Tag: pressure in fluids and atmospheric pressure

Questions Related to pressure in fluids and atmospheric pressure

 If the the height $(h)$ at which the atmospheric pressure is reduced to half. Find the value $\displaystyle h$.

physics pressure in fluids and atmospheric pressure common consequences of the atmospheric pressure atmospheric pressure and its consequences important points about atmospheric pressure

  1. $\displaystyle h=\dfrac{0.693RT}{Mg}$.

  2. $\displaystyle h=\dfrac{6.93RT}{Mg}$.

  3. $\displaystyle h=\dfrac{693RT}{Mg}$.

  4. None of these

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Correct Option: A
Explanation:

$\displaystyle P=P _{0}e^{-Mgh/RT} $
But $\displaystyle P=0.5P _{0}$
Hence, $\displaystyle 0.5P _{0}=P _{0}e^{-Mgh/RT} $
$\displaystyle 0.5=e^{-Mgh/RT} $


Take natural logarithm on both sides.
$\displaystyle \ln0.5=e^{-Mgh/RT} $
$\displaystyle -0.693 =  {-Mgh/RT}$
Hence, $\displaystyle \displaystyle h=\dfrac{0.693RT}{Mg} $

A fully loaded Boeing aircraft has a mass of $33\times 10^{5}$ kg. Its total wing area is $500 m^2$. It is in level flight with a speed of $960 km/h$. (a) Estimate the pressure difference between the lower and upper surfaces of the wings. (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. The density of air is $\rho = 1.2 kg m^{-3}$ and $g = 9.81 m^{-2}$.

physics pressure in fluids and atmospheric pressure common consequences of the atmospheric pressure atmospheric pressure and its consequences important points about atmospheric pressure

  1. $0.067$

  2. $0.075$

  3. $0.98$

  4. $0.88$

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Correct Option: B

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

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

  1. 9.4 m

  2. 4.9 m

  3. 0.49 m

  4. 0.94 m

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Correct Option: D
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 $\displaystyle P _1 = \frac{16}{17} = 0.94 m$

A capillary tube of radius $r$ is immersed in water and water rises to a height of $h$. Mass of water in the capillary tube is $5\times 10^{-3}kg$. The same capillary tube is now immersed in a liquid whose surface tension is $\sqrt{2}$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is $45^o$. The mass of liquid which rises into the capillary tube now is (in kg):

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

  1. $5\times 10^{-3}$

  2. $2.5\times 10^{-3}$

  3. $5\sqrt{2}\times 10^{-3}$

  4. $3.5\times 10^{-3}$

Show answer
Correct Option: A
Explanation:

Since the force on liquid is proportional to $Scos(\theta )$, in the latter case, force will be $\sqrt { 2 } cos(\dfrac { \pi  }{ 4 } )=1$ times the force in first case. Hence, since the same force is applied, mass of water rising in the tube will be same.

Four identical capillary tubes $a, b, c$ and $d$ are dipped in four beakers containing water with tube ‘$a$’ vertically, tube ‘$b$’ at $30^{o}$, tube ‘$c$’ at $45^{o}$ and tube ‘$d$’ at $60^{o}$ inclination with the vertical. Arrange the lengths of water column in the tubes in descending order.

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

  1. $d, c, b, a$

  2. $d, a, b, c$

  3. $a, c, d, b$

  4. $a, b, c, d$

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Correct Option: A
Explanation:

In capillary tube fluid always rise to the same vertical height as when the tube is perfectly vertical. So, the tube which is making greater angle with vertical will get more water in it.
So, order of lengths of water column will be $d > c > b > a$.

A capillary tube when immersed vertically in a liquid rises to 3 cm. If the tube is held immersed in the liquid at an angle of 60$^{o}$ with the vertical,the length of the liquid column along the tube will be:

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

  1. 2 cm

  2. 4.5 cm

  3. 6 cm

  4. 7.5 cm

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Correct Option: C
Explanation:


$h = \dfrac {2T cos \theta}{r \rho g}$
$h \alpha cos \theta$
for $\theta = 60^0 cos \theta = \dfrac {1}{2}$
$\therefore$ h is double $\Rightarrow h = 6 cm$.

A capillary tube is dipped in water vertically.Water rises to a height of 10mm. The tube is now tilted and makes an angle 60$^{o}$ with vertical.Now water rises to a height of:

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

  1. 10 mm

  2. 5 mm

  3. 20 mm

  4. 40 mm

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Correct Option: A
Explanation:

If capillary tube is tilted the vertical height of water in tube remains same but volume of the water increases in the tube. So, height of water column will be 10mm.

Water rises in a capillary upto a height h. If now this capillary is tilted by an angle of $45^{\circ}$, then the length of the water column in the capillary becomes

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

  1. 2h

  2. $\displaystyle \frac{h}{2}$

  3. $\displaystyle \frac{h}{\sqrt{2}}$

  4. $h\sqrt{2}$

Show answer
Correct Option: C
Explanation:

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


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

for all other parameters kept constant, if I change the angle of inclination to $ 45^o $

$ cos \theta $ will go from 1 to $ \dfrac {1}{\sqrt 2} $

Therefore, the height of capillary tube rise will also change from $ H to \dfrac{H}{\sqrt2} $

Two parallel glass plates are dipped partly in a liquid of density $'d'$ keeping them vertical. If the distance between the plates is $'x'$ Surface tension for liquid is $T$ & angle of contact is $\displaystyle \theta $ then rise of liquid between the plates due to capillary will be

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

  1. $\displaystyle \dfrac{T\cos \theta }{xd}$

  2. $\displaystyle \dfrac{2T\cos \theta }{xdg}$

  3. $\displaystyle \dfrac{2T}{xdg\cos \theta}$

  4. $\displaystyle \dfrac{T\cos \theta }{xdg}$

Show answer
Correct Option: B
Explanation:

weight of liquid of height 'h' = (area of tube x h) x g x d= (3.14/4)hdg${x}^2$

vertical component of surface tension force=(1/2)x(Txcircumference)x cosθ=3.14Txcosθ
therefore, (3.14/4)hgd${x}^2$=(1/2)Tx3.14xcosθ
h=(2Tcosθ)/(gdx)
θ
θθs=Tx3.14x

Two capillary tubes of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube is filled with water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is $7.3 \times 10^{-2} N/m$. Take the angle of contact to be zero and density of water to be $10^3 kg/m^3(g = 9.8 m/s^2)$

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

  1. 5 mm

  2. 10 mm

  3. 15 mm

  4. 20 mm

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Correct Option: A
Explanation:

$\varrho gh=\dfrac { 2T }{ R } $

${ h } _{ 1 }=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times 1.5\times { 10 }^{ 3 } } $
${ h } _{ 2 }=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times 3\times { 10 }^{ -3 } } $
So $\triangle h={ h } _{ 1 }-{ h } _{ 2 }$
             $=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times { 10 }^{ -3 } } \left( \dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 }  \right) $      
             $=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times 3\times { 10 }^{ -3 } } $
             $\boxed { \triangle h=5mm } $