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Chapter 10: Problem 23

Water rises to a height \(h\) in a capillary at the surface of earth. On the surface of the moon the height of water column in the same capillary will be (a) \(6 h\) (b) \(\frac{1}{6} h\) (c) \(h\) (d) Zero

Short Answer

Expert verified
The correct answer is (c).\n\n The height of the water column in the same capillary on the surface of the moon would be the same as on earth, which is \( h \). Capillary action, which is responsible for water rising in the tube, doesn't depend on gravity.

Step by step solution

01

Understand the problem

You are given a capillary action problem. On Earth, water rises to a certain height, \( h \), in a capillary tube. You need to find how high it would rise on the surface of the Moon.
02

Relate the problem with physical principles

Capillary action happens due to the adhesive and cohesive forces among molecules. It is not dependent on external gravity. Therefore, in terms of capillary action, there's no difference whether this happens on Earth or on the Moon even though the gravity is six times less on the Moon.
03

Choose the correct answer

From understanding the principles behind capillary action, it becomes clear that the height of water risen into the capillary tube on the surface of the Moon will be the same as on the surface of the Earth, which is \( h \). So, the correct answer is (c) \( h \)

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Most popular questions from this chapter

Water rises in a vertical capillary tube upto a height of \(2.0 \mathrm{~cm}\). If the tube is inclined at an angle of \(60^{\circ}\) with the vertical, then upto what length the water will rise in the tube (a) \(2.0 \mathrm{~cm}\) (b) \(4.0 \mathrm{~cm}\) (c) \(\frac{4}{\sqrt{3}} \mathrm{~cm}\) (d) \(2 \sqrt{2} \mathrm{~cm}\)

The lower end of a capillary tube is at a depth of \(12 \mathrm{~cm}\) and the water rises \(3 \mathrm{~cm}\) in it. The mouth pressure required to blow an air bubble at the lower end will be \(X \mathrm{~cm}\) of water column where \(X\) is \(\quad\) [CPMT 1989\(]\) (a) 3 (b) 9 (c) 12 (d) 15

The radii of two soap bubbles are \(R_{1}\) and \(R_{2}\) respectively. The ratio of masses of air in them will be (a) \(\frac{R_{1}^{3}}{R_{2}^{3}}\) (b) \(\frac{R_{2}^{3}}{R_{1}^{3}}\) (c) \(\left(\frac{P+\frac{4 T}{R_{1}}}{P+\frac{4 T}{R_{2}}}\right) \frac{R_{1}^{3}}{R_{2}^{3}}\) (d) \(\left(\frac{P+\frac{4 T}{R_{2}}}{P+\frac{4 T}{R_{1}}}\right) \frac{R_{2}^{3}}{R_{1}^{3}}\)

Several spherical drops of a liquid of radius \(r\) coalesce to form a single drop of radius \(R .\) If \(T\) is surface tension and \(V\) is volume under consideration, then the release of energy is (a) \(3 V T\left(\frac{1}{r}+\frac{1}{R}\right)\) (b) \(3 V T\left(\frac{1}{r}-\frac{1}{R}\right)\) (c) \(V T\left(\frac{1}{r}-\frac{1}{R}\right)\) (d) \(V T\left(\frac{1}{r^{2}}+\frac{1}{R^{2}}\right)\)

An air bubble in a water tank rises from the bottom to the top. Which of the following statements are true [Roorkee 2000] (a) Bubble rises upwards because pressure at the bottom is less than that at the top (b) Bubble rises upwards because pressure at the bottom is greater than that at the top (c) As the bubble rises, its size increases (d) As the bubble rises, its size decreases
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