Upthrust in Fluids, Archimedes' Principle and Floatation — Question 11

Given,

Weight of the solid in air (W₁) = 32 gf

Weight of the solid immersed completely in water (W₂) = 28.8 gf

$\text{Relative density of solid} = \big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{2}}\big) \\[1 em] = \big(\dfrac{32}{32 - 28.8}\big) \\[1 em] = \big(\dfrac{32}{3.2}\big) \\[1 em] = 10 \\[1.5em] \text{But, R.D. of solid} = \dfrac{\text {Density of solid in g m}^{-3}}{\text{1 g cm}^{-3}} \\[1 em] \Rightarrow 10 = \dfrac{\text {Density of solid in g m}^{-3}}{\text{1 g cm}^{-3}} \\[1 em] \Rightarrow \text {Density of solid} = 10 \text{ g m}^{-3} \\[1 em] \text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[1em] \therefore \text{Volume} = \dfrac{\text{Mass}}{\text{Density}} \\[1 em] = \dfrac{32}{10} \\[1 em] = 3.2 \text { cm}^{3}$

Hence, volume of solid = 3.2 cm³

Let weight of solid in liquid of density 0.9 g cm^-3 be W.

$\text{R.D. of solid} = \Big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{2}}\Big) \times \text{R.D. of liquid} \\[1em] \text{R.D. of liquid} = \dfrac{\text {Density of liquid in g cm}^{-3}}{\text{1.0 g cm}^{-3}} \\[1 em] = \dfrac{0.9}{1} \\[1 em] = 0.9$

Substituting the values in the formula for relative density of solid :

$10 = \Big(\dfrac{32}{32 − \text W}\Big) \times 0.9 \\[1 em] 10 = \dfrac{28.8}{32 − \text W} \\[1 em] 10 \times (32 - \text W) = 28.8 \\[1 em] 320 - 10\text W = 28.8 \\[1 em] 10\text W = 320 - 28.8 \\[1 em] \text W = \dfrac{291.2}{10} \\[1 em] \text W = 29.12 \text{ gf}$

Summarizing the answers:

(i) Volume of solid = 3.2 cm³

(ii) Relative density of solid = 10

(iii) Weight of solid in a liquid of density 0.9 g cm^-3 = 29.12 gf