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

According to Archimedes' Principle, the increase in buoyant force due to additional submergence equals the weight of the coin. The side of the cube is $10\ \text{cm}$, so base area $A = 100\ \text{cm}^2$.

Extra depth submerged $\Delta h = 0.3\ \text{cm}$. Additional volume of water displaced $\Delta V = A \times \Delta h = 100 \times 0.3 = 30\ \text{cm}^3 = 3 \times 10^{-5}\ \text{m}^3$.

The weight of the coin $W = \rho_{water} g \Delta V = 1000 \times 10 \times 3 \times 10^{-5} = 0.3\ \text{N}$.