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Chapter 7: Problem 899

Two similar wires under the same load yield elongation of \(0.1 \mathrm{~mm}\) and \(0.05 \mathrm{~mm}\) respectively. If the area of Cross-section of the first wire is \(4 \mathrm{~mm}^{2}\). Then what is the area of cross - section of the second wire? (A) \(6 \mathrm{~mm}^{2}\) (B) \(8 \mathrm{~mm}^{2}\) (C) \(10 \mathrm{~mm}^{2}\) (D) \(12 \mathrm{~mm}^{2}\)

Short Answer

Expert verified
The area of cross-section of the second wire is \(8 \mathrm{~mm}^{2}\) (Option B).

Step by step solution

01

Write down the relationship between elongation, length and area of cross-section

According to the relationship between elongation, length, and area of cross-section, we have: \[\frac{\text{Elongation}}{\text{Area of cross-section}} \propto \text{Length}\]
02

Write down the given information

We are given the following information: - Elongation of the first wire = 0.1 mm - Elongation of the second wire = 0.05 mm - Area of cross-section of the first wire = 4 mm² - We need to find the area of cross-section of the second wire.
03

Write equations for both wires

Using the relationship mentioned in Step 1, we can write equations for both wires as: \[\frac{0.1}{4} = kL \Rightarrow kL = 0.025 \,\,\,...(1)\] \[\frac{0.05}{A_2} = kL \Rightarrow kL = 0.05 \times A_2^{-1}\,\,\,...(2)\] Here, \(k\) is a proportionality constant and \(L\) is the length of the wire (which is the same for both the wires since they are similar). \(A_2\) is the area of cross-section of the second wire that we need to find.
04

Set up the equation to solve for the area of cross-section of the second wire

Since both equations (1) and (2) are equal to \(kL\), we can set them equal to each other: \[0.025 = 0.05 \times A_2^{-1}\]
05

Solve for the area of cross-section of the second wire

Now, isolate \(A_2\) by dividing both sides of the equation by 0.05: \[A_2^{-1} = \frac{0.025}{0.05} \Rightarrow A_2^{-1} = 0.5\] Invert both sides of the equation to find \(A_2\): \[A_2 = \frac{1}{0.5} \Rightarrow A_2 = 2\] Since the area of cross-section of the first wire is 4 mm² and the ratio of the areas of cross-section is 2, the area of cross-section of the second wire is: \[A_2 = 4 \times 2 = 8 \mathrm{~mm}^2\] So, the correct answer is (B) \(8 \mathrm{~mm}^{2}\).

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