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

A cylindrical container of radius R = 1 m, wall thickness 1 mm is filled with water up to a depth of 2 m and suspended along its upper rim. The density of water is 1000 kg/m3 and acceleration due to gravity is 10 m/s2. The self weight of the cylinder is negligible. The formula for hoop stress in a thin walled cylinder can be used at all points along the height of the cylindrical container.

If the Young's modulus and Poisson's ratio of the container material are 100 GPa and 0.3 respectively, the axial strain in the cylinder wall at mid-depth is

  1. 2 × 10-5

  2. 6 × 10-5

  3. 7 × 10-5

  4. 1.2 × 10-4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sigma_a=\frac{10^4 \times2\times1}{2\times10^{-3}\times100\times10^9}\bigg(\frac{1}{2}-0.3\bigg)\\ \sigma_a=\frac{10^4}{10^8}\bigg(\frac{1}{2}-0.3\bigg)=10^{-4}\times0.2=2\times10^{-5}$

Multiple choice

A cylindrical container of radius R = 1 m wall of thickness 1 mm is filled with water up to a depth of 2 m and suspended along its upper rim. The density of water is 1000 kg/m3 and acceleration due to gravity is 10 m/s2. The self weight of the cylinder is negligible. The formula for hoop stress in a thin walled cylinder can be used at all points along the height of the cylindrical container.

The axial and circumferential stress $(\sigma_d,\sigma_c)$ experienced by the cylinder wall at mid-depth (1 m as shown) are

  1. (10, 10) MPa

  2. (5, 10) MPa

  3. (10, 5) MPa

  4. (5, 5) MPa
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Multiple choice applications of gauss's law coulomb's law physics

A spherical shell of mass $m$ and radius $R$ filled completely with a liquid of same mass and set to rotate about a vertical axis through its centre has a moment of inertia ${I _1}$ about the axis$.$ The liquid starts leaking out of the hole at the buttom$.$ If moment of inertia of the system is ${I _2}$ when the shell is half filled and ${I _3}$ is the moment of inertia when entire water drained off$.$ then $:$

  1. $\dfrac{{{I _1}}}{{{I _2}}} \approx 1.5$
  2. $\dfrac{{{I _{ _1}}}}{{{I _2}}} \approx 0.67$
  3. $\dfrac{{{I _1}}}{{{I _3}}} \approx 1.6$
  4. $\dfrac{{{I _2}}}{{{I _3}}} \approx 1.4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Moment of inertia of a shell$,$ ${I _3} = \frac{2}{3}M{R^2}$ 

Moment of inertia of a completely filled sphere$,$ ${I _1} = \frac{2}{3}M{R^2} + \frac{2}{5}M{R^2} = \frac{{16}}{{15}}M{R^2}$ 
Moment of inertia of a half sphere$,$ ${I _2} = \frac{2}{3}M{R^2} + \frac{2}{5}\frac{M}{2}{R^2} = \frac{{26}}{{30}}M{R^2}$
Hence,
option $(C)$is correct answer.

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

What is the change in surface energy, when a mercury drop of radius $r$ splits up into 1000 droplets of radius $r$?

  1. $8pr^2t$
  2. $28pr^2t$
  3. $16pr^2t$
  4. $36pr^2t$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let bigger drop has radius $R$

Volume will remain same. 
 $\dfrac{4}{3}\pi R^3 = 1000\times \dfrac{4}{3} \pi r^3  \implies   r= \dfrac{R}{10}$

Change in surface energy-
 $A _2T - A _1T = 1000\times 4\pi r^2T-4\pi R^2T=4\pi[10R^2-R^2]T=36\pi R^2T$

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

Two soap bubbles have radii in the ratio of $4:3$. What is the ratio of work done to blow these bubbles?

  1. 4:3

  2. 16:9

  3. 9:16

  4. 3:4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work done to brust a soap bubble is change in surface energy which is equal surface Tension $\times $ Area.

Workdone $\alpha $ Area $\alpha $ ${ r }^{ 2 }$
hence $\dfrac { { w } _{ 1 } }{ { w } _{ 2 } } =\dfrac { { r } _{ 1 }^{ 2 } }{ { r } _{ 2 }^{ 2 } } ={ \left( \dfrac { 4 }{ 3 }  \right)  }^{ 2 }=\dfrac { 16 }{ 9 } $

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

W is the work done, when a bubble of volume V is formed from a solution. How much work is required to be done to form a bubble of volume 2V?

  1. $2W$
  2. $W$
  3. $2^{1/3}W$
  4. $2^{2/3}W$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the radius of the bubble with volume $V$ be $  r$.

Then the radius of the bubble  with volume $2V$ is $2^{1/3}r$.
Work is only because of change in surface energy.
Ratio of work done =  Ratio of surface energy =Rratio of surface area 
$\dfrac{W _1}{W} = \dfrac{4\pi (2^{1/3}r)^2}{4\pi r^2} \implies W _1= 2^{2/3}W$

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

A mercury drop of radius 1 cm is broken into 106 droplets of equal size. The work done is

$(T = 35 10^{-2} N/m)$

  1. $1.615\times 10^{-3}J$
  2. $4.35\times 10^{-4}J$
  3. $1.615\times 10^{-6}J$
  4. $4.35\times 10^{-8}J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the radius of smaller drop be r, so

$\dfrac{4\pi (1)^2}{3}=106\dfrac{4\pi r^3}{3}$
$r=0.21cm=0.21\times 10^{-2}m$
work done is $W=T(dA)=T(106\times 4\pi (0.21\times 10^{-2})^2-4\pi (1\times 10^{-2})^2)$
here $T=35\times 10^{-2}$
$W=1615\times 10^{-6}J$

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

Surface tension of a soap solution is $1.9 10^{-2} N/m.$ Work done in blowing a bubble of 2.0cm diameter will be

  1. $7.6 10^{-6} J$
  2. $15.2 10^{-6} J$
  3. $16 10^{-6} J$
  4. $2.5 10^{-6} J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work done = Change in surface energy

                   = Surface Tension $\times $ Area
                   $= 1.910 \times $ $4\pi \times { \left( 2\times { 10 }^{ -2 } \right)  }^{ 2 }$
workdone = $15.2\times { 10 }^{ -6 } J$.

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

The surface tension of a soap solution is $0.035 N/m$. the energy needed to increase the radius of the bubble from $4$ cm to $6$ cm is

  1. $1.75\times 10^{-3} J$
  2. $1.510^{-2} J$
  3. $310^{-3} J$
  4. $1.510^{-4} J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There are two free surface available in bubble.

Change in surface area $\Delta A = 2(4\pi R _f^2 - 4\pi R _i^2)$
Energy needed to change radius from 4 to 6 is 

$W=T \Delta A=T\times 2(4\pi R _f^2-4\pi R _i^2)$
$W=.035\times 2\times 4\pi((6\times 10^{-2})^2-(4\times 10^{-2})^2)$
$W=1.75\times 10^{-3}J$

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

Two soap bubbles of radii 4 cm and 3 cm respectively coalesce under isothermal conditions to form a single bubble. What is the radius of the new single bubble?

  1. $3 cm$
  2. $4 cm$
  3. $5 cm$
  4. $6 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given:
Radius of first soap bubble, $r _1 = 4 cm$
Radius of second soap bubble, $r _2 = 3 cm$
Let,
$P _1 = \dfrac{4T}{(r _1)^2}, V _1 = \dfrac{4}{3} \pi (r _1)^3$ be the excess pressure inside first soap bubble and volume of first soap bubble respectively.  
$P _2 = \dfrac{4T}{(r _2)^2}, V _2 = \dfrac{4}{3} \pi (r _2)^3$ be the excess pressure inside second soap bubble and volume of second soap bubble respectively. 
And $P = \dfrac{4T}{r^2}, V _2 = \dfrac{4}{3} \pi r^3$be the excess pressure inside new soap bubble, volume and radius of new soap bubble respectively. 
The two bubbles combine isothermally hence,
$PV = P _1 V _1 + P _2 V _2$

$\dfrac{4T}{r} \dfrac{4}{3} \pi r^3 = \dfrac{4T}{(r _1)} \dfrac{4}{3} \pi (r _1)^3 + \dfrac{4T}{(r _2)} \dfrac{4}{3} \pi (r _2)^3$

$r^2 = (r _1)^2 + (r _2)^2$
$r = \sqrt ((r _1)^2 + (r _2)^2)$
$r = \sqrt ((4)^2 +(3)^2)$
$r = \sqrt 25$
$r = 5 cm$
Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

A spherical water drop of radius $R$ is split up into $8$ equal droplets. If $T$ is the surface tension of water, then the work done in this process is-

  1. $4\pi R^2T$
  2. $8\pi R^2T$
  3. $48\pi R^2T$
  4. $2\pi R^2T$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let smaller drop has radius $r$

Volume will remain same. 
$\dfrac{4}{3}\pi R^3 = 8\times \dfrac{4}{3} \pi r^3  \implies   r= \dfrac{R}{2}$

Change in surface energy-
 $A _2T - A _1T = 8\times 4\pi r^2T-4\pi R^2T=4\pi[2R^2-R^2]T=4\pi R^2T$
Work done  = change in surface energy  = $4\pi R^2T$

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

A drop of oil is placed on the surface of water. Which of the following statement is correct?

  1. It will remain on it as a sphere

  2. It will spread as a thin layer

  3. It will partly be as spherical droplets and partly as thin film

  4. It will float as distorted drop on the water surface.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Adhesive force between oil and water molecules is greater than cohesive force between oil molecules. So the oil molecules do not mix with water molecules. As a result, oil spreads on the water surface forming a thin layer.

Multiple choice surface tension physics surface energy of a liquid work done in stretching a liquid surface: surface energy of a liquid properties of matter

One thousand small water droplets of equal size combine to form a big drop. The ratio of the final surface energy to the initial surface energy is: (Surface tension of water = 70 dyne/cm) 

  1. 10:1

  2. 1000:1

  3. 1:10

  4. 1:1000
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the radius of smaller radius be $r$ and bigger be $R$.

Since volume remains constant, thus  $\dfrac{4\pi R^3}{3}=1000\times \dfrac{4\pi r^3}{3}$
We get  $R=10r$
Ratio of final surface energy to initial is 
$\dfrac{W _f}{W _i}=\dfrac{T\times 4\pi R^2}{T\times 1000\times 4\pi r^2}$
$\dfrac{W _f}{W _i}=\dfrac{(10r)^2}{1000r^2}=1:10$