Q. 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 :

Solution:

Momentum per second carried by liquid per second is $\rho av^2$
net force due to reflected liquid = $2 \times [ \frac{1}{4} \rho av^2 ]$
net force due to stopped liquid = $\frac{1}{4} \rho av^2$
Total force = $\frac{3}{4} \rho av^2$
net pressure = $\frac{3}{4} \rho v^2$

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions