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(B) The voltmeter measures the potential difference across the parallel combination of the resistors connected to the switches. Let the $EMF$ of the b

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$V_2 > V_1 > V_3$

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(B) The voltmeter measures the potential difference across the parallel combination of the resistors connected to the switches. Let the $EMF$ of the battery be $E$ and the internal resistance be negligible.$1$. When only $S_1$ is closed,the circuit consists of a resistor $R$ in series with $3R$. The potential difference across the voltmeter is given by the voltage divider rule:$V_1 = E 	imes \frac{3R}{R + 3R} = E 	imes \frac{3}{4} = 0.75E$$2$. When only $S_2$ is closed,the circuit consists of a resistor $R$ in series with $6R$. The potential difference across the voltmeter is:$V_2 = E 	imes \frac{6R}{R + 6R} = E 	imes \frac{6}{7} \approx 0.857E$$3$. When both $S_1$ and $S_2$ are closed,the resistors $3R$ and $6R$ are in parallel. Their equivalent resistance $R_p$ is:$R_p = \frac{3R 	imes 6R}{3R + 6R} = \frac{18R^2}{9R} = 2R$The circuit now consists of $R$ in series with $2R$. The potential difference across the voltmeter is:$V_3 = E 	imes \frac{2R}{R + 2R} = E 	imes \frac{2}{3} \approx 0.667E$Comparing the values: $0.857E > 0.75E > 0.667E$,which means $V_2 > V_1 > V_3$.

In the circuit shown in the figure,the reading of the voltmeter is $V_1$ when only $S_1$ is closed,the reading of the voltmeter is $V_2$ when only $S_2$ is closed,and the reading of the voltmeter is $V_3$ when both $S_1$ and $S_2$ are closed. Then:

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