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Q10P

Expert-verifiedFound in: Page 15

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)**** ${\pi}\left({\text{r}}_{1}{\text{+r}}_{2}\right){[{h}^{2}+{({\text{r}}_{2}-{\text{r}}_{1})}^{2}]}^{1/2}$****, ${\pi}\left({\text{r}}_{1}{\text{+r}}_{2}\right){[{h}^{2}+{({\text{r}}_{2}-{\text{r}}_{1})}^{2}]}^{1/2}$(b)** **and(c) ${\pi}{h}({{\text{r}}_{1}}^{2}+{\text{r}}_{1}{\text{r}}_{2}+{{\text{r}}_{2}}^{2}){/}{3}$**** ****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)

**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 $A=\pi r(r+\sqrt{{h}^{2}+{r}^{2}}),$** ** and its dimension is ** ${L}^{2}.$**

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

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.

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

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

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.

$\pi h({{\text{r}}_{1}}^{2}+{\text{r}}_{1}{\text{r}}_{2}+{{\text{r}}_{2}}^{2})/3$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.

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

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

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