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Moment of inertia of big drop is $I$. If 8 droplets are formed from big drop, then moment of inertia of small droplet is
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Verified Answer
The correct answer is:
$\frac{I}{32}$
Moment of inertia of big drop is $I=\frac{2}{5} M R^{2}$ When small droplets are formed from big drop volume of liquid remain same
$$
\begin{aligned}
n \frac{4}{3} \pi r^{3} &=\frac{4}{3} \pi R^{3} \\
n^{1 / 3} r &=R \text { as } n=8 \Rightarrow r=\frac{R}{2}
\end{aligned}
$$
Mass of each small droplet $=\frac{M}{8}$
$\therefore$ Moment of inertia of each small droplet
$$
\begin{array}{l}
=\frac{2}{5}\left[\frac{M}{8}\right]\left[\frac{R}{2}\right]^{2} \\
=\frac{1}{32}\left[\frac{2}{5} M R^{2}\right]=\frac{I}{32}
\end{array}
$$
$$
\begin{aligned}
n \frac{4}{3} \pi r^{3} &=\frac{4}{3} \pi R^{3} \\
n^{1 / 3} r &=R \text { as } n=8 \Rightarrow r=\frac{R}{2}
\end{aligned}
$$
Mass of each small droplet $=\frac{M}{8}$
$\therefore$ Moment of inertia of each small droplet
$$
\begin{array}{l}
=\frac{2}{5}\left[\frac{M}{8}\right]\left[\frac{R}{2}\right]^{2} \\
=\frac{1}{32}\left[\frac{2}{5} M R^{2}\right]=\frac{I}{32}
\end{array}
$$
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