Question

In a large tank of cross-sectional area A, there is a small hole of area ‘a’ at the bottom. The time taken to emptying the tank from height h1 to h2 is

Answer 1

Given:

1. Cross-sectional area = A

2. Area of hole = a

3. Height of the liquid in the tank = H

To find:

Time taken to empty the tank from height H₁ to H₂.

Solution:

• We know that by Bernoulli's principle, at height H the rate of discharge of liquid through a hole at the bottom is √(2gh) x a.

• Let us consider that in time dt level of water in the tank decreases by dh.

• Now, equating volumes we get,

        A dh = dt √(2gh) x a

        dt = A dh / √(2gh) x a

• Integrating both the sides, we get,

        t = A / a x √(2g) (√H₁ - √H₂)

• Therefore, the time taken to empty the tank from height H₁ to H₂ is

        A / a x √(2g) (√H₁ - √H₂)

Answer 2

Given :

a=area at the bottom tank.

To find : t=??

Solution:

Following are the attachment of the question in the diagram form is given below

As we know that the flow rate is given as

Q=av

it means that

$\frac{d[V]}{dt}\ =av$

Where V =Volume of  cross-sectional area

v=velocity of cross-sectional area

Now,

$A \frac{dh}{dt} \ =\ a\sqrt{2gh}$

$\int\limits_{h1}^{h2} {\frac{dh}{\sqrt{h}}} =\int\limits^t_0 {\frac{a\sqrt{29}}{A}}\, dt$

$2\ [\sqrt{h2} \ -\sqrt{h1 \ ] } \ \ =\frac{a \sqrt{2g} \ t }{A}$

$t\ = \ \frac{2A\ [\sqrt{h2}\ - \sqrt{h1\ ]} }{a\sqrt{2g} }$