Question

A piece of alloy has a measured mass of 86g in the air and 73g when immersed in water. Find it's volume and it's density

Answer 1

The weight of the alloy in the air is:

$$P_{air} = mg = \rho Vg$$

where $m = 86g = 0.086kg$ is its real mass, $\rho$ is its density and $V$ is its volume. $g\approx 9.81m/s^2$ is the gravitational acceleration.

In the water due to the buoyancy the weight is:

$$P_{water} = P_{air} - F_b = m_{water}g$$

where $m_{water} = 73g = 0.073kg$ is the "effective" mass measured in water.

The buoyancy force is given as follows:

$$F_b = \rho_{water}Vg$$

where $\rho_{water} = 1000kg/m^3$ is the density of water.

Thus, obtain the volume:

$$mg - \rho_{water}Vg = m_{water}g\\ m - \rho_{water}V = m_{water}\\ V = \dfrac{m - m_{water}}{\rho_{water}}\\ V = \dfrac{0.086kg - 0.073kg}{1000kg/m^3} = 1.3\times 10^{-5}m^3$$

And the density:

$$\rho = \dfrac{m}{V} = \dfrac{0.086kg}{1.3\times 10^{-5}m^3} \approx 6615\space kg/m^3$$

Answer. Volume: $1.3\times 10^{-5}m^3$, density: $6615\space kg/m^3$.