It is a branch of mechanics which deals with study of fluids.

Fluid statics deals with study of fluids at rest while Fluid dynamics deals with study of fluids in motion.

An Ideal fluid is Incompressible and Non-Viscous. Incompressible means the density of liquid is constant and it is independent of variation of
pressure.

Non-viscous means no friction between the adjacent layers of fluid.

The force exerted by one part of liquid on other part is only perpendicular to the surface of contact.

It is defined as the ratio of mass and volume.
Let a substance has mass `(M)` and volume `(V)` , then its density is given by

`rho= M/V`

`S.l.` unit of density is `kg//m^3` . Its dimensional formula is `[M^1 L^(-3)]`

It is defined as the ratio of density of substance to the density of water at `4^o C` .

`rho_s =rho/rho_w`

It has no unit and hence, it is a dimensionless number.

Consider a number of substances of masses
`M_1,M_2,,M_3,......` having densities `rho_1,rho_2,rho_3........` respectively, are mixed together.

Total mass of mixture `= M_1 + M_2 + M_3 +.........`
Total volume of mixture `=M_1/rho_1 + M_2/rho_2 + M_3/rho_3 + .......`

Therefore, density of the mixture is given by

`rho_(max) = (M_1+M_2+M_3+..........)/(M_1/rho_1 + M_2/rho_2 +M_3/rho_3 +.............)`

For two substances, the density of the mixture isgiven by

`rho_(max) = (M_1+M_2)/(M_1/rho_1 +M_2/rho_2) = (rho_1rho_2(M_1+M_2))/(rho_1M_2+rho_2M_2)`

Consider a number of substances of volume `V_1 , V_2, V_3.....` and densities `rho_1, rho_2, rho_3 .....` respectively are mixed.

Total mass of mixture `= rho_1 V_1 + rho_2 V_2 + rho_3V_3 + ....`

Total volume of mixture `= V_1+ V_2 + V_3 + ...... `

Therefore, density of mixture is given by

`rho_(max) = (rho_1V_1 +rho_2V_2 +rho_3V_3+..........)/(V_1+V_2+V_3............)`

For two substances, the density of mixture is

`rho_(max) = (rho_1V_1 +rho_2V_2)/(V_1 +V_2)`

If a liquid is nor ideal, then its volume and hence density can change.

On increasing temperature of a liquid, its volume increases. Since, mass of liquid remains same, so its density decreases.

Let a liquid of mass `(m)` has density `( rho)` at temperature T. If temperature ofliquid is increased by `(Delta T)` . then its volume increases by `(Delta V)`

`Delta V = V gamma (Delta T)`

`gamma =` coefficient of volume expansion

`V =` Initial volume of liquid `=m/rho`

`Delta V =m/rho gamma (Delta T)`

`V' =` New volume of liquid

`V' = V +Delta V m/rho +m/rho gamma (Delta T)`

`V' = m/rho (1+gamma(Delta T))`

New density `(rho) = m/(V') = m/(m/rho (1+gamma(Delta T)))`

`rho' = (rho')/(1+gamma(Delta T))`

If temperature of liquid decreases, then `Delta T` is negative. Hence, density of liquid will increase.

For non-ideal liquid, if pressure increases, volume decreases and hence density will increase.
Consider a non-ideal liquid of mass `(m)` and density `( rho )` at pressure `(P)`.

Volume of liquid `=m/rho =V`

If pressure changes by `( Delta P)` and `(B)` is the bulk modulus of elasticity of liquid, then change in volume `(Delta V)` is given by

`B = -V(Delta P)/(Delta V)` or `-(V Delta P)/B`

`Delta =-(m Delta P)/(rho B)`

New volume `= V + Delta V = m/ rho - (m Delta P)/(rho B) = m/rho (1- (Delta P)/B)`

New density `rho' = m/(m/rho (1- (Delta P)/B)) = rho/(1-(DeltaP)/B)`

Consider a cylindrical vessel, rotating at constant angular velocity about its axis. lfit contains fluid then after an initial irregular shape, it will rotate with the tank as a rigid body. The acceleration of fluid particles located at a distance r from the axis of rotation will be equal to `omega^2r`, and the direction of the acceleration is toward the axis of rotation as shown in the figure. The fluid particles will be undergoing circular motion.

Lets consider a small horizontal cylinder of length dr and cross-sectional area A located y below the free surface of the fluid and r from the axis. This cylinder is accelerating in ground frame with acceleration `omega^2r` towards the axis hence the net horizontal force acting on it should be equal to the product of mass (dm) and acceleration.

`dm=Adrrho`

`P_2A- P_1A = (Adr rho)omega^2r`

If we say that the left face ofthe cylinder is y below the free surface of the fluid then the right surface is y + dy below the surface of liquid. Thus

`P_2 - P_1 = rhogdy`

Thus solving we get, `(dy)/(dr)=(romega^2)/g`

and, therefore, the equation for surfaces of constant pressure is

`y=(omega^2r^2)/(2g) + text(constant)`

This equation means that these surfaces of constant pressure are parabolic as shown in Fig.

The pressure varies with the distance from the axis of rotation, but at a fixed radius, the pressure varies
hydrostatically in the vertical direction as shown in fig.

`text(Force exerted by the jet on a stationary vertical plate :)`

Consider a jet of water coming out from the nozzle strikes the vertical plate

V = velocity of jet, d = diameter of the jet, a = area of x – section of the jet

The force exerted by the jet on the plate in the direction of jet.
`F_x =` Rate of change of momentum in the direction of force Rate of change of momentum in the direction of force

= (initial momentum – final momentum) / time

= (mass x initial velocity – mass x final velocity) / time

= mass/time (initial velocity – final velocity)

= mass/ sec x (velocity of jet before striking mass/ sec x (velocity of jet before striking – final velocity of jet after striking) final velocity of jet after striking)

`=ρaV (V -0)`

`=ρaV^2`