K the angular velocity of a body increases by `40%`, its kinetic energy of rotation increases by

Options

A. `80%`
B. `20%`
C. `96%`
D. None of these

Answer

To solve the problem, we need to find out how much the kinetic energy of a rotating body increases when its angular velocity increases by 40%.

### Step-by-Step Solution:

1. **Define Initial Conditions**:
- Let the initial angular velocity be \( \omega \).
- The initial kinetic energy \( K_i \) is given by the formula:
\[ K_i = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia.

2. **Calculate Final Angular Velocity**:
- If the angular velocity increases by 40%, the final angular velocity \( \omega_f \) can be expressed as:
\[ \omega_f = \omega + 0.4\omega = 1.4\omega \]

3. **Calculate Final Kinetic Energy**:
- The final kinetic energy \( K_f \) is given by:
\[ K_f = \frac{1}{2} I \omega_f^2 = \frac{1}{2} I (1.4\omega)^2 \]
- Simplifying this gives:
\[ K_f = \frac{1}{2} I (1.96\omega^2) = 0.98 I \omega^2 \]

4. **Calculate Change in Kinetic Energy**:
- The change in kinetic energy \( \Delta K \) is:
\[ \Delta K = K_f - K_i \]
- Substituting the expressions for \( K_f \) and \( K_i \):
\[ \Delta K = 0.98 I \omega^2 - \frac{1}{2} I \omega^2 \]
- Since \( K_i = \frac{1}{2} I \omega^2 \), we can express this as:
\[ \Delta K = 0.98 I \omega^2 - 0.5 I \omega^2 = (0.98 - 0.5) I \omega^2 = 0.48 I \omega^2 \]

5. **Calculate Percentage Increase in Kinetic Energy**:
- The percentage increase in kinetic energy is given by:
\[ \text{Percentage Increase} = \left( \frac{\Delta K}{K_i} \right) \times 100 \]
- Substituting the values:
\[ \text{Percentage Increase} = \left( \frac{0.48 I \omega^2}{0.5 I \omega^2} \right) \times 100 = \left( \frac{0.48}{0.5} \right) \times 100 = 96\% \]

### Final Answer:
The kinetic energy of rotation increases by **96%**.