Question

If earth shrinks to $\dfrac{1}{{64}}$ of its volume with mass remaining same, duration of day will be
(A) $1.5hr$
(B) $3hr$
(C) $4.5hr$
(D) $6hr$

Answer 1

We can use the moment of inertia concept to explain the problem. Initially the moment of inertia about its geographical rotational axis is the same as the spherical object. Originally, radius is R for volume V. When volume is decreased, using volume of the sphere formula, find how much the radii gets decreased and use it in angular momentum formula to find the change in duration.

Complete Step by step solution
It is given that the earth’s volume is reduced by $\dfrac{1}{{64}}$ of its original volume, which means that there will be a significant change in the radii due to the proportionality. Earth is spherical in shape, therefore its volume is given as ,
$\Rightarrow V = \dfrac{4}{3}\pi {r^3}$
Where , $r$ is the radius of the earth.
Now, the new volume is given as
$\Rightarrow {V'} = \dfrac{V}{{64}}$
Now, new volume of the sphere is given as
$\Rightarrow {V'} = \dfrac{4}{3}\pi {({r'})^3}$
$\Rightarrow \dfrac{V}{{64}} = \dfrac{4}{3}\pi {({r'})^3}$
On substituting $V$ in the above equation , we get
$\Rightarrow \dfrac{{\dfrac{4}{3}\pi {r^3}}}{{64}} = \dfrac{4}{3}\pi {({r'})^3}$
Simplifying common terms we get,
$\Rightarrow \dfrac{{{r^3}}}{{64}} = {({r'})^3}$
On further simplification we get,
$\Rightarrow \dfrac{r}{4} = {r'}$
Now, according to law of conservation of angular momentum, we can say that there is no external torque acting on earth and hence the angular momentum of the earth is constant
$\Rightarrow L = I\omega = Const$
Moment of inertia of spherical objects is given as $I = \dfrac{2}{5}M{R^2}$ and angular velocity is given as reciprocal to time period.
Now according to conservation of angular momentum, the angular momentum of earth before shrink is said to be equal to angular momentum post shrinking. Which means that,
$\Rightarrow {L'} = L$
$\Rightarrow {I'}{\omega '} = I\omega$
Substituting for the expressions we get
$\Rightarrow \dfrac{2}{5}M{(\dfrac{R}{4})^2} \times \dfrac{{2\pi }}{{{T'}}} = \dfrac{2}{5}M{R^2} \times \dfrac{{2\pi }}{T}$ (since, we found out that $\dfrac{r}{4} = {r'}$ )
Cancelling out the common terms we get,
$\Rightarrow {(\dfrac{R}{4})^2} \times \dfrac{1}{{{T'}}} = {R^2} \times \dfrac{1}{T}$
Squaring and cancelling out the radius term we get,
$\Rightarrow \dfrac{1}{{16{T'}}} = \dfrac{1}{T}$
We know that before shrinking the time period for one rotation of earth is 24 hours, substituting this we get
$\Rightarrow \dfrac{1}{{16{T'}}} = \dfrac{1}{{24}}$
$\Rightarrow \dfrac{{24}}{{16}} = {T'}$
$\Rightarrow {T'} = 1.5hr$
Therefore the new time period for earth to complete one revolution is $1.5hr$ .
Hence, option (A) is the right answer.

Note Law of conservation of angular momentum states that, when there is no external torque acting on the rotating body , then there won’t be any change in the state of angular momentum of the object, thus remaining constant throughout.