A capacitor of capacitance 900 µ F is charged by a 100V battery....

When a charged capacitor is connected to an uncharged capacitor, then there is loss of energy in the form of heat as charges flow from higher potential to lower potential.

So, heat $∆ H = \dfrac{1}{2}\dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}{\left({V}_{1} - {V}_{2}\right)}^{2}$

As the other capacitor is identical therefore charge is equally divided.

$= \dfrac{1}{2}\dfrac{{C}^{2}}{2C} ( 100 - 0{)}^{2}$

$= \dfrac{1}{2}\dfrac{900 \times {10}^{- 6}}{2} \times {10}^{4} = \dfrac{9}{4}$ Joule

$= 2 . 25$ Joule.

Hence, the value of $x = 225$.