Problem 861 - JEE Physics

Question

The temperature of a wire of length 1 meter and area of cross-sectional section \(1 \mathrm{~cm}^{2}\) is increased from \(0^{\circ}\) to \(100^{\circ} \mathrm{C}\). If the rod is not allowed to increase in length. What will be the force required ? \(\left[\alpha=10^{-5} /{ }^{\circ} \mathrm{C}, \mathrm{Y}=10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\right]\) (A) \(10^{3} \mathrm{~N}\) (B) \(10^{4} \mathrm{~N}\) (C) \(10^{5} \mathrm{~N}\) (D) \(10^{9} \mathrm{~N}\)

Step-by-Step Solution

Answer

The force required to prevent the wire from expanding is \(10^{4} N\). So, the correct answer is (B).

Step 1: 1. Convert area to the proper unit

Since the given area is in cm², we need to convert it to m²: Area = 1 cm² = 0.0001 m²

Step 2: 2. Calculate the change in length of the wire

Using the linear expansion formula, we can calculate the change in length: ΔL = α * L * ΔT Where: α = 10⁻⁵ /°C (coefficient of linear expansion) L = 1 m (initial length of the wire) ΔT = 100 °C (change in temperature) ΔL = (10⁻⁵ /°C)(1 m)(100 °C) = 0.001 m

Step 3: 3. Calculate the extension (strain) of the wire

Strain is the ratio of the change in length to the original length: Strain = ΔL / L Strain = 0.001 m / 1 m = 0.001

Step 4: 4. Use Young's modulus to find the stress

Now we can use the formula for Young's modulus: Y = stress / strain Where Y is the Young's modulus and is given as 10¹¹ N/m². We can rearrange the formula to find the stress: Stress = Y * strain Stress = (10¹¹ N/m²)(0.001) = 10⁸ N/m²

Step 5: 5. Find the force required to prevent the wire from expanding

Now we have the stress, we can find the force using the formula: Force = Stress * Area Force = (10⁸ N/m²)(0.0001 m²) = 10⁴ N The force required to prevent the wire from expanding is 10⁴ N. So, the correct answer is (B).

Key Concepts

Linear Expansion

When materials are subjected to temperature changes, they tend to change in size. In physics, this phenomenon is known as thermal expansion, and the linear expansion specifically refers to the change in length of an object. The formula used to calculate the change in length due to thermal expansion is:

• \( \Delta L = \alpha \times L \times \Delta T \)

Where:

• \( \Delta L \) is the change in length.
• \( \alpha \) is the coefficient of linear expansion.
• \( L \) is the original length of the object.
• \( \Delta T \) is the change in temperature.

In the exercise, the wire initially has a length of 1 meter. It is subjected to a temperature increase from \( 0^{\circ} \mathrm{C} \) to \( 100^{\circ} \mathrm{C} \). Utilizing the linear expansion coefficient \( \alpha = 10^{-5} / \degree \mathrm{C} \), we can calculate that the length would increase by 0.001 meters if allowed.

Young's Modulus

Young's modulus is a measure of the stiffness of a material. It is a mechanical property that describes the material's ability to withstand changes in length when under tension or compression. Mathematically, Young's modulus \( Y \) is defined as the ratio of tensile stress \( \sigma \) to tensile strain \( \epsilon \):

• \( Y = \frac{\text{Stress}}{\text{Strain}} \)

Where:

• Stress \( \sigma = \frac{F}{A} \) (force per unit area)
• Strain \( \epsilon = \frac{\Delta L}{L} \) (change in length per unit length)

The Young's modulus for steel, as given in the problem, is \( 10^{11} \text{ N/m}^2 \). This value indicates that steel is a very stiff material. In the context of the exercise, we used Young's modulus to derive the stress experienced by the material due to the constraint against thermal expansion.

Thermal Stress

Thermal stress arises when a material is prevented from expanding or contracting with changes in temperature. If a wire is heated, it naturally wants to expand, but if it is constrained, stress develops within the material. The thermal stress \( \sigma_t \) can be calculated using the relation from Young's modulus and strain:

• \( \sigma_t = Y \times \epsilon \)

From the exercise:

• Strain \( \epsilon \) was found to be 0.001
• Young's modulus \( Y \) is \( 10^{11} \text{ N/m}^2 \)

Therefore, the stress induced in the wire owing to the restriction of length change is \( 10^8 \text{ N/m}^2 \). This stress measures the internal force per unit area resisting the thermal expansion.

Strain and Stress

Strain and stress are fundamental concepts in understanding material behavior under loads. **Strain** \( \epsilon \), is a dimensionless number representing the deformation of a material. It is the ratio of the change in dimension (length, area, or volume) to the original dimension of the object involved:

• \( \epsilon = \frac{\Delta L}{L} \)

**Stress** \( \sigma \), on the other hand, is the force exerted per unit area within materials. It results from externally applied forces, uneven heating, or permanent deformation, and is given by:

• \( \sigma = \frac{F}{A} \)

In the exercise, stress was used to determine the force necessary to prevent the wire’s expansion, which was effectively calculated by multiplying the stress \( 10^8 \text{ N/m}^2 \) with the area of cross-section \( 0.0001 \text{ m}^2 \), yielding a force of \( 10^4 \text{ N} \). This demonstrates the interplay between stress, strain, and material response to prevent deformation.