1. Initial angular momentum and kinetic energy

The first wheel has:

• moment of inertia $I$
• angular speed $\omega$

The second wheel has:

• moment of inertia $3I$
• initial angular speed $0$

So initially, total angular momentum is $L_i = I\omega$

Initial kinetic energy is $K_i = \frac{1}{2}I\omega^2$

2. After coupling: conserve angular momentum

Since the shaft is frictionless externally, angular momentum is conserved.

Let the common angular speed after coupling be $\omega'$.

Total final moment of inertia: $I_{\text{final}} = I + 3I = 4I$

Applying conservation of angular momentum: $I\omega = (4I)\omega'$

Thus, $\omega' = \frac{\omega}{4}$

3. Final kinetic energy

$K_f = \frac{1}{2}(4I)\left(\frac{\omega}{4}\right)^2$

$K_f = 2I\cdot \frac{\omega^2}{16} = \frac{1}{8}I\omega^2$

4. Loss in kinetic energy

Initial kinetic energy: $K_i = \frac{1}{2}I\omega^2$

Final kinetic energy: $K_f = \frac{1}{8}I\omega^2$

Loss: $\Delta K = K_i - K_f = \left(\frac{1}{2}-\frac{1}{8}\right)I\omega^2 = \frac{3}{8}I\omega^2$

Fractional loss: $\frac{\Delta K}{K_i} = \frac{\frac{3}{8}I\omega^2}{\frac{1}{2}I\omega^2} = \frac{3}{4}$

5. Option check

• A: $0$ ❌
• B: $\frac{5}{6}$ ❌
• C: $\frac{1}{4}$ ❌
• D: $\frac{3}{4}$ ✅

So the correct answer is D.