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Chapter 7: Problem 890

A wire of diameter \(1 \mathrm{~mm}\) breaks under a tension of $100 \mathrm{~N}$. Another wire of same material as that of the first one, but of diameter \(2 \mathrm{~mm}\) breaks under a tension of \(\ldots \ldots \ldots\) (A) \(500 \mathrm{~N}\) (B) \(1000 \mathrm{~N}\) (C) \(10,000 \mathrm{~N}\) (D) \(4000 \mathrm{~N}\)

Short Answer

Expert verified
The tension at which the second wire breaks is 400 N. (Answer: D)

Step by step solution

01

Calculate the cross-sectional area of each wire

The cross-sectional area of a wire can be found using the formula for the area of a circle (\(A = πr^2\)), where \(r\) is the radius. Since the diameter is given, we need to find the radius for each wire (half of the diameter) and then calculate the cross-sectional areas. Wire 1 Diameter: 1mm, so Radius: \(0.5\ \mathrm{mm}\) Wire 2 Diameter: 2mm, so Radius: \(1\ \mathrm{mm}\) Cross-sectional Area of Wire 1: \(A1 = π(0.5^2)\ \mathrm{mm}^2\) Cross-sectional Area of Wire 2: \(A2 = π(1^2)\ \mathrm{mm}^2\)
02

Find the relationship between tensions and cross-sectional areas

Since both wires are made of the same material, the ratio of their tensions should be proportional to the ratio of their cross-sectional areas. \(\frac{T2}{T1} = \frac{A2}{A1}\) Where \(T1\) and \(A1\) are the tension and cross-sectional area for Wire 1, and \(T2\) and \(A2\) are the tension and cross-sectional area for Wire 2.
03

Substitute the known values and solve for T2

Now we can substitute the known values for Wire 1 (T1 = 100N) and the calculated cross-sectional areas (A1 and A2) into the equation and solve for T2, the tension for Wire 2. \(\frac{T2}{100\ \mathrm{N}} = \frac{π(1^2)\ \mathrm{mm}^2}{π(0.5^2)\ \mathrm{mm}^2}\) Simplify the equation by canceling π in the ratio of areas: \(\frac{T2}{100\ \mathrm{N}} = \frac{1}{0.25}\) Now solve for T2: \(T2 = 100\ \mathrm{N} \times \frac{1}{0.25}\) \(T2 = 100\ \mathrm{N} \times 4\) \(T2 = 400\ \mathrm{N}\) The tension at which the second wire breaks is 400N, which corresponds to option (D) in the exercise.

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Most popular questions from this chapter

The work done in blowing a soap bubble of \(10 \mathrm{~cm}\) radius is [surface tension of soap solution is \(\\{(3 / 100) \mathrm{N} / \mathrm{m}\\}\) ] (A) \(75.36 \times 10^{-4}\) Joule (B) \(37.68 \times 10^{-4}\) Joule (C) \(150.72 \times 10^{-4}\) Joule (D) \(75.36\) Joule

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