A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 12m/s over the top of the plate only, it is possible to keep the plate in a horizontal position, as illustrated in part (a) of the drawing. To what value of speed (in m/s) should the air be blown so that the plate is kept at ${30}^{\circ }$angle with respect to the vertical, as in part (b) of the drawing? (assume the velocity beneath the plate is almost zero)

Case (I)

Due to moving air at the top of the plate, pressure at the top is less than that at the bottom (due to Bernoulli’s theorem)
$\Delta p=\frac{1}{2}\rho v^2$
$F=\Delta p.A$(Where A is area of the plate;  F acts at the midpoint of the plate and directed upwards)
Mg acts as the midpoint of the plate and directed downwards.
$\therefore$ For equilibrium, balance moments about the hinge. 
$\frac{1}{2}\rho v^2.A.\frac{L}{2}=Mg.\frac{L}{2}\mathrm{......}\left(1\right)$
 

Case (II)

$F\text{'}=\frac{1}{2}\rho {\left(v\text{'}\right)}^{2}A$
$\therefore$ For equilibrium, balance moments about the hinge.
$\frac{1}{2}\rho {\left(v\text{'}\right)}^2.A.\frac{L}{2}=Mg.\frac{L}{2}\sin {30}^{\circ }\mathrm{.......}\left(2\right)$
$\begin{array}{l}\text{Dividing} \frac{\left(2\right)}{\left(1\right)}, \text{we get}\\ \Rightarrow v\text{'}=\frac{v}{\sqrt{2}}=\frac{12}{\sqrt{2}}=6\sqrt{2}{\text{\hspace{0.17em}ms}}^{\text{-1}}\end{array}$
Therefore, the correct answer is  (B)