A galvanometer of resistance \(50 \Omega\) is connected to a battery \(3 \mathrm{~V}\) along with a resistance \(2950 \Omega\) in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

Resistance of galvanometer, \(R_g=50 \Omega\)
Emf of battery, \(V=3 \mathrm{~V}\)
Resistance connected in series, \(R_s=2950 \Omega\)
Total resistance, \(R^{\prime}=R_g+R_s=50+2950=3000 \Omega\)
\(\therefore\) Current, \(I=\frac{V}{R^{\prime}}=\frac{3}{3000}=10^{-3} \mathrm{~A}\)
If the deflection has to be reduced to 20 divisions, then current, \(I^{\prime}=\frac{I}{30} \times 20=\frac{2}{3} \times 10^{-3} \mathrm{~A}\)
Let \(R_E\) be the effective resistance of the circuit, hence
\(3 =R_E I^{\prime} \)
\(\Rightarrow R_E =\frac{3}{I^{\prime}}=\frac{3}{\frac{2}{3} \times 10^{-3}}=4.5 \times 10^3=4500 \Omega\)
\(\therefore \text {Resistance to be added } =R_E-R_g \)
\(=4500-50=4450 \Omega\)