Question

A thin glass plate of area $A$ and mass $m$ is hinged along one of its sides. The speed with which the air should be blown parallel to its upper surface to hold the plate horizontal is $\gamma \sqrt{\frac{mg}{18\rho A}}$.
If the density of air is $\rho$. Then the value of $\gamma$ is

Answer 1

On Applying Bernoulli's theorem for above the plate $\left(2\right)$ and below the plate $\left(1\right)$, we have
$p_1=p_2+\frac{1}{2}\rho v^2\Rightarrow p_1-p_2=\frac{1}{2}\rho v^2$
Upward force is
$(p_1-p_2)A=\frac{1}{2}\rho v^{2}A\therefore \frac{1}{2}A\rho v^2=mg\Rightarrow v=\sqrt{\frac{2mg}{\rho A}}=6\sqrt{\frac{mg}{18\rho A}}\Rightarrow \gamma =6$