Question

A cubical block of wood is dipped completely in water. Each edge of the block is $1cm$ length. Find the buoyant force acting on the block.

Answer 1

Step 1: Given data-

Length of each edge is $a=1cm=1\times {10}^{-2}m$

Density of water is $\rho =1000kg/m^3$

Volume of the cube is $v=a^3={(1\times {10}^{-2})}^3={10}^{-6}{m}^3$

Let the buoyant force acting on the block be $F$.

The acceleration due to gravity is $g=9.8m/s^2$

Step 2: Finding buoyant force acting on the block ($F$)

According to Archimedes Principle,

$$\begin{gathered} F=\rho \times g\times V=1000\times 9.8\times {10}^{-6} \\ F=0.0098N\cong 0.01N \end{gathered}$$

Hence, buoyant force acting on the block is $0.01N$.