Two discs have moments of inertia $$I_1$$ and $$I_2$$ about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, $$\omega_1$$ and $$\omega_2$$ respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by:

The two discs are coupled coaxially, so no external torque acts about the common axis. Therefore, by the principle of conservation of angular momentum, the total angular momentum before contact equals the angular momentum after the discs have come to a common angular speed.

Initially, disc 1 has angular momentum $$I_1\omega_1$$ and disc 2 has angular momentum $$I_2\omega_2$$. Hence the total initial angular momentum is $$L_{\text{initial}} \;=\; I_1\omega_1 \;+\; I_2\omega_2.$$ After they come in contact and rotate together, let the common angular speed be $$\omega_f$$. The combined moment of inertia is $$I_1+I_2,$$ so the final angular momentum is $$L_{\text{final}} \;=\; (I_1+I_2)\,\omega_f.$$ Setting $$L_{\text{initial}}=L_{\text{final}}$$ we have $$(I_1+I_2)\,\omega_f \;=\; I_1\omega_1 + I_2\omega_2.$$ Now solving for the common angular speed, $$\omega_f \;=\; \dfrac{I_1\omega_1 + I_2\omega_2}{I_1+I_2}.$$

The kinetic energy of rotation is given by the formula $$K \;=\; \tfrac12 I\omega^2.$$ Using this, the initial kinetic energy of the system is $$K_{\text{initial}} \;=\; \tfrac12 I_1\omega_1^{\,2} \;+\; \tfrac12 I_2\omega_2^{\,2}.$$ The final kinetic energy, when both discs rotate together with angular speed $$\omega_f,$$ is $$K_{\text{final}} \;=\; \tfrac12 (I_1+I_2)\,\omega_f^{\,2}.$$

We substitute the value of $$\omega_f$$ obtained earlier: $$K_{\text{final}} \;=\; \tfrac12 (I_1+I_2) \left( \dfrac{I_1\omega_1 + I_2\omega_2}{I_1+I_2} \right)^{\!2} \;=\; \tfrac12 \dfrac{(I_1\omega_1 + I_2\omega_2)^{2}}{I_1+I_2}.$$

The loss in kinetic energy is the difference between the initial and final values: $$\Delta K \;=\; K_{\text{initial}} - K_{\text{final}} \;=\; \tfrac12 I_1\omega_1^{\,2} + \tfrac12 I_2\omega_2^{\,2} \;-\; \tfrac12 \dfrac{(I_1\omega_1 + I_2\omega_2)^{2}}{I_1+I_2}.$$

For easier algebra, we factor out $$\tfrac12$$ and bring everything over the common denominator $$I_1+I_2$$: $$\Delta K \;=\; \tfrac12\, \dfrac{(I_1+I_2)(I_1\omega_1^{\,2} + I_2\omega_2^{\,2}) - (I_1\omega_1 + I_2\omega_2)^{2}} {\,I_1+I_2\,}.$$ Now we expand the numerator term by term.

First expand the product: $$(I_1+I_2)(I_1\omega_1^{\,2} + I_2\omega_2^{\,2}) \,=\, I_1^{2}\omega_1^{\,2} \;+\; I_1I_2\omega_2^{\,2} \;+\; I_1I_2\omega_1^{\,2} \;+\; I_2^{2}\omega_2^{\,2}.$$ Next expand the square: $$(I_1\omega_1 + I_2\omega_2)^{2} \,=\, I_1^{2}\omega_1^{\,2} \;+\; I_2^{2}\omega_2^{\,2} \;+\; 2I_1I_2\omega_1\omega_2.$$ Substituting these back into the numerator gives $$$ \bigl[I_1^{2}\omega_1^{\,2}+I_1I_2\omega_2^{\,2}+I_1I_2\omega_1^{\,2}+I_2^{2}\omega_2^{\,2}\bigr] \;-\; \bigl[I_1^{2}\omega_1^{\,2}+I_2^{2}\omega_2^{\,2}+2I_1I_2\omega_1\omega_2\bigr]. $$$

Now we cancel identical terms and collect the remaining ones: $$I_1^{2}\omega_1^{\,2} \text{ and } I_2^{2}\omega_2^{\,2}$$ appear in both brackets and therefore cancel out. What is left is $$I_1I_2\omega_2^{\,2} + I_1I_2\omega_1^{\,2} - 2I_1I_2\omega_1\omega_2 \;=\; I_1I_2\bigl(\omega_1^{\,2} - 2\omega_1\omega_2 + \omega_2^{\,2}\bigr) \;=\; I_1I_2(\omega_1 - \omega_2)^{2}.$$ Hence the numerator simplifies neatly to $$I_1I_2(\omega_1 - \omega_2)^{2}.$$

Putting this back, we obtain $$\Delta K \;=\; \tfrac12\, \dfrac{I_1I_2(\omega_1 - \omega_2)^{2}}{I_1 + I_2}.$$ Thus the loss in kinetic energy is $$\Delta K \;=\; \frac{I_1I_2}{2\,(I_1+I_2)}\,(\omega_1 - \omega_2)^{2}.$$

Comparing with the options, this matches Option C.

Hence, the correct answer is Option C.