A piece of wire of resistance $20 Ω$ is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.
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We know that the resistance of a conductor is given by:
$\text{R}=\rho\frac{1}{\text{A}}=20\Omega$
Where $\rho=\text{resistivity}$
l = length of the conductor
A = area of cross-section of the conductor
Now the length is increased to twice the original length. Then let the new resistance be denoted by R'.
$\text{R}'=\rho\frac{2\text{I}}{\frac{\text{A}}{2}}=4\rho\frac{1}{\text{A}}$
$\text{R}'=4\text{R}=4\times20=80\Omega$
Thus, the new resistance will become four times.
$\text{R}=\rho\frac{1}{\text{A}}=20\Omega$
Where $\rho=\text{resistivity}$
l = length of the conductor
A = area of cross-section of the conductor
Now the length is increased to twice the original length. Then let the new resistance be denoted by R'.
$\text{R}'=\rho\frac{2\text{I}}{\frac{\text{A}}{2}}=4\rho\frac{1}{\text{A}}$
$\text{R}'=4\text{R}=4\times20=80\Omega$
Thus, the new resistance will become four times.
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