Tag: upthrust in fluids, archimedes' principle and floatation

Questions Related to upthrust in fluids, archimedes' principle and floatation

A piece of iron of density $7800kg/m^3$ and volume $100cm^3$ is completely immersed in water. Calculate the upthrust on the iron piece.

[Take, $g=10m/s^2$.]

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

  1. $1N$

  2. $2.8N$

  3. $7.8N$

  4. $6.8N$

Show answer
Correct Option: A

if a block of wood is floating in a river water, then the apparent weight of the floating block is

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

  1. equal to the weight of the displaced water

  2. zero.

  3. greater than the weight of the displaced water

  4. equal to the actual weight of the block

Show answer
Correct Option: B
Explanation:
Apparent weight in calculated as
Apparent weight$=$(weight of body)$-$(Buoyancy force)
Where
weight of body$=$(mass of body)$\times$(Gravitational force)
Buoyancy force$=$weight of fluid displaced
for the floating body:-
weight of body$=$Buoyancy force
$\therefore$ The body floating and
apparent weight equals to zero.
$\therefore$ Answer is B.

A body of mass 50 kg has a volume 0.049 $m^3$ and is kept on a table. The buoyant force on it due to air is.

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

  1. 0.588 N

  2. 0.49 N

  3. 58.8 N

  4. 49 kgf

Show answer
Correct Option: D
Explanation:

since the mass is in equilibrium. So, the net downward force =buoyant force

buoyant force=50x9.8=49kgf

Calculate the upthrust acting on a ball of volume $\displaystyle 0.2{ m }^{ 3 }$ immersed in a liquid having density $\displaystyle { 1kg }/{ { m }^{ 3 } }$.

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

  1. 0.2f

  2. 2f

  3. 0.2g f

  4. 2g f

Show answer
Correct Option: C
Explanation:

Upthrust = weight of the liquid displaced by the submerged part of the body Upthrust = mass of liquid displaced x Acceleration due to gravity Upthrust = volume of liquid displaced x density of liquid displaced x Acceleration due to gravity,Since volume of solid immersed is equal to the volume of the liquid displaced, Upthrust = volume of solid immersed x density of liquid displaced x Acceleration due to gravity Upthrust = 0.2 x 1 x g = 0.2g f.

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

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

  1. $42\ g\ cm^{-3}$

  2. $0.42\ g\ cm^{-3}$

  3. $0.58\ kg\ cm^{-3}$

  4. $600\ g\ cm^{-3}$

Show answer
Correct Option: B
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 $\rho _\textrm{wood} = 0.42 \textrm{ g cm}^{-3}$

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:

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

  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.

Show answer
Correct Option: A
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), 
$F _u-F _d=2mg-2U$
$F _u-F _d>0$
$F _u>F _d$

Mathematical proof of upthrust is based on 

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

  1. Definition of pressure

  2. Weight of object

  3. Pressure exerted by a column of fluid

  4. Viscosity

Show answer
Correct Option: C
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.

A cube of size 10 cm is floating in equilibrium in a tank of water. When a mass of 10 gm is placed on the cube, the depth of cube inside water increases by $\mathrm { g } = 10 \mathrm { m } / \mathrm { s } ^ { 2 }$ density of water $= 1000 \mathrm { kg } / \mathrm { m } ^ { 3 } )$

physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

  1. 0.1 m

  2. 1 mm

  3. 1 m

  4. 0.31 m

Show answer
Correct Option: B
Explanation:

Let $x$ be the initial depth upto which the cube is sinked in water.

Let $d$ be the density of the cube
Then
$x\times 10 \times 10\times 1 \times g$
$=10\times 10 \times 10\times 1 \times g\times d$........(1) 
$\Rightarrow x=10d$
Let $x^1$ be the new depth, then 
$x^1\times 100 \times g=1000\times d \times g+10g$........(2)
subtracting (1) and (2) we get
$\Rightarrow 100x^1=10x+10$
$x^2-x= \frac{1}{10cm}=1mm$
Hence,
option $B$ is correct answer.

A hydrogen balloon released on the moon would:

physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

  1. climb up with an acceleration of $9.8 \ m/s^2$

  2. climb up with an acceleration of $9.8 \times 6 \ m/s^2$

  3. neither climb nor fall

  4. fall with an acceleration of $9.8/6 \ m/s^2$

Show answer
Correct Option: D
Explanation:

As there is no atmosphere on the surface of the moon  so no bouyancy will act Hence it will fall with acceleration $9.8/6 m/s^2$

Statement I:- A block is immersed in a liquid inside a beaker,which is falling freely. Buoyant force acting on block is zero.
Statement II:- In case of freely falling liquid there is no pressure difference between any two points.

physics upthrust in fluids, archimedes' principle and floatation applications of floatation principle of floatation and its applications when do objects float on water?

  1. Statement I is true,statement II is true and statement II is a correct explanation for statement I.

  2. Statement I is true,Statement II is true and statement II is NOT the correct explanation for statement I.

  3. Statement I is true,Statement II is false.

  4. Statement I is false,Statement II is true.

Show answer
Correct Option: A