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A 6 volt battery is connected to the terminal of a three metre long wire of uniform thickness and resistance of 100 $\mathrm{ohm}$. The difference of potential between two points on the wire separated by a distance of $50 \mathrm{~cm}$ will be:
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The correct answer is:
1 volt
$$
\begin{gathered}
\therefore \text { Voltage on } 50 \mathrm{~cm}=\frac{6}{300} \times 50 \\
=1 \text { volt }
\end{gathered}
$$
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