Question

A wire of resistance $5\Omega$ is drawn out so that its length is increased by twice its original length, its new resistance is

• A. $45\Omega$
• B. $54\Omega$
• C. $20\Omega$
• D. $5\Omega$

Answer 1

Answer: (C)

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
$AL=A^{\prime }{L}^{\prime }=A^{\prime }\times (2L)$
$\Rightarrow A^{\prime }=\frac{A}{2}$
$\therefore \frac{R^{\prime }}{R}=\frac{\rho L^{\prime }/A^{\prime }}{\rho L/A}=\left(\frac{L^{\prime }}{L}\right)\left(\frac{A}{A^{\prime }}\right)$
$=\left(\frac{2I}{L}\right)\left(\frac{A}{A/2}\right)=4$
$\Rightarrow R^{\prime }=4R=4(5)=20\Omega$