Question

A tunnel is dug across the earth of mass $M$ and radius $R$ at a distance $R/2$ from the centre $\text{O}$ of the Earth as shown. A body is released from rest from one end of the smooth tunnel. The velocity acquired by the body when it reaches the centre $\text{C}$ of the tunnel is

• A. $\sqrt{\frac{3GM}{2R}}$
• B. $\sqrt{\frac{3GM}{4R}}$
• C. $\sqrt{\frac{3GM}{8R}}$
• D. $\sqrt{\frac{3GM}{16R}}$

Answer 1

The correct option is B $\sqrt{\frac{3GM}{4R}}$

Given,
Velocity at point $\text{A}$, $v_A=0$

Let, the velocity acquired at point $\text{C}$ be $v$.

Applying energy conservation principle between points $\text{A}$ and $\text{C}$, we have:


$P.E_A+K.E_A=P.E_C+K.E_C$

Substituting the values, we get

$\frac{-GMm}{R}+0=\frac{-GMm}{R^3}[\frac{3}{2}{R}^2-\frac{1}{2}{\left(\frac{R}{2}\right)}^2]+\frac{1}{2}m{v}^2$

$\Rightarrow \frac{-GMm}{R}=-\frac{11}{8}\frac{GMm}{R}+\frac{1}{2}m{v}^2$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{3}{8}\frac{GMm}{R}$

$\Rightarrow v=\sqrt{\frac{3GM}{4R}}$

Hence, option (b) is the correct answer.

Why this Question: This question will help students to understand the application of energy conservation principle inside a tunnel.