Question

Gravitational field at the centre of semicircle formed by a thin wire AB of mass m and length l is

• A. $\frac{GM}{l^2}$ along x - axis
• B. $\frac{GM}{{\pi l}^2}$ along y - axis
• C. $\frac{2\pi GM}{l^2}$
• D. $\frac{2\pi GM}{l^2}$

Answer 1

The correct option is D

$\frac{2\pi GM}{l^2}$

F = $\int$ dF cos$\theta$ = $\int$ $\frac{GMDM}{r^2}$ cos $\theta$

= $\frac{GM}{r^2}$ $\underset{\frac{\pi }{2}}{\overset{\frac{-\pi }{2}}{\int }}$ $\frac{m}{\pi r}$ . r d $\theta$ . cos $\theta$ d $\theta$

= $\frac{GM}{{\pi r}^2}$ $\underset{\frac{\pi }{2}}{\overset{\frac{-\pi }{2}}{\int }}$ cos $\theta$ d $\theta$

= $\frac{2GMm}{{\pi r}^2}$

= $\frac{2gMm}{\pi {\left(\frac{l}{\pi }r\right)}^2}$ = $\frac{2\pi GMm}{l^2}$

Gravitational Field = $\frac{F}{m}$