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ASTRONOMY NSEA 2016-2017 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 ASTRONOMY 2016-2017

Date of Examination 27th November 2016 Time 120 min.

[Ques. Paper Code : A422]

1. If 31xcos2 then cosec x =

(a) 3 (b) 3

2 (c) 32 (d)

23

Ans. [d]

Sol. 31xcos2

32xsin 2

32xsin

23ecxcos

2. Which of digits 1, 3, 4, 5, 7, 0 cannot appear at the ten's place of powers of 3 ? (a) 1, 3, 5 only (b) 1, 3, 4, 7 only (c) 1, 3, 5, 7 only (d) 0, 1, 3, 5 only Ans. [c] Sol. Clearly 1, 3, 5, 7 cannot appear at the ten's place of powers of 3. 3. A fish is swimming in still water. At a given instant there is a bird flying vertically above the fish. For the

bird the fish appears to be 15 m below the surface of water while for the fish the bird appears to be 20 m above the surface. Refractive index of water is 4/3. Actual distance between the bird and the fish is -

(a) 35 m (b) 40 m (c) 30 m (d) 25 m Ans. [a] Sol.

fish

y

x

bird

m15y

x = 20

y = 15

20x

3415y

3/420x

y = 20m x = 15m

Actual distance = x + y = 15 +20 = 35 m

Marks : 240

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CAREER POINT

4. The smallest integer n such that 01.0n1n is -

(a) 2499 (b) 2500 (c) 2501 (d) 2502

Ans. [b]

Sol. 100

1n1n

n100

11n

100

n2n10000

11n

n21009999

299.99n

n 2499.5

n = 2500

smallest

5. A uniform wire of resistance per unit length 1/m is bent in the form of an equilateral triangle. If effective

resistance between adjacent vertices is 2.4, length of each side of the triangle is-

(a) 1.8 m (b) 2.4 m (c) 3.6 m (d) 7.2 m

Ans. [c]

Sol.

R =

R =

R =

A B

3R2R AB

3

24.2

3

)1(24.2

= 3.6 m

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CAREER POINT

6. ABC is equilateral with each side being of unit length and P is an interior point, then the maximum product of the lengths AP, BP and CP is -

(a) 35

1 (b) 34

1 (c) 33

1 (d) 61

Ans. [c]

Sol. 31

)CPBPAP(3

CPBPAP

So max value of AP BP CP is 3

3CPBPAP

It happens when 3

160sin2

1RCPBPAP

7. The resultant of the forces P and Q is R. If Q is doubled then R gets doubled. If Q is reversed, even then R gets doubled. Then -

(a) 2:3:2:R:Q:P (b) 3:2:2:R:Q:P

(c) 2:3:3:R:Q:P (d) 3:3:2:R:Q:P

Ans. [a]

Sol. cosPQ2QPR 22 ..... (i)

cosPQ4Q4PR2 22

cosPQ2QPR2 22 ..... (ii)

For (1) & (2)

5R2 = 2P2 + 2Q2

P : Q : R

2:3:2

8. The unit's digit of 232015 × 72016 × 132017 is - (a) 1 (b) 3 (c) 7 (d) 9 Ans. [a] Sol. unit place of (23)2015 = 7

unit place of (7)2016 = 1

unit place of (13)2017 = 3

So unit place of product = 1

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CAREER POINT

9. A block of mass 5 kg is to be dragged along a rough horizontal surface having s = 0.5 and k = 0.3. The horizontal force applied for dragging it is 20 N. (g = 10 m/s2). Select correct statement/s.

(a) Frictional force acting on the block is 20 N (b) Block will be displaced (c) Block will move with acceleration 1m/s2 (d) Block will initially move and then stop. Ans. [a] Sol.

5kg

F= 20N

s = 0.5 k = 0.3 Limiting Fmax = s mg

Fmax = 0.5 × 5 × 10

Fmax = 25 N

a = 0 (F < Fmax )

Fr = F

Frictional Force Fr = 20 N

10. The negation of statement: ' f(x) is continuous for all real numbers x. ' is (a) f(x) is not continuous for all real numbers x (b) f(x) is not continuous for any real number x (c) f(x) is not continuous for every real number x (d) f(x) is not continuous for some real number x Ans. [b] Sol. Negation of the statement is

f(x) is not continuous for any real number x

11. A bullet moving with a speed of 72 m/s comes to a half in a fixed wooden block on travelling 9 cm inside it. If the wooden block (of the same type of wood) were to be 8 cm thick, the bullet would come out of the block with a speed.

(a) 9 ms–1 (b) 8 ms–1 (c) 24 ms–1 (d) 64 ms–1 Ans. [c] Sol. Assuming constant resistive force

as2uv 22 0 = (72)2 – 2 (a) (x1) (x1 = 9 cm)

1

2

x2)72(a

v2 = u2 +2as v2 = (72)2 – 2a (x2)

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CAREER POINT

v2 = (72)2 – )x(x)72(

21

2

v2 = (72)2

1

2

xx1

v2 = (72)2

981

v2 = (72)2

91

243

72v m/s

12. Let l be a vertical line and m a line that makes an angle of 6 with l. Consider the cone generated by rotating

m around the axis l. If plane L makes an angle of 15° with line l then the intersection of the plane and the

cone is –

(a) A parabola (b) A pair of straight lines

(c) An ellipse (d) A hyperbola

Ans. [d]

Sol.

30° m

L

Hyperbola

15°

30° m

L

Pair of straight line

15°

13. A piece of brass (an alloy of copper and zinc) weight 12.9 g in air. When completely immersed in water,

it weighs 11.3 g. What is the mass of copper contained in the alloy ? The density of copper and zinc are

8.9 g/cm3 and 7.1 g/cm3 respectively.

(a) 6.89 g (b) 4.54 g (c) 8.93 g (d) 7.61 g

Ans. [d]

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CAREER POINT

Sol. Let the mass of copper contained in alloy = x,

c = density of copper, zn = density of zinc, = density of water = 1 gm/cm3

(mg – vg) = m'g

(m– v) = m'

v = m – m'

v = (12.9 – 11.3)

v = 1.6 gm

6.1)x9.12(x

znc

6.1)x9.12(xznc

6.11.7

)x9.12(9.8

x

1.79.126.1

1.71

9.81x

1.7

54.11.79.89.81.7x

8.1

54.19.8x

x = 7.61 gm

14. The coefficient of x in the expansion of (1 + x)5 correspond to the - (a) 5th row of Pascal's triangle (b) 6th row of Pascal's triangle (c) 7th row of Pascal's triangle (d) 4th row of Pascal's triangle Ans. [c] Sol. Coefficient is 5C1 which corresponds to 6th row of Pascal's triangle.

15. Linked question (15-16) The adjacent figure shows a ramp (30°) holding a block of mass 50 kg. The block attached to a movable

pulley A with an inextensible massless string. The movable pulley is in turn held with the help of another fixed pulley B. The block kept on the ramp is to be raised through a vertical height of 10 cm. By what distance the string should be lowered down vertically, below E ?

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CAREER POINT

30°

Block of mass 50 kg

B

E A

(a) 20 cm (b) 5 cm (c) 40 cm (d) 10(31/2) cm Ans. [c] Sol.

=30°

50 kg

B

F A

F F E

T

m = x

10cm mgsin30° =30°

T

x sin 30 = 10 cm, x = 20 cm

xE = 2x

= 40 cm

16. Refer to figure in question 15, pulleys in the figure are massless and frictionless. Neglecting friction between

block and the ramp, the force that should be applied vertically downwards, at E, to side the block along the ramp without acceleration is (g = 10 ms–1)

(a) 65 N (b) 125 N (c) 175 N (d) 250 N Ans. [b]

Sol. mg sin 30° = T 2

mgT

T = 2F 2TF

4

10504

mgF

F = 125 N

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CAREER POINT

17. The figure shows a particular position of a Vernier callipers. The value of x in cm is -

3

Main scale (cm) x 4 5

0 10

Vernier Scale (a) 0.03 (b) 0.15 (c) 3.83 (d) 0.02 Ans. [a] Sol. Position of 2nd VSD = Reading + 2 × VSD

= (3.6 + 0.01 × 5) + 2 × 0.09

= 3.83

x = 3.83 – 3.8 cm

x = 0.03 cm

18. If A and B are two sets then the set A × B is given by - (a) {a × b | a A, b B} (b) {(a, b) | a B, b A} (c) {(a, b) | a A, b B} (d) {ab | a A, b B} Ans. [c] Sol. A × B = {(a, b) | a A, bB}

19. A lens is held directly above a pencil lying on a floor and forms an image of it. On moving the lens vertically through a distance equal to its focal length, it again forms image of same size as that of the previous image. If the length of the pencil is 5.0 cm, the length of the image is

(a) 10.0 cm (b) 15.0 cm (c) 20.0 cm (d) 12.5 cm Ans. [a] Sol.

x

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CAREER POINT

f1

u1

v1

fu

fuv

Ist virtual image

uvm

xf

fafu

fm

afx

IInd image real m = –a

)fx(f

fa

xfa

1a

aa

a11

1a

aff

fa

a – 1 = 1 a = 2

52H|a|HH

io

i Hi = 10 cm

20. If A = {2, 3}, B = {4, 5} then A × B = (a) {8, 15} (b) {8, 10, 12, 15} (c) {(2, 4), (3, 5)} (d) {(2, 4), (2, 5), (3, 4), (3, 5)} Ans. [d] Sol. A × B = {(2, 4), (2, 5), (3, 4), (3, 5)}

21. A balloon less dense than air is tied at the floor of a truck with a massless, inextensible and flexible string. The truck is observed to be taking a left turn. Select correct statement.

(a) The string will incline towards right (outward, w.r.t. person in the truck) (b) The string will incline towards left (inward, w.r.t. person in the truck) (c) The string will still be vertical as the balloon is less dense than air (d) Buoyant force on the balloon is equal to weight of the balloon Ans. [a] Sol.

r

B

mg Centrifugal

wrt truck

Wrt person in truck the string will incline towards right-outward.

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CAREER POINT

22. A function F from A to B is - (a) Relation F with (a, b), (c, b) F a = c (b) F A × B

(c) Relation F with (a, b), (a, c) F b = c (d) Relation F with (a, b), (b, c) F (a, c) F Ans. [c] Sol. As function is a relation from A to B

such that every element should have a unique image

so if (a, b), (a, c) F b = c

23. An ice cube with a steel ball bearing trapped inside it is floating above water in a glass. What will happen to the water level in the glass after the ice melts completely ?

(a) Rise (b) Fall (c) Will not change (d) Answer depends upon actual position of the steel ball Ans. [b] Sol.

v1

v2

When ice melts

ice v1 = vwater

vwater =

1icev

vsub g = (m1g + m2g)

vsub = )vv( 2ball1ice

2ball

1ice

sub vvv

Net sub0 vvv1

2ball

1ice

0 vvvv1

2water0 vvvv2

= 21ice

0 vvv

12

vv (level fall)

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CAREER POINT

24. Which of the following is a mathematically acceptable statement ? (a) It is an even number (b) 13th December is Saturday (c) Common donkey belongs to class orthopoda (d) Alexander was a great king Ans. [c] Sol. Common donkey belongs to class orthopoda.

25. Linked questions (25-29) A star can be considered as a spherical ball of hot gas of radius R. Inside the star, the density of the gas is r

at a radius r and mass of the gas within this region is Mr. The correct differential equation for variation of mass with respect to radius is (refer to the adjacent figure) –

Mr

r+dr

r

dMr

(a) 3r

r r34

drdM

(b) 2r

r r4dr

dM (c) 2

rr r

32

drdM

(d) 3r

r r31

drdM

Ans. [b] Sol.

dr

r

Volume of element = drr4 2

Mass of element = drr4dm r2

r2r4

drdm

26. A star in its prime age is said to be under equilibrium due to gravitational pull and outward radiation pressure (P). Consider the shell of thickness dr as in the figure of question (25). If the pressure on this shell is dP then the correct equation is (G is universal gravitational constant)

(a) r2r

rGM

drdP

(b) r2r

rGM

drdP

(c) r2r

rGM

32

drdP

(d) r2r

rGM

32

drdP

Ans. [a] Sol.

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CAREER POINT

dr

r

P P+dP

Excess force due to pressure = dP × 4r2

force due to gravitation = 4rr2 dr × 2r

rGM

2r

2r2

rdrGMr4r4dP

r2r

rGM

drdP

as r pressure decrease

r2r

rGM

drdP

27. In astronomy, order of magnitude estimation plays an important role. The derivative drdP can be taken as

difference ratio rP . Consider the star has a radius R, pressure at its centre is Pc and pressure at outer most

layer is zero. If the average mass is 2

M0 and average radius 2

R 0 then the expression for Pc is-

(a) 40

20

c RGM

23P

(b) 40

20

c RGM

43P

(c) 40

20

c RGM

32P

(d) 40

20

c RGM

23P

Ans. [a] Sol.

r

dr

mass of element (dm)

dF = (dm)E

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CAREER POINT

drr4R

34

MR

GMrdF 2

33

rdrR4

GM3r4

dF R

06

2

2

2

RGMR43P

22

6c

4

2

RGM

83P

40

20

c

2R4

GM83P

2R

R,2

MMputting 00

40

20

c RGM

23P

28. The value of mass and radius of sun are given by M0 = 2 × 1030 kg and R0 = 7 × 105 km respectively. The pressure at the centre is about (G = 6.67 × 10–11 m3 kg–1 s–2)

(a) 2 × 1014 N m–2 (b) 2 × 1015 N m–2 (c) 5 × 1014 N m–2 (d) 7 × 1015 N m–2 Ans. [c]

Sol. 40

20

c RGM

23P

Put M0 = 2 × 1030 kg, R0 = 7 × 105 km

Pc = 5 × 1014 N m–2

29. Assuming that the gas inside the sun behaves very much like the perfect gas, the temperature at the centre of

the sun is nearly (the number density of gas particles = HM

2 , Boltzmann constant kB = 1.4 × 10–23 JK–1 and

mass of proton MH = 1.67 × 10–27 Kg) (a) 3 × 107 K (b) 2 × 107 K (c) 4 × 107 K (d) 6 × 107 K Ans. [b]

Sol. KTVNP

30

0

R34M

P

HMP2

VN

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CAREER POINT

KTMR

34

2MP

H30

0

0

H30

KM2PM

3R4T

0

H30

KMPM

3R2T

T = 3023

142738

102104.11051067.1)107(14.3

32

T = 2×107 kelvin

30. At the earth's equator a satellite is observed passing directly overhead moving west to east in the sky. Exactly 12 hours later, satellite is again observed directly overhead. The altitude of the satellite is (Radius of earth = 6400 km)

(a) 1.82 × 107 m (b) 1.39 × 107 m (c) 3.59 × 107 m (d) 6.4 × 107 m Ans. [b] Sol. (WS – WE)t = 2

212

T1

T12

ES

121

241

Ts1

TS = 8 hr

Time period = TS = GM

r2 23

= 2

23

gR

r2

2

23

SgR

)hR(2T

Putting all value

h = 1.39 × 107 meter

31. Passage 31 to 33 Two stars, with masses M1 and M2 are in circular orbits around their common centre of mass. The star with

mass M1 has an orbit of radius R1 and the star with mass M2 has an orbit of radius R2. The correct relation is -

(a) 1

2

2

1

MM

RR

(b) 2

1

2

1

MM

RR

(c) 1

2

2

1

MM

RR

(d) 2

1

2

1

MM

RR

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CAREER POINT

Ans. [a] Sol.

M1 M2 CM

R1 R2

Summation of mass moment about centre of mass is equal to zero

2211 RMRM

1

2

2

1

RR

MM

or 1

2

2

1

MM

RR

32. The time period of each of the star is -

(a) 2

22

212

2

GMR)RR(4T

(b) )MM(G)RR(4T

21

321

22

(c) 1

22

212

2

GMR)RR(4T

(d) 1

12

212

2

GMR)RR(4T

Ans. [c] Sol.

M1 M2 R2R1

CM

1

211

221

21

RVM

)RR(MGM

221

211 )RR(

MGRV

1

1

VR2T

221

21

21

2

21

21

22 )RR(

MGRR4

VR4T

221

2

12

2 )RR(GM

R4T

as M1R1 = M2R2

1

2

2

1

MR

MR

1

2212

22

GM)RR(R4T

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CAREER POINT

33. The two stars in certain binary system move in circular orbits. The first star alpha has an orbital speed of 36.0 km.s–1. The second star, beta has an orbital speed of 12.0 km.s–1. The orbital period of first star is 137 days. The mass of the two stars are about

(a) 2.1 × 1030 and 6.8 × 1030 kg (b) 1.3 × 1030 and 3.9 × 1030 kg (c) 3.5 × 1030 and 9.2 × 1030 kg (d) 0.8 × 1030 and 2.3 × 1030 kg Ans. [d]

Sol. 3600241372

360024137v

r2

1

1

3600241371036r2

31

1

22

212

2

GMr)rr(4

T

2

2

1

1

rv

rv

21 r

12r36

r1 = 3r2

3r

3rr

GM4T 1

21

11

22

1627.GMr4T

1

31

22

1

1

vr2T

21

21

22

vr4T

1627.GM

r4v

r4

1

31

2

21

21

2

16GM27r

v1

1

121

kg108.0G2716vrM 30

211

1

M2 = 0.8 × 1030× 3 2.3 ×1030 kg Passage question (34-36) 34. Consider a spacecraft in an elliptical orbit around the earth. At the low point or perigee of its orbit, it is

400 km above the earth's surface. At the high point or apogee it is 4000 km above the earth's surface. The period of the space craft's orbit is (g = 9.8 ms–2 and R = 6400 km)

(a) 0.29 hr (b) 1.82 hrs (c) 2.21 hrs (d) 3.56 hrs Ans. [c]

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CAREER POINT

Sol. GM

a4T32

2

2

322

gRa4T

Rg

a2T2/3

4000 6400 6400 400

2a

V2

V1

Perigee

Apogee

Put a =

2440026400

=2

440012800

= 8600 km T = 2.21 hrs 35. The ratio of speed of the spacecraft at perigee to its speed at apogee is almost equal to (a) 10 : 1 (b) 3 : 2 (c) 2 : 3 (d) 1 : 10 Ans. [b] Sol. mv1r1 = mv2r2

6800

400,10rr

vv

1

2

2

1 = 1.5

= 10

5.1

= 23

36. The speed of the satellite at perigee is (a) 8576 m.s–1 (b) 57,307m.s–1 (c) 5876 m.s–1 (d) 7856 m.s–1 Ans. [c]

Sol. TE of satiable = 321 106800

GMmmv21

a2GMm

2

v1086002

GM106800

GM 21

33

= 21

523 v]286

1681[10g)106400(2

21

52 v172

16818.910642

v1 = 5876 ms–1

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CAREER POINT

37. Astronomers believe that a large percentage of the mass of the universe is dark matter. In one recent study, the transverse velocity of the large Magellanic cloud(LMC) was measured to be 200 km.s–1. The LMC is believed to orbit the centre of our galaxy at about 17 × 104 ly (1.6 × 1021m). Assuming a circular orbit, percentage of dark matter in our galaxy is about (independent estimate of visible matter is 2 × 1041 kg)

(a) 77 % (b) 82 % (c) 70 % (d) 80 % Ans. [b] Sol. 80% 38. The escape speed from jupiter is approximately 59.5 km.s–1 and its radius is about 12 times that of earth.

From this we may estimate the mean density of jupiter to be about (Radius of earth = 6400 km and escape speed from the earth is 11.2 km.s–1)

(a) 5 times that of earth (b) 0.2 times that of the earth (c) 2.5 times that of the earth (d) 0.4 times that of the earth Ans. [b]

Sol. vesc = RGM2 where M = 3R

34

vesc = 3

R4G2 2 vesc 2R

e

J

e

J

eesc

Jesc

RR

vv

122.115.59

e

J

2

2

e

J

)12(1

2.115.59

= 0.2

J = 0.2 time e 39. The orbit of planet mercury has the largest eccentricity of about 0.2 in the solar system. If the maximum

distance of mercury from the centre of the sun is about 69 million km, its minimum distance from sun is about

(a) 13.8 million km (b) 57.5 million km (c) 46 million km (d) 18 million km Ans. [c] Sol.

ae

Max. distance = a + ae

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CAREER POINT

Min. distance = a – ae a + ae = 69 mk a = 57.5 mk min. distance = a – ae = 46 Million km 40. As observed from a place in Australia the pole star (a) appears in the southern direction (b) appears at about 30º above the horizon (c) much brighter than that seen from India (d) can never be seen Ans. [d] Sol. pole star can be seen only from northern hemisphere as at is present above north pole of earth. Australia is in

southern hemisphere no pole star can be seen from Australia. 41. Evaporation of (sweat ) water is an essential mechanism in human beings for maintaining normal body

temperature. For human beings, heat of vaporization of water at a body temperature of 37ºC is nearly 2.3 MJ/kg and specific heat capacity is 3.5 kJ /(kg.K).On consuming a certain prescribed diet, the body temperature of an athlete of mass 82 kg is expected to increase by 2ºC. In order to prevent this, he drinks N bottles of mineral water (250 ml water in each ) at 37ºC. Assume that the entire amount of this water is given out as sweat, which vaporizes. N is nearly (density of water = 1000 kg.m–3)

(a) 1 (b) 2 (c) 3 (d) 4 Ans. [a] Sol. Q = msT = mL

82 × 3.5 ×103 ×2 = n

41 × 2.3 ×106

n = 2300

428.382

m = 1 42. The number n = 1 + 12 + 60 + 160 +240 + 192 + 64 is (a) A perfect square and a perfect cube (b) Neither a perfect square nor a perfect cube (c) A perfect cube but not a perfect square (d) A perfect square but not a perfect cube Ans. [a] Sol. n = 1 + 12 + 60 + 160 + 240 + 192 +64 = 729 = 272 = 93 43. A football (assumed to be a hollow sphere) of mass 200g and radius 16cm, is given horizontal kick 4 cm

above its centre. This imparts a speed of 8 cm/s to the football. Angular speed acquired by the fooball in radian/s is

(a) 9/16 (b) 15/16 (c) 3/4 (d) 16/15 Ans. [Bonus] Sol.

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CAREER POINT

4 cm

J

J = mv J × 4 = I mv × 4 = I

= Imv4

22 Rv6

mR)3/2(mv4

s/rad163

10)16(10108642

22

(Bonus)

44. For sets A,B we have (here XC denote complement of set X) (A × B)C = (a) AC × BC (b) BC × AC

(c) AC × B BC × A AC × BC (d) AC × B × BCAC × BC Ans. [b] Sol. (A × B)C = BC ×AC 45. A small marble (assumed to be a uniform solid sphere ) is released on one end of a parabolic mirror from a

vertical height of 70 cm. First half part of this mirror is rough on which the marble is released. Other half of the mirror is smooth. Throughout its motion the marble never slips. To what vertical height will it rise on the smooth surface?

(a) 98 cm (b) 70 cm (c) 63 cm (d) 50 cm Ans. [d]

Sol. mgh = 2mv107

v2 = 70g7

10gh7

10

70cm Smooth

h Rough

mghmv

21 2 " only translation energy will convert"

mgh70g7

10m21 Rotational K.E remains same

h = 50 m

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CAREER POINT

46. The number 312 + 29 + 3(3 × 64 + 65) + 26 is (a) A perfect square and a perfect cube (b) A perfect cube but not a perfect square (c) A perfect square but not a perfect cube (d) Neither a perfect square nor a perfect cube Ans. [c] Sol. 312 + 29 + 3(3×64 + 65) + 26 = (753)2 47. Radius and moment of inertia of a smooth pulley are 0.1 m and 1 kg.m2 respectively. A tangential force of

f = 40t – 10t2 sets the pulley into rotation . Direction of its rotation reverses after some time. The time duration after which the direction will reverse is

(a) 6s (b) 8s (c) 4s (d) 12 s Ans. [a] Sol. = f.r = (0.1) = 4t –t2

dtdw = 4t – t2

w = c3t–

2t4 32

at t = 0 ; w = 0

w = 2t2 – 3t3

w = 0 at t = 0 ; t = 5s 48. log10 0.01 + log0.1 10 + log10 0.001 + log0.1 0.001 = (a) log10.2 10.012 (b) log10 0.000001+ 3 (c) – 4 + log2 8 (d) None of the above Ans. [b] Sol. 3

103

10102

10 10log10log10log10log 11

= –2 –1 –3 + 3 = –3 49. A car is fitted with a rear view mirror of focal length 20cm. Another car, 2.8 m behind the first car is 15 m.s–1.

faster than the first car and approaching. The relative speed of image of the second car , with respect to first car at this instant, is

(a) 1/15 m.s–1 (b) 1/10 m.s–1 (c) 1/5 m.s–1 (d) 2/15 m.s–1 Ans. [a]

Sol. dtdv

uv

dtdv

2

2

201

2801

f1

u1

v1

v = 56/3

152802803

563

56

dtdv

s/m151

dtdv

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CAREER POINT

50. 3log9log16log31272

(a) Is defined but cannot be found (b) Is not defined

(c) Is defined and equals –31 (d) Is defined and equals

323

Ans. [d]

Sol. 3231

3283log9log16log

31272

51. An electric buggy is stationed exactly midway between two plane vertical walls parallel to each other. A man

standing adjacent to buggy blows whistle momentarily. Instantly the buggy starts running towards one of the walls with a velocity 35 m/s. The driver of the buggy records first two echoes of the whistle with a delay of exactly one second. Speed of sound in air at that temperature is 350 m/s. Distance between the walls must be

(a) 433.125 m (b) 866.25 m (c) 1732.5 m (d) 3465 m Ans. [c] Sol.

x x

wall -1 wall -2 at t = 0, whistle is blown.

and buggy start moving towards wall-1

Let driver records 1st echo at instant t.

then distance travelled by buggy = 35 t

distance travelled by sound = x + x – 35 t

i.e. (350) t = 2x – 35 t

385 t = 2x ...... (i)

at instant t + 1, driver record 2nd echo.

then during 0 to (t+1), distance travelled by sound which is reflected by wall-2

= x + x + 35 (t +1)

350 (t + 1) = 2x + 35t + 35

350t + 350 = 2x + 35t + 35

315t + 315 = 2x ....... (ii)

from equation (i) & (ii)

2x = 1732.5

hence distance between walls = 1732.5m 52. The number of points at which |x3 –1| is not differentiable is- (a) 3 (b) 2 (c) 1 (d) 0 Ans. [c]

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CAREER POINT

Sol. y = |x3 –1| = |x–1| |x2 + x + 1| = |x –1| (x2 + x + 1) is not diff. at x = 1 53. The circuit given below has a long straight wire AB placed horizontal along North – South direction . A

small magnetic needle can be held anywhere near this wire. Choose the correct statement.

A B(South)(North)

SN

(a) North pole of the magnetic needle will deflect towards East, if the compass is just above the wire (b) North pole of the magnetic needle will deflect towards West, if the compass is at exactly same level of

the wire (c) North pole of the magnetic needle will deflect towards East, if the compass is just below the wire (d) Magnetic needle will not deflect, if kept just below the wire Ans. [c] Sol. (i)decal North – south magnetic field below the wire into the plane Thus north pole move inwards that is east Below the wire, magnetic field is directed towards East hence North pole deflect towards East when placed

below the wire.

54. If DA,CD,BC,AB are unit vectors such that 2

1BC.AB then

(a) Points A, B, C, D are concyclic

(b) Quadrilateral ABCD has area 22

1

(c) Quadrilateral ABCD has half of the maximal area for quadrilateral with same perimeter

(d) The area determined by the vectors is 2

1

Ans. [a]

Sol. All the sides are of equal in magnitude then it can be either square or rhombus

2

1BC.AB so º

so it is rhombus it cyclic 55. INSAT series of satellites are launched by India for telecommunication. Such satellites appear stationary at a

particular point in the sky when observed from the earth . Consider the following statements : (i). The satellite always experiences gravitation of the earth (ii). The satellite does not need any fuel for its motion (iii). The satellite does not experience net force (iv). Such satellites have to be positioned vertically above the equator

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CAREER POINT

(a) Only (ii), (iii) & (iv) are correct (b) Only (i), (ii) & (iv) are correct (c) Only (i) & (iii) are correct (d) Only (i), & (ii) are correct Ans. [b] Sol. A geostationary satellite appears stationary with respect to Earth. It revolve in equatorial orbit. Gravitational

force by Earth on satellite provides required centripetal force of circular motion of satellite 56. The number 38(310 + 65) + 23(212 + 67) is (a) A perfect square and a perfect cube (b) Neither a perfect square 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 +38.25.35.+23.212 + 23.27.37

318 + 313.25 + 215 + 210.37

(36)3 + (25)3 + 3.(36)2. (25) + 3. (36)(25)2

= (36 + 25)3 57. The optical effects (phenomena) involved when we see a rainbow could be associated with (i) internal reflection (ii) dispersion (iii) total internal reflection (iv) deviation select the correct option (a) (ii), (iii) and (iv) (b) (i), (ii) and (iv) (c) (i) and (iv) (d) (iii) and (iv) Ans. [a] Sol. Rainbow formation involve total internal reflection, dispersion & deviation 58. Which of the following statements are true about periodic functions defined on the set of real numbers A: Sum of two functions with finite period is always a periodic function with finite period B. The period of a function that is sum of two periodic functions with finite period is least common multiple

of the period of two functions (a) A and B are correct (b) A is correct but B is incorrect (c) A is false but B is correct (d) A and B are false Ans. [d] Sol. For statement A, if f(x) = sin x, g(x) = {x} then f(x) + g(x) = sin x +{x} which is not periodic so statement 'A' is false for statement 'B' if f(x) = |sin x|, g(x) = |cos x| which are periodic with period ''

but period of f(x) + g(x) = |sin x| + |cos x| is 2

so statement (B) is also false

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CAREER POINT

59. Unaware about the fact that analog ammeters and voltmeters can also have zero error, a student recorded following readings while determining resistance by using ohm's law.

Obs No. Voltage/V Current/mA Obs. No. Voltage/V Current/mA

1 1.0 40 4 7.0 160 2 3.0 80 5 9.0 200 3 5.0 120

If the ammeter has no zero error, the zero error in the voltmeter is (a) – 1V (b) –1.5 V (c) 0.5 V (d) 1 V Ans. [d] Sol. Let end correction in voltmeter = a v i according to ohm's law

4080

a1a3

3 + a = 2 + 2a a = 1 60. The inverse function of the function sinx + cos x is

(a) sin–1x + cos–1 x (b) xcosxsin

1

(c)

2xsin 1 (d)

42xsin 1

Ans. [d] Sol. y = sinx + cosx

)4/xsin(2

y

x = sin–1

2y –/4

f–1(x) = sin–1

2x – /4

61. Particle A collides elastically (perfect) with another particle B which was at rest. They disperse in opposite

directions with same speeds. Ratio of their masses respectively must be (a) 1 : 2 (b) 1 : 3 (c) 1 : 4 (d) 2 : 3 Ans. [b] Sol. Let masses of A and B are m1 & m2

u

m1 m2

rest

v m1 m2

v

in elastic collision, relative velocity of separation = relative velocity of approach 2v = u

2uv

and Linear Momentum = constant m2v – m1v = m1 u (m2 – m1)v = m1 u

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CAREER POINT

(m2 – m1) 2u = m1 u

31

mm

2

1

62.

4

x

xcosxsinlim4

x=

(a) ∞ (b) – ∞ (c) 2

1 (d) 2

Ans. [d]

Sol. )

4x(

xcosxsinlim4

x

)

4xsin(

)4

xsin(2lim

4x

= 212

63. A long rowing boat put upside down, shown in adjacent figure, has to be weighed using only a single bathroom scale. The boat will sag if it is supported only in the middle, and so the scales must be put first at position A with a wooden support at B, and then at position B with the wooden support at A. The readings on the scale are 45 kg and 55 kg respectively. The distance of centre of mass (from point A) and mass of the boat are respectively

5 m 8 m 6 m A B

(a) 4.4 m, 120 kg (b) 4.4 m, 100 kg (c) 4.2 m, 100 kg (d) 4.2 m, 120 kg Ans. [b] Sol. Ist case :

8 m

mg

R2R1 = 450 N

A

Force balance Torque balance

R1 + R2 = mg R2 8 = mg x …(2)

450 + R2 = mg

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CAREER POINT

R2 = mg – 450 …(1)

by (1) (2)

8

mgx = mg – 450 …(3)

IInd case :

8

mg

R R

B

8–x

Force

R+ R mg

R + 550 = mg …(4)

Torque balance

R 8 = mg (8 – x) …(5)

by (4) (5)

8

)x–8(mg = mg – 550

by mg – 8

mgx = mg – 550

8

mgx = 550 …(6)

by (3) (6)

550 = mg – 450

1000 = mg

m = 100 kg

By eq. (6)

8

1000 x = 55

x = 100

855 = 4.4 m

64. The number of real solutions of the equation (x – 1) (3x – 2) = 1 is (a) 0 (b) 1 (c) 2 (d) More than 2 Ans. [c]

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CAREER POINT

Sol.

x = 1

(0, 1)

y = 2

y

x 0

3x – 2 = 1–x

1 .

Number of Intersection point = 2

65. A train of mass m is moving on a circular track of radius r with constant speed v. Length of the train is exactly equal to half the circumference of the circular track. Magnitude of its linear momentum is

(a) mv/ (b) 0.5 mv (c) 2mv/ (d) mv Ans. [c]

Sol.

Vy

V

Vx

d

x

y

dPy = dmVy

dPy = M d V sin

Py =

MV

2/

2/–dsin =

MV 2/

2/–]cos[– = 0

dPx = dmVx

= M d V cos

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CAREER POINT

Px =

MV

2/

2/–dcos =

MV 2/

2/–][sin =

MV [1 + 1] =

MV2

66. The number of integers a, b, c for which 2a2 + b2 – 8c = 7 is (a) 2 (b) Infinite (c) 0 (d) 4 Ans. [c] Sol. 2a2 + b2 – 8c = 7 b = odd 42

2 + 42 + 1

2a2 + 42 – 8c = 6 a = odd 4(2) + (2+1)

a2 + 22 – 4c = 3 c = odd

41 + 1 + 22 – 4c = 3 2a2 + b2 – 8c = 7

41 + 21 – 4c = 2

21 + 2 – 2c = 2 c = odd

21 + 23 – 2c = 2 2(21 + 1)2 + (22 – 1)2 – 8 (23 + 1) = 7

1 + 3 – c = 1 2(421 + 41 + 1) + 42

2 + 42 + 1 – 163 = 15

1 + 3 = 1 + c 821 + 42

2 + 81 + 42 – 163 = 12

221 + 2

2 + 21 + 2 – 43 = 3

Since, 1, 3 are even 221 + 21 – 43 + 2 (2 + 1) = 3

So, 'c' is odd

If a = odd the

a2 = 4 × even + 1

= 4 + 1

67. In SI units we use length, mass and time as fundamental quantities. Another intelligent world may not know these. However, three constants G (universal gravitational constant), c (speed of light in vacuum) and h (Planck's constant) are really universal and can be related to almost all the known interactions. In terms of these fundamental constants, the dimensions of time are

(a)

2

125

21

hcG – (b)

2

1

hcG 2–1 (c)

2

121

hcG –2 (d)

2

123

21

hcG –

Ans. [a] Sol. t = [Gx cy hz]

[M0 L0 T] = [M–1 L3 T–2]x [LT–1]y [ML2 T–1]z M0 L0 T = M–x + z L3x + y+2z T–2x–y–z – x + z = 0 …(1) x = z 3x + y + 2z = 0 …(2) 3x + y + 2x = 0 5x = – y – 2x – y – z = 1 …(3) – 2x + 5x – x = 1 2x = 1 x = 1/2 z = 1/2 y = – 5/2 t = [G1/2 c–5/2 h1/2]

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CAREER POINT

68. If p (x) = x (x + 1) (x + 2) … (x + 2001) – c then the maximum multiplicity of the roots of p (x) can be (a) 1 (b) 2 (c) 3 (d) 2001 Ans. [b] Sol. Multiplicity of roots is how many times a root is repeated. Ex. In (x – 2)3 (x – 3)2 (x – 4)5 x. 2 has multiplicity 3 3 has multiplicity 2 4 has multiplicity 5 0 has multiplicity 1 So, P(x) = x (x + 1) (x + 2) …. (x + 2001) – c

y = cc

Maximum multiplicity is two.

69. A train is running on a circular track of radius R with a constant speed. The driver is blowing siren of a constant frequency (n0) throughout the circular motion of period T. There is a listener on the diameter of the track at a distance R/2 from centre of the circle. At t = 0, the train siren is farthest from the listener. In the following graphs the frequency, as recorded by the listener, is plotted against time. Which of them is closest to the correct pattern?

(a)

t

n

n0

(b)

t

n

n0

T/2

(c)

t

n

n0

T/3 T

(d)

t

n

n0

Ans. [b] Sol.

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CAREER POINT

cos

A t = 0

P

DO Bt = T/3

Ct = T/2

R/2

at a general point P apparent frequency is

n = cosV–V

V

s n0

From AB, is increases so frequency is continuously from A to B and from B to C, is decreases so apparent frequency is decreases, but for t < T/2 freq. is always large than actual frequency. So graph (b) is correct.

70. If {x} denotes the fractional part of a real number then dx}x{2

0

2 =

(a) 322 (b)

31 (c) 1 –

32 (d) 1 +

32

Ans. [c]

Sol. I = 2

0

2 dx}x{

I = 2

0

22 dx]x[–x

I = 2

0

2dxx – 2

0

2 dx]x[

I = 2

0

3

3x

–

1

0

2 dx]x[ = 2

1

2 dx]x[

I = 3

22 – 0 – 2

0dx

I = 3

22 – 1–2

I = 3

22 – 2 + 1

I = –32 + 1

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CAREER POINT

71. Adjacent figure shows a block A, held by a spring balance D and submerged into a liquid in a beaker. The beaker is kept on a weighing balance E. Mass of the beaker plus the liquid is 2.5 kg. Balance D reads 2.5 kg and E reads 7.5 kg. Volume of the block is 0.003 m3. Consider the following statements

Figure

(i) The density of the liquid is 5000/3 kg/m3 (ii) The mass of block A is 7.5 kg (iii) The buoyant force is 5 kg. wt. (iv) If half the volume of the block is pulled out of the liquid, E would read 5 kg. Select correct option(s). (a) (i), (iii) and (iv) (b) (i), (ii) and (iv) (c) (i) and (iv) (d) (i), (ii), (iii) and (iv) Ans. [d] Sol. Let m = mass of liquid + beaker = 2.5 kg

T = reading of spring balance = 2.5 kg = 25N

R = reading of balance E = Normal reaction = 7.5 kg = 75 N

m = mass of block

mg

T Fb

mg

R Fb

for balance of block for balance of beaker + liquid

T + Fb = mg R = Fb + mg

25 + Fb = mg …(1) 75 = Fb + 25

Fb = 50 N

So,

buoyant force is = 5 kg.wt

by eq. (1)

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CAREER POINT

25 + Fb = mg

25 + 50 = mg

m = 7.5 kg

So, mass of block = 7.5 kg

Fb = 50

Vg = 50

(0.003) (10) = 50

= 3

5000 kg/m3

So, density of liquid = 3

5000 kg/m3

when half of block is pulled Fb became = 25 N

and New F.B.D of liquid + beaker

25 N

R Fb = 25

R = 50 N

= 5 Kg

So, all (i), (ii), (iii) & (iv) are correct.

72. AM-HM inequality for positive real numbers a, b, c states that 3

cba cacbab

abc3

. If a, b are positive

irrational numbers then

(a) ba2

ab9

a + b (b) ba2

ab9

1 (c) b2a

ab9

2a + b (d) ba2

ab18

a + 2b

Ans. [c] Sol. 2a + b = a + a + b

3

baa 3 2ba

3

bba 3 2ab

Multiply both,

9

)b2a)(ba2( ab

2a + b b2a

ab9

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CAREER POINT

73. Two identical stars with mass M orbit around their centre of mass in circular orbit. If radius of the orbit is R and the stars are always diametrically opposite, then consider the following statements

(i) Their binding force is equal to 2

2

R4GM

(ii) If the stars are heavier and closer, their orbital speed is greater.

(iii) The period of the orbit is T = GMR 3

(iv) The minimum energy required to separate the two stars to infinity is equal to R4

GM2

Select correct statement/s (a) Only (i) and (ii) (b) Only (i), (ii) and (iv) (c) Only (i), (iii) and (iv) (d) Only (i) and (iii) Ans. [b]

Sol.

M F com

F

M

Binding Force = G 2)R2(M.M = 2

2

R4M.G

for circular motion of any one star,

2

2

R4M.G =

RMV 2

V = R4

GM

V RM

when M is large and R is small orbital speed is high,

Time period T = V

R2 =

R4GM

R2 = GMR42

3 =

GMR4

3

Minimum Energy required to separate the stars = – (BE)

= G R2MM

= R4

GM2

(ii) and (iv) are correct Ans. (b)

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CAREER POINT

74. The number of positive integers n 50 such that 3 3 3 ...nnn is a natural number is

(a) Zero (b) 2 (c) 50 (d) 5 Ans. [b] Sol. 3 = n + , N

3 – = n

( – 1) ( + 1) = n

n = ( – 1) () ( + 1)

Possible value of for n 50

= 2, 3

75. A monkey is holding onto one end of a rope which passes over a frictionless pulley and at the other end is a plane mirror which has a mass equal to the mass of the monkey. At equilibrium the monkey is able to see her image in the mirror. How does the monkey see her image in the mirror as she climbs up the rope?

(a) The image of the monkey moves with double speed of that of the monkey. (b) The image of the monkey moves with half the speed of that of the monkey. (c) The image of the monkey moves as fast as the monkey. (d) The monkey will not be able to see her image. Ans. [c]

Sol.

TT

M

a

mg mg

a

M

Both monkey and mirror are have same mass

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CAREER POINT

NLM eq. T – mg = ma So, both are always move with same acceleration 76. If i = 1– then i2i is a (a) Purely imaginary number (b) A natural number (c) A real number (d) A complex number with non-zero real and imaginary parts Ans. [c]

Sol. i2i = i2

i 2e

= e–

which is a real number

77. A steel ball is dropped from a height of 1 m on to a hard non-conducting surface. Every time it bounces it reaches 80% of its previous height. All the losses in the energy are accounted only for increasing temperature. Nearly how much is the rise in temperature of the ball just after the third bounce? (g = 10 m/s2, specific heat capacity of material of the ball = 500 J/(kgK))

(a) 0.005 ºC (b) 0.01 ºC (c) 0.015 ºC (d) 0.02 ºC Ans. [b] Sol. Let mass of ball = m

Its initial energy = mgh = (m) (10) (1) = 10m

20% loss = (10m) × 0.2 = 2m …(1)

remaining energy = 8 m

now 20% loss in IInd collision = 8 m × 0.2 = 1.6 m …(2)

remaining energy = 6.4 m

now 20% loss in IIIrd collision = (6.4 m) (0.2) = 1.28 m …(3)

total loss = 2m + 1.6 m + 1.28 m

= 4.88 m

and

4.88 m = mst

t = s88.4 =

50088.4 = 0.0097ºC

After rounding off.

t = 0.01ºC

78. In a track event, a circumcircle and an incircle were drawn for a triangle having sides 50 m, 120 m and 130 m respectively. Ram and Sham were asked to walk on the in-circle and the circumcircle respectively. They started walking with same speed in the same direction (sense of rotation) from a point where they were closest. After how many rounds each, will they be closest again?

(a) Ram 4 and Sham 13 (b) Ram 13 and Sham 4 (c) Ram 5, Sham 15 (d) Ram 5, Sham 12 Ans. [a]

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CAREER POINT

Sol.

50

120

R

130

r

r = 20 ; R = 65

for Ram, 1 = 20V =

260V13 {V is there speed & V = r}

for Sham, 2 = 65V =

260V4

There for same position Ram will take 4 rounds & sham will take 13 rounds.

79. The angle between the two complex number a = ii and b = 1 is

(a) (b) 0 (c) 2 (d) –

2

Ans. [b]

Sol. a = ii = i

i 2e

= 2–e

b = 1

both are positive real numbers. So angle between them is zero.

80. The number of rectangles that can be formed by joining the points of 4 × 4 grid of equi spaced points is (a) 16 (b) 36 (c) 40 (d) 42 Ans. [b] Sol. 4C2 × 4C2 = 36

INDIAN ASSOCIATION OF PHYSICS TEACHERS · INDIAN ASSOCIATION OF PHYSICS TEACHERS NATIONAL STANDARD...

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