
A manometer connected to a closed tap reads \[4.5 \times {10^5}\,N{m^2}\]. When the tap is opened, the reading of the manometer falls to \[4.0 \times {10^5}\,N{m^2}\], then finds the velocity of the flow of water.
A. \[7\,m{s^{ - 1}}\]
B. \[8\,m{s^{ - 1}}\]
C. \[9\,m{s^{ - 1}}\]
D. \[10\,m{s^{ - 1}}\]
Answer
162.9k+ views
Hint: Before going to solve this question let us understand the streamlined flow. The streamlined flow is also referred to as laminar flow. When there is no turbulence and fluctuations in velocity then it is called streamlined flow. It is a path traced by a particle in a flow.
Formula Used:
Bernoulli’s equation for the unit mass of a liquid is,
\[\dfrac{P}{\rho } + \dfrac{1}{2}{v^2} = \text{constant}\]
Where, \[P\] is pressure, \[\rho \] is density and \[v\] is velocity.
Complete step by step solution:
When a manometer is connected to a closed tap it reads \[{P_1} = 4.5 \times {10^5}\,N{m^2}\]. When the value is opened, the reading of the manometer falls to \[{P_2} = 4.0 \times {10^5}\,N{m^2}\], then we need to find the velocity of the flow of water, that is \[v\]. As the liquid starts flowing, its pressure energy decreases, that is,
\[\dfrac{1}{2}{v^2} = \dfrac{{{P_1} - {P_2}}}{\rho } \\ \]
\[\Rightarrow {v^2} = \dfrac{{2\left( {{P_1} - {P_2}} \right)}}{\rho }\]
We know the density of water, that is \[1000\,kg{m^3}\]. Substitute the value of pressure and density in above equation, we get,
\[{v^2} = \dfrac{{2\left( {4.5 \times {{10}^5} - 4.0 \times {{10}^5}} \right)}}{{{{10}^3}}} \\ \]
\[\Rightarrow {v^2} = 100 \\ \]
\[\therefore v = 10\,m{s^{ - 1}}\]
Therefore, the velocity of the flow of water is \[10\,m{s^{ - 1}}\].
Hence, option D is the correct answer.
Additional information: Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. The Archimedes principle states that when an object is immersed in a fluid, it experiences some buoyant force that is equal in magnitude to the force of gravity on the displaced fluid. It is also referred to as the law of buoyancy. The weight of the displaced fluid is equal to the subtraction of the weight of an object in a vacuum and the weight of an object in a fluid.
Note: Here, one should notice that the Archimedes principle is only valid for fluids, where we observe the buoyant force. And, the displaced fluid is equal to the weight of the object immersed in the water. Here, gravity also plays an important role.
Formula Used:
Bernoulli’s equation for the unit mass of a liquid is,
\[\dfrac{P}{\rho } + \dfrac{1}{2}{v^2} = \text{constant}\]
Where, \[P\] is pressure, \[\rho \] is density and \[v\] is velocity.
Complete step by step solution:
When a manometer is connected to a closed tap it reads \[{P_1} = 4.5 \times {10^5}\,N{m^2}\]. When the value is opened, the reading of the manometer falls to \[{P_2} = 4.0 \times {10^5}\,N{m^2}\], then we need to find the velocity of the flow of water, that is \[v\]. As the liquid starts flowing, its pressure energy decreases, that is,
\[\dfrac{1}{2}{v^2} = \dfrac{{{P_1} - {P_2}}}{\rho } \\ \]
\[\Rightarrow {v^2} = \dfrac{{2\left( {{P_1} - {P_2}} \right)}}{\rho }\]
We know the density of water, that is \[1000\,kg{m^3}\]. Substitute the value of pressure and density in above equation, we get,
\[{v^2} = \dfrac{{2\left( {4.5 \times {{10}^5} - 4.0 \times {{10}^5}} \right)}}{{{{10}^3}}} \\ \]
\[\Rightarrow {v^2} = 100 \\ \]
\[\therefore v = 10\,m{s^{ - 1}}\]
Therefore, the velocity of the flow of water is \[10\,m{s^{ - 1}}\].
Hence, option D is the correct answer.
Additional information: Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. The Archimedes principle states that when an object is immersed in a fluid, it experiences some buoyant force that is equal in magnitude to the force of gravity on the displaced fluid. It is also referred to as the law of buoyancy. The weight of the displaced fluid is equal to the subtraction of the weight of an object in a vacuum and the weight of an object in a fluid.
Note: Here, one should notice that the Archimedes principle is only valid for fluids, where we observe the buoyant force. And, the displaced fluid is equal to the weight of the object immersed in the water. Here, gravity also plays an important role.
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