Problem 24 - Arihant AIEEE Physics

Question

A steel rod of diameter \(1 \mathrm{~cm}\) is clamped firmly at each end when its temperature is \(25^{\circ} \mathrm{C}\) so that it cannot contract on cooling. The tension in the rod at \(0^{\circ} \mathrm{C}\) is : \(\left(\alpha=10^{-5} /{ }^{\circ} \mathrm{C}, Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\right)\) (a) \(4000 \mathrm{~N}\) (b) \(7000 \mathrm{~N}\) (c) \(7400 \mathrm{~N}\) (d) \(4700 \mathrm{~N}\)

Step-by-Step Solution

Answer

The tension in the rod at \(0^{\circ} \mathrm{C}\) is approximately \(4000 \mathrm{~N}\), option (a).

Step 1: Identify Given Values

We are given a steel rod with the following parameters:- Diameter of the rod: \( 1 \text{ cm} \)- Initial temperature \( T_1: 25^{\circ} \text{C} \)- Final temperature \( T_2: 0^{\circ} \text{C} \)- Coefficient of linear expansion \( \alpha = 10^{-5}/{}^{\circ} \text{C} \)- Young's modulus \( Y = 2 \times 10^{11} \text{ N/m}^2 \).

Step 2: Calculate Change in Temperature

Determine the change in temperature:\[\Delta T = T_2 - T_1 = 0^{\circ} \text{C} - 25^{\circ} \text{C} = -25^{\circ} \text{C}\].

Step 3: Calculate Thermal Strain

Use the formula for thermal strain, which is the product of the coefficient of linear expansion and change in temperature:\[\epsilon = \alpha \times \Delta T = (10^{-5}/{}^{\circ} \text{C}) \times (-25^{\circ} \text{C}) = -2.5 \times 10^{-4}\].

Step 4: Calculate Thermal Stress

Since the rod is clamped and cannot contract, the thermal stress is equal to \( Y \times \text{strain} \) because stress and strain are related by Young's modulus \( Y \). Thus, we have:\[\sigma = Y \times \epsilon = (2 \times 10^{11} \text{ N/m}^2) \times (-2.5 \times 10^{-4})\].

Step 5: Simplify the Stress Calculation

Calculate the stress \( \sigma \):\[\sigma = 2 \times 10^{11} \times -2.5 \times 10^{-4} = -5 \times 10^{7} \text{ N/m}^2\].

Step 6: Calculate Tension in the Rod

To find the tension in the rod, multiply the stress by the cross-sectional area \( A \). First, calculate the cross-sectional area using the diameter:\[A = \pi \times \left(\frac{0.01}{2}\right)^2 = \pi \times (0.005)^2 = 7.85 \times 10^{-5} \text{ m}^2\].Then, multiply it by the absolute value of stress to find the tension (since tension is a magnitude and cannot be negative):\[T = |\sigma| \times A = 5 \times 10^{7} \times 7.85 \times 10^{-5} = 3925 \text{ N}\].

Step 7: Round and Identify Closest Answer

Since the calculated tension is approximately 3925 N, which is closest to \( 4000 \text{ N} \). Hence, the correct choice is (a).

Key Concepts

Coefficient of Linear Expansion

The Coefficient of Linear Expansion (\( \alpha \)) is a measure of how much a material expands per degree change in temperature. In the case of the steel rod, it's given as \( 10^{-5}/{}^{\circ} \text{C} \).

### How It Works

• Whenever the temperature of a material changes, its dimensions do too.
• In a solid, the change in length (\( \Delta L \)) is proportional to its original length (\( L_0 \)) and the change in temperature (\( \Delta T \)).
  • The formula is \( \Delta L = L_0 \times \alpha \times \Delta T \)
• The coefficient of linear expansion is crucial because it quantifies how temperature affects size.

### Application in the ExerciseFor this problem, the rod's temperature drops, and even though it cannot physically contract due to being clamped, we can calculate how much it "wants" to change using this coefficient. It guides us in computing thermal strain next.

Young's Modulus

Young's Modulus (\( Y \)) measures a material's ability to withstand changes in length under tension or compression. It links stress and strain through the equation:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
This constant describes the rod's stiffness—key in determining how much stress generates due to constraints like clamping.

### Key Takeaways

• Young's Modulus is extremely high for materials that don't stretch much under force, like steel (given as \( 2 \times 10^{11} \text{ N/m}^2 \)
• This means that even small strains can result in large stresses.

### Importance in the ProblemSince the rod cannot contract while cooling, Young's Modulus plays a vital role. It helps us calculate the stress using the strain derived from thermal expansion properties. Hence, knowing Young's Modulus is crucial in predicting how much force (tension) is exerted on the rod when temperature changes.

Thermal Strain Calculation

Thermal strain is the relative change in length due to temperature changes. For the steel rod, thermal strain (\( \epsilon \)) is calculated using the formula:

\[ \epsilon = \alpha \cdot \Delta T \]
This equation reveals how the rod's dimensions would change under different temperatures if it were free to expand or contract.

### Steps in the Context of the Problem

• Calculate the change in temperature (\( \Delta T \)) which was \( -25^{\circ} \text{C} \) as given by \( T_2 - T_1 = 0^{\circ} \text{C} - 25^{\circ} \text{C} \)
• Use the coefficient of linear expansion (\( \alpha \)) to find the strain.
• The resulting strain is \( \epsilon = -2.5 \times 10^{-4} \), indicating how much the material wants to contract per initial length.

### OutcomeAlthough the rod can't physically contract, this strain calculation lays the groundwork for determining stress (and ultimately tension) on the rod by fixing its potential expansion or contraction.