Upthrust in Fluids, Archimedes' Principle and Floatation — Question 16

Given,

Mass of the stone = 15.1 g

(a) Weight = mass x acceleration due to gravity
= 15.1 x 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 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 x g = V x 1 x g
⇒ V = 5.4 cm³

∴ Volume of piece of stone = 5.4 cm³

(c) From the relation,

$\text{Relative density of stone} = \big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{3}}\big) \\[1 em] = \big(\dfrac{15.1}{15.1 - 9.7}\big) \\[1 em] = \big(\dfrac{15.1}{5.4}\big) \\[1 em] = 2.79 \approx 2.8$

Hence,

Relative density of stone = 2.8

(d) From the relation,

$\text{Relative density of liquid} = \dfrac{\text W_{1} - \text W_{2} }{\text W_{1} − \text W_{3}}$

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,

$\text{Relative density of liquid} = \dfrac{15.1 - 10.9}{15.1 - 9.7} \\[0.5em] = \dfrac{4.2}{5.4} \\[0.5em] = 0.777 \approx 0.78\\[0.5em]$

Hence,

the relative density of the liquid = 0.777 = 0.78