91Ó°ÊÓ

The speed of an object is defined as the rate with which the distance of an object changes in unit time. The speed of the object is a scalar quantity.

Given data.

The diameter of the audio CD is $d = 12.0 \mathrm{cm}$.

The starting radius of the spiraling path is $r_1= 2.5 \mathrm{cm}$.

The ending radius of the spiraling path is $r_2= 5.8 \mathrm{cm}$.

The width of the spiraling path or the two spiral lines is $w = 1.6 × {10}^{-6} \mathrm{m}$.

The speed of the laser is $v = 1.2 \frac{\mathrm{m}}{\mathrm{s}}$.

Assumption.

Let the total length of the spiral path be $l$.

Let the maximum playing time for such a CD be $t$.

(a) Now, the total area inside the spiraling path is.

$\begin{array}{c}wl=\mathrm{π}{r}_2^2-\mathrm{π}{r}_1^2\\ l=\frac{\mathrm{π}\left(r_2^2-r_1^2\right)}{1.6×{10}^{-6} \mathrm{m}}\\ l=\frac{\mathrm{π}\left({\left(0.058\mathrm{m}\right)}^2-{\left(0.025\mathrm{m}\right)}^2\right)}{1.6×{10}^{-6} \mathrm{m}}\\ l=5.4\mathrm{km}\end{array}$

Therefore, the total length of the spiral is 5.4 km.

(b) Now, the maximum time for playing such a CD is

$\begin{array}{c}t = \frac{l}{v}\\ = \frac{5.4×{10}^3 \mathrm{m}}{1.2 \frac{\mathrm{m}}{\mathrm{s}}}\\ = 4482 \mathrm{s} ×\frac{1 \min }{60 \mathrm{s}}\\ = 75 \min \end{array}$

Therefore, the maximum time is 75 minutes for playing such a CD.