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 elongations of the steel and the brass wires.
Given data,
Diameter of both the wires, d = 0.25 cm
Hence, radius of both the wires = r = d/2
⇒ r = 0.25/2 cm
⇒ r = 0.125 cm.
Length of the steel wire, Ls = 1.5 m
Length of the brass wire, Lb = 1.0 m
From the given figure, the total force exerted on the steel wire:
Fs = (4 kg + 6 kg) × 9.8 ms-2
Here, 9.8 ms-2 is the acceleration due to gravity which acts along with the force attached to the wires.
⇒ Fs = 10 × 9.8 = 98 N.
Young’s modulus for steel:
Ys = 
………………… (1)
Where,
= Change in the length of the steel wire.
= Area of cross-section of the steel wire = πr2
⇒ As = π×(0.125× 10-2)2 m2
Young’s modulus for steel from standard table is, 2.0× 1011 Pa
Substituting the value is (1)
2.0× 1011 Pa = 
ΔLs = 
⇒ ΔLs = 1.49 × 10-4 m
Hence, elongation of the steel wire is 1.49 × 10-4 m.
From the given figure, the total force exerted on the brass wire:
Fb = 6 kg × 9.8 ms-2
Here, 9.8 ms-2 is the acceleration due to gravity which acts along with the force attached to the wires.
⇒ Fb = 6 kg × 9.8 ms-2 = 58.8 N.
Young’s modulus for steel:
Yb = 
………………… (1)
Where,
= Change in the length of the brass wire.
= Area of cross-section of the brass wire = πr2
⇒ Ab = π×(0.125× 10-2)2 m2
Young’s modulus for brass from standard table is, 0.91× 1011 Pa
Substituting the value is (1)
0.91× 1011 Pa = 
= 
⇒ ΔLb = 1.3 × 10-4 m
Hence, elongation of the steel wire is 1.3 × 10-4 m.