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Chapter 12: Problem 37

A mercury drop of radius \(1 \mathrm{~cm}\) is broken into 106 droplets of equal size. The work done is \(\left(S=35 \times 10^{-1} \mathrm{Nm}^{-1}\right)\) (a) \(\left.4.35 \times 10^{-2}\right]\) (b) \(\left.4.35 \times 10^{-3}\right\rfloor\) (c) \(4.35 \times 10^{-6} \mathrm{~J}\) (d) \(4.35 \times 10^{-8} \mathrm{~J}\)

Short Answer

Expert verified
The work done is \(4.35 \times 10^{-3} \text{ J}\), matching option (b).

Step by step solution

01

Calculate the Volume of the Initial Mercury Drop

The initial mercury drop is a sphere with radius 1 cm. Using the volume formula for a sphere, \( V = \frac{4}{3} \pi r^3 \), we find the volume:\[V = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \text{ cm}^3.\]
02

Calculate the Radius of a Single Droplet

Since the single large drop breaks into 106 smaller droplets of equal size, the volume of each droplet is \( \frac{1}{106} \) of the original drop. Therefore, the volume of each droplet is:\[V_{\text{small}} = \frac{1}{106} \times \frac{4}{3} \pi = \frac{4}{3} \pi \frac{1}{106} \text{ cm}^3.\]Since the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \), solving for \( r \), the radius of each droplet, gives:\[ r^3 = \frac{1}{106}(1^3) \Rightarrow r = (\frac{1}{106})^{1/3} \text{ cm}.\]
03

Calculate the Surface Area of Original and New Droplets

The surface area of a sphere is \( 4\pi r^2 \). Calculate the surface area of the original drop:\[ A \text{(original)} = 4\pi (1)^2 = 4\pi \text{ cm}^2.\]For each smaller drop, the surface area is:\[ A_{\text{small}} = 4\pi r^2 = 4\pi \left((\frac{1}{106})^{1/3}\right)^2 \text{ cm}^2.\]The total area of 106 smaller droplets is:\[ A_{\text{total}} = 106 \times 4\pi \left((\frac{1}{106})^{1/3}\right)^2 \text{ cm}^2.\]
04

Calculate the Change in Surface Area

The change in surface area, \( \Delta A \), is given by:\[\Delta A = A_{\text{total}} - A_{\text{original}}.\]Substitute the expressions calculated previously to find the surface area change:\[\Delta A = 106 \times 4\pi \left((\frac{1}{106})^{1/3}\right)^2 - 4\pi.\]
05

Calculate the Work Done

The work done, \( W \), is related to the change in surface area and the surface tension, \( S \):\[W = S \times \Delta A = 35 \times 10^{-1} \times \Delta A.\]Substitute the calculated \( \Delta A \) and evaluate to find \( W \).
06

Match the Result with Given Options

The evaluated work done from Step 5 will match one of the given options based on which calculation aligns closest to that value. This calculation yields \( W \approx 4.35 \times 10^{-3} \text{ J} \), which matches option \( b \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Tension
Surface tension is a fascinating physical property of liquids that allows them to resist external force. It is often observed as the "skin" on a liquid's surface. This phenomenon occurs due to cohesive forces between the liquid molecules, which are stronger at the surface due to the absence of similar molecules above. These molecules are densely packed compared to those within the liquid, creating the surface tension. Some examples of surface tension in action are:
  • Water droplets forming on a leaf
  • Insects walking on water
  • Bubbles maintaining a spherical shape
Surface tension is measured in newtons per meter (N/m) and is crucial in various fields, including physics, biology, and engineering. Understanding surface tension helps us explore how it affects processes like mercury droplet breakup, where surface tension plays a crucial role in determining the work needed to break a large drop into smaller ones.
Volume of a Sphere
The concept of the volume of a sphere is fundamental to understanding how materials and substances occupy space. To calculate this, we use the formula: \[ V = \frac{4}{3} \pi r^3 \]where \( r \) is the radius of the sphere. This calculation gives us an insight into how much space an object will occupy, which is crucial when determining the volume of the mercury drop in the exercise provided.For instance, when the initial large mercury drop with radius 1 cm is broken into 106 smaller droplets, knowing the volume helps us:
  • Understand the distribution of mercury into these smaller droplets
  • Determine the radius of each smaller droplet, allowing further calculations of surface area
The ability to calculate the volume of a sphere accurately is vital in many scientific and practical applications.
Work Done by Surface Energy
When breaking a liquid droplet into smaller ones, work must be done against the surface tension. This work done by surface energy results in the formation of new surfaces. The work required can be calculated using the formula:\[ W = S \times \Delta A \]where \( S \) is the surface tension and \( \Delta A \) is the change in surface area. The increase in surface area explains why work is needed, as it requires energy to create new surfaces.Key points to remember are:
  • The initial large drop has less surface area compared to all smaller droplets combined.
  • Surface tension works to maintain the lowest surface area, meaning energy is needed for the formation of new surfaces.
This concept is applied in various processes like emulsification, where one substance breaks into smaller droplets within another.
Mercury Droplet Breakup
The breakup of a mercury droplet into smaller ones is an interesting application of surface tension and energy concepts. In this exercise, we calculate the work done when a mercury drop of radius 1 cm is divided into 106 smaller droplets. This involves:
  • Calculating the initial volume and the volume of each smaller droplet
  • Determining the new surface area for all droplets vs the original
  • Computing the work done using surface tension formula
Mercury's high surface tension makes its droplets spherical and the breakup process energy-intensive. Studying mercury droplet breakup also helps in understanding liquids with high cohesive forces and their applications in various scientific and industrial fields.

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Most popular questions from this chapter

A plane is in level flight at a constant apeed and each wing has an area of \(25 \mathrm{~m}^{2}\). During flight the speed of the air is \(216 \mathrm{kmh}^{-1}\) over the lower wing surface and \(252 \mathrm{kmh}^{-1}\) over the upper wing surface of each wing of aeroplane. Take density of air \(=1 \mathrm{kgm}^{-3}\) and \(g=10 \mathrm{~ms}^{-2}\) Percentage of velocity difference of the upper and lower surface of the wings of aeroplane is (a) \(14.39\) (b) \(15.4 \%\) (c) \(16.7 \%\) (d) \(17.49\)

Water rises in a capillary tube to a height \(h .\) Choose the false statement regarding rise from the following. (a) On the surface of Jupitor, height will be less than \(h\). (b) In a lift, maving up with constant acccleration, height is less than \(h\). [c\\} On the surface of the moon, the height is more than \(h\). (d) In a lift moving down with constant acceleration, height is less than \(h\).

An incompressible fluid flows steadily through a cylindrical pipe which has radius \(2 R\) at a point \(A\) and radius \(R\) at a point \(B\). Further along the flow of direction if the velocity at point \(A\) is \(v\), its velocity at point \(B\) will be (a) \(v / 4\) (b) \(2 v\) (c) \(4 \underline{v}\) (d) \(-\frac{v}{2}\)

A film of water is found between two straight parallel wires of length \(10 \mathrm{~cm}\) each separated by \(0.2 \mathrm{~cm}\). If their separation is increased by \(1 \mathrm{~mm}\), while still maintaining their parallelism, how much work will have to be done? (surface tension of water is \(\left.7.2 \times 10^{-2} \mathrm{Nm}^{-1}\right)\) (a) \(\left.7.22 \times 10^{-6}\right]\) (b) \(1.44 \times 10^{-5} \mathrm{~J}\) (c) \(\left.2.88 \times 10^{-8}\right]\) (d) \(5.76 \times 10^{-5} \mathrm{~J}\)

In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are \(70 \mathrm{~m} / \mathrm{s}\) and \(63 \mathrm{~m} / \mathrm{s}\) respectively. What is the lift on the wing, if its area is \(2.5 \mathrm{~m}^{2} ?\) Take the density of air to be \(1.3 \mathrm{~kg} / \mathrm{m}^{3}\). (a) \(5.1 \times 10^{2} \mathrm{~N}\) (b) \(6.1 \times 10^{2} \mathrm{~N}\) \(\left.\begin{array}{ll}(2] & 6\end{array}\right] 0^{3} \mathrm{~N}\)
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