Q.
A liquid enter at point A1 with speed 3.5 m/s and leaves at point A2. Then find out the height attained by the liquid above point A2.
AIIMSAIIMS 2019
Solution:
A1V1 = A2V2
A1 = A2$\quad$$\quad$ $\Rightarrow $$\quad$$\quad$ V1 = V2 = 3.5 m/s
Maximum height achieved
Using Bernoulli theorem
$P_{atm}+\frac{1}{2}\rho\left(3.5\right)^{2}+\rho g.0=P_{atm}+\frac{1}{2}\left(0^{2}\right)+\rho gh$
$H=\frac{v^{2}}{2g}=\frac{3.5\times3.5}{20}=\frac{12.25}{20}=0.6125 m=61.25 cm$
