Question

In a potentiometer experiment of a cell of emf $1.25\, V$ gives balancing length of $30\, cm$. If the cell is replaced by another cell, balancing length is found to be $40\, cm$. What is the emf of second cell ?

• A. $\simeq$ 1.57 V
• B. $\simeq$ 1.67 V
• C. $\simeq$ 1.47 V
• D. $\simeq$ 1.37 V

Answer 1

Answer: (B)

Given,
First balancing length
$l_{1}=30\, cm$
Second balancing length
$l_{2} =40\, cm$
$E_{1} =1.25\, V $
$E_{2}$ =?
So, according to the principle of potentiometer
$E_{1} =Kl_{1}\,\,\,\,\,\dots(i)$
$E_{2} =K l_{2}\,\,\,\,\dots(ii)$
$\frac{E_{1}}{E_{2}} =\frac{Kl_1}{Kl_2}$
$\frac{1.25}{E_{2}} =\frac{30}{40}$
$\Rightarrow E_{2} =\frac{1.25 \times 40}{30}$
$\Rightarrow E_{2}=\frac{5}{3}$
$E_{2} =1.666\, V$
$\simeq 1.67\, V$