MHT CET · Physics · Rotational Motion
A solid sphere of mass 'M' and radius 'R' has moment of inertia 'l' about its diameter. It is recast into a disc of thickness 't' whose moment of inertia about an axis passing through its edge and perpendicular to its plane, remains 'I'. Radius of the disc will be
- A \(\mathrm{R}/\sqrt{19}\)
- B \(R / \sqrt{15}\)
- C \(2 \mathrm{R} / \sqrt{15}\)
- D \(2 R / \sqrt{19}\)
Answer & Solution
Correct Answer
(C) \(2 \mathrm{R} / \sqrt{15}\)
Step-by-step Solution
Detailed explanation
Moment of inertia of solid sphere is
\(\mathrm{I}_{\mathrm{s}}=\frac{2}{5} \mathrm{MR}^{2}\)
Moment of inertia of the disc is given by about the given axis is
\(\begin{aligned}
\mathrm{I}_{d} &=\frac{3}{2} \mathrm{Mr}^{2} \\
\therefore \mathrm{r}^{2} &=\frac{4}{15} \mathrm{R}^{2} \quad \therefore \frac{3}{2} \mathrm{Mr}^{2}=\frac{2}{5} \mathrm{MR}^{2} \\
\therefore \mathrm{r} &=\frac{2 \mathrm{R}}{\sqrt{15}}
\end{aligned}\)
\(\mathrm{I}_{\mathrm{s}}=\frac{2}{5} \mathrm{MR}^{2}\)
Moment of inertia of the disc is given by about the given axis is
\(\begin{aligned}
\mathrm{I}_{d} &=\frac{3}{2} \mathrm{Mr}^{2} \\
\therefore \mathrm{r}^{2} &=\frac{4}{15} \mathrm{R}^{2} \quad \therefore \frac{3}{2} \mathrm{Mr}^{2}=\frac{2}{5} \mathrm{MR}^{2} \\
\therefore \mathrm{r} &=\frac{2 \mathrm{R}}{\sqrt{15}}
\end{aligned}\)
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