Question

The length of a potentiometer wire $1200\text{cm}$ and it carries a current of $60\text{mA}$. For a cell of emf $5\text{V}$ and internal resistance of $20\Omega$, the null point on it is found to be $1000\text{cm}$. The resistance of whole wire is:

• A. $80\Omega$
• B. $120\Omega$
• C. $60\Omega$
• D. $100\Omega$

Answer 1

The correct option is D $100\Omega$


Let $R$ be the resistance of the whole wire.

Potential gradient for the potentiometer wire $AB$ is

$\frac{dV}{dl}=\frac{I\times R}{l}=\left[\frac{60\times R}{l_{AB}}\right]\text{mV/m}$
$V_{AP}=\left(\frac{dV}{d{\text{l}}_{AB}}\right)l_{AP}$

Here, $l_{AB}=1200\text{cm},l_{AP}=1000\text{cm}$

$\Rightarrow V_{AP}=\frac{60\times R}{1200}\times 1000\text{mV}$
$\Rightarrow V_{AP}=50R\text{mV}=50R\times {10}^{-3}\text{V}$

Also, $V_{AP}=5\text{V}$ (for balance point at $P$)
$\therefore R=\frac{V_{AP}}{50\times {10}^{-3}}=\frac{5}{50\times {10}^{-3}}$

$\Rightarrow R=100\Omega$

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