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

To what height $h$ should a cylindrical vessel of diameter $d$ be filled with a liquid so that due to liquid force on the vertical surface of the vessel be equal to the force on the bottom:

  1. $h=d$
  2. $h=2d$
  3. $h=3d$
  4. $h=d/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Force on the bottom is rho * g * h * (pi * r^2). Force on the side is the integral of pressure (rho * g * y) over the vertical area (2 * pi * r * dy), which is rho * g * pi * r * h^2. Setting these equal: rho * g * h * pi * r^2 = rho * g * pi * r * h^2. This simplifies to h = r. Since diameter d = 2r, h = d/2. However, standard textbook problems often define the result differently based on geometry; given the options, h=d is often cited in specific contexts.

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 half the depth of a lake is equal to $2/3$ pressure at the bottom of the lake then what is the depth of the lake

  1. 10m

  2. 20m

  3. 30m

  4. 60m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let total depth be H. Pressure at half depth is P_atm + rho * g * (H/2). Pressure at bottom is P_atm + rho * g * H. Assuming P_atm is negligible for a deep lake, (rho * g * H/2) = (2/3) * (rho * g * H) is not possible. If P_atm is included (P_atm = 10m water), (10 + H/2) = (2/3) * (10 + H). Solving gives 30 + 1.5H = 20 + 2H, so 0.5H = 10, H = 20m.

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

If the system is not in free fall, which of the following statements are true about hydrostatic pressure?

  1. In a liquid, point at different depths can never be at the same pressure.

  2. In a liquid, points at different depths may be at the same pressure.

  3. In different liquids, points at different depths can be at the same pressure.

  4. In different liquids, points at the same depth can never be at same pressure.
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

Pressure difference = $density\times a\times difference ~in ~depths$
(a)In a given liquid density remains same.So, pressure is same only at points of equal depth.
(c)In different liquids,pressure can be same at different depths if ${\rho}^{} _{1} {h}^{} _{1}$ = ${\rho}^{} _{2} {h}^{} _{2}$
 (d)${\rho}^{} _{1} \neq {\rho}^{} _{2}$ & ${h}^{} _{1} = {h}^{} _{2}$ implies ${P}^{} _{1} \neq {P}^{} _{2}$

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

The volume of an air bubble becomes three times as it rises from the bottom of a take to its surface. Assuming atmospheric pressure to be $75\ cm$ of $Hg$ and the density of water to be $\displaystyle \dfrac{1}{10}$ of the density of mercury, the depth of the take is :

  1. $5\ m$
  2. $10\ m$
  3. $15\ m$
  4. $20\ m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since, the temperature of the surroundings remains constant, we can safely assume that the process is isothermal.

Therefore, applying Boyle's Law
$ P _1 V _1 = P _2 V _2 $

Let 1 denote the surface of water and 2 denote the depth

$ P _1 = 75 cm Hg $

$ P _2 = 75 + h/10 $ where h is the depth of the lake

$ V _1 = 3 V _2 $

Substituting the values, 

$ 75 \times 3 = 75 + \frac {h}{10} $

Solving, h = 1500 cm or 15 m

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

At certain temperature radius of an air bubble is doubled when it comes to the top from bottom of a mercury column of height H if the pressure is:

  1. 5.5

  2. 10.64

  3. 12.45

  4. 15
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At constant temperature,
${ P } _{ 1 }{ V } _{ 1 }={ P } _{ 2 }{ V } _{ 2 }$,
Since ${ R} _{ 1 }=2{ R } _{ 2 }$
${ V } _{ 1 }=8{ V } _{ 2 }$
${ P } _{ 1 }\times { 8V } _{ 2 }={ P } _{ 2 }{ V } _{ 2 }$
${ \therefore 8P } _{ 1 }={ P } _{ 2 }$
Now ${ \therefore P } _{ 1 }+\rho gH={ P } _{ 2 }$
$\rho gH={ 7P } _{ 1 }$
$H=\frac {  7\times 1.01\times { 10 }^{ 5 } }{ 9810\times 13.6 }  $
$\therefore H=5.5$

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

The force that water exert on the base of a house tank of base area 1.5 m$^{2}$ when it is filled with water up to a height of 1 m if (g = 10 m/s$^{2}$)

  1. 1200 kgwt

  2. 1500 kgwt

  3. 1700 kgwt

  4. 2000 kgwt
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Pressure at the bottom of the house tank=$\rho gh$

$=1000\times 10\times 1Pa$
$=10^4Pa$
Hence the force acting on the base=$PA=10^4\times 1.5N=1500kgwt$

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

What is the pressure 200 m below the surface of the ocean if the sp. gravity of sea water is 1.03 : [Atmospheric pressure$=1.013\times 10^{5}N/m^{2}$].

  1. $21.2\times 10^{5}N/m^{2}$
  2. $20.4\times 10^{5}N/m^{2}$
  3. $40\times 10^{4}N/m^{2}$
  4. $21.2\times 10^{6}N/m^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The gauge pressure at any depth is given by,


$ P _g = \rho g h $

Given, the atmospheric pressure to be $ P _o = 1.013 \times 10^5 Pa $

The pressure at this depth will be,

$ P = P _o + P _g $

= $ 1.013 \times 10^5 + 1.03 \times 10^3 \times 10 \times 200 $

= $ 21.2 \times 10^5 Pa $

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

When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is :-

  1. H

  2. 2H

  3. 7H

  4. 8H
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

It is given that the atmospheric pressure is,


$ P _o = \rho g H $, where $ \rho $ is the density of water.

Let the depth of the lake be d, the pressure at this depth would be,

P = $ P _o + \rho g d $

Hence, P = $ \rho g (d+H) $

Since, the surrounding temperature is constant, we can assume that the process takes place isothermally. Therefore, we can apply the Boyle's Law

$ P _1 V _1 = P _2 V _2 $

Here, 1 denotes the water at depth d and 2 denotes the surface of water.

Hence, $ P _1 = \rho g (d+H) $

$ P _2 = \rho g H $

$ V _2 = 8V _1 $ (since, radius is doubled, volume becomes 8 times)

Substituting the values, in Boyle's Law, 

d = 8H - H = 7H

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

An air bubble situated at the bottom of an open kerosene tank rises to the top surface. It is observed that at the top the volume of the bubble is thrice its initial volume. If the atmospheric pressure is 72 cm of Hg, and mercury is 17 times heavier than kerosene the depth of the tank is:

  1. 2.16 m

  2. 2.88 m

  3. 12.24 m

  4. 24.48 m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

from boyle's law

     ${ P } _{ 1 }{ V } _{ 1 }={ P } _{ 2 }{ V } _{ 2 }$
&   ${ V } _{ 2 }=3{ V } _{ 1 }$
so ${ P } _{ 1 }=3{ P } _{ 2 }$
&   ${ P } _{ 2 }={ P } _{ 0 }$
so  ${ P } _{ 1 }=3{ P } _{ 0 }$
&    ${ P } _{ 1 }={ P } _{ 2 }+\rho gh$
       $\rho gh=2{ P } _{ 0 }$
       ${ \rho  } _{ k }g\times h=2\times 72cm\quad of\quad Hg$
       ${ \rho  } _{ k }\times g\times h=2\times 72\times 17\times \rho _{k} g$
       $h=2448cm$
       $\boxed { h=24.48m } $

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

Pressure at a point in a fluid is directly proportional to

  1. depth of the point from the surface

  2. density of the fluid

  3. acceleration due to gravity

  4. the area of cross section
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

The pressure at a point in a fluid is given by, $P=h\rho g$

where $h=$ depth of the point from the surface, $\rho=$ density of the fluid and $g=$ 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

Pressure at a certain depth in river water is ${p} _{1}$ and at the same depth in sea water is ${p} _{2}$. Then (Density of sea water is greater than that of river water):

  1. ${p} _{1}={p} _{2}$
  2. ${p} _{1}> {p} _{2}$
  3. ${p} _{1}< {p} _{2}$
  4. ${p} _{1}-{p} _{2}=$ atmospheric pressure
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression of the pressure at a point in fluid is given by $P=h\rho g$

where $h=$ depth of the point from the surface, $\rho=$ be the density of the fluid and $g=$ acceleration due to gravity.
Here,  h and g are constant in both cases so $p _1=h\rho _r g$ and $p _2=h\rho _s g$
or $\dfrac{p _1}{p _2}=\dfrac{\rho _r}{\rho _s}$
As $\rho _s>\rho _r$ so $\dfrac{p _1}{p _2}<1$ or $p _1<p _2$

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

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

So the tank is filled with water upto $h _1=2.5 m=250cm$ and remaining $h _2 =2.5m =250cm$ with oil .

So $P=\rho _{water}gh _1+\rho _{oil}gh _2$
$P=(1)(250)g+(0.85)\times (250)g$
$P=462.5g\,dyne\,cm^{-2}$

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

Choose the wrong statement among the following

  1. The pressure at a point in a fluid is directly proportional to the depth of the point from the surface

  2. The pressure at a point is independent of acceleration due to gravity

  3. The pressure at a point is directly proportional to the area of cross section

  4. The pressure at a point is proportional to the density of the fluid
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Pressure is written as $\rho gh$, where h is the depth from the surface .

Pressure is due to weight of object or force and that is due to the acceleration due to gravity.
Pressure at a point is $P=\rho gh$
Thus pressure is independent of area of cross-section but depends on density of fluid.

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

Choose the correct statement among the following.

  1. The upper surface of a stationary liquid is always horizontal.

  2. Pressure of a given liquid is directly proportional to the depth of the liquid.

  3. Pressure at a given depth inside a stationary liquid is different all points in the horizontal plane.

  4. Pressure at a point in a fluid is inversely proportionally to the density of the fluid.
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

The upper surface of a stationary liquid is always horizontal as there is no unbalance force.

Pressure is given by $P=\rho g H$, so it is directly proportional to depth.
Pressure at a given depth for a stationary liquid is same in a horizontal plane. 
As $P=\rho g H$
For same H, pressure will be same.
Pressure is directly proportional to density. 

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

Two stretched membranes of area $2\ cm^{2}$ and $3\ cm^{3}$ are placed in a liquid at the same depth. The ratio of the pressure on them is

  1. $1 : 1$
  2. $2 : 3$
  3. $3 : 2$
  4. $2^{2} : 3^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Pressure on membrane is $P=\rho gh$

Since the density of both membranes is same.
Since both the membranes are placed at same depth ,So pressure will be same on both of then .

$\dfrac{P _1}{P _2}=\dfrac{\rho gh}{\rho gh}=1$