BITSAT 2021 Paper

Let G be resistance of galvanometer and ig the current through it. Let V is maximum potential difference, then from Ohm's law

ig=‌

V

G+R

⇒R=‌

V

ig

−G
Given, G=10Ω,ig=0.01A
V=10‌ volt ‌
∴R=‌

10

0.01

−10=990Ω
Thus, on connecting a resistance R of 990Ω in series with the galvanometer, the galvanometer will become a voltmeter of range zero to 10V.
For the voltmeter, a high resistance is series with the galvanometer and so the resistance of a voltmeter is very high compared to that of galvanometer.