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A wire of resistance 4 $\Omega$ is stretched to twice its original length. The resistance of stretched wire would be:
A
16 $\Omega$
B
2 $\Omega$
C
4 $\Omega$
D
8 $\Omega$
Detailed Solution
$R = \frac{\rho l}{A} = \frac{\rho l^2}{Al}$; the volume $Al$ is constant, so $R \propto l^2$.
Doubling the length makes the resistance $4\times4 = 16\ \Omega$.
Doubling the length makes the resistance $4\times4 = 16\ \Omega$.