A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximu

Q: Why do we assume the block is exactly 100% submerged for the maximum weight?

If it were submerged any less, we could still add more weight. If it were submerged any more, water would spill over the top, violating the 'without fully submerging' condition.

Q: Does the shape of the block matter for the buoyant force?

No, Archimedes' principle states that the buoyant force depends only on the volume of the displaced fluid, not the shape of the object.

Q: Why did we cancel out gravity ($g$) in our equations?

Since gravity acts uniformly on both the mass of the block and the displaced water, it appears on both sides of the force balance equation and can be factored out, simplifying the math.

Q: What if the block was initially 50% submerged?

That would mean the block's density is exactly half that of water ($\rho_b = 0.5 \rho_w$). The steps to solve would be identical, just with a different initial fraction.

Q: Can we solve this without finding the block's density first?

Yes! You can think of the added mass as exactly replacing the weight of the additional water displaced when the block goes from 30% to 100% submerged. The extra displaced volume is 70% of $V$, so the added mass is simply $0.7 V \rho_w$.

Question

A cubical block of side $0.5 	ext{ m}$ floats on water with $30\%$ of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take, density of water = $10^3 	ext{ kg} / 	ext{m}^3$]

Options:

(A) 30.1 kg

(B) 46.3 kg

(C) 87.5 kg

(D) 65.4 kg

Sigma Percentile · Accepted Answer

### Analyzing the Initial State

Imagine a serene scene: a cubical block floating peacefully in a pool of water. The problem gives us a crucial piece of information right away—exactly **30% of the block's volume is submerged** beneath the surface.

Why does it float? This is the perfect time to invoke **Archimedes' Principle**, which states that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces.

Because the block is in equilibrium, its downward weight must be perfectly balanced by this upward buoyant force. Let's translate this physical reality into mathematics. The weight of the block is given by its volume $V$, multiplied by its density $ho_b$, and the acceleration due to gravity $g$.

$$W_b = V ho_b g$$

The buoyant force, on the other hand, depends only on the *submerged* volume. Since 30% is underwater, the submerged volume is $0.3V$. Multiplying this by the density of water $ho_w$ and gravity $g$ gives us the buoyant force.

$$F_{B1} = (0.3V) ho_w g$$

Equating the two, we get:

$$V ho_b g = 0.3V ho_w g$$

Notice how beautifully the volume $V$ and gravity $g$ cancel out from both sides! This leaves us with a simple, elegant relationship:

$$ho_b = 0.3 ho_w$$

The density of the block is exactly 30% of the density of water. This makes intuitive sense—an object that is 30% as dense as water will float with 30% of its volume submerged.

### The Master Equation for the Final State

Now, the problem introduces a challenge. We want to find the **maximum weight** (or mass $m$) we can place on top of the block without it sinking completely. "Without fully submerging" means we push the block down until its top surface is exactly flush with the water level. At this point, **100% of its volume is submerged**.

Let's set up a new force balance for this final state. The total downward force is now the weight of the block *plus* the weight of the new mass $m$.

$$W_{total} = V ho_b g + mg$$

This total weight is supported by a new, maximum buoyant force. Since the entire block is now underwater, the displaced volume is the full volume $V$.

$$F_{B2} = V ho_w g$$

Equating the total downward force to the new buoyant force gives us our master equation:

$$V ho_b g + mg = V ho_w g$$

### Final Calculation and Conclusion

We can immediately simplify our master equation by dividing every term by gravity $g$.

$$V ho_b + m = V ho_w$$

Rearranging to solve for our unknown mass $m$, we factor out the volume $V$:

$$m = V (ho_w - ho_b)$$

Remember our earlier discovery? We found that $ho_b = 0.3 ho_w$. Let's substitute that into our equation:

$$m = V (ho_w - 0.3 ho_w)$$

$$m = 0.7 V ho_w$$

*(Pro-Tip: You could have jumped straight to this step! The added mass $m$ is exactly responsible for pushing the remaining 70% of the block underwater. Therefore, the mass $m$ must equal the mass of the extra 70% of displaced water!)*

Now, it's just a matter of plugging in the numbers. The block is a cube with a side length of $0.5 	ext{ m}$, so its volume is:

$$V = (0.5)^3 = 0.125 	ext{ m}^3$$

The density of water is given as $10^3 	ext{ kg/m}^3$ (or $1000 	ext{ kg/m}^3$). Substituting these values into our equation for $m$:

$$m = 0.7 	imes 0.125 	imes 1000$$

$$m = 0.7 	imes 125$$

$$m = 87.5 	ext{ kg}$$

And there we have it! The maximum mass we can place on the block is exactly **87.5 kg**. Any more, and the block will sink beneath the waves.

Animated Solution for Physics - Properties of Solids and Liquids: A cubical block of side $0.5 m$ floats on water with $30%$ of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take, density of water = $1 0^{3} kg / m^{3}$ ]

Select Answer:

Visualized Solution

Visual Anchor: Initial State

Logic Bridge: Archimedes' Principle

Raw Setup: Initial Force Balance

Atomic Compute: Block Density

Visual Anchor: Final State

Logic Bridge: New Equilibrium

Raw Setup: Final Force Balance

Atomic Compute: Isolating $m$

Atomic Compute: Substituting $ρ_{b}$

Raw Setup: Known Values

Final Answer: Calculating $m$

The Way Forward: Denser Liquids

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Initial State

The Master Equation for the Final State

Final Calculation and Conclusion

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