Question 59 of 380

The resistance of a wire is $5\Omega$. Its new resistance in ohm, if stretched to $5$ times of its original length will be :

A$625$B$125$C$5$D$25$

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Answer: (B) $125$

The resistance of a wire $(R)$ is related to its resistivity $\left(\rho \right)$, length$(l)$ and cross-sectional area $\left(A\right)$ as:
$R=\rho \dfrac{l}{A}$

As the length of wire increases, the cross-sectional area decreases. The cross-sectional area in terms of volume $(V)$ and length $(l)$ is
$A=\dfrac{V}{l}$
$∴R=\rho \dfrac{{l}^{2}}{V}$
Taking the volume of wire constant,
$R\propto {l}^{2}$
So if ${R}^{'}$ is the resistance of wire after getting redrawn and $R$ is the initial resistance of wire,
$∴\dfrac{{R}^{'}}{R}=\dfrac{(5l{)}^{2}}{{l}^{2}}=25 \Rightarrow {R}^{'}=25\times R=125\Omega$

JEE Main Physics practice question on Current Electricity (Resistance and Resistivity), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.