Worked Examples-Calculation Part 2 - Archimedes Principle

Considered Formula;

$$\text {i)}\quad R.D = \frac {W_a - W_l}{W_a - W_w}$$ $$\qquad \text {Where}~~R.D \to Relative~density,\quad W_a \to Weight~in~air \\\qquad W_l \to Weight~in~liquid, \qquad W_w \to Weight~in~water$$ $$\text {ii)}\quad R.D = \frac {\rho_s}{\rho_w}$$ $$\qquad \text {Where}~~R.D \to Relative~density,\quad \rho_s \to Density~of~substance \\\qquad \rho_w \to Density~of~water$$

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 (W_a) = 1.2 N

Weight in water (W_w) = 0.7 N

Density of water (ρ_w) = 1g/cm³

a)        Relative density (R.D) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~R.D = \frac {W_a}{W_a - W_w}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {1.2}{1.2 - 0.7}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {1.2}{0.5}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {1.2 \times 10}{0.5 \times 10}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {12}{5}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 2.4$$

          The relative density = 2.4

b)        Density of glass (ρ_g) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~R.D = \frac {\rho_g}{\rho_w}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2.4 = \frac {\rho_g}{1}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\rho_g = 2.4 \times 1$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 2.4 g/cm^3$$

          The density of glass = 2.4 g/cm³

c)        Volume (V) = ?

                 Weight in air (W_a) = 1.2 N

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

$$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~W = mg$$ $$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~1.2 = m(10)$$ $$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac {10m}{10}= \frac {1.2}{10}$$ $$ \qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~m= 0.12 Kg = (0.12 \times 1000)g = 120 g$$

                 Then;

$$ \qquad~~~~~~~~~~~~~~~~~~~~~~~~~~~~\rho = \frac {m}{V}$$ $$ \qquad~~~~~~~~~~~~~~~~~~~~~~~~~~~~V = \frac {120}{2.4}$$ $$ \qquad~~~~~~~~~~~~~~~~~~~~~~~~~~~~V = \frac {120 \times 10}{2.4 \times 10}$$ $$ \qquad~~~~~~~~~~~~~~~~~~~~~~~~~~~~V = \frac {1200}{24}$$ $$ \qquad~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 50 cm^3$$

          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 (W_a) = 8.6 N

Weight Q in water (W_w) = 6.0 N

Weight Q in another liquid (W_l) = 5.4 N

Density of water (ρ_w) = 1000 Kg/m³

Relative density (R.D) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~R.D = \frac {W_a - W_l}{W_a - W_w}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {8.6-5.4}{8.6-6.0}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {3.2 \times 10}{2.6 \times 10}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {32}{26}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 1.2$$

The relative density = 1.2

Then

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~R.D = \frac {\rho_s}{\rho_w}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1.2 = \frac {\rho_g}{1000}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\rho_s = 1.2 \times 1000$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 1200~Kg/m^3$$

The density of liquid L= 1200 Kg/m³