If the radius of earth contracts 1/n of its present value, the length of the day will be approximately

The correct option is 24 h/n².

Step-by-step Explanation:

According to the law of conservation of angular momentum, in the absence of any external torque, the total angular momentum ($L$) of a rotating system remains constant. Since there is no external torque acting on the Earth during its contraction, its angular momentum is conserved.

The angular momentum of a rotating body is given by the relation:
$L=I\omega$
where:
- $I$ is the moment of inertia of the Earth.
- $\omega$ is the angular velocity of rotation.

Since the Earth is a solid sphere of mass $M$ and radius $R$, its moment of inertia about its axis of rotation is:
$I=\frac{2}{5}M{R}^2$

The angular velocity $\omega$ is related to the time period of rotation $T$ (which represents the length of a day) by:
$\omega =\frac{2\pi }{T}$

Substituting the expression for $\omega$ into the conservation of angular momentum formula ($L_1=L_2$):
$I_1{\omega }_1=I_2{\omega }_2$
$I_1\left(\frac{2\pi }{T_1}\right)=I_2\left(\frac{2\pi }{T_2}\right)$
Simplifying this, we get:
$\frac{I_1}{T_1}=\frac{I_2}{T_2}\Rightarrow T_2=T_1\left(\frac{I_2}{I_1}\right)$

Since the mass of the Earth ($M$) does not change during the contraction, the ratio of the moments of inertia depends only on the square of the radii:
$\frac{I_2}{I_1}=\frac{\frac{2}{5}M{R}_2^2}{\frac{2}{5}M{R}_1^2}={\left(\frac{R_2}{R_1}\right)}^2$

We are given that the radius of the Earth contracts to $\frac{1}{n}$ of its present value:
$R_2=\frac{R_1}{n}\Rightarrow \frac{R_2}{R_1}=\frac{1}{n}$

Now, substitute this ratio back into the equation for the new time period:
$T_2=T_1{\left(\frac{1}{n}\right)}^2=\frac{T_1}{n^2}$

Given that the original length of the day on Earth is $T_1=24\text{h}$, the new length of the day will be:
$T_2=\frac{24\text{h}}{n^2}$