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Chapter 7: Problem 986

Radius of a soap bubble is \(\mathrm{r}^{\prime}\), surface tension of soap solution is \(\mathrm{T}\). Then without increasing the temperature how much energy will be needed to double its radius. (A) \(4 \pi r^{2} T\) (B) \(2 \pi r^{2} T\) (C) \(12 \pi r^{2} T\) (D) \(24 \pi r^{2} T\)

Short Answer

Expert verified
The short answer is: \(\Delta E = 12\pi (\mathrm{r}^{\prime})^2 T\).

Step by step solution

01

Determine the initial and final surface areas of the bubble

The surface area of a sphere (soap bubble) is given by the formula \(A = 4\pi r^{2}\). Initially, the radius is \(\mathrm{r}^{\prime}\), so the surface area is \(A_{1} = 4\pi (\mathrm{r}^{\prime})^2\). When we double the radius, we have a new radius of \(2\mathrm{r}^{\prime}\). Thus, the new surface area is \(A_{2} = 4\pi (2\mathrm{r}^{\prime})^2 = 16\pi (\mathrm{r}^{\prime})^2\).
02

Calculate the initial and final surface energies of the bubble

Surface energy is proportional to the surface area, and the constant of proportionality is the surface tension, \(T\). The initial surface energy is \(E_{1} = T A_{1} = 4\pi (\mathrm{r}^{\prime})^2 T\), and the final surface energy is \(E_{2} = T A_{2} = 16\pi (\mathrm{r}^{\prime})^2 T\).
03

Determine the amount of energy required to double the radius

The energy required to double the radius will be the difference between the final and initial surface energies: \[\Delta E = E_{2} - E_{1} = 16\pi (\mathrm{r}^{\prime})^2 T - 4\pi (\mathrm{r}^{\prime})^2 T = 12\pi (\mathrm{r}^{\prime})^2 T\]
04

Identify the answer

The energy required to double the radius of the soap bubble is given by our result in Step 3, which is \(12\pi (\mathrm{r}^{\prime})^2 T\). Looking at the given options, the correct answer is: (C) 12\pi \(r^{2} T\).

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Most popular questions from this chapter

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

The surface tension of a liquid is \(5 \mathrm{~N} / \mathrm{m}\). If a thin film of the area \(0.02 \mathrm{~m}^{2}\) is formed on a loop, then its surface energy will be (A) \(5 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(2 \times 10^{-1} \mathrm{~J}\) (D) \(5 \times 10^{-1} \mathrm{~J}\)

When two soap bubbles of radius \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\left(\mathrm{r}_{2}>\mathrm{r}_{1}\right)\) coalesce, the radius of curvature of common surface is........... (A) \(r_{2}-r_{1}\) (B) \(\left[\left(r_{2}-r_{1}\right) /\left(r_{1} r_{2}\right)\right]\) (C) $\left[\left(\mathrm{r}_{1} \mathrm{r}_{2}\right) /\left(\mathrm{r}_{2}-\mathrm{r}_{1}\right)\right]$ (D) \(\mathrm{r}_{2}+\mathrm{r}_{1}\)

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) \(2 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\)

Cross-sectional area o if wire of length \(L\) is \(A\). Young's modulus of material is \(\mathrm{Y}\). If this wire acts as a spring what is the value of force constant? (A) (YA/L) (B) (YA/2L) (C) (2YA/L) (D) ( \(\mathrm{YL} / \mathrm{A})\)
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