A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long.
Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.
The formula for young’s modulus is given as:
Y = Stress/Strain
⇒ Y = (F/A)/Strain
Here, A = area of wire = π r2 = π (d/2)2 = 
F = Tension force
d = Diameter of the wire
⇒ Y = 
From the above equation, we can infer that young’s modulus,
Y α (1/d2) …………….. (1)
Young’s modulus for iron from standard table, Y1 = 190 × 109 Pa
Young’s modulus for copper from standard table, Y2 = 120× 109 Pa
Let,
Diameter of the iron wire be d1
Diameter of the copper wire be d2
From (1), the ratios of young’s modulus is,
⇒ 
⇒ 
⇒ 
⇒ 
Therefore, the ratio of their diameters is 1.25: 1