A tub is filled with water and a wooden cube \(10\) cm \(\times\) \(10\) cm \(\times\) \(10\) 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 \(1\) g/cm\(^3\) and density of wood as \(0.4\) g/cm\(^3\))

Let \(A\) be the base area of the wooden cube.

\(A = 10 \text{ cm} \times 10 \text{ cm} = 100 \text{ cm}^2\)

Let \(\Delta x\) be the increase in the submerged depth of the cube when the coin is placed on it.

\(\Delta x = 3.87 \text{ cm}\)

The additional volume of water displaced due to the weight of the metal coin is \(\Delta V = A \times \Delta x\).

\(\Delta V = 100 \times 3.87 = 387 \text{ cm}^3\)

By Archimedes' principle, the weight of the metal coin is balanced by the additional buoyant force, which is equal to the weight of the additional water displaced.

\(m g = \Delta V \rho_w g\)

\(m = \Delta V \rho_w\)

Substituting the values of \(\Delta V\) and the density of water \(\rho_w = 1 \text{ g/cm}^3\):

\(m = 387 \text{ cm}^3 \times 1 \text{ g/cm}^3 = 387 \text{ g}\)

Answer: \(387\)