Question

If the radius of the earth contracts to half of its present value without a change in its mass, then new duration of the day will be;

• A. $12\text{h}$
• B. $6\text{h}$
• C. $4\text{h}$
• D. $3\text{h}$

Answer 1

The correct option is B $6\text{h}$

If $R_1$ is the initial radius and $M$ is the mass of the earth revolving with angular speed ${\omega }_1$ about its own axis, then the present angular momentum of the earth with respect to its center is,

$L_1=I{\omega }_1=\frac{2}{5}M{R}_1^2{\omega }_1$

Let, the angular speed of the earth be ${\omega }_2$ when its radius becomes half of its initial radius.

New angular momentum of the earth,

$L_2=I{\omega }_2=\frac{2}{5}M{R}_2^2{\omega }_2$

$\Rightarrow L_2=\frac{2}{5}M{\left(\frac{R_1}{2}\right)}^2$

$\Rightarrow L_2=\frac{1}{10}M{R}_1^2{\omega }_2$

As the external torque is zero we can say that, $L_1=L_2$

$\frac{2}{5}M{R}_1^2{\omega }_1=\frac{1}{10}M{R}_1^2{\omega }_2$

$\therefore {\omega }_2=4{\omega }_1$

As, $T_1=\frac{2\pi }{{\omega }_1}$ and $T_2=\frac{2\pi }{{\omega }_2}$

$\frac{T_2}{T_1}=\frac{{\omega }_1}{{\omega }_2}=\frac{1}{4}$

$\Rightarrow T_2=\frac{T_1}{4}$

As, the actual duration of the day, $T_1=24\text{h}$ so,

$T_2=\frac{24}{4}=6\text{h}$

Hence, option (b) is the correct answer.