TUTORIAL: DENSITY AND UPTHRUST SIMPLE EXPLANATION

  DENSITY, UPTHRUST AND RELATIVE
                         DENSITY
             
Definition:
Density is defined as the mass per unit volume. It is represented by rho(¢) and measured in g/cm^3 or kg/m^3 depending on the value in which mass and volume is measured.
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Mathematically,
Density=
    mass(g or kg)/volume(cm^3 or m^3) i.e
            ¢ = m/v.

                    RELATIVE DENSITY
Definition:
Relative density is defined as the mass of a volume of a substance compare to the mass of an equal volume of water at 4^•c. It has no unit and it is denoted by R.D.

Mathematically expression,
R.D= mass of sample in air/apparent loss in weight of sample in water.

For a liquid,
R.D= loss in weight of body in air /loss in weight of body in water.

ARCHIMEDE'S PRINCIPLE
This law states that when a body is completly www.naijaexamrunz.blogspot.com immersed in a liquid, it will displace a volume of liquid which is equal to the volume of the object.

LOGIC OF ARCHIMEDES PRINCIPLE
First note all the underlisted points:
From ¢=m/v.
M=¢v i.e density × volume. Now,
 *volume of object = volume of liquid displaced*.

1)mass of liquid displaced =upthrust (g).
Upthrust maybe converted to Newton by multiply by 10.

2) mass of liquid displaced = volume of liquid×density of liquid. Also, it maybe
volume of object ×density of liquid.

3) If the body is not completely immersed, Archimedes principle is not obeyed so that maybe fraction of its volume is immersed e.g if it is said v/2 is immersed, then www.naijaexamrunz.blogspot.com
Mass of liquid displaced = v/2×density of liquid.

4) Object weight in air= Apparent weight + Upthrust.
(Apparent weight = Reading =Tension).

So, W=T+U. Where,
 W= mass in air(g/N).
  T= apparent weight(g/N).
  U= Upthrust(g/N)=mg i.e m ×10=¢vg=
        ¢v×10.

Worked Example:
A block of material if volume 20cm^3 and density of 2.5 g/cm^3 is suspended from a spring balance with half the volume of the block immersed in water. (www.naijaexamrunz.blogspot.com)What is the reading of the spring balance?( density of water is 1 g/cm^3) waec 1989.

Solution:
Given that:
object volume in air= 20cm^3,
Object density= 2.5 g/cm3.
Water density= 1g/cm^3.
Reading 'T'=?
From,
W=T+U.
W-U= T.
T=W-U. W=?, U=?.
Also, w=mg( object weight).
M=?
 mass of liquid displaced (upthrust)=density of liquid×v/2(since half immersed)= 1×20/2
     =10g.
mass of object (w)=density of object× v
                            =2.5×20=50g.
Recall,
T=w-u=50-10= 40g
T=40g.

2) An object weighs 10N in air and 7N in water. What is it weight when immersed in a liquid of relative density 1.5? (Waec 1990).

Solution:
Given that;
      r.d=1.5, w in air=10N, w in water
      =7N. T=?.
      R.d=w in air -T/w in water -T
      1.5=10-T/7-T
      10-T=1.5(7-T) cross multiply.
      10-T=10.5-1.5T
      -T+1.5T=10.5-10
       0.5T=0.5 therefore
       T=0.5/0.5
       T=1N.

Exercise:
A solid weighs 0.04N in air and 0.024N when fully immersed in a liquid of density 800kg/m^3. What is the volume of the solid?(g=10m/s^2) WAEC 1992.

                   END OF LECTURE.

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