In a circuit for finding the resistance of a galvanometer by half deflection method, a $6\ \text{V}$ battery and a high resistance of $11\ \text{k}\Omega$ are used. The figure of merit of the galvanometer is $60\ \mu\text{A}$ per division. In the absence of shunt resistance, the galvanometer produces a deflection of $\theta = 9$ divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of $\dfrac\theta2$, is closest to:
Answer: (D) $110\ \Omega$
**Galvanometer resistance.** Current for 9 divisions: $I = 9\times60\ \mu\text{A} = 540\ \mu\text{A}$.
$$\frac{6}{11000 + G} = 540\times10^{-6} \;\Rightarrow\; 11000 + G = 11111 \;\Rightarrow\; G = 111\ \Omega$$
**Shunt for half deflection.** In the half-deflection method $R \gg G$, so the total current hardly changes when the shunt is added. Half of it passes through the galvanometer when
$$S \approx G \approx 111\ \Omega$$
(An exact calculation including the small change in total current gives $S \approx 110\ \Omega$.) The closest option is $110\ \Omega$.
Solution by Sreeraj P, M.Sc Physics