A rectangular tank of height 10m filled with water, is placed near the bottom of a plane inclined at angle $30°$  with horizontal. At height h from the bottom a small hole is made (see figure) such that the stream coming from the hole, strikes the inclined plane normally. The value of h is $\frac{10x}{6}m$ . Then the value of x is

Let any time t fluid fills the portion BC and comes out with velocity v.
 $\mathcal{l}=1m,x=0.4m$
In rotating frame, centrifugal force on portion A ‘B.
 $F=\rho A\underset{ }{\overset{ }{\int }}{\omega }^{2}r.dr=\frac{\rho A{\omega }^2({\mathcal{l}}^2-x^2)}{ }$
Due to pressure difference between C and A’
 $\Delta P=\frac{1}{2}{\omega }^2({\mathcal{l}}^2-x^2-)$
Now  $\frac{1}{2}\rho V^2=\Delta P=\frac{\rho {\omega }^2}{2}({\mathcal{l}}^2-x^2)-\rho gx\mathrm{...}\left(1\right)$
From Bernoulli theorem,
Let it reaches upto a height h, then
 $$\begin{gathered} pgh=\frac{1}{2}p{v}^2=\frac{\rho {\omega }^2({\mathcal{l}}^2-x^2)}{2}-\rho gx \\ \rho \times 10\times h=\rho \times \frac{{\left(10\right)}^2}{2}(1^2-{\left(4\right)}^2)-4\times 10\times \rho \Rightarrow h=3.8m \end{gathered}$$