Tag: floating bodies

Questions Related to floating bodies

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. 

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):

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

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

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

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

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

Show answer
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

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

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

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

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

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

Show answer
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

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

  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

Show answer
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.

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

  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.

Show answer
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.