Question

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

Answer 1

Hint: The moment of inertia is a quantity which measures the torque required to change the rotation of an object, it depends on the radius or the distance of the rotating object from the axis of rotation. So if angular momentum is conserved, the angular velocity of rotation of earth will also change.
Formula used
$I = \dfrac{2}{5}m{r^2}$ (for a solid sphere)
${I_1}{\omega _1} = {I_2}{\omega _2}$
Where I represent the moment of inertia of a body, the subscripts 1 and 2 represent the initial and the final moments of inertia for the given body.
${\omega _1}$ and ${\omega _2}$ represent the initial and the final angular velocity of the body.
m is the mass of the body
r is the radius of the body.

Complete step by step solution:
Let the mass of earth be M and its initial radius be R.
The earth is a solid sphere, so the moment of inertia of earth is given by-
$I = \dfrac{2}{5}M{R^2}$
We know that the time taken by earth to complete one rotation around its axis is 24 hours, therefore-
Time period, $T = 24hr$
The angular velocity of an object is given by, $\omega = \dfrac{{2\pi }}{T}$
For earth, let ${\omega _1}$be the initial angular velocity be, then-
${\omega _1} = \dfrac{{2\pi }}{{24}}$
When the earth shrinks by a factor of n, let the new radius be $R'$
The new radius is given by,
$R' = \dfrac{R}{n}$
Now the new moment of inertia of the earth becomes-
${I_2} = \dfrac{2}{5}M{R'^2}$
${I_2} = \dfrac{{2M}}{5}{\left( {\dfrac{R}{n}} \right)^2}$
${I_2} = \dfrac{{2M{R^2}}}{{5{n^2}}}$
According to the conservation of angular momentum,
${I_1}{\omega _1} = {I_2}{\omega _2}$
${\omega _2} = \dfrac{{{I_1}{\omega _1}}}{{{I_2}}}$
On substituting the value of initial angular velocity ${\omega _1}$, initial and final moments of inertia in this equation, the final angular velocity is given by-
${\omega _2} = \dfrac{{\left( {\dfrac{2}{5}M{R^2}} \right) \times \left( {\dfrac{{2\pi }}{{24}}} \right)}}{{\left( {\dfrac{2}{{5{n^2}}}M{R^2}} \right)}}$
${\omega _2} = \dfrac{{2\pi {n^2}}}{{24}}$
Number of hours in a day is given by, $\dfrac{{2\pi }}{\omega }$
Therefore, the length of day in the new earth would be, $\dfrac{{24}}{{{n^2}}}$.

Option (C) is the correct option.

Note: If the radius of the earth is changed by a factor of n then the change in the rotation speed will be by a factor of n squared. If the earth is shrank, the days would be smaller and if the earth is enlarged then the days would be longer. This statement holds true because of the conservation of angular momentum.