A potentiometer wire has length $4\,\, m$ and resistance $8\,\,\Omega $. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\,\, V,$ so as to get a potential gradient $1\,\, m \,V$ per $cm$ on the wire is ............. $\Omega$
AIPMT 2015, Medium
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Required potential gradient $=1 \,\mathrm{mV} \,\mathrm{cm}^{-1}$
$=\frac{1}{10}\, \mathrm{Vm}^{-1}$
Length of potentiometer wire, $l=4\, \mathrm{m}$
So potential difference across potentiometer wire
$=\frac{1}{10} \times 4=0.4\, \mathrm{V}$ ....$(i)$
In the circuit, potential difference across $8 \,\Omega$
$=I \times 8=\frac{2}{8+R} \times 8$ ....$(ii)$
Using equation $(i)$ and $(ii),$ we get,
$0.4=\frac{2}{8+R} \times 8$ $\frac{4}{10}=\frac{16}{8+R}, \,\,8+R=40$
$\therefore \quad R=32 \,\Omega$
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