Correct option: D.Current sensitivity of galvanometer is $20\,\mu A$ $/ division$
d
Let full scale deflection of current $=1$
In case $1,$ when $\mathrm{R}=2400 \Omega$ and deflection of 40 divisions present.
$\therefore \quad \frac{40}{50} \mathrm{I}=\frac{V}{G+R}$
$\Rightarrow \quad \frac{4}{5} \mathrm{I}=\frac{2}{G+2400}\dots (1)$
In case $2,$ when $\mathrm{R}=4900 \Omega$ and deflection of 20 divisions present
$\therefore \quad \frac{20}{50} \mathrm{I}=\frac{V}{G+R}$
$\Rightarrow \quad \frac{2}{5} \mathrm{I}=\frac{2}{G+4900}\dots (2)$
From $(1)$ and $(2)$ we get,
$\frac{4}{2}=\frac{G+4900}{G+2400}$
$\Rightarrow 2 G+4800=G+4900$
$\Rightarrow \mathrm{G}=100 \Omega$
Putting value of G in equation (1), we get.
$\frac{4}{5} \mathrm{I}=\frac{2}{100+2400}$
$\Rightarrow \mathrm{I}=1 \mathrm{mA}$
Current sensitivity $=\frac{\mathrm{I}}{\text { number of divisions }}$
$=\frac{1}{50}$
$=0.02 \mathrm{mA} /$ division
$=20 \mu \mathrm{A} /$ division
Resistance required for deflection of 10 divisions
$\frac{10}{50} \mathrm{I}=\frac{V}{G+R}$
$\Rightarrow \frac{1}{5} \times 1 \times 10^{-3}=\frac{2}{100+R}$
$\Rightarrow \mathrm{R}=9900 \Omega$
