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

This problem involves the redistribution of charge and the consequent loss of energy when two capacitors are connected, a concept detailed in NCERT Example 2.10.

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

2. Final State: The switch is turned to position 2, connecting $C_1$ in parallel with an uncharged capacitor $C_2$ (implied to be $8 \mu\text{F}$ based on the answer option). Charge flows until they reach a common potential $V'$. $V' = \frac{\text{Total Charge}}{\text{Total Capacity}} = \frac{C_1 V}{C_1 + C_2}$

3. Final Energy: The total energy stored in the combination is: $U_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} (C_1 + C_2) \left( \frac{C_1 V}{C_1 + C_2} \right)^2 = \frac{1}{2} \frac{C_1^2 V^2}{(C_1 + C_2)}$

4. Energy Loss: $\Delta U = U_i - U_f = \frac{1}{2} C_1 V^2 - \frac{1}{2} \frac{C_1^2 V^2}{(C_1 + C_2)} = \frac{1}{2} C_1 V^2 \left( 1 - \frac{C_1}{C_1 + C_2} \right)$ $\Delta U = U_i \left( \frac{C_2}{C_1 + C_2} \right)$

5. Percentage Dissipated: $\% \text{ Loss} = \frac{\Delta U}{U_i} \times 100 = \left( \frac{C_2}{C_1 + C_2} \right) \times 100$ Substituting $C_1 = 2 \mu\text{F}$ and $C_2 = 8 \mu\text{F}$: $\% \text{ Loss} = \frac{8}{2 + 8} \times 100 = \frac{8}{10} \times 100 = 80\%$

This loss appears as heat and electromagnetic radiation [NCERT Class 12, Sec 2.15, Example 2.10].