A stone weighs 392.4 N in air and 196.2 N in water. Compute the volume of the stone and its specific gravity.

Stone Buoyancy Problem

Problem Statement

Given Data

Weight of stone in air, $ W_{air} = 392.4 \, \text{N} $
Weight of stone in water, $ W_{water} = 196.2 \, \text{N} $
Density of water, $ \rho_{water} = 1000 \, \text{kg/m}^3 $

Solution

(i) Volume of the Stone

The difference between the weight of the stone in air and its weight in water is equal to the buoyant force, which is the weight of the water displaced.

$$ \text{Weight of water displaced} = W_{air} - W_{water} $$ $$ \text{Weight of water displaced} = 392.4 - 196.2 $$ $$ \text{Weight of water displaced} = 196.2 \, \text{N} $$

The volume of water displaced is equal to the volume of the stone. We can find this volume from the weight of displaced water.

$$ V_{stone} = V_{displaced} = \frac{\text{Weight of water displaced}}{\rho_{water} \times g} $$ $$ V_{stone} = \frac{196.2}{1000 \times 9.81} $$ $$ V_{stone} = \frac{196.2}{9810} $$ $$ V_{stone} = 0.02 \, \text{m}^3 $$

(ii) Specific Gravity of the Stone

First, find the density of the stone. We can get the mass of the stone from its weight in air.

$$ M_{stone} = \frac{W_{air}}{g} $$ $$ M_{stone} = \frac{392.4}{9.81} $$ $$ M_{stone} = 40 \, \text{kg} $$

$$ \rho_{stone} = \frac{M_{stone}}{V_{stone}} $$ $$ \rho_{stone} = \frac{40}{0.02} $$ $$ \rho_{stone} = 2000 \, \text{kg/m}^3 $$

Specific gravity is the ratio of the stone's density to the density of water.

$$ S.G._{stone} = \frac{\rho_{stone}}{\rho_{water}} $$ $$ S.G._{stone} = \frac{2000}{1000} $$ $$ S.G._{stone} = 2.0 $$

Final Results:

Volume of the stone: $ 0.02 \, \text{m}^3 $.

Specific gravity of the stone: $ 2.0 $.

Explanation of Concepts

Archimedes' Principle: This principle states that the buoyant force on a submerged object is equal to the weight of the fluid it displaces. This buoyant force is also the reason for the apparent loss of weight when the object is submerged. Therefore, the difference between the weight in air and the weight in water gives us the weight of the displaced water.

Volume and Density: Once we know the weight of the displaced water, we can calculate its volume using the formula $V = W / (\rho g)$. Since the stone is fully submerged, the volume of water it displaces is equal to its own volume.

Specific Gravity: Specific gravity is a dimensionless quantity that compares the density of a substance to the density of a reference substance, which is usually water for solids and liquids. A specific gravity of 2.0 means the stone is twice as dense as water.