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Question

# Four point masses each of mass m are kept at four corners of a cube of side length ‘a’. Then

A
The gravitational potential energy of system is =32Gm2a
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B
The gravitational potential energy of system is =42Gm2a
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C
The magnitude of force on one of the particle due to the remaining masses is equal to 6Gm2a2
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D
The magnitude of force on one of the particle due to others is equal to 32Gm2a2
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Solution

## The correct option is D The magnitude of force on one of the particle due to others is equal to √32Gm2a2Gravitational potential energy of the system U=−Gm2r (for one face) For all 6 faces, U=−6Gm2r Here, r=a√2 U=−6 Gm2a√2=−3√2Gm2a F on A=−Gm2(a√2)2 due to one of the charges |F|=Gm2a32√2[√4a2+4a2+4a2] =√32Gm2a2

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