Initial State: The total resistance in the circuit is the sum of the galvanometer resistance ($G$) and the initial series resistance ($R_1$). $R_{total1} = G + R_1 = 50 + 2950 = 3000 \, \Omega$ The current ($I_1$) flowing through the galvanometer is given by Ohm's Law: $I_1 = \frac{V}{R_{total1}}$ This current produces a deflection of $\theta_1 = 30$ divisions. Since deflection is proportional to current ($I \propto \theta$), we have $I_1 = k \cdot 30$.

Final State: The deflection is reduced to $\theta_2 = 20$ divisions. The new current ($I_2$) is: $I_2 = k \cdot 20$ Taking the ratio of currents: $\frac{I_2}{I_1} = \frac{20}{30} = \frac{2}{3}$

Calculation: Let the new series resistance be $R_2$. The new total resistance is $R_{total2} = G + R_2$. Since voltage $V$ is constant, current is inversely proportional to resistance ($I \propto 1/R$): $\frac{I_2}{I_1} = \frac{R_{total1}}{R_{total2}} \Rightarrow \frac{2}{3} = \frac{3000}{50 + R_2}$ $2(50 + R_2) = 9000$ $50 + R_2 = 4500$ $R_2 = 4450 \, \Omega$ .