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In a laboratory experiment, a metallic wire of length L=2.5 \mathrm{~m} and diameter d=0.5 \mathrm{~mm} is subjected to a tensile force F=250 \mathrm{~N}. The extension x produced in the wire is measured as 1.5 \mathrm{~mm}. The cross-sectional area of the wire is given by A=\frac{\pi d^{2}}{4}.

The Young's modulus Y of the material is:

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

Young's modulus Y is defined by the formula:

Given:

Y=\frac{F \cdot L}{A \cdot x}

\begin{aligned} F & =250 \mathrm{~N} \\ L & =2.5 \mathrm{~m} \\ d & =0.5 \mathrm{~mm}=0.0005 \mathrm{~m} \\ x & =1.5 \mathrm{~mm}=0.0015 \mathrm{~m} \end{aligned}

The cross-sectional area A of the wire is:

A=\frac{\pi d^{2}}{4}

Substitute the values:

A=\frac{\pi \cdot(0.0005 \mathrm{~m})^{2}}{4} \approx 1.9635 \times 10^{-7} \mathrm{~m}^{2}

Now, substitute the values of F, L, A, and x into the formula for Young's modulus:

Y=\frac{250 \mathrm{~N} \cdot 2.5 \mathrm{~m}}{1.9635 \times 10^{-7} \mathrm{~m}^{2} \cdot 0.0015 \mathrm{~m}} \approx 212 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}

So, the Young's modulus Y of the material is approximately 212 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}.

Posted by

Sumit Saini

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