Question

In the arrangement shown in the figure when the switch $S_2$ is open, the galvanometer shows no deflection for $l=\frac{L}{2}$. When the switch $S_2$ is closed, the galvanometer shows no deflection for $l=\frac{5L}{12}$. The internal resistance $\left(r\right)$ of $6\text{V}$ cell, and the emf $E$ of the other battery are respectively

• A. $3\Omega$, $8\text{V}$
• B. $2\Omega$, $12\text{V}$
• C. $2\Omega$, $24\text{V}$
• D. $3\Omega$, $12\text{V}$

Answer 1

The correct option is B $2\Omega$, $12\text{V}$

$\text{Case I}:$ When $S_2$ is open at $l=\frac{L}{2}$.
Let us assume the point of contact of jockey of galvanometer be $C$. Since it shows no deflection for $l=\frac{L}{2}$, ​ thus no current will flow through the path $AC$. Therefore the $6\text{volt}$ cell will be in parallel with $\frac{L}{2}$ ​ arm of potentiometer wire. So, potential of both will be same.

So, no current flows through $6\text{V}$ battery.


Applying the condition of balanced Wheatstone bridge, we have

$\frac{6}{E}=\frac{\frac{L}{2}}{L}$

$\Rightarrow \frac{6}{E}=\frac{1}{2}$

$\Rightarrow E=12\text{V}$

$\text{Case II}$: When $S_2$ is closed at $l=\frac{5L}{12}$ same current flows in auxiliary circuit $PQRS$.


So, current passing the circuit, $i=\frac{6}{10+r}$

Potential difference across battery,
$=6-ir$
$=6-\frac{6r}{10+r}=\frac{60}{10+r}$

Equations for potentiometer:

$\frac{\frac{60}{10+r}}{12}=\frac{\frac{5l}{12}}{l}$

$\Rightarrow 60=50+5r$

$\therefore r=2\Omega$

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