Answer:
[tex]34.15 \mathrm{cm}^{3} \text { of "volume of iron" is needed to balance } 30 \mathrm{cm}^{3} \text { of "volume of copper". }[/tex]
Explanation:
Density of an object is defined as the mass occupied per unit volume. “Density” is represented by “ρ”. Mathematically, [tex]\rho=\frac{m}{v}[/tex] where, “m" mass of an object and "v" is volume of an object.
Given that,
[tex]\text { Volume of copper is } 30 \mathrm{cm}^{3}\left(v_{c}\right)[/tex]
[tex]\text { Density of copper is } 8.96 \mathrm{g} / \mathrm{cm}^{3}\left(\rho_{c}\right)[/tex]
[tex]\text { Density of iron is } 7.87 \mathrm{g} / \mathrm{cm}^{3}\left(\rho_{i}\right)[/tex]
[tex]\text { We need to find "Volume of iron" }\left(v_{i}\right) \text { that balances copper. The following steps are followed: }[/tex]
Step: 1
To find the mass of the copper:
[tex]\text { Take, } \rho_{c}=\frac{m_{c}}{v_{c}}[/tex]
[tex]8.96=\frac{m_{c}}{30}[/tex]
[tex]8.96 \times 30=m_{c}[/tex]
[tex]268.8=m_{c}[/tex]
Mass of the copper is 268.8 kg.
To balance the weight of the copper 268.8 kg [tex]\left(m_{c}\right)[/tex] = weight of the iron [tex]\left(m_{i}\right)[/tex] the volume required is
Step: 2
[tex]\text { Take, } \rho_{i}=\frac{m_{i}}{v_{i}}[/tex]
[tex]7.87=\frac{268.8}{v_{i}}[/tex]
[tex]v_{i}=\frac{268.8}{7.87}[/tex]
[tex]v_{i}=34.15 \mathrm{cm}^{3}[/tex]
[tex]\text { Therefore, the volume required is } 34.15 \mathrm{cm}^{3} .[/tex]
