Answer is (C)
$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eff}}}=\frac{\mathrm{V}}{1790+30}=\frac{2}{2000}=1 \times 10^{-3} \mathrm{~A}=1 \mathrm{~mA}$
This current provides full scale deflection (i.e. 20 division) In order to limit the deflection 10 divisions, the resistance needed to connect such that the current reduces can be obtained as
$\theta=\frac{\mathrm{ni} \mathrm{A} \mathrm{B}}{\mathrm{k}} \quad \therefore \quad \frac{\theta_{1}}{\theta_{2}}=\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}$
$\begin{array}{l}
\therefore \quad \theta_{1}: \theta_{2}=2 \\
\therefore \quad \mathrm{i}_{1}: \mathrm{i}_{2} \Rightarrow 2 \\
\therefore \quad \mathrm{i}_{2}=\frac{\mathrm{i}_{1}}{2}=\frac{1 \times 10^{-3}}{2}=5 \times 10^{-4} \mathrm{~mA} \\
\therefore \quad \mathrm{i}=\frac{\mathrm{v}}{\mathrm{R}_{\mathrm{eff}}+\mathrm{Rs}} \Rightarrow \mathrm{Rs}=\frac{\mathrm{V}}{\mathrm{i}}-\mathrm{R}_{\mathrm{eff}} \\
=\frac{2}{5 \times 10^{-4}}-2000 \\
=\frac{2}{5} \times 10^{-4}-2000 \\
=4 \times 10^{3}-2000 \\
=2000 \Omega
\end{array}$
$\therefore$ The resistance of $1970 \Omega$ is to be replaced by $1970+2000=3970 \Omega$
A galvanometer of resistance $30 \Omega$ is connected to a battery of emf $2 V$ with $1970 \Omega$ resistance in series. A full scale deflection of 20 divisions is obtained in the galvanometer. To reduce the deflection to 10 divisions, the resistance in series required is
A) $4030 \Omega$
B) $4000 \Omega$
C) $3970 \Omega$
D) $2000 \Omega$
A) $4030 \Omega$
B) $4000 \Omega$
C) $3970 \Omega$
D) $2000 \Omega$
A galvanometer of resistance $30 \Omega$ is connected to a battery of emf $2 V$ with $1970 \Omega$ resistance in series. A full scale deflection of 20 divisions is obtained in the galvanometer. To reduce the deflection to 10 divisions, the resistance in series required is
A) $4030 \Omega$
B) $4000 \Omega$
C) $3970 \Omega$
D) $2000 \Omega$
A) $4030 \Omega$
B) $4000 \Omega$
C) $3970 \Omega$
D) $2000 \Omega$