A potentiometer wire has length $10\, m$ and resistance $20\,\Omega $. A $2. 5\, V$ battery of negligible internal resistance is connected across the wire with an $80\,\Omega $ series resistance. The potential gradient on the wire will be
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(a)Potential gradient $x = \frac{e}{{(R + {R_h} + r)}} \cdot \frac{R}{L}$
$ \Rightarrow $ $x = \frac{{2.5}}{{(20 + 80 + 0)}} \times \frac{{20}}{{10}} = 5 \times {10^{ - 5}}\frac{V}{{mm}}$
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