Tag: pressure in fluids

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Multiple choice physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

A cube of wood floats in water, with $42$% of its volume is submerged, then the density of the wood is

  1. $42\ g\ cm^{-3}$
  2. $0.42\ g\ cm^{-3}$
  3. $0.58\ kg\ cm^{-3}$
  4. $600\ g\ cm^{-3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Archimedes Principle, the buoyant force on a body partially or fully immersed in a fluid is given by the weight of the fluid displaced.


Let the volume of wood be $V$
Thus, volume of wood submerged is $0.42V$

Thus, the buoyant force acting on the wood is $B = 0.42\rho gV$
Weight of wood is $\rho _\textrm{wood}gV$

Thus, in equilibrium, $0.42\rho gV = \rho _\textrm{wood}gV \Rightarrow \rho _\textrm{wood} = 0.42\rho$

As Density of water is $\rho = 1\textrm{ g cm}^{-3}$, we have $\rho _\textrm{wood} = 0.42 \textrm{ g cm}^{-3}$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

A: Diver dives in and reaches a depth of 100m.
B: Diver swims up from the depth of 100m to the surface.
Choose the correct alternative:

  1. A is easier than B

  2. B is easier than A

  3. A and B are equally hard

  4. A is easier than B if speed of descent and ascent is same and more.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Assume:
$F _d: \text{Force applied by diver during descent in downward direction}$
$F _u: \text{Force applied by diver during ascent in upward direction}$
$U: \text{Upthrust}$
$m: \text{Mass of the diver}$
$g: \text{Acceleration due to gravity}$
Uniform ascent and descent.

Diver has more density than water. Hence, weight of diver is more than the upthrust and without any effort, the diver sinks.
i.e. $mg>U..................(1)$

During upward motion, $F _u+U-mg=0............(2)$
During downward motion, $F _d-U+mg=0...............(3)$

From (1),(2) and (3), 
$F _u-F _d=2mg-2U$
$F _u-F _d>0$
$F _u>F _d$
Multiple choice physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

Mathematical proof of upthrust is based on 

  1. Definition of pressure

  2. Weight of object

  3. Pressure exerted by a column of fluid

  4. Viscosity
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Mathematical proof:
Consider a cylinder of cross section area $A$ and height $L$ completely submerged in water. Let depth of upper surface be $h$.
Using, pressure exerted by fluid column:
Force on the upper face of the cylinder = $hρgA$
Force on the lower face of the cylinder = $[h + L]ρgA$
Difference in force = $LAρg$

But $LA$ is the volume of liquid displaced by the cylinder, and $LrgA$ is the weight of the liquid displaced by the cylinder.

Therefore there is a net upward force on the cylinder equal to the weight of the fluid displaced by it.
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.
$P=hdg$
$h=\dfrac {p}{dg}=\dfrac {90000}{1000\times 10}=9m$.