Question

Two wires of same material have their lengths in ratio 1:4 and the ratio of the area of cross section is 1:2. Find the ratio of resistances of the wires?

Answer 1

Given :

▪ Ratio of lengths = 1:4

▪ Ratio of area of cross sections = 1:2

To Find :

▪ Ratio of resistances of the wires.

Concept :

✏ This question is completely based on the concept of resistivity (specific resistance).

✏ Resistivity is a materialistic property of metal, it is independent of the dimensions of the wire but depends upon the nature of the material of the wire.

✏ Both wires are made of the same matetial it means both have same resistivity.

Calculation :

$\dashrightarrow\sf\:{\rho}_1={\rho}_2\\ \\ \dashrightarrow\sf\:\dfrac{R_1A_1}{L_1}=\dfrac{R_2A_2}{L_2}\\ \\ \dashrightarrow\sf\:\dfrac{R_1}{R_2}=\dfrac{A_2}{A_1}\times \dfrac{L_1}{L_2}\\ \\ \dashrightarrow\sf\:\dfrac{R_1}{R_2}=\dfrac{2}{1}\times \dfrac{1}{4}\\ \\ \gray{\dashrightarrow}\underline{\boxed{\bf{\orange{R_1:R_2=1:2}}}}\:\gray{\bigstar}$

Answer 2

Solution -

In the above Question , we have the following information -

Two wires of same material have their lengths in ratio 1:4 and the ratio of the area of cross section is 1:2.

To find -

We have to find the ratio of resistances of the wires .

We know that -

$\sf{ Resistance = \rho \dfrac{ L }{ A } }$

The resistance in a wire is directly proportional to the length of the wire and is Inversely proportional to the cross Sectional area of the wire .

Two wires of same material have their lengths in ratio 1:4 and the ratio of the area of cross section is 1:2.

So,

Let the lengths of the wires by x and 4x respectively .

Let the ratio of cross sectional areas be y and 2y respectively .

$\sf{ \dfrac{Resistance_{1}}{Resistance_{2}} \propto \dfrac{Length_{1}} {Length_{2}} \times \dfrac{CSA_{1}}{CSA_{2}} } \\ \\ \sf{ => \dfrac{Resistance_{1}}{Resistance_{2}} = \dfrac{x}{4x} \times \dfrac{2y}{y} } \\ \\ \sf{ => \dfrac{Resistance_{1}}{Resistance_{2}} = \dfrac{1}{4} \times 2 } \\ \\ \sf{ => \dfrac{Resistance_{1}}{Resistance_{2}} = \dfrac{1}{2} }$

Hence , the required ratio of resistances of the wires is 1 : 2 .