Question

A wire X is half the diameter and half the length of a wire Y of similar material. The ratio of resistance of X to that of Y is
A. 8:1
B. 4:1
C. 2:1
D. 1:1

Answer 1

Hint:This problem can be directly solved by applying the formula of resistance in terms of the material and comparing the resistances of the two wires X and Y.
The resistance of a material is dependent on its resistivity, length and the area of cross-section.
The relation is given by the formula:
$R = \rho \dfrac{l}{A}$
where $\rho $= resistivity of the material, $l$ is the length of the material and A is the area of cross-section.

Complete step-by-step answer:
The resistance is a quantity which represents the obstruction offered by the material to the flow of electric current through it when its ends are maintained at a constant potential difference.

The value of resistance can be obtained by Ohm's law, which states that current through a conductor, $I$ is directly proportional to the voltage V, applied across its ends.
Resistance, $R = \dfrac{V}{I}$
However, the resistance of a material depends only on the property of the material known as resistivity and the dimensions.
Resistance, $R = \rho \dfrac{l}{A}$
where $\rho $= resistivity of the material, $l$ is the length of the material and A is the area of cross-section.
The resistance of wire X is, ${R_x} = \rho \dfrac{{{l_x}}}{{{A_x}}}$
The resistance of wire Y is, ${R_y} = \rho \dfrac{{{l_x}}}{{{A_y}}}$
The relationship given in the problem is,
Diameter, ${d_x} = \dfrac{1}{2}{d_y}$
Length, ${l_x} = \dfrac{1}{2}{l_y}$
The areas of the wires, given their diameters are –
${d_x} = \dfrac{1}{2}{d_y}$
Squaring and multiplying $\dfrac{\pi }{4}$ on both sides –
$\dfrac{\pi }{4}d_x^2 = \dfrac{1}{4} \times \dfrac{\pi }{4}d_y^2$
Since, the area of a circle of diameter, d is – $A = \dfrac{\pi }{4}{d^2}$
We have –
${A_x} = \dfrac{1}{4}{A_y}$
$ \Rightarrow \dfrac{{{A_y}}}{{{A_x}}} = 4$
Also,
${l_x} = \dfrac{1}{2}{l_y}$
$ \Rightarrow \dfrac{{{l_x}}}{{{l_y}}} = \dfrac{1}{2}$
By taking the ratios of the resistances of X and Y,
$\dfrac{{{R_x}}}{{{R_y}}} = \dfrac{{\rho \dfrac{{{l_x}}}{{{A_x}}}}}{{\rho \dfrac{{{l_y}}}{{{A_y}}}}}$
Note that the resistivity $\rho $ remains the same since the materials of the wires X and Y are the same.
$\dfrac{{{R_x}}}{{{R_y}}} = \dfrac{{{l_x}}}{{{A_x}}} \times \dfrac{{{A_y}}}{{{l_y}}}$
$\dfrac{{{R_x}}}{{{R_y}}} = \dfrac{{{l_x}}}{{{l_y}}} \times \dfrac{{{A_y}}}{{{A_x}}}$
Substituting,
$\dfrac{{{R_x}}}{{{R_y}}} = \dfrac{1}{2} \times 4 = \dfrac{2}{1}$
The ratios of the wires are 2:1.

Hence, the correct option is Option C.

Note:Always note that the quantity resistance is the dependent quantity and resistivity is the independent quantity in the above equation of resistance. Resistance depends on resistivity and the dimensions but the resistivity is not dependent on the dimensions. It only depends on the nature of the material and it only changes when there is a change in the temperature.