With the use of Archimedes' principle, state how you will find relative density of a solid denser than water and insoluble in it. How will you modify your experiment if the solid is soluble in water?

Relative density of a solid denser than water and insoluble in it

Procedure —

(i) Suspend a piece of the given solid with a thread from hook of the left pan of a physical balance and find it's weight W₁.

(ii) Now balance a wooden bridge over the left pan of balance and place a beaker nearly two-third filled with water on the bridge. Take care that the bridge and beaker do not touch the pan of balance,

(iii) Immerse the solid completely in water such that it does not touch the walls and bottom of beaker and find the weight W₂ of solid in water.

Observation —

Weight of solid in air = W₁ gf

Weight of solid in water= W₂ gf

Calculation —

Loss in weight of solid when immersed in water = (W₁ - W₂) gf

$\text{R.D.} = \dfrac{\text{ Weight of solid in air}}{\text{Loss in weight of solid in water}}$

or

$\text{R.D.} = \dfrac{W${1}}{W1 - W_2}

Relative density of a solid denser than water and soluble in it —

Procedure —

If solid is soluble in water, instead of water, we take a liquid of known relative density in which solid in insoluble and it sinks in that liquid. Then the process described above is repeated. Now

$\text{R.D.} = \dfrac{\text{Wt of solid in air}}{\text{Loss in wt of solid in liq}} \times \text{R.D. of liq}$