JEE Main 2019 — Current Electricity Question with Solution

Given that the galvanometer has $25$ divisions and the current required for deflection one division is $4\times {10}^{-4} \mathrm{A}$

current for full scale deflection $I=4\times {10}^{-4}\times 25 \mathrm{A}$
$={10}^{-2} \mathrm{A}$

To use this galvanometer as a voltmeter for a maximum voltage of  $2.5 \mathrm{V}$, we need to add a resistance $R$ in series to the resistance of the galvanometer of $G = 50 \mathrm{\Omega }$ such that,

$$\begin{gathered} I = \frac{2.5}{R+G} \\ \Rightarrow R = \frac{2.5}{{10}^{-2}} - 50 \\ =200 \mathrm{ohm} \end{gathered}$$