Question

A straight rod of length $L$, extends from $x=a$ to $x=L+a$. The linear mass density of the rod varies with x- coordinate is $\lambda =A+B{x}^2$. The gravitational force experienced by a point mass $m$ at $x=0$ is

• A. $Gm[\frac{A}{a}+BL]$
• B. $Gm[BL+\frac{A}{a+L}]$
• C. $Gm[A[\frac{1}{a}-\frac{1}{a+L}]+BL]$
• D. $Gm[\frac{BL}{a}+\frac{A}{a^2}]$

Answer 1

The correct option is C $Gm[A[\frac{1}{a}-\frac{1}{a+L}]+BL]$

Force experienced by m at $x=0$ due to a small mass dm of rod at $x=x$ is given by :

$dF=\frac{Gmdm}{x^2}$ $(AsF=\frac{GMm}{r^2})$

Now $dm=\lambda dx=(A+B{x}^2)dx$

$F={\int }_a^{L+a}\frac{Gm(A+B{x}^2)dx}{x^2}$

$\Rightarrow F=Gm(A{\int }_a^{L+a}\frac{1}{x^2}dx+B{\int }_a^{L+a}\frac{x^2}{x^2})$

$\Rightarrow F=Gm(A\frac{-1^{L+a}}{x_a}+B{x}_a^{L+a})$

$\Rightarrow F=Gm(A(\frac{1}{a}-\frac{1}{a+L})+BL)$