Question

Two cells of emf $E_1$ and $E_2(E_2>E_1)$ are connected in series in a secondary circuit of a potentiometer experiment for determination of emf. The balancing length is found to be 825 cm. Now when the terminals of cell of emf $E_1$ are reversed, then the balancing length is found to be 225 cm. The ratio of $E_1$ and $E_2$ is, then-

• A. 2:3
• B. 4:7
• C. 7:4
• D. None of these

Answer 1

Answer: (B)

Step 1, Given data

Balancing length in potentiometer experiment = 825 cm
When terminals of cell of emf $E_1$ are reversed, then the balancing length = 225 cm
here $E_2>E_1$

Step 2, Finding the ratio

We know that-
$E_2+E_1\propto l$

So we can write
So, $\frac{E_2+E_1}{E_2-E_1}=\frac{l_1}{l_2}=\frac{825}{225}$

Or,
$\frac{E_1+E_2}{E_2-E_1}=\frac{33}{9}=\frac{11}{3}$

Or, $3(E_1+E_2)=11(E_2-E_1)$
Or, $3E_1+3E_2=11E_2-11E_1$
Or,$3E_1+11E_1=11E_2+3E_2$
Or,$14E_1=8E_2$
Therefore,$\frac{E_1}{E_2}=\frac{8}{14}=\frac{4}{7}$

Hence the ratio is 4 : 7