Tag: variation of pressure with depth

Questions Related to 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: 

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

  1. 5.1 cm

  2. 1.7 cm

  3. 4 cm

  4. 2.9 cm

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Correct Option: A

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

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

  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.

Show answer
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?

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

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

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

  3. (i)remains same (ii) decreases

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

Show answer
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 :

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

  1. $5\ m$

  2. $10\ m$

  3. $15\ m$

  4. $20\ m$

Show answer
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

At certain temperature radius of an air bubble is doubled when it comes to the top from bottom of a mercury column of height H if the pressure is:

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

  1. 5.5

  2. 10.64

  3. 12.45

  4. 15

Show answer
Correct Option: A
Explanation:

At constant temperature,
${ P } _{ 1 }{ V } _{ 1 }={ P } _{ 2 }{ V } _{ 2 }$,
Since ${ R} _{ 1 }=2{ R } _{ 2 }$
${ V } _{ 1 }=8{ V } _{ 2 }$
${ P } _{ 1 }\times { 8V } _{ 2 }={ P } _{ 2 }{ V } _{ 2 }$
${ \therefore 8P } _{ 1 }={ P } _{ 2 }$
Now ${ \therefore P } _{ 1 }+\rho gH={ P } _{ 2 }$
$\rho gH={ 7P } _{ 1 }$
$H=\frac {  7\times 1.01\times { 10 }^{ 5 } }{ 9810\times 13.6 }  $
$\therefore H=5.5$

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$^{2}$)

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

  1. 1200 kgwt

  2. 1500 kgwt

  3. 1700 kgwt

  4. 2000 kgwt

Show answer
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$=1.013\times 10^{5}N/m^{2}$].

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

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

Show answer
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 :-

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

  1. H

  2. 2H

  3. 7H

  4. 8H

Show answer
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:

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

  1. 2.16 m

  2. 2.88 m

  3. 12.24 m

  4. 24.48 m

Show answer
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

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

  1. depth of the point from the surface

  2. density of the fluid

  3. acceleration due to gravity

  4. the area of cross section

Show answer
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.