Tag: pressure exerted by a liquid column

Questions Related to pressure exerted by a liquid column

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

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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 } $

Water rises up to a height $h _1$ in a capillary tube of radius $r$. The mass of the water lifted in the capillary tube is $M$. If the radius of the capillary tube is doubled, the mass of water that will rise in the capillary 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. $M$

  2. $2M$

  3. $\cfrac{M}{2}$

  4. $4M$

Show answer
Correct Option: B
Explanation:

Since we know that mass of water rise is proportional to volume of water.

Mass $\infty $ volume
$\dfrac { { M } _{ 1 } }{ { M } _{ 2 } } =\dfrac { { V } _{ 1 } }{ { V } _{ 2 } } =\dfrac { \pi { r } _{ 1 }^{ 2 }{ h } _{ 1 } }{ \pi { r } _{ 2 }^{ 2 }{ h } _{ 2 } } =\dfrac { { r } _{ 1 }^{ 2 }{ h } _{ 1 } }{ { r } _{ 2 }^{ 2 }{ h } _{ 2 } } \quad \rightarrow (1)$
and for capillary tube, we know that height $\alpha $ $\dfrac { 1 }{ radius } $
      So, $\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } \quad \rightarrow (II)$
     hence from (1) & (II)
     $\dfrac { { M } _{ 1 } }{ { M } _{ 2 } } =\dfrac { { r } _{ 1 }^{ 2 } }{ { r } _{ 2 }^{ 2 } } \times \dfrac { { r } _{ 2 } }{ { r } _{ 1 } } =\dfrac { { r } _{ 1 } }{ { r } _{ 2 } } $
     So ${ M } _{ 2 }=\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } \times { M } _{ 1 }=\dfrac { 2r }{ r } \times M=2M$
              $\boxed { { M } _{ 2 }=2M } $

In a surface tension experiment with a capillary tube water rises up to $0.1 m$. If the same experiment is repeated on an artificial satellite which is revolving around the earth. The rise of water in a capillary 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. $0.1 m$

  2. $9.8 m$

  3. $0.98 m$

  4. Full length of capillary tube

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Correct Option: D
Explanation:

If the experiment of capillary tube is performed in space then, it will rise to fall length of tube due to vaccum around it, i.e. no external pressure.

$5 g$ of water rises in the bore of capillary tube when it is dipped in water. If the radius of bore capillary tube is doubled, the mass of water that rises in the capillary tube above the outside water level 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. $1.5 g$

  2. $10 g$

  3. $5 g$

  4. $15 g$

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Correct Option: B

The height of water in a capillary tube of radius $2 cm$ is $4 cm$. What should be the radius of capillary, if the water rises to $8 cm$ in tube? 

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. $1cm$

  2. $2 cm$

  3. $3 cm$

  4. $4 cm$

Show answer
Correct Option: A
Explanation:

Since we know that height of capillary rise in inversely proportional to radius of capillary.

i.e.    height $\alpha $ $\dfrac { 1 }{ radius } $
         $\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } $
         $\dfrac { 4 }{ 8 } =\dfrac { { r } _{ 2 } }{ 2 } \Rightarrow \boxed { { r } _{ 2 }=1cm } $

Two capillary tubes of the same material but of different radii are dipped in a liquid. The heights to which the liquid rises in the two tubes are $2.2 cm$ and $6.6 cm$. The ratio of radii of the tubes 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. $1:9$

  2. $1:3$

  3. $9:1$

  4. $3:1$

Show answer
Correct Option: D
Explanation:

Since we know that height of capillary rise is inversely proportional to radii of tube, i.e.,

height $\propto \dfrac { 1 }{ radius } $

$\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } \Rightarrow \dfrac { 2.2cm }{ 6.6cm } =\dfrac { { r } _{ 2 } }{ { r } _{ 1 } } $

So, $\boxed { \dfrac { { r } _{ 1 } }{ { r } _{ 2 } } =3 } $

The height of water in a capillary tube of radius $2 cm$ is $4 cm$. What should be the radius of capillary, if the water rises to $8 cm$ in tube?

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. $1 cm$

  2. $0.1 cm$

  3. $2 cm$

  4. $4 cm$

Show answer
Correct Option: A
Explanation:

It is to be remembered that

height of a capillary rise $\propto \dfrac { 1 }{ radius\quad of\quad capillary } $
hence,
          $\dfrac { { h } _{ 1 } }{ { h } _{ 2 } } =\dfrac { { r } _{  2} }{ { r } _{ 1 } } $
          $\dfrac { 4 }{ 8 } =\dfrac { { r } _{ 2 } }{ 2 } \Rightarrow \boxed { { r } _{ 2 }=1cm } $

If the value of $g$ at a place is decreased by $2\%$. The barometric height of the mercury 

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

  1. Increases by $2\%$

  2. Decreases by $2\%$

  3. Remains unchanged

  4. Sometime increases and sometime decreases

Show answer
Correct Option: A
Explanation:
We know that,

$P=h\rho g$

Then,

$h=\dfrac{P}{\rho g}$

So, $h\propto \dfrac 1g$

If the value of $g$ decreased by $2\%$ then $h$ will increase by $2\%$.

The residual pressure of a vessel at ${27^0}C$ is  $1 \times {10^{ - 11}}N/{m^2}$. The number of molecules in this vessel is nearly:

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

  1. $2400$

  2. $2.4 \times {10^9}$

  3. ${10^{ - 11}} \times 6 \times {10^{23}}$

  4. $2.68 \times {10^{19}} \times {10^{11}}$

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Correct Option: B