Tag: applications of floatation

Questions Related to applications of floatation

A block of ice of total area A and thickness 0.5 m is floating in water. In order to just support a man of mass 100 kg, the area A should be (the specific gravity ofice is 0.9):

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

  1. $2.2m^{2}$

  2. $1.0m^{2}$

  3. $0.5m^{2}$

  4. None of these

Show answer
Correct Option: D
Explanation:

Let say $m _1=$ mass of the man = 100kg
and $m _2=$ mass of the ice $= 0.9 \times 1000V=900V$, where $V$ is the volume of the ice block.
For equilibrium,
Total downward weight = total upthrust
$100g +900 Vg=1000Vg \\Rightarrow V=1m^3$
Volume = Area $\times $ height
$\Rightarrow A=\frac{Volume}{Height}=\frac{1}{0.5}=2m^2$

The density of ice is $920kg/m^3$, and that of sea water is $1030kg/m^3$. What fraction of the total volume of an iceberg is outside the water?

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

  1. $0.107$

  2. $0.207$

  3. $0.307$

  4. $0.407$

Show answer
Correct Option: A
Explanation:

Let $V _L$ and $V _S$ be the volume of water displaced and volume of the ice respectively; $\rho _L$ and $\rho _S$ be the density of water and density of ice respectively. Since ice is floating on water, $F _B=W$
or $\rho _LV _Lg=\rho _SV _Sg$ or $\rho _LV _L=\rho _SV _S$
or $\frac{V _L}{V _S}=\frac{\rho _S}{\rho _L}$
This is the fraction of volume of the iceberg that is inside the water. Therefore, the fraction of volume of the iceberg that is outside the water is given by, 1 - 0.893 = 0.107
Hence, the fraction of the total volume of an iceberg is outside the water is 0.107.

An egg sinks when immersed in water contained in a vessel. On dissolving a lot of salt in the water,will the egg

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

  1. Develop cracks in the shell

  2. Break

  3. Rise and then float

  4. Remain where it is

Show answer
Correct Option: C
Explanation:

Answer is C.

An egg will sink in fresh water but it will float in very salty water; the density of the egg is greater than the density of fresh water but less than the density of the salty water.
The density of salty water can be a much as $10\%$ greater than that of fresh water i.e. up to $1.1 g/cm^{ 3 }$.
Hence, on dissolving a lot of salt in the water, the egg will rise and then float.

A body floats in water because of:

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

  1. No force is acting on it

  2. The buoyant force acting on it

  3. Gravitational pull

  4. Friction between body and the water

Show answer
Correct Option: B
Explanation:

When an object is floating, the net force on it will be zero. This happens when the volume of the object submerged displaces an amount of liquid whose weight is equal to the weight of the object.

Answer (B) the net force acting on this body is zero

Two solids A and B float in water. It is observed that A floats with half its volume immersed and B floats with $2/3$ of its volume immersed. Compare the densities of A and B:

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

  1. $4:3$

  2. $2:3$

  3. $3:4$

  4. $1:1$

Show answer
Correct Option: C
Explanation:

Given that $m _1=V _1/2 \times d \implies d _1=d/2$
and $m _2=2V _2/3 \times d \implies d _2=2d/3$
$\therefore d _1:d _2=3:4$

A piece of ice is floating in a concentrated solution of common salt (in water) in a pot. When ice melts completely, the level of solution will 

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

  1. Go up

  2. Remain the same

  3. Go down

  4. First go up then go down

Show answer
Correct Option: A
Explanation:

$\rho _{salt water0}>\rho _{ice}$

When ice floats, volume of ice in the salt water is given by
$V _{imm}\rho _{salt water}g=V _{ice}\rho _{ice}g$
$\implies V _{imm}=V _{ice}\dfrac{\rho _{water}}{\rho _{salt water}}<V _{ice}$

Hence when ice melts, whole of the volume would add to water and level of solution would rise.

An ice cube is floating in a glass of water. What happens to the water level when the ice melts?

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

  1. Rises

  2. Falls

  3. Remains the same

  4. First rises and then falls

Show answer
Correct Option: C
Explanation:

according to the Archimedes principle , the floating substance displaces some liquid, so when the ice melts ,there will be no change in the water level as the melted ice will occupy the same volume as it was occupying earlier.

A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four times the density of the material of the ball. The frictional force of the liquid on the rising ball is greater than the weight of the ball by a factor of

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

  1. $2$

  2. $3$

  3. $4$

  4. $6$

Show answer
Correct Option: B
Explanation:

Let $F _f = $ Force of friction  

$F _b = $ Force of bouyancy
$F _w = $ Weight of ball

Archimedes' principle:

$F _b = F _w + F _f$

Given: $F _b = (4P _B)gV$

$F _w = VP _Bg$

$F _f = F _b - F _w = 3P _B gV$

$\Rightarrow \dfrac{F _f}{F _w} = \cfrac{3P _B gV}{P _BgV} = 3 : 1 $

$\Rightarrow F _f = 3F _w$

An ice cube contains a glass ball. The cube is floating on the surface of water contained in a trough on the surface of water contained in a trough. What will happen to the water level, when the cube melts?

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

  1. It will remain unchanged

  2. It will fall

  3. It will rise

  4. First it will fall and then rise

Show answer
Correct Option: A
Explanation:

When the cube melts, the water level will remain the same.An floating object displaces an amount of water equal to its own weight. Since, water expands when it freezes, one ounce of frozen water has a larger volume than one ounce of liquid water.

A cylindrical block floats vertically in a liquid of density ${\rho} _1$ kept in a container such that the fraction of volume of the cylinder inside the liquid is $x _1$. then some amount of another immiscible liquid of density ${\rho} _2 ({\rho} _2 < {\rho} _1)$ is added to the liquid in the container so that the cylinder now floats just fully immersed in the liquids with $x _2$ fraction of volume of the cylinder inside the liquid of density ${\rho} _1$. The ratio ${\rho} _1 / {\rho} _2$ will be

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

  1. $\dfrac{1 - x _2}{x _1 - x _2}$

  2. $\dfrac{1 - x _1}{x _1 + x _2}$

  3. $\dfrac{x _1 - x _2}{x _1 + x _2}$

  4. $\dfrac{x _2}{x _1}-1$

Show answer
Correct Option: A
Explanation:

Let $V$ and $\rho$ be the volume and density of the cylindrical block.

Case 1 : When $x _1$ fraction of block's volume is immersed in liquid of density $\rho _1$
Using Archimede's principle :    Weight of cylindrical block = Weight of liquid displaced
$\therefore$  $\rho V g = \rho _1 x _1 V g$             ........(1)

Case 2 : When $x _2$ fraction of block's volume is immersed in liquid of density $\rho _1$ and $1-x _2$ fraction of block's volume is immersed in liquid of density $\rho _2$
Using Archimede's principle :    Weight of cylindrical block = Weight of liquid displaced
$\therefore$  $\rho V g = \rho _1 x _2 V g + \rho _2 (1-x _2) V g$             ........(2)

Equating (1)  and (2) we get    $\rho _1 x _1V g = \rho _1 x _2 V g + \rho _2 (1-x _2) V g$
OR    $\rho _1 x _1 = \rho _1 x _2 + \rho _2 (1-x _2)$
OR   $\rho _1 (x _1 - x _2) = (1-x _2)\rho _2$
$\implies$  $\dfrac{\rho _1}{\rho _2} = \dfrac{1-x _2}{x _1-x _2}$