An iron casting containing a number of cavities weighs 6000 Nin air and 4000 Nin water. What is the total volume of all the cavities in the casting? The density of iron (that is, a sample with no cavities) is $\mathbf{7}\mathbf{.}\mathbf{87}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$.

• The weight of iron casting in air, W=6000 N
• The weight of iron casting in water, $W_{eff}=4000N$
• The density of solid iron,${\rho }_{iron}=7.87\frac{g}{c{m}^3}or7.87\times {10}^3\frac{kg}{m^3}$

We can use the concept of Archimedes’ principle and expression of density. When an iron casting is fully submerged in water, then a buoyant force from the surrounding fluid acts on it. This force has a magnitude equal to the weight of the water displaced by the iron casting.

Formulae:

Force applied on body (or weight),$F_b=m_{f}g$ (i)

Density of a substance,$\mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{V}}$ (ii)

The volume of cavity is the difference between volume of casting and volume of iron. Hence,

$V_{cavity}=V_{casting}=V_{iron}$ (iii)

According to equation (ii), we get

The volume of iron as:

$$\begin{gathered} V_{iron}=\frac{m_{iron}}{{\rho }_{iron}} \\ =\frac{g{m}_{iron}}{g{\rho }_{iron}} \\ =\frac{W}{g{\rho }_{iron}} \end{gathered}$$

When the iron casting is submerged in water, the forces acting on it are$F_g-F_b$. Hence,the effective weight in water is given as:

$$\begin{gathered} W_{eff}=W-g{\rho }_w{V}_{casting} \\ {V}_{casting}=\frac{W-W_{eff}}{g{\rho }_w} \end{gathered}$$

Equation (iii) becomes and using the given values, we get

$$\begin{gathered} V_{cavity}=\frac{W-W_{eff}}{g{\rho }_w}-\frac{W}{g{\rho }_{iron}} \\ =\left(\frac{6000N-4000N}{9.8{\displaystyle \frac{m}{s^2}}\times 1000{\displaystyle \frac{kg}{m^3}}}\right)-\left(\frac{6000N}{9.8\frac{m}{s^2}\times 7.87\times {10}^3\frac{kg}{m^3}}\right) \\ =0.126m^3 \end{gathered}$$

Hence, the volume of cavity is$0.126m^3$