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(C) Let the internal resistance of the voltmeter be $R$. The resistor between $B$ and $C$ is $50 	ext{ k}\Omega$. When the voltmeter is connected in

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$50 	ext{ k}\Omega$

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(C) Let the internal resistance of the voltmeter be $R$. The resistor between $B$ and $C$ is $50 	ext{ k}\Omega$. When the voltmeter is connected in parallel with this resistor,the equivalent resistance $R'$ across $B$ and $C$ is given by:$\frac{1}{R'} = \frac{1}{R} + \frac{1}{50 	ext{ k}\Omega} = \frac{50 	ext{ k}\Omega + R}{50 R 	ext{ k}\Omega}$$R' = \frac{50 R}{50 + R} 	ext{ k}\Omega$The total resistance of the circuit is $R_{total} = 50 	ext{ k}\Omega + R' = 50 + \frac{50 R}{50 + R} = \frac{2500 + 50 R + 50 R}{50 + R} = \frac{2500 + 100 R}{50 + R} 	ext{ k}\Omega$.The total current $I$ in the circuit is $I = \frac{V}{R_{total}} = \frac{100}{\left( \frac{2500 + 100 R}{50 + R} ight)} = \frac{100(50 + R)}{2500 + 100 R} 	ext{ mA}$.The voltage across $B$ and $C$ is given as $V_{BC} = I R' = \frac{100}{3} 	ext{ V}$.Substituting the values:$\frac{100(50 + R)}{2500 + 100 R} 	imes \frac{50 R}{50 + R} = \frac{100}{3}$$\frac{5000 R}{2500 + 100 R} = \frac{100}{3}$$15000 R = 250000 + 10000 R$$5000 R = 250000$$R = 50 	ext{ k}\Omega$.

In the adjacent circuit,a voltmeter of internal resistance $R$,when connected across $B$ and $C$,reads $\frac{100}{3} \text{ V}$. Neglecting the internal resistance of the battery,the value of $R$ is:

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