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Pressure at a certain depth in liquid - class-XII

Attempted 0/81 Correct 0 Score 0

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


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):

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

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

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

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


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.

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


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 -

  1. $1: \sqrt{2}$

  2. $\sqrt{2}:1$

  3. 1:2

  4. 2:1


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:

  1. $8$cm

  2. $6$cm

  3. $10$cm

  4. $20$cm


Correct Option: D
Explanation:

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

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


Correct Option: B
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 $

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.

  1. $d, c, b, a$

  2. $d, a, b, c$

  3. $a, c, d, b$

  4. $a, b, c, d$


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:

  1. 2 cm

  2. 4.5 cm

  3. 6 cm

  4. 7.5 cm


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:

  1. 10 mm

  2. 5 mm

  3. 20 mm

  4. 40 mm


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

  1. 2h

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

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

  4. $h\sqrt{2}$


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

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

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

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

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


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 $10^3 kg/m^3(g = 9.8 m/s^2)$

  1. 5 mm

  2. 10 mm

  3. 15 mm

  4. 20 mm


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 

  1. $M$

  2. $2M$

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

  4. $4M$


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

  1. $0.1 m$

  2. $9.8 m$

  3. $0.98 m$

  4. Full length of capillary tube


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

  1. $1.5 g$

  2. $10 g$

  3. $5 g$

  4. $15 g$


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? 

  1. $1cm$

  2. $2 cm$

  3. $3 cm$

  4. $4 cm$


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

  1. $1:9$

  2. $1:3$

  3. $9:1$

  4. $3:1$


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?

  1. $1 cm$

  2. $0.1 cm$

  3. $2 cm$

  4. $4 cm$


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 

  1. Increases by $2\%$

  2. Decreases by $2\%$

  3. Remains unchanged

  4. Sometime increases and sometime decreases


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:

  1. $2400$

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

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

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


Correct Option: B

If pressure at the half depth of a lake is equal to $\dfrac{3}{4}$ times the pressure at its bottom, then find the depth of the lake . [Take g=$10 m/s^2]$

  1. $ \dfrac{P _{0}}{\rho g}\ $

  2. $ \dfrac{2P _{0}}{\rho g}\ $

  3. $ \dfrac{P _{0}}{2\rho g}\ $

  4. $ \dfrac{3P _{0}}{\rho g}\ $


Correct Option: A
Explanation:
Let depth of the lake be $h$ and pressure at bottom $= P$
Then $P=P _{0}+\rho gh\rightarrow (1)$    $(P _{0}=$ atmospheric pressure, $\rho $ = density of water)
At half depth $(h/2)$ pressure is $\dfrac{3P}{4}$ then :
$\dfrac{3P}{4}=P _{0}+\rho g\dfrac{h}{2}\rightarrow (2)$
On subtracting equation 2 from 1 we get :
$\dfrac{P}{4}=\rho g\dfrac{h}{2}$
$\Rightarrow P=2\rho gh$, substituting this value of $P$ in equation 1:
$2\rho gh=P _{0}+\rho gh$
$\Rightarrow h=\dfrac{P _{0}}{\rho g}\rightarrow $ Depth of the lake

A tank $4m$ high is half filled with water then filled to the top with a liquid of density $0.60 g/cc$ what is the pressure at the bottom of the tank due to these liquids? (take $g=10ms^{-2}$)

  1. $1.6 \times 10^3Nm^{-2}$

  2. $3.2\times 10^{-3}$

  3. $1.6 \times 10^4Nm^{-2}$

  4. $3.2 \times 10^4Nm^{-2}$


Correct Option: C

If the air density were uniform, then the height of the atmosphere above the sea level to produce a normal atmospheric pressure of 1.0 x 10$^{5}$ Pa is(density of air is 1.3 kg/m$^{3}$ , g $=$ 10m/s$^{2}$):

  1. 0.77 km

  2. 7.7 km

  3. 77 km

  4. 0.077 km


Correct Option: B
Explanation:

$\rho g h = 1 \times 10^5$
$\Rightarrow h = \dfrac {10^5}{\rho g} = \dfrac {10^5}{1.3 \times 10} = 7.7 km$

The pressure exerted by a liquid at depth $h$ is given by:

  1. $\displaystyle \dfrac{h}{dg}$

  2. $hdg$

  3. $\displaystyle \dfrac{h}{d}$

  4. $hg$


Correct Option: B
Explanation:

Answer is B.

The pressure exerted by a liquid at a point depends on its vertical depth and density of the liquid only. It is independent of the shape of the container. The pressure at the bottom of the three vessels of different shapes containing the same liquid acts equally in all directions.
Thus, the pressure exerted by the liquid of height $h$ is given as P=hdg, where $h$ is the height, $d$ is the density and $g$ is the acceleration due to gravity.

The pressure exerted by a liquid column of height h is given by (the symbols have their usual meanings).

  1. $\dfrac {h}{\rho g}$

  2. $h\rho g$

  3. $\dfrac {h}{\rho}$

  4. $hg$


Correct Option: B
Explanation:

The pressure exerted by a liquid column of height h is given by-

       $h\rho g$
Since, total mass =$\rho g$
      

At a depth of 1000 m in an ocean, what is the absolute pressure? Given density of sea water is $1.03 \times 10^3 kgm^{-3} ,\ g= 10ms^{-2}$

  1. 104 atm

  2. 100 atm

  3. 108 atm

  4. 110 atm


Correct Option: A
Explanation:

Given: $h=1000m  ,  d=1.03\times10^{3}kg/m^{3} ,  g=10m/s^{2}$ 

The absolute pressure is given by: absolute pressure = pressure of water + atmospheric pressure
$P=hdg+1atm=1000\times1.03\times10^{3}\times10+1atm=1.03\times10^{7}Pa+1atm$
$P=103atm+1atm=104atm$

A bubbles rises from the bottom of a lake $70m$ deep on reaching the surface its volume become (take atmospheric pressure equal to $10m$ of water)

  1. $4$ times

  2. $2$ times

  3. $10$ times

  4. $3$ times


Correct Option: B

The height of a barometer filled with a liquid of density $3.4\ g/cc$ under normal condition is approximately -

  1. $8\ m$

  2. $5\ m$

  3. $3\ m$

  4. $1\ m$


Correct Option: C
Explanation:

Given,

$\rho =3.4g/cc=3.4\times 10^3 kg/m^3$
$g=9.8m/s^2$
$P=1.01\times 10^5 Pa$
Pressure, $P=\rho gh$
$h=\dfrac{P}{\rho g}=\dfrac{1.01\times 10^5}{3.4\times 10^3\times 9.8}$
$h=3.03m$
The correct option is C.

A barometer tube reads $76 cm$ of mercury, If the tube is gradually inclined at an angle of $60^\circ$ with vertical, keeping the open end immersed in the mercury reservoir, the length of he mercury column will be:

  1. $152 \ cm$

  2. $76 \ cm$

  3. $38 \ cm$

  4. $38 \sqrt{3} cm$


Correct Option: C

A tank full of water has a small hole  at the bottom. If one-fourth of the tank is emptied in $t$ seconds and remaining three-fourths of the tank is emptied in $t _2$ seconds. Then the ratio $\frac{t _1}{t _2}$ is

  1. $\sqrt{3}$

  2. $\sqrt{2}$

  3. $\frac{1}{\sqrt{2}}$

  4. $\frac{2}{\sqrt{2}}-1$


Correct Option: C

A small hole is made at a height of $(1/\sqrt{2})m$ from the bottom of a cylindrical water tank. The length of the water column is $\sqrt{2}m$. Find the distance where the water emerging from the hole strikes the ground.

  1. $4m$

  2. $2m$

  3. $3m$

  4. $\sqrt{2}m$


Correct Option: C

 The average pressure of a liquid (density$\rho$) on the walls of the container filled upto height $h$ with the liquid is $\dfrac{1}{2}h\rho g$.

  1. True

  2. False


Correct Option: A

A spherical bubble of air has a radius of $1$mm at the bottom of a tank full of water. As the bubble rises it goes on becoming bigger on reaching the surface, the radius becomes $2$mm. The depth of tank is

  1. $142.80$m

  2. $72.35$m

  3. $723$m

  4. $100$m


Correct Option: B

The pressure at point in water is $10\ N/m^{2}$. The depth blow this point where the pressure becomes double is (Given density of water $=10^{3}\ kh\ m^{-3},\ g=10\ m\ s^{-2}$)

  1. $1\ mm$

  2. $1\ cm$

  3. $1\ m$

  4. $10\ cm$


Correct Option: D

Two vessels A and B are different shapes have the same base area and are filled with water upto same height as the force exerted between water on the base is FA for vessel A and F B for vessel B . The respective weight of the water filled in vessel are wA and wB. Then

  1. FA>FB , was>wB

  2. FA=FB, wA>wB

  3. FA=FB, wA

  4. FA>FB, wA=wB


Correct Option: A

The reading of a barometer containing some air above the mercury column is $73\ cm$ while that of a correct one is $76\ cm$. If the tube of the faulty barometer is pushed down into mercury until volume of air in it is reduced to half, the reading shown by it will be

  1. $70\ cm$

  2. $72\ cm$

  3. $74\ cm$

  4. $76\ cm$


Correct Option: A

A large container of negligeble mass and uniform cross-section area A has a small hole (of area a < < A) near its side wall at bottom. The container is open at the top and kept on a smooth horizontal floor . It contains a liquid of density $\rho $ and mass $m _0$ when liquid starts flowing horizontally at time t = 0. Find the speed of container when 75% of the liquid has drained out (Assume the liquid surface remains horizontal throughout the motion)

  1. $\left[ \frac { { m } _{ 0 }g }{ A\rho } \right] ^{ 1/2 }$

  2. $\left[ \frac { { 4m } _{ 0 }g }{ A\rho } \right] ^{ 1/2 }$

  3. $\left[ \frac { { m } _{ 0 }g }{ 2A\rho } \right] ^{ 1/2 }$

  4. $\left[ \frac { { 2m } _{ 0 }g }{ A\rho } \right] ^{ 1/2 }$


Correct Option: C

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


Correct Option: C

A large cylinderical vessel contains water to a height of $10m$ it is found taht the acting on the curved surface is equal to that at the bottom. If atmospheric pressure can supposed a semi column of $10m$, the radius of the vessel is

  1. $10m$

  2. $15m$

  3. $5m$

  4. $25m$


Correct Option: B

If the atmospheric pressure is 76 cm of Hg at what depth of water in a lake the pressure will becomes 2 atmospheres nearly.

  1. $862 cm$

  2. $932 cm$

  3. $982 cm$

  4. $1033 cm$


Correct Option: D
Explanation:

Pressure at that depth $= 2$ atmosphere $= 2 \times 76\, cm$ of $Hg$

Pressure is due to atmosphere $+$ pressure due to column of 
water $= 2 \times  76\, cm$ of $Hg$ 

$\implies 76\, cm$ of $Hg +$ depth $\times$ density of 
water $\times g = 2 \times  76 cm$ of $Hg$

Or 

$h \times d \times g = 76\, cm$ of $Hg$

Or 

$h = 76 \,cm \times  13600 \times \dfrac{g }{ 1000} \times g$  

Note: pressure due to $h$ meter of $Hg =h\times density\,of\,mercury \times g$$ 

Cancelling $g$ we have 

$h = 13.6 \times 76 = 1033.6 \,cm $ ( as cm is taken for atmosphere answer too comes in cm)

The pressure at the bottom of a lake, due to water is $4.9 \times 10 ^ { 6 } \mathrm { N } / \mathrm { m } ^ { 2 }$ . Whatis the depth of the lake? 

  1. 500$\mathrm { m }$

  2. 400$\mathrm { m }$

  3. 300$\mathrm { m }$

  4. 200$\mathrm { m }$


Correct Option: A

If the atmospheric pressure is 76 cm of Hg at what depth of water the pressure will becomes 2 atmospheres nearly.

  1. $826 cm$

  2. $932 cm$

  3. $982 cm$

  4. $1033 cm$


Correct Option: D
Explanation:

Let required depth be $h$


Pressure at that depth $= 2$ atmosphere $= 2\times  76\, cm$ of $Hg$

Pressure is due to atmosphere $+$ Pressure due to column of water $= 2 \times  76\, cm $ of $Hg$ 

$\implies 76 \,cm$ of $Hg +$ depth $\times$ density of water 

$h\times d\times  g = 2 \times 76 cm$ of $Hg$

Or 

$h \times  d \times  g = 76 \,cm$ of $Hg$

Or 

$h = \dfrac{76\, cm \times  13\times  g}{1000 \times g}$  ( Note: pressure due to $h$ meter of $Hg = h \times $ density of mercury $\times g$)

Cancelling $g$ we have $h = 13.6 \times  76 = 1033.6 \,cm$ ( as $cm$ is taken for atmosphere answer too comes in $cm$).

The depth of the dam is 240 m. The pressure of water is (Take $g=10 m/{ s }^{ 2 }$ density of liquid = $1000 kg/{ m}^{ 3})$

  1. $24\times { 10 }^{ 5 }N/{ m }^{ 2 }$

  2. $12\times { 10 }^{ 4 }N/{ m }^{ 2 }$

  3. $10\times { 10 }^{ 3 }N/{ m }^{ 2 }$

  4. None of these


Correct Option: A

The pressure on a swimmer $20$ m below the surface of water at sea level is

  1. $1.0$ atm

  2. $2.0$ atm

  3. $2.5$ atm

  4. $3.0$ atm


Correct Option: D
Explanation:

Given,

$P _0=1atm=1\times 10^5 Pa$
$h=20m$
$\rho=1000kg/m^3$
$g=10m/s^2$
The pressure on a swimmer $20m$ below the surface of water at sea level is
$P=P _0+\rho gh$
$P=1\times 10^5+1000\times 10\times 20$
$P=3\times 10^5$
$P=3atm$
The correct option is D.

The pressure at the bottom of a lake, due to water is $4.9 \times 10^{6} N/m^{2}$. What is the depth of the lake?

  1. 500m

  2. 400m

  3. 300m

  4. 200m


Correct Option: A
Explanation:
Given,

$P=4\times 10^6\,N/m^2$

$\rho=1000kg/m^2$

We have,

$P=\rho g h$

Then,

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

$=\dfrac{4\times 1066}{1000\times 9.8}=\dfrac{1000}{2}=500\,m$

A ball o mass m and density p is immersed in a liquid of density 3 p ar a depth h and released. to what height will the ball jump up above the surface of liquid ?(neglect the resistance of water and air)

  1. h

  2. 2h

  3. 3h

  4. 4h


Correct Option: A

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} $


Correct Option: D

A column of mercure of lenath $h = 10 \mathrm { cm }$ is contained in the middle of a narrow horizontal tube of length $1 \mathrm { m } ,$ closed at both ends. The air in both halves of the tube is under a pressure of $P _ { 0 } = 76 \mathrm { cm }$ of mercury. The tube is now slowly made vertical. The distance moved by mercury will be approximately

  1. $4.5$ $\mathrm { cm }$

  2. $3.0$ $\mathrm { cm }$

  3. $2.5$ $\mathrm { cm }$

  4. $1.2$$ $\mathrm { cm }$


Correct Option: B

The volume of an air bubble increases by $ \mathrm{x} \%  $ as it rises from the bottom of a lake to its surface. If the height of the water barometer is H, the depth of the lake is

  1. $

    \left(\dfrac{H+x}{100}\right)^{2}

    $

  2. $

    \dfrac{H x}{(100+x)}

    $

  3. $

    \dfrac{H x}{100}

    $

  4. $

    \dfrac{100 H}{x}

    $


Correct Option: C
Explanation:
We have,

$P _1V _1=P _2V _2$

$V _2=V _1+\dfrac{x}{100}V _1$

$P _1V _1=P _2(V _2+\dfrac{x}{100}V _1)$

$P _1=P _2(1+\dfrac{x}{100})$

But,

$P _2=1\,atm$

Then,

$P _1=P _2+\dfrac hH$

$1+\dfrac hH=1+\dfrac{x}{100}$

$h=\dfrac{xH}{100}$

A water tank is 20$\mathrm { m }$ deep. If the waterbarometer reads $10 \mathrm { m } ,$ the pressure at thebottom of the tank is

  1. 2 atmosphere

  2. 1 atmosphere

  3. 3 atmosphere

  4. 4 atmosphere


Correct Option: C

A cylindrical can open at the bottom end lying at the bottom of a lake $47.6\ \text{m}$ deep has $50\ \text{cm}^3$ of air trapped in it. The can is brought to the surface of the lake. The volume of the trapped air will become $($atmospheric pressure $= 70\ \text{cm}$ of Hg and density of Hg $= 13.6\ \text{g/cc)}$:

  1. $350\ \text{cm}^3$

  2. $300\ \text{cm}^3$

  3. $250\ \text{cm}^3$

  4. $22\ \text{cm}^3$


Correct Option: B
Explanation:

$P _o= 70\ \text{cm}$ of Hg $=70 \times 10^{-2} \times 13600 \times 9.8 =93296\ \text{Pa}$
Using Boyle's law: $P _1V _1 =P _2V _2$
$\Rightarrow (P _o+H \rho g) \times 50 \times 10^{-6}=P _o \times V _2$
$\Rightarrow (93296+47.6 \times 1000 \times 9.8) \times 50 \times 10^{-6}=93296 \times V _2$
$\Rightarrow (93296+466480) \times 50 \times 10^{-6}=93296 \times V _2$
$\Rightarrow V _2 =300 \times 10^{-6}\ \text{m}^3 =300\ \text{cm}^3$

A dam of water reservoir is built thicker at bottom than at the top because

  1. pressure of water is very large at the bottom due to its large depth.

  2. water is likely to have more density at the bottom due to its large depth.

  3. quantity of the water at the bottom is very large.

  4. none of the above.


Correct Option: A
Explanation:

The pressure applied to walls of the dam will be a function of the amount  of water that is over that particular point on the wall. So water pressure is very large at the bottom due to its large depth. That's why dams are constructed thicker at their bottoms than at their tops. So correct option is 'A'.

The height to which a cylindrical vessel be filled with a homogeneous liquid, to make the average force with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel, is equal to

  1. half of the radius of the vessel

  2. one-fourth of the radius of the vessel

  3. three-fourth of the radius of the vessel

  4. three eight of the radius of the vessel


Correct Option: A

A tank with a small hole at the bottom has been filled with water and kerosene (specific gravity 0.8). The height of water is 3m and that of kerosene 2m. When the hole is spend the velocity of fluid coming out from it is nearly .(take g=$10ms^{ -2 }$ and density of water = $10^{ 3 }kgm^{ -3 }$)

  1. ${ 10.7 }{ ms }^{ -1 }$

  2. ${ 9.8 }{ ms }^{ -1 }$

  3. ${ 8.5 }{ ms }^{ -1 }$

  4. ${ 7.6 }{ ms }^{ -1 }$


Correct Option: C

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. $5.5m{s}^{-1}$


Correct Option: C

To what height $h$ should a cylindrical vessel of diameter $d$ be filled with a liquid so that due to liquid force on the vertical surface of the vessel be equal to the force on the bottom:

  1. $h=d$

  2. $h=2d$

  3. $h=3d$

  4. $h=d/2$


Correct Option: A

If pressure at half the depth of a lake is equal to $2/3$ pressure at the bottom of the lake then what is the depth of the lake

  1. 10m

  2. 20m

  3. 30m

  4. 60m


Correct Option: B

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


Correct Option: A

If the system is not in free fall, which of the following statements are true about hydrostatic pressure?

  1. In a liquid, point at different depths can never be at the same pressure.

  2. In a liquid, points at different depths may be at the same pressure.

  3. In different liquids, points at different depths can be at the same pressure.

  4. In different liquids, points at the same depth can never be at same pressure.


Correct Option: A,C,D
Explanation:

Pressure difference = $density\times a\times difference ~in ~depths$
(a)In a given liquid density remains same.So, pressure is same only at points of equal depth.
(c)In different liquids,pressure can be same at different depths if ${\rho}^{} _{1} {h}^{} _{1}$ = ${\rho}^{} _{2} {h}^{} _{2}$
 (d)${\rho}^{} _{1} \neq {\rho}^{} _{2}$ & ${h}^{} _{1} = {h}^{} _{2}$ implies ${P}^{} _{1} \neq {P}^{} _{2}$

How is the reading of a barometer affected when it is taken to (i) a mine, and (ii) a hill?

  1. (i) increases, (ii) decreases.

  2. (i) decreases, (ii) increases.

  3. (i)remains same (ii) decreases

  4. (i) increases, (ii) remains same


Correct Option: A
Explanation:

(i) increases due to increase in height of atmosphere above it.
(ii) decreases. due to decrease in height of atmosphere above it.

The volume of an air bubble becomes three times as it rises from the bottom of a take to its surface. Assuming atmospheric pressure to be $75\ cm$ of $Hg$ and the density of water to be $\displaystyle \dfrac{1}{10}$ of the density of mercury, the depth of the take is :

  1. $5\ m$

  2. $10\ m$

  3. $15\ m$

  4. $20\ m$


Correct Option: C
Explanation:

Since, the temperature of the surroundings remains constant, we can safely assume that the process is isothermal.

Therefore, applying Boyle's Law
$ P _1 V _1 = P _2 V _2 $

Let 1 denote the surface of water and 2 denote the depth

$ P _1 = 75 cm Hg $

$ P _2 = 75 + h/10 $ where h is the depth of the lake

$ V _1 = 3 V _2 $

Substituting the values, 

$ 75 \times 3 = 75 + \frac {h}{10} $

Solving, h = 1500 cm or 15 m

The force that water exert on the base of a house tank of base area 1.5 m$^{2}$ when it is filled with water up to a height of 1 m if (g = 10 m/s$^{2}$)

  1. 1200 kgwt

  2. 1500 kgwt

  3. 1700 kgwt

  4. 2000 kgwt


Correct Option: B
Explanation:

Pressure at the bottom of the house tank=$\rho gh$

$=1000\times 10\times 1Pa$
$=10^4Pa$
Hence the force acting on the base=$PA=10^4\times 1.5N=1500kgwt$

What is the pressure 200 m below the surface of the ocean if the sp. gravity of sea water is 1.03 : [Atmospheric pressure$=1.013\times 10^{5}N/m^{2}$].

  1. $21.2\times 10^{5}N/m^{2}$

  2. $20.4\times 10^{5}N/m^{2}$

  3. $40\times 10^{4}N/m^{2}$

  4. $21.2\times 10^{6}N/m^{2}$


Correct Option: A
Explanation:

The gauge pressure at any depth is given by,


$ P _g = \rho g h $

Given, the atmospheric pressure to be $ P _o = 1.013 \times 10^5 Pa $

The pressure at this depth will be,

$ P = P _o + P _g $

= $ 1.013 \times 10^5 + 1.03 \times 10^3 \times 10 \times 200 $

= $ 21.2 \times 10^5 Pa $

When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is :-

  1. H

  2. 2H

  3. 7H

  4. 8H


Correct Option: C
Explanation:

It is given that the atmospheric pressure is,


$ P _o = \rho g H $, where $ \rho $ is the density of water.

Let the depth of the lake be d, the pressure at this depth would be,

P = $ P _o + \rho g d $

Hence, P = $ \rho g (d+H) $

Since, the surrounding temperature is constant, we can assume that the process takes place isothermally. Therefore, we can apply the Boyle's Law

$ P _1 V _1 = P _2 V _2 $

Here, 1 denotes the water at depth d and 2 denotes the surface of water.

Hence, $ P _1 = \rho g (d+H) $

$ P _2 = \rho g H $

$ V _2 = 8V _1 $ (since, radius is doubled, volume becomes 8 times)

Substituting the values, in Boyle's Law, 

d = 8H - H = 7H

An air bubble situated at the bottom of an open kerosene tank rises to the top surface. It is observed that at the top the volume of the bubble is thrice its initial volume. If the atmospheric pressure is 72 cm of Hg, and mercury is 17 times heavier than kerosene the depth of the tank is:

  1. 2.16 m

  2. 2.88 m

  3. 12.24 m

  4. 24.48 m


Correct Option: D
Explanation:

from boyle's law

     ${ P } _{ 1 }{ V } _{ 1 }={ P } _{ 2 }{ V } _{ 2 }$
&   ${ V } _{ 2 }=3{ V } _{ 1 }$
so ${ P } _{ 1 }=3{ P } _{ 2 }$
&   ${ P } _{ 2 }={ P } _{ 0 }$
so  ${ P } _{ 1 }=3{ P } _{ 0 }$
&    ${ P } _{ 1 }={ P } _{ 2 }+\rho gh$
       $\rho gh=2{ P } _{ 0 }$
       ${ \rho  } _{ k }g\times h=2\times 72cm\quad of\quad Hg$
       ${ \rho  } _{ k }\times g\times h=2\times 72\times 17\times \rho _{k} g$
       $h=2448cm$
       $\boxed { h=24.48m } $

Pressure at a point in a fluid is directly proportional to

  1. depth of the point from the surface

  2. density of the fluid

  3. acceleration due to gravity

  4. the area of cross section


Correct Option: A,B,C
Explanation:

The pressure at a point in a fluid is given by, $P=h\rho g$

where $h=$ depth of the point from the surface, $\rho=$ density of the fluid and $g=$ acceleration due to gravity. 

Pressure at a certain depth in river water is ${p} _{1}$ and at the same depth in sea water is ${p} _{2}$. Then (Density of sea water is greater than that of river water):

  1. ${p} _{1}={p} _{2}$

  2. ${p} _{1}> {p} _{2}$

  3. ${p} _{1}< {p} _{2}$

  4. ${p} _{1}-{p} _{2}=$ atmospheric pressure


Correct Option: C
Explanation:

The expression of the pressure at a point in fluid is given by $P=h\rho g$

where $h=$ depth of the point from the surface, $\rho=$ be the density of the fluid and $g=$ acceleration due to gravity.
Here,  h and g are constant in both cases so $p _1=h\rho _r g$ and $p _2=h\rho _s g$
or $\dfrac{p _1}{p _2}=\dfrac{\rho _r}{\rho _s}$
As $\rho _s>\rho _r$ so $\dfrac{p _1}{p _2}<1$ or $p _1<p _2$

A tank $5\ m$ high is half filled with water and then is filled to the top with oil of density $0.85\ g\ cm^{-3}$. The pressure at the bottom of the tank, due to these liquids is

  1. $1.85\ g\ dyne\ cm^{-2}$

  2. $89.25\ g\ dyne\ cm^{-2}$

  3. $462.5\ g\ dyne\ cm^{-2}$

  4. $500\ g\ dyne\ cm^{-2}$


Correct Option: C
Explanation:

So the tank is filled with water upto $h _1=2.5 m=250cm$ and remaining $h _2 =2.5m =250cm$ with oil .

So $P=\rho _{water}gh _1+\rho _{oil}gh _2$
$P=(1)(250)g+(0.85)\times (250)g$
$P=462.5g\,dyne\,cm^{-2}$

Choose the wrong statement among the following

  1. The pressure at a point in a fluid is directly proportional to the depth of the point from the surface

  2. The pressure at a point is independent of acceleration due to gravity

  3. The pressure at a point is directly proportional to the area of cross section

  4. The pressure at a point is proportional to the density of the fluid


Correct Option: A,B,D
Explanation:

Pressure is written as $\rho gh$, where h is the depth from the surface .

Pressure is due to weight of object or force and that is due to the acceleration due to gravity.
Pressure at a point is $P=\rho gh$
Thus pressure is independent of area of cross-section but depends on density of fluid.

Choose the correct statement among the following.

  1. The upper surface of a stationary liquid is always horizontal.

  2. Pressure of a given liquid is directly proportional to the depth of the liquid.

  3. Pressure at a given depth inside a stationary liquid is different all points in the horizontal plane.

  4. Pressure at a point in a fluid is inversely proportionally to the density of the fluid.


Correct Option: A,B
Explanation:

The upper surface of a stationary liquid is always horizontal as there is no unbalance force.

Pressure is given by $P=\rho g H$, so it is directly proportional to depth.
Pressure at a given depth for a stationary liquid is same in a horizontal plane. 
As $P=\rho g H$
For same H, pressure will be same.
Pressure is directly proportional to density. 

Two stretched membranes of area $2\ cm^{2}$ and $3\ cm^{3}$ are placed in a liquid at the same depth. The ratio of the pressure on them is

  1. $1 : 1$

  2. $2 : 3$

  3. $3 : 2$

  4. $2^{2} : 3^{2}$


Correct Option: A
Explanation:

Pressure on membrane is $P=\rho gh$

Since the density of both membranes is same.
Since both the membranes are placed at same depth ,So pressure will be same on both of then .

$\dfrac{P _1}{P _2}=\dfrac{\rho gh}{\rho gh}=1$

State True or False.
Pressure at a point in a liquid is inversely proportional to the height of the liquid column.

  1. True

  2. False


Correct Option: B
Explanation:

False statement.

The pressure exerted by a liquid depends on the height of the liquid column.
It can be defined as the weight of liquid column over an unit area. 
So pressure is define as $P=\rho g H$ 

A boy swims a lake and initially dives $0.5 m$ beneath the surface. When he dives $1 m$ beneath the surface, how does the absolute pressure change?

  1. It doubles

  2. It quadruples

  3. It slightly increases

  4. It cut to a half


Correct Option: C
Explanation:


When he goes from $0.5$ m to $1$ m , the pressure will slightly increases
Pressure at depth $0.5 m $ is $P _{0}+0.5dg$ , where $P _{0}$ is atmospheric pressure , $d$ is density
The pressure change when he divies to $1m$ is $P _{0}+dg$
So the pressure change slightly increases
Therefore option $C$ is correct

The pressure at the bottom of a tank of liquid is not proportional to:

  1. the acceleration

  2. the density of the liquid

  3. the area of the liquid surface

  4. the height of the liquid


Correct Option: C
Explanation:

Let the acceleration  of tank be $a$ $($ moving up $),$

Let the density of the liquid be $d$ and height of liquid be $h,$
The pressure at the bottom of tank is $P=P _{0}+dh(g+a),$
Therefore the pressure depends on acceleration, height of the liquid and the density of liquid.
It does not depend on the area of the liquid surface,
Therefore correct option is $C.$

The pressure on a swimmer 10 m below the surface lake is:(Atmospheric pressure=$1.01\times 10^5$ Pa,Density of water$=1000kg/m^3$ )

  1. $10\ atm$

  2. $5\ atm$

  3. $15\ atm$

  4. $2\ atm$


Correct Option: D
Explanation:
Given :  $h = 10 m$,  $\rho = 1000 kg/m^3$,  $g = 10 m/s^2$
Pressure on swimmer  $=$ pressure of atmosphere + pressure of water
                                      $ = 1  atm  + \rho gh \times 10^{-5} atm$
                                      $ = 1 atm + 1000\times 10 \times 10\times 10^{-5} atm$ 
                                      $ = 2atm$

What is the difference between the pressure on the bottom of a pool and the pressure on the water surface?

  1. $gh$

  2. $\dfrac{g}{h}$

  3. $0$

  4. $none$


Correct Option: A
Explanation:

The pressure difference is equal to the product of density , height and $g$

Given that $h$ is height difference between bottom and surface of a pool
$P _{b}-P _{s} = dgh$
Density $d$ of water is $1$
So we get $P _{b}-P _{s}=gh$
Therefore option $A$ is correct

Two containers $A$ and $B$ are partly filled with water and closed. The volume of $A$ is twice that of $B$ and it contains half the amount of water in $B$. If both are at the same temperature, the water vapour in the containers will have pressure in the ratio of

  1. $1 : 2$

  2. $1 : 1$

  3. $2 : 1$

  4. $4 : 1$


Correct Option: B

 $1m^3$ water is brought inside the lake upto $200 m$ depth from the surface of the lake. What will be change in the volume when the bulk modulus of elasticity of water is $22000 atm$?
(density of water is $1 \times 10^3 kg/m^3$ atmosphere pressure = $10^5 N/m^2$ and $g = 10 m/s^2$

  1. $8.9 \times 10^{-3} m^3$

  2. $7.8 \times 10^{-3} m^3$

  3. $9.1 \times 10^{-4} m^3$

  4. $8.7 \times 10^{-4} m^3$


Correct Option: C
Explanation:

$K = \dfrac{P}{\Delta V/V }$

$\therefore \Delta V = \dfrac{PV}{K}$

$P = h\rho g = 200 \times 10^3 \times 10 N/m^2$

$K = 22000 atm = 22000 \times 10^5 N/m^2$

V = 1$m^3$

$ \Delta V = \dfrac{200 \times 10^3 \times 10 \times 1}{22000 \times 10^5}=9.1 \times 10^{-4}m^3$

Three containers are used in a chemistry lab. All containers have the same bottom area and the same height. A chemistry student fills each of the containers with the same liquid to the maximum volume. Which of the following is true about the pressure on the bottom in each container?

  1. $P _1 = P _2 = P _3$

  2. $P _1 > P _2 > P _3$

  3. $P _1 < P _2 = P _3$

  4. $P _1 < P _2 > P _3$


Correct Option: A
Explanation:

Pressure applied on the bottom is equal to the force applied on the bottom per unit area of bottom

$\Rightarrow P=\frac{F}{A}$
$\Rightarrow P=\frac{dvg}{A}$
Where $d$ is density , $v$ is volume and $A$ is area
Given that for three containers , area is same and height is same. so the volume of three containers is same .
The density is also same for three containers.
So $d,v,g,A$ are same for all three containers
Therefore their pressures are same
So option $A$ is correct

The pressure at the bottom of a tank of water is $3P$ where $P$ is the atmospheric pressure. If the water is drawn out till the level of water is lowered by one fifth, the pressure at the bottom of the tank will now be:

  1. $2P$

  2. $(13/5)P$

  3. $(8/5)P$

  4. $(4/5)P$


Correct Option: B
Explanation:
If we ignore the atmospheric pressure, the pressure at the bottom is $2P$
We know that the pressure by a liquid column is given by $h\rho g$
$\therefore h\rho g=2P$
After the height getting lowered by one fifth, the height becomes four fifth. 
$\therefore \cfrac45h\rho g=\cfrac{2\times4}5P=\cfrac85P$
Now including the atmospheric pressure it becomes
$(\cfrac85+1)P=\cfrac{13}5P$ 

The force acting on a window of area 50 cm x 50 cm of a submarine at a depth of 2000 m in an ocean, the interior of which is maintained at sea level atmospheric pressure is (Density of sea water = 10$^3$ kg m$^{-3}$,g = 10 m s$^{-2}$)

  1. 5 x 10$^5$ N

  2. 25 x 10$^5$ N

  3. 5 x 10$^6$ N

  4. 25 x 10$^6$ N


Correct Option: C
Explanation:

Here, $h = 2000 m,$

          $ \rho= 10^3\, kg\, m^{-3},$ 
          $g = 10 \,m \,s^{-2}$
The pressure outside the submarine is
$P = P _a + \rho gh$
where $P _a$ is the atmospheric pressure. Pressure inside the submarine is $P _a$.
Hence, net pressure acting on the window is gauge pressure. Gauge pressure,
 $P _g = P - P _a = \rho gh $
$= 10^3\, kg\, m^{-3} \times 10 \,m\,s^{-2} \times 2000 \,m$
$ = 2 \times 10^7 Pa$
Area of a window is $A= 50 cm\times50 cm $
                                    $= 2500 \times 10^{-4}\, m^{2}$
Force acting on the window is
$F = P _gA $
$= 2 \times 10^7 \,Pa \times 2500 \times 10^{-4} m^2 $
$= 5 \times 10^6\,N$

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