A wire of resistance $4 \Omega$ is stretched to twice its original length. The resistance of stretched wire would be

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Aicharya · Accepted Answer

Let length and cross-section area of the wire be l and A respectively.
$\therefore$ Resistance $\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}$ where $\rho$ is the resistivity of the material
Given: $R=4 \Omega$
For $l^{\prime}=2 l$ and $A^{\prime}=\frac{A}{2}$, resistance of the wire be $R^{\prime}$ $\begin{aligned} & \therefore R^{\prime}=\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{2 l}{A / 2}=4 \rho \frac{l}{A}=4 R \\ & \therefore R^{\prime}=4 \times 4 \Omega=16 \Omega \end{aligned}$

A wire of resistance $4 \Omega$ is stretched to twice its original length. The resistance of stretched wire…

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