Question

In the figure, the potentiometer wire $AB$ of length $L$ and resistance $9r$ is joined to the cell $D$ of emf $\epsilon$ and internal resistance $r$. The cell $C^{\prime }s$ emf is $\epsilon /2$ and its internal resistance is $2r$. The galvanomater $G$ will show no deflection when the length $AJ$ is

• A. $\frac{4L}{9}$
• B. $\frac{5L}{9}$
• C. $\frac{7L}{18}$
• D. $\frac{11L}{18}$

Answer 1

The correct option is B $\frac{5L}{9}$

voltage drop on the balancing length of the wire $\Rightarrow$$i\rho x$
where $i$= current flowing in primary circuit

$\rho$ = resistance per unit length of the wire
x = balancing length of wire AB when deflection in galvanometer is zero
When there is no any deflection in the galvanometer $\Rightarrow$ voltage of secondary circuit $=$ voltage drop across the wire AJ

$\Rightarrow \frac{E}{2}=\left(\frac{E}{10r}\right)\left(\frac{9r}{L}\right)x$ where $x$ is the length of wire AJ

$x=\frac{5L}{9}$