Chapter 13, Problem 60A

Find the volume of this pin. All dimensions are in inches.
Volume of cylinder $=\pi \times R^2\times H$

Volume of cone $=0.2618\times D^2\times H$

To determine

The metal volume of given pin.

Answer to Problem 60A

Volume of given pin is $\text{9}{\text{.33 mm}}^3$.

Given information:

A metal pin is given as below indicating dimensions in millimeter.

Calculation:

Volume of the given pin will be volume of cylinder added with volume of cone.

As we know that volume of a cylinder is given by -

$V=\pi \times R^2\times H$

Here, radius of the cylinder is $0.65$ millimeter and height cylinder is $6.87$ millimeter. i.e. $R=0.65\text{ mm and }H=6.87\text{ mm}$.

So, volume of the cylinder will be -

$\begin{array}{l}V=\pi \times r^2\times H\\ \Rightarrow V=3.14\times {(}^{0.65}\times 6.87{\text{mm}}^3\text{ {Put }r=5.00\text{ mm,}h=3.20\text{ mm}}\\ \Rightarrow V=3.14\times 0.4225\times 6.87{\text{mm}}^3\\ \Rightarrow V=9.11{\text{mm}}^3\end{array}$

Now, as we know that volume of a cone is given by -

$V=0.2618\times d^2\times H$

Here, diameter of the given cone is $1.30$ millimeter and height is $0.50$ millimeter.

i.e. $d=1.30$ millimeter and $H=0.50$ millimeter.

So, volume of the cone will be -

$\begin{array}{l}V=0.2618\times d^2\times h\\ \Rightarrow V=0.2618\times {(}^{1.30}\times 0.50{\text{mm}}^3\text{ {Put }d=1.30\text{ mm, }h=0.50\text{ mm}}\\ \Rightarrow V=0.2618\times 1.69\times 0.50{\text{mm}}^3\\ \Rightarrow V=0.2212{\text{mm}}^3\\ \Rightarrow V=0.22{\text{mm}}^3\text{ {Rounding off to 2 decimal place}}\end{array}$

Since, volume of the given pin will be volume of cylinder added with volume of cone.

Hence, volume of given pin $=$$9.11+0.22{\text{ mm}}^3=9.33{\text{ mm}}^3$.

Explanation of Solution

Given information:

A metal pin is given as below indicating dimensions in millimeter.

Calculation:

Volume of the given pin will be volume of cylinder added with volume of cone.

As we know that volume of a cylinder is given by -

$V=\pi \times R^2\times H$

Here, radius of the cylinder is $0.65$ millimeter and height cylinder is $6.87$ millimeter. i.e. $R=0.65\text{ mm and }H=6.87\text{ mm}$.

So, volume of the cylinder will be -

$\begin{array}{l}V=\pi \times r^2\times H\\ \Rightarrow V=3.14\times {(}^{0.65}\times 6.87{\text{mm}}^3\text{ {Put }r=5.00\text{ mm,}h=3.20\text{ mm}}\\ \Rightarrow V=3.14\times 0.4225\times 6.87{\text{mm}}^3\\ \Rightarrow V=9.11{\text{mm}}^3\end{array}$

Now, as we know that volume of a cone is given by -

$V=0.2618\times d^2\times H$

Here, diameter of the given cone is $1.30$ millimeter and height is $0.50$ millimeter.

i.e. $d=1.30$ millimeter and $H=0.50$ millimeter.

So, volume of the cone will be -

$\begin{array}{l}V=0.2618\times d^2\times h\\ \Rightarrow V=0.2618\times {(}^{1.30}\times 0.50{\text{mm}}^3\text{ {Put }d=1.30\text{ mm, }h=0.50\text{ mm}}\\ \Rightarrow V=0.2618\times 1.69\times 0.50{\text{mm}}^3\\ \Rightarrow V=0.2212{\text{mm}}^3\\ \Rightarrow V=0.22{\text{mm}}^3\text{ {Rounding off to 2 decimal place}}\end{array}$

Since, volume of the given pin will be volume of cylinder added with volume of cone.

Hence, volume of given pin $=$$9.11+0.22{\text{ mm}}^3=9.33{\text{ mm}}^3$.