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

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

  1. $1 : 1$

  2. $2 : 3$

  3. $3 : 2$

  4. $2^{2} : 3^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 .

$\dfrac{P _1}{P _2}=\dfrac{\rho gh}{\rho gh}=1$

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

State True or False.
Pressure at a point in a liquid is inversely proportional to the height of the liquid column.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $P=\rho g H$ 

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

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?

  1. It doubles

  2. It quadruples

  3. It slightly increases

  4. It cut to a half
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

The pressure at the bottom of a tank of liquid is not proportional to:

  1. the acceleration

  2. the density of the liquid

  3. the area of the liquid surface

  4. the height of the liquid
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $C.$

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

The pressure on a swimmer 10 m below the surface lake is:(Atmospheric pressure=$1.01\times 10^5$ Pa,Density of water$=1000kg/m^3$ )

  1. $10\ atm$

  2. $5\ atm$

  3. $15\ atm$

  4. $2\ atm$
Reveal answer Fill a bubble to check yourself
D Correct answer
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$
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

What is the difference between the pressure on the bottom of a pool and the pressure on the water surface?

  1. $gh$

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

  3. $0$

  4. $none$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

  1. $1 : 2$

  2. $1 : 1$

  3. $2 : 1$

  4. $4 : 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The pressure of water vapour in a closed container at a constant temperature depends only on the temperature (saturated vapour pressure), not on the volume of the container or the amount of liquid present, provided some liquid remains.

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

 $1m^3$ 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 $g = 10 m/s^2$

  1. $8.9 \times 10^{-3} m^3$

  2. <span>$7.8 \times 10^{-3} m^3$</span>

  3. <span>$9.1 \times 10^{-4} m^3$</span>

  4. <span>$8.7 \times 10^{-4} m^3$</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
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$

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

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?

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

  2. $P _1 &gt; P _2 &gt; P _3$

  3. <span>$P _1 &lt; P _2 = P _3$</span>

  4. $P _1 &lt; P _2 &gt; P _3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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:

  1. $2P$

  2. $(13/5)P$

  3. $(8/5)P$

  4. $(4/5)P$
Reveal answer Fill a bubble to check yourself
B Correct answer
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
$(\cfrac85+1)P=\cfrac{13}5P$