Question

A point $P$ lies on the axis of a ring of mass $M$ and radius $a$, at a distance $a$ from its centre $C$. A small particle starts at rest from $P$ and reaches $C$ under gravitational attraction only. Its speed at $C$ will be

• A. zero
• B. $\sqrt{\frac{2GM}{a}(1-\frac{1}{\sqrt{2}})}$
• C. $\sqrt{\frac{2GM}{a}(\sqrt{2}-1)}$
• D. $\sqrt{\frac{2GM}{a}}$

Answer 1

The correct option is B $\sqrt{\frac{2GM}{a}(1-\frac{1}{\sqrt{2}})}$

Given, ring of mass $M$ and radius $a$.


Potential due to ring at point $P$: $V_p=\frac{-GM}{\sqrt{a^2+a^2}}=\frac{-GM}{\sqrt{2}a}$

Potential energy of mass ${}^{\prime }{m}^{\prime }$ at $P$ , $U_p=\frac{-GMm}{\sqrt{2}a}$

Similarly $V_C=\frac{-GM}{a}$ and $U_C=\frac{-GMm}{a}$

As we know that, gain in KE is equal to loss in PE

$\Rightarrow \frac{1}{2}m(v_C^2-v_P^2)=U_P-U_C$

$\frac{1}{2}m\left(v_C^2\right)=\frac{-GMm}{a}(\frac{1}{\sqrt{2}}-1)$

So, the speed at $C$ will be, $v_C=\sqrt{\frac{2GM}{a}(1-\frac{1}{\sqrt{2}})}$

Hence, option $\left(B\right)$ is correct.