Question

A potentiometer has five segments of wire, each of 1m length. A battery is used to pass a steady current through it. It is found that the e.m.f. of a cell can be balanced at 350 cm from the zero end. Now the battery is connected across the four segments of wire. Find the length of the potentiometer wire that will balance the e.m.f. of the given cells

Answer 1

$\green{\textbf{Answer}}$:-

let 'E ' is the $\green{\texttt{e.m.f of battery}}$ used in potentiometer circuit.

$\green{\texttt{length of potentiometer}}$ wire= 5m

5m = 500cm

$\green{\texttt{e.m.f of cell}}$=$\Large\frac{E}{500}$ ×350

= $\Large\frac{7E}{10}$ volt...............❶eqn

In second case,

$\green{\texttt{length of potentiometer}}$ wire = 4m

4m = 400cm

Now,

let $l_{1}$ is the $\green{\texttt{length of wire}}$ at which the balanced point is reached.

Then,

$\green{\texttt{e.m.f of cell}}$=$\Large\frac{E}{400}×l_{1}$ .........❷eqn

$\green{\texttt{comparing eqns}}$ ❶and ❷,

$\green{\texttt{we get}}$

$\Large\frac{7E}{10}$= $\Large\frac{E.l_{1}}{400}$

$l_{1}= 280cm$

so, the $\green{\textbf{length of potentiometer wire}}$ = 280cm

Answer 2

length of the potentiometer wire that will balance the e.m.f. of the given cells is

=280cm answer.

calculation is given on the attachment