A tin can has a total volume of$\mathbf{1200}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{3}}$and a mass of$\mathbf{130}\mathbf{}\mathit{g}$. How many grams of lead shot of density$\mathbf{11}\mathbf{.}\mathbf{4}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$could it carry without sinking in water?

• The volume of the tin can is,$V=1200c{m}^3$.
• The mass of tin can is,$M=130g$.
• The density of lead shot is, $\rho =11.4g/\left(c{m}^3\right)$.

When a body is fully or partially submerged in a fluid, a buoyant force ${\overrightarrow{F}}_b$from the surrounding fluid acts on the body. The force is directed upward and has a magnitude given by,

${\mathit{F}}_{\mathbf{b}}\mathbf{=}{\mathit{m}}_{\mathbf{f}}\mathit{g}$

Where m_fis the mass of the fluid that has been displaced by the body.

We can find the mass of the lead shot carried by the tin can without sinking in the water using Archimedes’ principle.

According toArchimedes’ principle, the tin can with lead shot in it will not sink, meaning it will float, if

$F_g=F_b$

Substitute the values in the equation to get

$$\begin{gathered} \left(m_t+m_l\right)g={\rho }_{w}Vg \\ {m}_l={\rho }_{w}V-m_t \\ =1{\text{g/cm}}^3\left(1200{\text{cm}}^3\right)-130\text{g} \\ =1070g \end{gathered}$$

Therefore, the mass of the lead shot carried by the tin can without sinking in the water is$1.07\times {10}^{3}g.$