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In a potentiometer experiment of a cell of emf \( 1.25 \mathrm{~V} \) gives balancing length of \( 30 \mathrm{~cm} \). If the
cell is replaced by another cell, balancing length is found to be \( 40 \mathrm{~cm} \). What is the emf of
second cell ?
Options:
cell is replaced by another cell, balancing length is found to be \( 40 \mathrm{~cm} \). What is the emf of
second cell ?
Solution:
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Verified Answer
The correct answer is:
\( \simeq 1.67 \mathrm{~V} \)
Given, for potentiometer experiment for cell of emf $E_{1}=1.25$, balancing length $L_{1}=30 \mathrm{~cm}$.
For cell of emf $E_{2}$, balancing length $L_{2}=40 \mathrm{~cm}$.
We know $E \propto L$
$\Rightarrow \frac{E_{1}}{E_{2}}=\frac{L_{1}}{L_{2}} \Rightarrow \frac{1.25}{E_{2}}=\frac{30}{40}$
$\Rightarrow E_{2}=1.25 \times \frac{4}{3}=1.666 \mathrm{~V}$
Therefore $E_{2} \sim 1.67 \mathrm{~V}$
Emf of second cell is $1.67$
For cell of emf $E_{2}$, balancing length $L_{2}=40 \mathrm{~cm}$.
We know $E \propto L$
$\Rightarrow \frac{E_{1}}{E_{2}}=\frac{L_{1}}{L_{2}} \Rightarrow \frac{1.25}{E_{2}}=\frac{30}{40}$
$\Rightarrow E_{2}=1.25 \times \frac{4}{3}=1.666 \mathrm{~V}$
Therefore $E_{2} \sim 1.67 \mathrm{~V}$
Emf of second cell is $1.67$
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