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 from P and reaches C under gravitational attraction only. Its speed at C will be

The correct answer is:
$\sqrt{\frac{2GM}{a}\left(1-\frac{1}{\sqrt{2}}\right)}$

Step-by-Step Derivation:

1. Gravitational Potential of a Ring:
The gravitational potential $V$ at a point on the axis of a uniform ring of mass $M$ and radius $a$, at a distance $x$ from its center, is given by the formula:
$V=-\frac{GM}{\sqrt{a^2+x^2}}$

2. Potential at Initial Point P:
The initial point $P$ lies on the axis of the ring at a distance $x=a$ from its center. Substituting $x=a$ into the potential formula:
$V_P=-\frac{GM}{\sqrt{a^2+a^2}}=-\frac{GM}{a\sqrt{2}}$

3. Potential at Center C:
At the center $C$ of the ring, the distance along the axis is $x=0$. Substituting $x=0$ into the potential formula:
$V_C=-\frac{GM}{\sqrt{a^2+0^2}}=-\frac{GM}{a}$

4. Conservation of Mechanical Energy:
Let a particle of mass $m$ start from rest at point $P$ (so its initial kinetic energy $K_P=0$) and reach the center $C$ with a speed $v$ (so its kinetic energy at $C$ is $K_C=\frac{1}{2}m{v}^2$).
By the principle of conservation of energy:
$E_P=E_C$
$K_P+m{V}_P=K_C+m{V}_C$
$0+m\left(-\frac{GM}{a\sqrt{2}}\right)=\frac{1}{2}m{v}^2+m\left(-\frac{GM}{a}\right)$

5. Solving for Speed v:
We can divide the entire equation by the mass of the particle $m$:
$-\frac{GM}{a\sqrt{2}}=\frac{1}{2}$
Rearranging the terms to isolate $v^2$:
$\frac{1}{2}$
$\frac{1}{2}$
Multiplying both sides by 2:
$v^2=\frac{2GM}{a}\left(1-\frac{1}{\sqrt{2}}\right)$
Taking the square root on both sides yields:
$v=\sqrt{\frac{2GM}{a}\left(1-\frac{1}{\sqrt{2}}\right)}$