1. Rolling motion down an incline

For a body rolling बिना slipping down an incline, the linear acceleration is

$a = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$

where:

• $I$ = moment of inertia about the center,
• $m$ = mass,
• $R$ = radius.

So, the body with smaller $\frac{I}{mR^2}$ gets larger acceleration and reaches first.

2. Compare hollow and solid cylinders

For a solid cylinder,

$I_{\text{solid}} = \frac{1}{2}mR^2$

Hence,

$a_{\text{solid}} = \frac{g\sin\theta}{1 + \frac{1}{2}} = \frac{2}{3}g\sin\theta$

For a hollow cylinder (thin-walled metal cylinder),

$I_{\text{hollow}} = mR^2$

Hence,

$a_{\text{hollow}} = \frac{g\sin\theta}{1 + 1} = \frac{1}{2}g\sin\theta$

Since

$\frac{2}{3}g\sin\theta > \frac{1}{2}g\sin\theta$

the solid cylinder reaches the bottom first, not the hollow cylinder.

So, Statement-1 is False.

3. Check Statement-2 using conservation of energy

Both start from the same height $h$ and roll without slipping.

Loss in potential energy = total kinetic energy at bottom:

$mgh = K_{\text{total}}$

Thus for each cylinder,

$K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

Since both have the same mass and descend through the same height, the total kinetic energy of both at the bottom is the same:

$K_{\text{total, hollow}} = K_{\text{total, solid}} = mgh$

So, Statement-2 is True.

4. Does Statement-2 explain Statement-1?

No. In fact, Statement-1 is false. Equal total kinetic energy does not mean equal speed or equal acceleration during descent. The division between translational and rotational kinetic energy depends on moment of inertia.

5. Final option

• Statement-1: False
• Statement-2: True

Therefore, the correct option is:

$\boxed{D}$