Questions
A mass m is at a distance a from one end of a uniform rod of length l and M. The gravitational force on the mass due to the rod is
a
GMma+l
b
GmMal+a
c
GMma
d
GmM2l+a
detailed solution
Correct option is B
Considering mass per unit length x=ml mass of a small portion dx=λdx at distance x. Force due to small portion Gmλdxx2, Total fore ∫aa+lGmλdxx2F=Gmλ-1xaa+l=Gmλ1a-1a+l=Gmλ1aa+l=GMmaa+lTalk to our academic expert!
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Where will be the centre of mass on combining two masses m and M (M > m)
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