The following question appeared in IIT-JEE 2008 question paper under Reasoning Type Questions.

**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.

**(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

The acceleration (*a*)* *of a body of mass *M* rolling down an inclined plane is given by

*a* = *g *sin*θ*/[1 + (*I*/*MR*^{2})]

where *θ* is the angle of the plane (with respect to the horizontal), *I* is the moment of inertia (about the axis of rolling) and *R* is the radius of the body.

Since the moments of inertia of solid cylinder and hollow cylinder are respectively *MR*^{2}/2 and *MR*^{2} the acceleration *a* is greater for the solid cylinder. Therefore, the *solid cylinder* will reach the bottom of the inclined plane *first*.

Since the cylinders have the same mass and are at the same height they have the *same* initial gravitational potential energy *Mgh*. This potential energy gets converted into translational and rotational kinetic energies (obeying the law of conservation of energy) when the cylinders roll down the incline and the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

A, B and C are three equal point masses (*m* each) rigidly connected by massless rods of length *L* forming an equilateral triangle as shown in the adjoining figure. The system is first rotated with constant angular velocity *ω* about an axis perpendicular to the plane of the triangle and passing through A. Next it is rotated with the same constant angular velocity *ω* about the side AB of the triangle. If *K*_{1} and *K*_{2} are the kinetic energies of the system in the first and the second cases respectively, the ratio *K*_{1}/*K*_{2} is

(a) 2/3

(b) 4/3

(c) 8/3

(d) 10/3

(e) 16/3

The rotational kinetic energy (*K*) is given by

*K* = ½ *Iω*^{2}

Since the angular velocity is the same in the two cases, the ratio of kinetic energies must be equal to the ratio of moments of inertia.

Thereore, *K*_{1}/*K*_{2} = *I*_{1}/*I*_{2} = 2*mL*^{2}/*m*(*L*sin60º)^{2}

[Note that in the second case the moment of inertia of the system is due to the single mass at C which is at distance *L*sin60º from the axis AB].

Thus *K*_{1}/*K*_{2} = 2/(√3/2)^{2} = 8/3.

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