Question

Explain Torricelli Law with derivation with neat and clean diagram & application.
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Answer 1

$\underline{\boxed{\bf{\pink{Torricelli's\:theorem:}}}}$

✴ The speed of liquid coming out through a hole at a depth h below the free surface is the same as that of a particle fallen freely through the height h under gravity. This is known as Torricelli's theorem. The speed of the liquid coming out is called the speed of efflux.

$\bigstar\bf\:Speed\:of\:Efflux$

▪ Consider a liquid of density $\rho$ filled in a tank of large cross-sectional area A$_1$. There is a hole of cross-sectional area A$_2$ at the bottom and the liquid flows out of the tank through the hole.

▪ Please see the attachment for better understanding. (A$_2$‹‹ A$_1$)

✒ Let V$_1$ and V$_2$ be the speeds of the liquid at A$_1$ and A$_2$. As both the cross sections are open to the atmosphere, the pressures there equals the atmosphere pressure Po. If the height of the free surface above the hole is H, Bernouli equation gives

$\bf\red{P_o+\dfrac{1}{2} \rho{V_1}^2+\rho gH={\rho}_o+\dfrac{1}{2}\rho{V_2}^2}\rightarrow (1)$

▪ By the equation of continuity

$\bf\blue{A_1V_1=A_2V_2}$

✒ Putting V$_1$ in terms of V$_2$ in equation (1),

$\bf\green{\dfrac{1}{2}\rho{\huge{(}}\dfrac{A_2}{A_1}{\huge{)}}^2{V_2}^2+\rho gH =\dfrac{1}{2}\rho{V_2}^2}$

or, $\sf\:{\huge{[}}1-{\huge{(}}\dfrac{A_2}{A_1}{\huge{)}}^2{\huge{]}}{V_2}^2=2gH$

✏ If, A$_2$<< A$_1$, this equation reduces to...

$\underline{\boxed{\bf{\purple{V_2=\sqrt{2gH}}}}}$

Answer 2

Let's consider that a column of fluid is passing outside through a container having a hole at the bottom.

Let the Area of cross section on the upper part of container be A and the area of hole be "a"

Now we shall Apply Bernoulli's Theorem at a point just outside the hole and just inside the hole.

Let atmospheric pressure be P, velocity with which the area of cross-section comes down be v1 and the speed of efflux be v2.

Most importantly , the reference line is considered at the level of hole:

Applying Bernoulli's Theorem

$P + \frac{1}{2} \rho {(v1)}^{2} + \rho gh = P + \frac{1}{2} \rho {(v2)}^{2} + \rho g(0)$

$= > \cancel P + \frac{1}{2} \rho {(v1)}^{2} + \rho gh = \cancel P + \frac{1}{2} \rho {(v2)}^{2}$

$= > \frac{1}{2} \rho {(v1)}^{2} + \rho gh =\frac{1}{2} \rho {(v2)}^{2}$

Now we need to replace any pne of the velocity terms:

Applying Equation of Continuity :

$= > A \times v1 = a \times v2$

Replacing v1 by v2 :

$= > \frac{1}{2} \rho { \{ \dfrac{a(v2)}{A} \}}^{2} + \rho gh =\frac{1}{2} \rho {(v2)}^{2}$

$= > \frac{1}{2} \rho( {v2)}^{2} \bigg \{1 - { (\dfrac{a}{A} ) }^{2} \bigg \} = \rho gh$

$= > v2 = \sqrt{ \dfrac{2gh}{ \bigg \{1 - {( \dfrac{a}{A}) }^{2} \bigg \} } }$

If a << A then , a² <<<< A² , hence (a/A)² is extremely small as compared to 1 , hence neglected :

$= > v2 = \sqrt{2gh}$

Definition :

The speed of efflux of a liquid from an orifice is equal to that of a body falling freely through a distance equal to the total height of the liquid at the orifice reference.

Application:

Toricelli's Equation has a classical application :

It helps us determine the velocity of efflux of water from beneath the dam by analysing the height of water stored behind the dam