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Q.

A uniform wire of resistance 100 Ω is melted and recast as a wire of length double that of the original. The resistance of the wire is

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a

150 Ω

b

250 Ω

c

300 Ω

d

400 Ω

answer is A.

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Detailed Solution

R∝l2

⇒R1R2=(l1l2)2

⇒100R2=(l2l)2⇒100R2=14

R2=400 Ω

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