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A steel wire of length L = 2.5 m  and diameter d = 1.0 mm is subjected to a tensile force F = 300 N, causing it to stretch. If the extension of the wire is e = 3.0 mm, calculate the Young’s modulus (Y) of the steel.

Option: 1

3.1795 \times 10^{11} \mathrm{~N} / \mathrm{m}^2.


Option: 2

3.1795 \times 10^9 \mathrm{~N} / \mathrm{m}^2.


Option: 3

4.1795 \times 10^{11} \mathrm{~N} / \mathrm{m}^2.


Option: 4

3.1795 \times 10^{10} \mathrm{~N} / \mathrm{m}^2.


Answers (1)

best_answer

Given:

  • Length of wire L=2.5 \mathrm{~m}
  •  Diameter of wire d=1.0 \mathrm{~mm}
  • Tensile force F=300 \mathrm{~N}
  • Extension of wire e=3.0 \mathrm{~mm}

Step 1: Calculate the cross-sectional area A of the wire:

            \begin{gathered} A=\frac{\pi d^2}{4} \\ A=\frac{\pi \times\left(1.0 \times 10^{-3}\right)^2}{4} \\ A=7.8539 \times 10^{-7} \mathrm{~m}^2 \end{gathered}

Step 2: Calculate the stress \sigma on the wire due to the tensile force:

            \begin{gathered} \sigma=\frac{F}{A} \\ \sigma=\frac{300 \mathrm{~N}}{7.8539 \times 10^{-7} \mathrm{~m}^2} \\ \sigma=3.8154 \times 10^8 \mathrm{~N} / \mathrm{m}^2 \end{gathered}

Step 3: Calculate the strain \varepsilon in the wire:

            \begin{gathered} \varepsilon=\frac{e}{L} \\ \varepsilon=\frac{3.0 \times 10^{-3} \mathrm{~m}}{2.5 \mathrm{~m}} \\ \varepsilon=1.2 \times 10^{-3} \end{gathered}

Step 4: Using Hooke's Law, Y=\frac{\sigma}{\varepsilon}, calculate the Young's modulus Y of the steel:

            \begin{aligned} & Y=\frac{3.8154 \times 10^8 \mathrm{~N} / \mathrm{m}^2}{1.2 \times 10^{-3}} \\ & Y=3.1795 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \end{aligned}

The Young's modulus (Y) of the steel is 3.1795 \times 10^{11} \mathrm{~N} / \mathrm{m}^2.

So, option A is correct

 

 

 

Posted by

Irshad Anwar

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