Question

One end of a wire 2m long and 0.2 cm^2 in cross section is fixed in a ceilign and a load of 4.8 kg is attached to the free end. Find the extension of the wire. Young modulus of steel `=2.0xx10^11Nm^-2`. Take `g=10ms^-2`.

Answer 1

We have
`y=stress/strain=(T/A)/(l/L)`
with symbols having their usual meanings. The extension is
`l=(TL)/(AY)`
As the load is in equilibrium after the extension, the tension in the wire is equal to the weight of the load
`=4.8kgxx10ms^-2=48N`
Thus, `l=((48N)(2m))/((0.2xx10^-4m^2)xx(2.0xx10^11Nm^-2))`
`=2.4xx10^-5m`