JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Fluid Mechanics

For an incompressible fluid the continuity equation $A_1{v}_1=A_2{v}_2$ holds, and since the cross-sectional area goes as the square of the diameter, the fluid moves slowest where the pipe is widest and fastest where it is narrowest. The ratio of the minimum to the maximum speed is therefore the ratio of the smaller area to the larger, $\frac{v_{\min }}{v_{\max }}={\left(\frac{d_{\min }}{d_{\max }}\right)}^2={\left(\frac{4.8}{6.4}\right)}^2={\left(\frac{3}{4}\right)}^2=\frac{9}{16}$.