Question

A $6\text{V}$ battery of negligible internal resistance is connected across a uniform wire of length $1\text{m}$. The positive terminal of another battery of emf $4\text{V}$ and internal resistance $1\Omega$ is joined to the point $A$ as shown in figure. The ammeter shows zero deflection when the jockey touches the wire at the point $C$. The length of $AC$ is equal to:

• A. $\frac{2}{3}m$
• B. $\frac{1}{3}m$
• C. $\frac{3}{5}m$
• D. $\frac{1}{2}m$

Answer 1

The correct option is A $\frac{2}{3}m$

If ammeter reading is zero, no current flows through $4V$ battery, Potential difference across $AC$ should be equal to $4V$,
$V_{AC}=4V$

As per the given condition, wire AB of length $1m$ is connected across $6V$ battery.
$\therefore$Potential gradient $\left(x\right)=\frac{V}{l}$
$x=\frac{6}{1}=6\frac{V}{m}$
Now,

$V_{AC}=4=x\times l_{AC}=6\times l_{AC}$
$\Rightarrow l_{AC}=\frac{2}{3}m$