Right answer is (d) 2.25 mJ
The explanation is: Common potential, V = \frac {(C_1V_1 + C_1V_2)}{(C_1 + C_2)}
V = \frac {[(5 \times 50) + (7 \times 40)] \times 10^{-6}}{(5 + 7) \times 10^{-6}}
V = \frac {(250 + 280)}{12} = 44.4 V
Energy lost by 5 μF capacitor = \frac {1}{2}C1V1^2 – \frac {1}{2}C1V^2 = \frac {1}{2}C (V1^2 – V^2)
\frac {1}{2} × (5 × 10^-6) × (50^2 – 40^2) = 0.00225
U = 2.25 × 10^-3 J
U = 2.25 mJ
Therefore, the energy lost is 2.25 mJ