Question

What volume of iron would be needed to balance a 30 cm3 sample of copper ? ( Density of iron is 7.87 g/cm3 and density of copper is 8.96 g/cm3)

Answer 1

`34.15 \mathrm{cm}^(3) \text { of`

Step-by-step explanation:

Density of an object is defined as the mass occupied per unit volume. “Density” is represented by “ρ”. Mathematically,
`\rho=(m)/(v)` where, “m" mass of an object and "v" is volume of an object.

Given that,

`\text { Volume of copper is } 30 \mathrm{cm}^(3)\left(v_(c)\right)`

`\text { Density of copper is } 8.96 \mathrm{g} / \mathrm{cm}^(3)\left(\rho_(c)\right)`

`\text { Density of iron is } 7.87 \mathrm{g} / \mathrm{cm}^(3)\left(\rho_(i)\right)`

`\text { We need to find`

Step: 1

To find the mass of the copper:

`\text { Take, } \rho_(c)=(m_(c))/(v_(c))`

`8.96=(m_(c))/(30)`

`8.96 * 30=m_(c)`

`268.8=m_(c)`

Mass of the copper is 268.8 kg.

To balance the weight of the copper 268.8 kg
`\left(m_(c)\right)` = weight of the iron
`\left(m_(i)\right)` the volume required is

Step: 2

`\text { Take, } \rho_(i)=(m_(i))/(v_(i))`

`7.87=(268.8)/(v_(i))`

`v_(i)=(268.8)/(7.87)`

`v_(i)=34.15 \mathrm{cm}^(3)`

`\text { Therefore, the volume required is } 34.15 \mathrm{cm}^(3) .`