Q 11-04-190JEE MainJEE Main 2019 (11 Jan, Shift 1)Medium
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. $25\%$ loses all of its momentum and $25\%$ comes back with the same speed. The resultant pressure on the mesh will be:
Answer: (B) $\dfrac34\rho v^2$
Mass hitting the mesh per second $= \rho Av$, where $A = \pi a^2$.
The part that stops ($25\%$) loses momentum $v$ per unit mass; the part that rebounds ($25\%$) changes momentum by $2v$ per unit mass; the rest is unaffected.
$$F = 0.25\rho Av\cdot v + 0.25\rho Av\cdot 2v = \frac34\rho Av^2$$
$$P = \frac FA = \frac34\rho v^2$$
Solution by Sreeraj P, M.Sc Physics