A point $\mathrm{P}$ lies on the axis of a ring of mass $M$ and radius $a$ at a distance $a$ from its centre $\mathrm{O}$. A small particle of mass $m$ starts from $\mathrm{P}$ and reaches $\mathrm{O}$ under gravitational attraction only. Its speed when it reaches $\mathrm{O}$ is

• A

  $\sqrt{\frac{2GM}{a}}$
• B

  $\sqrt{\frac{GM}{a}}$
• C

  $\sqrt{\frac{2GM}{a}\left(\sqrt{2}-1\right)}$
• Correct answer
  D

  $\sqrt{\frac{2GM}{a}\left(1-\frac{1}{\sqrt{2}}\right)}$

Solution

The gravitational potential at P, $V_P=-\frac{GM}{AP}$

$V_P=-\frac{GM}{\sqrt{a^2+a^2}}=-\frac{GM}{\sqrt{2a}}$ and gravitational potential at O, $V_O=-\frac{GM}{a}$

Let v be the velocity of the particle when it reaches O.

$\therefore$ K.E. at $O=\frac{1}{2}m{v}^2$

Gravitational P.E. = Gravitational potential × mass

$\therefore$ P.E. at $P=-\frac{GMm}{\sqrt{2}a}$ and P.E. at $O=-\frac{GMm}{a}$

By the principle of conservation of mechanical energy

$\frac{1}{2}m{v}^2-\frac{GMm}{a}=-\frac{GMm}{\sqrt{2}a}$

$\therefore v^2=2\left[\frac{GMm}{a}-\frac{GMm}{\sqrt{2}a}\right]=\frac{2GM}{a}\left[1-\frac{1}{\sqrt{2}}\right]$

$\therefore v=\sqrt{\frac{2GM}{a}\left(1-\frac{1}{\sqrt{2}}\right)}$

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