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 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 
$\dfrac{W _f}{W _i}=\dfrac{T\times 4\pi R^2}{T\times 1000\times 4\pi r^2}$
$\dfrac{W _f}{W _i}=\dfrac{(10r)^2}{1000r^2}=1:10$

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 spherical soap bubbles of a radii $r _1$ and $r _2$ in vacuum coalesce under isothermal conditions. The resulting bubble has the radius $R$ such that

  1. $R=r _1+r _2$
  2. $R^2 ={ r _1^2 + r _2^2}$
  3. $R=\dfrac{r _1+r _2}{r _2}$

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

When two soap bubbles coalesce in vacuum under isothermal conditions, the total number of moles of air is conserved. Since PV = nRT, and n is constant, P1V1 + P2V2 = PV. For a bubble, P = 4T/r, and V = (4/3)pi*r^3. So, (4T/r1)(4/3)pi*r1^3 + (4T/r2)(4/3)pi*r2^3 = (4T/R)*(4/3)pi*R^3. This simplifies to r1^2 + r2^2 = R^2.

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. $8\pi R^2T$
  2. $16\pi R^2T$
  3. $24\pi R^2T$
  4. $36\pi R^2T$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let Surface Tension be T.

Initial surface = $4\Pi { R }^{ 2 }$
Final Surface area = $1000\times 4\Pi { r }^{ 2 }$
& Initial Volume = final volume
$\dfrac { 4 }{ 3 } \Pi { R }^{ 3 }=1000\times \dfrac { 4 }{ 3 } \Pi { r }^{ 3 }$
R = 10r
So,
    Final surface area = $1000\times 4\Pi \times \dfrac { { R }^{ 2 } }{ 100 } =40\Pi { R }^{ 2 }$
So, Workdone = Change in Surface energy
                         = $T\Delta A=(40-4)\Pi { R }^{ 2 }T$
       $\boxed { Change\quad in\quad Surface\quad energy=36\Pi { r }^{ 2 }T } $.

Multiple choice

The volume of water in a piece of ice,

  1. Is equal to the surface area of the ice cube

  2. is equal to the volume of the ice below the water line

  3. is less than the volume of whole piece of ice

  4. Greater than (2*l*w)/(h), given the dimension of the ice
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Archimedes' principle states that a floating object displaces a weight of fluid equal to its own weight. Since ice is less dense than water, the volume of the displaced water is exactly equal to the volume of the submerged portion of the ice.

Multiple choice botany mineral nutrition in plants mechanism of absorption mineral absorption significance of transpiration

Which of the following statement is incorrect?

  1. Rise of water in a narrow tube is inversely proportional to the radius of the tube

  2. A water melon has over 92% water

  3. Dry plant material can imbibe ether but rubber cannot imbibe

  4. when water flows into thew cells and out of the cell and are in equilibrium, the cells are said to be flaccid
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice intermolecular forces: cohesive and adhesive forces surface tension properties of matter physics

A wire 10  cm long is placed horizontally on the surface of water. It is gently pulled up with a force of $1.4 \times 10^{-2}$ N to keep it in equilibrium. The surface tension of water is

  1. 50 dyne /cm

  2. 60 dyne/cm

  3. 70 dyne/cm

  4. 40 dyne/cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Surface tension T is force per unit length. Since the wire has contact on both sides of its surface on water, the effective length is 2l. Thus, T = F / (2l) = 1.4 * 10^-2 / (2 * 0.1) = 0.07 N/m, which equals 70 dyne/cm.

Multiple choice intermolecular forces: cohesive and adhesive forces surface tension properties of matter physics

The surface tension and vapour pressure of water at $20^oC$ is 7.28$ \times10^{ -2 }$ N/m and $2.33 \times { 10 }^{ 3 }$ pa, respectively. What is the radius of the smallest spherical water droplet which can from without evaporating at $20^oC$ ?

  1. $5\times { 10 }^{ -4 }m$
  2. $6.25\times{10}^{-5}m$
  3. $9\times{10}^{-2}m$
  4. $3\times{10}^{-5}m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice intermolecular forces: cohesive and adhesive forces surface tension properties of matter physics

 Work done in increasing the size of a soap bubble from a radius of $3\ cm$ to $5\ cm$ is nearly (Surface tension of soap solution$=0.03\ {Nm}^{-1}$):

  1. $0.2\pi\ mJ$
  2. $2\pi\ mJ$
  3. $0.4\pi\ mJ$
  4. $4\pi\ mJ$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Work done = Surface Tension * Change in Surface Area. A soap bubble has two surfaces, so Area = 2 * (4 * pi * r^2) = 8 * pi * r^2. Change in Area = 8 * pi * (r2^2 - r1^2) = 8 * pi * (0.05^2 - 0.03^2) = 8 * pi * (0.0025 - 0.0009) = 8 * pi * 0.0016 = 0.0128 * pi m^2. Work = 0.03 * 0.0128 * pi = 0.000384 * pi J = 0.384 * pi mJ, which is approximately 0.4 * pi mJ.

Multiple choice intermolecular forces: cohesive and adhesive forces surface tension properties of matter physics

If the resultant adhensive force between liquid and solid is smaller than the resultant cohesive then will_______

  1. partially wets the solid

  2. does not wets the solid

  3. completely wets the solid

  4. does not interact with solid
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When the adhesive force between a liquid and a solid is smaller than the cohesive force within the liquid, the liquid molecules prefer to bind to each other rather than the solid surface, causing it not to wet the solid.