Question

A particle would take time $t_1$ to move down a straight tube from the surface of earth (supposed to be homogeneous sphere) to its centre. If gravitational acceleration were to remain constant, time would be $t_2$. The ratio $t/t^{\prime }$ will be

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

Answer 1

The correct option is A $\frac{\pi }{2\sqrt{2}}$

Lets consider a distance of dr at a distance of r from center of earth.

now, $d{t}_1=\frac{dr}{-v}$ [since the velocity is towards the center of earth]

Energy is conserved , Initial = final energy

So, $\frac{-GMm}{R}=\frac{m{v}^2}{2}+\frac{-GMm}{2R^3}(3R^2-r^2)$

solving this, we get v = $(\frac{GM(R^2-r^2)}{R^3}{)}^{0.5}$

Integrating on both sides with limits from R to -R

$t_1=\pi (\frac{R^3}{GM}{)}^{0.5}=\pi \times (\frac{R}{g}{)}^{0.5}$

for $t_2$ , use S = $ut+\frac{a(t{)}^2}{2}$

So, $R=0+\frac{g(t{)}^2}{2}\to t=(\frac{2R}{g}{)}^{0.5}$

But, $t_2=2t=2(\frac{2R}{g}{)}^{0.5}$

Taking ratio, $\frac{t_1}{t_2}=\frac{\pi \times (\frac{R}{g}{)}^{0.5}}{2(\frac{2R}{g}{)}^{0.5}}=\frac{\pi }{8^{0.5}}=\frac{\pi }{2\sqrt{2}}$