Question

Question 9. 5. Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m.Young’s modulus of steel is 2.0 x 1011 Pa. Compute the elongations of steel and brass wires. (1 Pa = 1 N m2).​

Answer 1

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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, = 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:

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

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Answer 2

Explanation:

ATQ 

r = d/2 = 0.125 cm 

L1(steel) = 1.5 m 

L2(brass) = 2.0 m 

Force (steel) or F1 = (4+6)g = 10 x 9.8 = 98 N 

Young's modulus for steel 

⇒ Y1  = (F1/A1 ) / (ΔL1 / L1) 

⇒ ΔL1 = 9.8 x 15 / π(0.125 x 10⁻²)² x 2 x10¹¹ = 1.49 x 10⁻⁴ m

Similarly young's modulus for brass 

Y2 = (F2 / A2) / (ΔL2 / L2) 

And ΔL2 = 58.8 x 1.0 / π(0.125 x 10⁻²) ² x (0.91 x 10¹¹) = 1.3 x 10⁻⁴ m