Question

In given figure, the potentiometer wire $AB$ has a resistance of $5\Omega$ and length $10m.$ The balancing length $AM$ for the emf of $0.4V$ is:-

• A. $0.4m$
• B. $4m$
• C. $0.8m$
• D. $8m$

Answer 1

Answer: (D)

Emf of cell, $E=Xl=\frac{V}{l}=\frac{i R}{L}\times l$
$\therefore E=\frac{e}{\left(R_{1} + R_{2} + r\right)}\times \frac{R}{L}\times l$
$0.4=\frac{e}{\left(R_{1} + R_{2} + r\right)}\times \frac{R}{L}\times l$
$\therefore l=8m$