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

Step 1: Understand the principle of flotation.
When the coin is added, the block sinks further. The weight of the coin must be balanced by the additional buoyant force generated by the extra displaced volume of water.

Step 2: Identify the additional displaced volume.
Side of the cube $L = 10 \text{ cm}$. Area of base $A = L^2 = (10 \text{ cm})^2 = 100 \text{ cm}^2 = 100 \times 10^{-4} \text{ m}^2 = 10^{-2} \text{ m}^2$.
Additional dip $\Delta h = 0.3 \text{ cm} = 0.003 \text{ m}$.
Extra volume displaced $\Delta V = A \times \Delta h = 10^{-2} \times 0.003 = 3 \times 10^{-5} \text{ m}^3$.

Step 3: Calculate the weight of the coin using Archimedes' principle.
Weight of coin = Buoyant force of extra displaced water = $\rho_w \cdot \Delta V \cdot g$.
Taking $\rho_w = 1000 \text{ kg/m}^3$ and $g = 10 \text{ m/s}^2$.
Weight = $1000 \times (3 \times 10^{-5}) \times 10 = 10000 \times 3 \times 10^{-5} = 0.3 \text{ N}$.

Final Answer: $0.3 \text{ N}$. Correct option is (3).