Question

If the Earth shrinks such that it's density becomes $8$ times of the present value, then the new duration of the day in hours will be:

• A. $24$
• B. $12$
• C. $6$
• D. $3$

Answer 1

Answer: (C)

$6$

During the process of shrinking mass will not change only density will change. Hence mass can be equated for different cases:

Mass before shrinking $=$ mass after shrinking
$\Rightarrow \frac{4}{3}\pi r_1^3{\rho }_1=\frac{4}{3}\pi r_2^3{\rho }_2$, (Where $r_1$ ${\rho }_1$ and $r_2$ ${\rho }_2$ are the radius and density before and after shrinking)
$\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{r_2^3}{r_1^3}=\frac{1}{8}$, (As it given $\frac{{\rho }_1}{{\rho }_2}=\frac{1}{8}$)
$\Rightarrow \frac{r_2}{r_1}=\frac{1}{2}$.........(1)

Now, from conservation of angular momentum
$I_1{w}_1=I_2{w}_2$, (Where $I_1,{\omega }_1,T_1$ and $I_2,{\Omega }_2,T_2$ are moment of inertia, angular velocity and time period of rotation before and after shrinking)
$\Rightarrow I_1\frac{2\pi }{T_1}=I_2\frac{2\pi }{T_2}$
$\Rightarrow \frac{I_1}{I_2}=\frac{T_1}{T_2}$

$\Rightarrow \frac{T_1}{T_2}=\frac{m{r}_1^2}{m{r}_2^2}=4$, (As from equation (1) $\frac{r_2}{r_1}=\frac{1}{2}$)
$\Rightarrow T_1=4T_2$
So, time period will become $\frac{1}{4}$th of $T_1$ or $24hours$
So, duration will be $\frac{24}{4}=6hrs$