“There is no substitute for hard work.”

– Thomas A. Edison

You won’t be allowed to use equation tables and calculators to work out multiple choice questions (MCQ). (These are allowed in the case of free-response questions). You will have to remember certain basic formulas for solving multiple choice questions within the permitted time. The essential things you must remember in the section, ‘Fluid Mechanics’ are given below:

**1. Weight of floating body = weight of displaced liquid**

This can be modified as

*mass of floating body = mass of displaced fluid***2. Force of buoyancy (F**

_{buoy})=**weight of displaced fluid**

or,

**F**_{buoy}= V**ρg**where V is the volume of fluid displaced, ρ is the density of the fluid and g is the acceleration due to gravity.**3. Equation of continuity**in fluid dynamics, as applied to liquids which are supposed to be incompressible, is

**A**where A

_{1}v_{1}= A_{2}v_{2 }_{1}and A

_{2}are the cross section areas of the tube

**and v**

_{1}and v

_{2}are the velocities of the liquid at these sections respectively.

**4. Bernoulli’s equation (Bernoulli’s theorem)**is

**P +**

**ρgh + (½)ρv**where P is the pressure, ρ is the density, ‘h’ is the height (with respect to the reference level for estimating the gravitational potential energy) and ‘v’ is the velocity of the fluid. The symbol ‘y’ also may be used in place of ‘h’.

^{2}= constant
Note that Bernoulli’s theorem follows from the law of conservation of energy in the case of a small mass ‘m’ of a fluid throughout its flow and can be written as

PV+ mgh + (½ )mv

^{2}= constant
Another useful form of Bernoulli’s equation is

**P/**

**ρ**

**+**

**gh + (½)v**

^{2}= constant
In the case of a liquid flowing through a

*horizontal*pipe (constant gravitational potential energy), the pressure of liquid will be smaller at points where the velocity is greater and vice versa.**5.Torricelli’s Theorem:**

**Velocity of efflux (of liquid flowing out through a hole) =**

**√(2gh)**

where ‘h’ is the depth of the hole (fig).

(It is interesting to note that a particle dropped from a height ‘h’ will strike the ground with the above velocity)

The expression for the efflux velocity follows from Bernoulli’s theorem by considering the cases at the free surface of the liquid in the tank (fig) and at the hole:

**P + ρgH + 0 = P + ρg(H–h) + (½)ρv**

^{2}so that v =√(2gh),

where P is the atmospheric pressure. Note that the hole is open to the atmosphere, as is the free liquid surface in the tank.

**6. The horizontal range of liquid jet (fig), R= 2**

**√[h(H**

**–h)]**

Note that the above expression is obtained by considering the horizontal range (on the ground) of a particle projected horizontally from a height (H–h). Do it as an exercise.

**The range R will be**

*maximum*if h = H/2. [You may show this by putting dR/dh = 0 when R is maximum]. The maximum range is H.

In the next post we will consider some typical questions in this section. Of course, their solution also will be discussed.

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