Question

A piece of metal of density 7.8 * 10^3 weighs 20N in air, calculate it's weight when completely immersed in a liquid of density 8.3 * 10^3

Answer 1

Let's first find the mass and volume of the piece of metal:

$$m=\dfrac{W}{g}=\dfrac{20\ N}{9.8\ \dfrac{m}{s^2}}=2.04\ kg,$$ $$V=\dfrac{m}{\rho}=\dfrac{2.04\ kg}{7.8\cdot10^3\ \dfrac{kg}{m^3}}=2.61\cdot10^{-4}\ m^3.$$

The apparent weight of the piece of metal when it completely immersed in a liquid can be written as follows:

$$W_A=F_B-W_{air},$$ $$W_A=\rho_{liquid}V_{liquid}g-W_{air},$$ $$W_A=\rho_{liquid}Vg-W_{air},$$ $$W_A=8.3\cdot10^3\ \dfrac{kg}{m^3}\cdot2.61\cdot10^{-4}\ m^3\cdot9.8\ \dfrac{m}{s^2}-20\ N,$$ $$W_A=1.23\ N.$$