JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 26)
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
300
Explanation
Volume is constant so on length doubled
Area is halved so
$$R = \rho {l \over A}$$ and $$R' = \rho {{2l} \over {{A \over 2}}} = 4\rho {l \over A} = 4R$$
So percentage increase will be
$$R\% = {{4R - R} \over R} \times 100 = 300\% $$
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