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

\(\frac{10}{50} \mathrm{I}=\frac{V}{G+R}\)

\(\Rightarrow \frac{1}{5} \times 1 \times 10^{-3}=\frac{2}{100+R}\)

\(\Rightarrow \mathrm{R}=9900 \Omega\)