Two wires each of radius 0.2 cm and negligible mass, one made of steel and other made of brass are loaded as shown in the figure. The elongation of the steel wire is _____ x 10-6 m . [Young’s modulus for steel = 2 x 1011 Nm-2 and g = 10ms-2]
Show Hint
Remember the formula for elongation under tensile stress: \(∆L = \frac{FL}{AY}\) consistent units throughout your calculations
To find the elongation of the steel wire, we first calculate the force exerted by the weights and then use it to determine the elongation using Young’s modulus formula.
Step 1: Calculate Forces
The total weight supported by the steel wire is comprised of two weights:
Weight 1 = 2 kg, Weight 2 = 1.14 kg
Total mass = 2 kg + 1.14 kg = 3.14 kg
Total force, F = mass × gravity = 3.14 kg × 10 m/s2 = 31.4 N
Step 2: Calculate Cross-sectional Area
Radius, r = 0.2 cm = 0.002 m (converted to meters)
Area, A = πr2 = π(0.002 m)2 ≈ 1.25664 × 10-5 m2
Step 3: Calculate Elongation using Young's Modulus
Young’s modulus for steel, Y = 2 × 1011 N/m2
Elongation, ΔL = (F × L) / (A × Y)
Where L is the original length of the steel wire = 1.6 m.
ΔL = (31.4 N × 1.6 m) / (1.25664 × 10-5 m2 × 2 × 1011 N/m2)
ΔL ≈ 2.002 × 10-5 m or 20.02 × 10-6 m.
The elongation is 20 × 10-6 m, which lies within the range of [20, 20].
