Tag: principle of floatation and its applications

Questions Related to principle of floatation and its applications

Ice ______ in water, because the weight of water displaced by the immersed part of the ice is _____ its own weight 

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

  1. sinks, more than

  2. sinks, less than

  3. floats , equal to

  4. floats , less than

Show answer
Correct Option: C
Explanation:

According to Archimedes principle, A body immersed in water experiences an upward force equal to the mass of the fluid displaced by the body. If the weight of an object is greater than the weight of displaced fluid, it will float. If the two are equal, it is suspended, neither floating nor sinking. For example, when an object is placed in water, it will displace its own volume of water, and that water will push back against it proportionally, producing an upthrust.
Water has a weight density of $62$ pounds per cubic foot. It an object weighing $62$ pounds has a volume that displaces $2$ cubic feet of water, it will float. The displaced water will weigh $124$ pounds and the pressure of that water would be enough to keep the object floating.
Hence, Ice floats in water, because the weight of water is displaced by the immersed part of the ice is more than its own weight and the statement is true.

An ice-berg floating partly immersed in sea water of density $1.03 g/cm^3$. The density of ice is $0.92 g/cm^3$. The fraction of the total volume of the iceberg above the level of sea water is

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

  1. $8.1\%$

  2. $11\%$

  3. $34\%$

  4. $0.8\%$

Show answer
Correct Option: B
Explanation:

Let $v$ be the volume of the ice-berg outside the sea water and $V$ be the total volume of ice-berg. Then as per question
$0.92V = 1.03(V-v)$

or, $\dfrac vV = 1-\dfrac {0.92}{1.03}= \dfrac{11}{103} $
$\therefore  \dfrac vV \times 100 = 11 \times \dfrac {100}{103} \cong  11\%$

A wooden cylinder floats in water such that $4 cm$ of it is above water, if the same cylinder is made to float in alcohol (density $0.8gm^{-3}$), the length of cylinder above alcohol 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. $4cm$

  2. more than $ 4cm$

  3. less than $4cm$

  4. none of these

Show answer
Correct Option: C
Explanation:

C. less than 4cm  
Because more volume of alcohol needs to be displaced to displace the equal weight of water. since density of alcohol is lesser.

A wooden cube of side $10 \ cm$ has mass $700 \ g$. The part of it remains above the water surface while floating vertically on water surface is $X\ cm$. Find $X$.

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

  1. $3$

  2. $7$

  3. $0$

  4. Can not be detemined

Show answer
Correct Option: A
Explanation:

We know that the volume of part submerged in a liquid is given by, $V= \dfrac{Density\ of\ body}{Density\ of\ liquid}$
Density of cube $= \dfrac{Mass}{Volume}$$ =\dfrac{700 g }{ 10^{3}} = 0.7g/cm^{3}$
So, density of water $= 1 g/cm^{3}$
So, part of cube submerged in water $= \dfrac{Density\ of\ body}{Density\ of\ water} = 0.7/1 = 7/10$
$\therefore$ Part of cube above water $= 1 - 7/10 = 3/10$
i.e. $3cm$ of cube is above water.

A block of wood is so loaded that it just floats in water at room temperature. What change will occur in the state of floatation, if water is heated? 

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

  1. Floats with some part above water

  2. Just floats

  3. Sinks

  4. Floats completely above water

Show answer
Correct Option: C
Explanation:

Sinks.
On heating, the density of water decreases. So, upthrust on block decreases and weight of block exceeds upthrust due to which it sinks.

A block of wood is so loaded that it just floats in water at room temperature. What change will occur in the state of floatation, if some salt is added to 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. Floats with some part outside water

  2. Just floats

  3. Sinks in the water

  4. Floats completely above water

Show answer
Correct Option: A
Explanation:

Floats with some part outside water. 
On adding some salt to water, the density of water increases, so upthrust on block of wood increases and hence the block rises up till the weight of salty water displaced by the submerged part of block becomes equal to the weight of block.

Iron floats on the surface of mercury because its density is _____ the mercury

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

  1. more than

  2. less than

  3. same as

  4. cant say

Show answer
Correct Option: B
Explanation:
The substance having low density floats on substance with high density. 
Hence iron would float on mercury as it have lower density than mercury.
therefore, option (b) is correct.

Object having density less than that of the liquid in which they are immersed, _______on the surface of the liquid.

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

  1. Float

  2. Sink

  3. First sink and then float

  4. First float and then sink

Show answer
Correct Option: A
Explanation:

Float
We know according to Archimedes Principle, Object having density less than that of the liquid in which they are immersed, float on the surface of the liquid.
And Object having density greater than that of the liquid in which they are immersed, sink in the liquid.

A tin can has a volume of $1000cm^3$ and a mass of $100g$. What mass of lead shot can it carry without sinking in water $(\rho=1000kg/m^3)$?

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

  1. $900g$

  2. $100g$

  3. $1000g$

  4. $1100g$

Show answer
Correct Option: A
Explanation:
Volume of the floating tin = Volume of water displaced = $1000{ cm }^{ 3 }$.
weight of water displaced $= 1000\times 1 = 1000g = 10 N  =$ Upthrust.  Upthrust = max load + weight of tin 
That is, max load = upthrust - weight of tin.
It is given that the weight of tin is $100 g.$
Therefore, load $= 10 N - 1 N = 9 N$
Mass $M =$ Weight w/acceleration due to gravity $g$. Let us take $g=10m/{ s }^{ 2 }$.
So, $M=W/g = 9 N/10 = 0.9 kg = 900 g$.
Hence, the mass of lead shot the tin can carry without sinking in water is $900 g$.

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
$\Rightarrow A=\frac{Volume}{Height}=\frac{1}{0.5}=2m^2$