A $10 μ F$ condenser is charged to a potential of 100 volt. It is now connected to another uncharged condenser. The common potential reached is 40 volt. The capacitance of second condenser is :

2 μ F

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10 μ F

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15 μ F

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22 μ F

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Solution

The correct option is D $15 μ F$
Let $C_{2}$ be the capacitance of second capacitance.
here, $V_{C} = \frac{C_{1}}{C_{1} + C_{2}} V$
or $40 = \frac{C_{1}}{C_{1} + C_{2}} \times 100$
or $\frac{C_{1}}{C_{1} + C_{2}} = \frac{2}{5}$
or $2 C_{1} + 2 C_{2} = 5 C_{1}$
or $C_{2} = \frac{3}{2} \times 10 = 15 μ F$

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