Chapter 12.7, Problem 23IP

To determine

To find: The mass of the bead.

Answer to Problem 23IP

The mass of the bead is $4657\text{grams}$ .

Explanation of Solution

Given Information:

Consider the given figure.

  

Calculation:

In the figure, there are two cones and a cylinder. The total height of the bead is 10 centimeters and the height of the cylindrical section is 2 centimeters. So, the height of each cone will be 4 centimeters.

Write the expression to calculate the cone.

  ${\text{Volume}}_{\text{cone}}=\frac{1}{3}\pi r^{2}h$

Where,

• $r$ is the radius of the cone
• $h$ is the height.

Substitute the values in the above expression.

  $\begin{array}{c}{\text{Volume}}_{\text{cone}}=\frac{1}{3}\pi {\left(\frac{11}{2}\right)}^{2}4\\ \approx 126.7{\text{cm}}^3\end{array}$

Write the expression to calculate the volume of the cylindrical section.

  ${\text{Volume}}_{\text{cylinder}}=\pi r^{2}h$

Substitute the values in the above expression.

  $\begin{array}{c}{\text{Volume}}_{\text{cylinder}}=\pi r^{2}h\\ =\pi {\left(\frac{11}{2}\right)}^{2}2\\ \approx 190.1{\text{cm}}^3\end{array}$

Calculate the total volume of the bead.

  $\text{Total}\text{volume}=2{\text{Volume}}_{\text{cone}}+{\text{Volume}}_{\text{cylinder}}$

Substitute the values in the above expression.

  $\begin{array}{c}\text{Total}\text{volume}=2\left(126.7{\text{cm}}^3\right)+190.1{\text{cm}}^3\\ =443.5{\text{cm}}^3\end{array}$

To calculate the mass of the bead, multiply the total volume with the density of the pure silver.

  $\begin{array}{c}\text{Mass}=443.5\times 10.5\\ =4657\text{grams}\end{array}$

Conclusion:

Therefore, the mass of the bead is $4657\text{grams}$ .