Question

Assume there are two identical simple pendulum Clocks- 1 is placed on the earth and Clock-2 is placed on a space station located at a height $h$ above the earth surface. Clock-1 and Clock- 2 operate at time periods $4s$ and $6s$ respectively. Then the value of $h$ is (consider radius of earth $R_E=6400km$ and $g$ on earth $10m/s^2$ )

• A. $1200km$
• B. $1600km$
• C. $3200km$
• D. $4800km$

Answer 1

Answer: (C)

$t\propto \frac{1}{\sqrt{g}}\text{ and }g\propto \frac{1}{(R+h{)}^2}$
$\frac{t_1}{t_2}=\sqrt{\frac{g^{\prime }}{g}}=\sqrt{\frac{R^2}{(R+h{)}^2}}$
$\frac{t_1}{t_2}=\frac{4}{6}=\frac{R}{(R+h)}$
$\Rightarrow h=3200km$