Rotation 1 Question 2
2. A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass of $\frac{7 M}{8}$ and is converted into a uniform disc of radius $2 R$. The second part is converted into a uniform solid sphere. Let $I _1$ be the moment of inertia of the disc about its axis and $I _2$ be the moment of inertia of the new sphere about its axis.
The ratio $I _1 / I _2$ is given by
(2019 Main, 10 Apri II)
(a) 285
(b) 185
(c) 65
(d) 140
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Answer:
Correct Answer: 2. (d)
Solution:
- The given situation is shown in the figure given below
Density of given sphere of radius $R$ is
$$ \rho=\frac{\text { Mass }}{\text { Volume }}=\frac{M}{\frac{4}{3} \pi R^{3}} $$
Let radius of sphere formed from second part is $r$, then mass of second part $=$ volume $\times$ density
$$ \begin{aligned} \frac{1}{8} M & =\frac{4}{3} \pi r^{3} \times \frac{M}{\frac{4}{3} \pi R^{3}} \\ \therefore \quad r^{3} & =\frac{R^{3}}{8} \Rightarrow r=\frac{R}{2} \end{aligned} $$
Now, $I _1=$ moment of inertia of disc (radius $2 R$ and mass $\frac{7}{8} M$ ) about its axis
$$ =\frac{\operatorname{Mass} \times(\text { Radius })^{2}}{2}=\frac{\frac{7}{8} M \times(2 R)^{2}}{2}=\frac{7}{4} M R^{2} $$
and $I _2=$ moment of inertia of sphere
$$ \text { (radius } \frac{R}{2} \text { and mass } \frac{1}{8} M \text { ) about its axis } $$
$=\frac{2}{5} \times$ Mass $\times(\text { Radius })^{2}=\frac{2}{5} \times \frac{1}{8} M \times \frac{R}{2}{ }^{2}=\frac{M R^{2}}{80}$
$\therefore$ Ratio $\frac{I _1}{I _2}=\frac{\frac{7}{4} M R^{2}}{\frac{1}{80} M R^{2}}=140$
