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Question: The area of cross-section of steel wire is \(0.1c{m^2}\) and the young modulus of the steel wire is ...

The area of cross-section of steel wire is 0.1cm20.1c{m^2} and the young modulus of the steel wire is 2×1011Nm−22 \times {10^{11}}N{m^{ - 2}}. The force required to stretch by 0.1%0.1\% of its length is then
(A) 1000  N1000\;N
(B) 2000  N2000\;N
(C) 3000  N3000\;N
(D) 4000  N4000\;N

Explanation

Solution

To solve these questions we first need to understand the term Young modulus given. Young modulus can be defined as the ability of a material to resist changes in its length which can occur due to compression and tension in its length. It is also defined as the ratio of stress to strain.
Formula used:
Young modulus formula
Y=stressstrain=F/AΔl/lY = \dfrac{{stress}}{{strain}} = \dfrac{{F/A}}{{\Delta l/l}}
where FF is forced, AA is area and ll is length, Δl\Delta l is the change in length.

Complete step-by-step solution:
Here given that the young modulus of the steel wire is 2×1011Nm−22 \times {10^{11}}N{m^{ - 2}}, and cross-section area is also given as
A=0.1cm2=0.1×10−4m2A = 0.1c{m^2} = 0.1 \times {10^{ - 4}}{m^2} ………. (1)(1)
Here also given that the wire is stretch by 0.1%0.1\% which is the Δll\dfrac{{\Delta l}}{l},
Δll=0.1%\dfrac{{\Delta l}}{l} = 0.1\%
⇒Δll=0.1100=0.1×10−2\Rightarrow \dfrac{{\Delta l}}{l} = \dfrac{{0.1}}{{100}} = 0.1 \times {10^{ - 2}} ………. (2)(2)
Now we know that the young modulus YY is defined as the ratio of stretch and strain. Where stress is defined as the ratio of the force FF per unit area AA acting on the material while a strain is defined as the ratio of change is length Δl\Delta l to the length ll of the martial given.
Hence applying the definition the young modulus is given as
Y=stressstrain=FAΔllY = \dfrac{{stress}}{{strain}} = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}
Substituting the values of young modulus given as Y=2×1011Nm−2Y = 2 \times {10^{11}}N{m^{ - 2}}and the values from the equation (1)(1) and (2)(2), we get
2×1011Nm−2=F0.1×10−4m2×0.1×10−22 \times {10^{11}}N{m^{ - 2}} = \dfrac{F}{{0.1 \times {{10}^{ - 4}}{m^2} \times 0.1 \times {{10}^{ - 2}}}}
⇒F=2×1011Nm−2×0.1×10−4m2×0.1×10−2\Rightarrow F = 2 \times {10^{11}}N{m^{ - 2}} \times 0.1 \times {10^{ - 4}}{m^2} \times 0.1 \times {10^{ - 2}}
∴F=2×103N=2000N\therefore F = 2 \times {10^3}N = 2000N.
Hence the force required to stretch by 0.1%0.1\% of its length if the area of cross-section of steel wire is 0.1cm20.1c{m^2} and young modulus of the steel wire is 2×1011Nm−22 \times {10^{11}}N{m^{ - 2}} given by F=2000NF = 2000N.

Therefore option (B) is the correct answer.

Note: While solving this type of question one should ensure about the units used. Here in this question, the area was given in cm2c{m^2} which has to be converted to m2{m^2} which is a SI unit as young modulus is given in Nm−2N{m^{ - 2}} which is in the SI unit system.