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A solid cylinder P rolls without slipping from rest down an inclined plane attaining a speed $v_{p}$ at the bottom. Another smooth solid cyclinder $\mathrm{Q}$ of same mass and dimensions slides without friction from rest down the
inclined plane attaining a speed $v_{q}$ at the bottom. The ratio of the speeds $\left(\frac{v_{p}}{v_{q}}\right)$ is :
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inclined plane attaining a speed $v_{q}$ at the bottom. The ratio of the speeds $\left(\frac{v_{p}}{v_{q}}\right)$ is :
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Verified Answer
The correct answer is:
$\left(\sqrt{\frac{2}{3}}\right)$
$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$
$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} \cdot \frac{M R^{2} \omega^{2}}{2}$
$g h=\frac{1}{2} m v^{2}+\frac{1}{4} m v^{2}=\frac{3}{4} v^{2}$
$\begin{array}{l}V^{2}=\sqrt{\frac{4 g h}{3}} & V=\sqrt{\frac{4 g h}{3}} \\ V=\sqrt{2 g h} \\ & {V_{R}}=\sqrt{\frac{4}{3 \times 2}}=\sqrt{2 / 3}\end{array}$
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