1

Time : 120 Minutes Max. Marks : 240

INSTRUCTIONS TO CANDIDATES :

(i) Question paper has 80 multiple choice questions. Each question has four alternatives, out of which

only one is correct. Choose the correct alternative and fill the appropriate bubble, as shown.

Q. No. 22 A C D

(ii) A correct answer carries 3 marks whereas 1 mark will be deducted for each wrong answer.

Questions & Solutionsforforforforfor

INDIAN ASSOCIATION OF PHYSICS TEACHERS

NATIONAL STANDARD EXAMINATION

IN ASTRONOMY (NSEA) 2019-20

Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Ph.: 011-47623456

DATE : 24/11/2019 Paper Code : 41

1. If a and b are the roots of x2 – 3x + p = 0 while c and d

are the roots of x2 – 12x + q = 0. Also if a, b, c, d are

in GP then the value ( )

( )

q p

q p

, is

(a)17

15(b)

19

16

(c)35

32(d)

9

8

Answer (a)

Sol. Let the roots of x2 – 3x + p = 0 be a and ar and roots of

x2 – 12x + q = 0 be ar2 and ar3.

a + ar = 3, ar2 + ar3 = 12

r2 = 4

Now,

2 5 2 4

2 5 2 4

1 17

151

q p a r a r r

q p a r a r r

2. The date on which the Earth is at a minimum distance

from the Sun is around

(a) 3 January (b) 4 July

(c) 22 June (d) 21 September

Answer (a)

Sol. Earth comes closest to the sun every year around

January 3.

3. The plane of the circular coil is held in the east-west

direction. A steady current passed through the coil

produces a magnetic field (B) equal to 2 times the

horizontal component of the earth's magnetic field (BH)

at the place. Now the plane of the coil is rotated

carefully through an angle 45º about the vertical axis

through its diameter. What is the deflection of the

needle placed at the centre of the coil with respect to

BH?

(a)1 1

tan2

(b)

1 1tan

3

(c) tan–1 2 (d) tan–1 3

Answer (a)

Sol.

2H

B

W E

S

N

BR

B

H

45°

2

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The magnetic needle will point in the direction of

resultant magnetic field (BR) in horizontal direction

where is the angle made by BR

w.r.t. BH

2 sin45tan

2 cos45

H

H H

B

B B

1

(1 1) 2

H

H

B

B

1 1tan

2

4. If the ratio of the sum of first n terms of two different

AP series is (7n + 1) : (4n + 27), the ratio of their 10th

term is

(a)135

102(b)

134

103

(c)78

69(d)

103

89

Answer (b)

Sol.1 1

2 2

[2 ( 1) ]7 124 27

[2 ( 1) ]2

∵n

n

na n d

S n

nS na n d

1 1

2 2

1

7 12

1 4 27

2

na d

n

n na d

put n = 19

1 1

2 2

9 134

9 103

a d

a d

5. The solution to the set of simultaneous equations with

three variables x, y, z

x + y + z = 5; x2 + y2 + z2 = 5 and xy + yz + zx = 10 are

(a) All real

(b) One real, other two complex

(c) Two real, one complex

(d) All complex

Answer (d)

Sol. x + y + z = 5 …(1)

x2 + y2 + z2 = 5 …(2)

Eq. (1) represents a plane and eq. (2) represents a

sphere with centre at origin and radius 5. If ‘d’ be

the distance of the centre from the plane then

2 2 2

| 5 | 5

31 1 1

d

As5

53

(1) and (2) do not intersect

Hence given equations have no real solution.

6. An asteroid's closest approach to the Sun (perihelion)

is 2 AU, and farthest distance from the Sun (aphelion)

is 4 AU. The period of revolution of the asteroid and

the eccentricity of the orbit are respectively

(a) 4.2 Year, 0.40 (b) 4.8 Year, 0.50

(c) 5.2 Year, 0.33 (d) 6.0 Year, 0.20

Answer (c)

Sol. For earth, 2

32

0

0

41AUT

GM

…(i)

And for Asteroid, 2

32

0

43 AU

TGM

…(ii)

2

2

0

27T

T

T = (1 year) × 3 3 = 5.2 year

Clearly a = 3, and ae = 1

1

0.333

e

7. A circular coil carrying a definite current i produces a

magnetic field Bo = 2.83 T at the centre 'O' of the coil.

The magnetic field produced by the same current at a

point 'P' on the axis of the coil, where the angle OPM

as shown in figure is 45º, is

M

OP

45º

(a) 1.23 T (b) 1.00 T

(c) 1.78 T (d) 2.83 T

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Answer (b)

Sol. 45°

M

P

O

a

At axial point, magnetic field is given by

2

0

3/22 2

2

iaB

a x

At centre, x = 0

0

2

o

iB

a

Now, given that Bo = 2.83 T

And for the given point on the axis.

tan45

a

OP x a

2

0

3/22 2

2

P

iaB

a a

0

3/2

2.83T 1.00 T

2 2 8 8

oi B

a

8. A narrow U-tube of uniform cross-section, having a liquid

of density ‘’, is made to move with an acceleration 'a'

along the horizontal direction as shown in the

figure. What should be the value of 'a' so that the

difference in the level in the two arms of the tube is

3 cm (g = 10 ms–2)?

4 cm

a

(a) 6.5 ms–2 (b) 5.5 ms–2

(c) 13.3 ms–2 (d) 7.5 ms–2

Answer (d)

Sol. Since,

4 cm

3 c

m

1 2

a 1 2

�P P a

(where � = 4 cm)

0 0 �P gh P a

(where h = 3 cm)

�gh a

�

gha

210 37.5 m/s

4

9. The value of 4 sin 50 3 tan50 is

(a) –2 (b) –1

(c) 1 (d) 3

Answer (c)

Sol. 4sin50 3 tan50

2sin80 3 sin50

cos50

2sin80 2cos30 sin50

cos50

2sin80 (sin80 sin20 )

cos50

sin80 sin20

cos50

2sin30 cos50

cos50

= 1

10. The value of Z = cos10° cos30° cos50° cos70° is

(a)3

8(b)

5

16

(c)5

8(d)

3

16

Answer (d)

Sol. Z = cos10° cos30° cos50° cos70°

3cos10 cos(60 10 ) cos(60 10 )

2

3 1cos30

2 4

3

16

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11. The number of times a 5th magnitude star is brighter

than an 8th magnitude star is

(a) 15.85

(b) 3

(c) 20

(d) 6.4

Answer (a)

Sol. Brightness of star difference

2

2 1

1

2.50log

Bm m

B

or, 2

1

5 8 2.50log

B

B

or,2

3log

2.50 1

B

B

or,1.22

1

(10) 15.85 B

B

12. A uniform gravitational field (E) exists in a certain region

of space. Consider two parallel plane perpendicular to

the field E as shown in the figure. Three statements

are given below

(i) Work done by the force in moving a particle of

mass m from point A to B is 1

2mEd

(ii) Work done in moving a particle from point B to D

is zero

(iii) Work done in moving a particle from A to D is

3

2mEd

A

B

Y

E

X

d

30° 30°

d DC

d

Choose the correct statement(s)

(a) (i), (ii) and (iii)

(b) (i) and (ii) only

(c) (i) only

(d) (ii) and (iii) only

Answer (d)

Sol.

A B E

30°

d DC

1 2

(ii) Work done in moving a mass on an

equipotential surface from B to D

W = m[VD – V

B]

VB = V

D

W = 0

(iii) Work done in moving from A to D

W = m(dV)

= ����

m E dr

= mEd cos30°

3

2 mEd

13. A ball is projected with a speed of –110 3 ms at angle

of 60° from the foot of a wedge as shown in the figure.

The wedge also starts moving horizontally with a speed

of –110 3 ms . The time after which the ball hits the

inclined surface of the wedge is

60°30°

10 3 m/s

10 3 m/s

(a) 2 s (b) 2 s

(c) 3 s (d) 3 s

Answer ( b)

Sol.

30° 60°

10 3 m/s

10 3 m/s

y

x

Resolving velocity of ball

ˆ10 3 cos60ball

V i�

ˆ10 3 sin60 j

ˆ ˆ5 3 15i j

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ball /wedge ball wedge–V V V� � �

ˆ ˆ5 3 15i j

Relative to wedge :

30°

60°

10 3 m/s

15 m/s

5 3 m/s

2 10 3 sin (30)

cos (30)T

g

2 10 3

tan(30) 2 s10

14. In the previous question (number 13), what should be

the velocity of the wedge so that the ball hits the incline

perpendicular to it

(a) –18 3 ms in the same direction

(b) –15 3 ms in the opposite direction

(c) –116 3 ms in the same direction

(d) –115 3 ms in the opposite direction

Answer (a )

Sol.

5 3 m/s

15 m/s

Relative to wedge,

30°

15 m/s

5 3) m/s

60°30°

(v –

v = 15 cos 60°+(v – 5 3)cos 30°1

v = 15 sin 60°–(v – 5 3)sin 30°2

2(15 sin 60 ( 5 3) sin 30 )

cos30

vT

g

For particle to strike perpendicularly

15cos60 ( 5 3) cos30 sin(30 )v g T

15cos60 ( 5 3) cos30v

2(15sin60 ( 5 3)sin30sin(30 )

cos30

vg

g

Solving the above equation, we get 8 3 m/ sv

15. If 'p' is the perpendicular distance from the origin to

the straight line xcos– y sin= k cos2and 'q' is the

corresponding distance to the straight line xsec–ycosec= k, then the value of p2 + 4q2 is

(a) 2k2 (b) k2

(c) 4k2 (d) 3k2

Answer (b)

Sol.2 2 2

2 2

cos2cos 2

sin cos

kp p k

...(i)

and 2 2 2 2

2 2

sin .cos

sec cosec

kq q k

4q2 = k2sin22 ...(ii)

Adding (i) and (ii)

p2 + 4q2 = k2

16. If anda b are unit vectors and is the angle between

anda b then sin2

is equal to

(a) 1 (b)1| – |

2a b

(c) 0 (d)1| |

2a b

Answer (b)

Sol. ∵1– cos 1– .

sin2 2 2

a b

2 22 – 2 . | | | | –2 .

4 4

a b a b a b

1| – |

2a b

17. Which of the following statement is true for Saturn

and Jupiter ?

(a) Both rotate faster than the Earth

(b) Both rotate slower than the Earth

(c) Only one rotates rapidly while the other rotates

very slowly compared to Earth

(d) Their periods of rotation are linked to their period

of revolution

Answer (b)

Sol. Both saturn and jupiter rotate slower than the earth.

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NSEA 2019-20

18. In which of the following cases friction cannot be zero

for the disc having pure rolling?

F

R

Disc(i)

F

R

Disc(ii)

x

F

R

Disc(iii)

F

R

Disc(iv)

(a) (i), (iii) and (iv)

(b) (i) and (ii) only

(c) (i), (ii), (iii) and (iv)

(d) (i), (ii) and (iii)

Answer (a)

Sol. (i)

F

Since F is acting at height R from centre of mass

Friction will act in forward direction.

(ii)

F

o

x

If 2

cmI R

xMR

Then no friction will act on the disc.

(iii)

F

R

Here is non-zero due to F

Hence friction must be present to maintain pure

rolling by providing acm

(iv) FR

F will provide acm

and will be produced by friction

to maintain pure rolling.

Therefore, only in case (ii) friction can be absent

in pure roling.

19. A and B are two locations 120 m apart with a 40 m tall

wall midway between them. The minimum velocity and

the corresponding angle at which a ball be projected

from A so as just to clear the wall and strike at B are

(a) 28.63 ms–1 and 53° (b) 28.63 ms–1 and 37°

(c) 35.06 ms–1 and 53° (d) 28.28 ms–1 and 37°

Answer (c)

Sol. x = 60 m R = 120 m

y = 40 m

40 m

60 m60 mA B

tan 1x

y xR

6040 60 tan 1

120

140 60 tan

2

80 4tan

60 3

= 53°

2sin2u

Rg

2 4 3

25 5

1209.8

u

2 120 25 9.81225

24u

1225u

u = 35

20. The argument [arg (z)] of the complex number

1 3

2 2z i

is

(a)6

(b)

3

(c)2

3

(d)

4

3

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NSEA 2019-20

Answer (c )

Sol.1 3

–2 2

Z i

arg(Z) = –1 –13 / 2 2– tan – tan 3

1 3–2

21. The eccentricity of the ellipse represented by

4x2+ 9y2 – 16x = 20 is

(a)3

5(b)

5

3

(c)3

5(d)

1

2

Answer (b )

Sol. 4x2 + 9y2 – 16 x = 20

4(x2 – 4x + 4) + 9y2 = 36

2 2( – 2)

19 4

x y

4 51–

9 3e

22. Which of the following is TRUE for Retrograde motion

(apparent motion of planets in a direction opposite to

the normal)?

(a) Caused by epicycles

(b) Undergone only by superior planets

(c) Undergone only by inferior planets

(d) An effect due to the projection of planet's orbit

onto the sky

Answer ( a)

Sol. All planets undergo apparent retrograde motion hence

(b) and (c) are incorrect.

Projection of planet's orbit in the sky is cause of ecliptic

trace.

Retrograde motion is caused by epicycles. As a planet

moves around on its epicycle, the centre of epicycle

moves around the Earth in a deferent circle. When its

inside the deferent circle, the planet undergoes

retrograde motion.

23. A proton and a deuteron are projected into a uniform

magnetic field with velocities 50×106 ms–1 and 43.3 ×

106 ms–1 at angles 30° and 60° respectively with

respect to the direction of the magnetic field. Compare

the respective pitch of their helical paths.

(a) 1 : 2 (b) 2 : 1

(c) 1 : 1 (d) 4 : 1

Answer (c )

Sol.1 1 1 1 2

2 1 2 2 2

cos

cos

p m v q

p q m v

1 50 3 2

1 2 2 43.3

150 1

150 1

24. In a time of 4.606 days a radioactive sample loses

th4

5

of the amount present initially. The mean life of

the sample is

(a) 2.2 day (b) 2.86 day

(c) 2.46 day (d) 2.95 day

Answer ( b)

Sol. In 4.606 days, let the number of half lives be n

If t1/2

is half life.

then n × t1/2

= 4.606

and 1

5 =

1

2n

2n = 5

n log (2) = log (5)

n = 2.3219

Now, 2.3219 × t1/2

= 4.606

1/2

4.6061.9836 days

2.3219t

Mean Life = 1.44 t1/2

= 2.86 days

25. Assuming radius of the earth to be 6400 km, the

distance to the horizon visible from a 180 m tall building

is close to

(a) 56 km (b) 48 km

(c) 90 km (d) 64 km

Answer (b )

Sol. rr

R

Hh

S

Earth

Using Pythagoras theorem

(r + h)2 – (r2) = H2

putting r = 6400 × 103 m

and h = 180 m

H = 48000.337 m 48 km.

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NSEA 2019-20

26. Since none of the six trigonometric functions are

one-to-one, they are restricted in order to have their

unambiguous inverse functions. This range of the

inverse trigonometric functions is called the range of

the principal values. Range of principal value of

cot–1(x) is

(a) to2 2

(b) – to +

(c) 0 to (d) to 0 or 0 to2 2

Answer (c)

Sol. cot–1x : R (0, )

27. On which of the following planets would the Sun rise

in the west?

(a) Saturn (b) Mercury

(c) Venus (d) Jupiter

Answer (c)

Sol. Due to opposite spin of venus as compared to the

Earth, the Sun appears rise in the west on venus.

28. A prism of refracting angle 60° is made of a material of

refractive index 1.732. The angle of minimum deviation

produced by this prism is close to

(a) 30° (b) 38°

(c) 60° (d) 45°

Answer (c)

Sol.

sin2

sin2

mA

A

Given 1.732 3

60sin

23

60sin

2

m

60 3

sin2 2

m

60

602

m

60m

29. Two particles undergo simple harmonic motion given

by y1 = a sin 50 t and y

2 = a sin 40 t respectively.

Initially the particles are in phase. After what minimum

time will the particles be in phase again?

(a) 0.1 s (b) 0.2 s

(c) 0.25 s (d) 0.15 s

Answer (b)

Sol. Time periods of SHM(s) 1

2 1s

50 25T

and 2

2 1s

40 20T

The SHM(s) will be in phase again after a time,

1 2

1

1 1T

T T

1 1s

25 20 5

= 0.2 s

30. Bisectors of vertex angles A, B and C of a triangle

ABC intersect its circum-circle at the points D, E and

F respectively. The angle EDF is equal to

(a) 90° – 2

A(b) 45° +

2

A

(c) 90° – 2

B(d) A

Answer (a)

Sol.2 2

B CEDF

902

A

A

E

C

F

B

D

C/2 B/2

31. ABC is right angled at B. A square is constructed on

the side AC opposite to that of angle B. P is the centre

of this square. The angle PBC is equal to

(a) 30°

(b) 45°

(c) 60°

(d) None of the above

Answer (b)

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NSEA 2019-20

Sol. ∵ APCB is cyclic quadrilateral,

So, PBC = CAP = 45°

45°

45°P

A

B

C

32. A line through the three stars in Orion's belt appears

to point towards which one of the following star?

(a) Vega (b) Polaris

(c) Rigel (d) Sirius

Answer (d)

Sol. Line passing through the three stars in Orion's belt

appears to point towards 'Sirius'

33. In an experiment on photoelectric effect, the maximum

speed of photoelectron (vmax

) is measured for different

frequencies (f) of incident radiation. On a graph of vmax

versus f, the slope of the curve at any point gives

(a) Planck's constant (h)

(b) Planck's constant divided by electron charge (h/e)

(c) de-Broglie wavelength of the photoelectron

(d) Wavelength of incident radiation

Answer (c)

Sol. Photo electric equation

kmax

= hf –

2

max

1

2mv hf

2

max

2 2hfv

m m

Now, max

max

22 .

dv hv

df m

max

max max

2

.2

dv h h

df m v mv

=max

h

P

here slope of the curve at any point gives = photoelectron.

34. A dielectric slab of constant 0= 2 is inserted into a

parallel plate capacitor as shown in the figure. The

spring attached to the slab can set the slab into

oscillation of frequency . At equilibrium, end of the

slab is at the middle of the plates. If the slab is

pulled till the entire slab is inside the capacitor and

allowed to oscillate, the current flowing through the

circuit is

(a)0 sin( )2

b Vt

d

�

(b)0 cos ( )2

b Vt

d

�

(c)20 sin ( )

2

b Vt

d

�

(d)20 cos ( )

2

b Vt

d

�

Answer (a)

Sol.0 0

2( )

2 2

b bc x x x

d d

� �

Q = C V

dQ

dt�

x

l/2

At t = 0, x = cos( )2

t�

0 sin( )2

b Vt

d

��

35. If a, b, c are in G.P. and logc a, log

b c and log

ab are in

A.P., the common difference of the A.P. is

(a)1

2(b)

2

3

(c) 1 (d)3

2

Answer (d)

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NSEA 2019-20

Sol. Let log a = A, log b = B and log c = C

2

A CB

∵ and , ,

A C B

C B A are in A.P.

2, ,

2

A C A C

C A C A

are in A.P.

Let A

xC

2 1, ,1 2

xx

x x

are in A.P.

3 21 42 3 6 1 0

2 1

xx x x x

x x

2x2 + 5x – 1 = 0 3(x2 + x) = (x –1)2 ....(1)

Now common difference 1 2

2 1

xd

x x

2

2

( 1) 3

22( )

x

x x

36. ABCD is a line segment trisected by the points B and

C while P is any point on the circle whose diameter is

BC. If angles APB and CPD are and respectively,

the value of tan × tan is

(a)1

2(b)

1

4

(c)3

8(d)

3

2

Answer (b)

Sol. A DB O C 2r2r

P

90°–

90°– r r

Using cot m - n theorem in APO and in DPO

3cot = 2cot – cot

2cot = 4cot …(i)

and 3cot(90° – ) = 2cot – cot(90° – )

2cot = 4tan …(ii)

Multiplying (i) and (ii)

cot.cot = 4

1

tan tan4

37. A particle is projected with a velocity 20 ms–1, at angle

of 60° to the horizontal. Then radius of curvature of its

trajectory at a point where its velocity makes an angle

of 37° with the horizontal is close to (g = 10 ms–2)

(a) 16 m (b) 19.53 m

(c) 15.52 m (d) 25 m

Answer (b)

Sol.

37°P

v =

20 m

/s

60°

O x

v

vcos37° = vcos60°

4 120

5 2 v

50

m/s4

v

Radius of curvature at point P

2 2

cos37

N

v vr

a g

2500 5

16 10 4

= 19.53 m

38. A body is thrown up in a lift with a velocity u relative to

moving lift and its time of flight is t. The acceleration

with which the lift is moving will be

(a)2u

gt (b)

2ug

t

(c)2u

t(d) None

Answer (b)

Sol. Time of flight of body in lift = t

2 2 rel

rel

u ut a

a t...(i)

Now,rel l

a g a

Hence, 2 2

l l

u ug a a g

t t

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NSEA 2019-20

39. A particle of mass m, initially at rest, is acted upon by

a variable force F for a brief interval of time T. It begins

to move with a velocity u just after the force stops

acting. F, as a function of time, is shown in the graph.

The value of u is

F

O TTime

F0

(a)2

0

2

Fu

m

(b)2

8

Tu

m

(c)0

4

F Tu

m

(d)0

2

F Tu

m

Answer (c)

Sol. Area of F-t graph = change in momentum

010

2 2

F Tmu

0

4

F Tmu

0

4

F Tu

m

40. If the product of two roots of the equation

x4 –11x3 + kx2 + 269x –2001= 0 is –69, the value of k is

(a) 10 (b) 1

(c) –5 (d) –10

Answer (d)

Sol. x4 –11x3 + kx2 + 269x – 2001 = 0

Let roots of equation be a, b, c, d

then a + b + c + d = 11 ...(1)

ab + bc + cd + bd + ad + ac = k ...(2)

abc + abd + acd + bcd = –269 ...(3)

abcd = –2001 given let [ab = –69]

so [cd = 29]

from equation (3)

c + d = 6

In equation (2)

–69 + 29 + 6 (11 – 6) = k

k = –10

41. By a chord of the curve y = x3 we mean any line joining

two points on it. The number of chords which have

slope –1 is

(a) Infinite (b) 0

(c) 1 (d) 2

Answer (b)

Sol. Equation of any line

y x = 3

y x k + =

with slope = –1 is

x + y = k

This line can not intersect y = x3 is more than one

point.

Hence no chord with slope –1

42. The universe is estimated to be between ten and twenty

billion years old. This estimate is based on the value

of the constant(s)

(a) Speed of light (b) Fine structure constant

(c) Planck constant (d) Hubble constant

Answer (d)

Sol. Fact based.

43. Consider the charge configuration and a spherical

Gaussian surface as shown in the figure. While

calculating the flux of the electric field over any part of

the spherical surface the electric field will be due to

(a) q2

+q1

–q1

–q2

(b) Only the positive charges

(c) All the charges

(d) +q1 and –q

1

Answer (c)

Sol. Electric field at any point of the Gaussian surface is

due to all the charges. So, the flux of electric field over

any part of the surface will be due to all the charges.

44. The expression for electric potential at any point due

to an electric dipole p is (with usual notation)

(a)2

p rk

r

(b)

3

p rk

r

(c)2

.p rkr

(d)3

.p rkr

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NSEA 2019-20

Answer ( d)

Sol.3

.p rV k

r

� �

: Standard result

45. A function f(n) is defined by

–

–

4 – 4( )

4 4

n n

n nf n

for every

integer n.

If p and q are integers such that p > q, the sign of

f(p) – f(q) is

(a) Positive (b) Negative

(c) Indeterminate (d) Zero

Answer (a)

Sol. ∵

– 2

– 2 2

4 – 4 4 –1 2( ) 1–

4 4 4 1 4 1

n n n

n n n nf n

2 2

2 2( ) – ( ) 1– – 1–

4 1 4 1p qf p f q

= 2 2

2 2–

4 1 4 1q p

=

2 2

2 2

2(4 – 4 )

(4 1)(4 1)

p q

p q

Here, p > q = f(p) > f(q)

46. The first term of a descending A.P. series of 4 distinct

positive integers with greatest possible last term and

sum 2004 is

(a) 552 (b) 536

(c) 512 (d) 504

Answer (d)

Sol. Consider the four terms as:

a + 3d, a + d, a – d and a–3d

Sum = 4a = 2004 a = 501

If last term i.e., a – 3d is greatest then d should be

least.

As all terms are distinct integers then d = 1

First term = a + 3d = 504.

47. If the Earth is made to rotate in the opposite sense

(clockwise rather than counterclockwise), the length

of a solar day will be

(a) 23 hr 56 min (b) 24 hr

(c) 23 hr 52 min (d) 24 hr 4 min

Answer (c)

Sol. Right now, the solar day is about 4 minutes longer

than a sidereal day. If rotation of earth is reversed, it

would be 4 minutes shorter than a sidereal day.

48. In the case of a diatomic gas, the ratio of the heat

used in doing work for expansion of the gas to the

total heat given to it at constant pressure is

(a)2

5(b)

3

7

(c)2

7(d)

5

7

Answer (c)

Sol.5

,2

v

RC

7

2p v

RC C R

7 5

2 2

R RQ n T n T W W nR T

2

7

W

Q

49. Water coming out of the mouth of a tap and falling

vertically in streamline flow forms a tapering column,

i.e., the area of cross-section of the liquid column goes

on decreasing as it moves down, the most accurate

explanation for this is

(a) As the water moves down, its speed increases

and hence its pressure decreases. It is then

compressed by the atmosphere

(b) Falling water tries to reach a terminal velocity and

hence reduces the area of cross-section to

balance upward and downward forces

(c) The mass of water flowing past any cross-section

must remain constant. Also, water is almost

incompressible. Hence, the rate of volume flow

must remain constant.

(d) The surface tension causes the exposed surface

area of the liquid to decrease continuously

Answer (c)

Sol. vA = constant = Density

= speed

= (Area of cross-section)

v

A

vA = constant

As v increases, A decreases

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50. If x, y, z are positive integers with x + y + z =10, the

maximum value of xy + yz + zx + xyz is

(a) 69 (b) 59

(c) 64 (d) 61

Answer (a)

Sol. x, y, z N so the maximum value = 69

Occurs when x, y, z are 3, 3 and 4.

(xy + yz + zx + xyz)max

= 69

51. If a and b are positive real numbers and AB is a line

segment in a plane. The possible number of distinct

point C in the plane for which the triangle ABC will

have the lengths of medians and altitudes through C

as a and b respectively is

(a) 1 (b) 2

(c) 4 (d) Infinitely many

Answer (c)

Sol. AB is a given line segmant.

Draw two lines parallel to AB at distance b from it.

b

b

A BM

Now if a < b (No such triangle exist)

if a = b (only two triangles are possible)

if a > b (four triangles are possible)

52. The pair of planets, that is never visible at midnight is

(a) Mars and Neptune

(b) Venus and Neptune

(c) Neptune and Mercury

(d) Mercury and Venus

Answer (d)

Sol. Mercury and Venus are closer to the sun i.e., their

orbits are inside the Earth's orbit, Hence in midnight

they are opposite to sun hence both are never visible

at midnight.

Sun Mercury

Earth(Midnight)

Venus

53. A drop of water is broken into two droplets of equals

size. For this process, the correct statement is

(a) The sum of temperature of the two droplets

together is equal to the original temperature of

the drop

(b) The sum of masses of the two droplets is equal to

the original mass of the drop

(c) The sum of the radii of two droplets is equal to the

radius of the original drop

(d) The sum of the surface areas of the two droplets

is equal to the surface area of the original drop

Answer (b)

Sol. From conservation of mass, sum of masses of the

two droplets is equal to original mass of the drop.

54. A bullet of mass m moving horizontally with velocity v

strikes a wooden block of mass M suspended with a

string of length l. The bullet gets embedded into the

block as a result the block rises up to a height h. The

initial speed of wooden block and the embedded bullet

system is

(a) 2gh (b) 2M m

ghm

(c) 2m

ghM m

(d) 2M m

ghM

Answer (a)

Sol. After collision, let velocity of wooden block and bullet

system is v. Hence from mechanical energy

conservation after collision

h( + )M m

( + )M m

v

After Collision

l

reference

21

2 M m v M m gh

22 v gh

2 v gh

55. The number of diagonals in a regular polygon of 100

sides is

(a) 4950 (b) 4850

(c) 4750 (d) 4650

Answer (b)

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NSEA 2019-20

Sol. No. of sides = 100

No. of vertices = 100

No. of diagonals = 100C2 – 100

100 99–100

2

= 4850

56. Let a, b, c and p, q, r be all positive real numbers such

that a, b, c are in G.P. and ap = bq = cr. Then

(a) p, q, r are in G.P. (b) p, q, r are in A.P

(c) p, q, r are in H.P. (d) p2, q2, r2 are in G.P.

Answer (c)

Sol. b2 = ac

2lnb = ln a + ln c …(i)

given ap = bq = cr = (say)

ap =

log loge

p a

loglog

ea

p

Similarly log log

log , loge eb c

q r

from equation (i)

log log log2

q p r

2 1 1 q p r

p, q, r are in H.P.

57. If you look overhead at 6 p. m. (standard sunset time)

and notice that the moon is directly overhead, what

phase is it in?

(a) Last quarter

(b) First quarter

(c) Full moon

(d) 12th day from new moon

Answer (b)

Sol. In first quarter moon phase, at 12 noon moon rises in

east. At 3.00 pm moon is halfway up the sky between

eastern horizon. At 6.00 pm, moon is at the highest

point in the sky looking south. At 9.00 pm it is halfway

up the sky between western horizon and highest point

the moon can get looking south. At midnight (12.00)

moon sets.

58. The potential energy of a particle in a certain field has

the form 2

a bU

rr , where a and b are positive

constants and r is the distance from the centre of the

field. Find the value of r0 corresponding to the

equilibrium position of the particle. Is the equilibrium a

stable or unstable

(a)2a

b stable equilibrium

(b)a

b stable equilibrium

(c)2a

b unstable equilibrium

(d)a

b unstable equilibrium

Answer (a)

Sol. U = 2–

a b

rr

3 2

2 dU a b

dr r r

At equilibrium, 3 2

0 0

20 0 dU a b

dr r r

0

2 a

rb

2

2 4 3

6 2d U a b

dr r r

0

2 4

2 4 3 32

6 2

82 2ar r

b

d U a b b

dr aa a

b b

Since

2

2

d U

dr is positive i.e. potential energy is

minimum. Hence, equilibrium is stable.

59. The force acting on a particle is shown as a function

of the position in a one dimensional system. The

incorrect statement is

x (m)

2

–2

–1

F(N)

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NSEA 2019-20

(a) If total energy of the system is 1 J, motion is S.H.M

(b) If total energy of the system is 2J, motion is

periodic

(c) Work done on the particle as it moves from

x = 3 m to 2 m is 2J

(d) Work done on the particle as it moves from x = 3

m to 2 m is –2J

Answer (d)

Sol. From the graph, F = – kx for SHM

2 = –(k) (–1) k = 2 N/m

Now, 2 21 1

2 1 1J2 2kA E

If E = 2 J then F will not be directly proportional to x

beyond –1 < x < 1 i.e. motion is periodic but not SHM.

Now, when particle displaces from x = 3 m to x = 2 m

then ˆ ˆ(2 – 3) –s i i �

ˆ ˆ. (–2 ) (– ) 2 JW F s i i ��

Hence option (d) is incorrect.

60. Let Sk be the sum of an infinite G.P. whose first term

is k and common ratio is ( 0)1

kk

k

. Then the value

of 1

(–1)k

k

k S

is equal to

(a) loge4 (b) log

e2 – 1

(c) 1 – loge2 (d) 1 – log

e4

Answer (d*)

Sol. The question given appears to be mistyped we are

taking it as 1

(–1)k

k kS

( 1)

11

k

kS k k

k

k

Value of, 1 1

(–1) 1 1(–1) –

1

k

k

k kkS k k

1 1 1 1 1 1 1– 1– – – – – ...

2 2 3 3 4 4 5

1 1 1 1 1–1 2 – – ...

2 3 4 5 6

= –1 + 2 (1 – ln2) = 1 – ln 4

61. Which of the following is the closest to the distance in

kilometers of 30° extent of longitude on the surface of

the earth along the equator? How much is the

corresponding distance along a small circle (parallel

to the equator) at 60° N latitude?

(a) 3500 Km, 3000 Km (b) 3000 Km, 2000 Km

(c) 2500 Km, 1750 Km (d) 3300 Km, 1650 Km

Answer (d)

Sol. 1

302

360d R

= 3350 Km 3300 Km

d2 = 2R

1 ×

30

360 Km

60°

R

R1 = R cos 60°

d2 = 1675 Km 1650 Km

62. Any location on the surface of Earth as a sphere, is

determined by two coordinates-Latitude and Longitude.

Latitude of the North Pole is 90 degree. Its

corresponding longitude is .

(a) 0 degree

(b) 90 degree

(c) 180 degree

(d) None of these/indeterminate

Answer (d)

Sol. At 90° latitude, longitude circle converges to a point.

Corresponding longitude can have any value.

63. Two long wires AB and CD carrying current I1 and I

2

are perpendicular to each other as shown in the figure

(wires are insulated from each other)

A

DI2

I1

B

C

(a) Force on wire AB is towards left

(b) Force on wire AB is towards right

(c) Torque on wire AB is clockwise

(d) Torque on wire AB is anticlockwise

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Answer (d)

Sol. Force on AB will be shown by indicated arrows.

Torque on wire AB is anticlockwise

A

DI2

I1

B

C

X

64. A right circular cone of the height h and radius r is

suspended in liquid of density (). The density of cone

is 2 and its circular face is at a depth 3

h. The force

exerted by liquid on the curved surface of cone is

h

3

h

(a) Zero (b)

2

3

r h g

(c) r2hg (d)

22

3

r h g

Answer (d)

Sol.

2

2 13

RF F h g

= Net pressure force (Buoyant

force)

F2 : resultant force on curved surface

2

13

F R h g

F1

2

2

2

3 F R h g

65. A straight line through the point of intersection of the

line x + 2y = 4 and 2x + y = 4 meets the coordinates

axes at A and B. The locus of the midpoint of AB is

(a) 3(x + y) = 2xy (b) 2(x + y) = 3xy

(c) 2(x + y) = xy (d) (x + y) = 3xy

Answer (b)

Sol. B

A

M h, k ( )4 4,

3 3

Intersection point of x + 2y = 4 and 2x + y = 4 is

4 4,

3 3

Equation of line passing through 4 4,

3 3

whose slope

'm' is

4 4– –3 3

y m x

so coordinates of A are 4 4– , 0

3 3m

and –4 4

are 0,3 3

mB

Mid point of AB

2 2–

3 3h

m ...(i)

–2 2

3 3

mk ...(ii)

From (i) and (ii) locus of mid point of AB is

3xy = 2(x + y)

66. Let y be the solution of the differential equation

2

1– log

dy yxdx y x

satisfying y(1) = 1. Then y satisfies

(a) y = xy –1 (b) y = xy

(c) y = xy + 1 (d) y = xy + 2

Answer (b)

Sol.

2

.1– log

dy yxdx y x

2

1 1 log. –dx x

x dy yy

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2

1 1 1log

dxx

x dy y y

Let log x = t

1.dx dt

x dy dy

2

1 1dtt

dy y y

I.F = ln| |tdy ye e y

y.t = 2

1. lnydy y c

y

y (logx) = logy + c

for y(1) = 1 0c

y ln x = ln y

ln xy = ln y

yy x

67. The circle on the celestial sphere along which

declination of a star is measured is called

(a) Diurnal circle (b) Great circle

(c) Hour circle (d) Meridian

Answer (c)

Sol. The circle on the celestial sphere along which

declination of a star is measured is called hour circle.

68. A circular wire frame of radius R is dipped in a soap

solution of surface tension S. When it is taken out, a

thin soap film is formed inside the frame. If cross

sectional area of wire is A, then the stress developed

in the wire due to surface tension is

(a)SR

A

(b)

SR

A

(c)2SR

A(d)

2

SR

A

Answer (c)

Sol. Td = 2SR d2SRd

T = 2SR

Stress = 2SR

A

69. If two waves represented by y1 = 4 sint and

23sin

3y t

interfere at a point, the amplitude

of the resulting wave will be about

(a) 7 (b) 6

(c) 5 (d) 3.5

Answer (b)

Sol. A = 2 2

4 3 2 4 3 cos3

= 37 6

70. The value of

1

!lim

n

n

n

n is

(a) 1 (b)2

1

e

(c)1

2e(d)

1

e

Answer (d)

Sol.

1

Lim! n

n

n

n

1/

Lim !n

n n

n

n

1/

Lim ( 1)( 2)( 3)...........5.4.3.2.1

. . . .......... . . . .

n

n

n n n ny

n n n n n n n n n

Lim

1

1ln log

n

n

r

ry

n n

1

1

0

0

log .log 1xdx x x x 1 1

y ee

71. The value of the integral

5

1

– 3 1–x x dx is equal

to

(a) 4 (b) 8

(c) 12 (d) 16

Answer (c)

Sol. 5

1

| – 3 | |1– |x x dx

= 3 5

1 3

–1 3 – –1 – 3x x dx x x dx

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= 3 5

1 3

2 2 – 4dx x dx

= 25 – 20 – (9 – 12) + 2(2)

= 12

72. The definition of a nautical mile is

(a) One arc minute of latitude along any line of

longitude

(b) One arc minute of longitude along any line of

latitude

(c) One arc second of latitude along any line of

longitude

(d) One arc second of longitude along any line of

latitude

Answer (a)

Sol. A nautical mile is defined as one minute of latitude

along any line of longitude.

73. If the earth were to become a black hole, the minimum

radius to which it should be compressed is about

(mass remains constant, radius of earth = 6400 km)

(a) 1.8 m (b) 9.0 mm

(c) 3.6 cm (d) 1.8 km

Answer (b)

Sol. If 8

es3 10 m/sV then

0

es

2GMV

R

2 2

0 0 0

es 2

0

2 2 9.8

·

GM R RV

RR R

228 2 9.8 6400 1000

3 10R

R 8.9 × 10–3 m 9 mm

74. The in-centre of an equilateral triangle is (1, 1) and

the equation of one of the sides is 3x + 4y + 3 = 0.

Then the equation of the circum-circle of the triangle

is

(a) x2 + y2 – 2x – 2y – 2 = 0

(b) x2 + y2 – 2x – 2y – 14 = 0

(c) x2 + y2 – 2x – 2y + 2 = 0

(d) x2 + y2 – 2x – 2y + 14 = 0

Answer (b)

Sol.2 2

7 32

3 4

r OM

O

M

(1, 1)

= rs

23 32

4 2

aa

[where a is side of triangle]

4 3a

Now, 2sin

aR

A

4 3

23

2

R

R = 4

Circumcircle is (x – 1)2 + (y – 1)2 = 16

x2 + y2 – 2x – 2y – 14 = 0

75. How many odd numbers greater than 700000 can be

formed using the digits 4, 5, 6, 7, 8, 9, 0 repetitions

are allowed?

(a) 216 (b) 4920

(c) 9261 (d) 21609

Answer (d*)

Sol. If repetitions are allowed and odd numbers greater than

700000 is to be formed, then there are infinite number

of such numbers.

For 6 digit odd numbers greater than 700000, solution

is provided below.

N = (3) (3) (7)4 = 21609

76. The FALSE statement about planetary configurations

is,

(a) A superior planet can have an elongation between

0 and 180 degree

(b) Quadrature is when the elongation of a superior

planet is 90 degree

(c) Quadrature is when the elongation of an inferior

planet is 90 degree

(d) An inferior planet has zero elongation when in

conjunction

Answer (c)

Sol. Inferior planets can never be at quadrature.

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77. When an electron in a hydrogen like atom is excited

from a lower orbit to a higher orbit its

(a) Kinetic energy increases and potential energy

decreases.

(b) Both kinetic energy and potential energy

increases.

(c) Both kinetic energy and potential energy decrease

(d) Kinetic energy decreases and potential energy

increases

Answer (d)

Sol.0

n

v z

v

So, 2

2

0 2

1KE

2

zmv

And

2

0

2

2 EPE

z

So KE decreases and PE increases.

78. The number of divisors of 480 of the form 8n + 4(n 0),

where n is an integer is

(a) 3 (b) 4

(c) 8 (d) 10

Answer (b)

Sol. 480 = 4 × (23 × 31 × 51)

Number of divisors of form 8n + 4 = 4(2n + 1) is (2) (2)

= 4

79. The orbital period of Jupiter is 4333 mean solar days

and Jupiter’s mass is 1/1048 times the Sun’s mass.

The orbital period of a small body, of negligible mass,

moving in an elliptical orbit round the Sun with the

same major axis that of Jupiter is closest to

(a) 4329 days (b) 4333 days

(c) 4335 days (d) 4339 days

Answer (c)

Sol. mj M0

For Jupiter

3 22

0

0

1048 4

1049

R

TGM

…(i)

(M0 = Mass of Sun)

T0 = 4333 solar days

2 32

0

4 RT

GM…(ii)

2

2

0

1049

1048

T

T

2 2

0

1049

1048

T T

T = 4335 days

80. One Astronomical Unit (AU) is defined as the mean

distance between ths Sun and the Earth : 1 AU ~ 150

million kilometer. Even this unit is also small for

measuring large stellar distances which are quoted in

‘light year’ or ‘parsec’. A star one parsec away from

the Earth produces a parallax of one second of arc

when viewed from the Earth six months apart in its

orbit around the Sun, as shown in the accompanying

figure. There by one parsec is close to

Distant stars

Apparent parallaxmotion of near star

Parallax angle = 1arc second

Near star

1 P

ars

ec

1 AU

Earth's motion around Sun

p

(a) 205625 AU (b) 206265 AU

(c) 22620 AU (d) 226265 AU

Answer (b)

Sol. In figure is taken as parallex angle that is 1

rad180 60 60

And R

D

D

R

180 60 60

RD

∵ R = 1 AU

D 206265 AU

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