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 chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A boat floating in a tank is carrying passenger. If the passangers drink water, how will its affect the water level of the tank?

  1. It will go down

  2. It will rise

  3. It will remain unchanged

  4. It will depend on atmospheric pressure
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If passenger drinks water then the weight of boat get increase and the boat will displace the same amount of water in the tank resulting no net change in the water level.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Density of water is ___________?

  1. $1000kg/m^3$
  2. $1kg/m^3$
  3. $1000g/cm^2$
  4. $100kg/m^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Density of water is $1000kg/m^3$.

ρ (rho) = density, m = mass, V = volume. The SI unit of density is kg/m3. Water of 4 °C is the reference ρ = 1000 kg/m3 = 1 kg/dm3 = 1 kg/l or 1 g/cm3 = 1 g/ml.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A tank $2\ m$ high is half filled with water and then filled to the top with oil of density $0.80\ g/cc$. What is the pressure at the bottom of the tank due to these liquids (Take $g=10\ ms^{-2}$) is

  1. $1.80\times 10^{3}\ Nm^{-2}$
  2. $0.9\times 10^{3}\ Nm^{-2}$
  3. $1.8\times 10^{4}\ Nm^{-2}$
  4. $0.9\times 10^{4}\ Nm^{-2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Density of oil=0.80 g/cc=800 kg/m$^{3}$

Density of water=1 g/cc=1000 kg/m$^{3}$
Pressure of water=Height will be 1 m half filled
$\Longrightarrow hdg=1 m \times 1000 kg/m^{3} \times 10m/s^{2}=10000 Pa$
Pressure by oil$=hdg$
Total pressure$=(1 \times 8000 \times 10)=8000 Pa+10000 Pa = 18000 Pa=1.8 \times 10^{4} N/m^{2}$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Density of ice is $\sigma$ and that of water is $\rho$. What will be the decrease in volume when a mass $M$ of ice melts 

  1. $\dfrac {M}{\sigma - \rho}$
  2. $\dfrac {\sigma - \rho}{M}$
  3. $M\left [\dfrac {1}{\sigma} - \dfrac {1}{\rho}\right ]$
  4. $M\left [\dfrac {1}{\rho} - \dfrac {1}{\sigma}\right ]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of ice= Mass/Density= $M/P$

Volume of water= Mass/Density= $M/\sigma$
Decrease in volume= $V _i-V _w$
$=\cfrac {M}{P}-\cfrac {M}{\sigma}= M\left[\cfrac {1}{P}-\cfrac {1}{\sigma}\right]$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

By sucking through a straw, a boy can reduce the pressure in his lungs to $750\ mm$ of $Hg$ (density$ = 13.6\,\,g/c{m^3}$). Using a straw, he can drink water from a maximum depth of

  1. $13.6\ cm$
  2. $1.36\ cm$
  3. $0.136\ cm$
  4. $10\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given Pressure in lungs${P} _{lungs}=750$ mm of Hg

and atmospheric pressure${P} _{atm}=760$ mm of Hg
Now pressure difference $\Delta{P}=760-750=10$ mm of Hg=1 cm of Hg
Now
1 cm of Hg$={\rho} _{water}gh$
$1{\rho} _{Hg}g={\rho} _{water}gh$
$13.6g=gh$
$h=13.6 cm$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A body of density ${d _1}$ is counterpoised by $Mg$ of weights of density ${d _2}$ in air of density $d.$ Then the true mass of the body is

  1. $M$
  2. $M\left( {1-\dfrac{d}{{{d _2}}}} \right)$
  3. $M\left( {1 - \dfrac{d}{{{d _1}}}} \right)$
  4. $\dfrac{{M\left( {1 - d/{d _1}} \right)}}{{\left( {1 - d/{d _2}} \right)}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The true mass M_true is found by equating the forces: M_true * g - V * d * g = M * g - V * d2 * g, where V is volume. Substituting V = M_true / d1 leads to the correct buoyant force balance equation.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A block of mass 20 kg and volume $ { 10 }^{ 3 }{ cm }^{ 3 }$ is suspended vertically from a ceiling by  a wire . The liner mass density of the its length  is 50 cm. The wire is vibrating in its  fundamental mode and producing beats with a tuning fork f frequency block is just completely immersed in a liquid and vibrated in its fundamental mode , it produces the same number  of beats with  earlier . Density of the liquid is $\left( g={ 10 }{ m/s }^{ 2 } \right) $

  1. $3.8 gm/{ cm }^{ 3 }$
  2. $7.6 gm/{ cm }^{ 3 }$
  3. $1.9 gm/{ cm }^{ 3 }$
  4. $5.0 gm/{ cm }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When block is immersed in liquid, tension decreases by buoyant force. T' = T - ρ_liquid V g. Fundamental frequency f ∝ √T. Beats occur when frequencies differ from tuning fork. For equal beats, the tension ratio gives: (1 - ρ_liquid/20)/1 = 1/4. Solving gives ρ_liquid = 3.8 g/cm³. Option A is correct.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A metallic sphere with an internal cavity weighs 40 g weight in air and 20 g weight in water. If the density of the material with  cavity be 8 g per $c{m^3}$ then the volume of cavity is:

  1. zero
  2. 15 $c{m^3}$
  3. 5 $c{m^3}$
  4. 20 $c{m^3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$V _s=5cc$ As given

When it is placed in water net force$=$ new downward force
$\Rightarrow Hog-(V+V _s)g=20g\Rightarrow V+V _s=20\ \therefore V=20-V _s cc\V=20-5cm^2=15cm^2$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A body of uniform cross-sectional area floats in a liquid of dentisty thrice its value. The portion of exposed height will be:

  1. 2/3

  2. 5/6

  3. 1/6

  4. 1/3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For floating body: weight = buoyant force. Ah₁ρ_body g = Ah₂ρ_liquid g, where h₁ is total height and h₂ is submerged height. Given ρ_liquid = 3ρ_body, we get h₁ = 3h₂. Exposed height = h₁ - h₂ = 2h₂ = 2/3 h₁. Option A is correct.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

An orifice of radius $r$ is present at the bottom of a container $r$. The maximum height to which the container can be filled with a liquid of density $\rho$ for which liquid will not come out of the orifice is?

  1. $\dfrac {T}{r\rho g}$
  2. $\dfrac {3T}{r\rho g}$
  3. $\dfrac {2T}{r\rho g}$
  4. $\dfrac {4T}{r\rho g}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For liquid not to overflow or leak out through an orifice due to surface tension, the upward force created by surface tension must balance the downward hydrostatic pressure force. The pressure at depth h is rho*g*h, and the force due to surface tension along the perimeter of radius r is 2*pi*r*T. Equating the upward force (2*pi*r*T) to the downward force (rho*g*h * pi*r^2) yields the maximum height h = 2T/(r*rho*g).

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

When two liquid of same volume but different densities $\rho _{1}$ and $\rho _{2}$ are mixed together, then the density of the mixture is

  1. $\dfrac {p _{1}+p _{2}}{2}$
  2. $p _{1}+p _{2}$
  3. $\dfrac {2p _{1}p _{2}}{p _{1}+p _{2}}$
  4. $2p _{1}+2p _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Density \,\,of \,\,mixture = \dfrac{Total\,\, mass }{ Total \,\,volume}$

Let $1^{st}$ liquid have mass $M _1$, density $p _1$ and Volume $V$

and
$2^{st}$ liquid have mass $M _2$, density $p _2$ and Volume $V$
So,Density of mixture  $= \dfrac{M _1 +M _2}{2V}$$=$$\dfrac {p _1V+p _2V}{2V}$

                                 $\rho _m=\dfrac{p _1+p _2}{2}$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A tank  $5 { m }$  high is half filled with water and then is filled to the top with oil of density  $0.85 g / c m ^ { 3 } .$  The pressure at the bottom of the tank, due to these liquids is

  1. $1.85{ g }/{ cm }^{ { 2 } }$
  2. $89.25{ g }/{ cm }^{ { 2 } }$
  3. $462.5{ g }/{ cm }^{ { 2 } }$
  4. $500{ g }/{ cm }^{ { 2 } }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,

Total height of tank $=5\,m$

Density of water, $\rho _1=1\,g/cm^3$

Density of oil, $\rho _2=0.85\,g/cm^3$

We know that,

Half tank is filled with water and half tank is filled with oil.

So,  $h _1=h _2=\dfrac 52=2.5\,m=250\,cm$

Pressure at the bottom of the tank is

$P=h _1\rho _1g+h _2\rho _2g$

$P=g(250\times 1+250\times 0.85)$

$P=462.5\,g/cm^2$
Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Suppose that a student finds that $24 \ mL$ of a certain liquid weighs $36 \ g$. What is the density of this liquid in SI unit?

  1. $1500 \ kg m^{-3}$
  2. $1200 \ kg m^{-3}$
  3. $1550 \ kg m^{-3}$
  4. $1300 \ kg m^{-3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, mass of liquid, $m=36g=36\times 10^{-3}kg$;
density of liquid, $\rho=?$
Volume of liquid, $V=24mL=24\times 10^{-3}L$
$=24\times 10^{-3}\times 10^{-3}m^3$
$=24\times 10^{-6}m^3$
Always remember, $1L=10^{-3}m^3=10^3 cm^3$,
Also, $1mL=1cm^3$(also called $1 \ cc$)
Density, $\rho=\displaystyle \frac{Mass}{Volume}=\frac{m}{V}=\frac{36\times 10^{-3}}{24\times 10^{-6}}$
$=1500kg m^{-3}$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

The densities of three liquids are D, 2D and 3D. What will be the density of the resulting mixture if equal volumes of the three liquids are mixed? 

  1. 6D

  2. 1.4D

  3. 2D

  4. 3D
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

let $V$ be the volume of each liquid then the total volume of the mixture becomes $3V$. 

$\rho = \dfrac{total\ mass}{total\ volume}$

Therefore, the mass of the liquids can be written as:
$m _1=D\times V=DV$
$m _1=2D\times V=2DV$
$m _1=3D\times V=3DV$

the total mass of the liquids is
$M= DV+2DV+3DV=6DV$

Therefore, the density of the mixture is:
$\rho=\dfrac{6DV}{3V}$

$\rho = 2D$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Two non-mixing liquids of densities $\rho$ and $n \rho ( n >$ 1) are put in a container. The height of each liquids $h$ . A solid cylinder of length $L$ and density $d$ is put in this container. The cylinder floats with its its axis vertical and length $p L ( p < 1 )$ in the denser liquid. The density $d$ is equal to

  1. $\{ 1 + ( n - 1 ) p \} p$
  2. $\{ 1 + ( n + 1 ) p \} p$
  3. $\{ 2 + ( n + 1 ) p \} p$
  4. $\{ 2 - ( n + 1 ) p \} p$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} d=density\, \, of\, \, cylin{ { de } }r \ A=area\, \, of\, \, cross-sectional\, \, of\, \, cylinder \ U\sin  g\, \, law\, \, of\, \, floation, \ weight\, \, of\, \, cylinder=up\, thrust\, \, by\, \, two\, \, liquids \ L\times A\times d\times g \ =n\rho \times \left( { pL\times A } \right) g+\rho \left( { L-pL } \right) Ag \ d=np\rho +\rho \left( { 1-p } \right) =\left( { np+1-p } \right) \rho  \ d=\left{ { 1+\left( { n-1 } \right) p } \right} \rho  \end{array}$

Hence,
option $(A)$ is correct answer.