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A metallic wire has a resistance of $10 Ω$ . when the wire is stretched to twice its original length, resistivity of its material increases by 10%. Then the new resistance is

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$25 Ω$

$44 Ω$

$60 Ω$

$50 Ω$

answer is D.

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Detailed Solution

$R = ρ . \frac{l}{A} a n d l' A' = l A \Rightarrow A' = \frac{l A}{l'}$

$R' = ρ' . \frac{l'}{A'} = (1.10 ρ) \frac{l}{A} . {(\frac{l'}{l})}^{2} = 1.10 R {(\frac{l'}{l})}^{2}$

$\Rightarrow R' = 1.1 x 10 x {(2)}^{2} Ω = 44 Ω$

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