Question

A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.5 cm in diameter all the way through and perpendicular to one face, you find that the cube weighs 6.30 N. 1. What is the density of the metal? 2. What did the cube weigh before you drilled the hole in it?

Answer 1

1.

First, let's calculate the volume of the cube:

$V_1=6^3=216\text{ cm^^b3}$

Now, let's calculate the volume of the cylinder drilled:

$\begin{gathered} V_2=\frac{\pi d^2h}{4}\\ \\ V_2=\frac{3.14159\cdot2.5^2\cdot6}{4}\\ \\ V_2=29.45\text{ cm^^b3} \end{gathered}$

So the volume of the cube after being drilled is:

$\begin{gathered} V=V_1-V_2\\ \\ V=216-29.45\\ \\ V=186.55\text{ cm^^b3} \end{gathered}$

If the weight is 6.3 N, let's find the mass:

$\begin{gathered} W=m\cdot g\\ \\ 6.3=m\cdot9.8\\ \\ m=\frac{6.3}{9.8}\\ \\ m=0.643\text{ kg} \end{gathered}$

And the density of the metal is:

$\begin{gathered} d=\frac{m}{V}\\ \\ d=\frac{0.643}{186.55}\\ \\ d=0.0034468\text{ kg/cm^^b3}\\ \\ d=3.4468\text{ g/cm^^b3} \end{gathered}$

2.

To find the weight of the cube before being drilled, let's use the following rule of three:

$\begin{gathered} volume\rightarrow weight\\ \\ 216\text{ cm^^b3}\rightarrow x\text{ N}\\ \\ 186.55\text{ cm^^b3}\rightarrow6.3\text{ N}\\ \\ \\ \\ \frac{216}{186.55}=\frac{x}{6.3}\\ \\ x=\frac{216\cdot6.3}{186.55}\\ \\ x=7.295\text{ N} \end{gathered}$