Two capacitors of 2 µF and 3 µF are charged to 150 V and 120 V, respectively. The plates of capacitor are connected as shown in the figure. An uncharged capacitor of capacity 1.5 µF falls to the free end of the wire. Then,

Let ( + q) µC charge flows in the closed loop in clockwise direction. Then final charges on different capacitors are as shown in figure. 

Now, applying Kirchhoff's loop law, 

$\frac{360-q}{3}+\frac{300-q}{2}=\frac{q}{1.5}$

Solving the above equation, we get 

$\mathrm{q}=180\mathrm{\mu C}$

$U_i=Initial energy stored in the system =(\frac{1}{2}\times 2\times {150}^2+\frac{1}{2}\times 3\times {120}^2) \mu J=44100 \mu J$

$U_f=Final energy stored in the system =(\frac{{120}^2}{2\times 2}+\frac{{180}^2}{2\times 3}+\frac{{180}^2}{2\times 1.5}) \mu J = 19800 \mu J$

Therefore lost energy = (44100 - 19800 )$\mu J =24.3 mJ$