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

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$50 	ext{ k}\Omega$

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(C) Let the internal resistance of the voltmeter be $R$. The voltmeter is connected in parallel with the $50 	ext{ k}\Omega$ resistor between points $B$ and $C$.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}$$R' = \frac{50 R}{50 + R} 	ext{ k}\Omega$The total resistance of the circuit is $R_{total} = 50 	ext{ k}\Omega + R' = 50 + \frac{50 R}{50 + R} = \frac{2500 + 100 R}{50 + R} 	ext{ k}\Omega$.The total current $I$ in the circuit is:$I = \frac{V}{R_{total}} = \frac{100 	ext{ V}}{\frac{2500 + 100 R}{50 + R} 	ext{ k}\Omega} = \frac{100(50 + R)}{2500 + 100 R} 	ext{ mA}$.The voltage across $B$ and $C$ is given as $V_{BC} = \frac{100}{3} 	ext{ V}$.Using Ohm's law,$V_{BC} = I \cdot R'$:$\frac{100}{3} = \left( \frac{100(50 + R)}{2500 + 100 R} ight) \cdot \left( \frac{50 R}{50 + R} ight)$$\frac{1}{3} = \frac{50 R}{2500 + 100 R}$$2500 + 100 R = 150 R$$50 R = 2500$$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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