Given:
Step 1: Relationship Between Lateral and Longitudinal Strain
Poisson’s ratio is given by:
\[ \nu = \frac{\text{lateral strain}}{\text{longitudinal strain}} \]
Rearranging for longitudinal strain \( \varepsilon \):
\[ \varepsilon = \frac{\varepsilon_l}{\nu} = \frac{10^{-3}}{0.25} = 4 \times 10^{-3} \]
Step 2: Elastic Energy Density Formula
The elastic energy density \( U \) is given by:
\[ U = \frac{1}{2} Y \varepsilon^2 \]
Substituting values:
\[ U = \frac{1}{2} \times (2 \times 10^{11}) \times (4 \times 10^{-3})^2 \]
\[ U = \frac{1}{2} \times (2 \times 10^{11}) \times (16 \times 10^{-6}) \]
\[ U = \frac{32 \times 10^5}{2} = 16 \times 10^5 \text{ J/m}^3 \]
Answer: The correct option is D (16 × 105 J/m3).
The elastic energy density \( u \) is given by the formula: \[ u = \frac{1}{2} \sigma \varepsilon \] where: - \( \sigma \) is the stress, - \( \varepsilon \) is the strain. We know that stress \( \sigma \) is related to the Young's modulus \( Y \) and strain \( \varepsilon \) by: \[ \sigma = Y \cdot \varepsilon \] The lateral strain \( \varepsilon_l \) is related to the longitudinal strain \( \varepsilon \) by Poisson's ratio \( \nu \) as: \[ \varepsilon_l = - \nu \cdot \varepsilon \] Substituting the values: - \( Y = 2 \times 10^{11} \, \text{Nm}^{-2} \), - \( \nu = 0.25 \), - \( \varepsilon_l = 10^{-3} \), we can calculate the longitudinal strain \( \varepsilon \): \[ \varepsilon = \frac{\varepsilon_l}{\nu} = \frac{10^{-3}}{0.25} = 4 \times 10^{-3} \] Now, the stress \( \sigma \) is: \[ \sigma = Y \cdot \varepsilon = 2 \times 10^{11} \times 4 \times 10^{-3} = 8 \times 10^8 \, \text{N/m}^2 \] Finally, the energy density is: \[ u = \frac{1}{2} \cdot \sigma \cdot \varepsilon = \frac{1}{2} \cdot 8 \times 10^8 \cdot 4 \times 10^{-3} = 16 \times 10^5 \, \text{Jm}^{-3} \] Thus, the correct answer is (D).
Young's modulus is defined as the ratio of
The centre of mass of an extended body on the surface of the earth and its centre of gravity
A metallic rod breaks when strain produced is 0.2%. The Young's Modulus of the material is 7 x 109 N/m2 . The area of cross section to support a load of 104 N is