A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance l an

Q: Why does the coin displace more water when it is on top of the block than when it is submerged?

When the coin is on top of the floating block, the system must displace a volume of water whose weight equals the combined weight of the block and the coin. Thus, the coin effectively displaces its weight's worth of water. When the coin sinks, it is no longer supported by buoyancy and only displaces its own physical volume of water. Since the coin is denser than water, its weight's worth of water has a much larger volume than the coin's actual volume.

Q: Does the block rise or sink when the coin falls off?

The block rises because it no longer has to support the weight of the coin. Since the downward force on the block decreases, the required buoyant force also decreases, meaning a smaller volume of the block needs to be submerged. Thus, $l$ decreases.

Q: What if the coin was replaced by a plastic button with a density less than water?

If the plastic button falls off, it would float on the water surface instead of sinking. Since both the block and the button would still be floating, the total weight supported by buoyancy remains the same. Therefore, the total displaced volume of water would not change, and the water level $h$ would remain constant.

Q: Does the shape of the beaker affect the result?

No, the qualitative result that both $l$ and $h$ decrease is completely independent of the beaker's shape. However, the exact numerical change in the water level $\Delta h$ would depend on the cross-sectional area of the beaker.

Q: Is there any change in the buoyant force on the block itself?

Yes, the buoyant force on the block decreases because it no longer has to support the coin's weight. The buoyant force on the block changes from $(M+m)g$ to just $Mg$.

Question

A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance $l$ and $h$ are shown there. After sometime the coin falls into the water. Then

Options:

(A) $l$ decreases and $h$ increases

(B) $l$ increases and $h$ decreases

(C) Both $l$ and $h$ increase

(D) Both $l$ and $h$ decrease

Sigma Percentile · Accepted Answer

### Introduction: The Magic of Floating Objects

Have you ever wondered why a massive steel ship floats effortlessly on the ocean, while a tiny steel coin sinks like a stone to the bottom of a cup?

The answer lies in the beautiful and elegant **Archimedes' Principle**.

In this problem, we explore a classic physics puzzle that tests our conceptual understanding of buoyancy, density, and displaced volume.

Imagine a wooden block floating in a beaker of water, with a heavy coin resting on its top.

What happens to the submerged depth of the block ($l$) and the total water level ($h$) when the coin falls off and sinks to the bottom? Let's dive in and find out!

---

### Analyzing the Setup

Let us define our variables clearly to build a solid mathematical foundation:

* Let the mass of the wooden block be $M$.
* Let the mass of the coin be $m$.
* Let the density of water be $ho_w$.
* Let the density of the coin be $ho_c$.
* Let the cross-sectional area of the wooden block be $A$.

Initially, in **State 1**, the coin rests on top of the block. The entire system (block + coin) floats in equilibrium.

We are tracking two key parameters:
1. $l$: The submerged depth of the wooden block.
2. $h$: The total height of the water level in the beaker.

---

### State 1: The Combined Floating System

Since the combined system of the block and the coin is floating, the upward buoyant force must balance the total downward gravitational force.

According to Archimedes' Principle, the buoyant force is equal to the weight of the water displaced by the submerged part of the system:

$$	ext{Buoyant Force} = (M + m)g$$

$$ho_w V_{	ext{displaced}, 1} g = (M + m)g$$

Canceling $g$ from both sides, we find the total volume of water displaced in State 1:

$$V_{	ext{displaced}, 1} = \frac{M + m}{ho_w}$$

Since the coin is resting on top of the block and is completely out of the water, only the wooden block is submerged.

Therefore, the volume of water displaced is exactly equal to the submerged volume of the block:

$$V_{	ext{displaced}, 1} = A \cdot l$$

Equating the two expressions for the displaced volume, we get:

$$A \cdot l = \frac{M + m}{ho_w} \implies l = \frac{M + m}{ho_w A}$$

This gives us our first crucial result: the initial submerged depth $l$ is directly proportional to the combined mass of the block and the coin.

---

### State 2: The Coin Sinks

Now, let's look at what happens in **State 2** when the coin falls off the block.

Since the coin is made of metal, its density is much greater than that of water ($ho_c > ho_w$). Consequently, the coin sinks to the bottom of the beaker.

The wooden block, now freed from the extra weight of the coin, floats alone on the surface.

Let's find the new submerged depth of the block, which we will call $l'$.

Since the block floats alone, the buoyant force on it only needs to balance its own weight:

$$M g = ho_w (A \cdot l') g \implies l' = \frac{M}{ho_w A}$$

Comparing our two expressions for the submerged depth:

$$l = \frac{M + m}{ho_w A} \quad 	ext{and} \quad l' = \frac{M}{ho_w A}$$

Since $M + m > M$, it is mathematically clear that:

$$l' < l$$

Thus, the submerged depth of the block **decreases** ($l$ decreases). The block rises up in the water because it is carrying less load.

---

### The Water Level Puzzle: What Happens to $h$?

Now, let's tackle the more subtle and conceptual part of the problem: the total water level $h$.

The height of the water level in the beaker is directly determined by the **total volume of water displaced** by all objects in the beaker.

Let's calculate the total displaced volume in both states.

#### In State 1 (Coin on Block):

Since the entire system is floating, the total volume of water displaced is:

$$V_{	ext{total}, 1} = \frac{M + m}{ho_w} = \frac{M}{ho_w} + \frac{m}{ho_w}$$

#### In State 2 (Coin Sunk):

In this state, the block and the coin displace water independently:
1. The floating block displaces a volume of water equal to:  
   $$V_{	ext{block}} = \frac{M}{ho_w}$$
2. The sunken coin is fully submerged at the bottom, so it displaces a volume of water exactly equal to its own physical volume:  
   $$V_{	ext{coin}} = \frac{m}{ho_c}$$

Therefore, the total volume of water displaced in State 2 is:

$$V_{	ext{total}, 2} = \frac{M}{ho_w} + \frac{m}{ho_c}$$

---

### Comparing the Displaced Volumes

Let's find the difference between the total displaced volumes in the two states:

$$V_{	ext{total}, 1} - V_{	ext{total}, 2} = \left(\frac{M}{ho_w} + \frac{m}{ho_w}ight) - \left(\frac{M}{ho_w} + \frac{m}{ho_c}ight) = m \left(\frac{1}{ho_w} - \frac{1}{ho_c}ight)$$

Since the coin is denser than water ($ho_c > ho_w$), we have:

$$\frac{1}{ho_w} > \frac{1}{ho_c} \implies V_{	ext{total}, 1} > V_{	ext{total}, 2}$$

This is the key insight!

When the coin was floating on the block, it was forced to displace water equal to its **weight**.

But once it sank, it only displaced water equal to its **volume**.

Since the coin is denser than water, its weight's worth of water has a much larger volume than the coin's actual physical volume.

Therefore, the total volume of water displaced decreases when the coin sinks.

Since the total displaced volume decreases, the water level in the beaker must fall.

Thus, the height $h$ **decreases**.

---

### Conclusion

Both the submerged depth of the block ($l$) and the total water level ($h$) **decrease**.

This perfectly matches **Option (d)**.

This classic problem beautifully illustrates how buoyancy and density interact, showing that floating objects displace more water than they do when they sink!

Animated Solution for Physics - Properties of Solids and Liquids: A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance $l$ and $h$ are shown there. After sometime the coin falls into the water. Then

Select Answer:

Visualized Solution

Visualizing the Setup

Applying Archimedes' Principle

Initial State: Displaced Volume of Water

Submerged Depth $l$ in State 1

Final State: Coin Sinks to the Bottom

Submerged Depth $l^{'}$ in State 2

Comparing $l$ and $l^{'}$

Analyzing the Water Level $h$

Total Displaced Volume in State 1

Total Displaced Volume in State 2

Comparing Displaced Volumes

Concluding the Water Level $h$

Final Answer

The Way Forward

The Sigma Insight: Buoyancy and Archimedes' Principle

Introduction

Analyzing the Setup

State 1

State 2

The Water Level Puzzle

# In State 1 (Coin on Block)

# In State 2 (Coin Sunk)

Comparing the Displaced Volumes

Conclusion

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