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A solid oylinder and a solid sphere having same mass and same radius roll down on the same inclined plane. The fatio of the acceleration of the cylinder ' $\mathrm{a}_{\mathrm{c}}$ ' to that of sphere ' $\mathrm{a}_{\mathrm{s}}$ ' is
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The correct answer is:
$\frac {14}{15}$
For sphere: M.I. $\mathrm{I}_{\mathrm{S}}=\frac{2}{5} \dot{\mathrm{MR}}^2$
For cylinder: M.I. $\mathrm{I}_{\mathrm{C}}=\frac{1}{2} \mathrm{MR}^2$
Acceleration: $\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{MR}^2}}$
$\begin{aligned}
& \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{1+\frac{\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2}}{1+\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{MR}^2}}=\frac{\mathrm{MR}^2+\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2+\mathrm{I}_{\mathrm{c}}}=\frac{\mathrm{MR}^2+\frac{2}{5} \mathrm{MR}^2}{\mathrm{MR}^2+\frac{1}{2} \mathrm{MR}^2} \\
& \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{7}{5} \times \frac{2}{3}=\frac{14}{15}
\end{aligned}$
For cylinder: M.I. $\mathrm{I}_{\mathrm{C}}=\frac{1}{2} \mathrm{MR}^2$
Acceleration: $\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{MR}^2}}$
$\begin{aligned}
& \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{1+\frac{\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2}}{1+\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{MR}^2}}=\frac{\mathrm{MR}^2+\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2+\mathrm{I}_{\mathrm{c}}}=\frac{\mathrm{MR}^2+\frac{2}{5} \mathrm{MR}^2}{\mathrm{MR}^2+\frac{1}{2} \mathrm{MR}^2} \\
& \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{7}{5} \times \frac{2}{3}=\frac{14}{15}
\end{aligned}$
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