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Chapter 7: Problem 873

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) \(8.8 \mathrm{~N}\)

Short Answer

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

Step by step solution

01

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 = \(5^{\circ} \mathrm{C}\)
02

Calculate Δ𝐿th

Now we will plug in the values to find the increase in length due to thermal expansion: \(\Delta L_{th} = (2 \, \mathrm{m})(8 \times 10^{-6} \, \mathrm{C}^{-1})(5^{\circ} \mathrm{C})\) \(\Delta L_{th} = 8 \times 10^{-5} \, \mathrm{m}\)
03

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)
04

Calculate σ

Plugging in the values for \(F\) and \(A\): \(σ = \frac{200 \, \mathrm{N}}{10^{-6} \, \mathrm{m^2}}\) \(σ = 2 \times 10^8 \, \frac{\mathrm{N}}{\mathrm{m^2}}\)
05

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 = \(2 \times 10^8 \, \frac{\mathrm{N}}{\mathrm{m^2}}\)
06

Calculate ε

Now we will solve for \(ε\): \(ε = \frac{σ}{Y} = \frac{2 \times 10^8 \, \frac{\mathrm{N}}{\mathrm{m^2}}}{2.2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)}\) \(ε = 9.09 \times 10^{-4}\)
07

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 = \(2 \, \mathrm{m}\)
08

Calculate Δ𝐿st

Now we will plug the values to find the increase in length due to stress: \(\Delta L_{st} = (9.09 \times 10^{-4})(2 \, \mathrm{m})\) \(\Delta L_{st} = 1.82 \times 10^{-3} \, \mathrm{m}\)
09

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 \(\Delta σ = σ_{th} - σ_{st}\)
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).

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