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In the determination of the internal resistance of a cell with a potentiometer, the error in the measurement of the balancing length is $\pm 1 \mathrm{~mm}$. When the cell alone is connected in the circuit, the balancing length is obtained at $60 \mathrm{~cm}$ and when the cell is shunted with a resistance of $10 \Omega \pm 2 \%$, the balancing length is obtained at $50 \mathrm{~cm}$. The error in the determination of the internal resistance of the cell is
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The correct answer is:
$4.2 \%$
$\quad r=R\left(\frac{l_1}{l_2}-1\right)$
This is value of internal resistance here.
$$
\begin{aligned}
& \frac{\Delta r}{r} \times 100=\frac{\Delta R}{R} \times 100+\frac{\Delta\left(\frac{l_1}{l_2}\right)}{\left(\frac{l_1}{l_2}\right)} \times 100 \\
& =4+0.2=4.2 \%
\end{aligned}
$$
This is value of internal resistance here.
$$
\begin{aligned}
& \frac{\Delta r}{r} \times 100=\frac{\Delta R}{R} \times 100+\frac{\Delta\left(\frac{l_1}{l_2}\right)}{\left(\frac{l_1}{l_2}\right)} \times 100 \\
& =4+0.2=4.2 \%
\end{aligned}
$$
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