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 upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

A liquid mixture of volume $V$ has two liquids as its ingredients with densities $\alpha  \; and\; \beta $. If the density of the mixture is $\sigma $, then the mass of the first liquid in the mixture is :

  1. $\dfrac{\alpha V[\sigma \beta +1]}{\beta [\alpha +\sigma ]}$
  2. $\dfrac{\alpha V[\sigma -\beta ]}{ [\sigma +\beta]}$
  3. $\dfrac{\alpha V[\beta-\sigma ]}{ \beta-\alpha }$
  4. $\dfrac{\alpha V[1-\sigma\alpha ]}{ \beta[\alpha-\sigma ] }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let mass of liquid with density $\alpha =M _1$
mass of liquid with density $\beta =M _2$
Total volume$=V$
Net density of mixture$=\sigma$
Total mass$=M _1+M _2$
$\Rightarrow V\sigma =M _1+M _2$
$\Rightarrow M _2=V\sigma -M _1$ ......$(1)$

$\left[\because \dfrac{Total \, Mass}{v}=\sigma\right]$

$T=\dfrac{Total \,mass}{Total \, volume}=\dfrac{M _1+M _2}{\dfrac{M _1}{\alpha}+\dfrac{M _2}{\beta}}$ .......$(2)$
sub $(1)$ in $(2)$

$\Rightarrow \sigma =\dfrac{M _1+(v\sigma -M _1)}{\dfrac{M _1}{\alpha}+\left(\dfrac{v\sigma -M _1}{\beta}\right)}$

$\Rightarrow M _1=\dfrac{\alpha V(\beta -\sigma)}{\beta -\alpha}$.
Multiple choice chemistry materials liquid crystals liquid state the liquid state

A liquid of density $1.07\ g/ml$ is filled in the barometer, in place of mercury. What should be the length of liquid column, if another barometer filled with mercury, is measuring $75\ cm$ pressure?

  1. $75\ cm$
  2. $600\ cm$
  3. $9.375\ cm$
  4. $6000\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry materials liquid crystals liquid state the liquid state

A beaker is filled with a liquid of density $\rho $ upto a height pressure on the walls is :

  1. 0
  2. $h \rho g $
  3. $h \rho g/2 $
  4. $2 h \rho g $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The pressure at a depth h in a liquid is rho * g * h. The average pressure on the walls of a container filled to height h is the integral of pressure over the depth, which results in (1/2) * rho * g * h.

Multiple choice physics superposition of waves-2: stationary (standing) waves: vibrations of air columns determining wavelength and speed of sound resonance tube resonance and sonometer

As an empty vessel is filled with water, its fundamental frequency

  1. Increases

  2. Decreases

  3. Remains the same

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

   An empty vessel with a base is like a closed organ pipe , and fundamental frequency of a closed organ pipe is given by ,

       $n _{1}=v/4l$ ,
where $l=$ length of air column in pipe (height of pipe) ,
  from above we get,    
       $n _{1}\propto 1/l$ ,
when the empty vessel is filled with water , the length of air column $l$ in pipe decreases and therefore fundamental frequency $n _{1}$ increases as length of air column and fundamental frequency are inversely proportional to each other .

Multiple choice physics superposition of waves-2: stationary (standing) waves: vibrations of air columns determining wavelength and speed of sound resonance tube resonance and sonometer

In a resonance column experiment, the first resonance is obtained when the level of the water in tube is $20 cm$ from the open end. Resonance will also be obtained when the water level is at a distance of

  1. $40 cm$ from the open end.
  2. $60$ cm from the open end.
  3. $80$ cm from the open end.

  4. data insufficient
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Frequency of first resonance in closed pipe    $\nu = \dfrac{v}{4L _1}$      ....(1)
where $L _1 = 20 \ cm$
Frequency of next resonance in closed pipe  $\nu = \dfrac{3v}{4L _2}$      ....(2)
Equating (1) and (2), we get
$\dfrac{v}{4L _1} = \dfrac{3v}{4L _2}$
$\implies$  $L _2 = 3L _1  = 3\times 20 = 60 \ cm$

Multiple choice viscosity option b: engineering physics properties of matter physics

The viscous force on a small sphere of radius $R$ moving in a fluid varies as 

  1. $\propto \ R^2$
  2. $\propto \ R$
  3. $\propto \ (\dfrac{1}{R})$
  4. $\propto \ (\dfrac{1}{R})^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Stokes' Law, the viscous drag force F on a sphere of radius R moving at velocity v in a fluid is given by F = 6 * pi * eta * R * v. Thus, the force is directly proportional to the radius R.

Multiple choice viscosity option b: engineering physics properties of matter physics

A liquid rises in a capillary tube when the angle of contact is:

  1. $An\ acute\ one$
  2. $An\ obtuse\ one$
  3. $\pi/2\ radian$
  4. $\pi\ radian$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If angle of contact is acute, liquid rises in a capillary tube whereas if angle of contact is obtuse, liquid is depressed in a capillary tube.

Hence, option A is correct.

Multiple choice viscosity option b: engineering physics properties of matter physics

Two solid metal balls of radii $r$ $2r$ are falling with their terminal speeds in a viscous liquid.What is the ratio of drag force acting on these two balls?

  1. 1;2

  2. 1;4

  3. 1;8

  4. 4;1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Terminal velocity v_t is proportional to r^2. The drag force F = 6 * pi * eta * r * v_t. Substituting v_t proportional to r^2, we get F proportional to r * r^2 = r^3. The ratio of forces for radii r and 2r is (r/2r)^3 = 1/8.

Multiple choice viscosity option b: engineering physics properties of matter physics

When $200 ml$ of water is subjected to a pressure of $2 \times {10^8}pa,$ the decrease in its volume is $0.2 ml.$ the compressibility of water is -----

  1. $5 \times {10^{ - 8}}{m^2}{N^{ - 1}}$
  2. $5 \times {10^{ - 10}}{m^2}{N^{ - 1}}$
  3. $5 \times {10^{ - 12}}{m^2}{N^{ - 1}}$
  4. $None$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Compressibility K = -(1/V) * (dV/dP). Here, dV/V = 0.2 / 200 = 0.001. dP = 2 * 10^8 Pa. K = 0.001 / (2 * 10^8) = 0.5 * 10^-11 = 5 * 10^-12 m^2/N.

Multiple choice viscosity option b: engineering physics properties of matter physics

An air bubble of diameter 2mm rises steadily througha solution of density $1750 kg/m^3$at the rate of $0.35cm/s$.Calculate the coefficient of viscosity of the solution.The density of air is negligible. 

  1. 10

  2. 11

  3. 12

  4. 13
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The force of buoyancy B is equal to the weight of  the displaced liquid. Thus

$\Rightarrow B=\dfrac{4}{3}\pi r^36g$
This force is upward. The viscous force acting downward is $F=6\pi nrv$
The weight of the air bubble may be neglected as the density of air is small. for uniform velocity -
$\Rightarrow F=B$
$\Rightarrow 6\pi nrv=\dfrac{4}{3}\pi r^3 6g$
$\Rightarrow n=\dfrac{2r^36g}{9v}$
         $=\dfrac{2\times \left( 1\times 106{-3}m\right)^2\times \left( 1750kg/m^3\right)\times 9.8m/s^2}{9\times 0.35\times 10^{-2}m/s}$
         $=11\;poise$
This appears to be a highly viscous liquid.

Multiple choice viscosity option b: engineering physics properties of matter physics

A U-tube having identical limbs is partially filled with water. An immiscible oil having a density of 0.8 g/cc is poured into one side until the water rises by 25 cm on the other side. the level of oil will stand higher than the water level? 

  1. 6.25 cm

  2. 75 cm

  3. 22.5 cm

  4. 12.5 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice viscosity option b: engineering physics properties of matter physics

A small sphere of mass M and density $D _1$ is dropped in a vessel filled with glycerine. If the density of glycerine is $D _2$ then the viscous force acting on the ball will be in Newton.

  1. $M D _1 D _2$
  2. $Mg \displaystyle \left [ 1- \frac {D _2}{D _1} \right ]$
  3. $\displaystyle \frac {M D _1 g}{D _2}$
  4. $\displaystyle \frac {M}{g} \left ( D _1 + D _2 \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When sphere is in glycerine, three forces acts on it. which balances each other.

  • weight $(W)$
  • buoyant force $(F _B)$
  • viscous force $(F _V)$
and $W=F _B+F _V$
$\Rightarrow Mg=V _g D _2g +F _V$
$\Rightarrow Mg=V _s D _2g +F _V           \because V _g=V _s$
$\Rightarrow Mg=\frac{M}{D _1} D _2g +F _V             \because V _s=\frac{M}{D _1}$
$\Rightarrow F _V=Mg \left [ 1- \frac {D _2}{D _1} \right ]$

Multiple choice viscosity option b: engineering physics properties of matter physics

An air bubble of radius $1 \,cm$ is found to rise in a cylindrical vessel of large radius at a steady rate of $0.2 \,cm$ per second. If the density of the liquid is $1470 \,kg \,m^{-3}$, then coefficient of viscosity of liquid is approximately equal to

  1. $163$ poise
  2. $163$ centi-poise
  3. $140$ poise
  4. $140$ centi-poise
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice viscosity option b: engineering physics properties of matter physics

A capillary tube of area of cross-section A is dipped in water vertically. The amount of heat evolved as the water rises in the capillary tube up to height h is: (The density of water is $\rho$)

  1. $\dfrac{A\rho gh^2}{2}$
  2. $Agh^2\rho$
  3. $2Agh^2\rho$

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

The potential energy gained by the water column is m * g * h_cm = (A * h * rho) * g * (h/2) = (A * rho * g * h^2) / 2. This energy is provided by the surface tension work, and the heat evolved is the difference between the work done by surface tension and the potential energy gained.