Question

Water leaks out from an open tank through a hole of area 2 mm2 in the bottom. Suppose water is filled up to a height of 80 cm and the area of cross section of the tanks is 0.4 m2. The pressure at the open surface and at the hole are equal to the atmospheric pressure. Neglect the small velocity of the water near the open surface in the tank. (a) Find the initial speed of water coming out of the hole. (b) Find the speed of water coming out when half of water has leaked out. (c) Find the volume of eater leaked out using a time interval dt after the height remained is h. Thus find the decrease in height dh in terms of h and dt.
(d) From the result of park (c) find the time required for half of the water to leak out.

Answer 1

Explanation:

(a) The Bernoulli theorem ,

$P+\rho g h+\frac{1}{2} \rho v^{2}$

=$P^{\prime}+\rho g h^{\prime}+\frac{1}{2} \rho v^{\prime 2}$

Here $\mathrm{P}=\mathrm{P}^{\prime}$,  

h = 80 cm =0.80 m,

h'=0,  

v =0,  

hence  

$\rho g h=\frac{1}{2} \rho v^{\prime 2}$

$v^{\prime 2}=2 g h$

$v^{\prime}=\sqrt{(2 g h)}$

$=\sqrt{(2 * 10 * 0.8)}$

$=\sqrt{16}=4 m / s\left\{\text { Taking } g=10 \mathrm{m} / \mathrm{s}^{2}\right\}$

(b) Half the water leaks out,

$h=0.40 \mathrm{m}$

Now $v^{\prime}=\sqrt{(2 g h)}$

$=\sqrt{(2 * 10 * 0.40)}$

$=\sqrt{(20 * 0.40)}$

$=\sqrt{8} \mathrm{m} / \mathrm{s}$

(c) If the height remaining in the tank is h at any moment t, then dQ is the volume of the water leaked in a small interval dt.

dQ = Discharge rate × dt = Area ×s peed of water × dt  

$d Q=A \times v^{\prime} \times d t$

$=\left(2 m m^{2}\right) \sqrt{(2 g h)} d t$

And reduction in height dh = Water leaked out in time dt divided by the tank's open area

$d h=\frac{d Q}{0.4} m^{2}$

$d h=\frac{\left(2 m m^{2}\right) \sqrt{(2 g h)} d t}{0.40} m^{2}$

$d h=\frac{\left(2 m^{2} \times 10^{-6}\right) \sqrt{(2 g h)} d t}{0.40} m^{2}$  

$d h=\sqrt{(2 g h)}\left(2 \times 10^{-6} \times \frac{10}{4}\right) d t$  

$d h=\sqrt{(2 g h)} \times 5 \times 10^{-6} d t$

(d) From above i.e. $d h=(2 g h) \times 5 \times 10^{-6} d t$

 $d t=\frac{2 \times 10^{5}}{\sqrt{(2 g h)} d h}$

Integrating the height limits between h and h/2 we get

 $T=\int d t=\int\left\{\frac{2 \times 10^{5}}{\sqrt{(2 g h)} d h}\right\} d h$

$=\left\{\frac{2 \times 10^{5}}{\sqrt{(2 g})}\right\} \int h^{\frac{-1}{2}} d h$

$=\left[\frac{2 \times 10^{5}}{\sqrt{(2 g}} \times 2 \sqrt{h}\right]$

$=\frac{2 \sqrt{2 h} \times 10^{5}}{\sqrt{g}}$

putting the limits H = h to H = h/2 we get  

$T=\left[\left\{\frac{2 \sqrt{2} * 10^{5}}{\sqrt{g}}\right]\{\sqrt{h}-\sqrt{(h / 2)}\}\right.$

$=\left\{\left(\frac{\sqrt{2} \cdot 10^{5}}{\sqrt{g}}\right)\left(\frac{\sqrt{2}-1)}{\sqrt{2}}\right\} \sqrt{h} |\right.$

$=2(\sqrt{2}-1) \times 10^{5} \times \sqrt{\frac{h}{g}}$

Putting $\mathrm{h}=0.80 \mathrm{m}, \mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2} \mathrm{T}$

$=\frac{2 \times 0.414 \times 10^{5} \times \sqrt{\frac{0.8}{10}}}{3600}$

=6.50 hours