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A wire of length L=2.0 \mathrm{~m} and diameter d=0.5 \mathrm{~mm} is suspended vertically from a ceiling. A weight W=20 \mathrm{~N} is attached to the lower end of the wire, causing it to stretch. If the extension of the wire is e=2.5 \mathrm{~mm}, calculate the Young's modulus (Y) of the material.

Option: 1

8.1464 \times 10^8 \mathrm{~N} / \mathrm{m}^2


Option: 2

9.1464 \times 10^{10} \mathrm{~N} / \mathrm{m}^2.


Option: 3

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


Option: 4

8.1464 \times 10^{10} \mathrm{~N} / \mathrm{m}^2.


Answers (1)

best_answer

Given:

  • Length of wire L=2.0 \mathrm{~m}
  • Diameter of wire d=0.5 \mathrm{~mm}
  • Weight attached W=20 \mathrm{~N}
  • Extension of wire e=2.5 \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(0.5 \times 10^{-3}\right)^2}{4} \\ A=1.9635 \times 10^{-7} \mathrm{~m}^2 \end{gathered}

Step 2: Calculate the original volume V of the wire:
            \begin{gathered} V=A \cdot L \\ V=1.9635 \times 10^{-7} \mathrm{~m}^2 \times 2.0 \mathrm{~m} \\ V=3.927 \times 10^{-7} \mathrm{~m}^3 \end{gathered}

Step 3: Calculate the stress \sigma on the wire due to the weight:
            \begin{gathered} \sigma=\frac{W}{A} \\ \sigma=\frac{20 \mathrm{~N}}{1.9635 \times 10^{-7} \mathrm{~m}^2} \\ \sigma=1.0183 \times 10^8 \mathrm{~N} / \mathrm{m}^2 \end{gathered}

Step 4: Calculate the strain \varepsilon in the wire:
            \begin{gathered} \varepsilon=\frac{e}{L} \\ \varepsilon=\frac{2.5 \times 10^{-3} \mathrm{~m}}{2.0 \mathrm{~m}} \\ \varepsilon=1.25 \times 10^{-3} \end{gathered}
Step 5: Using Hooke's Law, Y=\frac{\sigma}{\varepsilon}, calculate the Young's modulus Y of the material:
              \begin{aligned} & Y=\frac{1.0183 \times 10^8 \mathrm{~N} / \mathrm{m}^2}{1.25 \times 10^{-3}} \\ & Y=8.1464 \times 10^{10} \mathrm{~N} / \mathrm{m}^2 \end{aligned}
The Young's modulus (Y) of the material is 8.1464 \times 10^{10} \mathrm{~N} / \mathrm{m}^2.
So, option D is correct

Posted by

Gaurav

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