Question

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?

Short answer

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

Step-by-step solution

Listing the given quantities

• 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, $\rho =11.4g/\left(c{m}^3\right)$.

Understanding the Archimedes principle

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.

Calculating the mass of the lead

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$1.07\times {10}^{3}g.$