Question

A legendary Dutch boy saved Holland by plugging a hole of diameter 1.20 cm in a dike with his finger. If the hole was 2.00 m below the surface of the North Sea (density $\mathbf{1030}\mathbf{}\mathbf{}\mathit{k}\mathit{g}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$), (a) what was the force on his finger? (b) If he pulled his finger out of the hole, during what time interval would the released water fill 1 acre of land to a depth of 1 ft? Assume the hole remained constant in size.

Short answer

(a) The force on his finger is$F=11.45N$

(b) The time interval would the released water fill 1 acre of land to a depth of 1 ft is$t=1.74\times {10}^{6}s$

Step-by-step solution

Step 1:

The sum of the pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume has the same value at all points along a streamline for an ideal fluid. This result is summarized in Bernoulli’s equation:

$P+\frac{1}{2}\rho v^2+\rho gy=cons\tan t$

Step 2:

Part(a):

Asweknowthattheforceisgivenbytheformula:

$F=P\times A$

Where $A$area and $P$is the pressure

$\begin{array}{rcl}F& =& P_0\times A\\ F& =& 1.013\times {10}^5\times \pi r^2\\ F& =& 1.013\times {10}^5\times \frac{22}{7}\times {\left(0.006m\right)}^2\\ F& =& 11.45N\\ & & \end{array}$

Step 3:

Part(b):

Given:

Diameterofhole(v),$D=1.2cm$

Radius(r),$r=0.6cm$

Height(h),$h=y_2-y_1=2m$

Density of sea water:$1030kg/m^3$

Applying Bernoulli’s equation, we get

$\begin{array}{rcl}P+\frac{1}{2}\rho v^2+\rho gy& =& cons\tan t\\ P_1+\frac{1}{2}\rho {v_1}^2+\rho g{y}_1& =& P_2+\frac{1}{2}\rho {v_2}^2+\rho g{y}_2\\ \because P_1& =& P_2=P_0\\ & & \end{array}$

Also neglect velocity at sea level since large diameter than the diameter of hole:

$\begin{array}{rcl}{P}_0+\frac{1}{2}\rho {v_1}^2+\rho g{y}_1& =& P_0+0+\rho g{y}_2\\ \frac{1}{2}\rho {v_1}^2+\rho g{y}_1& =& \rho g{y}_2\\ \frac{1}{2}{v_1}^2+g{y}_1& =& g{y}_2\\ {v_1}^2& =& \rho g\left(y_2-y_1\right)\\ v_1& =& \sqrt{2g\left(y_2-y_1\right)}\\ v_1& =& \sqrt{2gh}\\ v_1& =& \sqrt{2\times 9.8\times 2}\\ v_1& =& 6.26m/s\\ & & \end{array}$

Step 4:

Let R be the volume flow rate, then by using formula we can write:

$\begin{array}{rcl}R& =& A_1{V}_1=\pi r^2{V}_1\\ R& =& \pi \times {\left(0.06\right)}^2\left(6.26\right)\\ R& =& 7.08\times {10}^{-4}{m}^3/s\end{array}$

Volumeof1acrefoot$V=1.234\times {10}^3{m}^3$

Therefore;

$\begin{array}{rcl}R& =& \frac{V}{t}\\ t& =& \frac{V}{R}\\ t& =& \frac{1.234\times {10}^3}{7.08\times {10}^{-4}}\\ t& =& 1.74\times {10}^{6}s\\ & & \end{array}$