Problem 873 - JEE Physics

Question

3: A force of \(200 \mathrm{~N}\) is applied at one end of a wire of length \(2 \mathrm{~m}\) and having area of cross-section \(10^{-2} \mathrm{~cm}^{2}\), the other end of the wire is rigidly fixed. If of linear expansion of the wire \(\alpha=8 \times 10^{-6} /{ }^{\circ} \mathrm{C}\) and young's modulus \(\mathrm{Y}=2.2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) and its temperature is increased by \(5^{\circ} \mathrm{C}\) then the increase in the tension of the wire will be.......... (A) \(4.2 \mathrm{~N}\) (B) \(4.4 \mathrm{~N}\) (C) \(2.4 \mathrm{~N}\) (D) \(8.8 \mathrm{~N}\)

Step-by-Step Solution

Answer

The short answer to the increase in tension of the wire is \(8.8 \, \mathrm{N}\), which corresponds to option (D).

Step 1: Find the thermal expansion

To find the increase in length (\(\Delta L_{th}\)) due to thermal expansion, we use the formula: \(\Delta L_{th} = L \alpha \Delta T\) Where: \(L\) is the original length = \(2 \, \mathrm{m}\) \(\alpha\) is the coefficient of linear expansion = \(8 \times 10^{-6} \, \mathrm{C}^{-1}\) \(\Delta T\) is the change in temperature = \(5^{\circ} \mathrm{C}\)

Step 2: Calculate Δ𝐿th

Now we will plug in the values to find the increase in length due to thermal expansion: \(\Delta L_{th} = (2 \, \mathrm{m})(8 \times 10^{-6} \, \mathrm{C}^{-1})(5^{\circ} \mathrm{C})\) \(\Delta L_{th} = 8 \times 10^{-5} \, \mathrm{m}\)

Step 3: Find the stress

The stress, \(σ\) is the force per unit area: \(σ = \frac{F}{A}\) Where: \(F\) is the applied force = \(200 \, \mathrm{N}\) \(A\) is the area of cross-section = \(10^{-2} \,\mathrm{cm}^{2}\) = \(10^{-6} \,\mathrm{m}^{2}\) (by converting to SI units)

Step 4: Calculate σ

Plugging in the values for \(F\) and \(A\): \(σ = \frac{200 \, \mathrm{N}}{10^{-6} \, \mathrm{m^2}}\) \(σ = 2 \times 10^8 \, \frac{\mathrm{N}}{\mathrm{m^2}}\)

Step 5: Find the strain

To find the strain (\(ε\)), we will use the Young's Modulus formula: \(Y = \frac{σ}{ε}\) Where: \(Y\) is the Young's Modulus = \(2.2 \times 10^{11} \left(\mathrm{N} / \mathrm{m}^{2}\right)\) \(σ\) is the stress = \(2 \times 10^8 \, \frac{\mathrm{N}}{\mathrm{m^2}}\)

Step 6: Calculate ε

Now we will solve for \(ε\): \(ε = \frac{σ}{Y} = \frac{2 \times 10^8 \, \frac{\mathrm{N}}{\mathrm{m^2}}}{2.2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)}\) \(ε = 9.09 \times 10^{-4}\)

Step 7: Find the increase in length due to stress

Now by using the strain, we will find the increase in length due to stress (\(\Delta L_{st}\)): \(\Delta L_{st} = εL\) Where: \(ε\) is the strain = \(9.09 \times 10^{-4}\) \(L\) is the original length = \(2 \, \mathrm{m}\)

Step 8: Calculate Δ𝐿st

Now we will plug the values to find the increase in length due to stress: \(\Delta L_{st} = (9.09 \times 10^{-4})(2 \, \mathrm{m})\) \(\Delta L_{st} = 1.82 \times 10^{-3} \, \mathrm{m}\)

Step 9: Find the increase in tension

The wire is prevented from expanding by the rigid support, so the total expansion must be zero (\(\Delta L_{total} = 0\)). So, the increase in length due to stress must be equal and opposite to the increase in length due to thermal expansion: \(\Delta L_{st} = -\Delta L_{th}\) Now we will find the increase in tension (\(\Delta F\)) by using the increase in stress and the area of cross-section: \(\Delta σ = \frac{\Delta F}{A}\) where \(\Delta σ = σ_{th} - σ_{st}\)

Step 10: Calculate ΔF

We will now plug in the values to find the increase in tension: \(\Deltaσ = \frac{-\Delta L_{th}}{ε}\) \(\Deltaσ = \frac{8 \times 10^{-5}\, \mathrm{m}}{9.09 \times 10^{-4}}\) \(\Delta F = \Delta σ \times A = (8.8 \times 10 ^6 \, \frac{\text{N}}{\text{m}^2}) \times 10^{-6} \, \text{m}^2\) \(\Delta F = 8.8 \, \mathrm{N}\) The increase in tension of the wire is \(8.8 \, \mathrm{N}\), which corresponds to option (D).

Key Concepts

Thermal Expansion

When a material is heated, it expands. This change in size due to temperature is known as thermal expansion. For a linear expansion of a solid, the change in length is given by the formula:
\[ \Delta L_{th} = L \alpha \Delta T \]
where:

• \( \Delta L_{th} \) is the increase in length due to thermal change.
• \( L \) is the original length.
• \( \alpha \) is the linear expansion coefficient.
• \( \Delta T \) is the temperature change.

For instance, if a wire of length \(2 \, \mathrm{m}\) with a linear expansion coefficient of \(8 \times 10^{-6} \, \mathrm{C}^{-1}\) is heated by \(5^{\circ} \mathrm{C}\), its expansion can be calculated to determine the thermal stresses generated if it is restrained.

Stress and Strain

Stress and strain are related concepts in mechanics, describing how materials deform under force. Stress, \( \sigma \), is a measure of the force applied per unit area, calculated as:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the applied force and \( A \) is the cross-sectional area.
Strain, \( \varepsilon \), is the deformation experienced by the material, defined as the change in length divided by the original length:
\[ \varepsilon = \frac{\Delta L}{L} \]
Both stress and strain are essential for understanding how materials behave under loads, and they are connected through Young's Modulus \( Y \), a material property that relates stress and strain as follows:
\[ Y = \frac{\sigma}{\varepsilon} \]

Linear Expansion Coefficient

The linear expansion coefficient, \( \alpha \), is a measure of how a material's length changes with temperature. It's a constant specific to the material and is defined in the equation for thermal expansion as:
\[ \Delta L = L \alpha \Delta T \]
This coefficient helps predict the behavior of materials under thermal conditions, making it vital for engineering and construction where temperature variations are frequent. For instance, knowing \( \alpha \) allows engineers to ensure structures can withstand temperature-induced stresses without failure.

Applied Force

When a force is applied to a structure or material, it creates stress. The magnitude of this force impacts how much stress is generated, which is critical for ensuring safety and integrity of the object. Calculating stress involves dividing the applied force by the cross-sectional area, giving insight into potential deformations:
\[ \sigma = \frac{F}{A} \]
Understanding applied force is crucial when considering material limits, such as tensile strength, to prevent failures. Moreover, in the context of thermal expansion, an applied force helps balance or counteract expansion forces, as seen in materials with constrained movement.