Question

A cube of iron and a cube of wood, each have a volume of 2.1×10^-4th m3, are each placed in a large beaker of water. The density of the wood is 3.73×10^2 kg/m3 In the density of the iron is 7.86×10^3. Calculate the buoyant force on each.IronWood

Answer 1

Given:

The volume of wood is

`V=2.10*10^(-4)\text{ m}^3`

The density of wood is

`\rho_w=3.73*10^2\text{ kg/m}^3`

The volume of iron is

`V=2.10*10^(-4)\text{ m}^3`

The density of iron is

`\rho_(iron)=\text{ 7.86}*10^3\text{ kg/m}^3`

The density of water is

`\rho_(water)=\text{ 997 kg/m}^3`

To find the buoyant force on wood and on iron.

Step-by-step explanation:

The buoyant force can be calculated by the formula

`B=V*\rho_(water)* g`

Here, g =9.8 m/s^2 is the acceleration due to gravity.

On substituting the values, the buoyant force on the wood is

`\begin{gathered} B_(wood)=2.10*10^(-4)*997*9.8 \\ =\text{ 2.052 N} \end{gathered}`

On substituting the values, the buoyant force on the iron is

`\begin{gathered} B_(iron)=2.10\cdot10^(-4)*997*9.8 \\ =2.052\text{ N} \end{gathered}`