A $10\, m$ long wire of $20\,\Omega$ resistance is connected with a battery of $3\, volt$ $e.m.f.$ (negligible internal resistance) and a $10 \,\Omega$ resistance is joined to it is series. Potential gradient along wire in volt per meter is
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(c) Potential gradient $x = \frac{e}{{(R + {R_h} + r)}}.\frac{R}{L}$
$ = \frac{3}{{(20 + 10 + 0)}} \times \frac{{20}}{{10}} = 0.2$
$ = \frac{3}{{(20 + 10 + 0)}} \times \frac{{20}}{{10}} = 0.2$
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