To solve this problem, we need to understand the concept of conservation of angular momentum and the resulting energy loss when two rotating discs are brought into contact.$\\$Given:$\\$• Two discs with the same moment of inertia $I. \\$• Initial angular velocities $\omega_1$ and $\omega_2. \\$When the discs are brought into contact, they will rotate together with a common angular velocity $\omega_f. \\$By the conservation of angular momentum:$\\ I \omega_1 + I \omega_2 = 2I \omega_f \\$Solving for $\omega_f: \\ \omega_f = \frac{\omega_1 + \omega_2}{2} \\$The initial kinetic energy of the system is:$\\ K_i = \frac{1}{2} I \omega_1^2 + \frac{1}{2} I \omega_2^2 \\$The final kinetic energy of the system is:$\\ K_f = \frac{1}{2} (2I) \omega_f^2 = I \left(\frac{\omega_1 + \omega_2}{2}\right)^2 \\$Simplifying $K_f: \\ K_f = \frac{I}{4} (\omega_1 + \omega_2)^2 \\$The loss of energy $\Delta E$ is given by:$\\ \Delta E = K_i - K_f \\$Substitute the expressions for $K_i$ and $K_f: \\ \Delta E = \left(\frac{1}{2} I \omega_1^2 + \frac{1}{2} I \omega_2^2\right) - \frac{I}{4} (\omega_1 + \omega_2)^2 \\$Expand and simplify:$\\ \Delta E = \frac{1}{2} I \omega_1^2 + \frac{1}{2} I \omega_2^2 - \frac{I}{4} (\omega_1^2 + 2\omega_1\omega_2 + \omega_2^2) \\ \Delta E = \frac{1}{2} I \omega_1^2 + \frac{1}{2} I \omega_2^2 - \frac{I}{4} \omega_1^2 - \frac{I}{2} \omega_1 \omega_2 - \frac{I}{4} \omega_2^2 \\ \Delta E = \left(\frac{1}{2} I \omega_1^2 - \frac{I}{4} \omega_1^2\right) + \left(\frac{1}{2} I \omega_2^2 - \frac{I}{4} \omega_2^2\right) - \frac{I}{2} \omega_1 \omega_2 \\ \Delta E = \frac{1}{4} I \omega_1^2 + \frac{1}{4} I \omega_2^2 - \frac{I}{2} \omega_1 \omega_2 \\ \Delta E = \frac{1}{4} I (\omega_1^2 + \omega_2^2 - 2\omega_1 \omega_2) \\ \Delta E = \frac{1}{4} I (\omega_1 - \omega_2)^2 \\$Therefore, the expression for the loss of energy during this process is $\frac{1}{4} I (\omega_1 - \omega_2)^2. \\$This corresponds to Option 1.