Chapter 1, Problem 72P

(a)

To determine

The diameter of a viroid in $\text{m}$ .

Answer to Problem 72P

The diameter of a viroid in $\text{m}$ is $3.3\times {10}^{-8}\text{ m}$ .

Explanation of Solution

The viroid can be considered as circular loop containing $300$ bases as beads. It is given that the distance from one base to the next is $0.35\text{ nm}$ .

Find the circumference of the viroid.

$\begin{array}{c}C=300\times 0.35\text{ nm}\left(\frac{1\text{ m}}{{10}^9\text{ nm}}\right)\\ =300\times \left(0.35\times {10}^{-9}\text{ m}\right)\\ =1.05\times {10}^{-7}\text{ m}\end{array}$

Here, $C$ is the circumference of the circle

Write the equation for the circumference of a circle.

$C=\pi d$

Here, $d$ is the diameter of the circle

Rewrite the above equation for $d$ .

$d=\frac{C}{\pi }$ (I)

Conclusion:

Substitute $1.05\times {10}^{-7}\text{ m}$ for $C$ in equation (I) to find the value of $d$ .

$\begin{array}{c}d=\frac{1.05\times {10}^{-7}\text{ m}}{\pi }\\ =3.3\times {10}^{-8}\text{ m}\end{array}$

Therefore, the diameter of a viroid in $\text{m}$ is $3.3\times {10}^{-8}\text{ m}$ .

(b)

To determine

The diameter of a viroid in $\mu \text{m}$ .

Answer to Problem 72P

The diameter of a viroid in $\mu \text{m}$ is $3.3\times {10}^{-2}\mu \text{m}$ .

Explanation of Solution

Write the expression to convert distance in $\text{m}$ into $\mu \text{m}$ .

$d_{\left(\mu \text{m}\right)}=d_{\left(\text{m}\right)}\frac{{10}^6\mu \text{m}}{1\text{ m}}$ (II)

Here, $d_{\left(\mu \text{m}\right)}$ is the quantity in $\mu \text{m}$ and $d_{\left(\text{m}\right)}$ is the quantity in $\text{m}$ .

Conclusion:

Substitute $3.3\times {10}^{-8}\text{ m}$ for $d_{\left(\text{m}\right)}$ in equation (II) to find $d_{\left(\mu \text{m}\right)}$ .

$\begin{array}{c}{d}_{\left(\mu \text{m}\right)}=3.3\times {10}^{-8}\text{ m}\frac{{10}^6\mu \text{m}}{1\text{ m}}\\ =3.3\times {10}^{-2}\text{ m}\end{array}$

Therefore, the diameter of a viroid in $\mu \text{m}$ is $3.3\times {10}^{-2}\mu \text{m}$ .

(c)

To determine

The diameter of a viroid in $\text{in}$ .

Answer to Problem 72P

The diameter of a viroid in $\text{in}$ is $1.3\times {10}^{-6}\text{ in}$ .

Explanation of Solution

Write the expression to convert distance in $\text{m}$ into $\text{in}$ .

$d_{\left(\text{in}\right)}=d_{\left(\text{m}\right)}\frac{39.37\text{ in}}{1\text{ m}}$ (III)

Here, $d_{\left(\text{in}\right)}$ is the quantity in $\text{in}$

Conclusion:

Substitute $3.3\times {10}^{-8}\text{ m}$ for $d_{\left(\text{m}\right)}$ in equation (III) to find $d_{\left(\text{in}\right)}$ .

$\begin{array}{c}{d}_{\left(\mu \text{m}\right)}=3.3\times {10}^{-8}\text{ m}\frac{39.37\text{ in}}{1\text{ m}}\\ =1.3\times {10}^{-6}\text{ in}\end{array}$

Therefore, the diameter of a viroid in $\text{in}$ is $1.3\times {10}^{-6}\text{ in}$ .