JEE Challenger
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Rolling Motion of Hollow and Solid Cylinders Down Incline

STATEMENT-1
Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
and
STATEMENT-2
By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

Options

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

C

STATEMENT-1 is True, STATEMENT-2 is False

D

STATEMENT-1 is False, STATEMENT-2 is True

Correct

Step-by-Step Solution

To determine the correctness of the statements, let us analyze the physics of rolling motion down an inclined plane.

Analysis of STATEMENT-1:
Consider a body of mass MM, radius RR, and moment of inertia about its central axis IcmI_{\text{cm}} rolling without slipping down an inclined plane of angle θ\theta.

The linear acceleration of the center of mass down the incline is given by:
a=gsin⁡θ1+IcmMR2a = \frac{g \sin\theta}{1 + \frac{I_{\text{cm}}}{MR^2}}

  1. For the hollow cylinder (thin-walled pipe):
    Icm=MR2  ⟹  ahollow=gsin⁡θ1+1=12gsin⁡θI_{\text{cm}} = MR^2 \implies a_{\text{hollow}} = \frac{g \sin\theta}{1 + 1} = \frac{1}{2}g \sin\theta

  2. For the solid cylinder:
    Icm=12MR2  ⟹  asolid=gsin⁡θ1+12=23gsin⁡θI_{\text{cm}} = \frac{1}{2}MR^2 \implies a_{\text{solid}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{2}{3}g \sin\theta

Since asolid>ahollowa_{\text{solid}} > a_{\text{hollow}}, the solid cylinder accelerates faster.

Given that both cylinders start from rest and travel the same distance ss along the incline, the time taken to reach the bottom is:
t=2sat = \sqrt{\frac{2s}{a}}

Because asolid>ahollowa_{\text{solid}} > a_{\text{hollow}}, we have:
tsolid<thollowt_{\text{solid}} < t_{\text{hollow}}

Hence, the solid cylinder reaches the bottom first, not the hollow cylinder.
Therefore, STATEMENT-1 is False.


Analysis of STATEMENT-2:
During rolling without slipping, the point of contact is momentarily at rest, so static friction does no work. Consequently, mechanical energy is conserved throughout the motion.

Both cylinders have the same mass MM and are released from the same initial height hh. By the principle of conservation of mechanical energy:
Ktotal=ΔU=MghK_{\text{total}} = \Delta U = Mgh

where Ktotal=Ktrans+KrotK_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} is the total kinetic energy.

Since the mass MM and the vertical drop hh are identical for both cylinders, their total kinetic energies at the bottom of the incline must be identical:
Ktotal, hollow=Ktotal, solid=MghK_{\text{total, hollow}} = K_{\text{total, solid}} = Mgh

Therefore, STATEMENT-2 is True.


Conclusion:
STATEMENT-1 is False, and STATEMENT-2 is True.

This corresponds to Option D.