Question

Gravitational field at the center of the semicircle formed by thin wire AB of mass m and length l as shown in the figure is:-

• A. $\frac{Gm}{l^2}along+xaxis$
• B. $\frac{Gm}{\pi l^2}along+yaxis$
• C. $\frac{2\pi Gm}{l^2}along+xaxis$
• D. $\frac{2\pi Gm}{l^2}along+yaxis$

Answer 1

The correct option is D $\frac{2\pi Gm}{l^2}along+yaxis$

Mass of element $dm=\frac{m^1}{x}d\theta$

Gravitational field due to element $dg=\frac{Gdm}{R^2}=\frac{Gmd\theta }{\pi R^2}$

$dg\cos \theta$ of each element will be cancelled out

$\delta =\int dg\sin \theta$

$={\int }_0^{\pi }\left(\frac{\sin \theta }{{\pi R}^2}\right)\sin \theta$

$=\frac{Gm}{\pi R^2}{\int }_0^{\pi }\sin \theta d\theta$

$=\frac{2Gm}{\pi R^2}$

$=\frac{2\pi Gm}{l^2}$ $along+yaxis$