Question

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

• A. $Gml(b+\frac{a}{a(l+a)})$
• B. $\frac{Gm}{l(l+a)}\left\{a\right(l+a)+b{l}^3\}$
• C. $Gm(bl+\frac{a}{l+a})$
• D. $Gm\left\{b\right(l+a)+\frac{a}{l})$

Answer 1

Answer: (A)

$Gml(b+\frac{a}{a(l+a)})$

The gravitational force is $F={\int }_a^{l+a}\frac{Gm\lambda dx}{x^2}$
here, $\lambda =(a+b{x}^2)$
now, $F={\int }_a^{l+a}\frac{Gm(a+b{x}^2)dx}{x^2}=Gm{\int }_a^{l+a}(b+\frac{a}{x^2})dx$
or,$F=Gm\left[b\right(l+a-a)-a(\frac{1}{l+a}-\frac{1}{a}\left)\right]=Gml[b+\frac{a}{a(l+a)}]$