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
JEE MAIN 2019, Diffcult
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$F = \frac{1}{4} \times \rho A{v^2} + \frac{1}{4}2\rho a{v^2}$
$ \Rightarrow \frac{F}{A} = \frac{3}{4}\rho A{v^2} = \rho $
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