Question

A weight exerts force of 120N on steel wire of cross-sectional area 0.02cm^2.Find the extension produced if length of wire is 5m.(Y=2×10^13 N/m^2​)​

Answer 1

Answer:

15.01 μ

Explanation:

Given :

Force = 120 N

Cross section area = 0.02 cm² or 0.02 × 10⁻⁴ m²

Length of wire = 5 m .

Young's modulus = 2 × 10¹³ N m⁻²

We know :

Y = F / A × L / Δ L

Putting value here

2 × 10¹³ = ( 120 /  0.02 × 10⁻⁴ ) × ( 5 / Δ L )

2 × 10¹² = ( 12 /  2 × 10⁻⁶ ) × ( 5 / Δ L )

10¹² = ( 6 /  0.4 × 10⁻⁶ ) × ( 1 / Δ L )

10⁶ = ( 1 /  0.066 ) × ( 1 / Δ L )

Δ L = 15.01 × 10⁻⁶ m

Hence ,  extension produced if length is 15.01 μ