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Chapter 7: Problem 844

On increasing the length by \(0.5 \mathrm{~mm}\) in a steel wire of length $2 \mathrm{~m}\( and area of cross-section \)2 \mathrm{~mm}^{2}$ the force required is..... \(\left[\mathrm{Y}\right.\) for steel $\left.=2.2 \times 10^{11} \mathrm{~N} / \mathrm{m}\right]$ (A) \(1.1 \times 10^{5} \mathrm{~N}\) (B) \(1.1 \times 10^{4} \mathrm{~N}\) (C) \(1.1 \times 10^{3} \mathrm{~N}\) (D) \(1.1 \times 10^{2} \mathrm{~N}\)

Short Answer

Expert verified
The required force to increase the length of the steel wire by 0.5 mm is \(2.2 \times 10^2 N\).

Step by step solution

01

Understand the problem

We are given the original length, change in length, area of cross-section and Young's modulus of steel. We need to find the force required to cause the given increase in length.
02

Write the formula relating stress, strain, and Young's modulus

The formula relating stress, strain, and Young's modulus is: \(Stress = Young's~modulus × Strain\) where \(Stress = \frac{Force}{Area}\) and \(Strain = \frac{Change~in~length}{Original~length}\).
03

Substitute the given values

Substitute the given values in the formula: \(Force = Stress × Area = (Young's~modulus × Strain) × Area\) \(Force = (2.2 × 10^{11} N/m~ × \frac{0.5 × 10^{-3} m}{2 m}) × 2 × 10^{-6} m^2\)
04

Calculate the required force

Now, calculate the force: \(Force = (2.2 × 10^{11} × \frac{0.5 × 10^{-3}}{2}) × 2 × 10^{-6}\) \(Force = (2.2 × 10^{11} × 0.25 × 10^{-3}) × 2 × 10^{-6}\) \(Force = 1.1 × 10^{8} × 10^{-3} × 2 × 10^{-6}\) \(Force = 1.1 × 10^{5} × 2 × 10^{-6}\) \(Force = 1.1 × 2 × 10^{5-6}\) \(Force = 2.2 × 10^{5-6}\) \(Force = 2.2 × 10^{-1}\) \(Force = 2.2 × 10^{2} N\) So, the required force to increase the length of the steel wire by 0.5 mm is \(2.2 \times 10^2 N\) which corresponds to the answer choice (D).

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Most popular questions from this chapter

For a given material the Young's modulus is \(2.4\) times that of rigidity modulus. What is its poisson's ratio? (A) \(2.4\) (B) \(1.2\) (C) \(0.4\) (D) \(0.2\)

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) \(10^{9} \mathrm{~Pa}\)

The density \(\rho\) of coater of bulk modulus \(B\) at a depth \(y\) in the ocean is related to the density at surface \(\rho_{0}\) by the relation. (A) $\rho=\rho_{0}\left[1-\left\\{\left(\rho_{0} \mathrm{gy}\right\\} / \mathrm{B}\right\\}\right]$ (B) $\rho=\rho_{0}\left[1+\left\\{\left(\rho_{0} \mathrm{gy}\right\\} / \mathrm{B}\right\\}\right]$ (C) $\rho=\rho_{0}\left[1+\left\\{\left(\rho_{0} \mathrm{gyh}\right\\} / \mathrm{B}\right\\}\right]$ (D) $\rho=\rho_{0}\left[1-\left\\{\mathrm{B} /\left(\rho_{0} \mathrm{~g} \mathrm{y}\right\\}\right]\right.$

An iron bar of length \(10 \mathrm{~m}\) is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). If the coefficient of linear thermal expansion of iron is \(\left[\left\\{10 \times 10^{-6}\right\\} / \mathrm{C}\right]\) the increase in the length of bar is (A) \(0.5 \mathrm{~cm}\) (B) \(1.0 \mathrm{~cm}\) (C) \(1.5 \mathrm{~cm}\) (D) \(2.0 \mathrm{~cm}\)

Read the assertion and reason carefully and mark the correct option given below. (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (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. Assertion: The water rises higher in a capillary tube of small diametre than in the capillary tube of large diameter. Reason: Height through which liquid rises in a capillary tube is inversely proportional to the diameter of the capillary tube. (A) a (B) b (C) c (D) d
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