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Chapter 12: Problem 66

The aurface tension of soap solution is \(0.03 \mathrm{Nm}^{-1}\). The work done in blowing to form a soap bubble of surface area \(40 \mathrm{~cm}^{2}\), in joule is (a) \(1.2 \times 10^{-4}\) (b) \(2.4 \times 10^{-4}\) (c) \(12 \times 10^{-4}\) (d) \(24 \times 10^{-4}\)

Short Answer

Expert verified
(b) \(2.4 \times 10^{-4}\) J.

Step by step solution

01

Understand the Concept

The work done in forming a soap bubble is equal to the surface tension times the change in surface area. Since a soap bubble has two surfaces (inner and outer), the change in surface area is doubled.
02

Convert Units

Convert the surface area from square centimeters to square meters. Given area: 40 cm虏. Conversion: \(1 \text{ cm}^2 = 1 \times 10^{-4} \text{ m}^2\). Therefore, \(40 \text{ cm}^2 = 40 \times 10^{-4} \text{ m}^2\).
03

Calculate Work Done

Use the formula for work done: \(W = 2 \times T \times A\), where \(T\) is the surface tension and \(A\) is the area in square meters. Use \(T = 0.03 \text{ Nm}^{-1}\) and \(A = 40 \times 10^{-4} \text{ m}^2\).\[ W = 2 \times 0.03 \times 40 \times 10^{-4} = 2.4 \times 10^{-4} \text{ J} \]
04

Compare with Given Options

The calculated work done is \(2.4 \times 10^{-4} \text{ J}\). Comparing with the given options, the correct answer is (b) \(2.4 \times 10^{-4}\).

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Key Concepts

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

Work Done
When you perform work, you're essentially transferring energy. In the context of a soap bubble, work is needed to create and expand the bubble. The formula for calculating the work done on a soap bubble is:\[ W = 2 \times T \times A \]where:
  • \( W \) stands for the work done.
  • \( T \) is the surface tension, a measure of the force required to extend a liquid surface.
  • \( A \) represents the change in surface area.
It's important to note that a soap bubble has both an inner and outer surface. As a result, when the bubble is formed, the surface area changes for both surfaces, leading to double the effect. By understanding this, you recognize how the energy transformation is related to the bubble's tension and area expansion.
Soap Bubble
Soap bubbles are fascinating not just for their colors and patterns, but also for their scientific properties. These bubbles consist of a thin film of soapy water enveloping air inside. The surface tension of the liquid film holds the bubble together.Surface tension is a crucial aspect here. It dictates how much force is necessary to create the bubble's surface. Specifically, surface tension influences the bubble's stability and size. In this exercise, the bubble's surface tension is given as 0.03 Nm\(^{-1}\). This characteristic helps in calculating the work needed to form the bubble.
Understanding a soap bubble also involves appreciating that the shape is typically spherical. This shape minimizes surface area for a given volume, illustrating principles of physics and geometry in a captivating way. The beauty of soap bubbles lies in these intricate plays of force and form.
Surface Area Conversion
Converting surface area units is essential to align with the standard units used in physics problems, typically square meters. This exercise begins with a surface area in square centimeters (cm\(^2\)) and requires conversion into square meters (m\(^2\)) for accurate calculation.Here's how the conversion works:
  • 1 cm\(^2\) equals 1 \( \times \) 10\(^{-4}\) m\(^2\).
For a given area of 40 cm\(^2\), the conversion is calculated as follows:\[ 40 \text{ cm}^2 = 40 \times 10^{-4} \text{ m}^2 \]Understanding this conversion is vital, as it ensures the use of consistent units in equations. Physicists and engineers frequently convert units as a standard practice in calculations to maintain precision and accuracy.

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Most popular questions from this chapter

Water flowing out of the mouth of a tap and falling vertically in streamline flow forms a tapering column, Le., the area of cross-section of the liquid column decreases as it moves down. Which of the following is the most accurate explanation for this? (a) Falling water tries to reach a terminal velocity and hence, reduces the area of cross-section to balance upwand and downward forces (b) As the water moves down, its speed increases and hence, its pressure decreases. It is then compressed by atmosphere (c) The surface tension causes the exposed surface area of the liquid to decrease continuously (d) The mass of water flowing out per second through any cross-section must remain constant. As the water is almost incompressible, so the volume of water flowing out per second must remain constant. As this is equal to velocity \(x\) area, the area decreases as velocity increases

A tank is filled with water of density \(1 \mathrm{~g}\) per \(\mathrm{cm}^{\mathrm{3}}\) and oil of density \(0.9 \mathrm{~g} \mathrm{~cm}^{-3}\). The height of water layer is \(100 \mathrm{~cm}\) and of the oil layer is \(400 \mathrm{~cm}\). If \(g=980 \mathrm{cms}^{-2}\), then the velocity of efflux from an opening in the bottom of the tank is (a) \(\sqrt{900 \times 980} \mathrm{cms}^{-1}\) (b) \(\sqrt{1000 \times 980} \mathrm{cms}^{-1}\) (c) \(\sqrt{92 \times 980} \mathrm{cms}^{-1}\) (d) \(\sqrt{920 \times 980} \mathrm{cms}^{-1}\)

A uniform rod of density \(\rho\) is placed in a wide tank containing a liquid o \((\sigma>\rho)\). The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position, the rod makes an angle \(\theta\) with the horizontal. Then, \(\sin \theta\) is equal to (a) \(\frac{1}{2} \sqrt{\frac{\sigma}{\rho}}\) (b) \(\frac{1}{2} \frac{\sigma}{\rho}\) (c) \(\sqrt{\frac{\rho}{a}}\) (d) \(\sqrt{\frac{\rho}{d}}\)

If we dip capillary tubes of different radii \(r\) in water and the water rises to different height \(h\) in them, then we shall have constant (a) \(h / \underline{r^{2}}\) (b) \(h / r\) (c) \(h r^{2}\) (d) \(h r\)

An alloy of \(\mathrm{Zn}\) and Cu (i.e., brass) weights \(16.8 \mathrm{~g}\) in air and \(14.7 \mathrm{~g}\) in water. If relative density of \(\mathrm{Cu}\) and \(\mathrm{Zn}\) are \(8.9\) and \(7.1\) respectively then determine the amount of \(\mathrm{Zn}\) and \(\mathrm{Cu}\) in the alloy. (a) \(2 \mathrm{~g} .4 \mathrm{~g}\) (b) \(4 \mathrm{~g}, 2 \mathrm{~g}\) (c) 9.345g. \(7.455 \mathrm{~g}\) (d) \(0,3 \mathrm{~g}\)
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