An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is:

First, we recall the continuity equation for an ideal (incompressible) fluid flowing steadily through a pipe: $$A\,v=\text{constant},$$ where $$A$$ is the cross-sectional area of the pipe at any section and $$v$$ is the fluid speed at that section.

For a circular pipe the area is given by the well-known geometric formula $$A=\dfrac{\pi d^{2}}{4},$$ because the radius is $$d/2$$ and the area of a circle is $$\pi r^{2}.$$

Let us denote:

• $$d_{\text{max}} = 6.4\ \text{cm}$$ (the largest diameter),
• $$d_{\text{min}} = 4.8\ \text{cm}$$ (the smallest diameter),
• $$v_{\text{min}}$$ = fluid speed in the wider section (corresponding to $$d_{\text{max}}$$),
• $$v_{\text{max}}$$ = fluid speed in the narrower section (corresponding to $$d_{\text{min}}$$).

Using continuity, we write

$$A_{\text{max}}\,v_{\text{min}} = A_{\text{min}}\,v_{\text{max}}.$$

Substituting the area formula $$A=\dfrac{\pi d^{2}}{4}$$ into each term gives

$$\left(\dfrac{\pi d_{\text{max}}^{2}}{4}\right)v_{\text{min}} = \left(\dfrac{\pi d_{\text{min}}^{2}}{4}\right)v_{\text{max}}.$$

The common factors $$\pi/4$$ cancel out, leaving

$$d_{\text{max}}^{2}\,v_{\text{min}} = d_{\text{min}}^{2}\,v_{\text{max}}.$$

We need the ratio $$\dfrac{v_{\text{min}}}{v_{\text{max}}}.$$ Rearranging the previous relation, we obtain

$$\dfrac{v_{\text{min}}}{v_{\text{max}}} = \dfrac{d_{\text{min}}^{2}}{d_{\text{max}}^{2}}.$$

Now we substitute the given diameters:

$$\dfrac{v_{\text{min}}}{v_{\text{max}}} = \left(\dfrac{d_{\text{min}}}{d_{\text{max}}}\right)^{2} = \left(\dfrac{4.8\ \text{cm}}{6.4\ \text{cm}}\right)^{2}.$$

First simplify the fraction inside the brackets:

$$\dfrac{4.8}{6.4} = \dfrac{48}{64} = \dfrac{3}{4}.$$

Squaring this result gives

$$\left(\dfrac{3}{4}\right)^{2} = \dfrac{9}{16}.$$

Thus, the required ratio of the minimum velocity to the maximum velocity is

$$\dfrac{v_{\text{min}}}{v_{\text{max}}} = \dfrac{9}{16}.$$

Hence, the correct answer is Option A.