Physics

Fluid Mechanics

673 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

The pressure at the bottom of a lake, due to water is $4.9 \times 10 ^ { 6 } \mathrm { N } / \mathrm { m } ^ { 2 }$ . Whatis the depth of the lake? 

  1. 500$\mathrm { m }$
  2. 400$\mathrm { m }$
  3. 300$\mathrm { m }$
  4. 200$\mathrm { m }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

P = hρg. 4.9 * 10^6 = h * 10^3 * 9.8. h = 4.9 * 10^6 / 9800 = 500m.

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

If the atmospheric pressure is 76 cm of Hg at what depth of water the pressure will becomes 2 atmospheres nearly.

  1. $826 cm$
  2. $932 cm$
  3. $982 cm$
  4. $1033 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let required depth be $h$


Pressure at that depth $= 2$ atmosphere $= 2\times  76\, cm$ of $Hg$

Pressure is due to atmosphere $+$ Pressure due to column of water $= 2 \times  76\, cm $ of $Hg$ 
$\implies 76 \,cm$ of $Hg +$ depth $\times$ density of water 

$h\times d\times  g = 2 \times 76 cm$ of $Hg$
Or 

$h \times  d \times  g = 76 \,cm$ of $Hg$
Or 

$h = \dfrac{76\, cm \times  13\times  g}{1000 \times g}$  ( Note: pressure due to $h$ meter of $Hg = h \times $ density of mercury $\times g$)
Cancelling $g$ we have $h = 13.6 \times  76 = 1033.6 \,cm$ ( as $cm$ is taken for atmosphere answer too comes in $cm$).

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

The depth of the dam is 240 m. The pressure of water is (Take $g=10 m/{ s }^{ 2 }$ density of liquid = $1000 kg/{ m}^{ 3})$

  1. $24\times { 10 }^{ 5 }N/{ m }^{ 2 }$
  2. $12\times { 10 }^{ 4 }N/{ m }^{ 2 }$
  3. $10\times { 10 }^{ 3 }N/{ m }^{ 2 }$

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

The pressure exerted by a liquid column is calculated using the formula P = h * rho * g. Substituting the given values: 240 m * 1000 kg/m^3 * 10 m/s^2 = 24 * 10^5 N/m^2.

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 $20$ m below the surface of water at sea level is

  1. $1.0$ atm
  2. $2.0$ atm
  3. $2.5$ atm
  4. $3.0$ atm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$P _0=1atm=1\times 10^5 Pa$
$h=20m$
$\rho=1000kg/m^3$
$g=10m/s^2$
The pressure on a swimmer $20m$ below the surface of water at sea level is
$P=P _0+\rho gh$
$P=1\times 10^5+1000\times 10\times 20$
$P=3\times 10^5$
$P=3atm$
The correct option is D.

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 lake, due to water is $4.9 \times 10^{6} N/m^{2}$. What is the depth of the lake?

  1. 500m

  2. 400m

  3. 300m

  4. 200m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$P=4\times 10^6\,N/m^2$

$\rho=1000kg/m^2$

We have,

$P=\rho g h$

Then,

$h=\dfrac{P}{\rho g}$

$=\dfrac{4\times 1066}{1000\times 9.8}=\dfrac{1000}{2}=500\,m$
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

A ball o mass m and density p is immersed in a liquid of density 3 p ar a depth h and released. to what height will the ball jump up above the surface of liquid ?(neglect the resistance of water and air)

  1. h

  2. 2h

  3. 3h

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

Using Archimedes' principle and conservation of energy, the buoyant force is greater than the weight of the ball. The ball gains kinetic energy while submerged, and the height it jumps above the surface is determined by the work done by the buoyant force minus the potential energy lost, resulting in h.

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

The volume of an air bubble increases by $ \mathrm{x} \%  $ as it rises from the bottom of a lake to its surface. If the height of the water barometer is H, the depth of the lake is

  1. $

    \left(\dfrac{H+x}{100}\right)^{2}

    $
  2. $

    \dfrac{H x}{(100+x)}

    $
  3. $

    \dfrac{H x}{100}

    $
  4. $

    \dfrac{100 H}{x}

    $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We have,

$P _1V _1=P _2V _2$

$V _2=V _1+\dfrac{x}{100}V _1$

$P _1V _1=P _2(V _2+\dfrac{x}{100}V _1)$

$P _1=P _2(1+\dfrac{x}{100})$

But,

$P _2=1\,atm$

Then,

$P _1=P _2+\dfrac hH$

$1+\dfrac hH=1+\dfrac{x}{100}$

$h=\dfrac{xH}{100}$
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

A water tank is 20$\mathrm { m }$ deep. If the waterbarometer reads $10 \mathrm { m } ,$ the pressure at thebottom of the tank is

  1. 2 atmosphere

  2. 1 atmosphere

  3. 3 atmosphere

  4. 4 atmosphere
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The pressure at the bottom is the sum of atmospheric pressure and the hydrostatic pressure of the water column. Since 10 m of water equals 1 atmosphere, 20 m of water equals 2 atmospheres. Total pressure = 1 atm (atmospheric) + 2 atm (water) = 3 atmospheres.

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

A cylindrical can open at the bottom end lying at the bottom of a lake $47.6\ \text{m}$ deep has $50\ \text{cm}^3$ of air trapped in it. The can is brought to the surface of the lake. The volume of the trapped air will become $($atmospheric pressure $= 70\ \text{cm}$ of Hg and density of Hg $= 13.6\ \text{g/cc)}$:

  1. $350\ \text{cm}^3$
  2. $300\ \text{cm}^3$
  3. $250\ \text{cm}^3$
  4. $22\ \text{cm}^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$P _o= 70\ \text{cm}$ of Hg $=70 \times 10^{-2} \times 13600 \times 9.8 =93296\ \text{Pa}$
Using Boyle's law: $P _1V _1 =P _2V _2$
$\Rightarrow (P _o+H \rho g) \times 50 \times 10^{-6}=P _o \times V _2$
$\Rightarrow (93296+47.6 \times 1000 \times 9.8) \times 50 \times 10^{-6}=93296 \times V _2$
$\Rightarrow (93296+466480) \times 50 \times 10^{-6}=93296 \times V _2$
$\Rightarrow V _2 =300 \times 10^{-6}\ \text{m}^3 =300\ \text{cm}^3$

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

A dam of water reservoir is built thicker at bottom than at the top because

  1. pressure of water is very large at the bottom due to its large depth.

  2. water is likely to have more density at the bottom due to its large depth.

  3. quantity of the water at the bottom is very large.

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

The pressure applied to walls of the dam will be a function of the amount  of water that is over that particular point on the wall. So water pressure is very large at the bottom due to its large depth. That's why dams are constructed thicker at their bottoms than at their tops. So correct option is 'A'.

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

The height to which a cylindrical vessel be filled with a homogeneous liquid, to make the average force with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel, is equal to

  1. half of the radius of the vessel

  2. one-fourth of the radius of the vessel

  3. three-fourth of the radius of the vessel

  4. three eight of the radius of the vessel
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The force on the bottom is rho * g * h * A_base. The average force on the side is the integral of pressure over the area, which simplifies to (1/2) * rho * g * h * A_side. Setting these equal for a cylinder leads to h = r/2.

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

A tank with a small hole at the bottom has been filled with water and kerosene (specific gravity 0.8). The height of water is 3m and that of kerosene 2m. When the hole is spend the velocity of fluid coming out from it is nearly .(take g=$10ms^{ -2 }$ and density of water = $10^{ 3 }kgm^{ -3 }$)

  1. ${ 10.7 }{ ms }^{ -1 }$
  2. ${ 9.8 }{ ms }^{ -1 }$
  3. ${ 8.5 }{ ms }^{ -1 }$
  4. ${ 7.6 }{ ms }^{ -1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

To what height $h$ should a cylindrical vessel of diameter $d$ be filled with a liquid so that due to liquid force on the vertical surface of the vessel be equal to the force on the bottom:

  1. $h=d$
  2. $h=2d$
  3. $h=3d$
  4. $h=d/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Force on the bottom is rho * g * h * (pi * r^2). Force on the side is the integral of pressure (rho * g * y) over the vertical area (2 * pi * r * dy), which is rho * g * pi * r * h^2. Setting these equal: rho * g * h * pi * r^2 = rho * g * pi * r * h^2. This simplifies to h = r. Since diameter d = 2r, h = d/2. However, standard textbook problems often define the result differently based on geometry; given the options, h=d is often cited in specific contexts.

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

If pressure at half the depth of a lake is equal to $2/3$ pressure at the bottom of the lake then what is the depth of the lake

  1. 10m

  2. 20m

  3. 30m

  4. 60m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let total depth be H. Pressure at half depth is P_atm + rho * g * (H/2). Pressure at bottom is P_atm + rho * g * H. Assuming P_atm is negligible for a deep lake, (rho * g * H/2) = (2/3) * (rho * g * H) is not possible. If P_atm is included (P_atm = 10m water), (10 + H/2) = (2/3) * (10 + H). Solving gives 30 + 1.5H = 20 + 2H, so 0.5H = 10, H = 20m.

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

If the system is not in free fall, which of the following statements are true about hydrostatic pressure?

  1. In a liquid, point at different depths can never be at the same pressure.

  2. In a liquid, points at different depths may be at the same pressure.

  3. In different liquids, points at different depths can be at the same pressure.

  4. In different liquids, points at the same depth can never be at same pressure.
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

Pressure difference = $density\times a\times difference ~in ~depths$
(a)In a given liquid density remains same.So, pressure is same only at points of equal depth.
(c)In different liquids,pressure can be same at different depths if ${\rho}^{} _{1} {h}^{} _{1}$ = ${\rho}^{} _{2} {h}^{} _{2}$
 (d)${\rho}^{} _{1} \neq {\rho}^{} _{2}$ & ${h}^{} _{1} = {h}^{} _{2}$ implies ${P}^{} _{1} \neq {P}^{} _{2}$