In the adjacent shown 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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Internal resistance of voltmeter is $R.$

Therefore effective resistance across $B$ and $C, R^{\prime}$ is given by

$\frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{50}=\frac{50+R}{50 R}$

Or $R^{\prime}=\frac{50 R}{50+R}$

According to Ohm's law

$V^{\prime}=I R^{\prime}$

or $\frac{100}{3}=I \cdot \frac{50 R}{50+R}$

or $\frac{100}{3} \frac{50+R}{50 R}=I$

Now, total resistance of circult $R^{\prime \prime}=50+\frac{50 R}{50+R}$

OR $R^{\prime \prime}=\frac{2500+100 R}{50+R}$

Now, $V^{\prime \prime}=I R^{\prime \prime}$

$\Rightarrow 100=\frac{100}{3} \frac{50+R}{50 R} \frac{2500+100 R}{50+R}$

or $150 R=2500+100 R$

or $50 R=2500$

or $R=50 k \Omega$

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