Q 12-04-220JEE MainJEE Main 2019 (12 Jan, Shift 1)Medium
The galvanometer deflection, when key $K_1$ is closed but $K_2$ is open, equals $\theta_0$ (see figure). On closing $K_2$ also and adjusting $R_2$ to $5\ \Omega$, the deflection in galvanometer becomes $\dfrac{\theta_0}{5}$. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:
Answer: (B) $22\ \Omega$
$K_2$ open: $I_0 = \dfrac{E}{220 + G}$.
$K_2$ closed: total current $I = \dfrac{E}{220 + \frac{5G}{G + 5}}$, and the galvanometer gets
$$I_g = I\cdot\frac{5}{G + 5} = \frac{5E}{220(G + 5) + 5G}$$
Setting $I_g = \dfrac{I_0}{5}$:
$$25(220 + G) = 225G + 1100 \Rightarrow 200G = 4400 \Rightarrow G = 22\ \Omega$$
Solution by Sreeraj P, M.Sc Physics