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(D) Let the potential difference across the capacitor $C$ (connected between $X$ and $Y$) be $V_{XY} = 60 \ V$.The circuit consists of a capacitor $C$

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$30$

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(D) Let the potential difference across the capacitor $C$ (connected between $X$ and $Y$) be $V_{XY} = 60 \ V$.The circuit consists of a capacitor $C$ in parallel with a series combination of capacitors $2C$,$C$,and $2C$.Since the capacitors $2C$,$C$,and $2C$ are in series,the same charge $q$ flows through them.The potential difference across the series branch is $V_{XY} = 60 \ V$.The equivalent capacitance of the series branch is given by:$\frac{1}{C_{eq}} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{2C} = \frac{1+2+1}{2C} = \frac{4}{2C} = \frac{2}{C}$So,$C_{eq} = \frac{C}{2}$.The charge $q$ on each capacitor in the series branch is:$q = C_{eq} 	imes V_{XY} = \frac{C}{2} 	imes 60 \ V = 30C$.The potential difference between points $M$ and $N$ is the potential difference across the capacitor $C$ located between them.$V_{MN} = \frac{q}{C} = \frac{30C}{C} = 30 \ V$.

In the adjoining figure,the potential difference between $X$ and $Y$ is $60 \ V$. The potential difference between the points $M$ and $N$ will be: (in $V$)

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