Question

A steel wire is 4.0 m long and 2 mm in diameter. Young’s modulus of steel is 1.96 xx 10^(11) N//m^(2). If a mass of 20 kg is suspended from it the elongation produced will be-

Answer 1

Answer: Here's your answer

Explanation:

2.54 mm

1.27 mm

0.64 mm

0.27 mm

Answer 2

Given:

• A steel wire is 4.0m long
• 2mm in diameter
• young modulus of steel is 1.96×$10^{2} N/m^{2}$

To Find: If a mass of 20kg is suspended from it the elongation produced will be?

we have,

                           l = 4m

                          d = 2mm = 2×$10^{-3}$

                          E = 1.96×$10^{11} N/m^{2}$

                          σ = E×σ

                             = E×ΔL/L

∴ as mass (m) being suspended from there is,

                        m = 20kg

Then we will find the weight (w) with the help of mass (m),

                        w = mg = 20×9.8

                        σ = W/A = E×(ΔL/L)

                           = 20×9.8/$\frac{\pi }{4}$×(2×$10^{-3})^{2}$ = 1.96×$10^{11}$×(ΔL/L)

                           = ΔL = 0.00127m.

                              ΔL = 0.00127×1000 = 1.27mm.

hence the elongation produced will be 1.27mm.