Tag: pressure in liquids

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Questions Related to pressure in liquids

Multiple choice physics floatation pressure in fluids pressure in liquids introduction to pressure
The magnitude of buoyant force acting on an object immersed in a liquid depends on

  1. Volume of object immersed in the liquid.

  2. Density of the liquid.

  3. Both A and B

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Archimedes Principle, the magnitude of buoyant force experienced is equal to the weight of liquid displaced by it by being placed in it.

Hence $B=V _{immersed}\rho g$
Hence $B$ depends both on $V _{immersed}$ and $\rho$.
Correct answer is option C.

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

The pressure in a water pipe on the second floor of a building is 60,000 Pa, and on the third floor it is 30,000 Pa. Find the height of the second floor. (Density of water $=1000 kg m^{-3}, g=10 m s^{-2})$.

  1. 3 m

  2. 4 m

  3. 5 m

  4. 6 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Second floor :
$P _1=60,000 Pa, g=10 ms^{-2}$
$P _1=h _1dg$


$60,000=h _1\times 1000\times 10$

$h _1=\dfrac {60,000}{1000\times 10}=6 m$

[where $h _1=$ height of water tank above second floor]

Third floor :
$P _2=30,000 Pa, g=10 m s^{-2}$,


$\therefore 30,000=h _2\times 1000\times 19$

$\Rightarrow h _2=\dfrac {30000}{1000\times 10}=3 m$

[where $h _2=$ height of water tank above first floor]

$\therefore $ height of the second floor
$=h _1-h _2=6m-3m=3m$

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

The pressure in water pipe at the ground floor of a building is 120000 Pa, where as the pressure on a third floor is 30000 Pa. What is the height of third floor?
[Take $g=10 m s^{-2}$, density of water $=1000 kg m^{-3}]$.

  1. 9 m

  2. 10 m

  3. 11 m

  4. 12 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Difference in pressure of water at ground floor and third floor
$=(120000-300000)=900000 Pa$
Density of water $=1000 kg m^{-3}$
Let 'h' be the height third floor.
$P=hdg$
$h=\dfrac {p}{dg}=\dfrac {90000}{1000\times 10}=9m$.

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

The pressure of water on the ground floor is 50000 Pa and at the first floor is 20000 Pa. Find the height of the first floor. Take density of water is $10^3 kg m^{-3}$ and $g=10 ms^{-2}$

  1. 2 m

  2. 3 m

  3. 4 m

  4. 5 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Pressure on the ground floor $=$ Pressure on the first floor + hdg
$\Rightarrow 50000=20000+hdg$
$\Rightarrow 50000=20000+h\times 10^3\times 10$
$\Rightarrow h\times 10^4=50000-20000$
$\Rightarrow h=\frac {30000}{10000}m$
$\Rightarrow height=3 m.$

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

Calculate the pressure exerted by 0.8 m vertical length of alcohol of density $0.8 g cm^{-3}$. (Acceleration due to gravity $(g)=10 m s^{-2})$.

  1. 3200 Pa

  2. 6400 Pa

  3. 800 Pa

  4. 5000 Pa
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vertical length of the alcohol column
$(h)=0.8 m$
Density of alcohol $(d)=0.8 g cm^{-3}$
$=0.8\times 1000=800 kg m^{-3}$
$[\therefore 1 g cm^{-3}=1000 kg m^{-3}]$
$Pressure = hdg$
$=0.8\times 800\times 10$
$=6400 Pa$
Therefore, pressure exerted by the alcohol column is 6400 Pa.

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

Calculate the pressure exerted by water at the bottom of a lake of depth 6 m. (Density of water $=1000 kg m^{-3}, g=10 ms^{-2})$.

  1. $2\times 10^4 Pa$
  2. $4\times 10^4 Pa$
  3. $6\times 10^4 Pa$
  4. $8\times 10^4 Pa$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Height $(h)=6 m$,

$density=1000 kg m^{-3}$,
$g=10 m s^{-2}, Pressure = ?$
$P=hdg$
$=6\times 1000\times 10 Pa$
$=60000 Pa$
$=6\times 10^4 Pa.$