Question

A tub is filled with water and a wooden cube \(10\,\text{cm} \times 10\,\text{cm} \times 10\,\text{cm}\) is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by \(3.87\) cm. The mass of the metal coin is ____ gram. (Take water density \(=1\,\text{g/cm}^3\) and density of wood as \(0.4\,\text{g/cm}^3\)).}

Answer 1

Correct Answer: 387

Step 1: Understanding the Concept:
According to Archimedes' Principle, for a floating body, the weight of the body equals the buoyant force (weight of the displaced fluid). When an extra weight (coin) is added, the cube sinks deeper to displace an additional volume of water whose weight equals the weight of the coin.
Step 2: Key Formula or Approach:
1. Equilibrium condition: \(\Delta F_b = \Delta W\).
2. Additional Buoyant Force: \(\Delta F_b = (\text{Area} \times \Delta h) \cdot \rho_{\text{water}} \cdot g\).
3. Weight of coin: \(W_{\text{coin}} = m_{\text{coin}} \cdot g\).
Step 3: Detailed Explanation:
Given:
Area of the cube \(A = 10 \text{ cm} \times 10 \text{ cm} = 100 \text{ cm}^2\).
Increase in submerged depth \(\Delta h = 3.87 \text{ cm}\).
Density of water \(\rho_{\text{water}} = 1 \text{ g/cm}^3\).
The weight of the coin is supported by the extra buoyant force created by the additional submerged volume.
\[ \text{Mass of coin} = \text{Mass of additional water displaced} \]
\[ m_{\text{coin}} = \text{Volume}_{\text{additional}} \times \rho_{\text{water}} \]
\[ m_{\text{coin}} = (A \times \Delta h) \times \rho_{\text{water}} \]
\[ m_{\text{coin}} = (100 \text{ cm}^2 \times 3.87 \text{ cm}) \times 1 \text{ g/cm}^3 \]
\[ m_{\text{coin}} = 387 \text{ grams} \].
Step 4: Final Answer:
The mass of the metal coin is 387 grams.