Q134 / 160 · TS EAMCET 2021 (05 Aug Shift 2) · Physics

A point mass $' m'$ is located at a distance $r$ from a uniform thin rod of mass $M$ and length $L$ as shown in the figure. The magnitude of gravitational force of attraction is

A
$\frac{G M m}{r^{2}}$
B
$\frac{G M m}{(r + L)^{2}}$
Correct answer
C
$\frac{G M m}{r (r + L)}$
D
$\frac{G M m}{{(r + \frac{L}{2})}^{2}}$

Solution

Let us take an element at a distance $x$ from point mass as shown in figure below,

Force on $m$ due to element is given by, $d F = \frac{G m d m}{x^{2}}$
$\Rightarrow d F = \frac{G m λ d x}{x^{2}}$
$\Rightarrow \int_{0}^{F} d F = \frac{G m M}{L} \int_{r}^{r + L} \frac{d x}{x^{2}}$
$\Rightarrow F = \frac{G m M}{L} {[- \frac{1}{x}]}_{r}^{r + L}$
$\Rightarrow F = \frac{G m M}{L} [\frac{1}{r} - \frac{1}{r + L}]$
$\Rightarrow F = \frac{G m M}{L} [\frac{r + L - r}{r (r + L)}]$
$\Rightarrow F = \frac{G m M}{r (r + L)}$

Hence, the correct answer is option: 'C'.

A point mass 'm' is located at a distance r from a uniform thin rod of mass M and length L as shown in the figure. The magnitude of gravitational force of attraction is

Solution

Relevant Questions

A point mass $' m'$ is located at a distance $r$ from a uniform thin rod of mass $M$ and length $L$ as shown in the figure. The magnitude of gravitational force of attraction is