If the galvanometer \( G \) does not show any deflection in the circuit shown, the value of \( R \) is given by :

Using Kirchhoff's rules or bridge balance condition when no current flows through \( G \), the potential drop across \( R \) must equal the emf of the second battery (\( 2\text{ V} \)). Current through \( 400\ \Omega \) is \( (10 - 2)/(400 + R) \). Thus, \( I \times R = 2\text{ V} \), which yields \( R = 100\ \Omega \).