Two coaxial discs, having moments of inertia $I_1$ and $I_1/2$ are rotating with respective angular velocities ${\omega }_1$ and ${\omega }_1/2$, about their common axis. They are brought in contact with each other, and thereafter they rotate with a common angular velocity. If $E_f$ and $E_i$, are the final and initial total energies, then $(E_f-E_i)$ is

The correct option is A $-\frac{I_1{\omega }_1^2}{24}$
Initial kinetic energy of the given system,

$E_i=\frac{1}{2}{I}_1{\omega }_1^2+\frac{1}{2}\left(\frac{I_1}{2}\right){\left(\frac{{\omega }_1}{2}\right)}^2$

$E_i=\frac{9}{16}{I}_1{\omega }_1^2.........\left(i\right)$

$\because$ Net external torque on the system is zero, so using law of conservation of angular momentum.

Initial angular momentum = Final angular momentum

$I_1{\omega }_1+\left(\frac{I_1}{2}\right)\left(\frac{{\omega }_1}{2}\right)=I_1\omega +\frac{I_1}{2}\omega$

(assuming angular velocity after contact is $\omega )$

$\Rightarrow \frac{5}{4}{\omega }_1=\frac{3}{2}\omega$

$\therefore \omega =\frac{5}{6}{\omega }_1..........\left(ii\right)$

Now, final kinetic energy (after contact) is

$E_f=\frac{1}{2}{I}_1{\omega }^2+\frac{1}{2}\left(\frac{I_1}{2}\right){\omega }^2=\frac{3}{4}{I}_1{\omega }^2$

Putting the value of $\omega$ from eq. $\left(ii\right)$

$E_f=\frac{3}{4}{I}_1{\left(\frac{5}{6}{\omega }_1\right)}^2=\frac{25}{48}{I}_1{\omega }_1^2$

Hence, change in KE,
$\Delta E=E_f-E_i=\frac{25}{48}{I}_1{\omega }_I^2-\frac{9}{16}{I}_1{\omega }_1^2$

$\therefore E_f-E_i=-\frac{1}{24}{I}_1{\omega }_1^2$

Hence, option $\left(\text{A}\right)$ is correct.