Question

A metal wire of circular cross-section has a resistance $R_1$. The wire is now stretched without breaking so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now becomes $R_2$ then $R_2 : R_1$ is

• A. 1 : 1
• B. 1 : 2
• C. 4 : 1
• D. 1 : 4

Answer 1

Answer: (C)

As we know that, $R_{1}=\rho \frac{1}{a}=\rho \frac{1^{2}}{V}$
where, $l=$ length of wire
$a=$ area of cross-section of the wire
and $V=$ volume of the wire $R_{1} \propto l^{2}$
$\Rightarrow \frac{R_{1}}{R_{2}}=\left(\frac{l_{1}}{l_{2}}\right)^{2}=\left(\frac{1}{2}\right)^{2}$
$\Rightarrow R_{2}: R_{1}=4: 1$