A wooden block floating in a bucket of water has 4/5 of its volume submerged. When certain amount of

Q: Why do we take $V/2$ for both water and oil in the second case?

The problem explicitly states that in the final state, the block is 'just under the oil surface with half of its volume under water and half in oil'. This means exactly $50\%$ of the block's total volume $V$ is displacing water, and the other $50\%$ is displacing oil.

Q: Does the weight of the block change when oil is added?

No, the mass and the total volume of the wooden block remain constant throughout the process. Therefore, its downward weight $mg$ (or $V \rho_b g$) is unchanged.

Q: Why is the upthrust from oil added to the upthrust from water?

According to Archimedes' Principle, a body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. Since the block is displacing both water and oil simultaneously, both fluids exert an upward force on it. The total buoyant force is the sum of these individual forces.

Q: What exactly is relative density?

Relative density (also known as specific gravity) is the ratio of the density of a substance to the density of a given reference material. For liquids and solids, the reference material is almost always water. So, $\text{Relative Density} = \frac{\rho_{substance}}{\rho_{water}}$.

Q: Can we solve this without knowing the exact volume $V$ of the block?

Yes, absolutely! As you can see in the mathematical steps, the volume $V$ appears in every term of the equilibrium equations and cancels out completely. The final result depends only on the fractions of the volume that are submerged, not the absolute volume itself.

Question

A wooden block floating in a bucket of water has $\frac{4}{5}$ of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is

Options:

(A) 0.6

(B) 0.8

(C) 0.7

(D) 0.5

Sigma Percentile · Accepted Answer

This problem is a beautiful demonstration of Archimedes' Principle applied in two distinct scenarios. It tests your ability to set up equilibrium equations and relate the densities of different substances based on their floating behavior.

### The Initial Equilibrium

Imagine a wooden block floating peacefully in a bucket of water. According to Archimedes' Principle, any object floating in a fluid experiences an upward buoyant force (upthrust) equal to the weight of the fluid it displaces.

Since the block is in equilibrium, its downward weight is perfectly balanced by this upward upthrust. Let the total volume of the block be $V$ and its density be $ho_b$. The weight of the block is simply $W = V ho_b g$.

The problem states that $\frac{4}{5}$ of the block's volume is submerged in water. Therefore, the volume of water displaced is $\frac{4}{5}V$. The upthrust provided by the water is $F_{B1} = \left(\frac{4}{5}Vight) ho_w g$, where $ho_w$ is the density of water.

Equating the weight and the upthrust:

$$V ho_b g = \frac{4}{5} V ho_w g$$

Canceling out the common terms $V$ and $g$, we find a direct relationship between the density of the block and the density of water:

$$ho_b = \frac{4}{5} ho_w = 0.8 ho_w$$

This tells us that the wooden block is $80\%$ as dense as water.

### The Oil Invasion

Now, the scenario changes. Oil is poured into the bucket until the block is completely submerged, with exactly half of its volume in the water and the other half in the oil.

Even though the block is now fully submerged, it is still in equilibrium. However, the upward buoyant force is now a combined effort from both the water and the oil.

The volume of the block in water is $\frac{V}{2}$, so the upthrust from the water is $F_{Bw} = \left(\frac{V}{2}ight) ho_w g$.

Similarly, the volume of the block in oil is $\frac{V}{2}$, so the upthrust from the oil is $F_{Bo} = \left(\frac{V}{2}ight) ho_o g$, where $ho_o$ is the density of the oil.

### The Final Balance

Setting up the new equilibrium equation, the total weight of the block must equal the sum of the two upthrusts:

$$W = F_{Bw} + F_{Bo}$$

$$V ho_b g = \left(\frac{V}{2}ight) ho_w g + \left(\frac{V}{2}ight) ho_o g$$

Once again, we can divide the entire equation by $V g$ to simplify it:

$$ho_b = 0.5 ho_w + 0.5 ho_o$$

We already know from our initial analysis that $ho_b = 0.8 ho_w$. Substituting this into our new equation gives:

$$0.8 ho_w = 0.5 ho_w + 0.5 ho_o$$

Now, it's just a matter of simple algebra. Subtracting $0.5 ho_w$ from both sides yields:

$$0.3 ho_w = 0.5 ho_o$$

Solving for the density of the oil, $ho_o$:

$$ho_o = \frac{0.3}{0.5} ho_w = 0.6 ho_w$$

The question asks for the relative density of the oil, which is the ratio of its density to the density of water ($\frac{ho_o}{ho_w}$).

$$	ext{Relative Density} = \frac{ho_o}{ho_w} = 0.6$$

And there we have it! By carefully applying Archimedes' Principle to both states, we've elegantly deduced the relative density of the oil. **The correct option is (a).**

Select Answer:

Visualized Solution

$V_{s u b} = \frac{4}{5} V$

$W = F_{B 1}$

$V ρ_{b} g = \frac{4}{5} V ρ_{w} g$

$ρ_{b} = 0.8 ρ_{w}$

$V_{w a t er} = \frac{V}{2}, V_{o i l} = \frac{V}{2}$

$W = F_{B w} + F_{B o}$

$V ρ_{b} g = \frac{V}{2} ρ_{w} g + \frac{V}{2} ρ_{o} g$

$ρ_{b} = 0.5 ρ_{w} + 0.5 ρ_{o}$

$0.8 ρ_{w} = 0.5 ρ_{w} + 0.5 ρ_{o}$

$ρ_{o} = 0.6 ρ_{w}$

$\frac{ρ _{o}}{ρ _{w}} = 0.6$

$Think Beyond$

The Sigma Insight: Buoyancy and Archimedes' Principle

The Initial Equilibrium

The Oil Invasion

The Final Balance

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Select Answer:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=FB1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vρb​g=54​Vρw​g

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρb​=0.8ρw​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vwater​=2V​,Voil​=2V​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=FBw​+FBo​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vρb​g=2V​ρw​g+2V​ρo​g

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρb​=0.5ρw​+0.5ρo​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0.8ρw​=0.5ρw​+0.5ρo​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρo​=0.6ρw​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρw​ρo​​=0.6

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Think Beyond