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Worked Examples-Calculation Part 2 - Archimedes Principle

Considered Formula;

i)R.D=WaWlWaWw Where  R.DRelative density,WaWeight in airWlWeight in liquid,WwWeight in water ii)R.D=ρsρw Where  R.DRelative density,ρsDensity of substanceρwDensity of water

Example 1

A piece of glass weight in air 1.2N and 0.7N when completely immersed in water calculate it’s.

a)    Relative density

b)    Density of glass

c)     Volume of glass

(Density of water = 1g/cm3)

Solution

Data given

Weight in air (Wa) = 1.2 N

Weight in water (Ww) = 0.7 N

Density of water (ρw) = 1g/cm3

a)        Relative density (R.D) = ?

                            R.D=WaWaWw                                      =1.21.20.7                                      =1.20.5                                      =1.2×100.5×10                                      =125                                      =2.4

          The relative density = 2.4

b)        Density of glass (ρg) = ?

                            R.D=ρgρw                                2.4=ρg1                                  ρg=2.4×1                                      =2.4g/cm3

          The density of glass = 2.4 g/cm3

c)        Volume (V) = ?

                 Weight in air (Wa) = 1.2 N

                 Acceleration due to gravity (g) = 10 m/s2

                            W=mg                             1.2=m(10)                             10m10=1.210                             m=0.12Kg=(0.12×1000)g=120g

                 Then;

                            ρ=mV                             V=1202.4                             V=120×102.4×10                             V=120024                                      =50cm3

          The volume of glass= 50 cm3

Example 2

In an experiment to determine the relative density of a liquid L, a solid Q weighted as follows:
Weight Q in air = 8.6 N
Weight Q in water = 6.0 N
Weight Q in liquid = 5.4 N
What is the density of liquid L? (Density of water = 1000 Kg/m3)

Solution

Data given

Weight Q in air (Wa) = 8.6 N

Weight Q in water (Ww) = 6.0 N

Weight Q in another liquid (Wl) = 5.4 N

Density of water (ρw) = 1000 Kg/m3

Relative density (R.D) = ?

                                   R.D=WaWlWaWw                                      =8.65.48.66.0                                      =3.2×102.6×10                                      =3226                                      =1.2

The relative density = 1.2

Then

                            R.D=ρsρw                                1.2=ρg1000                                  ρs=1.2×1000                                      =1200 Kg/m3

The density of liquid L= 1200 Kg/m3

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