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 = 10$$^3$$ kg/m$$^3$$]

First, recall Archimedes’ Principle: “When a body floats in a fluid, the weight of the body equals the weight of the fluid displaced.” We translate this into the mathematical form

$$\text{Weight of block} \;=\; (\text{density of water}) \times (\text{volume submerged}) \times g.$$

We have a cube whose edge length is 0.5 m, so its total geometric volume is

$$V_{\text{cube}} \;=\; (0.5\ \text{m})^{3} \;=\; 0.5 \times 0.5 \times 0.5 \;=\; 0.125\ \text{m}^{3}.$$

At the moment the cube is floating with only 30 % of its volume under water. Therefore the submerged volume is

$$V_{\text{sub, now}} \;=\; 0.30 \times V_{\text{cube}} \;=\; 0.30 \times 0.125 \;=\; 0.0375\ \text{m}^{3}.$$

Using Archimedes’ Principle the weight of the cube itself equals the weight of 0.0375 m³ of water. Taking the density of water to be $$\rho_{\text{w}} = 10^{3}\ \text{kg m}^{-3},$$ we write

$$\text{Mass of cube}\; (m_{\text{cube}}) =\; \rho_{\text{w}} \, V_{\text{sub, now}} =\; 10^{3} \times 0.0375 =\; 37.5\ \text{kg}.$$

Now imagine loading the cube until it is just about to be completely immersed, but not more. In that limiting situation the entire volume 0.125 m³ is displaced. The corresponding maximum supporting (buoyant) mass is

$$m_{\text{max\,disp}} =\; \rho_{\text{w}} \, V_{\text{cube}} =\; 10^{3} \times 0.125 =\; 125\ \text{kg}.$$

This 125 kg must balance the combined mass of the cube plus the additional load. Writing this balance condition, we get

$$m_{\text{cube}} \;+\; m_{\text{load, max}} =\; m_{\text{max\,disp}}.$$

Substituting the numerical values,

$$37.5 \;+\; m_{\text{load, max}} =\; 125.$$

Solving for the unknown maximum load,

$$m_{\text{load, max}} =\; 125 \;-\; 37.5 =\; 87.5\ \text{kg}.$$

Hence, the correct answer is Option A.