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Question

# Four point masses each of mass m are at four corner 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 =6Gm2a2
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D
the magnitude of force on one of the particle due to other is 32Gm2a2
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Solution

## The correct options are A the gravitational potential energy of system is =−3√2Gm2a D the magnitude of force on one of the particle due to other is √32Gm2a2Gravitational potential energy of system is given by U=−GMmr Here, M=m and r=a√2 U=−6Gm2a√2=−3√2Gm2a So, option A matches correctly. →F on A=\frac{-Gm^2}{(a\sqrt2)^3}\left[a\hat i+a\hat k-a\hat j+a \hat k+a\hat i-a \hat j\right]|F|=\frac{Gm^2}{a^3 2\sqrt 2}\left[\sqrt{4a^2+4a^2+4a^2}\right]=\sqrt{\frac{3}{2}}\frac{Gm^2}{a^2}\$ So, option 4 matches.

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