A battery of emf 1.4V and internal resistance $2\Omega$ is connected to a resistance of $100\Omega$ resistance through an ammeter. The resistance of ammeter is $4/3\Omega .$ A voltmeter has also been connected to find the potential difference across the resistor. The ammeter reads 0.02A. What is the resistance of voltmeter.

Here we consider that R be the resistance of voltmeter and the voltmeter is connected in parallel with $100\Omega$ resistance.

The effective resistance with voltmeter will be $R\text{'}=100R/(100+R)$

Total resistance of the circuit is given as $R_T=2+\frac{4}{3}+\frac{100R}{(100+R)}\Rightarrow R_T=\frac{10}{3}+\frac{100R}{(100+R)}=\frac{1000+310R}{3(100+R)}$

Current in the circuit is given as $i=\frac{1.4}{\left\{\frac{1000+310R}{3(100+R)}\right\}}\Rightarrow 0.02=\frac{\left(1.4\right)\left\{3(100+R)\right\}}{1000+310R}\Rightarrow R=200\Omega$