91Ó°ÊÓ

1. The depth of the water in the tank, $\mathit{D}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{30}\mathit{m}$.

2. The cross-sectional area of the hole,role="math" localid="1657603615037" ${\mathit{A}}_{\mathbf{n}}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{5}\mathit{c}{\mathit{m}}^{\mathbf{2}}$.

3. The cross-sectional area of the stream = $1/2$cross-sectional area of the hole.

Find the velocity of the water flowing through the hole vh using Bernoulli’s principle. Then using the continuity equation, find the drainage rate of flow of water through the hole. Then, find the velocity of the water coming out of the hole vh using the continuity equation. Then, using the kinematic equation of motion, find the distance below the bottom of the tank at which the cross-sectional area of the stream is equal to the cross-sectional area of the hole. According to Bernoulli’s equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

Formulae are as follows:

1. $\mathit{p}\mathit{v}\mathbf{+}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{}\mathit{p}{{\mathit{g}}^{\mathbf{2}}}_{\mathbf{h}}\mathbf{+}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}$

2. $A_v$= constant

Where p is pressure, v is velocity, h is height, g is the acceleration due to gravity, h is height, and A is area and $P$is density.

The water is flowing out through the hole. Hence, the flow should obey the continuity equation.

Let,

$$\begin{gathered} {\mathit{A}}_{\mathbf{t}}=Thecross-sectionalareaofthe\tan k, \\ {\mathit{V}}_{\mathbf{t}}=Thevelocityofthewaterflowinginthe\tan k, \\ {\mathit{A}}_{\mathbf{h}}=Areaofthecross-sectionofthehole, \\ \mathbf{}{\mathit{V}}_{\mathbf{h}}=Thevelocityofthewaterflowingthroughthehole. \end{gathered}$$

So, according to the continuity equation,

$A_t{v}_t=A_h{V}_h$

Since $A_t$is very large compared to $A_{h,}{v}_t=\frac{A_h{v}_h}{A_t}$A will be very small as compared to $V_h$.

Hence, in further discussion, neglect the term $V_t$. The water flow follows Bernoulli’s principle. Consider the hole level as the reference level. Now, both the top of the tank and the hole are exposed to the atmosphere. The pressure P0 is the same for both.

Hence, the equation is,

$p_{0}v+1/2pgD+p=p_{0}v+1/2p{g^2}_h+\left(0\right)$

Simplifying,

$1/2{v^2}_t-gD1/2{v^2}_h$

And, since $v_{t^{}}$is negligible compared to$v_h$,

$$\begin{gathered} {V^2}_h=2gD \\ {V^2}_h=2×9.8×0.30 \\ {v}_h=2.42m/s \end{gathered}$$

Hence, the rate of flow of water is,

localid="1657604554074" $$\begin{gathered} A_{\mathrm{h}}{V}_{\mathrm{h}}=6.5×{10}^{-4}{m}^2×2.42m/s \\ =1.57×{10}^{-3}{m}^3/s \\ =1.6×{10}^{-3}{m}^3/s \end{gathered}$$

Hence, the drainage rate of flow of water through the hole is$1.6×{10}^3{m}^3/s$

The stream of water, after coming out of the hole, obeys the continuity equation. Hence,

${\mathit{A}}_{\mathbf{h}}{\mathit{v}}_{\mathbf{h}}\mathbf{=}{\mathit{A}}_{\mathbf{s}}{\mathit{v}}_{\mathbf{s}}$,

WhereAsandvsdenote the cross-sectional area and the speed of the stream, respectively.

So,
$$\begin{gathered} {\mathit{v}}_{\mathbf{s}}\mathbf{=}\frac{{\mathbf{A}}_{\mathbf{h}}}{{\mathbf{A}}_{\mathbf{s}}}{\mathit{v}}_{\mathbf{h}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}{\mathit{v}}_{\mathbf{h}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{×}\mathbf{2}\mathbf{.}\mathbf{42} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{84}\mathit{m}\mathbf{/}\mathit{s} \end{gathered}$$

The stream of water, after coming out of the hole, undergoes free-fall motion under gravity. Hence, use the kinematical equation to determine the speed of the stream.

$$\begin{gathered} {{\mathit{v}}^{\mathbf{2}}}_{\mathbf{s}}\mathbf{=}{{\mathit{v}}^{\mathbf{2}}}_{\mathbf{h}}\mathbf{+}\mathbf{2}\mathit{g}\mathit{h} \\ \mathit{h}\mathbf{=}\frac{{{\mathbf{v}}^{\mathbf{2}}}_{\mathbf{s}}\mathbf{-}{{\mathbf{v}}^{\mathbf{2}}}_{\mathbf{h}}}{\mathbf{2}\mathbf{g}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{84}\mathbf{)}}^{\mathbf{2}}\mathbf{-}{\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{42}\mathbf{)}}^{\mathbf{2}}}{\mathbf{2}\mathbf{×}\mathbf{9}\mathbf{.}\mathbf{8}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{17}\mathbf{.}\mathbf{56}}{\mathbf{2}\mathbf{×}\mathbf{9}\mathbf{.}\mathbf{8}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{98}\mathit{m} \end{gathered}$$

Hence, the distance below the bottom of the tank at which the cross-sectional area of the stream is equal to the cross-section area of the hole is$0.09m$