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ENTHUSIAST 2023
& LEADER COURSE
ENTHUSIAST & LEADER COURSE

PHYSICS GR # GRAVITATION

SOLUTION
SECTION-I
Single Correct Answer Type 8 Q. [3 M (–1)]
1. Ans. (A)

v0
Sol.
P1
R 3R v

Let speed at max height is v

Since, it is at max height, v is perpendicular to r
 By conservation of angular momentum
mv0R = mv (4R)
v0
v
4
 By conservation of mechanical energy
2
1 GMm 1  v 0  GMm
mv 20   m  
2 R 2  4  4R

8 GM
v0 
5 R

GM 8
g 2
,v 0  gR
R 5
2 Ans. (D)

P
r y

(R/2)
Sol.

Gravity at point P   gr 
R
Fcos N


mgr Fsin
F=
R
Fcos – N = 0
Fsin = ma

Physics / GR # Gravitation E-1/11

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mgr  y 
   ma
R r
2
g g R
a =  y = r2   
R R 2
a  f(r)
3. Ans. (C)
GM
Sol. ve = 2 r0

GM
v  v e  v 0   2 1 r0
4. Ans. (D)
Sol. Total mechanical energy = 0
GM 1
2 m  mv 2e  0
a 2
GM
ve = 2
a
5. Ans. (B)
Sol. Using conservation of angular momentum about O. vA
mVPrP = mVArA = mV0r0cos
3v 0 r0 A
VArA = VPrP =
5
using conservation of energy
rA
1  GMm  GMm 1
mVA2    mV02 v0 v0
2 rA r0 2 

9V02 r02 V02 r0 V02  r0  O

 50r 2   0  Let r  x  r0
rA 2 rP
A  A  P
 9x2 – 50x + 25 = 0 x = 5 or (5/9) vP

9r0 r0
 rA = ; rP =
5 5
VP rA
 V  r 9
A P

6. Ans. (D)
GM  2  3/ 2 GMm
Sol. Speed = , Time period =  GM  r , Total energy = –
r 2r
7. Ans. (A)
dA v0 r
Sol. Areal velocity = 
dt 2

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GM
& V0  , V0 = orbital speed
r
r = radius of orbit
8. Ans. (B)
M
V

2R 45° F2
F2 F1 V
R
Sol. M M

V
M

Net force on one particle

Fnet = F1 + 2F2 cos 45° = Centripetal force
GM 2  2GM2  MV2
   cos 45 
(2R)2  ( 2R)2  R

1 GM
V= (1  2 2)
2 R
Multiple Correct Answer Type 7 Q. [4 M (–1)]
9. Ans. (A, B, C)
Sol. For (A) : By COME total mechanical energy of the two objects
For (B) : By using reduced mass concept
1 2 G(4m)(m) (m)(4m) 4 10 Gm
µv rel – = 0 where µ = = m vrel =
2 r m  4m 5 r

G(m) (4m) 4Gm 2

For (C) total KE = –PE = =
r r
10. Ans. (A,C, D)
Extreme (v = 0)
R

Sol.
R
O Extreme (v = 0)

Let r is distance from center,

GMm
If r > R F (Not SHM)
r2
GMmr
r<R F (SHM)
R3
Physics / GR # Gravitation E-3/11
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 But motion is periodic to find velocity at center,

Apply COM E,
1 GMm 1 3 GMm
2
 
m 02 
2R
 mvC2 
2 2 R

2GM
vC 
R
11. Ans. (A,C)
Sol. VCM  0 , Mv1  2 Mv2  0

1 1 GM  2 M  1 2GM
Mv12  2 Mv22   0  relative velocity at t = T is
2 2 3R 3 R
12. Ans. (A,B,C,D)
13. Ans. (A, D)
Sol.  in both is same
so T1 = T2
and by vmax = A V1 > V2
14. Ans. (A,B,C,D)
GM
Sol. V 
r

VA r
 B 2
VB rA

GMm EA m A rB
E ,    12
2r EB mB rA

2GM uA
uA  , 2
r uB

2
T (r)3 / 2
GM
3/ 2
TA  rA  1
  
TB  rB  8
15. Ans. (B,C)
Sol. Initially parabolic path means escape velocity & hence (TE = 0)

Perigee (Perihelion)

r
a earth
C

apogee (Aphelion)

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a
rmin   a  c  
2
3a
rmax   a  c  
2
c 1 a
e  c
a 2 2
R 3R
rRh R 
2 2
3R
If rmin   a  3R
2
3R
and if rmax  a R
2
a 3R
If a = 3R, then, C  
2 2
3R
rmin   a  c    r1
2
9R
and rmax   a  c    r2
2
L = constant
mv1r1 = mv2r2

 3R   9R 
v1    v2  
 2   2  v1
v1 = 3v2
TE = conserved rmin = r1

1 GmM 1 GmM
mv12   mv 22 
2 r1 2 r2
rmax = r2
2
 v1 
v 2
GM   GM
3
1
    v2
2  3R  2  9R 
 2   2 
   

v12  1  GM  2 2 
2 1  9   R  3  9 
   

 4  GM  4 
v12   
9 R  9 
GM
v12 
R
1 GMm
TE  mv12 
2 r1

Physics / GR # Gravitation E-5/11

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1 GM GMm
 m 
2 R  3R 
 2 
 

 TE  GmM  1  2     GmM  1  
 finally  R 2 3  R  6 

GmM
So, change in TE required =
6R
Similarly solving for (a = R)
then we will get
GmM
TEfinal  
2R
GmM
So, change in TE 
2R
Alternate Method
For perihilion
3R
a(1 – e) = R  h 
2
 a = 3R
Also, for elliptical path
GMm GMm
T.E.   
2a 6R
GMm
 change in TE 
6R
For aphelion :
3R
a 1  e   R  h 
2
a=R
GMm GMm
& T.E. =  
2a 2R
GMm
 change in TE =
2R
Linked Comprehension Type (1 Para × 2Q.) [3 M (-1)]
(Single Correct Answer Type)
16. Ans. (B)
Sol. According to Keplar's Law
dA
= constant
dt
A1 t1
At 
A2 t2

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D A1 C
A2

B
A1 > A2
 t1 > t2
17. Ans. (C)
Sol. For a bounded system, TE < 0
 |U| > |KE|
Matching List Type (4 × 4) 1Q.[3 M (–1)]
18. Ans. (B)

m1 
 2
m2 
M  given
Sol. R1 1 

R1 R 2 4 
R1 m1
m2

GMm 1 m 1v12

R12 R1

GM GM
v12  , v 22 
R1 R2

v12 R 2
 4
v 22 R1

v1
(P) v  2
2

(Q) L = mvR
L1 m 1v1R1 1
  22  1
L2 m 2v 2 R 2 4

1
(R) K  mv 2
2
K1 m 1v12
 = 2 × (2)2 = 8
K 2 m 2v 22

Physics / GR # Gravitation E-7/11

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(S) T = 2R/V
T1 R1 v 2 R1 v 2 1 1 1
      
T2 v1 R 2 R 2 v1 4 2 8
SECTION-III
Numerical Grid Type (Ranging from 0 to 9) 2 Q. [4 M (0)]
19. Ans. 6
L total m1r12
Sol.  1
LB m2 r22
20. Ans. 7
Sol. Due to gravitational interaction connected masses have some acceleration.
Let both small masses are moving with acceleration 'a' towards larger mass M
a
M
m m

3 
Force eq. for mass nearer to larger mass
GMm Gm 2
  ma ... (i)
 3  2 2
Force eq. for mass away from larger mass
GMm Gm 2
  ma ... (ii)
 4 2 2
from equation (i) & (ii)
GM Gm GM Gm
2
 2
 2

9  16 2
M M
  m m
9 16
7M
 2m
144
7M  M 
m  k 
288  288 
K=7
SECTION-IV
Matrix Match Type (4 × 5) 1 Q. [8 M (for each entry +2(0)]
21. Ans. (A)-R, (B)-Q, (C)-Q (D)-P

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Perigee
vmax
rp

Sun

Sol. rA

vmin
Apogee

1
(A) v 
r
(B) rp < rA
1
(C) PE  
r
(D) L = constant
Subjective Type 5 Q. [4 M (0)]
3 G mM
22. Ans. A = –
2 R

M
R

C
Sol.

Wg = Ug
 m  Vg   m  Vg f  Vg i   m  V  VC 

  3GM   3GmM
Wg  m  0     
  2R   2R
2d 3 / 2
23. Ans. (a) T , (b) 2, (c) 2
3Gm
Sol. (a) The COM of the system divides the distance between the stars in the inverse ratio of thin masses. If
d1 & d2 are the distances of M & m from the COM. (COM = Centre of mass)
d d
d1   m & d2  M
Mm Mm
The stars will rotate in circles of radii d1 & d2 about their COM. The same force of attraction provides the
necessary contripetal force for their circular motion.
GmM
 = M12 d1  m22 d 2
d2

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2 Gm Gm M  m G M  m 
or 1  d 2 d = 2  =
1 d dm d3
G M  GM  M  m  G  M  m 
and 22 = 2 = 2   =
d d2 d  dM  d3

G M  m 
 1 = 2 =
d3
(b) From the fact that the moment of momentum is also angular momentum.
I M  Mv1  d1 M d12 M m2 m
  
I m  Mv 2  d 2 = m d 22 = m M 2 M

1 2
K M 2 Mv1 M 2 2
d M d12
(c)  =  1 1 =
K m 1 Mv 2 m 22 d 22 m d 22
2
2

KM M m 2 m
   
K m m M2 M
1/ 2
  2 1 
24. Ans. 2G (m1  m 2 )  
  a l 

25. Ans. (a) 2R, (b) 3 3  3 1 R

3M12R (M,R)

Sol. m

(a) speed will maximum at projection point

(b) To just reach the other planet small body must be able to reach the point where gravitation field due
to both planet is zero. At this point speed will be minimum and equal to zero.
G  3M GM
2
 2
 3  6R  x   x
x  6R  x 

 1  3 x  6 3R 
x
6 3  
3 1 R
2
 x2  R2 
26. Ans. 1   4R 2
 x 
 

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region near poles where

signal can not reach


R
Sol. ×
S

region near poles where

signal can not reach

R
sin  
x

R2
cos   1 
x2
Area required = 2(2 R2 (1 – cos ))
 x2  R2 
 4 R 2 1  
 x 
 

Physics / GR # Gravitation E-11/11