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Home » A disc of radius ‘ ‘ and thickness has moment of inertia ‘I’ about an axis passing through its centre and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter isA) B) C) D)
A disc of radius ‘ R ‘ and thickness \frac{\mathrm{R}}{6} has moment of inertia ‘I’ about an axis passing through its centre and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is
A) \frac{\mathrm{I}}{5}
B) \frac{\mathrm{I}}{6}
C) \frac{\mathrm{I}}{32}
D) \frac{\mathrm{I}}{64}
Physics
A disc of radius ‘ R ‘ and thickness \frac{\mathrm{R}}{6} has moment of inertia ‘I’ about an axis passing through its centre and perpendicular to its plane. Disc is melted and recast into a solid sphere. The moment of inertia of a sphere about its diameter is
A) \frac{\mathrm{I}}{5}
B) \frac{\mathrm{I}}{6}
C) \frac{\mathrm{I}}{32}
D) \frac{\mathrm{I}}{64}

Answer is (A)
Volume of disc is \mathrm{A} \cdot \mathrm{d}=\pi \cdot \mathrm{R}^{2} \times \frac{\mathrm{R}}{6}=\frac{\mathrm{R}^{3} \times \pi}{6}
Moment of inertia of disc is \mathrm{I}=\frac{1}{2} \mathrm{MR}^{2}
When the disk is remolded in solid sphere of volume V lowing radius r, then
\begin{array}{l} \frac{\pi R^{3}}{6}=\frac{4}{3} \times \pi r^{3} \Rightarrow \frac{R^{3}}{6} \times \frac{3}{4}=r^{3} \\ \therefore \quad r^{3}=\frac{R^{3}}{8} \Rightarrow r=\frac{R}{2} \end{array}
Moment of inertia of sphere is given by \frac{2}{5} \mathrm{~m} \cdot \mathrm{r}^{2}=\frac{2}{5} \times \mathrm{m} \cdot \frac{\mathrm{R}^{2}}{4}=\frac{\mathrm{MR}^{2}}{10}=\frac{\mathrm{MR}^{2}}{2} \times \frac{1}{5}=\frac{\mathrm{I}}{5}

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