Question

A ring of mass $M$ and radius $R$ is fixed in $y-z$ plane with its centre at origin $O$. From a point $P(\sqrt{3}R,0,0)$ along the axis of ring , a small particle of mass $m$ starts from rest and reaches $O$ under the influence of gravitational attraction alone. Its speed at $O$ will be.

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

Answer 1

The correct option is B $\sqrt{\frac{GM}{R}}$

The small mass will be attracted towards the ring due to gravitational force by the fixed ring. As it goes from point $P$ to origin $O$, its potential energy decreases and kinetic energy increases.

Let the speed of the mass be $v$ at $O$.

From law of conservation of mechanical energy, $$P{E}_i+K{E}_i=P{E}_f+K{E}_f$$$\Rightarrow -\frac{GMm}{\sqrt{R^2+r^2}}+0=-\frac{GMm}{R}+\frac{1}{2}m{v}^2$

$\Rightarrow -\frac{GMm}{\sqrt{R^2+(\sqrt{3}R{)}^2}}+0=-\frac{GMm}{R}+\frac{1}{2}m{v}^2$

On solving this, we get,

$v=\sqrt{\frac{GM}{R}}$

Hence, option (b) is the correct answer.
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Why this question? This question helps you to revise your concept of law of conservation of mechanical energy.