Considered Formula;
i)R.D=Wa−WlWa−Ww Where R.D→Relative density,Wa→Weight in airWl→Weight in liquid,Ww→Weight in water ii)R.D=ρsρw Where R.D→Relative density,ρs→Density of substanceρw→Density of waterExample 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=WaWa−Ww =1.21.2−0.7 =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=Wa−WlWa−Ww =8.6−5.48.6−6.0 =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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