1. Given data

• Initially charged capacitor: $C_1 = 5\,\mu F = 5 \times 10^{-6}\,F$
• Initial voltage on it: $V_0 = 220\,V$
• Second capacitor (initially uncharged): $C_2 = 2.5\,\mu F = 2.5 \times 10^{-6}\,F$

The first capacitor is disconnected from the battery and then connected to the second capacitor for charge redistribution.

2. Initial energy stored

Only the first capacitor has energy initially:

$U_i = \frac{1}{2} C_1 V_0^2$

$U_i = \frac{1}{2}(5 \times 10^{-6})(220)^2$

$(220)^2 = 48400$

$U_i = \frac{1}{2} \cdot 5 \times 10^{-6} \cdot 48400$

$U_i = 2.5 \times 10^{-6} \cdot 48400 = 0.121\,J$

3. Final common voltage after connection

When connected together for redistribution, total charge is conserved.

Initial charge on $C_1$:

$Q_{\text{total}} = C_1 V_0 = 5 \times 10^{-6} \cdot 220 = 1.1 \times 10^{-3}\,C$

Final common voltage $V_f$ is

$V_f = \frac{Q_{\text{total}}}{C_1 + C_2}$

$V_f = \frac{1.1 \times 10^{-3}}{(5+2.5)\times 10^{-6}} = \frac{1.1 \times 10^{-3}}{7.5 \times 10^{-6}}$

$V_f = 146.67\,V$

4. Final energy stored

$U_f = \frac{1}{2}(C_1 + C_2)V_f^2$

$U_f = \frac{1}{2}(7.5 \times 10^{-6})(146.67)^2$

Instead of calculating directly, use

$U_f = \frac{Q_{\text{total}}^2}{2(C_1+C_2)}$

$U_f = \frac{(1.1 \times 10^{-3})^2}{2(7.5 \times 10^{-6})}$

$= \frac{1.21 \times 10^{-6}}{1.5 \times 10^{-5}} = 0.08067\,J$

5. Energy change

$\Delta U = U_i - U_f$

$\Delta U = 0.121 - 0.08067 = 0.04033\,J$

Given,

$\Delta U = \frac{X}{100} \, J$

So,

$\frac{X}{100} = 0.04033$

$X = 4.033$

To the nearest integer,

$X = 4$

6. Comparison with stored answer

Stored correct answer = 4, which matches the derived answer.