JEE PYQ: Mechanical Properties Of Fluids Question 18
Question 18 - 2020 (07 Jan 2020 Shift 2)
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:
(1) $\frac{9}{16}$
(2) $\frac{\sqrt{3}}{2}$
(3) $\frac{3}{4}$
(4) $\frac{81}{256}$
Show Answer
Answer: (1) $\frac{9}{16}$
Solution
From the equation of continuity: $A_1 v_1 = A_2 v_2$ Here, $v_1$ and $v_2$ are the velocities at two ends of pipe. $A_1$ and $A_2$ are the area of pipe at two ends. $\Rightarrow \frac{v_1}{v_2} = \frac{A_2}{A_1} = \frac{\pi(4.8)^2}{\pi(6.4)^2} = \frac{9}{16}$
BETA
