Question

If the earth were to suddenly contract to $\dfrac{1}{n}th$ of its present radius without any change in its mass then the duration of the new day will be close to
A. $\dfrac{{24}}{n}hour$
B. $24n\,hour$
C. $\dfrac{{24}}{{{n^2}}}hour$
D. $24{n^2}hour$

Answer 1

Hint: First we have to assume the mass and radius of the earth. Then we know that reduction in the size or earth means reduction of radius, hence the radius is reduced to $\dfrac{1}{n}$ . Find the moment of inertia of two different situations then conserving their angular momentum we can solve this problem.

Formula Used:
Moment of inertia of a solid sphere,
$I = \dfrac{2}{5}M{R^2}$
Where, Mass of the sphere = M and Radius of the sphere = R.

Complete step by step answer:
As given in the problem, if the earth suddenly contracts to $\dfrac{1}{n}th$ of its present value.We need to calculate the time period of the new contracted earth.Let us first assume mass and radius of the earth be $M$ and $R$ respectively. $I_1$ be the moment of inertia of the earth before contraction. The shape of the earth is sphere hence the moment of inertia of the earth is,
$I_1 = \dfrac{2}{5}M{R^2}$

If the size of the earth contracts then the radius of the earth must contract. After the earth contract to $\dfrac{1}{n}th$ ,the new radius $R'$ will be,
$R' = \dfrac{R}{n}$
Hence the new moment of inertia of the earth,
$I_2 = \dfrac{2}{5}MR{'^2}$
Putting $R'$ value in the above formula we will get,
$I_2 = \dfrac{2}{5}M{\left( {\dfrac{R}{n}} \right)^2}$
$ \Rightarrow I_2 = \dfrac{2}{5}\dfrac{{M{R^2}}}{{{n^2}}}$

We know the value of $I_1$ comparing it with the above value we will get,
$I_2 = \dfrac{{I_1}}{{{n^2}}}$
After contraction also the angular momentum of the earth will remain the same.
Hence using conservation of angular momentum we will get,
Final angular momentum = Initial angular momentum
That is,
$I_1\omega_1 = I_2\omega_2 \ldots \ldots \left( 1 \right)$
Where , $\omega_1,\omega_2$ be the angular velocity of the rotation of earth and $\omega_1$ is the initial angular velocity.

Now, we know the time period of the earth to rotate in 24hour
$T_1 = 24hour$
Let the time period of the contract earth be $T_2$
We need to calculate this $T_2$ .
$\omega_1 = \dfrac{{2\pi }}{{T_1}}$
Putting $T_1$ value we will get,
$\omega_1 = \dfrac{{2\pi }}{{24}}$
Now putting $I_1,I_2$ and $\omega_1$ values in equation $\left( 1 \right)$ we will get,
$I_2\omega_2 = I_1\omega_1$
$ \Rightarrow \dfrac{{I_1}}{{{n^2}}} \times \dfrac{{2\pi }}{{T_2}} = I_1 \times \dfrac{{2\pi }}{{24}}$
Cancelling the common terms we will get,
$\dfrac{1}{{{n^2}}} \times \dfrac{1}{{T_2}} = \dfrac{1}{{24}}$
Rearranging the above values we will get,
$T_2 = \dfrac{{24}}{{{n^2}}}hour$

Therefore the correct option is $\left( C \right)$.

Note: We take the unit of $T_1$ in hour that's why the final answer will be in hour only. Always remember whatever changes in the radius or mass of the earth the angular momentum of the earth will always remain the same or we can say the angular momentum is conserved.