16 IN-CHAPTER EXERCISES

SOLUTIONS
Module - 2 / JEE-2023
IN-CHAPTER EXERCISES Physics Gravitation

EXERCISE-A
g
1. At equator: mg – N = mw2R  To fly off, N = 0  w
R
GM s GM e x
2. 2
 2 and x + y = 1.495 × 108 km and  3.24  105  Solve for x and y
x y y

3.(D) Equal and opposite forces

 Net zero force

GMm
4.(B) FNET  F '  2
 3 
 
 2 

GM M
5.(B) g  2
for any planet i.e. g 
R R2
g1 M1 R22 m  x3  y 2 mx
  
g 2 M 2 R12 n  y3  x2 ny

GM GM  R  x 
6.(B) g1  2 g2  g1 = g 2
 R  h R3
2
 h  x
 1    1    x = 2h (Binomial theorem)
 R   R 
2 2
 R  g  R 
g
7.(A) g  g   acceleration due to gravity at a height h above surface 100 
 Rh   Rh

R 1
 ; 10R = R + h ; h = 9R
R  h 10

8.(C) If T = constant

 1  2 
  constant   2 
g g1 g 2 4

Physics [JEE-2023/Module - 2]
 IN-CHAPTER EXERCISES 17

9.(A) Verify using law of conservation of energy. As it moves out of earths gravity, its energy is clearly zero.

1 6002
10. (A) S1  0  60 2  10  18000 m ; v = 0 + 10  60 = 600 m/s ; S2   18000 m  S1  S2  Smax  40 km
2 2g

EXERCISE-B
2
4 2
1. T2   r3  T  (3 Re / 2)3 / 2
GM gRe2
GMm 1 1 gR 2  m 1 1 GMm
2. EReq.  | E f  Ei |     
2 rf ri 2 2 R 3R 12 R

GMm 
3. Esatellite at earth's surface 
ri  GMm  1 2  ri  6400 km  2
E      r  7200 km  and GM  gR
1 GMm  (Req.) 2  rf ri 
Esatellite in orbit     f 
2 r f 
GM m  6400  5 GMm
 E   2 
2R  7200  9 R
GMm  GMm  GMm  1 1 
4. Loss in Energy     Ct  t   
2r  2R  2C  R r 

T2 1 1
5.(B) T 2  r3 ; 2
 ; T yrs .
1 64 8
6.(B) Using kepler law of periods,
2 3
 T1  r 
1
2
T r 3 ;     
 T2  r
 2

2
 2(4 R) 
  3
 4R  v 2
 3v    
 2( R)   R  ;  16  64 ; v   6v
  9v 2
 v1 

GMm
7.(B) Potential energy  ; as r increases potential energy increases
r

 n 1  n1
GM s m p m pV 2 GM s 2 R
8.(B) F   V  and T   t  R R 2   R 2
Rn R R n 1 V 




GMm GMm
9.(C) Total energy   E0 ; Potential energy   = 2E0
2r r

10.(D) Velocity changes in all elliplical path at every point. So velocity, momentum and kinetic energy change at every point.
Angular momentum remains constant about the sun.

[JEE-2023/Module - 2] Physics
 18 IN-CHAPTER EXERCISES

MISCELLANEOUS EXERCISE

1.(A) T 2  r 2  Take logrithm

2log T  log K  3Logr
2dT 3dr
Diff.  As T  T
T r
and Δr  r , ΔT   dT and Δr  dr
2ΔT 3Δr

T r
2.(D) LP = constant as the gravity force passes through P and hence  p  0
 mvR  mv1r
By conservation of Energy
1 2  GMm  1 2  GMm 
mv     mv1   
2    2  r 
Solve the get the value of R.

3.(A) After collision, their masses combine to become 2m whose velocity and KE) is zero. Hence its energy is purely potential i.e.
GM (2m)
r

4
G R3
3 3g
4.(A) g   
2 4RG
R
5.(C) Area velocity remains constant. If area is half time taken is half.
mgh
6.(C) ΔU 
h
1
R
7.(C) W  ΔU  U f  Vi

 GMM   GMM  3 GM 2
 3    3  l  
 2l    2 l

Physics [JEE-2023/Module - 2]
 IN-CHAPTER EXERCISES 19

8.(C) Just before exptosion

GM GM GM
v0   
r RR 2R
From conservation of linear momentum,
m m 2GM
mv0  0  v  v  2v0 
2 2 R
Increase in mechanical energy
 K f  Ki

1m 2 1 2 1 1
 v  mv0  m(2v0 )2  mv02
2 2 2 4 2
1 2 1  GM  1 GMm 1
 mv0  m    mgR
2 2  2R  4 R 4
9.(B) Ki  U i  K f  U f

1 2 GMm 2 2GM
 mv  2 00  v
2 2R R

10.(CD) At maximum distance (at A) kinetic energy is minimum. But angular momentum about centre of sun always remains constant.

[JEE-2023/Module - 2] Physics