nsea 2015 solutions by resonance

INDIAN ASSOCIATION OF PHYSICS TEACHERS

NATIONAL STANDARD EXAMINATION IN ASTRONOMY (NSEA) 2015-16

Examination Date : 22-11-2015 Time: 2 Hrs. Max. Marks : 000

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NATIONAL STANDARD EXAMINATION IN ASTRONOMY (Olympiad Stage-1) 2015-16 | 22-11-2015

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1. If high tides are observed on a particular date at a place in India, one may observe (a) low tides in the United States (b) low tides in Australia (c) high tides in both United States and Australia (d) low tides in India after half a month.

Sol. (d) High tides & low tides occur once in fortnight.

2. What is the value of x if x = 2 22sin2 cos2 sin cos

cos4 2sin cos

(a) 4sin (b) 2sin2 (c) cos2 (d) cos6 Sol. (c)

x (2sin2 cos2 )(sin2 ) (cos4 )( cos2 )

= sin4 sin2 cos2 cos4

= cos2

3. If

x a b c

a x b c 0

a b x c

then x is given by

(a) (a + b + c) (b) (a + b + c) (c) (ab + bc + ca) (d) (a2 + b2 + c2) Sol. (b)

x a b c

a x b c 0

a b x c

1 1 2 3C C C C

1 b c

(x a b c) 1 x b c 0

1 b x c

1 1 2 2 2 3R R R & R R R

0 x 0

(x a b c) 0 x x 0

1 b x c

x2 (x + a + b + c) = 0

x (a b c)

4. A particle of mass 1 kg is moving along a line y = x +2 (here x and y are in metre) with speed

2 ms-1. The magnitude of angular momentum of particle about the origin is

(a) 4kg m2s1 (b) 2 12 2 kgm s (c) 2 14 2 kg m s (d) 2kg m2s1

Sol. (b)





Angular momentum about origin = 1 (2) cos x 2

= 4 cos

= 2 12

4 2 2 kgm s2 2

5. If R is the radius of the earth and g is the acceleration due to gravity, the mean density of earth is

(a) 3 R

4gG

(b)

4 G

3gR

(c)

4RG

3gR (d)

3g

4 RG

Sol. (d)

As 2

GMg

R

2

3

M gR 3g

4V 4G RG R

3

6. Compared with others, the planets Earth and Venus have nearly same size and density. But

strength of the magnetic field of Venus is negligibly small compared to that of the Earth. This is due to (a) larger orbital speed of Venus (b) the absence of atmosphere around Venus (c) absence of metallic rocks inside Venus (d) very slow rotation of Venus

Sol. (c) Some physical attributes of venus is very similar to earth but absence of metallic rocks inside

makes its magnetic field weaker compared to earth 7. The sum of the cubes of three successive integers is always divisible by

(a) 4 (b) 8 (c) 9 (d) 12 Sol. (c)

Tn = n3 + (n + 1)3 + (n + 2)3 Tn + 1 = (n + 1)3 + (n + 2)3 + (n + 3)3 So Tn + 1 Tn = (n + 3)3 n3 (n + 3 n) ((n + 3)2 + n2 + n(n + 3)) 3(n2 + 9 + 6n + n2 + n2 + 3n) 3(3n2 + 9n + 9)





9(n2 + 3n + 3) Divided by 9 8. The circles x2 + y2 = 400 and x2 + y2 10x 24y + 80 = 0 are in same plane. Which of the

following statement is true? (a) They do not touch each other (b) They intersect each other in two points.

(c) They touch each other externally (d) They touch each other internally Sol. (b)

c1 (0, 0) r1 = 20

c2 (5, 12) r2 = 2 25 12 80 89

c1c2 = 13 r1 + r2 > 13

1 2r r 20 89

r1 + r2 > c1 c2 > 1 2r r

They intersect each other in two points.

9. A 40 kg slab S rests on frictionless floor and a 10 kg block B rests on the top of the slab. The

coefficient of friction between them is s 0.45 and k 0.4 . If the force applied on the block is

75N, the accelerations of the slab and the block respectively are (g = 10 ms-2).

(a) 3.5 ms-2, 1.0 ms-2 (b) 1.0 ms-2, 1.0 ms-2 (c) 3.5 ms-2, 3.5 ms-2 (d) 1.0 ms-2, 3.5 ms-2

Sol. (d)

B

S

s

k

0.45

0.4

m =

m = 75N

Friction require to move together = 75

4050

= 60N.

But fL (available) = 0.4 x 10g = 45 N

Relative motion takes place between slab and block

aslab = 240 1m / s

40

aBlock = 275 40 3.5 m / s

10

10. Select the wrong one (here N is newton, A is ampere, J is joule, m is meter, s is second and kg is

kilogram).

(a) 2stress

Nmstrain

(b) Surface tension = Jm-2

(c) Capacitance = A2s4Kg-1m-1 (d) Force = K gms-2





Sol. (c)

(a) 2

stress N(Correct)

strain m

(b) Surface tension =2

surfaceenergy J(Correct)

area m

(c)

2 2 2

2 2

2 4 1 2

2 4 1 1

Q Q Q A TC

V W / Q W ML T

A s kg m

A s kg m iswrong

(d) Force = ma = kg m.s2(Correct)

11. If cos20 - sin20 = k then cos40 is equal to

(a) 22k k (b) k (2 k)2 (c) 2k 2 k (d) 2k 2 2k k

Sol. (c)

k = cos 200 sin 200 k2 = cos2 200 + sin2 200 2 sin 200 cos 200 k2 = 1 sin 400 sin 400 = 1 k2

0 4 2cos40 1 (1 k 2k )

= 2 42k k

= 2k 2 k

12. Sun spot are used in the study of (a) rotation of the sun

(b) size of the sun (c) variation in the luminosity of Sun

(d) variation in the gravitational field around the Sun. Sol. (c)

Sun spots are temporary phenomena on the sun that appear visibly as dark spots compared to

surrounding regions

13. Which of the following statements are correct

(i) sin2 22 tan

1 tan

(ii)

2sin 2 2 tan cos

(iii) 2 2sin cos sin 2 (iv) tan 2

sin 2sin 2

(a) (i) & (iv) (b) (i), (ii) & (iii) (c) (i), (ii) & (iv) (d) All Sol. (c)

2

2tansin2

sec

2sin cos

(i) is true (ii) is true





(iii) cos2 sin2 False

(iv) sin2

sin2cos2 sec 2

is true

14. Five (5) parallel lines in a plane are intersected by set of four (4) parallel lines. The number of

parallelogram is (a) 12 (b) 20 (c) 25 (d) 60 Sol. (d)

Number of parallelogram is 5 42 2C . C

= 10 6 = 60 15. A horizontal platform at rest starts ascending from the ground with a constant acceleration of

1 ms-2. After 2 seconds a stone is thrown vertically upwards from the platform with a speed of 21ms-1 relative to the platform. The maximum height from the ground attained by the stone is (g = 10 ms-2)

(a) 2.8 m (b) 2.2 m (c) 2.4 m (d) 2.6 m Sol. (a)

at t = 2 sec plat-form is at 1

1 4 2m2

initial velocity of particle (u) = 2 + 2 = 4 m/s

From 2 m maximum height is 24 16

0.82 10 20

From ground ----- 2.8 m

16. The efficiency of carnots heat engine is 0.5 when the temperature of the source is T1 and that of

sink is T2. The efficiency of another carnots heat engine is also 0.5. The possible temperature of the source and sink of the second engine are respectively.

(a) T1 + 5, T2 5 (b) T1 + 10, T2 10 (c) 2T1, 2T2 (d) 2T1, 1T

2

Sol. (c)

2

1

T1

T

to have constant 2

1

T

T ratio should be constant i.e., (2T1, 2T2)

17. Two hypothetical stars A and B of same size have apparent magnitudes of 4.5 and 1.5 and real

magnitudes of 1.5 and 4.5 respectively. We conclude that (a) A is much hotter and closer to earth than B (b) B is much hotter and closer to earth than A (c) A is much hotter than B and B is much closer to earth than A (d) B is much hotter than A and A is much closer to earth than B Sol. (d) Lower the apparent magnitude higher is the brightness of star from earth hence A is nearer to

earth while real magnitude of B is less hence it is hotter compared to A





18. Find the value of

1 2 3

1 8 27

1 32 243

(a) 36 (b) 216 (c) 1296 (d) 61 Sol. (d)

2 2 1 3 3 1R R R & R R R

1 2 3

0 6 24

0 30 240

1 2 3

6 30 0 1 4

0 1 8

180 1(4) 720 6!

19. Find the value of tan1 2 + tan1 3

(a)

4 (b)

4 (c)

3

4 (d)

5

6

Sol. (c)

1 1

1

1

tan 2 tan 3

2 3tan

1 6

tan ( 1)

3

4 4

20. A physical quantity

4 2

1/34

a by

cd has four variables a, b, c and d. The percentage error in a, b, c

and d are 2 %, 3%, 4% and 5% respectively. The error in y will be (a) 6 % (b) 11 % (c) 12 % (d) 22 % Sol. (d)

y a b 1 c 4 d% 4 2 %

y a b 3 c 3 d

1 4

4(2) 2(3) (4) (5)3 3

= 8 + 6 + 8

= 22 %

21. Two identical drops of water fall through air with a terminal velocity 2 ms1. If the drops merge to

form a single drop, its terminal velocity is (a) 3.17 ms1 (b) 1.58 ms1 (c) 2.52 ms1 (d) 4 ms1 Sol. (a)

vr2

2

1 1

2 2

v r

v r





33 3 11 2

2

r4 4 12 r r

3 3 r 2

2

31

2

v 1

v 2

2

32 1v v (2) 3.17m / s

22. A part from the earth, Aurora phenomena is observed on which of the following planet (a) Venus (b) Mars (c) Mercury (d) Jupiter Sol. (d) Apart from earth Auroras have been observed on. Jupiter and Saturn 23. The mean of the five numbers is zero. Their variance is 2. If three of these numbers are 1, 1 and

2. The remaining numbers are (a) 5 & 3 (b) 4 & 2 (c) 3 & 1 (d) 2 & 0 Sol. (d) Let numbers be x & y

1 1 2 x y

05

x y 2

Variance =

2

2xi

(Mean)n

2 2 2 2 2( 1) (1) (2) x y

2 05

10 = 6 + x2 + y (x + y)2 = 4 x2 + y2 = 4 x + y = 2 Hence x = 2 & y = 0 24. Find the digit at the unit place of 272015 + 212015 232015 (a) 7 (b) 3 (c) 5 (d) 1 Sol. (a)

4 503 3 4 503 3 4 503 3(27) (21) (23)

The digit at unit place will be 3, 1, 7 respectively So, the unit place is 10 + 3 + 1 7 = 7 25. A cylindrical tube open at both ends has a fundamental frequency of 390 Hz tube is immersed,

vertically in water, the fundamental frequency of air column is (a) 130 Hz (b) 390 Hz (c) 520 Hz (d) 260 Hz Sol. (d)

l2

390 v 2l x 390 = v





'3l

4 4

' f v (3l) f = 2l x 390 f = 260 Hz

26. Three identical rods are joined as shown in the figure. The temperature at the junction is

(a) 48c (b) 44c (c) 40c (d) 36c. Sol. (d)

60 T 60 T 12 T

0R R 12

60 T + 60 T + 12 T = 0

108 = 3T

T = 360c

27. Sun is at a mean distance of about 27,000 light years from the centre of the milky way galaxy and

completes one revolution about the galactic centre in about 225 million years. The linear speed of Sun is

(a) 160 km s1 (b) 230 km s1 (c) 30 km s1 (d) 80 km s1 Sol. (b)

1 light year = 9.4607 1015m

Speed = dis tan ce

time

assuming path to be circular





v = 2 R

t

=

15

6

2 27000 9.4607 10

225 10 365 24 3600

= 230 km / sec

28. Find the value of 1cosx

tan1 sinx

(a) x

4 (b)

2 (c)

x

4 2 (d)

x

4 2

Sol. (c or d or cd)

1 cosxtan1 sinx

1

xsin

2 2tan

1 cos x2

1

2

x x2sin cos

4 2 4 2tan

x2cos

4 2

1 1 xtan tan4 2

=

x

4 2

29. If log 30 = 1.4771, log 40 = 1.6021, log 50 = 1.6990, log 45 = x then find x. (a) 1.6542 (b) 1.6232 (c) 1.6532 (d) 1.6832 Sol. (c) log(5 9) log 5 + 2 log 3 0.6990 + 0.4771 2 0.6990 + 0.9542 = 1.6532 30. The figure shows a pressure vs temperature (P T) graph for a given mass of an ideal gas

undergoing a process from A to B. In this process, volume of the gas

(a) is increasing (b) remains constant (c) is decreasing (d) cannot be predicted. Sol. (a)

P = MT + C

Pv = nRT

nRT

MT Cv





T

v nRMT C

2

dv (MT C)(1) T(M)nR

dT (MT C)

f(x, t) = f(x-vt, 0)

2

dv CnR ve

dT (MT C)

31. A window is 50 cm high. A stone starts falling from a height of 40 cm above the window. It crosses

the window in

(a) 1

S7

(b) 3

S7

(c) 5

S7

(d) 4

S7

Sol. (a)

v2 = v1 + at where AB = height of window

v1 = 2 9.8 0.4

v2 = 2 9.8 0.9

t = 2 1v v

a

=

2 9.8 0.9 2 9.8 0.4

9.8

=

1

7sec.

32. Light from the nearest star proxima centauri takes 4.24 light years to reach earth. The stellar

parallax of this star is about (a) 1.30 sec (b) 0.77 sec (c) 13.8 sec (d) 0.24 sec Sol. (c)

1 parsec (pc) = 3.0857 1016 m = 3.261 light year

1 light year = 9.4607 1015 m





Stellar parallax =

1

1.3009pc or 4.243light year

= 0.7687

33. The equation 2 2x y

125 a 49 a

represents

(a) an ellipse if a > 25 (b) a hyperbola if 25 < a < 49 (c) a hyperbola if a > 49 (d) an ellipse if 25 < a < 49 Sol. (b)

2 2x y

1(25 a) 49 a

25 a < 0, 4 a > 0 a > 25, 49 > a 25 < a < 49

34. The value of the determinant

x sin cos

sin x 1

cos 1 x

is

(a) Independent of (b) Independent of x (c) Independent of and x (d) none of these is true Sol. (a)

2

3 2 2

3

x( x 1) sin ( xsin cos ) cos ( sin x cos )

x x x(sin cos )

x

35. Two similarly charged spherical conductors are suspended by non conducting threads of length l

from a horizontal support. If this is taken to a place of zero gravitational effect, then angle between the threads and the separation between them are

(a) 0, 2l (b) 180, 2 l (c) 120, l /2 (d) 60, 3 / 2 l. Sol. (b)

In absence of gravity , only electric repulsive effect takes place acts

qqx

LL

36. A cylindrical object of volume 0.4 m3 floats in water with 20 % of its height seen above water. The

minimum force required to be applied on the object so that it just gets immersed in water is (Density of water = 103 kg ,m3 g = 10 ms2) (a) 32000 N (b) 800 N (c) 2400 N (d) 4800 N Sol. (b)





3v 0.4m=

0.8h

V = 0.4 = Ah (Given)

F = Extra Buoyancy force = ExtrasubmergedV g

= 103 x A x 0.2h x 10

= 103 x 0.4 x 0.2 x 10

= 800 N.

37. The International Space Station (ISS) launched in 1998 its satellite orbiting earth with perigee at

422 km and apogee at 425 km above mean sea level. Given mean radius of the earth is 6371 km and acceleration due to gravity is 9.8 ms2, the distance covered by ISS in one hour is about

(a) 27500 km (b) 766 km (c) 4150 km (d) 2660 km Sol. (a) Mean height is around423 km Distance from centre of earth 6371 + 423 6800 We know it takes around 1.5 hrs for near earth satellite

1 hr implies 2

3 of perimeter

2 2

2 R 2 6800km3 3

= 28000 km

Hence (a)

38. Find the 99th term of the sequence 7, 23, 47, 79, 119, . (a) 35999 (b) 379999 (c) 37799 (d) 39599 Sol. (d)

7, 23, 47, 79, 119, By method of difference

Tr = 4r2 + 4r 1 So, T99 = 4 992 + 4 99 1 = 39599

39. forty two cube with 1cm edge each are glued together to form a solid rectangular block. If the

perimeter of the base is 20 cm. Find the height of that rectangular block. (a) 2 cm (b) 3 cm (c) 6 cm (d) 7 cm Sol. (a)

Volume of the rectangle block = 13 x 42 cm3 = 42 cm3





2 (l + b) = 20

l + b =10 All sides will be integers

l b h = 42 3 x 7 h = 42 h = 2 40. Out of the following expressions

(i) a

yx t

(ii) y a x t (iii) 2y a x t (iv) 2 2y a x t The expression/s that represent/s progressive waves is/are (a) (i) and (ii) only (b) (ii) and (iii) only (c) (i), (ii) and (iii) only (d) all Sol. (c) f(x, t) = f(x vt, 0)

41. There are two open organ pipes of exactly the same length and material but of different radii.

Then, (a) narrow pipe has lower frequency (b) both the pipes have exactly the same frequency (c) wider pipe has lower frequency (d) either has lower frequency depending on the amplitude of sound waves. Sol. (c) end correction depends on radius large radius larger end correction larger wavelength smaller frequency (c) is correct. Read the following passage and answer the questions number 42 to 44 A star of mass greater than about 5 times the mass of the Sun (M0) dies as black hole. When the

nuclear burning stops, the star starts collapsing under its own gravity. On attaining a certain radius, its escape velocity becomes equal to the speed of light and it is called the Schwarzschild radius Rs. An event horizon is said to be formed when No information can be obtained by electromagnetic waves from within. For a non-rotating black hole, the Schwarzschild radius itself

forms the event horizon. ( . . / ,11 2 2G 6 674 10 Nm kg 8 1c 3 10 ms , . 300M 21 6 10 kg and

. 24earthM 5 972 10 kg )

42. The Schwarzschild radius of a star of mass 8M0 is about (a) 3 km1 (b) 256 km (c) 140 km (d) 480 km s-1 Sol. (b)

'

da 2L

dd 2 m





C = 2Gm

escape speedR

3 108 = 11 302 6.674 10 8 21.6 10

R

9 1016 = 191.6 6.674 10 21.6

R

R = 256 km 43. After a massive star attains the Schwarzschild radius Rs, on further gravitational contraction, the

radius of event horizon of a non rotating black hole (a) decreases (b) increases (c) remains constant (d) becomes infinite Sol. (c)

Rs is the radius at which escape velocity become C hence Rs is constant for a given star. 44. Suppose earth starts shrinking to become a black hole then the event horizon will (a) 8.9 km (b) 2.35 m (c) 0.89 cm (d) 12.45 cm Sol. (c)

x = 24m

R

(3 108)2 = 11 242 6.674 10 5.972 10

R

R = 13

16

8.857 10

10

R = 0.89 cm

45. 32 3251 15 is a multiple of (a) 765 (b) 2178 (c) 2826 (d) 2956 Sol. (c)

5132 1532 (5116 1516) (5116 + 1516) (518 158) (518 + 158) (5116 + 1516) (514 154) (514 + 154) (518 + 158) (5116 + 1516) (512 152) (512 + 152) (514 +154) (518 + 158) (5116 + 1516) (36) (66) (512 + 152) (514 + 154) (518 + 158) (5116 + 1516) Hence divisible by 2826

46. . .;0 6021 0 477110 4 10 3 and x10 72 find x.

(a) 1.85735 (b) 1.75725 (c) 1.87525 (d) 1.87255 Sol. (a)

100.6021 = 4 100.4771 = 3 10log 4 = 4 10log 3 = 3 log 4 = 0.6021

10x = 72 log 3 = 0.4771

10x = 72 x = log 72 = 2 log 3 + 3 log 2

= 2 0.4771 + 3 0.6021

2





= 0.9542 + 0.90315 = 1.85735

47. Find the smallest natural number n such that 265n 1 is a perfect square (a) 13 (b) 16 (c) 17 (d) 19 Sol. (b) 65n2 + 1 = k2

n2 2k 1

65

2(65 1) 1

65

2 265 2 65

65

= 265 2

So, minimum value of = 2

Hence n = 16 48. Two litres of milk purchased from a vendor, has a density of 1018 kgm-3. If the density of pure milk

from that source is 1030 kgm-3 and that of water used is 998 kg m-3, the amount of water mixed in he sample is

(a) 0.750 litre (b) 0.667 litre (c) 0.480 litre (d) 0.375 litre Sol. (a) water milk = 2L = V, m = 1018 kg/m

3

Density of milk = 1 = 1030, volume of milk = V2

Density of water = 2 = 998, volume of water = V2

mass of water + mass of milk = mass of (milk + water) 2 V2 + 1 V1 = Vfm

998 V2 + 1030 V1 = 2036 (i) V1 + V2 = 2 (ii) 998 V2 + 1030 [2 V2] = 2036 - 32 V2 + 2060 = 2036 V2 = 0.75 L 49. Particle A starts from origin with a constant acceleration 3ms-2. Its initial speed is zero. Particle B

starts from x = 48 m and moves with constant speed of 6ms-1. If both the particles move along x axis and start at the same instant, then identify the correct graph which depicts motion of both the particles.

(a) (b)

(c) (d)

Sol.

(a)

2Aa 3ms

, Au 0





x = 48, 1Bu 6ms

(Constant)

2A A A1

S u t a t2

B BS u t

2A1

x 0 (3)t2

xB 48 = 6t

xA = 1. 5t2 xB = 6t + 48

50. In the situation shown in adjacent figure, coefficient of friction between block B and wedge A is 0.5.

Wedge A is moved horizontally with uniform acceleration a. Then identify the correct statement

(take tan 0 3374

).

(a) Friction on the block is zero if a = g (b) Friction on the block is upward along the incline if a > g

(c) Friction on the block is downward along the incline if g

a2

(d) Friction on the block is kinetic in nature if a 2g

Sol. (d) If = g tan , tnen f = 0

3g

a4

for f = 0

If 3g

a4

then f is down the incline

3g

a4

then f is up the incline

(d) is correct. 51. Find he sum of all three-digit perfect square numbers (a) 9131 (b*) 10131 (c) 10231 (d) 11031 Sol. (b)

The sum of .........2 2 2 2 2 210 11 12 13 14 31

.......... ..........2 2 2 2 2 21 2 31 1 2 9

31 32 63 9 10 19

6 6

10416 285

10131

52. Which of the following statements are true

(i) sin cos sin .1 1 18 84

0 617 85

(ii) sin cos tan .1 1 18 84

0 7517 85





(iii) sin cos cot .1 1 18 84

0 7517 85

(iv) sin cos cos .1 1 18 84

0 8017 85

(a) (i) & (iv) (b) (i), (ii) & (iii) (c*) (i), (ii) & (iv) (d) All Sol. (c)

sin cos sin1 1 18 84

17 85

(say)

Taken sin both side sin sin cos sin sin1 1 18 8417 85

8 84 15 13

17 85 17 85

672 195

17 85

867

17 85

.0 6

Hence, ans is sin-1 (0.6), cos-1 (0.8) & tan-1 (0.75)

(i), (ii) & (iv) are true.

Read the following passage and answer the questions number 53 to 56

The Mars orbital Mission (MOM) space craft launched on 9th November 2013 reached the expected Martian orbit on 24th September 2014. Now it is revolving the planet Mars in a highly elliptic orbit. At the closest position the space craft is at a height of 421.7 km above the Martian surface, its farthest distance from the planets surface being 76993.6 km. The mean radius of Mars is about 3390 km. One of the two natural satellites of Mars is Phobos revolves the planet in a nearly circular orbit with a mean radius of 9377 km once every 7.66 hours (mass of mars = 6.39 x 1023 kg and G = 6.647 x 1011 N.m2/kg2).

53. The semi major axis of MOM's orbit is about (a) 42100 km (b) 38500 km (c) 40000 km (d) 36800 km Sol. (a)

2a = 421.7 + 2 x radius of mars + 76993.6 a = 42096.65 km 54. The semi minor axis is about (a) 21008 km (b) 17510 km (c) 4207 km (d) 19503 km Sol. (b)

ae a 421.7 3390 e = 0.909

2 2 2b a (1 e )

2b a (1 e )

b = 17545.7 km 55. The period of revolution of the space craft in hours is about (a) 36.4 (b) 26.5 (c) 72.8 (d) 53.0 Sol. (c)

2 3

1 1

2 2

T r

T r

From Keplers law





.

2 3

2 2

7 66 9377

T r

.

.2

2

58 670 0110

T

.

..

22

58 69T 73 031

0 011

56. The maximum speed of the space craft in its orbit is about (a) 2.15 kms1 (b) 1.85 kms1 (c) 7.89 kms1 (d) 1 kms1

Sol. (Bonus)

Maximum speed would be at A

maxGM 1 e

Va 1 e

e = 0.909 2

2

be 1

a

Where b & a are semi minor and semi major axes.

a = 42100 km.

By putting all values

max . / .V 4 31 km s

Ans : No option is correct hence BONUS.

57. If y = cos(x)

sec(x) then

dy

dx is given by

(a) sin (2x) (b) sin (2x) (c) sin(x) cosec(x) (d) sin(x)

cosec(x)

Sol. (b)

cos

cos

xy

1

x

cos2y x

cos sin

dy2 x x

dx

sin2x





58. The owner of a milk store finds that he can sell 980 litres of milk each week at Rs 42/litre and 1220

litres of milk each week at Rs 48/litre. Assuming the linear relationship between the selling price and the demand, how many litres could he sell weekly at Rs 51/litre.

(a) 1340 (b) 1430 (c) 1450 (d) 1540 Sol. (a)

Since we have to assume linear relationship

Z Ax B

Here A & B are constants.

Now, 980 = A (42) + B (1)

1220 = A (48) + B (2)

Solving (1) & (2)

A 40 B 700

Now finally Z 40 51 700

= 1340

59. Two hypothetical main sequence stars A and B have their radius in the ratio 1:2 and their surface

temperature in the ratio 2:1. Then the ratio of their luminosities respectively are in the ratio (a) 4:1 (b) 1:1 (c) 2:1 (d) 1:16 Sol. (a)

Luminosity 4L AT

421L 4 r 2T 2 42L 4 2r T

Ratio 2 4

1

2 42

L r 16 T 4

L 14r T

Ratio is 4 : 1

60. Find the value of if cos3 cos cos2

(a) (b) 2

(c)

2

2

(d)

4

Sol. (d)

cos3 cos cos2 2cos2 cos cos2 0

cos2 (2cos 1) 0

cos2 0 ,

1cos

2

22

3

4

61. Consider an isolated system of interacting particles. Then indentify the correct statement (i) Total mechanical energy of the system must be conserved. (ii) Total linear momentum of the system must be conserved.





(a) (i) is true and (ii) is false (b) (i) is false and (ii) is true (c) (i) and (ii) both are true (d) (i) and (ii) both are false Sol. (b)

For isolated system :

(i) Mechanical energy may not remain conserved. For interacting particles during collision energy

may be lost to heat energy.

(ii) Momentum of system will remain conserved.

62. A gaint wheel of mass M has a rim on which a person can walk. Consider a case where, in a person of mass m walks on the rim at a constant speed and the wheel rotates in opposite direction with constant angular speed. Further the speed of the person with respect to ground is zero. Then the net force exerted by axle on the wheel. (i) Is equal to (m + M) g (ii) ls greater than (m + M) g (iii) Depends on angular speed of the wheel (iv) Depends on radius of the wheel (a) (i) is true (b) (i) and (ii)are true (c) (ii) and (iii) are true (d) (ii), (iii) and (iv) are true Sol. (a)

For Man :

Since w.r.t. ground it is at rest.

N mg

So, no centripetal force or acceleration.

For wheel :





1N N Mg

1N M m g

Force exerted by axle on wheel = (M + m) g

No effect of angular speed or radius of wheel.

63. A particle is projected with a velocity, 20 ms1 at angle of 60. Then radius of curvature of its trajectory at a point where its velocity makes an angle of 37 with the horizontal is (g = 9.8ms2) (a) 16 m (b) 19.53 m (c) 15.62 m (d) 25 m Sol. (b)

cos cos0 0V 37 20 60

4

V 105

50

V4

m/s

2V

Ra





50 50 1 5

R4 4 10 4

6250

64 = .19 53m

64. A person standing on a plank is pulling a string as shown in the adjacent figure. Floor is rough and the motion is impending. Tension in the string is denoted by T and coefficient of friction

between plank and floor is . The tension in the string is

(a) mg

T1 cos sin

(b)

mgT

1 cos

(c) mg

T1 sin

(d)

mgT

1

Sol.

(a)

Friction

N Tsin mg

T Tcos (mg Tsin )

T(1 cos sin ) mg

mgT

1 cos sin

65. If and are the roots of the equation x2 = (x + 1). How many of the following statements are true.

(i) 2 = (ii) = 2 (iii) = 1 (iv) 3 3 = 0

(a) 1 (b) 2 (c) 3 (d) all Sol. (d)

2x x 1 0 2x , (where cube root of unity)

Let 2, ; obvious

66. The numbers a and b are chosen from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} such that a b

with replacement. Find the probability that a divides b

(a) 5

11 (b)

29

45 (c)

27

55 (d)

49

100 Sol. (c)

The total number of all possible chosen a & b such that a b is 10 + 9 + . + 1 = 55 &

for favorable case if a = 1 then b = 1, 2, 3..10





if a = 2 then b = 2, 4, 6, 8, 10 similarly if a = 10 then b = 10 Hence the total number of favorable case is 10 + 5 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 27 So the probability is 27/55 67. Assuming that straight line works as a plane mirror for a point. Find the image of the point (1, 2) in the line x 3y + 4 = 0 (a) (2, 1) (b) (1, 2) (c) (1.2, 1.4) (d) (1.2, 1.4) Sol. (c) The image of point (1, 2) w.r.t. line x 3y + 4 = 0 is

x 1 y 2 2(1 6 4)

1 3 10

x 1 y 2 1

1 3 5

6

x 1.25

3 7

y 2 1.45 5

68. There are two pendula as shown in the adjacent figure. Pendulum A is at rest in its equilibrium position while pendulum B is oscillating. When pendulum B passes through the equilibrium position, Consider the following two statements

(i) The net force experienced the bob at the equilibrium position is zero in both the cases. (ii) The acceleration of pendulum B in the tangential direction is zero (a) Statement (i) and (ii) are correct (b) Statement (i) is correct while Statement (ii) is wrong (c) Statement (ii) is correct while Statement (i) is wrong (d) Statement (i) and (ii) are wrong Sol. (c)

T mg 0 2mV

T mgR

equilibrium Not in equilibrium.

Hence statement (i) is wrong.

Now no net force is acting along the horizontal direction, therefore tangential acceleration is zero in

both cases.





69. An electrical circuit consisting of three identical bulbs and three switches is shown in the adjacent

figure. Let the glowing bulb be presented by one (1) and zero (0) when it is not glowing. The correct table that represent the circuit operation is

Ans. (Bonus) No figure is given.

70. In a polarization experiment a plane polarised light is allowed, to pass through a mica sheet and then through an analyser. The intensity of light is measured as function of orientation of mica sheet. The intensity attains minimum at a certain angle and then increases. The data obtained for three angles close to minima is given by.

Angle Intensity

-3.6 1.1

0 0.2

3.6 0.6

Assuming that variation of intensity (1) with respect to angle is a quadratic function, the angle at which the intensity of tight is zero is

(a) 0.7 (b) 0.9 (c) 0.3 (d) 0.5 Sol. (b)

Let I = a 2 +b + c

where is in radian

let 1 = + 3.6o 180

and 2 = - 3.6

o 180

let 1 = - 2 =





For 3 = 0 I = 0.2 C = 0.2

Now putting values of 1 2&

0.9 = a 2 - b + 0 .(1)

and 0.4 = a 2 +b .(2)

Adding 1 and 2

0.13 = 2a 2

2

2 2

0.13(180) 1300a

23.6 3.6 8

Putting value of a in eqn (1)

22

13000.9 b

8

22

13000.9 (3.6)3.6 b(3.6)

1808

83.50

b2

Now,

22

1300 83.500.2 0

28

2 21300 334 1.6 0

Now apply the quadratic formulae

2b b 4ac

2a

=

2 2334 334 4 1300 16

2x1300

Converting into degree

334 103236 9

x2x1300

334 321

9130

0

1

13x9 0.9

130





02334 321

9 45.36130

Correct answer is 0.90

71. A cylinder is closed at both ends and has insulating walls. It is divided into two compartments by

an adiabatic (insulating) partition that is perpendicular to the axis of the cylinder. Each compartment contains 1.00 mol of oxygen that behaves as an ideal gas with ratio of specific heat,

7y

5 . Initially, the compartments one and two have equal volumes and their temperatures are

550K and 250K The partition is then allowed to move slowly until the pressures on its two sides are equal. The final temperatures in the first and second compartments are respectively

(a) 473.5K & 264.5K (b) 498.9K & 284.2K (c) 492.9K & 270.2K (d) 250K & 550K

Sol. (b)

P1Vy = PV1y ...... (i) P2Vy = PV2y ..... (ii)

y

1 1

2 2

P V

P V

7/5

1

2

nR 550

VV

nR 250 V

V

1

2

V1.756

V ....... (iii)

V1 + V2 = 2V .......... (iv) Using (iii) and (iv)

1

2

V1.756

V

22

V V2.756

12 1.756

V2.756

then using TVy1 = constant for Partition number (1)

2/5

2/5

1

2 1.756550 V T

2.756

T1 = 499.17 K

and for partition number 2 y 1

y 1

2

2V250 V T

2.756

T2 = 283 K

72 The reminder that is obtained, if the number (100110011001)2 is divided by 91 is (a) 0 (b) 1 (c) 8 (d) 9 Sol. (a)





Here (100110011001)2

2

100 9 1100110011

291 1100110011

= So it is divisible by 91, Hence remainder = 0

73. If 3x = 7y = 63z then z (2y + x ) is given by

(a) x y (b*) xy (c) x2y (d ) none of these

Sol. (b)

x y z3 7 63 k

Let

x log3 ylog7 z log63 logk

logk 2logk logk

z(2y x)log63 log7 log3

2(log k) (2 log3 log7)

log 63 log7 log3

2(log k)xy

log7 log3

74. a and b are unit vectors. is the angle between a and b . Which of the given statement are

true.

(i) If a b is unit vector then 2

3

(ii) If a b is unit vector then

3

(iii) If a b is unit vector then 3

(iv) If a b is unit vector then

4

(a) (i) & (ii) (b) (iii) & (iv) (c) (i) & (iii) (d) none of above Sol. (a)

a b 1

22

a b 2 a b cos 1

1 1 2cos 1

1

cos2

2

3

and a b 1

22

a b 2 a b cos 1

1 1 2cos 1

1

sin2 2

3

(i), (ii) 75. When the cube root of 13683*0393208 is divided by 3, the remainder is 2. The digit in place of * is (a) 5 (b) 6 (c*) 7 (d) 9





Sol. (c)

Let 1/3(13683 0393208) 3n 2

313683 0393208 (3n 2)

9k 8 13683 0393200 should be divisible by ? 7

76. A glass slab is divided into three lenses l1, l2 and l3 as shown in the adjacent figure. If power of lens l1 is P1 = 0.25, the power of the lens l2.

(a) 0.5 D (b) 0.25 D (c) 0.75 D (d) 0.4 D Sol. (b)

Total power of slab P = 0.

Now, slab divided into three parts so, P1 + P2 + P3 = 0

11 1 2

1 1 1 1P 1 1

f R R R

.0 25

.3

3 1 2

1 21 1 1 1P 1 1

f R R 0 5R R

.0 5

Now, . .20 25 P 0 5 0

.2P 0 25D

77. An object (O) is placed at a distance of 30 cm from a convex lens of focal length 12 cm and real image is caught on a screen. A convex mirror is placed between lens and image screen. It's position, when adjusted forms a final image coincident with O. If the distance between lens and mirror is 10 cm, the focal length of the convex mirror is (a) 30 cm (b) 10 cm (c) 5 cm (d) 25 cm Sol. (c)

For convex lens 1 1 1

f v u

1 1 1

12 v 30





From above equation v 20cm

Now, distance of image form by convex lens is 10 cm from convex mirror. (because distance

between lens and mirror is 10 cm) To retrace path this point must be centre of curvature of mirror.

fm = R/2 = 10/2 = 5 cm.

78. An electrical heater drawing a current 5 amp is used for melting 5 Kg of ice (at 0). The temperature of the melt after 1000 sec (latent heat of melting of ice is 80 calg-1, specific heat of water is 1 calg-1 per degree Celsius and resistance of coil is 40 ). (a) 10 C (b) 30 C (c) 15 C (d) 0 C Ans. (d)

Heat required to melt ice H = ML = 45000 80 40 10 Cal

4 440 10 Cal 168 10 J

Now, 2H i R t

This will be equal to heat required to melt the ice, therefore 2 4i R t 168 10 J .

2 45 40 t 168 10 J

sec.t 1680 This time is greater than 1000, so temperature of liquid will be at 00.

79. A stone tied to string is whirled in a horizontal plane in a circular path at a rate 5rev s-1. If the same stone is whirled with same force by a string of half its length, the number of revolution for one second now is

(a) 180 revs (b)

110 revs (c) 150 revs (d)

140 revs

Sol. (c)

1 2F F

2 2

1 1 2 2mr mr

2 225

2

/2 50 rev s .

80. Consider a small balloon filled with an ideal gas which is submerged in water. Assuming that the temperature is the same everywhere in the water, the buoyant force on the balloon when it is at a depth d below surface, in terms of its volume at the surface V0, the atmospheric pressure P0, the density of water p0, and the acceleration due to gravity g.

(a) FB = 0 0

0

p v

pd

pg

(b) FB = 0 0

0

p v

dpg p (c) FB =

0

0 0

dpg p

p v

(d) FB =

0 0

0

p v

pgd

p

Sol. (a)

As temperature of liquid is same everywhere so the process is isothermal, therefore





1 1 2 2P V P V

, ,2 0 2 0 1 0P P V V P d g P

0 010

P VV

d g P

Now force of buoyancy at depth

0 0B 10

P VF V g g

d g P

0 0B0

P VF

Pd

g

nsea 2015 solutions by resonance

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