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Chapter 7: Problem 991

Two small drops mercury, each of radius \(\mathrm{r}\), coalesces the form a single large drop. The ratio of the total surface energies before and after the change is. (A) \(1: 2^{(1 / 2)}\) (B) \(2^{(1 / 3)}: 1\) (C) 2: 1 (D) \(1: 2\)

Short Answer

Expert verified
The ratio of the total surface energies before and after the change is \(\boxed{\text{(B)}\ 2^{(1 / 3)}: 1}\).

Step by step solution

01

Calculate the surface areas of the initial drops

Considering that both initial drops have a radius \(r\), we can find their surface areas using the formula for the surface area of a sphere: Surface area of a sphere = \(4\pi r^2\) Since there are two identical drops, the total surface area before coalescence is: Total surface area before coalescence = \(2(4\pi r^2) = 8\pi r^2\)
02

Calculate the volume of the final drop

The two initial drops coalesce to form a single large drop. The volume, therefore, remains constant throughout the process: Volume of initial drops = Volume of final drop Each initial drop has a volume given by the formula for the volume of a sphere: Volume of a sphere = \(\frac{4}{3}\pi r^3\) Since there are two identical drops, the total volume before coalescence is: Total volume before coalescence = \(2(\frac{4}{3}\pi r^3) = \frac{8}{3}\pi r^3\)
03

Calculate the radius of the final drop

Let the radius of the final drop be \(R\). We know that the volume of the final drop can be given as: Volume of final drop = \(\frac{4}{3}\pi R^3\) Now, equalizing the volumes of the drops before and after coalescence: \(\frac{8}{3}\pi r^3 = \frac{4}{3}\pi R^3\) Dividing both sides of the equation by \(\frac{4}{3}\pi\) and simplifying, we get: \(2r^3 = R^3\)
04

Find the surface area of the final drop

Using the radius \(R\) of the final drop, we can find the surface area using the formula for the surface area of a sphere: Surface area of the final drop = \(4\pi R^2\)
05

Calculate the ratio between the surface areas and surface energies

Now that we have the surface areas before and after coalescence, we can calculate the ratio of surface energies: \(\frac{\text{Total surface energy before coalescence}}{\text{Total surface energy after coalescence}} = \frac{8\pi r^2}{4\pi R^2}\) Since \(2r^3 = R^3\), we can rewrite the ratio in terms of \(r\): \(\frac{8\pi r^2}{4\pi (2^{1/3}r)^2}\) After simplifying, we get: \(\boxed{\text{(B)}\ 2^{(1 / 3)}: 1}\)

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Most popular questions from this chapter

A thin liquid film formed between a u-shaped wire and a light slider supports a weight of \(1.5 \times 10^{-2} \mathrm{~N}\) (see figure). The length of the slider is \(30 \mathrm{~cm}\) and its weight negligible. The surface tension of the liquid film is. (A) \(0.0125 \mathrm{Nm}^{-1}\) (B) \(0.1 \mathrm{Nm}^{-1}\) (C) \(0.05 \mathrm{Nm}^{-1}\) (D) \(0.025 \mathrm{Nm}^{-1}\)

Assertion and Reason: Read the assertion and reason carefully to mark the correct option out of the option given below (A) If both assertion and reason are true and reason is the correct explanation of the assertion. (B) If both assertion and reason are true but reason is not the correct explanation of the assertion. (C) If assertion is true but reason is false. (D) If assertion and reason both are false. Assertion: The stretching of a coil is determined by its shear modulus. Reason: Shear modulus change only shape of a body keeping its dimensions unchanged. (A) a (B) \(\mathrm{b}\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\)

What is the relationship between Young's modulus Y, Bulk modulus \(\mathrm{k}\) and modulus of rigidity \(\eta\) ? (A) \(\mathrm{Y}=[9 \eta \mathrm{k} /(\eta+3 \mathrm{k})]\) (B) \(\mathrm{Y}=[9 \mathrm{Yk} /(\mathrm{y}+3 \mathrm{k})]\) (C) \(\mathrm{Y}=[9 \eta \mathrm{k} /(3+\mathrm{k})]\) (D) \(\mathrm{Y}=[3 \eta \mathrm{k} /(9 \eta+\mathrm{k})]\)

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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) \(11.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\)
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