Question

A straight rod of length $L$ extends from $x = a$ to $x = L + a$. The gravitational force is exerts on a point mass $'m'$ at $x = 0$, if the mass per unit length of the rod is $A + Bx^2$, is given by:

• A. $Gm\left[A\left(\frac{1}{a + L} - \frac{1}{a}\right) - BL\right]$
• B. $Gm\left[A\left(\frac{1}{a} - \frac{1}{a + L}\right) + BL\right]$
• C. $Gm\left[A\left(\frac{1}{a + L} - \frac{1}{a}\right) + BL\right]$
• D. $Gm\left[A\left(\frac{1}{a} - \frac{1}{a + L}\right) - BL\right]$

Answer 1

Answer: (B)

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

Answer 2

$dm =\left(A+Bx^{2}\right)dx$
$dF = \frac{GMdm}{x^{2}}$
$=F = \int^{a+L}_{a} \frac{GM}{x^{2}}\left(A +Bx^{2}\right)dx$
$=GM \left[- \frac{A}{x}+Bx\right] ^{a+L}_{a}$
$=GM\left[A \left(\frac{1}{a} -\frac{1}{a+L}\right) +BL \right]$