Question

A cube of side 40 cm is floating with $\frac{1}{4}$th of its volume immersed in water. When a circular disc is placed on the cube, it floats with $\frac{2}{5}$th of its volume immersed in water. The mass of the disc is

• A. 6.4 kg
• B. 9.6 kg
• C. 3.2 kg
• D. 1.6 kg

Hint:

In buoyancy problems, the weight of the floating object (or combination of objects) is always equal to the weight of the fluid it displaces. The weight of the disc is simply the weight of the additional volume of water displaced when it was added.

Answer 1

The Correct Option is B

Step 1: Apply Archimedes' principle to the cube floating alone.
The buoyant force equals the weight of the object. Let $W_c$ be the weight of the cube and $V$ be the total volume of the cube. The weight of the displaced water is $(\frac{1}{4}V)\rho_w g$, where $\rho_w$ is the density of water. \[ W_c = \frac{1}{4}V\rho_w g. \]

Step 2: Apply Archimedes' principle to the cube with the disc.
Let $W_d$ be the weight of the disc. The total weight is $W_c + W_d$. The new volume of displaced water is $\frac{2}{5}V$. The buoyant force is $(\frac{2}{5}V)\rho_w g$. \[ W_c + W_d = \frac{2}{5}V\rho_w g. \]

Step 3: Solve for the weight of the disc.
Substitute the expression for $W_c$ from Step 1 into the equation from Step 2. \[ \frac{1}{4}V\rho_w g + W_d = \frac{2}{5}V\rho_w g. \] \[ W_d = \left(\frac{2}{5} - \frac{1}{4}\right)V\rho_w g = \left(\frac{8-5}{20}\right)V\rho_w g = \frac{3}{20}V\rho_w g. \]

Step 4: Calculate the mass of the disc.
The mass of the disc is $m_d = W_d/g$. \[ m_d = \frac{3}{20}V\rho_w. \] We are given the side of the cube is $L = 40$ cm. The volume is $V = L^3 = (40 \text{ cm})^3 = 64000 \text{ cm}^3$. The density of water is $\rho_w = 1 \text{ g/cm}^3$. \[ m_d = \frac{3}{20} (64000 \text{ cm}^3) (1 \text{ g/cm}^3) = 3 \times 3200 \text{ g} = 9600 \text{ g}. \] Converting to kilograms: \[ \boxed{m_d = 9.6 \text{ kg}}. \]