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Question

Mass M is distributed uniformly over a rod of length L. Find the gravitational field at P at a distance a on perpendicular bisector.
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  1. GM2a(L2+4a2)
  2. 2GMa(L2+a2)
  3. none
  4. 2GMa(L2+4a2)1/2

A
2GMa(L2+a2)
B
2GMa(L2+4a2)1/2
C
none
D
GM2a(L2+4a2)
Solution
Verified by Toppr

dm=MLdx

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

E=dEcosθ=GMaLL/2L/2dx(x2+a2)3/2

Put x=atanθ then dx=asec2θdθ

E=GMaLasec2θdθa3sec3θ=GMLacosθdθ

=GMLasinθ=GMxLax2+a2L/2L/2

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


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