MHT CET · Physics · Rotational Motion
Moment of inertia of big drop is \(I\). If 8 droplets are formed from big drop, then moment of inertia of small droplet is
- A \(\frac{I}{32}\)
- B \(\frac{I}{16}\)
- C \(\frac{I}{8}\)
- D \(\frac{I}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{I}{32}\)
Step-by-step Solution
Detailed explanation
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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