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One end of a wire 2m long and 0.2 cm^2 i...

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`.

Text Solution

Verified by Experts

We have
`Y=(stress)/(strai n)=(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 `=4.8kgxx10m//s^(2)=48N`
Thus, `l=((48N)(2m))/((0.2xx10^(-4)m^(2))(2.0xx10^(11)N//m^(2)))`
`=2.4xx10^(-5)m.`
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