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 50cm on the wire will be

(1) 1 V

(2) 1.5 V

(3) 2 V

(4) 3 V

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

$\frac{{V}_{1}}{{V}_{2}}=\frac{{l}_{1}}{{l}_{2}}$$\frac{6}{{V}_{2}}=\frac{300}{50}$

V2 =1 V.

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