Question

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. Question 9.4 machenical properties of solid

Answer 1

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 

Answer 2

Given,
diametre of wires = 0.25 cm
so, radius of wires ( r) = 1.25 × 10^-3 m

For steel wire
__________________


Tension in Steel wire ( F1) = 6g + 4g
= (10)g = 98 N
Length of steel wire (L1) = 1.5 m
Young's modulus (Y1) = 2.× 10¹¹ Pa
we know,
change in length (∆L1) = F1.L1/A1.Y1
= F1× L1/πr²×Y1
= 98×1.5/3.14 × (1.25×10^-3)²×2×10¹¹
= 1.5 × 10^-6 m

For brass wire
________________
Tension in brass wire(F2) = 6g= 6×9.8
= 58.8N
Length of brass wire (L2) = 1 m
Young's modulus (Y2) = 0.91 × 10¹¹

we know,
change in Length (∆L2) = F2×L2/A2×Y2
= 58.8×1/{3.14 × (1.25×10^-3)× 0.91 × 10¹¹}
= 1.3 × 10^-6 m