Two wires of the same material have lengths L and 2L cross-sectional areas 4A and A respectively. The ratio of their resistances would be
A. 1:1
B. 1:8
C. 8:1
D. 1:2

Hint:The resistivity is the property of the material so if different resistors are made of the same material then the resistivity of all the resistors will be equal. The resistance is proportional to the length and inversely proportional to the area of cross-section.

Formula used:
\[R = \dfrac{{\rho l}}{A}\]
where R is the resistance of the wire of length l and cross-sectional area A, \[\rho \] is the resistivity of the material of the wire.

Complete step by step solution:
If the length of the resistor is l and the area of cross-section is A. The wires are made of the same material. So, the resistivity of first wire is equal to the resistivity of the second wire,
\[{\rho _1} = {\rho _2}\]
\[\Rightarrow \dfrac{{{\rho _1}}}{{{\rho _2}}} = 1 \ldots \left( i \right) \\ \]
The length of the first wire is given as L and the area of cross-section is 4A.

Using the resistance formula, the resistance of the first wire will be,
\[{R_1} = \dfrac{{{\rho _1}L}}{{4A}}\]
The length of the second wire is given as 2L and the area of cross-section is A.
Using the resistance formula, the resistance of the second wire will be,
\[{R_2} = \dfrac{{{\rho _2}\left( {2L} \right)}}{A}\]

So, the ratio resistances of three wires will be,
\[{R_1}:{R_2} = \left( {\dfrac{{{\rho _1}L}}{{4A}}} \right):\left( {\dfrac{{{\rho _2}\left( {2L} \right)}}{A}} \right) \\ \]
On simplifying the ratio, we get
\[{R_1}:{R_2} = \left( {\dfrac{1}{4}} \right):\left( {\dfrac{2}{1}} \right) \\ \]
\[\Rightarrow {R_1}:{R_2} = \left( {\dfrac{1}{4} \times 4} \right):\left( {\dfrac{2}{1} \times 4} \right) \\ \]
\[\therefore {R_1}:{R_2} = 1:8\]
Hence, the required ratio of the resistances is \[1:8\].

Therefore, the correct option is B.

Note: We should be careful while using the ratio for the resistance. If we have given wires of different metals then the densities and the resistivity of wires would be different.