question_answer

A straight rod of length L extends from x = a to x = L + a. Find the gravitational force it exerts on a point mass m at x = 0 if the linear density of rod \[\mu =A+B{{x}^{2}}.\]

A) \[Gm\left[ \frac{A}{a}+BL \right]\]

B) \[Gm\left[ A\left( \frac{1}{a}-\frac{1}{a+L} \right)+BL \right]\]

C) \[Gm\left[ BL+\frac{A}{a+L} \right]\]

D) \[Gm\left[ BL-\frac{A}{a} \right]\]

Correct Answer: B

Solution :

[b] \[\because \]\[dF=\frac{Gm(\mu dx)}{{{x}^{2}}}\] \[F=Gm\int{(A+Bx\frac{dx}{x})}\] \[F=Gm\left[ A\left( \frac{1}{a}-\frac{1}{a+L} \right)+BL \right]\]