Tag: pressure at a certain depth in liquid

Questions Related to pressure at a certain depth in liquid

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

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

  1. 2 m/s

  2. 4 m/s

  3. 2.8 m/s

  4. 1 m/s

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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

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

  1. $10m$

  2. $15m$

  3. $5m$

  4. $25m$

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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.

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

  1. $862 cm$

  2. $932 cm$

  3. $982 cm$

  4. $1033 cm$

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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? 

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

  1. 500$\mathrm { m }$

  2. 400$\mathrm { m }$

  3. 300$\mathrm { m }$

  4. 200$\mathrm { m }$

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Correct Option: A

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

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

  1. $826 cm$

  2. $932 cm$

  3. $982 cm$

  4. $1033 cm$

Show answer
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 = $1000 kg/{ m}^{ 3})$

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

  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

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Correct Option: A

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

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

  1. $1.0$ atm

  2. $2.0$ atm

  3. $2.5$ atm

  4. $3.0$ atm

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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?

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

  1. 500m

  2. 400m

  3. 300m

  4. 200m

Show answer
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)

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. 3h

  4. 4h

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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

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

  1. q

  2. $ q^2 $

  3. $ \frac {1}{a} $

  4. $ \frac {1}{a^2} $

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Correct Option: D