A galvanometer has a current sensitivity of 1mA per division. A variable shunt is connected across the galvanometer and the combination is put in series with a resistance of $500\mathrm{\Omega }$ and cell of internal resistance $1\mathrm{\Omega }.$ It gives a deflection of 5 division for shunt of 5 ohm and 20 division for shunt of 25 ohm. Find the emf of cell and resistance of galvanometer.

Case I: ${\mathrm{I}}_{\mathrm{g}}=\frac{{\mathrm{I}}_1\mathrm{S}}{\mathrm{G}+\mathrm{S}}\Rightarrow {\mathrm{I}}_1=\frac{{\mathrm{I}}_{\mathrm{g}}(\mathrm{G}+\mathrm{S})}{\mathrm{S}}$

As the sensitivity is 1mA per division. There will be 5mA current for a deflection of 5 divisions. 

i.e. ${\mathrm{I}}_{\mathrm{g}}=5\mathrm{mA}$

$\therefore {\mathrm{I}}_1=\left(5\mathrm{mA}\right)\left(\frac{\mathrm{G}+5}{5}\right)={10}^{-3}(\mathrm{G}+5)......\left(1\right)$

Again ${\mathrm{R}}_{\mathrm{eff}}=\frac{\left(5\right)\mathrm{G}}{\mathrm{G}+5}+500+1=\frac{506\mathrm{G}+5\times 501}{\mathrm{G}+5}$

${\mathrm{I}}_1=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eff}}}=\frac{\mathrm{E}(\mathrm{G}+5)}{506\mathrm{G}+5\times 501}.....\left(2\right)$

$\left(1\right)=\left(2\right)\Rightarrow {10}^{-3}=\frac{E}{506\mathrm{G}+5\times 501}\pi$

$\Rightarrow \mathrm{E}=[506\mathrm{G}+5\times 501]{10}^{-3}\dots .\left(3\right)$

Similarly for the IInd case

${\mathrm{I}}_{\mathrm{g}}=20\mathrm{mA};\mathrm{S}=25\mathrm{\Omega }$

${\mathrm{I}}_2=20\times {10}^{-3}\left(\frac{\mathrm{G}+25}{25}\right)=\frac{4}{5}\times (\mathrm{G}+25){10}^{-3}\dots \left(4\right)$

Again ${\mathrm{R}}_{\mathrm{eff}}=\frac{\mathrm{G}\left(25\right)}{\mathrm{G}+25}+501=\frac{526\mathrm{G}+501\times 25}{\mathrm{G}+25}$

${\mathrm{I}}_2=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eff}}}=\frac{\mathrm{E}(\mathrm{G}+25)}{526\mathrm{G}+501\times 25}.....\left(5\right)$

$\left(4\right)=\left(5\right)\Rightarrow \frac{4}{5}\times {10}^{-3}=\frac{\mathrm{E}}{526\mathrm{G}+501\times 25}$

$\Rightarrow \mathrm{E}=[526\mathrm{G}+501\times 25]\frac{4}{5}\times {10}^{-3}\dots \left(6\right)$

On solving (3) & (6) we get $\mathrm{G}=882\mathrm{\Omega }$ and E = 47.13V