Physics - Current Electricity Question with Solution | TestHub

Let $I_g$ be the current required for full-scale deflection of the galvanometer.

When a resistance of $100\Omega$ is connected in series, the voltage range is $V$.
Using the formula for a voltmeter, we have:
$V=I_g(G+100)$

To double the range to $2V$, an additional resistance of $1000\Omega$ is connected in series with the existing arrangement. The new total series resistance becomes $100\Omega +1000\Omega =1100\Omega$.
The new voltage range is:
$2V=I_g(G+1100)$

Dividing the second equation by the first equation, we get:
$\frac{2V}{V}=\frac{I_g(G+1100)}{I_g(G+100)}$

$2=\frac{G+1100}{G+100}$

$2(G+100)=G+1100$

$2G+200=G+1100$

$G=1100-200$

$G=900\Omega$

Answer: $900\Omega$