Question 201 of 280
If the radius of earth is reduced to three-fourth of its present value without change in its mass then value of duration of the day of earth will be ______ hours 30 minutes.
Show answer and solution
Answer: 13
By conservation of angular momentum
$\begin{aligned}
& \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \\
& \left(\frac{2}{5} \mathrm{MR}^2\right) \frac{2 \pi}{\mathrm{T}_1}=\frac{2}{5} \mathrm{M}\left(\frac{3}{4} \mathrm{R}\right)^2 \frac{2 \pi}{\mathrm{T}_2} \\
& \frac{1}{\mathrm{~T}_1}=\frac{9}{16 \mathrm{~T}_2} \\
& \frac{1}{\mathrm{~T}_2}=\frac{9}{16} \times \mathrm{T}_1=\frac{9}{16} \times 24 \mathrm{hr}=\frac{27}{2} \mathrm{hr}=13 \mathrm{hr} 30 \mathrm{mins} .
\end{aligned}$
$\begin{aligned}
& \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \\
& \left(\frac{2}{5} \mathrm{MR}^2\right) \frac{2 \pi}{\mathrm{T}_1}=\frac{2}{5} \mathrm{M}\left(\frac{3}{4} \mathrm{R}\right)^2 \frac{2 \pi}{\mathrm{T}_2} \\
& \frac{1}{\mathrm{~T}_1}=\frac{9}{16 \mathrm{~T}_2} \\
& \frac{1}{\mathrm{~T}_2}=\frac{9}{16} \times \mathrm{T}_1=\frac{9}{16} \times 24 \mathrm{hr}=\frac{27}{2} \mathrm{hr}=13 \mathrm{hr} 30 \mathrm{mins} .
\end{aligned}$