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 upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

An ice-berg floating partly immersed in sea water of density $1.03 g/cm^3$. The density of ice is $0.92 g/cm^3$. The fraction of the total volume of the iceberg above the level of sea water is

  1. $8.1\%$
  2. $11\%$
  3. $34\%$
  4. $0.8\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $v$ be the volume of the ice-berg outside the sea water and $V$ be the total volume of ice-berg. Then as per question
$0.92V = 1.03(V-v)$

or, $\dfrac vV = 1-\dfrac {0.92}{1.03}= \dfrac{11}{103} $
$\therefore  \dfrac vV \times 100 = 11 \times \dfrac {100}{103} \cong  11\%$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A wooden cylinder floats in water such that $4 cm$ of it is above water, if the same cylinder is made to float in alcohol (density $0.8gm^{-3}$), the length of cylinder above alcohol will be 

  1. $4cm$
  2. more than $ 4cm$
  3. less than $4cm$

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

C. less than 4cm  
Because more volume of alcohol needs to be displaced to displace the equal weight of water. since density of alcohol is lesser.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A wooden cube of side $10 \ cm$ has mass $700 \ g$. The part of it remains above the water surface while floating vertically on water surface is $X\ cm$. Find $X$.

  1. $3$
  2. $7$
  3. $0$

  4. Can not be detemined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that the volume of part submerged in a liquid is given by, $V= \dfrac{Density\ of\ body}{Density\ of\ liquid}$
Density of cube $= \dfrac{Mass}{Volume}$$ =\dfrac{700 g }{ 10^{3}} = 0.7g/cm^{3}$
So, density of water $= 1 g/cm^{3}$
So, part of cube submerged in water $= \dfrac{Density\ of\ body}{Density\ of\ water} = 0.7/1 = 7/10$
$\therefore$ Part of cube above water $= 1 - 7/10 = 3/10$
i.e. $3cm$ of cube is above water.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

Iron floats on the surface of mercury because its density is _____ the mercury

  1. more than

  2. less than

  3. same as

  4. cant say
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The substance having low density floats on substance with high density. 
Hence iron would float on mercury as it have lower density than mercury.
therefore, option (b) is correct.
Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

Object having density less than that of the liquid in which they are immersed, _______on the surface of the liquid.

  1. Float

  2. Sink

  3. First sink and then float

  4. First float and then sink
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Float
We know according to Archimedes Principle, Object having density less than that of the liquid in which they are immersed, float on the surface of the liquid.
And Object having density greater than that of the liquid in which they are immersed, sink in the liquid.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A tin can has a volume of $1000cm^3$ and a mass of $100g$. What mass of lead shot can it carry without sinking in water $(\rho=1000kg/m^3)$?

  1. $900g$
  2. $100g$
  3. $1000g$
  4. $1100g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Volume of the floating tin = Volume of water displaced = $1000{ cm }^{ 3 }$.
weight of water displaced $= 1000\times 1 = 1000g = 10 N  =$ Upthrust.  Upthrust = max load + weight of tin 
That is, max load = upthrust - weight of tin.
It is given that the weight of tin is $100 g.$
Therefore, load $= 10 N - 1 N = 9 N$
Mass $M =$ Weight w/acceleration due to gravity $g$. Let us take $g=10m/{ s }^{ 2 }$.
So, $M=W/g = 9 N/10 = 0.9 kg = 900 g$.
Hence, the mass of lead shot the tin can carry without sinking in water is $900 g$.
Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A block of ice of total area A and thickness 0.5 m is floating in water. In order to just support a man of mass 100 kg, the area A should be (the specific gravity ofice is 0.9):

  1. $2.2m^{2}$
  2. $1.0m^{2}$
  3. $0.5m^{2}$

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

Let say $m _1=$ mass of the man = 100kg
and $m _2=$ mass of the ice $= 0.9 \times 1000V=900V$, where $V$ is the volume of the ice block.
For equilibrium,
Total downward weight = total upthrust
$100g +900 Vg=1000Vg \\Rightarrow V=1m^3$
Volume = Area $\times $ height
$\Rightarrow A=\frac{Volume}{Height}=\frac{1}{0.5}=2m^2$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

The density of ice is $920kg/m^3$, and that of sea water is $1030kg/m^3$. What fraction of the total volume of an iceberg is outside the water?

  1. $0.107$
  2. $0.207$
  3. $0.307$
  4. $0.407$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $V _L$ and $V _S$ be the volume of water displaced and volume of the ice respectively; $\rho _L$ and $\rho _S$ be the density of water and density of ice respectively. Since ice is floating on water, $F _B=W$
or $\rho _LV _Lg=\rho _SV _Sg$ or $\rho _LV _L=\rho _SV _S$
or $\frac{V _L}{V _S}=\frac{\rho _S}{\rho _L}$
This is the fraction of volume of the iceberg that is inside the water. Therefore, the fraction of volume of the iceberg that is outside the water is given by, 1 - 0.893 = 0.107
Hence, the fraction of the total volume of an iceberg is outside the water is 0.107.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

An egg sinks when immersed in water contained in a vessel. On dissolving a lot of salt in the water,will the egg

  1. Develop cracks in the shell

  2. Break

  3. Rise and then float

  4. Remain where it is
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Answer is C.

An egg will sink in fresh water but it will float in very salty water; the density of the egg is greater than the density of fresh water but less than the density of the salty water.
The density of salty water can be a much as $10\%$ greater than that of fresh water i.e. up to $1.1 g/cm^{ 3 }$.
Hence, on dissolving a lot of salt in the water, the egg will rise and then float.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

Two solids A and B float in water. It is observed that A floats with half its volume immersed and B floats with $2/3$ of its volume immersed. Compare the densities of A and B:

  1. $4:3$
  2. $2:3$
  3. $3:4$
  4. $1:1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that $m _1=V _1/2 \times d \implies d _1=d/2$
and $m _2=2V _2/3 \times d \implies d _2=2d/3$
$\therefore d _1:d _2=3:4$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four times the density of the material of the ball. The frictional force of the liquid on the rising ball is greater than the weight of the ball by a factor of

  1. $2$
  2. $3$
  3. $4$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $F _f = $ Force of friction  

$F _b = $ Force of bouyancy
$F _w = $ Weight of ball

Archimedes' principle:

$F _b = F _w + F _f$

Given: $F _b = (4P _B)gV$

$F _w = VP _Bg$

$F _f = F _b - F _w = 3P _B gV$

$\Rightarrow \dfrac{F _f}{F _w} = \cfrac{3P _B gV}{P _BgV} = 3 : 1 $

$\Rightarrow F _f = 3F _w$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A cylindrical block floats vertically in a liquid of density ${\rho} _1$ kept in a container such that the fraction of volume of the cylinder inside the liquid is $x _1$. then some amount of another immiscible liquid of density ${\rho} _2 ({\rho} _2 < {\rho} _1)$ is added to the liquid in the container so that the cylinder now floats just fully immersed in the liquids with $x _2$ fraction of volume of the cylinder inside the liquid of density ${\rho} _1$. The ratio ${\rho} _1 / {\rho} _2$ will be

  1. $\dfrac{1 - x _2}{x _1 - x _2}$
  2. $\dfrac{1 - x _1}{x _1 + x _2}$
  3. $\dfrac{x _1 - x _2}{x _1 + x _2}$
  4. $\dfrac{x _2}{x _1}-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $V$ and $\rho$ be the volume and density of the cylindrical block.

Case 1 : When $x _1$ fraction of block's volume is immersed in liquid of density $\rho _1$
Using Archimede's principle :    Weight of cylindrical block = Weight of liquid displaced
$\therefore$  $\rho V g = \rho _1 x _1 V g$             ........(1)

Case 2 : When $x _2$ fraction of block's volume is immersed in liquid of density $\rho _1$ and $1-x _2$ fraction of block's volume is immersed in liquid of density $\rho _2$
Using Archimede's principle :    Weight of cylindrical block = Weight of liquid displaced
$\therefore$  $\rho V g = \rho _1 x _2 V g + \rho _2 (1-x _2) V g$             ........(2)

Equating (1)  and (2) we get    $\rho _1 x _1V g = \rho _1 x _2 V g + \rho _2 (1-x _2) V g$
OR    $\rho _1 x _1 = \rho _1 x _2 + \rho _2 (1-x _2)$
OR   $\rho _1 (x _1 - x _2) = (1-x _2)\rho _2$
$\implies$  $\dfrac{\rho _1}{\rho _2} = \dfrac{1-x _2}{x _1-x _2}$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

A solid floats in a liquid in the partially submerged position:

  1. the solid exerts a force equal to its weight on the liquid

  2. the liquid exerts a force of buoyancy on the solid which is equal to the weight of the solid

  3. the weight of the displaced liquid equals the weight of the solid

  4. all of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that solid floats in a liquid, in the partially submerged position.

The weight of the displaced liquid will be equal to the weight of the solid, and the solid exerts a force equal to the weight of the liquid.
The weight of solid is equal to the buoyancy force exerted by the liquid on the solid.
Therefore option $D$ is correct.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

In a beaker containing liquid, an ice cube is floating. When ice melts completely, the level of liquid rises. Then the density of the liquid is:

  1. more than the density of ice

  2. less than the density of ice

  3. same as the density of ice

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

Given, That the ice cube is floating in the liquid.

Let the height of ice cube be $h,$
Given, If ice cube completely melts, the level of liquid raises. So initially the length of ice cube submerged in liquid be $l <h,$
Let the density of liquid be $d _{l}$ and density of ice cube be $d _{i}$
In equilibrium , $Mg=M _{l}g$
$\Rightarrow d _{i}Ahg=d _{l}Alg$
$\Rightarrow \frac{d _{l}}{d _{i}}=\frac{h}{l}>1$
$\Rightarrow d _{l} > d _{i}$
Therefore the density of liquid is more than the density of ice.
So option $A$ is correct.

Multiple choice physics pressure in fluids and atmospheric pressure pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

A capillary tube of radius $r$ is immersed in water and water rises to a height of $h$. Mass of water in the capillary tube is $5\times 10^{-3}kg$. The same capillary tube is now immersed in a liquid whose surface tension is $\sqrt{2}$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is $45^o$. The mass of liquid which rises into the capillary tube now is (in kg):

  1. $5\times 10^{-3}$
  2. $2.5\times 10^{-3}$
  3. $5\sqrt{2}\times 10^{-3}$
  4. $3.5\times 10^{-3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the force on liquid is proportional to $Scos(\theta )$, in the latter case, force will be $\sqrt { 2 } cos(\dfrac { \pi  }{ 4 } )=1$ times the force in first case. Hence, since the same force is applied, mass of water rising in the tube will be same.