Physics - Fluid Question with Solution | TestHub

Applying Bernoulli's equation between P and Q

$$\begin{array}{rlrl}& {\mathrm{p}}_0+\rho \mathrm{gh}+\frac{1}{2}\rho {\mathrm{v}}_1^2=& \rho \mathrm{gh}=\frac{\rho }{2}\left({\mathrm{v}}_2^2-{\mathrm{v}}_1^2\right)& {\mathrm{v}}_2^2-{\mathrm{v}}_1^2=2\mathrm{gh}\end{array}$$

From equation of continuity,

$$\begin{array}{rlrlr}& {\mathrm{Av}}_1={\mathrm{av}}_2=\frac{\mathrm{A}}{3}{\mathrm{v}}_2\Rightarrow & {\mathrm{v}}_2=3{\mathrm{v}}_1& 9{\mathrm{v}}_1^2-{\mathrm{v}}_1^2=2\mathrm{gh}& 8{\mathrm{v}}_1^2=2\mathrm{gh}\end{array}$$

$$\Rightarrow {\mathrm{v}}_1=\sqrt{\frac{\mathrm{gh}}{4}}=\frac{\sqrt{\mathrm{gh}}}{2}$$

Acceleration of top layer

$$\begin{array}{rlr}& \mathrm{a}=\frac{{\mathrm{dv}}_1}{\mathrm{dt}}=\frac{1}{2}\sqrt{\mathrm{g}}\times \frac{1}{2\sqrt{\mathrm{h}}}\times \frac{\mathrm{dh}}{\mathrm{dt}}& =\frac{1}{4}\sqrt{\frac{\mathrm{g}}{\mathrm{h}}}\times \left(-{\mathrm{v}}_1\right)=-\frac{1}{4}\sqrt{\frac{\mathrm{g}}{\mathrm{h}}}\times \frac{\sqrt{\mathrm{gh}}}{2}=-\frac{\mathrm{g}}{8}\mathrm{m}/{\mathrm{s}}^2\end{array}$$