JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Conservation of Momentum

The thrust on the mesh is the rate at which momentum is delivered to it. The mass arriving each second is $\dot{m}=\rho Av$. Half the liquid passes straight through and delivers no momentum; a quarter is stopped, delivering momentum at the rate $\frac{1}{4}\dot{m}v$; and a quarter rebounds at the same speed, reversing its momentum and delivering $\frac{1}{4}\dot{m}(2v)$. The total force is $F=\frac{1}{4}\rho Avv+\frac{1}{4}\rho Av(2v)=\frac{3}{4}\rho A{v}^2$, so the pressure is $F/A=\frac{3}{4}\rho v^2$.