
In a potentiometer arrangement, a cell of emf \[1.25\,{\text{V}}\] gives a balance point at \[35\,{\text{cm}}\] length of the wire. If the cell is replaced by another cell, the balance point shifts to \[63\,{\text{cm}}\], then the emf of second cell is
A. \[4.25\,{\text{V}}\]
B. \[2.25\,{\text{V}}\]
C. \[3.25\,{\text{V}}\]
D. \[1.25\,{\text{V}}\]
Answer
561k+ views
Hint: Use the formula for the balancing condition for a potentiometer. This formula gives the relation between emf of the first cell and its corresponding balancing length and emf of the second cell and its balancing length. Substitute all the values in the equation and calculate the emf of the second cell.
Formula used:
The balance condition for a potentiometer is given by
\[\dfrac{{{E_1}}}{{{L_1}}} = \dfrac{{{E_2}}}{{{L_2}}}\] …… (1)
Here, \[{E_1}\] is the emf of the first cell, \[{E_2}\] is the emf of the second cell, \[{L_1}\] is the balancing length for the first cell and \[{L_2}\] is the balancing length for the second cell.
Complete step by step answer:
We have given that for the first cell of emf \[1.25\,{\text{V}}\] the balancing length is \[35\,{\text{cm}}\].
\[{E_1} = 1.25\,{\text{V}}\]
\[{L_1} = 35\,{\text{cm}}\]
We have also given that the balancing length for the second cell is \[63\,{\text{cm}}\].
\[{L_2} = 63\,{\text{cm}}\]
We can determine the emf of the second cell using equation (1).
Rearrange equation (1) for the emf \[{E_2}\] of the second cell.
\[{E_2} = \dfrac{{{E_1}{L_2}}}{{{L_1}}}\]
Substitute \[1.25\,{\text{V}}\] for \[{E_1}\], \[35\,{\text{cm}}\] for \[{L_1}\] and \[63\,{\text{cm}}\] for \[{L_2}\] in the above equation.
\[{E_2} = \dfrac{{\left( {1.25\,{\text{V}}} \right)\left( {63\,{\text{cm}}} \right)}}{{35\,{\text{cm}}}}\]
\[ \Rightarrow {E_2} = 2.25\,{\text{V}}\]
Therefore, the emf of the second cell to have the given value of balancing length must be \[2.25\,{\text{V}}\].
So, the correct answer is “Option B”.
Note:
If one attempts to convert the units of balancing lengths of the two cells in the SI system of units, it is a total waste of time. This is because the units of the two balancing length terms get cancelled when the substitution is done. The only thing one should keep in mind that the units of two balancing lengths should be the same for getting them cancelled.
Formula used:
The balance condition for a potentiometer is given by
\[\dfrac{{{E_1}}}{{{L_1}}} = \dfrac{{{E_2}}}{{{L_2}}}\] …… (1)
Here, \[{E_1}\] is the emf of the first cell, \[{E_2}\] is the emf of the second cell, \[{L_1}\] is the balancing length for the first cell and \[{L_2}\] is the balancing length for the second cell.
Complete step by step answer:
We have given that for the first cell of emf \[1.25\,{\text{V}}\] the balancing length is \[35\,{\text{cm}}\].
\[{E_1} = 1.25\,{\text{V}}\]
\[{L_1} = 35\,{\text{cm}}\]
We have also given that the balancing length for the second cell is \[63\,{\text{cm}}\].
\[{L_2} = 63\,{\text{cm}}\]
We can determine the emf of the second cell using equation (1).
Rearrange equation (1) for the emf \[{E_2}\] of the second cell.
\[{E_2} = \dfrac{{{E_1}{L_2}}}{{{L_1}}}\]
Substitute \[1.25\,{\text{V}}\] for \[{E_1}\], \[35\,{\text{cm}}\] for \[{L_1}\] and \[63\,{\text{cm}}\] for \[{L_2}\] in the above equation.
\[{E_2} = \dfrac{{\left( {1.25\,{\text{V}}} \right)\left( {63\,{\text{cm}}} \right)}}{{35\,{\text{cm}}}}\]
\[ \Rightarrow {E_2} = 2.25\,{\text{V}}\]
Therefore, the emf of the second cell to have the given value of balancing length must be \[2.25\,{\text{V}}\].
So, the correct answer is “Option B”.
Note:
If one attempts to convert the units of balancing lengths of the two cells in the SI system of units, it is a total waste of time. This is because the units of the two balancing length terms get cancelled when the substitution is done. The only thing one should keep in mind that the units of two balancing lengths should be the same for getting them cancelled.
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