Worked Examples-Calculation Part 1-Archimedes Principle

Considered Formula;

$$\text {i)}\quad Up =W_a - W_l$$ $$\qquad \text {Where}~~Up \to Upthrust,\quad W_a \to Weight~in~air \\\qquad W_l \to Weight~in~liquid$$ $$\text {ii)}\quad R.D = \frac {W_a}{W_a - W_l}$$ $$\qquad \text {Where}~~R.D \to Relative~Density,\quad W_a \to Weight~in~air \\\qquad W_l \to Weight~in~liquid$$

Example 1

A block of iron weighs 20 N in air and 17 N when completely immersed in water. Calculate the Upthrust exerted by water

Solution

Weight in air (W_a) = 20 N
Weight in water (W_w) = 17 N
Upthrust (Up) = ? $$\qquad \qquad Up =W_a - W_w$$ $$~\qquad \qquad \quad =(20 - 17) N$$ $$~\qquad \qquad \quad =3 N$$

The Upthrust exerted by water = 3 N

Example 2

A body weighs 0.36 N in air and 0.18 N when immersed in water. Find the

i)        Upthrust exerted by water

ii)       Relative density of a body

Solution

Data given

Weight in air (W_a)  = 0.36 N

Weight in water (W_w) = 0.18 N

i)        Upthrust (Up) =?

$$\qquad \qquad Up =W_a - W_w$$ $$~\qquad \qquad \quad =(0.36 - 0.18) N$$ $$~\qquad \qquad \quad =0.18 N$$

The Upthrust exerted by water = 0.18 N

ii)        Relative density (R.D) =?

$$\qquad \qquad \qquad R.D = \frac {W_a}{W_a - W_w}$$ $$~\qquad \qquad \qquad~~~~~~~= \frac {0.36}{0.36 - 0.18}$$ $$~\qquad \qquad \qquad~~~~~~~= \frac {0.36}{ 0.18}$$ $$~\qquad \qquad \qquad~~~~~~~= 2$$

The The Relative density of a body = 2 N

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