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The potential gradient along the length of a uniform wire is 10 volt/metre $B$ and $C$ are the two points at $30 \mathrm{~cm}$ and $60 \mathrm{~cm}$ point on a meter scale fitted along the wire. The potential difference between $B$ and $C$ will be
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3 volt
Potential gradient $=$ Change in voltage per unit length
$\therefore 10=\frac{V_2-V_1}{30 / 100} \Rightarrow V_2-V_1=3$ volt
$\therefore 10=\frac{V_2-V_1}{30 / 100} \Rightarrow V_2-V_1=3$ volt
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