Q 12-02-154JEE MainJEE Main 2020 (2 Sep, Shift 1)Medium
A $5\ \mu\text{F}$ capacitor is charged fully by a $220\ \text{V}$ supply. It is then disconnected from the supply and is connected in series to another uncharged $2.5\ \mu\text{F}$ capacitor. If the energy change during the charge redistribution is $\dfrac{X}{100}\ \text{J}$ then value of $X$ to the nearest integer is ______.
Numerical value type. Enter your answer.
Answer: 4
When a charged capacitor is joined to an uncharged one, the energy lost is
$$\Delta U = \frac12\frac{C_1C_2}{C_1+C_2}V^{2} = \frac12\times\frac{5\times2.5}{7.5}\times10^{-6}\times220^{2} \approx 0.040\ \text{J}$$
So $\dfrac{X}{100} = 0.04$ and $X = 4$.
Solution by Sreeraj P, M.Sc Physics