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A battery of 6 volts is connected to the terminals of a three metre long wire of uniform thickness and resistance of the order of $100 Ω$ The difference of potential between two points separated by 50 cm on the wire will be

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1 V

1.5 V

2 V

3 V

answer is A.

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Detailed Solution

Here same current is passing throughout the length of the wire, hence $V \propto R \propto l$

$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow \frac{6}{V_{2}} = \frac{300}{50} \Rightarrow V_{2} = 1 V$

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