Question

A wooden cubic block of relative density $0.4$ is floating in water. Side of cubic block is $10\ \text{cm}$. When a coin is placed on the block the block dips $0.3\ \text{cm}$ in equilibrium. Weight of coin is

• (A) $0.2\ \text{N}$
• (B) $30\ \text{N}$
• (C) $0.3\ \text{N}$
• (D) $3\ \text{N}$

Hint:

Calculate the weight of the additional volume of water displaced when the block sinks by 0.3 cm.

Answer 1

The Correct Option is C

Step 1: Relate added weight to buoyancy. In equilibrium, a floating object displaces a weight of fluid equal to its own weight. When a coin is added, the extra buoyant force created by the additional submerged volume must equal the weight of the coin.

Step 2: Calculate the cross-sectional area. The block is a cube with side $a = 10\ \text{cm}$. The area of the base that is submerged in water is $A = a^{2} = 10 \times 10 = 100\ \text{cm}^{2}$.

Step 3: Find the additional volume of water displaced. The block dips further by $\Delta h = 0.3\ \text{cm}$. The extra volume of water displaced is $\Delta V = A \times \Delta h = 100\ \text{cm}^{2} \times 0.3\ \text{cm} = 30\ \text{cm}^{3}$.

Step 4: Convert volume to SI units. $\Delta V = 30 \times (10^{-2}\ \text{m})^{3} = 30 \times 10^{-6}\ \text{m}^{3} = 3 \times 10^{-5}\ \text{m}^{3}$.

Step 5: Compute the weight. Taking density of water $\rho = 1000\ \text{kg/m}^{3}$ and $g \approx 10\ \text{m/s}^{2}$, the weight $W = \rho g \Delta V = 1000 \times 10 \times 3 \times 10^{-5} = 0.3\ \text{N}$. Thus, the weight of the coin is 0.3 N.