A solid cylinder rolls down an inclined plane of height `3m` and reaches the bottom of plane with angular velocity of `2sqrt2 rad//s`. The radius of cylinder must be [take `g=10m//s^(2)`]

Options

A. 5 cm
B. 0.5 cm
C. `sqrt(10)` cm
D. `sqrt(5)` m

Explanation

To solve the problem, we will use the principle of conservation of energy. The potential energy at the top of the inclined plane will convert into kinetic energy at the bottom.

### Step-by-step Solution:

1. **Identify the initial potential energy (PE_initial)**:
The potential energy at the top of the incline can be calculated using the formula:
\[ PE_{\text{initial}} = mgh \]
where:
- \( m \) = mass of the cylinder (we will see that it cancels out),
- \( g = 10 \, \text{m/s}^2 \) (given),
- \( h = 3 \, \text{m} \) (height of the incline).

Thus,
\[ PE_{\text{initial}} = mg \cdot 3 \]

2. **Identify the final kinetic energy (KE_final)**:
At the bottom of the incline, the cylinder has both translational and rotational kinetic energy. The total kinetic energy can be expressed as:
\[ KE_{\text{final}} = KE_{\text{translational}} + KE_{\text{rotational}} \]
The translational kinetic energy is given by:
\[ KE_{\text{translational}} = \frac{1}{2} mv_{\text{cm}}^2 \]
The rotational kinetic energy is given by:
\[ KE_{\text{rotational}} = \frac{1}{2} I \omega^2 \]
For a solid cylinder, the moment of inertia \( I \) is:
\[ I = \frac{1}{2} m r^2 \]
Therefore,
\[ KE_{\text{rotational}} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \omega^2 = \frac{1}{4} m r^2 \omega^2 \]

3. **Relate translational velocity and angular velocity**:
Since the cylinder rolls without slipping, the relationship between the translational velocity \( v_{\text{cm}} \) and angular velocity \( \omega \) is:
\[ v_{\text{cm}} = r \omega \]
Therefore,
\[ KE_{\text{translational}} = \frac{1}{2} m (r \omega)^2 = \frac{1}{2} m r^2 \omega^2 \]

4. **Combine the kinetic energies**:
Now substituting back into the total kinetic energy:
\[ KE_{\text{final}} = \frac{1}{2} m r^2 \omega^2 + \frac{1}{4} m r^2 \omega^2 = \left(\frac{1}{2} + \frac{1}{4}\right) m r^2 \omega^2 = \frac{3}{4} m r^2 \omega^2 \]

5. **Set initial potential energy equal to final kinetic energy**:
By conservation of energy:
\[ mgh = KE_{\text{final}} \]
Substituting the expressions we derived:
\[ mg \cdot 3 = \frac{3}{4} m r^2 \omega^2 \]
Canceling \( m \) from both sides:
\[ g \cdot 3 = \frac{3}{4} r^2 \omega^2 \]

6. **Substituting known values**:
We know \( g = 10 \, \text{m/s}^2 \) and \( \omega = 2\sqrt{2} \, \text{rad/s} \):
\[ 10 \cdot 3 = \frac{3}{4} r^2 (2\sqrt{2})^2 \]
Simplifying:
\[ 30 = \frac{3}{4} r^2 \cdot 8 \]
\[ 30 = 6 r^2 \]
\[ r^2 = 5 \]
\[ r = \sqrt{5} \, \text{meters} \]

### Final Answer:
The radius of the cylinder must be \( \sqrt{5} \, \text{meters} \).