Physics

Fluid Mechanics and Hydraulics

361 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

How much work is done by an agent fcn forcing 3$\mathrm { m } ^ { 3 }$ of water through a pipe of radius 2$\mathrm { cm }$ , It the difference in pressure at the two ends of the pipe is $10 ^ { 4 } \mathrm { N } \mathrm { m } ^ { 2 } \mathrm { ? }$

  1. $3 \times 10 ^ { 6 } J$
  2. $2 \times 10 ^ { 6 } J$
  3. $4 \times 10 ^ { 5 } J$
  4. $1 \times 10 ^ { 5 } 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
A Correct answer
Explanation

The volume of water flowing out per second is given by the product of the hole area and the speed of efflux, Q = A * sqrt(2gh). For the square hole of side L, area is L^2 and depth is y. For the circular hole of radius R, area is pi * R^2 and depth is 4y. Equating the two flow rates: L^2 * sqrt(2gy) = pi * R^2 * sqrt(2g(4y)). Simplifying this gives R = L / sqrt(2pi).

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

Water is falling in a cylindrical tank at the rate of $ \pi m^3 / s. $ If the radius of the tank is 2 m, the rate of increases in the level of water in the tank is

  1. 1 m/s

  2. 0.25 m/s

  3. 0.5 m/s

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

The rate of change of volume in the tank is given by dV/dt = A * (dh/dt) = pi * r^2 * (dh/dt). Given dV/dt = pi m^3/s and radius r = 2 m, we have pi = pi * (2^2) * (dh/dt), which yields dh/dt = 1 / 4 = 0.25 m/s.

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

The water flowing from a garden hose fills a container $ 3 \pi $ litre in one minute.Then speed of the water coming from that pipe with opening of radius 1 cm is 

  1. $ 4 ms^{-1} $
  2. $5 ms^{-1} $
  3. $ 1 ms^{-1} $
  4. $ 0.5 ms^{-1} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The volume flow rate is Q = 3 pi litres / minute = 3 pi * 10^-3 m^3 / 60 s = (pi / 20) * 10^-3 m^3/s. Also, Q = A * v = pi * r^2 * v, where r = 1 cm = 10^-2 m. Equating the two expressions: pi * (10^-2)^2 * v = (pi / 20) * 10^-3, which simplifies to v = 0.5 m/s.

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 pressure in fluids and atmospheric pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

Which of the following are the common consequences of atmospheric pressure in our daily life.

  1. Sucking a drink with a straw

  2. Filling a syringe with a liquid

  3. Filling ink in a fountain pen

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Liquid flows from higher pressure region to the lower pressure region.

Thus sucking a drink with a straw, filling a syringe with a liquid and filling ink in a fountain pen are the consequences of atmospheric pressure.