Question

If the earth shrinks such that its density becomes $ 8 $ times its present value then the new duration of the day in hours will be: (assume that there is no change in axis of rotation of Earth)
(A) $ 24 $
(B) $ 12 $
(C) $ 6 $
(D) $ 3 $

Answer 1

Hint To solve this question, firstly we need to find the new radius of the Earth using the new density and keeping the mass constant on both the cases. Using the new radius we find the moment of inertia for the new case and then finally apply the principle of conservation of angular momentum to find the new duration of the day.

Formula Used: The formulae used in this problem are,
 $\Rightarrow L = I\omega $
Here, $ L $ is the angular momentum, $ I $ is the moment of Inertia, and $ \omega $ is the angular velocity.
 $\Rightarrow \omega = \dfrac{{2\pi }}{T} $
Here, $ T $ is the time period.

Complete step by step answer
Now, Since we are given that the new density is 8 times the old density and the mass of Earth is constant, we can use this condition to find the relation between new and old radius.
We know, density of a body is given as,
 $\Rightarrow \rho = \dfrac{M}{V} $
Where, M=Mass of the Earth, V = Volume of the Earth.
Since, Earth is spherical; we get the Volume of Earth as,
 $\Rightarrow V = \dfrac{4}{3}\pi {R^3} $
Now, if the new parameters are represented by the subscript ‘new’, then, we get the new density as,
 $\Rightarrow {\rho _{new}} = \dfrac{{{M_{new}}}}{{{V_{new}}}} $
Since Mass of Earth is constant in both the cases, thus, we get,
 $\Rightarrow
  {M_{new}} = M \\
   \Rightarrow {\rho _{new}}{V_{new}} = \rho V \\
 $
Now, putting in the values, we get,
 $ \Rightarrow
  {M_{new}} = M \\
   \Rightarrow (8\rho )\left( {\dfrac{4}{3}\pi R_{new}^3} \right) = \rho \left( {\dfrac{4}{3}\pi R_{}^3} \right) \\
 $
This gives the relation between old and new radius of Earth as,
 $ \Rightarrow
  R_{new}^3 = \dfrac{{{R^3}}}{8} \\
   \Rightarrow {R_{new}} = \dfrac{R}{2} \\
 $
Now, According to Principle of Conservation of Angular Momentum, if no external torque is applied, the initial and final angular momentum of a rotating system is conserved,
So, the new Angular Momentum and old Angular Momentum will remain same, i.e.
 $ \Rightarrow
  {L_{new}} = L \\
   \Rightarrow {I_{new}}{\omega _{new}} = I\omega \\
 $
So, now putting in the relation between angular velocity and Time period in the above formula we get,
 $\Rightarrow {I_{new}}\left( {\dfrac{{2\pi }}{{{T_{new}}}}} \right) = I\left( {\dfrac{{2\pi }}{T}} \right) $
Now, we know that, Moment of Inertia of a Sphere is given as
 $\Rightarrow I = \dfrac{2}{5}(M{R^2}) $
And the old duration of day was 24 hours, So, we the above equation as,
 $\Rightarrow \left( {\dfrac{2}{5}MR_{new}^2} \right)\left( {\dfrac{{2\pi }}{{{T_{new}}}}} \right) = \left( {\dfrac{2}{5}M{R^2}} \right)\left( {\dfrac{{2\pi }}{{24}}} \right) $
So, now simplifying and putting in the value of $ {R_{new}} $ obtained above, we get,
 $\Rightarrow
  \left( {\dfrac{{R_{new}^2}}{{{T_{new}}}}} \right) = \left( {\dfrac{{{R^2}}}{{24}}} \right) \\
   \Rightarrow \left( {\dfrac{{{{\left( {\dfrac{R}{2}} \right)}^2}}}{{{T_{new}}}}} \right) = \left( {\dfrac{{{R^2}}}{{24}}} \right) \\
 $
Hence, we get the new duration of the day as,
 $\Rightarrow
  \left( {{T_{new}}} \right) = 24\left( {\dfrac{{{R^2}}}{{4{R^2}}}} \right) \\
   \Rightarrow {T_{new}} = \dfrac{{24}}{4} = 6hours \\
 $
$ \therefore $ Option (C) is correct out of the given options.

Note
Here since shrinking the Earth, the density was a variable so we kept the mass of Earth constant. However, if the question had kept the Earth to be of same size then we would have kept the volume to be constant and in that case mass would have been taken as variable.