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JEE Mains · Physics · STD 11- Mechanical properties of solids
  • JEE Mains 2026
  • Medium
  • MCQ

Figure represents the extension (\(\Delta l\)) of a wire of length \(1\) meter, suspended from the ceiling of the room at one end with a load \(W\) connected to the other end. If the cross-sectional area of the wire is \(10^{-5}\) m\(^2\) then the Young's modulus of the wire is __________ N/m\(^2\).

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Verified Solution

Answer & Solution

Correct Answer

(C) \(1.0 \times 10^{10}\)

Step-by-step Solution

Detailed explanation

From the given graph, we can observe that for a load \(W = 20\) N, the extension is \(\Delta l = 2 \times 10^{-4}\) m.

The formula for Young's modulus \(Y\) is:

\(Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{W / A}{\Delta l / L} = \dfrac{W L}{A \Delta l}\)

Given:
Length of the wire, \(L = 1\) m
Cross-sectional area, \(A = 10^{-5}\) m\(^2\)

Substituting the values into the formula:

\(Y = \dfrac{20 \times 1}{10^{-5} \times 2 \times 10^{-4}}\)

\(Y = \dfrac{20}{2 \times 10^{-9}}\)

\(Y = 10 \times 10^9 = 1.0 \times 10^{10}\) N/m\(^2\)

Answer: \(1.0 \times 10^{10}\)
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