Question

A potentiometer wire is $10m$ long and a potential difference of $6{\text{V}}$ is maintained between its ends. The emf of a cell which balances against a length of $180cm$ of the potentiometer wire is:
(A) \[1.8{\text{V}}\]
(B) $1.1{\text{V}}$
(C) \[1.08{\text{V}}\]
(D) \[1.2{\text{V}}\]

Answer 1

Hint: To solve this question, we need to find out the potential gradient of the given potentiometer wire from the given values of the potential difference, and the length of the potentiometer wire. Then using the value of the potential gradient, we can determine the value of the emf of the cell which gets balanced against the given length of the potentiometer wire.

Formula used: The formula used to solve this question is given by
$k = \dfrac{V}{l}$, here $k$ is the potential gradient of a potentiometer wire, $l$ is the length of the potentiometer wire against which a cell of potential difference $V$ gets balanced.

Complete step-by-step solution:
According to the question, the length of the potentiometer is equal to $10m$, across which a potential difference of $6{\text{V}}$ is maintained. So the potential gradient of the given potentiometer wire is given by
$k = \dfrac{V}{l}$..................(1)
Substituting $V = 6{\text{V}}$ and $l = 10m$ in the above expression, we get
$k = \dfrac{6}{{10}}$
$ \Rightarrow k = 0.6{\text{V}}{{\text{m}}^{ - 1}}$................(2)
Now, according to the question, a cell balances against a length of $180cm$ of the same potentiometer wire. Let the emf of this cell $E$. So we have
$l = 180cm$
We know that $1cm = 0.01m$. So the value of the given length can be written as
$l = 1.8m$
Substituting (2), (3) and $V = E$ in (1) we get
$0.6 = \dfrac{E}{{1.8}}$
\[ \Rightarrow E = 0.6 \times 1.8\]
On solving we get
$E = 1.08{\text{V}}$
Thus, the emf of the cell which balances against the given length of the potentiometer wire is equal to \[1.08{\text{V}}\].

Hence, the correct answer is option C.

Note: Do not forget to convert the value of the given length of the potentiometer wire, against which the second cell gets balanced, into the SI unit. It is given in centimetres. Therefore, it has to be converted to metres.