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A solid sphere of mass 'M' and radius 'R' is rotating about its diameter. A disc of same mass and radius is also rotating about an axis passing through its centre and perpendicular to the plane but angular speed is twice that of the sphere. The ratio of kinetic energy of disc to that of sphere is
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The correct answer is:
$5: 1$
Kinetic energy is given by :
For sphere $\mathrm{k}_{\mathrm{s}}=\frac{1}{2} \mathrm{I}_{\mathrm{s}} \omega_{\mathrm{s}}^{2}$
For $\operatorname{disc} \mathrm{k}_{\mathrm{d}}=\frac{1}{2} \mathrm{I}_{\mathrm{d}} \omega_{\mathrm{d}}^{2}$
$\frac{\mathrm{k}_{\mathrm{d}}}{\mathrm{k}_{\mathrm{s}}}=\frac{\mathrm{I}_{\mathrm{d}} \omega_{\mathrm{d}}^{2}}{\mathrm{I}_{\mathrm{s}} \omega_{\mathrm{s}}^{2}}=\frac{\frac{1}{2} \mathrm{MR}^{2}(2 \omega)^{2}}{\frac{2}{5} \mathrm{MR}^{2} \omega^{2}}=5$
For sphere $\mathrm{k}_{\mathrm{s}}=\frac{1}{2} \mathrm{I}_{\mathrm{s}} \omega_{\mathrm{s}}^{2}$
For $\operatorname{disc} \mathrm{k}_{\mathrm{d}}=\frac{1}{2} \mathrm{I}_{\mathrm{d}} \omega_{\mathrm{d}}^{2}$
$\frac{\mathrm{k}_{\mathrm{d}}}{\mathrm{k}_{\mathrm{s}}}=\frac{\mathrm{I}_{\mathrm{d}} \omega_{\mathrm{d}}^{2}}{\mathrm{I}_{\mathrm{s}} \omega_{\mathrm{s}}^{2}}=\frac{\frac{1}{2} \mathrm{MR}^{2}(2 \omega)^{2}}{\frac{2}{5} \mathrm{MR}^{2} \omega^{2}}=5$
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