INDIAN ASSOCIATION OF PHYSICS TEACHERS - … that she prepared (i) HCl solution by dissolving 73 g of hydrochloric acid in one litre of water and (ii) sodium hydroxide solution by

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JUNIOR SCIENCE NSEJS 2017-2018 EXAMINATION

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-5151200 Website : www.careerpoint.ac.in, Email: [email protected]

CAREER POINT

INDIAN ASSOCIATION OF PHYSICS TEACHERS NATIONAL STANDARD EXAMINATION IN JUNIOR SCIENCE (NSEJS) 2017-2018

Total time : 120 minutes (Total Marks : 240) [CODE JS531]

19-11-2017

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ONLY ONE OUT OF FOUR OPTIONS IS CORRECT

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 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. According to Archimede's principle: loss of weight = bouyant force (FB)

Since, solid is immersed in water, so FB density of liquid. As density of sea water is more in compare to density of tap water, loss of weight of solid is more in sea water.

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2. The ratio of atoms present in 4g of magnesium and 4 g of sulphur is (Mg = 24; S = 32) (a) 1 : 1 (b) 2 : 1 (c) 3 : 2 (d) 3 : 4 Ans. [d] Sol. Given mass of Mg = 4 g Atomic weight of Mg = 24 g

Moles = 424

= 16

No. of atoms = AN6

Given Mass of S = 4g Atomic weight of S = 32 g

Moles = 324 = 1

8

No. of atoms = AN8

Ratio of atoms of sulphur is to magnesium is

A

A

N8

N6

=86 = 3

4 3 : 4

3. If Z = 10 the valency of the element is ............ (a) zero (b) one (c) two (d) three Ans. [a] Sol. Atomic No. (Z) = 10 Its configuration is 2, 8. It is the configuration of noble gas. Hence its valency is zero. 4. The average atomic mass of an element X is 80u. The percent of isotopes 79X35 and 82X35 in the sample is - (a) 90.99 and 9.01 (b) 80.8 and 19.2 (c) 66.67 and 33.34 (d) 50 and 50 Ans. [c] Sol. Let the % of 79X35 be x and that of other be (100 – x).

Average atomic mass = 79 x 82 (100 x)100

80 = 79x 8200 82x100

8000 = 8200 + 79x – 82x 82x – 79x = 8200 – 8000 3x = 200

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x = 2003

= 66.66 %

i.e. percentage of 79X35 is 66.66 & that of 82X35 is (100 – 66.67) = 33.34 5. An aqueous solution used to preserve biological specimen is - (a) Methane (b) Methanol (c) Methanal (d) Methonoic acid Ans. [c] Sol. An aqueous solution used to preserve biological specimen is methanal

HCHO formaldehyde or methanal.

6. The molecular formulae of some organic compounds are given below, which of these compounds contains a

Ketone group ? (a) C3H6O2 (b) C3H6O (c) C3H4O (d) C3H8O Ans. [b] Sol. The molecular formula of a ketone containing 3 carbon atom is C3H6O H3C– C

||O

–CH3

7. '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 is an alloy of aluminium with copper, manganese and magnesium. 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 starts when the pH of the mouth is lower than 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. Ag.)aq()NO(Cu.)aq(AgNO)s(Cu)Blue(23

)colouress(3

It is a displacement reaction. Cu(NO3)2 solution is blue in colour. Silver will deposit on the copper piece.

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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 Green Pale Blue Dark Blue Violet Strong acid Weak acid Neutral Weak alkali Strong alkali

(a) Green (b) Yellow (c) Pale Blue (d) Violet Ans. [c] Sol. Millimoles of HCl = 0.1 × 10 = 1

Millimoles of NaOH = 0.067 × 15 = 1.005

So, concentration of OH– = 1.005 125

[OH–] = 2 × 10–4

pOH = 4 – log 2 = 3.7

pH + pOH = 14

pH = 14 – pOH = 14 – 3.7 = 10.3

The pH range of 8-11 is of weak base and it gives pale blue colour.

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 (c) III (d) IV

Ans. [c]

Sol. (I) CuZnSOCuSOZn 44

Reaction occur because Zn is more reactive than Cu.

(II) 342 )SO(AlZn Reaction does not occur because Al is more reactive than Zinc.

(III) Ag2)NO(ZnAgNO2Zn 233

It is a displacement reaction Zn is more reactive than Ag, So It displaces the Ag from AgNO3

(IV) Pb2)NO(ZnPbNO2Zn 233

Reaction occur because Zn is more reactive than Pb Now according to question (III) is not correct.

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 (b) 700 K (c) 550 K (d) 500 K Ans. [d]

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Sol. P1 = P P2 = P [Open vessel] V1 = V V2 = V

n1 = n n2 = n – 25

(n) = 3n5

T1 = 300 K T2 = ?

1 1 1 1

2 2 2 2

P V n RTP V n RT

n1T1 = n2T2

n × 300 = 35

× n × T2

T2 = 300 53

T2 = 15003

= 500 K

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 (b) 10 mL (c) 46 mL (d) 5 mL Ans. [b]

Sol. Molarity HCl =

7336.5

1 litre = 2

2 Na + 2H2O 2NaOH + H2

2moles 2 moles

2346.0 = 0.02 mole 0.02 mole

HCl + NaOH NaCl + H2O Mole 1 1

0.02 mole of NaOH required 0.02 mole HCl Since molarity of HCl = 2 M M × V = no. of moles 2 × V = 0.02

V = 0.022 100

V = 1100

Litre

V = 10 ml

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14. What would be the atomic number of the next halogen element, if discovered in future ? (Periodic table

provided Page-2)

(a) 103 (b) 115 (c) 117 (d) 121

Ans. [c]

Sol. Atomic number of the next halogen element is 117.

Atomic no. of At (Astatine) = 85

Atomic no. of next halogen = 85 + 32 = 117 15. A white solid known to be a compound of sodium, gives 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 (d) Sodium nitrate

Ans. [b]

Sol. 2)CarbonateSodium(232

)ebicarbonatSodium()solidWhite(

3 COOHCONaNaHCO2

CO2 is a odourless and colourless gas

When Na2CO3 dissolved in water it will give alkaline solution.

When we add this solution in alum solution white gelatinous ppt of Al(OH)3 is obtained

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. Assume

Caustic soda (NaOH) is a monoacidic base, Calcium hydroxide Ca(OH)2 is diacidic base, Hydrated alumina

Al(OH)3 is triacidic base.

For neutralization with one equivalent of phosphoric acid (tribasic acid) each time

(Molesbase × (V.F.) base = Eq. of acid).

The ratio of moles of bases required will be

NaOH : Ca(OH)2 : Al(OH)3

1 : 0.5 : 0.33

17. 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 (b) one-fourth that of oxygen

(c) four times that of oxygen (d) twice that of oxygen

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Ans. [d] Sol. Let mass of SO2 be x & that of O2 be y. Since, volume is same therefore no of moles of SO2 & O2 are also same.

Now no. of moles Mass.MolMassGiven

Now Mol. Mass of O2 = 32 g & Mol. Mass of SO2 = 64 g

Moles of O2 32y

Moles of SO2 64x

Since, no. of moles is same.

32y

64x

32 x = 64 y x = 2y therefore, the wt. of SO2 is 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] Sol. C12 H22 O11 = 342 gm.

mole = GMM

massGiven = g342g17.1 = 0.005 mole

No. of carbon atoms = 0.005 × 6.02 × 1023 × 12 = 3.61 × 1022 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) & (ii) (b) (ii) & (iii) (c) (iii) & (iv) (d) (i) & (iv) Ans. [d] Sol. (i) CO2 : MgO : N2O : H2O (Acidic) (Basic) (Neutral) (Amphoteric) (ii) SO2 : NO : CO : Al2O3 (Acidic) (Neutral) (Neutral) (Amphoteric) (iii) P2O5 : ZnO : NO : Al2O3 (Acidic) (Amphoteric) (Neutral) (Amphoteric) (iv) SO3 : CaO : N2O : PbO (Acidic) (Basic) (Neutral) (Amphoteric) Therefore, the correct order is in (i) & (iv).

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20. 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. [b] Sol. The pH will become less that 7 due to increases in carbon content as the combustion of meteors will occur

when it enters the earth's atmosphere due to friction. Due to meteorite shower temperature of water body increases as a result pH decreases. 21. The positions 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 2 3 4 5 6 7

1 2 3 4 5

Block a

Block b

The accelerations 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 accelerations are zero. Ans. [d]

Sol. Block - a 2.0

42.026v1

2.0

42.0

610v2

Block - b 2.0

112.0

011v1

2.0

112.01122v2

aA = 0 ab = 0

22. In rural areas, an indigenous way of keeping kitchen materials cool is to put them in a box and wrap the box with wet blanket ; the blanket is kept wet as tap is allowed to drip in to its corner. Choose the correct statement :

(a) This method works because the water from the tap is cold. If one uses room temperature 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) This 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.

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Ans. [d]

Sol. This 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.

23. In the adjacent circuit what is the current flowing from N to K ?

30

30

60

60

10 30 60

120 V

L

M

K A B

D C

J

H

N

G

(a) 3 A (b) 2 A (c) 1 A (d) 0.5 A Ans. [c]

Sol.

120 V

10 10 20

.Amp340

120I

60

60

60

I '

I '

I '

I

30

30

30

10

Amp133

3I'I

24. If x, v and t represent displacement (m), velocity (m/s) and time (s) respectively for a certain particle then which

pair of the following figures can be best correlated to each other.

x

t Fig. I

v

t Fig. II

v

t Fig. III

v

t Fig. IV

(a) I & II (b) I & III (c) I & IV (d) None

Ans. [B]

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Sol. Slope of x – t gives v which is constant in OA & AB both. But slope of AB is less than OA so correct v – t

graph will be fig. III.

x

t

A

B

O

v

t 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 engineers for this aircraft for a take-off ?

(a) 1800 m (b) 2000 m (c) 2200 m (d) 2400 m Ans. [b] Sol. u = 0

80185288

hrkm288v m/s

t = 50 sec

502800t

2vuS

= 2000 m

26. An empty office chair is at rest on a floor. Consider the following forces : 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] Sol. Due to air net force will be negligible and in upward direction.

27. The ability of eye of focus both near and distant objects, by adjusting its focal lengths, is called (a) Myopia (b) Presbyopia (c) accommodation of eye (d) Tyndall effect Ans. [c] Sol. Power of accommodation of eye

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28. In bringing an -particle towards another -particle, the electrostatic potential energy of the system ______ .

(a) increases (b) decreases (c) remains unchanged (d) becomes zero

Ans. [a]

Sol. We need to do work against electrostatic repulsion force.

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 statements is correct :

A B

G

N S

C D

G (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

(b) 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

(c) looking from end A, current in coil AB will be clockwise and looking from end D, the direction of

current in coil CD will be anticlockwise

(d) looking from end A, current in coil AB will be clockwise and looking from end D, the direction of

current in coil CD will be clockwise

Ans. [a]

Sol. By 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 highest

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 an 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. After leaving from hand, there will be only gravitation force on the body, which always acts in downward

direction.

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31. A large truck collides head-on with a small compact car. During the collision. (a) the truck exerts a greater force on the car than the car than the car exerts 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 exerts 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. From Newton's third law. 32. A common hydrometer has a uniform scale and its stem is graduated downwards from 0 to 20. While floating

in water, it reads 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 read 10 is ________ .

(a) 0.7 g/cm3 (b) 0.85 g/cm3 (c) 1.17 g/cm3 (d) 2.8 g/cm3 Ans. [c] Sol. When hydrometer is floating in water, Let its x length is inside water. In all three cases given in question buoyant force will balance weight of hydrometer. FB = w g (x + 20) A = 1.4 g(x) A = l g (x + 10) A …….. (i) x + 20 = 1.4 x x = 50 Now on using value of x in equation (i) 50 + 20 = l (50 + 10)

l 6070

= 1.166 g/cm3 1.17 g/cm3

33. 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 ? (A is refractive index of A) (a) A is maximum and velocity of light is maximum in medium A. (b) A is minimum and velocity of light is maximum in medium A. (c) A is maximum and velocity of light is minimum in medium A. (d) A is minimum and velocity of light is minimum in medium A. Ans. [b]

Sol. rsin

1A A > B > C

v1

A vA < vB < vC

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

(a) a1 v2 = a2 v1 (b) a1 v1 = a2 v2 (c) 1

222

21 vava (d) 2

22211 vava

Ans. [b] Sol. From continuity equation : a1v1 = a2v2

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35. As shown in adjacent figure, two plane mirrors M1 and M2 are inclined to each other at an angle 70°

(angle M1OM2). Incident ray AB makes an angle of incidence on M1. This ray after reflection at B on M1

and further at C on M2 travels along the direction CD, such that path CD is parallel to M1. Then angle is_:

O

M2

E

A

M1

D

B

C

(a) 45° (b) 50° (c) 55° (d) 60° Ans. [b] Sol.

O

M2

E

A

M1

D

B

C

70°

70°

90° – = 180° – (2 × 70°)

90° – = 180° – 140° = 40°

= 90° – 40° = 50°

36. A copper disc of radius a0 has a hole of radius b0 at the centre, at 0°C. The disc is now heated and maintained at 200°C. The new radii of disc and hole are at and bt respectively. For the heated disc it can be concluded that :

b0

a0

(a) a0 < at, b0 < bt and density of disc decreases (b) a0 < at, b0 > bt and density of disc decreases (c) a0 < at, b0 < bt and density of disc increases (d) a0 < at, b0 > bt and density of disc increases Ans. [a] Sol. On heating expansion occurs so every linear dimension increases and since mass is same, density will

decreased.

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37. A concave mirror of radius of curvature 1 m is placed at the bottom of a water tank. 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. When object is at infinity, rays come parallel to the principal axis, converges at focus of the concave mirror

after reflection and image will form there. Rays entering the water surface will not suffer any deviation as they strike the surface normally.

F 100 cm

50 cm

So, image will form at focus which is

f = 2R =

2100 = 50 cm above the concave mirror inside water

38. The equivalent resistance of two resistances in series is 'S'. These resistances are now joined in parallel. The

parallel equivalent resistance is 'P'. If S = n P. Then the minimum possible value of n is : (a) 2 (b) 3 (c) 4 (d) 5 Ans. [c] Sol. S = nP (R1 + R2)2 = nR1R2

Minimum occurs for R1 = R2

(R1 + R1)2 = 21nR

21R4 = 2

1nR n = 4 39. An electron and -particle enter a region of uniform magnetic field (of induction B) with equal velocities.

The direction of B is perpendicular and into the plane of the paper. Then qualitatively identify the direction of paths of electron and the -particle.

I II

III

(a) I for -particle, II for electron (b) I for electron, II for -particle (c) I for -particle, III for electron (d) I for electron, III for -particle Ans. [c] Sol. From Fleming's left hand rule, positive charge will move in left direction and negative charge in right

direction.

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40. Two wave pulses I and II have the same wavelength. 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_______.

a

P

I R

2aII

(a) a (b)

a

(c) a (d)

a

Ans. [a] Sol. Resultant amplitude is given by

cosaa2aaA 2122

21

= 180°

180cosa2a2)a2(aA 22

= aa4a4a 222

a

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 (b) Fruit juice and Saccharomyces (c) Fruit juice and Lactobacillus (d) Salt solution and Saccharomyces Ans. [b] Sol. Alcohol is produced due to fermentation of fruit juices by Saccharomyces. 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] Sol. Euglena shows characters of both plants and animals. Plant characters - Photosynthesis Animal characters -Motility, presence of flagellum

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43. Which amongst the following are not plastids :

(a) Leucoplasts (b) Chromoplasts (c) Amyloplasts (d) Tonoplasts

Ans. [d]

Sol. Tonoplast is not a plastid, it is the membrane of vacuole.

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 (d) Fragmentation

Ans. [c]

Sol. Binary fission as in this one cell gives rise to two daughter cells.

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]

Sol. Synthesis and diffusion of auxins promotes phototropism

46. What would be the minimum required length of codon to encode 400 amino acids, if there existed three

purines and pyrimidines eeach ?

(a) 3 (b) 4 (c) 5 (d) 6

Ans. [b]

Sol. Minimum required length of codon to encode 400 amino acids is 4.

47. A 'life-style' disorder among these is :

(a) Hypertension (b) Presbyopia (c) Herpes (d) Scurvy

Ans. [a]

Sol. Hypertension or high blood pressure is a life-style disorder as certain habits like smoking, obesity and lack

of physical activity play a major role in development of Hypertension

48. Metamerism is a characteristic of :

(a) Hirudinaria (b) Taenia (c) Asterias (d) Pila

Ans. [a]

Sol. Hirudinaria belongs to phyllum-Annelida and its body is segmented showing Metamerism.

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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] Sol. Lysosomes are cleaners of the cell. 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] Sol. During twilight, at the time of dawn and dusk, light is dim so photosynthesis is less. Rate of photosynthesis = Rate of respiration. so, plants may not exchange gases at twilight. 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] Sol. So, small portions of food will move is small interval of time due to the length of esophagus being less now.

due to this person might feel full earlier. This is the reason why he has 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' GAGCUAUCCGAG 5' 5' TTCCATATGTTA 3' 5' CUCGAUAGGCUC 3' (c) 3' ACGTCCGCTGCG 5' (d) 3' ATTAGCTAGCAA 5' 5' TGCAGGCGACGC 3' 5' TAATCGATCGTT 3' Ans. [c] Sol. There exist a double bond between A = T, A = U and a tripple bond between C ≡ G. So, the maximum

temperature required for melting will be in the one which has maximum number of tripple bonds. 53. In a self-pollinated plant, what would be minimum number of meiotic divisions required for setting 400

seeds ? (a) 100 (b) 200 (c) 400 (d) 500 Ans. [d] Sol. For the formation of 400 seeds 400 pollen grains and 400 egg cells are required. 400 pollen grains will be

formed from 100 meiotic divisions. 400 egg cells will be formed from 400 meiotic divisions as each embryo sac contains 1 egg cell. So, formation of 400 seeds would require 500 meiotic divisions.

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CAREER POINT

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] Sol. Large number of minute and smooth pollens are pollinated by Air. 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. [a] Sol. Green revolution, as Wheat, Rice, Soyabean has led to increase in C in atmosphere (CO2) which is a green

house gas. 56. A mammal adapted to desert conditions is likely to have large : (a) Nostrils (b) Pinnae (c) Muzzle (d) Nails Ans. [c] Sol. Large muzzle increases the surface area of animal body. So, it lets heat escape from the body of desert

animals. Though, pinnae are also larger in desert animals but it is not so in case of all desert mammals like camels.

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] Sol. Canines are present in carnivores for tearing flesh and premolars and molars are present in Herbivores for

chewing and grinding. Presence of both of these in humans shows its omnivorous nature. 58. In a dihybrid cross, what is the proportion of organisms with dihybrid genotype ? (a) 2/16 (b) 6/16 (c) 4/16 (d) 9/16 Ans. [d] Sol. Out of 16 chances, there are 4 chances of plants with dihybrid genotype. 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]

Sol. Organic acids as its respiratory quotient is more than one. RQ = 2

2

CO eliminatedO consumed

.

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CAREER POINT

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. Due to high latent heat of vapourization, panting acts as a method of Thermoregulation. 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 (d) 101 Ans. [a] Sol. Minimum four digit number = 1000 & maximum four digit number = 9999. Now,

1000 –909

101 9

91 First number (a1) = 1000 – 91 + 99 = 1008 &

9999 –909

101 99

909 –909

× Last number (an) = 9999 – 2 = 9997 Here d = 101 an = a1 + (n – 1) d 9997 = 1008 + (n – 1) (101) 8989 = (n – 1) 101 n – 1 = 89 n = 90

62. If x = ( 21 – 20 ) and y = ( 18 – 17 ), then - (a) x = y (b) x < y (c) x > y (d) x + y = 0 Ans. [b]

Sol. x = 21 – 20

1x

= 121 20

1x

= 21 2021 20

1x

= 21 + 20

y = 18 – 17

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CAREER POINT

1y

= 118 17

1y

= 18 1718 17

1y

= 18 + 17

Here, 21 + 20 > 18 + 17

1x

> 1y

x < y 63. What is the sum of all odd numbers between 500 and 600 ? (a) 26000 (b) 27000 (c) 27500 (d) 29500 Ans. [c] Sol. Here a1 = 501, an = 599, d = 2, n = 50

Sn = n2

[2a + (n – 1)d]

or

= n2

[a1 + an]

= 502

[501 + 599]

= 25 × 1100 = 27,500

64. 1 12

+ 1 16

+ 1 112

+ 1 120

+ 1 130

+ ….. + 1 1380

= _________ .

(a) 20.25 (b) 20.05 (c) 19.95 (d) 19.85 Ans. [c]

Sol. 1 12

+ 1 16

+ 1 112

+ 1 120

+ … + 1 1380

= 1 + 12

+ 1 + 16

+ 1 + 112

+ 1 + 120

+ … + 1 + 1380

= 1 + 11 2

+ 1 + 12 3

+ 1 + 13 4

+ 1 + 14 5

+ … + 1 + 119 20

= 1 × 19 + 1 1 1 1 1...1 2 2 3 3 4 4 5 19 20

= 19 + 1 1 1 1 1 1 1 1 11 ....2 2 3 3 4 4 5 19 20

= 19 + 1120

= 19 + 1920

= 19 + 0.95 = 19.95

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CAREER POINT

65. A train is running at a speed of 54 km/hr. It is not stopping at a certain station. It crosses the person showing

green flag in 20 seconds and crosses the platform in 36 seconds. What is the length of the train ? (a) 240 m (b) 300 m (c) 320 m (d) 360 m Ans. [b] Sol. Speed = 54 km/hr

= 54 × 518

m/sec

= 15 m/sec Now, Case-I : When train passes through platform

(Platform Train)

Total distance

= 15 × 36 m = 540 m

Case-II : When train passes through person

(Length of train)Distance = 15 × 20 = 300 m

& length of platform = 540 – 300 = 240 Now, length of train = 300 m 66. In triangle ABC, segment AD, segment BE and segment CF are altitudes. If AB X AC = 172.8 cm2 and

BE X CF = 108.3 cm2 then AD X BC = _______. (a) 136.8 cm2 (b) 132.4 cm2 (c) 129.2 cm2 (d) 128.6 cm2 Ans. [a] Sol.

B D C

E F

A

AB × AC = 172.8 cm2 BE × CF = 108.3 cm2 AD × BC = ?

Area of ABC = 12

× AB × CF …(1)

Also, Area of ABC = 12

× AC × BE …(2)

& Area of ABC = 12

× BC × AD …(3)

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CAREER POINT

Now Eqn (1) × Eqn (2)

(Area of ABC)2 = 12

× 12

× AB × CF × AC × BE

Area of ABC = 1 1 AB AC BE CF2 2

Area of ABC = 12

172.8 108.3

Area of ABC = 12

18714.24

Area of ABC = 12

× 136.8 cm2 …(4)

From (3) & (4) , we get

12

× BC × AD = 12

× 136.8

BC × AD = 136.8 cm2

67. Diagonals of a quadrilateral bisect each other. Therefore the quadrilateral must be a _____________ .

(a) parallelogram (b) rhombus (c) rectangle (d) square

Ans. [a]

Sol. Diagonals of a parallelogram bisects each other. Also, rhombus, rectangle & square are parallelogram.

68. If (a + b + c + d) = 4, then

1(1 a)(1 b)(1 c)

+ 1(1 b)(1 c)(1 d)

+ 1(1 c)(1 d)(1 a)

+ 1(1 d)(1 a)(1 b)

= __________ .

(a) 0 (b) 0.25 (c) 1 (d) 4

Ans. [a]

Sol. a + b + c + d = 4

Now,

1(1 a)(1 b)(1 c)

+ 1(1 b)(1 c)(1 d)

+ 1(1 c)(1 d)(1 a)

+ 1(1 d)(1 a)(1 b)

= (1 d) (1 a) (1 b) (1 c)(1 a)(1 b)(1 c)(1 d)

= 4 (a b c d)(1 a)(1 b)(1 c)(1 d)

= 4 4(1 a)(1 b)(1 c)(1 d)

= 0

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CAREER POINT

69. The sum of two numbers is 13 and the sum of their cubes is 1066. Find the product of those two numbers :

(a) 26 (b) 27 (c) 28 (d) 29

Ans. [d]

Sol. Let the number be a & b

Here, a + b = 13 …(1)

(a)3 + (b)3 = 1066 …(2)

Use (1) in (2),

a3 + (13 – a)3 = 1066

a3 + 2197 – a3 – 39a (13 – a) = 1066

39a (13 – a) = 2197 – 1066

39ab = 1131

ab = 29

Product of the number = 29

70. By which smallest number we should divide 198396198 to get a perfect square ?

(a) 14 (b) 18 (c) 22 (d) 28

Ans. [c]

Sol. 198396198 = 2 × 3 3 × 11 11 × 11 × 7 7 × 13 13

The number should be divided by 2 × 11 i.e. 22 to make it a perfect square.

71. What will be the remainder if the number (7)2017 is divided by 25 ?

(a) 1 (b) 7 (c) 18 (d) 24

Ans. [b]

Sol. Remainder =

2572017

= 257.7 2016

= 25

)49.(7 1008

= 25

)1–50.(7 1008

Remainder = 7

72. If ABCD is a cyclic quadrilateral, AB = 204, BC = 104, CD = 195, DA = 85 and BD = 221, then AC = _____________ .

(a) 210 (b) 220 (c) 225 (d) 240 Ans. [b]

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CAREER POINT

Sol. AB = 204 BC = 104 CD = 195 DA = 85 BD = 221

D C

AB

By ptoleny's theorem AC × BD = AB × CD + BC × AD AC × 221 = (204 × 195) + (104 × 85)

AC = 221

884039780

AC = 220 73. If x2 + xy + xz = 135, y2 + yz + xy = 351 and z2 + xz + yz = 243, then x2 + y2 + z2 = ____________ . (a) 225 (b) 250 (c) 275 (d) 300 Ans. [c] Sol. x2 + xy + xz = 135 ....(1) y2 + yz + xy = 351 ....(2) z2 + xz + yz = 243 ....(3) on adding x2 + y2 + z2 + 2 (xy + yz + xz) = 729 (x + y + z)2 = 729 x + y + z = 27 from (1) x (x + y + z) = 135

x = zyx

135

= 27

135 = 5 x = 5

from (2) y (y + z + x) = 351

y = xzy

351

= 27351 = 13 y = 13

from (3) z (z + x + y) = 243

z = yxz

243

= 27243 = 9 z = 9

x2y2 + z2 = 25 + 169 + 81 = 275

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CAREER POINT

74. What is the radius of the circumcircle of a triangle whose sides are 30 cm, 36 cm and 30 cm.

(a) 15 cm (b) 16 cm (c) 17 cm (d) 18 cm

Ans. [Bonus]

Sol. a = 30

b = 30

c = 36

Radius of circumcircle = 4

abc

30 30

36

r

r r

Now = )c–s()b–s()a–s(s & s = 2

363030 = 48

= )12()18()18(48 = 432

radius of circumcircle = 4324

363030

= 18.75

75. On seventy first 'Independence day' there was Tuesday. After how many years there will be Tuesday on

'Independence day' ?

(a) 4 yrs. (b) 5 yrs. (c) 6 yrs. (d) 7 yrs.

Ans. [c]

Sol. 71st Independence day 2017

day Tuesday

or 15-Aug-2017 Tuesday

for same calendar we have to work on odd number of days.

Year 2017 2018 2019 2020 2021 2022 Total

Odd day 1 1 1 2 1 1 7

In 2023, the independence day will occur on Tuesday i.e. after 6 year.

76. If p + q + r = 2, p2 + q2 + r2 = 30 and pqr = 10, the value of (1 – p) (1 – q) (1 – r) will be - (a) –18 (b) –24 (c) –27 (d) –35 Ans. [b]

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CAREER POINT

Sol. p + q + r = 2 .... (1) p2 + q2 + r2 = 30 .... (2) pqr = 10 .... (3)

(1 – p) (1 – q) (1 – r) = ?

(1 – r) (1 – q – p + pq)

1 – q – p + pq – r + rq + rp – pqr

1 – (p + q + r) + pq + qr + pr – pqr

1 – 2 + pq + qr + pr – 10 .... (4)

from (1)

p + q + r = 2

Squaring

p2 + q2 + r2 + 2 (pq + qr + pr) = 4

30 + 2 (pq + qr + pr) = 4

pq + qr + pr = –13

From (4)

(1 – p) (1 – q) (1 – r) = 1 – 2 – 13 – 10 = – 24

77. The mean of the following frequency distribution is _____________ .

Class interval 0-10 10-20 20-30 30-40 40-50

Frequency 4 6 8 10 12

(a) 25 (b) 28 (c) 30 (d) 32 Ans. [c] Sol.

CI Mid point (xi) fi xifi 0-10 5 4 20

10-20 15 6 90 20-30 25 8 200

30-40 35 10 350 40-50 45 12 540

if 40 i ix f 1200

i i

i

x f 1200x 3040f

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CAREER POINT

78. If the roots of the equation 2x bx

ax c

= m 1m 1

are equal and of opposite signs, then the value of 'm'

is _____________ .

(a) a ba b

(b) a ba b

(c) aba b

(d) a bab

Ans. [a]

Sol. 1m1m

caxbxx2

(m + 1) x2 – b (m + 1) x = a(m – 1) x + c

(m + 1) x2 + (–bm – b – am + a) x – c = 0

Acc. to question sum of roots = 0

01m

aambbm

–m (a + b) – b + a = 0

babam

79. If 1xx

= 5, then 33

1xx

– 5 22

1xx

+ 1xx

= _______________ .

(a) 0 (b) 5 (c) –5 (d) 10

Ans. [a]

Sol. 5x1x

31x 125

x

33

1 1 1x 3.(x). x 125x xx

11015125x1x 3

3

23225x1x 2

2

x1x

x1x5

x1x 2

23

3

= 110 – 5 × 23 + 5

= 110 – 115 + 5

= 0

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CAREER POINT

80. If x2 – 3x + 2 is a factor of x4 – px2 + q, then p, q are : (a) 2, 3 (b) 4, 5 (c) 5, 4 (d) 0, 0 Ans. [c] Sol. P(x) = x2 – 3x + 2 = x2 – 2x – x + 2 = x(x – 2) – 1 (x – 2) P(x) = (x – 1) (x – 2)

(x – 1) & (x – 2) are also the factors of f(x) = x4 – px2 + qAccording to Question f(1) = 0

1 – p + q = 0

p – q = 1 .......(1) f(2) = 0

16 – 4p + q = 0

4p – q = 16 .......(2) Eqn (1) – (2) –3p = –15

p = 5

5 – q = 1

q = 4 p, q are 5, 4

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INDIAN ASSOCIATION OF PHYSICS TEACHERS - … that she prepared (i) HCl solution by dissolving 73 g of hydrochloric acid in one litre of water and (ii) sodium hydroxide solution by