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Question

A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M as shown in the figure.The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction.When the point mass nearer to M is at distance r=3l from M, the tension in the rod is zero for m=k(M288). The value of k is


A
7.00
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B
7.0
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C
7
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Solution

Since both particles of mass m are connected by a rod, so the acceleration of both the masses will be the same, let’s say a.


For m close to M, FBD is as follows.


Here,FM force of attraction on m due to mass M placed distance (r) from M.
And Fm is the force of attraction on m due to the another mass m placed at distance l from it.

From the Newton's second law of motion, The net force on the m can be written as,

GMmr2Gm2l2=ma...(1)

For the other m, FBD is as follows,


Here, FM force of attraction on m due to mass M placed distance (r+l) from M.
And Fm is the force of attraction on m due to the another mass m placed at distance l from it.

Hence,
GMm(r+l2)+Gm2l2=ma...(2)

Solving Equation (1) and (2),

GMmr2Gm2l2=GMm(l+r)2+Gm2l2

Given, r=3l

GMm(3l)2Gm2l2=GMm(l+3l)2+Gm2l2

GMm9l2Gm2l2=GMm(16l2)+Gm2L2

M9m=M16+m

2m=M9M16

2m=7M144

m=7M288

Given, m=kM288,

k=7

Thus, Accepted value of k is 7.

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