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Chapter 7: Problem 880

A copper wire of length \(4 \mathrm{~m}\) and area of cross-section $1.2 \mathrm{~cm}^{2}\( is stretched with a force of \)4.8 \times 10^{3} \mathrm{~N}\(. If young's modulus for copper is \)1.2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$. What will be the length increase of the wire? (A) \(1.33 \mathrm{~mm}\) (B) \(1.33 \mathrm{~cm}\) (C) \(2.66 \mathrm{~mm}\) (D) \(2.66 \mathrm{~cm}\)

Short Answer

Expert verified
The length increase of the copper wire is \(\Delta L = 1.33\,\text{mm}\), which corresponds to answer (A).

Step by step solution

01

Write down the given values

First, we need to write down all the given information: - Length of the copper wire, \(L = 4\,\text{m}\) - Area of cross-section, \(A = 1.2\,\text{cm}^2 = 1.2 \times 10^{-4}\,\text{m}^2\) (converted to square meters) - Stretching force, \(F = 4.8 \times 10^3\,\text{N}\) - Young's modulus for copper, \(Y = 1.2 \times 10^{11}\,\text{N/m}^2\) - Length increase of the wire, \(\Delta L = ?\)
02

Rearrange the Young's modulus formula to solve for the length increase

We will rearrange the Young's modulus formula to solve for the length increase, \(\Delta L\). \(Y = \frac{F \cdot L}{A \cdot \Delta L}\) Rearrange for \(\Delta L\): \(\Delta L = \frac{F \cdot L}{A \cdot Y}\)
03

Substitute the given values into the formula

Now, we can substitute all the given values into the rearranged formula: \(\Delta L = \frac{(4.8 \times 10^3\,\text{N}) \cdot (4\,\text{m})}{(1.2 \times 10^{-4}\,\text{m}^2) \cdot (1.2 \times 10^{11}\,\text{N/m}^2)}\)
04

Calculate the length increase

Calculate the length increase, \(\Delta L\): \(\Delta L = \frac{1.92 \times 10^4\,\text{N} \cdot \text{m}}{1.44 \times 10^7\,\text{N} \cdot \text{m}^2} = 1.33 \times 10^{-3}\, \text{m}\) Convert the result into millimeters: \(\Delta L = 1.33 \times 10^{-3}\, \text{m} \cdot \frac{1000\,\text{mm}}{1\,\text{m}} = 1.33\,\text{mm}\) So the length increase of the copper wire is \(\Delta L = 1.33\,\text{mm}\), which corresponds to answer (A).

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Most popular questions from this chapter

When liquid medicine of density \(\mathrm{S}\) is to be put in the eye. It is done with the help of a dropper as the bulb on the top of the dropper is pressed a drop forms at the opening of the dropper we wish to estimate the size of the drop. We dirst assume that the drop formed at the opening is spherical because the requires a minimum increase in its surface energy. To determine the size we calculate the net vertical force due to surface tension \(\mathrm{T}\) when the radius of the drop is \(\mathrm{R}\). When this force becomes smaller than the weight of the drop the drop gets detached from the dropper. If $\mathrm{r}=5 \times 10^{-4} \mathrm{~m}, \mathrm{p}=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}=10 \mathrm{~ms}^{-2} \mathrm{~T}=0.11 \mathrm{~N} \mathrm{~m}^{-1}$ the radius of the drop when it detaches from the dropper is approximately (A) \(1.4 \times 10^{-3} \mathrm{~m}\) (B) \(3.3 \times 10^{-3} \mathrm{~m}\) (C) \(2.0 \times 10^{-3} \mathrm{~m}\) (D) \(4.1 \times 10^{-3} \mathrm{~m}\)

The force required to separate two glass plates of area $10^{-2} \mathrm{~m}^{2}\( with a film of water \)0.05 \mathrm{~mm}$ thick between them is (surface tension of water is \(70 \times 10^{-3} \mathrm{~N} / \mathrm{m}\) ) (A) \(28 \mathrm{~N}\) (B) \(14 \mathrm{~N}\) (C) \(50 \mathrm{~N}\) (D) \(38 \mathrm{~N}\)

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Read the assertion and reason carefully to mark the correct option out of the option given below. (a) If both assertion and reason are true and the reason is the correct explanation of the reason. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If the assertion and reason both are false. (e) If assertion is false but reason is true. Assertion: Specific heat capacity is the cause of formation of land and sea breeze. Reason: The specific heat of water is more then land. (A) a (B) \(b\) (C) \(c\) (D) \(\mathrm{d}\) \((\mathrm{E}) \mathrm{e}\)

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