Tag: pressure at a certain depth in liquid

Questions Related to pressure at a certain depth in liquid

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. 

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

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

  1. $1 : 1$

  2. $2 : 3$

  3. $3 : 2$

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

Show answer
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.

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

  1. True

  2. False

Show answer
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?

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

  1. It doubles

  2. It quadruples

  3. It slightly increases

  4. It cut to a half

Show answer
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:

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

  1. the acceleration

  2. the density of the liquid

  3. the area of the liquid surface

  4. the height of the liquid

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

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

  1. $10\ atm$

  2. $5\ atm$

  3. $15\ atm$

  4. $2\ atm$

Show answer
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?

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

  1. $gh$

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

  3. $0$

  4. $none$

Show answer
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