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 fluid pressure pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

A $20$cm long capillary tube is dipped in water. The water rises up to $8$cm. If the entire arrangement is put in a freely falling elevator, the length of water column in the capillary tube will be:

  1. $8$cm
  2. $6$cm
  3. $10$cm
  4. $20$cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a freely falling lift, gravitational pull is zero hence the capillary tube will be filled completely.

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

Four identical capillary tubes $a, b, c$ and $d$ are dipped in four beakers containing water with tube ‘$a$’ vertically, tube ‘$b$’ at $30^{o}$, tube ‘$c$’ at $45^{o}$ and tube ‘$d$’ at $60^{o}$ inclination with the vertical. Arrange the lengths of water column in the tubes in descending order.

  1. $d, c, b, a$
  2. $d, a, b, c$
  3. $a, c, d, b$
  4. $a, b, c, d$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In capillary tube fluid always rise to the same vertical height as when the tube is perfectly vertical. So, the tube which is making greater angle with vertical will get more water in it.
So, order of lengths of water column will be $d > c > b > a$.

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 when immersed vertically in a liquid rises to 3 cm. If the tube is held immersed in the liquid at an angle of 60$^{o}$ with the vertical,the length of the liquid column along the tube will be:

  1. 2 cm

  2. 4.5 cm

  3. 6 cm

  4. 7.5 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


$h = \dfrac {2T cos \theta}{r \rho g}$
$h \alpha cos \theta$
for $\theta = 60^0 cos \theta = \dfrac {1}{2}$
$\therefore$ h is double $\Rightarrow h = 6 cm$.

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 is dipped in water vertically.Water rises to a height of 10mm. The tube is now tilted and makes an angle 60$^{o}$ with vertical.Now water rises to a height of:

  1. 10 mm

  2. 5 mm

  3. 20 mm

  4. 40 mm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If capillary tube is tilted the vertical height of water in tube remains same but volume of the water increases in the tube. So, height of water column will be 10mm.

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

Water rises in a capillary upto a height h. If now this capillary is tilted by an angle of $45^{\circ}$, then the length of the water column in the capillary becomes

  1. 2h
  2. $\displaystyle \frac{h}{2}$
  3. $\displaystyle \frac{h}{\sqrt{2}}$
  4. $h\sqrt{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression for the capillary rise in a tube is given by, 


H = $ \dfrac {2T cos \theta}{\rho g r} $

for all other parameters kept constant, if I change the angle of inclination to $ 45^o $

$ cos \theta $ will go from 1 to $ \dfrac {1}{\sqrt 2} $

Therefore, the height of capillary tube rise will also change from $ H to \dfrac{H}{\sqrt2} $

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

Two parallel glass plates are dipped partly in a liquid of density $'d'$ keeping them vertical. If the distance between the plates is $'x'$ Surface tension for liquid is $T$ & angle of contact is $\displaystyle \theta $ then rise of liquid between the plates due to capillary will be

  1. $\displaystyle \dfrac{T\cos \theta }{xd}$
  2. $\displaystyle \dfrac{2T\cos \theta }{xdg}$
  3. $\displaystyle \dfrac{2T}{xdg\cos \theta}$
  4. $\displaystyle \dfrac{T\cos \theta }{xdg}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

weight of liquid of height 'h' = (area of tube x h) x g x d= (3.14/4)hdg${x}^2$

vertical component of surface tension force=(1/2)x(Txcircumference)x cosθ=3.14Txcosθ
therefore, (3.14/4)hgd${x}^2$=(1/2)Tx3.14xcosθ
h=(2Tcosθ)/(gdx)
θ
θθs=Tx3.14x

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

Two capillary tubes of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube is filled with water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is $7.3 \times 10^{-2} N/m$. Take the angle of contact to be zero and density of water to be $10^3 kg/m^3(g = 9.8 m/s^2)$

  1. 5 mm

  2. 10 mm

  3. 15 mm

  4. 20 mm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\varrho gh=\dfrac { 2T }{ R } $

${ h } _{ 1 }=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times 1.5\times { 10 }^{ 3 } } $
${ h } _{ 2 }=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times 3\times { 10 }^{ -3 } } $
So $\triangle h={ h } _{ 1 }-{ h } _{ 2 }$
             $=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times { 10 }^{ -3 } } \left( \dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 }  \right) $      
             $=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times 3\times { 10 }^{ -3 } } $
             $\boxed { \triangle h=5mm } $

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

Water rises up to a height $h _1$ in a capillary tube of radius $r$. The mass of the water lifted in the capillary tube is $M$. If the radius of the capillary tube is doubled, the mass of water that will rise in the capillary tube will be 

  1. $M$
  2. $2M$
  3. $\cfrac{M}{2}$
  4. $4M$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since we know that mass of water rise is proportional to volume of water.

Mass $\infty $ volume
$\dfrac { { M } _{ 1 } }{ { M } _{ 2 } } =\dfrac { { V } _{ 1 } }{ { V } _{ 2 } } =\dfrac { \pi { r } _{ 1 }^{ 2 }{ h } _{ 1 } }{ \pi { r } _{ 2 }^{ 2 }{ h } _{ 2 } } =\dfrac { { r } _{ 1 }^{ 2 }{ h } _{ 1 } }{ { r } _{ 2 }^{ 2 }{ h } _{ 2 } } \quad \rightarrow (1)$
and for capillary tube, we know that height $\alpha $ $\dfrac { 1 }{ radius } $
      So, $\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } \quad \rightarrow (II)$
     hence from (1) & (II)
     $\dfrac { { M } _{ 1 } }{ { M } _{ 2 } } =\dfrac { { r } _{ 1 }^{ 2 } }{ { r } _{ 2 }^{ 2 } } \times \dfrac { { r } _{ 2 } }{ { r } _{ 1 } } =\dfrac { { r } _{ 1 } }{ { r } _{ 2 } } $
     So ${ M } _{ 2 }=\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } \times { M } _{ 1 }=\dfrac { 2r }{ r } \times M=2M$
              $\boxed { { M } _{ 2 }=2M } $

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

In a surface tension experiment with a capillary tube water rises up to $0.1 m$. If the same experiment is repeated on an artificial satellite which is revolving around the earth. The rise of water in a capillary tube will be

  1. $0.1 m$
  2. $9.8 m$
  3. $0.98 m$

  4. Full length of capillary tube
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If the experiment of capillary tube is performed in space then, it will rise to fall length of tube due to vaccum around it, i.e. no external pressure.

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

$5 g$ of water rises in the bore of capillary tube when it is dipped in water. If the radius of bore capillary tube is doubled, the mass of water that rises in the capillary tube above the outside water level is

  1. $1.5 g$
  2. $10 g$
  3. $5 g$
  4. $15 g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The height of water in a capillary tube is inversely proportional to the radius (h = 2T / rρg). If the radius is doubled, the height is halved. Since mass m = πr^2 h ρ, and h is proportional to 1/r, then m is proportional to r^2 * (1/r) = r. Doubling the radius doubles the mass.

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

The height of water in a capillary tube of radius $2 cm$ is $4 cm$. What should be the radius of capillary, if the water rises to $8 cm$ in tube? 

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

Since we know that height of capillary rise in inversely proportional to radius of capillary.

i.e.    height $\alpha $ $\dfrac { 1 }{ radius } $
         $\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } $
         $\dfrac { 4 }{ 8 } =\dfrac { { r } _{ 2 } }{ 2 } \Rightarrow \boxed { { r } _{ 2 }=1cm } $

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

Two capillary tubes of the same material but of different radii are dipped in a liquid. The heights to which the liquid rises in the two tubes are $2.2 cm$ and $6.6 cm$. The ratio of radii of the tubes will be

  1. $1:9$
  2. $1:3$
  3. $9:1$
  4. $3:1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since we know that height of capillary rise is inversely proportional to radii of tube, i.e.,

height $\propto \dfrac { 1 }{ radius } $

$\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } \Rightarrow \dfrac { 2.2cm }{ 6.6cm } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } $

So, $\boxed { \dfrac { { r } _{ 1 } }{ { r } _{ 2 } } =3 } $

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

The height of water in a capillary tube of radius $2 cm$ is $4 cm$. What should be the radius of capillary, if the water rises to $8 cm$ in tube?

  1. $1 cm$
  2. $0.1 cm$
  3. $2 cm$
  4. $4 cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

It is to be remembered that

height of a capillary rise $\propto \dfrac { 1 }{ radius\quad of\quad capillary } $
hence,
          $\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{  2} }{ { r } _{ 1 } } $
          $\dfrac { 4 }{ 8 } =\dfrac { { r } _{ 2 } }{ 2 } \Rightarrow \boxed { { r } _{ 2 }=1cm } $

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

A tank $4m$ high is half filled with water then filled to the top with a liquid of density $0.60 g/cc$ what is the pressure at the bottom of the tank due to these liquids? (take $g=10ms^{-2}$)

  1. $1.6 \times 10^3Nm^{-2}$
  2. $3.2\times 10^{-3}$
  3. $1.6 \times 10^4Nm^{-2}$
  4. $3.2 \times 10^4Nm^{-2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The tank is 4m high and half filled with water, meaning 2m of water and 2m of the other liquid. The pressure at the bottom is the sum of the hydrostatic pressures due to both liquid columns, calculated as P = h1*d1*g + h2*d2*g. Substituting the given values yields 1.6 x 10^4 N/m^2.