A cord is wound round the circumference of wheel of radius `r`. The axis of the wheel is horizontal and fixed and moment of inertia about it is `I`. A weight `mg` is attached to the end of the cord and falls from rest. After falling through a distance `h`, the angular velocity of the wheel will be.

Options

A. `sqrt((2 gh)/(I + mr))`
B. `[(2 mgh)/(I + mr^2)]^(1/(2))`
C. `[(2 mgh)/(I +2 m)]^(1/(2))`
D. `sqrt(2 gh)`

Explanation

To solve the problem, we will use the principle of conservation of energy. The potential energy lost by the falling mass will be equal to the kinetic energy gained by both the mass and the rotating wheel.

### Step-by-Step Solution:

1. **Identify the Initial and Final Energies**:
- Initially, the mass \( m \) has potential energy given by:
\[ PE_{\text{initial}} = mgh \]
- The mass starts from rest, so its initial kinetic energy is:
\[ KE_{\text{initial}} = 0 \]

2. **Determine the Final Energies**:
- After falling a distance \( h \), the potential energy becomes:
\[ PE_{\text{final}} = 0 \]
- The kinetic energy of the mass as it falls is:
\[ KE_{\text{mass}} = \frac{1}{2} mv^2 \]
- The kinetic energy of the wheel (rotational kinetic energy) is:
\[ KE_{\text{wheel}} = \frac{1}{2} I \omega^2 \]

3. **Apply Conservation of Energy**:
- According to the conservation of energy:
\[ PE_{\text{initial}} = KE_{\text{mass}} + KE_{\text{wheel}} \]
- Substituting the expressions we have:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

4. **Relate Linear and Angular Velocity**:
- The linear velocity \( v \) of the falling mass is related to the angular velocity \( \omega \) of the wheel by the equation:
\[ v = r\omega \]
- Therefore, we can express \( \omega \) in terms of \( v \):
\[ \omega = \frac{v}{r} \]

5. **Substitute \( \omega \) into the Energy Equation**:
- Substitute \( \omega \) into the energy equation:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \left(\frac{v}{r}\right)^2 \]
- This simplifies to:
\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \frac{I v^2}{r^2} \]

6. **Factor Out \( v^2 \)**:
- Factor out \( v^2 \):
\[ mgh = \frac{1}{2} v^2 \left( m + \frac{I}{r^2} \right) \]

7. **Solve for \( v^2 \)**:
- Rearranging gives:
\[ v^2 = \frac{2mgh}{m + \frac{I}{r^2}} \]

8. **Find \( \omega \)**:
- Now, substituting back to find \( \omega \):
\[ \omega = \frac{v}{r} = \frac{1}{r} \sqrt{\frac{2mgh}{m + \frac{I}{r^2}}} \]

### Final Result:
Thus, the angular velocity \( \omega \) of the wheel after the mass has fallen through a distance \( h \) is given by:
\[ \omega = \frac{1}{r} \sqrt{\frac{2mgh}{m + \frac{I}{r^2}}} \]