A) 300%
B) 200%
C) 100%
D) 50%
Correct Answer: A
Solution :
[a] The new length is 2l if the original length is l. clearly, the new cross-sectional area is \[\alpha /2\] if a is the initial cross-sectional area. This is because the volume of the wire has to remain constant. Now, \[R\alpha =\rho \frac{21}{9/2}=4R\] This increase in resistance is \[4R-R=3R.\] The percentage increase in resistance is \[\frac{3R}{R}\times 100=\] 300%
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