#### 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 10^{8} N/m^{2};Y for steel (Ysteel) = 20 x 10^{10} N/m^{2}; π = 3.142)

#### Solution

Given : g = 9.8 m/s^{2;} Elastic limit of steel is 2.4 x 10^{8} N/m^{2};Y for steel (Ysteel) = 20 x 10^{10} N/m^{2}; 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} m^{2}.