A cube 5.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.30 N. (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

• The side length of the cube is,$a=5.0\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.05\mathrm{m}$
• The diameter of the drill is,$D=2.0\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.02\mathrm{m}$
• The weight of the cube after the hole is drilled was,$W=6.3\mathrm{N}$

The density of a substance is defined as the amount of mass contained inside a unit volume.

The density can be given as,

$P=\frac{m}{V}$………………..(1)

Where m and V are the mass and volume of the substance.

The mass of the metal can be evaluated by,

$$\begin{gathered} m=\left(\frac{W}{g}\right) \\ =\frac{6.3\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =0.64\mathrm{N}/(\mathrm{m}/{\mathrm{s}}^2) \\ =0.64\mathrm{kg} \end{gathered}$$

The volume of the cube after the drill can be evaluated as,

role="math" localid="1655791407187" $$\begin{gathered} V=V_{cube}-V_{hole} \\ ={(0.05)}^3-\mathrm{\pi }{\left(0.01\mathrm{m}\right)}^2\left(0.05\mathrm{m}\right) \\ =1.09\times {10}^{-4}{\mathrm{m}}^3 \end{gathered}$$

Now, the density of metal can be evaluated using equation (1),

role="math" localid="1655791898368" $$\begin{gathered} p=\frac{0.64kg}{1.09\times {10}^{-4}{\mathrm{m}}^3} \\ =5.9\times {10}^3\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Thus, the density of the metal is$5.9\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$.

The mass of the cube can be evaluated by rearranging equation (1),

$m=p{V}_{cube}$......................(2)

Substitute the values in equation (2),

$$\begin{gathered} m=\left(5.9\times {10}^3\mathrm{kg}/{\mathrm{m}}^3\right){\left(0.05\mathrm{m}\right)}^3 \\ m=0.74\mathrm{kg} \end{gathered}$$

The weight of the drill hole can be evaluated by,

$$\begin{gathered} W=mg \\ =\left(0.74\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =7.25\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =7.25\mathrm{N} \end{gathered}$$

Thus, the weight of the cube is 7.25N