An ideal battery of $4\, V$ and resistance $R$ are connected in series in the primary circuit of a potentiometer of length $1\, m$ and resistance $5\,\Omega $ . The value of $R$, to give a difference of $5\, mV$ across $10\, cm$ of potentiometer wire, is: ................ $\Omega$
JEE MAIN 2019, Medium
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$i=\frac{4}{5+R}$
$V_{A B}=i(5)=\frac{20}{5+R}$
$V_{A P}=\frac{V_{A B}}{L}(0.1)=\frac{20}{5+R}\left(\frac{0.1}{1}\right)=\frac{2}{5+R}$
Now, $\frac{2}{5+R}=5 \times 10^{-3}$
$\Rightarrow \quad R=395\, \Omega$
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