A solid cylinder starts rolling from a height h on an inclined plane. At some instant t, the ratio of its rotational K.E. and the total K.E. would be :

Options

A. `1:2`
B. `1:3`
C. `2:3`
D. `1:1`

Explanation

To solve the problem of finding the ratio of the rotational kinetic energy to the total kinetic energy of a solid cylinder rolling down an inclined plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Kinetic Energies**:
- The rotational kinetic energy (RKE) of the solid cylinder is given by:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
- The total kinetic energy (TKE) is the sum of the translational kinetic energy and the rotational kinetic energy:
\[ \text{TKE} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v_{cm}^2 \]

2. **Substituting the Moment of Inertia**:
- For a solid cylinder, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} m r^2 \]
- Substitute \( I \) into the expression for RKE:
\[ \text{RKE} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \omega^2 = \frac{1}{4} m r^2 \omega^2 \]

3. **Relate Angular Velocity to Linear Velocity**:
- For pure rolling motion, the relationship between angular velocity \( \omega \) and linear velocity \( v_{cm} \) is:
\[ \omega = \frac{v_{cm}}{r} \]
- Substitute this into the RKE:
\[ \text{RKE} = \frac{1}{4} m r^2 \left(\frac{v_{cm}}{r}\right)^2 = \frac{1}{4} m r^2 \cdot \frac{v_{cm}^2}{r^2} = \frac{1}{4} m v_{cm}^2 \]

4. **Substituting into Total Kinetic Energy**:
- Now substitute RKE into the TKE:
\[ \text{TKE} = \frac{1}{4} m v_{cm}^2 + \frac{1}{2} m v_{cm}^2 \]
- Combine the terms:
\[ \text{TKE} = \frac{1}{4} m v_{cm}^2 + \frac{2}{4} m v_{cm}^2 = \frac{3}{4} m v_{cm}^2 \]

5. **Finding the Ratio**:
- Now we can find the ratio of RKE to TKE:
\[ \text{Ratio} = \frac{\text{RKE}}{\text{TKE}} = \frac{\frac{1}{4} m v_{cm}^2}{\frac{3}{4} m v_{cm}^2} \]
- The \( m v_{cm}^2 \) terms cancel out:
\[ \text{Ratio} = \frac{1/4}{3/4} = \frac{1}{3} \]

### Conclusion:
The ratio of the rotational kinetic energy to the total kinetic energy of the solid cylinder is \( \frac{1}{3} \).