Two wires with equal lengths are made of pure copper. The diameter of wire A is three times the diameter of wire B. When 8 kg masses are hung on the wires, wire B stretches more than wire A. You make careful measurements and compute Young's modulus for both wires. What do you find? a. YA > YBb. YA = YBc. YA < YB

Answer: c. YA < YB

Explanation:

The formula for Young's modulus is = Tensile stress / Tensile strain

Tensile stress = Force x Length

Force = mass x acceleration due to gravity

= 8kg x 10m/s

= 80kgm/s

Tensile stress = 80kgm/s x 2m = 160kgm2/s

Tensile strain = Area x change in length

Area = pi x D2 / 4; Pi = 3.14

Change in length = L2 - L1 (New length - Initial length)

Given parameters:

Length of wire A = Length of wire B, (let's use 2meters for the calculation)

For wire A, Diameter = 3 x Wire B diameter

Assuming Diameter of wire B = 1meter

Therefore, diameter of wire A = 1 x 3 = 3meters

It is said that wire B stretches more than wire A when the man of 8kg is placed on both

For wire B, let's assume new length is = 4m

For wire A let's assume new length is = 3m.

(i) Tensile strain of wire A =

Area of wire A = 3.14 x (32) / 4 = 7.065m2

Change in length = 3m - 2m = 1m.

Therefore, tensile strain = 7.065m2 x 1m = 7.065m3

Young's modulus for wire A (YA) = 160kgm2/s divided by 7.065m3

= 22.64Pa.

(ii) Tensile strain of wire B =

Area of wire B = 3.14 x (12) / 4 = 0.785m2

Change in length = 4m - 2m = 2m

Therefore, tensile strain = 0.785m2 x 2m = 1.57m3

Young's modulus for wire B (YB) = 160kgm2/s divided by 1.57m3

= 101.91Pa.

From the calculations above, we see that YA is less than YB (YA < YB). This is true given that wire A has a greater diameter than wire B which in turn impacts the Area of the wire since the diameter is directly proportional to area and the area is inversely proportional to the young's modulus.