Fluid Statics
Pascals Law
Learning Objectives:

State Pascal’s law.

Apply Pascal’s Principle to describe pressure behavior in static fluids
The normal stresses at any point in a fluid element at rest are directed towards the point from all directions and they are of the equal magnitude.
Units and scales of Pressure Measurement:
Pascal \((N/m^2)\) is the unit of pressure. Pressure is usually expressed with reference to either absolute zero pressure (a complete vacuum) or local atmospheric pressure.

Absolute pressure: It is the difference between the value of the pressure and the absolute zero pressure.

Gauge pressure: It is the difference between the value of the pressure and the local atmospheric pressure.
At sealevel, the international standard atmosphere has been chosen as = 101.32 \(kN/m^2\)
Solved Example: 61101
1 GPa is equivalent of:
A. $10^9 N/mm^2$
B. $10^6 N/m^2$
C. $10^3 N/mm^2$
D. $10^3 N/m^2$
Correct Answer: C
Solved Example: 61102
Liquids transmit pressure equally in all the directions. This is according to:
A. Boyle's law
B. Archimedes principle
C. Pascal's law
D. Newton's formula
Boyle's law states that the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature; i.e., in equation form, pv = k, a constant.
Archimedes' principle states: An object immersed in a fluid experiences a buoyant force that is equal in magnitude to the force of gravity on the displaced fluid.
There is no such thing called Newton's formula. We have Newton's laws and Newton's equation of motion.
Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container. The pressure at any point in the fluid is equal in all directions.
Correct Answer: C
Solved Example: 61103
A hydraulic press has a ram of 15 cm diameter and plunger of 1.5 cm. It is required to lift a weight of 1000 kg. The force required on the plunger is equal to: (NPCIL Stipendiary Trainee ME 2018)
A. 10 kg
B. 100 kg
C. 1000 kg
D. 10,000 kg
Correct Answer: A
Solved Example: 61104
The hydraulic lift is based on which of the following? (SSC Scientific Assistant Physics Nov 2017 Shift II)
A. Bernoulli's principle
B. Pascal's Law
C. Archimedes' principle
D. Boyle's law
Correct Answer: B
Solved Example: 61105
A hydraulic press has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 30 kN. Find the force required at the plunger. (SSC JE ME Oct 2020 Evening)
A. 975 N
B. 1075 N
C. 675 N
D. 875 N
Correct Answer: C
Solved Example: 61106
As per Pascal's law: (SSC JE CE Oct 2020 Evening)
A. Intensity of pressure at a point in a moving fluid is equal in the direction of applied force.
B. Intensity of pressure at a point in a static fluid is equal in all directions.
C. Intensity of pressure at a point in a moving fluid is equal in all directions
D. Intensity of pressure at a point in a static fluid is equal in the direction of applied force.
Correct Answer: B
Solved Example: 61107
_____ gives an idea about the pressure in a static fluid. (SSC JE ME Oct 2020 Evening)
A. Grashof's law
B. Weber's law
C. Pascal's law
D. Newton's law
Correct Answer: C
Solved Example: 61108
Which of the following statements is INCORRECT in the context of static fluid pressure? (SSC JE ME Mar 2021 Evening)
A. At any particular depth, it changes with direction.
B. It exerts force normal to the walls of the container.
C. It is independent of shape or area of the container.
D. It is directly proportional to depth of fluid.
Correct Answer: A
Solved Example: 61109
A hydraulic press has a ram of 300 mm diameter and a plunger of 45 mm diameter. When the force applied at the plunger is 50 N, the weight lifted by the hydraulic press will be nearly: (ESE Civil 2020)
A. 2133 N
B. 2223 N
C. 2316 N
D. 2406 N
Correct Answer: B
Pressure Measurement
Learning Objectives:

Apply the principles of manometer to calculate pressure.

Differentiate between absolute pressure and gauge pressure.
Piezometer Tube:
The direct proportional relation between gauge pressure and the height h for a fluid of constant density enables the pressure to be simply visualized in terms of the vertical height, \[h = \frac{P}{\rho g}\]
The height h is termed as pressure head corresponding to pressure p. For a liquid without a free surface in a closed pipe, the pressure head \(\displaystyle h = \frac{P}{\rho g}\) at a point corresponds to the vertical height above the point to which a free surface would rise, if a small tube of sufficient length and open to atmosphere is connected to the pipe.
Such a tube is called a piezometer tube, and the height h is the measure of the gauge pressure of the fluid in the pipe. If such a piezometer tube of sufficient length were closed at the top and the space above the liquid surface were a perfect vacuum, the height of the column would then correspond to the absolute pressure of the liquid at the base. This principle is used in the well known mercury barometer to determine the local atmospheric pressure.
Barometer:
Barometer is used to determine the local atmospheric pressure. Mercury is employed in the barometer because its density is sufficiently high for a relative short column to be obtained. and also because it has very small vapour pressure at normal temperature. High density scales down the pressure head(h) to represent same magnitude of pressure in a tube of smaller height.
Manometers:
A manometer is also frequently used to measure the pressure difference, in course of flow, across a restriction in a horizontal pipe.
Inclined Tube Manometer:
For accurate measurement of small pressure differences by an ordinary utube manometer, it is essential that the ratio \(\displaystyle \frac{r_{m}}{r_{w}}\) should be close to unity. This is not possible if the working fluid is a gas; also having a manometric liquid of density very close to that of the working liquid and giving at the same time a well defined meniscus at the interface is not always possible. For this purpose, an inclined tube manometer is used.
If the transparent tube of a manometer, instead of being vertical, is set at an angle \(\displaystyle \theta\) to the horizontal, then a pressure difference corresponding to a vertical difference of levels x gives a movement of the meniscus s = x/sin q along the slope.
If \(\displaystyle \theta\) is small, a considerable magnification of the movement of the meniscus may be achieved.
Angles less than 5\(^\circ\) are not usually satisfactory, because it becomes difficult to determine the exact position of the meniscus.
Inverted Tube Manometer:
For the measurement of small pressure differences in liquids, an inverted Utube manometer is used.
Solved Example: 61201
Differential manometer is used to measure: (Based on JKSSB JE CE Oct 2021Shift I)
A. Pressure in pipes, channels etc.
B. Atmospheric pressure
C. Very low pressure
D. Difference of pressure between two points
Correct Answer: D
Solved Example: 61202
Pitot tube is used for measurement of: (ISRO RAC 2017)
A. Pressure
B. Flow
C. Velocity
D. Discharge
Correct Answer: C
Solved Example: 61203
1 bar pressure is nearly equal to:
A. 1 mm Hg
B. 750 mm Hg
C. 100 mm Hg
D. 13.4 mm Hg
1 bar = 1 $\times$ 10$^5$ N/m$^2$ \[750\ mm\ Hg = 750 \times 10^{3} \times 13600 \times 9.81 = 100062\ N/m^2\]
Correct Answer: B
Solved Example: 61204
Mercury is often used in barometer because: (Gujrat Engg Services 2017 Civil Part II)
A. It has good viscosity
B. The height of barometer will be less
C. Its vapour pressure is so low that it may be neglected
D. Both (B) and (C)
Mercury is much heavier than water (13.6 times) hence the column height required is only 760 mm as compared to 10.33m of water column. Also, it has very less vapor pressure hence the reading will be accurate. Mercury has in fact less viscosity, hence it is faster to react to the changes in pressure.
Correct Answer: D
Solved Example: 61205
Barometer is used to measure: (CTET Sept 2014 Paper II)
A. Pressure in pipes, channels etc.
B. Atmospheric pressure
C. Very low pressure
D. Difference of pressure between two points
Barometer is used to measure atmospheric pressure. To measure difference of pressure between two points, Utube manometer is used.
Correct Answer: B
Solved Example: 61206
Select the correct statement:
A. Local atmospheric pressure depends upon elevation of locality only
B. Standard atmospheric pressure is the mean local atmospheric pressure at sea level
C. Local atmospheric pressure is always below standard atmospheric pressure
D. A barometer reads the difference between local and standard atmospheric pressure
Correct Answer: B
Solved Example: 61207
Which of the following is correct?
A. Absolute pressure = Gauge pressure + Atmospheric pressure
B. Gauge pressure = Absolute pressure + Atmospheric pressure
C. Atmospheric pressure = Absolute pressure + Gauge pressure
D. Absolute pressure = Gauge pressure  Atmospheric pressure
Absolute pressure is measured with reference to perfect vacuum, whereas gauge pressure is measured with reference to atmospheric pressure. Hence the difference between absolute pressure and gauge pressure is equal to atmospheric pressure.
Correct Answer: A
Solved Example: 61208
The pressure in Pascals at a depth of 1 m below the free surface of a body of water will be equal to:
A. 1 Pa
B. 981 Pa
C. 9810 Pa
D. 98,100 Pa
\[P = h \rho g = 1 \times 1000 \times 9.81 = 9810\ Pa\]
Correct Answer: C
Solved Example: 61209
The absolute pressure exerted on a diver at 30 m below the free surface of the sea will be :(Take barometric pressure = 101 kPa, specific weight of seawater = 10.3 $kN/m^3$
A. 410 kPa
B. 309 kPa
C. 101 kPa
D. 187.8 kPa
\[P = P_{atm} + h\rho g = 101 \times 10^3 + 30 \times 10.3\times 10^3 =\ 410\ kPa\]
Correct Answer: A
Solved Example: 61210
One of the differences between absolute pressure and gauge pressure is: (Prof. Prashant More's notes of Fluid MechanicsII)
A. Difference between gauge pressure and absolute pressure changes based on viscosity.
B. Absolute pressure can be positive or negative but gauge pressure is always positive.
C. Absolute pressure can be calculated by adding atmospheric pressure to gauge pressure.
D. Gauge pressure is measured from perfect vacuum whereas absolute pressure is measured from atmospheric pressure.
Let us study these options individually.
Option A: Difference between gauge pressure and absolute pressure is the atmospheric pressure.
Option B: Absolute pressure can be positive or negative but gauge pressure can be positive or negative.
Option C: Already correct.
Option D: Gauge pressure is measured from atmospheric pressure whereas absolute pressure is measured from perfect vacuum.
Correct Answer: C
Solved Example: 61211
A lake has maximum depth of 60 m. If the mean atmospheric pressure in the lake region is 91 kPa and the unit weight of the lake water is 9790 $N/m^3$, the absolute pressure (in kPa, round off to two decimal places) at the maximum depth of the lake is: (GATE Civil 2021)
A. 587.4
B. 678.4
C. 778.4
D. 785.4
Absolute pressure at the maximum depth of the lake \begin{align*} &= P_{atm} + \rho g h\\ &= 91 + \dfrac{9790 \times 60}{1000}\\ &= 678.4\ kPa \end{align*}
Correct Answer: B
Hydrostatic Forces on Submerged Surfaces
Learning Objectives:

Apply the principles of hydrostatics to derive an expression for buoyancy.

Determine the magnitude, direction and location of the resultant hydrostatic force acting on any submerged surface.
Center of Pressure is defined as the point of application of the resultant pressure. The depth of center of pressure of an immersed surface from free surface of the liquid is given by:

For vertically immersed surface, \[y_{CP} = y_C + \dfrac{I_{xC}}{y_C A}\] y\(_C\) = slant distance from liquid surface to the centroid of area
y\(_C\) = \(\dfrac{h_C}{\sin \theta}\)
h\(_C\) = vertical distance from liquid surface to centroid of area
y\(_{CP}\) = slant distance from liquid surface to center of pressure
\(\theta\) = angle between liquid surface and edge of submerged surface
I\(_{xC}\) = moment of inertia about the centroidal xaxis
This equation indicates that the centre of the pressure is always below the centroid of the submerged plane.
Solved Example: 61301
The center of pressure location and the toal pressure force on a 1 m $\times$ 1 m gate immersed vertically at a depth of 2 m below the free water surface will be:
A. 2.5m, 21819 N
B. 2.5m, 44819 N
C. 2.53m, 21234 N
D. 2.53m, 24525 N
\begin{align*} S_{c} &=\dfrac {I_{G}}{A\overline {x}}+\overline {x}\\ &=\dfrac {\dfrac {bd^{3}}{12}}{A\overline {x}}+\overline {x}\\ &=\dfrac {\left( 1\right) \left( 1\right) ^{3}}{12\left( 1\right) \left( 1\right) \left( 2.5\right) }+2.5\\ &=\dfrac {1}{30}+2.5\\ &=2.53\ m \end{align*} \[\mathrm{Total}\ P =wA\overline {x} =9810\left( 1\right) \left( 1\right) \left( 2.5\right) =24525\ N\]
Correct Answer: D
Solved Example: 61302
The total pressure on the surface of a vertical sluice gate 2 m wide x 1 m depth with its top 2 m surface being 0.5 m below the water level will be: (Based on CIL MT Mech 2020)
A. 4905 $N/m^2$
B. 9810 $N/m^2$
C. 14715 $N/m^2$
D. 19600 $N/m^2$
\[\bar{h} = 0.5 + \dfrac{1}{2} = 1\] \[ P = \rho g \bar{h} = 1000 \times 9.81 \times 1 = 9810\ N/m^2\]
Correct Answer: B
Solved Example: 61303
A completely submerged body, with center of gravity G and center of buoyancy B, will be stable if: (GATE ME 2014)
A. G above B
B. G below B
C. G coincident with B
D. Independent of G and B
Correct Answer: B
Solved Example: 61304
Metacentric height is given as the distance between: (NPCIL Stipendiary Trainee ME 2018)
A. The center of gravity of the body and the meta center
B. The center of gravity of the body and the center of buoyancy
C. The center of gravity of the body and the center of pressure
D. Center of buoyancy and metacentre
Correct Answer: A
Solved Example: 61305
The center of gravity of the volume of the liquid displaced by an immersed body is called: (SSC JE ME Paper 10 Dec 2020)
A. Metacenter
B. Center of pressure
C. Center of buoyancy
D. Center of gravity
Correct Answer: C
Solved Example: 61306
In an immersed body, centre of pressure is: (NPCIL SA/ST ME GJ Nov 2019 Shift I)
A. At the centre of gravity
B. Above the centre of gravity
C. Below be centre of gravity
D. Could be above or below e.g. depending on density of body and liquid
Correct Answer: C
Solved Example: 61307
The centre of gravity of the volume of the liquid displaced by an immersed body is called:
A. Centre of gravity
B. Centre of pressure
C. Metacentre
D. Centre of buoyancy
Correct Answer: D
Solved Example: 61308
“Any object, wholly or partly immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object”. This is known as the _____________.
A. Bernoulli’s Principle
B. Torricelli’s Principle
C. Archimedes’ Principle
D. Pascal’s Principle
Correct Answer: C
Solved Example: 61309
Resultant pressure of the liquid in the case of an immersed body acts through: (ISRO VSSC Tech. Asst. Mech Feb 2015)
A. Centre of gravity
B. Centre of pressure
C. Metacentre
D. Centre of buoyancy
Correct Answer: B
Solved Example: 61310
It is the upward force on an object produced by the surrounding fluid in which it is fully or partially immersed. (RRB ALP Aug 2018 Shift I)
A. Archimedes’ force
B. Fluid pressure
C. Buoyancy
D. Weight reaction
Correct Answer: C
Solved Example: 61311
For a body floating in a liquid the normal pressure exerted by the liquid acts at:
A. bottom surface of the body
B. C.G. of the body
C. Metacentre
D. All points on the surface of the body
Correct Answer: D
Solved Example: 61312
Choose the wrong statement.
A. Any weight, floating or immersed in a liquid, is acted upon by a buoyant force
B. Buoyant force is equal to the weight of the liquid displaced
C. The point through which buoyant force acts, is called the center of buoyancy
D. Center of buoyancy is located above the center of gravity of the displaced liquid.
Correct Answer: D
Solved Example: 61313
According to the principle of buoyancy a body totally or partially immersed in a fluid will be lifted up by a force equal to:
A. The weight of the body
B. More than the weight of the body
C. Less than the weight of the body
D. Weight of the fluid displaced by the body
Correct Answer: D
Solved Example: 61314
When a body floating in a liquid, is displaced slightly, it oscillates about: (DSSSB JE CE Oct 2019 Shift I)
A. C.G. of body
B. Center of pressure
C. Center of buoyancy
D. Metacentre
Correct Answer: D
Solved Example: 61315
Buoyant force is:
A. Resultant force acting on a floating body
B. Equal to the volume of liquid displaced
C. Force necessary to keep a body in equilibrium
D. The resultant force on a body due to the fluid surrounding it
Correct Answer: D
Solved Example: 61316
The two important forces for a floating body are: (SSC JE ME March 2017)
A. Buoyancy, gravity
B. Buoyancy, pressure
C. Buoyancy, inertial
D. Inertial, gravity
Correct Answer: A
Archimedes Principle and Buoyancy
Learning Objectives:

Define buoyant force.

State Archimedes’ principle.

Understand why objects float or sink.

Understand the relationship between density and Archimedes’ principle

The buoyant force exerted on a submerged or floating body is equal to the weight of the fluid displaced by the body.

A floating body displaces a weight of fluid equal to its own weight; i.e., a floating body is in equilibrium.
The center of buoyancy is located at the centroid of the displaced fluid volume. In the case of a body lying at the interface of two immiscible fluids, the buoyant force equals the sum of the weights of the fluids displaced by the body.
Solved Example: 61401
The point in the immersed body through which the resultant pressure of the liquid may be taken to act is known as: (SJVNL JE Mech 2018)
A. Meta center
B. Center of pressure
C. Center of buoyancy
D. Center of gravity
Correct Answer: B
Solved Example: 61402
The upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. This definition is according to: (Based on SSC CPO Tier I July 2017 Shift I)
A. Equilibrium of a floating body
B. Archimedes' principle
C. Bernoulli's theorem
D. Metacentric principle
Correct Answer: B
Solved Example: 61403
The pressure in the air space above an oil (sp. gr. 0.8) surface in a tank is 0.1 $kg/cm^2$. The pressure at 2.5 m below the oil surface will be:
A. 2 metres of water column
B. 3 metres of water column
C. 3.5 metres of water column
D. 4 m of water column
Correct Answer: B
Solved Example: 61404
A piece of wood having weight 5 kg floats in water with 60% of its volume under the liquid. The specific gravity of wood is: (ISRO URSC Tech Asst. Mech Mar 2019)
A. 0.83
B. 0.6
C. 0.4
D. 0.3
Correct Answer: B
Solved Example: 61405
Flow of water in a pipe about 3 metres in diameter can be measured by:
A. Orifice plate
B. Venturi
C. Rotameter
D. Pitot tube
Correct Answer: D
Solved Example: 61406
Which of the following instrument can be used for measuring speed of an aeroplane? (Based on UP Jal Nigam Civil 2016)
A. Venturimeter
B. Orifice plate
C. Hot wire anemometer
D. Pitot tube
Correct Answer: D
Solved Example: 61407
A rock of weight 10 N suspended by a string is lowered into water, displacing water of weight 3 N. Determine the tension in the string.
A. 13 N
B. 7 N
C. 10 N
D. 3 N
Correct Answer: B
Solved Example: 61408
If the buoyancy of an object exceeds its weight, the object:
A. Tends to rise
B. Tends to sink
C. A or B
D. None of the above
Correct Answer: A
Solved Example: 61409
Choose the wrong statement.
A. The center of buoyancy is located at the center of gravity of the displaced liquid
B. For stability of a submerged body, the center of gravity of body must lie directly below the center of buoyancy
C. If c.g. and center of buoyancy coincide, the submerged body must lie at neutral equilibrium for all positions
D. All floating bodies are stable.
Correct Answer: D
Solved Example: 61410
An open vessel of water is accelerated up an inclined plane. The free water surface will:
A. Be horizontal
B. Make an angle in direction of inclination of inclined plane
C. Make an angle in opposite direction to inclination of inclined plane
D. Any one of above is possible
Correct Answer: C
Solved Example: 61411
Center of pressure on an inclined plane is: (SSC JE CE March 2017 Evening)
A. At the centroid
B. Above the centroid
C. Below the centroid
D. At metacentre
Correct Answer: C
Solved Example: 61412
The line of action of the buoyant force acts through the centroid of the: (ISRO Scientist Civil 2015)
A. Submerged body
B. Volume of the floating body
C. Volume of the fluid vertically above the body
D. Displaced volume of the fluid
Correct Answer: D
Solved Example: 61413
The time oscillation of a floating body with increase in metacentric height will be: (CIL MT Mech 2020)
A. Same
B. Higher
C. Lower
D. Lower/higher depending on weight of body
Correct Answer: C
Solved Example: 61414
In an immersed body, center of pressure is: (NPCIL SA/ST ME GJ Nov 2019)
A. At the center of gravity
B. Above the center of gravity
C. Below be center of gravity
D. Could be above or below c.g. depending on density of body and liquid
Correct Answer: C
Solved Example: 61415
The normal stress is same in all directions at a point in a fluid: (SSC JE ME March 2017 Morning)
A. Only when the fluid is friction less
B. Only when the fluid is incompressible and has zero viscosity
C. When there is no motion of one fluid layer relative to an adjacent layer
D. Irrespective of the motion of one fluid layer relative to an adjacent layer
Correct Answer: C