1. Given data

• First disc: $I_1 = 0.1\,\text{kg m}^2$, $\omega_1 = 10\,\text{rad s}^{-1}$
• Second disc: $I_2 = 0.2\,\text{kg m}^2$, $\omega_2 = 5\,\text{rad s}^{-1}$

They rotate in the same direction and then stick together.

2. Use conservation of angular momentum

Since no external torque acts about the common axis,

$L_{\text{initial}} = L_{\text{final}}$

Initial angular momentum:

$L_i = I_1\omega_1 + I_2\omega_2$

Substitute values:

$L_i = (0.1)(10) + (0.2)(5) = 1 + 1 = 2\,\text{kg m}^2\text{/s}$

When they stick together, total moment of inertia is

$I = I_1 + I_2 = 0.1 + 0.2 = 0.3\,\text{kg m}^2$

Let final angular speed be $\omega$. Then,

$I\omega = 2$

$0.3\omega = 2$

$\omega = \frac{2}{0.3} = \frac{20}{3}\,\text{rad s}^{-1}$

3. Find final kinetic energy

Rotational kinetic energy of combined system:

$K = \frac{1}{2}I\omega^2$

$K = \frac{1}{2}(0.3)\left(\frac{20}{3}\right)^2$

$K = 0.15 \cdot \frac{400}{9}$

$K = \frac{60}{9} = \frac{20}{3}\,\text{J}$

4. Check options

The obtained value is

$\boxed{\frac{20}{3}\,\text{J}}$

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They match.