Explanation

(d) $q_{i} = C_{i}V = 2V = q$ (say)
This charge will remain constant after switch is shifted from position 1 to position 2.
$U_{i} = \frac{1}{2}\frac{q^{2}}{C_{i}} = \frac{q^{2}}{2 \times 2} = \frac{q^{2}}{4}$
$U_{i} = \frac{1}{2}\frac{q^{2}}{C_{i}} = \frac{q^{2}}{2 \times 10} = \frac{q^{2}}{20}$
$\therefore$Energy dissipated$\ \left( \frac{q^{2}}{5} \right)is\ $80% of the initial stored energy $\left( = \frac{q^{2}}{4} \right)$.