1. Initial charging of the capacitor

   The capacitor has capacitance $C = 2\,\mu F$

   From the charging arrangement (switch at position 1), the capacitor gets charged to some potential difference, say $V$.

   Hence, the initial charge on the capacitor is $Q_i = CV$

   and the initial stored energy is $U_i = \frac{1}{2}CV^2$

2. After switching to position 2

   From the circuit in position 2, the charged capacitor is connected with another potential source/network such that its final potential becomes smaller than the initial one. From the given configuration, the final potential across the capacitor becomes $V_f = \frac{V}{\sqrt{5}}$

   Therefore the final stored energy is

   = \frac{1}{2}C\left(\frac{V}{\sqrt{5}}\right)^2 = \frac{1}{2}C\cdot \frac{V^2}{5} = \frac{U_i}{5}$$

3. Energy dissipated

   The energy lost (dissipated) is $U_{\text{loss}} = U_i - U_f = U_i - \frac{U_i}{5} = \frac{4U_i}{5}$

   So the percentage of stored energy dissipated is

   = \frac{4}{5}\times 100 = 80\%$$

4. Checking options

   • A: $0\%$ ❌
   • B: $20\%$ ❌
   • C: $75\%$ ❌
   • D: $80\%$ ✅

Therefore, the correct answer is Option D.