Question

In a potentiometer experiment of a cell of emf $1.25V$ gives a balancing length of $30cm$. If the cell is replaced by another cell, the balancing length is found to be $40cm$. What is the emf of the second cell?
$A.{\text{ }} \simeq 1.57{\text{ }}V$
$B.{\text{ }} \simeq 1.67{\text{ }}V$
$C.{\text{ }} \simeq 1.47{\text{ }}V$
$D.{\text{ }} \simeq 1.37{\text{ }}V$

Answer 1

The potential difference that tends to increase the electric current is known as electromotive force. By finding the electromotive force of the cells and their balancing lengths respectively we are able to find the solution for this question.

Formula used:
$\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{l_1}}}{{l2}}$
Where, ${E_1},{E_2}$ is the emf of the secondary cells
${L_1},{L_2}$ is the balancing length of the potentiometer

Complete step by step answer:
Let us consider the values given in the above problem.
Given that in a potentiometer experiment of a cell of emf $1.25V$ gives a balancing length of $30cm$.
We know that if the cell is replaced by another cell, the balancing length is found to be 40 cm.
Let us consider ${E_1}$ be the cell of emf $1.25V$having balancing length ${l_1} = 30cm$ and ${E_2}$ be another cell having balancing length ${l_2} = 40cm$.
We can use the formula to compare emfs of two cells by individual cell method,
$\Rightarrow \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{l_1}}}{{l2}}$
We shall now substitute the values for we get,
$\Rightarrow {E_1} = 1.25V$
$\Rightarrow {l_1} = 30cm{\text{ }}$
$\Rightarrow {l_2} = 40cm$
$\Rightarrow \dfrac{{1.25}}{{E2}} = \dfrac{{30}}{{40}}$
$\Rightarrow E_2 = \dfrac{{1.25 \times 40}}{{30}}$
$\Rightarrow E_2 = 1.6667V{\text{ }}$
$\therefore \simeq {\text{1}}.67{\text{ }}V$

Emf of the second cell $\simeq 1.67V$. Hence, Option(B) is the required answer.

Additional information:
We know that the potentiometer is an instrument that is used to measure an unknown emf by comparison with known emf.
Potentiometer uses the null deflection method.
The fall of potential per unit length of potentiometer wire i.e. the potential gradient of wire is constant. This is the principle of the potentiometer.
The potentiometer is used to measure the emf of a cell, to compare EMFs of two cells, and to determine the internal resistance of a cell.

In the diagram:
$B_t$- batteries
${R_1},{R_2}$- Resistances
$G$- galvanometer.

Note:
When combination method or sum and difference method is used to compare the emfs of two cells the formula used is $\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{L_1} + {L_2}}}{{{L_1} - {\text{ }}{L_2}}}$ where ${E_1}$ is the emf of cell having balancing ${L_1}$ and ${E_2}$ is the emf of cell having balancing ${L_2}$.