Question

A straight rod of length $l$ extends from $x=\alpha$ to $x=\ell +\alpha .$ If the mass per unit length is $(a+b{x}^2).$ The gravitational force It exerts on a point mass $m$ placed at $x=0$ given by

• A. $\mathrm{Gm}(a(\frac{1}{\alpha }-\frac{1}{\alpha +\ell })+bl)$
• B. $\frac{\mathrm{Gm}(a+b{x}^2)}{l^2}$
• C. $\mathrm{Gm}(a(\frac{1}{\alpha }+\frac{1}{\alpha +l})+bl)$
• D. $\mathrm{Gm}(a(\frac{1}{\alpha +l}-\frac{1}{\alpha })+bl)$

Answer 1

The correct option is A $\mathrm{Gm}(a(\frac{1}{\alpha }-\frac{1}{\alpha +\ell })+bl)$

REF.Image.

solution

Consider a differential part of

the rod.so,

$df=\frac{G\left(dm\right)m}{x^2}$

$df=G\frac{[a+b{x}^2]dxm}{x^2}$

${\int }_0^{f}df={\int }_{\alpha }^{\alpha +l}Gm\left[\frac{a+b{x}^2}{x^2}\right]dx$

$\Rightarrow F=\left[GM\right[\frac{-a}{x}+bx]{]}_{\alpha }^{\alpha +l}$

$\Rightarrow F=Gm\left[\right[\frac{1}{\alpha }-\frac{1}{\alpha +l}]a+bl].$

option - A is correct.