Chapter 12.4, Problem 23PPS

To determine

To find: the mass in grams of the bead

Answer to Problem 23PPS

The mass of the bead is approximately 4654.34 grams.

Explanation of Solution

Given:

  

Pure silver's mass is 10.5 grams per cubic cm.

Calculation:

The figure is the combination of two cones whose diameter is 11 cm and height is

  $\begin{array}{c}\frac{10-2}{2}=\frac{8}{2}\text{cm}\\ =4\text{cm}\end{array}$

and a cylinder whose diameter is 11 cm and height is 2 cm.

The objective is to find the approximate mass in gram of the bead.

To find the mass first find the volume of the bead.

The required volume is the sum of the volume of two cones and the volume of a cylinder.

The volume V of a cone is given by the formula

  $V=\frac{1}{3}B\cdot h$

where $B=\pi r^2$ is the area of the circle and h is the height the cone.

The volume V of the cylinder is given by the formula

  $V=B\cdot h$

where $B=\pi r^2$ is the area of the circle and h is the height the cylinder.

First find the volume of the cone.

Since radius is half the diameter of a circle, so radius of the circle is $\frac{11}{2}$ cm that is

5.5 cm

The area B of the circle is given

  $\begin{array}{c}B=\pi r^2\\ =\pi {\left(5.5\right)}^2\\ =30.25\pi \\ \approx 30.25\left(3.14\right)\left[Since\pi \approx 3.14\right]\\ =94.985\end{array}$

Put $B\approx 94.985$ and $h=4$ in the volume formula to find the volume of the cone.

Thus, the volume $V_1$ of the cone is

  $\begin{array}{c}{V}_1=\frac{1}{3}B\cdot h\\ \approx \frac{1}{3}\left(94.985\right)\left(4\right)\\ =\frac{379.94}{3}\\ \approx 126.65\end{array}$

Therefore, the volume of the cone is approximately $126.65{\text{cm}}^3$ .

Put $B\approx 94.985$ and $h=2$ in the volume formula to find the volume of the cylinder.

Thus, the volume $V_2$ of the cylinder is

  $\begin{array}{c}{V}_2=B\cdot h\\ \approx \left(94.985\right)2\\ \approx 189.97\end{array}$

Therefore, the volume of the cylinder is approximately $189.97{\text{cm}}^3$ .

The volume V of the figure is

  $\begin{array}{c}V=2V_1+V_2\\ \approx 3\left(126.65\right)+189.97\\ =253.3+189.97\\ =443.27\end{array}$

Therefore, the volume of the figure is approximately $443.27{\text{cm}}^3$ .

Since pure silver's mass is 10.5 grams per cubic cm, so the mass of the bead

  $10.5\times 443.27\text{grams}\approx 4654.34\text{grams}$

Therefore, the mass of the bead is approximately 4654.34 grams.

Conclusion:

Therefore, the mass of the bead is approximately 4654.34 grams.