Physics

Fluid Mechanics and Hydraulics

376 Questions

Fluid mechanics and hydraulics questions address the principles of fluid flow, pipe resistance, and open channel dynamics. The topics include Bernoulli equation, Navier-Stokes equation, and hydrograph calculations. These concepts are crucial for civil and mechanical engineering competitive examinations.

Fluid flow equationsOpen channel flowPipe frictionHydraulic jumpHydrograph analysis

Fluid Mechanics and Hydraulics Questions

Multiple choice physics pressure in liquids and gases pressure dependence on force and area concept of pressure pressure on surface

The area of cross-section of the pump plunger and the press plunger of hydraulic press are $0.03 m^2$ and $9 m^2$ respectively. How much is the force acting on the pump plunger of the hydraulic press overcomes a load of 900 kgf?

  1. 2 kgf

  2. 3 kgf

  3. 4 kgf

  4. 5 kgf
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Load=900 kgf, Effort (E)=?$
Area of cross section of the press plunger $=a _1=9 m^2$
Area of cross - section of the pump plunger $=a _2=0.03 m^2$
By Pascal's law,
Pressure exerted by pump plunger $=$ pressure exerted on press plunger.
i.e., $\displaystyle \frac {L}{E}=\frac {a _1}{a _2}\Rightarrow \frac {900 kgf}{E}=\frac {9 m^2}{0.03 m^2}$
$\displaystyle \Rightarrow E=\frac {900 kgf\times 0.03 m^2}{9 m^2}=3 kgf$.

Multiple choice physics pressure in liquids and gases pressure dependence on force and area concept of pressure pressure on surface

A Liquid of density p is coming out of a hosepipe of the radius with horizontal speed $\upsilon $  and hits a mesh.  50 % of the liquid passes through the mesh unaffected.  25 % loses all of its momentum and 25 % comes back with the same speed.  The resultant pressure on the mesh will be: 

  1. $\frac{3}{4}p\upsilon ^{2}$
  2. $p\upsilon ^{2}$
  3. $\frac{1}{2}p\upsilon ^{2}$
  4. $\frac{1}{4}p\upsilon ^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Force on mesh = change in momentum per unit time. 50% pass through (no change), 25% stop (delta p = mv), 25% rebound (delta p = 2mv). Total force = 0.25(mv) + 0.25(2mv) = 0.75mv. Pressure = Force/Area = 0.75 * rho * v^2.

Multiple choice physics pressure in liquids and gases pressure dependence on force and area concept of pressure pressure on surface

A hole is made at the bottom of a tank filled with water. If total pressure at the bottom of the tank is three atmosphere, then the velocity of efflux at the bottom is (1 atm = $10^5 N/m^2$)

  1. $\sqrt{400} m/s$
  2. $\sqrt{200} m/s$
  3. $\sqrt{600} m/s$
  4. $\sqrt{500} m/s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$\begin{array}{l} P=1000kg/{ m^{ 3 } } \ P=3atm=3\times { 10^{ 5 } }N/{ m^{ 2 } } \ { P _{ 0 } }=1atm={ 10^{ 5 } }N/{ m^{ 2 } } \end{array}$
Pressure due to liquid column is 
$P - {P _0} =\rho gh$
velocity of flux $V = \sqrt {2gh} $
$\begin{array}{l} V=\sqrt { 2\dfrac { { \left( { p-{ P _{ 0 } } } \right)  } }{ \rho  }  }  \ =\sqrt { \dfrac { { 2\left( { 3\times { { 10 }^{ 5 } }-{ { 10 }^{ 5 } } } \right)  } }{ { 1000 } }  }  \ =\sqrt { \dfrac { { 4\times { { 10 }^{ 5 } } } }{ { 1000 } }  }  \ =\sqrt { 400 } m/s \end{array}$
Hence,
option $A$ is correct answer.

Multiple choice physics pressure in liquids and gases pressure dependence on force and area concept of pressure pressure on surface

A person blows into open-end of a long pipe. As a result, a high pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe,

  1. a high pressure pulse starts traveling up the pipe, it the other end of the pipe is open.

  2. a low pressure pulse starts traveling up the pipe, if the other end of the pipe is open.

  3. a low pressure pulse starts traveling up the pipe, if the other end ot the pipe is closed.

  4. a high pressure pulse starts traveling up the pipe, if the other end of the pipe is closed.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A pressure wave undergoes a phase change of $ \pi$ radians on 

reflection from open end of no phase change occurs on reflection from 
closed end.

Multiple choice power work and power work, energy and power physics energy and its forms

A pump ejects $12000kg$ of water at speed of $4m/s$ in $40$ second. Find the average rate at which the pump is working

  1. $0.24KW$
  2. $2.4KW$
  3. $24KW$
  4. $24W$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Force = change in momentum / time = (m * v) / t = (12000 * 4) / 40 = 1200 N. Power = Force * velocity = 1200 * 4 = 4800 W = 4.8 kW. The provided answer 2.4 kW suggests the calculation might be (1/2) * m * v^2 / t = (0.5 * 12000 * 16) / 40 = 2400 W = 2.4 kW, which is the rate of kinetic energy delivery.

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

The pump tube of hydraulic press is of cross section area $a$ and is connected with a press tube of cross section area $A$ such that :

  1. $a>A$
  2. $A>a$
  3. $A=a$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the construction of hydraulic press, release valve and press foot valves are there.

Hydraulic press consist of a pump tube of cross sectional area $a$ connected with a press tube of cross sectional area $A$ such that $A>a$.
The lower end of the pump tube is connected with a smaller tube which is further connected with an underground water tank.
At the junction of the pump tube and the smaller tube is fitted $a$ "pump foot valve" which opens in the upward direction only.

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

Discharge of centrifugal pump is

  1. inversely proportional to diameter of its impeller
  2. inversely proportional to $diameter^2$ of its impeller

  3. directly proportional to (diameter)2 of its impeller

  4. directly proportional to diameter of its impeller
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Discharge of centrifugal pump is inversely proportional to square of the diameter of its impeller, thus option B is correct.

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

Slip of a reciprocating pump is defined as

  1. Difference of theoretical discharge and actual discharge

  2. ratio of actual discharge to theoretical discharge

  3. ratio of theoretical discharge to actual discharge

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Slip of a reciprocating pump is defined as  Difference of theoretical discharge and actual discharge.


Hence, option A is correct.

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

To avoid vaporization in the pipe line, the pipe line over the bridge is laid such that it is not more than

  1. 2.4m above hydraulic gradient

  2. 6.4m above hydraulic gradient

  3. 10m above hydraulic gradient

  4. 1.4m above hydraulic gradient
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To avoid vaporization in the pipe line, the pipe line over the bridge is laid such that it is not more than 6.4m above hydraulic gradient

Multiple choice physics pressure in fluids and atmospheric pressure hydraulic equipment applications of pascals law examples of hydraulic press

A hydraulic press has a ram of 15 cm diameter and plunger is 1.5 cm. It is required to lift a weight of 1 tonne. The force required on plunger is

  1. 10kg

  2. 100 kg

  3. 1000 kg

  4. 1 kg
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Weight on the ram  $F  =  mg=1000\times 10 = 10000$

Using Pascal's law      $\dfrac{F}{A } = constant$              where $A = \pi D^2$
$\therefore$      $\dfrac{10000}{(15)^2 } =\dfrac{F _p}{(1.5)^2}$              $\implies F _p = 100$ N $= 10\ kgf$

Thus, force required on plunger is $10\ kgf$

Multiple choice physics machines levers and pulleys pulley pulleys

It is easy to draw water from a well when the rope passes over a pulley than drawing it directly. Because:

  1. pulley adds more force

  2. pulley decreases the weight of the bucket

  3. pulley has a mechanical advantage of more than one

  4. pulley changes the direction of force
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A fixed pulley does not reduce the weight or add force; it simply changes the direction of the force, allowing the user to use their body weight to pull down instead of lifting up.

Multiple choice physics machines levers and pulleys pulley pulleys

It is easy to draw water from a well when the rope passes over a pulley than drawing it directly. Because :

  1. Pulley adds more force

  2. Pulley decreases the weight of the bucket

  3. Pulley has a mechanical advantage of more than one

  4. Pulley changes the direction of force
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A single fixed pulley does not provide a mechanical advantage greater than one, nor does it reduce the weight of the load. Its primary purpose and advantage in this context is changing the direction of the applied force, making it more convenient to pull downward.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

Neglecting air resistance, the upward velocity of the water in the stream of a particular fountain is given by the formula $v = -32t + 28$, where $t$ is the number of seconds after the water leaves the fountain. While going upward, the water slows down until at the top of the stream, the water has a velocity of $0$ feet per second. How long does it take a droplet of water to reach the maximum height?

  1. $0.825$ seconds
  2. $0.925$ seconds
  3. $0.875$ seconds
  4. $0.975$ seconds
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $v = -32t + 28$
It is mentioned that at the maximum height, the velocity of water is $0$ feet per second.
Therefore, final velocity $(v) = 0$. 
$\Rightarrow 0=-32t+28$

$\Rightarrow 32t=28$      
$\Rightarrow t=\cfrac { 28 }{ 32 } =0.875$ seconds

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A jet water issues from a nozzel with a velocity of $20 m/s$ and it impinges normally on a flat plate moving away from it at $10 m/s$. If the cross-sectional area of the jet is $0.02 m^2$ and the density of water is taken as $1000 kg/m^3$, then the force developed on the plate will be

  1. $ 10 N$
  2. $ 100 N$
  3. $1000 N$
  4. $2000 N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A jet water issues from a nozzel with a velocity$=20m/s$

It impinges normally on a flat plate moving away from it$=10m/s$
Cross-sectional area of the jet $=0.02m^2$
The density of water is taken$=1000kg/m^3$
The force developed on the plate will be
Hydrostatic force o the bottom surface of the tank will be
$F _b=\rho g(l\times b)\times h\F _b\rho g\times(1\times2)\times 2\ F _b=4\rho g(lbh)\rightarrow(1)$
Hydrostatic force on vertical surface will be
$F _v=\rho g(l\times h)\times\cfrac{h}{2}\F _v=\rho g(2\times 2)\cfrac{lh^2}{2}\F _v=2\rho g(lh^2)\rightarrow(2)$
Ratio will be
$\cfrac{F _b}{F _V}=\cfrac{4\rho g(lbh)}{2\rho g(lh^2)}\ \cfrac{F _b}{F _V}=1\F _b=1\times10\ \quad=10N$