Answer
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Hint: In this question use the concept that potentiometer helps establish the relationship between the potential drop (say emf) and the length across which the potential drop is to be measured. This will help understand the principle on which potentiometer works.
Complete Step-by-Step solution:
Principle of potentiometer:
The principle of potentiometer is that, for a wire having uniform area of cross section, the potential drop in the wire is directly proportional to its length.
The above principle is valid only when the potentiometer is used in comparing the EMF of two cells.
For example:
Consider two potentiometers the length of the first potentiometer is 10 cm and another potentiometer is 20 cm if the EMF of the first potentiometer is 5 V then what is the EMF of another potentiometer.
So according to above information we have,
$ \Rightarrow EMF \propto {\text{length}}$
$ \Rightarrow \dfrac{{{{\left( {EMF} \right)}_1}}}{{{{\left( {EMF} \right)}_2}}} = \dfrac{{{L_1}}}{{{L_2}}}$
$ \Rightarrow \dfrac{5}{{{{\left( {EMF} \right)}_2}}} = \dfrac{{10}}{{20}}$
$ \Rightarrow {\left( {EMF} \right)_2} = 10$ V.
So the EMF of another cell is 10 V.
So this is the required answer.
Note – There is always a great bit of confusion between a potentiometer and a rheostat. A potentiometer is a three terminal device used for current control and a rheostat is an electromechanical device designed so that the resistance values can easily be changed. A rheostat is used in potentiometer assembly so as to vary the resistance across specific lengths of potentiometer.
Complete Step-by-Step solution:
Principle of potentiometer:
The principle of potentiometer is that, for a wire having uniform area of cross section, the potential drop in the wire is directly proportional to its length.
The above principle is valid only when the potentiometer is used in comparing the EMF of two cells.
For example:
Consider two potentiometers the length of the first potentiometer is 10 cm and another potentiometer is 20 cm if the EMF of the first potentiometer is 5 V then what is the EMF of another potentiometer.
So according to above information we have,
$ \Rightarrow EMF \propto {\text{length}}$
$ \Rightarrow \dfrac{{{{\left( {EMF} \right)}_1}}}{{{{\left( {EMF} \right)}_2}}} = \dfrac{{{L_1}}}{{{L_2}}}$
$ \Rightarrow \dfrac{5}{{{{\left( {EMF} \right)}_2}}} = \dfrac{{10}}{{20}}$
$ \Rightarrow {\left( {EMF} \right)_2} = 10$ V.
So the EMF of another cell is 10 V.
So this is the required answer.
Note – There is always a great bit of confusion between a potentiometer and a rheostat. A potentiometer is a three terminal device used for current control and a rheostat is an electromechanical device designed so that the resistance values can easily be changed. A rheostat is used in potentiometer assembly so as to vary the resistance across specific lengths of potentiometer.
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