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, we will follow these steps:

### Step 1: Calculate the initial angular momentum of each disc
The angular momentum \( L \) of a rotating object is given by the formula:
\[ L = I \cdot \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

For the first disc:
- Moment of inertia \( I_1 = 0.1 \, \text{kg m}^2 \)
- Angular velocity \( \omega_1 = 10 \, \text{rad/s} \)

Calculating the angular momentum of the first disc:
\[ L_1 = I_1 \cdot \omega_1 = 0.1 \cdot 10 = 1 \, \text{kg m}^2/\text{s} \]

For the second disc:
- Moment of inertia \( I_2 = 0.2 \, \text{kg m}^2 \)
- Angular velocity \( \omega_2 = 5 \, \text{rad/s} \)

Calculating the angular momentum of the second disc:
\[ L_2 = I_2 \cdot \omega_2 = 0.2 \cdot 5 = 1 \, \text{kg m}^2/\text{s} \]

### Step 2: Apply conservation of angular momentum
When the two discs stick together, the total angular momentum before they stick together must equal the total angular momentum after they stick together.

Total initial angular momentum:
\[ L_{\text{total}} = L_1 + L_2 = 1 + 1 = 2 \, \text{kg m}^2/\text{s} \]

Let \( \omega \) be the common angular velocity after they stick together. The combined moment of inertia \( I_{\text{total}} \) is:
\[ I_{\text{total}} = I_1 + I_2 = 0.1 + 0.2 = 0.3 \, \text{kg m}^2 \]

Using conservation of angular momentum:
\[ L_{\text{total}} = I_{\text{total}} \cdot \omega \]
\[ 2 = 0.3 \cdot \omega \]

Solving for \( \omega \):
\[ \omega = \frac{2}{0.3} = \frac{20}{3} \, \text{rad/s} \]

### Step 3: Calculate the kinetic energy of the combined system
The kinetic energy \( K \) of a rotating system is given by:
\[ K = \frac{1}{2} I \omega^2 \]

Substituting for \( I_{\text{total}} \) and \( \omega \):
\[ K = \frac{1}{2} (I_1 + I_2) \omega^2 = \frac{1}{2} (0.3) \left(\frac{20}{3}\right)^2 \]

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

Now substituting this back into the kinetic energy formula:
\[ K = \frac{1}{2} (0.3) \left(\frac{400}{9}\right) = \frac{0.3 \cdot 400}{2 \cdot 9} = \frac{120}{9} = \frac{40}{3} \, \text{J} \]

### Final Answer
The kinetic energy of the combined system is:
\[ \boxed{\frac{40}{3} \, \text{J}} \]