Answer: Option B
:
B
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 2ndwire$A=2m{m}^2$
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.