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Q.
A wire of resistance $5\, \Omega$ is drawn out so that its length is increased by twice its original length, its new resistance is
ManipalManipal 2020
Solution:
Resistance of a wire is given by $R=\frac{\rho L}{A}$
where, $\rho=$ resistivity, $L=$ length of the wire
$A=$ area of cross- section of the wire
Equating initial and finial volumes of the wire, we can write
$A L=A' L'=A' \times(2 L)$
$\Rightarrow A'=\frac{A}{2}$
$\therefore \frac{R'}{R}=\frac{\rho L' / A'}{\rho L / A}=\left(\frac{L'}{L}\right)\left(\frac{A}{A'}\right)$
$=\left(\frac{2 I}{L}\right)\left(\frac{A}{A / 2}\right)=4$
$\Rightarrow R'=4 R=4(5)=20\, \Omega$