Torricelli’s law: Theorem and Law Derivation

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Torricelli’s law or Torricelli theorem is related to fluid dynamics which determines the speed of fluid that is flowing from the opening to the extent of fluid above the opening. It creates a relationship between the height of the fluid and its velocity of leaving the container. It states that the exit velocity of the fluid is equal to the velocity with which the body is falling from the height. Torricelli’s law determines the speed of efflux or the speed of fluid outflow.

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Table of Content

Torricelli’s’ theorem
Torricelli’s law derivation
Points to remember
Sample questions

Key Terms: Fluid, Speed, Velocity, Height, Bernoulli’s equation, Torricelli’s barometer, Mercury, Pressure

Torricelli’s Theorem

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The theorem explains the relationship between the height of the liquid in the container and the velocity of the fluid leaving the hole. Suppose there is a container full of water with a hole through which water gushes out, the velocity of the water which leaves from the hole is equal to the velocity of the liquid if it is dropped from a height. It is assumed that the fluid is non-viscous, compressible, and has a laminar flow which makes it an ideal fluid.

Check Important Relationship Between Pressure and Density

Torricelli’s Law Derivation

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A₁ = area of the slit

V₁ = velocity of the flow of liquid

A₂ = area of the free surface of the fluid

V₂ = velocity of the fluid at the free surface

Using equation of continuity,

Av = constant

A₁V₁ = A₂V₂

Using Bernoulli’s equation at the hole,

P_a + (1/2)ρv₁₂ + ρgy₁ …. (i)

Where, P_a = atmospheric pressure

Y₁ = height of the slit from the base

Using Bernoulli’s equation at the surface,

P + ρgy₂ … (ii)

Solving (i) and (ii)

P_a + (1/2)ρv₁₂ + ρgy1 = P + ρgy₂

(1/2) ρv₁₂ = (p –p_a) + ρg(y₂ – y₁)

(p –p_a) ρgh (where h = (y₂ – y₁))

V₁₂ = 2/ρ [(p-p_a) + ρgh

Speed of efflux (V₁) = √2/ρ[(P –P_a) + ρgh]

Also Read:

Hydrostatic Paradox

Pascal’s Law

Continuity Equation

Things to Remember

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In a soda bottle with a hole in the bottom, the exit velocity decreases as the height of the fluid in the soda bottle decreases.
Bernoulli’s equation is used in deriving Torricelli’s law
The distance between the orifice(slit) and the free surface is denoted by the letter “h”
The flow of the direction of the fluid affects the speed of the efflux.
V = 2gh1/2
The velocity by which the fluid will come out of the slit (v1)= √2/ρ[(P –P_a) + ρgh]
The formula of the speed of efflux from a container and that of a freely falling body is identical.
Speed of freely falling body (v₁) = √2gh
The pressure of the fluid at a free surface affects the velocity with which the fluid comes out of the container.
Torricelli’s barometer was made of mercury

Also Check:

Gravitational Potential Energy

Boyle’s Law

Weightlessness

Sample uestions

Ques. A Torricelli’s barometer contains mercury and wine. Calculate the height of the wine column for normal atmospheric pressure if the density of the wine is 984kgm^-3.

Ans: density of mercury (ρ₁) = 13.6 x 103 kg/m³

Density of wine (ρ₂) = 984kgm^-3

Height of mercury column ( h₁) = 0.76m

Height of wine column (h₂) = h₂

Acceleration due to gravity (g) = 9.8 m/s²

The pressure in the mercury column = the pressure in the wine column

ρ₁ h₁ g = ρ₂ h₂ g

h₂ = ρ₁ h₁ / ρ₂

= (13.6 x 103 x 0.76) / 984

=10.5m

Ques. The measurement of the hole in the side of a wide cell is 15cm below the surface of the liquid. Calculate the velocity of the water that is coming out of the hole.

Ans: according to torricelli’s law (v₁) = 2gh

= √2 x 9.8 x 15 x 10^-2 m/s

=1.7m/s

Ques. The density of the atmosphere at sea level is 1.29kg/m³ which does not change with height. What is the height of the atmosphere?

Ans: ρ = mv

Ρgh = 1.29kg/m³ x 9.8ms² x hm

= 1.01 x 105 pa

h = 7989m = 8km

Ques. (a)What is the absolute pressure of an ocean at the depth of 1000m? (b) calculate the gauge pressure. (c) if the area of the window of the submarine is 20cm x 20cm. find the force acting on the window at this depth. The density of seawater is 1.03 x 103 kg m^-3, g = 10ms^-2.

Ans: (a). h = 1000m , ρ= 1.03 x 103 kg m^-3

P₂ – P₁ = ρgh

P = P_a + ρgh

= 1.01 x 105 p_a

+ 1.03 x 103 kg m^-3 x 10ms^-2 x 1000m

=104.01 x 105 pa

= 104atm

(b). gauge pressure

P – P_a = ρgh = P_g

P_g = 1.03 x 103 kg m^-3 x 10ms² x 1000m

= 103 atm

(c). the pressure inside the submarine

P = P_a + ρgh

The pressure inside the submarine

P_a

Gauge pressure is defined as the net pressure acting on the window

P_g = ρgh

Area of the window(A) = 0.04m²

Force acting on the window = PgA

=103 x 105p_a x 0.04m²

=4.12 x 105N

Ques. A tube with a height of 100cm is filled with water. Q is a hole that is located at a distance of 10cm above the ground. Calculate the horizontal distance (x)

Ans: distance between the hole and the surface of water = 100cm – 10 cm

=90cm = 0.9m

Acceleration due to gravity = 10m/s2

The speed of the flow of water at the hole

V₂ = 2gh

V = √2gh

= √2(10)(0.9) = √2(9)

= 3√2 m/s

Ques. A large tub is filled with water and there is a hole. calculate the velocity of water out of the hole if g = 10ms^-2.

Ans: height (h) = 85cm – 40cm = 45cm = 0.45 m

Acceleration due to gravity = 10m/s²

The free fall motion formula to find out the water velocity

Vt₂ = 2gh

= 2(10)(0.45) = 9

Vt =√9 = 3m/s

Ques. The height of the water in tank is 1m (g = 10ms^-2) where there is a leak in the wall. Calculate the speed of water coming out of the tank

Ans: height (h) = 1m – 0.20m = 0.8m

Acceleration due to gravity (g) = 10m/s²

vt² = 2gh

=2 (10)(0.8) = 16

vt = √16

= 4m/s

Physics Related Links:

Bulk Modulus Formula	Motion of Celestial Bodies in Space
Shearing Stress	Elastic Limit

CBSE CLASS XII Related Questions

1.
Read the following paragraph and answer the questions that follow.
In an experiment with convex lens of focal length f, the screen is fixed at a distance D from the object. A student slowly moves the lens away from the object towards the screen and finds that she is able to form sharp image of the object for two positions of the lens. The distance between these two positions of the lens is d.

2.
A charged particle $+q$ in an electric field $\vec{E}$ experiences a force in the direction of the electric field. As a result, its kinetic energy changes. Similarly, the charged particle also experiences a force when it moves in a magnetic field $\vec{B}$. But this magnetic force is perpendicular to both velocity $\vec{v}$ of the charged particle and the magnetic field $\vec{B}$, so it cannot change the kinetic energy of the charged particle. Consider two charged particles 1 and 2 of masses $m$ and $\frac{m}{2}$ having charges $-q$ and $+2q$ respectively. They are accelerated from rest through the same potential difference $V$ and acquire kinetic energy $K_1$ and $K_2$. Then they enter in a region of uniform magnetic field $\vec{B}$ perpendicular to their velocities.

3.
Two air-filled capacitors of capacitances $C_1$ and $C_2$ are connected in parallel with a dc battery. After the capacitors are fully charged, a slab of dielectric constant K is inserted between the plates of each capacitor. How will the (i) charge on each capacitor and (ii) energy stored in the capacitor affected after the slab is introduced.

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4.
Capacitors are manufactured with certain standard capacitances and working voltages. However, these standard values may not be the ones that are actually needed in a particular application. Two or more capacitors can be grouped in series or in parallel to achieve desired capacitance and voltage. When connected in series, the total capacitance decreases while the voltage rating increases, whereas in parallel connections, the total capacitance increases and maintains the same voltage rating. A capacitor stores energy in the electric field between its plates and stored energy is proportional to the square of the voltage and capacitance $U = \frac{1}{2}CV^2$, where symbols have their usual meanings.
Two capacitors, one of $3 \ \mu$F and the other of $6 \ \mu$F, are connected in series in the circuit as shown in the figure, for a long time. }

5.
Consider the nuclear reaction $ X \to Y + Z $. Let $ M_x $, $ M_y $, and $ M_z $ be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

$ (M_x - M_z)<M_y $
$ (M_x - M_y)<M_z $
$ M_x>(M_y + M_z) $
$ M_x<(M_y + M_z) $

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6.
Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

The total charge of the two spheres is conserved.
Both spheres attain the same potential.
The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2)}{(r_1 + r_2)}$
The final potential of the system equals $\frac{1}{4\pi\epsilon_0} \frac{(q_1 + q_2) (r_1 + r_2)}{r_1 r_2}$

View Solution

Torricelli’s law: Theorem and Law Derivation

Torricelli’s Theorem

Torricelli’s Law Derivation

Things to Remember

Sample uestions

CBSE CLASS XII Related Questions

5.
Consider the nuclear reaction \( X \to Y + Z \). Let \( M_x \), \( M_y \), and \( M_z \) be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

6.
Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

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Torricelli’s law: Theorem and Law Derivation

Torricelli’s Theorem

Torricelli’s Law Derivation

Things to Remember

Sample uestions

CBSE CLASS XII Related Questions

5.Consider the nuclear reaction \( X \to Y + Z \). Let \( M_x \), \( M_y \), and \( M_z \) be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

6.Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

Comments

SUBSCRIBE TO OUR NEWS LETTER

5.
Consider the nuclear reaction \( X \to Y + Z \). Let \( M_x \), \( M_y \), and \( M_z \) be the masses of the three nuclei X, Y, and Z respectively. Then which of the following relations hold true?

6.
Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?