An ideal fluid flows (laminar flow) through a | vedclass

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(D) According to the equation of continuity for an ideal fluid,the product of the cross-sectional area and the velocity of the fluid remains constant:

Vedclass · Accepted Answer

$\frac{9}{16}$

Vedclass · Answer

(D) According to the equation of continuity for an ideal fluid,the product of the cross-sectional area and the velocity of the fluid remains constant: $A_1 v_1 = A_2 v_2$.This implies that the velocity is inversely proportional to the area: $v \propto \frac{1}{A}$.Since the area $A = \pi r^2 = \pi (d/2)^2$,we have $A \propto d^2$,where $d$ is the diameter.Therefore,the ratio of the minimum velocity to the maximum velocity is given by the ratio of the minimum area to the maximum area:$\frac{v_{\min}}{v_{\max}} = \frac{A_{\min}}{A_{\max}} = \left( \frac{d_{\min}}{d_{\max}} \right)^2$.Substituting the given values $d_{\min} = 4.8 \; cm$ and $d_{\max} = 6.4 \; cm$:$\frac{v_{\min}}{v_{\max}} = \left( \frac{4.8}{6.4} \right)^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16}$.

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

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