Q 11-09-128JEE MainJEE Main 2020 (7 Jan, Shift 2)Easy
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 fluid in this pipe is:
Answer: (A) $\dfrac{9}{16}$
Continuity: $Av$ = constant, so $v \propto \dfrac{1}{d^2}$. The speed is least where the pipe is widest:
$$\frac{v_{\min}}{v_{\max}} = \left(\frac{d_{\min}}{d_{\max}}\right)^2 = \left(\frac{4.8}{6.4}\right)^2 = \frac{9}{16}$$
Solution by Sreeraj P, M.Sc Physics