A potentiometer wire of length $100\, cm$ has a resistance of $10\, ohm.$ It is connected in series with a resistance and an accumulator of emf $2\,V$ and of negligible internal resistance. A source of emf $10\, mV$ is balanced against a length of $40\, cm$ of the potentiometer wire. What is the value of external resistance :- ................. $\Omega$
Medium
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$\mathrm{i}=\frac{2}{10+\mathrm{R}}$
$\mathrm{x}=\frac{\mathrm{V}}{\ell}=\frac{2 \times 10}{(\mathrm{R}+10)} \cdot \frac{1}{100}$
$\mathrm{V}_{1}=\mathrm{x} \ell \quad \Rightarrow \quad 10 \times 10^{-3}=\frac{2 \times 10}{(\mathrm{R}+10)} \times \frac{40}{100}$
$\mathrm{R}+10=\frac{8}{10 \times 10^{-3}}$
$\Rightarrow \quad \mathrm{R}+10=800 \Rightarrow \mathrm{R}=790\, \Omega$
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