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Chapter 7: Problem 987

The amount of work done in blowing a soap bubble such that its diameter increases from \(d\) to \(D\) is \((T=\) Surface tension of solution) (A) \(4 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) (B) \(8 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) (C) \(\pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) (D) \(2 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\)

Short Answer

Expert verified
The short answer is: The amount of work done in blowing a soap bubble such that its diameter increases from \(d\) to \(D\) is \(4 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\).

Step by step solution

01

Understanding the problem

The work done in blowing a soap bubble can be calculated using the formula: Work = Surface tension × Change in surface area. We are given the initial and final diameters, d and D, respectively, and we need to find the change in surface area.
02

Calculate the surface area of the initial and final bubbles

The surface area of a sphere can be calculated using the formula: A = 4πr². To find the surface areas of the soap bubble initially and finally, we need to convert diameters into radii. The radii for initial and final bubbles can be represented as r₁ = d/2 and r₂ = D/2. Now, calculate the initial and final surface areas: - Initial surface area: A₁ = 4πr₁² = 4π(d/2)² - Final surface area: A₂ = 4πr₂² = 4π(D/2)²
03

Calculate the change in surface area

We can find the change in surface area by subtracting the initial surface area from the final surface area: ΔA = A₂ - A₁ = 4π(D/2)² - 4π(d/2)² To factor out 4π, we get: ΔA = 4π[(D/2)² - (d/2)²]
04

Calculate the work done

Using the formula, Work = Surface tension × Change in surface area, we can now find the work done: Work = T × ΔA = T × 4π[(D/2)² - (d/2)²] Comparing this expression to the given options, we find that it matches with option (A): \(4 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) Therefore, the correct answer is (A).

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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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