NSEJS IJSO STAGE 1 PREVIOUS YEAR QUESTION PAPER WITH SOLUTION 2016 to 2019 National Standard Exam for Junior Science

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NSEJS-2016 (IJSO STAGE-I) Date of Examination : 20th November, 2016 PAPER CODE - JS531

SOLUTIONS 1.

Rod AB of radius 2r is joined with rod BC of radius r. They are of same material and are of same length. The combination carries a current I. Choose the correct statement. A

B

C

I

(a) VAB = 4VBC

(b) Current per unit in AB and BC are equal

(c) Resistance of AB is greater than of BC Ans. (d) Sol.

I

(d) VBC = 4VAB

VAB IR AB Irl / p(2 p)2 1 VBC = IR BC = Irl / pr 2 = 4 or VBC = 4VAB

2.

The statement “a is not less than 4” is correctly represented by (a) a < 4

(b) a > 4

(c) a ³ 4

(d) a £ 4

Ans. Sol. "a is not less than 4" Þ a is greater or equal to 4 Þ a³4 So option (c) is correct 3.

A chemist mixes two ideal liquids A and B to form a homogeneous mixture. The densities of the liquids are 2.0 g/mL for A and 3 g/mL for B. When she drops a small object into the mixture, she finds that the object becomes suspended in the liquid that is it neither sinks to the bottom nor does it float on the surface. If the mixture is made of 40% A and 60% B, by volume, what is the density of the object?

(a) 2.60 g/mL (b) 2.50 g/mL Ans. (a) Sol. density of mixture = density of object =

(c) 2.40 g/mL

(d) 1.50 g/mL

VA r A + VBrB 0.4 ´ 2.0 + 0.6 ´ 3 = VA + VB 1

= 0.8 + 1.8 = 2.6 g/mL 4.

How many different compounds can have the formula C3 H4 ?

(a) One (b) Two Ans. (c) Three Sol. Structures are : CH3 – C º CH

(c) Three

(d) Four

CH2 = C = CH2 and 1

5.

In the figure shown, the current carrying loop is fixed, where as current carrying straight conductor is free to move. Then straight wire will (ignore gravity) (a) remain stationary I1

(b) move towards the loop

I2

(c) move away from the loop (d) rotate about the axis perpendicular to plane of paper Ans. (b) Sol. Due to current carrying loop, magnetic field on straight wire will be in outward direction so a force towards right will be exerted on this wire. 6.

Two friends A and B watched a car from the top of their buildings. Angle of depression for A was 10° more than angle of depression for B, then (a) A's apartment is taller than B's apartment (b) B's apartment is taller than A's apartment (c) A's apartment and B's apartment have same height (d) We cannot compare the heights of the two apartments

Ans. (d) Sol. Let 'H1' and 'H2' i.e. the height of two buildings, then tan q =

H2 H1 , tan (q + 10°) = b a

A

D

H1

Þ H2 = b tan q,

B

H1 = a tan(q + 10°)

H2 q + 10° a

q E

b

C

Now, we can compare the height of two building only when a and b are known. so option (d) is correct 7.

How many times would a red blood cell pass through the heart during one complete cycle? (a) Once

(b) Twice

(c) 4 times

(d) 72 times

Ans. (b) Sol. In double circulation, blood passes twice from heart. 8.

A gene has two alleles P(dominant) and p(recessive). The homozygous recessive combination leads to death in the embryo stage. If two individuals with genotype Pp are mated, out of the offspring that survive to adulthood, what is the probability of the genotype to be Pp?

(a) 0.75 Ans. (d) Sol.

Pp

pp

(b) 0.33

×

Pp

(c) 0.5

(d) 0.67

Pp

Pp

pp

As mention in question homozygous recessive will die. So pp will die then Pp will be

2

2 = 0.67. 3

9.

A convex mirror of focal length f produces an image of size equal to Then the object distance is (a) nf

(b)

f n

1 times the size of the object. n

(c) (n + 1)f

(d) (n – 1)f

Ans. (d) Sol.

v 1 u = , v= u n n Writing mirror formula, 1 1 1 + = v u f n 1 1 - = u u f 1 1 (n – 1) = u f \ u = (n – 1)f or,

10.

Total surface area of a sphere S with radius

2 + 3 cm is

(a) 400p (5 + 2 6) sqmm

(b) p ( 2 + 3)2 sqcm

(c) 2p ( 2 + 3)2 sqcm

(d) 40p (5 + 2 6) sqmm

Ans. (a) Sol. r =

2+ 3

S.A of sphere = 4p ( 2 + 3 )2 = 4p (5 + 2 6 ) cm2 = 400 p (5 + 2 6 ) mm2 So option (a) is correct. 11.

There are many elements in the periodic table that are named after the country, where they were first made or obtained. For example, the Latin name for copper was coined by the Romans because their chief source of copper was from the Island of Cyprus. However, there is one country in the world which was named after an element (the Latin name). A long time ago, it was believed that this country had mountains full of a valuable element, however all expeditions to find these mountains failed. But the name stuck on. The element in question is used for many applications today, and many of its compounds are used as catalysts. The ions of this metal have very good anti-microbial property and finds application in water purification. The element is

(a) Sodium (b) Gold Ans. (c) Silver Sol. Argentina is named after silver "Argentuns"

(c) Silver

(d) Francium

3

12.

All of these species have the same number of valence electrons as nitrate ion, except (a) Carbonate ion

(b) Bicarbonate ion

(c) NF 3

(d) SO3

Ans. (c) NF 3 Sol. No. of valence e–s in nitrate ion is NO3– = 24. (a) CO 3 2–

13.

:

Valence e– = 24

(b) HCO 3 – :

Valence e– = 24

(c) NF 3

:

Valence e– = 26

(d) SO3

:

Valence e– = 24

The angle between the hour arm and the minute arm of a clock at 2:10 a.m. is (a) Zero

(b) 4°

(c) 5°

(d) 6°

Ans. (c) Sol.. Angle between hour hand and min hand is given by

11M - 60H 2

Where M = min. and H = Hour So, angle between hour hand and Min hand of 2 : 10 AM is

11 ´ 10 - 2 ´ 60 2

= 5º

Option is (c) correct 14.

A craft teacher reshapes the wax from a cylinder of candle with section diameter 6 cm and the height 6cm into a sphere. The radius of this sphere will be (a) r = 6

3 / 2 cm

(c) r = 3 3 3 / 2 cm

(b) r = 6 cm (d) r = 6 3 2 cm

Ans. (c) Sol. Volume of wax in cylindrial shape = volume of wax is spherical shape = pr2h = 4/3pR3 =

22 4 × (3)2 × 6 = pR3 7 3

Þ R3 =

32 ´ 6 ´ 3 4

Þ R = 33

3 cm 2

Option (c) is correct 15.

Plants absorb nitrates from the soil, which are most essential to produce (a) Proteins

(b) Carbohydrates

(c) Fats (d) Cell wall Ans. (a) Sol. Plant absorb nitrates from the soil which is essential to produce protein. 4

16.

The dry mass (mass excluding water) of a seed in the process of germination (a) increases over time until the first leaves appear (b) decreases over time until the first leaves appear (c) stays constant until the first leaves appear

(d) first increases and then decreases until the first leaves appear. Ans. (b) Sol. During the time of germination the dry mass of seed will decrease, because it will do respiration. 17.

A point object O is kept at origin. When a concave mirror M1 placed at x = 6 cm, image is formed at infinity. When M1 is replaced by another concave mirror M2 at same position, image is formed at x = 30 cm, then ratio of the focal length of M1 to that of M2 is (a)

3 4

(b)

4 3

(c) 5

(d)

1 5

Ans. (a) Sol. Since image is formed at ¥ fM1 = –6

For M2, 1 1 1 - = (30 - 6) 6 fM2 1 1 1 or 24 - 6 = f M2 1- 4 1 = 24 fM2

Þ fM2 = –8 cm fM1 6 3 \ f =8=4 M2

18.

The number 38 (310 + 65) + 23(212 + 67) is (a) A perfect square and a perfect cube (b) Neither a perfect squre nor a perfect cube (c) A perfect cube but not a perfect square (d) A perfect square but not a perfect cube

Ans. (c) Sol. 38(310 + 65) + 23(212 + 67) 318 + 38.35.25 + 215 + 210.37 318 + 313.25 + 37210 + 215 318 + 37.25 (36 + 25) + 215 (36)3 + 3(36)(25) (36 + 25) + (25)3 (36 + 25)3 which is perfect cube. It is a perfect cube but not a perfect square So, option (c) is correct. 5

19.

Ans. Sol. 20.

Ans. Sol.

21.

Melting point of a substance is 10°C. What does this mean? (a) The substance is a liquid at 10°C. (b) The substance is a solid at 10°C. (c) There is an equilibrium between solid phase and liquid phase at 10°C (d) The substance is 50% solid and 50% liquid at 10°C. (c) Melting opint is defiend as the temperature at which solid and liquid phase of a substance are in state of dynamic equilbrium at a fixed pressure. The following substances have approximately same molecular mass. Which is likely to have the highest boiling point? (a) n-butane (b) Isobutane (c) n-butanol (d) Isobutanol (c) n- butanol. B.P. of n – Butanol = 117.7°C B.P. of iso – Butanol = 108°C B.P. of n – Butane = –1°C B.P. of iso – Butane = –11.7°C U-tube contains some amount of mercury. Immiscible Liquid X is poured in left Immiscible liquid Y is poured in the right arm. length of liquid X is 8 cm. length Y is 10 cm and upper levels of X and Y are equal. If density of Y is 3.36 g.cm–3 and l3.6 g.cm–3 then density of X is X

arm of

Y liquid of

Hg is

(a) 0.8 g-cm–1 (b) 1.2 g-cm–1 (c) 1.4 g-cm–1 (d) 16 g-cm–1 Ans. (a) Sol. PA = PB P0 + rxg(8) + rHgg(2) = rY × G × 10 + P0 8 cm X Y 10 cm 2 cm 8rX = 10rY – 2rHg = 10 × 3.36 – 2 × 13.6 = 33.6 – 27.2 B A 8rX = 6.4

6.4 = 0.8 g cm–3 8 Let the number of rectangles formed by 6 horizontal and 4 vertical lines be n and those formed by 5 vertical and 5 horizontal lines be m then we have (a) n = m (b) n ³ m + l (c) m ³ n (d) m > n + 5 (d) Number of rectangles formed by 6 horizontal and 4 vertical lines is 6C × 4C = 90 2 2 \ n = 90 Also number of rectangles formed by 5 vertical and 5 horizontal lines is 5C × 5C = 100 2 2 Þ m = 100 \m>n+5 so, option (d) is correct. In a human cell undergoing Meiosis, what are the total number of cellular DNA molecules present during Prophase-l (a) 23 (b) 46 (c) 69 (d) 92 (d) During the interphase the process of DNA replication will occur. That's why the DNA content in prophase-I will be doubled. rX =

22.

Ans. Sol.

23.

Ans. Sol. 6

24.

During gaseous exchange in the alveoli, what happens to nitrogen? (a) There is no net nitrogen exchange, as nitrogen is filtered out by the alveoli. (b) The nitrogen is absorbed by the alveolus to form amino acids. (c) The nitrogen is filtered out by the alveolus, as the nitrogen molecule is too large to cross the gaps in the capillaries (d) There is no net nitrogen exchange, as the blood is saturated with nitrogen Ans. (d) Sol. There is no net nitrogen exchange as, the blood is saturated with nitrogen. 25. The effective resistance between A and D in the circuit shown in the adjacent figure is 5W

A

10W

5W

10W C

(b) 10 W

(a) 5W Ans. (d) Sol. From the fig. 5W

A

P

10W

10W 5W

D

(c) 15 W

(d) 20 W

10W

A 10W

10W 10W

26.

5W

B

Q

5W

5W

10W

Þ D

5W

D

Equivalent resistance Between P and Q is 10W Req. = 5 + 10 + 5 = 20 W If ABCD is a rhombus and ÐABC = 60° then (a) The points A,B,C,D are concyclic (b) The quadrilateral has exactly half the area of the square with same sides as ABCD

3 AB2 2 (d) The diagonals of the quadrilateral ABCD are equal and bisect each other at right angle Ans. (c) Sol. Here A AM = sin 60º × a D

(c) The quadrilateral has area

=

a

3a 2

B

\ Area of rhombus = Base × height = a × =

=

60º

M

C a

3a 2

3a 2 2

3 (AB)2 2

so, option (c) is correct. 7

27.

Identify the overall change in the following set of reactions: (1) Carbon dioxide ® carbonic acid (H2 CO3 ) (2) Ethanol (alcohol) ® Ethanal (aldehyde) (3) Ethanal (aldehyde) ® Ethanol (alcohol) (4) Sulphuric acid ® Sulphur trloxlde (SO3 ) Choose the correct option which best describes these conversions (a) oxidation, oxidation, reduction, reduction (b) hydration, oxidation, reduction, dehydration (c) reduction, dehydration, hydration, oxidation (d) reduction, reduction, oxidation, oxidation Ans. (b) Sol. (1) CO2 + H2O ® H2CO3 : Hydration (2) C2H5OH ® CH3CHO : Oxidation (3) CH3CHO ® CH3CH2 OH : Reduction (4) H2SO4 ® SO3 + H2O : Dehydration 28.

An element with atomic number 44, is below which element in the periodic table? (a) Calcium (b) Iron (c) Argon (d) Magnesium Ans. (b) Fe Sol. At. No. 44 "Ruthenium" is below Fe in the peridic table 29. Three bulbs B1 , B2 and B3 having rated powers 100 W, 60 W and 60 W at 250 V are connected in a circuit as shown in the adjacent figure. If W1, W2 and W3 are the output powers of the bulbs B1 , B2 and B3 respectively, then B1

B2 B3

250V

(a) W1 > W2 = W3 (b) W1 > W2 > W3 Ans. (d) Sol. Since power rating is provided.

(c) W1 < W2 = W3

B1

i1

(d) W1 < W2 < W3

B2 B3

i2 250V

R B1 < R B2 = R B3

From the given circuit current through B3 will be more than that through B1 and B2. (i1 < i2) So, after comparing power from equation i2R. We get W1 < W2 < W3 , which is option (d). 30.

8

If a, b > 0 then (a) a + b £

ab

(b) a + b >

(c) a + b ³

2ab

(d) None of the above inequalities will hold

ab

Ans. (d) Sol. If a, b > 0 Then AM ³ GM Þ

a+b ³ ab 2

Þ a + b ³ 2 ab Þ a + b > 31.

Ans. Sol. 32.

Ans. Sol.

33.

ab

So option (b) is correct. Which of the following is true about ATP (a) It is a derivative of one of the nitrogenous bases that form DNA (b) It splits into ADP and phosphate and the energy produced is used by muscle cells to contract (c) It is produced in both aerobic and anaerobic conditions. (d) All of the above (d) ATP is derivative of Adenosine, ATP is also used in muscles contraction, ATP is also produced in both aerobic & anaerobic condition. That's why the answer will be all of the above. Which of the following is true regarding communication in neurons (a) Free electrons are moved along the plasma membrane of the axon and control the expression of neurotransmitters (b) A chemical signal travels along the axon and is converted into an electric impulse at the synapse (c) An electric impulse travels along the length of the axon. The electric impulse is converted to a chemical signal at the synapse. (d) An electrical signal is converted to a chemical signal by the Myelin sheath before it reaches the synnapse (c) An electric impulse travels along the length of the axon. The electric impulse is converted to a chemical signal with the help of neurotransmitter “Acetyl choline” at synapse. Following diagram shows refraction of parallelo beam of light through a spherical surface. Identify the correct ray diagram Denser

(a)

Rarer

Rarer

(b)

Denser

(c)

Denser

Rarer

(d)

Ans. (b) Sol. As rays bend towards normal in denser medium so, option (b) is correct. 9

34. Ans. Sol.

35.

Ans. Sol. 36.

Ans. Sol. 37.

Tenth term in the sequence 12, 18, 20, 28, ... is (a) 336 (b) 63 (c) 216 (d) 68 (b) 12,18,20,28...... 22 × 3, 32 × 2, 22 × 5, 22 × 7,............. These are the number with 6 factors so next possible terms with 6 factors are 25 × 1, 22 × 11, 32 × 5, 52 × 2, 22 × 13, 32 × 7, ............. \ 10th term is 63. So option (b) is correct. An electron pair donor is a Lewis base and an electron pair acceptor is a Lewis acid. Which among the following statements, is correct? (a) NH3 is a Lewis acid, because nitrogen has only 6 electrons around it (b) BF3 is a Lewis base, because fluorine has 8 electrons around it. (c) NF3 is a Lewis base, because nitrogen has a lone electron. (d) BCI3 is a Lewis acid because it has only 6 electrons around it. (d) (d) most appropriate (c) Option has lone e– not lone pair of e–s. Greenhouse gases absorb (and trap) outgoing infared radiation (heat) from Earth and contribute to global warming. A molecule that acts as a greenhouse gas, generally has a permanent dipole moment and sometimes for other reasons. Going only by the condition of permanent dipole moment, in the list of gases given below, how many can be potential greenhouse gases? Water, Sulphur dioxide, Boron trifluoride, carbon monoxide, carbon dioxide, nitrogen, oxygen, methane, hydrogen sulphide, ammonia. (a) Five (b) Six (c) Seven (d) Four (a) five Water, SO2, CO, H2S and NH3 will have permanent dipole moment In the diagram M1 and M2 are two plane mirrors at right angles to each other. O is a luminous point object. Consider two images formed due to first reflection at M1 and M2. The area of the triangle formed by the object and two images is M1

2 cm O 2 cm M2

(a) 4 cm2

(b) 2 cm2

(c) 8 cm2

(d) 16 cm2

Ans. (c)

1 Sol. Area of DOI1I2 = × I1 × OI2 2 =

2 cm 2 cm

1 ×4×4 2

= 8 cm2 10

O

2 cm I2

2 cm

I1

38.

The probability of a point within an equilateral triangle with side 1-unit lying outside its in-circle (inscribed circle) is (a) 1 –

1 2( 3)

(b) 1 –

p

(c) l –

3 3

p

(d) 1 –

2 3

Ans. (b)

2p 3 3

A

D Sol. Radius of in-circle = S 3 2 (1) 1 4 = = 2 3 3 2

C

B

\ Probability of a point lying outside the circle =

3 2 (1) - p(r 2 ) p 4 = 1– 3 3 3 2 (1) 4

So option (b) is correct. 39. Penicillin cannot be used to treat influenza because: (a) It only helps to bring the temperature down, and does not reduce the infection (b) The penicillin is broken down by the organism (c) Viruses do not have cell walls (d) Reproduction of protozoans is not affected by penicillin Ans. (c) Sol. Antibiotics are not used in viral disease, because virus do not have cell wall. 40. Thin cuboidal strips are made by slicing a potato. They are all made to be exactly 8 cm long and 2 mm wide. Each strip is placed in sugar solutions of different concentration. After soaking it for 5 hours, their lengths are measured again. The following graph shows the results of the experiment. What concentration of sugar solution is isotonic with the contents of the cells of the potato. 8.5 8.4 8.3 8.2 8.1 8 7.9 7.8 7.7 7.6 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

(a) 0.2 (b) 0.4 (c) 0.6 (d) 0.1 Ans. (b) Sol. In the mention graph, 8 cm length of the slice will be on 0.4 concentrate solution [Isotonic] which will be similar to the initial slice length 8 cm.

11

41.

A fisherman of height h is standing on the bank of a lake. A fish in the water perceives his height as h'. Then (a) h' > h

(b) h' < h

(c) h' = h

(d) h' > h or h' < h depending on position of fish

Ans. (a) Sol.

h'

h

From Fig. itself h' = mh 42.

A triangle has perimeter 316 and its sides are of integer length. The maximum possible area for such a triangle is achieved for (a) Single triangle

(b) Two triangles

(c) Three triangles

(d) More than three triangle

Ans. (a) Sol. a + b + c = 316 and a, b, c Î I Now, area is maximum only if a = b = c But here sides are integers, so it is not possible as 3a = 316 Þ a =

316 = 105.33 3

\ Area is Max. only when a = b = 105 and c = 106. So option (a) is correct. 43.

Hennig Brand, one of the many alchemists was in pursuit of the philosopher's stone. Brand's method is believed to have consisted of evaporating large quantities of urine to leave a black residue that was then left for a few months. The residue was then heated with sand, driving off a variety of gases and oils. The final substance to be driven off, was condensed as a white solid, which he called as "cold fire" as it was luminousand glowed in the dark and also caught fire on slight warming and producing a large quantity of light. It has also been called as the "Bearer of light". Which element is "cold fire"? (a) Lithium

(b) Tungsten

(c) Phosphorous

(d) Cesium

Ans. (c) Sol. Phosphorous is called cold fire. 44.

When solid KOH is mixed with solid NH4Cl, a gas is produced. Which gas is it? (a) Chlorine

(b) hydrogen

Ans. (d) ammonia Sol. KOH + NH4Cl ® KCl + H2O + NH3 ­ 12

(c) hydrogen chloride

(d) ammonia

45.

Object A is completely immersed in water. True weight of object A is WA. Weight of water with beaker is WB. Let B be the buoyant force. W1 and W2 are scale readings of spring balance and weighing scale respectively. Spring balance

A Weighing scale ....... gm

(a) W1 = WA

(b) W1 = WA + B

(c) W2 = WB

(d) W2 = WB + B

Ans. (d) Sol. For spring balance kx = WA – B

B

kx

Þ WA – B

....... (1)

For weighing machine

Spring balance

N = B + WB Þ W2 = WB + B

wB

.......(2)

A

Weighline B

mg = wA Hence ‘d’ is correct 46. Number of numbes less than 40 having exaclty four divisors is (a) 15 (b) 12 (c) 11 Ans. (d) Sol. No having 4 divisor should be is the form of a11 × a21 or a13 \ Possible Numbers are

6=2×3 10 = 2 × 5 14 = 2 × 7 22 = 2 × 11 26 = 2 × 13 34 = 2 × 17 38 = 2 × 19

N

Scale

(d) 14

15 = 3 × 5 21 = 3 × 7 33 = 3 × 11 39 = 3 × 13 35 = 5 × 7 8 = 23 27 = 33

So, option (d) is correct. Antibodies play an important role in defending the body agains infections by which of the following meachanism: (a) They engulf the bacteria and make them harmelss (b) They bind to the surface of pathogens, so taht they can be easily indentified and removed by other cells of the immune system (c) They enter the pathogen and prevent cell division (d) They are highlty reactive and chemically react with the DNA of the pathogen Ans. (b) Sol. Antibody, Binds to the surface of pathogen, so that they can be easily identified & removed by other cells of Immune system. 47.

13

48.

The figure shows a food we, where A, B, C, D etc. are different species. And the direction of the arrows symbolizes the direction of flow of natrients An increase in the population of which species is likely to decrease the population of species. A

B

C

D

E

F G

(a) Species D (b) Sepecies F (c) Species G (d) Species E Ans. (a) Sol. As mention in the figure, “C” is eaten by “A”, “E” is eaten by both “C” & “D”, that's why if we increase the number of “D”, the number of “E” will be decrease similarly “C” & “A” also decrease. 49. Velocity of a particle moving along a straight line varies with time as shown in the figure. Net forces acting on the aprticle are F1, F2, F3, F4 and F5 in the intervals OA, AB, BC, CD and DE respectively. Indentify the correct statement V A

D t B

(a) F1 increases with time (c) F1 and F2 are in opposite directions

C

E

(b) F5 is initially positive and then becomes negative (d) F3 is negative

v

st =

c

D

A

F1

=c on

Sol.

+v e

Ans. (c)

F5 = ve

F2 = ve

F

O

F4 = +ve B

F3 = 0

C

Q Slope of the velocity time gives us acceleration and F = ma option ‘c’ is correct. 50. If set X consists of three elements then the number of elements in the power set of power set of X is (a) 3 3 (b) 23 (c) 38 (d) 28 Ans. (d) Sol. Let X = {a, b c} Then No. of elements in the power set of A = 23 = 8 \ No. of element in the power set of power set of A = 28 So, option (d) is correct. 14

51.

Ans. Sol. 52.

Ans. Sol. 53.

The heat of neutralisation of CH3COOH, HCOOH, HCN and H2S are – 55. 2, – 56.07 – 2.8 and – 3.34 kJ per equaivalent respectively. The increasing order of strength of these acids is (a) HCOOH < CH3COOH < H2S < HCN (b) H2S < HCN < HCOOH < CH3COOH (c) HCN < H2S < CH3COOH < HCOOH (d) CH3COOH < HCOOH < HCN < H2S (c) HCN < H2S < CH3COOH < HCOOH Higher the heat of neutralisation stronger will be the acid To prevent the formation of oxides, peroxides, and superoxides, alkali metals are sometimes stored in an unreactive atomosphere. Which of the following gases should not be used for lithium : (a) Ne (b) Ar (c) N2 (d) Kr (c) N 2 As Li can react with N2 forming explosive Li3N. A wooden block (W) is suspended by using a coard from aheavy steel ball (B). The entire system is dropped from a height. Neglecting air resistance, the tension in the cord is (a) Zero (b) The difference in the masses of B and W (c) The differences in the weights of B and W (d) The weights of B

Ans. (a) Sol. For wooden block

B T

g

W – T = mg Þ T=0

W

54.

In a n- sided regular polygon, the radius of the circum-circle is equal in length to the shortest diagonal. the number of values of n < 60 for which this can happen is (a) 0 (b) 1 (c) 10 (d) 2 Ans. (b) Sol. DOCE is an equilateral triangle. So ÐCOE = 60° DOCP @ DOEP A B So ÐDOE = 30° G Angle made at centre by the side of regular polygon = 30°

360° So, = 30° n

R

C

D

55. Ans. Sol. 56. Ans. Sol.

P

O R R

F

360° E n= = 12 30 So, option (b) is correct. Which of the following does NOT contain living cells? (a) Bone tissue (b) Xylem sieve tubes (c) Phloem (d) epidermis (b) Xylem sieve tubes do not contain living cells. If DNA was made of 6 nucleotides instead of 4, what are the total number of triplet codons possible? (a) 24 (b) 18 (c) 64 (d) 216 (d) If we have 4 nitrogen base then we have 64 codons (triplet) 43 Þ 64, As in question there are 6 nucleotide, then 63 = 216. 15

57.

A circus performer of weight W is standing on a wire as shown in the figure. The tension in the wire is

(a) Approximately

w 4

(b) Approximately

w 2

(c) Much more than

w w (d) Much less than 2 2

Ans. (c) Sol. From equilibrium condition W = 2T sinq q

W or T = 2sin q

Ans. Sol.

59.

Ans. Sol.

T W

Since q is small, T will be much more than 58.

q T

W . 2

Number of integers n such that the number 1 + n is a divisor of the number 1 + n2 is : (a) 0 (b) 1 (c) 4 (d) 2 (c) n2 + 1 = n2 – 1 + 2 = (n – 1)(n + 1) + 2 Now, n2 + 1 is divisible by n + 1 only when 2 is divisible by n + 1 So, n = –3, –2, 0, 1 So, option (c) is correct. When 1 gram of a mixture of aluminium and zinc was treated with HCl, a gas was liberated. At the end of the reaction the volume of the liberated gas was found to be 524 cm3 , under STP conditions. The individual weight of aluminium and zinc in the mixture respectively are : (a) 0.2 g and 0.8 g (b) 0.8 g and 0.2 g (c) 0.5 g and 0.5 g (d) 0.322 g and 0.678 g (a) Al = xg Zn = 1 – x g Al + 3HCl ® AlCl3 +

3 H 2 2

x 3 x ´ moles 27 2 27 Zn + 2HCl ® ZnCl2 + H2 1- x 65.3

1- x moles. 65.3

x (1 - x) 524 + moles = (give) e) 18 65.3 22400 Þ Solving this x = 0.21 gm and Zn = 0.79 gm 60. Choose that element which is most different from the other three. (a) Carbon (b) Nitrogen (c) Silicon Ans. (a) Sol. Carbon. due to is special character forming organic chemistry. Total H2produced =

16

(d) Phosphorous

61.

In the following diagrams O is point object and I is its image formed by a concave mirror. Identify the diagram. In which position of image I is nearly correct.

C

F

(a)

C

(b)

O

F

I C

I

O

(c)

F O

C

(d)

F O

I

I

Ans. (a) Sol. From ray diagram, it is clear that option ‘a’ is correct.

C

F

q q O I

62.

If for x, y > 0 we have

1 1 + = 2 then the minimum value of xy is x y

(a) 2

(b) 1

(c) 4

(d)

(c) Crab

(d) Octopus

2

Ans. (b) Sol. If

1 1 + =2 x y

Now, AM ³ GM

1 1 + 1 x y ³ Þ 2 xy

Þ

1 2 £ xy 2

Þ

1 £1 xy

Þ xy ³ 1 Hence least value of xy = 1 So, option (b) is correct. 63.

Which of these is a mollusc? (a) Lobster

(b) Scorpion

Ans. (d) Sol. “Octopus” Belongs to mollusc 17

64.

What is the mechanism used by the kindly to remove waste products from the body? (a) Nephrons convert nitrogenous waste to uric acid and pass it out as urine (b) Nephrons actively transport uric acid and other nitrogenous waste A into the proximal and distal convoluted tubules, from where it is collected to form urine. (c) The blood is filtered to retain cells and large plasma proteins within the blood. The remaining filtrate passes through the proximal and distal convoluted tubules, where needed substances are reabsorbed by active transport (d) Nephrons filter out the nitrogenous waste which is passed through the proximal and distal convoluted tubules and collected by the collecting duct as urine.

Ans. (c) Sol. In kidney blood cells & plasma will not enter in PCT & the remaining filterate is passed through the Nephron, & all useful materials are re-absorbed actively. 65.

Two bodies A and B are charged with equal magnitude of charge but A with positive charge and B with negative. If MA and MB are masses before charging and MIA and MIB are the masses after charging the (m0 is some constant mass) (a) MIA = MA + m0 and MIB = MB – m0

(b) MIA = MA – m0 and MIB = MB + m0

(c) MIA = MIB

(d) MIA = MA –

m0 and MIB = MB + m0 2

Ans. (b) Sol. If a particle is positively charged then electron leaves the object. If particle is negatively charged, extra electron comes into it. Q M IA = M A – m 0 , M IB = M B + m 0 66.

The number of natural numbers n £ 30 for for which (a) 30

(b) Zero

n + n + n + ..... is a natural number is

(c) 6

(d) 5

Ans. (d) Sol. Let K = Þ K=

n + n + n + ..... n+K

Þ K =n+K 2

Þ K2 – n – K = 0 Þ K=

1 + 1 + 4n is a natural no only 2

When 4n + 1 is a perfect square Q n = 2, 6, 12, 20, 30 So, option (d) is correct. 67.

Elements A, B and C have atomic numbers X, X + 1 and X + 2, respectively. 'C' is an alkali metal 'A' reacts with another element 'Y' to form the compound 'AY'. 'A' and 'Y' belong to the same group. 'AY' possesses an (a) ionic bond

(b) covalent bond

Ans. (b) Covalent bond Sol. AY is an interhalogen compound Ex ICl 18

(c) metallic bonding

(d) coordinate bond

68.

Air has three major components : nitrogen, oxygen and argon. Given that one mole of air at sea level is made up of 78% nitrogen, 21% oxygen, and 1% argon, by volume. What is the density of air? Assume that gases behave in an ideal manner. (Atomic mass of argon is 40 amu) (a) 14.62 g/L (b) 1.3 g/L (c) 29 g/L (d) 0.65 g/L Ans. (b) 1.3 gm/L Sol. Molar mass of air =

78 ´ 28 + 21 ´ 32 + 1´ 40 = 28.96 gm 100

mass volume 1 moles of air will occupy 22.4 litre. 28.96 gm \ d= = 1.29 gm/L. 22.4 lit A conductor of length L has a varying cross section with area 2A at P and A at Q as shown in the figure. If it carries a steady current I, then density =

69.

P Q

L

(a) Net charge per unit volume near P is more than net charge per unit volume near Q. (b) Net charge per unit volume near Q is less than net charge per unit volume. (c) Current per unit area near P is more than current per unit area near Q. (d) Current per unit area near P is less than current per unit area near Q. 70.

The number of natural numbers n £ 30 for which

(a) Three Ans. (a) Sol. Let K =

71.

Ans. Sol. 72.

Ans. Sol.

n + n + n + ...... is a prime number is (c) Nine (d) Two

(b) Zero

n + n + n + .... Þ K =

1 + 1 + 4n 2

Here, K is prime only when n = 2,6,20 So, option (a) is correct. Rhodoferax fermentans is a species of photosynthetic bacteria. From your knowledge about bacteria in general, identify the components that CANNOT be present in this organism. (a) Chloroplasts (b) ATP (b) Ribosomes (d) Cell wall (a) Bacteria do not have membrane bound cell - organelles, that's why chloroplast will be absent in bacteria. If atmospheric humidity decreases, transpiration rate (a) Decreases because the concentration gradient between the mesophyll and the atmosphere decreases. (b) Stays the same because active transport does not depend on humidity (c) Increases because of the higher concentration gradient between the air spaces of the mesophyll and the atmosphere (d) Decreases because the concentration of water vapour decreases (c) If humudity decreases the transpiration rate will increase, because there will be more water in mesophyll cell as compared to atmosphere. 19

73.

Vessels A and B are made of conducting material. Both contain water. Vessel A floats in B. Vessel B is now heated at a uniform rate, then (a) Water in A boils first. (b) Water in A boils some time after water in B starts boiling. (c) Water in both A and B start boiling simultaneously. (d) Water in A does not boil.

Ans. (d) Sol. As water in vessel A gets heat via conduction through the vessel, so when temp. reaches 100°C on both inside and outside of vessel A, conduction stops and the water inside, does not get latent heat for boiling whereas the vessel B gets continuous heat from the burner for boiling. So water in A does not boil. 74.

The number of squares formed by 5 vertical and 4 horizontal lines (all are equispaced) is (a) 60

(b) 20

(c) 40

(d) 46

Ans. (d) Sol. No. of square formed by 5 vertical and 4 horizontal lines = 4 × 3 + 3 × 2 + 2 × 1 = 20 So, option (d) is correct. 75.

If 0.50 mole of a monovalent metal (M+1 ) halide is mixed with 0.2 mole of a divalent metal (L+2 ) phosphate, the maximum number of moles of M3PO4 that can be formed is (a) 0.25

(b) 0.30

(c) 0.16

(d) 0.20

Ans. (c) 0.166 mole of M3PO4 Sol. 6MX + L3(PO4)2

® 2M3PO 4 + 3LX2

MX is acting as Limiting meganet. 76.

Every major city in India has a pollution control board to monitor air and water pollution. The following data is from three different localities in Bangalore city from the year 2015.

Locality Annual average of SO2 in the air (volume/volume) P 16.3 mL/m 3 Y 16.3 ppb (m 3/m 3) Z 16.3 ppm (m3/m3) ppb stands for parts per billion and ppm stands for parts per million. These are different units to express concentration. They are very similar to percentage (which is actually parts per hundred). Based on the above data, which place will you choose to live in? (a) All localities are equally polluted, so I have no preference. (b) P is the more polluted than Y and Z, hence I will live in either Y or Z. (c) Locality Y is least polluted, hence I will live in Y. (d) Z and Y are more polluted than P, hence I will live in P. 20

Ans. (c) Locality Y is least polluted, hence I will live in Y.

ml 16.3ml =16.3 ppm ml/ml 3 = m 106 ml 16.3 m3 Y = 16.3 ppb = 109 m3

Sol. P = 16.3

Z = 16.3 ppm =

16.3 m3 = 16.3 ppm 106 m3

77.

A body thrown vertically up reaches a maximum height and returns back. Its acceleration is (a) Downward during both ascent and descent. (b) Downward at all positions except at the highest point, where it is zero. (c) Upward during both ascent and descent. (d) Downward during ascent and upward during descent. Ans. (a) Sol. In both cases acceleration will be in downward direction and equal to ‘g’ if we neglect air friction. 78. The number of integers a, b, c for which a2 + b2 – 8c = 3 is (a) 2 (b) Infinite (c) 0 (d) 4 Ans. (c) Sol. If a2 + b2 – 8c = 3 so a2 + b2 = 8c + 3 Case I : If both a = even, b = even a = 2m, b = 2n 4m2 + 4n2 = 8c + 3 LHS = Even , But RHS = odd Hence no solution Like wise for both a = odd , b = odd., There is no solution. Case II :If any one number is odd, Let a = 2m, b = 2n + 1 so, (2m)2 + (2n + 1)2 = 8c + 3 4 m2 + 4n2 + 4n + 1 = 8c + 3 4(m2 + n2 + n) = 8c + 2 LHS is divisible by 4 But RHS is not divisible by 4. Hence no solution. So, option (c) is correct. 79. Which of the following is NOT produced by the endoplasmic reticulum? (a) Lipids (b) Proteins (c) Monosaccharides (d) Hormones Ans. (c) Sol. Monosaccharides, do not produced by endoplasmic reticulum. 80. Vaccines prevent infections by pathogens by: (a) Presenting the body's immune system with antigens in a controlled manner, so that it is prepared to counter the pathogen producing it when it attempts to infect the body (b) Affecting the reproductive cycle of the invading pathogen (c) Binding to antigens on the surface of the pathogen and inactivating it (d) Affecting the metabolic pathways of the pathogen Ans. (a) Sol. Vaccines stimulates formation of antibody in our body to counters with the upcoming pathogen. 21

NSEJS-2017 (IJSO STAGE-I) Date of Examination : 19th November, 2017 PAPER CODE - JS531

SOLUTIONS 1.

Rajiv, Nikhil, Shubha and Nilima wanted to establish a relationship between loss in weight of a solid with weight of water displaced by immersing it in tap water and sea water. After performing their experiment, they noted their observations for the same solid as follows : Rajiv : Loss of weight of solid is more in tap water. Nikhil : Loss of weight of solid is more in sea water. Shubha : Loss of weight of solid is equal in the tap water and the sea water. Nilima : Loss of weight of solid may be more in tap water or sea water, depending upon how deeply it is immersed, identify the correct observation. (a) Nikhil

(b) Nilima

(c) Shubha (d) Rajiv Ans. (a) Sol. dsea water > dtap water So, weight of water displaced in sea water > in tap water. 2.

The ratio of atoms present in 4 g of magnesium and 4 g of sulphur is (Mg = 24 ; S = 32) (a) 1 : 1

(b) 2 : 1

(c) 3 : 2 Ans. (d)

(d) 3 : 4

4 ´1 4 24 Sol. Ratio = 4 = 3 32 3. If Z = 10 the valency of the element is..............

(a) zero (c) two Ans. (a) Sol. Z = 10 is for neon (noble gas) Hence, valency is zero. 4.

(b) one (d) three

The average atomic mass of an element X is 80 u. The present of isotopes sample is (a) 90.99 and 9.01

(c) 66.67 and 33.34 Ans. (c)

79

X35 and

82

X35 in the

(b) 80.8 and 19.2 (d) 50 and 50

(79 ´ x) + 82 ´ (100 - x) 100 x = 66.67% % 79X35 = 66.67% & % 82X35 = 33.34%

Sol. 80 =

1

5.

An aqueous solution used to preserve biological specimen is (a) Methane

(b) Methanol

(c) Methanal Ans. (c) Sol. Biological specimen is stored in HCHO. 6.

The molecular formulae of some organic compounds are given below, which of these compounds contains a Ketone group ? (a) C3 H6 O 2

(b) C3 H6 O

(c) C3 H4 O Ans. (b) Sol. Ketone has D.U. = 1 7.

(d) Methanoic acid

(d) C3 H8 O

'Duralumin' is an alloy of aluminium with (a) iron, manganese and magnesium

(b) copper, manganese and magnesium

(c) copper, chromium and magnesium (d) iron, nickel and magnesium Ans. (b) Sol. Duralumin contains : Al - 95%, Cu - 4%, Mg - 0.5%, Mn - 0.5% 8.

Tooth decay starts when the pH around tooth is around (a) 7.5

(b) 7

(c) 6.5 (d) 5.5 Ans. (d) Sol. Tooth decay occurs in acidic medium, pH = 5.5. 9.

What will happen if a copper piece is dipped in aqueous solution of silver nitrate for quite some time? (i) Solution will remain colourless (ii) Solution will turn blue (iii) Silver will deposit on the copper piece (iv) Bubbles of brown gas will be formed around copper piece (a) i and iv

(b) ii and iv

(c) ii and iii (d) iii and iv Ans. (c) Sol. Copper being more reactive than silver will displace it, so silver will bet deposited & solution turns blue because of formation of Cu2+. Cu + AgNO3 ® Cu2+ + Ag¯ + NO3– Blue

10.

Neeta mixed 10 mL of 0.1 M HCl solution with 15 mL of 0.067 M NaOH solution. She checked the pH of the resulting solution using pH paper. The colour obtained was Red

Orange

Yellow

Strong acid ¬¾¾ Weak acid

(a) Green

2

Neutral

Pale blue

Dark blue

15 × 0.067 1.005 mmol

Violet

Weak alkali ¬¾¾ Strong alkali

(b) Yellow

(c) Pale blue Ans. (c) Sol. HCl + NaOH ® NaCl + H2O 10 × 0.1 1mmol

Green

(d) Violet

Solution is basic, [OH–] =

0.005 = 2 × 10–4 25

pOH = 3.7 pH = 10.3 Pale blue – Dark blue Appropriate answer (c) 11.

(I) Zn + CuSO4(aq) ® Reaction occurs (II) Zn + Al2(SO4)3 (aq) ® Reaction does not occur (III) Zn + AgNO3 (aq) ® Reaction does not occur (IV) Zn + PbNO3(aq) ® Reaction occurs Which of the above statements is not correct ?

(a) I (b) II Ans. (c) Sol. (I) Zn + CuSO4(aq) ® ZnSO4 + Cu – True

(c) III

(d) IV

(II) Zn + Al2(SO4)3 (aq) ® No reaction – True (III) Zn + AgNO3 ® Zn(NO3)2 + Ag – False (IV) Zn + Pb(NO3)2 ® Zn (NO3) + Pb – True 12.

An open vessel contains air at 27°C. The vessel is heated till two-fifth of the air in it has been expelled. Assuming the volume of the vessel remains constant, find the temperature to which the vessel has to be heated ?

(a) 750 K Ans. (d) Sol. T = 300 K; n1T1 = n2T2

(b) 700 K

(c) 550 K

(d) 500 K

3 × n × T2 5 Þ T2 = 500 K

n × 300 =

13.

A teacher wanted to give acid base titration to her students. For that she prepared (i) HCl solution by dissolving 73 g of hydrochloric acid in one litre of water and (ii) sodium hydroxide solution by dissolving 0.46 g of sodium metal in one litre of water. Find the volume of the hydrochloric acid solution required for complete neutralisation of sodium hydroxide solution. (Cl = 35.5 ; Na = 23.0 ; O = 16.0)

(a) 20 mL Ans. (b) Sol. [HCl] =

(b) 10 mL

(c) 46 mL

(d) 5 mL

73 = 2M 36.5 1

Na + H2O ¾® NaOH + 1H2

0.46 23 [NaOH] = 0.02 M 0.02 × 1 = 2 × v v = 0.01 L = 10 mL 3

14.

What would be the atomic number of the next halogen element, if discovered in future ?

(a) 103 (b) 115 (c) 117 Ans. (c) Sol. After Astatine next element will have atomic number = 85 + 32 = 117. 15.

(d) 121

A white solid known to be a compound of sodium, given rise to water vapour and a colourless gas on heating. The residual white powder is dissolved in water and when the solution is added to alum solution, a white gelatinous precipitate is obtained. The original solid was : (a) Sodium carbonate

(b) Sodium bicarbonate

(c) Sodium hydroxide Ans. (b)

(d) Sodium nitrate

D Sol. NaHCO3 ¾¾ ® Na2CO3 + H2O + CO2

Na2CO3 solution alkaline Al3+ + OH– ¾® Al(OH)3 (white gelatinous ppt) 16.

Harsha was trying to neutralize phosphoric acid using various bases. Those available were caustic soda, lime water and hydrated alumina. If Harsha took 1 equivalent of phosphoric acid each time, what will be the ratio for moles of each of the above bases required for complete neutralization ? (a) 1 : 1 : 1

(b) 1 : 0.5 : 0.33

(c) 1 : 2 : 3

(d) 1 : 0.33 : 0.5

Ans. (b) Sol. Caustic soda - NaOH

n-factor 1

No. of moles 1

Lime water - Ca(OH)2

2

1 2

Hydrated Alumina - Al(OH)3

3

1 3

1 eq of phosphoric acid = 1 eq. of NaOH = 1 eq. of lime water = 1 eq. of Al(OH)3 1: 17.

1 1 : 2 3

A flask containing SO2 gas was weighed at a particular temperature and pressure. The flask was then flushed and filled with oxygen gas at the same temperature and pressure. The weight of SO2 in the flask will be about : (a) same as that of oxygen

(c) four times that of oxygen Ans. (d) Sol.

w1 w 2 = m1 m 2 w1 = 2w2 (SO2) (CO2)

4

w1 w 2 = 61 32

(b) one-fourth that of oxygen (d) twice that of oxygen

18.

Arun needs 1.71 g of cane sugar (C12H22O11) to sweeten his tea. What would be the number of carbon atoms consumed through sugar in the tea ?

(a) 3.66 × 1022 (b) 7.2 × 1021 (c) 5 × 1021 (d) 6.6 × 1022 Ans. (a) 1.71 Sol. × 12 × 6.023 × 1023 = 3.61 × 1022 342 19. Choose the correct sets which represent the oxides as Acidic : basic : neutral : amphoteric respectively (i) CO2 : MgO : N2O : H2O

(ii) SO2 : NO : CO : Al2O3

(iii) P2O5 : ZnO : NO : Al2O3

(iv) SO3 : CaO : N2O : PbO

(a) i and ii

(b) ii and iii

(c) iii and iv Ans. (d) Sol. (i) CO2 - Acidic MgO - Basic N2O - Neutral H2O - Amphoteric (iv)SO3 - Acidic CaO - Basic N2O - Neutral PbO - Amphoteric 20.

(d) i and iv

During a meteorite shower a few meteorites fell into a water body having pH around 7. The pH of the water body was measured after meteorite shower and found to be (a) > 7 (b) < 7 (c) = 7

(d) no change in pH of water due to the meteorite shower Ans. (a) Sol. Meteorites are mainly composed of metal oxides hence solution formed is basic. pH > 7 21.

The position of two blocks at successive 0.20-second time intervals are represented by the numbered squares in the figure below. The blocks are moving towards right. 1

Block a Block b

1

2

2

3

4

3

5

7

6

4

5

The acceleration of the blocks are related as follows : (a) acceleration of ‘a’ is greater than acceleration of ‘b’. (b) acceleration of ‘a’ equals acceleration of ‘b’. Both accelerations are greater than zero. (c) acceleration of ‘b’ is greater than acceleration of ‘a’ (d) acceleration of ‘a’ equals acceleration of ‘b’. Both acceleration are zero Ans. (d) Sol. Velocity is constant for the blocks a and b, so acceleration will be zero. 5

22.

In rural areas, an indigenous way of keeping kitchen materials cool is to put them in a box and wrap the box with weiblanket; the blanket is kept wet as tap is allowed to drip in to its comer. Choose the correct statement: (a) This method works because the water from the tap is cold. If one uses room temperatufe water, it will not work. (b) Method will work only if the box is a bad Conductor of heat. If one uses tin box, it will not work. (c) Method doesn't work

(d) method works because the latent heat necessary for evaporation of water in the blanket is taken from the box so the box and its content remain cool' Ans. (d) Sol. A wet blanket is wraped around a box so latent heat necessary for evaporation of water is the blanket is taken from the box so the box and its content remain cool. 23. In the adjacent circuit what is the current flowing from N to K? 30W

A

30W D

30W

60W

B J 10W G H C N

K

60W

L M

60W

120 V

(a) 3A Ans. (c)

(b) 2A

(c) 1A 60W

30W 10W

30W

60W K

N

Sol.

(d) 0.5 A

30W

60W V=120 V 10W

10W

20W

i V=120 V

120 = 3A 40 \ in NK current will be 1A. If x, v and t represent displacement (m), velocity (m/s) and time 9s) respectively for a certain particle then which pair of the following figures can be best correlated to each other. i=

24.

x

x

Fig.I

(a) I & II 6

t

x

Fig.II

(b) I & III

t

x

Fig.III

(c) I & IV

t

Fig.IV

(d) None

t

Ans. (b) Sol. Slope of x - t graph gives velocity Slope = v = tan q if q will greater than v will be more. 25.

The take off speed of Airbus A340 is 288 km/hr. From the taxi track it comes to the main runway and waits for a while for the final clearance from Air Traffic control. The aircraft then achieves this speed within 50 seconds. Neglecting the effect of the wind direction and friction, what should be the minimum length of main runway decided by civil engineeis for this aircraft for a take-off?

(a) 1800 m Ans. (b) Sol. u = 0, v = 288 × a=

(b) 2000 m

(c) 2200 m

(d) 2400 m

5 = 80 m/s 18

80 - 0 8 = m/s2 50 5

u = 0, v = 80 m/s, a =

8 m/s2, s = ?, t = 50 sec 5

1 1 8 × × at2 = = 2500 = 2000 m 2 2 5 An empty office chair is at rest on a floor. Consider the following forces :

s= 26.

I- A downward force of gravity, II-An upward force exerted by the floor, III-A net downward force exerted by the air. Then, which of the force(s) is (are) acting on the office chair? (a) I only (b) I and II (c) I, II and III (d) None of the forces. (Since the chair is at rest there are no forces acting upon it.) Ans. (b) N

Sol.

mg

But in air net force will be upward byount force. 27.

The ability of eye to focus both near and distant objects, by adjusting its focal length, is called (a) Myopia

(b) Presbyopia

(c) accommodation of eye

(d) Tyndall effect

Ans. (c) Sol. Change in focal length of eye is known as accomodation of eye. 28.

In bringing a a-particle, the electrostatic potential energy of the system _________. (a) increases

(b) decreases

(c) remains unchanged (d) becomes zero 7

Ans. (a) Sol. When a particle are brought together electrostatic potential energy increase. 29.

A magnet is placed between two coils AB and CD as shown. It is being moved in the direction as shown by the arrow, then which of the following statement is correct. (a) looking from end A, current in coil AB will be anticlockwise and looking from end D, the direction of current in coil CD will be anticlockwise. A B C D (b) looking from end A, current in coil AB will be clockwise and looking frojm end D, the direction of current in coil CD will be clockwise

N

S

G

G

(c) looking from end A, curent in coil AB will be clockwise and looking from end D, then direction of current in coil CD will be anticlockwise (d) looking from end A, current in coil AB will be anticlockwise and looking from end D, the direction of current in coil CD will be clockwise Ans. (a) Sol. Using Lenz's law. 30.

A boy throws a steel ball straight up. Consider the motion of the ball only after it has left the boy's hand but before it touches the ground and assume that forces exerted by the air are negligible. For these conditions, the force(s) acting on the ball is (are) : (a) a downward force of gravity along with a steadily decreasing upward force. (b) a steadily decreasing upward force from the moment it leaves the boy's hand until it reaches its height point; on the way down there is a steadily increasing downward force of gravity as the object gets closer to the earth. (c) constant downward force of gravity along with un upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity.

(d) constant downward force of gravity only. Ans. (d) Sol. In free fall only gravity acts. 31.

A large truck collides heat-on with a small compact car. During the collision : (a) the truck exerts a greater force on the car than the car exertes on the truck. (b) the car exerts a greater force on the truck than the truck exerts on the car. (c) the truck exerts a force on the car but the car does not exert a force on the truck.

(d) the truck exerts the same force on the car as the car exerts on the truck. Ans. (d) Sol. According to Newton's IIIrd law for every action there is always equal and opposite reaction. 32.

A common hydrometer has a uniform scale and its stem is graduated downwards from 0 to 20. While floating in water, it read 0 and while floating in a liquid of density 1.40 g/cm3, it reads 20. Then the density of the liquid in which it will reacd 10 is (a) 0.7 g/cm3

(c) 1.17 g/cm3 Ans. (c)

8

(b) 0.85 g/cm3 (d) 2.8 g/cm3

Sol. x

20 rw

rl = 1.4

10

x

Weight of body = A (20 + x) × rw

in water.

Weight of body = A × x × 1.4

in liquid.

x

r

Equating A × x × 1.4 = A × (20 + x) × rw 20 + x = 1.4x Þ 0.4 x = 20 x = 50 Similarly in unknown liquid Þ A × (10 + x) × r = A × x × 1.4 (10 + x) × r = 1.4 × x 60r = 1.4 × 50 r= 33.

1.4 ´ 50 7 = = 1.17 g/cm3 6 60

For the same angle of incidence, the angle of refraction in three different media A, B, C are 15°, 25° and 35° respectively. Then which statement is correct : (µA is refractive index of A) (a) µA is maximum and velocity of light is maximum in medium A. (b) µA is minimum and velocity of light is maximum in medium A. (c) µA is maximum and velocity of light is minimum in medium A. (d) µA is minimum and velocity of light is minimum in medium A.

Ans. (c) Sol. As we known by Snell's law A, B, C have

r = 15°, 25°, 35° respectively

sin i × 1 = µ sin r sin i = constant for all.

µµ

1 (Q sine is increasing function) sin r

air

i

r

Hence as sin r increase µ decreases \ µA > µB > µC and µ =

c 1 Þµµ v v

vA < vB < vC. 9

34.

A liquid, whose density doesn't change during the motion, is flowing steadily through a pipe of varying cross sectional area as shown in the adjacent figure. If a1, a2 are the cross sectional areas, v1 , v2 are the values of velocities (or speeds) at L and H respectively, then the correct relation between a1, a2 and v1, v2 is : H

L

(b) a1v1 = a2v2 (c) a12v2 = a23v1 (a) a1v2 = a2v1 Ans. (b) Sol. Equation of continuity states that for steady flow A1V1 = A2V2 35.

(d) a1 v12 = a2v22

As shown is adjacent figure two plane mirros. M1 and M2 are inclined to each other at the angle 70° (angle M1OM2). Incident ray AB makes an algne of incidence q on M1. This ray after relfection at B on M1 and further at C on M2 travesl along the direction CD, such that path CD is parallel to M1. Then angle q is _______. M2

C

D A

E

O

(b) 50° (a) 45° Ans. (b) Sol. 70° + 70° + 90° – q = 180°

M1 B

(c) 55º 70°

D

70°

q = 230° – 180° q = 50° 36.

(d) 60°

70° 90-q

q

90-q

A copper disc of radius a0 has a hole of radius b0 at the centre, at 0°C. The disce is now heated and mainteained at 200°C. The new radii of disc and hole are a1 and b1 respectively. for the heated disc it can be concluded that. a0 b0

(a) a0 < a1, b0 < b1and density of disc decreases (b) a0 < a1, b0 > b1and density of disc decreases (c) a0 < a1, b0 < b1and density of disc increases (d) a0 < a1, b0 > b1and density of disc increases Ans. (a) Sol. On expansion of Annular disc. Inner and outer both will expand uniformly On increase in volume, density decreases 10

37.

A concave mirror of radius of curvature 1m is placed at the bottom of a water tahnk. The mirror forms an image of the sun when it is directly overhead. If the depth of water in the tank is 80 cm, then the distance of the image formed is _______(refractive index of water is 1.33) (a) on surface of water

(b) 110 cm above mirror

(c) 50 cm above mirror

(d) image cannot be formed

Ans. (c) Sol. R = 100 cm f = –50 cm

From sun u = ¥

u=¥ v=?

1 1 1 = + f v u 1 1 = 50 v

v= –50 cm 38.

80 cm

//////////////////////////// //////

–

I

The equivalent resistance of two resistances in series is 'S'. These resistance are now joined in parallel. The parallel equivalent resistance is 'P'. If S = n P. Then the minimumpossible value of n is (a) 2

(b) 3

(c) 4

(d) 5

Ans. (c) Sol. Let r1 = r2 = r

(where r1 and r2 are two resistors)

In series

s = r1 + r2 = 2r ...... (1)

In parallel

p=

r1r2 r2 r = = r1 + r2 2r 2

......(2)

equation (1) / equation (2)

s 2r = =4 p r/2 39.

An electron a a-particle enter a regionof uniform magnetic field (of induction B) with equal velocities. The direction of B is perpendicualr and into the plane of the paper. The qualitatively identify the direction of paths of electron and the a-particle. II I

(a) I for a- particle, II for electron

III

(b) I for electron, II for a- particle

(c) I for a- particle, III for electron (d) I for electron, III for a- particle Ans. (c) Sol. According to fleming left hand rule we can find force on charge particle. 11

40.

Two wave pulses I and II and the same wavelenth. They are travelling in the directions as shown by the single headed arrows. The resultant sketch of the two wave pulses at some instant of time when P coincides with R is _________. I

a 2a II

a

(a)

(b)

a

(d)

(c)

a

a

Ans. (a) Sol. Due to superposition principle of wave having same wavelength & frequency, amplitude is simply added. Logic of standing waves. 41.

Ravi mixed two substances A and B in a vessel and left it as it is. After few hours he detected an alcoholic smell emanating from the vessel. Identify what A and B are: (a) Salt solution and Lactobacillus

(c) Fruit juice and Lactobacillus Ans. (b) Fruit juice and Saccharomyces

(b) Fruit juice and Saccharomyces (d) Salt solution and Saccharomyces

Sol.

Saccharomyces Fruit juice ¾¾¾¾¾ ® Ethanol + CO 2

42.

Which amongst the following shows the characters of both plants and animals: (i) Anabaena

(ii) Paramecium

(iii) Euglena

(iv) Amoeba

(a) i and iv (b) iii (c) ii (d) i and iii Ans. (b) - (iii) Sol. Euglena shows mixotrophic nutrition so it shows character of both plants and animals. 43.

Which amongst the following are not plastids:

(a) Leucoplasts (b) Chromoplasts (c) Amyloplasts (d) Tonoplasts Ans. (d) Tonoplasts Sol. Tonoplast is the outer membrane of plant vacuole. Remaining three are plastids. 44.

During a study the number of cells was recorded to increase as follows: 64 ® 128 ® 256 ® 512 ® 1024. This represents:

(a) Budding (b) Meiosis (c) Binary fission Ans. (c) Binary fission Sol. In Binary fission each parent cell divides into two daughter cells.

12

(d) Fragmentation

45.

A plant kept in a box with only a hole for entry of light shows bending, the process called phototropism. It occurs due to: (a) Synthesis and diffusion of cytokinin in the leaves (b) Breakdown of auxin in the shoot (c) Synthesis and diffusion of abscisic acid (d) Synthesis and diffusion of auxin in the shoot

Ans. (d) Synthesis and diffusion of auxin in the shoot. Sol. Auxin accumulates in shaded area of shoot tip, and causes bending of shoot tip towards light i.e. phototropism. 46.

What would be the minimum required length of codon to encode 400 amino acids, if there existed three purines and pyrimidines each? (a) 3

(b) 4

(c) 5

(d) 6

Ans. (b) 4 Sol. 3 purines and 3 pyrimidines each so total 6 nitrogen base pairs. We require to encode 400 amino acids. 6n > 400 Þ 6 × 6 × 6 × 6 = 1296 codons So, n = 4 Minimum required length of codon would be 4. 47.

A 'life- style' disorder among these is: (a) Hypertension

(b) Presbyopia

(c) Herpes

(d) Scurvy

Ans. (a) Hypertension Sol. Life style disorders like stress, improper diet, lack of sleep etc. is responsible for causing hypertension. Presbyopia is due to hardening of lens of eye. Herpes is caused due to Herpes simplex virus. Scurvy is caused due to deficiency of Vitamin C. 48.

Metamerism is a characteristic of (a) Hirudinaria

(b) Taenia

(c) Asterias

(d) Pila

Ans. (a) Hirudinaria Sol. Metamerism is the characteristic feature of phylum Annelids. eg. Hirudinaria (Leech) 49.

Health is all about 'eating-fasting' balance. When you fast for extended periods, your cells clean out and recycle the intracellular garbage. The organelles responsible for this are: (a) Microtubules

(b) Microfilaments

(c) Golgi Apparatus

(d) Lysosomes

Ans. (d) Lysosomes Sol. Lysosomes contain digestive enzymes and help in cleaning of cells. 50.

A plant may not exchange CO2 or O2 with air at: (a) twilight

(b) mid-night

(c) late hours in the morning

(d) noon

Ans. (a) twilight Sol. The (light) compensation point is the amount of light intensity where the rate of photosynthesis is equals to the rate of respiration that is at twilight. 13

51.

If a small part of the esophagus of a person is excised, the consequence would be the person will have to eat _________. (a) larger portion of food with large time interval (b) small portions of food at small time interval (c) small portions of food at large time interval

(d) majorly subsist on liquid diet Ans. (b) small portions of food at small time interval Sol. Because small part of esophagus is excised and the patient need to eat small portions of food at small time interval 52.

When heated, the hydrogen bonds between the complementary strands of DNA break and the 2 strands separate in a process called melting. Which of the following pieces of DNA will require maximum temperature for melting? (a) 3' AAGGTATACAAT 5'

(b) 3' GAGCUAUCCGAG5'

5' TTCCATATGTTA 3'

5'CUCGAUAGGCUC 3'

(c) 3' ACGTCCGCTGCG 5'

(d) 3' ATTAGCTAGCAA 5'

5' TGCAGGCGACGC 3' 5' TAATCGATCGTT 3' Ans. (c) 3' ACGTCCGCTGCG 5', 5' TGCAGGCGACGC 3' Sol. 3' ACGTCCGCTGCG 5', 5' TGCAGGCGACGC 3' DNA fragments with more G/C base pairs will require higher temperature for melting. Option (c) has 9 G/C base pairs which is the maximum among all other options. 53.

In a self-pollinated plant, what would be minimum number of meiotic divisions required for setting 400 seeds?

(a) 100 Ans. ( d) 500

(b) 200

(c) 400

Sol. For pollen grain

1 microspore mother cell meiosis uuuuuuuuur 4 microspores.

(d) 500

1 megaspore mother cell meiosis uuuuuuuuur 1 megaspore.

Total number of meiosis = 100 for pollen grains, 400 for megaspore = 500 meioisis to produce 400 seeds. 54.

If a flower is producing a large number of minute and smooth pollen, the agency for cross pollination is most likely to be:

(a) Air (b) Water (c) Insects (d) Bats Ans. (a) Air Sol. Wind pollinated pollen should be small is size, light in weight, large in number and non sticky. 55.

To meet the increasing demand for food, there have been several 'revolutions'. Which of the following revolutions is likely to have contributed most to global warming?

(a) Green (b) White (c) Blue (d) Silver Ans. (b) White Sol. White revolution has major time contributed to increases milk production but the live stock produces numerous polluting gases specially methane which is a potent gas for global warming.

14

56.

A mammal adapted to desert conditions is likely to have large:

(a) Nostrils (b) Pinnae (c) Muzzle (d) Nails Ans. (b) Pinnae Sol. The large ears of the desert animals are used in cooling and dissipating heat. 57.

Which of the following feature indicates omnivorous feeding of human species? (a) Presence of canines as well as premolars and molars (b) Presence of appendix (c) Presence of 11th and 12th pair of ribs

(d) Presence of opposable thumb Ans. (a) Presence of canines as well as premolars and molars Sol. Presence of canines indicates carnivorous nature and presence of both pre molars and molars are indicative of hebivorous nature so humans has both the teeth present so that is suggestive of omnivorous nature. 58.

In a dihybrid cross, what is the proportion of organisms with dihybrid geonotype?

(a) 2/16 (b) 6/16 (c) 4/16 (d) 9/16 Ans. (c) 4/16 Sol. In a dihybrid cross the proportion of organism with dihybrid genotype (RrYy) is 4/16. 59.

If the cell is using less oxygen molecules than the molecules of carbon dioxide evolved in respiration, the substrate for respiration has to be: (a) simple sugars

(b) organic acids

(c) fatty acids (d) cholesterol Ans. (b) organic acids Sol. Respiratory quotient (RQ) is the ratio of volume of CO2 produced to the volume of O2 consumed in respiration. RQ > 1 of organic acids. RQ < 1 of fats RQ = 1 of simple sugars. 60.

Panting is a means of thermoregulation in dogs. This is due to: (a) high specific heat of water

(b) high vapour pressure of water

(c) high latent heat of vapourization (d) high specific gravity of water Ans. (c) Sol. Panting is a means of thermoregulation in dogs due to high latent heat of vaporization. 61.

How many four digit numbers are there such that when they are divided by 101, they have 99 as remainder ?

(a) 90 (b) 98 (c) 100 Ans. (a) Sol. Number is of the form 101 K + 99 Then first four digit number = 101 × 9 + 99 = 1008 Last four digit number = 101 × 98 + 99 = 9997 So number of such four digit numbers is = 90 62.

If x =

(

)

21 – 20 and y =

(a) x = y Ans. (b)

(

(d) 101

)

18 – 17 , then

(b) x < y

(c) x > y

(d) x + y = 0

15

Sol. x = Þ x

21 – 20 , y =

18 – 17

( 21 - 20 )( 21 + 20 )

21 + 20

Similarly, y =

1 21 + 20

1 18 + 17

1 ,y= 21 + 20

Þ x=

=

1 18 + 17

Þ x 4 is such that a(n – 2), a(n), a(n + 3) forms an arithmetic progression. The sum of digits of n is (a) 2

(b) 3

(c) 4

(d) 5

Ans. (b) Sol. For any n-sided regular polygon, Interior angle a(n) =

(n - 2) ´ 180° n

Now, a(n – 2), a(n), a(n + 3) are in A.P

Þ

(n - 4) ´ 180° (n - 2) ´ 180° (n + 1) ´ 180° , , are in AP n-2 n n +3

Þ

2(n - 2) n - 4 n + 1 = + n n -2 n +3

Þ

2n - 4 n 2 - n - 12 + n 2 - n - 2 = n n2 + n - 6

Þ (2n – 4)(n2 + n – 6) = n(2n2 – 2n – 14) Þ 2n3 – 2n2 – 16n + 24 = 2n3 – 2n2 – 14n Þ 2n = 24 Þ n = 12 Q sum of the digits of n = 3 56.

The sum of 5 numbers in geometric progression is 24. The sum of their reciprocals is 6. The product of the terms of the geometric progression is (a) 36

(b) 32

(c) 24

Ans. (b) Sol. Let the terms of GP be

a a , , a, ar, ar2 r2 r

Where a is the first term and r is the common ratio ATQ,

a a + + a + ar + ar2 = 24 r2 r

æ1 1 ö a ç 2 + + 1 + r + r 2 ÷ = 24 r r è ø 1 1 24 + + 1 + r + r2 = 2 r r a

Also,

=

24

r2 r 1 1 1 + + + + 2 =6 a a a ar ar

1æ 2 1 1 r + r +1+ + 2 ç aè r r

ö ÷=6 ø

.....(1)

(d) 18

=

1 24 ´ =6 a a

(from (1))

24 =6 a2

a2 =

24 6

a2 = 4 a = ±2 Product of terms :

a a ´ × a × ar × ar r2 r

= a5 = (±2)5 = ±32 only 32 is given in options 57.

Digits a and be are such that the product 4a1´ 25b is divisible by (in base 10). The number of ordered pairs (a, b) is (a) 15

(b) 8

(c) 6

(d) 4

Ans. (Bonus) Sol. Number is : 4a1 ´ 25b 36 = 22 × 32 Hence we have to check for divisibility by 4 and 9. Product of two numbers, A × B is divisible by 4 and 9, if Case 1 : A is devisible by 36 and B can be any number, Case 2 : B is divisible by 36 and A can be any number, Case 3 : A is divisible by 4 and B is divisible by 9, Case 4 : A is divisible by 9 and B is divisible by 4, Case 5 : A is divisible by 2 and B is divisible by 18, Case 6 : A is divisible by 18 and A is divisible by 2, Case 7 : A is divisible by 3 and B is divisible by 12, Case 8 : A is divisible by 12 and B is divisible by 3, Case 9 : A is divisible by 6 and B is also divisible by 6. Case 1, Case 3, Case 5, Case 6, Case 8, Case 9 are neglected because 4a1 is odd and hence cannot be divisible by any even number. Case 2 : 25b , 5b must be divisible by 4, so be can be 2 or 6 only. 2 + 5 + b = 7 + b must be divisible by 9. for b = 2, 7 + 2 = 9 is divisible by 9. for b = 6, 7 + 6 = 13 not divisible by 9. So b = 2, a can be any digit from {0, 1 ......9} So (a, b) = (0, 2), (1, 2), (2, 2) ....... (9, 2) Þ total 10 pairs. 25

Case 4 : 4a1 : 4 + a + 1 = 5 + a must be divisible by 9. so a can be 4 only. 25b : 5b must be divisible by 4.

So b can be 2 or 6 only. So (a, b) can be (4, 2) and (4, 6) But (4, 2) is already counted in case (2) So (4, 6)

....... 1 Pair

Case 7 : 4a1 : 4 + a + 1 = 5 + a must be divisible by 3, So a can be 1, 4, 7. 25b : 5b must be divisible by 4 so b can be 2 or 6, also 2 + 5 + b = 7 + b must be divisible by 3, b can

be 2, 5, 8. Hence be can be 2 only. So (a, b) can be (1, 2), (4, 2), (7, 2) But all of the these have correctly been considered. Overall total 10 + 1 = 11 Pairs will be there. 58.

The integer closest to (a)

101009 - 1 3

111...1 - 222...2 , where there are 2018 ones and 1009 twos, is (b)

101009 - 1 9

(c)

102018 - 1 3

(d)

102018 - 1 9

Ans. (a) Sol. Using gemetric progression for sum of 'n' term 111 ...... 1 + 2018 time =

102018 - 1 9

æ 101000 - 1 ö 2 ´ 222 ..... 2 + 1009 have = ç ÷ 9 è ø

Now,

59.

=

111......1 - 222...2

=

102018 - 1 2 1009 - (10 - 1) 9 9

=

1 102018 - 2x101009 + 1 3

=

1 1 (101009 - 1)2 = (101009 - 1) 3 3

In a triangle ABC, a point D on AB is such that AD : AB = 1 : 4 and DE is parallel to BC with E on AC. Let M and N be the mid points of DE and BC respectively. What is the ratio of the area of the quadrilateral BNMD to that of triangle ABC?

(a) 1/4 Ans. (d) 26

(b)9/32

(c) 7/32

(d) 15/32

Sol.

Qud.[BNMD] =? D[ABC]

A

Now, since DADE ~ DABC(DE || BC) D

[ADE] AD2 \ = [ABC] AB2

æ AD ö =ç ÷ è AB ø æ1ö =ç ÷ è4ø =

Also,

xM x

E

h 2

y B

y N

C

2

1 16

[ADE] 1 = [ADE] + [DBEC] 16

16[ADE] = [ADE] + [DBEC] 15[ADE] = [DBEC] \

[DBEC] 15 = ADE 1

and [ADE] =

....(1)

[ABC] 16

.....(2)

Also, let h be the height of trapezium BCED and let DM = ME = x and BN = NC = y \ ar of trapezium [DMNB] =

and ar of trapezium [CNME] =

\ [DMNB] = [CNME] =

1 (x + y)h 2 1 (x + y)h 2

1 [BCED] 2

or [DBEC] = 2[BNMD] Put in (1), we get 2[BNMD] 15 = 1 [ABC] 16 32[BNMD] 15 [BNMD] 15 Þ\ = = [ABC] [ABC] 32 1 27

60.

é 10 2 ù é102 ù é102 ù é 10 2 ù The number of distinct integers in the collection ê , , ,..... ú ê ú ê ú ê ú , where [x] denotes the ë 1 û ë 2 û ë 3 û ë 20 û

largest not exceeding x, is (a) 20

(b) 18

Ans. (d) Sol.

é 102 ù é102 ù é102 ù ê ú, ê ú , ..... ê ú ë 1 û ë 2 û ë 20 û é 102 ù ê ú = 100 ë 1 û

é 102 ù ê ú= 9 ë 11 û

é 102 ù ê ú = 50 ë 2 û

é 102 ù ê ú= 8 ë 12 û

é 102 ù ê ú = 33 ë 3 û

é 102 ù ê ú= 7 ë 13 û

é 102 ù ê ú = 25 ë 4 û

é 102 ù ê ú= 7 ë 14 û

é 102 ù ê ú = 20 ë 5 û

é 102 ù ê ú =6 ë 15 û

é 102 ù ê ú = 16 ë 6 û

é 10 2 ù ê ú= 6 ë 16 û

é 102 ù ê ú = 14 ë 7 û

é 10 2 ù ê ú= 5 ë 17 û

é 102 ù ê ú = 12 ë 8 û

é 10 2 ù ê ú= 5 ë 18 û

é 102 ù ê ú = 11 ë 9 û

é 10 2 ù ê ú= 5 ë 19 û

é 102 ù ê ú = 10 ë 10 û

é 10 2 ù ê ú= 5 ë 20 û

\ distinct integers = 15 28

(c) 17

(d) 15

61.

True coelom is not present in animals of: (a) Platyhelminthes

(b) Annelida

(c) Echinodermata

(d) Arthropoda

Ans. (a) Sol. Platyhelminthes is an example of acoelomate animals. 62.

The intracellular organelle that is responsible for formation of acrosomal vesicle is: (a) Endoplasmic reticulum

(b) Golgi apparatus

(c) Mitochondrion

(d) None of the above

Ans. (b) Sol. Golgi apparatus Acrosomal vesicle form during spermiogenesis by golgibody. 63.

The genetically modified (GM) brinjal in India has been developed for: (a) Enhancing shelf life

(b) Insect-resistance

(c) Drought-resistance

(d) Enhancing mineral content

Ans. (b) Sol. The main goal behind genetically modified brinjal is to make it insect resistance. 64.

A scientist observed few cells under a microscope with following characters: i. Cells divided by binary fission or fragmentation, or budding ii. Cells moved with the help of flagella iii. Ether lipids were observed in cell membranes iv. Peptidoglycans were noted in the cell walls Which of the following category do the cells belong to? (a) Archaea

(b) Plant cells

(c) Unicellular eukaryotes

(d) Cyanobacteria

Ans. (a) Sol. Petidoglycan cell wall present in monera means cyanobacteria, archaea etc. But if we compare cyanobacteria and archaea flagella absent in cyanobacteria and present in Archaea. According to question scientist observed cell moved with the help of flagella and flagella absent in cyanobacteria so answer will be (a). 65.

Character(s) of acquired immunity is (are): (a) Differentiation between self and non-self

(c) Retains memory

(b) Specificity of antigen

(d) All the above

Ans. (d) Sol. The immunity acquired by an organism after birth is known as acquired immunity. Which has following characteristics (a) It can differentiate between self and non-self antigen. (b) It is specific for specific antigen. (c) It retains the memory of first encounter with antigen 29

66.

Instead of using chemical fertilizers in a paddy field, a farmer thought of employing nitrogen fixation technique. Amongst the following which would be beneficial for his cause? (a) Glycine max - Rhizobium

(b) Cycas-Nostoc

(c) Casuarina - Frankia

(d) Azolla-Anabaena

Ans. (d) Sol. Azolla (Fern) and Anabaena (B.G.A.) both show symbiotic relationship and perform N2 fixation. This type of relationship is beneficial for paddy field. It increases efficiency of nitrogen use and reduce the water pollution. So Azolla and Anabaena symbiotic relation replace the uses of chemical fertilizers. So among all the four option the pair of Azolla - Anabaena is suitable to grow in paddy field as Azolla is a water fern and paddy grows in fields filled with water. 67.

An action potential in the nerve fibre is produced when positive and negative on outside and inside of the axon membrane are reversed because: (a) all potassium ions leave the axon (b) more potassium ions enter the axon as compared to sodium ions leaving it (c) more sodium ions enter the axon as compared to potassium ions leaving it (d) all sodium ions enter the axon

Ans. (c) Sol. The action potential in nerve fibre is produced due to more permeability of voltage gated sodium channels which causes large influx of Na+ which reverses the polarity of axon membrane. 68.

A geneticist was studying the pathway of synthesis of an amino acid 'X' in an organism. The presence (either synthesized de novo or externally added) of 'X' is a must for the survival of that organism. She isolated several mutants that require 'X' to grow. She tested whether each mutant would grow when different additives, P, Q, R, S and T were used.'+' indicates growth and '-' indicates the inability to grow in the mutants tested. Find out the correct sequence of additives in the biosynthetic pathway of 'X'.

Organisms Wild-type Mutant 1 Mutant 2 Mutant 3 Mutant 4

P + – – – –

Q + – + – +

Additives R + – + + +

S + – + –

T + + + + +

(a) P ® Q ® R ® S ® T

(b) P ® R ® S ® Q ® T

(c) T ® P ® Q ® S ® R

(d) P ® S ® Q ® R ® T

Ans. (d) Sol. '×' is an aminoacid which is required for the growth of organism / Survival. '+' indicates growth '–' indicates inability to grow (So number of '+' increase/number of mutant increase that increase the survival of additives.) P ® S ® (+)

30

(+, +)

Q (+, +, +)

® R ® T (+, +, +, +) (+, +, +, +, +)

69.

In a case of mammalian coat color, the principal gene identified is 'C which codes for a tyrosinase enzyme. In case of rabbits four different phenotypes are observed Full Color > Chinchilla > Himalayan > Albino (in order of the expression of gene 'C and its alleles). In a progeny obtained after crossing two rabbits, the percentages of Chinchilla, Himalayan and Albino rabbits were 50, 25 and 25 respectively. What must have been the genotypes of the parent rabbits? (a) CchCchX Cchc (b) CchCh X Cchc (c)Cchc X Ch c (d) ChCh X CchCch Ans. (c) Sol. Cch ¾® Chinchilla Cch ¾® Himalayan cc ¾® Albino Cchc CchCh

Cchc

Chinchilla

Chc Chc

cc

Chinchilla Himalaya Albino 50

70.

X

25

25

It was observed in a group of tadpoles of a mutant frog reared in a laboratory that their development was arrested at a particular stage. The exact tissue that was affected by the mutation is unknown. The development was then resumed and accelerated by injecting the tadpoles with the extracts prepared from various tissues of the wild type frogs. The observations of the experiment are given below.

Experiment No. Tissue Extract Observations 1 Anterior lobe of pituitary Development resumed 2 Posterior lobe of pituitary Development did not resume 3 Thyroid gland Development resumed 4 Anterior lobe of pituitary + Thyroid gland Development resumed 5 Anterior + posterior lobe of pituitary Development resumed 6 Posterior lobe of pituitary + Thyroid gland Development did not resume From the above observations, find out the tissue that is affected by the mutation. (a) Anterior lobe of pituitary (b) Posterior lobe of pituitary (c) Thyroid gland (d) Both pituitary and thyroid gland Ans. (a) Sol. Anteror pituitary gtand has tropic action which stimualate thyroid gland in turns it screte thyroxine, which stimnlate metamorphosis, mutated Anterior lobe of pituitary would not produc TSH so development is arrested. 71. Identify the odd ones from each group (A and B) based on same criterion. Group A Salmon Bullfrog Platypus Bull shark

Group B Alpine salamander Spiny anteater Common toad Crocodile

(a) Platypus, Alpine Salamander (b) Bull shark, Alpine salamander (c) Bullfrog, Crocodile (d) Platypus, Common toad Ans. (b) Sol. The criteria that is used to distinguish members of group A as well members of groups B is ovoviviparity. In group a bull shark is ovo-viviparous while other three members are oviparous. In group B alpine salamander is ovo-viviparous while other three members are oviparous. 31

72.

A patient was administered a chemical agent called Guanfacine hydrochloride after the patient showed the symptoms like shortness of breath and headache. Guanfacine hydrochloride is a known stimulant of central a2-adrenergic receptors of the medulla regulating the sympathetic nervous system. The patient in this case must be suffering from______. (a) Hypertension

(b) Hyperstimulation

(c) Hyperpolarization

(d) None of the above

Ans. (a) Sol. Drug guanfacine hydrochloride is used to treat the cases of hypertension and this drug binds to the a2 -adrenergic receptors present in medulla oblongata. 73.

A bacterial dsDNA molecule, 2988 bp in length, was found to have the following composition: The respective values of X and Y are:

Strand I Strand II (a) 1400 and 590

T 348 650

(b) 590 and 1400

C X

A

(c) 590 and 590

G 1400 Y (d) None of the above

Ans. (c) Sol. According to chargraff's rule A+T =1 C+G

(Strand I) T = (Strand II) A = 348 (Strand II) T = (Strand I) A = 650 (Strand I) G = (Strand II) C = 1400 \ Value of Y = total nmber of base in strand II – Total number of T, C, A in strand II = 2988 – (348 + 650 + 140) = 590 Value of X = Total nimber of base in strand I – Total number of T, A, G, in strand I = 2988 – (348 + 650 + 1400) = 590 74.

What would be the length of a polypeptide translated from mRNA which is encoded by 2988 bp of a bacterial gene? (a) 989

(b)992

(c)995

(d)998

Ans. (c) Sol. Number of bases in m-RNA = 2988 Number of codon in m-RNA =

2988 3

(as codon is made up of 3 bases) Þ 996 codons Out of 996 codon, one codon is stop codon that does not code for any amino acid. So the length of polypeptide translated will be 995 amino acid. 32

75.

A student recorded the data for five types of cells as given below:

Character P Cell wall Centrioles Chloroplast Mitochondrion Nucleus Plasma membrane RNA/DNA Vacuoles

P + – – – – + + +

Q + – + + + + + +

R – – – + + -

S – + – + + + + +

T + _ _ + + + + +

The five cell types P, Q, R, S and T are: (a) P - Bacterium, Q - Plant, R- Virus, S - Animal, T - Fungus (b) P - Bacterium, Q - Plant, R- Virus, S - Fungus, T - Animal (c) P - Fungus, Q - Plant, R- Bacterium, S - Animal, T - Virus (d) P - Plant, Q - Bacterium, R- Virus, S - Animal, T - Fungus Ans. (a) Sol. According to question cell wall present in bacteria, plant and fungus but Nucleus and membrane bound cell organelles present in both plant and fungus and absent in Bacteria (underdeveloped nucleus) so accurate order will be P-Bacterium, Q-Plant, R-Virus, S-Animal, T-Fungus 76.

An environment conservation group performed a survey of some diverse locations in the country and represented it as under:

Which amongst these sites should be included as a biodiversity hotspot? (a) Site A (b)Sitc B (c)Site C (d)Site D Ans. (a) Sol. To qualify as a biodiversity hotspot on myers 2000 edition of the hosrpot-map a region must contain atleast 1500 species of vasular plants as endonics. 33

77.

A bacterium has a generation time of 50 minutes. A culture containing 108 cells per mL is incubated for 300 minutes. What will be the number of cells after 300 minutes? (1) 64 × 103 cells

(b) 6.4 × 108 cells

(c) 64 × 109 cells

(d) 6.4 × 109 cells

Ans. (d) Sol. Generation time

= 50 minutes

Initial

= 108 cells /ml

Cell count Incubation time

= 300 minutes

[F = I × 2n]

(F = final number of bacteria) (I = Initial number of bacterial = 108)

n = number of generation n= n=

300 =6 50

F = 108 × 26 = 108 × 64 = 6.4 × 109 78.

The blood grouping system is an example of 'multiple allelism. In order to find out the gene products of various gene variants, different enzymes (codes used for the purpose of experimentation are X and Y) from four blood samples were assayed. The enzymes were quantified and the information obtained from these experiments is given in percentages in the following table. indicates presence of an enzyme and indicates the absence of that enzyme from the blood sample. The standard codes for dominant and recessive alleles are considered. Identify the blood groups of subjects and choose the correct option of their genotypes from given options. (In table: + means present, – means absent) Subjects

Ramesh

Ali

Sophia

Balwinder

Enzymes

P/A

%

P/A

%

P/A

%

P/A

%

X

+

50

+

50

+

100

–

–

Y

–

–

+

50

–

–

+

100

(a) IAi, ii, IBi, IAIB

(b) IAi, IAIB, IAIA, IBIB

(c) IBi, IAIB, ii, IBi

(d) IBi, ii, IAIB, IAi

Ans. (b) Sol. Let X enzyme is coded by IA Let Y enzyme is coded by IB In case of Ramesh 50% X is formed which shows that alleles are in heterozygous condition = IAi In case of Ali 50% X and 50% y is formed which shows the presence of both dominant allels IAIB. In case of Sofia 100% of × indicate the genotype IAIA In case of Balwinder 100% of Y indicates the genotyfie IBIB. 34

79.

In an experiment, a scientist discovered a darkly stained chromatin body on the periphery of nucleus of epithelial cells obtained from an eight year old boy. This is indicative of a particular syndrome. Find out the best possible chromosome combination of their parents from the options given below; which have the highest probability of producing the child under investigation. 'A' indicates autosome. 'X' and 'Y' represent the sex chromosomes. (a) 22AA+XY, 22AA+XXX

(c) 22AA+XY, 22AA+XX

(b) 22AA+XXY, 22AA+XXX

(d) 22AA+XXY, 22AA+XX

Ans. (a) Sol. In the given options, option (b) and (d) indicates having kleinfelter's syndrome so the male will be infertile. Option (c) shows normal genotype so the option A has highest probabilty of producing the child under investigation. 80.

A millionaire Mr. Jim, died recently. Two women, Mary and Lou, claiming to have a child by Jim approached the police demanding a share in his wealth. Fortunately Jim's semen sample was cryopreserved. The scientists used DNA fingerprinting technique to study the three highly variable chromosome regions. The results obtained are shown in the adjoining figure:

Jim

Mary

Marry's child

Lou

Lou's child

After studying the DNA profile, which of the alleged heirs are children of Jim? (a) Mary's child

(b) both are children of Jim

(c) Lou's child

(d) none are children of Jim

Ans. (b) Sol. From the given DNA profile it can be observed that the some DNA bands of Jim are matching with some DNA bands of Mary's Lou's child so we can conclude that both the child's belong to Jim.

35

NSEJS-2019 (IJSO STAGE-I) Date of Examination : 17th November, 2019 PAPER CODE - 54

SOLUTIONS 1.

A group of students was studying development of an organism under controlled laboratory conditions. Following observations were made by them. i.

The larvae had a rod-like supporting structure that separated the nervous system and the gut.

ii. A prominent central cavity was present in the transverse section of the part of the nervous system of the larvae; while the adults had cerebral ganglia as the main component of the nervous system. iii. The eyes were prominently seen in larvae. iv. The tails were absent in the adults, which the larvae had. v.

A lot of phagocytic activity was observed before conversion of larvae into adults.

vi. The adults had a cuticular exoskeleton. The organism under study must be belonging to: (a) Amphibia

(b) Pisces

(c) Protochordata

(d) Arthropoda

Ans. (c) Sol. All the observations are of protochordata (Protochordata include urochordata and cephalochordata, urochordata has retrogressive metamorphosis while cephalochordata has progressive metamorphosis) 2.

In case of mice coat colour, two genes are responsible for colour of the hair. Gene 'A' is responsible for distribution of pigments on shaft of hair. Wild type allele of 'A' produces a yellow band on dark hair shaft (agouti), whereas recessive allele produces no yellow band. There is another allele of A, known as AY, which is embryonic lethal in homozygous condition only. In an experiment, two yellow mice were crossed to obtain a progeny of 6 pups. What would be the most probable number of agouti mice among them? (a) 0

(b) 2

(c) 4

Ans. (b)

Sol.

Yellow y

AA

AA

×

Yellow

×

AA

y

AA

y

y

AA

Yellow agauti

living

y

y

AA

In homozygous condition dead

Probability of agouti is 1/3 out of 3 ® 1 agouti so out of 6 ® 2 agouti

1

(d) None of the above

3.

A stain was developed by a group of scientists to stain a particular cell organelle. The stain was tested on various tissues derived from an autopsy sample from a mammal. The organelles were counted. The results showed maximum number of the organelles in cells of brain, lesser in cells of heart, least in mature sperms and absent in erythrocytes. Identify the organelles from following options. (a) Nissl bodies

(b) Mitochondria

(c) Golgi bodies

(d) Endoplasmic reticulum

Ans. (b) Sol. The organelle which is obtained after stain is mitochondria which is abudantly present in brain cell lesser in heart, least in mature sperms and completely absent in erythrocytes. 4.

Pinus sylvestris grows at low temperatures in Russia. The plant survives under such freezing conditions due to the presence of: (a) Saturated lipids in plasma membrane

(b) Glycoproteins in plasma membrane

(c) Glycolipids in plasma membrane

(d) Polyunsaturated lipids in plasma membrane*

Ans. (d) Sol. The plants which grows at low temperature basicaly have unsaturated lipid in their plasma membrane. For maintaining flexibility of plasma membrane. 5.

In an experimental setup, certain pathogen caused a disease in primates with nasal congestion, sore throat and fever being the common symptoms. The scientists injected an extract from blue-green mold as the first line of action. However, the symptoms did not subside. The possible causative agents of the disease were listed out as follows. i. a virus

ii. a fungus

iii. a conjugation deficient bacterium

iv. a tapeworm

Choose the correct option from the following that indicate the pathogen. (a) i, ii

(b) i, iii

(c) ii, iv

(d) iii only

Ans. (b) Sol. Nasal congestion also called sinusitis has sore throat and fever which is caused by virus, conjugation deficient bacterium has no role in it. (Conjugation in bacteria is a process in which plasmid are transferred by themselves alone or along with other DNA element from one cell to another cell through conjugation tube) 6.

A process is represented in the adjacent figure. The arrows indicate the flow of a biochemical reaction. The arrowhead points to the product, while the base of the arrow indicates the template biomolecule. What do P, Q, R, and S represent?

R

X

(a) P : Replication, Q : Translation, R : Transcription, S : Reverse Transcription (b) P : Transcription, Q : Replication, R: Reverse Transcription, S : Translation (c) P : Reverse Transcription, Q : Replication, R : Translation, S : Transcription (d) P : Reverse Transcription, Q : Replication, R : Transcription, S : Translation Ans. (d)

Transcription Sol.

RNA

Reverse Transcription

DNA

Translation Protein

Replication

2

S

P

Y Q

Z

7.

The whooping cranes were on the verge of extinction with only 21 individuals in wild in 1941. After conservation measures, the cranes are now included in the endangered category by IUCN. The highlight of the conservation efforts is the reintroduction of the whooping cranes in wild. This was possible due to raising of the young cranes in absence of their parents by biologists dressed in crane costumes. Aircraft Guided bird migration technique was used for teaching the captive-bred cranes to follow the scientists to learn the migratory route. What type of animal behaviour might be responsible for these captive-bred cranes to follow the crane costume dressed scientists? (a) Cognitive learning

(b) Habituation

(c) Operant conditioning

(d) Genetic Imprinting

Ans. (d) Sol. Imprinting : A type of behavior that includes both learned and innate components is called imprinting. it is become an important component of efforts to save endangered species. 8.

A 4 µm long bacterial cell was magnified and drawn to a dimension of 6 cm. How many times has it been magnified? (a) 1.5 × 103

(b) 15 × 104

(c) 1.5 × 104

(d) 1.5

Ans. (c) Sol. 4µm long bacterial cell = 4 × 10–6 m magnification to a dimension of 6 cm = 6 × 10–2 m magnification = 9.

6 ´ 10-2 magnifying dimension = = 1.5 × 104 4 ´ 10-6 actual dimention

In the baking industry, when the dough is prepared, various ingredients are mixed together with the flour. At one instance, the dough was fermented, but failed to rise sufficiently during the baking process. Choose the correct cause(s) from following possibilities. i.

The salt was mixed before the fermentation process was completed

ii. The sugar was added in excess iii. Yeast granules were not activated prior to mixing with the flour. (a) i, iii

(b) iii only

(c) i, ii, iii

(d) i, ii

Ans. (a) Sol. In process of fermentation, if the dough failed to rise sufficiently during backing process, this might be due to inactivation of yeast. Salt addition to the better can be done immidiately after fermenting. 10.

Given below are four statements. 1. Prokaryotic cells are unicellular while eukaryotes are multicellular. II. Histones are present in eukaryotes and absent in prokaryotes. III. The nucleoid contains the genetic material in prokaryotes and eukaryotes. IV. Prokaryotic flagellum is composed of flagellin while eukaryotic flagellum is composed of tubulin. Identify which amongst these are false. (a) I and II

(b) III and IV

(c) II and III

(d) I and III

Ans. (d) Sol. because prokaryotic cell are unicellular but eukaryotic are not only multicellular, it may be unicellular also. Only prokaryotic genetic material is called nucleoid.

3

11.

The students of a college were working on regeneration using Planaria (Platyhelminthes) and Asterias (Echinodermata). Planaria was cut in three pieces, namely, a piece with head, with tail and the middle piece. Asterias (bearing five arms) was cut in such a way that after separation, six pieces were obtained, namely, an arm with a portion of the central disc, four pieces cut from tips of each of the remaining arms and the remaining body. The animals were allowed to regenerate completely. How many Planaria and Asterias respectively will be obtained after the completion of regeneration in both? (a) 1, 1

(b)3,2

(c) 3, 6

(d)l,2

Ans. (b) Sol. Planaria can be cut into pieces, and each piece can regenerate into a complete organism over the course of a few weeks. Most species of sea stars must split part of their central disc along with a limb for regeneration to occur. It is very unlikely that a severed limb will be able to regenerate into a full-grown starfish unless it is already attached to at least a portion of the central disc. 12.

Fecundity in animal world is the maximum possible ability of an individual to produce offsprings during its entire lifetime. Following factors were checked for their effect on fecundity of different animal models. i.

Availability of food during breeding season

ii. Mode of fertilization iii. Population density Which of these factor(s) can regulate fecundity? (a) i, ii

(b) ii, iii

(c) i, ii, iii

(d) None of the above

Ans. (c) Sol. Fecundity means maximum possible ability of an individual to produce offsprings during life time. Availability of food, mode of fertilization and population density regulate the fecundity. 13.

In case of peppered moths, pale and dark moths are observed. Pale variety is known to be the wild type variety. During industrial revolution, industrial melanism led to prevalence of dark variety around the cities and pale variety continued to be in majority in areas away from the industries. After enforcement of regulations for controlling pollution, reappearance of pale moths in majority was observed around cities again. Driving force(s) for these adaptive changes is/are: i.

Increased pollution around industries

ii. A stable transposition of a gene in moths iii. Limitations of the vision of birds to differentiate dark moths on darkened barks and pale moths in presence of lichens iv. Ability of lichens to grow on barks in less polluted areas only. (a) i, iv

(b) i, iii, iv

(c) i, ii

(d) i, ii, iii and iv

Ans. (b) Sol. It is the example of progressive/ directional Natural selection, in which only one extremity is selected, when environment condtion change it is shited towards another extremity, pale moth is selected after controlling of pollutions and its reappearance observed in cities again. 14.

Four different human body fluid samples were subjected to quantification of hydrogen ion concentration. mEq/L is the unit of measurement for hydrogen ion concentration. The results of the experiment were as follows: Sample A: 1.6 X 102 units

Sample B: 4.5 X 10–5 units

Sample C: 1 X 10–3 units

Sample D: 3 X 10–2 units

Identify the samples in sequence from A to D. 4

(a) Gastric HCl, Venous blood. Intracellular Fluid, Urine (b) Venous blood, Intracellular Fluid, Gastric HCl, Urine (c) Urine, Gastric HCl, Venous blood, Intracellular Fluid (d) Intracellular Fluid, Urine, Gastric HCl, Venous blood Ans. (a) Sol. PH = –log10 [H+] According to this formula correct answer is (a) 15.

Any damage or injury to a particular area causes nociceptors to release some chemicals, which carry the signal to the higher centres in the nervous system for the processing and a subsequent action. However, there is a difference in the way in which the stimulus is received which is related to the acuity of the detection. Fingertips are more sensitive as compared to the forearm. Following reasons for the observed phenomenon were suggested. i.

The receptive fields in the fingertip are smaller

ii. The number of nociceptors per receptive field in the forearm is lesser iii. The amount of prostaglandins released by the nociceptors per receptive field is more in fingertips The most probable reason(s) for this may be: (a) i

(b) i, iii

(c) ii, iii

(d) i, ii, iii

Ans. (d) Sol. Due to any external stimulus (injury) nociceptors release prostaglandins which is more in fingertips per receptive field that's why fingertips are more sensitive as compared to forearm Rate of photosynthesis in hydrophytes depends on various parameters. The adjacent graph shows the effect of one parameter (while keeping all the others constant) on the rate of photosynthesis. Rate of photosynthesis is plotted on Y axis. Identify the parameter which is plotted along X axis: Rate of Photosynthesys

16.

Parameter (a) light intensity

(b) wavelength

(c) temperature

(d) CO2 concentration

Ans. (b) Sol. According to figure wavelength of light affect the photosynthesis. 17.

An organism has 27 pairs of homologous chromosomes. In each daughter cell after completion of mitosis and in each gamete after completion of meiosis II, ________ and ________ chromosomes would be present respectively. (a) 27 and 27

(b) 54 and 27

(c) 108 and 54

Ans. (b) Sol. A cell have 27 pairs of chromosome. 2n = 54 chromosomes.

5

(d) 54 and 108

2n = 54

In mitosis : 2n 2n = 54

In meiosis II : 2n

M

sis eio

I

is ios Me

II n = 27

n n = 27

= 54 n

is ios e M

II n = 27

n = 27 So ans is 54 and 27 respectively. 18.

Gymnosperms are called naked seed bearing plants' because they lack: (a) Male gamete

(b) Ovule

(c) Ovary

(d) Seeds

Ans. (c) Sol. Gymnosperm are called ‘naked seed bearing plant’ because they lacks ovary. 19.

Rahul sprayed a chemical 'X' on a plant with rosette habit. After few days, he found the internodal distances to have increased suddenly. The chemical 'X' might be: (a) Ethylene

(b) Abscisic acid

(c) Auxin

(d) Gibberellic acid

Ans. (d) Sol. Gibberellic acid initiate the internodal distance. 20.

On a study tour, plants with leathery leaves with thick cuticle, vivipary, salt glands, apogeotropic roots, and stomata limited to abaxial surface were observed. The plants might be: (a) Bromeliads

(b) Cycads

(c) Mangroves

(d) None of the above

Ans. (c) Sol. Mangroves plant have the properties of thick cuticle, vivipary, apogeotropic root and so on. 21.

An element Y is a white translucent solid at room temperature and exhibits various allotropic forms. Some compounds of element Y find application in agriculatural industry. Y forms two solid oxides which dissolve in water to form comparatively weak acids. The element Y is : (a) Sulphur

(b) Nitrogen

(c) Phosphrous

(d) Carbon

Ans. (c) Sol.

P

15

Allotropic forms white P, red P, black P, P2 fertilizers : Ammonium phosphate P4O6 + 6H2O ¾® 4H3PO3 P4O10 + 6H2O ¾® 4H3PO4 22.

How many sigma bonds are present between any two carbon atoms in fullerenes? (a) 1

(b) 2

(c) 3

Ans. (a) 6

(d) 4

Sol. Number of sigma bonds between any two atoms will be equal to one. 23.

Four gram of mixture of calcium carbonte and sand is treated with excess of HCl and 0.880 g of carbon-di-oxide is produced. What is the percentage of calcium carbonate in original mixture? (a) 40%

(b) 50%

(c) 55%

(d) 45%

Ans. (b) Sol. CaCO3 + 2HCl ¾® CaCl2 + H2O + CO2 CO2 = moles =

0.88 = 0.02 moles 44

CaCO3 amount = 0.02 × 100 = 2g % of CaCO3 in the mixture is, 24.

2 × 100 = 50% 4

A student was studying reactions of metals with dilute NaOH at room temperature. The student took dilute NaOH in four different test tubes and added Copper powder to test tube A, Zinc dust to test tube B, Aluminium powder to test tube C and Iron powder to test tube D and observed effervescence in. (a) Test tubes A & B

(b) Test tubes B & C

(c) Test tubes C & D

(d) Test tubes A & D

Ans. (b) Sol. For the given infomation, Zn and Al can only react will aqeous NaOH because of their amphoteric nature. Zn + 2NaOH ¾® Na2ZnO2 + H2 ­ Al + NaOH + H2O ¾® NaAlO2 + 1.5 H2 ­ 25.

A magician performed following act: He dipped Rs. 50 note in a 50% solution of alcohol in water and held it on the burning flame, but the note did not burn. The reason behind this is(a) The alcohol kept on dousing the fire (b) Air required for burning was not available (c) The Rs 50 note failed to reach ignition temperature (d) The Rs. 50 note is fire proof

Ans. (c) Sol. Conceptual question 26.

Gammaxene insecticide powder is prepared by the reaction given in the adjacent box. If 78 g of benzene when reacted with 106.5 g of chlorine, how much Gammaxene would be formed?

(a) 140 g

(b) 154.5 g

(c) 145.5 g

Ans. (c) 7

(d) 160 g

Sol.

H UV

+ 3 Cl2 Benzene

78g

Cl

Cl

H

H

Cl H H Cl Gammaxene

Cl

106.5 g (L.R.)

Amount of gammaxene formed is, 27.

Cl H

291 × 106.5 = 145.5 g 3 ´ 71

Arrange following solutions in increasing hydronium ion concentration. The solutions are : (P) 0.1 M HCl (Q) 0.1 M H2SO4 (R) 0.001 M NH4OH (S) 0.001 M Ca(OH)2 (a) P > Q > R > S

(b) Q > P > S > R

(c) S > R > Q > P

(d) S > R > P > Q

Ans. (Bonus) Sol. Correct option should be Q > P > R > S 28.

A zinc rod was dipped in 100 cm3 of 1M copper chloride solution. After certain time the molarity of Cu 2+ ions in the solution was found to be 0.8 M. If the weight of zinc rod, 20 g, then the molarity of chloride ions is ____. (a) 2M

(b)1.5M

(c) 1 M

(d) 0.5 M

Ans. (a) Sol. CuCl2 (aq) ¾® Cu2+(aq) + 2Cl– (aq) 1M

1M

2M

the actual reaction is, Cu+2 + Zn ¾® Zn+2 + Cu. (There will be no change in concentration of Cl– ions) 29.

Which of the following polymeric material will be ideal for remolding? (a) Polythene and Melamine

(b) Polyvinyl chloride and Polythene

(c) Melamine and Bakelite

(d) Bakelite and Polyvinyl chloride

Ans. (b) Sol. Polyvinyl chloride and polyethene are thermoplastics. 30.

When four dilute solutions of (I) vinegar, (II) common salt, (III) caustic soda and (IV) baking soda are tested with universal indicator which will be the correct observation (a) I- Green, II - Violet, III - Blue, IV - Red (b) I - Green, II - Blue, III -Violet IV- Red (c) I - Red, H - Green, III - Violet, IV - Blue (d) I-Red, II- Violet, III - Green, IV - Blue

Ans. (c) Sol. Vinegar – acidic, NaCl – Netural, NaOH - Strong Base, Baking Soda-weak base.

8

31.

Substance X is white crystalline solid which melts after 10 seconds on burner flame. It is soluble in water and insoluble in CCl4 It is a poor conductor of electricity in molten state as well as in the form of aqueous solution, hence we conclude that substance X is (a) an ionic compound

(b) a non polar covalent compound

(c) a polar covalent compound

(d) a pure element

Ans. (c) Sol. Informative question. 32.

In a beaker 50 ml of a normal HC1 solution was taken and NH3 gas was passed through it for some time. The contents of the beaker were then titrated, which required 60 ml of semi normal NaOH solution. How much ammonia was passed through the beaker? (a) 0.85 g

(b) 0.34 g

(c)0.51 g

(d) 0.4 g

Ans. (b) Sol. HCl taken = 1 × 50 NaOH consumed =

= 50 m.xmol

1 × 60 = 30 m.mol 2

NH3 Reacted

= 20 m.mol.

Amount of NH3 passed is,

20 ´ 17 1000

W = 0.34 g 33.

Which is the correct order of metals with reference to their melting point in increasing order? (a) Hg, Ga, Li, Ca

(b) Ca, Li, Ga, Hg

(c) Hg, Li, Ga, Ca

(d) Hg, Ga, Ca, Li

Ans. (a) Sol.

Element Melting point Hg –38.83 °C Ga Li Ca

34.

29.76 °C 180.5 °C 842 °C

Sodium tungstate has formula Na 2WO4, lead phosphate has formula Pb3(PO4)2, formula for lead tungstate should be: (a) PbWO4

(b) Pb2(WO4)3

(c) Pb3(WO4)2

(d) Pb3(WO4)4

Ans. (a) Sol. Na2WO4 ¾® WO42–, Na+ Pb2(PO4)2 ¾® Pb2+, PO43– Hence, the compound will be Pb (WO4) 35.

What is the ratio of reducing agent to oxidizing agent, if the following reaction is correctly balanced? NH3 + O2 ¾® NO + H2O (a) 4 : 5

(b) 5 : 4

(c) 5 : 3

Ans. (a) Sol. Balanced Equation is 4 NH3 + 5O2 ¾® 4NO + 6H2O Ratio of Reduing agent to Oxidizing agent is 4 : 5. 9

(d) 3 : 5

36.

Which of the following is iso-structural with CO2? (a) NO2

(b) N2O4

(c) NO

(d) N2O

Ans. (d) Sol. O = C = O –

37.

+ N=N=O

In one litre of pure water, 44.4 g of calcium chloride is dissolved. The number of ions in one mL of the resultant solution is : (a) 7.23 × 1023

(b) 7.23 × 1020

(c) 4.82 × 1023

(d) 4.82 × 1020

Ans. (b) Sol. Molarity =

44.4 1 × = 0.4 molar 111 1

CaCl2(aq) ¾®

Ca2+(aq)

0.4 m mol

0.4 m mol

+

2Cl–(aq) 2 × 0.4 m mol

Ions = 1.2 × 10 × 6.023 × 10 = 7.23 × 1020 –3

38.

23

Which of the following species is / are isoelectronic with Neon? (i) N3–

(ii) Mg2+

(iii) K+

(iv) Ca2+

(a) only (iv)

(b) only (ii)

(c) both (i) and (ii)

(d) both (i) and (iii)

Ans. (c) Sol.

39.

Species No. of electrons Ne 10 e– N 3– 10 e– Mg 2+ 10 e– Which of the following gases will have equal volume at STP, if the weight of gases is 14.0 g? (i) N2O

(ii) NO2

(iii) N2

(iv) CO

(a) (i) & (ii)

(b) (ii) & (iii)

(c) (i) & (iii)

(d) (iii) & (iv)

Ans. (d) Sol. For the given amount of gas (14g) at S.T.P. as volume is same, molecular mass should be same. Molecular mass : N2 = 28 ; CO = 28 40.

Which of the following are not ionic? (i) AlCl3

(ii) CaCl2

(iii) MgCl2

(iv) LiCl

(a) (i) and (iv)

(b) (i) and (ii)

(c) (ii) and (iii)

(d) (iii) and (iv)

Ans. (a, d) Sol. AlCl3, MgCl2, LiCl are covalent in nature. 41.

Apples dropping from apple trees were observed by many many people before Newton. But why they fall, was explained by Isaac Newton postulating the law of universal gravitation. Which of the following was explained by Isaac Newton postulating statements best describes the situation. (a) The force of gravity acts only on the apple (b) The apple is attracted towards the surface of the earth (c) Both earth and apple experience the same force of attraction towards each other (d) Apple falls due to earth's gravity and hence only (a) is true and (c) is absurd 10

Ans. (c) Sol. Both earth and apple will attract each other with force

FG =

A rectangular metal plate, shown in the adjacent figure has a charge of 420 mC assumed to be uniformly distributed over it. Then how much is the charge over the shaded area? No part of metal plate is cut. (Circles and the diagonal are shown for clarity only. p = 22/7)

14 cm

42.

G m 1m 2 r2

28 cm (a) 45 µC

(b) 450 µC

(c) 15 µC

(d) 150 µC

Ans. (a) Sol. Area of shaded region =

14 ´ 28 14 ´ 28 22 2 - pr 2 = ´ 7 = 42 cm2 2 2 7

\ charge of shaded area = 42 × 43.

420 mC = 45 mC 14 ´ 28

A piece of wire P and three identical cells are connected in series. An amount of heat is generated in a certain time interval in the wire due to passage of current. Now the circuit is modified by replacing P with another wire Q and N identical cells, all connected in series. Q is four times longer in length than P. The wire P and Q are of same material and have the same diameter. If the heat generated in second situation is also same as before in the same time interval, then find N. (a) 4

(b) 6

(c) 16

Ans. (b) Sol. Given ; Power in circuit - I = Power in circuit - II

11

(d) 36

44.

9e2 N 2 e2 Þ R = R P Q

9 N2 = Þ R P 4R P

Þ N2 = 36

Þ N=6

ëéQ l Q = 4 l P Þ R Q = 4R P ûù

A piece of ice is floating in water at 4° C in a beaker. When the ice melts completely, the water level in the beaker will (a) rise

(b) fall

(c) remains unchanged

(d) unpredictable

Ans. (a) Sol. Since ice at 0° melts completely in water at 4°C. Equilibrium temperature will be slightly less than 4°C. Density of water is maximum at 4°C. \ Hence volume of water increases as ice melts completely.

m

So water level will rise. Mathematically

V1 =

V2 =

m rw m r'w

4°C

b. If an = an – bn for n ³ 1 the the value of (a) 2

(b) 3

(c) 4

Ans. (d) Sol. x2 – 5x + 3 = 0 Þ x2 + 3 = 5x then, a2 + 3 = 5a Þ a8 + 3x6 = 5a7

......(1)

and b8 + 3b6 = 5b7

......(2)

Subtracting equation (2) from equation (1) 3(a6 – b6) + a8 – b8 = 5(a7 – b7) Þ

3a 6 + a8 =5 a7

19

(d) 5

3a 6 + a 8 is a7

62.

In the given figure, two concentric circles are shown with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at points B, C, D and A. What is the ratio of the perimeter of the outer circle to that of quadrilateral ABCD?

P

A

S (a)

p 4

(b)

B

Q

O

C

D

R

3p 2

(c)

p 2

(d) p

Ans. (c) Sol. Let radius of inner circle be r, then AB = 2r Þ AC = 2r = PQ = QR PQ2 + QR 2 = 2 PQ = 2 2r

so, PR = Þ OR =

PR = 2r 2

Required Ratio is 2p( 2 r) : 4 2 r Þ p:2 63.

In rectangle ABCD, AB=5 and BC=3. Points F and G are on the line segment CD so that DF = 1 and GC=2. Lines AF and BG intersect at E. What is the area of DAEB? (a) 10 sq. units

(b) 15/2 sq. units

(c) 25/2 sq. units

(d) 20 sq. units

Ans. (c) E

Sol. DEFG ~ DEAB Þ

EM 2 = EN 5

D

F

2 M

EN - MN 2 = Þ EN 5

Þ 1-

1

G

2

C

3

3 2 = EN 5

Þ EN = 5 ar (DABE) = 64.

A

1 25 ×5×5= sq. units 2 2

N 5

B

The number of triples (x,y,z) such that any one of these numbers is added to the product of the other two, the result is 2, is (a) 1

(b) 2

(c) 4

Ans. (b) Sol. xy + z = 2

.....(1)

yz + x = 2

.....(2)

zx + y = 2

.....(3) 20

(d) infinitely many

equation (1) – equation (2) Þ y(x – z) + z – x = 0 Þ (x – z) (y – 1) = 0 Either x = z or y = 1 Case-1 x = z Þ x(y + 1) = 2

.....(4) (from equation 1)

Þ x2 + y = 2 Þ x2 = 2 – y .....(5) (from equation 3) and equation (4)/(5) Þ

x2 (y + 1)2 4 = 2 2-y x

Þ (y2 + 2y + 1) (z – y) = y y2 – 3y + 2 = 0 Þ (y – 1)2 (y + 2) = 0 When y = 1 we get x = z = 1 and when y = –2 we get x = z = –2 Case-2 y = 1 x+z=2

(from equation 1)

xz = 1

(from equation 3)

On solving these equation we get x = z = 1 Therefore possible triples are (1, 1, 1) & (–2, –2, –2) 65.

What is the product of all the roots of the equation (a) – 64

5|x | +8 = x 2 – 16 ?

(b) – 24

(c) 576

(d) 24

Ans. (a) Sol.

5 x + 8 = x 2 - 16

Þ 5|x| + 8 = x2 –16 Þ |x|2 – 5|x| – 24 = 0 Þ (|x| – 8) (|x| + 3) = 0 Þ |x| = 8

As |x| ¹ –3

Þ x=±8 \ Products of the roots = –64 66.

How many positive integers N give a remainder 8 when 2008 is divided by N. (a) 12

(b) 13

(c)14

Ans. (d) Sol. Given : 2008 º 8 (mod N) Þ 2008 = NK + 8, K Î N Þ NK = 24 × 53 \ Number of factors = 20 Leaving factors 1, 2, 4, 5, 8 as divisor > remainder So number of required positive integers = 20 – 5 = 15 21

(d) 15

67.

LCM of two numbers is 5775. Which of the following cannot be their HCF? (a) 175

(b) 231

(c) 385

(d) 455

Ans. (d) Sol. Let a and b be the two number The, LCM (a, b) = 5775 = 52 × 3 × 7 × 11 Here, possible products are 231, 175, 385 \ 455 is not possible. 68.

If a, b, c are distinct real numbers such that a + (a) ± 2

(b)

1 1 1 = b + = c + evaluate abc. b c a

(c)

2 –1

3

(d) ±1

Ans. (d) Sol.

a+

1 1 1 =b+ =c+ b c a

Then, a – b =

1 1 b-c – = c b bc

b–c=

1 1 c -a - = a c ac

c–a=

1 1 a-b - = b a ab

Þ (a – b)(b – c)(c – a) =

é ë

(a - b)(b - c)(c - a) a 2 b 2c 2

Þ (a – b)(b – c)(c – a) ê1 -

1 ù =0 a b2c 2 úû 2

So, either (a – b)(b – c)(c – a) = 0 or 1 -

1

( abc )2

=0

But a = b, b = c, c = a is not possible as a, b, c are distinct. So, a2b2c2 = 1 \ abc = ± 1 69.

Mr. X with his eight children of different ages is on a family trip. His oldest child, who is 9 years old saw a license plate with a 4-digit number in which each of two digits appear two times. "Look daddy" she exclaims. "That number is evenly divisble by the age of each of us kids! "That's right," replies Mr. X, "and the last two digits just happen to be my age" Which of the following is not the age of one of Mr. X's children? (a) 4

(b) 5

(c) 6

Ans. (b)

22

(d) 7

Sol. age of oldest child = 9 years \ age of 7 younger childs Î {1, 2, ...., 8} The 4 digit number is divisible by all the ages of 8 childrens. And in {1, 2, ..., 9}, there are 4 even numbers. So atleast 3 childs will be even aged, so that 4 digit number will be surely divisible by all even number so option (a) 4 and (c) 6 will be neglected. Let's check option (b) i.e., 5. If none of children has age 5, then smallest number divisible by rest of the numbers is 9 × 8 × 7 = 504 and checking its multiples we get 504 × 11 = 5544 a four digit number with same two digits. So, age of X = 44 age of children = 1, 2, 3, 4, 6, 7, 8, 9. So answer is option (b) 5. 70.

How many numbers lie between 11 and 1111 which divided by 9 leave a remainder 6 and when divided by 21 leave a remainder 12? (a) 18

(b) 28

(c) 8

(d) None of these

Ans. (a) Sol. 33 is one of the number that satisfies the condition Required number will be k × LCM (21, 9) + 33 Þ 63k + 33 Þ 11 < 63k + 33 < 1111 -22 1078