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(C) Let the internal resistance of the voltmeter be $R$. When the voltmeter is connected across $B$ and $C$,it is in parallel with the $50 \ k\Omega$

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$50$

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

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

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