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Question

Two lead balls of masses m and 5m having radii R and 2R are separated by 12R. If they attract each other by gravitational force, the distance covered by small sphere before they touch each other is:

A
10 R
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B
7.5 R
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C
9 R
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D
2.5 R
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Solution

The correct option is B 7.5 R
Effective distance =9R
Distance travelled by smaller mass=x
x=(m2m1+m2)(9R)

The gravitational force between masses m and 5m is given by:
FG=G(m)(5m)(12Rx)2=5Gm2(12Rx)2
The acceleration of mass m is given by:
a(m)=FGm=5Gm(12Rx)2
The acceleration of mass 5m is given by:
a(5m)=FG5m=Gm(12Rx)2
Distance travelled by mass m is given by:
x=12a(m)t2
x=125Gm(12Rx)2t2 ................................(1)
The distance covered by mass 5m is given by:
9Rx=12a(5m)t2
9Rx=12Gm(12Rx)2t2 .................... (2)
Dividing 1 by 2:
x9Rx=125Gm(12Rx)2t212Gm(12Rx)2t2=5
x=45R5x
6x=45R
x=7.5R

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