A wooden block with a coin placed on its top floats in water in a container. The vertical distance $l$ is the depth of the block that lies submerged below the water surface, and $h$ is the height of the water level in the container (the height of the water column, measured from the bottom of the container up to the free water surface). After some time the coin slides off the block and falls into the water, sinking to the bottom of the container (the coin is denser than water). Considering the block's submerged depth $l$ and the water level $h$ afterwards, which of the following statements are correct? (More than one option may be correct.) (a) $l$ decreases. (b) $h$ decreases. (c) $l$ increases. (d) $h$ increases.

A wooden block with a coin placed on its top floats in water in a container. The vertical distance l is the de

NCERT Exemplar · Q8

l

(a)

l

decreases.

(b)

h

decreases.

(c)

l

increases.

(d)

h

increases.

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Concept understanding — Archimedes Principle

Archimedes' Principle

Push a beach ball underwater in a swimming pool and feel it fight back — the deeper it goes, the harder it pushes up. Or try lifting a heavy stone while it's submerged: it feels far lighter than it does on land. Something about being surrounded by water makes objects seem to lose weight. That "something" is Archimedes' Principle.

The Intuition

A fluid is made of molecules pressing on everything around them, and pressure increases with depth. So the fluid beneath a submerged object pushes up on it harder than the fluid above pushes down — because there is more fluid weighing down from above onto the fluid below. This net upward push is the buoyant force.

The key insight: this upward force doesn't depend on the object's shape or material. It depends only on how much fluid the object displaces — the volume of fluid it pushes out of the way to make room for itself.

Note

Displaced fluid = the volume of the object that is submerged. Fully dunk a 1-litre bottle, and you displace 1 litre of water.

The Precise Statement

F_b = \rho_{\text{fluid}} \, V_{\text{submerged}} \, g

where $F_b$ is the buoyant force, $\rho_{\text{fluid}}$ is the fluid's density, $V_{\text{submerged}}$ is the submerged volume of the object, and $g$ is the acceleration due to gravity.

Archimedes' Principle: any object, wholly or partly immersed in a fluid, experiences an upward buoyant force equal to the weight of the fluid it displaces.

Why This Matters

Observation	What's happening
A ship floats	Its hull shape displaces a huge volume of water. The weight of that displaced water equals the ship's own weight, so it floats.
A stone feels lighter underwater	The water pushes up with a force equal to the weight of water it displaces, partly cancelling gravity.
A helium balloon rises	The balloon displaces air; the weight of that displaced air exceeds the weight of the helium inside, giving a net upward force.

What Decides Sink or Float: Density

Object density greater than fluid density $\Rightarrow$ it sinks (weight exceeds buoyant force).
Object density less than fluid density $\Rightarrow$ it floats (weight is less than buoyant force).
Object density equal to fluid density $\Rightarrow$ it hovers at any depth. …

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When the coin (denser than water) drops off the floating block and sinks, the block no longer has to support the coin's weight, so it rises and its submerged depth $l$ decreases. The coin at the bottom displaces only its own small volume rather than a volume of water equal to its weight, so the total displaced volume falls and the water level $h$ decreases too. Hence $l$ decreases and $h$ decreases.

Set-up

Let the block have mass $M$ and the coin mass $m$ , and let $\rho_w$ and $\rho_c$ be the densities of water and the coin, with $\rho_c > \rho_w$ (a metal coin sinks). Take the water's cross-sectional area in the container as $A$ .

Effect on the block's submerged depth $l$

While the coin rides on top, the block floats and by Archimedes' principle the buoyant force equals the total weight:

\rho_w \, g \, V_{sub} = (M + m) g \;\Rightarrow\; V_{sub} = \frac{M+m}{\rho_w}.

After the coin falls off, the block alone floats:

V_{sub}' = \frac{M}{\rho_w} < V_{sub}.

Since the block now displaces less water, it sits higher and its submerged depth $l$ decreases — option (A) is correct, and (C) is wrong.

Effect on the water level $h$

Compare the total volume of water displaced (which sets the water level) before and after.

Before: the floating system displaces a volume equal to the weight of block + coin divided by $\rho_w$ : $V_1 = \frac{M+m}{\rho_w} = \frac{M}{\rho_w} + \frac{m}{\rho_w}.$ …

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Set-up

Effect on the block's submerged depth lll

Effect on the water level hhh

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Effect on the block's submerged depth $l$

Effect on the water level $h$

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