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Chapter 7: Problem 883

A wire of length \(L\) and radius \(\mathrm{r}\) is rigidly fixed at one end on stretching the other end of the wire a force \(\mathrm{F}\) the increase in its lengths is \(\ell\). If another wire of same material but of length $2 \mathrm{~L}\( and radius \)2 \mathrm{r}\( is stretched with a force of \)2 \mathrm{~F}$, the increase in its length will be........... (A) \(\varepsilon\) (B) \(2 \ell\) (C) \(\mathrm{C} / 2\) (D) 4

Short Answer

Expert verified
The increase in length for the second wire is 2 times the increase in length for the first wire. So, the correct option is \( (B) 2l \).

Step by step solution

01

Understand the given values

The first wire has: - Length: L - Radius: r - Force: F - Increase in length: \(l\) The second wire has: - Length: 2L - Radius: 2r - Force: 2F - Increase in length: We have to find this value
02

Young's modulus formula

Young's modulus is a measure of the stiffness of a material, and it is defined as the ratio of stress (force per unit area) to strain (relative deformation). In general, it's denoted as Y and is given by the formula: \(Y = \frac{\text{Stress}}{\text{Strain}}\) For wires, we can rewrite this formula using stress and strain in terms of force, cross-sectional area (A), and displacements (∆L): \(Y = \frac{F/A}{\Delta L/L}\) Rearranging the formula for increased length (∆L): \(\Delta L = \frac{FL}{AY}\)
03

Determine Force and Area relationship

Here, we must determine the relationship between the cross-sectional area (A) and the radius (r). Since the wire has a circular cross-section, the area A can be expressed as: \(A = \pi r^2\) Now, for the second wire, the radius is doubled, meaning: \(A' = \pi (2r)^2 = 4\pi r^2 = 4A\) The force on the second wire is also doubled: \(F' = 2F\)
04

Calculate the increased length for the second wire

Using the formula \(\Delta L = \frac{FL}{AY}\), we will plug in the values of F', A', and L': \(\Delta L' = \frac{F'L'}{A'Y} = \frac{2F \cdot 2L}{4AY}\) Now, we notice that the given values (F, L, A) are already present in the first wire, which means we can substitute the value of the first wire into the second: \(\Delta L' = \frac{2F \cdot 2L}{4AY} = 2 \cdot \frac{FL}{AY} = 2 \Delta L\) Therefore, the increase in length for the second wire is:
05

Conclusion

The increase in length for the second wire is 2 times the increase in length for the first wire. So, the correct option is (B) 2l.

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Most popular questions from this chapter

A rubber ball when taken to the bottom of a \(100 \mathrm{~m}\) deep take decrease in volume by \(1 \%\) Hence, the bulk modulus of rubber is $\ldots \ldots \ldots . . .\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]$ (A) \(10^{6} \mathrm{~Pa}\) (B) \(10^{8} \mathrm{~Pa}\) (C) \(10^{7} \mathrm{~Pa}\) (D) \(10^{9} \mathrm{~Pa}\)

Two drops of the same radius are falling through air with a steady velocity for \(5 \mathrm{~cm}\) per sec. If the two drops coakesce the terminal velocity would be (A) \(10 \mathrm{~cm}\) per sec (B) \(2.5 \mathrm{~cm}\) per sec (C) \(5 \times(4)^{(1 / 3)} \mathrm{cm}\) per sec (D) \(5 \times \sqrt{2} \mathrm{~cm}\) per sec

The surface tension of a liquid is \(5 \mathrm{~N} / \mathrm{m}\). If a thin film of the area \(0.02 \mathrm{~m}^{2}\) is formed on a loop, then its surface energy will be (A) \(5 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(2 \times 10^{-1} \mathrm{~J}\) (D) \(5 \times 10^{-1} \mathrm{~J}\)

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