Question

a graduated cylinder is filled to the 40.00 ml mark with a mineral oil. the masses of the cylinder before and after the addition of the mineral oil are 124.966 g and 159.446 g, respectively. in a separate experiment, a metal ball bearing of mass 20.133 g is placed in the cylinder and the cylinder is again filled to the 40.00 ml mark with the mineral oil. the combined mass of the ball bearing and mineral oil is 50.952 g. calculate the density and radius of the ball bearing.
part 1 of 2
round your answer to 3 significant digits.
density of the ball bearing = \\( \frac{\square}{\mathrm{cm}^{3}} \\)
part 2 of 2
round your answer to 3 significant digits.
radius of the ball bearing = \\( \square \mathrm{cm} \\)

Explanation:

Step1: Calculate the mass of the ball bearing

The mass of the ball bearing \(m\) is the combined mass of the ball bearing and mineral oil minus the mass of the mineral oil.
\(m = 50.952\space g-159.446\space g = 30.506\space g\)

Step2: Calculate the volume of the ball bearing

The volume of the ball bearing \(V\) is the volume of the displaced mineral oil. The volume of the displaced mineral oil is the difference in the volume of the mineral oil before and after adding the ball bearing.
\(V=40.00\space mL - 12.4966\space mL=27.5034\space mL = 27.5034\space cm^{3}\) (since \(1\space mL = 1\space cm^{3}\))

Step3: Calculate the density of the ball bearing

The density formula is \(
ho=\frac{m}{V}\).
Substitute \(m = 30.506\space g\) and \(V = 27.5034\space cm^{3}\) into the formula:
\(
ho=\frac{30.506\space g}{27.5034\space cm^{3}}\approx1.11\space g/cm^{3}\)

Step4: Calculate the radius of the ball bearing

The volume of a sphere \(V=\frac{4}{3}\pi r^{3}\).
We know \(V = 27.5034\space cm^{3}\), then \(r^{3}=\frac{3V}{4\pi}\)
Substitute \(V = 27.5034\space cm^{3}\) and \(\pi\approx3.14\) into the formula:
\(r^{3}=\frac{3\times27.5034}{4\times3.14}=\frac{82.5102}{12.56}\approx6.57\)
\(r=\sqrt[3]{6.57}\approx1.87\space cm\)

Answer:

Density of the ball bearing \(= 1.11\space g/cm^{3}\), Radius of the ball bearing \(= 1.87\space cm\)