A solid cylinder of mass `M` and radius `R` rolls down an inclined plane of height `h` without slipping. The speed of its centre when it reaches the bottom is.

Options

A. `sqrt((2 gh))`
B. `sqrt((4//3)gh)`
C. `sqrt((3//4)gh)`
D. `sqrt((4g//h)`

Explanation

To find the speed of the center of mass of a solid cylinder of mass \( M \) and radius \( R \) rolling down an inclined plane of height \( h \) without slipping, we can use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify Initial and Final Energy States:**
- Initially, the cylinder is at rest at a height \( h \), so its initial potential energy (PE) is given by:
\[ PE_{\text{initial}} = Mgh \]
- At the bottom of the incline, the cylinder has both translational kinetic energy (KE) and rotational kinetic energy (KE).

2. **Write the Expression for Final Kinetic Energy:**
- The total kinetic energy when the cylinder reaches the bottom is the sum of translational and rotational kinetic energy:
\[ KE_{\text{final}} = KE_{\text{translational}} + KE_{\text{rotational}} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 \]
- For a solid cylinder, the moment of inertia \( I \) about its central axis is:
\[ I = \frac{1}{2} M R^2 \]
- Since the cylinder rolls without slipping, the relationship between linear speed \( v \) and angular speed \( \omega \) is:
\[ v = \omega R \quad \Rightarrow \quad \omega = \frac{v}{R} \]

3. **Substitute \( I \) and \( \omega \) into the Kinetic Energy Equation:**
- Substitute \( I \) and \( \omega \) into the kinetic energy equation:
\[ KE_{\text{final}} = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2 \]
- Simplifying the rotational kinetic energy term:
\[ KE_{\text{final}} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2 \]

4. **Apply Conservation of Energy:**
- Set the initial potential energy equal to the final kinetic energy:
\[ Mgh = \frac{3}{4} M v^2 \]
- Cancel \( M \) from both sides (assuming \( M \neq 0 \)):
\[ gh = \frac{3}{4} v^2 \]

5. **Solve for \( v^2 \):**
- Rearranging gives:
\[ v^2 = \frac{4}{3} gh \]

6. **Take the Square Root to Find \( v \):**
- Finally, taking the square root gives:
\[ v = \sqrt{\frac{4}{3} gh} \]

### Final Answer:
The speed of the center of mass of the cylinder when it reaches the bottom is:
\[ v = \frac{2}{\sqrt{3}} \sqrt{gh} \]