The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ____________ %.
Answer (integer)
300
Solution
<p>Volume is constant so on length doubled</p>
<p>Area is halved so</p>
<p>$R = \rho {l \over A}$ and $R' = \rho {{2l} \over {{A \over 2}}} = 4\rho {l \over A} = 4R$</p>
<p>So percentage increase will be</p>
<p>$R\% = {{4R - R} \over R} \times 100 = 300\%$</p>
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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