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Chapter 7: Problem 902

A steel wire of \(1 \mathrm{~m}\) long and \(1 \mathrm{~mm}^{2}\) cross sectional area \(1 \mathrm{~s}\) hung from rigid end when weight of \(1 \mathrm{~kg}\) is hung from it then change in length will be........... [ $\left.\mathrm{Y}=2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\right]$ (A) \(0.5 \mathrm{~mm}\) (B) \(0.25 \mathrm{~mm}\) (C) \(0.05 \mathrm{~mm}\) (D) \(5 \mathrm{~mm}\)

Short Answer

Expert verified
The change in length of the steel wire is approximately \(0.05\,\text{mm}\).

Step by step solution

01

List the given information

We are given the following information - Original length of wire \(L = 1\,\text{m}\) - Cross-sectional area of wire \(A = 1\,\text{mm}^2 = 1 \times 10^{-6}\,\text{m}^2\) - Weight of \(1\,\text{kg}\) (force applied) \(F = 1\,\text{kg} \times 9.81\,\text{m/s}^2 = 9.81\,\text{N}\) - Young's modulus \(Y = 2 \times 10^{11}\,\text{N/m}^2\) Now, we will plug the given information into the formula to calculate the change in length.
02

Calculate the change in length using the formula

We will use the formula \(\Delta L = \frac{FL}{AY}\) to calculate the change in length. \[\Delta L = \frac{9.81\,\text{N} \times 1\,\text{m}}{1 \times 10^{-6}\,\text{m}^2 \times 2\times 10^{11}\,\text{N/m}^2}\]
03

Simplify and find the change in length

Now, simplify the equation: \[\Delta L = \frac{9.81}{2 \times 10^5} \approx 4.905 \times 10^{-5}\,\text{m}\] To convert the change in length into millimeters, multiply by 1000: \[\Delta L \approx 4.905 \times 10^{-5}\,\text{m} \times 1000 = 0.04905\,\text{mm}\] Looking at the given options, the closest answer is (C) \(0.05\,\text{mm}\)

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Most popular questions from this chapter

When \(100 \mathrm{~N}\) tensile force is applied to a rod of $10^{-6} \mathrm{~m}^{2}\( cross-sectional area, its length increases by \)1 \%$ so young's modulus of material is \(\ldots \ldots \ldots \ldots\) (A) \(10^{12} \mathrm{~Pa}\) (B) \(10^{11} \mathrm{~Pa}\) (C) \(10^{10} \mathrm{~Pa}\) (D) \(10^{2} \mathrm{~Pa}\)

A square frame of side \(\mathrm{L}\) is dipped in a liquid on taking out a membrance is formed if the surface tension of the liquid is \(\mathrm{T}\), the force acting on the frame will be. (A) \(2 \mathrm{TL}\) (B) \(4 \mathrm{TL}\) (C) \(8 \mathrm{TL}\) (D) \(10 \mathrm{TL}\)

A tank is filled with water up to a height \(\mathrm{H}\). Water is allowed to come out of a hole P in one of the walls at a depth \(\mathrm{D}\) below the surface of water express the horizontal distance \(\mathrm{x}\) in terms of \(\mathrm{H}\) and \(\mathrm{D}\). (B) $\left.\mathrm{x}={ }^{\alpha} \sqrt{[}\\{\mathrm{D}(\mathrm{H}-\mathrm{D})\\} / 2\right]$ (D) \(\mathrm{x}=4[\mathrm{D}(\mathrm{H}-\mathrm{D})]\)

The temperature at which the vapour pressure of a liquid becomes equals of the external pressure is its. (A) Melting point (B) sublimation point (C) Critical temperature (D) Boiling point

Cross-sectional area o if wire of length \(L\) is \(A\). Young's modulus of material is \(\mathrm{Y}\). If this wire acts as a spring what is the value of force constant? (A) (YA/L) (B) (YA/2L) (C) (2YA/L) (D) ( \(\mathrm{YL} / \mathrm{A})\)
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