A flywheel is in the form of a uniform circular disc of radius `1 m` and mass `2 kg`. The work which must be done on it to increase its frequency of rotation from `5` to `10 rev//s` is approximately

Options

A. `1.5 xx 10^(2) J`
B. `3.0 xx 10^(2) J`
C. `1.5 xx 10^(3) J`
D. `3.0 xx 10^(3) J`

Explanation

To solve the problem of finding the work done on a flywheel to increase its frequency of rotation from 5 to 10 revolutions per second, we can follow these steps:

### Step 1: Calculate the Moment of Inertia (I)
The moment of inertia \( I \) for a uniform circular disc is given by the formula:
\[ I = \frac{1}{2} m r^2 \]
Where:
- \( m = 2 \, \text{kg} \) (mass of the disc)
- \( r = 1 \, \text{m} \) (radius of the disc)

Substituting the values:
\[ I = \frac{1}{2} \times 2 \times (1^2) = \frac{1}{2} \times 2 \times 1 = 1 \, \text{kg m}^2 \]

### Step 2: Convert Frequencies to Angular Velocities
The initial frequency \( f_i = 5 \, \text{rev/s} \) can be converted to angular velocity \( \omega_i \) using the formula:
\[ \omega_i = 2 \pi f_i \]
Calculating \( \omega_i \):
\[ \omega_i = 2 \pi \times 5 = 10 \pi \, \text{rad/s} \]

The final frequency \( f_f = 10 \, \text{rev/s} \) can be converted to angular velocity \( \omega_f \):
\[ \omega_f = 2 \pi f_f \]
Calculating \( \omega_f \):
\[ \omega_f = 2 \pi \times 10 = 20 \pi \, \text{rad/s} \]

### Step 3: Calculate the Change in Rotational Kinetic Energy
The work done on the flywheel is equal to the change in its rotational kinetic energy, which is given by:
\[ W = \Delta KE = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2 \]
Substituting the values of \( I \), \( \omega_f \), and \( \omega_i \):
\[ W = \frac{1}{2} \times 1 \times (20 \pi)^2 - \frac{1}{2} \times 1 \times (10 \pi)^2 \]

Calculating each term:
\[ W = \frac{1}{2} \times 1 \times (400 \pi^2) - \frac{1}{2} \times 1 \times (100 \pi^2) \]
\[ W = 200 \pi^2 - 50 \pi^2 = 150 \pi^2 \]

### Step 4: Calculate the Numerical Value of Work Done
Using \( \pi \approx 3.14 \):
\[ W \approx 150 \times (3.14^2) \approx 150 \times 9.86 \approx 1479 \, \text{J} \]

### Step 5: Round the Result
Rounding \( 1479 \, \text{J} \) gives approximately:
\[ W \approx 1.48 \times 10^3 \, \text{J} \]

### Conclusion
The work done to increase the frequency of rotation from 5 to 10 rev/s is approximately:
\[ \boxed{1.5 \times 10^3 \, \text{J}} \]