A battery of internal resistance one ohm and $emf$ $3\,volt$ sends a current through $1\,metre$ of uniform wire of resistance $5\,\Omega $. The pole of the cell of $emf$ $1.4\,volt$ are connected to two points on the wire so that no current passes through this cell. Then, the potential gradient of the wire is
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Current through the wire of $5$ $\mathrm{ohm}$
$=\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}}=\frac{3}{1+5}=0.5 \mathrm{\,A}$
Potential cifference actioss the wire of $5 \Omega=0.5 \times 5=2.5 \mathrm{\,V}$
Length of wire $=1 \mathrm{\,m}$
$\therefore $ Potential gradient $=2.5 \mathrm{\,V} / \mathrm{m}$
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