JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 4)
A liquid of density $$\rho $$ is coming out of a hose pipe of radius a with horizontal speed $$\upsilon $$ 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 :
$${3 \over 4}\rho {v^2}$$
$${1 \over 4}\rho {v^2}$$
$${1 \over 2}\rho {v^2}$$
$$\rho {v^2}$$
Explanation
Momentum per second carried by liquid per second is $$\rho $$av2
net force due to reflected liquid = 2$$ \times $$$$\left[ {{1 \over 4}\rho a{v^2}} \right]$$
net force due to stopped liquid = $${{1 \over 4}\rho a{v^2}}$$
Total force = $${{3 \over 4}\rho a{v^2}}$$
net pressure = $${{3 \over 4}\rho {v^2}}$$
net force due to reflected liquid = 2$$ \times $$$$\left[ {{1 \over 4}\rho a{v^2}} \right]$$
net force due to stopped liquid = $${{1 \over 4}\rho a{v^2}}$$
Total force = $${{3 \over 4}\rho a{v^2}}$$
net pressure = $${{3 \over 4}\rho {v^2}}$$
Comments (0)
