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 $R_1\propto l^2$
$\Rightarrow \frac{R_1}{R_2}={\left(\frac{l_1}{l_2}\right)}^2={\left(\frac{1}{2}\right)}^2$
$\Rightarrow R_2:R_1=4:1$