Question:
The stress versus strain graphs for wires of two materials A and B are as shown. If \(Y_A\) and \(Y_B\) are the Young's moduli of the materials, then:
The stress versus strain graphs for wires of two materials A and B are as shown. If \(Y_A\) and \(Y_B\) are the Young's moduli of the materials, then:
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If graph is strain vs stress, then slope \(=\frac{1}{Y}\).
Higher slope $\Rightarrow$ Lower Young’s modulus.
Updated On: Apr 17, 2026
- \( Y_B = 2Y_A \)
- \( Y_A = Y_B \)
- \( Y_B = 3Y_A \)
- \( Y_A = 3Y_B \)
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The Correct Option is C
Solution and Explanation
Concept:
\[
Y = \frac{\text{Stress}}{\text{Strain}}
\]
Hence, slope of stress vs strain graph:
\[
\text{slope} = \frac{\text{Strain}}{\text{Stress}} = \frac{1}{Y}
\]
So, slope is inversely proportional to Young’s modulus.
Step 1:From the graph:}
•Line A makes \(60^\circ\)
•Line B makes \(30^\circ\)
Step 2:Slopes:} \[ \text{slope}_A = \tan 60^\circ = \sqrt{3}, \quad \text{slope}_B = \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Step 3:Ratio of slopes:} \[ \frac{\text{slope}_A}{\text{slope}_B} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 \]
Step 4:Using inverse relation:} \[ \frac{1/Y_A}{1/Y_B} = 3 \Rightarrow \frac{Y_B}{Y_A} = 3 \] \[ \Rightarrow Y_B = 3Y_A \]
Step 1:From the graph:}
•Line A makes \(60^\circ\)
•Line B makes \(30^\circ\)
Step 2:Slopes:} \[ \text{slope}_A = \tan 60^\circ = \sqrt{3}, \quad \text{slope}_B = \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Step 3:Ratio of slopes:} \[ \frac{\text{slope}_A}{\text{slope}_B} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 \]
Step 4:Using inverse relation:} \[ \frac{1/Y_A}{1/Y_B} = 3 \Rightarrow \frac{Y_B}{Y_A} = 3 \] \[ \Rightarrow Y_B = 3Y_A \]
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