Physics - Current Electricity Question with Solution | TestHub

Internal resistance of voltmeter $=R$. Effective resistance across $B$ and $C$ $$\frac{1}{R^{\prime }}=\frac{1}{R}+\frac{1}{50}=\frac{50+R}{50R}$$ or $R^{\prime }=\left(\frac{50R}{50+R}\right)$
According to Ohm's law, $V^{\prime }=I{R}^{\prime }$ $$\begin{array}{l}\text{ or }\frac{100}{3}=I\cdot \left(\frac{50R}{50+R}\right)\\ \text{ or }\frac{100}{3}\left(\frac{50+R}{50R}\right)=I...(i)\end{array}$$ Now, total resistance of circuit $$\begin{array}{l}{R}^n=50+\frac{50R}{50+R}\\ \text{ or }{R}^{\prime \prime }=\frac{(2500+100R)}{(50+R)}\end{array}$$ Now, $V^{\prime \prime }=I{R}^{\prime \prime }$ $$\Rightarrow 100=\frac{100}{3}\left(\frac{50+R}{50R}\right)\frac{2500+100R}{(50+R)}$$ $$\Rightarrow 150R=2500+100R\text{ or }R=50\mathrm{k}\Omega$$