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Chapter 7: Problem 875

The length of a wire is \(1.0 \mathrm{~m}\) and the area of cross-section is \(1.0 \times 10^{-2} \mathrm{~cm}^{2}\). If the work done for increase in length by \(0.2 \mathrm{~cm}\) is \(0.4\) joule. Then what is the young's modulus? Of material of the wire ? (A) \(2.0 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (B) \(4.0 \times 10^{10}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (C) \(2.0 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (D) \(4.0 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

Short Answer

Expert verified
The Young's modulus of the material of the wire is approximately \(2.0 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\).

Step by step solution

01

Find stress

The work done (W) is given as 0.4 J. We can use the formula for work: Work = Force × Distance. Let's find the force applied on the wire, F: 0.4 J = F × 0.2 cm Now, convert the distance from cm to meters: 0.2 cm = 0.002 m So, we get: 0.4 J = F × 0.002 m Solve for the force, F: F = 0.4 J / 0.002 m = 200 N Now, we need to find the stress. Stress is given by: Stress = Force / Area The area of cross-section is given as 1.0 × 10^{-2} cm². We need to convert it to square meters: 1.0 × 10^{-2} cm² = 1.0 × 10^{-6} m² Now, we can find the stress on the wire: Stress = 200 N / (1.0 × 10^{-6} m²) = 2.0 × 10^8 N/m²
02

Find strain

Now, we need to find the strain on the wire. Strain is given by: Strain = Change in length / Original length Change in length = 0.2 cm = 0.002 m (already converted to meters) Original length = 1.0 m Now, find the strain: Strain = 0.002 m / 1.0 m = 0.002
03

Find Young's modulus

Young's modulus (Y) is given by: Young's modulus = Stress / Strain Now, we can find the Young's modulus of the wire: Y = (2.0 × 10^8 N/m²) / 0.002 = 1.0 × 10^{11} N/m² From the given options, the closest value is: (C) \(2.0 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) So, the answer is (C).

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Most popular questions from this chapter

The work done increasing the size of a soap film from $10 \mathrm{~cm} \times 6 \mathrm{~cm}\( to \)10 \mathrm{~cm} \times 11 \mathrm{~cm}\( is \)3 \times 10^{-4}$ Joule. The surface tension of the film is (A) \(1.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (B) \(3.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (C) \(6.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (D) \(11.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\)

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) \(0.2\)

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) \(38 \mathrm{~N}\)

A soap bubble of radius \(r\) is blown up to form a bubble of radius $2 \mathrm{r}\( under isothermal conditions if the \)\mathrm{T}$ is the surface tension of soap solution the energy spent in the slowing is. (A) \(3 \pi \mathrm{Tr}^{2}\) (B) \(6 \pi \mathrm{Tr}^{2}\) (C) \(12 \pi \mathrm{Tr}^{2}\) (D) \(24 \pi \mathrm{Tr}^{2}\)

Water is flowing continuously from a tap having an internal diameter $8 \times 10^{-3} \mathrm{~m}\(. The water velocity as it leaves the tap is \)0.4 \mathrm{~m} / \mathrm{s}\(. The diameter of the water stream at a distance \)2 \times 10^{-1} \mathrm{~m}$ below the tap is close to (A) \(5.0 \times 10^{-3} \mathrm{~m}\) (B) \(7.5 \times 10^{-3} \mathrm{~m}\) (C) \(9.6 \times 10^{-3} \mathrm{~m}\) (D) \(3.6 \times 10^{-3} \mathrm{~m}\)
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