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 option b: engineering physics buoyancy floatation fluid pressure

oil bath (density of oil$=0.85\times { 10 }^{ 3 }kg/m^{ 3 })$ has a spherical cavity of diameter $26\times { 10 }^{ -6 }$ m at a depth of 0.2 face tension of oil is $26\times { 10 }^{ -3 }$ N/m and the pressure of air over the surface of oil is 76 cm of mercury, the 

  1. $1.03\times 105N/m^{ 2 }$
  2. $1.17\times { 10 }^{ 5 }N/m^{ 2 }$
  3. $3.07\times { 10 }^{ 5 }N/m^{ 2 }$
  4. $1.07\times { 10 }^{ 5 }N/m^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

The height to which a cylindrical vessel be filled with a homogeneous liquid, to make the average force with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel is equal to:

  1. half of the radius of the vessel.

  2. radius of the vessel.

  3. one-fifth of the radius of the vessel.

  4. three-fourth of the radius of the vessel.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $h$ is the height of liquid in cylinder, $r$ be the radius of the cylinder and $\rho$ be the density of the liquid. then we have
weight of the liquid $=\pi { r }^{ 2 }h\rho g          ....... (I)$
Mean pressure on the wall $=\frac{1}{2} \rho g h $
force on the wall $=\frac{1}{2} \rho g h \times 2 \pi r h=\pi r \rho g h^2          ....... (II)$
On equating $(I)$ and $(II)$ we have
$\pi { r }^{ 2 }h\rho g=\pi r \rho g h^2$
$\Rightarrow r=h$
i.e. the liquid should be filled up-to a height equal to the radius of the cylinder.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

By sucking through a straw, a student can reduce the pressure in his lungs to $750mm$ of $Hg$ (Density $=13.6g/{cm}^{3}$). Using the straw, he can drink water from a glass up to a maximum depth of

  1. $10cm$
  2. $75cm$
  3. $13.5cm$
  4. $1.36cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,
$\rho _H=13.6g/cm^3$
$\rho _w=1g/cm^3$
$P _0=760mm\,  of\,  Hg$
$P _l=750mm\,   of\,  Hg$
The pressure difference between the lungs of student and the atmosphere is given by
$\Delta P=P _0-P _l$
$\Delta P=760-750=10mm \,  of\,   Hg$
$\Delta P=1cm\,  of\,   Hg$
This pressure can be used for the drinking water.
$1cm\,   of\,   Hg=$ Pressure difference due to water column
$\rho _H gh _H=\rho _w gH$
$1\times 13.6\times g=1\times g\times H$
$H=13.6cm $
The correct option is C.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

The hydrostatic pressure on a diver $100m$ below the surface of an ocean is

  1. $1$ atm
  2. $2$ atm
  3. $11$ atm
  4. $20$ atm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Hydrostatic pressure P = P_atm + rho*g*h. P_atm = 1 atm. rho*g*h = 1000 * 10 * 100 = 10^6 Pa. Since 1 atm approx 10^5 Pa, rho*g*h = 10 atm. Total pressure = 1 + 10 = 11 atm.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A tank of height H is fully filled with water.If the water rushing from a made in the tank below the free surface,strikes the floor at maximum horizontal distance then depth of the hole from the free surface must be.

  1. $ \left( \frac { 3 }{ 4 } \right) H $
  2. $ \left( \frac { 2 }{ 3 } \right) H $
  3. $ \left( \frac { 1 }{ 4 } \right) H $
  4. $ \left( \frac { 1 }{ 2 } \right) H $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A conical portion is cut out of a solid hemisphere of radius R and the remaining portion is held in a liquid of density $ \rho $ through a string as shown in the figure.What is the net force exerted by the liquid on the body.

  1. $ \frac {1}{6} \pi R^3 \rho g $
  2. $ \frac {1}{3} \pi R^3 \rho g $
  3. $ \frac {1}{4} \pi R^3 \rho g $
  4. $ \frac {1}{2} \pi R^3 \rho g $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

Mark out the correct statement(s)

  1. Net force acting on the base of the vessel > weight of the liquid inside the vessel
  2. Net force acting on the base of the vessel $=$ weight of the liquid inside the vessel
  3. Net pressure force acting on the liquid $=$ weight of the vessel

  4. Both (a) and (c) are correct
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Weight of the liquid inside the vessel,

$W=\rho(A _1\times \dfrac{5}{100}+A _2\times \dfrac{1}{100})g=1N$
So, $F>W$
Net force on the liquid is zero.

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A uniform solid cylinder of density $ 0.8 g/cm^3 $ floats in equilibrium in a combination of teo non-mixing liquid A and B with its axis vertical. the densities of liquid A ad B with its axis vertical. the densities of liquid A and B are $ 0.7 g /cm^3 $ and $ 1.2 \times gm/cm^3 $. the height of liquid A is $ h _A = 1.2 cm $ and the length of the part of cylinder immersed in liquid B is $ h _B = 0.8 cm $ then the length of the cylinder in air is

  1. 0.21 m

  2. 0.25 cm

  3. 0.35 m

  4. 0.4 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

A cylindrical vessel of $100\ cm$ height is kept filled upto the brim. It has four holes $1, 2, 3, 4$ which are respectively at heights of $27\ cm, 30\ cm, 50\ cm$ and $80\ cm$ from the horizontal floor. The water falling at the maximum horizontal distance from the vessel comes from

  1. Hole number $4$
  2. Hole number $3$
  3. Hole number $2$
  4. Hole number $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

Two metal plates $'A'$ and $'B'$ having the same breadth but different lengths $\ell 1$ and $\ell _2 $ respectively are placed at same depth inside water such that their breadth is held exactly in vertical positions. Then, the ratio of the pressure acting on $'A'$ and $'B'$ by water is ____.

  1. $1:1$
  2. $\ell _1:\ell _2$
  3. $\ell _2:\ell _1$
  4. $\ell _1 b:\frac{\ell _2}{b}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

To what height h should a cylindrical vessel of diameter d be filled with a liquid so that the total 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
D Correct answer
Explanation

If we fill the cylinder upto a height h, the force exerted on at the bottom of the cylinder would be equal to F = PA


$ F = \rho gh \times \pi d^2/4 $

Similarly, the average force exerted along the sides of the cylinder will be because of half the height filled for the cylinder.

Therefore, $ F = \rho g h/2 \times \pi d h $

Equating the 2 forces, and solving for h, gives h = d/2

Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

The height of liquid in a cylindrical vessel of diameter $d$ so that the total force on the vertical surface of the vessel be equal to the force on the bottom, will be:

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

Force on bottom = P * Area = (rho * g * h) * (pi * (d/2)^2). Force on vertical surface = Integral of (rho * g * y) * (pi * d) dy from 0 to h = (rho * g * h^2 / 2) * (pi * d). Setting these equal: h/4 = h^2/2d => h = d/2.