DENSITY OF AN IRREGULAR SOLID:

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 DENSITY OF AN IRREGULAR SOLID: 

What is irregular solid?: It is the shape of three dimensional object whose sides can not be measured and volume can not be determined by mathematical formula e.g.  stone boulder, wooden piece

What is meniscus?:

A meniscus is the formation  of a curved  surface of substance  like water, or other liquid  when comes in contact with the vertical wall of  glass cylinder or other material, due to force of molecular attraction between the two materials. Water inside a glass cylinder forms a concave surface. Water level shall be read with the bottom line of the curve i.e. lower meniscus  as shown in FIG-1.  

What is Density?: Density is the measure of  the compactness of mass in a body. It is termed as  the ratio of the mass (M) and volume (V) of the object , Density, d=M/V

Volume, V of a three dimensional body is the space occupied by the object.

M=Mass of the object

Density of an immersible object: 

Let us take a stone sample and weighed=W in gm

Required quantity of water is taken in a graduated cylinder with known initial volume: V₁ ml

Put the stone sample into the cylinder, being heaver density, the sample will immerse, air bubble if any to be shaken out. Water in the cylinder will rise up and the level is noted :V₂ ml

Water displaced equal to the volume of stone = Difference of the two readings i.e the volume of stone sample, V= V_{2 -} V₁

All values are known to find the numerical value of density. It is known that 1ml of water=1cm3,

Hence,  Density of stone, d=W/(V₂-V₁) in gm /cc

Density of a Floating object: 

method-: 

1. Let us take a graduated glass cylinder with initial water level V1

2. The object  is put gently on the water surface in cylinder. It will float being lighter than water. Now the object is just  immersed with a stick so that stick does not go inside water as far as possible, water will rise up and volume is recorded = V_{2. Volume of the object =Volume of water displaced = V2-V1}

Mass of the object in air is recorded accurately=M. 

Density= M / (V₂-V₁)

FIG-1

Method-2: 

1.Required quantity of water is taken in a graduated glass cylinder, say initial volume of water, V1 =800 ml=800CC in FIG-1

2. An iron piece is immersed into the cylinder tied with a thin plastic thread,air bubble if any is to be shaken out, reading of the water level is  recorded , V2= 900ml= 900 CC FIG-2

3. Now the object, tied with an iron piece is immersed  (Iron piece acts as a sinker of the object)  as shown in fig.3 water level is recorded, V_{3= 960ml= 960CC}

Water displaced due to immersion of the iron piece i.e volume of iron piece, V2 - V1= 100CC.

      FIG-2

Water displaced due to immersion of the iron piece and object i.e combined volume , V3-V1= 160CC.

Hence volume of the floating object, 160-100=60ml=60CC

Let the mass of the floating object=48gm, 

Hence, Density of the object= 48/60=0.8 gm/cc.

FIG-3

Method-3: Let us take some quantity of water in a small container up to the brim. An iron piece is slowly left on the surface. It will immerse and displace water equal to the volume of iron piece in Fig-4 Displaced water is collected in a non-sticky container carefully with no loss. 

FIg-4

Let us weigh the collected water, W1=100gm, As 1gm water occupies volume 1 cubic cm.  Hence volume of iron piece, V1= 100cc.

Now the container is again filled up to the brim and  the floating object is tied with the same iron piece with a thin plastic thread and placed gently on the surface.

 

The object alongwith iron piece will immerse. The displaced water is again collected and weighed, W2=180gm (say). As explained above volume of the iron and the object, V2=180cc. as in Fig-4.

Volume of the object, V=180-100=80cc.

Let us take the mass of the object, W=68gm

Density of the object, d=W/V = 68/80=0.85 gm/cc.

Remarks: At least 03 readings shall be taken to get the average density.

DENSITY OF A HOLLOW SPHERE WITH KNOWN OUTER RADIUS( R) AND INNER RADIUS(r)

Let us take some quantity of water in  a container  up to the brim. A hollow cylinder is left gently on the surface of water.  It will float and  displace some quantity water equal to the submerged part of the sphere. The displaced water is collected in a non-sticky container with great care so that water is not lost. let us take  the mass of  collected water, M=450gm. Hence volume of the sphere under water=450cc

We know from Archimedes principle , For a floating body weight of hollow sphere (Material)= weight of the displaced liquid

Hence Weight of the hollow sphere,W=M x g=450 x g.............1

Again, weight of the hollow sphere, W= Volume of the solid material(Vs) x Density of the material(ρ) x g or W=Vs x ρ x g.....................2

Equating 1 and 2, We can write M x g= Vs x ρ x g

or Density of the material,ρ =M/Vs

Let us consider outer Rad of the sphere, R=5cm and Inner Rad, r=4.8cm

Hence density of the sphere  material can be determined following the above procedure.

Also See the Following Link

Sp.Gr. of soils,                       Dry Density of Soils,                   Dry Density by core cutter