91Ó°ÊÓ

The term speed may be defined as the ratio of distance and time.

Consider the given data as below.

Density of water, $\mathrm{p}=1000\mathrm{kg}/{\mathrm{m}}^{3^{}}$

Pressure, ${\mathrm{p}}_1=1000\mathrm{kg}/{\mathrm{m}}^3$

Pressure, ${\mathrm{p}}_1=4.2×{10}^5\mathrm{Pa}$

Height, ${\mathrm{h}}_1=3.5\mathrm{m}$

Height, ${\mathrm{h}}_2=1\mathrm{m}$

Gravity, $\mathrm{g}=2.981\mathrm{m}/{\mathrm{s}}^2$

Using the Bernoulli’s equation

localid="1668311202703" $$\begin{gathered} {\mathrm{p}}_1+{\mathrm{Òϲ\mu ³ó}}_1+\frac{1}{2}{\mathrm{ÒÏ\pm ¹}}_1^2={\mathrm{p}}_2+{\mathrm{Òϲ\mu ³ó}}_2+\frac{1}{2}{\mathrm{ÒÏ\pm ¹}}_2^2 \\ \mathrm{And}\mathrm{the} \\ {\mathrm{v}}_2=\sqrt{\frac{2}{\mathrm{ÒÏ}}â‹\dots \left({\mathrm{p}}_1−{\mathrm{p}}_2\right)+2\mathrm{g}\left({\mathrm{h}}_1−{\mathrm{h}}_2\right)} \\ \mathrm{Put}\mathrm{the}\mathrm{given}\mathrm{values} \\ {\mathrm{v}}_2=\sqrt{\frac{2}{1000}â‹\dots \left(4.2×{10}^5−1×{10}^5\right)+2.981(3.5−1)} \\ =26.2\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed out of the hose is ${\mathrm{v}}_2=26.2\mathrm{m}/\mathrm{s}$.

When the pressure increases the volume also increase so

$\begin{array}{r}\mathrm{p}\left(4−{\mathrm{h}}_2\right)\mathrm{A}={\mathrm{p}}_1\left(4−{\mathrm{h}}_1\right)\mathrm{A}\\ \mathrm{p}=\frac{{\mathrm{p}}_1\left(4−{\mathrm{h}}_1\right)}{4−{\mathrm{h}}_2}\\ \mathrm{The}\mathrm{pressure}\mathrm{can}\mathrm{be}\mathrm{calculated}\mathrm{as}\\ \mathrm{p}=4.2×{10}^5×\frac{4−3.5}{4−2}\\ =1.05×{10}^5\mathrm{Pa}\end{array}$

Now apply Bernoulli equation

localid="1668311230158" $$\begin{gathered} v_2=\sqrt{\frac{2}{\mathrm{ÒÏ}}â‹\dots \left(p−p_1\right)+2g\left(h_1−h_2\right)} \\ {v}_2=\sqrt{\frac{2}{1000}\left(1.05×{10}^5−1×{10}^5\right)+2.981(3.5−1)} \\ =5.44\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of flow for h = 3.00 m and forh = 2.00 mis${\mathrm{v}}_2=5.44\mathrm{m}/\mathrm{s}$.

Now consider $v_2=0$

From the Bernoulli equation

localid="1668311326821" $$\begin{gathered} \sqrt{\frac{2}{\mathrm{p}}\left({\mathrm{p}}_1−{\mathrm{p}}_2\right)+2\mathrm{g}\left({\mathrm{h}}_1−{\mathrm{h}}_2\right)}=0 \\ \mathrm{Put}\mathrm{the}\mathrm{given}\mathrm{values} \\ 4.2×{10}^5\frac{0.5}{4−\mathrm{h}}=9.81×1000(1−\mathrm{h}) \\ 9.8{\mathrm{h}}^2=149×\mathrm{h}+229.2 \\ \mathrm{h}=\frac{1}{2}\left(15.2Â\pm \sqrt{(15−2{)}^2−4.23.39}\right) \\ =(7.6Â\pm 5.86)\mathrm{m} \end{gathered}$$

But h < 4m you can conclude that right value of the height is h = 1.74 m .