1. Given data

• Disc 1: mass $M$, radius $R$, initial angular speed $\omega_1$
• Disc 2: mass $M$, radius $\dfrac{R}{2}$, initially at rest
• Both are coaxial and finally rotate together with angular speed $\omega_2$

We need the percentage loss in rotational kinetic energy.

2. Moment of inertia of each disc

For a solid disc about its central axis, $I = \frac{1}{2}MR^2$

So for disc 1, $I_1 = \frac{1}{2}MR^2$

For disc 2, $I_2 = \frac{1}{2}M\left(\frac{R}{2}\right)^2 = \frac{1}{2}M\cdot \frac{R^2}{4} = \frac{1}{8}MR^2$

3. Apply conservation of angular momentum

Since no external torque acts about the common axis, $I_1\omega_1 + I_2\cdot 0 = (I_1+I_2)\omega_2$

$\frac{1}{2}MR^2\omega_1 = \left(\frac{1}{2}MR^2 + \frac{1}{8}MR^2\right)\omega_2$

$\frac{1}{2}MR^2\omega_1 = \frac{5}{8}MR^2\omega_2$

Thus, $\omega_2 = \frac{\frac{1}{2}}{\frac{5}{8}}\omega_1 = \frac{4}{5}\omega_1$

4. Initial rotational kinetic energy

Only disc 1 is rotating initially: $K_i = \frac{1}{2}I_1\omega_1^2 = \frac{1}{2}\cdot \frac{1}{2}MR^2\omega_1^2 = \frac{1}{4}MR^2\omega_1^2$

5. Final rotational kinetic energy

Both discs rotate together, so total moment of inertia is $I_f = I_1 + I_2 = \frac{1}{2}MR^2 + \frac{1}{8}MR^2 = \frac{5}{8}MR^2$

Hence, $K_f = \frac{1}{2}I_f\omega_2^2$

Substitute $I_f = \frac{5}{8}MR^2$ and $\omega_2 = \frac{4}{5}\omega_1$: $K_f = \frac{1}{2}\cdot \frac{5}{8}MR^2 \left(\frac{4}{5}\omega_1\right)^2$

$K_f = \frac{5}{16}MR^2 \cdot \frac{16}{25}\omega_1^2$

$K_f = \frac{1}{5}MR^2\omega_1^2$

6. Energy lost

$\Delta K = K_i - K_f = \frac{1}{4}MR^2\omega_1^2 - \frac{1}{5}MR^2\omega_1^2$

$\Delta K = \left(\frac{1}{4}-\frac{1}{5}\right)MR^2\omega_1^2 = \frac{1}{20}MR^2\omega_1^2$

7. Percentage loss

$p = \frac{\Delta K}{K_i}\times 100$

$p = \frac{\frac{1}{20}}{\frac{1}{4}}\times 100 = \frac{4}{20}\times 100 = 20$

So, $\boxed{p=20}$