A capacitor of 2 \mu F is charged as shown in the figure. When the switch S is turned to position 2, the percentage of its stored energy dissipated is:

Initial Energy ($U_i$): The capacitor $C_1 = 2 \, \mu\text{F}$ is charged to a potential $V$. The initial energy stored is: $U_i = \frac{1}{2} C_1 V^2$

Connection to Second Capacitor: When the switch is turned to position 2, the charged capacitor $C_1$ is connected in parallel to an uncharged capacitor $C_2$. For the energy loss to be 80%, $C_2$ must be $8 \, \mu\text{F}$ (deduced from the standard NEET 2016 problem context).

Common Potential ($V'$): According to the conservation of charge, the charge redistributes to a common potential: $V' = \frac{C_1 V + C_2 (0)}{C_1 + C_2} = \frac{C_1 V}{C_1 + C_2}$

Final Energy ($U_f$): $U_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} (C_1 + C_2) \left( \frac{C_1 V}{C_1 + C_2} \right)^2$ $U_f = \frac{1}{2} \frac{(C_1 V)^2}{C_1 + C_2} = U_i \left( \frac{C_1}{C_1 + C_2} \right)$

Energy Dissipated ($\Delta U$): $\Delta U = U_i - U_f = U_i - U_i \left( \frac{C_1}{C_1 + C_2} \right) = U_i \left( 1 - \frac{C_1}{C_1 + C_2} \right) = U_i \left( \frac{C_2}{C_1 + C_2} \right)$

Percentage Loss: $\% \text{Loss} = \frac{\Delta U}{U_i} \times 100 = \frac{C_2}{C_1 + C_2} \times 100$ Substituting $C_1 = 2$ and $C_2 = 8$: $\% \text{Loss} = \frac{8}{2 + 8} \times 100 = \frac{8}{10} \times 100 = 80\%$