Question

If the earth becomes a thin shell of the same radius the length of a day becomes
A. 24 hr
B. 6 hr
C. {\raise0.7ex\hbox{7} \\!\mathord{\left/ {\vphantom {7 2}}\right.} \\!\lower0.7ex\hbox{2}}\,hr
D. 40 hr

Answer 1

Hint
The given question is based on the concept of the relation between the time period and the moment of inertia. Using the formulae for the relation between the time period and the moment of inertia, we can consider two cases and equate them. Finally, by substituting the given value we can find the value time period in the second case.
In this solution we will be using the following formula,
$\Rightarrow T = 2\pi \sqrt {\dfrac{I}{{mgl}}}$
Where $T$ is the time period, $I$ is the moment of inertia
$m$ is the mass, $l$ is the length and $g$ is the acceleration due to gravity.
\Rightarrow I = {\raise0.7ex\hbox{2} \\!\mathord{\left/ {\vphantom {2 5}}\right.} \\!\lower0.7ex\hbox{5}}M{R^2} for solid sphere and
\Rightarrow I = {\raise0.7ex\hbox{2} \\!\mathord{\left/ {\vphantom {2 3}}\right.} \\!\lower0.7ex\hbox{3}}M{R^2} for spherical shells.

Complete step by step answer
Firstly, let us derive the relation between the time period and the moment of inertia.
Let us consider a pendulum for derivation purposes.
The time period of a pendulum depends on the moment of inertia of the body around the pivot point and the distance from the pivot to the body’s centre of mass.
Thus, the time period of the pendulum is given by,
$\Rightarrow T = 2\pi \sqrt {\dfrac{I}{{mgl}}}$
Hence we can write,
$\Rightarrow T \propto \sqrt I$
Thus, we have obtained the relation between the time period and moment of inertia.
We will be using this relation for further calculation.
The moment of inertia of a solid sphere is given as by the formula,
\Rightarrow I = {\raise0.7ex\hbox{2} \\!\mathord{\left/ {\vphantom {2 5}}\right.} \\!\lower0.7ex\hbox{5}}M{R^2}
Where $M$ is the mass and $R$ is the radius of the sphere.
The moment of inertia of a thin spherical shell given as,
\Rightarrow I = {\raise0.7ex\hbox{2} \\!\mathord{\left/ {\vphantom {2 3}}\right.} \\!\lower0.7ex\hbox{3}}M{R^2}
Where $M$ is the mass and $R$ is the radius of the shell.
As the time period is proportional to the square root of the moment of inertia, we can express the above equations in terms of a ratio.
So, we have,
$\Rightarrow \dfrac{{{T_{sphere}}}}{{{T_{shell}}}} = \dfrac{{\sqrt {{I_{sphere}}} }}{{\sqrt {{I_{shell}}} }}$
Substituting the values,
\Rightarrow \dfrac{{{T_{sphere}}}}{{{T_{shell}}}} = \dfrac{{\sqrt {{\raise0.7ex\hbox{2} \\!\mathord{\left/ {\vphantom {2 5}}\right.} \\!\lower0.7ex\hbox{5}}M{R^2}} }}{{\sqrt {{\raise0.7ex\hbox{2} \\!\mathord{\left/ {\vphantom {2 3}}\right.} \\!\lower0.7ex\hbox{3}}M{R^2}} }}
The $M$ and $R$ gets cancelled from the numerator and the denominator and we are left with,
$\Rightarrow \dfrac{{{T_{sphere}}}}{{{T_{shell}}}} = \sqrt {\dfrac{3}{5}}$
We know that the length of a day is 24 hours when the earth is a solid sphere. So, we substitute this value in the above equation to obtain the value of the length of the day when the earth becomes a thin shell of the same radius.
So, we have,
$\Rightarrow \dfrac{{24}}{{{T_{shell}}}} = \sqrt {\dfrac{3}{5}}$
Hence taking ${T_{shell}}$ to the RHS and rest of the terms to the LHS we get,
$\Rightarrow {T_{shell}} = \sqrt {\dfrac{5}{3}} \times 24$
Calculating this gives a value of,
$\Rightarrow {T_{shell}} = 30.98 hrs$
$\therefore$ The length of the day, if the earth becomes a thin shell of the same radius is approximately 30 hours, thus, none of the options is correct.

Note
The moment of inertia is the measure of rotational inertia of the body. In simple words, the moment of inertia is a physical quantity that describes how easily a body can be rotated about a given axis. There are two different formulae for the moment of inertia of the sphere. They are, for a solid sphere, the moment of inertia is I = {\raise0.7ex\hbox{2} \\!\mathord{\left/ {\vphantom {2 5}}\right.} \\!\lower0.7ex\hbox{5}}M{R^2}, and for the hollow sphere, also called a thin shell, the moment of inertia is I = {\raise0.7ex\hbox{2} \\!\mathord{\left/ {\vphantom {2 3}}\right.} \\!\lower0.7ex\hbox{3}}M{R^2}.