Current Electricity

JEE Main level, numerical answer, medium

133 of 380

Question 133 of 380

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 _____ %

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Answer: 300

As volume is constant, so, length doubled and area reduces to half.
Resistance of a wire is given by R=ρlAR=\dfrac{\rho l}{A}, where, ρ\rho is resistivity of a material.
Here, R1=ρlA{R}_{1}=\dfrac{\rho l}{A} and R2=ρ(2l)A2=4R{R}_{2}=\dfrac{\rho \left(2l\right)}{\dfrac{A}{2}}=4R
So percentage increase in the resistance will be =4R−RR×100=\dfrac{4R-R}{R}\times 100%=300%

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