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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-9 Mechanical properties of Solids



Q.5:- 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.

 

 

Answer:-

Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m
Diameter of the wires, d = 0.25 m
Hence, the radius of the wires, r = d/2  =  0.125 cm
Length of the steel wire, L1 = 1.5 m
Length of the brass wire, 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:
Y1 = (F1/A1) / (ΔL1 / L1)
Where,
ΔL1 = Change in the length of the steel wire
A1 = Area of cross-section of the steel wire = πr12
Young’s modulus of steel, Y1 = 2.0 × 1011 Pa
∴ ΔL1 = F1  × L1 / (A1  × Y1)
= (98  × 1.5) / [ π(0.125  × 10-2)2  × 2  × 1011]   =   1.49  × 10-4 m



Total force on the brass wire:
F2 = 6 × 9.8 = 58.8 N
Young’s modulus for brass:
Y2 = (F2/A2) / (ΔL2 / L2)
Where,
ΔL2 = Change in the length of the brass wire
A1 = Area of cross-section of the brass wire = πr12
∴ ΔL2 = F2  × L2 / (A2  × Y2)
= (58.8 X 1) / [ (π  × (0.125  × 10-2)2  × (0.91  × 1011) ] = 1.3  × 10-4 m
Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m.