A circular disc of mass M and radius R is rotating about its axis with angular speed $$\omega_1$$. If another stationary disc having radius $$\frac{R}{2}$$ and same mass M is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed $$\omega_2$$. The energy lost in the process is p% of the initial energy. Value of p is

For a uniform circular disc, the moment of inertia about the central axis is given by the formula $$I=\frac12\,M R^{2}.$$

We have one disc of mass $$M$$ and radius $$R$$ already rotating with angular speed $$\omega_1.$$ For this disc

$$I_1=\frac12\,M R^{2}.$$

The second disc is identical in mass but has radius $$\dfrac{R}{2}$$ and is initially at rest. Using the same formula

$$I_2=\frac12\,M\left(\frac{R}{2}\right)^{2} =\frac12\,M\left(\frac{R^{2}}{4}\right) =\frac18\,M R^{2}.$$

When the stationary disc is dropped coaxially, the two discs interact frictionally until they spin together with a common angular speed $$\omega_2.$$ Because the external torque about the axis is zero, angular momentum is conserved:

$$I_1\,\omega_1 = \bigl(I_1+I_2\bigr)\,\omega_2.$$

Substituting the values of the moments of inertia,

$$\frac12\,M R^{2}\,\omega_1 =\left(\frac12\,M R^{2}+\frac18\,M R^{2}\right)\omega_2 =\left(\frac58\,M R^{2}\right)\omega_2.$$

Canceling the common factor $$M R^{2}$$ and solving for $$\omega_2$$ gives

$$\omega_2=\frac{\frac12}{\frac58}\,\omega_1 =\frac12\cdot\frac85\,\omega_1 =\frac45\,\omega_1.$$

The initial rotational kinetic energy of the system is only that of the first disc:

$$E_i=\frac12\,I_1\,\omega_1^{2} =\frac12\left(\frac12\,M R^{2}\right)\omega_1^{2} =\frac14\,M R^{2}\,\omega_1^{2}.$$

After both discs rotate together, the total kinetic energy is

$$E_f=\frac12\bigl(I_1+I_2\bigr)\omega_2^{2} =\frac12\left(\frac58\,M R^{2}\right)\left(\frac45\,\omega_1\right)^{2}.$$ Simplifying step by step,

$$E_f=\frac12\cdot\frac58\,M R^{2}\cdot\frac{16}{25}\,\omega_1^{2} =M R^{2}\,\omega_1^{2}\cdot\frac{1}{2}\cdot\frac{5}{8}\cdot\frac{16}{25} =M R^{2}\,\omega_1^{2}\cdot\frac{80}{400} =\frac15\,M R^{2}\,\omega_1^{2}.$$

The mechanical energy lost in the process is therefore

$$\Delta E = E_i - E_f =\left(\frac14-\frac15\right)M R^{2}\,\omega_1^{2} =\left(\frac5{20}-\frac4{20}\right)M R^{2}\,\omega_1^{2} =\frac1{20}\,M R^{2}\,\omega_1^{2}.$$

To find the percentage loss, compare $$\Delta E$$ with the initial energy $$E_i$$:

$$p=\left(\frac{\Delta E}{E_i}\right)\times100 =\left(\frac{\frac1{20}\,M R^{2}\,\omega_1^{2}} {\frac14\,M R^{2}\,\omega_1^{2}}\right)\times100 =\left(\frac1{20}\times\frac4{1}\right)\times100 =\frac4{20}\times100 =20\%.$$

So, the answer is $$20$$.