Question

A wooden cubical block of relative density 0.4 is floating in water. Side of cubical block is $10 \text{ cm}$. When a coin is placed on the block, it dips by $0.3 \text{ cm}$, weight of coin is:

• A. $0.1 \text{ N}$
• B. $0.2 \text{ N}$
• C. $0.3 \text{ N}$
• D. $0.4 \text{ N}$

Answer 1

The Correct Option is C

To determine the weight of the coin placed on the cubical block, we need to analyze the additional buoyant force exerted due to the extra dip in water caused by the coin.

Concept and Calculations

The principle of flotation states that the weight of the floating body (wooden block plus coin) is equal to the weight of the liquid displaced.

Step 1: Find the volume of water displaced by the additional dip

The additional volume of water displaced due to the dip is the product of the area of the top surface of the cube and the additional depth by which the cube dips.

• Side of the cube, \(a = 10 \text{ cm}\)
• Additional dip, \(h = 0.3 \text{ cm}\)
• Volume of water displaced, \(V = a^2 \cdot h = (10 \text{ cm})^2 \cdot 0.3 \text{ cm} = 30 \text{ cm}^3\)

Step 2: Calculate the buoyant force

The buoyant force, which equals the weight of the coin, is the weight of the displaced water:

• Density of water, \(\rho_{\text{water}} = 1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3\)
• Weight of displaced water (buoyant force), \(F_b = \rho_{\text{water}} \cdot V \cdot g\)
• \(F_b = 1 \text{ g/cm}^3 \times 30 \text{ cm}^3 \times 9.8 \text{ m/s}^2 \times 10^{-3} \text{ kg/g}\)
• \(F_b = 0.294 \text{ N}\)

Conclusion

The weight of the coin (approximated) is \(0.3 \text{ N}\), which is best matched with the given option.

Correct Answer

The correct option is: $0.3 \text{ N}$