A wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance becomes respectively

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1.21 times, same

1.1 times, 1.1 times

1.2 times, 1.1 times

both remain same

answer is A.

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Detailed Solution

R $\propto$ l² (volume constant)
$\frac{R_{1}}{R_{2}} = {(\frac{l_{1}}{l_{2}})}^{2} = {(\frac{l}{1.1 l})}^{2}$
R₂ =1.21 R
Specifies resistance depends on temperature and material hence remains same

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