• # question_answer A ring takes ${{t}_{1}}$ second in rolling down an inclined plane without slipping. The same ring ${{t}_{2}}$ second in sliding down without rolling on a similar but frictionless inclined plane. Then the ratio ${{t}_{1}}:{{t}_{2}}$ is A)  $1:\sqrt{2}$            B)  $\sqrt{2}:1$C)  $1:2$             D)  $2:1$

${{t}_{1}}=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}\left( 1+\frac{{{K}^{2}}}{{{r}^{2}}} \right)}$ For ring $=\frac{{{K}^{2}}}{{{r}^{2}}}=1$ so ${{t}_{1}}=\frac{1}{\sin \theta }\sqrt{\frac{4h}{g}}$ In sliding $\frac{{{K}^{2}}}{{{r}^{2}}}=0$ ${{t}_{2}}=\frac{I}{\sin \theta }\sqrt{\frac{2h}{g}}$      so   ${{t}_{1}}:{{t}_{2}}=\sqrt{2}:1$