Tag: introduction to pressure

Questions Related to introduction to pressure

The magnitude of buoyant force acting on an object immersed in a liquid depends on
physics floatation pressure in fluids pressure in liquids introduction to pressure
  1. Volume of object immersed in the liquid.

  2. Density of the liquid.

  3. Both A and B

  4. None

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Correct Option: C
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.

The property of a fluid to exert a buoyant force on an object immersed in it is known as _______.

physics floatation pressure in fluids pressure in liquids introduction to pressure

  1. Adhesion

  2. Buoyancy

  3. Cohesion

  4. none of these

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Correct Option: B
Explanation:

The property of fluid to exert a buoyant force on an object immersed in it known as Buoyancy .

so  option (B) is correct

When a body is partially or completely immersed in a fluid at rest, it experiences an upthrust which is equal to the weight of the fluid displaced by it.

physics floatation pressure in fluids pressure in liquids introduction to pressure

  1. True

  2. False

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Correct Option: A
Explanation:

The given statement is the statement of archimedes principle , so it is true 

hence option (A) is correct

The upward force exerted by the floating body on the fluid is known as upthrust.

physics floatation pressure in fluids pressure in liquids introduction to pressure

  1. True

  2. False

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Correct Option: B
Explanation:
Upthrust is the upward force acting on the floating body and is exerted by the fluid .
so given statement is false

hence option (B) is correct

How much is the hydrostatic pressure exerted by water at the bottom of a beaker? Take the depth of water as 45 cm. (density of water $10^3 kg m^{-3})$.

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

  1. 2410 Pa

  2. 3410 Pa

  3. 4410 Pa

  4. 5410 Pa

Show answer
Correct Option: C
Explanation:

$Height=45 cm=0.45 m$,
$gravity=9.8 ms^{-2}$
$Density=1000 kg m^{-3}$
$Pressure=hdg$
$=0.45\times 9.8\times 1000Pa=4410 Pa$.

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 $=1000 kg m^{-3}, g=10 m s^{-2})$.

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

  1. 3 m

  2. 4 m

  3. 5 m

  4. 6 m

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Correct Option: A
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
$=h _1-h _2=6m-3m=3m$

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 $=1000 kg m^{-3}]$.

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

  1. 9 m

  2. 10 m

  3. 11 m

  4. 12 m

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Correct Option: A
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.
$P=hdg$
$h=\dfrac {p}{dg}=\dfrac {90000}{1000\times 10}=9m$.

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 $g=10 ms^{-2}$

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

  1. 2 m

  2. 3 m

  3. 4 m

  4. 5 m

Show answer
Correct Option: B
Explanation:

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

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

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

  1. 3200 Pa

  2. 6400 Pa

  3. 800 Pa

  4. 5000 Pa

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Correct Option: B
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.

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

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

  1. $2\times 10^4 Pa$

  2. $4\times 10^4 Pa$

  3. $6\times 10^4 Pa$

  4. $8\times 10^4 Pa$

Show answer
Correct Option: C
Explanation:

Height $(h)=6 m$,

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