Physics

Fluid Mechanics and Hydraulics

376 Questions

Fluid mechanics and hydraulics questions address the principles of fluid flow, pipe resistance, and open channel dynamics. The topics include Bernoulli equation, Navier-Stokes equation, and hydrograph calculations. These concepts are crucial for civil and mechanical engineering competitive examinations.

Fluid flow equationsOpen channel flowPipe frictionHydraulic jumpHydrograph analysis

Fluid Mechanics and Hydraulics Questions

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

A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water. The quater of water flowing out per second from both holes are the same. Then radius R, is equal to :

  1. $\dfrac { L }{ \sqrt { 2\pi } } $
  2. $2\pi L$

  3. L
  4. $\dfrac { L }{ 2\pi } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice physics option b: engineering physics buoyancy floatation fluid pressure

The efflux velocity of a liquid of density $1500 kg m^{-3} $ from a tank in which the pressure of liquid is $1000pa$ above the atmosphere is :

  1. $115 ms^{-1} $
  2. $11.5 ms^{-1} $
  3. $0.115 ms^{-1} $
  4. $1.15 ms^{-1} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The velocity of a liquid is given as,

$v = \sqrt {2gh} $

$v = \sqrt {2g \times \frac{{\Delta P}}{{\rho g}}} $

$v = \sqrt {2 \times \frac{{1000}}{{1500}}} $

$v = 1.15\;{\rm{m/s}}$

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 capillaries of same length and radii in the ratio 1: 2 are.connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination is 1 m of water, the pressure difference across first capillary is

  1. 9.4 m

  2. 4.9 m

  3. 0.49 m

  4. 0.94 m
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Here, $l _1 = l _2 = 1m$ and $\displaystyle \frac{r _1}{r _2} = \frac{1}{2}$

As $V = \displaystyle \frac{ \pi P _1 r _1^4 }{8 \eta l} = \frac{ \pi P _2 r _2^4 }{8 \eta l}$ or $\displaystyle \frac{ P _1 }{P _2 } = \left( \frac{ r _2 }{ r _1} \right)^4 = 16$

$\therefore P _1 = 16 P _2$

Since, both tubes are connected in series, hence pressure difference across the combination is
$P = P _1 + P _2$ $\Rightarrow$  $\displaystyle 1 = P _1 + \frac{P _1}{16}$
or $\displaystyle P _1 = \frac{16}{17} = 0.94 m$
Multiple choice physics fluid pressure pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

A liquid is allowed to flow in a tube of truncated cone shape. Identify correct statement from the following.

  1. The speed is high at the wider end and low at the narrow end

  2. The speed is low at the wider end and high at the narrow end

  3. The speed is same at both ends in a stream line flow

  4. The liquid flows with uniform velocity in the tube
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an incompressible liquid equation of continuity $ Av = constant$

or, $ A \propto \cfrac{1}{v}$
Therefore at the wider end speed will be low and at the narrow end speed will be hgih.

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

If a capillary tube is tilted to $45^{\circ}$ and $60^{\circ}$ from the vertical then the ratio of length $l _{1}$ and $l _{2}$ of liquid columns  in it will be -

  1. $1: \sqrt{2}$
  2. $\sqrt{2}:1$

  3. 1:2

  4. 2:1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


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


where $ \theta$ is the contact angle and not the angle of tilt.


Thus, for all parameters constant,


at $ 60^o $


$ l _1 = H cos 60^0 $ = H/2


At $ 45^o $


$ l _2 = H cos 45^0 $ = $ H/\sqrt 2 $


Therefore the ratio of the 2 lengths is given by,


$ l _1/l _2 = \dfrac {1}{\sqrt 2} $


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

Water rises in a vertical capillary tube upto a length of $10cm.$ If the tube is inclined at $45^o$, the length of water risen in the tube will be,

  1. $10 cm$
  2. $10 \sqrt2 cm$
  3. $\displaystyle \dfrac {10}{\sqrt2}$

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

The vertical rise in the level of liquid is constant.
Hence for an inclined tube, the effective length is $ \displaystyle\dfrac {h}{\cos\theta} $
So, the length of water risen in the tube will be $ 10 \sqrt 2 cm $

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

A large vessel with a small hole at the bottom is filled with water and kerosene. The height of the water column is 20 cm and that of the kerosene is 25 cm. the velocity with which water flows out the hole is

  1. 2 m/s

  2. 4 m/s

  3. 2.8 m/s

  4. 1 m/s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Torricelli's law: v = sqrt(2gh_eff). Pressure at the hole is due to water and kerosene. P = h_w * ρ_w * g + h_k * ρ_k * g. h_eff = h_w + h_k * (ρ_k/ρ_w). h_eff = 0.2 + 0.25 * (0.8) = 0.2 + 0.2 = 0.4m. v = sqrt(2 * 10 * 0.4) = sqrt(8) = 2.82 m/s.

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

Water is being poured into a vessel at a constant rate $ qm^2/s $. There is small aperture of cross-section area 'a' at the bottom of the vessel.The maximum level of water level of water in the vessel is proportional to

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

At steady state, the rate of inflow q equals the rate of outflow, which is given by a * v = a * sqrt(2 * g * h). Thus, q = a * sqrt(2 * g * h). Solving for h gives h = q^2 / (2 * g * a^2), meaning h is proportional to 1/a^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 cylindrical tank having cross-sectional area $A$ is filled with water to a height of $2.0m$. A circular hole of cross-sectional area $a$ is opened at a heigh of $75cm$ from the bottom. If $\cfrac{a}{A}=\sqrt{0.2}$, the velocity with which water emerges from the ole is ($g=9.8m{s}^{-2}$)

  1. $4.9m{s}^{-1}$
  2. $4.95m{s}^{-1}$
  3. $5.0m{s}^{-1}$
  4. $5.5m{s}^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Torricelli's law, v = sqrt(2 * g * h), where h is the height of the water column above the hole. Here h = 2.0m - 0.75m = 1.25m. v = sqrt(2 * 9.8 * 1.25) = sqrt(24.5) = 4.95 m/s. Given the options, 5.0 m/s is the closest approximation.

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

Water flows into a large tank with flat bottom at the rate of $ 10^{-4} m63s^{-1} $. water is also leaking out of a hole of area $ 1cm^2 $ at its bottom. if the height of the water in the tank remains steady , then this height is: 

  1. 5.1 cm

  2. 1.7 cm

  3. 4 cm

  4. 2.9 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Steady state implies inflow rate = outflow rate. Inflow = 10^-4 m^3/s. Outflow = a * sqrt(2 * g * h). 10^-4 = 10^-4 * sqrt(2 * 10 * h). 1 = sqrt(20 * h). 1 = 20h, h = 0.05 m = 5 cm.

Multiple choice physics simple machine common machines terms related to machines introduction to simple machines

A pump ejects $12000\ kg$ of water at the speed of $4\ { m }/{ s }$ in $40\ second.$ Find the average rate at which the pump is working

  1. $2.4kw$
  2. $2.5kw$
  3. $2.3kw$
  4. $1.7kw$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$F=ma=12000\times \dfrac {4}{40}=1200\ N$
$A=F/m=1200/1200=1/10\ m/s^2$
$s=ut +\dfrac {1}{2}at^2=0+\dfrac {1}{2}\times \dfrac {1}{10}\times 40\times 40=80\ m$
then, $W=Fs=80\times 1200=96000\ J$
rate of working, $P=\dfrac {W}{t}$
$P=\dfrac {96000}{40}=2400W=2.4\ kW$
Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

In an open pipe pressure at the ends of the pipe is

  1. minimum

  2. maximum

  3. zero

  4. depending on temperature, it can be maximum or minimum.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an open pipe, the ends are pressure nodes (displacement antinodes) because they are open to the atmosphere, meaning the pressure variation is zero relative to atmospheric pressure.