A solid cube of side 50cm and mass 75kg floats in a liquid with $\frac{1}{3}$ of its height above the liquid surface. The relative density of the liquid is?

1. Analyze the Question

We are given a cube with a side length ($L$) of 50 cm and a mass ($m$) of 75 kg. It floats in a liquid such that $\frac{1}{3}$ of its height is above the surface. We need to find the relative density of the liquid.

2. Explain the Concept

According to the Law of Flotation, a floating object displaces a weight of liquid equal to its own weight.

• Weight of the object ($W$) = $mg$
• Upthrust (Weight of displaced liquid) = $V_{submerged} \times \rho_{liquid} \times g$

Since the cube floats, $W = \text{Upthrust}$. Also, Relative Density (RD) = $\frac{\text{Density of liquid}}{\text{Density of water}}$. Note that the density of water ($\rho_w$) is $1000 \, \text{kg/m}^3$.

3. Show the Solution

Step 1: Determine the volume of the cube. Side $L = 50 \, \text{cm} = 0.5 \, \text{m}$. Total Volume ($V_{total}$) = $L^3 = 0.5^3 = 0.125 \, \text{m}^3$.

Step 2: Determine the submerged volume. If $\frac{1}{3}$ of the height is above the liquid, then $\frac{2}{3}$ of the height is submerged. $V_{submerged} = \frac{2}{3} \times V_{total} = \frac{2}{3} \times 0.125 = 0.0833 \, \text{m}^3$.

Step 3: Apply the Law of Flotation. $mg = V_{submerged} \times \rho_{liquid} \times g$ The $g$ cancels out: $m = V_{submerged} \times \rho_{liquid}$ $75 = 0.0833 \times \rho_{liquid}$ $\rho_{liquid} = \frac{75}{0.0833} \approx 900 \, \text{kg/m}^3$.

Step 4: Calculate Relative Density. $\text{RD} = \frac{\rho_{liquid}}{\rho_{water}} = \frac{900}{1000} = 0.90$.

4. Identify the Answer

Comparing our result to the options provided, the correct value is 0.90. This matches option D.

5. Tip

When dealing with floating objects, remember the shortcut formula:

First, find the density of the cube: $\rho_{cube} = \frac{\text{mass}}{\text{volume}} = \frac{75}{0.125} = 600 \, \text{kg/m}^3$. Relative density of cube = 0.6. Then, $\text{RD}_{liquid} = \frac{0.6}{2/3} = 0.6 \times 1.5 = 0.9$. This is a much faster way to solve it in the exam hall!