If the earth were to suddenly contract to 1/n^th of its present radius without any change in its mass, the duration of the new day will be nearly

The correct option is C

24/n²hr

Step1: Given data

Earth was to suddenly contract to ${\frac{1}{n}}^{th}$ of its present radius without any change in its mass

Step2: Formula used

Moment of inertia of a solid sphere $I=\frac{2}{5}M{R}^2$

Expression for angular velocity, $\omega =\frac{2\pi }{T}$

Step3: New time period

Let us take, the mass of the earth=$M$, and the radius of the earth=$R$

The time period of rotation of the earth $\left(T\right)=24hr$

Before shrinking, $I_1=\frac{2}{5}M{R}^2$

After shrinking, the radius of the earth becomes $R\text{'}=\frac{R}{n}$

$I_2=\frac{2}{5}MR{\text{'}}^2=\frac{2}{5}M{R}^2\times {\left(\frac{1}{n}\right)}^2\left[AsR\text{'}=\frac{R}{n}\right]$

Putting value of $I_1$

$I_2=\frac{I_1}{n^2}$

Let us take the time period of rotation of the earth after shrinking be $T_{2}hr$

Step 4: Conservation of angular momentum

Using conservation of angular momentum

$$\begin{gathered} I_2{\omega }_2=I_1{\omega }_{1}where{\omega }_1,{\omega }_{2}areangularvelocity \\ \frac{I_1}{n^2}\times \frac{2\pi }{T_2}=I_1\times \frac{2\pi }{24} \\ {T}_2=\frac{24}{n^2}hr \end{gathered}$$

If the earth were to suddenly contract to 1/n^th of its present radius without any change in its mass, the duration of the new day will be $24/n^{2}hr$

Hence, option C is the correct answer.