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 —
(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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Answer

Mass of the stone = 15.1 g

(a) Weight = mass × acceleration due to gravity

= 15.1 × g

= 15.1 gf

Hence, Weight of the piece of stone in air (W1) = 15.1 gf

(b) W1 = 15.1 gf

Weight of the stone when immersed in water(W3) = 9.7 gf

Upthrust on the stone = loss in weight when immersed in water

= Weight in air (W1) – Weight in water (W3)

= 15.1 – 9.7

= 5.4 gf

Let the volume of the piece of stone be V.

From the relation, Upthrust on stone = volume of stone x density of water x acceleration due to gravity

5.4 × g = V × 1 × g

⇒ V = 5.4 cm3

∴ Volume of a piece of stone = 5.4 cm3

(c) From the relation,

RD of stone = 15.1\over 15.1 - 9.7 = 15.1\over 5.4 = 2.79 ≈ 2.8

Relative density of stone = 2.8

(d) From the relation

RD of liquid = W_1 - W_2\over W_1 - W_3

where,

W1 is the weight of piece of stone in air,
W2 is the weight of piece of stone in liquid,
W3 is the weight of piece of stone in water

Substituting the values in the formula above we get,

RD of liquid = 15.1 - 10.9\over 15.1 - 9.7 = 4.2\over 5.4 = 0.777 ≈ 0.78

Hence,

the relative density of the liquid = 0.777 = 0.78

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