Question

The length of potentiometer wire is $10$m and is connected in series with an accumulator. The e.m.f. of a cell balances against $250$ cm length of wire. If the length of potentiometer wire is increased by $1$m, calculate the new balancing length of wire.

Answer 1

In case of potentiometer, we know that potentiometer is a device used to find unkown value of EMF

$\epsilon =kl$

where $\epsilon =EMF$

k=potential Gradient$=\frac{V}{L}$

where V=voltage across potentiometer

L=length of potentiometer

l=point where null point occurs

According to $1st$ case

$\epsilon =\frac{V}{L}\times \frac{250}{100}----\left(1\right)$

According to $2nd$ case

Potential gradient will change but value of EMF does not change

$\epsilon =\frac{Vl}{L+1}----\left(2\right)$

Divide $\left(1\right)$ and $\left(2\right)$

$1=\frac{L+1}{L}\times \frac{250}{100l}$

and it is given that L$=10m$

$l=\frac{11}{10}\times \frac{250}{1000}$

$l=\frac{11}{4}$

$l=2.75m$

$l=275cm$