Question

A scale reads 274 N when a piece of brass is hanging from it. What does it read (in N) when it is lowered so that the brass is submerged in water?

Answer 1

The weight of the piece of brass in air is $F_a = \rho_b V g$, where $\rho_b$ is the brass density, $V$ is its volume, and $g$ is the acceleration of gravity. By the Archimedes law, the weight of the same piece of brass submerged in water is $F_w = \rho_b V g - \rho_w V g$, where $\rho_w$ is the density of water. The last formula can be rewritten as

$$F_w = \rho_b V g \left( 1 - \frac{\rho_w}{ \rho_b} \right) = F_a \left( 1 - \frac{\rho_w}{ \rho_b} \right) \, .$$

Substituting here the corresponding densities $\rho_b \approx 8.5\, \text{g} / \text{cm}^3$ and $\rho_w \approx 1\, \text{g} / \text{cm}^3$, we obtain the weight of the brass in water: $F_w \approx 242\, \text{N}$.

Answer: 242 N.