A tub is filled with water and a wooden cube $10\ \text{cm}\times10\ \text{cm}\times10\ \text{cm}$ is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by $3.87$ cm. The mass of the metal coin is ______ gram.
(Take water density as $1\ \text{g/cm}^3$ and density of wood as $0.4\ \text{g/cm}^3$)
Numerical value type. Enter your answer.
Answer: 387
Without the coin the cube floats $4$ cm deep ($0.4\times10$ cm), so an extra $3.87$ cm still keeps the top above water.
The coin's weight equals the weight of the extra water displaced:
$$m=\rho_w\times(10\times10\times3.87)=1\times387=387\ \text{g}$$
Solution by Sreeraj P, M.Sc Physics