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A wire of resistance $4 \Omega$ is stretched to twice its original length. The resistance of stretched wire would be
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The correct answer is:
$16 \Omega$
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 \(R^{\prime}\)
\(\begin{aligned}
& \therefore R^{\prime}=\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{2 l}{A / 2}=4 \rho \frac{1}{A}=4 R \\
& \therefore R^{\prime}=4 \times 4 \Omega=16 \Omega
\end{aligned}\)
\(\therefore\) 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 \(R^{\prime}\)
\(\begin{aligned}
& \therefore R^{\prime}=\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{2 l}{A / 2}=4 \rho \frac{1}{A}=4 R \\
& \therefore R^{\prime}=4 \times 4 \Omega=16 \Omega
\end{aligned}\)
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