Question

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

The length AC is equal to

• A. 2/3 m
• B. 1/3 m
• C. 3/5 m
• D. 1/2 m

Answer 1

Answer: (A)

2/3 m

we have,

$E=\frac{Irl}{L}$

$l=\frac{EL}{Ir}$

$l=\frac{\left(4\right)\left(1\right)}{\left(6\right)\left(1\right)}$.......$\because I=\frac{6}{1}=6A$

$l=\frac{2}{3}$