Physics

Fluid Mechanics

637 Questions

Fluid mechanics is a core physics topic that evaluates the principles of liquid pressure, buoyancy, density, and viscosity through complex numerical problems. The questions require calculating the volume of submerged objects, understanding hydraulic jumps, and applying fundamental fluid statics principles. It is a highly scoring subject for candidates preparing for technical and engineering competitive exams.

Liquid pressure and depthBuoyancy and densityVolume expansionHydraulic jump calculationsSurface tension mechanics

Fluid Mechanics Questions

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

State True or False.
Pressure at a point in a liquid is inversely proportional to the height of the liquid column.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

False statement.

The pressure exerted by a liquid depends on the height of the liquid column.
It can be defined as the weight of liquid column over an unit area. 
So pressure is define as $P=\rho g H$ 

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

A boy swims a lake and initially dives $0.5 m$ beneath the surface. When he dives $1 m$ beneath the surface, how does the absolute pressure change?

  1. It doubles

  2. It quadruples

  3. It slightly increases

  4. It cut to a half
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


When he goes from $0.5$ m to $1$ m , the pressure will slightly increases
Pressure at depth $0.5 m $ is $P _{0}+0.5dg$ , where $P _{0}$ is atmospheric pressure , $d$ is density
The pressure change when he divies to $1m$ is $P _{0}+dg$
So the pressure change slightly increases
Therefore option $C$ is correct

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

The pressure at the bottom of a tank of liquid is not proportional to:

  1. the acceleration

  2. the density of the liquid

  3. the area of the liquid surface

  4. the height of the liquid
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the acceleration  of tank be $a$ $($ moving up $),$

Let the density of the liquid be $d$ and height of liquid be $h,$
The pressure at the bottom of tank is $P=P _{0}+dh(g+a),$
Therefore the pressure depends on acceleration, height of the liquid and the density of liquid.
It does not depend on the area of the liquid surface,
Therefore correct option is $C.$

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

The pressure on a swimmer 10 m below the surface lake is:(Atmospheric pressure=$1.01\times 10^5$ Pa,Density of water$=1000kg/m^3$ )

  1. $10\ atm$
  2. $5\ atm$
  3. $15\ atm$
  4. $2\ atm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given :  $h = 10 m$,  $\rho = 1000 kg/m^3$,  $g = 10 m/s^2$
Pressure on swimmer  $=$ pressure of atmosphere + pressure of water
                                      $ = 1  atm  + \rho gh \times 10^{-5} atm$
                                      $ = 1 atm + 1000\times 10 \times 10\times 10^{-5} atm$ 
                                      $ = 2atm$
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

What is the difference between the pressure on the bottom of a pool and the pressure on the water surface?

  1. $gh$
  2. $\dfrac{g}{h}$
  3. $0$
  4. $none$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The pressure difference is equal to the product of density , height and $g$

Given that $h$ is height difference between bottom and surface of a pool
$P _{b}-P _{s} = dgh$
Density $d$ of water is $1$
So we get $P _{b}-P _{s}=gh$
Therefore option $A$ is correct

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

 $1m^3$ water is brought inside the lake upto $200 m$ depth from the surface of the lake. What will be change in the volume when the bulk modulus of elasticity of water is $22000 atm$?
(density of water is $1 \times 10^3 kg/m^3$ atmosphere pressure = $10^5 N/m^2$ and $g = 10 m/s^2$

  1. $8.9 \times 10^{-3} m^3$
  2. $7.8 \times 10^{-3} m^3$
  3. $9.1 \times 10^{-4} m^3$
  4. $8.7 \times 10^{-4} m^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$K = \dfrac{P}{\Delta V/V }$

$\therefore \Delta V = \dfrac{PV}{K}$

$P = h\rho g = 200 \times 10^3 \times 10 N/m^2$

$K = 22000 atm = 22000 \times 10^5 N/m^2$

V = 1$m^3$

$ \Delta V = \dfrac{200 \times 10^3 \times 10 \times 1}{22000 \times 10^5}=9.1 \times 10^{-4}m^3$

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

Three containers are used in a chemistry lab. All containers have the same bottom area and the same height. A chemistry student fills each of the containers with the same liquid to the maximum volume. Which of the following is true about the pressure on the bottom in each container?

  1. $P _1 = P _2 = P _3$
  2. $P _1 > P _2 > P _3$
  3. $P _1 < P _2 = P _3$
  4. $P _1 < P _2 > P _3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Pressure applied on the bottom is equal to the force applied on the bottom per unit area of bottom

$\Rightarrow P=\frac{F}{A}$
$\Rightarrow P=\frac{dvg}{A}$
Where $d$ is density , $v$ is volume and $A$ is area
Given that for three containers , area is same and height is same. so the volume of three containers is same .
The density is also same for three containers.
So $d,v,g,A$ are same for all three containers
Therefore their pressures are same
So option $A$ is correct

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

The pressure at the bottom of a tank of water is $3P$ where $P$ is the atmospheric pressure. If the water is drawn out till the level of water is lowered by one fifth, the pressure at the bottom of the tank will now be:

  1. $2P$
  2. $(13/5)P$
  3. $(8/5)P$
  4. $(4/5)P$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
If we ignore the atmospheric pressure, the pressure at the bottom is $2P$
We know that the pressure by a liquid column is given by $h\rho g$
$\therefore h\rho g=2P$
After the height getting lowered by one fifth, the height becomes four fifth. 
$\therefore \cfrac45h\rho g=\cfrac{2\times4}5P=\cfrac85P$
Now including the atmospheric pressure it becomes
$(\cfrac85+1)P=\cfrac{13}5P$ 

Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

The force acting on a window of area 50 cm x 50 cm of a submarine at a depth of 2000 m in an ocean, the interior of which is maintained at sea level atmospheric pressure is (Density of sea water = 10$^3$ kg m$^{-3}$,g = 10 m s$^{-2}$)

  1. 5 x 10$^5$ N
  2. 25 x 10$^5$ N
  3. 5 x 10$^6$ N
  4. 25 x 10$^6$ N
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $h = 2000 m,$

          $ \rho= 10^3\, kg\, m^{-3},$ 
          $g = 10 \,m \,s^{-2}$
The pressure outside the submarine is
$P = P _a + \rho gh$
where $P _a$ is the atmospheric pressure. Pressure inside the submarine is $P _a$.
Hence, net pressure acting on the window is gauge pressure. Gauge pressure,
 $P _g = P - P _a = \rho gh $
$= 10^3\, kg\, m^{-3} \times 10 \,m\,s^{-2} \times 2000 \,m$
$ = 2 \times 10^7 Pa$
Area of a window is $A= 50 cm\times50 cm $
                                    $= 2500 \times 10^{-4}\, m^{2}$
Force acting on the window is
$F = P _gA $
$= 2 \times 10^7 \,Pa \times 2500 \times 10^{-4} m^2 $
$= 5 \times 10^6\,N$

Multiple choice physics pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

A cubical vessel sealed vessel with edge $L$ is placed on a cart, which is moving horizontally with an acceleration `a' as shown iin figure. The cube is filled with an ideal fluid having density $\rho$. Find the gauge pressure at the centre of the cubical vessel.

  1. $\dfrac{Ldg}2$
  2. $\dfrac{Ld(g+a)}2$
  3. $\dfrac{Lda}2$
  4. $\dfrac{Ld(g-a)}2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an accelerating frame, the effective gravity is the vector sum of g and -a. At the center of a cube of side L, the depth from the free surface is L/2. Thus, pressure P = rho * g_eff * depth = rho * sqrt(g^2 + a^2) * L/2. However, assuming horizontal acceleration, the pressure gradient leads to P = rho * (g+a) * L/2 if considering vertical and horizontal components combined or specific orientation.

Multiple choice physics pressure in fluids and atmospheric pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

While filling a syringe with liquid which of the following is true for pressure inside the barrel and the atmospheric pressure acting on the liquid?

  1. pressure inside the barrel is less than atmospheric pressure

  2. pressure inside the barrel is more than atmospheric pressure

  3. pressure inside the barrel is same as atmospheric pressure

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the plunger is pulled up in the barrel, the pressure inside the barrel is much less than atmospheric pressure. As a result atmospheric pressure forces the liquid to rise up in the syringe.

Multiple choice physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

A bubble is at the bottom of a lake of depth $h$. As the bubble comes to the sea level, its radius increase three times. If atmospheric pressure is equal to $l$ metre of a water column, then $h$ is equal to

  1. $26\ l$
  2. $l$
  3. $25\ l$
  4. $30\ l$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Boyle's Law P1V1 = P2V2. Initial pressure P1 = P_atm + rho*g*h. Final pressure P2 = P_atm. Since radius triples, volume increases 27 times. P_atm + rho*g*h = 27 * P_atm. With P_atm = rho*g*l, we get h = 26l.

Multiple choice physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

A narrow glass tube, $80$ cm long and opens at both ends, is half immersed in mercury, now the top of the tube is closed and is taken out of mercury A column of mercury $20$ cm long remains in the tube. Find the atmospheric pressure

  1. $20$ cm of air column
  2. $60$ cm of Hg column
  3. $60$ cm of air column
  4. $20$ cm of Hg column
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Boyle's Law for the trapped air: P_initial * V_initial = P_final * V_final. Initially, the tube is half-immersed (40 cm air at P_atm). Finally, 20 cm Hg remains, so air column is 60 cm. Solving gives P_atm = 60 cm Hg.

Multiple choice physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

Two communicating cylindrical vessel contain mercury. The diameter of one vessel is n times larger than the diameter of the other.A column of water of height h is poured into the vessel.The mercury level will rise in the right - hand vessel (s=relative density of mercury and p = density of water ) by 

  1. $ \frac { n^ h }{ (n\quad +\quad 1)^ 2s } $
  2. $ \frac { h }{ (n^ 2\quad +\quad 1)s } $
  3. $ \frac { h }{ (n\quad +\quad 1)^ 2s } $
  4. $ \frac { h }{ (n^{ 2 }s) } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

It is given that the diameter of one vessel is n times larger than the diameter of other. So,

$ \pi {{r}^{2}}{{h} _{1}}=\pi {{(nr)}^{2}}{{h} _{2}} $

$ {{h} _{1}}={{n}^{2}}{{h} _{2}} $

Since, pressure at point A is equal to pressure at B

$ \rho gh=({{h} _{1}}+{{h} _{2}})\rho 'g $

$ \because s=\dfrac{\rho '}{\rho } $

$ \rho gh=({{n}^{2}}{{h} _{2}}+{{h} _{2}})s\rho g $

$ {{h} _{2}}=\dfrac{h}{({{n}^{2}}+1)s} $