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Question

Physics Question on mechanical properties of solids

Steel wire of length ' LL ' at 40C40^{\circ} C is suspended from the ceiling and then a mass ' mm ' is hung from its free end. The wire is cooled down from 40C40^{\circ} C, to 30C30^{\circ} C to regain its original length 'LL'. The coefficient of linear thermal expansion of the steel is 105/C10^{-5} /{ }^{\circ} C, Young's modulus of steel is 1011N/m210^{11} \,N / m ^{2} and radius of the wire is 1mm1 \,mm. Assume that LL \gg diameter of the wire. Then the value of ' mm ' in kgkg in nearly.

A

1

B

2

C

3

D

5

Answer

3

Explanation

Solution

Youngs modulus We know that E=F/AΔL/LE=\frac{F / A}{\Delta L / L} ΔLL=FAE\Rightarrow \frac{\Delta L}{L}=\frac{F}{A E} \ldots (i) Also ΔLL=αΔT\frac{\Delta L }{ L }=\alpha \Delta T\ldots (ii) FAE=αΔT\frac{F}{A E}=\alpha \Delta T mg=(αΔT)AE\Rightarrow m g=(\alpha \Delta T) A E m=αΔTgAE\Rightarrow m=\frac{\alpha \Delta T}{g} A E =105×10×π×106×101110=\frac{10^{-5} \times 10 \times \pi \times 10^{-6} \times 10^{11}}{10} =π3=\pi \approx 3