A copper wire and an aluminium wire have length's in the ratio 3:2.diameters n the ratio 2:3 AND FORCES APPLIED IN THE RATIO 4:5 find the ratio of increase in length of the two wires.(Y FOR COPPER =1.1*1O^11

N/m^2,Y for aluminum=0.70*10^11 N/m^2).
ans:
1)189:110

X = Length increase = L*Force/(Area)(Y)

Y is Young's modulus

X(copper)/X(aluminum) = (3/2)*(4/5)/[(2/3)^2*(1.1/.7]
= 1.718
which happens to equal 189/110

To find the ratio of increase in length of the two wires, we'll use the Young's modulus formula:

Young's modulus (Y) = (Stress)/(Strain)

The stress is the force applied divided by the cross-sectional area, and the strain is the increase in length divided by the original length.

Let's denote the original length of the copper wire as Lc, the original length of the aluminum wire as La, the increase in length of the copper wire as ΔLc, and the increase in length of the aluminum wire as ΔLa.

Given:
Length ratio (Lc:La) = 3:2
Diameter ratio (Dc:Da) = 2:3
Force ratio (Fc:Fa) = 4:5
Young's modulus of copper (Yc) = 1.1 * 10^11 N/m^2
Young's modulus of aluminum (Ya) = 0.70 * 10^11 N/m^2

Step 1: Calculate the cross-sectional areas

The cross-sectional area of a wire is proportional to the square of its diameter.

Let's assume the diameter of the copper wire is 2x.
Then the diameter of the aluminum wire is 3x.

The cross-sectional area of the copper wire (Ac) is A = πr^2 = π(x)^2
The cross-sectional area of the aluminum wire (Aa) is A = πr^2 = π(1.5x)^2 = π(2.25x^2)

Step 2: Calculate the stress

The stress is the force divided by the cross-sectional area.

The stress on the copper wire (σc) is Fc/Ac = (4x)/(π(x)^2) = 4/πx
The stress on the aluminum wire (σa) is Fa/Aa = (5x)/(π(2.25x^2)) = 5/(2.25πx)

Step 3: Calculate the strain

To find the strain, we need to divide the increase in length by the original length.

The strain on the copper wire (εc) is ΔLc/Lc
The strain on the aluminum wire (εa) is ΔLa/La

Step 4: Calculate the ratio of increase in length

Finally, let's find the ratio of increase in length of the two wires.

Ratio of increase in length (ΔLc:ΔLa) = (σc/εc):(σa/εa)
= (4/πx):(5/(2.25πx))
= (4/π):(5/(2.25π))

Simplifying the ratio, we get:
= (4*2.25):(π*5)
= 9:5

Therefore, the ratio of increase in length of the two wires is 9:5.

To find the ratio of increase in length of the two wires, we can use the formula:

ΔL/L = F/(A*Y)

Where:
ΔL = change in length
L = original length
F = applied force
A = cross-sectional area of the wire
Y = Young's modulus

First, let's find the ratio of cross-sectional areas of the two wires using their diameters:

Ratio of diameters = 2:3
Ratio of radii = (2/2):(3/2) = 1:1.5
Ratio of areas = (1^2):(1.5^2) = 1:2.25

Since the lengths are in the ratio 3:2, we can assume the original lengths as 3x and 2x respectively, where x is a constant.

For the copper wire:
Original length = 3x
Applied force = 4

For the aluminum wire:
Original length = 2x
Applied force = 5

Now, let's calculate the change in length for each wire:

ΔL_copper = (4/(A_copper*Y_copper)) * L_copper
ΔL_aluminium = (5/(A_aluminium*Y_aluminium)) * L_aluminium

Substituting the values:
ΔL_copper = (4/(1*1.1*10^11)) * 3x = (4/3.3*10^11) * 3x = 4x/1.1*10^11
ΔL_aluminium = (5/(2.25*0.70*10^11)) * 2x = (5/3.9375*10^11) * 2x = 10x/7*10^11

Now, let's find the ratio of increase in length:

Ratio of increase in length = ΔL_copper/ΔL_aluminium
= (4x/1.1*10^11)/(10x/7*10^11)
= (4x * 7*10^11)/(1.1*10^11 * 10x)
= (4 * 7*10^11)/(1.1 * 10^11 * 10)
= (28 * 10^11)/(11 * 10^11)
= 28/11
= 189/77

Simplifying the ratio gives the final answer:
189:77