Two discs of moment of inertia $$I_1 = 4$$ kg m$$^2$$ and $$I_2 = 2$$ kg m$$^2$$ about their central axes & normal to their planes, rotating with angular speeds 10 rad s$$^{-1}$$ & 4 rad s$$^{-1}$$ respectively are brought into contact face to face with their axes of rotation coincident. The loss in kinetic energy of the system in the process is _________ J.

Using conservation of angular momentum:

$$I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega$$

$$4(10) + 2(4) = 6\omega \Rightarrow \omega = \frac{48}{6} = 8$$ rad/s.

Loss in KE = Initial KE - Final KE:

$$= \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 - \frac{1}{2}(I_1+I_2)\omega^2$$

$$= \frac{1}{2}(4)(100) + \frac{1}{2}(2)(16) - \frac{1}{2}(6)(64)$$

$$= 200 + 16 - 192 = 24$$ J.

Therefore, the answer is $$\boxed{24}$$.