Upthrust in Fluids, Archimedes' Principle and Floatation — Question 8

Given,

Volume of body = 100 cm³

Weight of the body in air (W₁) = 1 kgf = 1000 gf

Let weight of the body in water be W₂

$\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \\[0.5em] = \dfrac{1000}{100} \\[0.5em] = 10 \text { g cm}^{-3}$

Hence,

Density of body = 10 g cm^-3

$\text{R.D. of body} = \dfrac{\text {Density of body in g cm}^{-3}}{\text{1.0 g cm}^{-3}} \\[0.5em] = \dfrac{10}{1} \\[0.5em] = 10$

Therefore,

Relative Density of body = 10

$\text{R.D. of body} = \dfrac{\text W_{1}}{\text W_{1} − \text W_{2}} \\[1 em] \Rightarrow 10 = \dfrac{1000}{1000 − \text W_{2}} \\[1 em] \Rightarrow 10 \times (1000 − \text W_{2}) = 1000 \\[1 em] \Rightarrow 10000 - 10\text W_{2} = 1000 \\[1 em] \Rightarrow 10\text W_{2} = 9000 \\[1 em] \Rightarrow \text W_{2} = \dfrac{9000}{10} \\[1 em] \Rightarrow \text W_{2} = 900 \text { gf}$

Hence,

(i) Weight in water = 900 gf

(ii) Relative density of body = 10