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 :
(a) the weight of the piece of stone in air,
(b) the volume of the piece of stone,
(c) the relative density of stone,
(d) the relative density of the liquid.

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Solution

a. The mass of stone is 15.1 g. Hence, its weight in air will be W_a = 15.1 gf

b. 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.

c. 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

$R. D space o f space s t o n e equals fraction numerator W subscript a over denominator W subscript a minus W subscript w end fraction equals fraction numerator 15.1 over denominator 15.1 minus 9.7 end fraction R. D space o f space s t o n e equals fraction numerator 15.1 over denominator 5.4 end fraction equals 2.8 d right parenthesis r e l a t i v e space d e n c i t y space o f space l i q u i e d space i s R. D space o f space l i q u i d space i s equals fraction numerator W subscript a minus W subscript l over denominator W subscript a minus W subscript w end fraction equals fraction numerator 15.1 minus 10.9 over denominator 15.1 minus 9.7 end fraction equals fraction numerator 4.2 over denominator 5.4 end fraction R. D space o f space l i q u i d space i s equals 0.777$

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