Question:
The figure shows stress versus strain graphs of two materials A and B. If $Y_A$, $Y_B$ are the Young's moduli of materials respectively, then:
The figure shows stress versus strain graphs of two materials A and B. If $Y_A$, $Y_B$ are the Young's moduli of materials respectively, then:
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On a Stress-Strain graph, a steeper slope indicates a more rigid material with a larger Young's Modulus.
Since line A is steeper than line B, $Y_A$ must be larger than $Y_B$, quickly eliminating options (B) and (C).
Since line A is steeper than line B, $Y_A$ must be larger than $Y_B$, quickly eliminating options (B) and (C).
Updated On: Jul 22, 2026
- $Y_A = \sqrt{2} Y_B$
- $Y_B = \sqrt{3} Y_A$
- $Y_B = \sqrt{2} Y_A$
- $Y_A = \sqrt{3} Y_B$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We are given a stress-strain graph for two materials, A and B.
We need to find the relationship between their Young's moduli based on the angles of their linear regions with the strain (X) axis.
Step 2: Key Formula and Approach:
According to Hooke's Law:
\[ \text{Stress} = Y \times \text{Strain} \] \[ Y = \frac{\text{Stress}}{\text{Strain}} \] On a Stress (Y-axis) vs Strain (X-axis) plot, the slope of the linear region represents the Young's Modulus:
\[ Y = \tan\theta \] where $\theta$ is the angle made by the curve with the horizontal strain axis.
Step 3: Detailed Explanation:
• Identify the angles from the graph:
For material A, the angle with the strain axis is $\theta_A = 45^\circ$.
For material B, the angle with the strain axis is $\theta_B = 30^\circ$.
• Calculate Young's Modulus for A ($Y_A$):
\[ Y_A = \tan(45^\circ) = 1 \]
• Calculate Young's Modulus for B ($Y_B$):
\[ Y_B = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
• Find the relationship:
\[ \frac{Y_A}{Y_B} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \] \[ Y_A = \sqrt{3} Y_B \]
Step 4: Final Answer:
The correct relationship is $Y_A = \sqrt{3} Y_B$, which corresponds to Option (D).
We are given a stress-strain graph for two materials, A and B.
We need to find the relationship between their Young's moduli based on the angles of their linear regions with the strain (X) axis.
Step 2: Key Formula and Approach:
According to Hooke's Law:
\[ \text{Stress} = Y \times \text{Strain} \] \[ Y = \frac{\text{Stress}}{\text{Strain}} \] On a Stress (Y-axis) vs Strain (X-axis) plot, the slope of the linear region represents the Young's Modulus:
\[ Y = \tan\theta \] where $\theta$ is the angle made by the curve with the horizontal strain axis.
Step 3: Detailed Explanation:
• Identify the angles from the graph:
For material A, the angle with the strain axis is $\theta_A = 45^\circ$.
For material B, the angle with the strain axis is $\theta_B = 30^\circ$.
• Calculate Young's Modulus for A ($Y_A$):
\[ Y_A = \tan(45^\circ) = 1 \]
• Calculate Young's Modulus for B ($Y_B$):
\[ Y_B = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
• Find the relationship:
\[ \frac{Y_A}{Y_B} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \] \[ Y_A = \sqrt{3} Y_B \]
Step 4: Final Answer:
The correct relationship is $Y_A = \sqrt{3} Y_B$, which corresponds to Option (D).
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