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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 \propto l^{2}

$\Rightarrow \frac{R_{1}}{R_{2}} = {(\frac{l_{1}}{l_{2}})}^{2}$

$\Rightarrow \frac{100}{R_{2}} = {(\frac{l}{2 l})}^{2} \Rightarrow \frac{100}{R_{2}} = \frac{1}{4}$

$R_{2} = 400 Ω$

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