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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

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

When liquid medicine of density \(\mathrm{S}\) is to be put in the eye. It is done with the help of a dropper as the bulb on the top of the dropper is pressed a drop forms at the opening of the dropper we wish to estimate the size of the drop. We dirst assume that the drop formed at the opening is spherical because the requires a minimum increase in its surface energy. To determine the size we calculate the net vertical force due to surface tension \(\mathrm{T}\) when the radius of the drop is \(\mathrm{R}\). When this force becomes smaller than the weight of the drop the drop gets detached from the dropper. If $\mathrm{r}=5 \times 10^{-4} \mathrm{~m}, \mathrm{p}=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}=10 \mathrm{~ms}^{-2} \mathrm{~T}=0.11 \mathrm{~N} \mathrm{~m}^{-1}$ the radius of the drop when it detaches from the dropper is approximately (A) \(1.4 \times 10^{-3} \mathrm{~m}\) (B) \(3.3 \times 10^{-3} \mathrm{~m}\) (C) \(2.0 \times 10^{-3} \mathrm{~m}\) (D) \(4.1 \times 10^{-3} \mathrm{~m}\)

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) \(5 \times 10^{-1} \mathrm{~J}\)

A material has poisson's ratio \(0.50\). If uniform rod of it suffers longitudinal strain of \(2 \times 10^{-3}\). Then what is percentage change in volume ? (A) \(0.6\) (B) \(0.4\) (C) \(0.2\) (D) 0

When there is no external force, the shape of liquid drop is determined by (A) Surface tension of liquid (B) Density of Liquid (C) Viscosity of liquid (D) Temperature of air only
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