Answer:
Option C
Explanation:
According to question
Net current through the galvanometer is given by
$l=\frac{V}{R_{eff}}=\frac{2}{1970+30}=\frac{2}{2000}= 10^{-3} A$
As we know, this current provides full scale deflection (i.e, 20 div) . In order to limit the deflection to 10 divisions , the resistance needed to connect such that the current reduces can be obtained as
$\theta= \frac{nlAB}{k}(i.e, \theta \propto l)$
[symbols have their usual meanings ]
$\Rightarrow $ $\frac{\theta_{1}}{\theta_{2}}=\frac{l_{1}}{l_{2}}=2$
$\Rightarrow$ $l_{2}=\frac{l_{1}}{2}=\frac{10^{-3}}{2}=5 \times 10^{-4} A$
Again $l= \frac{V}{R_{eff}+R_{5}}$
$\Rightarrow$ $R_{5}= \frac{V}{l}-R_{eff}$
$=\frac{2}{5\times10^{-4}}-2000$
= $4 \times 10 ^{3}-2000$=2000 $\Omega$
So, the resistance of 1970 $\Omega$ is to be replaced by 1970+2000=3970 $\Omega$
