Calculate the water pressure and the thrust at the bottom of a tank whose length, width and depth are $2\,m$, $1.5\,m$, $1\,m$ respectively. Density of water is $1000\,kg\,{m^{ - 3}}$.
Answer
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Hint: The water pressure at the bottom of a tank is the force exerted by the mass of water on the surface per unit area of the surface. The thrust is the perpendicular force exerted by water downwards, on the surface.
Complete step-by-step answer:
Liquid pressure(P) depends upon
Height of the water column (depth) which is denoted by ‘h’
Density of liquid $(\rho )$
Acceleration due to gravity (g) which equals to $9.8\,m/{s^2}$
Thus, by using pressure formula
$P\, = \,h\, \times \,\rho \, \times \,g$
$P\, = \,1\, \times \,1000\, \times \,9.8$
$P\, = \,9800\,N/{m^2}$
For calculating the thrust acting on the surface, we will first calculate the area of the surface.
Area $ = $ length $ \times $ width
Area $ = \,2\, \times \,1.5\,{m^2}$
Area $ = \,3\,{m^2}$
Now we know that
${\text{Pressure}}\,{\text{ = }}\dfrac{{{\text{Force(Thrust)}}}}{{{\text{Area}}}}$
${\text{Thrust(Force)}}\,{\text{ = }}\,{\text{Pressure}}\, \times \,{\text{Area}}$
Now Substituting values of Pressure and Area from above, we get
${\text{Thrust}}\,{\text{ = }}\,9800\, \times \,3\,N$
${\text{Thrust}}\,{\text{ = }}\,29400\,N$
So the Pressure at the bottom of the tank is $P\, = \,9800\,N/{m^2}$ and ${\text{Thrust}}\,{\text{ = }}\,29400\,N$.
Additional Information: Liquid pressure is also called hydrostatic pressure. Hydrostatic pressure is the pressure that is exerted by a fluid at equilibrium at a given point within the fluid, due to gravity. This pressure is independent of the shape of the container. The pressure due to a liquid column is the force exerted by a liquid column downwards per unit surface area. This pressure is directly proportional to height of the liquid column, higher is the column, larger is the pressure. Thrust is a kind of force, necessarily perpendicular to the surface area.
Note: This is not the total pressure exerted by liquid on the container as the atmosphere above the liquid also exerts atmospheric pressure.
Therefore,
${P_{Total}}\, = \,{P_{hydrostatic\,}}\, + \,{P_{atm}}$
The units of pressure must be checked. The units are $N/{m^2}$, which is also called Pa [Pascal], torr, atm.
Complete step-by-step answer:
Liquid pressure(P) depends upon
Height of the water column (depth) which is denoted by ‘h’
Density of liquid $(\rho )$
Acceleration due to gravity (g) which equals to $9.8\,m/{s^2}$
Thus, by using pressure formula
$P\, = \,h\, \times \,\rho \, \times \,g$
$P\, = \,1\, \times \,1000\, \times \,9.8$
$P\, = \,9800\,N/{m^2}$
For calculating the thrust acting on the surface, we will first calculate the area of the surface.
Area $ = $ length $ \times $ width
Area $ = \,2\, \times \,1.5\,{m^2}$
Area $ = \,3\,{m^2}$
Now we know that
${\text{Pressure}}\,{\text{ = }}\dfrac{{{\text{Force(Thrust)}}}}{{{\text{Area}}}}$
${\text{Thrust(Force)}}\,{\text{ = }}\,{\text{Pressure}}\, \times \,{\text{Area}}$
Now Substituting values of Pressure and Area from above, we get
${\text{Thrust}}\,{\text{ = }}\,9800\, \times \,3\,N$
${\text{Thrust}}\,{\text{ = }}\,29400\,N$
So the Pressure at the bottom of the tank is $P\, = \,9800\,N/{m^2}$ and ${\text{Thrust}}\,{\text{ = }}\,29400\,N$.
Additional Information: Liquid pressure is also called hydrostatic pressure. Hydrostatic pressure is the pressure that is exerted by a fluid at equilibrium at a given point within the fluid, due to gravity. This pressure is independent of the shape of the container. The pressure due to a liquid column is the force exerted by a liquid column downwards per unit surface area. This pressure is directly proportional to height of the liquid column, higher is the column, larger is the pressure. Thrust is a kind of force, necessarily perpendicular to the surface area.
Note: This is not the total pressure exerted by liquid on the container as the atmosphere above the liquid also exerts atmospheric pressure.
Therefore,
${P_{Total}}\, = \,{P_{hydrostatic\,}}\, + \,{P_{atm}}$
The units of pressure must be checked. The units are $N/{m^2}$, which is also called Pa [Pascal], torr, atm.
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