Question

Potentiometer wire $\text{PQ}$ of $1\text{m}$ length is connected to a standard cell $E_1$. Another cell $E_2$ of emf $1.02\text{V}$ is connected with a resistance $r$ and a switch $S$ as shown in circuit diagram. With switch $S$ open, null position is obtained at a distance of $51\text{cm}$ from $\text{P}$. Calculate the e.m.f. of cell $E_1$.

• A. $2.2\text{V}$
• B. $2\text{V}$
• C. $1.8\text{V}$
• D. $1.02\text{V}$

Answer 1

The correct option is B $2\text{V}$

Given:
$E_2=1.02\text{V};l=1\text{m};l_1=51\text{cm}$
Since, point $\text{A}$ is null point, so no current will pass through internal resistance $r$. It means potential drop across $\text{PA}$ will be equal to $E_2.$

Let $K$ be the potential gradient in wire $\text{PQ}$.

$\therefore E_2=K\times \left(\text{PA}\right)=K{l}_1$

$K=\frac{E_2}{l_1}=\frac{1.02}{0.51}=2\text{V/m}$

So, potential drop across wire $\text{PQ}$ will be

$V_{PQ}=E_1=Kl=2\times 1$

$\therefore E_1=2\text{V}$

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