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In a potentiometer experiment the balancing length with a cell is \(560 \mathrm{~cm}\). When an external resistance of \(10 \Omega\) is connected in parallel to the cell, then the balancing length changes by \(60 \mathrm{~cm}\). Find the internal resistance of the cell.
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Verified Answer
The correct answer is:
\(1.2 \Omega\)
Given, \(l_1=560 \mathrm{~cm}, R=10 \Omega\),
\(\Rightarrow \quad \begin{aligned}
& l_2=l_1-60 \\
& l_2=560-60=500 \mathrm{~cm}
\end{aligned}\)
\(\therefore\) Internal resistance of cell is given by
\(\begin{aligned}
& r=\left(\frac{l_1-l_2}{l_2}\right) R=\left(\frac{560-500}{500}\right) 10 \\
& =\frac{60}{500} \times 10=\frac{6}{5}=1.2 \Omega
\end{aligned}\)
\(\Rightarrow \quad \begin{aligned}
& l_2=l_1-60 \\
& l_2=560-60=500 \mathrm{~cm}
\end{aligned}\)
\(\therefore\) Internal resistance of cell is given by
\(\begin{aligned}
& r=\left(\frac{l_1-l_2}{l_2}\right) R=\left(\frac{560-500}{500}\right) 10 \\
& =\frac{60}{500} \times 10=\frac{6}{5}=1.2 \Omega
\end{aligned}\)
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