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Question: A force of \(1kgm{m^{ - 2}}\) is applied to a wire of Young’s modulus \({10^{11}}N{m^{ - 2}}\). Find...

A force of 1kgmm−21kgm{m^{ - 2}} is applied to a wire of Young’s modulus 1011Nm−2{10^{11}}N{m^{ - 2}}. Find the percentage increases in length.

Explanation

Solution

Here in this question, we are given the Young's modulus of the wire and the stress pressure we apply on the ends of the wire which is a tension type stress as the wire length has increased. We use the formula for Young's modulus and strain (ratio of increases in length to the length) to find out the length increased.

Complete answer:
Now the force in 1kgmm−21kgm{m^{ - 2}} we need to convert it into standard form
1kgmm−2=1kg(10−3)2m2=106kgm−21kgm{m^{ - 2}} = \dfrac{{1kg}}{{{{\left( {{{10}^{ - 3}}} \right)}^2}{m^2}}} = {10^6}kg{m^{ - 2}}
Stress force=106kgm−2 = {10^6}kg{m^{ - 2}}
Young’s modulus =stressstrain=1011kgm−2 = \dfrac{{stress}}{{strain}} = {10^{11}}kg{m^{ - 2}}
Now by substituting the value of stress in the Young’s modulus equation we get
stressstrain=1011kgm−2\dfrac{{stress}}{{strain}} = {10^{11}}kg{m^{ - 2}}
106strain=1011\dfrac{{{{10}^6}}}{{strain}} = {10^{11}}
strain=1061011=10−5strain = \dfrac{{{{10}^6}}}{{{{10}^{11}}}} = {10^{ - 5}}
Now strain is ratio of increases in length to the original length =Δll = \dfrac{{\Delta l}}{l}
So, percentage increase =Δll×100=strain×100 = \dfrac{{\Delta l}}{l} \times 100 = strain \times 100
Percentage increases =10−5×100=10−3%= {10^{ - 5}} \times 100 = {10^{ - 3}}\%
Percentage increase in length = 0.001%

Note:
Here we use the strain of the wire to determine the ratio of increased length of the wire. The stress at the ends of the wire is in the form of tension, we can assume this as the wire length has increased which is generally due to stretching. Some material even has a negative Young's modulus, in such material compression stress across the ends causes increase in the length from sideways.