JEE Main 2026 — Mechanical Properties of Fluids Question with 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.

$mg=\Delta V{\rho }_{w}g$

$m=\Delta V{\rho }_w$

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: $387$