91Ó°ÊÓ

• 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,

$P=\frac{m}{V}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(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" $$\begin{gathered} V=V_{cube}-V_{hole} \\ ={(0.05)}^3-\mathrm{π}{\left(0.01\mathrm{m}\right)}^2\left(0.05\mathrm{m}\right) \\ =1.09×{10}^{-4}{\mathrm{m}}^3 \end{gathered}$$

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

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

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

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

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

Substitute the values in equation (2),

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

The weight of the drill hole can be evaluated by,

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

Thus, the weight of the cube is 7.25N