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(C) From figure-$1$,the voltmeter is connected in parallel with resistance $R$. Let the resistance of the voltmeter be $R_v$.The equivalent resistance

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$8$

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(C) From figure-$1$,the voltmeter is connected in parallel with resistance $R$. Let the resistance of the voltmeter be $R_v$.The equivalent resistance of the parallel combination of $R$ and $R_v$ is $R_{eq} = \frac{R \cdot R_v}{R + R_v}$.The total resistance of the circuit is $R_{total} = 2 + \frac{R \cdot R_v}{R + R_v}$.The voltage across the voltmeter (reading $V$) is given by the voltage divider rule:$V = 10 \cdot \frac{R_{eq}}{R_{total}} = 10 \cdot \frac{\frac{R \cdot R_v}{R + R_v}}{2 + \frac{R \cdot R_v}{R + R_v}} = 10 \cdot \frac{R \cdot R_v}{2(R + R_v) + R \cdot R_v}$.From figure-$2$,as $R_v \rightarrow \infty$,the reading $V \rightarrow 8 \, V$.Substituting $R_v \rightarrow \infty$ in the expression for $V$:$V = 10 \cdot \frac{R \cdot R_v}{2R + 2R_v + R \cdot R_v} = 10 \cdot \frac{R}{\frac{2R}{R_v} + 2 + R} = \frac{10R}{2 + R}$.Given $V = 8 \, V$ at the limit:$8 = \frac{10R}{2 + R} \Rightarrow 16 + 8R = 10R \Rightarrow 2R = 16 \Rightarrow R = 8 \, \Omega$.

In the given figure-$1$,the resistance of the shown voltmeter is variable. The variation of its reading with respect to its resistance is shown in figure-$2$. The value of $R$ is ............... $\Omega$.

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