Tag: pressure exerted by a liquid column

Questions Related to pressure exerted by a liquid column

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

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

  1. $1 : 2$

  2. $1 : 1$

  3. $2 : 1$

  4. $4 : 1$

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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 $g = 10 m/s^2$

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

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

Show answer
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?

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

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

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

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

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

Show answer
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:

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

  1. $2P$

  2. $(13/5)P$

  3. $(8/5)P$

  4. $(4/5)P$

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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$^{-2}$)

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

  1. 5 x 10$^5$ N

  2. 25 x 10$^5$ N

  3. 5 x 10$^6$ N

  4. 25 x 10$^6$ N

Show answer
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 $
$= 5 \times 10^6\,N$