the choices given below the list. [JEE Main 2015]
List I
List II
A. FranckHertz experiment
1. Particle nature of light
B. Photoelectric experiment
2. Discrete energy levels of atom
C. DavissonGermer experiment
3. Wave nature of electron 4. Structure of atom
A B (a) 1 4 (c) 2 1
C 3 3
A (b) 2 (d) 4
B 4 3
C 3 2
Exp. ( c ) (A) FranckHertz experiments is associated with discrete energy levels of atom. (B) Photoelectric experiment is associated with particle nature of light. (C) DavissonGermer experiment is associated with wave nature of electron.
21. The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10−4 T.
367
Dual Nature of Radiation If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to [JEE Main 2014]
(a) 1.8 eV
(b) 1.1 eV
(c) 0.8 eV
(d) 1.6 eV
/ The problem is based on frequency dependence of photoelectric emission. When incident light with certain frequency (greater than on the threshold frequency) is focus on a metal surface, then some electrons are emitted from the metal with substantial initial speed.
Exp. (b) When an electron moves in a circular path, then mv r 2e 2 B2 m2 v 2 Radius, r = ⇒ = eB 2 2 (mv )2 r 2e 2 B2 KEmax = ⇒ =(KE)max 2m 2m Work function of the metal (W), i.e., W = hν − KEmax r 2e 2 B2 1 r 2eB2 1.89 − φ = eV = eV 2m 2 2m [hν → 1.89 eV, for the transition on from third to second orbit of Hatom] . × 10−19 × 9 × 10−8 100 × 10−6 × 16 = . × 10−31 2 × 91 . ×9 16 = 189 . − 0.79 = 11 . eV . − φ= 189 . 2 × 91
22. The anode voltage of a photocells kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of photocell varies as follows [JEE Main 2013] I
I
(a)
(b) λ
λ
I
I
(c)
(d) λ
λ
Exp. ( d ) As λ is increased, there will be a value of λ above which photoelectron will be cease to come out, so photocurrent will becomes zero.
23. This question has Statement I and Statement II. Of the four choices given the statements, choose the one that describes the two statements. [AIEEE 2012] Statement I DavissonGermer experiment established the wave nature of electrons. Statement II If electrons have wave nature, they can interfere and show diffraction. (a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is false (c) Statement I is true, Statement II is true; Statement I is the correct explanation of Statement I (d) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I
Exp. ( c ) Davisson and Germer experimentally established wave nature of electron by observing diffraction pattern while bombarding electrons on Nicrystal.
24. This question has Statement I and Statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement I A metallic surface is irradiated by a monochromatic light of frequency ν > ν 0 (the threshold frequency). The maximum kinetic energy and the stopping potential are K max and V 0, respectively. If the frequency incident on the surface is doubled, both the K max and V 0 are also doubled. Statement II The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light. [AIEEE 2011] (a) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I (c) Statement I is false, Statement II is true (d) Statement I is true, Statement II is false
Exp. ( c )
368
JEE Main Chapterwise Physics If
Kmax = eV0 = h(ν − ν0 ) ν′ = 2 ν
∴
′ Kmax = eV0′ = h(2ν − ν0 )
∴
= 2Kmax + hν0 ′ > 2 Kmax and Kmax
crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure). [AIEEE 2008]
ν′0 > 2 ν0
From graph, we can predict P that Kmax and ν do not Kmax correspond linear graph from origin. O
Outgoing electrons
Incoming electrons
Aliter
i
90° – I ν
90°
d
25. If a source of power 4 kW
produces 1020 , photons/second, the radiation belong to a part of the spectrum called [AIEEE 2010] (a) Xrays (c) microwaves
(b) ultraviolet rays (d) γrays
As power = Energy × time Q energy = power 4 × 103 = hf × 1020 time
Exp. ( a ) or
f=
4 × 103 20
10
× 6.023 × 10−34
f = 6.64 × 1016 Hz The obtained frequency lies in the band of Xrays.
26. The surface a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is [AIEEE 2009] (hc = 1240 eVnm) (a) 3.09 eV (b) 1.42 eV (c) 151 eV (d) 1.68 eV
Exp. ( b ) Q ⇒ ⇒ ⇒
KEmax = eV0 1 mv 2 = eV0 = 1.68 eV 2 hc 1240 eV nm hν = = = 3.1 eV 400 nm λ 3.1 eV = W0 + 1.6 eV [from Einstein equation, E = W0 + Kmax ] W0 = 1.42 eV
æ Directions
Q. Nos. 27 and 29 are based on the following paragraph. Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from
90° + 90°
Crystal plane
27. Electrons accelerated by potential V are diffracted from a crystal. If d = 1 Å and i = 30°,V should be about (h = 6.6 × 10−34 Js, me = 9.1 × 10−31 kg, e = 1.6 × 10−19 C
[AIEEE 2008]
(a) 2000 V
(b) 50 V
(c) 500 V
(d) 1000 V
Exp. ( b ) For constructive interference, 2 d cos i = nλ =
h 2 meV
On substituting values, we get ~ 50 V V—
28. If a strong diffraction peak is observed when electrons are incident at an angle i from the normal to the crystal planes with distance d between them (see figure), deBroglie wavelength λ dB of electrons can be calculated by the relationship (n is an integer) [AIEEE 2008] (a) d sin i = nλ dB (c) 2d sin i = nλ dB
(b) 2d cosi = nλ dB (d) d cosi = nλ dB
Exp. ( b ) Expression is given by 2d cos i = nλdB .
29. In an experiment, electrons are made to pass through a narrow slit of width d comparable to their deBroglie wavelength. They are detected on a screen at a distance D from the slit (see figure). [AIEEE 2008]
369
Dual Nature of Radiation Exp. (a) y=0 d
D
Which of the following graphs can be expected to represent the number of electrons N detected as a function of the detector position y ( y = 0 corresponds to the middle of the slit) ? y
hν0 = 6.2eV, eV0 = 5 eV From Einstein’s photoelectric equation hν = hν0 + eV0 = 6.2 + 5 = 11.2 eV hc ⇒ = 11.2 eV λ hc or λ= = 1108.9 Å 112 . Which belongs to ultraviolet region.
32. The time taken by a photoelectron to come out after the photon strikes is approximately (a) 10−4 s
y
−16
(c) 10 (a)
d
N
(b)
d
N
y
(b) 10−10 s
[AIEEE 2006]
(d) 10−1 s
s
Exp. (b) The photoelectric effect is an instantaneous phenomenon (experimentally proved). It takes approximate time of the order of 10−10 s.
y
33. The anode voltage of a photocell is kept (c)
N
d
(d)
d
N
fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows [AIEEE 2006]
Exp. ( d ) As diffraction pattern has to be wider than slit width, so (d) is the correct option.
(a)
ν (a) c
(b) hνc
(c)
hν
c2
hν (d) c
h hν = λ c
31. The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in [AIEEE 2006] (a) (b) (c) (d)
ultraviolet region infrared region visible region Xray region
(c)
I O
O
λ
(d) None of these λ
Exp. ( d )
Exp. ( d ) The momentum of the photon, p =
λ
O
30. Photon of frequency ν has a momentum associated with it. If c is the velocity of light, the momentum is [AIEEE 2007]
(b) I
I
According to the photoelectric effect in a photocell, if a light of wavelength λ is incident on a cathode, then electrons are emitted, which constitute the photoelectric current. Photocell is based on the principle of photoelectric effect. As the wavelength of light changes, there is no change in number of electrons emitted and hence, no change in current (plate current of photocell). Thus, the two wavelength of incident light and plate current are independent to each other. Plate current depends on intensity of light used. Hence option (d) is true.
370
JEE Main Chapterwise Physics
34. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity, then [AIEEE 2005]
and ⇒
(a) its velocity will decrease (b) its velocity will increase (c) it will turn towards right of direction of motion (d) it will turn towards left of direction of motion
⇒
Exp. (a)
or
When E, v and B are all along same direction, then magnetic force experienced by electron is zero while electric force is acting opposite to velocity of electron, so velocity of electron will decrease. QLorentz force = 0 BE + ∴ θ = 0 − (ν)e qvBsinθ = 0 ⇒
35. A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed 1/2 m away, the number of electrons emitted by photocathode would [AIEEE 2004] (a) (b) (c) (d)
decrease by a factor of 4 increase by a factor of 4 decrease by a factor of 2 increase by a factor of 2
36. If the kinetic energy of a free electron doubles, its deBroglie wavelength changes by the factor [AIEEE 2005]
Exp. ( c )
(b) 2
(c)
K1 K1 λ2 = = [Q K 2 = 2 K1 ] 2 K1 K2 λ1 λ λ2 1 or λ 2 = 1 = 2 λ1 2
37. According
to Einstein’s photoelectric equation, the plot of the kinetic energy of the emitted photoelectrons from a metal Versus the frequency of the incident radiation gives a straight line whose slope [AIEEE 2004]
(a) depends on the nature of the metal used (b) depends on the intensity of the radiation (c) depend both on the intensity of the radiation and the metal used (d) is the same for all metals and independent of the intensity of the radiation
Einstein’s photoelectric equation is KEmax = hν − φ
We know intensity depends on source distance. I2 (r1 )2 as I ∝ 1 ∴ = 2 I1 (r2 ) r 2 2 I2 (1) = ⇒ I1 (1 / 2 )2 I2 = 4 I1 Now, since number of electrons emitted per second is directly proportional to intensity, so number of electrons emitted by photocathode would increase by a factor of 4.
1 2
Thus,
mv = 2 mK h 1 ⇒ λ∝ λ= 2 mK K
Exp. ( d )
Exp. ( b )
(a)
h mv (mv )2 1 K = mv 2 = 2 2m
We know that, λ =
1 2
(d)
2
…(i)
KE v c=φ
The equation of line is …(ii) y = mx + c Comparing above two equations, we get m = h, c = − φ Hence, slope of graph is equal to Planck’s constant (nonvariable) and does not depend on intensity of radiation. But interception ‘c’ depends on work function ‘φ’.
38. The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately [AIEEE 2004]
371
Dual Nature of Radiation (a) 540 nm (c) 310 nm
(b) 400 nm (d) 220 nm
Exp. ( a )
Exp. (c) 6.6 × 10−34 × 3 × 108 hc hc = φ0 ⇒ λmax = = φ0 λ0 4 × 1.6 × 10−19 = 310 nm
We know, Einstein’s equations hf = hf0 + 1 mv2 2
Energy of incident light
39. A charged oil drop is suspended in uniform
field of 3 × 104 Vm −1 so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10−15 kg and g = 10 ms −2) [AIEEE 2004] −18
(a) 3.3 × 10
C
(c) 1.6 × 10−18 C
−18
(b) 3.2 × 10
C
(d) 4.8 × 10−18 C
Kinetic energy Work function
⇒ and ∴
2 hf1 2 hf0 − m m 2 hf2 2 hf0 2 v2 = − m m 2h [f1 − f2 ] v12 − v 22 = m v12 =
41. Formation of covalent bonds in compounds
Exp. ( a )
exhibits
In steady state, F = qE
(a) (b) (c) (d)
[AIEEE 2002]
wave nature of electron particle nature of electron both wave and particle nature of electron None of the above
Exp. ( a ) mg
Electric force on drop = Weight of drop i.e., qE = mg mg 9.9 × 10−15 × 10 q= = ⇒ E 3 × 104 = 3.3 × 10−18 C
40. Two identical, photocathodes receive light of frequencies f1 and f 2. If the velocities of the photoelectrons (of mass m) coming out are respectivelyv1 and v 2, then [AIEEE 2003]
2h ( f1 − f 2 ) m 1/ 2 2h (b) v1 + v 2 = ( f + f 2 ) m 1 2h (c) v12 + v 22 = ( f1 + f 2 ) m 1/ 2 2h (d) v1 − v 2 = ( f 1 − f 2 ) m (a) v12 − v 22 =
Formation of covalent bonds due to the wave nature of particles is done in compounds.
42. Sodium and copper have work functions 2.3 eV and 4.5 eV, respectively. Then, the ratio of the wavelengths is nearest to (a) 1 : 2 (c) 2 : 1
(b) 4 : 1 (d) 1 : 4
[AIEEE 2002]
Exp. ( c ) hc λ [here, they are interested in asking threshold wavelength] where, h = Planck’s constant, c = velocity of light. WNa λCu Therefore, = WCu λNa λNa WCu 4.5 or = = 2 (nearly) = λCu WNa 2.3 Work function, W =
372
JEE Main Chapterwise Physics
18 Atoms and Nuclei 1. Radiation coming from transitions n = 2 to +
n =1 of hydrogen atoms fall on He ions in n = 1 and n = 2 states. The possible transition of helium ions as they absorb energy from the radiation is [JEE Main 2019, 8 April ShiftI] (a) n = 2 → n = 3 (c) n = 2 → n = 5
(b) n = 1 → n = 4 (d) n = 2 → n = 4
Exp. (d)
2. A nucleus A, with a finite deBroglie wavelength λ A , undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same directions as that of A, whileC flies in the opposite direction with a velocity equal to half of that of B. The deBroglie wavelengths λ B and λ C of B and [JEE Main 2019, 8 April ShiftII] C respectively λA , λA 2 λ (d) λ A , A 2
(a) 2λ A , λ A
Deexcitation energy of hydrogen electron in transition n = 2 to n = 1is 1 1 1 1 E = 13.6 × 2 − 2 eV = 13.6 2 − 2 n n 1 2 f i = 102 . eV Now, energy levels of helium ion’s (He + ) electron are (For helium, Z = 2) n=4 10.2 eV 13.6(Z)2 En = ——— n2
n=3
(c) λ A , 2 λ A
Exp. (b) Let m be the mass of nuclei B and C. So, the given situation can be shown in the figure below
– 3.4 eV
A
– 6.04 eV
2m
– 13.6 eV
Initial
n=2 – 54.4 eV n=1
So, a photon of energy 10.2 eV can cause a transition n = 2 to n = 4 in a He + ion. Alternate Solution For He + ion, when in n = 1state, 1 1 102 . = 13.6 × 2 2 2 − 2 ⇒ n = 1 1 n Thus, no transition takes place. Similarly, when in n = 2 state, 1 1 102 . = 13.6 × 2 2 2 − 2 ⇒ n = 4 2 n
(b)
pA
m
C
B
vB vC= — 2
m vB
Final
Now, according to the conservation of linear momentum, Initial momentum = Final momentum ⇒ pA = pB + pC or 2mv A = mv B + mvC mv B 2 mv A = mv B − 2 1 …(i) 2 v A = v B ⇒ v B = 4v A ⇒ 2 v and vC = B = 2 v A …(ii) 2 So, momentum of B and C respectively, can now be given as
373
Atoms and Nuclei pB = mB v B = m4v A = 2(2 mv A ) [using Eq. (i)] or pB = 2 pA …(iii) and pC = mC vC = m2 v A [using Eq. (ii)] or pC = pA …(iv) From the relation of deBroglie wavelength, h i.e. λ= p where, p is momentum and h is Planck’s constant. h h So, for A, λ A = or pA = …(v) pA λA h h [using Eq. (iii)] For B, λ B = = pB 2 pA
hc Z 2 me 4 1 1 = − λ 8h2 E02 m2 n2
or
1 1 1 2 ∝ − Z λ m2 n2 For first case, λ = 660nm, m = 2 and n = 3 1 1 1 ∝ 2 − 2 Z2 ∴ 660 nm (2 ) (3) ⇒
⇒
1 1 1 5 2 Z ∝ − Z 2 or 660 nm 4 9 36
…(i)
3. The ratio of mass densities of nuclei of 40Ca
For second case, transition is from n = 4 to n = 2, i.e. m = 2 and n = 4 1 1 1 2 ∴ ∝ − Z λ (2 )2 (4)2 1 1 1 ∝ − Z 2 ⇒ λ 4 16 1 3 2 …(ii) or ∝ Z λ 16 From Eqs. (i) and (ii), we get λ 5 16 = × 660nm 36 3 80 λ= × 660 nm ⇒ 108 = 488.9 nm
[JEE Main 2019, 8 April ShiftII]
5. A He+ ion is in its first excited state. Its
From Eq. (v), λ B can be written as λ h = A λB = h 2 2× λA h h [using Eq. (iv)] Similarly, for C, λC = = pC pA Similarly, from Eq. (v), λ C can be written as h = λA λC = h λA
and 16 O is close to (a) 5
(b) 2
(c) 0.1
(d) 1
Exp. (d) Mass density of nuclear matter is a constant quantity for all elements. It does not depends on element’s mass number or atomic radius. ∴ The ratio of mass densities of 40 Ca and 16 O is 1 : 1.
4. Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2 nd Balmer line (n = 4 to n = 2) will be [JEE Main 2019, 9 April ShiftI]
(a) 889.2 nm (c) 642.7 nm
(b) 388.9 nm (d) 488.9 nm
Exp. (d) Expression for the energy of the hydrogenic electron states for atoms of atomic number Z is given by Z 2 me 4 1 1 (Here, m < n) − E = hν = 8h2 E02 m2 n2
ionisation energy is [JEE Main 2019, 9 April ShiftII]
(a) 54.40 eV (c) 48.36 eV
(b) 13.6 eV (d) 6.04 eV
Exp. (b) Energy of a hydrogen atom like ion by Bohr’s Z2 model is En = − 13.6 2 n where, Z = atomic number and n = principal quantum number. For a He + ion in first excited state, n = 2, Z = 2 4 ∴ E2 = − 13.6 × = − 13.6 eV 4
6. Two radioactive materials A and B have
decay constants 10 λ and λ, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of A to that ofB will be 1/e after a time [JEE Main 2019, 10 April ShiftI]
374
JEE Main Chapterwise Physics (a)
1 11 λ
(b)
11 10 λ
(c)
1 9λ
(d)
1 10 λ
Exp. (c) Given, λ A = 10λ and λ B = λ Number of nuclei (at any instant) present in material is N = N0e − λt So, for materials A and B, we can write −λ t NA e A = −λ t = e − ( λ A − λ B )t NB e B NA 1 Given, = NB e
E4= – 0.85 × 9 eV
n=4 n=3
E3
… (i) E2
n=2
… (ii)
Equating Eqs. (i) and (ii), we get 1 = e − ( λ A − λ B )t e ⇒ e −1 = e − ( λ A − λ B )t Comparing the power of ‘e’ on both sides, we get or (λ A − λ B ) t = 1 1 t = ⇒ λA − λB By putting values of λ A and λ B in the above equation, we get 1 1 t = ⇒t = 9λ 10 λ − λ ++
7. In Li , electron in first Bohr orbit is excited to a level by a radiation of wavelength λ. When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of λ? [Take, h = 6.63 × 10− 34 Js; c = 3 × 108 ms− 1] [JEE Main 2019, 10 April ShiftII]
(a) 9.4 nm (c) 10.8 nm
So, L+ + electron is in it’s 3rd excited state. Now, using the expression of energy of an electron in nth energy level, 13.6Z 2 eV En = − n2 where, Z is the atomic number. ∴Energy levels of L+ + electron are as shown
(b) 12.3 nm (d) 11.4 nm
Exp. (c) Number of spectral lines produced as an excited electron falls to ground state (n = 1) is, n(n − 1) N= 2 In given case, N = 6 n(n − 1) ∴6 = 2 ⇒ n=4
l E1= – 13.6 × 9 eV
n=1
So, energy absorbed by electron from incident photon of wavelength λ is hc ∆E = λ hc ⇒(13.6 × 9 − 0.85 × 9) = λ hc λ= ⇒ 9(13.6 − 0.85) 1240 eV nm λ= ⇒ = 10.8 nm 9 × 12.75 eV
8. Two radioactive substances A and B have
decay constants 5 λ and λ, respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of 2 1 the number of nuclei to become will be e [JEE Main 2019, 10 April ShiftII]
(a) 2 / λ (c) 1 / 4 λ
(b) 1 / 2 λ (d) 1/ λ
Exp. (b) Number of active nuclei remained after time t in a sample of radioactive substance is given by N = N0e − λt where, N0 = initial number of nuclei at t = 0 and λ = decay constant. Here, at t = 0, Number of nuclei in sample A and B are equal, i.e. N0 A = N0 B = N0 Also, λ A = 5λ and λ B = λ
375
Atoms and Nuclei So, after time t, number of active nuclei of A and B are NA = N0e −5λt and NB = N0e − λt If
⇒
NA 1 = , then NB e 2 NA N e −5 λt 1 = 0 − λt = 2 NB N0e e e( λ
− 5 λ )t
= e −2
9. An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n corresponding to its initial excited state is [for photon of wavelength λ, energy
(a) n = 4 (c) n = 7
1240 eV 1 − 1 = 2 30 4 . × 13.6 × 4 m 1240 eV Given, E = λ (in nm)
1 − 1 = 074980 . ≈ 075 . m2 1 or = 1 − 075 . = 025 . m2 1 ⇒ =4 m2 = 025 . Hence, m = 2 So, by putting the value of m in Eq. (ii), we get 1 1 1240 13.6 × 4 × 2 − 2 = eV 2 n 108. 5 ⇒
Comparing the power of e on both sides, we get 4λt = 2 1 t = ⇒ 2λ
1240 eV ] E= λ ( in nm)
⇒
⇒ ⇒ or
[JEE Main 2019, 12 April ShiftI]
(b) n = 5 (d) n = 6
⇒ ⇒
Exp. (b) Change in energy in transition from n to m stage is given by (n > m), E Z2 En = − 0 2 n Here, Z =2 hc 1 1 ...(i) ∆En = + 13.6 × 4 2 − 2 = m λ n n E1,λ1
1240 1 − 1 = 4 n2 108. 5 × 13.6 × 4 1 1 − = 021 . 4 n2 1 = 025 . − 021 . = 0.04 n2 1 = 25 n2 = 0.04 n2 = 25 n=5
10. The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths λ 1 / λ 2 of the photons emitted in this process is [JEE Main 2019, 12 April ShiftII]
(a) 20/7 (c) 7/5
(b) 27/5 (d) 9/7
Exp. (a) m λ2 n=1
Let it start from n to m and then m to ground. So, in first case, 1 1 hc ...(ii) 13.6 × 4 × 2 − 2 = m n 108.5 nm and in second case, 1 1 hc 13 . 6 × 4 × 2 − 2 = 1 m 30 . 4 nm
Wavelength λ of emitted photon as an electron transits from an initial energy level ni to some final energy level nf is given by Balmer’s formula, 1 1 1 = R 2 − 2 λ ni nf where, R = Rydberg constant. In transition from n = 4 to n = 3, we have 1 1 1 = R 2 − 2 3 λ1 4 7 =R 9 × 16
…(i)
376
JEE Main Chapterwise Physics 10 min
In transition from n = 3 to n = 2, we have 1 1 1 = R 2 − 2 2 λ2 3 5 =R 9 × 4 So, from Eqs. (i) and (ii), the ratio of
→
…(ii) λ1 is λ2
11. Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å). The deBroglie wavelength of this electron is [JEE Main 2019, 12 April ShiftII] (b) 6.6 Å (d) 9.7 Å
Exp. (d) By Bohr’s IInd postulate, for revolving electron, nh nh Angular momentum = ⇒ mvrn = 2π 2π nh ⇒ Momentum of electron, p = mv = 2 πrn deBroglie wavelength associated with electron is h 2 πrn λn = = p n n = 3, rn = 4.65 Å (2 × π × 4.65) λn = ≈ 97 . Å 3
Given, ∴
12. Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If initially a sample has equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be [JEE Main 2019, 12 April ShiftII]
(a) 3 : 8
(b) 1 : 8
(c) 8 : 1
(d) 9 : 8
Exp. (d) For substance A, halflife is 10 min, so it decays as 10 min N 10 min N 10 min N 0A N0 A → → 0 A → 0 A 4 2 8 (initial number of nuclei at t = 0) (Active nuclei remained after 10 min)
16
10 min
→
nuclei remained after 60 min)
9 × 16 λ1 7 R 20 = = 9 × 4 λ2 7 5R
(a) 3.5 Å (c) 12.9 Å
N0 A
N0 A 32
10 min
→
N0 A 64
(number of active
∴ For substance A, number of nuclei decayed in 60 min is N 63 N0 A N0 A − 0 A = 64 64 Similarly, for substance B, halflife is 20 min, so it’s decay scheme is 20 min N 20 min N 20 min N 0B 0B 0B N0 B → → → 2 4 8 So, number of nuclei of B decayed in 60 min is N0 B 7 N0 B − = N0 B 8 8 Hence, ratio of decayed nuclei of A and B in 60 min is 63 N0 64 A = 9 [QN0 A = N0 B ] 7 8 N0 B 8 Alternate Solution Number of active nuclei remained after 60 min can also be calculated as N N′ = T / t 2 1/ 2 where, T = 60 min So, for nuclei A, N N N N0′ A = 060A = 06A = 0 A 64 2 2 10 Similarly, for nuclei B, N N N N0′ B = 060B = 03B = 0 B 8 2 2 20
13. Surface of certain metal is first illuminated
with light of wavelength λ 1 = 350 nm and then by light of wavelength λ 2 = 540 nm. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to 1240 (energy of photon = eV) λ (inn m ) [JEE Main 2019, 9 Jan ShiftI]
(a) 5.6
Exp. (c)
(b) 2.5
(c) 1.8
(d) 1.4
377
Atoms and Nuclei Let maximum speed of photo electrons in first case is v1 and maximum speed of photo electrons in second case is v 2 Assumption I if we assume difference in maximum speed in two cases is 2 then v1 = v and v 2 = 3v According to Einstein’s photo electron equation Energy of incident photon = work function + KE hc 1 i.e. = φ0 + mv 2 2 λ Where hc = 1240 eV, λ is wavelength of light incident, φ 0 is work function and v is speed of photo electrons. When λ1 = 350 nm hc 1 ∴ = φ 0 + mv 2 350 2 hc 1 …(i) or − φ 0 = mv 2 350 2 when λ 2 = 540 nm hc 1 = φ0 + m(3v 2 ) ∴ 540 2 hc 1 …(ii) − φ0 = ( mv 2 ) × 9 ∴ 540 2 Now, we divide Eq. (i) by Eq. (ii), we get hc 1 − φ0 mv 2 1 350 = 2 = hc 1 2 − φ 0 ( mv ) × 9 9 540 2 hc hc or − φ0 9 − φ0 = 350 540 9 1 or − 8φ 0 = hc 350 540 9 × 540 − 350 1 or φ 0 = × 1240 . eV. = 37 350 × 540 8 No option given is correct. Alternate Solution Assumption II If we assume velocity of one is twice in factor with second then. Let v1 = 2 v and v 2 = v We know that from Einstein’s photoelectric equation, energy of incident radiation = work function + KE hc 1 or = φ + mv 2 2 λ Let when λ1 = 350 nm then v1 = 2 v and when λ1 = 540 nm then v 2 = v ∴ Above Eq. becomes 1 hc = φ + m(2 v 2 ) 2 λ1
or and or
1 hc − φ = m × 4v 2 2 λ1 1 hc = φ + mv 2 2 λ2 1 hc − φ = mv 2 2 λ2
…(i)
…(ii)
Now, we divide Eq. (i) by Eq. (ii), we get hc − φ 1 m × 4v 2 λ1 =2 =4 1 hc −φ mv 2 2 λ2 hc 4hc or −φ= − 4φ λ1 λ2 or
φ= =
1 4 1 − hc 3 λ 2 λ1 4 × 350 − 540 1 × 1240 350 × 540 3
or φ = 1.8 eV According the assumption II, correct option is (c).
14. A sample of radioactive material A, that has
an activity of 10 mCi (1 Ci = 3.7 × 1010 decays/s) has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for halflives of A and B would, then be respectively [JEE Main 2019, 9 Jan ShiftI]
(a) 20 days and 10 days (b) 5 days and 10 days (c) 10 days and 40 days (d) 20 days and 5 days
Exp. (d) Activity of a radioactive material is given as R = λN where, λ is the decay constant and N is the number of nuclei in the radioactive material. For substance A, R A = λ A NA = 10 mCi For substance B, ...(i) R B = λ B NB = 20mCi As given in the question, NA = 2 NB ...(ii) ⇒ R A = λ A (2 NB ) = 10 mCi
378
JEE Main Chapterwise Physics ∴Dividing Eq. (ii) and Eq.(i), we get RA λ (2 NB ) = A RB λ B (NB ) =
λ 1 10 or A = λB 4 20
So, for substance A, ln 2 ln2 (T1/ 2 )A = ⇒ ln2 = λA λA ...(iii)
As, halflife of a radioactive material is given as 0.693 T1/ 2 = λ ∴ For material A and B, we can write 0.693 (T1/ 2 )A λ λA = B = 0.693 λ A (T1/ 2 )B λB Using Eq. (iii), we get (T1/ 2 )A 4 = (T1/ 2 )B 1 Hence, from the given options, only option (d) satisfies this ratio. Therefore, (T1/ 2 )A = 20 days and (T1/ 2 )B = 5 days
15. In given timet = 0, Activity of two radioactive substances A and B are equal. After time t , R the ratio of their activities B decreases RA −3t according to e . If the half life of A is In 2, the halflife of B will be
λA = 1 According to the given question, R at time t, B = e −3t RA Using Eqs. (i), (ii) and (iii). we get λ N e − λ Bt RB = e −3t = B 0 B − λ t RA λ A N0 A e A e −3t = e( λ A − λ B )t − 3 = λA − λB ⇒ λB = λA + 3 λB = 1 + 3 = 4 The halflife of substance B is ln2 ln2 = (T1/ 2 )B = λB 4
(a) 4 ln 2 (c)
ln 2 2
Exp. (b) Activity of radioactive material is given as R = λN where, λ is the decay constant N is the number of nuclei in the radioactive material. For substance A, R A = λ A NA = λ A N0 A (initially NA = N0 A ) For substance B, R B = λ B NB = λ B N0 B (initially NB = N0 B ) At t = 0, activity is equal, therefore …(i) λ A N0 A = λ B N0 B The halflife is given by 0.693 ln2 T1/ 2 = = λ λ
…(iii)
⇒
…(iv)
16. The magnetic field associated with a light wave is given at the origin, by B = B 0 [sin ( 3.14 × 10 7 ) ct + sin (6.28 × 10 7 )ct ]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photoelectrons? (Take,c = 3 × 108 ms −1 andh = 6.6 × 10−34 Js) [JEE Main 2019, 9 Jan ShiftII]
[JEE Main 2019, 9 Jan ShiftII]
ln 2 (b) 4 (d) 2 ln 2
…(ii)
(a) 7.72 eV (c) 8.52 eV
(b) 6.82 eV (d) 12.5 eV
Exp. (a) According to question, the wave equation of the magnetic field which produce photoelectric effect B = B0 [(sin(314 . × 107ct ) + sin(628 . × 107ct )] Here, the photoelectric effect produced by the angular frequency (ω) = 628 . × 107 c ⇒
ω = 628 . × 107 × 3 × 108 ω = 2 π × 107 × 3 × 108 rad/s
Using Eqs. (i) hω h × 2 π × 107 × 3 × 108 = hν = 2π 2π hν =12.4 eV Therefore, according to Einstein equation for photoelectric effect E = hν = φ + KEmax
…(i)
379
Atoms and Nuclei ⇒
KEmax = E − φ (where, φ = workfunction = 47 . eV) KEmax = 12.4 − 47 . = 7.7 eV or KEmax = 7.7 eV
17. Using a nuclear counter, the count rate of emitted particles from a radioactive source is measured. At t = 0, it was 1600 counts per second and t = 8 s, it was 100 counts per second. The count rate observed as counts per second att = 6 s is close to [JEE Main 2019, 10 Jan ShiftI]
(a) 400
(b) 200
(c) 150
(d) 360
Exp. (b) Here given, at t = 0, count rate or initial activity is A0 = 1600 s − 1. At t = 8 s, count rate or activity is A = 100 s −1 So, decay scheme for given sample is T1 / 2
T1 / 2
T1 / 2
T1 / 2
1600 → 800 → 400 → 200 → 100 So, 8 s = 4T1/ 2 where, T1/ 2 = Halflife time. ⇒ T1/ 2 = 2 s ∴From above decay scheme, we see that activity after 6 s is 200 counts per second.
18. Consider the nuclear fission Ne20 → 2He4 + C12 Given that the binding energy/nucleon of Ne20 , He4 and C12 are respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement. [JEE Main 2019, 10 Jan ShiftI]
(a) Energy of 3.6 MeV will be released. (b) Energy of 12.4 MeV will be supplied. (c) 8.3 MeV energy will be released. (d) Energy of 11.9 MeV has to be supplied.
Exp. (*) Energy absorbed or released in a nuclear reaction is given by; ∆Q = Binding energy of products − Binding energy of reactants. If energy is absorbed, ∆Q is negative and if it is positive then energy is released. Also, Binding energy = Binding energy per nucleon × Number of nucleons.
Here, binding energy of products = 2 × (B.E . of He 4 ) + (B.E.of C12 ) = 2 (4 × 7.07 ) + (12 × 7.86) = 150.88 MeV and binding energy of reactants = 20 × 8.03 = 160.6 MeV So, ∆Q = (B.E.)Products − (B.E.) reactants = 150.88 − 160.6 = −972 . MeV. As ∆Q is negative ∴ energy of 9.72 Mev is absorbed in the reaction. ∴ No option is correct.
19. A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 Å. The radius of the atom in the excited state in terms of Bohr radius a 0 will be (Take hc = 12500 eVÅ) [JEE Main 2019, 11 Jan ShiftI]
(a) 4a 0
(b) 9a 0
(c) 16a 0
(d) 25a 0
Exp. (c) We know that net change in energy of a photon in a transition with wavelength λ is ∆E = hc / λ. Here, hc = 12500 eV Å and λ = 980 Å ∆E = 12500 / 980 = 12.76 eV ⇒ En − E1 = 12.76 eV Since, the energy associated with an electron in nth Bohr’s orbit is given as, − 13.6 …(i) En = eV n2 …(ii) ⇒ En = E1 + 12.76 eV − 13.6 = + 12.76 = −0.84 (1)2 Putting this value in Eq. (i) − 13.6 = 16 ⇒ n = 4 n2 = ⇒ − 0.84 and radius of nth orbit, rn = n2 a0 ⇒ rn = 16 a0
20. In a hydrogen like atom, when an electron jumps from the Mshell to the Lshell, the wavelength of emitted radiation is λ. If an electron jumps from Nshell to the Lshell, the wavelength of emitted radiation will be [JEE Main 2019, 11 Jan ShiftII]
27 (a) λ 20
Exp. (c)
25 (b) λ 16
(c)
20 λ 27
(d)
16 λ 25
380
JEE Main Chapterwise Physics For hydrogen or hydrogen like atoms, we know that 1 1 1 …(i) = RZ 2 2 – 2 λ n1 n2 where, R is Rydberg constant and Z is atomic number. When electron jumps from M  shell to the L  shell, then ( for L  shell) n1 = 2 (for M  shell) n2 = 3 ∴Eq (i) becomes 1 1 1 5 …(ii) = RZ 2 2 − 2 = RZ 2 ] 2 λ 3 36 Now, electron jumps from Nshell to the L  shell, for this (for L  shell) n1 = 2 (for N  shell) n2 = 4 ∴Eq (i) becomes 1 1 1 3 …(iii) = RZ 2 2 − 2 = RZ 2 2 λ′ 4 16
So, for nth orbit, ⇒ m2 v n2 = mkrn2
Therefore,
1 in a central potential field U (r ) = kr 2. If 2 Bohr’s quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as [JEE Main 2019, 12 Jan 2019, ShiftI]
(d) rn ∝ n , E n ∝
rn2 ∝ n
So, rn ∝ n Energy of particle is En = PE + KE 1 1 = krn2 + mv n2 2 2 1 1 = krn2 + krn2 2 2 = krn2
[using Eq. (i)]
So, energy, En ∝ rn2 ⇒
En ∝ n
22. In a radioactive decay chain, the initial
(a) A = 202; Z = 80 (c) A = 200; Z = 81
(b) A = 208; Z = 82 (d) A = 208; Z = 80
Exp. (b) An αparticle decay ( 42 He) reduces, mass number by 4 and atomic number by 2. ∴ Decay of 6αparticles results 232 90 Th
2
n 1 n
6a
− 24 208 → 232 90 − 12 Y = 78 Y
A βdecay does not produces any change in mass number but it increases atomic number by 1. ∴Decay of 4βparticles results,
Exp. (c) dU As, for conservative fields F = − dr
4 b 208 208 78 Y → 82 X
∴ In the end nucleus A = 208, Z = 82
23. If the series limit frequency of the Lyman
∴Magnitude of force on particle is dU d 1 2 F= = ⇒ kr dr dr 2 ⇒ F = kr This force is acting like centripetal force. mv 2 = kr ∴ r
4π r rn4 ∝ n2
⇒
1
(c) rn ∝ n , E n ∝ n
Q v = nh n 2 πmr
= mkrn2
[JEE Main 2019, 12 Jan ShiftII]
21. A particle of mass m moves in a circular orbit
(b) rn ∝ n 2 , E n ∝
2 2
nucleus is 232 90 Th. At the end, there are 6 αparticles and 4 βparticles which are emitted. If the end nucleus is AZ X , A and Z are given by
Now, we divide Eq (ii) by Eq (iii) λ′ 5 3 20 = RZ 2 ÷ RZ 2 = 16 27 λ 36 20 or λ′ = λ 27
(a) rn ∝ n , E n ∝ n
n2 h2
⇒
series is ν L , then the series limit frequency of the Pfund series is [JEE Main 2018] (a) 25 νL
...(i)
Exp. (d)
(b)16 νL
(c)
νL 16
(d)
νL 25
381
Atoms and Nuclei / Series limit occurs in the transition n2 = ∞ to n1 = 1 in Lyman series and n2 = ∞ to n1 = 5 in pfund series. For Lyman series, n2=∞ hνL = Eg= E0 12 1 = 13.6 eV n 1=1
hνL = 13.6 In pfund series
Fractional loss of energy of neutron −Kf + Ki = Ki for neutron v2 1 1 − mv12 + mv 02 − 0 + v 02 2 = 2 = 9 2 1 v0 mv 02 2 8 1 = − + 1 = = 0.88 = 0.89 9 9
1 ∞
…(i) n 2= ∞
m 13.6 1 = 2 ∞ 5
hνp = E0 12 5 n2 = 5
hνp =
13.6
…(ii)
52
From Eqs. (i) and (ii), we get 25hνp = hνL ν νp = L 25
∴
v0
12m
m
v1
+ 12m
v2
Similarly, for neutroncarbon atom collision; Momentum conservation gives; v 0 = v1 + 12 v 2 and e = 1 ⇒ v 0 = v 2 − v1 11 So, v1 = v0 13 121 + 1 = 0.28 ∴Loss of energy = − 169 So, Pd = 0.89 and Pc = 0.28
25. Some energy levels of a molecule are shown
24. It is found that, if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is Pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is Pc . The values of [JEE Main 2018] Pd and Pc are respectively (a) (.89, .28) (b) (.28, .89) (c) (0, 0)
in the figure. The ratio of the wavelengths r = λ 1 / λ 2 is given by [JEE Main 2017 (Offline)] –E –4/3E
λ1 –2E
(d) (0, 1)
Exp. (a)
–3E
2 3 1 (c) r = 3
3 4 4 (d) r = 3
(a) r =
NeutronDeuterium collision; m
2m
m
2m
v0
v=0
v1
v2
Momentum conservation gives; …(i) mv 0 = mv1 + 2 mv 2 ⇒ v 0 = v1 + 2 v 2 Collision given is elastic . So, coefficient of restitution, e = 1 v − v1 Velocity of separation ∴ e = 1= ⇒ 1= 2 v0 − 0 Velocity of approach …(ii) ⇒ v 0 = v 2 − v1 On adding Eqs. (i) and (ii), we get 2v0 = v2 2 v 0 = 3v 2 ⇒ 3 So, from Eq. (i), we get 4v v v1 = v 0 − 2 v 2 = v 0 − 0 ⇒ v1 = − 0 3 3
λ2
(b) r =
Exp. (c) hc ∆E ∴ So, ratio of wave lengths We have, λ =
4 E − E λ1 hc / ∆E1 ∆E2 3 1 = = = = 2E − E 3 λ 2 hc / ∆E2 ∆E1
26. A radioactive nucleus A with a halflife T ,
decays into a nucleus B. At t = 0, there is no nucleus B. After sometimet , the ratio of the number of B to that of A is 0.3. Then, t is given by [JEE Main 2017 (Offline)]
382
JEE Main Chapterwise Physics log 1.3 loge 2 T (c) t = log 1.3
(b) t = T log 1.3
(a) t = T
(d) t =
T loge 2 2 log 1.3
Exp. (a) Decay scheme is , N atoms of B A, B
A No at t=0
Given, So,
Let N atoms decays into B in time t
NB 3 = 0.3 = 10 NA
⇒
NB 30 = NA 100
N0 = 100 + 30 = 130 atoms
By using
N = N0 e − λt
We have,
100 = 130e − λt
⇒
1 = e − λt 13 .
⇒
No – N atoms of A
∴
log 13 . = λt loge 2 T loge 2 log 13 . = ⋅t T T ⋅ log (13 . ) t = loge 2
27. Halflives of two radioactive elements A and B are 20 min and 40 min, respectively. Initially, the samples have equal number of nuclei. After 80 min, the ratio of decayed numbers of A and B nuclei will be [JEE Main 2016 (Offline)]
(a) 1 : 16 (c) 1 : 4
28. As an electron makes a transition from an excited state to the ground state of a hydrogen like atom/ion [JEE Main 2015] (a) its kinetic energy increases but potential energy and total energy decrease (b) kinetic energy, potential energy and total energy decrease (c) kinetic energy decreases, potential energy increases but total energy remains same (d) kinetic energy and total energy decrease but potential energy increases
Exp. (a)
If T is halflife, then λ = ⇒
For radioactive element B, N ⇒ Number of B nuclides 22 N 3 decayed =N− = N 4 4 ∴ Ratio of decayed numbers of A and B nuclei will be (15/16)N 5 = (3/ 4)N 4 NB after 80 min. =
(b) 4 : 1 (d) 5 : 4
Exp. (d) Given, 80 min = 4 halflives of A = 2 halflives of B. Let the initial number of nuclei in each sample be N. For radioactive element A, N NA after 80 min = 4 2 ⇒ Number of A nuclides decayed N 15 =N− = N 16 16
As we know that kinetic energy of an electron is 2 Z KE ∝ n When the electron makes transition from an excited state to the ground state, then n decreases and KE increases. We know that PE is lowest for ground state. As TE = − KE. TE also decreases.
29. Hydrogen (1H1 ), deuterium (1H 2 ), singly
ionised helium ( 2He 4 )+ and doubly ionised lithium( 3Li8 )+ + all have one electron around the nucleus. Consider an electron transition form n = 2 to n = 1. If the wavelengths of emitted radiation are λ 1 , λ 2 , λ 3 and λ 4 respectively for four elements, then approximately which one of the following is correct? [JEE Main 2014] (a) 4λ1 = 2 λ 2 = 2 λ 3 = λ 4 (b) λ1 = 2 λ 2 = 2 λ 3 = λ 4 (c) λ1 = λ 2 = 4λ 3 = 9λ 4 (d) λ1 = 2 λ 2 = 3λ 3 = 4λ 4
Exp. (c) For hydrogen atom, we get 1 1 1 = R Z 2 2 − 2 λ 1 2 1 3 = R(1)2 4 λ1
383
Atoms and Nuclei 1 3 = R(1)2 4 λ2
⇒
Exp. (a) According to given data, mass of neutron and proton are equal which do not permit the breaking up of neutron and proton. But if we take standard mass of neutron as 1.6750 × 10−27 kg, then Energy released = Mass defect × c 2
1 3 = R(2 )2 4 λ3 1 3 = R(3)2 4 λ4 1 1 1 1 = = = λ1 4λ 3 9λ 4 λ 2
∴
= (mn − mp − me ) × c 2 [Qenergy released mass of reactant − mass of product × c 2 ]
30. In a hydrogen like atom electron make transition from an energy level with quantum number n to another with quantum number (n − 1). If n >> 1, the frequency of radiation emitted is proportional to [JEE Main 2013] (a)
1 n
(b)
1 n2
(c)
1
(d)
n 3 /2
1 n3
Exp. (d) ∆E = hν where, ∆E is energy of radiation, h is Planck's constant and ν is frequency. 1 ∆E 1 ⇒ ν= =k − 2 2 h n (n − 1) 2k 1 k2 n = 2 ≈ ∝ 3 n n (n − 1)2 n3
=
1.66 × 10−27 × 931.5 MeV [Q1amu = 931.5 MeV] 16 × 10−31 × 931.5 MeV = 1.66 × 10−27 1.6 × 0.9315 MeV = 1.66 ≈ 0.9 MeV which is close to option (a).
33. A diatomic molecule is made of two masses m1 andm 2 which are separated by a distance r. If we calculate its rotational energy by applying Bohr’s rule of angular momentum quantisation, its energy will be given by (n is an integer) [AIEEE 2012]
31. Hydrogen atom is excited from ground state
(a)
to another state with principal quantum number equal to 4. Then, the number of spectral lines in the emission spectra will be
(c)
[JEE Main 2013]
(a) 2
(b) 3
(c) 5
(d) 6
Exp. (d) In emission spectrum, number of bright lines is given by n(n − 1) 4(4 − 1) = =6 2 2
32. Assume that a neutron breaks into a proton and an electron. The energy released during this process is (mass of neutron = 1.6725 × 10−27 kg, mass of proton = 1.6725 × 10−27 kg, mass of electron = 9 × 10−31 kg) [AIEEE 2012] (a) 0.73 MeV (c) 6.30 MeV
(b) 7.10 MeV (d) 5.4 MeV
(1.6750 × 10−27 −1.6725 × 10−27 − 9 × 10−31 )
(m1 + m2 )2 n 2h2 2 m12m 22 r 2 2 2 2n h
(b) (d)
(m1 + m2 )r 2
n 2h2 2(m1 + m2 )r 2 (m1 + m2 )n 2h2 2m1m2 r 2
Exp. (d) Rotational kinetic energy of the two body system rotating about their centre of mass is 1 RKE = µω2 r 2 2 m1m2 where, µ = = reduced mass m1 + m2 nh and angular momentum, L = µωr 2 = 2π 2 nh 2 1 1 RKE = µω2 r 2 = µ ⋅ r ∴ 2 2 2 2 πµr = =
n2 h2 8 π 2µ r 2
=
n2 h2 2 µ r2 2 2
(m1 + m2 )n h 2 m1m2 r 2
384
JEE Main Chapterwise Physics
34. After absorbing a slowly moving neutron of mass m N (momentum ~ 0), a nucleus of mass M breaks into two nuclei of masses m1 and 5m1 ,(6m1 = M + m N ), respectively. If the deBroglie wavelength of the nucleus with mass m1 is λ , then deBroglie wavelength of the other nucleus will be [AIEEE 2011]
(a) 25λ
(c) λ / 5
(b) 5λ
(d) λ
Exp. (d) deBroglie wavelength, λ =
∴
λ1 = λ 2 = λ
37. The halflife of a radioactive substance is 20 min. The approximate time interval (t 2 − t 1 ) between the timet 2 when 2/3 of it has decayed and time t 1 when 1/3 of it had decayed is [AIEEE 2011] (a) 14 min (b) 20 min (c) 28 min (d) 7 min
h h = mv p
where, p = momentum By conservation of momentum, p1 + p2 = 0 or p1 = p2 ⇒
1 1 1 1 ∆E = 13.6Z 2 2 − 2 Q ∆E = Rhc 2 − 2 n n n n 1 1 2 2 1 2 1 = 13.6(3) 2 − 2 = 108.8 eV 1 3
Exp. (b)
1 2 N0 = N0 3 3 2 1 N2 = N0 − N0 = N0 3 3 n N1 1 = ⇒ n=1 N2 2 N1 = N0 −
h h = p1 p2 h Q λ = p
35. Statement I A nucleus having energy E1
∴ t 2 − t 1 = one halflife = 20 min
decays be β − emission to daughter nucleus having energy E 2, but β − rays are emitted with a continuous energy spectrum having end point energy E 1 − E 2.
æ Directions
Statement II To conserve energy and momentum in βdecay, atleast three particles must take part in the transformation. [AIEEE 2011]
38. The binding energy per nucleon for the
(a) Statement I is false, Statement II is true (b) Statement I is true, Statement II is false (c) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I (d) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I
Exp. (c) In particle situation, atleast three particles take place in transformation, so Energy of βparticle + Energy of third particle = E1 − E2 Hence, energy of βparticle ≤ E1 − E2
36. Energy required for the electron excitation inLi
2+
from the first to the third Bohr orbit is
(a) 36.3 eV (c) 122.4 eV
Exp. (b)
(b) 108.8 eV (d) 12.1 eV
[AIEEE 2011]
(Q. Nos. 38 to 39) are based on the following paragraph. A nucleus of mass M + ∆m is at rest and decays into two daughter nuclei of equal mass M /2 each. Speed of light is c.
parent nucleus is E 1 and that for the daughter nuclei is E 2. Then, [AIEEE 2010]
(a) E 2 = 2 E 1 (c) E 2 > E 1
(b) E 1 > E 2 (d) E 1 = 2 E 2
Exp. (c) After decay, the daughter nuclei will be more stable, hence binding energy per nucleon will be more than that of their parent nucleus.
39. The speed of daughter nuclei is ∆m (a) c M + ∆m (c) c
∆m M
[AIEEE 2010]
(b) c
2∆m M
(d) c
∆m M + ∆m
Exp. (b) M M v1 − v 2 2 2 ...(i) v1 = v 2 1 M 1 M ...(ii) ∆mc 2 = ⋅ v12 + ⋅ ⋅ v 22 2 2 2 2 M ∆mc 2 = v12 2
By conservation of momentum, 0 =
385
Atoms and Nuclei 2 ∆m 2 ∆mc 2 = v12 ⇒ v1 = c M M
42. The transition from the state n = 4 to n = 3 in
40. A radioactive nucleus (initial mass number A and atomic number Z ) emits 3αparticles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be [AIEEE 2010] A − Z −8 (a) Z −4 A − Z − 12 (c) Z −4
A−Z −4 (b) Z −8 A−Z −4 (d) Z −2
Exp. (b)
a hydrogen like atom results in ultraviolet radiation. Infrared radition will be obtained in the transition from [AIEEE 2009] (a) 2 → 1
(b) 3 → 2
(c) 4 → 2
(d) 5 → 4
Exp. (d)
1 1 IR corresponds to least value of 2 − 2 n2 n1 i.e., from Paschen, Brackett and Pfund series. Thus, the transition corresponds to 5 → 4.
æ Directions
In positive beta decay, a proton is transformed into a neutron and a positron is emitted. p+ → n0 + e + Number of neutrons initially was A − Z. Number of neutrons after decay ( A − Z ) − 3 × 2 (due to alpha particles) + 2 × 1 (due to positive beta decay) The number of protons will reduce by 8. [as 3 × 2 (due to alpha particles) + 2 (due to positive beta decay)] Hence, atomic number reduces by 8. So, the ratio of number of neutrons to that of A−Z−4 protons = Z−8
41. The below is a plot of binding energy per nucleon E b , against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions (i) A + B → C + ε (ii) C → A + B + ε (iii) D + E → F + ε and (iv) F → D + E + ε
B
Eb
C
D
E
A
F M
where, ε is the energy released. In which reactions ε is positive? [AIEEE 2009] (a) (i) and (iv) (c) (ii) and (iv)
(b) (i) and (iii) (d) (ii) and (iii)
Exp. (a) 1st reaction is fusion and 4th reaction is fission.
Q. No. 43 is AssertionReason type question. The question contains two statements : Statement I (Assertion) and Statement II (Reason). The question also has four alternative choices, only one of which is correct answer. You have to select the correct choice.
43. Statement I Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. Statement II For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei, it decreases with increasing Z . [AIEEE 2008] (a) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (b) Statement I is true, Statement II is false (c) Statement I is false, Statement II is true (d) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I Fe
Exp. (b) Statement I is correct and Statement II is wrong which can be directly concluded from binding energy/ nucleon curve.
BE A
A
44. Suppose an electron is attracted towards the origin by a force k /r , where k is a constant and r is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the nth orbital of the electron is found to be rn and the kinetic energy of the electron to be Tn . Then, which of the following is true? [AIEEE 2008]
386
JEE Main Chapterwise Physics 1 , rn ∝ n 2 n2 1 (c) Tn ∝ , rn ∝ n n
(a) Tn ∝
(b) Tn is independent of n , rn ∝ n 1 (d) Tn ∝ , rn ∝ n 2 n
Exp. (b) mv 2 k = rn rn nh mvrn = 2π
[given]
45. The halflife period of a radioactive element X is same as the mean life time of another radioactive elementY .Initially they have the same number of atoms. Then, [AIEEE 2007] X will decay faster than Y Y will decay faster than X Y and X have same decay rate initially X and Y decay at same rate always
Exp. (b) T1/ 2 ( X ) = τ(Y ) 0.693 1 = λX λY λX λY = 0.693 λY > λX
⇒ or ⇒
So, Y will decay faster than X.
46. If Mo is the mass of an oxygen isotope 8O17 , Mp and Mn are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is [AIEEE 2007] (a) ( Mo − 8M p )c 2 (c) Mo c
2
(b) ( Mo − 8M p − 9Mn )c 2 (d) ( Mo −17Mn )c 2
Exp. (b) Binding energy, BE = (Mnucleus − Mnucleons )c 2 = (Mo − 8M p − 9M n )c 2
47. Which of the following transitions in hydrogen atoms emit photons of highest frequency? [AIEEE 2007] (a) n = 2 to n = 6 (c) n = 2 to n =1
n2 ν
n1
48. In gamma ray emission from a nucleus,
[from Bohr’s theory]
On solving, rn ∝ n and Tn is independent of n.
(a) (b) (c) (d)
Frequency of emitted photon is proportional to change in energy of two energy levels, i .e., 1 1 ν = RcZ 2 2 − 2 n2 n1
(b) n = 6 to n = 2 (d) n =1 to n = 2
Exp. (c) Emission spectrum would rises when electron makes a jump from higher energy level to lower energy level.
(a) both the neutron number and the proton number change [AIEEE 2007] (b) there is no change in the proton number and the neutron number (c) only the neutron number changes (d) only the proton number changes
Exp. (b) In gamma ray emission, the energy is released from nucleus, so that nucleus get stabilised (to get the minimum energy state).
49. An alpha nucleus of energy (1 / 2 ) mv 2 bombards a heavy nuclear target of charge Ze. Then, the distance of closest approach for the alpha nucleus will be proportional to [AIEEE 2006]
(a) v
2
1 (b) m
1 (c) 4 v
(d)
1 Ze
Exp. (d) Since, nuclear target is heavy, it can be assumed safely that it will remain stationary and will not move due to the Coulombic interaction force. At distance of closest approach, relative velocity of two particles is v. Here, target is considered as stationary, so αparticle comes to rest instantaneously at distance be closest approach. Let required distance be r, then from workenergy theorem, 1 Ze × 2 e mv 2 0− =− 2 4 πε0 r 1 1 or r ∝ 2 or r ∝ Ze 2 r∝ ⇒ m v
50. If the binding energy per nucleon in 37Li and 4 2 He
nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction p + 37Li → 2 42 He energy of proton must be (a) 28.24 MeV (c) 1.46 MeV
[AIEEE 2006]
(b) 17.28 MeV (d) 39.2 MeV
387
Atoms and Nuclei Exp. (b) From the equation, Energy of proton + (7 × 5.60) = 2 × [4 × 7.06] Energy of proton = 17 .28 MeV ∴
53. Starting with a sample of pure 66 Cu, 7/8 of it decays into Zn in 15 min. The corresponding halflife is [AIEEE 2005] (a) 10 min (b) 15 min (c) 5 min
(d) 7
51. The energy spectrum of βparticles [number N ( E ) as a function of βenergy E ] emitted from a radioactive source is [AIEEE 2006]
Exp. (c) N = N0 (1 − e − λ t ) ⇒ 1 / 8 = e −λ t
(b)
E0
(d) E0
E
N (E)
(c)
E
N (E)
E0
E
E0
E
Exp. (c) Energy spectrum of emitted βparticles from radioactive source is drawn as N(E)
27 54. If radius of the 13 Al nucleus is estimated to
be 3.6 fm, then the radius of be nearly (a) 6 fm
(b) 8 fm
Exp. (a)
E
(c) 4 fm
nucleus
[AIEEE 2005]
(d) 5 fm
R = R 0 ( A)1/ 3
R Al 3 (27 )1/ 3 = = 1/ 3 5 R Te (125) 5 R Te = × 3.6 or R Te = 6 fm 3
or
When 37Li nuclei are bombarded by protons, and the resultant nuclei are 84 Be,the emitted (a) alpha particles (c) gamma photons
125 52 Te
R Al R ( A )1/ 3 R Al ( A )1/ 3 = 0 Al 1/ 3 or = Al 1/ 3 R Te R Te ( ATe ) R 0 ( A Te )
E0
particles will be
or 8 = e λ t
3 ln 2 = λt [Qloge mn = n loge m] 3 × 0.693 or λ= 15 0.693 Halflife period, t 1/ 2 = × 15 3 × 0.693 0.693 Qt = 1/ 2 λ = 5 min
or
52.
N0 − N = e −λ t N0
or
N(E)
N (E)
∴ (a)
1 min 2
[AIEEE 2006]
(b) beta particles (d) neutrons
55. The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy? [AIEEE 2005] n=4 n=3
Exp. (c) The nuclear reaction can be represented as 7 1 8 A 3 Li + 1H → 4 Be + Z X Applying conservation of atomic number (charge), 3 + 1= 4 + Z ⇒ Z=0 Applying conservation of atomic mass, 7 + 1= 8 + A ⇒ A=0 Thus, the emitted particles are γphotons ( 00 X ).
n=2
I
(a) III (c) I
Exp. (a)
II
III
IV
(b) IV (d) II
n=1
388
JEE Main Chapterwise Physics 1 1 E = Rhc 2 − 2 n n 1 2 1 1 E( 4 → 3 ) = Rhc 2 − 2 3 4 7 = Rhc = 0.05 Rhc 9 × 16
58. The binding energy per nucleon of deuteron (12H) and helium nucleus ( 42 He) is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is [AIEEE 2004]
3 . Rhc = Rhc = 02 16
(a) 13.9 MeV (c) 23.6 MeV
1 1 E( 2 → 1) = Rhc 2 − 2 (2 ) (1)
X (n , α ) → 37Li. Which of the following is the nucleus of element X ? [AIEEE 2005] (c)
9 5B
(d)
11 4 Be
Exp. (b) A Z
X + 10 n → 73 Li + 42 He
57. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2 : 1. The ratio of their nuclear sizes will be [AIEEE 2004]
(a) 2
:1 (b) 1 : 3
(c) 3
1/ 2
:1
(d) 1 : 2
Exp. (d) Law of conservation of momentum gives m v m1v1 = m2 v 2 ⇒ 1 = 2 m2 v1 4 3 But m = πr ρ or m ∝ r 3 3 ∴
m1 r13 v 2 = = m2 r23 v1
The binding energy per nucleon of helium ( 42 He) = 7 MeV ∴ Total binding energy = 4 × 7 = 28 MeV Hence, energy released in the above process = 28 − 2 × 2 . 2 = 28 − 4.4 = 23.6 MeV
through 180° by a fixed uranium nucleus. The distance of the closest approach is of the order of [AIEEE 2004]
Thus, it is Boron 105 B.
1/ 2
= 1.1 MeV ∴Total binding energy of one deuteron nucleus = 2 × 1.1 = 2 .2 MeV
59. An αparticle of energy 5 MeV is scattered
It implies that, A + 1 = 7 + 4 ⇒ A = 10 and Z + 0=3+2 ⇒ Z =5
1/ 3
+ 12 H → 42 He + energy
The binding energy per nucleon of a deuteron(12 H)
56. A nuclear transformation is denoted by 10 5B
2 1H
As given
1 1 8 E( 1 → 3 ) = Rhc 2 − 2 = − Rhc = −0.9 Rhc 9 ( 3 ) ( 1 ) Thus, III transition gives most energy. I transition represents the absorption of energy.
(b)
(b) 26.9 MeV (d) 19.2 MeV
Exp. (c)
3 . Rhc = Rhc = 075 4
12 6C
1/ 3
r1 : r2 = 1 : 2 1/ 3
or
1 1 E( 4 → 2 ) = Rhc 2 − 2 2 4
(a)
r1 1 = r2 2
⇒
1/ 3
(b) 10−10 cm (d) 10−15 cm
(a) 1 Å (c) 10−12 cm
Exp. ( c ) According to law of conservation of energy, Kinetic energy of αparticle = Potential energy of αparticle at distance of closest approach 1 1 q1 q 2 i.e., mv 2 = 2 4 πε0 r 5 MeV =
∴
9 × 109 × (2 e ) × (92 e ) r Q 1 mv 2 = 5 MeV, given 2
9 × 109 × 2 × 92 × (1.6 × 10−19 )2
⇒
r=
or
r = 5.3 × 10−14 m ≈ 10−12 cm
5 × 106 × 1.6 × 10−19
389
Atoms and Nuclei 60. The manifestation of band structure in solids is due to (a) (b) (c) (d)
[AIEEE 2004]
Heisenberg’s uncertainty principle Pauli’s exclusion principle Bohr’s correspondence principle Boltzmann’s law
According to Pauli’s exclusion principle, the electronic configuration of number of subshells existing in a shell and number of electrons entering each subshell is found. Hence, on the basis of Pauli’s exclusion principle, the manifestation of band structure in solids can be explained.
61. When U 238 nucleus originally at rest, decays by emitting an alpha particle having a speed u , the recoil speed of the residual nucleus is [AIEEE 2003]
4u (a) 238 4u (c) 234
4u (b) − 234 4u (d) − 238
Exp. ( c ) 238 U 234 X
Exp. ( a ) N0 λ = 5000, N λ = 1250
Given,
N = N0 e − λt = N0 e −5 λ where, λ is decay constant per minute. N λ = N0 λ e −5 λ
Exp. (b)
v
(a) 0.4 ln 2 (b) 0.2 ln 2 (c) 0.1 ln 2 (d) 0.8 ln 2
[before decay]
4 2 He
u [after decay]
Apply conservation of linear momentum. 0 = 4u − 234v [Q Mu = m1v1 + m2 v 2 ] 4u ⇒ v= 234 The residual nucleus will recoil with a velocity of 4u unit. 234
Note If they will ask the recoil velocity, then answer 4u 4u remains same i.e., and not − as the word ‘recoil’ 234 234 itself is signifying the direction of motion of residual nucleus.
62. A radioactive sample at any instant has its disintegration rate 5000 disintegrations/min. After 5 min, the rate is 1250 disintegrations/min. Then, the decay constant (per minute) is [AIEEE 2003]
1250 = N0 λe −5 λ
⇒ ∴
N0 λ
N0 λe −5 λ
=
5000 =4 1250
e 5λ = 4
or or
5λ = 2 loge 2
or
λ = 0.4 ln 2
63. A nucleus with Z = 92 emits the following in a sequence α , α , β − , β − , α , α , α , α ; β − , β − , α , β + , β + , α. The Z of the resulting nucleus is [AIEEE 2003] (a) 76 (c) 82
(b) 78 (d) 74
Exp. ( b ) and Since, 8 αparticles 4β − particles 2β + particles are emitted, so new atomic number [α → 42He and β → − 10 n]. Z′ = Z − 8 × 2 + 4 × 1 − 2 × 1 = 92 − 16 + 4 − 2 = 92 − 14 = 78
64. Which of the following cannot be emitted by radioactive substances during their decay? (a) Protons (c) Helium nuclei
(b) Neutrinos [AIEEE 2003] (d) Electrons
Exp. (a) Protons cannot be emitted by radioactive substances during their decay.
65. In the nuclear fusion reaction, 2 1H
+ 13 H → 42 He + n
given that the repulsive potential energy between the two nuclei is 7.7 × 10−14 J, the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann’s constant, k = 1.38 × 10−23 J/K] [AIEEE 2003]
(a) 107 K (c) 103 K
(b) 105 K (d) 109 K
390
JEE Main Chapterwise Physics
Exp. (b)
or
Exp. (a)
3 T = 7.7 × 10−14 J 2 2 × 7.7 × 10−14 = 3.7 × 109 K T= 3 × 1.38 × 10−23
E = − Z2
(a) (c)
14 7N 40 18 Ar
Exp. (b)
(b) (d)
E2 = − 32 ×
13 .6 4
= −30.6 eV Ionisation energy for first excited state of Li 2+ is 30.6 eV.
According to problem,
atoms given, thus electrons present in the outermost orbit will be away from the nucleus and the electrostatic force experienced by electrons due to nucleus will be minimum. Therefore, the energy required to liberate electrons from outer orbit will be minimum in case of 133 55 Cs.
67. The wavelengths involved in the spectrum of deuterium (12D) are slightly different from that of hydrogen spectrum, because [AIEEE 2003]
(a) sizes of the two nuclei are different (b) nuclear forces are different in the two cases (c) masses of the two nuclei are different (d) attraction between the electron and the nucleus is different in the two cases
Exp. (c) Q Masses of the two nuclei are different.
68. If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li2+ is [AIEEE 2003] (b) 13.6 eV (d) 122.4 eV
eV
[Q n = 2, for Ist excited state ]
[AIEEE 2003] 133 55 Cs 16 8O
Since, ionisation potential is minimum energy required to eject the etectron from the outermest orbit. As 133 55 Cs has larger size among the four
(a) 30.6 eV (c) 3.4 eV
n2
For first excited state,
66. Which of the following atoms has the lowest ionisation potential?
13.6
69. If 13.6 eV energy is required to ionise the hydrogen atom, then the energy required to remove an electron from n = 2 is [AIEEE 2002] (a) 10.2 eV (b) zero
(c) 3.4 eV
(d) 6.8 eV
Exp. (c) Energy required to remove an electron from nth orbit is 13.6 En = − 2 n Here, n = 2 13.6 Therefore, E2 = − 2 = − 3.4 V 2
70. If N 0 is the original mass of the substance of
halflife periodt 1/2 = 5 yr, then the amount of substance left after 15 yr, is [AIEEE 2002] (a)
N0 8
(b)
N0 16
(c)
N0 2
(d)
N0 4
Exp. (a) N0 is the initial amount of substance and N is the amount left after decay. Thus,
1 N = N0 2
n
n = number of halflives = 3
Therefore,
t 15 = =3 t 1/ 2 5
N 1 N = N0 = 0 2 8
19 Electronic Devices 1. The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit. 200 W IZ
2. A common emitter amplifier circuit, built using an n p  n transistor, is shown in the figure. Its DC current gain is 250, RC = 1 kΩ and VCC = 10 V. What is the minimum base current forVCE to reach saturation? [JEE Main 2019, 8 April ShiftII]
800 W
9V
The current I z through the Zener is (a) 10 mA (b) 17 mA (c) 15 mA (d) 7 mA [JEE Main 2019, 8 April ShiftI]
RB
RC VCC
VB
Exp. (a) Given circuit is Zener diode circuit 200Ω
(a) 40µA (c) 100µA
i2 iz
i1
Vz=5.6 V
(b) 10µA (d) 7µA
Exp. (a) 800Ω
+ – 9V
where,potential drop across 800 Ω resistance = potential drop across Zener diode = 5.6 V 5.6 V So, current, i 2 = = = 7mA R 800 Now, potential drop across 200 Ω resistance = 9 − 5.6 = 3.4 V V 3.4 Current, i1 = = = 17 mA R 200 So, current, i z = i1 − i 2 = 17 − 7 = 10 mA
For a common emitter npn transistor, DC current gain is I βDC = C IB At saturation state, VCE becomes zero. ∴ VCC − IC RC = 0 V IC ≈ CC RC 10 = = 10−2 A 1000 Hence, saturation base current, I 10−2 IB = C = = 40µA βDC 250
392
JEE Main Chapterwise Physics
3. An npn transistor is used in common emitter configuration as an amplifier with 1 kΩ load resistance. Signal voltage of 10 mV is applied across the baseemitter. This produces a 3 mA change in the collector current and 15µA change in the base current of the amplifier. The input resistance and voltage gain are [JEE Main 2019, 9 April ShiftI] (a) 0.67 kΩ , 200
(b) 0.33 kΩ, 1.5
(c) 0.67 kΩ, 300
(d) 0.33 kΩ, 300
Exp. (c) Truth table for given combination of logic gates is Y1 A Y B Y2
Exp. (c) Given, load resistance, R L = 1kΩ Input voltage, Vin = 10 mV = 10 × 10−3 V −6
Base current, ∆IB = 15µA = 15 × 10
A
Collector current, ∆IC = 3 mA Input resistance, V 10 × 10−3 Rin = in = = 0.67 kΩ ∆IB 15 × 10−6 R and voltage gain = β × L Rin =
∆IC × R L 3 mA 1 kΩ = × ∆IB × Rin 15µA 0.67 kΩ
3 × 10−3 1 × 103 = −6 3 15 × 10 0.67 × 10 1000 × 3 × 3 (Q0.67 ~ = 2 / 3) = 300 = 15 × 2 Alternate Solution V ∴Voltage gain = output Vinput = =
Y1 = A
0
0
1
1
0
1
0
0
1
1
0
1
1
0
1
1
1
0
0
1
Y2 = B Y = Y1 ⋅ Y2
Output Y resembles output of an OR gate. So, given combination acts like an OR gate. Alternate Solution The given logic gate circuit can be drawn as shown below A
Y1=A Y=Y1× Y2
B
Here,
R L × ∆IC Vin 1 × 10 3 × 3 × 10−3 −3
10 × 10
= 300
[JEE Main 2019, 9 April ShiftII]
A Y
Y 2= B
Y = Y1 ⋅ Y2 = A ⋅ B
∴Y = A + B = A + B
[Q x = x]
This represents the boolean expression for OR gate.
5. Four point charges −q , + q , + q and −q are placed on Y axis at y = −2d , y = −d , y = +d and y = +2d , respectively. The magnitude of the electric field E at a point on the X axis at x = D, with D >> d , will behave as [JEE Main 2019, 9 April ShiftII]
B
(a) NOR (c) OR
B
Using deMorgan’s theorem, i.e. x⋅ y = x + y
4. The logic gate equivalent to the given logic circuit is
A
(b) NAND (d) AND
1 (a) E ∝ D 1 (c) E ∝ 2 D
(b) E ∝
1
D3 1 (d) E ∝ 4 D
393
Electronic Devices Exp. (d)
–q
Given charge distribution is as shown below
+q
–q y=2d – d p
d r
y =d + +q
P
θ
+q
d
–q
So, we can view above point charges as combination of pair of dipoles or a quadrupole. By symmetry, the field components parallel to quadrupole cancels and the resultant perpendicular field is D 2q 1 − E= 4 πε0 D2 (D2 + d 2 )3 / 2 − 3/ 2 2 1 − 1 + d D2 4 πε0 D2
2q
3 − d2 2
1 + 2 D
3 d2 ≈ 1 − 2 D2 (using binomial expansion)
We have, E= ∴
E1
D
y=–2d – – q
As,
E1 θ1 θ1
E2
d
E2
d
P
θ2 θ2 Enet
y=–d + +q
E2
E1
x=D
=
d
E∝
3 qd 2 4 πε0 D4 1
D4 Note Dependence of field for a point charge is 1 E∝ 2 r For a dipole, it is 1 E∝ 3 r For a quadrupole, it is 1 E ∝ 4 ..... etc. r Alternate Solution The given distribution of charges can be shown as the figure below
Electric field at point P, E = E1 cos θ1 + E1 cos θ1 − E2 cos θ2 − E2 cos θ2 = 2 E1 cos θ1 − 2 E2 cos θ2 2 kq 2 kq cos θ1 − cos θ2 = 2 (d + D2 ) (2d )2 + D2 D As, cos θ1 = 2 (d + D2 )1/ 2 D Similarly, cos θ2 = 1 [(2d )2 + D2 ] 2 = 2 kqD[(d 2 + D2 ]− 3 / 2 − (4d 2 + D2 )− 3 / 2 ] =
− 3/ 2 − 3/ 2 d2 2 kqD 4d 2 1 1 − + + D3 D2 D2
As D > > d , then by applying binomial approximation, we get 3 4d 2 2 kq 3 d2 = 2 1 − − 1 − 2 2 2 D 2 D D 2 2 2 kq 9d 9kqd = 2 2 = D 2 D D4 1 ⇒ E∝ 4 D
6. An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is 100 Ω and the output load resistance is 10 kΩ. The common emitter current gain β is [JEE Main 2019, 10 April ShiftI]
(a) 102
(b) 6 × 102
(c) 104
(d) 60
394
JEE Main Chapterwise Physics
Exp. (a) Given, AP = 60 dB (in decibel) Power gain in decibel can be given as Output power AP = 10log10 Input power P P ⇒ 60 = 10 log10 out ⇒ log10 out = 6 Pin Pin Pout 6 … (i) ⇒ = 10 = AP Pin Also, given Rout = 10 kΩ, Rin = 100 Ω ∴Power gain of a transistor is given by R AP = β 2 out Rin where, β is current gain. R 100 ⇒ β 2 = AP × in = 106 × Rout 10 × 103 4
Now, when VB = 8 V i1
R1
A
iL
iZ + –
8V
RL B
Potential drop across R1 = 8 − 6 = 2 V V 2 = R1 1 × 103
So, current through R1 is i1 =
= 2 × 10−3 A
⇒ β = 10 or β = 10 2
Zener breakdown voltage,VZ = 6 V So, across R L , potential drop is always 6 V. So, current through load resistance is V 6 iL = Z = = 15 . × 10−3 A R L 4 × 103
2
7. The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is R L = 4 kΩ. The series resistance of the circuit is Ri = 1 kΩ. If the battery voltage V B varies from 8V to 16V, what are the minimum and maximum values of the current through Zener diode? [JEE Main 2019, 10 April ShiftII]
So, current through Zener diode is i Z = i1 − i L = 2 × 10−3 − 1.5 × 10−3 = 0.5 × 10−3 A = 0.5 mA Similarly, when VB = 16 V VR1 = 16 − 6 = 10 V 10 = 10 × 10−3 A i1 = ∴ 1 × 103 Hence, i Z = i1 − i L = 10 × 10−3 − 1.5 × 10−3 = 8.5 × 10−3 A = 8.5 mA
Ri
8. The truth table for the circuit given in the figure is VB
[JEE Main 2019, 12 April ShiftI]
RL Y
A B
(a) 1.5 mA, 8.5 mA (c) 0.5 mA, 8.5 mA
(b) 1 mA, 8.5 mA (d) 0.5 mA, 6 mA
Exp. (c) R1
+ –
VZ = 6V
RL=4 k Ω
A B Y 0 0 1
(a) 0
1
1
(b) 0
1
0
1
0
1
1
0
0
1 1 1 A B Y
In given voltage regulator circuit,
VB
A B Y 0 0 1
0 (c) 0 1 1
0 1 0 1
Exp. (c)
1 1 0 0
1 1 0 A B Y 0 (d) 0 1 1
0 1 0 1
0 0 1 1
395
Electronic Devices Given circuit is
Y
A B
X
Let the intermediate state X of OR gate is shown in figure. Clearly, …(i) Y = AX Here, …(ii) X = A+ B Y = A ( A + B) = AA + AB ∴ (Q AA = A) = A + AB = A (1 + B) (Q1 + B = 1) ⇒ Y = A So, truth table shown in option (c) is correct. Alternate Solution We can solve it using truth table A 0 0 1 1
X = A+ B 0 1 1 1
B 0 1 0 1
Y = AX 1 1 0 0
9. The transfer characteristic curve of a transistor, having input and output resistance 100 Ω and 100 kΩ respectively, is shown in the figure. The voltage and power gain, are respectively [JEE Main 2019, 12 April ShiftI]
(400, 20)
In CE configuration, I I Current gain, β = out = c Iin Ib
…(i)
Voltage gain, V I × Rout R AV = out = c = β × out Vin Ib × Rin Rin
…(ii)
and power gain, I 2 × Rout P R = β 2 × out AP = out = c2 Pin Rin Ib × Rin
…(iii)
Given, Rin = 100 Ω and Rout = 100 × 103 Ω From Eq. (i), we get 5 mA 10 mA 15 mA 20 mA or β= or or 100 µA 200 µA 300 µA 400 µA
⇒
AV = 50000 = 5 × 104
…(v)
From Eqs. (iii) and (iv), we get R 100 × 103 Power gain, AP = β 2 × out = (50)2 × Rin 100 = 2500 × 1000 AP = 2.5 × 106
⇒
10. Figure shows a DC voltage regulator circuit,
with a Zener diode of breakdown voltage = 6 V. If the unregulated input voltage varies between 10 V to 16 V, then what is the maximum Zener current? [JEE Main 2019, 12 April ShiftII]
(300, 15) Ic (mA)
5 × 10−3
…(iv) = 50 100 × 10−6 From Eqs. (ii) and (iv), we get 100 × 103 R Voltage gain, AV = β × out = 50 × Rin 100 ⇒ β=
Is
(200, 10)
Rs=2kΩ IL
(100, 5) Ib (µA)
IZ RL=4kΩ
(a) 2.5 × 10 , 2.5 × 10
(b) 5 × 10 , 5 × 10
(c) 5 × 10 , 5 × 10
(d) 5 × 10 , 2.5 × 10
4
4
6
5
4
4
6 6
Exp. (d) Given curve is between Ic and Ib as output and input currents, respectively. So, it is transfer characteristics curve of a common emitter (CE) configuration.
(a) 2.5 mA (c) 7.5 mA
(b) 1.5 mA (d) 3.5 mA
Exp. (d) In given voltage regulator circuit breakdown of Zener occurs at 6 V.
396
JEE Main Chapterwise Physics IS
12. At 0.3V and 0.7 V, the diodes Ge and Si become conductor respectively. In given figure, if ends of diode Ge overturned, the change in potentialV 0 will be
RS=2kΩ IL
10 V to 16 V
[JEE Main 2019, 9 Jan ShiftII] Ge
IZ RL=4kΩ
V0
After breakdown, voltage across load resistance (R L = 4kΩ ) is, V = VZ = 6V ∴Load current after breakdown, V 6 = 15 . × 10−3 A IL = Z = R L 4000 When unregulated supply is of 16 V, potential drop occurring across series resistance (RS = 2 kΩ ) is; VS = V − VZ = 16 − 6 = 10 V So, current across series resistance is V 10 IS = S = = 5 × 10−3 A RS 2 × 103
Si
12 V
(a) 0.2 V
(b) 0.6V
(c) 0.4 V
5 kΩ
(d) 0.8V
Exp. (c) Initially Ge and Si are both forward biased. So, current will effectively pass through Ge diode with a voltage drop of 0.3 V. Ge
So, current across Zener diode is IZ = IS − IL = 5 × 10−3 − 1.5 × 10−3 = 3.5 × 10−3 A
V0
= 3.5 mA
11. Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If for an n  type semiconductor, the density of electrons is 1019m −3 and their mobility is 1.6 m 2 (Vs), then the resistivity of the semiconductor (since, it is an ntype semiconductor contribution of holes is ignored) is close to [JEE Main 2019, 9 Jan ShiftI]
(a) 2 Ωm
(b) 0.2 Ωm (c) 0.4 Ωm (d) 4 Ωm
12 V
5 kΩ
∴ Initial output voltage, V0 = 12 − 0.3 = 11.7 V If Ge is reversed biased, then only Si diode will work. In this condition, output voltage V0 = 12 − 07 . = 11. 3 V 0.7 V
Exp. (c) Since, it is an ntype semiconductor and concentration of the holes has been ignored. So, its conductivity is given as σ = ne e µ e where, ne is the number density of electron, e is the charge on electron and µ e is its mobility. Substituting the given values, we get σ = 1019 × 1.6 × 10− 19 × 1.6 = 2.56 1 1 As, resistivity, ρ = = σ 2.56 or ρ = 0.39 ~ − 0.4 Ωm
Si
V0 Si 12 V
∴ Change in output voltage = 11.7 − 11. 3 = 0. 4 V
5 kΩ
397
Electronic Devices 13. To get output ‘1’ at R, for the given logic gate circuit, the input values must be [JEE Main 2019, 10 Jan ShiftI]
Exp. (d) In the circuit, let the current in branches is as shown in figure below A
X P
I2
By Kirchhoff’s node law, … (i) I1 = I2 + I3 Now, when diode conducts, voltage difference between points A and B will be VAB = 120 − 50 = 70 V V 70 So, current I1 = AB = 5 kΩ 5 × 103
(b) X = 1 , Y = 0 (d) X = 0, Y = 1
Exp. (b) The given circuit can be drawn as shown in the figure given below NOT
A1 OR
R
NOR
NOT
…(ii) I1 = 14 mA Since, diode and 10 kΩ resistor are in parallel combination, so voltage across 10 k Ω resistor will be 50 V only. 50 50 ⇒ = I3 = 10 kΩ 10 × 103
A4
Y
NAND A2
⇒ I3 = 5 mA ∴From Eqs. (i), (ii) and (iii), we get 14 mA = I2 + 5 mA or current through diode, I2 = 14 mA − 5mA = 9 mA
A3
Truth table for this given logic gate is Given inputs in the A1 options
A2
A3
10 kΩ
50 V
120 V
Q
X
I3
I1 B
R
Y
(a) X = 0, Y = 0 (c) X = 1, Y = 1
5 kΩ
A4
R
… (iii)
15. In the given circuit, the current through zener diode is close to
X
Y
0
0
1
1
1
1
0
1
0
0
1
0
0
1
1
1
0
0
1
1
0
0
1
1
0
1
1
0
So to get output R = 1, inputs must be X = 1and Y = 0.
14. For the circuit shown below, the current
R1
500 Ω
R2
1500 Ω
12 V V2 = 10 V
R2
[JEE Main 2019, 11 Jan ShiftI]
(a) 6.0 mA (c) 0
(b) 6.7 mA (d) 4.0 mA
Exp. (c)
through the Zener diode is [JEE Main 2019, 10 Jan ShiftII] 5 kΩ
120 V
(a) 14 mA
50 V
(b) zero
(c) 5 mA
10 kΩ
(d) 9 mA
Key Idea When the applied reverse voltage (V ) reaches the breakdown voltage of the Zener diode, then only a large amount of current is flown through it, otherwise it is approximately zero. In the given situation, if we consider that Zener diode is at breakdown. Then, potential drop across 1500 Ω resistances will be10 V. So potential drop at 500 Ω resistor will be 2 V.
398
JEE Main Chapterwise Physics ∴Current in R1 =
2 = 4mA = I1 (say) 500
(c) AB + AB (d) AB + AB
(b) AB
(a) AB
Exp. (a)
Current in each 10 2 R2 = = = 13.33 mA = I2 (say) 750 150 ⇒ I1 < I2 which is not possible. So, Zener diode will never reach to its breakdown. ∴Current flowing through a reverse biased Zener diode = 0.
Truth table for given circuit is A NAND
NAND
Y3
Y
NAND
Y1
16. The circuit shown below contains two ideal diodes, each with a forward resistance of 50 Ω. If the battery voltage is 6 V, the current through the 100 Ω resistance (in ampere) is [JEE Main 2019, 11 Jan ShiftII] D1
D2
A
B
Y1
Y2
Y3
Y
0
0
1
1
1
0
75Ω
1
0
1
1
0
1
0
1
1
1
1
0
1
1
0
1
1
0
100Ω
(b) 0.020
(c) 0.030
This is the same output produced by A ⋅ B gate or
(d) 0.036
A
Exp. (b)
AND
D1
150 Ω
D2
75 Ω
So, given circuit is equivalent to Boolean expression A ⋅ B. Alternate Solution Using the Boolean algebra, output of the given logic circuit can be given as
Open 6V
Y
B
In this circuit, D1 is forward biased and D2 is reversed biased.
Y2= A.(A.B)
A
100 Ω Y1= A.B
Resistance of D1 is 50Ω. ∴Net resistance of the circuit, Rnet = 50 + 150 + 100 = 300 Ω ∴Current through the 100Ω resistance V 6 = 0.020 A = = Rnet 300
B
[JEE Main 2019, 12 Jan ShiftI]
Y
Y2 = A ⋅ ( A ⋅ B)
Using deMorgan’s principle, x ⋅ y = x + y and x + y = x ⋅ y ⇒
A
Y=Y2.Y3
Y3= B+(A.B)
Here
17. The output of the given logic circuit is
B
Y2
150Ω
6V
(a) 0.027
OR
B
Y2 = A + ( A ⋅ B)
[Q x = x]
…(i) = A + ( A. B) Similarly, Y3 = B + A + B = 1 + A [Q x + x = 1] …(ii) [Q x + 1 = 1] ⇒ Y3 = 1 As, Y = Y2 ⋅ Y3
399
Electronic Devices Now, we apply Kirchhoff’s voltage rule in baseemitter closed loop, we get VBB = IB R B + VBE
Using Eqs. (i) and (ii), we get Y = ( A + A ⋅ B)(1) = ( A + A ⋅ B) + 1 = A ⋅ ( A ⋅ B) + 0 = A ⋅ ( A + B) = A⋅ A + A⋅ B
[Q x + 0 = x] [Q x x = 0]
RB
[JEE Main 2019, 12 Jan ShiftII] IC
VBB
⇒ VBB = (25 × 10−6 × 100 × 1000) + 1 = 3.5 V
19. The reading of the ammeter for a silicon
RC
B
v0 IE
VBB
[JEE Main 2018]
200Ω
E vi
IE + –
diode in the given circuit is
C
RB
E
IB
18. In the figure, given thatV BB supply can vary
from 0 to 5.0 V, VCC = 5 V, β DC = 200, R B = 100 k Ω, RC = 1 k Ω and V BE = 1.0 V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively
C
B
VCC 3V
(a) 25µA and 2.8 V (c) 20µA and 3.5 V
(b) 25µA and 3.5 V (d) 20µA and 2.8 V
Exp. (b) Transistor saturation occurs when VCE = 0. Now, for closed loop of collector and emitter by Kirchhoff’s voltage rule, we have IC C B
E
RC
IE VCC
VCE = VCC − IC RC ⇒ 0 = VCC − IC Rc V 5 ⇒ = 5 × 10−3 A IC = CC = RC 1 × 103 I Now, βDC = 200 (given) ⇒ C = βDC = 200 IB ⇒ ⇒
−3
IC 5 × 10 = 200 200 IB = 2.5 × 10−5 = 25 µA IB =
(a) 0 (c) 11.5 mA
(b) 15 mA (d) 13.5 mA
Exp. (c) / Potential drop in a silicon diode in forward bias is around 0.7 V. In given circuit, potential drop across200 Ω resistor is 3 − 07 . ∆V I = net = 200 R ⇒ I = 0.0115 A ⇒ I = 115 . mA
20. In a common emitter amplifier circuit using an npn transistor, the phase difference between the input and the output voltages will be [JEE Main 2017 (Offline)] (a) 90°
(b) 135°
(c) 180°
(d) 45°
Exp. (c) vi
vo t (Input)
t (Output)
In a CE npn transistor amplifier output is 180° out of phase with input.
400
JEE Main Chapterwise Physics
21. If a, b, c and d are inputs to a gate and x is its output, then, as per the following time graph, the gate is [JEE Main 2016 (Offline)] a.
b.
c.
d.
x.
(a) NOT
(b) AND
(c) OR
(d) NAND
Exp. (c) Output of OR gate is 0 when all inputs are 0 and output is 1 when atleast one of the inputs is 1. Observing output x It is 0 when all inputs are 0 and it is 1 when atleast one of the inputs is 1. ∴The gate is OR.
a
b
c
d
x
0
0
0
0
0
0
0
0
1
1
0
0
1
0
1
0
0
1
1
1
0
1
0
0
1
0
1
0
1
1
0
1
1
0
1
0
1
1
1
1
1
0
0
0
1
23. For a common emitter configuration, if α [JEE Main 2016 (Offline)]
(c) α =
(d) α =
β 1− β β2
1 + β2
As, we know, In case of a commonemitter I configuration, DC current gain, α = c . Ie
I
(b)
β 1+ β
(b) α =
Exp. (a, c)
characteristics are as given below, in the order (a),(b),(c),(d). [JEE Main 2016 (Offline)]
V
Zener diode works in breakdown region. So, Simple diode → (a) Zener diode → (b) Solar cell → (c) Light dependent resistance → (d)
1 1 (a) = + 1 α β
22. Identify the semiconductor devices whose
(a)
Exp. (a)
and β have their usual meanings, the incorrect relationship between α and β is
Alternative Solution OR Gate
I
(a) Simple diode, Zener diode, Solar cell, Light dependent resistance (b) Zener diode, Simple diode, Light dependent resistance, Solar cell (c) Solar cell, Light dependent resistance, Zener diode, Simple diode (d) Zener diode, Solar cell, Simple diode, Light dependent resistance
V
Where, Ic is collector current and Ie is emitter current I and AC current gain, β = c . Ib where, Ib is base current. Also, Ie = Ib + Ic Dividing whole equation by Ic , we get I Ie 1 1 β ⇒ = +1 ⇒ α= = b +1 ⇒ α β 1+ β Ic Ic
24. A red LED emits light at 0.1 W uniformly I
I Dark
(c)
Resistance V
Illuminated
(d)
V Intensity of light
around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is [JEE Main 2015] (a) 1.73 V/m (c) 5.48 V/m
(b) 2.45 V/m (d) 7.75 V/m
Exp. (b) Consider the LED as a point source of light. Let power of the LED is P.
401
Electronic Devices Exp. (a) r
To increase frequency of light emitted from LED (Light Emitting Diode) potential barrier to diode is increased, so for same value of current, higher value of voltage is required for higher frequency.
P
Intensity at r from the source P I= 4 πr 2 1 As we know that I = ε0 E 20 c 2 From Eqs. (i) and (ii), we can write P 1 = ε0 E02 c 4 πr 2 2 . × 9 × 109 2 × 01 2P or E02 = = 2 1 × 3 × 108 4 πε0 r c E02
or
= 6 ⇒ E0 =
...(i)
27. Truth table for system of four NAND gates as shown in figure is
...(ii)
Y B
(a)
6 = 2.45 V/m
25. The forward biased diode connection is [JEE Main 2014]
+2V
(a)
–2V
(c) (b) (c)
–3V
–3V
2V
4V
[AIEEE 2012]
A
A
B
Y
A
B
Y
0 0 1 1
0 1 0 1
0 1 1 0
0 0 1 1
0 1 0 1
0 0 1 1
A
B
Y
A
B
Y
0 0 1 1
0 1 0 1
1 1 0 0
0 0 1 1
0 1 0 1
1 0 0 1
(b)
(d)
Exp. (a) –2V
(d)
+2V
A.(A.B)
A A .B
Y
Exp. (a) For forward bias, pside must be a higher potential than nside. So, + 2V
+
– 2V
–
[JEE Main 2013] R YG B
O
(b) V
R G Y R V
O
O
V
(c)
(d)
I
O
V
Y = { A ⋅ ( A ⋅ B)} ⋅ [B ⋅ ( A ⋅ B)] [using DeMorgan's rules] [break the line, change the sign]
26. The IV characteristic of an LED is
I
B.(A.B)
= ( A + A ⋅ B) ⋅ (B + A ⋅ B)
is forward biased.
(a)
B
R Y G B
I
Red Yellow Green Blue
= ( A + A ⋅ B) + (B + ( A ⋅ B)) = A ⋅ ( A ⋅ B) + B ⋅ ( A ⋅ B) = A ⋅ ( A + B) + B⋅ ( A + B) = A ⋅ B + B⋅ A [Q A ⋅ A = B ⋅ B = 0] A
B
A
B
A⋅ B
B⋅ A
Y
0 0 1 1
0 1 0 1
1 1 0 0
1 0 1 0
0 0 1 0
0 1 0 0
0 1 1 0
It is the truth table of XOR gate.
402
JEE Main Chapterwise Physics
28. The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as a [AIEEE 2011] (a) OR gate (c) NOR gate
(b) NOT gate (d) AND gate
Given figure is half wave rectifier. [since, diode allows the signal to pass during positive half only i .e., forward bias]
31. In the adjacent circuit, A and B represent two inputs andC represents the output.
Exp. ( c )
A
A
C
y
y′
B
B
y′ = A + B and y = y′ = A + B i .e., output of a NOR gate.
29. The combination of gates shown below yields
[AIEEE 2010] A
The circuit represents (a) NOR gate (c) NAND gate
[AIEEE 2008]
(b) AND gate (d) OR gate
Exp. (d) X
If we give the following inputs to A and B, then corresponding output is shown in truth table
B
(a) OR gate (c) XOR gate
(b) NOT gate (d) NAND gate
Exp. (a) Truth table for given combination is A 0 1 0 1
B 0 0 1 1
A
B
C
0 0 1 1
0 1 0 1
0 1 1 1
The above table is similar to OR gate.
X 0 1 1 1
32. If in a pn junction diode, a square input signal of 10 V is applied as shown. [AIEEE 2007]
5V
This comes out to be truth table of OR gate.
RL
30. A pn junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit.
D V
R
~
Then, the output signal across R L will be [AIEEE 2009]
I
10 V
I
(a)
(a)
(b) t
(b)
t
I
I
(c)
–10 V
(d) t
Exp. (c)
–5 V
5V t
(c)
(d) –5 V
403
Electronic Devices (a) 49
Exp. (d) During − ve cycle, diode will not allow the signal to pass. 5V
(b) 50
(d) 48
Exp. (a) β = IC / IB IE = IC + IB I β= C IE − IC 5.488 = = 49 5.60 − 5.488
and ∴ R L Vo
Vi
(c) 51
35. If the lattice constant of this semiconductor
–5 V
is decreased, then which of the following is correct? [AIEEE 2006]
Fig. (i)
Ec
Conduction band width Band gap Vo
Vi
Eg
Vo
Vi
Ev
Valence band width Fig. (ii)
Fig. (iii)
For Vi < 0, the diode is reverse biased and hence offer infinite resistance, so circuit would be like as shown in Fig. (ii) and Vo = 0. For Vi > 0, the diode is forward biased and circuit would be as shown in Fig. (iii) and Vo = Vi. Hence, the option (d) is correct.
33. Carbon, silicon and germanium have four valence electrons each. At room temperature, which one of the following statements is most appropriate? [AIEEE 2007] (a) The number of free conduction electrons is significant in C but small in Si and Ge (b) The number of free conduction electrons is negligibly small in all the three (c) The number of free electrons for conduction is significant in all the three (d) The number of free electrons for conduction is significant only in Si and Ge but small is C
Exp. (d) The number of free electrons for conduction is significant only in Si and Ge but small in C, as C is an impurity.
34. In a commonbase mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor (β ) will be [AIEEE 2006]
(a) (b) (c) (d)
All E c , E g , E v increase E c and E v increase but E g decreases E c and E v decrease but E g increases All E c , E g , E v decrease
Exp. (c) If lattice constant of semiconductor is decreased, then Ec and Ev decrease but Eg increases.
36. A solid which is not transparent to visible light and whose conductivity increases with temperature is formed by [AIEEE 2006] (a) (b) (c) (d)
ionic bonding covalent bonding van der Waals’ bonding metallic bonding
Exp. (b) Covalent bonding. [they are simply semiconductors having 4th group]
37. If the ratio of the concentration of electrons to that of holes in a semiconductor is
7 and 5
7 , then what is the 4 ratio of their drift velocities? [AIEEE 2006] the ratio of currents is
(a)
5 8
Exp. (c)
(b)
4 5
(c)
5 4
(d)
4 7
404
JEE Main Chapterwise Physics
∴ Here, ∴
I = ne Avd n × (vd )e Ie = e Ih nh × (vd )h ne 7 Ie 7 = , = nh 5 Ih 4 (v ) 5 7 5 7 7 (vd )e or d e = × = = × (vd )h 7 4 4 4 5 (vd )h
38. The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit? [AIEEE 2006] 4Ω
∆1
∆2
3Ω
2Ω
Exp. (d) For forward bias, p is − ve, n is − ve. p
n
For reverse biasing of an ideal diode, the potential of nside should be higher than potential of pside. Only option(d) is satisfying the criterion for reverse biasing.
40. The
electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm, is incident on it. The band gap in (eV) for the semiconductor is [AIEEE 2005]
12 V
(a) 1.1 eV
(b) 2.5 eV
(c) 0.5 eV
(d) 0.7 eV
Exp. (c) (a) 1.71 A
(b) 2.00 A (c) 2.31 A
(d) 1.33 A
Eg = hν =
hc 6.63 × 10−34 × 3 × 108 eV = −9 λ . × 10−19 2480 × 10 × 16 [Q1 eV = 1. 6 × 10−19 J]
Exp. (b) In the given, circuit diode D1 is reverse biased while D2 is forward biased, so the circuit can be redrawn as 4Ω
π 4 (c) zero
39. In the following, which one of the diodes is reverse biased?
[AIEEE 2006]
R
R –10 V
+5V
+5 V
(c)
R –10 V
(d)
(d)
π 2
Exp. (c) In common base amplifier, the input signal is amplified but remain in phase with output signal.
42. In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be [AIEEE 2005] (a) 50 Hz
–12 V
(b)
(b) π
(a)
2Ω
[Qfor ideal diodes, reverse biased means open and forward biased means short] Apply KVL to get current following through the circuit 12 −12 + 4 I + 2 I = 0 or I = = 2A 6
(a)
41. In a common base amplifier, the phase difference between the input signal voltage and output voltage is [AIEEE 2005]
12 V
+ 10 V
= 0.5 eV
(b) 25 Hz
(c) 100 Hz (d) 70.7 Hz
Exp. (c) Given, ∴
f = 50 Hz 1 T= 50
For full wave rectifier, T1 =
R
∴
T 1 = 2 100
f1 = 100 Hz
Q f = 1 T
405
Electronic Devices 43. When npn transistor is used as an amplifier, (a) (b) (c) (d)
[AIEEE 2004]
electrons move from base to collector holes move from emitter to base electrons move from collector to base holes move from base to emitter
Exp. (d)
44. For a transistor amplifier in common emitter configuration for load impedance of 1 k Ω (h fe = 50 andhoe = 25 µ A/V), the current gain is [AIEEE 2004] (b) – 15.7 (d) – 48.78
Exp. (d) For a transistor amplifier in common emitter configuration, current gain hfe Ai =− 1 + hoe R L where, hfe and hoe are hybrid parameters of a transistor. 50 Ai = − ∴ 1 + 25 × 10−6 × 1 × 103 = − 4878 .
[AIEEE 2004]
(a) the depletion region is reduced and barrier height is increased (b) the depletion region is widened and barrier height is reduced (c) both the depletion region and barrier height are reduced (d) both the depletion region and barrier height are increased
Exp. (c) When pend of pn junction is connected to positive terminal of battery and nend to negative terminal of battery, then pn junction is said to be in forward bias. In forward bias, the more numbers of electrons go from nregion to pregion and more numbers of holes go from pregion to nregion. Therefore, major current due to both types of carriers takes place through the junction, causing more recombination of electron hole pairs, thus causing reduction in height of depletion region and barrier potential.
47. The difference in the variation of resistance
45. A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of [AIEEE 2004, 03]
(a) (b) (c) (d)
46. When pn junction diode is forward biased, then
When forward bias is applied on npn transistor, then it works as an amplifier. In forward biased npn transistor, electrons move from emitter to base and holes move from base to emitter.
(a) – 5.2 (c) – 24.8
resistance increases with decrease in temperature or viceversa. Since, copper is pure conductor and germanium is a semiconductor, hence due to decrease in temperature, resistance of conductor decreases while that of semiconductor increases.
each of them increases each of them decreases copper decreases and germanium increases copper increases and germanium decreases
Exp. (c) We know that resistance of conductor is directly proportional to temperature (i.e., R ∝ ∆t ), while resistance of semiconductor is inversely 1 proportional to temperature i .e., R ∝ . ∆t Therefore, it is clear that resistance of conductor decreases with decrease in temperature or viceversa, while in case of semiconductor,
with temperature in a metal and a semiconductor arises essentially due to the difference in the [AIEEE 2003] (a) crystal structure (b) variation of the number of charge carriers with temperature (c) type of bonding (d) variation of scattering mechanism with temperature
Exp. (b) Metal has number of free electrons while semiconductor has not at room temperature. The difference in the variation of resistance with temperature in metal and semiconductor is caused due to difference in the variation of the number of charge carriers with temperature.
406
JEE Main Chapterwise Physics
48. In the middle of the depletion layer of reverse biased pn junction, the [AIEEE 2003] (a) (b) (c) (d)
electric field is zero potential is maximum electric field is maximum potential is zero
Exp. (a) Due to the reverse biasing, the width of depletion region increases and current flowing through the diode is almost zero. In this case, electric field is almost zero at the middle of the depletion region.
49. At absolute zero, Si acts as (a) nonmetal (c) insulator
(b) metal (d) None of these
Exp. ( c ) At absolute zero, the energy band is large (> 3 eV) and hence Si acts as insulator or in other words, we can say that there is no free charge carriers.
50. The part of a transistor which is most heavily doped to produce large number of majority carriers is [AIEEE 2002] (a) emitter (b) base (c) collector (d) can be any of the above three
Exp. (a) Because emitter has largest size and maximum number of charge carriers.
51. The energy band gap is maximum in [AIEEE 2002]
(a) metals (c) insulators
(b) superconductors (d) semiconductors
Exp. ( c ) Metals have minimum energy gap and insulators have maximum energy gap and semiconductors have energy gap lying between insulators and metals.
52. By increasing the temperature, the specific resistance of semiconductor (a) (b) (c) (d)
a
conductor
and
a
increases for both [AIEEE 2002] decreases for both increases, decreases respectively decreases, increases respectively
Exp. ( c ) Increases, decreases respectively. If we increase the temperature, the specific resistance (resistivity) of conductor increases while that of semiconductor decreases.
20 Communication Systems 1. The wavelength of the carrier waves in a modern optical fibre communication network is close to (a) 2400 nm (c) 600 nm
d hT
(b) 1500 nm (d) 900 nm [JEE Main 2019, 8 April ShiftI]
Exp. (b) In optical fibre communication network, the signals are transmitted by laser light operating in range of 1310 nm1550 nm. So, the closest value is 1500 nm.
2. In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70 m, then the minimum height of the transmitting antenna should be (Radius of the earth = 6.4 × 106 m)
hR
Given, d = 50 km = 50 × 103 m, hR = 70 m, R = 6.4 × 106 m Then, distance between transmitting and receiving antenna, i.e. d = 2 RhT + 2 RhR 50 × 103 = 2 R ( hT + hR ) = 2 × 6.4 × 106 ( hT + ⇒
[JEE Main 2019, 8 April ShiftII]
(a) 20 m
(b) 32 m
(c) 40 m
(d) 51 m
Exp. (b)
or
70 )
50 × 10 − 8.37 3577.7 = 13.98 − 8.37 = 5.61 hT = 315 . ≈ 32 m
hT ≈
3
3. A signal A cosωt is transmitted usingv 0 sinω 0t
Key Idea In line of sight communication, distance d between transmitting antenna and receiving antenna is given by d=
2RhT +
2RhR
Here in figure, hR and hT is the height of receiving and transmitting antenna, respectively.
as carrier wave. The correct amplitude modulated (AM) signal is [JEE Main 2019, 9 April ShiftI] (a) (v 0 sin ω0t + A cosωt (b) (v 0 + A )cosωt sin ω0t (c) v 0 sin[ω0 (1 + 001 . A sin ωt )t ] A A (d) v 0 sin ω0t + sin(ω0 − ω)t + sin(ω0 + ω)t 2 2
408
JEE Main Chapterwise Physics
Exp. (d) Given, modulating signal, Am = A cosωt Carrier wave, Ac = v 0 sinω0t In amplitude modulation, modulated wave is given by Ym = [ A0 + Am ]sinω0t where, A0 is amplitude of the carrier wave (given as v 0 ) ∴ Ym = [v 0 + A cos ωt ] sinω0t = v 0 sinω0t + A sinω0t cos ωt A = v 0 sinω0t + [sin(ω0 + ω)t + sin(ω0 − ω)t ] 2 A A = v 0 sinω0t + sin(ω0 − ω)t + sin(ω0 + ω)t 2 2
Ac = 400 V Range of frequency in case of amplitude modulation is (fc − fm ) to (fc + fm ). ∴Bandwidth = 2 fm = 2 × 100 × 106 Hz = 2 × 108 Hz and modulation index, 100 A MI = m = = 0.25 400 Ac
6. Given below in the left column are different modes of communication using the kinds of waves given in the right column. [JEE Main 2019, 10 April ShiftI] A. Optical fibre communication
P. Ultrasound
B.
Radar
Q. Infrared light
4. The physical sizes of the transmitter and
C.
Sonar
R. Microwaves
receiver antenna in a communication system are [JEE Main 2019, 9 April ShiftII]
D.
Mobile phones
S. Radio waves
(a) proportional to carrier frequency (b) inversely proportional to modulation frequency (c) independent of both carrier and modulation frequency (d) inversely proportional to carrier frequency
Exp. (d) Size of antenna is directly proportional to the wavelength of the signal. Also, the speed at which signal moves = carrier frequency × wavelength 1 ⇒ fλ = c ⇒ λ ∝ f 1 ∴Size of antenna ∝ . f Note Minimum size of the antenna is λ / 4.
5. A message signal of frequency 100 MHz and peak voltage 100 V is used to execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V. The modulation index and difference between the two side band frequencies are [JEE Main 2019, 10 April ShiftI]
(a) 0.25 ;1 × 108 Hz
(b) 4 ;1 × 108 Hz
(c) 0.25 ; 2 × 108 Hz
(d) 4 ; 2 × 108 Hz
Exp. (c) Given, fm = 100 MHz = 100 × 106 Hz, Am = 100 V,
From the options given below, find the most appropriate match between entries in the left and the right column. (a) (b) (c) (d)
AQ, BS, CR, DP AS, BQ, CR, DP AQ, BS, CP, DR AR, BP, CS, DQ
Exp. (c) In optical fibre communication, infrared light is used to transmit information from one point to another. RADAR (Radio detection and ranging) is a detection system that uses radio waves to determine range, angle or velocity of objects. SONAR (Sound navigation and ranging) is also a detection system that uses ultrasound to detect under water objects, submarines, etc. Mobile phone is a portable telephone that can make and receive calls, which make use of microwave. ∴Correct sequence is A → Q, B → S, C → P and D → R
7. In an amplitude modulator circuit, the carrier wave is given by C( t ) = 4 sin(20000 πt ) while modulating signal is given by, m( t ) = 2 sin(2000 πt ). The values of modulation index and lower side band frequency are [JEE Main 2019, 12 April ShiftII]
(a) 0.5 and 10 kHz (c) 0.3 and 9 kHz
(b) 0.4 and 10 kHz (d) 0.5 and 9 kHz
409
Communication Systems Exp. (d) Given, carrier wave, C(t ) = 4sin(20000 πt ) Modulating signal, m(t ) = 2 sin(2000 πt ) So, carrier wave’s amplitude and frequency are Ac = 4 V, ωc = 20000 π = 2 π × 104 rad/s ω fc = c = 104 Hz = 10 kHz ⇒ 2π and modulating signal’s amplitude and frequency are Am = 2 V, ωm = 2000 π = 2 π × 103 rad/s ω ⇒ fm = m = 103 Hz = 1kHz 2π So, modulating index is A 2 m = m = = 0.5 Ac 4 and lower side band frequency is, fLSB = fc − fm = 10 − 1 = 9 kHz
frequency of signal of wavelength 800 nm can be used as bandwidth. How many channal of 6MHz bandwidth can be broadcast this? (c = 3 × 108 m / s, h = 6.6 × 10−34 J s) [JEE Main 2019, 9 Jan ShiftII]
(b) 386 . × 106 (d) 487 . × 105
Here, Signal wavelength, λ = 800 nm = 8 × 10−7 m Frequency of source is c 3 × 108 As, = 3.75 × 1014 Hz f= = λ 8 × 10−7 ∴Total bandwidth used for communication = 1% of 375 . × 1014
= =
…(i)
total bandwidth available for communication bandwidth of TV signal 375 . × 1012 6 × 106
= 0.625 × 106 = 625 . × 105
[JEE Main 2019, 10 Jan ShiftI]
(a) 65 km (c) 40 km
(b) 80 km (d) 48 km
Exp. (a) Maximum distance of transmission is given by d = 2 RhT + 2 RhR where, hT = height of transmitter, = 140 m, hR = height of receiver = 40 m and R = radius of earth = 6.4 × 106 m. Substituting values, we get d = 2 × 6.4 × 106 ( 140 +
40 )
10. The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license, what broadcast frequency will you allot ? [JEE Main 2019, 10 Jan ShiftII] (a) 2000 kHz (c) 2900 kHz
(b) 2250 kHz (d) 2750 kHz
Exp. (a)
Exp. (c)
= 375 . × 1012 Hz So, number of channel for signals
m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower in LOS (Line of Sight) mode? (Take, radius of earth = 6.4 × 106 m).
= 65 km
8. In communication system, only one percent
(a) 3.75 × 106 (c) 6.25 × 105
9. A TV transmission tower has a height of 140
For a given carrier wave of frequency fc with modulation frequency fm , the bandwidth is calculated by fupper = fc + fm … (i) flower = fc − fm To avoid overlapping of bandwidths, next broadcast frequencies can be f1 = fc ± 2 fm , f2 = fc ± 3fm So, next immediate available broadcast frequency is f1 = fc + 2 fm and f1′ = fc − 2 fm Given, fm = 250 kHz and also that fm is 10% of fc , i.e. fc = 2500 kHz. So, f1 = 2500 + (2 × 250) = 3000 kHz and f′1 = 2500 − (2 × 250) = 2000 kHz
410
JEE Main Chapterwise Physics C m = ( Ac + Am sinωmt ) ⋅ sinωc ⋅ t
11. An amplitude modulates signal is given by v (t ) = 10[1 + 0.3 cos(2.2 × 104t )] sin(5.5 × 105t ). Here, t is in seconds. The sideband 22 frequencies (in kHz) are Take, π = 7 [JEE Main 2019, 11 Jan ShiftI]
(a) 892.5 and 857.5 (c) 178.5 and 171.5
(b) 89.25 and 85.75 (d) 1785 and 1715
Exp. (b)
(as from given graph, Tc = 8 × 10–6 s)
v(t )=10 [1+ 0.3cos(2.2 × 104 t )] [sin(5.5 × 105 t )] and
Upper band angular frequency ων = (2.2 × 104 + 5.5 × 105 ) rad / s
= 2 .5 π × 105 s −1 2π 2π ωm = = Tm 100 × 10−6 = 2 π × 104 s −1
Similarly, lower band angular frequency. ωL = (5.5 × 105 − 2.2 × 104 ) rad / s = 528 × 10 rad / s 3
∴ Side band frequency are, ω 572 kHz ~ − 91kHz fu = u = 2π 2π ω 528 and kHz ~ − 84 kHz fL = L = 2π 2π Thus, the approximate side band frequencies are 89.25 and 85.75.
12. An amplitude modulated signal is plotted below V(t)
t
100 µs
8 µs
Which one of the following best describes the above signal? [JEE Main 2019, 11 Jan ShiftII] (a) (b) (c) (d)
...(v)
(as from given graph, Tm = 100 × 10−6 s)
= 572 × 103 rad / s
10 V 8V
…(i)
For the given graph, maximum amplitude, Ac + Am = 10 …(ii) and minimum amplitude, Ac − Am = 8 …(iii) From Eqs. (ii) and (iii), we get …(iv) Ac = 9V and Am = 1V QFor angular frequency of message signal and carrier wave, we use a relation 2π 2π ωc = = Tc 8 × 10−6
[1 + 9 sin( 2 π × 104 t )]sin( 2.5 π × 105 t )V [9 + sin( 2 π × 104 t )]sin( 2.5 π × 105 t )V [9 + sin( 4 π × 104 t )]sin( 5 π × 105 t )V [9 + sin( 2.5 π × 105 t )]sin( 2 π × 104 t )V
Exp. (b) Equation of an amplitude modulated wave is given by the relation,
When we put values of Ac , Am, ωc and ωm in Eq. (i), we get C m = [9 + sin(2 π × 104 t )]sin(2 .5 π × 105 t ) V
13. A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index? [JEE Main 2019, 12 Jan ShiftI]
(a) 0.4 (c) 0.6
(b) 0.5 (d) 0.3
Exp. (c) Modulation index is given by − Amin A V − Vmin µ = max = max Amax + Amin Vmax + Vmin 160 − 40 120 = = 0.6 = 160 + 40 200
14. To double the covering range of a TV transmission tower, its height should be multiplied by [JEE Main 2019, 12 Jan ShiftII] (a)
2
(b) 4
(c) 2
(d)
1 2
Exp. (b) Range of TV transmitting tower d = 2 hR where, h is the height of the transmission tower. when range is doubled, ∴ 2d = 2 2 hR = 2(4h)R So, height must be multiplied with 4.
411
Communication Systems 15. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilised for transmission. How many telephonic channels can be transmitted simultaneously, if each channel requires a bandwidth of 5 kHz? [JEE Main 2018]
(a) 2 × 103
(b) 2 × 104
(c) 2 × 10
(d) 2 × 106
5
Exp. (c) Only 10% of 10 GHz is utilised for transmission. ∴ Band available for transmission 10 = × 10 × 109 Hz = 109 Hz 100 Now, if there are nchannels each using 5 kHz then, n × 5 × 103 = 109 ⇒ n = 2 × 105
16. In amplitude modulation, sinusoidal carrier frequency used is denoted by ωc and the signal frequency is denoted by ωm . The bandwidth ( ∆ωm ) of the signal is such that ∆ωm n1 ). The refractive index of cladding can be either changing abruptly or gradually changing (graded index fibre). (iii) There is a very little transmission loss through optical fibres. (iv) There is no interference from stray electric and magnetic fields to the signals through optical fibres.
21 Practical Physics 1. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet, if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line? [JEE Main 2016 (Offline)] (a) 0.75 mm (c) 0.70 mm
(b) 0.80 mm (d) 0.50 mm
Exp. (a) Given that the screw gauge has zero error. 0.5 So, least count of a screw gauge = mm. 50 Thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line, we have 0.5 mm = 0.50 mm + (25) × 50 = 0.50 mm + 0.25 mm = 075 . mm
2. A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? [JEE Main 2014] (a) A meter scale (b) A vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in 1 cm
(c) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm (d) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm
Exp. (a) If student measures 3.50 cm, it means that there is an uncertainly of order 0.01 cm. For vernier scale with 1 MSD = 1mm and 9 MSD = 10 VSD LC of vernier calliper = 1MSD − 1VSD 9 1 = 1 − 10 10 1 cm = 100
3. A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading 0 mm Circular scale reading 52 divisions Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is (a) 0.052 cm (c) 0.005 cm
(b) 0.026 cm [AIEEE 2011] (d) 0.52 cm
Exp. (a) Diameter of wire, d = MSR + (CSR × LC) = 0 + 52 × = 0.052 cm
1 = 0.52 mm 100
QLC = 1 100
415
Practical Physics 4. An experiment is performed to find the
z = 3 + 35 ×
refractive index of glass using a travelling microscope. In this experiment, distances are measured by [AIEEE 2008]
= 3.38 mm
(a) a vernier scale provided on the microscope (b) a standard laboratory scale (c) a meter scale provided on the microscope (d) a screw gauge provided on the microscope
Exp. ( a ) If an experiment is performed to find the refractive index of glass using a travelling microscope, distances are measured by a vernier scale which is provided on the microscope.
5. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of – 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is [AIEEE 2008] (a) 3.32 mm (c) 3.67 mm
(b) 3.72 mm (d) 3.38 mm
Exp. ( d ) Diameter = Main scale reading + Circular scale reading × LC + Zero error
1 + 0.03 2 × 50
6. A working transistor with its three legs marked P, Q and R is tested using a multimeter. No conduction is found between P and Q. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P orQ, some conduction is seen on the multimeter. Which of the following is true for the transistor? [AIEEE 2008] (a) It is an npn transistor with R as base (b) It is a pnp transistor with R as collector (c) It is a pnp transistor with R as emitter (d) It is an npn transistor with R as collector
Exp. (a) Since, no conduction is found when multimeter is connected across P and Q, it means either both P and Q are ntype or ptype. So, it means R is base. When R is connected to common terminal and conduction is seen when other terminal is connected to P or Q, so it means transistor is npn with R as base. p P
n
n R
Q
Practice Set 1 Instructions This test consists of 30 questions to be completed in 3 hrs. Each question is allotted 4 marks for correct response. 1 marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted according as per instructions.
1. For the following diagram [used to measure
3. A 1200 kg car is coasting down a 30° hill as
the length of a small metal piece by using vernier callipers], determine the length of the metal piece. Least count of the vernier callipers is 0.1 mm.
shown in figure. At a time when the car’s speed is 12 m/s, the driver applies the brakes. What constant force F (parallel to the road) must result, if the car is to stop after travelling 100 m?
10
15
20
mm
0 Vernier scale
10
Object
(a) 18 mm (c) 12.6 mm
(b) 15.7 mm (d) None of these
2. Consider the situation shown in figure. The uniform 0.60 kN beam is hinged at P. Find the tension in the tie rope. Give your answer to two significant figures.
0m 100 sin 30°
30°
(a) 6.7 kN (c) 100 N
(b) 6.7 N (d) 20 N
4. A plane is travelling eastward at an air speed of 500 km/h. But a 90 km/h, wind is blowing southward. What is speed of the plane relative to the ground?
P
40°
(a) 235 km/h
(b) 158 km/h
(c) 258 km/h
(d) 508 km/h
5. A skier starts from rest and slides 9.0 m down
3L/4
800 N
(a) 2280 N (b) 2.3 kN (c) 2.28 kN (d) 228 N
a slope in 3.0 s. In what time after starting will be skier acquire a speed of 24 m/s? Assume that the acceleration is constant. (a) 12 s
(b) 15 s
(c) 2 s
(d) 12.5 s
Note All the Practice Sets (110) available for free online practice see detailed instructions back side of the title.
417
Practice Set 1 6. As shown in figure, a uniform solid sphere
11. When a force of 500 N pushes on a 25 kg box
rolls on a horizontal surface at 20 m/s and then rolls up the incline. If friction losses are negligible, what will be the value of h, where the ball stops?
as shown in figure, the acceleration of the box up the incline is 0.75 m/s 2. The coefficient of kinetic friction between box and incline is 500N
h
(b) 24.5 m (c) 29 m
(d) 90 m
7. A wind tunnel is to be used with a 20 cm high model car to approximately reproduce the situation in which a 550 cm high car is moving at 15 m/s. What should be the wind speed in the tunnel? (a) 2.41 km/s (c) 0.41 km/s
(b) 3.41 km/s (d) 5.41 km/s
Two identical blocks P and Q have mass m each. They are attached to two identical springs initially unstretched. Now, the left spring (along with P) is A compressed by and the right spring (along with Q) is 2 compressed by A. Both the blocks are released simultaneously. They collide perfectly inelastically. Initially time period of both the blocks was T. A
8. The time period of oscillation of combined mass is (a)
T 2
(b)
(c) T
(d)
A 4
(b)
A 2
(c)
(d)
3A 4
10. What is energy of oscillation of the combined mass? (a)
1 kA 2 2
(b)
1 2 kA 4
(c)
1 2 kA 8
FW 40°
(a) 0.41
(b) 0.3
(c) 0.25
(d) 1.5
steel beam (crosssectional area 45 cm 2) is put in place with its ends cemented in pillars. If the sealed ends cannot move, what will be the compressional force in the beam when the temperature is 25°C ? For this kind of steel, and α = 1.1 × 10−5 ° C −1 11 2 Y = 2.0 × 10 N/m . (a) 2.5 × 103 N
(b) 3.5 × 105 N
(c) 2.5 × 105 N
(d) 3.5 × 103 N
æ Directions : Question. 13 is AssertionReason type questions. Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is correct answer. You have to select the correct choice in the codes (a), (b), (c) and (d) in the given below.
tornado is alarmingly high. Statement II If no external torque acts on a body, its angular velocity remains conserved.
T 2
2A 3
FN
Ff
13. Statement I The speed of whirlwind in a
2T
9. The amplitude of combined mass is (a)
x
40°
12. When a building is constructed at –10°C, a
æ Directions : Question Nos. 8 to 10 are based on following paragraph.
A 2
y
g
m
g
(a) 20 m
64 0.
kg
m
38
30°
25
77 0.
321N
40° 3N
v = 20 m/s
(d)
1 kA 2 16
(a) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I (b) Statement I is true, Statement II is false (c) Statement I is false, Statement II is true (d) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I
418
JEE Main Chapterwise Physics
14. A 2.0 mg droplet of liquid nitrogen is present
19. A voltmeter is to deflect full scale for a
in a 30 mL tube as it is sealed off at very low temperature. What will be the nitrogen pressure in the tube when it is warmed to 20°C? Express your answer in atmospheres. (M for nitrogen is 28 kg/K mol).
potential difference of 5.000 V across it and is to be made by connecting a resistor R x in series with a galvanometer. The 80.00 Ω galvanometer deflects full scale for a potential of 20.00 mV across it. Find R x .
(a) 0.057 atm (c) 5.7 atm
(a) 12.2 kΩ (c) 19.92 kΩ
(b) 0.57 atm (d) 0.0057 atm
(b) 20.54 kΩ (d) None of these
15. A beverage cooler is in the shape of a cube,
20. The lights on a car are inadvertently left on.
42 cm on each inside edge. Its 3.0 cm thick walls are made of plastic(kT = 0.050 W/mK). When the outside temperature is 20°C, how much ice will melt inside the cooler each hour?
They dissipate 95.0 W. About how long will it take for the fully charged 12.0 V car battery to run down, if the battery is rated at 150 Ah?
(a) 2.5 g (c) 0.38 g
(b) 2.5 kg (d) 0.38 kg
16. A steam engine operating between a boiler temperature of 220°C and a condenser temperature of 35.0°C delivers 8.00 HP. If its efficiency is 30.0 per cent of that for a Carnot engine operating between these temperature limits, how many calories are absorbed each second by the boiler? (a) 12.7 kcal/s (c) 2.54 cal/s
(b) 12.7 cal/s (d) 2.54 kcal/s
17. In the Bohr model of the hydrogen atom, an
electron (q = − e ) circles a proton (q ′ = e ) in an orbit of radius 5.3 × 10−11 m. The attraction of the proton for the electron furnishes the centripetal force needed to hold the electron in orbit. What is the electron’s speed ? The electron mass is 9.1 × 10−31 kg. (a) 25.0 × 106 m/s
(b) 22.5 ×103 m/s
(c) 2.2 ×106 m/s
(d) 4.2 × 103 m/s
18. Two capacitors, 3.0 µF and 4.0 µF, are individually charged across a 6.0 V battery. After being disconnected from the battery, they are connected together with a negative plate of one attached to the positive plate of the other. What is the final charge on capacitors? (a) 2.6 µ C, 3.4 µC (c) 2.6 µC, 2.6 µC
(b) 3.4 µC, 2.6 µC (d) 3.4 µC, 3.4 µC
(a) 18.9 h (c) 22.5 h
(b) 16.0 h (d) 29.0 h
21. At a certain place on the planet, the earth’s
magnetic field is 5.0 × 10−5 T, directed 40° below the horizontal. Find the force per metre of length on a horizontal wire that carries a current of 30 A northward. (a) 2.45 × 10−4 N −2
(c) 9.6 × 10
N
(b) 2.45 × 10−2 N (d) 9.6 × 10−4 N
22. A 5.0 µF capacitor is charged to a potential difference of 20 kV between plates. After being disconnected from the power source, it is connected across a 7.0 MΩ resistor to discharge. What is the initial discharge current and how long will it take for the capacitor voltage to decrease to 37% of the 20 kV ? (a) 20 s (c) 19 s
(b) 35 s (d) 14 s
23. How much electrical potential energy does a proton lose as it falls through a potential drop of 5 kV? (a) − 8 × 10−16 J
(b) − 2.4 × 10−16 J
(c) − 1.5 × 10−14 J
(d) 19.5 × 10−14 J
24. What is the minimum value of the refractive index for a 45.0° prism which is used to turn a beam of light by total internal reflection through a right angle? (a) 1.67 (c) 1.41
(b) 1.5 (d) 2.0
419
Practice Set 1 25. A lens has a convex surface of radius 20 cm and a concave surface of radius 40 cm and is made of glass of refractive index 1.54. Compute the focal length of the lens (a) 74 cm
(b) 23 cm (c) 15 cm
28. Given below are symbols for some logic gates.
(d) 29 cm
1
2
3
4
26. Red light falls normally on a diffraction grating ruled 4000 lines/cm and the second order image is diffracted 34.0° from the normal. Compute the wavelength of the light. (a) 229 nm (c) 699 nm
(b) 137 nm (d) 250 nm
(a) 1 and 2 (c) 3 and 4
27. When a hydrogen atom is bombarded, the atom may be raised into a higher energy state. As the excited electron falls back to the lower energy levels, light is emitted. What are the three longest wavelength spectral lines emitted by the hydrogen atom as it returns to the n = 1 state from higher energy states? Give your answers to three significant figures. (a) (b) (c) (d)
The XOR gate and NOR gate respectively are
122 nm, 102 nm, 96.9 nm 122 nm, 112 nm, 96.9 nm 145 nm, 102 nm, 83.7 nm 145 nm, 112 nm, 83.7 nm
(b) 2 and 3 (d) 1 and 4
29. If an AM transmitter shows an antenna current of 8 A when only carrier is transmitted and 9 A when current is modulated using sine wave. Then, the modulation index is (a) 65.8% (c) 40.2%
(b) 72.9% (d) 0
30. The bandwidth required for an FM signal, if modulating frequency is 2 kHz maximum deviation is 10 kHz, is (a) 32 kHz (c) 15 kHz
and
(b) 10 kHz (d) 40 kHz
Answers 1. (c) 11. (a) 21. (d)
2. (b) 12. (b) 22. (b)
3. (a) 13. (b) 23. (a)
4. (a) 14. (a) 24. (c)
5. (a) 15. (d) 25. (a)
6. (c) 16. (a) 26. (c)
7. (c) 17. (c) 27. (a)
8. (c) 18. (a) 28. (b)
9. (a) 19. (c) 29. (b)
10. (d) 20. (a) 30. (a)
Solutions 1. Reading of vernier calliper = Main scale divisions + Least count × Vernier scale divisions = MSD + LC × VSD As 6th VSD is coinciding with one of the MSD and MSD is 12th, hence reading = 12 + 6 × 01 . = 12.6 mm
2. The reaction forces acting on the beam are shown in
represented by its components, FRH and FRV . The torque equation about P as axis is FT
FRH P
40°
FRV 600N
figure, where the force exerted by the hinge is 800N
420
JEE Main Chapterwise Physics 3 L (F )(sin 40° ) − (L)(800 N)(sin 90° ) T 4 L − (600 N) (sin 90° ) = 0 2 We take the axis at P, so that FRH and FRV do not appear in the torque equation. Solution of this equation yields FT = 2280 N or to two significant figures FT = 2.3 kN.
3. From workenergy theorem, The change in total energy of the car (KE + PE) is equal to the work done on it by the breaking force F. This work is Fs cos 180° because F retards the car’s motion. We have, 1 m(v f2 − v i2 ) + mg (hf − hi ) = Fs(−1) 2 where, m = 1200 kg vf = 0 v i = 12 m/s hf − hi = (100 m) sin 30° s = 100 m With these values, the equation yields F = 67 . kN
6. The rotational and translational KE of the sphere at the bottom will be changed to PEG when it stops. Therefore, we write 1 mv 2 + 1 Iω2 = (mgh)end 2 start 2 2 But for a solid sphere, I = mr 2 . Also, ω = v/r. The 5 above equation becomes 2
v 1 1 2 mv 2 + (mr 2 ) = mgh r 2 2 5 1 2 1 2 v + v = (9.81 m/s 2 ) h 2 5 Using v = 20 m/s gives h = 29 m
or
7. We want the Reynolds number NR to be the same in
4. The plane’s resultant velocity is the sum of two velocities, 500 km/h—East and 90 km/h— South. These component of velocities are shown in figure. The plane’s resultant velocity is then R = (500 km/h)2 + (90 km/h)2 = 508 km/h N 500 km/h α
We can now use this value of a for the longer trip, from the starting point to the place, where vtx = 24 m/s. For this trip, v i x = 0, v f x = 24 m/s, a = 2.0 m/s 2 . Then, from v f = v i + at , v − v ix 24 m/s t = fx = = 12 s a 2.0 m/s 2
E 90 km/h
R
The angle α is given by 90 km/h tan α = = 018 . 500 km/h from which α = 10°. The plane’s velocity relative to the ground is 508 km/h at 10° south to east.
5. We must find the skier’s acceleration from the data concerning the 3.0 s trip. Taking the direction of motion as the + xdirection, we have t = 3.0 s, v ix = 0 and x = 9.0 m. 1 Then, x = v ix t + at 2 gives 2 2x 18 m a= 2 = = 2.0 m/s 2 t (3.0 s)2
both cases, so that the situation is similar in both the cases. That is, we want ρvD ρvD NR = = η tunnel η air Both ρ and η are the same in the two cases, so we have D vt Dt = v a Da from which vt = v a a Dt 550 = (15 m/s) = 0.41 km/s 20
8. They will collide at their mean positions because m . k = 2 k.
time period of both are same and that is 2 π After collision, combined mass is 2 m and keff Hence, time period remains unchanged. 2m m T = 2π ⇒ T = 2π 2k k
9. From conservation of linear momentum, we can see that velocity of combined mass just after collision is ωA v= . 4
⇒ Before collision
Just after collision
Since, this is the velocity at mean position.
421
Practice Set 1 Hence,
or
10.
E=
v = ω′ A ′ ωA = ω′ A ′ 4 A A′ = because ω′ = ω = 4
ω A 1 1 (2 m)v 2 = (2 m) 4 2 2 =
14. We use pV = m RT to find p = mRT M
2k k or m 2m
2
k 2k mω2 A 2 kA 2 = = Q ω2 = m 2 m 16 16 2
or
1 A kA 2 E = (2 k ) = 4 2 16
11. Since, the box moves up the incline, the friction force (which always acts to retard the motion) is directed down the incline. Let us first find Ff by writing ΣFx = ma x . From figure (using sin 40° = 0.643) F − Ff −mg sin θ = ma 383 N − Ff − (0.64)(25)(9.81) N = (25 kg)(0.75 m/s 2 ) from which Ff = 207 N We also need FN . Writing ΣFy = may = 0 and using cos 40° = 0766 . , we get FN − 321 N − (0.77)(25)(9.81) N = 0 or FN = 510 N F 207 Then, µk = f = = 0.41 FN 510
12. ∆L = α ∆T = (1.1 × 10−5 °C −1 )(35°C ) L0
where, ∆L = Change in length [Q ∆L = L0 α ∆t ] and L0 = original length = 3.85 × 10−4 F/ A ∆L So, F = YA Q Young’s modulus, Y = ∆ L / L0 L0 11 2 −4 2 −4 = (2.0 × 10 N/m ) (45 × 10 m ) (3.85 × 10 )
=
−6
(2.0 × 10
kg)(8314 J/kmolK)(293 K)
(28 kg/kmol)(30 × 10−6 m3 )
= 5800 N/m2 10 . N/atm = (5800 N/m2 ) 5 2 × 10 1.01 N/m = 0.057 atm
15. We have with the ice inside at 0°C ∆Q/∆t = kT A ∆T/L ∆Q 20° C = (0.050 W/mK )(6)(0.42 m)2 0.030 m ∆t = 35.3 J/s = 8.43 cal/s In one hour, ∆Q = (60)2 (8.43) = 30.350 cal. To melt 1.0 g of ice requires 80 cal, so the mass of ice melted in one hour is 30.350 cal m= = 0.38 kg 80 cal/g
16. Actual efficiency = (0.30) (Carnot efficiency) 308 K = (0.30) 1 − . = 0113 493 K Q η = 1− T2 where, η = efficiency T1 T2 = sink and T1 = saurce. Output work But the relation efficiency = Input heat Output work /s ∴ Input heat/s = Efficiency 1.00 cal/s (8.00 HP)(746 W/HP) 4.184 W = 0113 .
= 3.5 × 105 N
13. In a whirlwind in a tornado, the air from nearby regions gets concentrated in a small space thereby decreasing the value of its moment of inertia considerably. Since, Iω = constant, so due to decrease in moment of inertia of the air, its angular speed increases to a high value. If no external torque acts, then dL = 0 or L = constant τ = 0 or dt or Iω = constant As in the rotational motion, the moment of inertia of the body can change due to the change in position of the axis of rotation, the angular speed may not remain conserved.
MV
17. FE = k
= 12.7 kcal/s qq ′
[as q ⇒ e
r2
er r
= (9.0 × 109 N m2 /C 2 ) −8
= 8.2 × 10
N = 82 N
and q → e]
e
r
(1.6 × 10−19 C)2 (5.3 × 10−11 m)2
422
JEE Main Chapterwise Physics The force found is the centripetal force, mv 2 /r. Therefore, mv 2 8.2 × 10−8 N = r from which
(8.2 × 10−8 N)(r ) m
v= =
−8
(8.2 × 10
19. When the galvanometer is deflecting full scale, the −11
N)(5.3 × 10
m)
9.1 × 10−31 kg
= 2.2 × 106 m/s Aliter Velocity in nth orbit, v n = v1 =
⇒
c 1 137 n
c 1 ⋅ = 2.2 × 106 m/s 137 1
18. The situation is shown in figure. Before being connected, their charges are q 3 = CV = (3.0 × 10−6 F)(6.0 V) = 18 µC q 4 = CV = (4.0 × 10−6 F)(6.0 V) = 24 µC These charges partly cancel when the capacitors are connected together. Their final charges are given by q3 = 18µC +
−
+
− +
−
+
+
−
+
A +
−
+
−
q′4 (b) After
q 3′ + q 4′ = q 4 − q 3 = 6.0 µC Also, the potentials across them are now the same, so that V = q / C which gives q 3′ q 4′ = [Q V1 = V2 ] 3.0 × 10−6 F 4.0 × 10−6 F or
q 3 ′ = 075 . q4 ′
from which
R x = 19.92 kΩ
20. As an approximation, assume the battery maintains 12.0 V until it goes dead. Its 150 Ah rating means it can supply the energy equivalent of a 150 A current that flows for 1.00 h (3600 s). Total output energy = (power)(time)= (VI)t = (12.0 V × 150 A ) (3600 s) = 6.48 × 106 J
Northward i.e., the direction in which a compass needle points. Therefore, the situation is that shown in figure. The force on the wire is FM = (30 A )(L)(5.0 × 10−5 T) sin 40° B
V
When R x is connected in series with the galvanometer, we wish I to be 2.500 × 10−4 A for a potential difference of 5.000 V across the combination. Hence, V = IR becomes 5.000 V = (2.500 × 10−4 A)(80.00 Ω + R x )
21. Nearly everywhere, the earth’s field is directed
(a) Before q′3 −
current through it is V 20.00 × 10−3 V I= = = 2.500 × 10−4 A 80.00 Ω R
The energy consumed by the lights in a time t is Energy dissipated = (95 W)(t ) Equating these two energies and solving for t, we find t = 6.82 × 104 s = 18.9 h
q4 = 24µC −
Substitution in the previous equation gives 075 . q 4 ′ + q 4 ′ = 6.0 µC or q 4 ′ = 3.4 µC Then, q 3 ′ = 075 . q 4 ′ = 2.6 µC
So that
[QF = iBl sin θ]
FM = 9.6 × 10−4 N L Bk 40° Bv B
30 A North
The right hand rule indicates that the force is into the page, which is West.
22. The loop equation for the discharging capacitor is Vc − IR = 0 where, Vc is the potential difference across the capacitor. At the first instant, Vc = 20 kV, so
423
Practice Set 1 I=
20 × 103 V Vc = = 2.9 mA R 7.0 × 106 Ω
The potential across the capacitor, as well as the charge on it, will decrease to 0.37 of its original value in one time constant. The required time is RC = ( 7.0 × 106 Ω )(5.0 × 10−6 F ) = 35 s
23. The proton carries a positive charge. It will therefore move from regions of high potential the regions of low potential if left free to do so. Its change in potential energy as it moves through a potential different V is Vq. In our case, V = − 5 kV. Therefore, change in PEE = Vq = (−5 × 103 V)(1.6 × 10−19 C) = − 8 × 10−16 J
24. The ray enters the prism without deviation, since it strikes side AB normally. It then make an incidence angle of 45.0° with normal to side AC. The critical angle of the prism must be smaller than 45.0°, if the ray is to be totally reflected at side AC and thus turned through 90°. A 45°
27. We are interested in the following transitions n = 2 → n = 1: ∆E( 2, 1) = − 3.4 − (−13.6) = 10.2 eV
n = 3 → n = 1:
∆E( 3, 1) = − 1.5 − (−13.6) = 12.1 eV
n = 4 → n = 1:
∆E( 4, 1) = − 0.85 − (−13.6) = 12.8 eV To find the corresponding wavelengths, we can use hc . ∆E = hf = λ For example, for the n = 2 to n = 1transition, hc λ= ∆E( 2, 1) =
(6.63 × 10−34 Js) (2.998 × 108 m/s) (10.2 eV)(1.60 × 10−19 J/eV)
= 122 nm The other lines are found in the same way to be 102 nm and 96.9 nm. These are the first three lines of the Lyman series.
28. XOR ⇒ Exclusive OR NOR ⇒ NOT + OR Symbols 2 and 3 represent XOR and NOR gates respectively.
45° C
B
29. Modulation index ma is given by 2 I ma = 2 tot − 1 Icarrier
1 Minimum ni = . = 141 sin 45.0°
25. First, notice that r1 > 0 and r2 > 0 because both surfaces have their centres of curvature to the right. Consequently, 1 1 1 = (n − 1) − f r1 r2 1 1 0.54 cm . − 1) = (154 − = 20 40 40 or
f = + 74 cm
Since, f turns out to be positive, the lens is converging.
26. From the grating equation mλ = a sin θ2 1 cm (0.559) a sin θ2 λ= = 4000 2 2 = 6.99 × 10−5 cm = 699 nm
where,
Itot = 9 A, Icarrier = 8 A
∴
9 2 ma = 2 − 1 8 81 − 64 ma = 2 8 = 72.9% (approx)
Hence, option (b) is true. 30. M f = δ = 10 = 5 fm 2 For M f = 5, Highest J coefficient is J8 . Thus 8th pair sidebands is the farthest from the carrier to be included here. Thus, bandwidth D = 2 × 8 × 2 = 32 kHz
Practice Set 2 Instructions For instructions refer to Practice Set 1.
1. A car and a truck are moving with same kinetic energy. To stop them brakes exerting equal force are applied simultaneously. Then, which of the following statement is correct? (a) The truck stops earlier as it covers less distance (b) The car stops earlier as it covers less distance (c) The car and the truck stop simultaneously covering equal distances (d) The car stops earlier though it covers the same distance as the truck before coming to stop
2. A DC voltage with appreciable ripple expressed asV = V1 + V 2 cos ωt is applied to a resistorR. The amount of heat generated per second is (a)
V12 + V 22 2R
(b)
2V12 + V 22 2R
(c)
V12 + 2V 22 2R
(a) a
(b) 2a
(c) zero
(d)
a 2
5. A car accelerates from rest at a constant rate
α for some time after which it decelerates at a constant speedβ to come to rest. If the total time elapsed is t , the maximum velocity acquired and total distance travelled by the car are given by α2 + β2 1 t , (a) αβ 2
α2 + β2 2 t αβ
α2 − β2 1 t , (b) αβ 2
α2 − β2 2 t αβ
α + β 1 α + β 2 (c) t, t αβ 2 αβ αβ 1 αβ 2 (d) t, t α + β 2 α + β
6. A body slides down a plane inclined at an
(d) depends on whether V1 is greater than or less than V 2
3. The work function for the surface of aluminium is 4.2 eV. How much potential difference will be required to just stop the emission of maximum energy electrons emitted by light of 2000 Å? (a) 1.51 V (c) 2.99 V
difference between two points (5a , 0) and ( −3a , 4a ) will be
(b) 1.99 V (d) None of these
4. A charge of 20µC is situated at the origin of an xycoordinate system. The potential
angle θ to the horizontal. The coefficient of friction µ down the plane varies in direct proportion to the distance moved down the plane. The body will move down the plane with a
(a) (b) (c) (d)
constant acceleration g sin θ constant acceleration ( g sin θ − µg cos θ) constant acceleration (µg cos θ − g sin θ) variable acceleration that first decreases, then becomes negative
7. Unit of LCR is (a) ohm × sec (c) ohm × sec 2
(b) ohm s −1 (d) No unit
425
Practice Set 2 8. Two factories are sounding their sirens at 800 Hz each. A man goes from one factory to the other at a speed of 2 ms −1. The speed of sound is 320 ms −1. Therefore, the number of beats heard by the person in one second will be (a) 2 (c) 8
(b) 4 (d) 10
9. A container open to atmosphere contains air (assumed to be an ideal gas) at temperature 27°C. The temperature is now raised to 227°C. The ratio of number of atoms in the container now and at the beginning is 3 (a) 5
5 (b) 3
3 (c) 4
12. A ray of light is
φ α
(i) Assuming that the Q R prism is immersed in air, find the largest value for angle φ, so that the ray is totally reflected at face PR. (ii) Find φ if the prism is immersed in water (µ =1.33). Choose the correct answer from the following code.
(a) (i) 28.9° (b) (i) 48.9° (c) (i) 41.1° (d) (i) 36°
4 (d) 5
æ Directions Q. no. 10 and 11 are based on following paragraph. A block of mass 2 kg is suspended by a spring of force constant k = 10 N/m. An another spring of same value of force constant is 1 m below it. Initially both the springs were unstretched. Mass is released from rest.
P
incident normally on the face PQ of a glass prism ( µ = 1.52 ) as shown in the figure
(ii) (ii) (ii) (ii)
48.9° 28.9° 61.1° 30°
13. Four charges Q each are located at four vertices of a regular tetrahedron of side L. The potential energy of the system is (a)
4Q 2 πε0 L
2
(b)
6Q 2 4 πε0 L
(c)
Q2 πε0 L
(d)
8Q 2 4 πε0 L
14. The moment of inertia of a body about a
given axis is 1.2 kg m 2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad  s −2 must be applied about that axis for a duration of
2 kg 1m
(a) 4 s
(b) 2 s
(c) 8 s
(d) 10 s
15. Light passes from a denser medium 1 to a rarer medium 2. When the angle of incidence is θ, the reflected and refracted rays are mutually perpendicular. The critical angle will be
10. What will be the maximum extension in the
(b) sin −1 (tan θ) (d) sin −1 ( secθ)
16. A person looking through telescope T just
upper spring? (a) 3.26 m
(b) 1.67 m
(c) 2.82 m
(d) None of these
11. In the above problem, equilibrium position of the block is at ... m from where it was released. (a) 1.0 (c) 1.2
(a) sin −1 (cot θ) (c) sin −1 (cos θ)
(b) 2.0 (d) 1.5
sees the point A on the rim at the bottom of a cylindrical vessel when the vessel is empty. When the vessel is completely filled with a liquid ( µ = 1.5), he observes a mark B at the centre of the bottom without moving the telescope or the vessel. The diameter of the base of the vessel is 10 cm. The height of the vessel is (a) 10 cm
(b) 15 cm (c) 4.85 cm (d) 8.45 cm
426
JEE Main Chapterwise Physics
17. A particle A has charge +q and particle B has charge +4q with each of them having same mass m. When allowed to fall from rest through the same electric potential v difference, the ratio of their speeds A will vB become (a) 2 :1
(b) 1 : 2
(c) 1 : 4
(d) 4 :1
18. A radio transmitter operates at 880 kHz and its power is 10 kW. The number of photons emitted per second is (a) 1.72 ×1031
(b) 1.72 ×1032
(c) 3.44 ×1031
(d) None of these
19. The whistle of a railway engine is heard in winter at much longer distances. This is due to (a) decrease in the velocity of sound in winter (b) decrease in the density of air with respect to height from the surface of the earth (c) cold air absorbs much small energy from sound waves (d) increase in the density of air with respect to height from the surface of the earth
20. The position vector of an electron is r = 5i + 4j − 3k . To an observer moving along x direction with speed0.6 c, the magnitude of position vector is (a) 5 2
(b)
34
(c)
41
(d) 5
21. Magnus effect is related to (a) (b) (c) (d)
magnetic effect of current production of eddy currents Bernoulli’s effect None of the above
22. The displacement of a particle is given by
23. In
Young’s double slit experiment, constructive interference is produced at a certain point P. The intensities of light at P due to the individual sources are 4 and 9 units. The resultant intensity at point P will be (a) 13 units (c) 97 units
24. Substances for which permeability µ is slightly greater than 1, are called (a) Diamagnetic (c) Ferromagnetic
electric charge. S is momentarily connected to a distant uncharged sphere T which has a radius r. The ratio of surface charge density of S to that of T is (a)
R r
(b)
r R
(c)
r 2R
(d)
(r + R ) R
26. In post office box, the graph of galvanometer deflection versus resistance R (pulled out of resistance box) for the Deflection ratio 100 : 1 is given as shown (due to unsuitable values of R, galvanometer shows 5 326 deflection). The two 320 R (Ω) consecutive values of –3 R are shown in the figure. The value of unknown resistance would be (a) 3.2 Ω
(b) 3.24 Ω (c) 3.206 Ω (d) 3.237 Ω
27. The forward bias characteristics of two diodes D1 and D2 are shown, the knee voltages for D1 and D2 are respectively (approx) I (mA) D1
D2
1000 750
Then, the incorrect statement is
250
The amplitude of motion is 12 The angular frequency is ω The velocity at t = 0 is 3ω The initial phase of motion is π/3
(b) Paramagnetic (d) Nonmagnetic
25. An isolated sphere S of radius R carries an
π y = 2 sin (ωt ) + 2 sin ωt + 3 (a) (b) (c) (d)
(b) 25 units (d) 5 units
500
0.3
(a) 0.4 V and 0.7 V (c) 0.6 V and 0.8 V
0.6
0.9
1.2
V
(b) 0.6 V and 0.9 V (d) 0.4 V and 0.9 V
427
Practice Set 2 æ Directions Q. no. 28 and 29 are AssertionReason type questions. Each of these question contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is correct answer. You have to select the correct choice in the codes (a), (b), (c) and (d) in the given below
(a) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (b) Statement I is true, Statement II is false (c) Statement I is false, Statement II is true (d) Statement I is true, Statement II is true Statement II is a correct explanation of Statement I
28. Statement I An empty test tube dipped into water in a beaker appears silvery, when viewed from a suitable direction.
29. Statement I The electric field due to a dipole on its axial line at a distance r is E . Then, electric field due to the same dipole on the equatorial line and at the same E distance will be . 2 Statement II Electric field due to dipole varies inversely as the square of distance.
30. The output current of a 60% modulated AM generator is 1.5 A. To what value will the current rise, if the generator is additionally modulated by another audio wave of modulation index 0.7? (a) (b) (c) (d)
0.64 A 1.64 A 2.34 A 5.32 A
Statement II Due to refraction of light, the substance in water appears silvery.
Answers 1. (d) 11. (d) 21. (c)
2. (b) 12. (b) 22. (d)
3. (b) 13. (b) 23. (b)
4. (c) 14. (b) 24. (b)
5. (d) 15. (b) 25. (b)
6. (d) 16. (d) 26. (b)
7. (c) 17. (b) 27. (a)
8. (d) 18. (a) 28. (b)
9. (a) 19. (b) 29. (b)
10. (c) 20. (c) 30. (b)
Solutions 1. Momentum, p = 2 mE As energy E is the same, car will have lesser momentum than the truck. Final momentum is zero for both. Hence, the change in momentum is less for the car. Now, change in momentum = Impulse = F × t Since, the force F is the same, it is clear that the car stops earlier. Again, E=F×s where, s is the distance travelled before coming to rest. This is the same for both the car and the truck as E and F are same. 2 (V + V cos ωt )2 2. Heat generated, Q = V = 1 2
R R V12 + V22 cos 2 ω t + 2 V1V2 cos ωt = R
Over a cycle, (cos 2 ωt )average =
1 2
Q cos 2 ωt = 1 − cos 2ωt 2
(cos ωt )average = 0 (cos 2ωt )average = 0 and (cos ωt ) = 0 1 2 V2 2 R 2 V12 + V22 Q= 2R
Q=
⇒ ⇒
V12 +
3. From equation of photoelectric effect, we have Ek = hν − W =
hc −W λ
as ν = c …(i) λ
W = 4.2 eV = 4.2 × 1.6 × 10−19 = 6.7 × 10−19 J
428
JEE Main Chapterwise Physics From Eq. (i), we get 6.62 × 10−34 × 3 × 108 Ek = − 6.72 × 10−19 2000 × 10−10
⇒ Now,
⇒ Ek = 9.9 × 10−19 − 672 . × 10−19
⇒
⇒ Ek = 3.18 × 10−19 J Hence, the stopping potential E 3.18 × 10−19 = 199 . V V0 = k = e 1.6 × 10−19
αβ v= t α + β S = S1 + S 2 =
v 2 1 1 + 2 α β
=
1 αβt α + β 2 α + β αβ
=
1 2
2
y (−3a,4a) 4a 20µC
αβ 2 t α + β
6. The force on the body moving down the plane
(5a,0) 3a
…(ii)
Also, total distance travelled,
r1 = 5 a 20 K V1 = 5a
x′
v2 v2 v2 = − β 2 β 2β
t1 + t 2 = t v v + =t α β
⇒
4. The distance of point (5 a ,0) from origin ⇒
S2 =
x
⇒
F = mg sin θ − µN F = mg sin θ − µ mg cos θ µN
N
y′
The distance of point (−3a, 4a) from origin r2 =
9 a2 + 16 a2
mg sin θ
= 5a 20K V2 = 5a
⇒ Hence,
θ
V1 − V2 = 0
5. Let the car accelerate for time t 1, travelling distance S1 and acquiring maximum velocity v. 1 1 Then, S1 = αt 12 [from the equation,S = ut + at 2 ] 2 2 and ⇒
v = αt 1
[from the equation, v = u + at ] 1 v2 S1 = α × 2 2 α =
v2 2α
…(i)
After this car decelerates for timet 2 to come to rest. 1 Hence, S 2 = vt 2 − βt 22 2 1 [from the equation, S = ut − at 2 ] 2 and
0 = v − βt 2 [from the equation, u = v − at ]
θ
mg cos θ
mg
Acceleration down the plane a = (g sin θ − µg cos θ) As µ varies in direct proportion to the distance moved down the plane, it is clear that first the acceleration decreases, becomes zero and then reverses. 7. L and RC are time constants. Each has unit second. R L So, LCR = (RC )(R ) R = sec × sec × ohm = ohm × sec 2 v − vL 8. Apparent frequency is given by n1 = n v − vs where, n is frequency of source. 320 − 2 × 800 = 795 Hz n1 = 320 − 0 [minimum frequency]
429
Practice Set 2 n2 =
320 + 2 × 800 = 805 Hz 320 − 0
In this case, sin α =
[maximum frequency] So, number of beats heard per second is nb = n2 − n1 = 805 − 795 = 10 beats/s
=
pV = nkT
9. ⇒ ⇒
Q3
From conservation of mechanical energy, decreases in potential energy of block is equal to increases in elastic potential energy of both the springs. 1 1 mgh = kx2 + k ( x + 1)2 2 2 1 1 (2 )(10)( x + 1) = × 10 × x2 + × 10 × ( x + 1)2 2 2
L Q0
10 x2 − 10 x − 15 = 0
or
2 x2 − 2 x − 3 = 0 x=
2±
U=
∴ Maximum extension of upper spring = 1 + x = 2.82 m
⇒ ⇒
11. At equilibrium position, net force on the block should
⇒
be zero. So, let it be at distance y from where it was released. Then, mg = ky + k ( y − 1) or ∴
20 = 10 y + 10( y − 1) or 20 y = 30 y = 1.5 m
⇒
15.
6 Q2 4 π ε0 L
1 2 Iω 2 1 3000 1500 = × 1.2 ω2 ⇒ ω2 = = 2500 2 1.2 ω = 50 rads −1 ω [Q ω = 0 + αt] t = α 50 t = =2 s 25 sin (90° − θ) = = cot θ sin θ E=
14.
x = 182 . m
Q1
Q0Q1 (Q0 + Q1 ) Q2 (Q0 + Q1 + Q2 )Q3 + + 4πε0 L 4 π ε0 L 4 π ε0 L 2 Q2 3 Q2 Q2 + + U =0+ 4 π ε0 L 4 π ε0 L 4 π ε0 L
4 + 24 4
L
U = 0+
2
∴
Q2
L
or 20 x + 20 = 5 x + 5 x + 5 + 10 x
or
µ w 133 . = = 0.8750 152 . µg
= Total work done in bringing all four charges (Q0 = Q1 = Q2 = Q3 = Q ) in their position
10. Let lower spring compresses maximum by x metre.
∴
a
13. The potential energy of the system
n2 T1 273 + 27 300 3 = = = = n1 T2 273 + 227 500 5
2
a
1 = gµ w µg
α = 611 .° φ = 90° − α = 90° − 611 . ° = 28.9°
Hence,
n1kT2 = n2 kT2
w
2 µ1
12. If the ray is to be totally reflected, then α must be equal to or greater than the critical angle. Since, α + φ = 90°, φ is a maximum when α is a minimum allowed value. Hence, α should be equal to the critical angle. 1 Now, [Qα = C] sin α = µ ⇒
sin α =
1 152 .
1 Qsin C = µ
⇒ sin α = 0.6579 ⇒ α = 411 .° Hence, φ = 90° − α = 90° − 411 . ° = 48.9°
°– 90
Medium 2 (Rarer)
Refracted ray
90°
Medium 1 (Denser) Incident ray
θ
θ
θ
Reflected ray
430
JEE Main Chapterwise Physics Also, ⇒
1 sin C 1 1 sin C = = = tan θ µ θ cot 2 1 2 µ1
=
16. Suppose the position of the telescope is at T when the point A is visible on the rim as shown in the figure. Now, when the vessel is filled with a liquid, point B is observed at the same arrangement due to refraction of the ray at point D. BC In ∆BCA, sin i = BD BC 5 = = 2 2 (BC ) + (CD) 25 + h2
Now,
v 2 = u 2 + 2 ax
⇒
v A2 = 0 + 2 aA x and v B2 = 0 + 2 aB x
⇒
v A2
=
v B2
aA v 1 1 = ⇒ A = aB 4 vB 2
18. Number of photons emitted per second n= =
Q n = P E
P hν 10 × 1000 6.6 × 10−34 × 880 × 103
= 172 . × 1031
19. In winter, the density of air near the ground is more,
E
producing total internal reflection. Hence, the sound is heard loud and to a greater distance.
Te les
co pe
T r
aA =
Since, x is the same for both the particles.
C = sin−1 (tan θ)
⇒
1 aB 4
∴
20. Only xcomponent of radius vector contracts
D
L = L0
i h
1−
v2 c2
= 5 1 − (0.6)2
= 5 × 0.8 = 4 r = 4i + 4 j − 3k [Q only xcomponent will be altered]
A 5 cm
Again, ∴
B
∠EDT = ∠ADC = ∠r AC sin r = AD AC 10 = = 2 2 ( AC ) + (CD) 100 + h2
Now, applying Snell’s law, sin i 1 = wµa = sin r aµ w ⇒
or
 r  = (4)2 + (4)2 + (−3)2 =
C 5 cm
5 25 + h
2
×
100 + h2 10 100 + h2 25 + h2
or
17. Acceleration = Force Mass F qE a= = m m
=
1 15 .
=
16 9
h = 8.45 cm
41
u+v
21. u
v−u
u v
[Qµ a = 1]
According to Bernoulli’s theorem, pressure difference on a ball thrown spinning exerts a force on the ball due to which a swing is produced in the ball. This effect is known as Magnus effect. 22. y = 2 sin ωt + 2 sin ωt + π 3 π π = 2 sin ωt + 2 sin ωt cos + cos ωt sin 3 3 1 3 = 2 sin ωt + 2 sin ωt + cos ωt 2 2 = 3 sin ωt +
3 cos ωt
431
Practice Set 2 1 2 3 sin ωt ⋅ + 2 3 cos ωt ⋅ 2 3 2 π π = 12 sin ωt cos + cos ωt sin 6 6 π y = 12 sin ωt + 6
From the given graph, the galvanometer shows zero deflection at
=3×
R = 323.75 = 324 Ω
27. The forward voltage when current in circuit starts increasing rapidly is the knee voltage. As it is clear from the graph, current starts increasing rapidly when voltage is about 0.4 and 0.7 V.
y = a sin(ωt + φ), we get a = 12
Amplitude,
28. Rays of light travelling from water to air at the surface of test tube strikes at an angle of incidence greater than the critical angle for waterair interface. Therefore, they suffer total internal reflection and reach the eye. Thus, the surface of testtube appears silvery.
π v = 12 cos ωt + ⋅ ω 6 π v = 12 cos ⋅ ω 6
Q At t = 0,
2 3 × 3ω 3 ×ω= = 3ω 2 2 π Initial phase, φ= 6 π It is not equal to . 3
S = 3.24 Ω
So,
Comparing it with standard equation
Angular velocity = ω dy Velocity, v = dt
[upto 3 significant digits] Q S = R 100
Empty test tube Eye
= 12 ×
29. We know that for an electric dipole,
amplitude at point P is2 + 3 = 5. So, resultant intensity at point P will be
and
(5)2 = 25 units
Hence,
24. S and T will have same potential. Thus,
…(ii)
Eequatorial =
E 2
Reason is false as electric field due to dipole varies 1 inversely as cube of distance i.e., E ∝ 3 . r
Q1 Q2 Q R = ⇒ 1 = r 4 π ε 0 R 4 π ε 0 r Q2
balanced conditions P R = , where S is unknown resistance. Q S Q P [Q = 100, given] S= ×R P Q
Eequatorial
Eaxial = Eequatorial 2
placed in a magnetic field, the magnetic flux density is slightly greater than that in the free space. Thus, the relative permeability for them is slightly greater than 1.
26. For post office box (Wheatstone bridge), under
…(i)
From Eqs. (i) and (ii), we get
24. We know that when a paramagnetic substance is
Again, if σ1 and σ 2 are the surface densities of spheres S and T, then Q1 4 πR 2 σ1 R σ r = = ⇒ 1 = 2 Q2 σ2 R r 4 πr σ 2
1 2 p 4 πε0 r 3 1 p = 4 πε0 r 3
Eaxial =
23. Amplitudes of the sources are 2 and 3. The resultant
30.
1 + m12 2 1.5 1.5 I Ic = = = 1.18 (0.6)2 m12 1+ 1+ 2 2 I = Ic
Inet = Ic 1 +
Q m = 60 1 100
m12 m2 + 2 2 2
0.49 1.5 1 + 0.18 + = 1.64 A 2 1.18 Hence, option (b) is true. =
Practice Set 3 Instructions For instructions refer to Practice Set 1.
1. The length of the string of a simple pendulum is measured with a meter scale to be 92.0 cm, the radius of the bob plus the hook is measured with the help of vernier calliper to be 2.17 cm. Mark out the correct statement. (a) Least count of meter scale is 0.1 cm (b) Least count of vernier calliper is 0.01 cm (c) Effective length of simple pendulum is 94.2 cm (d) All of the above
2. Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k 1 and k 2 , respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that ofB is (a)
k1 k2
(b)
k1 k2
(c)
k2 k1
(d)
k2 k1
3. As the object moves from infinity to focus, then which is true, about the image formed by single concave mirror? (a) Always real and speed of image continuously increases (b) Always real and speed is initially smaller and finally larger than object speed (c) Initially real and moving with speed smaller than object speed but later on image becomes virtual and moving with speed of object (d) Always virtual and speed is less than object speed
4. When a glass capillary tube of radius 0.015 cm is dipped in water, the water rises to a height of 15 cm within it. Assuming contact angle between water and glass to be
0°, the surface tension of water [ ρwater = 1000 kg/m 3 , g = 9.81 m/s 2] (a) 0.11 N/m
(b) 0.7 N/m
(c) 0.072 N/m
(d) None of these
is
5. Two bodies of different masses have been released from the top of tower. One is thrown in horizontal direction while other is dropped, then which will reach the ground first? (a) (b) (c) (d)
The body which has been thrown horizontally The body which has been dropped Both will reach the ground simultaneously Depends on the velocity with which the first body has been projected horizontally
6. A cylinder of radius R made of a material of thermal conductivity K 1 is surrounded by a cylindrical shell of inner radius R and outer radius 2 R made of a material of thermal conductivity K 2. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. Find the effective thermal conductivity of the system. (a) K 1 + K 2 (c)
(3K1 + K 2 ) 4
K 1K 2 (K1 + K 2 ) (K1 + 3K 2 ) (d) 4 (b)
æ Directions Q. Nos. 7 to 9 are based on following paragraph. A solid sphere of mass M and radius R is surrounded by a spherical shell of same mass M and radius 2 R as shown in figure. A small particle of mass m is released from rest from a height h(< < R ) above the shell. There is a hole in the shell.
m h A B
R
2R
433
Practice Set 3 7. In what time will it enter the hole at A? 2
hR GM
(a) 2 (c)
2
(b)
hR 2 GM
2hR GM
(d) None of these
8. What time will it take to move from A to B? R2 GMh R2 (c) < GMh (a) =
(b) >
R2 GMh
m 1 = 1 kg and then comes to rest inside a second plate of massm 2 = 2.98 kg. It is found that the two plates, initially at rest, now move with equal velocities. The percentage loss in the initial velocity of bullet when it is between m 1 and m 2 (neglect any loss of material of the bodies, due to action of bullet) will be (a) 20%
(d) None of these
9. With what approximate speed will it collide
(b) 25%
14. Given, force =
(c) 30%
(d) 45%
α
density + β 3
What are the dimensions of α , β ? (a) [ML2 T −2 ], [ML−1/3 ]
at B ? (a)
13. A 20 g bullet pierces through plate of mass
2GM (b) R
GM 2R
(c)
3GM (d) 2R
GM R
10. An
AC source producing emf is e = e 0 [cos (100 πs−1 )t + cos (500 πs−1 )t ] connected in series with a capacitor and resistor. The steady state current in the circuit is found to be i = i1 cos[(100 πs−1 )t + φ1] + i2 cos[(500 π s−1 )t + φ 2]. (a) i1 > i 2 (b) i1 = i 2 (c) i1 < i 2 (d) The information is insufficient to find the relation betweeni1 andi 2
11. A block of wood has a mass of 25 g. When a
5 g metal piece with a volume of 2 cm 3 is attached to the bottom of the block, the wood barely floats in water. What is the volumeV of the wood? (a) 28 cm 3 (c) 48 cm 3
(b) 35 cm 3 (d) 12 cm 3
(b) [M2 L4 T −2 ], [M1/3 L−1 ] (c) [M2 L−2 T −2 ], [M1/3 L−1 ] (d) [M2 L−2 T −2 ], [ML−3 ]
15. If electric potential due to some charge distribution is given by V = 3/r 2, when r is radial distance, then find electric field at (1, 1, 1). (a) (c)
2 (i + j + k ) 3 3 (d) 2( i + j + k )
2 3
(b)
2 8( i + j + k )
16. In the diagram, a plot between δ (deviation) versus i (angle of incidence) for a triangular prism is given. From the observed plot, some conclusions can be drawn. Mark out the correct conclusions. δ
δ0 δm
12. A stone is swinging in a horizontal circle 0.8 m in diameter, at 30 rev/min. A distant light causes a shadow of the stone to be formed on a nearby wall. What is the amplitude of the motion of the shadow? What is the frequency? (a) 0.4 m, 1.5 Hz (c) 0.8 m, 0.5 Hz
(b) 0.4 m, 0.5 Hz (d) 0.2 m, 0.5 Hz
i0
π/2
i
(a) The range of deviation for which two angles of incidence are possible for same deviation is δ 0 − δm (b) The curve is unsymmetrical abouti 0 (c) For a given δ, i is unique (d) Both (a) and (b) are correct
434
JEE Main Chapterwise Physics 20. All the accelerations shown in figure are with
17. Mark the correct option (a) When the branch of the circuit containing battery is open, then potential difference across the terminals of battery is same as emf (b) When two terminals of a battery are shorted, then potential difference across the terminals of battery is zero (c) If external resistance of the circuit changes, then potential difference across the terminals of the battery changes but emf remains same (d) All of the above
18. Two clocks are being tested against a standard clock located in a national laboratory. At 12 : 00 : 00 noon by the standard clock, the readings of the two clocks are given as follows. Monday Tuesday Wednesday Thursday Friday Saturday Sunday
Clock I 12 : 00 : 05 12 : 01 : 15 11 : 59 : 08 12 : 01 : 50 11 : 59 : 15 12 : 01 : 30 12 : 01 : 19
Clock II 10 : 15 : 06 10 : 14 : 59 10 : 15 : 18 10 : 15 : 07 10 : 14 : 53 10 : 15 : 24 10 : 15 : 11
If you are doing an experiment that requires precision time interval measurements, which of the two clocks will you prefer? (a) Clock I (b) Clock II (c) Both have same precision time interval measurements (d) None of the above
19. The emf and internal
4.3 V 1 Ω resistance of the battery shown in figure are 4.3 A V and 1Ω, respectively. 50 Ω The external resistance 2Ω R is 50 Ω. The resistance V of the voltmeter and 200 Ω ammeter are 200 Ω and 2 Ω , respectively. Find the readings of the two meters.
(a) 0.1 A, 2 V (c) 0. 4 A, 1 V
(b) 0.1 A, 4 V (d) 0.4 A, 4 V
respect to ground, find acceleration of B. (a) 3 m/s 2 upward (b) 5 m/s 2 upward (c) 3 m/s 2 downward (d) None of the above
21. Three dielectric slabs of thickness d/4, d/7 and d/2 having dielectric constants 2, 8/7 and 4 respectively are inserted between the plates of a parallel plate capacitor having plate separation d and plate area A. The remaining space is filled with a conducting medium. Find the capacitance of the system. (a)
8 ε0 A d
(b)
8 ε0 A 3d
(c)
26 ε0 A 35d
(d)
12 ε0 A 35d
22. The electron emitted from an electron gun are having 2 eV of kinetic energy, then they are accelerated by applying a potential ofV volts and are allowed to fall on atomic hydrogen gas. It has been observed that red light of wavelength 656.3 nm has been emitted by the hydrogen. Assume that total energy of electron has been given, then V has the value given by (a) 2 V (c) –2 V
(b) 0.11 V (d) – 0.11 V
23. A body dropped from a height H reaches the ground with a speed of 1.2 gH . Calculate the work done by airfriction. (a) 2.8 mgH (c) 1.3 mgH
(b) –1.3 mgH (d) –0.28 mgH
24. The ratio of the acceleration due to gravity at the bottom of a deep mine and that on the surface of the earth is 978/980. Find the depth of the mine, if the density of the earth is uniform throughout and the radius of the earth is 6300 km. (a) 12.86 km
(b) 13.0 km
(c) 25.38 km
(d) 90.9 km
435
Practice Set 3 25. A sky wave with a frequency 55 MHz is
27. Statement I If the halflife of a radioactive
incident on Dregion of the earth’s atmosphere at 45°. The angle of refraction is (electron density for Dregions is 400 electron/cm 3)
substance is 40 days, then 25% substance decays in 20 days.
(a) 60°
(b) 45°
(c) 30°
(d) 15°
æ Directions Q. Nos. 26 and 27 are AssertionReason type questions. Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is correct answer. You have to select the correct choice in the codes (a), (b), (c) and (d) in the given below
(a) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (b) Statement I is true, Statement II is false (c) Statement I is false, Statement II is true (d) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
26. Statement I The rocket works on the principle of momentum.
conservation
n
of
linear
Statement II Whenever there is the change in momentum of one body, the same change occurs in the momentum of the second body of the same system but in opposite direction.
1 N = N0 2 time elapsed n= halflife period
Statement II where,
28. The voltage gain of an amplifier without feedback is 100. If a negative feedback is introduced with a feedback fraction β = 0.1, then the gain of the feedback amplifier is (a) 9.09 (c) 100.1
(b) 10 (d) 90.0
29. In frequency modulated wave, (a) (b) (c) (d)
frequency varies with time amplitude varies with time both frequency and amplitude vary with time both frequency and amplitude are constant
30. The output current of a 60% modulated AM generator is 1.5 A. If the generator is additionally modulated by another radio wave of modulate index 0.7, the new modulation index is (in percentage) (a) 20% (c) 80%
(b) 92% (d) 55%
Answers 1. (c) 11. (a) 21. (b)
2. (d) 12. (b) 22. (d)
3. (b) 13. (b) 23. (d)
4. (a) 14. (c) 24. (a)
5. (c) 15. (b) 25. (b)
6. (d) 16. (d) 26. (d)
7. (a) 17. (d) 27. (c)
8. (c) 18. (b) 28. (a)
9. (d) 19. (b) 29. (a)
10. (c) 20. (d) 30. (b)
Solutions 1. Effective length of the pendulum is (92.0 + 2.17 ) cm
gt 12 2 gt 22 h= 2 t1 = t 2 h=
So,
= 94.2 cm after rounding off to 3 significant digits.
2. Since, maximum velocity is same, so maximum KE i.e., TE would be same. Also, maximum PE would also be same. Let A1 and A 2 be the amplitudes of two bodies A and B, respectively. k1 A12 k2 A 22 A1 k2 Then, = ⇒ = 2 2 A2 k1
∴
So, image speed is smaller in beginning when the object is moving from infinity to centre of curvature and increases thereafter.
4. When liquid rises to height h, then
[for dropping]
6. This can be considered as a parallel combination of two, one the inner cylinder and the other surrounding cylinder. 2
3. As the object moves from infinity to centre of curvature, the image formed by a concave mirror would be real and is moving from focus to centre of curvature, but as the object crosses centre of curvature and moves towards focus the image is still real but moves from centre of curvature towards infinity and when the object is at focus, the real image would be formed at infinity.
[for horizontal throwing]
R T1
1
2R
T2
A1 = πR 2 , A 2 = π (4R 2 − R 2 ) = 3 πR 2 L1 = L 2 = L Heat is flowing only along the length of tube. ∴ Heq = H1 + H2 K A (T − T2 ) K 2 × A 2 (T1 − T2 ) = 1 1 1 + L L ⇒ ⇒
h
Keq × 4 = K1 + 3 K 2 (K + 3 K 2 ) Keq = 1 4
7. Acceleration due to gravity near the surface of shell can be assumed to be uniform (h Z 2 , therefore i1 < i 2
So,
β = [M L−3 ]1/ 3 = [M1/ 3L−1 ]
∴
11. Let volume of wood be V cm3 , then total volume of displaced water is (V + 2 ) cm3 , then for translational equilibrium,
= [M2 L−2 T−2 ]
Wood
15. Metal
(V + 2 ) ρ g = (25 g + 5 g ) where, all the quantities are in CGS unit and ρ is the density of water. ⇒
(V + 2 ) × 1 = 30 ∴V = 28 cm3
12. The amplitude is the radius of the circle 0.8 m = 0.4 m 2 The frequency of the shadow is the same as that of the circular motion, so ω = 30 rev/min = 0.5 rev/s = π rad/s ω π and v= = = 0.5 Hz 2π 2π R=
13. The situation is shown in figure. m1
m2
m u
u1 v
v
V=
3 r2
dV E = − r dr ∂ 3r =− ∂ r r 2 r 3 × ( −2 ) r 6 =− = r r r4 r3 2 = (i + j + k ) 3
16. The range of deviation for which two angles of incidence are possible for same deviation is δ 0 − δ m and the given curve is symmetrical about i 0 .
17. In the option (a), the circuit is open, as a result no current flows through it, so the potential difference across internal resistor is zero and hence terminal potential difference is same as emf of battery. For option (b), the terminal potential difference gets zero as battery is shorted although emf remains same. In option (c), discharging of battery takes place, current has been withdrawn from battery, whose value is depending on value of resistance, which is turn changes the value of terminal potential difference but emf remains same.
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JEE Main Chapterwise Physics
18. A simple look at the data shows that average
21. Here, three slabs are in series.
reading of clock I is much closer to the standard time that average reading of clock II. Therefore, zero error in clock II is much larger than the zero error in clock I. But zero error can always be corrected by applying necessary correction.
8
Now, over the seven days, range of variation of clock I is 12 : 01 : 50 − 11 : 59 : 08 = 162 s and range of variation of clock II is 10 : 15 : 24 − 10 : 14 : 53 = 31 s. Therefore, precision of clock II is much better than the precision of clock I. We will prefer clock II.
Ceq =
19. In the given circuit, 200 Ω and 50 Ω are in parallel, hence equivalent resistance is 4+1 1 1 1 5 = + = = R 200 50 200 200 ⇒ R = 40 Ω The equivalent circuit is as follows in this circuit, then 40 Ω, 2 Ω and 1 Ω resistors are connected in series, hence equivalent resistance is Req = 40 + 2 + 1 = 43 Ω 4.3 V
1Ω
⇒
d 7
d 2
1242 656.3
= − 0.11 eV V = − 011 . V
23. The forces acting on the body are (i) force of gravity, (ii) airfriction According to workenergy theorem, total work done on the body = Gain in KE 1 1 . mgH W = mv 2 = m (1.2 gH )2 = 072 2 2 As work done by gravity, W1 = mgH ∴ Work done by friction, W2 = W − W1 = 0.72 mgH − 1 mgH = − 0.28 mgH
24. The ratio of the acceleration due to gravity at bottom and by deep mine is given by.
a AP = − a BP a AG = a AP + a PG 3 = a AP − 2 ⇒ a AP = 5 m/s 2
g ′ 978 d d = = 1− Qat depth, g ′ = g 1 − g 980 R Re 978 2 d or = 1− = 980 980 R 2 R 2 × 6300 or d= = = 12.86 km 980 980
2 m/s2
P
neff = n0
25.
80.5 N 1 − ν2
= 1 1−
3 m/s2
a BG = a BP + a PG = − 5 − 2 = − 7 m/s 2
d 4
eV = 1.89 − 2
⇒ ∴
From Ohm’s law, V = I R V 4.3 = 0.1 A I= = ⇒ R 43 Hence, voltmeter reading = 4.3 − 0.1 × 3 = 4 V
∴
4
ε0 A 8ε A ε0 A = = 0 d /4 d /7 d /2 d + d + d 3d + + 8 8 8 2 8 / 7 4
2 + eV =
2Ω
20. Consider downward direction as positive
8 7
22. Total energy of the electron
A 40Ω
2
Also, ⇒
neff
80.5 × (400 × 106 )
=1 (55 × 106 )2 sin i ⇒ sin i = sin r = sin r
r = i = 45°
439
Practice Set 3 26. Since, the rocket fuel is undergoing combustion, the
29. In frequency modulation, frequency of a carrier
gases produced in this process leave the body of the rocket with large velocity and produce upthrust to the rocket.
signal varies in accordance with the modulating signal. The amplitude of the carrier wave is fixed while its frequency is changing.
Let us assume that the fuel be undergoing combustion at the constant rate, then rate of change of momentum of the rocket will be constant. Since, more and more fuel will be burnt, the mass of rocket will go on decreasing, so it will lead to increase the velocity of the rocket more and more rapidly.
When the instantaneous amplitude of the modulating voltage is large, the instantaneous carrier frequency is higher. It is lower when the modulating voltage is small. Hence, frequency
27. Here, or
1 N = N0 2 N 1 = N0 2
varies with time.
30. Output current by generator is
t /T
I = Ic
t /T
…(i)
N is fraction of N0 atoms left after time t. Here, T = 40 days and N 25 1 = = . N0 100 4
Ic =
1 1 = 4 2 or or
t / 40
N in Eq. (i), we get N0 2
1 1 or = 2 2
t = 80 days
28. The gain of feedback amplifier Af =
=
A 100 100 = = = 9.09 1 + βA 1 + 01 . × 100 11
I m12 2 1.5
1+
= ma =
2
=
(0.6) 2
Inet = Ic = 1 +
t / 40
t =2 40
m12 2
1+
where, T is the halflife period and
Putting the values of T and
1+
1.5 1.18
m12 m2 + 2 2 2
0.49 1.5 1 + 0.18 + = 1.64 A 2 1.18 m12 + m22
= (0.6)2 + (0.7)2 =
36 + 0.49
=
0.85 = 0.92 or 92%
Practice Set 4 Instructions For instructions refer to Practice Set 1.
1. A particle of mass 10 kg starts from point A, with an initial velocity of 3 m/s towards negative xaxis, it has been acted by a force of 10 N towards positive xaxis. Find the distance travelled by particle in 4 s. (a) 4 m
(b) 0.5 m
(c) 4.5 m
(d) 5 m
2. Find the charge on 5 µF capacitor. 2µF
4µF
(a) 5.45 µC (c) 10.9 µC
quickly pour boiling water into it ? O
6V
(b) 16.37 µC (d) 18 µC
3. A uniform slender rod 1m long is initially standing vertically on a smooth l C horizontal surface. It is struck by a sharp horizontal blow at the top end, with the blow directed at right angles to the rod axis. As a result, the rod acquires an angular velocity of 3.00 rad/s. What is the translational velocity of the centre of mass of the rod after the blow? (a) 0.25 m/s (c) 0.5 m/s
Here, EI represents the rate at which electrical energy is converted to nonelectrical energy (c) In both the above cases, the term I 2r represents the rate of dissipation of energy in the internal resistance of battery (d) None of the above
5. Why does a glass sometimes break, if we
5µF
6V
(b) During the charging of battery, the battery consumes energy at the rate of VI = EI + I 2r .
(b) 0.64 m/s (d) 1.2 m/s
4. Mark the incorrect option. (a) During the discharging of battery, the battery supplies electrical energy at the rate of VI = EI − I 2r . Here, EI represents the rate at which nonelectrical energy is converted to electrical energy
(a) Hot water expands, pushing the glass out (b) The hot water cools when it touches the glass, shrinking and pulling the glass in (c) The glass becomes hot and expands causing the molecules to break (d) The inside of the glass expands faster than the outside of the glass, causing the glass to break
6. Find the equivalent capacitance between A and B. 2 µF A
B 2µF
4µF
2µF
4 µF 3
(d) 6µF
2 µF
2 (a) µF 3
(b) 1µF
(c)
7. A 0.5 kg ball is dropped from rest at a point 1.20 m above the floor. The ball rebounds straight upward to a height of 0.7 m. What is the magnitude and the direction of the impulse of the net force applied to the ball during the collision with the floor? (a) 4.28 Ns, upward (c) 8.56 Ns, upward
(b) 4.28 Ns, downward (d) 8.56 Ns, downward
441
Practice Set 4 8. A body is moving with uniform speed v in a
13. A girl is seated on the top of a hemispherical
horizontal circle in anticlockwise direction as shown in figure. The motion starts from point A, find the change in velocity in second quarter of revolution.
mound of ice. She is given a very small push and starts sliding down the ice. She leaves the ice at a point whose height is
N
Girl R
W
A
E
(a) S
(a)
2 v NW
(b)
2 v NE
(c)
2 v SW
(d)
2 v SE
cylinder at temperatureT . If we supply some heat to it, then N /3 moles of gas dissociates into atoms while temperature remains constant. Heat supplied to the gas is NRT 6
(b)
5 NRT 2
(c) 5.6 NRT (d)
8 NRT 3
10. Electrons having KE 15 eV is collided with hydrogen atom and 80.6% of it is used to excite the electrons from its ground state. Find the number of emitted wavelengths. (a) 1
(b) 2
(c) 3
(d) 4
æ Directions Q. Nos. 11 and 12 are based on following paragraph. Work function of metal A is equal to the ionisation energy of hydrogen atom in first excited state. Work function of metal B is equal to the ionisation energy of He + ion in second orbit. Photons of same energy E are incident on both A and B. Maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B.
11. Value of E (in eV) is (a) 20.8
(b) 32.2
(c) 24.6
(d) 23.8
12. The difference in maximum kinetic energy of photoelectrons from A and from B (a) (b) (c) (d)
(b)
R 3
(c)
R 5
(d)
2R 3
14. A person of mass m is
9. N moles of an ideal diatomic gas are in a
(a)
2R 5
increases with increase in E decreases with increase in E first increases then decreases with increase in E remains constant
standing on a structure made up of pulley, strings and platform as shown in figure. Find the force exerted by the person on the rope, so that the system (person + structure) remains in equilibrium.
m M
(a) mg (b) (m + M ′ + M ) g ( M + M ′ + m) g (c) 2 (d) None of the above
M'
15. Find the distance between the object and its doubly magnified real image by a concave mirror of focal length f . (a) 2 f (c)
5f 2
(b)
3f 2
(d) 3 f
16. A point source of light is taken away from the experimental set up of photoelectric effect, then which is the most appropriate statement? (a) Saturation photocurrent remains same, while stopping potential increases (b) Saturation photocurrent and stopping potential both decrease (c) Saturation photocurrent decreases while stopping potential remains same (d) Saturation photocurrent decreases and stopping potential increases
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JEE Main Chapterwise Physics
17. The pitch of a screw gauge is 0.5 mm and
20. Statement I A car battery is of 12 V. Eight
there are 50 divisions on circular scale. When there is nothing between the two ends (studs) of screw gauge, 45th division of circular scale is coinciding with screw gauge and in this situation, zero of main scale is not visible. When a wire is placed between the studs, the linear scale reads 2 divisions and 20th division of circular scale coincides with reference line. For this situation, mark the correct statement(s).
dry cells of 1.5 V connected in series can give 12 V. Still such cells are not used in starting a car.
(a) Least count of the instrument is 0.01 mm (b) Zero correction for the instrument + 0.45 mm (c) Thickness of wire is 1.65 mm (d) All of the above
is
18. A man in an empty swimming pool has a telescope focussed at 4’O clock sun. When the swimming pool is filled with water, the man (now inside the water with his telescope undisturbed) observes the setting sun. Find the refractive index of water, if sun rises and sets at 6’O clock. (a)
4 3
(b)
2 3
(c)
8 5
(d)
2 5
19. A 25 kg bear slides from rest, 12 m down a pine tree, moving with a speed of 5.6 m/s just before hitting the ground. What is the average frictional force that acts on the sliding bear? (a) 210 N (c) 110 N
(b) 180 N (d) 170 N
æ Directions Question Nos 20 to 22 are AssertionReason type questions. Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is correct answer. You have to select the correct choice in the codes (a), (b), (c) and (d) in the given below
(a) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (b) Statement I is true, Statement II is false (c) Statement I is false, Statement II is true (d) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
Statement II It is easier to start a car engine on a warm day than on a rainy day.
21. Statement I The minimum inductance that can be obtained by combining three inductors of 2H, 3H and 6H is 1H. Statement II Minimum inductance is obtained when the inductors are connected in series.
22. Statement I A small body of mass 0.1 kg is undergoing SHM of amplitude 1.0 m and period 0.2 s. The maximum value of the force acting on it is 98.7 N. Statement I Maximum force acting on it is given by F = mω 2r .
23. A Carnot air conditioner takes energy from the thermal energy of a room at 70°F and transfers it to the outdoors, which is at 96°F. For each joule of electric energy required to operate the air conditioner, how many joules are removed from the room? (a) 15 J
(b) 10 J
(c) 20 J
(d) 10.5 J
24. A UHF television loop antenna has a diameter of 11 cm. The magnetic field of a TV signal is normal to the plane of the loop and at one instant of time, its magnitude is changing at the rate 0.16 T/s. The magnetic field is uniform. What emf is induced in the antenna? (a) 1.52 × 10−3 V (b) 0.35 × 10−2 V (c) 0.54 × 10−3 V (d) 1.39 × 10−2 V
25. What direct current will produce the same amount of thermal energy, in a particular resistor, as an alternating current that has a maximum value of 2.60 A? (a) 2.35 A (c) 1.20 A
(b) 1.84 A (d) 1.29 A
443
Practice Set 4 26. A 1.0 kg copper rod rests on two horizontal
(a)
rails 1.0 m apart and carries a current of 50 A from one rail to the other. The coefficient of static friction between rod and rails is 0.60. What is the smallest magnetic field (not necessarily vertical) that would cause the rod to slide? (a) 0.10 T
(b) 0.20 T
(c) 0.50 T
(c)
region i.e., the breakdown region for which the circuit diagram is as shown in the figure. R
VZ
A
0
0
0
0
0
0
1
0
1
1
0
0
1
1
1
1
1
1
1
1
A
B
Y
A
B
Y
0
0
1
0
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
0
1
1
0
0
0
1 0
(b)
(d)
modulated with a carrier wave of frequency 2.51 MHz. The upper and lower side band frequencies are respectively
B
Here, take V Z = 7 V and R = 10 kΩ. For potential difference equal to 8 V across AB, what is the current through microammeter ? (c) 10µA
Y
Y
29. A signal wave of frequency 12 kHz is
+ A µA –
(a) 1000µA (b) 1 mA
B
B
(d) 0.30 T
27. A zener diode is operating in its normal
E
A
A
(a) (b) (c) (d)
2512 kHz and 2508 kHz 2522 kHz and 2488 kHz 2502 kHz and 2498 kHz 2522 kHz and 2498 kHz
(d) 100µA
28. The output of a 2 input NAND gate is fed to a NOT gate. The truth table for the output of the combination for all possible inputs of A and B is
30. An oscillator is producing FM waves of frequency 2 kHz with a variation of 10 kHz. The modulation index is (a) 0.20
(b) 5.0
(c) 0.67
(d) 1.5
Answers 1. (d) 11. (d) 21. (b)
2. (b) 12. (d) 22. (d)
3. (c) 13. (d) 23. (c)
4. (d) 14. (c) 24. (a)
5. (d) 15. (b) 25. (b)
6. (b) 16. (c) 26. (a)
7. (a) 17. (d) 27. (d)
8. (d) 18. (b) 28. (a)
9. (a) 19. (a) 29. (d)
10. (c) 20. (a) 30. (b)
Solutions 1. This is a question based on the concept that when
2. We will solve this question by using point potential
acceleration and initial velocity are in opposite direction, then particle comes to rest momentarily at some instant and after that direction of velocity becomes same as that of acceleration. Upto this instant, distance and displacement are equal. F 10 = = 1 m/s 2 a= m 10
theory.Consider point O at zero potential and then mark the potentials of various junctions as shown in figure. Apply KCL at junction having potential V1. 2µF
4µF
6V
5µF
u = 3 m/s, t = 0 a = 1 m/s2
v=0 O
A 6V
s1 t = 3 s, t = 4 s B
C s=4m s1 = 4.5 m
t=0 A
Let velocity becomes zero after t second, then applying v = u + at ⇒
V1
6V
0 = 3 − 1× t
⇒ t =3s Distance travelled in 3 s = Displacement in 3 s 1 s1 = ut + at 2 ⇒ 2 1 = 3 × 3 − × 1 × 32 2 9 = m = 4.5 m 2 [ Since, the above equation is a vector equation and gives us displacement and not distance] Displacement of particle in 4 s, 1 s = ut + at 2 2 1 = 3 × 4 − × 1 × 42 2 = 12 − 8 = 4 m From the diagram, distance travelled in 4 s = AB + BC = 4.5 + 0.5 = 5 m
O 0V
6V
5 × (V1 − 0) + 2 × (V1 − 6) + 4 × (V1 − 6) = 0 ⇒ (5 + 2 + 4) V1 = 12 + 24 36 V ⇒ V1 = 11 36 So, charge on 5 µF = × 5 µC = 16.37 µC 11
3. Let the magnitude of the impulse delivered by the blow be represented by p. From the translational and rotational forms of Newton’s second law, p pl and Ic ωc = vc = M 2 Here, vc is the rightward velocity of the centre of mass after the blow and ωc is the (positive clockwise) angular velocity of the rod after the blow about the point C. Ml 2 , we find that Using the fact that Ic = 12 6p ωc l = M (ωc l ) Therefore, vc = 6 Since, ωc = 3.00 rad/s and l = 100 . m, we find vc = 0.5 m/s
4. During discharging, battery supplies electric power to external circuit, while during charging the charger supplies electrical energy to the battery. In both the cases, some part of the energy is dissipated as heat in internal resistance of the battery.
445
Practice Set 4 5. When hot water is poured into the glass, the inner portion of the glass comes into the contact of hot water immediately while of outer portion after sometimes as heat is conduced through glass. As a result, the inside of the glass expands faster than the outside of the glass and hence causes the glass to break.
8. The diagram is clearly showing the change in velocity vector ∆v = v 2 − v1 v1 ⇒ v2
v2
6. Equivalent circuit can be drawn as shown in figure.
–v1
1 1 1 = + QC p = C1 + C 2 and C s C1 C 2
E √2v
⇒ v2 A
B 2 µF
2 µF
2 µF
–v1
The middle capacitor is shorted and hence no charge flows through it. So, the equivalent circuit can be redrawn as A
⇒
S
9. Heat supplied = Change in internal energy ∆U = U f − U i 5RT N 3RT N + ×2 × U f = N − × 3 2 3 2
B 2 µF
2 µF B
A
5NRT 8 + NRT = NRT 3 3 NRT NRT Ui = 5 × ∴ ∆Q = 2 6 =
1 µF
and
7. The velocity of ball just before collision = 2 g × 1.2 = 2.4 g m/s in downward direction.
SE E
10. Change in kinetic energy of electron
1.2m 0.7m
Ground
The velocity of ball just after collision = 2 g × 07 . = 14 . g m/s in upward direction. Impulse = pf − pi = m [ 1.4 g − (− 2.4 g )]
= 4.28 Ns in the direction of final momentum, i.e., upward.
80.6 × 15 = ∆E = 12.09 =  E3 − E1  100 So, electron can make the hydrogen atom to go into 2nd excited state, i.e, n = 3 energy state. From there 3 transitions are possible, i.e, from 3 to 2, 3 to 1 and 2 to 1. 11. We know that, En = −132.6 in nth orbit. n WA = Ionisation energy of electron in 2nd orbit of hydrogen atom = 3.4 eV. WB = Ionisation energy of electron in 2nd orbit of He + ion = 13.6 eV Now, given that K A = 2 K B or ∴
( E − WA ) = 2 ( E − WB ) E = 2 WB − WA = 23.8 eV
12. (E − WA ) − (E − WB ) = WB − WA = 10 . 2 eV = constant
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JEE Main Chapterwise Physics
13. The free body diagram for the girl is shown in the figure.
N
15.
v m = − u
θ
v = 2u
1 1 1 From mirror formula, + = v u f
mg
⇒
1 1 1 + = 2u u f 3 1 = ⇒ 2u f
⇒ …(i)
u=
3f 2
16. As the source of light is taken away, the intensity of light at the location of experimental setup decreases.
N=0 v2 So, g cos θ = R Conservation of energy gives 1 0 = mv 2 − mgR (1 − cos θ) 2
∴ …(ii)
Photocurrent ∝ Intensity
So, photocurrent decreases. hc φ Now, V0 = − λ 0e e which is independent of intensity.
v = 2 gR (1 − cos θ) 2
…(iii)
17. Least count , Pitch Divisions on circular scale 0.5 = = 0.01 mm 50
LC =
From Eqs. (ii) and (iii), we have g cos θ = 2 g (1 − cos θ) 2 cos θ = 3
⇒
v u
θ
Net inward force = mg cos θ − N mv 2 Also, mg cos θ − N = R where, v is the speed of the girl. At the point where girl leaves
or
−2 =
[–ve magnification because image is real] ⇒
R
⇒
Thus, the required height = R cos θ =
2R 3
14. The net force on the system can be calculated directly after drawing the free body diagram. T
Zero error is negative in nature and is e = − 45 × LC = − 0.45 mm Zero correction = − Zero error = 0.45 mm Reading of the screw gauge, d = 2 × 0.5 + 20 × 0.01 + Zero correction (0.45 mm) = 165 . mm
18. When no water is there in the swimming pool, the T
T T/2
T/2
T m M
telescope is along the line PQ. When water has been filled into the pool, the person is able to see the setting sun, i.e, refracted ray will be along QP while incident ray is along RP, as shown in figure. That is from the object (sun) to the telescope.
T/2
Sun 12'O clock noon
T/2
Q Sun 4'O clock evening
M' (M + M' + m) g
T = (M + M ′ + m) g [Q upward force = downward force] T (M + M ′ + m) g = 2 2
So, the required force, F =
Sun 6'O clock morning
30° P 60°
R Sun 6'O clock evening
447
Practice Set 4 Let µ be the refractive index of water, then Q µ1 = 1 P
when the inductors are connected in parallel. Hence, 1 1 1 1 1 1 1 6 = + + = + + = L p L1 L 2 L 3 2 3 6 6 ⇒
i R
r = 60°
21. We know that minimum inductance can be obtained
µ2 = µ
L p = 1H
22. The maximum force is given by 2
2π F = mω2 r = m × × r = 0.1 × T
2
2 π × 1.0 0.2
= 0.1 × (10 π )2 = 98.7 N sin 90° µ = sin 60° 1 ⇒
µ=
1 2 = sin 60° 3
19. The change in the mechanical and thermal energy must sum to zero. Since, the change in thermal energy is ∆Eth = f L , where f is the magnitude of the average frictional force. ∆K + ∆U f=− L (K f − K i ) + (U f + U i ) =− L 1 mv 2 − 0 + (− mgL − 0) =− 2 L 1 × 25 × (5.6)2 − 25 × 9.8 × 12 2 =− 12 3.92 × 105 − 2.94 × 103 =− 12 = 210 N
20. To start a car, a very high current is required. A car battery has very low internal resistance, so that it can provide the required high current. When eight dry cells are joined in series, the internal resistance of the combination becomes very high. Due to this high internal resistance, small current will be drawn from it. Hence, such cells cannot be used to start a car. On a warm day, the internal resistance of car battery decreases and so large current can be drawn from the battery. But on a chilly day, reverse process occurs.
23. K =
TC 294 or K = = 19.6 TH − TC 309 − 294
The coefficient of performance is the energy QC drawn from the cold reservoir as heat divided by the work done Q  K= C W  Thus,
 QC  = K  W  = (19.6)(1.0) = 20 J
24. The magnetic field is normal to the plane of the loop and is uniform over the loop. Thus, at any instant, the magnetic flux through the loop is given by φB = AB = πr 2 B, where A (= πr 2 ) is the area of the loop. According to Faraday’s law, the magnitude of the emf in the loop is dφ dB e = B = πr 2 dt dt = π (0.055 m)2 (0.16 T/s) = 152 . × 10−3 V
25. The average rate with which thermal energy is generated in resistance R when the current is alternating is given by 2 Pav = Irms R
I2 R 2 The rate of thermal energy generation in the same resistor when the current is direct is given by P = i 2 R, where i is the current. Set the two rates equal to each other and solve for i. You should get I 2.60 A ,= i = = 184 . A 2 2 or
Pav =
448
JEE Main Chapterwise Physics
26. Suppose the magnetic
FB
B
N field makes the angle θ θ with the vertical. The θ diagram to the right shows f the view from the end of the sliding rod. The forces are also shown. Notice that the magnetic force mg makes the angle θ with the horizontal. When the rod is on the verge of sliding, the net force acting on it is zero and the magnitude of the frictional force is given by f = µ sN, where µ s is the coefficient of static friction. The magnetic field is perpendicular to the wire, so the magnitude of the magnetic force is given by FB = ILB, where I is the current in the rod and L is the length of the rod. The vertical component of Newton’s second law yields …(i) N + ILB sin θ − mg = 0 and the horizontal component yields …(ii) ILB cos θ − µ s N = 0 Solve the Eq. (ii) and substitute the resulting expression into the Eq. (i), then solve for B. You should get µ s mg B= IL(cos θ + µ s sin θ)
The minimum value of B occurs when cos θ + µ s sin θ is a maximum. Set the derivative of cos θ + µ s sin θ equal to zero and solve for θ. You should get θ = tan−1 ( µ s ) = tan−1 (0.60) = 31°. Now, evaluate the expression for the minimum value of B. 0.60 (10 . )(9.8) Bmin = (50)(10 . )(cos 31° + 0.60 sin 31° ) = 0.10 T The magnetic field makes an angle of 31° with the vertical.
27. Write KVL equation, − VBA + IR + VZ = 0 [as diode is operating in breakdown region] ⇒ IR = VBA − VZ = 8 − 7 1V ⇒ I= = 100 µA 10 kΩ
28. If the output of two inputs NAND gate is used as the input of NOT gate as shown below, we get back AND gate. Here, the Boolean expression for output Y ′ is Y ′ = A ⋅ B and Y = A ⋅ B = A ⋅ B The truth table of this combination will be as shown below, which is a truth table of AND gate.
A
B
Y′
Y
0
0
1
0
1
0
1
0
0
1
1
0
1
1
0
1
29. Upper side band (USB) frequency = fc + fm Given, fc = 2.51 MHz = 2.51 × 106 Hz = 2510 kHz fm = 12 kHz ∴ Upper side band (USB) frequency = (2510 + 12 ) kHz = 2522 kHz Lower side band (LSB) frequency = fc − fm = 2510 − 12 = 2498 kHz 30. Modulation index is given by β = ∆f fm where, ∆f is frequency deviation, fm is the highest modulation frequency. Given, ∆f = 10 kHz = 10 × 103 Hz fm = 2 kHz = 2 × 103 Hz ∴
β=
10 × 103 2 × 103
=5
Practice Set 5 Instructions For instructions refer to Practice Set 1.
1. A carpet is to be installed in a room whose
6. Satellite dishes do not have to change
length is measured to be 12.71 m (four significant figures) and whose width is measured to be 3.46 m (three significant figures). Find the area of the room.
directions in order to stay focussed on a signal from a satellite. This means that the satellite always has to be found at the same location with respect to the surface of earth. For this to occur, the satellite must be at a height such that its revolution period is the same as that of earth, 24 h. At what height must a satellite be to achieve this?
(a) 43.97 m 2 (c) 43.98 m 2
(b) 43.9766 m 2 (d) 44.0 m 2
2. If the charge of 10µC and −2 µC are given to two plates of a capacitor which are connected across a battery of 12 V, find the capacitance of the capacitor. (a) 0.33µF (b) 0.5 µF
(c) 0.41µF
(d) 066 . µF
3. A passenger at the rear of a train travelling at 15 m/s relative to earth throws a baseball with a speed of 15 m/s in the direction opposite the motion of the train. What is the velocity of the baseball relative to earth? (a) Zero (c) 30 m/s
(b) 15 m/s (d) −15 m/s
T2 (a) 2 GMe 4π
1/ 3
T2 (c) 2 GMe 4π
1/ 2
(a) 21.5 m/s (c) 4.9 m/s
(b) 14.9 m/s (d) 20.3 m/s
5. A body when projected vertically up covers a total distance s, during its time of flight. If we neglect gravity, then how much distance the particle will travel during the same time? Will it fall back? (a) s, Yes (c) 2 s , Yes
(b) s, No (d) 2 s , No
T (d) 2 GMe 4π
1/ 3
7. Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of the principal quantum number n? v
C A
4. A sled and its rider together weigh 800 N. They move down on a frictionless hill through a vertical distance of 10.0 m. Use conservation of mechanical energy to find the speed of the sled at the bottom of the hill, assuming the rider pushes off with an initial speed of 5.00 m/s. Neglect air resistance.
T2 (b) 2 GMe 4π
D
B
n
(a) A
(b) B
(c) C
(d) D
8. Water with a mass of 2.0 kg is held at constant volume in a container while 10.0 kJ of energy is slowly added by a flame. The container is not well insulated and as a result 2.0 kJ of energy leaks out to the surroundings. What is the temperature increase of the water? (a) 0.28°C (c) 0.96°C
(b) 27°C (d) 1.27°C
450
JEE Main Chapterwise Physics
æ Directions Q. no. 9 to 11 are based on following paragraph. We have two radioactive nuclei A and B. Both convert into a stable nucleus C. Nucleus A converts into C after emitting two αparticles and three βparticles. Nucleus B converts into C after emitting one αparticle and five βparticles. At time t = 0, nuclei of A are 4 N 0 and that of B are N 0. Half life of A (into the conversion of C) is 1 min and that of B is 2 min. Initially number of nuclei of C are zero.
9. If atomic numbers and mass numbers of A and B are Z 1 , Z 2 , A1 and A 2, respectively. Then, (a) Z 1 − Z 2 = 6 (b) A1 − A 2 = 4 (c) Both (a) and (b) are correct (d) Both (a) and (b) are wrong
10. What are number of nuclei of C , when number of nuclei of A and B are equal? (a) 2 N 0
(b) 3 N 0
(c)
9N 0 2
(d)
5N 0 2
11. At what time rate of disintegrations of A and B are equal? (a) 4 min
(b) 6 min
(c) 8 min
(d) 2 min
12. When two particles q and −q are placed at a distance r, then Coulomb’s force of attraction between them has magnitude F. If another charge particle having charge +3q is placed near to −q , then the force exerted by −q on q will be (a) 4F , repulsive (c) F , attractive
(b) 2 F , attractive (d) 2 F , repulsive
13. Find the equivalent thermal resistance of combination of rods as shown in figure. Every rod has same length l and crosssectional area Thermal A. conductivities are mentioned in figure. 1 2
K
3
4K
2K T2
T1
l (a) 4KA
7l (b) 4KA
7l (c) 12 KA
l (d) 12 KA
14. If the blood vessels in a human being acted as simple pipes (which they do not in actual), what would be the difference in blood pressure between the blood in a 1.80 m tall man’s feet and in his head when he is standing? Assume the specific gravity of blood to be 1.06. (a) 16.23 kPa (c) 17.3 kPa
(b) 18.7 kPa (d) 15.9 kPa
15. In J J Thomson experiment, the potential difference of 320 V is accelerating the electron. The electron beam is entering a region having uniform magnetic field 6 × 10−5 T acting perpendicular to it. Find the value of electric field in this region so that the electron does not experience any deflection. (Take me = 9.1 × 10−31 kg) (a) 6.4 V/m (c) 640 V/m
(b) 64 V/m (d) 6.4 ×104 V/m
16. A wire of length 100 cm is connected to a cell of emf 2V and negligible internal resistance. The resistance of the wire is 3 Ω. The additional resistance required to produce a potential difference of 1 mV/cm is (a) 47 Ω
(b) 57 Ω
(c) 60 Ω
(d) 55 Ω
17. A 20 g bullet is fired horizontally with a speed of 600 m/s into a 7 kg block sitting on a table top; the bulletb lodges in the block B.If the coefficient of kinetic friction between the block and the table top is 0.4, what is the distance the block will slide? b
(a) 0.5 m
B
(b) 1.2 m
µ = 0.4
(c) 0.37 m (d) 0.85 m
æ Directions Q. no. 18 is AssertionReason type questions. Each of these questions contains two statements : Statement I (Assertion) and Statement II (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice in the codes (a), (b), (c) and (d) in the given below
(a) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I (b) Statement I is true, Statement II is false
451
Practice Set 5 (c) Statement I is false, Statement II is true (d) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
18. Statement I When the range of projectile is maximum the time of flight is the largest. Statement II Range is maximum when angle of projection is 45°.
19. A uniform rope, of mass m per unit length, hangs vertically from a support so that the lower end just touches the table top. If it is released, then at the time a length y of the rope has fallen, the force on the table is equivalent to the weight of the length ky of the rope. Find the value of k. (a) 1 (c) 3
(b) 2 (d) 3.5
20. The two blocks shown in figure have equal
masses and µ s = µ k = 0.3 for both blocks. WedgeW is fixed and block A is given initial speed of 1 m/s down the plane. How far will it move before coming to rest if inclines and strings are quite long ? [Take g = 10 m/s 2]
/s 1
m
A
(a) 0.45 m (c) + 4.8 m
37°
W
(a) 0.75 m/s
(b) 0.50 m/s
(c) 0.25 m/s
(d) 2.25 m/s
23. A small conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied, the current induced in the loop is (a) clockwise (b) anticlockwise (c) zero (d) clockwise or anticlockwise depending on whether the current is increased or decreased
24. An inductor coil joined to a 6 V battery draws a steady current of 12 A. This coil is connected to a capacitor and an AC source of rms voltage 6 V in series. If the current in the circuit is in phase with the emf, the rms current will be
(b) 0.9 m (d) Question is irrevalant
21. As shown in figure, a uniform solid sphere rolls on a horizontal surface at 20 m/s. It then rolls up the incline shown. What will be the value of h, where the ball stops?
(c) 8 A
(d) 9.87 A
length of 1.0 × 10−4 N/m, are strung parallel to one another above earth’s surface, one directly above the other. The wires are aligned in a NorthSouth direction so that earth’s magnetic field will not affect them. When their distance of separation is 0.10 m, what must be the current in each in order for the lower wire to levitate the upper wire? Assume that the wires carry the same currents, travelling in opposite directions. (a) 2.7 A
v = 20 m/s
(b) 0.1 A
(c) 3.5 A
(d) 7.1 A
26. Two converging lenses are placed 20.0 cm h 30°
(a) 28.6 m (c) 6 m
(b) 12 A
25. Two wires, each having a weight per unit
m 53°
attached to a spring with a spring constant of1 0 × 103 N/m. The spring is compressed to a distance of 2.0 cm and the block is released from rest. Calculate the speed of the block as it passes through the equilibrium position, x = 0, if the surface is frictionless.
(a) 16.9 A
B
m
22. A 1.6 kg block on a horizontal surface is
(b) 8.6 m (d) 18.6 m
apart, as shown in figure. If the first lens has a focal length of 10.0 cm and the second has a focal length of 20.0 cm, locate the final image formed of an object 30.0 cm infront of the first lens.
452
JEE Main Chapterwise Physics 30.0 cm
t1 t2
20.0 cm
t3
t5
t6
t4 A (Inputs) Object
B t1 t2
f1 = 10.0 cm f2 = 20.0 cm
(a) 6.67 cm left (c) 15.0 cm left
t5
t6
t5
t6
(a)
(b) 6.67 cm right (d) 15.0 cm right
t1
27. Calculate the minimum thickness of a
soapbubble film (n = 1. 33) that will result in constructive interference in the reflected light, if the film is illuminated by light with a wavelength in free space of 602 nm. (a) 98 nm (b) 113 nm (c) 125 nm (d) 25 nm
t2
signal at 100 MHz, transmitted from a ground transmitter at a height of 300 m located at a distance of 100 km. Then, signal is coming via space wave signal is coming via sky wave signal is coming via satellite transponder None of the above
29. The output waveform (Y ) of AND gate for the following inputs A and B given below is
t3
t4
(b) t1
t2
t3
t4
t5
t6
t1
t2
t3
t4
t5
t6
(c)
28. A ground receiver station is receiving a
(a) (b) (c) (d)
t3 t4
(d)
30. A load resistor of 2 kΩ is connected in the collector branch of an amplifier using CE mode. β = 50 and input resistance of the transistor 500 Ω. If the input current is changed by 50 µA, then the power gain is (a) 102 (c) 104
(b) 103 (d) 106
Answers 1. (d) 11. (b) 21. (a)
2. (b) 12. (c) 22. (b)
3. (a) 13. (c) 23. (c)
4. (b) 14. (b) 24. (b)
5. (d) 15. (b) 25. (d)
6. (a) 16. (b) 26. (a)
7. (b) 17. (c) 27. (b)
8. (c) 18. (c) 28. (c)
9. (b) 19. (c) 29. (b)
10. (c) 20. (a) 30. (c)
Solutions 1. If we multiply 12.71 m by 3.46 m, we will get an 2
answer of 43.9766 m . We have only three significant figures in our least accurate measurement, so we should express our final answer as 44.0 m2 .
Note that in the answer given, we used a general rule for rounding off numbers, which states that the last digit retained is to be increased by 1, if the first digit dropped was equal to 5 or greater.
453
Practice Set 5
where, Me is earth’s mass and r is the satellite’s distance from the centre of earth. Also, we find the speed of the satellite to be d 2 πr …(ii) v= = T T
2. Charge of capacitor is the charge on facing surfaces of the plates of capacitor q − q 2 [10 − (− 2 )] = 6 µC Q = 1 = 2 2 Potential difference across the capacitor = 12 V Q (6 × 10−6 ) So, C= = F = 0.5 µF V 12
where, T is the orbital period of the satellite. Solving Eqs. (i) and (ii) simultaneously for r yields T2 r = 2 GMe 4π
3. We first write our knowns and unknowns with appropriate subscripts. vte = + 15 m/s [velocity of the train relative to earth] v bt = − 15 m/s [velocity of the baseball relative to the train] We need v bc , the velocity of the baseball relative to earth. v bc = v bt + vtc = + 15 − 15 = 0
4. The initial energy of the sledriderearth system
7. Velocity in nth orbit, v n = 2.2 × 10 ms −1 6
where,
8. Recall that an isovolumetric process is one that takes place at constant volume. In such a process, the work done is equal to zero because there is no change in volume. Thus, the first law of thermodynamics gives ∆U = Q [QQ = ∆U + W ] This indicates that the net energy Q added to the water goes into increasing the internal energy of the water. The net energy added to the water is Q = 10.0 − 2.0 = 8.0 kJ Because Q = mc∆T, the temperature increase of the water is 8.0 × 103 Q = 0.96° C ∆T = = mc (2.0)(4.186 × 103 )
If we set the origin of our coordinates at the bottom of the incline, the initial and final y coordinates of the sled are yi = 10.0 m and yf = 0. Thus, we get 1 2 1 v i + gyi = v f2 + 0 2 2 v f2 = v i2 + 2 gyi
Z1 − 2 × 2 + 3 × 1 = Z 2 − 2 × 1 + 5 × 1 = Zc
9.
v f = 14.9 m/s
5. Let particle be projected with speed u, so total time of flight
∴ ∴
2u T= g
n v1 = 2.2 × 106 m / s
For hydrogen, which implies velocity is inversely proportional to n. Hence, the plot is a rectangular hyperbola.
includes kinetic energy because of the initial speed. 1 1 ∴ mv i2 + mgyi = mv f2 + mgyf 2 2 1 2 1 or v i + gyi = v f2 + gyf 2 2
= (5.00)2 + 2 (9.80)(10.0)
1/ 3
Z1 − Z 2 = 4 A1 − 4 × 2 = A2 − 1 × 4 = Ac A1 − A2 = 4
1 min 2 N → 1 min 10. At A, 4N0 → 0 1 min N0 → 1 min N0 N0 → 2 4 N0 2 min N0 2 min At → B, N0 → 2 4 N After 4 min, NA = NB = 0 4 N N 9N0 NC = (4N0 + N0 ) − 0 + 0 = ∴ 4 4 2
and s = 2 × Maximum height u2 u2 =2 × = g 2g If there is no gravity, then s′ = u × T =
2u2 = 2s g
If gravity is not there, it will never fall back.
[Qinitial − final = NC ]
6. The force that produces the centripetal acceleration of the satellite is the gravitational force, so M m mv 2 G e2 = r r
11. Given, …(i)
or
RA = RB
∴ λ A NA = λ B NB
ln 2 ln 2 − λ Bt (4N0e − λ At ) = (N0 ) e TA TB
454
JEE Main Chapterwise Physics
∴ ∴
e( λ A − λ B ) t = 8 (λ A − λ B )t = ln 8 = 3 (ln 2 ) ln 2 − ln 2 t = 3 ln(2 ) or t = 6 min 1 2
Let current in circuit be I. ∴
polarity is placed, the force exerted will be attractive, also force between two charge particles is independent of presence of any other charge from the principle of superposition. Hence, option (c) is true. l l l 13. R1 = , R2 = , R3 = KA 2 KA 4KA RR 7l Req = 1 2 + R 3 = R1 + R 2 12 KA
14. Specific gravity = Density of substance (ρ) Density of water(ρw )
[Q ρ = ρw × 1000] ∆p = ρgh = 1060 × (9.8)(1.8) = 18.7 kPa
15. Electric potential energy = Kinetic energy mv 2 i.e., qV = 2 2 qV and qvB = qE ⇒ E = vB v= ⇒ m 2 qV ×B E= m After substituting these values, we get E = 640 V/m
16. Let a resistance R be connected in series with the battery and wire. 100 cm, 3Ω
2 ⇒ R = 57 Ω R+ 3
17. By conservation of momentum, the momentum of
12. From Coulomb’s law, since charge of opposite
Q 1 = 1 + 1 and R = R + R S 1 2 R R1 R 2 P as rods 1 and 2 are in parallel and equivalent in series with 3
. =I×3=3× 01
the block bullet system just after the interaction is p = mv. where, m is the mass of bullet and v is its velocity before striking the block. Hence, the kinetic energy of the system just after the lodging of bullet into the block is p2 K= 2 (M + m) =
(mv )2 2 (M + m)
The friction force does work Wf = − f s = − µ k ( m + M ) g s In stopping the block, where s is the distance traversed by blockbullet system on the tabletop. From workenergy theorem, ∆K = Wf ⇒ ∴
0 − K = − µ k (m + M ) g s s = 0.37 m
18. The horizontal range, u 2 sin 2 θ g 2 u sin θ Time of flight, T = g R=
Range is maximum when θ = 45°. So, sin 2 θ = sin 90° = 1 u2 Rmax = g Time of flight is maximum when θ = 90°. So, sin θ = sin 90° = 1 2u ∴ Tmax = g
19. The descending part of the rope is in free fall. It has R
speed v = 2 gy at the instant all its points have
2V
Voltage drop across wire = 1 × 10−3 × 100 = 0.1 V
descended a distance y. The length of the rope which lands on the table during an interval dt following this instant is vdt. The increment of momentum imparted to the table by this length in coming to rest is m(vdt ) v. Thus, the rate at which momentum is transferred to the table is
455
Practice Set 5 y v F
dp = mv 2 = (2 my) g dt and this is the force arising from stopping the downward fall of the rope. Since, a length of rope y of the weight (my) g , already lies on the tabletop, the total force on the tabletop is (2 my) g + (my) g = (3 my) g or the weight of a length 3y of rope. So, k=3
20. Find acceleration 2 ma = mg sin 53° − µmg cos 53° − mg sin 37 ° − µmg cos 37 ° Then, use v 2 = u 2 + 2 as,
v = 0, u = 1m/s.
On solving, we get s = 0.45 m
21. The rotational and translational kinetic energy of the ball at the bottom will be changed to gravitational potential energy, when the sphere stops. We therefore write Mv 2 Iω2 = (Mgh)end + 2 start 2 For a solid sphere,
I=
2 Mr 2 5
v Also, ω = . Then, above equation becomes r 2
2 Mr 2 v = Mgh 5 r 1 2 1 2 or v + v = (9.8) h 2 5 Using v = 20 m/s, gives h = 28.6 m. Note The answer does not depend upon the mass of the ball or the angle of the incline. 1 1 Mv 2 + 2 2
22. The initial elastic potential energy of the compressed spring is 1 PE s = kxi2 2 Because the block is always at the same height above earth’s surface, the gravitational potential energy of the system remains constant. Hence, the initial potential energy stored in the spring is converted to kinetic energy at x = 0 i.e., 1 2 1 kxi = mv f2 2 2 Solving for v f gives 1.0 × 103 k (2.0 × 10−2 ) xi = m 1.6 = 0.50 m/s
vf =
23. The angle between magnetic field and area vector is 90°, so the flux associated with coil is zero. Although magnetic field is changing but flux is remaining constant equal to zero, so emf induced and hence current in the loop is equal to zero. V 24. Resistance of coil = DC = 6 = 0.5 Ω IDC 12 In an AC circuit, the current is in phase with emf. This means that the net reactance of the circuit is zero. The impedance is equal to the resistance, i.e., Z = 0.5 Ω. RMS voltage 6 RMS current = = = 12 A Z 0.5
25. If the upper wire is to float, it must be in equilibrium under the action of two forces : the force of gravity and magnetic repulsion. The weight per unit length here 10 . × 10−4 N/m must be equal and opposite the magnetic force per unit length. Because the currents are the same, we have F1 mg µ 0 I 2 = = l l 2 πd ⇒
. × 10−4 = 10
(4 π × 10−7 ) (I 2 ) (2 π )(010 . )
We solve for the current to find I = 7.1 A.
456
JEE Main Chapterwise Physics
26. First we make ray diagrams roughly to scale to see, where the image from the first lens falls and how it acts as the object for the second lens. The location of the image formed by the first lens is found via the thinlens equation 1 1 1 + = ⇒ v1 = + 15.0 cm 30.0 v1 10.0 The image formed by this lens becomes the object for the second lens. Thus, the object distance for the second lens is 20.0 cm − 15.0 cm = 5.00 cm . We again apply the thinlens equation to find the location of the final image (see figure). 1 1 1 ⇒ v 2 = − 6.67 cm + = 5.00 v 2 20.0 Thus, the final image is 6.67 cm to the left of the second lens.
I2 I1 O1
F1
F1
F2
10.0 cm
F2
15.0 cm 6.67 cm
30.0 cm
20.0 cm Lens 1
27. For constructive interference,
λ 2 nt = , we have 2 λ 602 t = = = 113 nm 4n (4)(133 . )
Lens 2
For t 4 to t 5 ; A = 0, B = 0; hence Y = 0 For t 5 to t 6 ; A = 1, B = 0; hence Y = 0 For t > t 6 ; A = 0, B = 1; hence Y = 0 Based on the above, the output waveform for AND gate can be drawn as given below t2
28. Maximum distance covered by space wave
t3 t4
t1
communication
t5
t6
= 2Rh = 2 × 6.4 × 10 × 300 = 62 km 6
Since, receivertransmitter distance is 100 km, this is ruled out for signal frequency. Further fc for ionospheric propagation is
∆V0 Q β = ∆Vi
30. ∆V0 = β ∆Ib(RL )
fc = 9 (Nmax )1/ 2 = 9 × (1012 )1/ 2 = 9 MHz
Given, β = 50, ∆I = 50 µ A = 50 × 10−6 A
So, the signal of 100 MHz (7fc ) comes via the satellite mode.
∴
R L = 2 kΩ = 2000 Ω
29. For t ≤ t 1; A = 0, B = 0; hence Y = 0 For t 1 to t 2 ; A = 1, B = 0; hence Y = 0 For t 2 to t 3 ; A = 1, B = 1; hence Y = 1 For t 3 to t 4 ; A = 0, B = 1; hence Y = 0
∆V0 = 50 (50 × 10−6 ) × 2000 ∆V0 = 5 V
∴
Ap = β 2 ⋅
RL 2000 = (50)2 × = 104 ri 500
Hence, option (c) is true.