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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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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.

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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,
Pa + (1/2)ρv12 + ρgy1 …. (i)
Where, Pa = atmospheric pressure
Y1 = height of the slit from the base
Using Bernoulli’s equation at the surface,
P + ρgy2 … (ii)
Solving (i) and (ii)
Pa + (1/2)ρv12 + ρgy1 = P + ρgy2
(1/2) ρv12 = (p –pa) + ρg(y2 – y1)
(p –pa) ρgh (where h = (y2 – y1))
V12 = 2/ρ [(p-pa) + ρgh
Speed of efflux (V1) = √2/ρ[(P –Pa) + ρgh]
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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 –Pa) + 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
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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 = Pa + ρgh
= 1.01 x 105 pa
+ 1.03 x 103 kg m-3 x 10ms-2 x 1000m
=104.01 x 105 pa
= 104atm
(b). gauge pressure
P – Pa = ρgh = Pg
Pg = 1.03 x 103 kg m-3 x 10ms2 x 1000m
= 103 atm
(c). the pressure inside the submarine
P = Pa + ρgh
The pressure inside the submarine
Pa
Gauge pressure is defined as the net pressure acting on the window
Pg = ρgh
Area of the window(A) = 0.04m2
Force acting on the window = PgA
=103 x 105pa 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
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