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 (W _{1}) = 15.1 gf**

(b) W_{1} = 15.1 gf

Weight of the stone when immersed in water(W_{3}) = 9.7 gf

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

= Weight in air (W_{1}) – Weight in water (W_{3})

= 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 cm^{3}

∴ **Volume of a piece of stone = 5.4 cm ^{3}**

(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,

W_{1} is the weight of piece of stone in air,

W_{2} is the weight of piece of stone in liquid,

W_{3} 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

**the relative density of the liquid = 0.777 = 0.78**