Question

Two bars of different materials and same size are subjected to the same tensile force. If the bars have unit elongation in the ratio of \(2:5\), then the ratio of modulus of elasticity of the two materials will be

• A. \(2:5\)
• B. \(5:2\)
• C. \(4:3\)
• D. \(3:4\)

Hint:

If two members experience the same stress, \[ E=\frac{\text{Stress}}{\text{Strain}} \] implies \[ \boxed{E\propto\frac1{\text{Strain}}.} \] Greater strain means lower Young's modulus, and vice versa.

Answer 1

The Correct Option is B

Concept: The modulus of elasticity (Young's modulus) is given by \[ E=\frac{\text{Stress}}{\text{Strain}}. \] Since both bars have the same cross-sectional area and are subjected to the same tensile force, \[ \text{Stress}=\frac{P}{A} \] is the same for both bars. Hence, \[ E\propto\frac1{\text{Strain}}. \] Since unit elongation is the strain, \[ E\propto\frac1{\text{Unit Elongation}}. \]

Step 1: Write the given strain ratio. The ratio of unit elongations is \[ \epsilon_1:\epsilon_2=2:5. \]

Step 2: Use the inverse relationship. Since \[ E\propto\frac1{\epsilon}, \] we have \[ E_1:E_2 = \frac1{2}:\frac1{5} = 5:2. \] Therefore, \[ \boxed{E_1:E_2=5:2.} \] Hence, the correct option is \[ \boxed{(B)\;5:2.} \]