Question

Keeping the mass of earth constant, if its radius is reduced to its of its $\frac{1}{4}$ of its present value, then the period of revolution of the earth about its own axis and passing through the centre (in hours) is (assume the earth to be a . solid sphere and its initial period of rotation as 24h)

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

Answer 1

Answer: (D)

Assuming earth to be a solid sphere, its moment of inertia $\quad I=\frac{2}{5} MR ^2$ As no external torque is acting on earth, thus its angular momentum is constant, i.e, Iw $=$ constant
$\therefore \frac{2}{5} MR ^2 \times \frac{2 \pi}{ T }=$ constant $\quad$ where $T$ is the time period of earth's rotation
$\Longrightarrow \frac{ R _1^2}{ T _1}=\frac{ R _2^2}{ T _2} \quad(\because M =$ constant $)$
Given : $R _1= R \quad R _2=\frac{ R }{4} \quad T _1=24 hrs$
$\therefore \quad \frac{ R ^2}{24}=\frac{\frac{ R ^2}{16}}{ T _2} \quad \Longrightarrow T _2=\frac{24}{16}=1.5 hrs$