Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongation of the steel and the brass wires.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongation of the steel and the brass wires.
Given,
Diameter of the wires,d = 0.25 m
Hence, the radius of the wires, 
= 0.125 cm
Length of the steelwire, L1 = 1.5 m
Length of the brasswire, L2 = 1.0 m
Total force exerted on the steel wire:
F1 =(4 + 6) g = 10 × 9.8 = 98 N
Young’s modulus for steel:
Where,
ΔL1= Change in the length of the steel wire
A1 =Area of cross-section of the steel wire
Young’s modulus of steel, Y1 = 2.0 × 1011 Pa
Total force on the brass wire:
F2 =6 × 9.8 = 58.8 N
Young’s modulus for brass:
Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m
Given,
Diameter of the wires,d = 0.25 m
Hence, the radius of the wires, = 0.125 cm
Length of the steelwire, L1 = 1.5 m
Length of the brasswire, L2 = 1.0 m
Total force exerted on the steel wire:
F1 =(4 + 6) g = 10 × 9.8 = 98 N
Young’s modulus for steel:
Where,
ΔL1= Change in the length of the steel wire
A1 =Area of cross-section of the steel wire
Young’s modulus of steel, Y1 = 2.0 × 1011 Pa
Total force on the brass wire:
F2 =6 × 9.8 = 58.8 N
Young’s modulus for brass:
Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m
