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(C) Let the resistance of the voltmeter be $R_v$. The voltmeter is connected in parallel with the $160 \Omega$ resistor. The equivalent resistance of

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$160 \Omega$

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(C) Let the resistance of the voltmeter be $R_v$. The voltmeter is connected in parallel with the $160 \Omega$ resistor. The equivalent resistance of this parallel combination is $R_p = \frac{160 R_v}{160 + R_v}$.The total resistance of the circuit is $R_{eq} = R_p + 20 = \frac{160 R_v}{160 + R_v} + 20$.The total current in the circuit is $I = \frac{V}{R_{eq}} = \frac{10}{\frac{160 R_v}{160 + R_v} + 20}$.The voltage across the parallel combination (voltmeter reading) is $V_p = I 	imes R_p = 8 \,V$.Substituting $I$ and $R_p$: $\frac{10}{\frac{160 R_v}{160 + R_v} + 20} 	imes \frac{160 R_v}{160 + R_v} = 8$.Let $x = \frac{160 R_v}{160 + R_v}$. Then $\frac{10x}{x + 20} = 8 \implies 10x = 8x + 160 \implies 2x = 160 \implies x = 80 \Omega$.Now, $\frac{160 R_v}{160 + R_v} = 80 \implies 160 R_v = 80(160 + R_v) \implies 160 R_v = 12800 + 80 R_v \implies 80 R_v = 12800 \implies R_v = 160 \Omega$.

In an experiment on a circuit as shown in the figure, the voltmeter shows $8 \,V$ reading. The resistance of the voltmeter is,

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