Two solid cylinders $P$ and $Q$ of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder $P$ has most of its mass concentrated near its surface, while $Q$ has most of its mass concentrated near the axis. Which statement(s) is(are) correct?

Step 1: Analyze the mass distribution and moment of inertia.

Let both cylinders have mass $m$ and radius $R$, and define the dimensionless parameter $k$ representing the mass distribution.

$$k = \frac{I}{mR^2}$$

Since cylinder $P$ has its mass concentrated near the surface and cylinder $Q$ has its mass concentrated near the axis, the moment of inertia of $P$ is greater than that of $Q$.

$$k_P > k_Q$$

Step 2: Compare the linear accelerations.

The acceleration $a$ of a body rolling down an incline of angle $\theta$ without slipping is given by the rolling formula.

$$a = \frac{g\sin\theta}{1 + k}$$

Since $k_P > k_Q$, the denominator for $P$ is larger, resulting in a smaller acceleration.

$$a_P < a_Q$$

Thus, statement (2) is false.

Step 3: Compare the times taken to reach the bottom.

For a distance $s$ along the incline starting from rest, the time $t$ is related to the acceleration.

$$s = \frac{1}{2}at^2$$ $$t = \sqrt{\frac{2s}{a}}$$

Since $a_P < a_Q$, the time taken by $P$ is greater than that of $Q$.

$$t_P > t_Q$$

Thus, statement (1) is false.

Step 4: Compare the translational kinetic energies at the bottom.

Using conservation of energy, the loss in potential energy equals the gain in total kinetic energy.

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

Using the rolling without slipping condition $v = \omega R$, express the total energy in terms of $k$.

$$mgh = \frac{1}{2}mv^2(1+k)$$

Solve for the translational kinetic energy.

$$\text{KE}_{\text{trans}} = \frac{1}{2}mv^2$$ $$\text{KE}_{\text{trans}} = \frac{mgh}{1+k}$$

Since $k_P > k_Q$, the translational kinetic energy of $P$ is less than that of $Q$.

$$(\text{KE}_{\text{trans}})_P < (\text{KE}_{\text{trans}})_Q$$

Thus, statement (3) is false.

Step 5: Compare the angular speeds at the bottom.

Express the linear velocity at the bottom of the incline.

$$v = \sqrt{\frac{2gh}{1+k}}$$

The angular speed is related to the linear velocity.

$$\omega = \frac{v}{R}$$ $$\omega = \frac{1}{R}\sqrt{\frac{2gh}{1+k}}$$

Since $k_Q < k_P$, the angular speed of $Q$ is greater than that of $P$.

$$\omega_Q > \omega_P$$

Thus, statement (4) is true.

Answer: Only statement (4) is correct: Cylinder $Q$ reaches the bottom with a larger angular speed than cylinder $P$ ($\omega_Q > \omega_P$).