Date: 20/03/2022 Test Booklet Code

42

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Answers & Solutions

Time : 75 Minute Max. Marks : 120
for
International Olympiad Qualifiers
Astronomy (IOQA) 2021-22
(Part I)

INSTRUCTIONS TO CANDIDATES

(1) There are 32 objective type questions. Out of 32 questions, 24 questions in Part A-1 and 8 questions

in Part A-2. All questions are compulsory.

(2) In Part A-1 (Q. No. 1 to 24), each question has four alternatives out of which one is correct.

(3) In Part A-2 (Q. No. 25 to 32), each question has four alternatives out of which any number of

alternative(s) (1,2,3 or 4) may be correct. You have to choose all correct alternative(s).

(4) For Part A-1, each correct answer carries 3 marks whereas 1 mark will be deducted for each wrong

answer.

(5) For Part A-2, each correct answer carries 6 marks if all the correct answers are marked and no

incorrect. No negative marks in this part.

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 IOQA-2021-22

PART A-1
1. Globular clusters in our galaxy are primarily found
(a) in the spiral arms.
(b) distributed throughout the disk including regions between the spiral arms.
(c) in the bulge at the center of our galaxy.
(d) in the halo of our galaxy.
Answer (d)
Sol. Globular clusters in our galaxy are primarily found in the halo of our galaxy.
 1
2. Consider the function n =1 . What is the coefficient of x5 in its expansion near x = 0?
1− xn
(a) 7 (b) 5
(c) 3 (d) 10
Answer (a)
 1 1 1 1 1
Sol.  =   
(1 − x ) 1 − x 2 1 − x 3 (1 − x 4 )
....
r −1 1 − x
n

as x → 0 binomial expansion for any index can be applied

i.e.,
required coefficient of x5 in (1 – x)–1 (1 – x2)–1 (1 – x3)–1 ....
 (1 + x + x2 + x3 + x4 + x5 ...) (1 + x2 + x4 + ....) (1 + x3 + ....) (1 + x4 + ....) (1 + x5 + ...)
 1+1+2+1+1+1=7
 coefficient of x5 is 7.
3. If a and b are roots of x2 – 6x + p = 0 and c and d are roots of x 2 – 24x + q = 0, and if a, b, c, d are in geometric
progression then the value of the product pq is
(a) 192 (b) 64
(c) 24 (d) 1024
Answer (d)
Sol. Let b = ar, c = ar2, d = ar3
given
a+b=6 ... (1), c + d = 24 ...(2)
ab = p = a2r and cd = q = a2r5
by (1) and (2)
a(1 + r) = 6 and ar2 (1 + r) = 24
 r2 = 4  r=±2
if r = 2 if r = –2
a=2 a = –6
 b = 4, c = 8, d = 16 b = 12, c = –24, d = 48
 pq = abcd
= 1024

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 IOQA-2021-22

4. A student measures the displacement x from the equilibrium of a stretched spring and reports it to be 100 m
with a 1% error. The spring constant K is known to be 500 Nm–1 with 0.5% error. The percentage of error in the
1
estimate of the potential energy V = Kx 2 is
2
(a) 0.8% (b) 2.5%
(c) 1.5% (d) 3.0%
Answer (b)
1 2
Sol. V = Kx
2
V K x
  100 =  100 + 2  100
V K x
= (0.5) + 2 × 1
= 2.5
5. Two satellites are in the same geosynchronous orbit (assumed to be circular), but in diametrically opposite
positions. One satellite descends into a lower circular orbit and catches up with the other after 8 complete orbits.
Neglect the time of descent into lower orbit. If the radius of the geosynchronous orbit is 40000 km, the radius of
the lower, faster orbit is about (using 7.52/3  3.84 )
(a)  32400 km (b)  34000 km
(c)  36000 km (d)  38400 km
Answer (d)
Sol. T2  R3
2 3
T  R 
 1 = 1
 T2   R2 
2/3
T 
 R2 = (R1 )  2 
 T1 
( 40,000 ) ( 7.5 )2/3
= = 38400 km
( 8 )2/3
6. The ratio of the masses of the Earth and Mars is 10 and the ratio of the radii of the Earth and Mars is 2. If two
persons jump with the same velocity and angle of the surface of each of the planets, the ratio of maximum height
reached at Earth to that reached at Mars is
1 2
(a) (b)
5 5
5
(c) (d) depends on the ratio of the masses of two people
2
Answer (b)
Sol. As it is person who is making the jump so the trajectory can be considered as a simple projectile with constant
acceleration due to gravity.
He u 2 sin2  2g
=  2 m2
Hm 2g e u sin 

 R2   Mm 
=  e   2 
 Me   Rm 
 4 2
= =
 10  5

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 IOQA-2021-22

7. Kepler's laws for planetary orbits are derived using Newton's law of gravitational force.

These laws state that

First: All planets move in elliptical orbit with sun at one of the foci,

Second: The line joining the planet to the sun sweeps equal area in equal time interval,

Third: Square of the period of revolution of a planet is proportional to cube of semi-major axis. Newton's law of
gravitation receives corrections from relativity theory, with the modified law of force between bodies of masses
M and m at a distance r given by

GMm  A ˆ
F =− 2 
1+ 2 r
r  r 

where A is some constant. With this correct form of law of gravitation, for which of the Kepler's laws you can be
sure that it (these) will remain unchanged.

(a) First and Second (b) First

(c) Second (d) All laws will change

Answer (c)

Sol. According to Kepler's Second law

dA L
Areal velocity =
dt 2m

As even after correction the force formula remains of a central force that means the net angular momentum is
still conserved about any point connecting two masses.

 L = constant

dA
 = constant
dt

 The second law will hold good.

8. A schematic Hertzsprung-Russel diagram for stars in the solar neighbourhood is shown below.

L
The radius R of a star, its luminosity L and surface temperature T are related as R  .
T2

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 IOQA-2021-22

R
If for a star = 20 and T = 3000 K (using Tsun = 6000 K), then the star is a
Rsun

(a) Blue Supergiant

(b) Blue Giant

(c) Red Supergiant

(d) Red Giant

Answer (d)

Sol. L = kR2T4
Where k is some constant
2 4
Lstar  Rstar   Tstar 
=   
Lsun  Rsun   Tsun 

Lstar 1
= 400  = 25
Lsun 16
 the star is a Red Giant
9. Statement I: We cannot see what is near the centre of Galaxy.

Statement II: There is a super-massive black hole at the centre of Galaxy.

(a) If statement I is true and statement II is true and also if the statement II is a correct explanation of
statement I

(b) If statement I is true and statement II is true but the statement II is not a correct explanation of statement I

(c) If statement I is true but the statement II is false

(d) If statement I is false but statement II is true

Answer (b)

Sol. Statement I is true as we cannot see into the centre of galaxy as the region is surrounded with huge dust and
gas clouds.
Statement II is true as there is a super-massive Blackhole at the centre of galaxy but is not the correct explanation
of statement I.
10. Statement I: Hydrogen gas is not found in large amount in the atmosphere of terrestrial planets.

Statement II: Speed of Hydrogen molecules was higher than the escape velocity on the terrestrial planets.

(a) If statement I is true and statement II is true and also if the statement II is a correct explanation of
statement I

(b) If statement I is true and statement II is true but the statement II is not a correct explanation of statement I

(c) If statement I is true but the statement II is false

(d) If statement I is false but statement II is true

Answer (a)

Sol. Statement-I is true that hydrogen gas is not found in abundance in the atmosphere of terrestrial planets as is the
case for jovian planet and statement-II is correct explanation for the observation in statement-I as lighter atoms
escaped from these planets due to their higher speeds.

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 IOQA-2021-22

11. Two balls of masses m1 and m2 respectively (m1 > m2) are projected towards each other from initial separation
d at t = 0 such that their motion is in the same plane. Initial velocity of each ball is v facing towards each other at
an angle 45° from the horizontal. The two balls undergo completely inelastic collision, and fuse to form a single
body of mass m1 + m2. The total time t (counting from t = 0) after which the fused body will fall to the ground
(neglecting air resistance) is

v 2 m1v 2
(a) (b)
g g ( m1 + m2 )

(c)
v
(d)
(m1 + m2 ) v
g 2 m1g 2

Answer (a)
Sol.

Both the masses have same vertical component of velocity of centre of mass at initial moment
m1vmv
+ 2
 v cm = 2 2 = v
y
m1 + m2 2
As both the bodies fuse together they will move with the velocity of centre of mass
2v
 time taken = 2 =v 2
g g

dy
12. The solution for + y = Ae− x for y(1) = 5 is (here A is constant)
dx

1 −x
(a)
5
e (b) 5e + A(x − 1) e−x

(c) 5e − A(x − 1) e−x (d) 5e e−x

Answer (b)

dy
Sol. + y = Ae − x (Linear D.E.)
dx

I.F. = e 
1 dx
= ex

  d ( ye ) =  Adx
x

 yex = Ax + C

 y (1) = 5

 5e = A + C  C = 5e – A

 Solution of D. E yex = Ax + 5e – A

 y = [A(x – 1) + 5e] e–x

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 IOQA-2021-22

13. A 1-inch telescope is pointed towards Sirius (m = – 1.4) and another bigger one towards Castor (m = 1.6), m
being the apparent magnitude, but both the telescopes deliver equal amount of energy per second to the CCD
detector at the eyepiece. The objective diameter of the bigger telescope is (using log 10 2 = 0.3).

(a) 2 - inch (b) 4 - inch

(c) 8 - inch (d) 16 - inch

Answer (b)

Sol. We know apparent magnitude,

M = 5 log D + 2, where

D is the diameter of telescope

– 1.4 = 5 log D1 + 2 (i)

For castor

1.6 = 5 log D2 + 2 (ii)

Equation (ii) – equation (i)

D 
3 = 5 log  2 
 D1 

D2
 0.6 = log
D1

D2
 log4 = log  D2 = 4D1 = 4 inch
D1

14. Consider a circle circumscribing an equilateral triangle of side L. The ratio of the area of the circle to that of the
equilateral triangle is

 4
(a) (b)
2 3 3 3

 6
(c) (d)
3 3

Answer (b)

3 2
Sol. Area of triangle = L =  (say)
4
L L  L L
Radius of circumcircle = R = 2
=
3L 3
4.
4
L2
Area of circle = R 2 =
3

L2 4 4
 Ratio = 2
=
3 3L 3 3

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 IOQA-2021-22

15. A light ray having frequency f and wavelength  enters from air into water. After entering into water, its

(a) f remains unchanged,  decreases

(b) f increases and  decreases

(c) f decreases,  remains unchanged

(d) f and  both decrease

Answer (a)

Sol. When light rays travel from one medium to another medium its frequency does not change.
 air
As w = ,  is refractive index of water

as  > 1
 w < air
16. A lens of diameter d and focal length f is used to project image of an object on a screen. The object is kept at a
distance u from the lens and consists of two points separated by distance r0 in the plane perpendicular to the
principal axis. The wavelength of light used is . The two points will be resolved in the image if

1.22 r0 1.22 f

(a) u  (b) d 
d r0

r0 d 1.22 ur0
(c) u  (d) d 
1.22  f

Answer (c)

Sol. We know

2sin 
Resolving power of lens =
1.22 

 is small

d
 sin  
2u

d
 R.P.=
1.22 u

1 d
 
r0 1.22 u

dr0
 u
1.22 

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 IOQA-2021-22

17. Assume that the density () of the earth has following dependence on the distance r from the centre

(r) = 0 for r < r0,

1
(r )  for r0  r  R,
r

R being earth’s radius and 0 is constant density of the central core. For any point r between r0 and R, the
gravitational acceleration g(r) will have the following form (A and B being constants in the following)

B
(a) A + (b) Ar
r2

A
(c) + Br (d) A + Br
r

Answer (a)

Sol. Given  = 0 for r < r0

1
 for r0  r  R
r

4 3
m1 = r 
3 0 0

For r0  r  R

0 r0
=
r

r 0 r0 r2 −r2 
 m2 = r0 4r 2dr = 40 r0  
0
r  2 

 1 1
 m1 + m2 = 40 r03  −  + 20 r0 r 2
3 2

G(m1 + m2 ) 40 r03  1 

 g (r ) = =  −  + 20 r0
r2 r2  6

B
= A+
r2

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 IOQA-2021-22

2 K
18. Given f(x) is a continuous function such that for x  1, f(x) = 5e−( x −1) and for x  1, f ( x ) = .
( x + 4)2

The value of K is

1
(a) (b) 5
5

(c) 25 (d) 125

Answer (d)

5e −( x −1)2 ; x  1

Sol. f ( x ) =  K
 ; x 1
 ( x + 4)
2

For continuity LHL = RHL i.e.,

2
LHL = lim f ( x ) = 5e −(1−1) = 5
x →1−

K
and RHL = lim f ( x ) =
x →1+ (5)2
K
 = 5  K = 125
25
19. The geometric albedo (A) of the solar system objects is related to their absolute magnitude (M) and diameter (D)
2
 1329  10−M /5 
as A =   . Initially it was thought that the radius of Phobos, one of the satellites of Mars, is 7 km
 D 
assuming the albedo of Phobos is same as that of Mars i.e., about 0.15. But later Mariner spacecraft’s
photographs revised the radius to 10 km. The correct albedo of Phobos, therefore, is

(a) 0.10

(b) 0.05

(c) 0.07

(d) 0.12

Answer (c)

Sol. AD2 = constant as absolute magnitude does not changes so

0.15 × 142 = A × 202
 A = 0.0735
20. The area in the region 1  x  2 between the line y = 2x + 10 and the curve y = 3x2 is

(x and y are in cm and area is in cm2)

(a) 6

(b) 5

(c) 7

(d) 8

Answer (a)

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 IOQA-2021-22

Sol.

2
 A =  ((2x + 10) − 3 x 2 )dx
1
2
= x 2 + 10x − x 3 1 = (4 + 20 − 8) − (1 + 10 − 1)
= 16 – 10 = 6
21. The summer triangle refers to the three stars
(a) Vega, Altair and Deneb (b) Regulus, Antares and Sirius
(c) Sirius, Procyon and Betelgeuse (d) Pollux, Caster and Regulus
Answer (a)
Sol. Unlike many asterisms, the Summer Triangle is actually an amalgamation of stars from three separate
constellations. Three stars make up the triangle. Deneb, Vega, and Altair.
22. The three points (1, 0, 5), (2, 3, 1) and (4, 9, r) are collinear. The value of r is
(a) 12 (b) –14
(c) –7 (d) 0
Answer (c)
Sol. Points are P(1, 0, 5), Q(2, 3, 1) and R(4, 9, r)

For collinearity coefficient of i , j and k in PQ and PR shall be in same ratio.

 PQ = i + 3j − 4k

and PR = 3i + 9j + (r − 5)k

1 3 −4
 = =
3 9 r −5

 r – 5 = –12
 r = –7
23. Astronomers have discovered 200 stars having parallax of 0.10  with fair completeness. Supposing the
distribution of stars around us is pretty much homogeneous and isotropic, the number of stars having parallaxes.
0.025 or more is
(a) 3200 (b) 25600
(c) 12800 (d) 6400
Answer (a)
Sol. According to the data we are viewing stars at 4 times more distance which will consist of 16 times more stars so
the number of stars having parallaxes 0.025 = 16 × 200 = 3200 stars.

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 IOQA-2021-22

24. The value of (x, y) for which z = 3x + 4y + 10 is a maximum under the constraints x + y  40, 2x + 3y  90,
x, y  0 is

(a) (0, 30) (b) (0, 40)

(c) (40, 0) (d) (30, 10)

Answer (d)

Sol.

(x, y) z

(0, 0) 10

(40, 0) 130

(0, 30) 130

(30, 10) 140

 zmax at (30, 10)

PART A-2

25. Analysis of the spiral galaxy NGC 1357 spectra reveals a strong emission line at 6606 Å. Knowing the H
emission line is at 6560 Å.
Use c = 3 × 105 km/s and Hubble’s constant = 70 (km/s)/Mpc. Choose correct option(s)
(a) The velocity of NGC 1357 is about 2 × 103 km/s.

(b) NGC 1357 is an example of blue-shifted galaxy.

(c) The galaxy is at a distance of 30 Mpc from us.

(d) It is an old galaxy having little to no star formation regions.

Answer (a, c, d*)

Sol. Apparent wavelength is more than the wavelength of emitted light so, it is red shifted
 v
=
 c
46 v
 =
6560 c
 46 
 v = 3  105   3
 = 2.10 × 10 km/s
 6560 
v = H0D
2.10  103
 D= = 30 Mpc
70

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 IOQA-2021-22

26. The Hubble plot below depicts two alternate universe A (solid line) and B (dashed curve).

The quantity z determines the amount of redshift (negative z for blue shift) in the absorption and emission spectra
of the galaxies. From the Hubble plot above, we can deduce that
(a) Both the universe A and B are contracting but at different rates.
(b) The galaxies in the universe A are blue shifted while those in B are red shifted.
(c) In the universe A, farther galaxies are approaching us at faster velocities than closer galaxies.
(d) The z – value is larger for the farther galaxies in B compared to those closer.

Answer (b, c, d)
Sol. A is contracting while B is expanding. The galaxy in A are blue shifted whereas B is red shifted. Speed increases
with increase in distance for A, so option (c) is also correct. Also option (d) is correct as speed is more for further
galaxy so is the blue shift.
27. Electromagnetic waves also undergo Doppler effect just as sound waves. Use the expression of Doppler effect
valid for small velocities (of source, or observer, they both give the same result for small velocities), replacing
sound speed by speed of light c. A hydrogen atom moving along x axis with velocity v undergoes transition of
electron from 1st excited level (n = 2) to the ground state, emitting radiation which travels along x axis. This
radiation is absorbed by another hydrogen atom at rest in its ground state causing it to get excited to n = 3 level.
The value of v is approximately.
(a) 5.55 × 104 km/sec (b) 9.45 × 104 km/sec
(c) 0.315 c (d) 0.185 c
Answer (a, d)
 1 3Rc
Sol. Frequency of emitted photon, v 0 = Rc 1 –  =
 4 4

vs
1+
v = v0 c
vs
1–
c

vs
1+
3Rc c = 8 Rc
=
4 vs 9
1–
c
 For vs << c
3Rc  v s  8Rc
= 1 + =
4  c  9

 v  32
 1 + s  =
 c  27
5
 vs = c
27

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 IOQA-2021-22

28. A metal rod moving through a magnetic field may get induced e.m.f. (depending on the direction of the magnetic
field and the orientation of the rod) due to the fact that
(a) Current flowing through the rod leads to a force on the rod due to magnetic field
(b) Magnetic field applies force on electrons in the rod
(c) Changing magnetic field produces electric field
(d) Electrons have a magnetic dipole moment which feels force due to magnetic field.
Answer (b)
Sol. Free electrons moving with velocity in magnetic field experience force in magnetic field leading to polarisation of
charge at ends which leads to induced emf.

29. If sin   = 1 − 2sin2  then  can be
3

 13 
(a) (b)
12 12

11 25
(c) − (d)
12 12

Answer (a, b, c, d)

3
Sol. = 1 − 2sin2  = cos 2
2

 cos2 = cos
6


 2 = 2n 
6


  = n  , nZ
12
 13  11  25
i.e., All , , − and are acceptable.
12 12 12 12

a b
30. Let the 2 × 2 matrix A =   , where a, b, c, d  R, with A = 0. Then,
2
 c d 

(a) A = 0 (b) a + d = 0 and det A = 0

(c) A–1 does not exist (d) a + d = 0 and b, c can be arbitrary
Answer (b, c)

a b  a b 
Sol. A 2 =   
c d c d

a2 + bc ab + bd 0 0 
=  2
= 
ac + cd bc + d  0 0 
i.e. a2 + bc = 0, b(a + d) = 0, c(a + d) = 0 and bc + d2 = 0
 a + d = 0, |A| = 0 and A–1 does not exist.

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 IOQA-2021-22

31. A light beam passes through a transparent medium of thickness L. After passing through the medium, the
intensity of the light beam reduces because

(a) Each photon loses some energy in the medium

(b) Some photons get absorbed in the medium

(c) Some photons get scattered by the medium in different directions

(d) Some photons get reflected back as the light enters the transparent medium

Answer (b, c, d)

Sol. When light passes through medium every possibility occurs except possibility depicted in option (a). Photons
may get absorbed, reflected back and scattered.

32. Consider a pyramid with square base of length L and triangular faces as equilateral triangle. For this pyramid,
the true statement(s) is/are

L
(a) The height of the pyramid is
2

(b) The area of the pyramid surface (including the base) is ( 3 + 1)L2 .

(c) The volume of the pyramid is L3.

(d) The angle between the base and the side is tan−1 2 .

Answer (a, b, d)

Sol.

L 2 L
OA = =
2 2

L2 L2
OE 2 = H 2 = AE 2 − OA2 = L2 − =
2 2

L
H=
2

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 IOQA-2021-22

 3 2 2
Surface area = 4  L +L
 4 

= ( 3 + 1)L2

1
Volume =  base area × height
3

1 2 L L3
= L  =
3 2 3 2

OE L / 2
tan  = = = 2
OL L/2

❑ ❑ ❑

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