Question

A point particle is held on the axis of a ring of mass $m$ and radius rat a distance $r$ from its centre $C$. When released, it reaches $C$ under the gravitational attraction of the ring. Its speed at $C$ will be

• A. $\sqrt{\frac{2Gm}{r}\left(\sqrt{2}-1\right)}$
• B. $\sqrt{\frac{Gm}{r}}$
• C. $\sqrt{\frac{2Gm}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$
• D. $\sqrt{\frac{2Gm}{r}}$

Answer 1

Answer: (C)

Let 'M be the mass of the particle
Now $E_{initial} = E_{final}$
$i.e. \frac{GMm}{\sqrt{2}r}+0=\frac{GMm}{r} + \frac{1}{2}MV^{2}$
or, $\frac{1}{2}MV^{2} = \frac{GMm}{r}\left[1-\frac{1}{\sqrt{2}}\right]$
$\Rightarrow \frac{1}{2}V^{2} = \frac{Gm}{r}\left[1-\frac{1}{\sqrt{2}}\right]$
or, $V = \sqrt{\frac{2Gm}{r}\left(1-\frac{1}{\sqrt{2}}\right)}$