Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are `0.1` kg `-m^(2)` and 10 rad `s^(-1)` respectively while those for the second one are`0.2` kg ` -m^(2)` and `5"rad s"^(-1)` respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The Kinetic energy of the combined system is:

Options

A. `2/3` J
B. `10/3` J
C. `5/3 ` J
D. `20/3` J

Explanation

To solve the problem of two uniform circular discs rotating independently and then getting stuck together, we will follow these steps:

### Step 1: Identify Given Values
- For Disc 1:
- Moment of Inertia, \( I_1 = 0.1 \, \text{kg m}^2 \)
- Angular Velocity, \( \omega_1 = 10 \, \text{rad/s} \)

- For Disc 2:
- Moment of Inertia, \( I_2 = 0.2 \, \text{kg m}^2 \)
- Angular Velocity, \( \omega_2 = 5 \, \text{rad/s} \)

### Step 2: Calculate Total Moment of Inertia
The total moment of inertia \( I \) of the combined system when the discs are stuck together is given by:
\[ I = I_1 + I_2 = 0.1 + 0.2 = 0.3 \, \text{kg m}^2 \]

### Step 3: Apply Conservation of Angular Momentum
According to the conservation of angular momentum, the initial angular momentum before they get stuck together is equal to the final angular momentum after they stick together.

The initial angular momentum \( L_i \) is:
\[ L_i = I_1 \omega_1 + I_2 \omega_2 \]
Substituting the values:
\[ L_i = (0.1 \times 10) + (0.2 \times 5) = 1 + 1 = 2 \, \text{kg m}^2/\text{s} \]

The final angular momentum \( L_f \) after they stick together is:
\[ L_f = I \omega \]
Where \( \omega \) is the common angular velocity after they stick together.

Setting \( L_i = L_f \):
\[ 2 = 0.3 \omega \]

### Step 4: Solve for Common Angular Velocity \( \omega \)
Rearranging the equation gives:
\[ \omega = \frac{2}{0.3} = \frac{20}{3} \, \text{rad/s} \]

### Step 5: Calculate Kinetic Energy of the Combined System
The kinetic energy \( KE \) of the combined system is given by:
\[ KE = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 0.3 \times \left(\frac{20}{3}\right)^2 \]

Calculating \( \left(\frac{20}{3}\right)^2 \):
\[ \left(\frac{20}{3}\right)^2 = \frac{400}{9} \]

Now substituting back into the kinetic energy equation:
\[ KE = \frac{1}{2} \times 0.3 \times \frac{400}{9} = \frac{0.3 \times 400}{18} = \frac{120}{18} = \frac{20}{3} \, \text{J} \]

### Final Answer
The kinetic energy of the combined system is:
\[ \boxed{\frac{20}{3} \, \text{J}} \]