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Chapter 8: Problem 50

In a surface tension experiment with a capillary tube, water rises up to \(0.1 \mathrm{~m}\). If the same experiment is repeated on an artificial satellite, which is revolving around the earth, water will rise in the capillary tube up to a height of (A) \(0.1 \mathrm{~m}\) (B) \(0.2 \mathrm{~m}\) (C) \(0.98 \mathrm{~m}\) (D) Full length of tube

Short Answer

Expert verified
In an artificial satellite revolving around the earth, the effective gravitational acceleration becomes zero, implying weightless conditions. In these conditions, water would try to rise up to an infinite height in the capillary tube. However, due to the tube's length limit, the water will rise up to the full length of the tube (D).

Step by step solution

01

Recall the Jurin's Law formula for capillary rise

Jurin's Law states that the height 'h' to which a liquid rises or falls in a capillary tube in equilibrium is given by the formula: \( h = \dfrac{2 \times S \times \cos \theta}{\rho \times g \times r} \) Where: \(S\) = Surface tension of the liquid \(\theta\) = Contact angle between the solid and the liquid \(\rho\) = Density of the liquid \(g\) = Acceleration due to gravity \(r\) = Capillary tube radius
02

Consider the gravitational force on the satellite

Since the satellite is moving around the earth, it experiences a centripetal force towards the center of the earth. This force equals the gravitational force acting on the satellite: \(F_g = m \times g_s\) Where: \(F_g\) = Gravitational force \(m\) = Mass of the satellite \(g_s\) = Effective gravitational acceleration on the satellite Also, the centripetal force acting on the satellite is given by: \(F_c = \dfrac{m \times v^2}{R}\) Where: \(F_c\) = Centripetal force \(v\) = Orbital velocity of the satellite \(R\) = Distance from the center of the earth to the satellite Since \(F_g = F_c\), we get: \( m \times g_s = \dfrac{m \times v^2}{R} \) Thus, the effective gravitational acceleration on the satellite is: \( g_s = \dfrac{v^2}{R} \)
03

Compare the capillary rise on earth and the satellite

Since we're only concerned with the height 'h' to which the liquid rises in the capillary tube, and since all other factors in the Jurin's Law formula remain the same (surface tension, contact angle, density, and radius), we can compare the ratio of gravitational accelerations (\(g_{earth}\) and \(g_{satellite}\)) to find the ratio of heights: \( \dfrac{h_{satellite}}{h_{earth}} = \dfrac{g_{earth}}{g_{satellite}} \) Given that on the earth's surface \(h_{earth} = 0.1\) m, we can solve for \(h_{satellite}\): \( h_{satellite} = h_{earth} \times \dfrac{g_{earth}}{g_{satellite}} \)
04

Apply the condition of weightlessness on the satellite

In an orbiting satellite that is in free fall, the effective gravitational acceleration becomes zero (\(g_s = 0\)). This implies weightlessness: \( h_{satellite} = h_{earth} \times \dfrac{g_{earth}}{0} \) This indicates that in weightless conditions, water would try to rise up to an infinite height in the capillary tube. However, in reality, the water would not rise infinitely due to the tube's limit on length. So, the answer is: (D) Full length of the tube

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Most popular questions from this chapter

The surface tension of a liquid is \(5 \mathrm{~N} / \mathrm{m}\). If a film is held on a ring of area \(0.02 \mathrm{~m}^{2}\), its total surface energy is about (A) \(5 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(2 \times 10^{-1} \mathrm{~J}\) (D) \(3 \times 10^{-1} \mathrm{~J}\)

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