A 5μ F capacitor is charged fully by a 220V supply. It is then...

${C}_{1} = 5\muF{V}_{1} = 220Volt$

${C}_{2} = 2 . 5\muF{V}_{2} = 0$

Heat loss;

$\DeltaH = {U}_{1} - {U}_{t} = \dfrac{1}{2}\dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}{\left({v}_{1} - {v}_{2}\right)}^{2}$

$= \dfrac{1}{2} \times \dfrac{5 \times 2 . 5}{( 5 + 2 . 5 )} ( 220 - 0{)}^{2}\muJ$

$= \dfrac{5}{2 \times 3} \times 22 \times 22 \times 100 \times {10}^{- 6}J$

$= \dfrac{5 \times 11 \times 22}{3} \times {10}^{- 4}J = \dfrac{55 \times 22}{3} \times {10}^{- 4}J$

$= \dfrac{1210}{3} \times {10}^{- 4}J = \dfrac{1210}{3} \times {10}^{- 3}J = 4 \times {10}^{- 2}$

According to questions

$\dfrac{x}{100} = 4 \times {10}^{- 2}$

so, $x = 4$