MHT CET · Physics · Rotational Motion
A solid metallic sphere of radius ' \(R\) ' having moment of inertia ' I ' about diameter is melted and recast into a solid disc of radius ' \(r\) ' of a uniform thickness. The moment of inertia of a disc about an axis passing through its edge and perpendicular to its plane is also equal to ' I '. The ratio \(\frac{r}{R}\) is
- A \(\frac{1}{\sqrt{2}}\)
- B \(\frac{2}{\sqrt{5}}\)
- C \(\frac{2}{\sqrt{10}}\)
- D \(\frac{2}{\sqrt{15}}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{\sqrt{15}}\)
Step-by-step Solution
Detailed explanation
M.I. of the solid sphere about a diameter
\(\mathrm{I}=\frac{2}{5} \mathrm{MR}^2...(i)\)
M.I. of the disc about an axis through its edge and perpendicular to its plane is
\(\begin{array}{ll}
& I=\frac{3}{2} \mathrm{Mr}^2 ...(ii)\\
\therefore \quad & \frac{2}{5} \mathrm{MR}^2=\frac{3}{2} \mathrm{Mr}^2 \\
\therefore \quad & \mathrm{r}=\frac{2}{\sqrt{15}} \mathrm{R}
\end{array}\)
...[From(i) and (ii)]
\(\mathrm{I}=\frac{2}{5} \mathrm{MR}^2...(i)\)
M.I. of the disc about an axis through its edge and perpendicular to its plane is
\(\begin{array}{ll}
& I=\frac{3}{2} \mathrm{Mr}^2 ...(ii)\\
\therefore \quad & \frac{2}{5} \mathrm{MR}^2=\frac{3}{2} \mathrm{Mr}^2 \\
\therefore \quad & \mathrm{r}=\frac{2}{\sqrt{15}} \mathrm{R}
\end{array}\)
...[From(i) and (ii)]
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