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Chapter 12: Problem 31

A vessel whose bottom has round holes with diameter of \(1 \mathrm{~mm}\) is filled with water. Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is (Surface tension of water is \(75 \times 10^{-3} \mathrm{Nm}^{-1}\) and \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(3 \mathrm{~cm}\) (b) \(0.3 \mathrm{~cm}\) (c) \(3 \mathrm{~mm}\) (d) \(3 \mathrm{~m}\)

Short Answer

Expert verified
The maximum height is 3 cm. Option (a) is correct.

Step by step solution

01

Understanding the Problem

We have a vessel with round holes through which water can leak. Surface tension is the only force holding water from leaking.
02

Find the Critical Pressure

First, we compare the pressure due to the water column with the pressure from surface tension. The maximum pressure that surface tension can hold is \( \frac{2T}{r} \), where \( T \) is the surface tension and \( r \) is the radius of the holes.
03

Convert Diameter to Radius

The diameter of the hole is given as \(1\, \text{mm}\), thus the radius \( r \) is \( \frac{1}{2} \times 1\, \text{mm} = 0.5\, \text{mm} = 0.5 \times 10^{-3}\, \text{m} \).
04

Calculate Pressure by Surface Tension

Substitute \( T = 75 \times 10^{-3} \text{Nm}^{-1} \) and \( r = 0.5 \times 10^{-3}\, \text{m} \) into the formula for pressure. Hence, the maximum pressure due to surface tension \( P_t = \frac{2 \times 75 \times 10^{-3}}{0.5 \times 10^{-3}} = 300\, \text{Pa} \).
05

Hydrostatic Pressure Balance

The hydrostatic pressure from the water column is given by \( P_h = \rho g h \), where \( \rho \) is the density of water (approximately \( 1000\, \text{kg/m}^3 \)) and \( g = 10\, \text{ms}^{-2} \). Set \( P_h = 300\, \text{Pa} \).
06

Calculate the Maximum Height

Using the equation \( \rho g h = 300 \), solve for height \( h \). Thus, \( h = \frac{300}{1000 \times 10} = 0.03 \), which is \( 0.03\, \text{m} = 3\, \text{cm} \).
07

Select the Correct Answer

The maximum height of water without leakage is \(3\, \text{cm}\). Therefore, the correct answer is (a) \(3\, \text{cm}\).

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Key Concepts

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

Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. Imagine a column of water; the deeper you go, the more water is above you, applying pressure downward. This pressure increases with depth because more fluid presses down from above. This is important when filling a vessel with water, as the water's own weight can cause it to leak through holes. The concept of hydrostatic pressure is expressed through the formula: \( P_h = \rho gh \), where \( \rho \) represents the fluid's density, \( g \) is the gravitational acceleration, and \( h \) is the height of the fluid column. It's why, in our problem, the height of the water that doesn't cause leaking is crucial and is calculated to ensure pressure balance with the surface tension at the holes.
Density of Water
The density of water is a fundamental concept in physics and plays a crucial role in calculations involving fluids. It tells us how much mass of water is present in a given volume. The standard density of water is approximately \( 1000 \, \text{kg/m}^3 \). This means one cubic meter of water has a mass of 1000 kilograms. The density helps us understand how water behaves under different conditions, such as pressure changes. It is a key factor in the hydrostatic pressure formula, as it directly affects the pressure with which water pushes down. In this exercise, knowing the density allows us to accurately calculate the maximum height to which the vessel can be filled without water leakage.
Physics Problem Solving
Physics problem solving requires a step-by-step approach to understand and apply concepts. Here, let's think about how we methodically solve the problem of determining the maximum fill height of water in a vessel to prevent leakage.
  • First, understand the forces at play: surface tension and hydrostatic pressure.
  • Next, translate all physical properties into usable data, such as converting the diameter of the hole into a radius.
  • Utilize relevant formulas, like those for pressure due to surface tension and hydrostatic pressure.
  • Equate these pressures to solve for the unknown, in this case, the height of the water column.
By breaking down the problem, organizing data, and logically applying concepts and formulas, physics problems become a sequence of manageable steps. This systematic approach not only finds solutions but also builds a deeper understanding of the involved physical concepts.

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Most popular questions from this chapter

An aeroplane of \(\operatorname{mass} 3 \times 10^{4} \mathrm{~kg}\) and total wing area of \(120 \mathrm{~m}^{2}\) is in a level flight at some height. The difference in pressure between the upper and lower surfaces of its wings in kilo pascal is \(\left(g=10 \mathrm{~ms}^{-2}\right.\) ) (a) \(2.5\) (b) \(5.0\) (c) \(10.0\) (d) \(12.5\)

A body floats in a liquid contained in a beaker. If the whole system falls under gravity, then the upthrust on the body due to liquid is IUP SEE 2009] (a) equal to the weight of the body in air (b) equal to the weight of the body in liquid (c) zero (d) equal to the weight of the immersed part of the body

The rate of steady volume flow of water through a capillary tube of length \(l\) and radius \(r\) under a pressure difference of \(p\), is \(V\). This tube is connected with another tube of the same length but \(\underline{\text { half the }}\) radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is \(p\) ) (a) \(\frac{v}{16}\) (b) \(\frac{v}{17}\) (c) \(\frac{16 \mathrm{~V}}{17}\) (d) \(\frac{17 \mathrm{~V}}{16}\)

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A glass tube \(80 \mathrm{~cm}\) long and open at both ends is half immersed in mercury. Then the top of the tube is closed and it is taken out of the mercury. A column of mercury \(20 \mathrm{~cm}\) long then remains in the tube. The atmospheric pressure (in \(\mathrm{cm}\) of \(\mathrm{Hg}\) ) is (a) 90 (b) 75 (c) 60 (d) 45
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