Question

Keeping the mass of earth as constant, if its radius is reduced to $\frac{1}{4}$th of its initial value, then the period of revolution of earth about its own axis, and passing through the centre, in $hrs$, is:

(Assume earth to be a solid sphere and its initial period of rotation as $24hrs$)

• A. $12$
• B. $3$
• C. $6$
• D. $1.5$

Answer 1

Answer: (D)

$1.5$

Assuming earth to be a solid sphere, its moment of inertia $I=\frac{2}{5}M{R}^2$

As no external torque is acting on earth, thus its angular momentum is constant, i.e, $Iw=constant$

$\therefore$ $\frac{2}{5}M{R}^2\times \frac{2\pi }{T}=constant$ where $T$ is the time period of earth's rotation

$⟹$ $\frac{R_1^2}{T_1}=\frac{R_2^2}{T_2}$ $(\because M=constant)$

Given : $R_1=R$ $R_2=\frac{R}{4}$ $T_1=24hrs$

$\therefore$ $\frac{R^2}{24}=\frac{\frac{R^2}{16}}{T_2}$ $⟹T_2=\frac{24}{16}=1.5hrs$