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Chapter 7: Problem 988

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) \(24 \pi \mathrm{Tr}^{2}\)

Short Answer

Expert verified
The energy spent in expanding the soap bubble of radius r to form a bubble of radius 2r under isothermal conditions is \(12 \pi Tr^2\), which matches option (C).

Step by step solution

01

Understand the problem and its relevant formula

First, let's understand the problem. We have a soap bubble of radius r, and we want to expand it to form a bubble with radius 2r under isothermal conditions, meaning the temperature remains constant during the entire process. Now, we want to find the energy spent in doing this. The key concept to understand here is surface tension (T) of the soap solution. The work done against the surface tension force will be equal to the energy spent in expanding the bubble. The surface tension work is given by the formula: Work done = T * (change in surface area) Now, let's figure out the change in surface area for the given bubble expansion.
02

Find the initial and final surface area of the soap bubble

The bubble is a sphere, and the surface area of a sphere is given by: Surface area = 4 * π * r^2 For our bubble, the initial surface area (A₁) is: A₁ = 4 * π * r^2 And, after expansion, the final surface area (A₂) is: A₂ = 4 * π * (2r)^2 = 4 * π * 4 * r^2 = 16 * π * r^2 Now that we have both the initial and final surface areas, we can find the change in surface area (∆A).
03

Calculate the change in surface area of the soap bubble

To find the change in surface area, ∆A, we simply need to subtract the initial surface area (A₁) from the final surface area (A₂). ∆A = A₂ - A₁ ∆A = (16 * π * r^2) - (4 * π * r^2) ∆A = 12 * π * r^2
04

Calculate the energy spent

Now that we have the change in surface area, we can find the energy spent in expanding the bubble. Recall our formula for work done against surface tension: Work done = T * (change in surface area) Energy spent = T * ∆A Energy spent = T * (12 * π * r^2) Now let's match this result with the given options: (A) 3 * π * Tr^2 (B) 6 * π * Tr^2 (C) 12 * π * Tr^2 (D) 24 * π * Tr^2 Our result matches option (C). Therefore, the energy spent in expanding the soap bubble is 12 * π * Tr^2.

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Most popular questions from this chapter

When more than \(20 \mathrm{~kg}\) mass is tied to the end of wire it breaks what is maximum mass that can be tied to the end of a wire of same material with half the radius? (A) \(20 \mathrm{~kg}\) (B) \(5 \mathrm{~kg}\) (C) \(80 \mathrm{~kg}\) (D) \(160 \mathrm{~kg}\)

In a capillary tube water rises by \(1.2 \mathrm{~mm}\). The height of water that will rise in another capillary tube having half the radius of the first is (A) \(1.2 \mathrm{~mm}\) (B) \(2.4 \mathrm{~mm}\) (C) \(0.6 \mathrm{~mm}\) (D) \(0.4 \mathrm{~mm}\)

The compressibility of water \(4 \times 10^{-5}\) per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere will be.......... (A) \(4 \times 10^{-5} \mathrm{CC}\) (B) \(4 \times 10^{-5} \mathrm{CC}\) (C) \(0.025 \mathrm{CC}\) (D) \(0.004 \mathrm{CC}\)

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) \(3.6 \times 10^{-3} \mathrm{~m}\)

A large number of water drops each of radius \(r\) combine to have a drop of radius \(\mathrm{R}\). If the surface tension is \(\mathrm{T}\) and the mechanical equivalent at heat is \(\mathrm{J}\) then the rise in temperature will be (A) \((2 \mathrm{~T} / \mathrm{rJ})\) (B) \((3 \mathrm{~T} / \mathrm{RJ})\) (C) \((3 \mathrm{~T} / \mathrm{J})\\{(1 / \mathrm{r})-(1 / \mathrm{R})\\}\) (D) \((2 \mathrm{~T} / \mathrm{J})\\{(1 / \mathrm{r})-(1 / \mathrm{R})\\}\)
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