If error in measuring the length of a wire is 2 % and error in measuring its radius is 1 %, find the maximum possible error in its resistance

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4 %

2 %

2.5 %

3 %

answer is A.

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Detailed Solution

R=ρ.lπr2⇒R∝lr-2∴∆RRmax×100 %=∆ll × 100 %+2.∆rr × 100 % =2 %+2 x 1 %=4 %

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