Angular Momentum & its Conservation question from JEE Main 2024 (30 Jan, Shift 1)

Answer: 250

1. $I = \frac{1}{2}MR^2 = \frac{1}{2}(5)(2)^2 = 10\ \text{kg m}^2$ for each disc.
2. Conservation of $L$: $I\omega = 2I\omega' \Rightarrow \omega' = 5\ \text{rad/s}$.
3. $K_i = \frac{1}{2}(10)(10)^2 = 500\ \text{J}$, $K_f = \frac{1}{2}(20)(5)^2 = 250\ \text{J}$
4. Energy dissipated $= 500 - 250 = 250\ \text{J}$
5. Answer: 250 J