Question

A tunnel is dug inside the earth across one of its diameters. Radius of earth is $R$ and its mass is $M$. A particle is projected inside the tunnel with velocity $\sqrt{\frac{2GM}{R}}$ from one of its ends then maximum velocity attained by the particle in the subsequent motion is (assuming tunnel to be frictionless)

• A. $\sqrt{\frac{GM}{R}}$
• B. $\sqrt{\frac{3GM}{R}}$
• C. $\sqrt{\frac{2GM}{R}}$
• D. $\sqrt{\frac{5GM}{R}}$

Answer 1

The correct option is B $\sqrt{\frac{3GM}{R}}$

Given velocity at one the end of the tunnel,

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

The maximum velocity will be achieved at the centre of the tunnel. Let the velocity at the centre of the tunnel be $v$.

Now, applying energy conservation principle between these two points, we get

$\frac{1}{2}m{u}^2+\frac{-GMm}{R}=\frac{1}{2}m{v}^2+\frac{-3GMm}{2R}$

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

$\Rightarrow \frac{1}{2}m{v}^2=\frac{3}{2}\frac{GMm}{R}$

$\therefore v=\sqrt{\frac{3GM}{R}}$

Hence, option (b) is the correct answer.