Question

A mass of 1 kg is hung from a steel wire if radius 0.5 mm and length 4 m. Calculate the extension produced. What should be the area of cross-section of the wire so that elastic  limit is not exceeded? Change in radius is negligible

(Given : g = 9.8 m/s^2; Elastic limit of steel is 2.4 x 10⁸ N/m²;Y for steel (Ysteel) = 20 x 10¹⁰ N/m²; π = 3.142)

Answer 1

Given : g = 9.8 m/s^2; Elastic limit of steel is 2.4 x 10⁸ N/m²;Y for steel (Ysteel) = 20 x 10¹⁰ N/m²; π = 3.142

To find:  Extension in length (l)
             Area of cross section (A)
Formulae:

           i. Y = FL/Al
           ii. Elastic limit = F/A

`l=(FL)/(AY)=(MgL)/(pir^2Y)`

`l=(1xx9.8xx4)/(3.14xx(0.5xx10^-3)^2xx20xx10^10)`

l = 2.495x 10^-4 m
From formula (ii),

`A=(Mg)/("Elastic limit")`

`=(1xx9.8)/(2.4xx10^8)`

`A=4.o83xx10^-8 m^2`

The extension produced in length is 2.495x 10^-4 m and the area of cross section of the wire should be 4.083 x 10^-8 m².