Question:

The adjacent graph shows the extension $(\Delta l)$ of a wire of length 1 m suspended from the top of a roof at one end with a load W connected to the other end. If the cross-sectional area of the wire is $10^{-6} m^2 , $ calculate the Young's modulus of the material of the wire.

Updated On: Jul 7, 2022
  • $2 \times 10^{11} \, Nm^{-2}$
  • $2 \times 10^{-11} \, Nm^{-2}$
  • $3 \times 10^{-12} \, Nm^{-2}$
  • $2 \times 10^{-13} \, Nm^{-2}$
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The Correct Option is A

Solution and Explanation

From the graph $l = 10^{-4}m ,F = 20 \, N$ $\hspace35mm A = 10^{-6} m^2 , L = 1 \, m$ $\therefore \hspace30mm Y = \frac{FL}{Al}$ $\hspace40mm = \frac{20 \times 1}{10^{-6} \times 10^{-4}}$ $\hspace40mm = 20 \times 10^{10} = 2 \times 10^{11} \, Nm^{-2}$
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Notes on Hooke's Law

Concepts Used:

Hooke’s Law

Hooke’s Law states that for small deformities, the stress and strain are proportional to each other. Thus,

Stress ∝ Strain

Stress = k × Strain … where k is the Modulus of Elasticity.

When a limited amount of Force or deformation is involved then concept of Hooke’s Law is only applicable . If we consider the fact, then we can deviate from Hooke's Law. This is because of their extreme Elastic limits.