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

If pressure at the half depth of a lake is equal to $\dfrac{3}{4}$ times the pressure at its bottom, then find the depth of the lake . [Take g=$10 m/s^2]$

  1. $ \dfrac{P _{0}}{\rho g}\ $
  2. $ \dfrac{2P _{0}}{\rho g}\ $
  3. $ \dfrac{P _{0}}{2\rho g}\ $
  4. $ \dfrac{3P _{0}}{\rho g}\ $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let depth of the lake be $h$ and pressure at bottom $= P$
Then $P=P _{0}+\rho gh\rightarrow (1)$    $(P _{0}=$ atmospheric pressure, $\rho $ = density of water)
At half depth $(h/2)$ pressure is $\dfrac{3P}{4}$ then :
$\dfrac{3P}{4}=P _{0}+\rho g\dfrac{h}{2}\rightarrow (2)$
On subtracting equation 2 from 1 we get :
$\dfrac{P}{4}=\rho g\dfrac{h}{2}$
$\Rightarrow P=2\rho gh$, substituting this value of $P$ in equation 1:
$2\rho gh=P _{0}+\rho gh$
$\Rightarrow h=\dfrac{P _{0}}{\rho g}\rightarrow $ Depth of the lake
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 exerted by a liquid at depth $h$ is given by:

  1. $\displaystyle \dfrac{h}{dg}$
  2. $hdg$
  3. $\displaystyle \dfrac{h}{d}$
  4. $hg$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Answer is B.

The pressure exerted by a liquid at a point depends on its vertical depth and density of the liquid only. It is independent of the shape of the container. The pressure at the bottom of the three vessels of different shapes containing the same liquid acts equally in all directions.
Thus, the pressure exerted by the liquid of height $h$ is given as P=hdg, where $h$ is the height, $d$ is the density and $g$ is the acceleration due to gravity.

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 exerted by a liquid column of height h is given by (the symbols have their usual meanings).

  1. $\dfrac {h}{\rho g}$
  2. $h\rho g$
  3. $\dfrac {h}{\rho}$
  4. $hg$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The pressure exerted by a liquid column of height h is given by-

       $h\rho g$
Since, total mass =$\rho g$
      

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

At a depth of 1000 m in an ocean, what is the absolute pressure? Given density of sea water is $1.03 \times 10^3 kgm^{-3} ,\ g= 10ms^{-2}$

  1. 104 atm

  2. 100 atm

  3. 108 atm

  4. 110 atm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $h=1000m  ,  d=1.03\times10^{3}kg/m^{3} ,  g=10m/s^{2}$ 

The absolute pressure is given by: absolute pressure = pressure of water + atmospheric pressure
$P=hdg+1atm=1000\times1.03\times10^{3}\times10+1atm=1.03\times10^{7}Pa+1atm$
$P=103atm+1atm=104atm$

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 of a barometer filled with a liquid of density $3.4\ g/cc$ under normal condition is approximately -

  1. $8\ m$
  2. $5\ m$
  3. $3\ m$
  4. $1\ m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,

$\rho =3.4g/cc=3.4\times 10^3 kg/m^3$
$g=9.8m/s^2$
$P=1.01\times 10^5 Pa$
Pressure, $P=\rho gh$
$h=\dfrac{P}{\rho g}=\dfrac{1.01\times 10^5}{3.4\times 10^3\times 9.8}$
$h=3.03m$
The 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 average pressure of a liquid (density$\rho$) on the walls of the container filled upto height $h$ with the liquid is $\dfrac{1}{2}h\rho g$.

  1. True

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

The pressure at the surface is 0 and at depth h is hρg. The average pressure on the wall is (0 + hρg) / 2 = 1/2 * hρg.

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

Two vessels A and B are different shapes have the same base area and are filled with water upto same height as the force exerted between water on the base is FA for vessel A and F B for vessel B . The respective weight of the water filled in vessel are wA and wB. Then

  1. FA>FB , was>wB

  2. FA=FB, wA>wB

  3. FA=FB, wA<wb< div=""></wb<>

  4. FA>FB, wA=wB
Reveal answer Fill a bubble to check yourself
A 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

The reading of a barometer containing some air above the mercury column is $73\ cm$ while that of a correct one is $76\ cm$. If the tube of the faulty barometer is pushed down into mercury until volume of air in it is reduced to half, the reading shown by it will be

  1. $70\ cm$
  2. $72\ cm$
  3. $74\ cm$
  4. $76\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial state: P_atm = P_air + 73. So P_air = 76 - 73 = 3 cmHg. When volume is halved, P_air becomes 6 cmHg. New reading = 76 - 6 = 70 cm.

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

A large container of negligeble mass and uniform cross-section area A has a small hole (of area a < < A) near its side wall at bottom. The container is open at the top and kept on a smooth horizontal floor . It contains a liquid of density $\rho $ and mass $m _0$ when liquid starts flowing horizontally at time t = 0. Find the speed of container when 75% of the liquid has drained out (Assume the liquid surface remains horizontal throughout the motion)

  1. $\left[ \frac { { m } _{ 0 }g }{ A\rho } \right] ^{ 1/2 }$
  2. $\left[ \frac { { 4m } _{ 0 }g }{ A\rho } \right] ^{ 1/2 }$
  3. $\left[ \frac { { m } _{ 0 }g }{ 2A\rho } \right] ^{ 1/2 }$
  4. $\left[ \frac { { 2m } _{ 0 }g }{ A\rho } \right] ^{ 1/2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a classic problem of a container moving due to the reaction force of fluid efflux. The force F = dm/dt * v_exit. Integrating the momentum equation leads to the result.