A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance $\ell$ and $h$ are shown here. After some time the coin falls into the water. Then

Step 1: Analyze the initial floating condition.

Initially, the wooden block of mass $m$ floats with the coin of mass $c$ on top of it. The total weight is balanced by the buoyant force.

$$(m + c)g = \rho A d_i g$$

Here, $\rho$ is the density of water, $A$ is the cross-sectional area of the block, and $d_i$ is the initial immersed depth $\ell$.

$$\ell = d_i = \frac{m + c}{\rho A}$$

Step 2: Determine the change in the immersed depth $\ell$.

After the coin falls into the water, only the wooden block floats. The new immersed depth $d_f$ is determined by the weight of the block alone.

$$m g = \rho A d_f g$$ $$d_f = \frac{m}{\rho A}$$

Comparing the initial and final immersed depths:

$$d_f < d_i$$

Thus, the immersed depth $\ell$ decreases.

Step 3: Formulate the initial total water volume equation.

Let $V_w$ be the volume of water and $A_c$ be the cross-sectional area of the container. The initial total volume below the water surface is the sum of the water volume and the submerged volume of the block.

$$A_c h_i = V_w + \frac{m + c}{\rho}$$

Step 4: Formulate the final total water volume equation.

After the coin falls, it sinks to the bottom because its density $\rho_c$ is greater than the density of water $\rho$. The final volume below the water surface includes the submerged volume of the block and the actual volume of the coin.

$$A_c h_f = V_w + \frac{m}{\rho} + \frac{c}{\rho_c}$$

Step 5: Calculate the change in the water level $h$.

Subtract the initial volume equation from the final volume equation to find the change in height $\Delta h$.

$$h_f - h_i = \frac{1}{A_c} \left( \frac{c}{\rho_c} - \frac{c}{\rho} \right)$$

Since the coin sinks, its density is greater than that of water.

$$\rho_c > \rho$$ $$\Delta h < 0$$

Thus, the total water depth $h$ decreases.

Answer: Both $\ell$ and $h$ decrease.