Question

If the radius of the earth suddenly decreases by half of its present value. Then the time duration of one day will be

• A. 6 h
• B. 8 h
• C. 12 h
• D. 24 h

Answer 1

Answer: (A)

6 h

Answer 2

As we know in the absence of external torque, angular momentum is constant so,
$L _{\text {initial }} = L _{\text {final }}$
$m_{1}\left( r _{1} \times v _{1}\right) =m_{2}\left( r _{2} \times v _{2}\right)$
since, $v =r \omega$
$\Rightarrow m_{1} \omega_{1} r_{1}^{2}=m_{2} \omega_{2} r_{2}^{2}$
For earth mass remain constant (i.e., $m_{1}=m_{2}$ )
$\Rightarrow \omega_{1} r_{1}^{2}=\omega_{2} r_{2}^{2} \left(\because \omega=\frac{2 \pi}{T}\right)$
$\Rightarrow \frac{r_{1}^{2}}{T_{1}}=\frac{r_{2}^{2}}{T_{2}}$
Given, $T_{1}=24\, h,\, r_{1}=R_{1}$
and $r_{2}=\frac{R_{1}}{2} \Rightarrow \frac{T_{2}}{24}=\frac{R_{1}^{2}}{R_{1}^{2} 4}$
$\Rightarrow T_{2}=6\, h$