JEE MAIN - Physics (2022 - 25th July Morning Shift - No. 5)
A solid cylinder and a solid sphere, having same mass $$M$$ and radius $$R$$, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be :
$$\sqrt{\frac{5}{3}}$$
$$\sqrt{\frac{4}{5}}$$
$$\sqrt{\frac{3}{5}}$$
$$\sqrt{\frac{14}{15}}$$
Explicació
$$a = {{g\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}$$
$$v = \sqrt {{{2Sg\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}} $$
$$ \Rightarrow {{{v_c}} \over {{v_{ss}}}}\sqrt {{{1 + {{K_{ss}^2} \over {{R^2}}}} \over {1 + {{K_c^2} \over {{R^2}}}}}} = \sqrt {{{1 + {2 \over 5}} \over {1 + {1 \over 2}}}} $$
$$ \Rightarrow \sqrt {{{{7 \over 5}} \over {{3 \over 2}}}} = \sqrt {{{14} \over {15}}} $$