Question

In a potentiometer experiment, two cells connected in series get balanced at 9 cm length on the wire. Now the connections of terminals of the cell of the lower emf is reversed, then the balancing length is obtained at 3 cm. the ratio of the emfs of the two cells will be
\[\left( {\text{A}} \right){\text{ }}1:2\]
\[\left( {\text{B}} \right){\text{ }}2:1\]
\[\left( {\text{C}} \right){\text{ }}1:4\]
\[\left( {\text{D}} \right){\text{ }}4:1\]

Answer 1

Hint:Here we have to find the ratio, try to think of a relation between emf and length.
We use the principle of potentiometer; the emf of the cell is directly proportional to the length at which it is balanced.
Also, by using the given data we will get the required answer.

Formulae used:
$E \propto l$
Here, E= emf of the cell and l= the length at which the emfs get balanced

Complete step by step solution:
According to the principle of potentiometer, the emf of the cell is directly proportional to the length at which it is balanced.
$E \propto l$
Now, let \[{E_1}\] be the emf of the first cell and \[{E_2}\] be the emf of the second cell.
We need to find $\dfrac{{{E_1}}}{{{E_2}}}$
For case one, the cells are connected in series with the negative of cell 1 connected to the positive of cell 2.
In this case, the resultant emf will be the sum of the given emfs.
$E = {E_1} + {E_2}$
$\therefore {E_1} + {E_2} \propto l$
Here, length is equal to 9 cm
$\therefore {E_1} + {E_2} \propto 9$ ------1
For case 2, the cells are connected in series with the negative of cell 1 connected to the negative of cell 2.
In this case, the resultant emf will be the difference of the given emfs.
$E = {E_1} - {E_2}$
Here, length is equal to 3 cm
$\therefore {E_1} - {E_2} \propto 3$ --------2
Now we divide equation 1 by 2
$\dfrac{{{E_1} + {E_2}}}{{{E_1} - {E_2}}} \propto \dfrac{9}{3}$
On dividing the terms we get,
$ \Rightarrow \dfrac{{{E_1} + {E_2}}}{{{E_1} - {E_2}}} \propto 3$
Now we have to do cross multiply, so that we have to removing the proportionality sign and multiply with the constant and we get
${E_1} + {E_2} = 3k\left( {{E_1} - {E_2}} \right)$
${E_1} + {E_2} = 3k{E_1} - 3k{E_2}$
Bringing similar terms to one side and adding we get,
$ - 2k{E_1} = - 4k{E_2}$
We can write it as in the form of ratio we get,
$\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{4k}}{{2k}}$
Cancelling the term and we get
$\therefore \dfrac{{{E_1}}}{{{E_2}}} = \dfrac{2}{1}$

Thus, the correct answer is option B.

Additional information: The potentiometer can also be used to find the internal resistance of the cell and the potential difference across resistors. In 1841, the side wire potentiometer was invented by Johann Christian Poggendorff, a German physicist who is also the inventor of the electrostatic motor.

Note:For potentiometer questions there is only one relation, which is the working principle so only that formula is used in each question. Be it internal resistance of a cell or potential difference across resistors, there is only one formula.