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. 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.

Step 1: Analyze Statement-2 using the conservation of energy.

Both cylinders start from rest at the same height $h$ and have the same mass $m$, meaning they lose the same potential energy.

$$\Delta U = mgh$$

Since they roll without slipping, there is no energy loss due to non-conservative forces.

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

Since $m$, $g$, and $h$ are identical for both cylinders, their total kinetic energies at the bottom are identical, making Statement-2 true.

Step 2: Express the total kinetic energy of a rolling body.

The total kinetic energy is the sum of translational and rotational kinetic energies.

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

For rolling without slipping, the linear velocity $v$ and angular velocity $\omega$ are related by the radius $R$.

$$v = \omega R$$

Substitute $\omega = \frac{v}{R}$ into the kinetic energy equation.

$$K_{\text{total}} = \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right)$$

Step 3: Relate potential energy loss to kinetic energy gain.

Equate the potential energy lost to the total kinetic energy gained.

$$mgh = \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right)$$

Solve for the final linear velocity $v$.

$$v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}}$$

Step 4: Determine the acceleration down the incline.

The acceleration $a$ of a body rolling down an incline of angle $\theta$ depends on its mass distribution.

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

For the hollow cylinder, the moment of inertia is $I = mR^2$.

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

For the solid cylinder, the moment of inertia is $I = \frac{1}{2}mR^2$.

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

Step 5: Compare the accelerations to evaluate Statement-1.

Compare the acceleration values.

$$a_{\text{solid}} > a_{\text{hollow}}$$

Since the solid cylinder has a greater acceleration, it reaches the bottom first, making Statement-1 false.

Answer: Statement-1 is False, Statement-2 is True.