Tag: fluid pressure

Questions Related to fluid pressure

What should be 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,

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $d$

  2. $2d$

  3. $4d$

  4. $\dfrac{d}{2}$

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Correct Option: D

To what height should a cylindrical vessel be filled with a homogeneous liquid to make the force with which the liquid pressure on the sides of the vessel equal to the force exerted by the liquid on the bottom of the vessel?

physics option b: engineering physics buoyancy floatation fluid pressure

  1. Equal to the radius.

  2. Less than radius

  3. More than radius

  4. Four times of radius

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Correct Option: A
Explanation:
If  $h$ is the height of liquid in cylinder, $r $ be the radius of the cylinder and $ ρ$ be the density of the liquid.

Then we have
Weight of the liquid $=\pi r^2h \rho g$........................................(I)
Mean pressure on the wall $=\dfrac12 \rho  gh$

The total force on the wall =  $ 2\pi rh \times \dfrac12 \rho  gh= \pi rh^2\rho g $....................................(2)

On equating (I) and (2) we have
$\pi r^2h \rho g=\pi rh^2\rho g $
$r=h$
$\therefore$ The liquid should be filled up-to a height equal to the radius of the cylinder.

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 :

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $115 ms^{-1} $

  2. $11.5 ms^{-1} $

  3. $0.115 ms^{-1} $

  4. $1.15 ms^{-1} $

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Correct Option: D
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}}} $

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

A small hollow vessel open to atmosphere having a small circular hole radius $R\ mm$  in its base is immersed in a tank of water. To what depth should the base of vessel be immersed in water so that water will start coming into the vessel through the hole. ($TT$ is surface tension of water) ($\rho=$density of water).

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $\dfrac {2T}{\rho g R}$

  2. $\dfrac {T}{\rho g R}$

  3. $\dfrac {T}{4\rho g R}$

  4. $\dfrac {4T}{\rho g R}$

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Correct Option: A

A cylindrical vessel filled with water up to the height H becomes empty in time $ t _0 $ due to a small  hole at the bottom of the vessel. if water is filled to a height 4 H it will flow out in time 

physics option b: engineering physics buoyancy floatation fluid pressure

  1. $ t _0 $

  2. $ 4t _0 $

  3. $ 8t _0 $

  4. $ 2t _0 $

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Correct Option: B

A tank with a square base of area 2 m$^2$ is divided into two compartments by a vertical partition in the middle. There is a small hinged door of face area 20 cm$^2$ at the bottom of the partition. Water is filled in one compartment and an acid of relative density 1.53 x 10 kg m$^{-3}$ in the other, both to a height of 4 m. The force necessary to keep the door closed is (Take g = 10 m s$^{-2}$)

physics option b: engineering physics buoyancy floatation fluid pressure

  1. 10 N

  2. 20 N

  3. 40 N

  4. 80 N

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Correct Option: C
Explanation:

The situation is as shown in the figure.
For compartment contain water,
$h = 4 m, \rho _w = 10^3\, kg \,m^{-3}$
Pressure exerted by the water at the door at the bottom is
$P _w=\rho _w hg  $

$=10^3\,kg \,m^{-3} \times 4 \,m \times 10 \,m s^{-2}$
$= 4 \times 10^4\,N\,m^{-2}$
For compartment containing acid.
$\rho _a =1.5 \times 10^3\, kg\, m^{-3}, h = 4 \,m$
Pressure exerted by the acid at the door at the bottom is
$P _a=\rho _ahg $
$= 1.5 \times 10^3\,kg \,m^{-3} \times 4\,m \times 10\,m\,s^{-2} $
$= 6 \times 10^4\,N\,m^{-2}$
$\therefore$ Net pressure on the door =$P _a - P _w = (6  \times 10^4 - 4 \times 10^4) N \,m^{-2}$

$= 2 \times 10^4 \,N \,m^{-2}$
Area of the door $= 20 cm^2 = 20 \times 10^{-4} m^2$
$\therefore$ Force on the door$= 2 \times 10^4 N m^{-2} \times 20 \times 10^{-4} m^2 = 40 N$
Thus, to keep the door dosed the force of $40 N$ must be applied horizontally from the water side.

A vessel, whose bottom has round holes with diameter 0.1 mm, is filled with water. The maximum height up to which water can be filled without leakage is:

physics option b: engineering physics buoyancy floatation fluid pressure

  1. 100 cm

  2. 75 cm

  3. 50 cm

  4. 30 cm

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Correct Option: D
Explanation:

Sln :
For equilibrium,
Total upward force by surface tension 
= Weight of the water in tube
$\Rightarrow \, \pi \, \times \, D \, \times$ surface tension (circumference)
$= \, \pi(D/2)^2 \, \times \, h \, \times \, density \, \times \, g(cross \, section)$
where D (diameter) = 0.1 mm = 0.01 cm
Density of water = 1 $\times \, 10^{-3} \, gcm^3$
$\Rightarrow \, \pi \, \times \, (0.01) \, \times \, 75 \, \times \, 10^{-3}$
$= \, \pi \times \, \left(\dfrac{0.01}{2} \right)^2 \, \times \, h \, \times \, 1 \, \times \, 10^{-3} \, \times \, 1000$
$\therefore \, h = \, \dfrac{0.75 \, \times \, 0.01 \, \times \, 4}{0.01 \, \times \, 0.01}$ = 0.3 m = 30 cm

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

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

  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

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Correct Option: B
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.

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 -

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

  1. $1: \sqrt{2}$

  2. $\sqrt{2}:1$

  3. 1:2

  4. 2:1

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Correct Option: A
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} $


A $20$cm long capillary tube is dipped in water. The water rises up to $8$cm. If the entire arrangement is put in a freely falling elevator, the length of water column in the capillary tube will be:

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

  1. $8$cm

  2. $6$cm

  3. $10$cm

  4. $20$cm

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Correct Option: D
Explanation:

In a freely falling lift, gravitational pull is zero hence the capillary tube will be filled completely.