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Fluid Mechanics

JEE Main level, numerical answer, medium

32 of 192

Question 32 of 192

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 $^3$ )

Show answer and solution

Answer: 387

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

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