The following bodies,
(1) a ring
(2) a disc
(3) a solid cylinder
(4) a solid sphere,
of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ___________. [Mark the body as per their respective numbering given in the question]

Solution

Physics sir · 3 steps

On a rolling race, the body that stores the least energy in spinning wins, and that is the solid sphere.

1. Rolling acceleration. For a body rolling without slipping down an incline of angle θ, write I = mk². Its acceleration along the incline depends only on the ratio k²/R², not on mass or radius.

   a = (g sin θ)/(1 + k²/R²), t = √(2s/a)

2. Compare the bodies. All four start from rest and cover the same length s. So the one with the smallest k²/R² has the largest a and the shortest time.

   Body	I	k²/R²	a
   (1) Ring	mR²	1	1/2g sin θ
   (2) Disc	1/2mR²	1/2	2/3g sin θ
   (3) Solid cylinder	1/2mR²	1/2	2/3g sin θ
   (4) Solid sphere	2/5mR²	2/5	5/7g sin θ, largest

3. So. The solid sphere has the smallest k²/R² and the largest acceleration. It reaches the bottom first, so the answer is 4.

The common slip: 1. Taking the largest k²/R² as the fastest gives 1, but the ring puts the most energy into spinning and actually arrives last.

The idea to keep. In a rolling race from rest, a smaller k²/R² means more of the energy goes into moving forward. Order of arrival: solid sphere, then disc and cylinder together, then the ring.

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