Question

If the radius of the earth becomes $n$ times its present value, without change in mass, then the duration of the day becomes,

Answer 1

Hint: In a system if there is no net external torque acting then the angular momentum of the system is always conserved. Like an extended object like earth, the point of the object is in rotation then the angular momentum is given by the moment of inertia. The formula of the conservation of the angular momentum is needed here.
Formula used:
The conservation of the angular momentum:
$ \Rightarrow {I_1}{\omega _1} = {I_2}{\omega _2}$
Where, $I$ is the inertia and $\omega $is the angular momentum.

Complete step by step answer:
Consider the given diagram as earth. The earth is now rotating along with the angular momentum. It has the mass of $m$ and the radius of $R$.

The product of the mass and the velocity is known as momentum. The property of any rotating body given by the product of the inertia and the angular velocity is known as the angular momentum.
For extended objects like the earth, the momentum is given by the product of inertia and velocity.
To calculate the duration of the day an angular momentum formula is needed.
Inertia is defined as the state of rest or a uniform motion unless or otherwise the object is disturbed. The inertia value is given as,
$ \Rightarrow I = \dfrac{2}{5}m{R^2}$
Where,
$I$ is the inertia, $m$ is the mass, and $R$is the radius.
Consider the values given. The value of the mass $m$is given without any changes. The value of radius is given as $n$. It can be taken as $nR$.
Consider the angular momentum has two values that are initial and final angular momentum.
The initial angular momentum is, ${L_1} = {I_1}{\omega _1}$
The final angular momentum is, ${L_2} = {I_2}{\omega _2}$
The conservation of the angular momentum:
The initial angular momentum= The final angular momentum
That is,
$ \Rightarrow {L_1} = {L_2}$
Substitute the values of the angular momentum.
$ \Rightarrow {I_1}{\omega _1} = {I_2}{\omega _2}$
Substitute the value of the inertia in the equation.
$ \Rightarrow \dfrac{2}{5}m{R^2}{\omega _1} = \dfrac{2}{5}m{R^2}{\omega _2}$
The value of the radius of the final angular momentum is given as $n$. It can be taken as $nR$.
$ \Rightarrow \dfrac{2}{5}m{R^2}{\omega _1} = \dfrac{2}{5}m{n^2}{R^2}{\omega _2}$
Cancel out the common terms.
$ \Rightarrow {\omega _1} = {n^2}{\omega _2}$
The velocity $\omega $ is given as $\dfrac{{2\pi }}{{24hrs}}$. The value of velocity differs for the initial and final velocity.
For initial velocity, the velocity $\omega $ is given as $\dfrac{{2\pi }}{T}$.
For final velocity, the velocity $\omega $ is given as $\dfrac{{2\pi }}{{24hrs}}$ .
$ \Rightarrow \dfrac{{2\pi }}{T} = {n^2}\dfrac{{2\pi }}{{24hrs}}$
Canceling out the common terms we get,
$ \Rightarrow T = 24{n^2}$
Therefore, when the radius of the earth becomes $n$times its present value, without change in mass, then the duration of the day becomes $T = 24{n^2}$.

Note:
Angular momentum quantum number will always be synonymous with the azimuthal quantum number. The atomic orbital has the quantum number that decides the angular momentum, orbital shape, etc. The typical value ranges from \[0\] to \[1\].