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 :

Solution

Physics sir · 2 steps

Same drop, same mass: the body that spends less energy on spinning reaches the bottom faster.

1. Energy conservation. There is no slipping, so friction does no work and Mgh turns fully into kinetic energy. Write I = Mk² and use ω = v/R. Then Mgh = 1/2Mv²(1 + k²/R²).

   v² = 2gh/(1 + k²/R²)

   Body	k²/R²	1 + k²/R²	v²
   Solid cylinder	1/2	3/2	4gh/3
   Solid sphere	2/5	7/5	10gh/7

2. So. v_c²/vₛ² = (4/3)/(10/7) = 28/30 = 14/15, so v_c/vₛ = √(14/15). The sphere is slightly faster, and the answer is (d).

The idea to keep. For rolling from rest, v² = 2gh/(1 + k²/R²). Mass and radius drop out. A smaller k²/R² means less energy goes into rotation, so that body is faster.

Answer from the NTA final answer key. Solution by PrepWiser.