Question

A tunnel is dug along a chord of the earth at a perpendicular distance $\frac{R}{2}$ from the earth's centre. The wall of the tunnel may be assumed to be frictionless. The particle is released from the one end of the tunnel. The pressing force by the particle on the wall and the accelaration of the particle varies with $x$ (distance of the particle from the centre) according to

• A.
• B.
• C.
• D.

Answer 1

Answer: (B), (C)

$\text{Force Analysis}:$
Let the particle of mass $m$ be at a point which lies at a distance $x$ from the centre of the earth as shown in the figure.

Gravitational field at point $\text{P}$will be, $$E_P=\frac{GM}{R^3}x$$So, the force on the particle placed at point $\text{P}$ will be,

$F=m{E}_P=\frac{GMm}{R^3}x$

Now, the pressing force on the particle,
$N=F\cos \theta$

Substituting $\cos \theta =\frac{\frac{R}{2}}{x}$ from the figure,
$\Rightarrow N=\frac{GMm}{R^3}x\times \frac{\frac{R}{2}}{x}$

$\Rightarrow N=\frac{GMm}{2R^2}=\text{Constant}$ or independent of $x$.


So, option (b) is correct answer.

$\text{Acceleration Analysis}:$
Let $a$ be the acceleration of the particle along the tunnel. So,

$ma=F\sin \theta$

Let the distance of particle from $B$ be $b$.

$\Rightarrow ma=\frac{GMm}{R^3}x\times \frac{b}{x}$

$\Rightarrow a=\frac{GM}{R^3}b$

Substituting the value of $b$ from figure,

$\Rightarrow a=\frac{GM}{R^3}\sqrt{x^2-\frac{R^2}{4}}$

Between $\text{ABC}$ the motion of partcle is $\text{SHM}$ , with acceleration $a$.

At point $\text{B}$, $x=\frac{R}{2}$, $a=0$, As this is the mean position of $\text{SHM}$.

At $\text{A}$ or $\text{C}$, $x=R$, $a=\text{maximum}$

As, these are the extreme positions of $\text{S.H.M.}$

Hence, options (b) and (c) are the correct answers.