Q75 / 75 · JEE Main 2026 (02 April Shift 1) · Physics

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$^3$)

Correct answer: 387

Solution

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: $387$

Hence, the correct answer is 387.

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