A wooden cubical block of relative density 0.4 is floating in water. Side of the cube is 10cm. When a coin is placed on the block, it dips by 0.3cm. Find the weight of the coin.

To find the weight of the coin, we can analyze the equilibrium of the wooden cubical block floating in water before and after the coin is placed on top of it.

According to Archimedes' principle, a floating body displaces an amount of fluid whose weight is equal to the weight of the floating body.

Let:
- $L$ be the side of the wooden cube, so $L=10\text{cm}=0.1\text{m}$.
- $A$ be the cross-sectional area of the face of the cube, where $A=L^2={\left(0.1\text{m}\right)}^2=0.01{\text{m}}^2$.
- ${\rho }_w$ be the density of water, which is approximately $1000{\text{kg/m}}^3$.
- $g$ be the acceleration due to gravity, taken as $10{\text{m/s}}^2$.
- $d$ be the extra depth by which the block dips when the coin is placed, so $d=0.3\text{cm}=0.003\text{m}$.

When the coin is placed on the block, the total weight supported increases by the weight of the coin ($W_{\text{coin}}$). This additional downward force is balanced by the additional upward buoyant force resulting from the extra volume of water displaced by the block as it dips further.

The additional volume of water displaced by dipping an extra depth $d$ is given by:

$\Delta V=A\times d$

The weight of this additional displaced water (additional buoyant force) is:

$F_B={\rho }_w\times \Delta V\times g={\rho }_w\times A\times d\times g$

For the block to remain in equilibrium, the weight of the coin must equal this additional buoyant force:

$W_{\text{coin}}={\rho }_w\times A\times d\times g$

Substituting the given values into the equation:

$W_{\text{coin}}=1000\text{kg}/{\text{m}}^3\times 0.01{\text{m}}^2\times 0.003\text{m}\times 10{\text{m/s}}^2$

$W_{\text{coin}}=10\times 0.003\times 10\text{N}$

$W_{\text{coin}}=0.3\text{N}$

Thus, the weight of the coin is 0.3N.