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- GM2a(L2+4a2)
- 2GMa(L2+a2)
- none
- 2GMa(L2+4a2)1/2

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Solution

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dm=MLdx

dE=GMLdx(x2+a2)dEnet=dEcosθ=GMadxL(x2+a2)3/2

E=∫dEcosθ=GMaL∫L/2−L/2dx(x2+a2)3/2

Put x=atanθ then dx=asec2θdθ

E=GMaL∫asec2θdθa3sec3θ=GMLa∫cosθdθ

=GMLasinθ=GMxLa√x2+a2∫L/2−L/2

=GM(2L)aL(L2+4a2)1/2=2GMa(L2+4a2)1/2

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