Question

Two wires are made of the same material. Wire 1 has length that is 1.35 times the length of wire 2 and diameter that is 0.91 times the diameter of wire 2. What is the ratio of the resistance of wire 1 to the resistance of wire 2

Answer 1

Answer: 1.63 : 1

Step-by-step explanation:

let length of wire 2 be 10 m, so that length of wire 1 will be 13.5 m

Also, let the diameter of wire 2 be 10 m, so that diameter if wire 1 will be 9.1 m

Area, A = πd²/4, so that, area of wire 2 will be

A2 = π*10²/4

A2 = 25π

A1 = π*9.1²/4

A1 = 20.7025π

Since they are both made if the same material, we can agree that their resistivity is the same, thus,

ρ = R1*A1/L1 = R2*A2/L2

R1*A1*L2 = R2A2L1, so that

R1/R2 = A2L1/A1L2

R1/R2 = L1/L2 * A2/A1 when we substitute, we have

R1/R2 = 13.5/10 * 25π/20.7025π

R1/R2 = 337.5π / 207.025π

R1/R2 = 1.63 : 1

Answer 2

1.117935:1

Step-by-step explanation:

Since the wires are of the same material, they will have the same resistivity
`\rho`.

The cross-sectional area of the of a wire is given by;

`A=\pi(d^2)/(4)................(1)`

where d is the diameter of the wire.

Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;

`\rho=(RA)/(l)..................(2)`

Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;

`(R_1A_1)/(l_1)=(R_2A_2)/(l_2)..................(3)`

Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;

`(R_1)/(R_2)=(l_1A_2)/(l_2A_1)..................(4)`

Given;

`l_1=1.35l_2`

`(R_1)/(R_2)=(1.35l_2A_2)/(l_2A_1)\\(R_1)/(R_2)=(1.35A_2)/(A_1).............(5)`

We then use equation (1) to fine the ratio of the area
`A_1` to
`A_2`

bearing in mind that
`d_1=0.91d_2`

This ratio gives 0.8281. Substituting this into equation (5), we get the following;

`(R_1)/(R_2)= 1.35*0.8281=1.117935`