Question

A straight rod of length $L$ extends from $x=0$ to $x=L$. The variation of linear mass density of the rod with $x$ co-ordinate is $\lambda =a_0+b_0{x}^2$. What is the gravitational force experienced by a point mass $m$ at $x=-a$ ?

• A. $Gm(\frac{a_0}{a}+b_{0}L)$
• B. $Gm[a_0(\frac{1}{a}-\frac{1}{a+L})+b_{0}L+b_0{a}^2(\frac{1}{a}-\frac{1}{a+L})+2a{b}_{0}ln\left(\frac{a+L}{a}\right)]$
• C. $Gm(b_{0}L+\frac{a_0}{a+L})$
• D. None of these

Answer 1

The correct option is A $Gm(\frac{a_0}{a}+b_{0}L)$

NL of gravitation

Small mass dm at the centre of rod having length dx

length of rod =L

mass of rod =M K=linear mass density

$K=a_0+b_0{x}^2$

$L\to M$ unit length $\to \frac{M}{L}$

$dx\to \frac{M}{L}dx$, $dm\to kdx.....$

$DF=\frac{Gm\left(dm\right)}{x^2};df=\frac{GM\left(kdx\right)}{x^2}$

$dF=Gm\frac{(a_0+b_0{x}^2)}{x^2}dx$

by in integration from $x=1$ to $x=aM$

$\int df=Gm{\int }_a^{a+L}(\frac{a_0}{x^2}+b_0)dx$

$F=Gm{[\frac{-a}{x}+bx]}_a^{a+L}$

$F=Gm(b_{0}L+\frac{a_0}{a+L})$