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Found in: Page 15

Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

Figure P1.10 shows a frustum of a cone. Match each of theexpressions(a) ${\pi }\left({\text{r}}_{1}{\text{+r}}_{2}\right){\left[{h}^{2}+{\left({\text{r}}_{2}-{\text{r}}_{1}\right)}^{2}\right]}^{1/2}$, ${\pi }\left({\text{r}}_{1}{\text{+r}}_{2}\right){\left[{h}^{2}+{\left({\text{r}}_{2}-{\text{r}}_{1}\right)}^{2}\right]}^{1/2}$(b) and(c) ${\pi }{h}\left({{\text{r}}_{1}}^{2}+{\text{r}}_{1}{\text{r}}_{2}+{{\text{r}}_{2}}^{2}\right){/}{3}$ with the quantity it describes: (d) the total circumference of the flat circular faces, (e) the volume, or (f) the area of the curved surface.

The matches are given by:

(a) and (f)

(b) and (c)

(c) and (e)

See the step by step solution

Step 1: A concept and analysis of the given data

The analysis of a relationship between different physical quantities by using the units of measurements and dimensions is called dimensional analysis. It is used to examine the correctness of an equation.

The formula of a circumference is $\text{C = 2}\pi \text{r,}$ and its dimension is $L.$

The formula of a volume is $V=\pi {r}^{2}\frac{h}{3},$ and its dimension is ${L}^{3}.$

The formula of an area is $A=\pi r\left(r+\sqrt{{h}^{2}+{r}^{2}}\right),$ and its dimension is ${L}^{2}.$

Step 2: Find the dimensions of equation (a)

The given equation is $\pi \left({\text{r}}_{1}{\text{+r}}_{2}\right){\left[{h}^{2}+{\left({\text{r}}_{2}-{\text{r}}_{1}\right)}^{2}\right]}^{1/2}.$

In the above formula, $\pi$ has no dimension.

The dimensionsof ${\text{r}}_{1}$, ${\text{r}}_{2},$ and $h$ is $L$

Replace for ${\text{r}}_{1}$, ${\text{r}}_{2},$ and $h$ in the above equation to get its dimension.

$\begin{array}{c}Dimension=\left(\text{L+L}\right){\left[{L}^{2}+{\left(L-L\right)}^{2}\right]}^{1/2}\\ \text{\hspace{0.17em}}=L{\left({L}^{2}\right)}^{1/2}\\ =L\left(L\right)\\ ={L}^{2}.\end{array}$

Hence, the dimension of equation (a) is equal to the dimension of the area of the curved surface.

Step 3: Find the dimensions of equation (b)

The equation given is $\text{2}\pi \left({\text{r}}_{1}{\text{+r}}_{2}\right).$

In the above formula, $\pi$ has no dimension.

The dimensionsof ${\text{r}}_{1}$, ${\text{r}}_{2},$ and $h$ is $L.$

Replace $L.$ for ${\text{r}}_{1}$ , and ${\text{r}}_{2}$ in the above equation to get its dimension.

$\begin{array}{c}\text{Dimenstion =}\left(L\text{\hspace{0.17em}+ L}\right)\\ =\text{\hspace{0.17em}}L.\end{array}$

Hence, the dimension of equation (b) is equal to the dimension of the circumference of the flat circular base.

Step 4: Find the dimensions of equation (c)

$\pi h\left({{\text{r}}_{1}}^{2}+{\text{r}}_{1}{\text{r}}_{2}+{{\text{r}}_{2}}^{2}\right)/3$Consider the given equation as shown below.

In the above formula, $\pi$ has no dimension.

The dimensionsof ${\text{r}}_{1}$, ${\text{r}}_{2},$ and $h$ is $L$

Replace for ,${\text{r}}_{1}$ ${\text{r}}_{2},$ and $h$ in the above equation to get its dimension.

$\begin{array}{c}Dimension=L\left({L}^{2}+LL+{L}^{2}\right)\\ =L\left({L}^{2}\right)\\ ={L}^{3}.\end{array}$

Hence, the dimension of the equation (c) is equal to the dimension of the volume.