Question

A block of wood of side 40 cm floats in water in such a way that its lower face is 5 cm below the free surface of water. What is the weight of the block?

• A. 64 kg
• B. 16 kg
• C. 8 kg
• D. cannot be determined as density of wood is not given

Hint:

Floating body: Weight = weight of displaced liquid.

Answer 1

The Correct Option is B

To find the weight of the block, we apply the principle of buoyancy, also known as Archimedes' principle. According to this principle, the buoyant force is equal to the weight of the fluid displaced by the submerged part of the object.

1. Given:
   • Side of the block = 40 cm = 0.4 m
   • Depth submerged below the water surface = 5 cm = 0.05 m
2. The volume of water displaced by the block (submerged volume) is given by: \(V = \text{side}^2 \times \text{depth submerged} = (0.4)^2 \times 0.05 \, \text{m}^3\)
3. Calculating the volume: \(V = 0.4 \times 0.4 \times 0.05 = 0.008 \, \text{m}^3\)
4. The weight of the water displaced is equal to the weight of the block. Using the formula for the buoyant force: \(F_b = \rho \times V \times g\)
   • \(\rho = 1000 \, \text{kg/m}^3\) (density of water)
   • \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)
5. Substituting the values: \(F_b = 1000 \times 0.008 \times 9.8 = 78.4 \, \text{N}\)
6. Since \(1 \, \text{N} = 0.10197 \, \text{kg}\), the weight of the block in kilograms is approximately: \(\frac{78.4}{9.8} = 8 \, \text{kg}\)

Therefore, the weight of the block is approximately 8 kg. However, the correct given answer seems to be 16 kg, which might indicate there's an error in the options or additional information is missing. Double-checking the provided details in the problem could be helpful.