FLUID AT REST

DENSITY AND RELATIVE DENSITY

The concept of density.

Figure below Shows two identical flasks one filled with water to 250cm3 mark and the other filled with kerosene to the same 250cm3 mark, when measured in electronic balance the flask filled with water is found to be heavier than that filled with kerosene why? The answer is in finding the mass per unit volume of kerosene and water in respective flasks.

The mass of the flask filled with water is 330g, and the mass of flask filled with kerosene 280g, the empty flasks were measured and found to be 80g therefore mass of water only is 250g and kerosene only is 200g.

Mass per unit volume of water is 250g/ 250cm3 this is 1g/cm3.

Mass per unit volume of kerosene is 200g/ 250cm3this is 0.8g/cm3.

The results 1g/cm3 and 0.8g/cm3 are the densities of water and kerosene respectively.

Therefore the density of a substance is the mass per unit volume of a given substance.

The SI unit of density is kilogram per meter cubic (kg/m3) also gram per centimeter cubic (g/cm3). The symbol for density is rho (ρ) ρ=mass/volume.

Example 1.
A block of ice with volume 5.5m3 has a mass of 5060kg find the density of ice.

Solution

Volume of block=5.5m3

Mass of block=5060kg

Density=mass /volume

=5060/5.5m3.

=920kg/m3.

The density of ice is 920kg/m3.

Example 2.

A silver cylindrical rod has a length of 0.5m and radius of 0.4m,find the density of the rod if its mass is 2640kg.

Solution

Mass of cylinder=2640kg

Volume of cylinder= πr²h

=3.14 x 0.4 2 x 0.5

=0.2512m3

Density=mass/volume

=10509kg/m3.





Example3.


A stone has a mass of 112.5g.when the stone totally immersed in water contained in measuring cylinder displaced water from 50cm3 to 95cm3.find the density of the stone.


Solution


Mass of the stone=112.5g


Volume of stone=95cm3-50cm3=45cm3





Density=mass/volume


=2.5g/cm3.






Example 4.


Beaker contain 262.5cm3 of a certain liquid weigh 410g,if the mass of an empty dry beaker is 200g,find the density of the liquid.


Solution


Mass of liquid=410g-200g=210g


Volume of liquid=262.5cm3.


Density=mass/volume


=0.8g/cm3.



DENSITY BOTTLE


The density bottle (pycnometer) consists of ground glass stopper with a fine hole through it.





The function of the fine hole in a stopper is that, when the bottle is filled and the stopper is inserted, the excess liquid rises through the hole and runs down outside the bottle, by this way the bottle will always contain the same volume of whatever the liquid is filled in provided the temperature remains constant.



density bottle


the bottle is used to measure density and relative density, relative density is comparison of one density to another, thus a density of a given volume of a substance to a density of equal volume of referenced substance, for example a ratio of a density of a given volume of substance to a density of an equal volume of water, this is referred to a relative density of a given substance or Specific gravity of a given substance. The term specific gravity is used when the reference substance is water.


Measurement of density of liquid by relative density bottle


ü The mass of bottle is found when dry and empty


ü The bottle is then filled with the liquid density is to be determined


ü The stopper is then inserted causing the liquid to overflow


ü The bottle is dried up by using blotting paper


ü The mass of the liquid and the bottle is found


ü Density is found from the collected data




Mass of empty bottle=m1g


Volume of liquid in the bottle=V


mass of bottle and the liquid=m2g


mass of liquid only=(m2-m1)g


density= mass /volume


density= .(m2-m1)g/V


The volume of the bottle is known, usually 25ml, 30ml or 50ml


Example 1


A 30ml density bottle was filled with kerosene and found to weigh 86g.if the mass empty dry bottle was 62g, find the density of kerosene.






Solution


Mass of empty bottle=62g


Mass of bottle and kerosene=86g


Mass of kerosene only=86g-62g=24g


Density= mass /volume


=24g/30ml


=0.8g/cm3.






Determination of densities of granules and sand





To find the density of sand or granules such as lead shots a density bottle is used as follows


1. Find the mass of empty dry density bottle mo


2. Put some granules and find the mass m1 =( mass of empty bottle + mass of granules)


3. Pour water in the bottle until it is full and find mass m2 = ( mass of bottle + mass of granules + mass of water on top of granules)



4. Find the mass of bottle filled with water only m3 =( mass of bottle + mass of water)




The mass of sand = (m1-m0) g


Mass of water above the sand = (m2-m1) g


Mass of water filling the bottle = (m3-m0) g




Since density of water is 1g/cm³






Volume of sand = [(m1+m3)-(mo+m2)]/1g/cm³


= [(m1+m3)-(mo+m2)] cm³



Density=mass/volume








Example 1


Given the data below find the density of granules


Mass of empty dry density bottle =18g


Mass of density bottle and granules=131g


Mass of density bottle and granules together with water on top =171g





Mass of density bottle full of water=68g

RELATIVE DENSITY

Relative density of a substance is the ratio of the density of substance to the density of water.


Or




Relative density of a substance is the ratio of mass of any volume of substance to the mass of an equal volume of water.




Example 1



A globe of steel has a mass of 12g and a volume of 15.2cm³,find its relative density.







To measure relative density of liquid by density bottle






ü Find mass of empty bottle –m0


ü Find mass of bottle and liquid-m1


ü Empty the bottle and rinse it with water





ü Fill the bottle with water and find mass m2








Mass of liquid= (m1-m0)g





Mass of equal volume of water= (m2-m0)g








Since comparison of density is done with water (referenced substance) the other name of the ratio is specific gravity of a given substance. Because the density of water is 1g/cm³.Relative density has no units it is simply a number or ratio.


Example



The mass of density bottle is 19g when dry and empty, 45g when filled with water and 40g when full of liquid x. calculate the density of the liquid x.




Determinations of relative density by eureka can method


ü Find the mass m₀ g of solid


ü Fill the eureka can and let water overflow until last drop


ü Place under the spout of overflow can a clean dry beaker of mass m₁ g.


ü Lower the solid slowly with thin thread until it is totally immersed


ü Obtain the mass of water that overflow from the eureka can and the beaker itself m₂ g.







The volume of water overflows into a beaker is equal to the volume of solid


Mass of solid = m₀ g


Mass of beaker and water=m₂ g


Mass of beaker=m ₁g.


Mass of water only = (m₂-m₁)g






EXAMPLE

A certain piece of metal has a mass of 282.5 g,if when the block was totally immersed in overflow can displaced water in a beaker of mass 20 g.if the mass of water and the beaker was 45 g,find the relative density of the metal.

ARCHIMEDES' PRINCIPLE

A plastic cup containing air (literally empty) when immersed in water upside down it will tend to come up, likewise the piece of wood or bee wax will come up if totally or partially immersed in water









In both cases objects rise up due to upthrust force which is greater than the weight (mg) of objects. If the weight would be greater than the upthrust then the object would sink but would be much lighter if measured when immersed in water.


Upthrust force is also known as buoyant force


Buoyant force is an upward force exerted by a fluid to a body which is partially or totally immersed in it and it is equal in magnitude to the weight of fluid displaced by a body.

ARCHIMEDES’ PRINCIPLE (LAW OF BUOYANCY)

States that, when a body is partially or totally immersed in fluid, it experiences an upthrust force which is equal to the weight of fluid displaced.






Experiment to show upthrust is equal to the weight of the fluid displaced





A is a real weight (weight of an object in air) 4.5 N


B is apparent weight (weight of an object in fluid) 3.0 N


C is apparent loss of weight (weight of fluid displaced) 1.5 N


Measuring the weight of displaced fluid is equal to apparent loss of weight; this apparent loss is the buoyancy force on an object which is upthrust.



Conclusion


Apparent loss of weight is equal to the weight of displaced fluid.



Real weight = apparent weight + apparent loss


Since apparent loss is equal to upthrust


Real weight = apparent weight + upthrust


Upthrust =real weight – apparent weight.




Example 1.


A body weighs 5.5 N when measured in air and 3 N when completely immersed in kerosene, what is upthrust experienced by a body.



Solution


Weight of body in air (Real weight) =5.5 N


Apparent weight of a body when immersed in kerosene = 3 N.


But upthrust = real weight – apparent weight loss


Upthrust = 5.5 N – 3 N


= 2.5 N.


Example 2.


A body weighs 3 N when totally immersed in water contained in an overflow can, if weight of displaced water was 2.6 N, what is the real weight of an object.




Solution


Apparent weight = 3 N


Upthrust = 2.6 N (according to Archimedes’ principle)


But


Upthrust = real weight – apparent weight


2.6 N = Real weight – 3 N.


Real weight = 2.6 N + 3 N


= 5.6 N.

DETERMINATION OF RELATIVE DENSITY OF SOLID BY ARCHIMEDES’ PRINCIPLE

If mass is multiplied by gravitational acceleration ‘g’ becomes weight (times 10 above and below which does not change magnitude but changes the physical quantity from mass to weight)




Weight of a given volume of substance: - is the weight of substance in air.


Weight of an equal volume of water: - is the weight of water displaced (liquid has to be water because the density of solid object is compared to the density of water, meaning of relative density)




Other formulae will be




Example 1.


A metal block weighs 108 g in air and 68 g in water, find the relative density of metal. What possibly this metal can be? (Use the table of densities).





Example 2


A body weighs 135 g in air displaced 54 g of water when completely immersed, find the relative density and volume of a body.






Volume of a body is equal to the volume of displaced water.


Density= mass/volume


Therefore 1 g/cm 3= 54 g/volume


Volume = 54 cm 3.

DETERMINATION OF RELATIVE DENSITY OF OBJECT THAT FLOATS BY ARCHIMEDES’ PRINCIPLE

CLASS WORK

AIM: To determine the relative density of an object that floats by Archimedes’ principle.


REQUIREMENTS: floating object, sinker (piece of metal or stone), spring balance, string, and water in a beaker


PROCEDURES:


1. Tie a sinker with a string.


2. Immerse a sinker totally in water and measure its weight (apparent weight) W 1


3. Tie a floating object just before the sinker and immerse totally the sinker leaving the floating object in air then measure weight W 2 (weight of floating object in air + apparent weight of a sinker)



4. Immerse totally both the sinker and the object in water and measure the weight W 3












Example 1


In an Experiment to find the relative density of floating object the followings were observed,


I. Weight of a sinker in air = 64 g


II. Weight of a sinker when totally immersed in water = 49 g


III. Weight of a sinker totally immersed and floating object in air =56 g



IV. Weight when both sinker and floating object totally immersed in water = 46 g

DETERMINATION OF RELATIVE DENSITY OF LIQUIDS BY ARCHIMEDES’ PRINCIPLE

If an object of given volume is totally immersed in different liquids it will always displace volume of given liquid equals to its own volume but the masses of these equal volumes of liquids displaced will differ owing to their densities(concentration of particles)


We can simply know mass of liquid displaced by noticing the apparent loss of mass of an object when totally immersed in liquid; this is according to Archimedes’ principle.


Remember, when mass is multiplied by ‘g’ (gravitational acceleration) it becomes weight also:-

CLASS WORK

AIM: TO FIND RELATIVE DENSITY OF ‘X’


REQUIREMENTS: Liquid x in a beaker, water in a beaker, object that can sink, and spring balance.


PROCEDURES:


1. Measure the sinking object in air and record its weight W 1


2. Measure the sinking object when totally immersed in X and record its apparent weight W 2



3. Measure the sinking object when totally immersed in water and record its apparent weight W 3

THE LAW OF FLOATATION

FLOATATION

Ships, canoes and ferry boats move on the upper part of the water bodies, we say they float.it may happen that due to accident or over loading they sink. Example Titanicwent to the bottom of North Atlantic Ocean after hitting larger ice- berg.


A balloon filled with lighter gas such as hydrogen will keep rising up when released from the hand and a hot air balloon will keep rising as well with the people in carriage once the air in balloon has been heated and expanded.


Therefore floatation can be defined as the tendency of an object to rise up to the upper levels of the fluid or to stay on the surface of the fluid.


The opposite of floatation is sinking and can be defined as the tendency of an object to go to the lower levels of the fluid.









MV.Liemba floats on lake Tanganyika

THE LAW OF FLOATATION

The law of floatation states that, a floating body displaces its own weight of the fluid in which it floats.


This means if a log of 200kg (2000N) floats in water displaces 200kg (2000N) of water, if the same log is placed in other liquid and be able to float it will displace the same 200kg of fluid in which it floats.

CONDITIONS FOR OBJECTS TO FLOAT

I. The average density of the object should be less than the density of the fluid in which the object has to float.


Example, a ship is very heavy but it floats because it is hollow inside it contains air, this causes its average density to be lower than that of water.


II. The upthrust force of the fluid on the object must be equal total weight of the object.(law of floatation)


Example, a coin will sink to the bottom when placed on the surface of water, this is because the upthrust of water on coin is less than its weight


III. The volume of object submerged must be large so as to displace large amount of fluid.

RELATIONSHIP BETWEEN UPTHRUST AND REAL WEIGHT OF FLOATING BODY

We know that,


Real weight = apparent weight + apparent loss in weight


Apparent loss in weight = weight of fluid displaced = upthrust


Therefore,


Real weight =apparent weight + upthrust


But when an object floats its apparent weight is zero


Apparent weight = 0


Then,


Real weight = 0 + upthrust


Real weight = upthrust.


Therefore relationship between upthrust and real weight of floating object is such that upthrust is equal to real weight, and since upthrust is equal to weight of fluid displaced then floating body displaces its own weight of the fluid in which it floats (law of floatation)


In order an object to float its real weight must be equal to upthrust.


If an object floats the volume of fluid displaced is equal to the volume of an object submerged and the percentage of the volume of an object submerged is equal to the relative density of an object with respect to the density of fluid in which it floats.


Thus,


Example


An ice cube of density 0.9g/cm3 floats in freshwater of density 1.0g/cm3 what fraction of volume of ice is submerged? If the same ice is floating in sea water of the density 1.3g/cm³ what is the percentage volume of ice will be submerged?



Solution


Percentage volume of an object submerged in fresh water,




Percentage volume of an object submerged in sea water

APPLICATION OF LAW OF FLOATATION IN EVERY DAY LIFE

In transportation by water ways


The law of floatation is applied in all vessels which travel by waterways that include ships, submarines and ferry boats.


In transportation by air ways


It is also applied in some vessels which travel by air ways such as hot air balloon and air ship.


In decoration


Balloons of different colors and shapes are filled with lighter gas so that will float in air.


In measurement of specific gravity of liquids


Hydrometer is an instrument which is used to measure specific gravity of liquids, in its operation it employs the law of floatation.


Other uses include making of bulges and transportation of logs down the river.




Ships

Ship is like an iron bowl floating in water, although iron is denser than water the bowl does not sink. Being hollow it contains air causing its average density to be lower than that of water.

Sand can be poured in the bowl but it will keep floating, in excess mass of sand the bowl will sink, in ship we call it overloading.

water, so to make it be able to float it is hollow and filled with air making the average density lower than density of water even when it is loaded with her cargo.


Ship travels in different densities of water, sea water or fresh water, (hot or cold)


So by changing water ways ship can displace more or less water in which it floats that means it may gain or lose upthrust. If the upthrust becomes lower than its own weight the ship sinks.


So for safety loading of the ship under different sea conditions plimsol lines are provided.


Plimsol lines are lines which show maximum height of the ship that should be under water.


Plimsol lines are also referred as plimsol marks.










Pic of marksxxxxxxxxxx


Submarine


Submarine is ship which can travel deep under water and also at the surface of water, it owes its ability to dive and come up to law of floatation.it contains ballast tanks at either sides which are filled with water when it has to dive and are emptied when it has to come up. The tanks are filled and emptied by special pumps.




submarine at different levels from sea surface













submarine at the surface and underwater






Hot air balloon


To watch animals at Serengeti helicopters is noisy it would scare the animals,plane is faster it would pass without clear vision but hot air balloon will do.


Hot air balloon consist of three parts,the balloon,the burner and the basket.


The burner uses the propane gas to heat up air in the balloon as the air in the balloon gets hot it expands,its density becomes lower than the surrounding air so it raises.


To get the balloon down the pilot can open a parachute valve at the top of the balloon causing cold air to enter in the balloon and decrease the temperature inside the balloon so the density,also the pilot can let the air inside the balloon to cool itself by burning less fuel.









hot air balloon






Air ship


Air ship uses lighter gas such as helium, hydrogen or hot air to rise up, and maneuvering is done by means of propeller attached to it,example at cabin.


The lighter gas gives it lifting and the engine is used to propel it .



airship


Balloons


Balloons used for decoration are filled with lighter gas such as hydrogen or helium so that they stay in the air,when filled by blowing air to them the can not rise because the air we blew in them denser or equals to the density of air when it is air itself(if not contaminated by exhaled air).






Transportation of logs by the river










Bulge


A bulge is floating platform which can be used for several activities for example recreational,transportation,research or working platform,bridge etc.













The hydrometer