JEE PYQ: Mechanical Properties Of Fluids Question 32
Question 32 - 2019 (11 Jan 2019 Shift 1)
A liquid of density $\rho$ is coming out of a hose pipe of radius $a$ with horizontal speed $v$ and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:
(1) $\frac{1}{4}\rho v^2$
(2) $\frac{3}{4}\rho v^2$
(3) $\frac{1}{2}\rho v^2$
(4) $\rho v^2$
Show Answer
Answer: (2) $\frac{3}{4}\rho v^2$
Solution
Mass per unit time of the liquid = $\rho a v$ Momentum per second carried by liquid: Net force due to bounced back liquid, $F_1 = 2 \times \frac{1}{4}\rho a v^2$ Net force due to stopped liquid, $F_2 = \frac{1}{4}\rho a v^2$ Total force, $F = F_1 + F_2 = \frac{1}{2}\rho a v^2 + \frac{1}{4}\rho a v^2 = \frac{3}{4}\rho a v^2$ Net pressure $= \frac{3}{4}\rho v^2$
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