KVPY SX SB Exam 20141 Question Paper

If perfect rolling (solid cylinder P)
According to energy conservation law
$\mathrm{mgh}=\frac{1}{2}\mathit{m}{\mathit{v}}_{\mathit{P}}^2+\frac{1}{2}\mathit{I}{\left(\frac{{\mathit{v}}_{\mathit{P}}}{\mathit{R}}\right)}^2$
Here, I → moment of inertia, R → Radius
$\mathit{I}=\frac{\mathit{m}{\mathit{R}}^2}{2}$
$\mathit{\omega }=\frac{{\mathit{v}}_{\mathit{P}}}{\mathit{R}}$
$\mathrm{mgh}=\frac{1}{2}\mathit{m}{\mathit{v}}_{\mathit{P}}^2+\frac{1}{2}\mathrm{}\frac{\mathit{m}{\mathit{R}}^2}{2}\mathrm{}\frac{{\mathit{v}}_{\mathit{P}}^2}{{\mathit{R}}^2}$
$\mathrm{mgh}=\frac{1}{2}\mathit{m}{\mathit{v}}_{\mathit{P}}^2\left[1+\frac{1}{2}\right]=\frac{1}{2}\mathit{m}{\mathit{v}}_{\mathit{P}}^2\times \frac{3}{2}$
$\mathrm{mgh}=\frac{3}{4}\mathit{m}{\mathit{v}}_{\mathit{P}}^2$
${\mathit{v}}_{\mathit{P}}^2=\frac{4}{3}\mathit{g}\mathit{h}$.... (1)
If sliding without friction
(solid cylinder Q)
According to energy conservation law
$\mathrm{mgh}=\frac{1}{2}\mathit{m}{\mathit{v}}_{\mathit{Q}}^2$
⇒ ${\mathit{v}}_{\mathit{Q}}^2=2\mathit{g}\mathit{h}$ .... (2)
from equation (1) and (2)
$\frac{{\mathit{v}}_{\mathit{Q}}^2}{{\mathit{v}}_{\mathit{P}}^2}=\frac{2\mathit{g}\mathit{h}}{\left(\frac{4}{3}\mathit{g}\mathit{h}\right)}=\frac{3}{2}$
$\frac{{\mathit{v}}_{\mathit{Q}}}{{\mathit{v}}_{\mathit{P}}}=\sqrt{\frac{3}{2}}$