If the angular velocity of a body increases by `20%` then its kinetic energy of rotation will increase by

Options

A. `20%`
B. `30%`
C. `44%`
D. `66%`

Explanation

To solve the problem of how the kinetic energy of rotation changes when the angular velocity increases by 20%, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Kinetic Energy of Rotation**:
The kinetic energy (K) of a rotating body is given by the formula:
\[ K = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

2. **Initial Conditions**:
Let the initial angular velocity be \( \omega_1 \) and the initial kinetic energy be \( K_1 \):
\[ K_1 = \frac{1}{2} I \omega_1^2 \]

3. **Final Conditions**:
If the angular velocity increases by 20%, the new angular velocity \( \omega_2 \) can be expressed as:
\[ \omega_2 = \omega_1 + 0.2 \omega_1 = 1.2 \omega_1 \]

4. **Calculating Final Kinetic Energy**:
The final kinetic energy \( K_2 \) when the angular velocity is \( \omega_2 \) is:
\[ K_2 = \frac{1}{2} I \omega_2^2 = \frac{1}{2} I (1.2 \omega_1)^2 \]
Simplifying this:
\[ K_2 = \frac{1}{2} I (1.44 \omega_1^2) = 1.44 \left(\frac{1}{2} I \omega_1^2\right) = 1.44 K_1 \]

5. **Finding the Increase in Kinetic Energy**:
The increase in kinetic energy \( \Delta K \) is given by:
\[ \Delta K = K_2 - K_1 = 1.44 K_1 - K_1 = (1.44 - 1) K_1 = 0.44 K_1 \]

6. **Calculating Percentage Increase**:
The percentage increase in kinetic energy is calculated as:
\[ \text{Percentage Increase} = \frac{\Delta K}{K_1} \times 100 = \frac{0.44 K_1}{K_1} \times 100 = 44\% \]

### Final Answer:
The kinetic energy of rotation will increase by **44%**.