Oswaal NTA CUET (UG) Chapterwise Question Bank Physics (For 2024 Exam) [2024 ed.] 9359582352, 9789359582351

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CUET (UG) COMMON UNIVERSITY ENTRANCE TEST

CHAPTER-WISE

QUESTION

BANK Includes SOLVED PAPERS (2021

-

2023)

PHYSICs

Section ll(DomainSpecific Subject) as per the Latest Notification & Examination Scheme issued by NTA on 6 March 2024

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Chapter-1-Electrostatics

-

1

Chapter-2 - Current Electricity 15

Chapter-3- Magnetic Effects

of

Currentand Magnetism

Chapter-4-Electromagnetic Induction and Alternating Currents Chapter-5-Electromagnetic

Waves

-

14

26

27 - 38 39 - 52 53

- 61

62

- 80

81

- 91

Chapter-8- Atom and Nuclei

92

-106

Chapter-9- Electronic Devices

107

-124

Chapter-10-Communication Systens

125

-136

Chapter-6-Optics Chapter-7-Dual Nature

of

Matter and Radiation

CUET (UG) Exam Paper 2023* National Testing Agency Held on 26th May 2023

PHYSICS

[This includes Questions pertaining toDomain Specific Subject onlyi Max. Marks: 200

Time allowed: 45 Minutes

General Instructions: The test is of 45 Minutes duration. 1, The test contains 50questions Out of which 40 questions need to be attempted. Marking Scheme of the test: 3. a. Correct answer or the most appropriate answer: Five marks (+5). b. Any incorrectly marked option will be given minus one

c

Unanswered/Marked for Review will be given zero •**.io...*...ei..

1,

A

beam of electron is used in Young's double slits experiment. The slit width is d. When veloity of electron is increased, then: (1) No interference is observed (2) Fringe width increases (3) Fringe width decreases (4) Fringe width remains same 2. Match List - I with List - II. List -I

(4)

Simple

Compound

()

Microscope Refracting Telescope

Reflecting Telescope Choose

below:

List-IIK

Microscope

(B)

(D)

mark (-1). mark (0).

the

)

A concave mirror is |used as an objective

()

Two convex lenses both

of small focal length (I) Two convex lenses, one of small focal length other of large focal |length (IV) |Single convex lens of small focal length correct answer from the options given

(A-IV), (B)-(I), (C-), (D-() (A-IV), (B-(II), (C)-(III), (D)-() (A)-M), (B)-IV), (C)-(III), (D)-) (AJ-IV), (B)-(III), (C-(I), (D-() 3. Match List - I with List - II

(B)

(II)

(C)

(II)q>0

(D)

(IV)

(Field lines)

(2) (A)-(IV), (B)-(), (3) (A)-(II), (B)-(0),

(C-), (D-() (C)-V), (D)-(W) (4) (A)-), (B)-(I), (C)-(IV), (D-() 4. The emfs and resistances in the given circuit have the following values

(A)

E

w Ej = 4.2V, E, = 1.9 V

I=

q In parallel- C, = C+C, +...+C,

3P

V=+,t...+V,

Equipotential surfaces: > An equipotential surface is a surface with a constant value of potential at all points on the surface. It requires no work to move the charge on such surface. > Electric field is always perpendicular to the equipotential

Combination of capacitors

Capacitance of parallel plate capacitor: > Parallel plate capacitor is a capacitor with two identical plane parallel plates separated by a small distance where space between them is filled by dielectric

Scan to know

more about

this topic

medium.

> Capacitance of parallel-plate capacitor with area A separated by a distance d is,

surface.

> Spacing among equipotential surfaces allows to locate regions of strong and weak electric field. > Equipotential surfaces never intersect cach other. If they intersect, then the intersecting point of two equipotential surfaces results in two values of electric potential, which is impossible. Electrie potential energy of a system of charges: > Potential energy

of a system

of

two charges-U=

1

9924 993,993

Electric potential energy in an external field: > Potential energy due to a single charge-U = qV() >Potential energy due to two charges in an external field 1

U-ar(i)+4(a) Potential energy

of

a

> If we have a number of slabs of same area as the plates of the capacitor with thickness t,,4,,tq,.. . and dielectric constants K, K,, K3,...inserted between the plates, the capacitance of the capacitor is given by,

C=

4142

> Potential energy of a system of three charges

U=

Parallel plate capacitor

d

9192

K K, K, -t.... >

If

a

single slab of thickness and dielectric constant 'K is introduced between the plates having separation 'd' then the capacitance of the capacitor is given by,

+)

1

> If a single slab of conductor of thickness

between the plates, having separation of the capacitors is given by, C=

dipole in an external field

U(e) =-pE cos0 Where,

p=2ag and 0 is the angle between electric field and

dipole.

Conductors and insulators: >Conductors- Materials through which charges can move freely. Examples- metals, semi-metals such as carbon, graphite, antimony and arsenic.

>Insulators- Materials in which the electrical current does not tow easily. Examples- Plastics and glass.

K,A K(d-i) +i

C=.

(d-1)

is introduced then the capacitance

d-t

1

Energy stored in a capacitor:

>U-0or-cv 2 C

> The energy density of in

a region with electric field is,

Oswaal CUET (UG) Chapterwise

6

Question Bank PHYSICa

OBJECTIVE TYPE QUESTIONS IAJAMULTIPLE CHO1CE QUESTIONS I| 1. Match List -I with List List II List -I configuration) (Charge (Field lites) (U)

(A)

q0

(D)

(IV)

Lst -I (Physical Quantity) (A) Lincar chargc density

Uncqual charges

(B) Choose the correct answer from theoptions given below:

(C)

(1) (A-), (B)>-(), (C)-(1), (D)-(1V) (2) (A)-(IV), (B)-(||I), (C)-(), (D)-(|1)

(3)

(A).

(B-(),

(C)-(1V),

(D)-()

(4) (A}-O), (B)-(II), (C)-(IV), (D)-(I1) 2.

(CUET 2023, 26th May) radius 6.0 cm is held in conductor of metallic A spherical air. The maximum charge it can hold if dielectric strength of air is 20 kV/cm will be

Take-9x10'Nm'c* (4) 8 iC (CUET 2023, 26h May) Two capacitors of capacitances 8 uF and 20 uF are connected in series with a battery. The voltage across the 8 uF capacitor is 5 V. The total battery voltage is 5V (2) 10 V (4) 125V 7V (CUET 2023, 26th May) is Electrical capacity of earth approximately

(1) 0.8 uC

3.

(1)

(3)

(2) 8.0uC

4. (1) 57.6 mF

(3) 80 uC

(2)

(D)

(1) (2) (3) (4)

List- II (S.I. Unit) newton(metre)/ columb

()

Elcctric dipole moment Polarisation vector

E, E> Ec E, > Eg > Ec (CUET 2023, 26h May)

Match List - I with List I.

7.

(C)

(2) (4)

E

(1) Ep> Ec> (3) Eç> Bp > Ea

Electric dipole

(II) coulomb/metre

(II)

coulomb/(metre)

(IV) coulomb - metre answer correct from the options given below: Choose the EIectric flux

(AJ-(II), (B)-(I), (C)-(II), (D)-(1V) (A)-(IV), (B)-(II), (C)-(I), (D)-()

(A)-(1),(B)-(IV),(C)-), (D)-(IIn) (A)-(II), (B)-(IV), (C)-(I), (D)-(I)

(CUET 2023, 26h May) q are + 8. Three charges 44, 2 and placed along x-axis at points, x =0, /2 and l respectively. The value of to make the net force on q to be zero Q

)

(1) 4q

-

(3)

(4) -g

-24

(CUET 2023, 26th May) one 9. Which of the following correctly represents the variation of electric field and electric potential with a distance r from a point charge?

E,VV (1)

(2)

1F

(4) 9 x 109 mF (CUET 2023, 26th May) a B A, points Three in uniform Electric field E of and Clie 5. 2 x 10 NIC as shown in figure. The potential difference (3) 711 uF

between points

A

and Cis

5

cm',

VE

EV

E

E,V

(4)

(3)

13cm

of radius R' is having surface charge density a cm2, Electric field at axial distance 2R' is

10. A circular lamina (1) 100 V (3) 80 V 6.

(2) 60 V

(4) Zero

(CUET 2023, 26th May) For a given charged conducting hollow sphere arrange the electric field at the given points A, B and C in decreasing order

(1)

(3)

4nE,R

(4)

e

4€o 2E August) 5'h 2022, (CUET

11. Two-point charges +5 and -5 uC are placed at O (0 mm, 0 mm) and P (3 mm, 4 mm) respectively force on +>

C

uC is

ELECTROSTATICS

7

P(3, 4)

(0 1s($-4j)x10*N

() -18(+4j)×10"N

)

(4)

18(3$+4)x10'N

12. Three charges Q,

1.8(-si+4j)x10'N

(CUET 2022, 5" August) tq and tq are placed at the vertices of an

equilateral triangle of side l, as shown in the figure. If the net electrostatic energy of the system is zero, the Q is equal to

(2) In a charge free region, clectric field lines can be taken to be continuous without any break. (3) Two ficld lincs never cross cach other. (4) Electrostatic ficld lines form closed loops. 18. Select the correct statement from the following A. Gauss's law is true for any closed surface, no matter what its shape or size. B. If thc net clectric flux through a closed surface is zero, we can conclude that the total charge contained in the closed surface is also zero. C. Gauss's law is not based on the inverse square dependence on distance contained in the Coulomb's law. D. The electric ficld due to a charge outside the Gaussian surface contributes zero net flux through the surface. Choose the correct answer from the options given below (2) A and C only (1) A only (4) B, C and D only (3) A, B and D only (CUET 2022, 8" August) constant K has the same dielectric 19. A slab of material of area as the plates of a parallel plate capacitor but has a and placed between plates of parallel

thickness

tq (1)

(2) Zero

2

(3)

2

plate capacitor, where d is the separation of the plates. The ratio of capacitances of capacitor with dielectric and with

(4) -g

(CUET 2022, 5th August) capacitor circuit charged by a battery to a potential difference Vo as shown in the 3Co figure. When a dielectric slab inserted into both the capacitors, new potential difference will be

13. Consider

(1) (2)

(3)

14.

(1) 15.

,

(1)

20.

(1) (2) (3) (4)

4K K (4) (3) K+3 +1 3K +1 4K Charged bodies are considered as point charges if the linear size of the charged bodies is Much smaller than the separation between them Greater than the separation between them Equal to the separation between them Independent of separation between them 3K +1

(2)

4

3

(CUET 2022, 8th August)

K 8, 5K

(4)

8

K (CUET 2022, 5th August) Electric field at the surface ofa conducting shell of radius a ' is measured as X. Electric field at distance 3r fom the centre of the shell is 3

air as medium

a

(2)

(3) 6

9

21. In the Fig. below, two positive charges q2 and q3 fixed along they axis, exert a net electric force in the + x direction on a charge g fixed along the x axis. If a positive charge is added at (x, 0), the force ong

(4) X

(CUET 2022, 6th August) at (0 mm, 0 mm). Another + 10 uC is placed A charge mm, O mm) to (0 mm, 3 (3 from is moved uC charge -5 agency is mm). Work done by the external + (3) 150 J (4) -300 (2) -150 J 0J (CUET 2022, 6th August) a N. If the Two charges in air experiences a force of 20 space between them is filled with mediumn of dielectric constant K= 4, the new force will be (4) 2 N (3) 4 N 20 N (2) 5 N

(x, o)

J

(1) 16.

(1)

(CUET 2022, 8" August) Which o of the following statements is wrong? at (1) Electric field lines start from positive charges and end negative charges. 17.

(a)

(1) (2) (3) (4)

(b)

shall increase along the positive xX-axis. shall decrease along the positive x-axis. shall point along the negative x-axis. shall increase but the direction changes because of the intersection of with q, and q3. [NCERT Exemplar, Q.1.1 Page No. 1,] 22. A point positive charge is brought near an isolated conducting sphere (Fig.). The electric field is best given by

Oswaal CUET

8 (2)

E

(UG) Chapterise

Question Bank

on

(1)

only.

(ii)

(iv) (2) fig (ii)

contributions

INCERT Exemp. 25. A point charge Q.1.4 Page No. tq, is placed at a distance 3| conducting planc. The ficld at a point d from an isolated on the planc is P the other side of (1) directed perpendicular to the planc and away from (2) directed perpendicular the plane. (3) directed radially away to the plane but towards the plane. from the point charge. (4) directed radially towards the point charge. A INCERT Exemp. Q.1.6 26. A capacitor Page No. 3) of 4 uF is connccted as The intcrnal resistance shown in the circuit. of the battery is 0.52. of charge on the capacitor The amount plates will be

(i)

(1) fig (i)

PHYSICS

the eontributionLHS of the above from equation will have a contribution all charges while q on a the RHS will from q, and q4 only. (3) E on the have contributionLHS of the above cquation will have a contributionfrom all charges whilcq on the from RHS q1, 93 and qs only. will haye (4) BothE on the LHSand q on the RHSwill have from q, and 44

10 2

(3) fig (iii)

23. The Elcctric flux through the surface

(4) fig (iv)

2 S2

(1) 0

(i)

(i)

(ii)

(1) (2) (3) (4)

(iv)

in Fig. (iv) is the largest. in Fig. (ii) is the least. in Fig. (ii) is same as Fig. (iii) but is smaller than Fig. (iv) is the same for all the figures. [NCERT Exemp.Q.1.3 Page No. 2] 24. Five charges q1,42, 43, q4, and qs are fixed at their positions shown in Fig. S is a Gaussian surface. The Gauss's law is given by

E• ds

(2) 4 HC

(3) 16 uC (4) 8 uC INCERT Exemp.Q.2.1 Page No. 10] 27. Apositively charged particle is released from rest in a uniform clectric field. The electric potential energy the of charge (1) remains a constant because the electric field is uniform. (2) increases because the charge moves along the electric field. (3) decreases because the charge moves along the electric field. (4) decrcases because the charge moves opposite to the electric field |NCERT Exemp.Q.2.2 Page No. 10) 28. Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B. 30V 40V

30V I

= 9

Which of the following statements are

Gaussian Surface

14

B

A

correct? 10V 20V 30V 40V 50V

Fig. I (1) (2) (3) (4)

(1) E on the LHS of the above cquation contribution from g, gs and g: while q on will have a the RHS will have a contribution from g, and q4 only.

10V 20V 50V Fig. II

10

20V

40V

50V

Fig. III

The work done in Fig. (i) is the greatest. The work done in Fig. (ii) is least. The work done is the same Fig. (i), Fig. (ii) and Fig. (ii). The work done in Fig. (iii)) is greater than Fig. (ii)but equal to that in Fig. (i). [NCERT Exemp. Q.3 Page No. 11]

29. Equipotentials at a great distance from a collection of e,charges whose total sum is not zero are approximately (1) spheres. (2) planes. (3) paraboloids (4) ellipsoids. [NCERT Exemp. 0.5 Page No. 11|

ELECTROSTATICS

9

30. A parallel plate capacitor is made in series. One

of two dielectric blocks

of the blocks has thickness d, and dielectric constant k, and the other has thickness d, and dielectric constant k, as shown in Fig. This arrangement can be thought as a diclectric slab of thickness d (= d + d) and effective dielectric constant k. The k is

(1)

(3)

kd

+

d,

k,d

+d, kk,(d, +d,) (kd, + kyd,)

(2)

kd, +k,d,

Reason (R): Negative gradient of electric potential is electric field. any 9. Assertion (A): Work done in moving a charge between the independent of a is field two points in uniform electric path followed by the charge between these two points. Reason (R): Electrostatic forces are non-conservative. 10. Assertion (A): Two parallel metal plates having charge +0 and -0 are facing at a distance between them. The plates are now immersed in kerosene oil and thé clectric potential between the plates decreases. Reason (R): Diclectric constant of kerosene oil is less than I.

(4)

1.

[C] COMPETENCY BASED QUESTIONS

2k,k, +k,

k,

[NCERT Exemp.Q.6 Page No. 12]

BJ ASSERTION REASON QUESTIONS Question Nos. 1 to 10 consist of two statements -Assertion and Reason. Answer these questions by selecting the appropriate option given below: (1) Both (A) and (R) are true and (R) is the correct explanation

Read the following text and answer the following same: questions on the basis of the or Faraday shield is an cage A Faraday Cage: Faraday enclosure made of conducting mate rial. The fields within so the a conductor cancel out with any external fields, zero. Faraday is These enclosure the within field electric cages act as big hollow conductors you can put things in to shield them from electrical fields. Any electrical shocks the cage receives, pass harmlessly around the outside of the cage.

of (A). (2) Both (A) and (R) are true and (R) is not the correct explanation of (A). (A)is true but (R) is false. (4) (A) is false but (R) is true. 1. Assertion (A): Circuit containing capacitors should be handled very carefully even when the power is off. Reason (R): The capacitors may break down at any time. 2. Assertion (A): In a non-uniform electric field, a dipole will have translatory as well as rotatory motion. a dipole Reason (R): In a non-uniform electric field, as as a torque. well experiences force cross each other. 3. Assertion (A): Electric lines of force at a point is the field electric Reason (R): The resultant at that point. fields electric the superposition of different radii 4. Assertion (A): If two spherical conductors of their electric then densities, have the same surface charge field intensities will be equal. = Total charge Reason (R): Surface charge density Area a conductor, the electric field a 5. Assertion (A): In cavity of (3)

is zero.

6.

reside only at its a Reason (R): Charges in conductor surface. square law is not obeyed, Assertion (A): When the inverse Gauss 1law shows deviation. leads to Gauss Reason (R): The conservation of charges law.

7.

8.

in an electric field Assertion (A): A negative charge field. electric moves opposite direction of the a force acts in the a charge Reason (R): On negative field. opposite direction of the electric is always normál to field Assertion (A): Electric equipotential surfaces and along the direction ofdecreasing order of potential.

Which of the following material can be used to make a Faraday cage? (3) Copper (4) Wood (2) Glass (1) Plastic 2. Example of a real-world Faraday cage is (2) plastic box (1) car 1.

(3)

lighting rod

(4)

metal rod

3. What is the electrical force inside a Faraday cage when it is struck by lightning? (1) The same as the lightnig (2) Half that of the lightning (3) Zero (4) A quarter of the lightning 4. Ifisolated point charge tq is placed inside the Faraday cage. Its surface must have charge equal to (3) -q (4) +2q (2) tq (1) Zero 5. A point charge of 2uC is placed at centre of Faraday cage in the shape of cube with surface of 9 cm edge. The number of electric field lines passing through the cube normally will be (1) 1.9 x 10 Nm/C entering the surface. (2) 1.9 x 10 Nm/C leaving the surface. (3) 2.0 x 10° Nm/C leaving the surface. (4) 2.0 x 10 Nm²/C entering the surface.

II. Read the given text and answer any four of the following questions on the basis of the same: Super capacitor: Super capacitor is a high capacity capacitor with a capacitance value much higher than normal capacitors but with lower voltage limits. Such

OswaalCUET (UG) Chapterwise Question Bank PHYSICS

10 capacitors gap betwecn electrolytic capacitors bridges the and rechargeable batteries. capacitors train, cranc, clcvator such encrgy In automobile, bus, short term are uscd for regencrative braking, power delivery. storage or burst-mode many advantages over batteries: contain Super capacitors have weight and generally don't they are very low can charged be mctal. They harmful chemicals or toxic without cver times number of and discharged innumerable wearing out. well that super capacitors aren't The disadvantage is rate of discharge storage. The cnergy suited for long-term lithium-ion than higher super capacitors is significantly as 10-20% of their charge batteries; they can lose as much 6.

(1) (3) 7.

(1)

(2)

per day due to self-discharge. Capacity of super capacitor is (2) medium. very low. (4) may have any value. very high. a between Super capacitor makes bridge battery. rechargeable and clectrolytic capacitor capacitor. clectrolytic use and battery single

(3) clectrolytic capacitor and dynamo.

(4) electrolytic and non-electrolytic capacitors. 8. Supcr capacitors can be charged and discharged (1) few number of times. (2) oncc only. (3) several number of times but less than rechargeable batteries (4) several number of times much more than rechargeable batteries. 9. Self-discharge rate of Super capacitors (1) 10-20% of their charge per day (2) -2% of their charge per day (3) 0% of their charge per day (4) 1009% of their charge per day 1

capacitors are used for degenerative braking. regenerative braking. small appliances. long time charge storage.

10. Super

(1) (2) (3)

(4)

ANSWER KEY CHOICE QUESTIONS 7.(4)8. 5.(3)6.(1)

[AJ MULTIPLE 1.

(3)|2.

3. (3)

(3)

11. (3)

12. (3)

21. (1)

22. (1)

1. (1)

1.(3)

2. (1)

|2. (1)

13.(3) 23. (4)

4. (3) 14. (3) 24. (2)

15.(1)

|25. (1)

16. (2)

(4)

17. (4)

26. (4)

27. (3)28.

|3. (4)

BASSERTIONREASON QUESTIONS |7.(1)8. 6. (2) 5. (1) 4. (2)

3. (3)

jC COMPETENCY BASED QUESTIONS 6. (3) 7.(1)8. 4. (3) |5.(3)

9.(3) 19. (2)

18. (3) (3)

29.

(1)30.

10. (2)

20. (1) (3)

(1)

9. (3)

10. (3)

(4)

9. (1)

10. (2)

ANSWERS WITH EXPLANATIONS JAJ

MULTIPLE CHOICE QUESTIONS

Option (3) is correct. Explanation: Electric field lines always point away from a positive charge and towards a negative charge. In fact, electric fields originate at a positive charge and terminate at a negative 1.

3. Option (3) is correct. 20 Explanation: Cpotdl = 8x =5.7luF 8+20 Q=CV=8x5=40uF

40

charge.

For Unequal charges the field lines will not be proportionate at the two sides of the charge, that is the number of field lines will diffe. 2. Option (3) is correct. Explanation: Consider that the charge is Q. The dielectric strength of air is given. The maximum charge which can be given without ionizing the air around it is given by, kQ

Or, Or,

20x10 =

9x10' x@ 20x 10' x36 9x10"

Q=80x10C=80uC

C

5.71

=7V

4. Option (3) is correct.

Explanation: Here,

C= 4reoR 1

4nE09y10° R=6400 km =6.4 C=

6.4x106

9x10

x 10°m

-0.711x10F-71 luF

5. Option (3) is correct. Explanation: The line joining B to C is perpendicular to clectric field, so potential at B= potential at C Distance AC=5 cm Potential difference between A and C= Ex (AB) 2x 10 x (4 x 102) = 80 volt.

ELECTROSTATICS

11

6. Option (1) is

correct. Explanation:There are no charges inside the hollow conducting sphere, allcharges reside on its surface. So, clectric ficld inside the hollow conducting sphere is zero. Electric ficld decrcases exponentially outside the conductor. 7. Option (4) is correct. Eplanation: Lincar charge density (2) is tlhe quantity of charge per unit length, measured in coulonb per meter (Cm') at any point on a linc charge distribution. The SI unit for clectric dipole moment is the coulonb-meter (Cm). Polarization vector, P is cqual to the bound charge per unit arca or cqual to the surfacce density of bound charges (because surface charge density is charge per unit area), S.I. unit is Cm". The Sl unit of clcctric flux is volt meter (V-m), this is also cqual to newton-meter squared per coulomb (Nm'Cm). 8. Option (4) is correct. Explanation: The force exerted on the charge q by charge is

given as:

F(9.0)

12. Option (3) is correct. Explanation:

Net electrostatic energy

Given,

U

U0

kq', 2kq2-0

Or, So,

2

Co

correct.

13. Option (3) is

Explanation: On inserting diclectric, charge on the combination of capacitor docs not change. Equivalent capacitance Co+ 3Co 4Co Ncw potential difference be V1

3Co H Vo

=

K2

Charge = 4CoVo

LI14J The force exerted on the charge g by charge 4q is given as:

So, V, =

The net force is zero, s0,

F(9.0) +F

(9,4q)= 0

4kql4k'-o

8

5K

14. Sol. Option (3) is

correct.

Explanation: For shell,

4g

2->-/2->

Q+q=0

Therefore, the charge of Q should be to remain zero. 9. Option (3) is correct.

Explanation:

- q for the net force on q

Or,

E,

E=l9 4E, r V=

(3r)°

X E,

Or,

(r)*

X E, = 9

1_g

15. Option (1) is

correct.

Explanation:

So, E will come down faster then

V

with an increase of r as

E=P

(Omm, 3mm)

Option (2) is correct. Explanation: 10.

E=(l-cos

0),where cos

=

2R

V4R²+

(Omm,

R²

F

KDe.;-)

Omm)O

O

and

Q2

Fait

Explanation:

Fmed

- 1s(6+4j)x10

Omm)

Since, OA =OB, potentials at 4 and B are same and potential difference between A and B is 0. So, work done is also 0. 16. Option (2) is correct.

9x10 x5x10* x5x 104(4j+ 3)x10 F

A

+10 ucOmm,

Putting the value of cos

11.Option (3) is correct. Explanation: Force between two-point charges

B

N

Fnsd

(g) -() K4

12 Or,

Oswaal CUET (UG)

20

Fd

5 N

correct.

Explanation: Electrostatic field lines don't form closed loops since it is conservative in nature. 18. Option (3) is correct. Explanation: Gauss's law is true for any closed surface, no matter what its shape or size. So, option (C) is correct. Gauss's law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface. So option (B) is correct. Flux is due to charge enclosed. The clectric field due to a charge outside the Gaussian surface contributes nothing through the surface. So, option (D) is correct. Gauss's law is based on the inverse square dependence on distance contained in the Coulomb's law. So, option (C) is incorrect. 19. Option (2) is correct. Explanation:

E

Now, as a positive charge is placed on the x axis, the fre between q and is also attractive and is in the + x dircctir

=4

Fmcd

17. Option (4) is

Chapterwise Question Bank PHYSCS

d C=

d-t+ K d

d-+4 4K d

3d 4

F1,3

Fad

(k, 0)

Hence, the net force on charge g shall increase along the

+x direction. 22. Option (1) is correct. Explanation: Thc electric field lines come out of positive are alwavs charge and go into negative charge. The field lines a to Now, when positive charge normal the conducting surface. is brought ncar an isolated conducting sphere, without touching it, then the free clectrons in the sphere gets attracted towards the positive charge. As a result, the left surface of the sphere becomes charged with negative charge, whereas the ríght

surface becomes charged positively. It should be emphasised that neither type of charge is free to exit the metal sphere. Therefore, they are located on the sphere's cxterior. 23. Option (4) is correct. Explanation: According to Gauss law of clectrostatics, the total electric flux out of a closcd surface is cqual to the charge cnclosed divided by the permittivity. Hence, the electric flux through a surface does not depend on its shape, size or area. Therefore, the clectric flux through all the surfaces is same as the charge enclosed is same.

Co=

d,

F1,2

24. Option (2) is correct. Explanation: On the LHS of the cquation, the electric field is due to all the charges present both inside as well as outside the Gaussian surfacc. On the RHS of cquation, the flux is due to all the charges enclosed in the Gaussian surface, i.e., 4) and q4.

4K

d

4K 3K +1

20. Option (1) is correct. Explanation: When the linear size of charged bodies

is much

may be smaller than the separation between them, the size ignored and the charged bodies are considercd as point charges. 21. Option (1) is correct. g is in +x Explanation: As the net electric force on charge direction, hence the nature of force betwcen q1, 42 and gz, 43 should be attractive which means qshould be negative.

25. Option (1) is correct. Explanation: Free electrons are drawn to a point positive charge when it is positioncd close to an isolated conducting plane. As a result, a little amount of negative charge forms on the planc's surface in the direction of the positive charge, and a similar positive charge forms on the other side of the plane. The clcctric ficld lines are parallel to the surface and away from positive charge. As a result, option (1) is true since the field at point P on the other side of the plane is directed away from and perpendicular to the planc. 26. Option (4) is correct. Explanation: At steady state capacitor gets fully charged and offers infinite resistance in DC circuit and acts as open circuit, hence no current flows through the capacitor and 102 resistance. So, the potential difference across the capacitor is equal to potential differencc across 2 S2 resistance. Current through 2 SQ resistance, 2.5

=1A 2+0.5 R+r Potential difference across 2 2 resistance,

V= IR =1x2=2V

ELECTROSTATICS

13

So, potential difference across capacitor is also 2V. Therefore,

the charge on the capacitor,

Q=CV=2juFx 2V =8uC 27. Option (3) is correct.

Explanation: Elcctric potential energy decreases in the direction of the field. And as the particle is positively charged it will move along the direction of the electric ficld hence its electric potential energy will decreasc. 28. Option (3) is correct. Explanation: In all the threc figures, V = 20V and V= 40V Work done in carrying a charge q form A to B is W= q(Vg-V) Hence, work done is same in all figures. 29. Option (1) is correct. Explanation: From a large distance, a collection of charges whose sum is not zero can be considered as a point charge. The equipotential surfaces due to a point charge are spherical. 30. Option (3) is correct. Explanation: This system as shown in the figure can be considered as two capacitors say C,and C, connected in series. We know, C=* and C, = Kde d, d, Now,

Ca C

C,

d d, k,A6, k,deo d d,

kk,Eg4

kd, + k,d,

...)

We can write equivalent capacitance as

C=

ke,A d, + d,

..(i)

On comparing (i) and (ii), we get,

kk,(4, + d, ) k= kd, + k,d,

[B]ASSERTION REASON QUESTIONS 1. Option (1) is correct. Explanation: Even when power is off capacitor may have stored charge which may discharge through human body and

thus one may get a shock. 2. Option (1) is correct. a Explanation: When an electric dipole is placed in uniform electric field at an angle with the field, the dipole experiences

a torque. as couple The torque produced by two parallel forces qE acting T=qE (2a sin 0) on both the ends of In case of non-uniform field, force acting be a combination will there So, the dipole will not be equal. way, dipole will have both of couple and a net force. In this rotational as well as linear motion. explains the So, both assertion and reason are true. Reason also assertion. 3. Option (4) is correct. never cross each other. If Explanation: Electric lines of force we get two directions they cross cach other, then at that point, the assertion is false. So, not possible. is of electric field, which a vector sum of the The resultant electric field at a point reason is true. electric fields at that point. So, the

4. Option (2) is correct. Explanation: If a be the surface charge density of the two spheres of radius r and R, then clectric fields for the two spheres are respectivcly: = K4n

K4nR's R

= K4nG

So, electric field intensities are cqual. The assertion is true. = Surface charge density is charge per unit area Total chargelarea. So, reason is also true. But the reason does not explain the assertion.

5. Option (1) is correct. Explanation: The charge enclosed by the Gaussian surface is surrounding the cavity is zero. Hence, the electric field also a reside in conductor zero. So, the assertion is true. Charges no charge. So, the only at its surface. So, in cavity there is reason is also true and properly explains the assertion. 6. Option (2) is correct. Explanation: The total electric flux out of

a

closed surface

equal to the charge enclosed divided by the permittivity, according to Gauss law. Inverse square dependency of the electric field on distance is the basis of Gauss theorem. Therefore, both assertion and reason are correct but reason is not correct explanation of assertion. 7. Option (1) is correct. Explanation: The direction of electric field is from equipotential surface with high electric potential energy to low electric potential energy. So, the negative charge gets attracted is

towards the equipotential surface with high electrostatic energy because of strong Coulomb attraction and hence, moves in direction opposite to the electric field. 8. Option (1) is correct. Explanation: E= dV dr The electric field is always perpendicular to the equipotential surface. Therefore, the assertion is true. Negative gradient of electric potential is electric field. So, direction of electric field must be in the direction of the decreasing order of electric potential. Therefore, the reason is also true and is the correct explanation of assertion too. 9. Option (3) is correct. Explanation: Work done in moving a charge between any two points in a uniformelectric field =charge x potential difference. So, it is independent of the path followed by the charge. Hence, the assertion is true. Electrostatic forces are conservative type, Hence, the reason is false. 10. Option (3) is correct. Explanation: Electric field for parallel plate capacitor in vacuum =E= s/e) Electric field in dielectric E'= c/Eo Since, the value of K for Kerosene oil is greater than 1, then E'R= Scan to know more about this topic

9

At

Instantaneous current I= lim

Aq

Electric current

dq

dt Drift velocity: > Drift velocity is an average velocity which is obtained by certain particles like electrons due to the presence of an electric

P/

A

Where, p=resistivity of the material l= length of the wire A = Area of cross-section of the wire > The SI unit of resistance is ohm (2). >p is called resistivity of the material which depend on the nature of the material of theconductor but not on its dimensions.

I-V characteristics: > Ohm's law and their I-V characteristic curve are straight lines passing through the origin.

field.

> Drift velocity, v,

Where, relaxation time, Here, e = charge

).= mean free path

eV

eE

=-t=

Current

mL

R, When electric current is set up in a conductor, electrons drift through the conductor with velocity given by

v

vneA

or I=neAv,

Where. I = Electric current through the conductor, n= Number of free electrons per unit volume A= Area of crosS-section of the wire e= Charge of an electron

> Semiconductors, p-n junction diodes do not obey Ohm's law and their I–V curves are non-linear. urrent(mA)

Mobility: > Mobility is the drift velocity of an electron when applied electric field in unity.

lo

E

(uA)

etEIm Current

E

Ohm's law:

Reverse through a

conductor is directly proportional the Current / V potential difference applicd across the ends of the conductor provided the physical conditions such as temperature, hechanical strain, etc. remain unchanged.

V= IR

Or, Where,

Voltage (V

R

is a constant called resistance

of

the conductor.

breakdown 3 Electric energy and power: > Electrical energy due to conduction of charged particles in a conductor causing electric current () is

E=Vxlxt= I' xRxt =t

R

Oswaal CUET (UG) Chapterwise Question Bank PHYSICS

16 resistance

sB

W

to

E=

E-

when

potential

,

E,+En,i=

cell

Ez

externally.

the

terminals

Vy-

between

not

= emf

the

Internal

V,=

the

difference

equal

ir

connected

r=

E+

of

=

entering

to

equal currents

junction,

Level currents

+ E

thejunction.

is junction

E,

i,+

Third

of

any

i Resistane

thetheleaving

i,+i,=i,+i,

=

are

is the It

sum

R=rtr;

R+

the

+

45

they

sum

At

Map

of

grouping

Level

Mind

grouping

EMF

Second

along

eries uternal maintains

zero.

sum potential a

Rule

the

is 0

algebraic

1unction

loop =

parallel

Se

difference

terminals.

difference

two

closed

of

Level

The all

Loop

Cellb

which

Trace

2E

the

Rule

of

First

Kirchhoffs

Grouping

device potential

Rules

its between

a

cell

is It

Electric

a

Electricity Current balanced

G

R

Wheatstone

through

bridge

Bridge

for

current

wheatstone

R

Condition

0

>

Ig

R

Current

No i.e.,

7j)I

(TElectric Temperature

7,

T

(T- at

at

temperature

temperature

of

Temperature

Resistance

coefficient

Resistance

a

DependenceResistance

7))

resistance

a

+

+

R[l p=[1 Resistors

density

charge

Conductivity

R=

of

=

R=

a=

R

velocity Resistivity

area, Current

of

Mobility

flow

(ds

an

of

Law

Grouping combination

Drift

t

Ohm's

through

of

i=

Rate

resistors

K+R,

IR

or

parallel

V=

1

two

For Area flow

time

J between

section

of

neA

combination

JE o=

cross=Angle

As

and

1

+Rn collision

-| AS=Area

p=am

e

current g=Conductivity

cose

In

--Ry=R,+R+R,+

resistors,

T=Average

vector

R+R series

0

In

two

For

Ry=

Rey

CuRRENT

EucTRICIIY

Where,

17

I m electrical energy P

Potential difference scross the ende of the conductor Curent throuph the conductor

is brawn fus ditferenc is leos thAn the end. When

I RResistance of the wire

> The SI unit of eneryy is joule (U). > Power is the work done per unit

be time.

cinet

comes in esies

resistare

R,tten tte

A

tero

t,

&

yferta

ezrfa

curtet in the crast

will be

I,p+nr P-t'R > The unit of power is joule per second (J/s) or watt (W). Electrical resistivity and conductivity: Resistivity is the resistance of a conductor having unit Jength and unit area of cross-scction, m

ner > Conductivity is the reciprocal resistivity. of

> Parales conbination of th cels are connectd in parallel resistance R, then current throvs

m

>

Wn acnn a

te eitze npves by

m cd

fm cells having ínternal reistancn hhho

are cuneced in pralled acxss E,E,,E,,..„, rcpetively the current

resistance R, then the given by

throuyh

exterral revitaen

ne't

Gr

m

Temperture dependence of resistance: > With small change in temperature, resistivity varies with temperature as: R= R,(1+ ast) Wherc, a temperature cocfficient of resistance. The temperature cocfficient for conductors is positive i.e., resistance increases as the temperature rises. > The temperature cocfficient for the insulators and semiconductors negative i.c., their resistance decreases as the temperature increases, Internal resistance of a cell: > Cell is a devicc that maintains the potential difference between the two electrodes as a result of chemical reaction. > Internal resistance is the resistance of clectrolyte which resists the flow of current when connected to a circuit.

Kirehhoffs Rule:

Sean

t kam

> First ule The alycbraic um of currert at a junction is zero ie., EI- 0.Thís inplies ths tupe that the total current entering a junction is equal to the total curent Seaving the junction, Thís follows law of conervtisn of charge. Kirchbs > Second rule- Around any coed loop in as a circuit, sum of the emfs and the potentíal differences across all elements ís zero, ie., ZV= 0 or ZV-ZIR. This follows law of conservation of enery. Wheatstone bridge: > It is a circuít having four resístances P, Q, R galvanomcer (G) and a battery connected as shown.

nd

S, 2

Potential diference and emf of a cell: The emf and terminal potential difference ofa cell- Let the emfof a cell is E' and its internal resistance is r. If an external resistance R is connected across the cell through a key, then IR R. potential difference across the external resistance This across cell, the is equal to the termninal potential difference

R

E=V+Ir

> Balanced conditson: PIO RIS Meter bridge: > It measures unknown resistance. It works on the principle of balanced condition ofWheatstone bridge. > The rcsistance of the unknown wire can be calculated from:

100-/

Oswaal CUET (UG) Chapterwise Question Bank PHYSICS

18

Resistancc Box

cad

100-/1

D

2

BatteryBE

lumlu

-Metre Scale R Variable K Resistance

G

K1 K

Potentiometer: > The potentiometer is basically a long piece of uniform wire

across which a standard cell is connected. > Principle- The basic principle of the potentiometer is that the potential drop across any section of the Scan to know more about wire will be directly proportional to the a this topic length of the wire, provided the wire has uniform cross-sectional area and a uniform current flows through the wire. > Comparison of emf of two cells

>Measuremcnt of internal resistance of a cell: RB B

Battery

N,

EMF R

LVariable

Resistance

r=l

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS

The emfs and resistances in the given circuit have the

1. 2

following values

3. The S.I. unit of electron mobility is (3) Vs/m? (2) m²/Vs (1) m/Vs

(4) Vsm?

(CUET 2023, 26h May) S50 division deflection in a A current 10-7 A produces of 4. galvanometer. The figure of merit would be (2) 50 10'A/diy (1) 50x10A/div x

E

Ww R E

=4.2

V,

and

E,= 1.9 V

r=2.0 2, r,=

1.6

2,

2x 10A/div

(4) 2×10*A/div (CUET 2023, 26th May) S. In the given circuit potential difference between point A

(3)

R=6.0 2

potentiometer is supplied with a constant voltage ofs

2S2 W

39

V.

potentiometer wire will be (2)

V

A

A cell of emf 1.4 V is balanced by the voltage drop across 280 cm of the potentiometer wire. The total length of the

(1) 5m (3) 10 m

12

12

(CUET 2023, 26th May)

A

(Va -V)=

42

What is the current in the circuit? (2) 635 mA (1) 383 mA (4) 958 mA (3) 240 mA 2.

B

4m

(4) 8 m

(CUET 2023, 26th May)

(1) (3)

v v 3

B

32 (2)

(4)

(CUET 2022, 5th August)

19

CURRENT ELECTRICITY

6.

Current through the galvanometer in the circuit given below is B

1002

209

(2) 0.125 x 10!8

(1) 1.25 x 10!8 (3) 1.25 x 1019 13.

(4) 12.5x 1019

,

(CUET 2022, 8h August) and respective internal Two cells of emf's E, and are in parallel as shown in r, connected r, resistanccs and figure. The effective emf will be

D

(2)

(1) (1) 6.13 mA (3) 6.69 mA

(2) 33.9 mA (4) O mA (CUET 2022, 5th August) 7. An ac source of emf t) = Vsinot is put across a pure capacitor. The value of angular frequency of instantaneous power is

(1) 0

l

(2) o

(3)

20

of emf

R

(1)

15.

16.

2 S2 (3) 2A, 2A

1iE,

(CUET 2022, 23rd August) and internal resistance r are connected with external resistor R as shown in fig. The value of current (/) flowing in the circuit will be

14. Three cells, each

(1) (3)

(2) 1A, 2A

E,h

(4)

(CUET 2022, 5th August) 8. In a meter bridge, null point is found at a distance of 20 cm form the end A, then the resistance of 1092 is replaced by another resistance of 2002 the null now is at (1) 20cm (2) 30cm (4) 40cm (3) 15cm (CUET 2022, 6h August) 9. Changing current through 1V cell and through 29 resistor respectively is

(1) 2A, 1A

(4)

(3)

%+t

(4) 1A, IA

(CUET 2022, 6th August) mobility of charge carriers increases with (1) Increase in average collision time interval (2) Increase in the electric field (3) Increase in the mass of the charge carriers (4) Decrease in the charge of the mobile carriers (CUET 2022, 6h August) x 10° m/s in a conductor 11. The drift velocity of electrons is 2 x 1028 per m². If in which the number of free electrons is 6 x m, then the the cross section of the conductor is 5 10 current flowing through the conductor is (Given e= 1.6 x 10-1 coulomb) (2) 48 Amp (1) 96 Amp (4) 4.8 Amp (3) 9.6 Amp (CUET 2022, 8'h August) resistance of 0.5 S is emnf 2 internal V and 12. A battery of connected across a resistance of 9.5 SQ. The number of electrons passing through the resistance in l second,

(1)

3e

r+R R

10. The

2

(3)

(4) Zero R+3r (CUET 2022, 23rd August) When a cell is connected across external resistance 5 2, a current of 0.25 A flows through the circuit. If the external resistance is replaced by 2 S2, a current of 0.5 A flows through it. The emf of the cell in the circuit is 0.75V (2) 1 V 1.2s5 V (4) 1.5 V (CUET 2022, 23rd August) Two identical cells of emf and internal resistancerwhen connected in series or in parallel across external resistance Rgive the same value of current. Then the relation between r and R is

R+3r

(2)

Rr

(2)

r+2R 3R 2

17.

(1) (2) (3) (4)

18.

(CUET 2022, 23rd August) Which of the following characteristics of electron determines the current in a conductor? Drift velocity alone Thermal velocity alone Both drift and thermal velocity Neither drift nor thermal velocity (CUET 2022, 30*h August) AB is a potentiometer wire. If the value ofR is increased then in which direction will the balance point shift?

Oswaal CUET (UG) Chapterwise Question Bank PHYSICe

20

(2) The battery

(1) Towards A

R

(2) 0.564 mm (4) 1.128 x 10 m

(CUET 2022, 30h August) is correct? statements one following the 20. Which of measure current to the is ina circuit. used (1) Potentiometer measure resistance of to internal is the (2) Potentiometer used a cell. (3) Potentiometer is used to measure the resistance ofa circuit. (4) Potentiometer is used to measure the potential difference across a resistor. 21. Consider a current carrying wire (current ) in the shape of a circle. Note that as the current progresses along the wire, an exact the direction ofj (current density) changes in manner, while the current I remain unaffected. The agent that is essentially responsible for is (1) source of emf. on the (2) electric field produced by charges accumulated surface of wire. segment of wire which a (3) the charges just behind given way by repulsion. push them just the right Page No. 16] (4) the charges ahead. [NCERT Exemp. Q.3.1, &, (&, > &;) and internal and 22. Two batteries of emf are connected in parallel resistances and r, respectively as shown in fig.

,

r

resistance will be opposite faces. The acrose battery is connected (1) Maximum when the faces. cmx l cm 2 across 10 cmnx1m battery is connected (2) Maximum when the faces. (3) Maximum

when

the

battery

is

connected acrose

1

xcm faces.

10 cm

irrespective of the three faces. [NCERT Exemp. Q.3.5, Page No. 171 characteristics of electrons 26. Which of the following a conductor? determines the current in

(4)

(1) (2) (3) (4)

27. (1) (2) (3)

(4)

B

Samne

Drift velocity alone. Thermal velocity alone. Both drift velocity and thermal velocity.h Neither drift nor thermal velocity. [NCERT Exemp. Q.3.6, Page No. 17 Kirchoff's junction rule is a reflection of Conservation of current density vector. Conservation of charge. a particle The fact that the momentum with which charged as a vector) the (as a unchanged is approaches junction charged particle leaves the junction. a The fact that there is accumulation ofcharges at junction. [NCERT Exemp. Q.3.7, Page No. 18] stands for a

28. Consider a simple circuit shown in fig. (1) The equivalent emf Epg Of the two cells is between

1s1

across the wire potential and adjusted so that the slightly cxcceds 10V. cm of wire itself should have a (3) The first portion of 50 potential drop of 10V. resistanoes used for comparing (4) Potentiometer is usually No. 1 Page Exemp. Q.3.4, cree and not voltages. [NCERT a rectangular 10 cm and length rod metal of 25. A to a battery acrose section of lcm x-cm is connected

(2) Towards B (3) No change (4) Will remain fixcd at the mid of wirc AB (CUET 2022, 30"August) a 19. Nichrome wire of lcngth 10 m is uscd to make hcating consumes power of 160 W when there is a coil. This coil across it. Find the diameter of potential difference of 40 V of Nichrome is this wire. Resistivity 10Nm. (1) 1.128 mm (3) 0.10 mm

a of potentiomcter can havedrop voltage of

and

variable resistance R'. R' can vary from Ro to infinity. r is internal resistance of the battery (rx

FNe

25 x


Electromagnetic induction is the rxexs of gencrating the electric current with a changing magnetic field. If the

magnetic ficid is changing. the changing NBA cos9, magnctic fux will be where, is the angle between the magnetic feld and normal to the planc.

Electromagnetic

Induction: Faraday's Laws and Lenz Law

Faraday's law, induced emf and current: > Faraday'% First Law: Whenever a conductor is placed in a as varying magnctic field, an emf is induced which is known induced etmf and if the conductor circuit is closed, current is also inducod which is called induced current. > Faraday's Second Law: The induced emf is cqual to the rate of change of magnctic flux linkage where flux linkage is the product of the number of turns in the coil and flux associated with the coil. Induced cmf

=

E=-where dt

,

èa is the magnetic flux

-fBdA through the circuit and is represented as a area in circuit and With N loops of similar being the fux through cach loop. induced emf is

g

c=N

Joduced current > Whena magnet is moved towardsa loop of wire connected 10 zn ammetcr, thc ammeter shows current induced in the loop. > Whena magnct is held stationary, there is no induced current in the loop. > When a magnet is moved away frorm the loop, the opposite current is induccd in the loop.

Ln's Law: > The irection of an induccd emf always opposes the change in the

magnctic flux which causes the cmf.

> It explains the ncgative sign in Faraday's

ule,

t

m

do,

a showing that the polarity of induced emf tends to produce flux. cause, in opposes magnetic i.e,, change the cuTent that As per conservation of energy, induced emf opposes its cause, making mechanical work to continue with the process which gets convertod into elecrical energy.

Eddy current: Eddy currents arc loops of electric current induced within conductors by a changing magnetic ficld in the conductor according to Faraday's laws of electromagnetic induction or by the relative nvaion ofa conductor in a magnetic field. > The magnitude of Pddy current is O proportional to the strength of the magnetic field,

proportional to the arca of the loop. O proportional to the rate of chaige of fux, o inverscly proportional to the resistivity of the material. Applications of eddy current: o Electromagnctic damping o Induction furnace o Domestic elcctric meter o Non-destructive testing Eddy current also produces undesirable heat and oss of power. This may be reduccd by larninating thc metal core. Mutual and Self Inductance: Scan t kaow > Mutual inductance: Mutual inductance more about this topic is the phenomenon of production of induced emf in a circuit, when the current in the neighbouring circuit changes. > The mutual induction between two coils depends on: Metual Indactance o The nurnber of turns of primary and secondary coils. o The shape, size or geomctry of the two coils, ie., the area of cross-section and the length of the coils. > Coefficient of mutual induction: For instantancous current / in the primary coil, if the magnetic flux linked with the secondary coil be then O

...) where M is the constant of proportionality. It is called cocfficient of mutual induction. The induced emfe in the secondary coil is given by

do dt

...(iü)

dt

The negative sign is in accordance with the Lenz's law, ie., the induced emf in the sccondary coil opposes the variation of current in the primary coil. Taking magnitude of induced emf from Eq. (ü).

MTall dt) Unit of M

=

VAs

If n,n, be the nunnber of turns per unit length in prinary and secondary coils and r be their radi, then the coefficient of mutual induction is given as MHori

Self-Induction:

Self inductance is the phenomenon of protuction ofinduced emf in a circuit, when the current in the sane circuit changes. For instantancous current l in the coil, if the magnetíc flux , linked is then: « lor

LI

(eemSO

130 (9n)

aswrde

Uonsn SISAH

ot sin

length

within solenoid

o

in

by

or

=NBA

unit

induced magnetic

of

per a.c.generator,

magnetc

Faraday's

field

induction

a

the

loop turns currentchanging

in

conductor

to

according electromagnetic

each

7

(Hg:)

of

n'

|

NN,nn

=

Clerk

of

James

Current

dt

of

relative

Loops

the

Michael

=Mi

M=

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by

conductors

L= n=

in induced

of

a

mathematically

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a

electric

r=Radius

=Flux

electromagnetic

an

NNnh

of

Number

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field

and Emf

Eddy

it.

discovered

M,

solenoid

described

inductance

inductionMaxwell

1831,

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long

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induction

of

Mutual

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omagElectronetic

induced

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1.Whenever

area

dt

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Mind

E=-NBds

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the

each

a

is = emf

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is

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turns,

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the

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coil,

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41

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ota

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thrgh

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42

Oswaal CCUET (UG)

where L is the constant of proportionality. It is called the cocfficient of self-induction. The induced emf in the coil is given by

d. --

Average power

Chapterwise Question Bank

=V=[sin

=P

PHYSICs

(20or)] =0

[Since average of sin2ot over a complete cycle is zerol +Vn

dt dt The negative sign is in accordance with the Lenz's law, i.., the induced emf opposes the variation of current in the coil. Taking the magnitude of the induced emf from Eq. (i),

+I 180°

90°:

360

L=s/(dldt) The self inductance of a cicular coil is given by:

-V

Waveform of pure inductive circuít

The self inductance of a solenoid of length / is given by,

=HrAl

=

p,°V

901

Here, n =NIl= Number of turnsper unit length and V=Al =Volume of the coil. Note: If two coils of inductance L and L, are coupled togeth er, then their mutual inductance is given by

M=kJ44 where, k is called the coupling constant. The value of k lies between 0 and 1. For perfectlycoupled coils, k= 1, it means that the magnetic flux of primary coil is completely linked with the secondary

Phasor diagram of pure inductive circuit Inductive reactance (X)is the resistance offered by inductor.X, = oL = 2nfL >AC applied to pure capacitive circuit:

an

coil.

Alternating currents, peak and RMS value of alternating current/voltage: Alternating current: > Altenating curent represented by sine curve or cosine curve as I = Insin ot or I= Locos ot where, Io is peak value of current and Iis instantaneous value of current. > Frequencyf, is defined as the number of cycles completed per second. Unit: Hertz (Hz). In India, the frequency of a.c. is

V=Vasin ot

V= V,sin

ot

I=1, sin

[which shows current leads the voltage by)

2

50 Hz.

Peak and rms value of alternating current/voltage: > ms value or effective value or virtual value of a.c. is represented as I,m Lef or I;

of*

Average power

=P,=sin 2

(20or) =0 (Since, average

of sin 20t over a complete cycle is zero]

=0.7071,

I.n V2

> rms voltage value is represented as:

90°

0

180°

360°

ot

,

and Vo are the peak value of altermating curent and voltage respectively.

Reactance and impedance: > AC voltage applied to pure inductive circuit:

V=Vnsin ot 90°

V= Vsin ot

Capacitive reactance capacitor [which shows current lags the voltage byJ

2

(X)

is the resistance offered by a

oC

2rC

EECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

43

Impedance: > Impedance is the comprchensive expression of all forms of opposition to clectron flow, including resistance and reactance, where an alternating current after passing through it produces a voltage drop which will be out of phase betwecn 0 and 90 with current given as,

L

Z=VR+x? Impedance of circuit, R = Resistance, X =

where, Z

Rcactance Unit of reactance and impedance is Ohm. LC oscillations (qualitative treatment only): > An LC circuit, also called a resonant circuit, tank circuit, or funed circuit, is an clectrical circuit consisting of an inductor and a capacitor.

> LC circuits are used either for generating signals at a particular frequency, or picking out a signal at a particular frequency from a more complex signal. > In an LC circuit there is no dissipation of energy due to resistance. > An LC circuit, oscillatíng at its natural resonant frequency can store electrical energy. >If an inductor is connected across a charged capacitor, the voltage across the capacitor drives a current through the inductor, building up a magnetic field around it. The voltage across the capacitor falls to zero as the charge is used up by the current flow. At this point, the energy stored in the magnetic feld of the inductor induces a voltage across the inductor. This induced voltage causes a current to begin to recharge the capacitor. Thus, the energy required to charge the capacitor is extracted from the magnetic field. When the magnetic field is

C

u) LCR series circuit, resonance V=V, sin

ot

completely dissipated the current stops and the charge again gets stored in the capacitor. The cycle repeats.

C

> In an LCR circuit: > Inductive Reactance =X= oL E

C

Capacitive Reactance = =

X=X-X¢ Impedance

oL-

X

=

1

Scan to know more about this topic

oC

1

oC

=Z =/R² +X?

LC circuit

I.TR+X,- X¢) Ifo is the angle between voltage and current then,E X -X¢ tano =

B

R

is positive, when X>Xc.

AM

C

is negative, when X LCR circuit:

44

Oswaal CUET (UG) Chapterwise Question

Bari

Resonance: > When current in a series LCR becomes maximum then the circuit is called as series resonant circuit. > The necessary condition for resonance in LCR series circuit is: Ve= V Xçand Z= R which gives

S N

X-

1

o-orf= LC 2rLC

4

ctelss

D

B,

R

R Z This is the maximum current at resonance. > The sharpness of tuning at resonance is measured by factor or quality factor of the circuit given as

R

ABCD -Armature

Wattless Current: > If the instantaneous values of the voltage and current in an ac circuit are given by

V=V,sinot

I=1,sin(ot-¢) where is the phase difference between voltage and the current. Then, the instantaneous power

P, =Vxl =VI,sinot -sin (ot-¢) or average power

cos

P

2

SN- Permanent magnet Rj, R, x Slip ríngo B1, B:- Carbon bushOts R.

- External cireuit

where, cos

X

voltages. It is based on the phenomenon of mutual induction. It comprises of two sets of coils which are insulated on one another on soft-iron core. > One coil is called primary (input coil) having Np turns while other coil is secondary (output coil) having N, turns, so we

Effective power Apparent power

> The value of power factor varies from 0 to 1. > The current in an 4C circuit said to be Wattless Current when the average power consumed in such circuit corresponds to zero. This happens in case of pure inductive and capacitive circuits.

AC generator and transformer: AC Generator: > An AC generator is an electrical machine which converts mechanical energy into alternating electrical energy. >It works on the principle of electromagnetic induction where coil rotates in uniform magnetic field and sets an induced emf givenas:

this tople

Transformers

sht oad

> Transformation Ratio:

N

N,

,

is defned as the transformation ratio.

> In step-up transformer N,

>

N,,V, >V, and 1, > Step-down transformer: N, < N, this,

V,

Energy losses in transformers are due to: 1. Flux leakage 2. Resistance of windings

3. Eddy currents 4. Hysteresis

> Efficiency

of

transformer=1=

I,

>,

45

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

OBJECTIVE TYPE QUESTIONS TAJ 1.

MULTIPLE CHOICE QUESTIONS

The instantaneous magnetic flux linked with a coil is given by =(15t -100² + 40) Wb. The emf induced in

the coil at time t =1 sis (1) + 150 V (2) 155 V

-

– 150V (4) +

(3)

a Current in an alternating circuit is ahead of voltage by

6.

155 V

(CUET 2023, 7th June) 2. A transformer has 4000 turns in primary and 100 turns in secondary coils. The primary coil is connected to a 240 V supply and secondary to a bulb of 1.2 Q resistance. The current in the bulb is (1) 0.50A (3) 0.20 A (2) 0.25 A (4) 5.00 A (CUET 2023, 7th June) 3. For an LCR circuit 14

D

3

(2) LC circuit (4) L circuit

(1) C circuit (3) RC circuit 7.

(1) (2) (3) (4) 8.

(CUET 2022, 5h August) copper, up is placed on a table top. A circular coil made of Magnetic field is normally outwards. Now, the magnetic field is gradually decreased. The direction of induced current as seen from the top side is Clock wise Anti clock wise Curent will not be induced (CUET 2022, 5th August) Information is insufficient a inductance of 0.005 H. The mutual Two coils have current changes in the first coil according to the equation .

, I=1,sinaot where I, =10 A and o=100rad /s The maximum value of emf in the second coil is: 6r (4) 12r (2) (1) 2. 5(3) (CUET 2023, 5h August) 9. An ac source of emf V() = Vo sinot is put across a pure capacitor. The value of angular frequency of instantaneous

A

B

power is:

() Cand D represent R and Z, respectively. (i) A, B, C, D, represent R, Xo, X, and Z, respectively. (iii) Cand D represent Z and R, respectively. (iv) A andB represent R and X;, respectively. (v) For o = o', the phase difference between current and voltage is zero. (R, Z, X, X, 0, o' have their usual meaning) Choose the correct answer from the options given below: (2) (ii), (iv) only (1) (), (i) only (4) (iii), (v) only (3) (ii), (v) only (CUET 2023, 7th June) 4. The current flowing through an inductor of self inductance Lis continuously increasing. The correct graph, showing

di the variation of induced emf(e) versus

(2)

(1) 0

o

(4)

2

(CUET 2022, 5th August) a a solenoid, a current across 10. When 100 V dc is applied 1 a V, it. A 100 flows When 50 Hz ac is applied in of across the same solenoid, the current drops to 0.5 A. The inductance of the solenoid is (4) 0.93H (1) 0.55H (3) 1 H (2) 0.86 H 11. Faraday's law in integral form is (The symbols have their usual meaning)

-do

(1) Ed (3)

dt

[Bds =0

e L

(1)

(3) 2o

12. Consider two inductor circuits. Assume that the current is increasing with time equally in both the circuits.

dt

RC

.The alternating circuít is

of

phase difference

= 10 mH

L

= 20 mH

(2)

The correct graph for energy stored in inductor with current is

dl/dt |20 mH

e

(3)

(2)

(1)

10 mH

4)

U

U 4.

20 mH

d/dt (CUET 2023, 7tà June) 5. Two solenoids of equal number of turns having their lengths and radii in the same ratio 1:3. The ratio of their self inductances Lj:Ly will be (4) 9:1 (3) 1:9 1:27 (1) (2) 1:3

(3)

10

mHs

10 mH

(4)

-20

mH

(CUET 2023, 6 August)

46

sOswaal CUET (UG) Chapterwise

inductance 1000 mH is connected in series with resistance, a variable capacitance and an AC source of frequency 2.0 kHz. What value of the capacitance result in drawing the maximum current from the circuit? (2) 63 nF (3) 63 mF (4) 63 F (1) 63 uF 14. In a solenoid, if number of turns per unit length is doubled, then self- inductance will become (1) Half of its initial value (2) Double of its initial value 13. An inductor

of

interpretation

of

of induced

current is given in The Correct figure.

N

its initial value (4) 4 times of its initial value

15. The current in a coil falls from 5.0 to 0.0 A in 0.1 s. If average emf of 200 V is induced, the value of self inductance of coil is (1) 2 H (3) 3 H (4) 1 H (2) 4 H 16. A bulb and an iron core inductor are connected to an AC source through key as shown in figure.

S

-N

(2)

1

times

BankPHYSICS P

19. Given figures show a plane coil moving with v with respect to N-pole of a bar vekity magnet.

a

(3)

Question

(3)

S

(4)

(CUET 2023, 23d August) 20. A coil of 200 turns, area 0.20 is rotated in a unifor magnetic field of 0.4 G perpendicular to the axis of the coi at the rate of 7 rps. The maximum emf induced in the coil m

The bulb glows with certain brightness. Now iron rod is taken out of the inductor. Then the brightness of bulb. (1) Increases (2) Decreases (3) Is unchanged (4) First increases then decreases (CUET 2022, 23rd August) 17.Match List-I with List-II

List-I5

(Physical quantity) (A) Self-Inductance (B) Magnetic flux (C) Impedance (D) Induced emf

SList-II

(SI unit)

()

Weber

()

Volt

(I) Henry IV) Ohm

Choose the correct answer from the options given below: (1) (A)-(III), (B)-), (C)-(D, (D)-(IV) (2) (A)- (III), (B) (), ()-(I), (D) -(I) (3) (A)-), (B)-(), (C)- (I), (D)- () (4) (A)-), (B)- (II), (C)- (ITD, (D)-(V (CUET 2022, 23rd August)

-

18. A 50Qresistance and an inductance of-H are connected 31

ín series with power supply of 220 volt AC of 50 Hz. Choose the correct statement.

(1) Current leads the potential difference by tan (2) Potential difference leads the current by 90°

(3) CurTent leads the potential difference by tan (4) Potential difference leads the current by tan

is (1) 3.52 x 10-2 y (3) 6.24 x 10-2 V

(2) 4.28 x 10-2 y (4) 7.04 >x 10-2y (CUET 2022, 26" August)

21. An LCR series circuit containing resistance Rj, inductance C gives resonance at the same L and capacitance a frequencyfas second similar combination R, L, and Ch. If the two circuits are connected in series, then the frequency of the combined circuit is (1)

2f

(2)

4f

(3)f

(CUET 2022, 30th August) 22. Which amongst the following is not responsible for energy loss in transformer due to heat? (1) Flux leakage (2) Resistance of the windings (3) Eddy currents (4) Hysteresis (CUET 2022, 30 August) 23. In ideal transformer back emf (e,) in primary coil is equal to the applied voltage (V) across primary coil. If this were (1) (2) (3) (4)

not so then. Primary current would be zero Secondary current would be zero Primary current would be infinite Secondary current would be infinite

(CUET2022, 30th August) 24. Parameter that remains unchanged in a transformer is (1) voltage. (2) current. (3) frequency. (4) none of these. 25. There are two coils A and B as shown in figure. A curent starts flowing in B as shown, when A is moved towards A B and stops when A stops moving. The current in counter clockwise. B is kept stationary when A moves. can infer that 1S

e

bAB

47

HECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

there is a constant current in the clockwise direction in A. (2) there is a varying current in A. 3) therc is no current in A. (4) there is a constant current in the counter clockwise direction in 4. [NCERT Exemp. Q.6.4, Page No. 34] 26. The sclf-inductance L of a solenoid of length / and area of cross-section A, with a fixcd number of turns Nincreases as (1) / and A increasc. (2) I decreascs and A incrcases. decreases, (3) /increases and (4) both l and A dccrease. [NCERT Exemp. Q.6.6, Page No. 34] 27. When voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220 V. This mcans (1) Input voltage cannot be AC voltage, but a DC voltage. (2) Maximum input voltage is 220 V. (3) The meter reads not v but (v) and is calibrated to read

)

4

The pointer of the meter is stuck by some mechanical [NCERT Exemp. Q.7.3, Page No. 41] defect. 28. Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication? (1) R=20 S2, L = 1.5 H, C= 35 uF (2) R= 25 2, L=2.5 H, C=45 uF (3) R= 15 2, L= 3.5 H, C= 30 uF (4) R=25 2, L= 1.5 H, C= 45 uF 29. The rms current in a circuit connected to S50 Hz ac source

(4)

The value of the current the circuit after the instant the current is zero, is is 15 A.

15

(3)

15

A

(2)

15/2A

(4)

8A

30. Which of the following is not a part of an A.C. generator? (2) Carbon brush (4) Split ring commutator [Indigenous(since there was no question from generator)]

(1) Armature (3) Slip ring

BJ ASSERTION REASON QUESTIONS

Question Nos. 1 to 10 consist of two statements-Assertion and Reason. Answer these questions by selecting the appropriate option given below: 1. Both assertion (A) and reason (R) are true, and reason (R) is the correct explanation of assertion (A). 2. Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A). 3. Assertion (A) is true, but reason (R) is false. 4. Assertion (A) is false, but reason (R) is truc. 1. Assertion (A): Lenz's law does not violate the principle of conservation of energy. Reason (R): Induced e.m.f. never opposes the change in the magnetic flux that causes the e.m.f. 2. Assertion (A): Ifhe number of turns ofa coil is increased, it becomes more difficult to push a bar magnet towards the coil. Reason (R): The difficulty faced is according to the Lenz's law.

3.

a Assertion (A): When the number of turns in coil is doubled, the cocfficient of self-inductance of the coil

becomes four times. Reason (R): Coefficient of self-inductance is proportional to the square of number of turns. as a 4. Assertion (A): A step-up transformer cannot be uscd transformer. step-down Reason (R): A transformer works only in one direction. a.c. to 5. Assertion (A): Capacitor blocks dc. and allows pass.

Reason (R): Capacitive reactance is inverscly proportional to frequency. 6. Assertion (A): VRMS value of an alternating voltage V =4N2sin3 14: is 4 volt.

Reason (R): Peak value of the alternating voltage V= 4/2 sin 3141 is 4/2 volt. resonance, the current becomes 7. Assertion (A): At minimum in a series LCR circuit. Reason (R): At resonance, voltage and current are phase in a series LCR circuit. 8. Assertion (A): Quality factor of a series LCR circuit is

increases in a Reason (R): As bandwidth decreases, resonant LCR circuit. 9. Assertion (A): A transformer does not work on d.c. Reason (R): d.c. neither changes direction nor magnitude. 10. Assertion (A): When two identical loops of copper and aluminium are rotated with the same speed in the same magnetic field, the induced e.m.f. will be the same. Reason (R): Resistance of the two loops are equal. CI COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5. Bottle Dynamo: A bottle dynamo is a small generator to generate electricity to power the bicycle light. It is not a dynamo. Dynamo generates d.c. but a bottle dynamo generates a.c. Newer models are now available with a rectifier.

The available d.c. can power the light and small electronic gadgets. This is also known as sidewall generator since, it operates using a roller placed on the sidewall of bicycle tyre. When the bicycle is in motion, the dynamo roller is engaged and clectricity is generated as the tyre spins the roller. When engaged, a dynamo requires the bicycle rider to exert more effort to maintain a given speed than would otherwise be necessary when the dynamo is not present or disengaged. Bottle dynamos can be completely disengaged during day when cycle light is not in use. In wet conditions, the dhtime roller on a bottle dynamo can slip against the surface of the tyre, which interrupts the electricity generated. This causes the lights to go out intermittently. 1. Why bottle dynamo is not a dynamo ? (1) It generates a.c. only (2) It generates d.c. only (3) It looks like a bottle (4) It requires no fuel to operate

48 2.

Oswaal CUET (UG) Chapterwise Question Bank

Can you recharge the battery of your mobilc phonc with the help of bottle dynamo?

(1) Yes (2) No (3) Yes, when a rcctificr is uscd (4)) Yes, when a transformer is uscd 3. Bottle gencrator gencrates clcctricity (1) when fuel is pourcd in the bottlc. (2) when cycle is in motion. (3) when it is mountcd propcrly. (4) when wind blows. 4. Bulb of bicycle light glows (1) with a.c. supply only. (2) with d.c. supply only. (3) with both a.c. and d.c. supply. (4) only when a.c. supply is rectificd. 5. Which onc of the following is not an advantages of newer model of bottle dynamo? (1) Works intermittently when its roller slips on the tyre (2) Small electronic gadgets can be charged (3) Can be casily disengaged during day time (4) Requires no fuel II. Based on following passage answer questions from 6-10. At power plant, a transformer increases the voltage of generatcd power by thousands of volts so that it can be sent of long distances through high-voltage transmission power lines. At substations, transformers lower the voltage ofincoming power to make it acceptable for high-volume delivery to ncarby cnd-users. Electricity is sent at extremely high voltage because it limits so-called line losses. Very good conductors of clectricity also offer some resistance and this resistance becomes considerable over long distances causing considerable loss. Another option of minimising loss the use of wires of super-conducting material. Super-conducting materials are capable of conducting without resistance. Still it is impractical to have superconductors at normal temperature. Transformers generate waste heat when they are in operation and oil is the coolant of choice. Flash point

PHHYSICS

very important parameter of transformer oil. Flash point of an oil is the temperature at which the oilignites spontancously. This must be as high as possible (not less than 160°C from the point of safety). at which the temperature Fire point oilflaashes is the is

a

and continuously burns. This must be very high for the chosen oil (not less than 200°C). (Please take care: Certain portions of the passage have been deletcd) Which of the following statement is true for long distance

6.

transmission of electricity? (1) Step-down transformer is used at generating station and step-up transformer is used at destination substation. (2) Step-down transformers are used at generating station and destination substation. (3) Step-up transformers are used at generating station and destination substation. (4) None of the above 7. Super-conducting transmission line has the following advantages: (1) Resistance being zero, there is no power loss. (2) There is no requirement of costly step-up and step-down transformers. (3) Cable is capable of sending more electricity. (4) All of the above 8. Why does stepping up voltages reduce power loss? (1) Since resistance of conductor decreases with increase of voltage

(2) (3) (4) 9.

Since current decreases with increase of voltage Both of the above None of the above Oil transfers heat fromn transformer winding the process by of

(1) (3) 10. (1)

convection. (2) conduction. radiation. (4) All of these Flash point of an oil is the temperature at which the oil flashes and continuously burns. (2) the temperature at which the oil ignites spontaneously. (3) the temperature at which the oil starts boiling. (4) The temperature at which the oil forms fumes.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1.

(4)

2. (4)

3.(3)

4. (1)

11. (2)

12. (2)

13. (2)

21.(3)

22. (1)

23. (3)

6. (3)

14. (4)

5.(2) 15.(2)

16. (1)

(2)8. 17. (1)18.

24. (3)

25. (4)

26. (2)

27. (3)

7.

(2)

(4) 28. (3)

|9.(3) 19.(3) 29. (1)

30. (4)

9. (1)

10. (3)

9.(1)

10. (2)

10. (1)

20. (4)

B] ASSERTION REASONQUESTIONS 1.(3)

2. (1)

|3.(1)s|

4.

(2)

5.(1)

6. (1)

7.(4)8.(1)

[C] COMPETENCY BASED QUESTIONS 1.(1)

2. (3)

3.

(2)4.

(3)

5.(1)6.

(4)

|7.(4)

8.(2)

49

EECTROMAGNETIC INDUCTION AND ALTERNATING CÜRRENTS

ANSWERS WITH EXPLANATIONS FAI MULTIPLE

CHOICE QUESTIONS

Option (4) is correct. Explanation: p=15-100/

7. Option (2) is correct. Explanation: Magnetic field decreases outwards gradually. so According to Lenz's law, current will flow anticlockwise that decrease is opposed. 8. Option (2) is correct. Explanation: emf induced in the second coil is given by

1,

+40

0r,

d =8=-dp/ dt = -(15r dt E=-(45-2001)

At

t=ls

Induced emf

-100r +40)

e=-(45x1-200×1) =155Ve Option (4) is correct. Explanation: NJN,= VJVp = V240 100/4000 Or,

V,=6V

Or,

le

Or,

lel

dt

5r

In a capacitor

I()

= 4sin

of+

= lhcosot

Instantaneous Power = V)I) =

phase.,0.S

Vo sin

ot lo cos ot

- oosin 2ot 2

e-doldt

20ie

Angular frequency = So, 10. Option (1) is correct. Explanation: When 100 V dc is applied:

d LI dt dI

L

i

R

dt

dIetos dt

100 V

It is the equation of a straight line passing through origin. 5. Option (2) is correct.

Explanation:

HNA

HN'

R=

Here,

00100Q 1

When 100 V ac is applied:

L

100

2

thgth

Or,

.

L:L,=

1)

100r |elmax =0.005 Or, lelmax So, 9. Option (3) is correct. V() = Vosinot Explanation:

resonance condition. So, voltage and current are in 4. Option (1) is correct.

= Self inductance= L

(Since, maximum value cos is x 10 x

X.

lelL

sin ot)

lel=L lo cosot L l, o

graph A represents R. Graph B represents Graph Crepresents X Graph D represents Z. At o = o', the impedance value is minimum. Hence this is the

e=-L

L

lelmax

Or,

So,

or.

=

dt

Or,

Current in the bulb = VYR=6/1.2 = SA 3. Option (3) is correct. Explanation: In LCR Circuit: o. Ris independent of

e=:

dt

a

2.

Explanation:

di

£= -L

100

1:3

V,

50

Hz3

Z=100 0.5

Option (3) is correct. Explanation: 6.

I

lc

2002

Z= J(oL)? + R²

Vc

+ R²

200 =

(2L)³

Or,

200 =

J(2rx s0L)°

Or,

200 = J(100rL) + 100²

T/3

In RC circuit current leads voltage.

at

Or,

200 =

100(rL)? +1

+100

CUET (UG) Chaçtstvia

50

Wn

tan|2012 Or,

L,0.5511ist

Or, 11.Option (2) is

Explanation:

correct,

In LR circuit potentíal differene lesds the curtene

ee

Induced cmí

Hence, the potential difference leads the currerA

s

do dt

will be anti-clockwise so that an M

This is Faraday's law in integral form. 12, Option (2) is correct. Explanation: Energy stored in an Inductor is proportional to P. So, the graphs will be upward parabolic. 13. Option (2) is correct. Explanation: Current is maximum at resonance.

Or,

oL =

the motion.

20, Option (4) is correct,

Explanaton: Given, N (n0,

A

20)

0,2 m

Magnetic field B = 04G = 047114T Frequency of oscillation =7

)

oC

Magnetic fiux,

=2f

1

14z rad's,

BA co) cos

OgBA

ot

e=-dodt e= NBAn cos ot

Induced emf, COS

Here,

(2/)°L C=

of turns)

Area

C=

Or,

D=1

e= 200 04 10*1021 1%

21. Option (3) Ís correct. Explanation: Equivalent capacitance ín series,

(2xrx2x10')? x 100x10

= 63 nP 14. Option (4) is correct. Explanation: Coefficient of self inductance, is directly proportional to square of number of turns in coil, So, when number of turns per unit length is doubled, the self- inductance becomes 4 times of its initial value. 15. Option (2) Is correct.

C=

CC

G+Cq

Equivalent ínductance in series L = L + Equivalent resistance in series R =R, +R,

Frequency of combined circuit Ocombined

Explanation: Or, Or,

Or,

200 50L

L= 4H

16.Option (1) Is correct. Explanatlon: When the iron rod is removed, the inductance of the inductor decreases. So, the total circuit resistance decreascs. Hence, the current increascs which results in the increase in the brightness of the bulb. 17.Option (2) is correct. Explanation: SI unit of self-inductance is Henry. SI unit ofmagnetic flux is Weber.

SIunit of impedance is Ohm. Sl unit of induced emf is Volt. 18.Option (4) is correct. Explanation:

g(5

19, Option (3) Ís correct. Explanatont According to Lenz's law when thee towards the N-pole of the maypet, the urert Coil will be such that it opposesthe tim,

And

At resonance

b

Or,

Wcombined =

VI,cC,

+LCC,b

Frequency of individual circuit is cqual.

. Putting in above cquation, Wcombincd

Or,

Ocombined

i tan

VCC, + LCC

G+G

V4C(C+C)

Wcombincd

tan )

eonbincd Scombined

f

51

ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS 22. Option (1) is correct. Explanation: Flux leakage is responsible for voltage drop. It is not responsiblc for energy loss due to heat. 23. Option (3) is correct. Explanation: In any coil, applied AC voltage causes current to flow which induces a back emf. Current gets limited by the

back emf.

If this were not so, there were nothing to limit the current and thus primary current would become infinite. 24. Option (3) is correct. Explanation: The voltage and current vary in a transformer but the frequency remains constant. 25.Option (4) is correct.

Explanation: When coil A moves towards coil B with constant velocity, so rate of change of magnetic flux due to coil B in coil will be constant that gives constant current in coil A in same direction as in coil B by Lenz's law. 26. Option (2) is correct. Explanation: As we know that, A

L=4 As L is

constant for

a

N°A

As u, and N are constant here so, to increase L for a coil, arca must be increased and /must be decreased. Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/ permcability of the medium. 27.Option (3) is correct. b Explanation: The voltmeter in 4C circuit reads value < > rms < and meter is calibrated to which is multiplied value

In other words, voltmeter connected to by v2 to get the AC main read root mean square value of AC voltage, i.c.,

V.

by.

correct.

-C-R circuit is given

Q-D L

For Q to be high, R should be low, L should be high and C should be low. Therefore, option (c) is most suitable. 29. Option (1) is correct. Explanation: I = I,sinaot

OT.

or.

of

opposes the change in the magnetic flux that causes the e.m.f. So, the reason is incorrcct. 2. Option (1) is correct. Explanation: As it is tricd to push a bar magnet towards a coil, magnetic flux increases. According to Faraday's law, c.m.f. is induccd. As the number of turns incrcases, induced e.m.f. increases. According to Lenz's law, induced e.m.f. always opposes the change in magnctic flux that causes the induction of e.m.f. So, the induced e.m.f. willoppose the motion of the bar magnet towards the coil. As the number of turns increases, opposition increases. Hence, both Assertion and Reason are correct. 3. Option (1) is correct.

I=x2xsin (2rf)

Iel5x/2xsin 2x1x50 600

I=15x/Zxsin

15

Lc N'

Hence, both A and R are correct. 4. Option (2) is correct. Explanation: Step-up transformer can be used as a step down transformer or vice versa. The Assertion is false. So, the transformer is not a uni-dírectional device. The Reason is false. 5. Option (1) is correct. 1

Explanation: Capacitive reactance

=

2nfC

So, as f (frequency) increases, reactance decreases. For dc, frequency = 0, hence capacitor offers infinite reactance. So, it blocks d.c. For a.c, frequency 0, hence capacitor offers low reactance and allows a.c to pass. Hence, Assertion and Reason both are correct. 6. Option (1) is correct. Explanation: Comparíng with the general expression of an alternating voltage V= Vsinot value of the given alternating voltage

aPeak

Tuning of an L-C-R circuit depends on quality factor of the circuit. Tuning will be better when quality factor of the circuit is high. As, quality factor (0) of an L-C-R circuit is given by,

or,

1. Option (3) is correct. Explanation: Lenz's law is based on the principle conservation of energy. So, the assertion is correct. Induced cmf. always

So,

A

Explanation: Quality factor (0) of an

commutator. It is a part of d.c generatof. [B] ASSERTION REASON QUESTIONS

Explanation: L= pA

coil,

L«A and

28. Option (3) is

30. Option (4) is correct. Explanation: An a.c gencrator does not have split ring

V,

=4/2 v

So, A and R both are correct. 7. Option (4) is correct. Explanation: At resonance, X=X, so the circuit impedance becomes minimum and resistive and hence the current becomes maxímum. So, the Assertion is incorrect. At resonance, X, = Xc, so the circuít impedance becomes resistive. In resistive circuit voltage and current are always in same phase. Hence, Reason is correct.

Option (1) is correct. Explanation: Quality factor of a series LCR circuít 8.

ís

.

Assertion is correct.

Quality factor is also defined as

-

Resonant frequency Bandwidth

So, as bandwídth decreases, Qincreases. So, Recason is correct.

Oswaal CUET (UG) Chapterwise Question Bank PHYSCS

52

5. Option (1) is

9. Option (1) is correct.

Explanation: Transformer has two coils, If current fluctuates

in one coil, c.m.f. is induced in the other coil. For d.c. supply current does not change, so there is no induced c.m.f., Hence both A and R are correct. 10. Option (3) is correct. Explanation: Induced e.m.f. in a rotating loop in a magnethe field depends on the area of the loop, number of turns, spccd of rotation and magnetic ficld strength. It does not depend on the material of the coil. So, when two identical loops of copper and aluminium arc rotated with the same spccd in the same magnetic field, the induccd c.m.f., will be thc same. So, the Assertion is correct. Resistance of the two loops cannot be cqual. Resistance of copper loop is less than that of the aluminium loop. So, thc Reason is incorrect.

[C] COMPETENCY BASED QUESTIONS correct. Explanation: Dynamo gencrates d.c. But bottle dynamo generates a.c. So, it is not a dynamo in that sense. But, it generates clectricity for bicycle light. 2. Option (3) is correct. Explanation: Newer models of bottle generators arc now available with a rectifier. d.c. available from such bottle generator can be used directly for charging mobile 1. Option (1) is

phone.

Otherwise with the old models, a rectifier is to be attachcd to convert a.c. to d.c. 3. Option (2) is correct. Explanation: Bottle gencrator is also known as sidewall generator since it operates using a roller placed on the sidewall of bicycle tyre. When the bicycle is in motion, the dynamo roller is engaged and clcctricity is generated as the tyre spins the roller. 4. Option (3) is correct. Explanation: Normal lamps work with both a.c. and d.c. So, bottle generators of older model or newer model can bc dircctly used for bicycle lamp.

correct.

wet conditions, the roller on a bottec dynamo new modcl) can slip against the (old model surface clectricity of the gencrated. Thiscauses tyre, which interrupts the is lights to go out intermittently. This not an ndvantage. the 6. Option (4) is correct. Eyplanation: At power plant, a step-up transformer increases powcr by thousands of volts, so the voltage of gencrated that can be sernt of long distancesthrough high-voltage transmission it

Explanation: In or

power lincs. At substations, step-down transformcrs lower the voltage it acceptable for high-volume of incoming power to make to cnd-uscrs. delivery ncarby

7. Option (4)is correct. materials arec Eyplanation: Super-conducting capable of So, conducting without resistance. this climinates the line loss and the cable is capablc of sending more clcctricity than conventional cablc. Using super-conducting cables, ont get rid of the nced of costly transformcrs. 8. Option (2) is correct. Explanation: At gencrating station, normally voltage i stepped up to around thousands of volts. Powcr losses increase with the square of current. Thercforc, keeping voltage high. current bccomes low and the loss is minimized. 9. Option (1) is correct. Explanation: Transformers gencrate waste hcat when they are in operation and oil is the coolant of choice. It transfers the heat through convection to the transformer housing. neh 10. Option (2) is corrcct. l Explanation: Flash point is a very important paramcter of transformer oil. Flashpoint ofan oil is the temperature at which the oil ignites spontancously. This must be as high as possible (not less than 160°C from the point n of safety).i tx

tat

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CHAPTER

5

ELECTROMAGNETIC WAVES field

Revision Notes Nni

Sur

ispiacemert current:

Time andspace vaning electric field and magnetic the form of sine wave, perpendicular to each other, propagating to both the as a transverse wave in a direction perpendicular wave. as electromagnetic a ñelds with speedof light is known

Electric feld

Direction of

propagation of electromagnetic wave

Magnetic field

Characteristics of EM waves:

Le's eomiãer tuzt z capacitor being charged from a cell (E) throngh R 2n the 2ssociated wires to C. This current is knon as conduction cituge ne of the curet LO The cherges get accumulated on the plates between. in develops ffeld electric 2 v2rving Caaicr 2nd tine it Ez megnic neede is placed in between the plates, defects z magnetic feld has also developed there. it inöcztes tat is the Bur no cherge iows bereen the plates. Then, how developed? Tagneic feid to define a Srietist Jznes Clerk Maxwell feit the need fow any of without type of current which develops plates the between eieconAcoording to him 2 curent fows produces curent this and time vzring electric feld due 1o This curent feld in the region between the plates. megetic he is called displacement current. A

Ir îs curtcapacor

aEt

te

>

Displaceet

CUTeL

dt and E, is the wbere o, =JE-dA is the electric fux pmitiiy of free space. field law, line integral of magnetic As per Ampere Mzxwell x (sum of conduction Ho to equal 20nd 2ny closed path iscurrent through that path.) displacement

curTet 2d

fields EM waves propagate as electric and magnetic directions. oscillating in mutually perpendicular a > EM waves travel in vacuum along straight line with thex x assunmed as 3 velocity 2.997924591 10 m/s which is often

m's. > EM waves are not affected by electric and magnetic ficlds. > Relation between electric and magnetic field components: 10

Bo =

where

Elc 1

c=

Transverse nature of electromagnetic waves: In electromagnetic wave, electric and magnetic field vectors are perpendicular to each other in the direction of propagation of wave which shows its transverse nature. An electromagnetic wave travelling along x direction is in the form: = E sin (kr -ot + ) É(r.)

B(x) =

B

sin (kr - ot +)

where E and B are electric field vector and magnetic field vector respectively. and B are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is in the direction of xB.

Electromagnetic

d,

responsible for This proves that chznge in electric field E, is the induction of magnetic field.

Electromagnetic waves and their characteristics (qualitative ideas onty:

3x10 m/s

spectrum

(radio

waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays)

including elementary facts about their uses Electromagnetic spectrum shows the entire distribution of electromagnetic waves according to frequency and wavelength.

Scan to know

more about this topic

Electromagnetic spectrum

Chapterwise COuestion Oswaal CCUET (UG)

54

wave

radar

nm

nm

10

700

oven,

2>

signal >

10>

À>

nm

Infrared:10

400

night light

micro

Visible

Micro

Use:

Total

bone

Eo

>

:

Use:

Use

to

treat

10

detect

: Yrays:

X-rays

:

:

Use UV

Use

Use:

an as

to

Use:

•

•

•

•

2

mutually

Level

Third

(as

long

;E

volume

a

to referred

•

time

Ho

0.01

B

c=

=

nm >

rays: Radio

and

electric cancer

perpendicular

waves

radio,

4,

electromagnetic

>0.01

destroying

observe

>

waves:

10 >>

break

:700

vision

and

nm

bacteria

world

>

wave

TV

:

energy

400

10 >

=l

per

nm

nm

Bank PHYSICS

Over

B}

H%

1

2

Map Level

Mind

2

Second

unit

Ug

of

the Trace

•

is

Level

fields

Combination

First

density

magnetic Electromagnetic

Spectrum

Energy

as current

Naves

if

field, current,

electric

and

electric

magnetic Displacement

field

displacement

Current

changing

an

magnetic

dt

exists

lectror

a there

as

results causes

well

If Conduction

Current

flow

called

to

often

current,

History

waves

due

of current

is

Characteristics

conduction

are

particle

Electromagnetic

charge

1=7

dt The

of

fields constant

charged magnetic

(c) material

propagation

light

is

accelerated

ratio

of

transmit controlled

Heinrich

& electric

the need became

to

person receive

Hertz

signal.

radio

for with medium

not 1886

nature

speed

any

be

and

in Transverse

Oscillating

Produced

phase

Travel

•

aEo =

Do

in

first and

In

their

•

•

B

(c)

55

BECTROMAGNETIC WAVES

Frequency and wavelength of various electromagnetic

ndiations: EM Wave

nameWavelength (m)

Frequency (Hz)

Radio

>lx10-!

UV rays > IR rays > Microwave (2) Microwave > IR rays > UV rays > Gamma rays (3) UVrays > Gamma rays > Microwave > IR rays (4) IR rays > UV rays > Microwave> Gamma rays 19. In vacuum, the physical property which remains same for microwave of wavelength Imm and UV radiation 1600 is

(1) Wavelength

(2) Frequency (3) Speed (4) None of these 20. In vacuum, the wavelength of the electromagnetic wave of te frequency S x 10!9 Hz is (1)

6x 102 x

(3) 1.6

10"

m

3 m

(2) 3 x 10 m (4)15 x 102

m

Question Bank

PHHYSICS

linearly polarised clectromagnetic wave given as E= E,icos(kz - or) is incident normally on a perfectly = z Assuming that reflecting infinite wall at the wä of the wall is optically inactive, the reflected wave mateial given as (2) E, = 1). (1) E, =-E, i cos(kz E,i cos(kz +or). 21.

A

a.

(3) E, =-E,i cos(kz + or).

(4) E, = E,i sin(z-o)

22. In the electromagnetic spectrum, X-rays fall in between (1) Visible light and microwaves (2) Ultra violet and infrared rays (3) Gamma rays and ultra violet (4) Infrared and radio waves 23. The ratio of contributions made by the electric feld and magnetic field components to the intensity of an EM wave is (2) c': 1(3)

1:1

(4)

Ve:l [NCERT Exemp. Q.8.6, Page No. 48) 24. A plane electromagnetic wave propagating along r direction can have the following pairs of.E and B (2) Ey, B,.s () E, By (3) B, Ey (4) None of these 25. A charged particle oscillates about its mean equilibrium position with a frequency of 10 Hz. The electromagnetic waves produced: (1) will have frequency of 10 Hz. (2) will have a wavelength of 0.3 m. (3) fall in the region of radio waves. (4) All of the above 26. The source of electromagnetic waves can be a charge (1) moving with a constant velocity. (2) in a circular orbit. (1)

c:l

(3) falling in an electric field. (4) Both (2) and (3) 27. X-rays are produced by

(1) Klystron valve (2) Excitation of atoms or inner shell electron (3) Nuclear origin

(4) Oscillating circuit 28. We consider the radiation emitted by the human body. Which of the following statements is true? (1) The radiation emitted is in the infrared region (2) The radiation is emitted only during the day (3) The radiation is emitted during the summers and absorbed during the winters (4) The radiation emitted lies in the ultraviolet region and hence is not visible 29. Which of the following rays are not electromagnetic waves? (1) X-rays (2) y-rays (3) B-rays (4) Heat rays 30. A wave travelling the + positive x-direction having displacement along y-direction as Im, wavelength 2rm and frequency of l/t Hz is represented by (1) y= sin (x- 2) (2) y= sin (2rx - 2u) (3) y=sin (10x-20r) (4) y= sin (2x + 2ri)

B] ASSERTION REASONQUESTIONS

Question Nos. 1 to 10consist oftwo statements-Assertion and Reason. Answer these questions by selecting the approprate option given below:

ELECTROMAGNETIC

Roth (A) and (R) are truc, and (R) is the correct explanation

of (A). (3) (4)

2.

3.

57

WAVES

Roth (A) sand (R) are truc, and (R) is not the correct explanation of (A). (A) is true but (R) is false. (A) is false but (R) is truc. Assertion (A): Electromagnetic radiation cxerts pressure. Reason (R): Elcctromagnetic waves carry momentum and energy Assertion (A): Electromagnetic wave docs not require any medium to travel. Reason (R): Clectromagnetic wave cannot travel through any medium. Assertion (A): Dipole oscillations produce electro-magnetic waves.

Reason (R): Accelerated charge produce electro-magnetic waves.

Assertion (A): X-ray travels with the speed of light. sti Reason (R): X-ray sis an electromagnetic wave. 5. ÁSsertion (A): Microwaves are considered suitable for

4.

radar system.

Reason (R): Microwaves are of shorter wavelength. 6. Assertion (A): Ozone layer is very important for human survival. Reason (R): It blocks the infrared radiation. 7. Assertion (A): Electromagnetic waves are longitudinal in

nature. Reason (R): Electric field, magnetic field and the direction of propagation of electromagnetic waves are mutually perpendicular to each other. 8. Assertion (A): Gamma rays are electromagnetic waves having largest wavelength. Reason (R): Gamma rays also have highest frequency. as "Heat 9. Assertion (A): Ültraviolet rays are also known wave". is Reason (R): In electromagnetic spectrum, ultraviolet wave. micro positioned between visible light and waves to 10. Assertion (A): Ratio of frequencies of ultraviolet 1. waves greater than is infrared Frequency of UV rays is greater than infrared

Reason (R): waves.

[C]COMPETENCY BASED QUESTIONS 1-5. Based on following passage answer questions from electromagnetic Microwave oven: The spectrum of as microwaves. These radiation contains a part known energy smaller than visible waves have frequency and oven our microwave In larger. light and wavelength such up. items warm food All it objective is to cook food or as a water contain etc., as fruit, vegetables, meat, cereals, say we mean when that a constituent. Now, what does it temperature certain object has become warmer? When the energy of the random motion of of a body rises, the increases and the molecules travel atoms and molecules or vibrate or rotate with higher energies. The frequency gigalhertz of rotation of water molecules is about 2.45 frequency, this microwaves of (GHz). If water receives equivalent to its molecules absorb this radiation, which is energy with this share heating up water. These molecules up food. One the heating molecules, neighbouring food

in a

non-metal containers should use porcelain vessels and danger of getting a shock microwave oven because of the Metals may also melt from accumulated clcctric charges. container remains unaffected from heating. The porcelain rotate molccules vibrate and and cool, because its large cannot absorb and thus with much smaller frequencies microwaves. on the method, the vessel In the conventional heating gets inside food the then burner gets heated first and cnergy In vessel. the from heated because of transfer of water to energy is dircctly delivered the microwave oven, by the entire food. is shared smolecules which microwave has frequency and light visible to 1. As compared energy (1) more than visible (2) less than visible light (3) equal to visible light energy is more (4) Frequency is less but temperature of a body rises 2. When the motion of atoms and molecules random energy of the (1) the increases. atoms and molecules (2) the energy of the random motion of decreases. atoms and molecules (3) the energy of the random motion of remains same. becomes (4) the random motion of atoms and molecules streamlined. 3. The frequency of rotation of water molecules is about (3) 2.45 GHz(4) 2.45 THz (1) 2.45 MHz (2) 2.45 kHz 4. Why should one use porcelain vessels and non-metal containers in a microwave oven? (1) Because it will get too much hot (2) Because it may crack due to high frequency (3) Because it will prevent the food items to become hot (4) Because of the danger of getting a shock from accumulated electric charges. 5. In the microwave oven, (1) Energy is directly delivered to water molecules which is shared by the entire food. (2) The vessel gets heated first and then the food grains inside. (3) The vessel gets heated first and then the water molecules çollect heat from the body of the vessel. (4) Energy is directly delivered to the food grains. II. Based on following passage answer questions from 6-10. LASER: Electromagnetic radiation is a natural phenomenon found in almost all areas of daily life, from radio waves to sunlight to x-rays. Laser radiation - like all light -is also a form of clectromagnetic radiation. Electromagnetic radiation that has a wavelength between 380 nm and 780 nm is visible to the human eye and is commonly referred to as light. At wavelengths longer than 780 nm, optical radiation is termed infrarcd (IR) and is invisible to the eye. At wavelengths shorter than 380 nm, optical radiation is termed ultraviolet (U) and is also invisible to the eye. The term "laser light" refers to a much broader range of the electromagnetic spectrum that just the visible spectrum, anything between 150 nm up to 1000 nm (i.e., from the UV up to the far IR). The term laser is an acronym which stands for light amplification by stimulated emission of radiation".

lightacrtoSO)

Oswaal CUET(UG)

58

6.

(1) (2) (3) (4) 7.

(1)

Chapterwise QuestionBank

absorbing onc photon (2) release of twos photons clectron comes back from higher to lower cncrgy level. a when electron moves fromlower (3) absorption of photon t, energy level. higher above. the (4) None of ofLASER? 8. What is the range of amplitude nm (2) 700 rm-11000 nm – (1) 150 nm 400 (4) None of the ahe (3) Both the above 9. Lithotripsy is application (1) An industrial application (2) A medical application (3) Laboratory application control Process by

Einstein explaincd the stimulated emission. In an atom, electron may move to higher energy level by absorbing a photon. When the clectron comes back to the lower energy level it releases the same photon. This is called spontancous emission. This may also so happen that the Cxcited electron absorbs another photon, relcases two photons and returns to the lower energy state. This is known as stimulated emission. Laser emission is therefore a light emission whose energy is used, in lithotripsy, for targeting and ablating the stone inside human body organ. Apart from medical usage, laser is used for optical disk drive, printer, barcode reader etc. What is the full form of LASER? light amplified by stimulated emission of radiation light amplification by stimulated emission of radiation radiation light amplification by simultaneous emission of of radiation emission synchronous by light amplified The "'stimulated emission" is the process of comes back from higher release of a photon when electron energy level. to lower

(4) 10. LASER is used in

(1) (2) (3) (4)

Optical disk drive Transmitting Satellite signal Radio communication Ionisation

ANSWER KEY QUESTIONS [A] MULLTIPLE CHOICE 1. (3)

2. (2)

11. (4)

12. (1)

21. (2)

22. (4)

7. (4)

10. (4)

6. (3)

9. (3)

5. (3)

8. (2)

4. (2)

18. (1)

13. (3)

14. (2)

17. (1)

20.

16. (3)

19.(3)

15. (4)

23. (3)

27.(3)

30. (1)

26. (4)

29. (3)

25.(4).

28. (1)

24. (2)

3. (2)

o

1. (1)

1. (2)

2. (3)

[B]ASSERTION

3.(1)4. bGRSC 3.

| 2. (1)

(3)

(1)

(1)

REASON QUESTIONS

5.(1)

6. (3)

|7. (4)8.(4)

9.(2)

10. (1)

9.(2)

10. (1)

cOMPETENCY BASED QUESTIONS

5.(1)

|4. (4)

6. (2)

8. (3)

7. (2)

ANSWERS WITH EXPLANATIONS 4.

CHOICE QUESTIONS [AJMULTIPLE

Option (3) is correct. waves are made of photons. Explanation: Electromagnetic

Option (2) is correct.

=3x10°/W2.25

Ba Or,

10

3x10

Option (2) is correct. ultraviolet rays. Explanation: Photocells are used to detect Ionisation chambers are used to detect X-rays. Bolometers are used to detcct infrared rays. 6. Option (3) is correct. electromagnetic waves are Evplanation: Speeds of all 5.

ms-!

=C

B

. Ratio

in air.

Bo3x10 -=1.66x10 So,

BA

9x10 -=3x10-2 T

x4

correct. 3. Option (2) is Explanation:

9x104

3x10 =

=cl/us

=3x10°/3 =

as2

Explanation: C= avg

1.

carry charge. Hence, they do not correct. is (3) 2. Option medium Explanation: Speed of light in the

B,=

1.66 x 10

sin (0.25 x

10

+ 2.5 x

10!!)

P

of speeds

=

1l:1

S

59

ELECTROMAGNETIC WAVES

7.

Optlon (3)

is

20.Option (1)

correct.

are relatcd Explanation: Magnctron Klystron or Gunn diodes to microwaves, 8, Optlon (4) Is correct. Explanation: We can check all the given cquations by checking their corrCctncss using Dimensional analysis. Option (4) is not dimensionally corect. Hence, option (4) is correct. 9. Option (2) Is correct.

Explanation: i,

Eyplanation: For clectromagnetic wave Eo

cBo

E,

6x10

m

- E(lcos[k(-z)-ot+ a]2--i andi

-i

--Ei cos[n-(k+ or)] --E,i[-cos{(k+or))]

6.3 3x10* 2.1 x 10%T

Option (3) is correct. Explanation: In increasing order of frequency: Radio waves < Visible light < UV rays < y-rays 11. Option (4) Is correct. wavec, electric and magnetic Explanation: For clcctromagnctic to cach other. ficld vectors arc perpendicular 12. Option (4) Is correct. Explanation:

3x10

=clv= 5x105

after waves renains identical. wave, we have z=-z,i=-i and Thereforc, for the reflected wave. additional phase of n in the incident wave is given by clectromagnetic Therefore, the reflected =

Co

10.

correct.

21.Option (2) is correct. a wave changes by 180° or r radian Explanation: The phase of medium. But the type of a getting reflectcd from denser

B

B

is

=

E,icos(k+ ot)

22.Option (3)

is

correct.

spectrum in increasing order of Explanation: The clectrormagnctic waves Image formation by convex mirror for different positions of object: Position of the object

Position of the image

Size of the image

Nature of the image

At infinity

At the focus F, behind the mirror Between P and F, behind the mirror

Highly diminished, point sized

Virtual and erect

Diminished

Virtual and erect

Between infinity and the pole P of the mirror

63

OPTICS]

(air)] point]

adjustment]

point] distance

H;=4H,=I

lens

near

near least

at

of

in are

(image

=

at

Magnification

[image

at

they

image

M-2

d

are

P7

they

when

[For

P

for

final

M=1+

se

when

image

M-1+

lof.

combination

at

[

lens

combination

at

image

h,

formula)

If

infinity]

[normal

lens

of

infinity]

u

urated

contact)

D

111

For

M Lateral

fe

M=-

M=

Microsco?

using

mpound PIcTOSCope

Hypermetropia:

sightedness.

bylens. Omple

Corrected

Telescop

small =i+i-A

smaller

Lightelectricalit

convex

where

Instruments

Marker's

and

if

of lens

Ominimum=(u-1)4,

deviation

Lens

intobrain

of Defects

inverted

Power

on

to

image

sent

converted

=2i-A

a real,

of

far

retina.

Formula

A

fi=

&

eye

Optical

is

i]

using

inverted.

Nearsightedness.

lens.

by

and

Corrected Ominimum

Human

Angle

Forms

sizeis

again signalsignal

concave

Myopia:

eye

in

is

Oispersio polychromatic

its

of

Splitting

constituent

Level colours.

into

Prism

Third

optics-/

light

Mlap Level \Mind

Second

and

surface

Reflection

the

Light

ray

Trace reflecting

Level

of

reflected

Scattering

First

Conventions

ray the Incident

Spherical

coplanar

Refraction

Surface

to

Sign Internal

normal

beamwhenthe

reflection

are

Magnification

Critical

on

Angle

Refraction

R-v

R-u

a

light

Spherical

Mirror

comparable

R

of

by

case

different

Lateral

in

to

directions

which Total

phenomenon

ve.

size

centre origin

direction

direction

light.

taken-

many opticallyincident

angle,back

optical

from

the

origin

and

X-axis

ve.

in

taken

of are case

medium

as are axis

image

are

depth

the

ray

distances distances

opposite

in

real Lateral

#v

Pole

All All of All

of •

•

angle

i

sin

r

denser

angle entire

with

wavelength

interacts

Scattering

passes

is

sin •Incident

the

sin

to

ray

i.e.,

process

When

optical

is rarer

light

to

of it

of

denser

the of

of

is

fibre.

to

(i) When denser

redirected

refraction

light

ray

=1/m

(0:) 90o called

is

in

light

fom

for

TILR.

greater

rarer

R

medium.

then

apparent

to

is

It

than passes

depth

distances

incident

lenses Priniple

•

from medium,

Magnification

the

which

if

shift

=

incident

as taken

the

reflected

critical

measured

measured

mirror

particle optically

again,

the ray

in

incident

angle

used

t

measured

of

this in

are

Oswaal CUET (UG) Chapterwise Question Bank PHYSIca

64

maxima

22 are the

difference

with

light

if

maxima

fringe)

incoherent

they

waves

coherent

tanp

fringe)

sources

central

2DA

=

constant

bright

otherwise

the

of sources

phase

amplitude.

~A,)

Two

(dark

the

of

by

be resultant

light.

between

emitted

to

initial

called remains

said

of

2

width

Law

maxima

are

Linear

material

Width

bsin

bsin

A=(A,

parallel.

Brewesters

Angular

of

=

analyzer.

cellulose

packed

(2n+1)

nà

sOurce

time,

central

width

an

in

as axis

crystals

(for

well

a synthetic

form same

optic

as

herapathite Iperference,

vibration

interference,

perpendicular

polarizer

to or

particular

wave

slit

superimpose

lower

of

of

single

or

Polarisation

Restricting destructive

•For

Two

of to For Interference

• Light

of

Optics-/|

fringe fringe

bright

dark

Young'sDouble

Experiment

AQ=n1

For

For

•

• Level

Slit

Principle

Huygens Third

Resolving

Power

dark

Map

slits

Level

fringe between

nth

interference

on

_(21+1)2D between

central

light

power

Fringe-width

Second

the

d

of

distance

power4usin

1.222

Based

2d

Mind

of

of

microscope;

Telescope:: Resolving

Trace

Resolving

and

Level

Distance

fringe

First

primary

d=

source wavelets

•

the the between

secondary

on is

or

•

phase

wavefront

wavefront

consecutive,bright

point

between

particles between

scCreen

fringe

some

fringe

Each

distance

naD central Distance

a

of

D=distance

and

all

fringes.

bright

and

spherical

of

SOurce

h

in

vibrating

Locus

the

is It

two dark

AEM

For •

as

Used

the

A=A,

wave

acetate

with

light

light

I-(Vit

of

Large

direction

by

+A,

waves

a

tiny

a

in

greater

with

sheet

propagation

SOurce

constructive

of

of

direction

the

Diffraction

Coherent

their

65 Relraction of light: Retaction is deviation of light path when it obliquely travels to another. mNn onc medium

Refraction of Light The incident ray, thc reffacted and the nomal to te ileriace of two transparent media, nt the point of incidehce, Laws of

y

coplanar, i.c., they all lic in the same plane. Snell's law: The ratio of sine of angle of incidence to the cine of angle of reftraction is a constant, for the light of n given ur and for the given pair of media. This constant value is alled the refractive index of the second medium with respect

are

sin i to the first medium, sin r > Ifthe first medium is free space, then the refractive index is known as the absolutc refractive index of the second medium, The absolute refractive index of a medium is cxpressed as Velocity of light in free spacc H2 Velocity of light in the medium

Conditlons for totat fnternal reflection arei

The light shoutd travel from denser mediurn to the rarer inedium. Angle of incidence shoutd be latyer than the criticaf angle.

Optleat nbres Optical Shre is the technology ASsOciated with data a transmisaion sing light pulses travelling along with long or is nbre which usually nade of plastic glass. It works on the principle of total internal reflection. Cadding ws

Cort Light out

Light in

Refractlon at spherical surfaces: > If the rays arc incident from a medium of refractíve index H, to another medium of refractive index z, then R

> if light ray bends away from normal after refraction, then second medium is optically rarer. If the light ray bends towards Dormal, then second medium is optically denser. Total Internal refection and its applications, optical fAbres: > Wben light travels from an optically denser medium to a rarer medium at the interface, it is reflccted back into the same nedium when angle of incidence is greater than critical angle. This reflection is called total internal reflection. >Citical angle the angle of incidence for which angle of

where, R= Radius of curvature of spherical surface and object is placed at rarer medium. u=Object distance from spherical surface v= Image distance from spherical surface

Lenses:

>A lens is a picce of transparent glass bounded by two surfaces out of which at least once surface is spherical. There are two types of lenses: o Convex lens: A convex lens is one which is thinner at refraction is 90°. edges and thick at centre. o Concave lens: A concave lens is one which is thick at cdges and thin at centre. > Image formation in conver lens for different positions of object: Position of the image Nature of the Image Position of the object Relattye stze of the image At infinity at focus F2 Highly diminished, point sized Real and inverted Beyond 2F

Between F2 and 2F2

At2 F; Between 2F, and 2F,

at 2F2

Diminished Same sized Enlarged Infinitely enlarged

Beyond 2F, At infinity F and optical On the same side of the Enlarged lens as object cntre > Image formation in concave lens for different positions of object: of the imageReltye Postion ofthe objectPositon size of the image

FocusF, Between focus At

At infinity

At focus

Highly diminished point sized

F

Between infinity and the Between focus and optical Diminished optiçal centre O of the lens centre 0 Thin lens formula: Magnification: Thin lens is a lens whose thickness is negligible compared Lo the radii of curvature of the lens surfaces. Relation between objct distance (u), image distance (v) and focal length () of Power of lens: a thin lens.

Real and inverted Real and inverted Real and inverted Real and inverted Virtual and erect

|

Nature of the image

Virtual and erect

F

Virtual and erect

m=is

> The power of a lens is defined as the reciprocal of its focal length, represented by the letter P and exprcessed as

Lens maker's formula: where

u=

The SI unit of power is dioptre when focal

length is in metre.

Ovesl CUET (UG) Chaptervise Oustin

66 Combination of thin lenses in contact and combination of a lens and a mirror: > When two or more lenses are combined, the equívalent focal length and power of the combination is

P =P+P+t.....

> When lens and mirror are used coaxially, the image formation is considered one after another in steps. The image formed by the 1 lens facing the object serves as the object

(may be real or virtual) for the miror on the 2nd lens. In each case, lens appropriate lens or mirror formula to be applied with proper sign convention. Refraction and dispersion of light through a prism: > Refraction through Prism:

ite=5+A Here, i=e

Scatteríng of light - blue colour of the sky appearance of the sun at suríse and sunset: tl% Scatteríng: Scattering of igt is the phenonenon by which a beam of light is redírected in many different directiuns when it interacts with particle of size comparahle to the wavelength of the incident light. Blue colour of sky: Sunlight in carth's atmosphere is scattered by the gas molecules and the tiny particles floating in air. Ble light is scattered more than the other colours because it is the visible light of smallest wavelength which is comparable to the molecules and floating tíny particles. So, the sky ppan ti Reddish appearance of the sun 2t sunri and During sunrise and sunset sun stays near the horizon. So te rays travel longer distance through the atmosphere. On the wy most of the blue light is scattered away since its wveleng is the least in the visible range and cormparable to the molecules zrd floating tíny particles in the atrmosphere. Theret light is scattered the least since its wavelength is lzrzest in visible range and hence not cotparable to the gzs molecies and floating tiny particles in the ztrmosphere. So, the red igt reaches our eyes preinireraty and therefore appezrs red z sunrise zd sunst. Optical istruments: Human eye, image formztion, zsd 2ccommodation: > Human eye: Vision is the most precious sense of humea being Our vísual system is cornpsei of highly sophísticated brain and modest optical instrument - eye. Ciiary musdes >

ut:

e

p B

Refracted ray always bends towards the base. Angle of deviation, ô = (i-r)+ (e-r)

> At minimum deviation, i =e(-) and r =n(),

then

sin[(A+6.)/2]

Crystalline lens

sin[4 12]

Aqueeous

For thinprism, 6,,

=(H21-1)A Angle of incidence vs. angle of deviation graph:

humour Pupil Iris

-Retina

Optic nerve

(6) 604

Cornea

deviation

50

of

Angle

40t

i=ei

30 e

40 80 Angle of incidence ) Dispersion through prism: In a medium, lights of different colours travel with different welocities and hence refracted in different angles. Thus white light assing through a prism ges split ito its constituent colours and orm a spectrum. This phenomenon is known as dispersion. 20

Angular dispersion: The angle subtended between the directions of emergent iolet and red rays of light is known as the angular dispersion. Angular dispersion =6,-õp (4-#)A Dispersive power: Dispersive power is defined as the ratio of angular dispersion mean deviation (deviation yellow light) by the prism. of -1

Vitreous humour Different parts of human eye: o Cornea: Clear ome shaped front portion eye through of which light enters into eye. o Iris: The coloured portion behind the cornea o Pupil: A dark hole at the middle of Iris. o Pupil: A dark hole at the middle of Iris. o Aqueous humour: Fluid present in bctween conea znd iis o Lens: Just behind the iris; focus the incoming light on the retina. o Ciliary Muscle: Changes the curvature of the lens to focus the incoming rays on retina. o Retina: Photosensitive screen on which the incorning light is focused and image is formed. The image is real, inverted, smaller than object. Retina is made of photoreceptor celis rods and cones which converts light into electrical energy. Rods perceives black and while and enable night vision. Cones perceive colour and provides details the vision. of o Optic nerve: Carries the electrical signal to brain. Range of clear vision of human eye: 25 cm (near point) to infinity (far point).

67

OPTICS >Image formation by human eye: With the help of convex lens an inverted real and smaller in size image is formed on the retina. Photoreceplor cclls convert light signal into the clectrical signal. The signal is carricd brain Derves

by the optical

where image is again inverted and we sce an ercct imagc. w

Inverted Real Image

Convex lens

Brain

> Hypermetropia:

this condition, the Hypermetropia is far-sightedness. In appcar blurred. ncar objects distant objccts appcar clear whercas size, (ii) Dccrcase of Causes: (i) Reduction of cycball curvature of cyc lens. convex corrective lens of appropriate Correction: Use of a power.

aEye

N

Erect Image

Optic nerve Accommodation: The ability to change the curvature of the eye lens with the help of ciliary muscle to focus far and distant object. Correction of eye defects (myopia and hypermetropia) using lenses: > Defects of human eye: () Myopia, (ii) Hypermetropia > Myopia: Scan to know Myopia is near-sightedness. In this more about condition,the near objects appear clear this topic whereas distant objects appear blurred. Causes: (G) Elongation of eyeball size, () Increase of curvature of eye lens. Correction: Use of a concave corrective Defects of eye lens of appropriate power. >

Hypermetropic eye without correction: Image

shifts close to Myopic eye without correction: Far point cye.

Concave lens Myopic eye after correction

object at

near point forms behind retina.

N

Hypermnetropic eye without correction: Near point shifts away from eye.

N Myopic eye without correction: Image of object at infinity forms in front of retina.

of

N

Convex lens Hypermetropic eye after correction. Microscopes and astronomical telescopes (refiecting and 2 refracting) and their magnifying powers:st e7t > Microscope: An optical instrument which helps us to see and study micro objects or organisms. > Simple microscope:

Eye focussed

on near

pointois

Image at near point of distinct vision

Oswaal CUET (UG) Chapterwise

68

Question Bank PHYSICS

Magnification: When image forms at near point of distinct vision:

o

m=1+

D

o When image forms at infinity:

m= D

Eye focussed at infinity

Image at infinity

> Compound microscope: Objective

u=fe

Eyepiece

VE

Normal Adjustment: Image at Infinity

oasqF. B

B

-Do

u,

Eye piece

-D

Objective

Image at near point of distinct vision

B

Fo

--D,

> Magnification:

o

When image forms at near point of distinct vision:

-) o

When image forms at infinity:

-a2)

m:

> Reflecting telescope: Objective mirror

Secondary mirror

Eyepiece

> Telescope: Itis an optical instrument which helps us to see and study far off objects magnified and resotved (with clarity).

-4)

Magnification m =

69

OPTICS

Refracting telescope:

>

Objective

-fo

Eyepiece

Fo

Fe

Fo

A'

R

Eye

Objective lens

piece

Normal Adjustment: Lmare at Infinity

D

e

Image at near point of distinct vision >

Magnification: o When image forms at near point of distinct vision:

L M

N

M

m=

o When image forms at infinity: m

XriTmTmtiinftTTmY

=

A

Wave optics: Wave front and Huygens' Principle

> Wavefront: Wavefront is locus of all points in which light waves are in same phase. Propagation of wave energy is perpendicular to the wavefront. > Shapes of wavefronts:

Source Point source

Wavefront

Line source

Spherical wavefront Cylindrical wavefront

Plane source Point source very far away

Plane wavefront Plane wavefront

> Refraction of plane wavefront: wavefront PQ is a line of separation of two media. A plane to incident the perpendicular on it. The lines AB is incident rays. wavefront represent incident

becomes the The wavefront arrives at point A first and the B end of by source of secondary wavelets. Time required = t the wavefront to reach the line of separation is and BB' refractive index Vt (V = velocity of light in the medium of A' from and AA' = Vyt reaches H). By that time wavelets (Vy

>Huygens' Principle: o Every point of a wavefront becomes secondary source of

= velocity

of light in the medium of refractive index u).

Arc with radius AA' is drawn. Tangent A'B' is the refracted wavefront. The lines perpendicular to the refracted wavefront are refracted rays.

light.

o These secondary sources give their own light waves. own wave called Within small time, they produce their same secondary wavelets. These secondary wavelets have speed and wavelengths as waves by primary sources.sta common tangential surface on all these O At any instant, a direction. wavelets gives new wavefront in forward a Reflection, and refraction of plane wave at plane surface using wavefronts: a surface. >Reflection of plane wavefront: XY is reflecting MP on and LA lines it. The A plane wavefront AP is incident represent which are perpendicular to the incident wavefrontAP, incident rays.

R

p A

A'

B

source

of the The wavefront arives at point A first and becomes wavefront the end of Secondary wavelets. Time required by the P of surface is tand PP = vt (y= velocity light n the medium). Arc with radius PP is drawn. Tangent A'P is the to the reflected wavefront. The lines A'L' and PM perpendicular

to reach the reflecting

reflected wavefront are reflected rays.

A

70

Oswaal CUET (UG) Chapterwise Question Bank PHYSIKS G

Proof of laws of reflection and refraction using Huygens' Principle: > Proof of law of reflection:

Ip S

Incident wavefront AB

Reflected

-D

wavefront

C!

Ma The figure drawn here shows the reflected wavefront CE corresponding to the given incident wavefront AB. It is seen that AEAC and ABAC are congruent. This is law

SjP

td

> Condition for P to be a bright spot:

Zi=Lr of reflection.

nh,

> Proof of law of refraction: Medium1

> Central maxima forms at O. Here, path difference = 0 > AtP. which is at x distance from path difference = 0, S.p

i..,

nD.

Xnth bright

>Condition for

Incident wavefront

(2n

+l)ie.,

P

xnth dark

> Fringe width = ß =

Jnoisas Medium 2 A E

Refraçtedot

wavefront

urnis

to be

=0,

Dt

a dark spot: xd =0,

(2n+1)DA

D2

.

27.,

...

32. ****

2d

DA

d

Coherent sources, and sustained interference of light: > Coherent sources: Two sources are said be coherent sources

The figure drawn here shows the refracted wavefront CE corresponding to the given incident wavefront AB. It is seen that

ACAC taeiDs

Sin

sin

r=

AE AC V

=

sinr

AC

P21

This is Snell's law of refraction.

Interference:

> Interference is a phenomenon in which waves from two coherent sources combined by adding their intensities with due consideration of their phase difference. The resultant wave will have greater intensity (constructive interference) or lower intensity (destructive interference) if the two waves are in phase or out of phase respectively. > Condition for constructive Interference: Phase difference between the waves should be even multiple of T. Hence Path difference = 0, ,.....n2. Here, n=0,1, 2, 3.....

> Condition for destructive interference: Phase difference between the waves should be odd multiple of Hence Path (2n+1)2 difference = Here, n = 1,2,3..... 2'2 Young's double hole experiment and expression for fringe

.

width:

if they produce two waves having same frequency. same waveform and have time independent constant phase different between them. > Sustained interference of light: Sustained interference of light indicates that the positions of the maxima and minima of light intensity remains constant throughout the screen. > Conditions for sustained interference: (i)) The two interfering waves must be coherent which means the two light waves must have a constant phase difference or must be in the same phase. (ii) The two waves must have the same wavelength. There may be little difference in amplitudes. (iii) The two sources should be very narrow. (iv) The two sources mnust lie very close to each other. > Intensity distribution pattern:

Interference pattern on screen dark bright dark bright dark bright dark bright dark bright dark

Intensity distribution

71

OPTICS

nifraction due to a single slit, width of central maximum: > Difraction: It is the bending of light around the corners of

> Resolving power of telescopc

of the obstacle.

> Resolving power of microscope

an obstacle or aperture

=

into the region where we expect shadow

To P

From S

To C

> Path difference between ray from L and ray from N=N -a sin® (where width of the opening = a and = angle of elevation of point P from principal axis.)

> For first maxima: a sin

=, i..,

0=

(sin

value of 0)

> Pisadark point when path difference =h, 2.,

for small

1.222 =

2/sin

Polarization, plane polarized light: is > Polarization: Polarization phenomenon by which vibrations of light waves arc restricted in a particular planc. > Plane polarized light: A polarized light which vibrates in single plane, perpendicular to direction of

Scan to know more about this topic

Polarization of light propagation is known as planc polarized light. they are Light waves are electromagnetic in nature i.c., waves. unpolarized as In as magnetic composed ofelectric well light, the orientation of these fields is along various planes. a polarizing filter the Passing unpolarized light through is a vibrations may be restricted in single plane and such light as known plane polarized light. POLARIZING|

......., n

FILTER

32 >P is a bright point when path difference = 2 (2n +1)a 2

> Fringe width =

DA

a

> Width of central bright fringe = 2DA a

> Intensity distribution pattern: Diffraction pattern on screen

neion

Dark

Bright Dark

central maximum double the width Dark

Bright Dark Bright There is no gain or loss ofenergy in interference or diffraction, which is consistent with the principle of conservationof energy. Energy is only redistributed in these phenomena. Resolving the power of microscopes and astronomical telescopes: > Resolving power: The resolving power of an optical instrument is measured by its ability to differentiate two lines or points in an object. The greater the resolving power, the smaller the minimum distance between two lines or points that can still be distinguished.

Plan polarized Unpolarized Light Light Brewster's law: > Brewster's law gives the exact angle of incidence for different materials and wavelengths of light for which reflected ray is completely polarized. > The law states that maximum polarization of an unpolarized light ray occurs when the refractive index of the reflector material is equal to the tangent of the angle of incidence of the ray.

Uses of plane polarized light and Polaroids: Uses of plane polarized light: > Used to reduce glare of bright light - sunglass, windows of aeroplane etc. > In number of digital applications, digit, letters, figures are formed by LCD through polarization. > Polarized LASER beam is used for playing CD. > Used for photography. > Used for recording and reproducing 3D pictures. > Used for study of asymmetries in molecules and crystals. > Used for determination of refractive index using Brewster's law.

Polaroids: Polaroid is a large sheet of synthetic material packed with tiny herapathite crystals in cellulose acetate with their optic axis parallel so that it transmits light vibrations only in one direction. The layers of the crystal are mounted between two glass sheets for protection. This acts as a sheet of the polarizer and known as polaroids. It can be used as a polarizer as well as an analyzer.

72

Oswaal CUET (UG) Chapterwise Question Bank PHYSICs

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1.

(1) (2)

(3) (4) (1) (3) 2.

(1)

A converging lens is used to form image on a screen. When the lower half of the lens is covered by an opaque screcn then, the correct statements are: Intensity of Image will decrease. Iatensity of Image will increase. Complete Image will be forned.irnd Half the Image will disappear. Choose the correct answer from the options given below. (A), (B) and (C) Only (2) (A), (C) and (D) Only (A) and (C) Only (4) (B) and (C) Only (CUET 2022, 7th June) In Young's double slit experiment, the ratio of slit widths is 4:1.The intensity ratio in the interference pattern would be:

1:2

(2)

1:3

(3)

9:1

(4)

1:9

(CUET 2022, 7lh June) 3. Huygen's wave theory of light could not explain the phenomenon of: (1) interference (2) diffraction (3) rectilincar propagation

of light (4) photoelectric effect (CUET 2022, 7th June) 4. Two sources are producing waves as given in the options below. Which two sources are called coherent? (1) of same velocity (2) of equal wavelength (3) having a constant phase difference (4) having wavefront of same shape (CUET 2022, 7h June) 5. The radii of curvature of the faces of a double convex lens of focal length 12 cm made up of glass (u = 1.5) are 10 cm and 'p'cm respectively. The value ofp is: (1) 10 cm (2) 15 cm (3) 30 cm (4) 20 cm 6. The essential condition for total internal reflection to occur is: (1) angle of incidence i> critical angle i (2) angle of incidence i > critical angle ie (3) angle of incidence i= critical angle i, (4) angle of incidence i= 90° (CUET 2022, 7th June) 7. A student has drawn the following courses of rays through a glass prism. The one which represents the position of minimum derivation is: (1)

(4)

(CUET 2022, 7h June) The Brewster's angle for air to glass interface when unpolarised light is incident on a glass (" = 1.5), such that reflected and refracted rays are perpendicular to cach other, is nearly: (1) iz =30° (2)i= 45° (3) ig = 570 (4) ip = 470 8.

(CUET 2022, 7"

ssonh Nowadays optical

August)

fibres are cxtensively used for 9. transmitting audio and video signals through long distances. The optical fibres work on (2) Refraction (1) Double refraction Reflection (4) (3) Total internal reflection (CUET 2022, 5" August) 10. A star is scen using a telescope whose objective lens has a diameter of 250 cm. The wavelength of light coming from the star is 500 nm. The limit of resolution of telescope is (1) 1.2 x 10- radians (2) 2.4 x 10- radians x 1.5 (3) 10- radians (4) 3.9 x 10radians (CUET 2022, 6h August) 11. Match List-I and List-II

List-I

List-II

Parabolic mirors are used B. Compound micro- II. Only one convex lens is scope used C. Telescope III. Objective of large focal (reflecting) length and aperture is used D. Telescope IV. Objective of small focal (refracting) length and aperture is used Choose the correct answer from the options given below: (1) A-III, B-II, C-I, D-IV (2) A-II, B-II, C-IV, D-I (3) A-I, B-III, C-II, D-IV (4) A-II, B-IV, C-I, D-III (CUET 2022, 6'h August) 12. A convex mirror of focal length f is placed at a distance d from a convex lens of focal length f,. A beam light rays of coming from infinity falling on convex lens-convex mirror combination and returns back to infinity. Distance is d (1) (+) (2) 6-) (3) (2-2) (9) (+ 25) 13. A short pulse of white light is incident from air to a glass slab at normal incidence, after travelling through the slab, the first colour to emerge is (1) Blue (2) Green (3) Violet (4) Red (CUET 2022, 6'h August) A.

Simple microscope

I.

14.

=143 1

2l=1.39

45

II

=121 1V

h 90°

45

73

DPTICS There are four light rays incident on a right angled prism. The refractive index of prism material for the rays is 1.43, L39, 1.38, 1.21 respectively. The ray that suffers total

internal reflection is

II

I(2)

(3) III (4) IV (CUET 2022, 6h August) .. hollow double concave lens is madc of very thin transparent material. It can be filled with air or cither of the two liquids L, or L, having refractive indices n, and n, respectively (n, > n > 1). The lens will diverge a parallel beam of light if it is filled with (1) Air inside and placed in air (2) Air inside and immersed in (3) L, inside and immersed in Lo (4) L, inside and immersed in (CUET 2022, 6th August) 16. Two prisms made of material of same refractive index and having angles of thc prism 65° and 60 respectively are combined to form a glass block as shown. If for any ray passing across the block the minimum angle of deviation in 3° then refractive index for the material of the prism is

(1)

refractive index 1.5 is 22. The Brewster's angle for glass of (1) 57.3º

67.3

(CUET 2022, 8" August) on a thin convex lens. The 23. A planc wavcfront is incident refraction is nature of wavefront immediately after (2) Spherical and converging (1) Planc wavefront (4) Cylindrical and diverging (3) Spherical and diverging (CUET 2022, 8t August) appearance of the sun during sunrise and Z4.

(1)

L

(2)

L

60Y

(4)

(3) 60

(2) 45°

(3) (4)

The reddish sunset is due to more and hence its scattering is Wavclength of red light is more. energy and hence reaches our eyes Red light has higher cven after travelling long distances. scattering is least hence Due to its longer wavelength, its cycs. our travels longer and reach Because it does not follow Rayleigh scattering.

(CUET 2022, 8h August) ray parallel to the x-axis is incident 25. A parallel beam of light x as shown in the on a parabolic reflecting surface =2by focal point F. passes through figure. After reflecting it surface? reflecting the What is the focal length of

65\

0 (4) 1.35 (3) 1.6 light when of wavelength experiment In Young's double slit mm. 0.1 Calculate the is nm width fringe the is used 600 immersed in a fringe width when the entire apparatus

(1) 1.5 17.

(2) 2.0

medium of refractive indcx light in air (1) 0.01l mm (3) 0.075 mm 18.

by keeping the source of (2) 1.05 mm (4) 0.0075 mm

(CUET 2022, 8" August) to the Among the following which one is not due reflection? phenomenon of total internal

(1) Mirage

(2) Spectacular

brilliance of diamond

(3) Optical fibres (9) Dispersion of light a convex miror of 19. An object is placed 12 cm in front of distance is focal length 18 cm, the value of the image (4) 6 cm (3) 7.2 cm (1) 18 cm (2) 12 cm (CUET 2022, 8th August)

an object. The image of 20. A screen is placed 90 cm from a convex lens at two the object on the screen formed by cm. The focal length of by 20 different locations separated the lens is

(1) 10.5 cm

(2) 15.4 cm

(3) 18.5 cm (4) 21.4 cm

(CUET 2022, 8'h August)

a telescope where objective has

The resolution of wavelength 4000 diameter of 200 inches when the light of A is coming from the star is (2) 96.06 x 10- rad (1) 9.606 x 10 rad x (4) 9.006 10rad (3) 0.9606 x 10- rad (CUET 2022, 8" August) Z1.

1

(4)

(3)

(2)

(1)

46 86 25 wavelengths 5000 Å and two consisting of light beam 26. A of obtain interference fringes in Young's is used 6000 double slit experimnent. The least distance from the central maxima, here the bright fringes due to both wavelengths coincide, will be (If separation between slits =lmm and separation between slits and screen is m) (3) 2 mm (4) lmm (2) 3 mm (1) 4 mm are true for refraction statements 27. Which of the following of white light through a glass prism at minimum deviation position of prism. () The angle of prism becomes zero. () Angle, of refraction at first refracting surface is equal to angle of refraction at second refracting surface (ii) The refracted ray inside the prism is parallel to the base of the prism. (iv) Angle of emergence becomes 90 (v) Angle of incidence is equal to angle of emergence. Choose the correct answer from the options given below: (1) (ii), (ii) and (v) only (b) (1), (üi) and (iv) only (3) (ii), (iii) and (iv) only (4) (), (iv) and (v) only 28. Match List-I with List-II 1

r

r.

List-I (A) Convex

List-II

miror

(B) Total intemal reflection

)

Accommodation

() Reflecting type

Oswaal CUET (UG) Chapterwise Question Bank PHYSICS

74

Assertion (A): According to Huygen's theoryno backward wavefront is possible. Reason (R): Amplitude of secondary wavelets is (1 + cos0), wheret 0 is theangle betwcen proportional the ray at the point of consideration and direction f secondary wavelet. 7. Assertion (A): Wavefront emitted by a point source light in an isotropic medium is spherical. Reason (R): Isotropic medium has same refractive indn. in all directions. 8. Assertion (A): When a light wave travels from rarer x denser medium, its speed decreases. Due to this reduction of speed the energy carried by the light wave reduces. Reason (R): Energy of wave is proportional to the frequency. 9. Assertion (A): No interference pattern is detected when two coherent sources are too close to each other. Reason (R): The fringe width is inversely proportional to the distance between the two slits. 10. Assertion (A): Difraction takes place with all types of waves. Reason (R): Difiraction is perceptible when the wavelength of the wave is comparable to the dimension of the diffracting device. 6.

(C) Ciliary muscles

(III) Optical fiber

(D) Cassegrain

(IV) Used as a rear view

telescope

miror

Choose the correct answer from the options given below: (1) (A)- (IV), (B)- (), (C)- ([), (D)-(I) (2) (A) - (IV), (B) – (I), (C) (), (D)- (1) (3) (A)- (D, (B) – (I), (C)- (), (D)- (V) (4) (A) - (), (B) (III), (C) (IV, (D) -() (CUET 2022, 23rd August) rays 29. A converging beam of is incident ona concave lens. After passing through the lens, the rays converge at a distance of 15 cm from the lens on the other side. If the lens is removed, the converging point of rays decreases by 5 cm. The focal length of lens is (1) -10 cm (2) -20 cm (3) -30 cm (4) -5 cm 30. A beam oflight converges ata point M. Now a concave lens of focal length 32 cm is placed in the path of convergent bean 24 cm from M. Now the beam will converge at: (1) 56 cm from lens (2) 8 cn from lens (3) +96 cm from lens (4) -32 cm from lens (CUET 2022, 23rd August)

t

-

-

-

[B]ASSERTIONREASON

QUESTIONS

Question Nos. 1 to 10 consist of two statements -Assertion and Reason. Answer these questions by selecting the appropriate option given below: 1. Both assertion (A) and reason (R) are true, and reason (R) is the correct explanation of assertion (A). Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A). 3. Assertion (A) is true, but reason (R) is false. 4. Assertion (A) is false, but reason (R) is true. 1. Assertion (A): A convex mirror cannot form real images. Reason (R): Convex mirror converges the parallel rays that are incident on it. 2. Assertion (A): The focal length of a concave miror is f and an object is placed at a distance x from the focus. The

[C]COMPETENCYBASED QUESTIONS I.

2

magnification produced by the mirror is

Reason (R): Magnification = size of image / size of object.

1,1 3. Assertion (A): The mirror formula+*=

20 cm and 30

cn. If refractive index of glass is

3

2

then

the focal length of the lens will be: (2) -24 cm (1) 24 cm (3) 120 cm (4) -120 cm 2. Find the focal length of an equiconvex lens whose each face has radius of curvature 12 cm. Given refractive index

of glass is

3

(1) 0 (2) Infinite Ch (3)-12 cm (4) 12 cm 3. A convex lens has 30 cm focal length in air. What is

focal length in a liquid whose refractive index is

is valid for

Assertion (A): A diamond of refractive index V6 is immersed in a liquid of refractive index 3. If light travels from diamond to liquid, total internal reflection will take place when angle of incidence is 30°. Reason (R): p = 1sin C, where is the refractive index of diamond with respect to the liquid. 5. Assertion (A): A double convex air bubble is fomed within a glass slab. The air bubble behaves like a converging lens. Reason (R): Refractive index of glass is mnore that the refractive index of air. 4.

1.

1

mirors of small aperture. Reason (R): Laws of reflection of light is valid for only plane surface and not for large spherical surface.

Based on following passage answer questions from 1-5. of derived focal length using surface of suitable radii of curvature. Lens maker formula helps finding various unknowns and is widely applied. (CUET 2022, 5lh August) The radii of curvature of faces ofa double concave lens are It is useful to design lenses

(Given refractive index glass is

its

4

?

3

) 2

(1) -120 cm (2) +120 cm (4) -60 cm (3) 60 cm 4. If a convex lens made of a material of refractive index n, is immersed in a liquid of refractive index n, such that the incident beam of light willemerge as shown

n=n,

(A)

(B)

75

OPTICS

3)

waves meeting at point P is Path differencc between the

7.

given by

lxd 2D mm distance 20 In YDSE 20h bright fringe is obtained the 5th between distance from central bright point. The xd

xD

(1) (4)

8.

(3)

2xD

(4)

d

is bright fringe and 3rd dark fringe

Find the power ofa concave lcns whose focal lcngth

E

(1)-2.5D x 10-D (3) +25

T

Sth

is 40 cm.

(2) +2.5D (4) -2.5 x10D

bright

3rd dark CBP Screen mm (4) 8 mm 2.5 (3) mm mm (2) 2 (1) two consecutive bright 9. In YDSE the distance between mm. be the fringe width would What fringes is given by 3 of refractive index 1.2? a medium ifYDSE is performed in mm (3) 2.5 mm (4) 0.5 3 mm (2) 1.5 mm 1

Based on following passage answver questions from 6-10. In Young's double slit cxperiment as shown in figure, interferencse of light waves was observed on the screen. Thomas Young made two pinholes S, and (very close to S, each other) on an opaque screen in front of parent source S.

and S, behaved as coherent sources, and produced interference patten on the screen which has alternate bright fringe and dark fringes. This experiment proved Huygen's wave theory of Light. S,

(1)

path difference SS, 10. In YDSE as shown, the

- SS, =

bright point. This causes the shift in position of central is point bright central new position of The |P Power)

|X

O

Parent sOurce

Bright

(CBP)

(Central

D 6. In YDSE experiment position fringes is given by

(1)

3)

D

of bright fringes and dark

D

(1)

i, then there is no refraction. Total light is reflected back. 7. Option (3) is correct. Explanation: For minimum deviation, angle of incidence = angle of emergence and the ray inside the prism is parallel to the base. 15

8. Option (3) is correct.

Explanation: From Brewster's law tan(ig)

=

ig =

tanu = tan(1.5)

9. Option (3) is

Focal length=f

d=fs-Rh correct.

So, 1

57°

correct.

Explanation: Refractive Index of inner core of optical fibre is optically denser than the outer cladding Hence repeated total internal reflection takes place between these two media and thus the signal is sent through long distance. 10. Option (2) s correct. Explanation: Limit of resolution of telescope is given by,

of

only onc convex lens. In compound microscope, objective has small focal length and aperture since smaller the focal length higher the magnification. In reflccting type telescope parabolic mirrors are used to avoid aberration and to make the instrument light. Image formed is also brighter. In refracting type telescope objective has large focal length and aperture since large focal length increases the magnifying power and large aperture help in collecting large amount of light coming from the object so that a bright image is formed. 12. Option (3) is correct.

For ray I,

-i,=

sin

in4437

sin

i

For ray II,

90

=

For ray II, ;

ie=

For ray IV,

i,=

sin sin

1

= 46.43°

1.38

sin55.730 .21

So, only for ray I, the angle of incidence is greater than critical angle. So, only this ray will suffer total internal reflection. 15. Option (4) is correct.

Explanation:

77

ks sunNNing

will

have like a divging lona if the the matial of the lens is than hat of mtum

22,Optlon

me

the liyunt l

having

cmnwnatively higher eftactive insid The lons shoukl be inmmersel in liquid mmywrativcty lower ng neftactive inles. L, ptkon (2) is corvt, tinimum angle of deviation

t

0,

ytan)

tanO3) n36.3" In

The value in olose to 57,3". 23,

ati:

(-S-(-1)

() correet.

Kylanaton: Brewter's ongle

Optton (2)

s correct.

Kylanatlon:

60-3

Su-S3 Su

.6

13Optin (3) ts corret.

AanaiN:

Incident wavefhont (Plane) 24. Optlon (3)

Bmerging wavefront [opherical and converging] Is

correct.

sunsct, the sunrays have to Aylanation: During sunrise and the atmosphere. The blue

0.1 4

pass through greater distance through most; the rod colour is less scattered sun during liglt gets scattered thewe find reddish colour of the and reaches us. Thus, sunrise and sunsct. 25, Optlon (2) Is correct. Bplanatlon: Given, x 2by

0.075 mm)h

3 18, Option

(4) is correct.

Alenanion: Dispersion is the phenomenon of splitting of

visible light into its component colouns,

Dissersion of light is caused due the change of speed of light aY of each wavelength bya different amount, This is not due to total intemal reiection. 19, Option (3) is correct.

So,

25

parabola Comparing with the general cquation of a focal length) (where

y'

4ar,

Elanation: 2b 12

So,

8b

30

1

Focal length of the reflccting surface

12x18 y=7.2 cm 20. Option (4) is

correct. D

Eplanation:

4D

screen = 90 cm Where, D = distance between object and 20 lens between two positions of

f=

PaetbtA

Sn6n -6 n6

Lenst distance from central maxima

[90°-20°)

-

[4x90]

cmi

Resolution

=

6x5000 x 10-10 x

2

21.4

21.Option (1) is correct.

86

26. Option (2) is correct. Explanation: If nh bright fringe of 5000 A wavelength coincides with (n - 1)h bright fringe of 6000 Å then nx 5000 (n-1)6000

or,

cmalse

I-Distance

Erplanation:

og

is

30 x

27. Option (1) is

1.222

10 m3

mm

correct.

Explanation: For minimum deviation, x

10-10

a- 200 inch

200 x 2.54

)e

100

Resolution

fe

1.22.3 a

1.22 x 4000 x 10-!0 200x 2.54

100

9.6 x 10

rad

s

lx10

78

Oswaal CUET (UG) Chapterwise Question Bank PHYCIOR

The refracted ray inside the prism. is parallel to the base of the prism. 28. Option (2) is correct. Explanation: Parabolíc reflector is used in Cassegrain telescope. Optical fibre utilises the total nternal reflection phenomenon. Ciliary muscles control the curvature of eye lens. This is known as accommodation. Convex miror is used as rear-view mirror since it has large field of view and produces erect image. 29. Option (3) is correct.

Explanation:

5 cm

15

cm

O is the virtual object. I is the real image.

1

Or,

h

Magnification (m)

1

15

10

1

size of image

size

of object

f+x

x

So, both the assertion and reason are true and reason is the correct explanation of assertion. 3. Option (3) is correct. Explanation: The mirror formula is derived under the consideration that the incident rays are paraxial which means that the rays lie very close to the principal axis. Hence the mirror aperture is considered to be small. So, the assertion is true. Laws of reflection are valid for any surface plane or spherical. Hence the reason is false. 4. Option (4) is correct. Explanation: Refractive index of diamond with respect to the

pair 1

ff+x)

x

-=/2

So, critical angle for the diamond-liquid

media is sin

For total internal reflection, -45.Fort

liquid is

Using lens formula,

V=

of

angle of incidence should be greater than critical angle. Since, angle of incidence is 30°, total internal reflection cannot take So, the assertion is false but the reason is true. place. f=-30 bt 13) 30. Option (3) is correct. 5. Option (4) is correct. Explanation: Given focal length of concave lens = 32 cm Explanation: Speed of light is slower in glass compared to object is 24 from point M; U = 24 cm that in air. Hence, the refractive index of glass is more than that of air. So, the reason is true. When a double convex air bubble is formed within a glass slab, the refractive index of the medium of the bubble is less than the refractive index of M k-32 cm> the surrounding medium. Hence, the lens will not behave like f2 a converging lens. It will behave like a diverging lens. So, the 24 cm(S abg0.03 assertion is false. 6. Option (1) is correct. Explanation: According to Huygen's theory each and every 1 l point on a wavefront is the source of secondary wavelets. Secondary wavelets do not proceed backward. So, the assertion -1 is true. Kirchhoff explained that amplitude of secondary wavelets is 32 1 proportional to (1 + cos0), where 0 is the angle between the ray at the point of consideration and direction of secondary 32 24 wavelets. In the backward direction = 180; so 1 + cose 0; so, the secondary wavelets do not proceed backward. Hence, 13+4 reason is also true and is correct explanation 96 96 of assertion.T13 V=96 cm from lens 7. Option (1) is correct. Explanation: If a medium has same refractive index at every B] ASSERTION REASONQUESTIONS point in all directions, then the wavefront obtained from a 1. Option (3) is correct. source point in such a medium is spherical since wave travels Explanation: Convex miror always form virtual image. So, in all direction with same speed. Such a medium is known as true. the asscrtion ís isotropic medium. Parallel rays incident on convex mirror do not actually meet. 8. Option (4) is correct. They get reflected in such a manner that their extensions meet at a point. So, the rcason is false. Explanation: When a light wave travels from rarer to denser mediumn, its speed decreases. But this reduction 2. Option (1) is correct. of speed does not imply the loss of energy carried by the light wave. So, Explanation: u=f+x the assertion is false. Energy of wave is proportional to the Using miror formula, frequency of the wave which remains same in very medium. Hence there is no loss of energy. So, the reason is true. Or,

cmt

ien3

+l

79 Optioa (1) is correct.

Elenetien: berent

No

SOurces

interference pattem is detected wbea two are too close to cach othr. The assertion is

Fringe widhi is proportional to

When

d

becomes toosmall,

ine width boomes too large. So, the rcason is also truc of assertion sd is corTect explanation correct. eOptioa (2) is

without any devintion. 5. Option (1) is cerrect.

Eplanation:

(-04)

f=-25

So,

D

6. Option (1) is coerrect.

Eplanation: For bright fringe to be

formed

niD

tsle

re

=

(in meter)

P

etion:

Difiraction is spreading of waves around It takes place with all types of waves (mechanical, chanical, tranverse, longitudinal) and with very small uing particles (2tom, ncutron, clectron cte.) which show like property. Sa, the assertion is true. DEation is percepible when the wavelength of the wave is oonparable to the dimension of the iffracting device. Tbe sis also true but reason is not the correct erplarnation of

ight tays wili pass through

no reflection, no refraction. The

±

(n=0,

1,+2,...)

For dark fringe to be formed

(2n + 1)D

2d

ICICOMPETENCY BASED QUESTIONS 1. Optien (2) is correct. Eglereion: Using ens maler 's formula

(n=0,+ 1,+2,...)

G3

7. Option (2) is

correct.

Esplenation:

}-6\)

X

Parent source Se

f=-24 cm

S,

2 Opion (4) is correct. Eqglenetion: Using lens maker's formula

Path difference = SN In AS,S,N,

Or,

--(] }-0-() f=

So,

In AKP0, 0

12 cm

SN= D correct. Explanation: For 20th bright fringe

8. Option (3) is

1

(Whea the lens is in air)

20 mm

)

the liquid)

cm = Sivia 4 x 30 120 4. Option (4) is correct. Eplanation: Tbe refractive index of the material of the lens same, there will be d that of the surrounding mediun being

So.

D sine

being small, tan0

3 Option (2) is correct. Eplanation: Using Lens Maker's Formula,

(When the lens is in Taking ratio,

d

So,

AD =lmnm d

For Sth bright fringe,

=5x|=

For 3rd dark fringe, AD

Smm

OswaalCUET (UG) Chapterwise Questiori Bark PHYstCa

80 10.

Option (4) is correct.

25

Explanation:s

2f

mm bright fringe and 5th thc between distance So, the = 2.5 mm is fringe

2.5

3r

S,P-SPm

5-2.5

Option (3) is correct. two consecutive Explanation: Fringe width= Distancc betwecn bright fringe 9.

a When the system is immersed in medium

D

xd=

D3 For

n central maxima, putting

-2.5 mm

Pe

xd

Total path differcnce

- 3mm (given)

Then B' =

(given)

3 dark

1.22

1

1

15

10

n2. (For

=0

0 AD

3d

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DUAL NATURE OF MATTER AND

RADIATION

Revision Notes CRUX POINTS:

Scan to know more about Photoelectric effect this topic Photoelectric effect: When electron emission occurs by illumination of metal by the light of suitable frequcncy, it is known as photoelectric emission. Here, emitted Photoelectric clectrons are called photoclectrons. When effect light falls on the metal surface, free electrons absorb energy from light and if this energy is more than the work function of metal, the Scan to know electron escapes from the surface. more about this topic Hertz and Lenard's observations: > In 1888, Lenard observed that when ultraviolet light falls on zinc metal, metal becomes positively charged. With the discovery of electrons, it was established oelectric Photoel effe that this is due to emission of electrons. The current produced by these photoclectrons is called photoelectric current. > The frequency of the incident light below which ernission of electrons does not take place, is called its threshold frequency. > Intensity of light has linear relationship with photoelectric current at a potential higher than the stopping potential. > For a given frequency of the incident radiation, the stopping potential is independent of its intensity. current

to

I>l,>1,

Stopping potential

Retarding potential

Collector plate potential

>

> Photoemission current starts only at certain minimum Irequency of light known as threshold frequency. Below this frequency, photoemission does not take place in spite of the

increase in the intensity

of light.

Photoelectric Current

V3>V2>V1

Saturation

current -Vo1 -Vo2 -Vo3

+-Retarding potential

Collector plate potential->

Einstein's photoelcctric equation Particle nature of light > Wave theory failed to explain the photoelectric effect: o According to wave theory, higher amplitude means higher energy but experiments show that even larger amplitude (higher intensity) of light below threshold frequency docs not

show photoclectric effect. o According to wave theory, same intensity of different colour should have same energy but experiments show that cnergy depends upon frequency, not on amplitude. O According to wave theory, wavefront should take some was found time to give energy to clectron but cxperimentally, it that ejection of electron is instantaneous. > In 1900, Max Planck stated that clectromagnctic energy can be emited onlyin quantized form. E= hv (where, k is Planck's constant). > Based upon this postulate, Einstein established quantum theory of radiation and was able to explain photoclectric phenomenon. > It states that light encrgy packets are known as photons (Particle nature of light). > In photoelectric effect, an clectron absorbs a quantum of energy (E = hv) of radiation. If this absorbed quantum of energy exceeds, the minimum energy necded for the electron to escape from the metal surface (work function oo), the clectron is emitted with maximum kinetic energy. K.E. = hy oo (where, oo is the work function of the metal). > At stopping potential, the kinetic energy of the cjected electron is zero. Below this potential, the electrons cannot be ejected. Hence, maximum kinetic energy of an electron is calculated by K•Epmas =eV, where, Vo is the stopping potential. > Work function of metal, @, = hy, where, y, is the cut-off frequency or threshold frequency. > Maximum speed of emitted photoelectrons can be calculated as

-

m

Matter waves - Wave nature of particles > de-Broglie's postulate is based upon the symmetry of nature. If radiation has dual nature, then matter should also have dual nature. > According to his hypothesis, moving particles of matter should display wave nature under suitable conditions, He named the wave as matter wave. It is the third type of wave. It is different from mechanical wave and electromagnetic wave.

> de-Broglie proposed that the wavelength 2. known as de Broglie wavelength is associated with momentum of particle p as = h Hence, de-Broglie's wavelength of particle, A = m > Wavelength of electron-wave: Kinetic energy of clectron at potential V is K.E, =cV.

Question

Oswaal CUET (UG) Chapterwise

82

pattern.

Bank PHYSICS

scattered

peak theoretically

Anode intensity wavelerngth

diffraction

of

phenomenon

metal

work

Intensity

small

with

to

ejected

sufficient

called

on are

photoelectric

energy electrons

metal,

of is

momentum

or

light

be to as so

surface,

the

(6)

p=

,

verified

Cathode

of

used

It

= nm.

by as

Kma

Ejected

from

50°.

hc the

function

called

accelerating

tally metal

electron

When

slit. single

wavelength

particle

=

observing

must

photoelectrons

Minimum

with

the

are

the

electrons

wavelength

bring

equal

incident

J2m

given it

to

de-Broglie

Kmax

the

voltage.

effect. associated

different

angle

0.165

eVo=

an out

is

was •

scattering calculated

was

for Nickel electron

plotted

relation

ata

of

of

Photoelectric

Broglie

nature

Experment

exhibit

electrons

Observation

behaviour

Davisson

Germer

and Lenard's

Effect

wavelength.

wavelength

space was

54V

beam

at observed

Inter-atomic

diffracted

wave

wave

Hertz

academic

calculated

Electron

electron

can

of

awater

matter wave-like beam

was

The

technical

quantum

while

e.g,

All

average

waves

be

•

•

•

can like

office.

1905 forward

an

grade

in light

after

put Einstein,

of

Matter

II patent

a

Contribution

as a working

careertheory

Radicction

in

Dual,

officer

equation

nature NatureRadiation Einstein's

nature

wave

Photoelectric

definite

particle

Dual

Level

of

by it

Third

wavelength,

explained

and Map

as

potential

energy

"photons".

Elc

well

Level

c

be small as can

called

diffractionsufficiently

Second

onto

=

definite

character

hv

having

wave

out come

each

as

on

will

of

ie.,

electrons

V=

maximum

(v-vo) Threshold

-,

come depends

may electron

surface

of

behaves

When

Light

=0,

Threshold

ejected

linear

=

=

.>g

Electron

Electrons

Kmar


E

Kmar

the •

eVo

E-=

out

just

Energy

•

metal

Momentum

light

Light

Level

with

are

particdes

Trace

out come

Interference

has

used

each

and

frequency

kinetic

momentum

the

Wavelength

stopping

photon

particle.

isof

definite

= photon

=

both

Mind

energy

KE.

If

=

ie., no

If •

bg

=

of

,

Here,

=

2g=

=

Vo

Kmax

First

aAL

NATURE OF MATTER

AND RADIATION

Duting this valuc ofkinetic energy equation,

83

in de-Broglie wavelength

constant (h= 6.62607015 x 1034 J/Hz, it becomes

1.227

h

nm

V2meV Ry putting the value of mass of clectron (m = 9.1093837 x 103 kg), its charge (e =1.60217663 x 10-19C, and Planck's

Davisson-Germer experiment: Experimental arrangement:

From this equation, we can infer that the wavelength of the particle is inversely proportional to the mass of the particle and its velocity. Hence, heavier particles have shorter wavelengths. Scan to know more about this topic

+

A

F

Davisson and Germer experiment

Nickel

Electron Beam

Targete Scan to know more about this topic

Electron+

LT.

gun

Diffracted electron

Movable

beam

Vacuum Chambe

collector To

de Broglie wavelengths

galvanometer

> During this experiment the wave nature of electron was verified by observing the difraction pattern. > In the experimental set up, inter-atomic space of Nickel was taken as single slit. Scattered clectron beam was collected by

movable collector and its intensity was plotted. > The experiment was performed by varying the accelerating voltage from 44 V to 68 V. A strong peak of the intensity of scattered electrons appeared for accelerating voltage 54 Vat a scattering angle = 50°.

incident incident

direction

direction

50

incident

....e....*.**.esn**.

direction

between scattred

dscattered

between

scattred

Angle

and g

Angle

and

Intensity of scattered beam

bah

atacceleratingaa

potential 48V

eh

Intensity of scattered beam at acelerating potential 54V

> This experiment concluded that the electrons exhibit atoms diffraction when they are scattered from crystals whose are spaced appropriately. diffraction >Electron wavelength was calculated using as the 1st values Tormula of wave optics. Putting experimental space nickel of inter-atomic dark diffraction angle 50° and the nm. d=0.91Å, obtained =0.165

Intensity of scattered beam at accelerating potential 64V

> Wavelength obtained from theoretical calculation using de I.227

Broglie formula

V=

nm

0.165 nm by putting

$4V.

OBJECTIVE TYPE QUESTIONS IA]MULTIPLE CHOICE QUESTIONS

of an electron having kinetic The de Broglie wavelength energy 56 eV is: (2) 16.4 A (1) 0.022 nm (4) 0.0164 nm

(3) 0.164 nm 2

The

(CUET 2023, 7th June)

of: wavelength of matter wave is independent

(3) velocity (4) mass (2) charge 3. According to Einstein's photoclectric equation: (2) hy= Kmax (1) Kmax = hvKmay hv (3) Kmax = hv+ (4) (where all symbols have their usual meaning) (CUET 2023, 7th June) current of 4. Agraph for variation photo with anode potential is given for different intensities of incident radiation:

(1) momentum

,

,=

-o

t

Oswaal CUET (UG) Chapterwise Question Bank

84

PHYSICS

Vs Vs

Photo Current

10-15

x

Slope=4.15

Vo

(V)

Collector plate potential Choose the correct order of anode intensities. (2) 7;-l,=y () I>,>, (4) ,I (3) , Vo). So. h

(1) Slope for all metals is same > work function of B (2) Slopc for all metals is same > work function ofA (3) Slope for all metals is same > work function of B (4) Slope for all nmetals is same > work function ofA

Here

work function ofA

Here work function ofB

-

Here work function

of

A

h. Hlere work function

of

B

A.

(CUET 2022, 26th August) 15, Photoclectric cmission occurs only when the incident ligbt

has more than a certain minimum.

(1) Intensity (3) Specd 16.

(2) Angle of Incidence (4) Frequeney

(CUET2022, 26 August) The work function for a metal surfaco is 4.I4eV The threshold wavelengtth fur this metal surfaco is

(1). 4125A

(2) 2062.SA

(3)

3000 4)

6000A

85

DUAL NATUREOF MATTER AND RADIATION

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly (1) 1.2 nm (3) 1.2 × 10° nm 18.

(1)

(2) (3) (4) 19.

(2) 1.2 x 10 nm (4) 1.2 x 10 nm [NCERT Exemp. Q. 11.2, Page 68] Kinetic cnergy ofelectrons emitted in photoclectric effect is directly proportional to the intensity of incident light. inversely proportional to the intensity of incident light. independent of the intensity of incident light. independent of the frequency of light. Threshold wavelength of a photoelectric emission from a material is 600 nm. Which of the following illuminating

source will emit photoclectrons? (1) 400 W, infrared lamp (2) 10 W, ultraviolet lamp (3) 100 W, ultraviolet lamp (4) Both (B) & (C) 20. Photoclectrons emitted from a metal have (1) different speeds starting from zero to certain maximum. (2) same kinetic energy. (3) same frequency. (4) Both (2) & (3) 21. At stopping potential, the kinetic energy of emitted photoelectron is (1) minimum. (2) maximum. (4) cannot be predicted (3) zero. 22. A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to

(2) HI2

(1) H

H2

(3) H° () [NCERT Exemp. Q. 11.1, Page 681

vvi

an initial velocity 23. An electron is moving with and is in a magnetic field B= B,j. Then, its de Broglie

wavelength (1) remains constant. (2) incrcases with time. (3) decreases with time. (4) incrcases and decreases periodically. (NCERT Exemp. Q. 11.6, Page 69] 24. Anelectron (mass m) withan initial velocity v =v,i(v, >0) = constant in an electric field E=-E,i (E, at Broglie wavelength time t is given by is L

0). Its de

>

(2)

(1) m

v

o

(4) (3) 2o 25. Consider a beam of electrons (cach clectron with energy E) incident on a metal surface kept in an evacuated chamber. Then, (1) no electrons will be emitted as only photons can enit electrons. (2) electrons can be emitted but all with an energy, Eç (3) electrons can be emitted with any energy, with a maximum of Eg- (G is the work function). (4) clectron can be emitted with energy, with a maximum of Eg (NCERT Exemp. Q. 113, Page 68]

26. Consider the figure given below. Suppose the voltage applied to A is increased. The diffracted becam will have the maximum at a value of 0 that (2) will be the same as the carlier value (1) will be larger than the carlier value, (4) will depend on the target. (3) will be less than the carlier value

+

onHI S

A

Nickel

Electron Beam

Target

Electron

gun

:L.T

Diffracted electron

ops 27. Two particles A, and A, of masses m,

m, (m> m) have

the same de Broglie wavelength. Then, (1) their momenta are the same. (2) their energies are the same. energy of (3) energy of A, is more than the (4) Both (1) and (3)

u

NCERT Exemp. Q. 114, Page 69)

28. The momentum of a photon is 3.3 x 10 frequency is

(1) Az.

Vacuum Chamber

beam

Movable collector

(2) (3)

kg

10 3xHze

6x 10 Hz 7x 102 Hz

(4) 1.5x 10

Hz

sd

:

ms. Its

86 29. (1) (2) (3) (4) 30.

Oswaal CUET (UG) Chapterwise Question The rest mass of photon is Infinite Zero More than zero Equal to that of an clectron Photon in motion has mass

(1) 0

CICOMPETENCY

st (2)

I.

s

9)

hy

S

(4) hv

[B] ASSERTION REASON OUESTIONS Question Nos. 1 to 10 consist of two statements -Assertion and Reason. Answer these questions by selecting the appropriate option given below: l. Both assertion (A) and rcason (R) arc true, and reason (R) is the correct explanation of assertion (A). 2. Both assertion (A) and rcason (R) are true, but reason (R) is not the correct explanation of assertion (A). 3. Assertion (A) is true, but reason (R) is false. 4. Assertion (A) is false, but reason (R) is true. 1. Assertion (A): The photoelectrons produced by a t monochromatic light beam incident on a metal surface have a spread in their kinetic energies. Reason (R): The energy of electrons emitted from inside the metal surface, is lost in collision with the other atoms in the metal. 2. Assertion (A): Energy of moving photon varies inversely as the wavelength. x Reason (R): Energy of the particle = Mass (Speed of light) 3. Assertion (A): Ifthe frequency of an incident photon is twice the threshold frequency then two electrons are emitted. Reason (R): According to Einstein's equation, Energy of photon = Maximum kinetic energy of photo electron work function. 4. Assertion (A): de Broglie wavelength is significant for microscopic particles. Reason (R): de Broglie wavelength is inversely proportional to the mass of a particle when velocity is kept constant. 5. Assertion (A): Ifa proton and electron are moving with same

velocity, then wavelength of de Broglie wave associated with electron is longer than that associatèd with proton. Reason (R): The wavelength of de Broglie wave associated with a moving particle is inversely proportional to its mass. wave 6. Assertion (A): Photoelectric effect demonstrates nature of radiation. Reason (R): The number of photoelectrons ejected is proportional to the intensity of incident radiation. 7. Assertion (A): Photosensitivity of a material increases when its work function decreases. Reason (R): Work function = hvg 8. Assertion (A): In photoelectric emission, all the emitted photoelectrons have the same kinetic energy. Reason (R): In photoelectric emission, incident photons transfer its whole energy. 9. Assertion

(A): Electron microscope offers higher resolution than optical microscope. Reason (R): Electron beam as a wave has much smaller wavelength compared to visible light. 10. Assertion (A): According to de Broglie, a wave is associated with cach moving particle which is called matter waves. Statement II: Heavier particles have larger wavelengths.

Bank PHysi

BASED QUESTIONS

fronm Based on following passage answer questions a technological application is photocell A Photocell: a device whose electri the photoclectric effect. It is A photocell consists of light. arc affected by properties metal plate C (emitt photo-sensitive semi-cylindrical in an evacuate supported a (collector) A and wirc loop to the external cire glass or quartz bulb. It is connectcd ammeter (u B micro and having a high-tension battery as shown in the Figure.

Incident light

Emitter.

ItCollector

A photo cell to enter it A part of the bulb is left clean for the light on the emitte falls wavelength suitable When light of C. photoelectrons are emitted. These photoelectrons ar drawn to the collector A. Photocurrent of the order ofa fe a microamperes can be normally obtained from photo cel A photocell converts a change in intensity of illumination S into a change in photocurrent. 1. Photocell is an application of (2) photoelectric effect (1) thermoelectric effect (4) None of the above (3) photoresistive effect 2. Photosensitive material should be connected to (1) Negative terminal of battery s (2) Positive terminal of (3) Either positive or negative terminal of battery (4) ground 3. Which of the following statement is true? (1) The photocell is totally painted black. (2) A part of the photocell is left clean.

batteryti

The photo cell is completely transparent. A part of the photocell is made black. The photocurrent generated is in the order of Ampere (2) Milliampere Microampere (4) None of these A photocell converts a change in of incident light into a change in (1) Intensity, photo-voltage (2) Wavelength, photo-voltage (3) Frequency, photo-current (4) Intensity, photo-current Based on following passage answer questions from 6-10. A photon is the smallest discrete amount or quantum of elcctromagnetic radiation. It is the basic unit of all light. According to Einstein, photons have energy equal to ther frequency times Planck's constant. The intensity of the light corresponds to the number of photons. The basic properties of photons are: ) They have zero mass and rest energy. They only exISt moving particles. () They are elementary particles despite lacking rest mas. (iül) They have no electric charge. (3) (4) 4. (1) (3) 5.

I.

s

DUAL

NATURE

(tv) Tbey

MATTER

AND RADIATION

87

are stable.

They çarry cneEy and momentum which are dependernt on the frequency.

.nThey can have interactions

with other as clectrons, such as the Compton cffect. particles such si0 Theycan be destroyed or crcated many by natural processcs, for instance when radiation is absorbed or cmitted. wi) In frece spacc, they travel at the specd of light, K Photons have energy cqual to their frequency times constant Rydberg's (2) Planck's constant () (3) Avogadro's constant (4)) Boltzmann constant 7. The intensity of the light corresponds to () Number of photons (2) Speed of photons (3) Energy of photons (4) Frequency of photons

Charge of a photon is

8.

(2) +e (4) None of the above (3) 0 9. Which of the following statements is wrong? (1) Photons only exist ss moving particles. (2) Photons carry energy and momentum. (3) Mass of photon is equal to the mass of electron. (4) Photons travel at the spced of light. 10. Which of the following staternents is wrong? (1) Photons can neither be destroyed not created. (2) Photons can have interactions with other particles. (3) Photons are clementary particles. (4) Photon is the basic unit of all light. (1)-e

ANSWER KEY [A]MULLTIPLE CHO1CE QUESTIONS 1.(3) 11. (3)

2. (2) 12. (1)

21.(3)

22. (4)

3.(1) 13.(4) 23. (1)

4. (3)

5.(2)

6. (4)

|7.(3)

8.(4)

9.(2)

10. (2)

14. (2)

15. (4)

16. (3)

19.(4)

20.(1)

25. (4)

26. (3)

17.(2) |27. (4)

18. (3)

24. (1)

28. (4)

29. (2)

30.(3)

8.(4)

9.(1)

10. (3)

(B] ASSERTION REASON QUESTIONS

1.(1)

3.(4)

2. (2)

4.(1)

5.(1)

7.

6. (2)

(1)

IC| COMPETENCY BASED QUESTIONS 1.(2)

3.(2)

2. (1)

|4. (3)

|5.(4)

8.(3)

|7.(1)

6. (2)

|9.(3)

10. (1)

ANSWERS WITH EXPLANATIONS Photo

[AJMULTIPLE CHOICE QUESTIONS 1.

cCurrent

Option (3) is correct.

Explanation:

h

=mE 6.6x104 2x9.1lx10-*56x1.6x10"

Or,

2=0.164 nm correct.

2. Option (2) is

Eplanation: ),

=

h

h

p So, it depends on mass, velocity and momentum. But, it is independent of charge. Option (1) is correct. on the a Explanation: Einstein theorised that when photon falls to the is energy transferred of the photon surface of a metal, the remove clectron the to energy is used clectron. A part of this to the cjected clectron as from the metal and the rest is given = hy kinetic energy. their usual mcaning. have Where all symbols 4. Optlon (3) Is correct. as intensity Explanatlon: For a fixed collector plate potential, increases of incident light increases, emission ofphotoelectrons and the photo current increases. photo current also As collector plate potential increases, a constant value. This current reaches Incrcases and ultimately is increased. does not increase further unless the intensity

K

my

-o

Collector plate

"I,>l,>, potential 5. Option (2) is correct.

Explanation: de-Broglie wavelength of a moving particle, h my

,

Since, >hy s0 v, < iy and the velocity of the particke becomes greater at a point nearer to the origin on an clliptical orbit. 6. Option (4) is correct. Explanation: The photon of bue light has higher energy as compared to red light. So, the red light emits electrons of reduced kinetic energy than that of blue light and hence the stopping potential also decreases.

s

7. Option (3) correct. Explanation: de -Broglie wavelength of electron 1.23

nm

123

,

VI00

=0.123 nm

Oswaal CUET (UG) Chapterwise Question Bank PHYSICe

88 So, the wavelength is comparable to the wavelength - nm. Range of wavelength of X-rays is (0.01 10)

of X-rays.

hv-

W

hv

Option (4) is correct.

8.

=

Explanation: We know that

eVo

hv,

of straight line, y = mx +c

On comparing it with equation

J2mgV

With V remaining constant

e mq

C

(constant)

-e

of metal B is more than metal function of metal B is more than metal A thercfore, work

A

Since, (mq),< (mg), < (mg)a So, Ag 9. Option (2) is correct.

As threshold frequcncy

Explanation:

is the minimum frequency emission. to photoclectric initiate is which required

mass of 2c. m

clcctron. So, wavelength of clectron > wavelength of proton. So, the assertion is also correct and reason is the correct explanation of assertion. 6. Option (2) is correct. Explanation: Photoelectric effect does not demonstrate particle nature of radiation. It demonstrates the wave nature of radiation. So, assertion is false.

The number of photoelectrons cjectcd is not proportional to tf frequency of radiation but intensity of incident radiation. s reason is also true. 7. Option (1) is correct. Explanation: Work function =hv Where h = Planck's constant and vo is the threshold frequenc Hence the rcason is correct. Increase of photoscnsitivity mean it is casier to cjcct photoclcctrons from a material, ic., energ requirement is less. It means that the work function of th material is less. So, photosensitivity of a material increase when its work function decreases. The assertion is also corree but the reason is not the correct explanation of assertion. 8. Option (4) is correct. Explanation: In photoclectric emission, incident photons transfer its whole energy. So, the reason is corrcct. Allthe electrons do not come out from the surface. Electrons coming out from inside the mctal surface, face collisions with the other atoms in the metal and in the process lose of cnergy. So, they cannot have same kinetic energy. Hence, the assertion is incorrect. 9. Option (1) is correct. Explanation: Resolution is inverscly proportional to the wavelength. Wavelength of visible light is about S000 times larger compared to the wavelength of clectron beam. So, clectron microscope offers much higher resolution than an optical microscope. Resolution of optical microscope is less than 200 nm, Resolution of clectron microscope may go upto 0.I nm. 10. Option (3) is correct. Explanation: According to de Broglie hypothesis, moving particles of matter should display wave nature under suitable conditions. This wave is named as matter wave. Hence the assertion is true. de Broglie's wavelength of particle is given

by à =

my

So,

m.s

So, heavier particles have shorter wavelengths. Hence, the reason is false.

JCICOMPETENCY BASED QUESTIONS correct. Explanation: Photocell is a technological application of the 1. Option (2) is

photoclectric effect. 2. Option (1) is correct. Explanation: Photosensitive material used as emitter should be connected to negative terminal of the battery so that the emitted electrons are repelled by emitter and collected by collector. 3. Option (2) is correct. Explanation: A part of the bulb is left clean for the light to enter in it. 4. Option (3) is correct. Explanation: Photocurrent of the order of a few microampere can be normally obtained from a photo cell. 5. Option (4) is correct. Explanation: A photocell converts a change in the intensity of illumination into a change in photocurrent.

DUAL

NATURE OF MATTER AND RADIATION

correct. 6. Option (2) is Explanation: Photons have energy equal to their frequency times F Planck's constant. E= hy. 7. Option (1) is correct. Explanation: Intensity per 8.

unit area. Option (3) is

of light

depends on photons per second

correct.

91 9. Option (3) is correct. Explanation: Photons have zero rest mass. 10. Option (1) is correct. Explanation: Photons can be destroyed or created by many or natural processes, for instance when radiation is absorbed emitted,

Explanation: Photon is an uncharged particle.

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8

ATOM AND NUCLEI All positively charged particles are together at one location

Revision Notes

at centre called nucleus.

Alpha - particie scattering experiment > Experimental set-up:

Size

of

m Size

of

the nucleus is calculated to be about 10-1 m to 104 one atom is of the order of 10-10 m.

Rutherford's model of atom: > Rutherford overtuned Thomson's model in 1911 with

Vacuun

alpha-particle scattering experiment in which he demonstrated that the atom has a tiny and heavy, positively charged nucleus > In his model, the atom is made up of a central charge, the nucleus, surrounded by orbiting electrons in circular orbits at the speed of light.

Gold foil

Screen

Source of a-particles Bi was taken as collimated source Radioactive element of alpha particles. Thin foil of highly malleable heavy metal gold was used as target. The detector was made from ZnS. > Alpha-particle trajectory:

Target nucleus

> Impact parameter: It is the perpendicular distance between the direction of the given a-particle and the centre of the nucleus. It is represented by b. > Distance of closest approach: It is the distance between centre of nucleus and the a-particle where it stops and reflects back. It is represented by 'd.' This distance gives an approximation of nucleus size. 1

2eZe

d= 4nE, K

> Experimental observations: Most of the a-particles passed roughly in a straight line (within 1°) without deviation. A very small number of a-particles were deflected. (1 out of 8000) > Conclusions: Most of thespace in the atom is mostly empty (only 0.14% Scatters more than 1°).

> Limitations of Rutherford model: o It could not explain the stability of

Scan to know the atom. Rutherford applied classical more about are So, the electrons electromagnetism. this topic attracted to the nucleus. As the electron orbits, it accelerates. Because of the acceleration, the electrons should fall inward toward the nucleus. But, that would Nuclei make any atom highly unstable! But this does not happen since atoms in the natural world are stable. o It could not explain the nature of energy spectrum. Bohr model > Bohrpresented his atomic model with the three following

postulates: O An electron can revolve in certain stable orbits without emission of radiant energy. These orbits are called stationary states of the atom. o Electron revolves around nucleus only in those orbits for which the angular momentum is the integral multiple of h2, where, h is Planck's constant. Hence, angular momentumn, L = nh/2n > An electron mnay make a transition from one of its specified non-radiating orbit to another of lower energy. When it does so, a photon is radiated having energy equal to energy difference between initial and final state.

ATOM AND NUCLET

93

wavelength

-13623

tube through

KZe

E

by sealed

1216 6563

different

region]

series

region]

Amax

in

of

each

components

radiation

gas

(U.V.

in

[visible

2,3,4..

,

series

5...

=

yax

Anoy

6... [IR

4,

4,5,

n=3, A, series

series

=3648

n,==912

Wavelength

=

=

region]

Anax

yy0y

,

8......

n,=5,6,7...

region]

(IR

IR

emits

Paschan

Lyman

appear.

Balmer

1,

n,= min

n=2

n;=3 hmin

=

=

Brackett

n=4 Amin

Amin

KZe

=23850

14592

n= 8208 series

n

me

&e3h

13.6

2r, Energy

=5 min

n

=

Energy

fund

E,

Total

n;

P

z n,=6,7,

Å

Hydrogen

prism

2

E=

=

region]

=

passed

a heatedwhen

U,

40533

18761

2r,

+

74618

Å

given

Kinetic

-eV

4ehn

Z

Ze

-27.2 n'

-me

K

Z*

is orbits

Hydrogen

energy

ev

also

model

model

releases stationary

are

higher

first

1

of

planetary

the

=-13.6ev

each

plum-pudding

prepared

in

prepared

tottal

from level,

Potentia

energy

n'

theory.

a

proposed

it

atom

of

=

levels.

n=1,2,3..

electron

=E-E,

stationary

quantum

Model

136

jumps energy

number

Bohr

E,

Rutherford

Bohr

as

on Niels

known

of

by

Energy

Given

lower

energy

10

Where

•These called•When

to

=

AE

energy.

level

Z=

Atoms n

*Peyy

me'

atom.1913,

1898,atom 1911,

2.2

atomic

based

JJ

x

Velocity of electr oro in Bohr's orbit

nth

electron

Thomson

s

Ener

Level

orbits

model

Energy

Hydrogen Spectrum

Energy

model.

E,h atom

Limitations

of

InofIn of In

T

=

Zeeman's

spectrum Contribution

=0.53

multi

K,

of atoms/ions

effect.

angle

system.

angle

explain explain Postulates

scattering

small

Rutherford's

Stark's scattering

complex

Model

to Fails

experiment

a

awbacks

with

electron

Doesn't

and

a-particle

or foil

by deflected

undeviated

integer

gold

10m electrons

on

orbits.

were

thin

Postulates

the explain

particles parameter

/2

bombarded

8000 particle

a-

of

2

Doesn't stability

a-

of 1

atomic

Impact

s

in nucleus

Atoms

Level

or

nx Angular

=

of Size

I Irace the Mind Map p

Third

Electron

revolve.

atoms.

2r

h

nucleus

•

Number.

known

positively

out

hydrogen spectrums

as

revolves

have

Second Level

of

where,

nucdeus

a

First Level

like

n

momentum

around

b=

1

•

fermi

core which

principal

is

stationary

charged Doesn't

of of Most

around

spectra

central,

Ze'cot

a-

=

explain

4nem

9

particles

atom

an

called

the

passed

electron

massive,

Quantum

[Het,

mvT,

is It

the •

Lit*]

explains hydrogen

also

It •

Oswaal CUET (UG) Chapterwise Question Bank PHYSCa

94 MeV

more

c²

of

x -M]

200

or two

N

m

+

3,n'

No.

sBa+sK+

-

-Z)m]

X

V=

+(A neutron

=B.EJA

nucleus

R,=

R=

Level

+

of

of

of

mass

mass

mass

10

=

=

M=

B.E.=

m,

the kgm

X

of

Trace

Ax1.67

radioactivity

rx(1.2x

Level

10"

volume

mass

First

x

Ut

= a

AMe

AM= m,=

Second

nucleus

heavy nucleon

(Zm,

Mind

N=

Mass

of

(A

105xA

nucleus

A

[Zm,

Map

-dN/dt

RA 1.2

VcA

4

nucleus

NoeAt

=

1615

(mass

into

M

Third

(independent

no.)

-Z)m,

proton,

Level

A)

Radius

2.9

nuclei.

Volume

of

of

Splitting

per

Law

Nuclear

Density

lighter

B.E,

•

0.693

Nuclear

e.g.,

1/,

NudearD

•

comparatively

t=

emission

Nucleus

heavy

radiations.

and is

Radioactivity

energy

Defect

fission

of

=0.693/. No(1/2)n

elements

Disintegration

into

Size

that showed

of

of

Nuclear

mc²

i

Mean

by

BindingEnergy

mass

t/ty2)

y

and

E= elements

Mass

form

t2

higher ß

relation

a,

N= =

(n

of

another

Einstein

a

Mass-Energy

Relations

gave

an Nucle charge.

Chadwick

-

particles

Contribution

neutrons

electrical

James

wave. power.

Composition

elementary

and

discovered

any

of

Nuclei

power. atomic

e.m.

Properties

Nuclear

1932,

rays penetrating

fusion

of

Y

devoid

In

trequency

p,

of

ionizing

a,

Types

in

change

no. d-particle

(energy)

Isotopes

Highest

B-particle

mass

Least

High

No.

the neutrons.

a

in

•

form

Isotones

MeV

•

•

neutrons

Isobars

to Number

Number protons

having

No.

& neutrons

Mass

ion. power.

particle

of of name

H+}H}H+

nucleus

A

two

No.(A)

of

Same

e.g

No.

of Mg

e.&,C",,N4

Combining

whose number

•

heavy

e-grNa Same

Highest

Doubly

Different

of

no.

•

4. atomic

no. ionizing

in

Least

electron.

mass

of in

Moderate

decrease

•

mass

change

Decay

and

•

no.

one

and

Less

An •

by

ß

Decay

e.g.,

Atoms

by

penetrating

by a

ionized

nucleus

mass

N=

ionizing

one

same

„X, •

Atomic

Z=

is

Nuclei

,C",

no.

particle

atomic 4.

nuclei atomic

Z+

consist

Common

N=

1

by power.

penetrating

Different nucleons

power.

lighter

No.(Z)

C4

and

2

Helium

atomic

and causes

ettv+

(N) neutrons

mass

protons

causes

power.

(Z) and

0.42

nuclei

No.(A)

•

•

•

•

ATOM AND

NUCLEL

95 Increasing energy of orbits

=3

energy level to higher > When electron jumps from lower cnergy. energy level it absorbs Scan to know > Emission atomic spectra: When more about an excited electron comes down from this topic energy level, higher energy level to lower radiation of particular wavelength is emitted. This atomic spectra is known as emission spectra. energy > Absorption atomic spectra: If Binding and fission clectrons are excited, they absorb the radiation corresponding to their emission energy lcvel. Black spectra and transit from lower to higher spectra. This lines appear in the same places of the emission as spectra. type of spectra is known absorption > Hydrogen spectrum: > Lyman series: Lines emitted by Scan to know more about transitions of the electron from an outer orbit

n=2

Aphotonis emitted with eergy E=hv

Speed of electron in V,

n

=

orbit.

e n 4ne,

(h/2r)

Radius ofn" orbit,

of

-}

4TE e

For innermost orbit n = 1; the value radius ao

of

r,

x

Tme Total energy

of

an electron

E,

is known as Bohr's

=5.29 10-"m in

1

0.53A

orbit,

=ev

to the n=

This is in UV range. Longest wavelength = 1216 Åite

> Limitation of Bohr's atomic model: o Bohr's model is for hydrogenic

atoms. It does not hold true for a multi electron model. o This model fails to explain the effect

of magnetic field on the spectra of atoms (Zeeman effect). o This model fails to explain the effect of electric field on the spectra of atom (Stark

lonization

Scan to know more about this topic

> Balmer series: Lines emitted by transitians ofthe electrons of quantum number n > 2 to the n=2artit. from an outer orbit

Bohr's atomic model

5,..

Longest wavelength =6566.4 Shortest wavelength 3648 This is in visible range. > Paschen series: Lines emitted by transitions of the electrons from an outer orbit of quantum number n >3 to the n=3 orbit.

n=4,

5, 6,

..s2

Longest wavelength = 18761.14 Shortest wavelength 8208

This is in far infrared region.

> Brackett series: Lines emitted by transitions of the electrons Energy (eV) 0.00

S excited state, excited state,

3Cxcited state,

excited state,

-3.40

n2

atom

n=3,4,

effect).

Energy levels & Hydrogen spectrum Level

Spectral lines of Hydrogen

=912

-13.6

n'

this topic

l orbit.

... n=2, 3, 4, 5,

Shortest wavelength

Negative value shows that electron is bound to nucleus.

Ist excited state,

quantum number n >

from an outer orbit of quantum number n >4 to the n=4 orbit.

n=$, 6,.... Longest wavelength = 40533.33 A & 2sG = Shortest wavelength 14592 This is in far infrared region. > Pfund series: Lines emitted by transitions of the electrons from an outer orbit of quantum number n>4 to the n=4 orbit.

n=6,7,,8. Longest wavelength =74618.1 A Shortest wavelength = 22800 A This is in far infrared region.

Ground State n=]

-13.60

Composition and size of nucleus > The atomic nucleus is the small, dense region consisting of protons (positively charged particles) and neutrons (electrically neutral particles) at the centre of an atom.

Owaal CUET (UG) Chapterwise Question Bank PHYSICe

96 > The diameter of the nucleus is in the range of 1.70 fm for Hydrogen to about l1.7 fm for Uranium. >A nucleus of nass number A has a radius where, Ro 12 x 10-15 m R= > Nuciear matter density = 2.3 x 10!" kg

R4S

m

> The composition of a nucleus is described using the folowing terms and symbols: Z Atomic number = Number of protons (cqual the number of clectrons) N= Neutron mumber - Number of neutrons A= Atomic mass number = (2 + M) = Total number of

protons and neutrons An ztom is represented as X, where X= Symbol of element, A=Mass number, Z= Atomic number. Atomic miss > The atomic mass is the mass of an atom. Atomic mass is often expressed in the unifîed atomic mass unit (u). as of themass of a carbon-12 atom at rest >luis defined h in its ground state. The protons and neutrons of the nucleus account for Dearly all ofthe total mass ofatom, with the electrons znd muclear binding energy making minor contributions. Thus, the numeric value of the atomic mass when expressed in u is Deariy the same value as the masS number. Isotopes, Isobars, Isotones > Isotopes : The atoms of an element having the same atomic nber (z is same) but different atomic mass numbers (ue to the ifferent nunber of neutrons) are said to be isotopes. Examples: > Isobars: The atoms of different elements having the same m2ss number but different atomic numbers are said to be

;c, c,

c

isobars.

Examples:Fe,

Ni

> Isotones: The atoms of different elements having different mass numbers and atomic numbers such that their difference is sarne are said to be isotones. It means they have same number of neutons. Examples: 0, "N,

C

Radioactivity

- alpha, beta, and gamma particles/rays,

2nd their properties

> Radioactivity: When atoms of some elements become very hezvy, neutrons are unable to bind and some nucleons keep on leaving the nucleus. These elements are known as radioactive elements and the process of spontaneous ejection of nucleons 2nd radiztions is known as radioactivity. > There are 3 types of radioactive decay in nature: O a-decay: In this decay, a-particles eject out. o Bdecay: In this decay, electrons or positrons eject out. o y-decay: In this decay, high energy photons are emitted.

> Properties of a-rays:

o These are doubly ionized helium atoms. o Afier a-decay, a nucleus transforms into

a different

X

t+(He

Mass number of daughter nucleus reduces by 4 and atomic number reduces by 2. o

U

Example: Th+jHe o Deflected by clectric and magnetic field. o Velocity is about 10% of the velocity of light. o Affect photographic plates.

power when

externally exposed to.

> Properties of f-rays:

o These are fast moving electrons or positrons.

o In B decay, an clectron is emitted from nucleus alone with antincutrino. [Therc is no clcctron in the nucleus. A neutron splits into a proton and an clectron.] o After B*-decay, a nucleus transforms into a different muclee X

,4Y+e

+ v

o Mass number of daughter nucleus remains same. Atomie number increases by 1. Example: S+fe + o B* decay, a positron is emitted from nucleus along with neutrino. [In this process, a proton splits into a neutron and a positron.] o After B-decay, a nucleus transforms into a different nucleus.

P

X Y+fe

+v

o Mass number of daughter nucleus remains same. Atomic number decreases by 1. o Example: C 9B+e+ o Deflected by electric and magnetic field. o Velocity is about 90% of the velocity of light. o Affect photographic plates. o Have moderate ionizing power. o Have moderate penetration range. o Have some biological damaging power when extemally

exposed to.

> Properties of y-rays: o When an excited nucleus makes a transition from lower

energy state, y radiation is emitted. These are photons of very short wavelength

o

10-m).

o

(10-m to

Carry no charge.

o Neither deflected by magnetic field nor by electric field. o Velocity is equal to that of the velocity of light.

o Affect photographic plates. o Have very weak ionizing power. o Have very long penetration range. o Have very high biological damaging

power when externally exposed to. Radioactive decay law > dNIdt =-W (a= disintegration constant) Integrating both sides, N() = Nge > Alternative fornm of radioactive decay:

R=

ucleus.

o

o Have very high ionizing power. o Have very short penetration range. o Have very little biological damaging

Roe

(Ro = Radioactive decay rate at

=0 andR=

radioactive

decay rate at any subsequent time ) SI unit of radioactive decay is becquerel. I becquerel is I decay per second. I curie= 3.7 x 10° Bq > Half life time (T): It is the time period in which both and R reduce to half of initial values.

In2 =0.693/

ATOM AND NUCLEI

97

life (t): is the time valucs. initial lof It

Mean

at

which both Nand

Relation between half life time and

>

mnean

R

reduceto

life:

= tln2

so from the formula

of half

MeV.

life time and radioactive decay,

NIN, = (1/2y"

E= me

Where E is the equivalent energy of mass m. c is the velocity of light in vacuum. > This enables the understanding of nuclear masses and the interaction of nuclei with each other (fusion, fission, etc.) and evolution of energy. According to law of conservation of energy, the initial energy and final energy of such reactions are same provided the mass-energy is taken into consideration. Mass defect

AM= [Zm,

+

(4-Z )m,]-M

Binding energy per nucleon and its variation with mass number > Nuclear binding energy of a nucleus that quantity of energy which when given to nucleus, its nucleons will become free and will leave the nucleus. > Binding energy, E, = AMc? > Binding energy per nucleon (E, A): The average energy required to remove an individual nucleon from nucleus is the binding energy per nucleon. It is the ratio of the binding energy E; of a nucleus to the number of the nucleons. > Variation of Binding energy per nucleon (E, lA) with mass number (4): (MeV)

a

per nucleon is quite

Nuclear fission and fusion > Ifa nucleus of lower binding energy is converted into higher

are twO methods binding energy, then energy is released. There energy into higher nucleons binding lower of converting energy nucleons. binding > Fission: When a heavy nucleus (low binding energy per nucleon) is broken into two lighter nuclei (higher binding energy per nucleon), the process is known as fission. In this process huge amount of energy is released. Example: When a neutron bombards a Uranium target, the Uranium nucleus breaks into two nearly equal size fragments releasing huge amount of energy. Ba + yKr +3,n

n+U

on+U

> The difference in mass of a nucleus and its constituents, AM, is called the mass defect.

SFeMo

to

>For A 170, binding energy

low.

Mass-energy relation b Einstein's theory of special relativity established the fact hat mass is another forn of energy. Also, it is possible to convert mass-energy into other forms of energy. According to Einstein, mass-energy cquivalence relation is given by,

nucleon

more stable is the > Higher the binding energy per nucleons, element. mass number are in the > Most of the atoms where atomic energy per nucleon is fairly range 30 Fusion: When two light nuclei (of low binding energy per nucleon) fuse and form one nucleus of higher binding energy per nucleon, the process is known as fusion. In this process huge amount of energy is released. }H + H +4.03 MeV Example: H+}H }He +tn+3.27 MeV process gives more energy than fission process. Fusion Source of energy of stars is fusion process.

¡H+H

Nuclear force: > This high binding energy per nucleon counters the repulsive force betvween protons and bind both protons and neutrons into the tiny nuclear volume. > The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses. But it is a short-range force.

Heo IREN

per energy H

Binding

150

100

50

200

250

Mass number (A)

OBJECTIVE TYPE QUESTIONS [A]

MULTIPLE CHOICE QUESTIONS

1.

Match List - I with List - II

08List -I (Scientist) (A) Rutherford (B) A.H. Becquerel (C) Niels Bohr

s

List -II (Phenomenon) () Nuclear Model of atom atom (II) Plum-Pudding Model of (II) Radioactivity

(D) J.J. Thomson

() Hydrogen atom model Choose the correct answer from the options given below: (1) (A)-0), (B)-(III), (C)-(IV), (D)-(I) (2) (A)-(II), (B)-(), (C)-(1), (D-(IV) (3) (A)>-(TV), (B)-(I), (C)-(II), (D-) (4) (A)-(I), (B)-(ID, (C)-(I), (D)-(T) (CUET 2023, 7th June)

98 The correct scquence 4, atomic spectral series is :

2.

B,

C, D, E with reference to

(A) Lyman series

Series () Lyman

Pfund Series (iü)) Paschen Series (iv)

(o) Bracket series (E) Paschen series

Choose

(4) (), (iii), (v), (ii), (iv) (CUET 2022, 6" August) mass in the ratio 1:27 the number 9. Two nuclei have ratio of their radii and densitics should be respectively (2) 1:9 and 1:1 (1) 1:1 and 1:3 1:1 (4) 1:1 and 1:9 (3) 1:3 and (CUET 2022, 6" August) 10. The Bohr model for the spectra of an H-atom

--

A.

B.

C. D.

(1) (3)

11.

of initial amount remains unchanged after 2 hours? (1) (2) 45 min

(3) 30 min (4) 15 min (CUET 2023, 7th June) The variation of the number of undecayed nuclei with time is best represented by (Where No -initial number of atoms) No

ea

(1)

N

(2) time

time

No (4) tume

6. (1) (2)

(3) (4) 7. (1) (2) (3) (4)

8.

12.

1

time

(CUET 2022, 5 August) Bohr model of atom is valid for Only hydrogen atom Only one electron atom Only one electron atoms and ions (CUET 2022, 5th Augut) All the atoms Which of the following is correct for muclear forces? Nuclear forces are much weaker than Coulomb force. The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometre. Nuclear forces are charge dependent. Nuclear forces are long range forces. (CUET 2023, 5h August) spectrum has mainly five atom The study of hydrogen series. Choose the correct sequence of these series in ncreasing order of their shortest wavelength.

Will not be applicable to hydrogen in the molecular form a Will not be applicable as it is for He-atom. Is valid only at room tempcrature. Predicts continuous as well as discrete spectral lines. Choose the correct answer from the options given below: (2) A, B,C only A, B only (4) B, C, D only A, B, D only (CUET 2022, 6'H August) In a Geiger Marsden experiment, an alpha particle of energy 6 meV hits a nucleus of Z= 72. The distance of closest approach nearly is (2) 45fm (3) 25fm (4) 30fm 35um (CUET 2022, 8m August) Arrange the following in increasing order of binding energy per nucleon

A. Li B. Fe C. O D. N E. H Choose the correct answer fromn the options given below: (1) E 1 (1) 0 wave is 17. Number of side bands of an amplitude modulated

(1) 10%

(3) 5% 12. Which

1

(1) (3)

1

(2) 2

3

(4) 0

130

Oswaal CUET (UG)

Chaptervise

Ouestion

Bank

PHYSICS

18. The last stage of production ofAM waves using squarc law device is a

(1) Band pass filter (2) Band reject filter (3) Low pass filter (4) High pass filter 19. The last stage of AM wave detector using square law device is a (1) Band pass filter (2) Band reject filter (3) Low pass filter (4) High pass filter 20. Three waves A, B and C of frequencies 1,600 kllz, 5 MHz and 60 MHz. respcctivcly are to be transmitted from one place to another. Which of the following is the most appropriatc mode of communication? (1) A is transmitted via space wave while B and C are transmitted via sky wave. (2) A is transmitted via ground wavc, B via sky wave and C Via spacc wave. (3) B and C are transmitted via ground wave while A is transmitted via sky wave. (4) B is transmitted via ground wave while A and C are transmitted via space wave. |NCERT Exemp. Q. 15.1, Page 98] 21. A 100 m long antenna is mounted on a 500 m tall building. The complex can become a transmission tower for waves with 2 (1) ~400 m. (2) - 25 m. (3) - 150 m. (4) ~ 2,400 m. |NCERT Exemp. Q. 15.2, Page 98] 22. A speech signal of 3 kHz is used to modulate a carrier signal of frequency 1 MHz, using amplitude modulation. The frequencies of the side bands will be (1) 1.003 MHz and 0.997 MHz. (2) 3.001 kHz and 2,997 kHz. (3) 1,003 kHz and 1,000 kHz. (4) 1 MHz and 0.997 MHz. [NCERT Exemp. Q. 15.4, Page 99] 23. A message signal offrequency is superposed on a carrier wave of frequency o, to get an amplitude modulated wave (AM). The frequency of the AM wave will be (1) om (2) (3) (Omto)2 (4) (0,m - o)2 (NCERT Exemp. Q. 15.5, Page 99] 24. 1-V characteristics of four devices are shown in figure.

o,

o

(i)

1

(ii)

(1)

Carrier USB

LSB

Frequency (2)

Carrier LSB

USB

Frequency (3)

vt baseband

Carrier USB

Frequency

0

(4) Df

Carrier baseband USB

Frequency 26. A basic communication system consists (4) transmitter. of (B) information source, () user of information, (D) channel and (E) receiver.

Choose the correct sequence in which these are aranged in a basic communication system: (1) ABCDE. (2) BADEC. (3) BDACE. (4) BEADC. |NCERT Exemp. Q. 15.8, Page 100] 27. Audio sine waves of 3 kHz frequency are usei to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true? (1) The bandwidth required for amplitude modulation is 1.5MHz. (2) The bandwidth required for amplitude modulation is kHz. (3) The bandwidth required for amplitude modulation is 3 MHz. (4) The bandwidth requirement is infinite. 28. If a TV telecast has to cover 200km, the height of the antenna should be (1) 6250m (2) 3.12Sm (3) 15.625m (4) 3125m 29, Transducer converts (1) signals of various physical forms to electrical signals, and vice versa. (2) signals of various physical forms to electrical signals ony (3) signals of electrical signals to various physical forms ony (4) None of the above. 6

(iii) I|

Identify devices that can be used for modulation: (1) (() and (iii).

(2) only (ii). (3) (ii) and some regions of (iv). (4) All the devices can be used. (NCERT Exemp. Q. 15.6, Page 99) 25, Frequency spectrum of amplitude modulated wave is represented by

OMMUNICANON SYSTEMS

figure shows ovemodulated AM wave?

Fig. (a)

Fig. (b)

(1) Fig. (a) (3) Fig.

(c)

Fig. (c) (2) Fig. (b) (4) Fig. (a) and (c)

131 tower used for Assertion (A): When the height of the the maximum by is increased 21o, communication LOS distance coverage decreases by 10%. Reason (R): Maximum distance coverage through LOS communication decrcases as the height of tower incrcases. 9. Assertion (A): In undermodulated amplitude modulation, zero, nor will then the carricr amplitude will ncither fall to it rise to twice its unmodulated level. Reason (R): Amplitude modulation with modulation index 100% is called perfect modulation. 10. Assertion (A): Medium usecd for ground wave propagation. sky wave propagation, space wave propagation air. Reason (R): Unguided media transport clectromagnetic waves without using a physical conductor.

8.

[C| COMPETENCY BASED QUESTIONS I.

IRIASSERTION REASON QUESTIONS Oestion Nos. 1 to 10consist of two statements –Assertion and Retson. Answer these questions by selecting the appropriate option given below: .

1.

Both assertion (A) and reason (R) are true, and reason (R) is the

correct explanation of assertion (A).

Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A). 3 Assertion (A) is true, but reason (R) is false.

Assertion (A) is false, but reason (R) is true.1. 1. Assertion (A): A radio wave having frequency greater than 30MHz is transmitted through sky wave propagation. Reason (R): lonosphere reflects electromagnetic waves having frequencies greater than certain critical frequency. 2. Assertion (A): For speech signals, frequency range 300 Hz to 3100 Hz is considered adequate. Reason (R): The bandwidth of the speech signal is 3400 Hz. 3. Assertion (A): A very large antenna of impractical size is required to transmit an electromagnetic wave of frequency 20KHz. Reason (R): The antenna should have a size comparable to the wavelength of the signal. 4. Assertion (A): The square law device used in an amplitude a half modulated wave detector basically behaves like wave rectifier. Reason (R): The envelope detector used in an amplitude modulated wave detector is basically a low pass RC filter. above 5. Assertion (A): A communication satellite is placed 4.

the ionosphere.

6.

I

Reason (R): Radio wave reflection from satellite is sky wave propagation. Assertion (A): The amplitude modulated signal consists sinusoidal of the carrier wave of frequency o, and four waves cach with a frequency slightly different from oe, known as side bands. are cqual and Reason (R): The amplitudes of side bands wave. less than the amplitude of the carrier waves are suitable for Assertion (A): High frequency ground wave propagation. curvature Reason (R): Ground wave propagates along the of the earth.

Based on following passage answer questions from 1-5. A village is situated at a distance 50km from a TV transmission station. The TV transmission antenna is 32m high whereas the receiving antenna in the village is 30m high. Villagers are unable to watch the TV programmes. What should be the minimum height of the receiving antenna to receive the signal in the village? (4) 56m (3) 32m (2) 40m What is the type of wave propagation taking place in this TV signal transmission? Ground wave propagation (2) Sky wave propagation (4) None of these Space wave propagation wave suitable if a flying type What propagation of aircraft desire to receive the TV signal? Ground wave propagation (2) Sky wave propagation (4) None of these Space wave propagation What may be the alternate wave propagation to receive the TV signal in the village? Ground wave propagation Sky wave propagation Optical fibre cable communication Both (2) and (3) Why is it not possible to receive the signal by ground wave propagation technique?

(1) 70m

2.

(1) (3) 3. (1) (3) 4. (1)

(2) (3) (4) 5.

(1) Ground wave propagation is suitable for low and medium frequency only. (2) Attenuation is high for ground wave propagation and increases with increase in frequency. (3) Large antenna, whose dimension is proportional to the wavelength of the wave, is required for ground wave. (4) All of the above. II. Based on following passage answer questions from 6-10. Modulation index is a measure of extent of modulation done on a carrier signal. t describes how much the modulated variable of the carrier signal varies around its unmodulated level. It is defined diflerently in each modulation scheme. InAmplitude modulation, it is defined as the ratio of the amplitude of modulating signal to that of the carrier signal.

m= Aelm

It is also defined as m = (Amat - dmin) /(Ama Amin) Its value is kept less than l to avoid overmodulation which leads to distortions in the modulated signal and makes it

t

very hard to demodulate and extract the modulating signal.

Oswaal CUETr(UG)

132 A caricr wave frequency 1.5 MHz and amplitude 50 V is modulated by a sinusoidal wave of frequency 10 kHz. producing S0% amplitude modulation. The amplitude of the AM wave is (1) Amar = 50V, A,min = 25V (2) Amar = 50V, A,min = 50V A,min (3) Amax = 75V, Amin = 25V (4) Amax 25V

Chaptervíse Question Bank PHYSICS

6.

(3) 1.49MHZ 1.5MHz 1.51MHz

=25V, MHz and amplitude 50

v A carrier wave frequency 1.5 is modulated by a sinusoidal wave of frcquency 10 kHz producing 50% amplitude modulation. The frequcncy

7.

spcctrum of the AM wave is

(4)

1.49MHZ 1.5MHz 1.51MHz

carrier wave frequency 1.5 MHz and amplitude s0 V is modulated by a sinusoidal wave of frequency 10 kHz The bandwidth is (1) 20kHz (2) 20MHz

(1)

A

8.

1.49MHZ 1.5MHz 1.51MHz

(3) 1kHz (4) 0.02kHz

(2)

1.49MHZ 1.5MHz 1.51MHz

9.

A carrier wave frequency 1.5 MHz and amplitude 50 V is modulated by a sinusoidal wave following diagrams correctly displays 60% amplitude modulation?

of frequency

10 kHz. Which

,60VL

,80VL Level with nct modulationli

Level with no

modulation 30

20

(2)

(1)

witV

Level modulation

.60V

Level with nol modulation 10v

N

(4)

(3)

10. When the amplitude modulation as shown below occur, the modulation index is

(1) =100%

(2) > 100%

(3) < 100%

(4) =0%

ANSWER KEY [A]MULTIPLE CHOICE QUESTIONS (3)

2. (1)

3. (4)

4. (2)

5. (2)

6. (3)

7. (4)

8. (3)

12. (2)

9. (4)

10. (1)

11. (2)

13. (3)

14. (3)

15. (3)

16. (2)

17. (2)

18. (1)

21.(1)

19. (3)

20. (2)

22. (1)

23. (2)

24. (3)

25. (2)

26. (2)

27. (2)

28. (4)

29. (1)

30. (2)

8. (2)

9. (1)

10.(1)

1.

(BIASSERTION REASON QUESTIONS 1. (2)

|2. (2)

3.(1)

4. (1)

5. (3)

6. (4)

7. (3)

C] COMPETENCY BASED QUESTIONS 1.

(1)

2.(3)

3. (3)4.

(4)

(4)6.

5.

(3)

7. (1)

8. (1)

|9.(1)10.

(2)

of

the

COMMUNICATION SYSTEMS

133

OBJECTIVE TYPEQUESTIONS MULTIPLE CHOICE QUESTIONS Option (3) is correct.

JAJ 1.

Or. Or.

J2 x6400 x(32/1000)

d, 45.5

20,2

(4) is

Option Explanation: 3.

J2Rh,

J2 Rh,

Eyplanation: dy

+

25.3

Or,

- 113Jh, 0,0484 km

Iy 2. Option (1) is +

J2 x6400 × h,

Explanathon: Or, 0.70

J12800 × h,

A,,

48,4 m

correct.

A,/

24

0.7 x 24

16.8 V

correct. Communication system

Information

Transmitter

SOurcc

Transmitted

Signal

Channel

Reccived Signal

Receiver| Messago Signal

Option (2) is correct Explanation: Wavelength transmission depends on the length of the antenna. If2 is the wavelength and 7 is the length of the antenna, then the relation between

User of information

4.

,m4/

4 x 5

2.

them is

200 m

Option (2) is correct Explanation: 5.

Amplifter

I Stage

Option (3) is correct Explanation: In IF stage the original information bearing signal is extracted from carrier wave. 7. Option (4) is correct Explanation: The Amplitude Modulated (AM radio) uses frequency range is 535 to 1605 kHz. Lowest frequency band. Sky wave propagation, uses frequency range 3 MHz to 30 MHz. FM radio broadcast uses the frequency range 88 MHz to 108 kHz. For satellite communication uses the frequency range 5,925 to 6.425 GHz for uplink and 3.7 to 4.2GHz for downlink. Highest frequency band. 8. Option (3) is correct Explanation: Modulation index 6.

Y.

Or,

75 100

12

V

9 V

Option (4) is correct. Explanation: T.V. transmission is either line of sight signal propagation or by reflection of signal from satellite. 10. Option (1) is correct. Explanation: Modulation is a process of superimposition of signal, on high low frequency signal, known as modulating as trequency signal known carrier signal. 9.

I1, Option (2) is

Explanation:

mar Vmin

Or,

30/2 = 15 mV 10/2 = Sim

Amplifieror

12. Option (2) is correct. Explanation: Waves having frequency range 3 -30 MIHz are suitable for sky wave propagation. 13. Option (3) is correct. Explanation: The values of critical frcquency are found to be 4 MHZ, 5 MHZ and 6 and 8 MHz for D, E, F and F, ayers of ionosphere respectively. IHence, the wave penctrates all the layers of ionosphere except F. 14. Option (3) is correct. Explanation: Optical fibre works on the principle of total

internal reflcction. 15. Option (3) is correct. Explanation: Frequency of carrier signal should be higher than the frequency of modulating signal. 16. Option (2) is correct Explanation: Amplitude modulated waves having modulation index greater than l are over modulated and distorted. 17. Option (2) is correct. Explanation: The cquation of an AM wave is

V=Vsino,

m, cos(o, 2

-o.)! - -cos(o,

It shows that there are two side bands:

2

+

o,)!.

cos(a,-o,)! and

mcos(o, + ao.y. Explanation:

V

m= (Vmr - Vmin )/(m m=(15- 5)/(15+ 5)

ma 0.5

-

18. Option (1) is correct.

correct.

V.

Detector

50%

+

min)

V

Square Law Device

Band Pass Filter

Vam

134

Oswaal

CUET (UG)

Chapterwise Question

19. Option (3) is

Bark PHYSKS According to the prooblem, frequency of the | MHz and frequency of speech signal ccarrier MHz = 0.003 MHz sigraliss.

correct.

=

Explanation:

|Square

lavw

So,

device

and

R.Mešsage

Vam

signal

t

Low pass filter

20. Option (2) is correct. Explanation: The radio waves emitted from a transmitter antenna can reach the receiver antenna by the following mode of operation: () Ground wave propagation (Frequency range: Fewhundred KHz to 2MHz) (i1) Sky wave propagation (Frequency range: 3-30 MHz) (ii) Space wave propagation (Frequency range: Above 40 MHz) So, A is transmitted via ground wave, B via sky wave and C via sky wave.

400m

= 4L =

Explanation: Side band frequencies of amplitude modulated wave are f. +f) and f-fn). fe+sm) is the frequency of upper side band (USB). is the frequency

- 0.003) =0.997MHz.

of

lower side band (LSB).

(1+0.003))=\03

MIH Option (2) is correct. Explanation: The process changing wave in accordance with the theamplitude amplitude (AF) signal of the audio ofcariet as is known amplitude modulation freqensy In AM, frequency of (AM). the carrier wave we can say remains that the frequency of unchanged modulated wave the frequency of carrier wave. G Now, is equal according frequency of carrier wave to to isf,. theprobem, Thus, the amplitude modulated wave 1 also hasfrequency 24. Option (3) is correct. o Explanation: square law modulator isthe device produce modulated waves by the application of whichcan signal and the carrier wave. the messag of

A

Square law modulator Characteristics shown

is

by (i)

used for modulation and (iii) corresponds

purpose.

devices. And by (i1) to linea: shows non-linear corresponds to square law device relations. Some part uH of (iv) also follo square law.

Vam

=Vsino / +

of amplitude modulated wayeie

m

Scos(o, 2

22. Option (1) is correct.

f-f)

(1

23.

25.Option (2) is correct. Explanation: The equation

21. Option (1) is correct.

Explanation: Length of antenna= m Wavelength of the wave which canL= 100 be transmitted =

the frequencies of side bannds are

-o.)-cos(o, +o,

2 It has the carrier wave, LSB and USB. Frequency of LSB is less frequency of USB is more than carrier wave frequency and Hence, LSB will be situatedthan the carrier wave frequency the left side and USB will be situated at the right side of the carrier.

26. Option (2) is correct Explanation: A basic communication system consists an of information source, a transmitter, a communication system is the set-up used in the transmission link (channel) and a receiver or a and reception of information The whole system consists from one place to another. of several elements in a sequence. It can be represented as the diagram given Information| Source (B)

Message

Signal

below:

Transmitter Transmitted Signal (A)

27. Option (2) is correct.

Explanation: Bandwidth requirement modulation is 2f., =2 x 3KHz = 6KHz. 28. Option (4) is correct.

for

amplitude

Explanation: d= V2RA Or, Or,

h=I2R (200 x10°) 2 X6400x10

h=3125 29. Option (1) is correct Explanation: At the transmitting end, signal from information source may not be in electrical form. So, transducer converts it into electrical form. At the receiving end, it required to convert the electrical m

signal into various physical forms. Transducer does this conversion.

Channel

(D)

Received Signal+Receiver (E)

30. Option (2) is correct.

Message| User of Signal

Information (C)

Explanation: An over-modulated AM wave is a signal that has been modulated so significantly that the carrier wave clips of distorts. This can happen when: ()) The modulation index is greater than (i) The modulating signal voltage is much greater than the carrier voltage (iii) The instantaneous level of the modulating signal exceeds the value necessary to produce 100% modulation of he carrier

[B] ASSERTION REASON QUESTIONS 1. Option (2) is correct. Explanation: The frequency range for sky wave propagati

3-30 MHz. Hence, A radio wave having frequency greater u 30MHz cannot be transmitted through sky wave propdg The assertion is incorrect.

cOMMUNICATION SYSTEMS

135

Lonosphere refleets electromagnetic r than certain

waves having frequencies critical iequency. Electronmagnetic wave tavine frequency greater than the critical frequency penetrates ionosphere. Hence, the reason is also incorreet 2. Option (2) is correct. Fenlanation: For specch signals, frequcncy range 300 1lz to 3100 Hz is considered adeqate. The assertion is correct. The bandvwidth of the speech signal is 3 1 00 Hz–30011z 2800 Hz. So, the reason is incorrect, 3. Option (1) is correct. Eyplanation: The antenna should have a size comparable to he wavelength of the signal. It should be at least of size of one fourth of the wavelength. So, the reason is correct. Eor electromagnetic wave of frequency 20KHz is 15km, So, the size of antenna should be at least 1Skm/4 3.75 km which is impractical. So, the assertion is also correct. 4. Option (1) is correet. Explanation: The output of the square law device used in an amplitude modulated wave detector is as shown:

ime Square Law Device

time

So. it basically

bchaves like a half wave rectifier. Assertion is correct. The envelope detector used in an amplitude modulated wave detector is basically a low pass RC filter as shown: R envelop detector Square law

device

Vam

signal

Low pass filter So, reason is also correct. 5. Option (3) is correct.

Explanation: Communication satellite is placed approximately at a height 36000 km from the surface of carth whereas ionosphere is stretched upto 650km approximately. So, assertion is correct. Radio wave reflection from satellite is space wave propagation since it has to penetrate ionosphere. So, reason is incorrect 6. Option (4) is

correct.

as: Explanation: An amplitude modulated wave is represented mV cos(o, + o,) mV cos(o, Vam = V 0)1–

sin

o+

2 So, it has the carrier wave of frequency o, and two side bands .

is frequency slightly different from o, Hence, the assertion incorrect. The amplitude of each side bands is V2. Hence, the carrier amplitudes are equal and less than the amplitude of the wave. Hence, the reason is correct. 7. Option (3) is correct. curvature of Explanation: Ground wave propagates along the the earth. Hence, the reason is correct. medium Ground wave propagation is suitable for low and As the MHz. 2 to KHz Irequency ie., from few hundred

of

frequency increases, attenuation inereases, Hence, high frequency vwaves are not suitable for ground wave propagation. So, the assertion is incorrect. 8. Option (2) is correct. Explanation: Maximunm distance coverage dn 2Rh So, So, masimunm distance coverage through IOS communication inereases as the height of tower inereases, So, thc reason is incorrect. Sincc,

So. 1.21/%

.21 =11

dJd,

So, maximum distance coverage inereases by 10%. lence, the assertion is also incorrect. 9. Option (1) is correet. Explanation: Modulation inde, in an amplitude modulation, indicates the maximum and minimum variation of carricr amplitude during modulation. When nodulation indes is less than 100% then the carrier amplitude neither falls to zero, nor rises to twice its unmodulated level. Hence, it is called undermodulatcd. So, the assertion is correct. When modulation index is 100%, then the carrier amplitude falls to zero, and rises to twice its unmodulated level during modulation. When modulation index is more than 100%, then the modulated wave gets distorted and it is called overmodulated. So, amplitude modulation with modulation index 100% is the perfect modulation since there is maximum variation of carrier

amplitude without any distortion. Hence, reason is also correct. 10. Option (1) is correet. Explanation: Ground wave propagation and sky wave propagation takes place in troposphere. Space wave propagation involves ionosphere. There is physical conductor carrying the signal. Hence, the air medium is called unguided medium. Hence, the assertion and reason both are correct.

BASED QUESTIONS

|C|COMPETENCY 1. Option (1) is

correct.

Explanation: The maximum distance coverage = S0 km Maximum distance coverage = 2Rih, + J2Ri, Or. 50000 = V2x 6400 x 1000x 32 +

Or, 50000 - 20238.6 +

2x

J2x 6400x 1000 x

,

6400x 1000x 4,

Or, 885740929.96 = 2x6400 x 1000 x h, .:. h=69.2 m 70 m

2. Option (3) is correct

Explanation: Propagation of radio waves in a straight line from transmitting antenna to the receiving antcnna is known as space wave propagation. Space waves are used for line-of sight (LOS) communication as well as satellite communication. 3. Option (3) is correct Explanation: Propagation of radio waves in a straight line from transmitting antenna to the receiving antenna is known as space wave propagation. Space waves are used for linc-of sight (LOS) communication as well as satellite communication.

Oswaal CUET (UG) Chaptervise

136 Option (4) is correct. Evplanation: Ground wave propagation will not be suitable since high frequency wave will get absorbed by carth crust 4.

casily. Sky wave propagation, i.c., by reflecting the signal from ionosphere is possiblc. Signal may be propagated through high bandwidth optical fibre cable. 5. Option (4) is correet. 6. Option (3) is correct. Explanation: Amplitude of unmodulated carrier wave is 50V. Modulation index = 50% So, the variation amplitude unmodulated wave is 50V 50% of

of 50V

-

So, Amar = 50V + 25V=75V, A,min = 50V 25V = 25V 7. Option (1) is correct. Explanation: In the frequency spectrum, there are the carrier wave and two side bands.

fUSRtm

×

QuestionPark Bar PWYSICS

1049

MHz

10"+ 10* L51 MI Amplitudes of the side bands are cqual amplitude of the carrier wavc. 8. Option (1) is correct. 1.5

and

Explanation: Band width = 2 fm2 x 10KHz 20KH 9. Option (1) is correct. Explanation: Amplitude of unmodulated carrier wave is S5V, Modulation index =60% So, the variation of amplitude unmodulated wave is 50V of 50V So. A.,= 50V + 30V 80V, Amin = 50V-30V 20V 10. Option (2) is correct. Explanation: When the modulation index is greater the 100%, then the amplitude modulated wave is distorted

±%

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