$\rho_0 T_0=\rho T$

$\Rightarrow 1.2 \times 300=\rho(360) \therefore \rho=1$

Between $A$ & $B$

$\mathrm{P}_0+\frac{1}{2} \rho \mathrm{V}_0^2=\mathrm{P}+\frac{1}{2} \rho^2+\rho \mathrm{gh}$

$\frac{\pi \mathrm{D}^2}{4} \mathrm{~V}_0=\frac{\pi \mathrm{d}^2}{4} \mathrm{~V} \ldots \ldots \ldots \text { (2) }$. . . . .

Between $B$ & $C$

$\mathrm{P}+\frac{1}{2} \rho^2=\mathrm{P}_0-\rho_0 \mathrm{~g}(\mathrm{H}+\mathrm{h})+\frac{1}{2} \rho^2 \text {. }$

from $(1)$ & $(2)$ :

$\Rightarrow P_0+\frac{1}{2} \rho\left(V \frac{d^2}{D^2}\right)^2=P+\frac{1}{2} \rho^2+\rho g h$

$\Rightarrow \rho_0 g(H+h)=\frac{1}{2} \rho V^2\left[1-\frac{d^4}{D^4}\right]+\rho g h$

$\Rightarrow V^2 \simeq \frac{2 \rho_0}{\rho} g(H+h)-2 g h$

$=2 \times 1.2 \times 10 \times 10-2 \times 10 \times 9$

$=240-180=60 \quad \therefore V=\sqrt{60 m} \mathrm{~m} / \mathrm{s}$

$Q_m=\rho \frac{\pi d^2}{4} V=1 \times \frac{\pi}{4} \times 10^{-2} \times \sqrt{60}=60.80$

(image)

$\mathrm{P}=\text { constant }$

$\Rightarrow \rho_2 \mathrm{~T}_2=\rho \mathrm{T}$

$1.2 \times 300=\rho \times 360$

$\rho=1 \mathrm{~kg} / \mathrm{m}^3$

$\Delta \mathrm{P}=\rho_2 \mathrm{~g}(\mathrm{~h}+\mathrm{H})-\rho \mathrm{gh}$

$=1.2 \times 10 \times 10-1 \times 10 \times 9$

$=120-90=30 \mathrm{~N} / \mathrm{m}^2$