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Question: Steel wire of length ‘L’ at \({{40}^{0}}C\)is supposed from the ceiling and then a mass ‘m’ is hung ...

Steel wire of length ‘L’ at 400C{{40}^{0}}Cis supposed from the ceiling and then a mass ‘m’ is hung from its free end. The wire is cooled down from 400C{{40}^{0}}C to 300C{{30}^{0}}C to regain its original length ‘L’. The coefficient of linear thermal expansion of the steel is 10−5/0C{{10}^{-5}}{{/}^{0}}C, young’s modulus of steel is 1011N/m2{{10}^{11}}N/{{m}^{2}} and radius of the wire is 1mm. Assume that L>>L >> diameter of the wire. Then the value of ‘m’ in kg is nearly
A. 3
B. 2
C. 9
D. 5

Explanation

Solution

To solve this question, we will use the concept of the young’s modulus and the dependence of length or size of material on temperature. Obtain the expression for Young’s modulus and the change in length of the wire with temperature. Put the given values and make required arrangements to find the answer.

Complete answer:
Given I the question that,
The coefficient of linear thermal expansion of the steel wire is α=10−5/0C\alpha ={{10}^{-5}}{{/}^{0}}C
The young’s modulus of the steel wire is, Y=1011N/m2Y={{10}^{11}}N/{{m}^{2}}
The original length of the wire is L at Ti=400C{{T}_{i}}={{40}^{0}}C
The wire expands in length after we hang a mass m from its free end.
At Tf=300C{{T}_{f}}={{30}^{0}}C, the wire regains its original length L with the mass m.
The radius of the wire is r=1mm=10−3mr=1mm={{10}^{-3}}m
The young modulus of any wire can be defined as the ratio of the stress and the strain produced in the wire.
Y=stressstrain Y=mgAΔLL Y=mgLAΔL \begin{aligned} & Y=\dfrac{\text{stress}}{\text{strain}} \\\ & Y=\dfrac{\dfrac{mg}{A}}{\dfrac{\Delta L}{L}} \\\ & Y=\dfrac{mgL}{A\Delta L} \\\ \end{aligned}
Now, the change in length can be expressed as,
ΔL∝LΔT ΔL=αLΔT \begin{aligned} & \Delta L\propto L\Delta T \\\ & \Delta L=\alpha L\Delta T \\\ \end{aligned}
Putting this value on the expression of young’s modulus, we get,
Y=mgLAαLΔT=mgAαΔT m=YAαΔTg \begin{aligned} & Y=\dfrac{mgL}{A\alpha L\Delta T}=\dfrac{mg}{A\alpha \Delta T} \\\ & m=\dfrac{YA\alpha \Delta T}{g} \\\ \end{aligned}
The value of acceleration due to gravity, g=9.8ms−2g=9.8m{{s}^{-2}}
The cross-sectional area of the wire is, A=πr2=π×(10−3)2 A=10−6πm2 \begin{aligned} & A=\pi {{r}^{2}}=\pi \times {{\left( {{10}^{-3}} \right)}^{2}} \\\ & A={{10}^{-6}}\pi {{m}^{2}} \\\ \end{aligned}
The change in temperature of the wire is, ΔT=Ti−Tf ΔT=400C−300C ΔT=100C \begin{aligned} & \Delta T={{T}_{i}}-{{T}_{f}} \\\ & \Delta T={{40}^{0}}C-{{30}^{0}}C \\\ & \Delta T={{10}^{0}}C \\\ \end{aligned}
Putting the given values on the expression for mass, we get,
m=1011×10−6π×10−5×1010 m=π m≈3kg \begin{aligned} & m=\dfrac{{{10}^{11}}\times {{10}^{-6}}\pi \times {{10}^{-5}}\times 10}{10} \\\ & m=\pi \\\ & m\approx 3kg \\\ \end{aligned}
So, the mass m will be nearly 3kg.

The correct option is (A).

Note:
Young’s modulus or the modulus of elasticity of a material can be defined as a mechanical property which measures the stiffness of the material. It gives us the ability of the material to regain its original shape when the object is subjected under a force to deform its shape.