A straight rod of length l extends from x=a to x=l+a. If the mass per unit length is (a+bx2), the gravitational force it exerts on a point mass m placed at x=0 is given by
Gml(l+a){a(l+a)+bl3}
Gm(bl+al+a)
Gml(b+aa(l+a))
Gm{b(l+a)+al)
A
Gml(b+aa(l+a))
B
Gml(l+a){a(l+a)+bl3}
C
Gm(bl+al+a)
D
Gm{b(l+a)+al)
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Solution
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The gravitational force is F=∫l+aaGmλdxx2 here, λ=(a+bx2) now, F=∫l+aaGm(a+bx2)dxx2=Gm∫l+aa(b+ax2)dx or,F=Gm[b(l+a−a)−a(1l+a−1a)]=Gml[b+aa(l+a)]
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