Question

A piece of stone of mass 15.1 g is first immersed in a liquid and it weighs 10.9 gf. Then on immersing the piece of stone in water, it weighs 9.7 gf. Calculate:

1. The weight of the piece of stone in air,
2. The volume of the piece of stone,
3. The relative density of stone,
4. The relative density of the liquid.

Answer 1

1. The mass of stone is 15.1 g. Hence, its weight in air will be W_a = 15.1 gf
2. When stone is immersed in water its weight becomes 9.7 gf. So, the upthrust on the stone is 15.1 - 9.7 = 5.4 gf, Since the density of water is 1 g cm^-3, the volume of stone is 5.4 cm³.

3. Weight of stone in liquid is W_l = 10.9 gf

   Weight of stone in water is Ww = 9.7 gf

   Therefore, the relative density of stone is 

`"R.D"_("stone") = (W_a)/(W_a - W_w) = (15.1  "gf")/(15.1 - 9.7  "gf")`

`"R.D"_("stone") = 15.1/5.4 = 2.8`

Relative density of liquid is 

`"R.D"_("liquid") = (W_a - W_l)/(W_a - W_w) = (15.1 - 10.9)/(15.1 - 9.7) = 4.2/5.4`

`therefore "R.D"_("Stone") = 0.7777 ≈ 0.78`