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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:
- The weight of the piece of stone in air,
- The volume of the piece of stone,
- The relative density of stone,
- The relative density of the liquid.
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Solution
1) The mass of stone is 15.1 g. Hence, its weight in air will be Wa = 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 Wl = 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")`
= `15.1/5.4`
= 2.8
4) 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`
∴ R.DStone = 0.7777 ≈ 0.78
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