Question

A cell can be balanced against $110cm$ and $100cm$ of potentiometer wire respectively, with and without being short-circuited through a resistance of $10\Omega $. Its internal resistance is:
A. $1.0\Omega $
B. $0.5\Omega $
C. $2.0\Omega $
D. Zero

Answer 1

Hint:In this question, we are given the length of the potentiometer with and without being short circuited. We will use the relation between these lengths and the emf of the cell and the terminal potential difference to determine our answer. We will use the emf as the function of both internal resistance and resistance used to short circuit which will lead us to the final answer.

Formula used:
$\dfrac{E}{V} = \dfrac{{{l_1}}}{{{l_2}}}$
where, $E$ is the emf of the cell, $V$ is the terminal potential difference, ${l_1}$ is the length of the potentiometer with short circuit and ${l_2}$ is the length of potentiometer without short circuit.
$E = I(R + r)$
where, $E$ is the emf of the cell, $I$ is the current,$R$ the resistance through which the potentiometer is short circuited and $r$ is the internal resistance.
$V = IR$
where, $V$ is the terminal potential difference, $I$ is the current and $R$ the resistance through which the potentiometer is short circuited.

Complete step by step answer:
We are given that
${l_1} = 110cm$, ${l_2} = 100cm$ and $R = 10\Omega $
We know that$\dfrac{E}{V} = \dfrac{{{l_1}}}{{{l_2}}}$
Now, we will put $E = I(R + r)$and $V = IR$
$
\dfrac{{I(R + r)}}{{IR}} = \dfrac{{{l_1}}}{{{l_2}}} \\
\Rightarrow 1 + \dfrac{r}{R} = \dfrac{{{l_1}}}{{{l_2}}} \\
\Rightarrow 1 + \dfrac{r}{{10}} = \dfrac{{110}}{{100}} \\
\Rightarrow \dfrac{r}{{10}} = \dfrac{{110}}{{100}} - 1 \\
\Rightarrow \dfrac{r}{{10}} = \dfrac{{10}}{{100}} \\
\therefore r = \dfrac{{100}}{{100}} = 1\Omega $
Thus, the internal resistance is $1\Omega $.

Hence, option A is the right answer.

Note:We have used the working principle of potentiometer to solve this question. This principle states that the potential across any portion of the wire which is directly proportional to the length of the wire that has a uniform cross-sectional area and current flow is constant. It is important to remember that in the emf of a potentiometer, we need to consider both the resistances as we have done here.