Chapter 20, Problem 47E

(a)

To determine

The number of bits per centimeter stored on a track.

Answer to Problem 47E

The number of bits stored per centimeter is $1592\text{bits}\cdot {\text{cm}}^{\text{-1}}$.

Explanation of Solution

Write the expression for the number of bits stored on a track.

    $n=\frac{U}{l}$        (1)

Here, $n$ is the number of bits, $U$ is the transfer rate and $l$ is the length covered by the track.

Write the expression for the length of the track.

    $l=\omega r$        (2)

Here, $\omega$ is the spin rate and $r$ is the radius of the track.

Conclusion:

Substitute $300\text{rev}{\text{min}}^{\text{-1}}$ for $\omega$ and $5\text{cm}$ for $r$ in equation (2).

    $\begin{array}{c}l=\left(300\text{rev}{\text{min}}^{\text{-1}}\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1\min }{60\text{s}}\right)\right)\left(5\text{cm}\right)\\ =50\pi \text{cm}{\text{s}}^{\text{-1}}\end{array}$

Substitute $250\text{kilobits}{\text{s}}^{\text{-1}}$ for $U$ and $50\pi \text{cm}{\text{s}}^{-1}$ for $l$ in equation (1).

    $\begin{array}{c}n=\frac{250\text{kilobits}{\text{s}}^{\text{-1}}\left(\frac{{10}^3\text{bits}}{1\text{kilobits}}\right)}{50\pi \text{cm}{\text{s}}^{\text{-1}}}\\ =1592\text{bits}\cdot {\text{cm}}^{\text{-1}}\end{array}$

Thus, the number of bits stored per centimeter is $1592\text{bits}\cdot {\text{cm}}^{\text{-1}}$.

(a)

To determine

The number of bits stored per unit length of the track.

Answer to Problem 47E

The number of bits stored per centimeter is $4406\text{bits}\cdot {\text{cm}}^{\text{-1}}$.

Explanation of Solution

Write the expression for the number of bits stored on a track.

    $n=\frac{U}{l}$

Write the expression for the length of the track.

    $l=\omega r$

Conclusion:

Substitute $3633\text{rev}{\text{min}}^{\text{-1}}$ for $\omega$ and $5\text{cm}$ for $r$ in equation (2).

    $\begin{array}{c}l=\left(3633\text{rev}{\text{min}}^{\text{-1}}\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1\min }{60\text{s}}\right)\right)\left(10.5\text{cm}\right)\\ =3994.7\text{cm}{\text{s}}^{\text{-1}}\end{array}$

Substitute $17.6\text{megabits}{\text{s}}^{\text{-1}}$ for $U$ and $3994.7\text{cm}{\text{s}}^{\text{-1}}$ for $l$ in equation (1).

    $\begin{array}{c}n=\frac{17.6\text{megabits}{\text{s}}^{\text{-1}}\left(\frac{{10}^6\text{bits}}{1\text{megabits}}\right)}{3994.7\text{cm}{\text{s}}^{\text{-1}}}\\ =4406\text{bits}\cdot {\text{cm}}^{\text{-1}}\end{array}$

Thus, the number of bits stored per centimeter is $4406\text{bits}\cdot {\text{cm}}^{\text{-1}}$.