Question

Water leaks out from an open tank through a hole of area $2m{m^2}$in the bottom. Suppose water is filled up to a height of $80cm$and the area of the cross section of the tank is $0.4{m^2}$. 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 the water has leaked out.
(c) Find the volume of water leaked out during a time interval dt after the height remained is h.Then find the decrease in height dh in terms of h and dt.
From the result of part.
(d) Find the time required for half of the water to leak out.

Answer 1

To solve such types of questions you need to apply Bernoulli's theorem and principle of continuity between the top and bottom of the tank. You can also get the answer if you simply use formulas but it's better if you use basics.

Complete step by step answer:
Applying Bernoulli’s theorem at top and bottom we have
${P_{top}} + \rho g{h_{top}} + \dfrac{1}{2}\rho {v_{top}}^2 = {P_{bottom}} + \rho g{h_{bottom}} + \dfrac{1}{2}\rho {v_{bottom}}^2$
$\Rightarrow \rho g{h_{top}} = \dfrac{1}{2}\rho {v_{bottom}}^2(\because {P_{top}} = {P_{bottom}}\& {v_{top}} \approx 0)$and assuming ${v_{bottom}} = 0$
$\Rightarrow {v_{bottom}} = \sqrt {2g{h_{top}}}$
Putting values we have
${v_{bottom}} = \sqrt {2 \times 10 \times \dfrac{{80}}{{100}}} = 4m/s$
So for (a) answer will be $4\;m/s$
for (b) height will be reduced to half so velocity will be $= \sqrt {2 \times 10 \times \dfrac{{80}}{{2 \times 100}}} = 2\sqrt 2 m/s$
when height is h the velocity $v = \sqrt {2gh}$
for time dt volume water leaked will be ($dV = a \times v.dt$) where a is area of cross section of the hole
also change in volume can be written as ($dV = Adh$) where A is area of cross section of the tank
Thus, $dV = Adh = a \times vdt \Rightarrow \dfrac{{dh}}{{\sqrt {2gh} }} = \dfrac{a}{A}dt$
After integration we have,
$\int { - \dfrac{{dh}}{{\sqrt {2gh} }} = \int {\dfrac{a}{A}} } dt \Rightarrow - \sqrt {\dfrac{{2h}}{g}} = \dfrac{a}{A}t + C$
At $t = 0$height $h = H$so $C = - \sqrt {\dfrac{{2H}}{g}}$
Therefore total time taken $T = \dfrac{A}{a}(\sqrt {\dfrac{{2H}}{g}} - \sqrt {\dfrac{H}{g}} )$
Putting values we have $T = \dfrac{{0.4}}{{2 \times {{10}^{ - 6}}}}(\sqrt {\dfrac{{2 \times 0.8}}{{10}}} - \sqrt {\dfrac{{2 \times 0.4}}{{10}}} ) = (8 - 4\sqrt 2 ) \times {10^4}s$
So for (c) answer will be $\dfrac{{dh}}{{\sqrt {2gh} }} = \dfrac{a}{A}d$
And for (d) answer will be $(8 - 4\sqrt 2 ) \times {10^4}s$

Note: Most of the students have confusion in finding the time taken to empty half of the tank but if you apply basics properly and apply integration you will get solutions for such types of questions very comfortably.