Upthrust in Fluids, Archimedes' Principle and Floatation — Question 13

Given,

Weight of the solid in air (W₁) = 1.5 kgf

Weight of the solid in liquid (W₂) = 0.9 kgf

Density of the liquid = 1.2 x 10³ kg m^-3

$\text{R.D. of liquid} = \dfrac{\text {Density of liquid in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \\[0.5em] = \dfrac{1.2 \times 10^3}{1000} \\[0.5em] = 1.2$

$\text{R.D. of solid} = \Big(\dfrac{\text W_{1}}{\text W_{1} − \text W_{2}}\Big) \times \text{R.D. of liquid} \\[0.75em] = \Big(\dfrac{1.5}{1.5 - 0.9}\Big) \times 1.2 \\[0.75em] = \Big(\dfrac{1.5}{0.6}\Big) \times 1.2 \\[0.75em] = 3.0$