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A hole is made at the bottom of a tank filled with water (density \(\left.=10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right)\). If the total pressure at the bottom of the tank is \(3 \mathrm{~atm}\left(1 \mathrm{~atm}=10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\), then the velocity of efflux is (1) \(\sqrt{400} \mathrm{~m} / \mathrm{s}\) (2) \(\sqrt{200} \mathrm{~m} / \mathrm{s}\) (3) \(\sqrt{600} \mathrm{~m} / \mathrm{s}\) (4) \(\sqrt{500} \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The velocity of efflux is \(\sqrt{400} \text{ m/s}\).

Step by step solution

01

Determine Pressure Due to Water Column

The total pressure at the bottom of the tank is given as 3 atm, which is equal to \(3 \times 10^{5} \text{ N/m}^2\). Out of this, 1 atm is the atmospheric pressure exerted on the water surface. Thus, the pressure due to the water column alone is \(3 \times 10^{5} - 1 \times 10^{5} = 2 \times 10^{5} \text{ N/m}^2\).
02

Apply Bernoulli’s Equation for Efflux Velocity

Using Torricelli's theorem, the velocity of the efflux \(v\) from a hole can be found using the equation derived from Bernoulli's equation: \(v = \sqrt{2gh}\), where \(h\) is the height of the water column. Since \(P = \rho gh\), \(h\) can be found from \(2 \times 10^{5} = 10^{3} \times 9.8 \times h\), leading to \(h = \frac{2 \times 10^{5}}{9.8 \times 10^{3}} \approx 20.41 \text{ m}\).
03

Calculate Efflux Velocity

Now, we substitute \(h\) into the equation for velocity. Therefore, \(v = \sqrt{2 \times 9.8 \times 20.41}\). Thus, \(v \approx \sqrt{400} \text{ m/s}\).

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Key Concepts

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

Bernoulli's Equation
Bernoulli's equation is a principle that helps us understand the behavior of fluids, such as water or air, in motion. It states that in a streamline flow, the total mechanical energy of the fluid remains constant. This energy is a combination of three parts:
  • Pressure Energy: This is related to the pressure exerted by the fluid.
  • Kinetic Energy: This deals with the speed of the fluid.
  • Potential Energy: This is dependent on the height or elevation of the fluid above a reference point.
Bernoulli's equation is expressed as:\[ P + \frac{1}{2} \rho v^2 + \rho gh = \,\text{constant} \]Where, \(P\) is the pressure energy, \(\rho\) is the fluid density, \(v\) is the velocity, \(g\) is the acceleration due to gravity, and \(h\) is the height. In practical terms, this equation is incredibly useful for solving various fluid mechanics problems, including understanding how water flows out of a tank when a hole is introduced.
For example, in the exercise, knowing the pressure at the bottom helps us determine the velocity at which water exits the tank using the concepts embedded within Bernoulli's equation.
Torricelli's Theorem
Torricelli's Theorem is an important principle derived from Bernoulli's equation, specifically used to calculate the speed of liquid flowing out of an opening. It essentially transforms the energy principles from Bernoulli's equation into a more focused formula:\[ v = \sqrt{2gh} \]Here, \(v\) denotes the velocity of the efflux, \(g\) is the gravitational acceleration, and \(h\) is the height of the fluid above the hole. Torricelli discovered that the velocity of fluid flowing out from an open tank or container is the same as if the liquid were falling freely from the height \(h\).
By applying Torricelli's theorem, we can see that with the calculated height of around 20.41 meters from the given pressure, we find how fast the water shoots out from a hole at the bottom of the tank. Torricelli's theorem simplifies the procedure significantly and is an elegant way of relating fluid motion to the gravitational effects on the liquid column.
Pressure Calculation
In fluid mechanics, pressure calculation is a vital part of understanding how fluids behave under different conditions. Pressure, in simple terms, is the force applied perpendicular to the surface of an object per unit area.
For the problem at hand, pressure acts in determining the flow and velocity of water. Total pressure at the tank's bottom includes both atmospheric pressure and the pressure due to the weight of the water column. In our scenario, it's mentioned as 3 atm, which means:
  • 1 atm is the atmospheric pressure, always present.
  • The remaining 2 atm is due to the water's weight above the hole.
This breakdown allows us to find the pressure due solely to the water, helping translate that 'stored energy' into the height of the water column, \(h\). The height \(h\) was crucial to obtaining the velocity of efflux using Torricelli's theorem. Without correctly calculating pressure contributions, the right height and subsequent velocity wouldn't be accurately determined. Thus, understanding pressure calculations equips us to tackle various fluid dynamics problems proficiently.

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Most popular questions from this chapter

If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid for calculation of pressure, effective \(g\) is used. A closed box with horizontal base \(6 \mathrm{~m}\) by \(6 \mathrm{~m}\) and a height \(2 \mathrm{~m}\) is half filled with liquid. It is given a constant horizontal acceleration \(g / 2\) and vertical downward acceleration \(g / 2\). What is the value of vertical acceleration of box for given horizontal acceleration \((g / 2)\), so that no part of the bottom of the box is exposed? (1) \(g / 2\) upwards (2) \(\mathrm{g} / 4\) downwards (3) \(g / 4\) upwards (4) not possible

A goldsmith desires to test the purity of a gold ornament suspected to the mixed with copper. The ornament weighs \(0.25 \mathrm{~kg}\) in air and is observed to displace \(0.015\) litre of water when immersed in it. Densities of gold and copper with respect to water are, respectively, \(19.3\) and \(8.9\). The approximate percentage of copper in the ornament is (1) \(5 \%\) (2) \(10 \%\) (3) \(15 \%\) (4) \(25 \%\)

A solid cone of height \(H\) and base radius \(H / 2\) floats in a liquid of density \(\rho\). It is hanging from the ceiling with the help of a string. The force by the fluid on the curved surface of the cone is \(\left(P_{0}=\right.\) atmospheric pressure \()\) (1) \(\pi H^{2}\left(\frac{P_{0}}{4}+\frac{\rho g H}{3}\right)\) (2) \(\pi H^{2}\left(\frac{P_{0}}{4}+\frac{\rho g H}{6}\right)\) (3) \(\frac{\pi H^{2}}{4}\left(\frac{P_{0}}{4}+\rho g H\right)\) (4) \(\frac{\pi H^{2}}{4}\left(P_{0}+\rho g H\right)\)

A block of wood is floating in water in a closed vessel as shown in the figure. The vessel is connected to an air pump. When more air is pushed into the vessel, the block of wood floats with (neglect compressibility of water) (1) larger part in the water (2) smaller part in the water (3) same part in the water (4) at some instant it will sink

A vessel contains oil (density \(=0.8 \mathrm{~g} / \mathrm{cm}^{3}\) ) over mercury (density \(\left.=13.6 \mathrm{~g} / \mathrm{cm}^{3}\right)\). A uniform sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in \(\mathrm{g} / \mathrm{cm}^{3}\) is (1) \(3.3\) (2) \(6.4\) (3) \(7.2\) (4) \(12.8\)

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