Question

1 kg piece of copper is drawn into a wire of cross-sectional area 1 mm², and then the same wire is remoulded again into a wire of cross-sectional area 2 mm². Find the ratio of the resistance of the first wire to the second wire.

• A. 2:1
• B. 4:1
• C. 8:1
• D. 16:1

Answer 1

The correct option is B 4:1

Volume of a solid is the product of length and cross-sectional area. $V=AL$
Since volume is same in both the cases, the length of second wire will be half of the length of 1st wire.
Accordingly, let length of 1st wire = L
area of 1st wire = A = 1$m{m}^2$
So resistance $R=\rho \frac{L}{A}=\rho \frac{L}{1}=\rho L$

Length of the second wire $=\frac{L}{2}$ and
area of 2nd wire $A=2m{m}^2$
So resistance of 2nd wire $R^{\prime }=\rho \frac{\frac{L}{2}}{A}$
$R^{\prime }=\rho \frac{L}{2\times 2}$
$R^{\prime }=\frac{1}{4}\left(\rho L\right)$
$R^{\prime }=\frac{1}{4}\left(R\right)$
$\frac{R}{R^{\prime }}=\frac{4}{1}$
R : R' = 4 : 1
we get the resistance of the 1st wire is 4 times the value of resistance of the 2nd wire.