A tub is filled with water and a wooden cube 10 cm 10 cm 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 )

Step-by-step solution

Let A be the base area of the wooden cube. A = 10 cm 10 cm = 100 cm ^2 Let x be the increase in the submerged depth of the cube when the coin is placed on it. x = 3.87 cm The additional volume of water displaced due to the weight of the metal coin is V = A x . V = 100 3.87 = 387 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 = V _w g m = V _w Substituting the values of V and the density of