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Step-by-Step Solution
Step 1: Identify the Relevant Physical Principle
For an ideal (incompressible) fluid in a steady (laminar) flow through a pipe, the equation of continuity applies. Mathematically, it is stated as
$A_1 V_1 = A_2 V_2,$
where
$A_1$ and $A_2$
are the cross-sectional areas of the pipe at two different points, and
$V_1$ and $V_2$
are the corresponding fluid velocities.
Step 2: Expressing Velocities in Terms of Cross-sectional Areas
From the equation of continuity:
$V_1 / V_2 = A_2 / A_1.$
Step 3: Relate Cross-sectional Areas to Diameters
The cross-sectional area $A$ of a circular pipe is proportional to the square of its diameter $d,$ specifically
$A = \pi \times \left(\dfrac{d}{2}\right)^2.$
Because the factor of $\pi/4$ is common and cancels when taking ratios, one can write:
$A_1 / A_2 = (d_1 / d_2)^2.$
Step 4: Substitute the Given Diameters
The maximum diameter is 6.4 cm (call this $d_\text{max}$), and the minimum diameter is 4.8 cm (call this $d_\text{min}$). Here, we want the ratio of the minimum velocity to the maximum velocity, i.e.
$\dfrac{V_\text{min}}{V_\text{max}} = \dfrac{A_\text{max}}{A_\text{min}} = \left(\dfrac{d_\text{max}}{d_\text{min}}\right)^2 = \left(\dfrac{6.4}{4.8}\right)^2.$
Step 5: Calculate the Numerical Value
Compute:
$\left(\dfrac{6.4}{4.8}\right)^2 = \left(\dfrac{64}{48}\right)^2 = \left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9}.$
However, the question specifically asks for the ratio of minimum and maximum velocities. By continuity, if the diameter is smaller, the velocity is higher, and vice versa. Hence for the ratio $\dfrac{V_\text{min}}{V_\text{max}},$ we actually use
$\dfrac{d_\text{min}}{d_\text{max}} = \dfrac{4.8}{6.4}.$
Thus:
$\dfrac{V_\text{min}}{V_\text{max}} = \left(\dfrac{4.8}{6.4}\right)^2 = \left(\dfrac{48}{64}\right)^2 = \left(\dfrac{3}{4}\right)^2 = \dfrac{9}{16}.$
Final Answer
The ratio of the minimum to the maximum velocities is
$\dfrac{9}{16}.$
