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A uniform metal rod (fixed at both ends) of \(2 \mathrm{~mm}^{2}\) cross- section is cooled from \(40^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\). The co-efficient of the linear expansion of the rod is \(12 \times 10^{-6}\) per degree \(\&\) it's young modulus of elasticity is \(10^{11} \mathrm{~N} / \mathrm{m}^{2}\). The energy stored per unit volume of the rod is: (1) \(2880 \mathrm{~J} / \mathrm{m}^{3}\) (2) \(1500 \mathrm{~J} / \mathrm{m}^{3}\) (3) \(5760 \mathrm{~J} / \mathrm{m}^{3}\) (4) \(1440 \mathrm{~J} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The energy stored per unit volume of the rod is 2880 J/m鲁.

Step by step solution

01

- Note down given data

Record all the given data. Cross-sectional area, A = 2 mm虏 = 2 脳 10鈦烩伓 m虏 Initial temperature, T鈧 = 40掳C Final temperature, T鈧 = 20掳C Coefficient of linear expansion, 伪 = 12 脳 10鈦烩伓 per 掳C Young's modulus, Y = 10鹿鹿 N/m虏
02

- Calculate the temperature change

Calculate the change in temperature 螖T = T鈧 - T鈧 螖T = 40掳C - 20掳C 螖T = 20掳C
03

- Compute strain due to temperature change

Use the formula for strain, 系 = 伪螖T Plug in the values, 系 = (12 脳 10鈦烩伓) 脳 20 系 = 2.4 脳 10鈦烩伌
04

- Determine stress in the rod

Use the relationship between stress and Young's modulus, 蟽 = Y 脳 系 Plug in the values, 蟽 = (10鹿鹿) 脳 (2.4 脳 10鈦烩伌) 蟽 = 2.4 脳 10鈦 N/m虏
05

- Calculate the energy stored per unit volume

Use the formula for energy stored per unit volume, U = 陆 脳 蟽 脳 系 Substitute the calculated values, U = 陆 脳 (2.4 脳 10鈦) 脳 (2.4 脳 10鈦烩伌) U = 陆 脳 5.76 脳 10鲁 U = 2.88 脳 10鲁 J/m鲁

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

thermal strain
When a material undergoes a temperature change, it either expands or contracts. Thermal strain is the measure of this deformation relative to the original dimensions of the material.
Thermal strain can be mathematically defined by the formula: \[ \epsilon = \alpha \Delta T \] Here:
  • \( \epsilon \) represents the thermal strain,
  • \( \alpha \) is the coefficient of linear expansion (which shows how much the material expands or contracts per degree change in temperature),
  • and \( \Delta T \) is the change in temperature.
For example, if a metal rod with a coefficient of linear expansion of \(12 \times 10^{-6} \) per degree Celsius is cooled from 40掳C to 20掳C, the resulting thermal strain can be calculated by: \[ \epsilon = (12 \times 10^{-6} ) \times 20 = 2.4 \times 10^{-4} \] This result helps us understand the amount of deformation experienced by the rod due to the temperature change.
Young's modulus
Young's modulus is a measure of the stiffness of a material. It provides the ratio of stress to strain in a material within its elastic limit. Mathematically, it can be expressed as: \[ Y = \frac{\sigma}{\epsilon} \] Here:
  • \(Y\) is Young's modulus,
  • \(\sigma\) represents the stress,
  • and \(\epsilon\) is the strain.
For the metal rod example, we know the Young's modulus is \( 10^{11} \, \text{N/m}^2 \) and the calculated strain is \(2.4 \times 10^{-4} \). Hence, the stress (\(\sigma\)) can be determined using: \[ \sigma = Y \times \epsilon = (10^{11}) \times (2.4 \times 10^{-4}) = 2.4 \times 10^7 \, \text{N/m}^2 \] This stress value gives us insight into the internal forces acting on the material when it is subjected to thermal strain.
energy stored per unit volume
When a material is deformed, energy is stored within it. This stored energy per unit volume can be calculated using the formula: \[ U = \frac{1}{2} \sigma \epsilon \] Where:
  • \(U\) is the energy stored per unit volume,
  • \(\sigma\) is the stress,
  • and \(\epsilon\) is the strain.
For our example, the stress (\(\sigma\)) is \(2.4 \times 10^7 \, \text{N/m}^2 \) and the strain (\(\epsilon\)) is \(2.4 \times 10^{-4} \). Plugging these values into the formula, we get: \[ U = \frac{1}{2} (2.4 \times 10^7) (2.4 \times 10^{-4}) = \frac{1}{2} (5.76 \times 10^3) = 2.88 \times 10^3 \, \text{J/m}^3 \] This is the energy stored per unit volume of the rod due to the thermal contraction. Understanding this concept is crucial for studying how materials store and release energy under different conditions.

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