Question

A thin ring has a mass M and radius 'R'.P is a point on its axis at a distance $\sqrt{3}$R from its centre . Then the intensity of gravitational field at point P is

• A. $\frac{GM}{4R^2}$
• B. $\frac{GM}{4\sqrt{2}{R}^2}$
• C. $\frac{3GM}{4R^2}$
• D. $\frac{\sqrt{3}GM}{8R^2}$

Answer 1

The correct option is D $\frac{\sqrt{3}GM}{8R^2}$

Consider ring uniform mass $M$. let radius be $a$.

Now, $dV\left(r\right)=-\frac{G.dm}{z}=\frac{-Gdm}{\sqrt{a^2+r^2}}$

$\Rightarrow V\left(r\right)=-\int \frac{Gdm}{\sqrt{a^2+r^2}}=\frac{-G}{\sqrt{a^2+r^2}}\int dm$

$\Rightarrow V\left(r\right)=\frac{-GM}{\sqrt{a^2+r^2}}$

Now, $a=R,r=\sqrt{3}R$

$\Rightarrow V=\frac{-GM}{\sqrt{R^2+3R^2}}=\frac{GM}{2R}$

Now, $E=\int \frac{Gdm}{z^2}\cos \theta =\frac{G\cos \theta }{z^2}\int dm=\frac{MG\cos \theta }{z^2}=\frac{GMT}{(r^2+a^2{)}^{3/2}}$

$\Rightarrow E=\frac{+GMR\times \sqrt{3}}{(\sqrt{R^2+3R^2}{)}^3}=\frac{+GM\times \sqrt{3}R}{(2R{)}^3}=\frac{+\sqrt{3}GM}{8R^2}$