Question

A potentiometer wire has $10\text{m}$ length and $10$ $\Omega$ resistance. A battery of $E$ $=$ $2\text{volt}$ and a rheostat is connected to it. If the potential gradient is $1$ $\mu$$\text{V}$/$\text{mm}$, the value of resistance in rheostat in $\text{ohm}$ will be

• A. $1.99$
• B. $19.9$
• C. $199$
• D. $1990$

Answer 1

The correct option is D $1990$

Given:
$E=2\text{V};l=10\text{m}$
$K=1\mu \text{V/mm}={10}^{-3}\text{V/m},R=10\Omega$

The potential drop across the wire will be

$V=K\times l=1\times {10}^{-3}\times 10=0.01\text{V}$

So, The current passing through circuit will be :

$i=\frac{V}{R}=\frac{0.01}{10}=0.001\text{A}$

Let $R^{\prime }$ be the resistance of rheostat, and current $i$ also passes through the rheostat, so

$i=\frac{E}{R+R^{\prime }}\Rightarrow 0.001=\frac{2}{10+R^{\prime }}$

$\therefore R^{\prime }=2000-10=1990\Omega$

Hence, option $\left(d\right)$ is correct.