Solution:

`"Surface Tension :"` Force on unit length of an imaginary line drawn on the surface of the liquid is called surface tension. Its S.l. unit is `Nm^(-I)` and its dimension is `[ML^0 'r-2].` Surface Energy. Energy possessed by the surface of the liquid is called surface energy. Change in surface energy is the product of surface tension and change in surface area under constant temperature.

`"Surface Energy."` Energy possessed by the surface of the liquid is called surface energy. Change in surface energy is the product of surface tension and change in surface area under constant temperature.

Let `S =` Surface tension of soap solution
`d =` Length of the wire `PQ`
`=` length of wire `PQ`

Surface tension acts on both the free surfaces of film.
Hence, total inward force on wire `PQ`

Increase in area of the film `PQ Q_1 P_1`

`= DeltaA = 2 (l xx)`

.. Work done in stretching film is
W = Force applied x Distance moved

`= (S xx 2l) xx x = S xx (2l xx)`


`= S xx DeltaA (·: 2lx=M)`

This work done is stored in the film as its surface energy.

`E=W=SxxDeltaA`
`=> S= WDeltaA`

If increase in area is unity then, `DeltaA = 1`

S= W
:. Surface tension of a liquid is numerically equal to surface energy of the liquid surface.

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Solution:

`"Coefficient of Viscosity."` Coefficient of viscosity is defined as the viscous force acting in unit area of a layer having unit velocity gradient. It is measured in `Nsm^(-2)` and has dimension of `ML^(-I)T^(-1.)`

`"Stoke's Formula."` The viscous drag experienced by a spherical ball moving through vertical column of highly dense liquid is given by `F = 6pietarv` where r-radius of the ball, 11-coefficient of viscosity of the liquid and v-terminal velocity attained by the ball.


`"Terminal Velocity."` The constant velocity with which a body drops down after initial acceleration in a dense liquid or fluid is called terminal velocity. This is attained when the apparent weight is compensated by the viscous force. It is given by


`v = 2/9 (r^2g)/(eta) (p - eta)` where p and sigma are denseness

of the body and liquid respectively, `eta` is the coefficient of viscosity of the liquid and r is the radius of the spherical body. Consider a lengthy column of a dense liquid like glycerine. As the ball or spherical ball is dropped in it, the forces experienced are,

(i) weight `= mg = 4/3 pia^3` pg where `p -` density of ball

(ii) upthrust, `U = 4/3 pia^3 p'g` where `p' -` density of liquid


(iii) viscous force `F_v = 6pieta ap' v_t`
where `v_1 -` terminal velocity

When terminal velocity is attained,
acceleration should be zero and the
net force should be zero.

`:.mg - U - F_v = 0`

`=> 4/3 pia^3 pg - 4/3 pia^3 ptg - ppieta av = 0`

`v_t = (4/3) pia^3 g (p-p')/(6pietaa)`

`= 2/9 (a^2g(p-p'))/(eta)`
Thus, terminal velocity depends upon
(i) square of radius of the body.
(ii) coefficient of viscosity of the medium.
(iii) density of the body and the medium.

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Solution:

(a) Streamline is the actual path followed by the procession of particles in a steady flow, which may be straight or curved such that tangent to it at any point indicates the direction of flow of a liquid at that point.

(b) Two properties of streamlines are- (i) Two streamlines can never cross each other.

(ii) The greater is the crowding of streamlines at a place, the greater will be the velocity of liquid particles at that place and vice-versa.

Volume of liquid entering per second at `A= a_1 u_1` Mass of liquid entering per second at `A= a'_1u_1p_1` Similarly, mass of liquid leaving per second at `B = a_2u_2p_2` If there is no loss of liquid in tube and the flow is steady then, Mass of liquid entering per second at A = Mass of liquid leaving per second at B

`a_1v_1p_1 = a_2U_2P_2`

If liquid is incompressible then,

`p_1 = p_2`

`a_1v_1 = a_2v_2`
av = constant
This is the equation of continuity.

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Solution:

According to Bernoulli's theorem, for an incompressible, non-viscous liquid having streamlined flow, the sum of pressure head, velocity head and gravitational head is a constant,

`i.e P/pg + (v^2)/(2g) + h`

Consider an incompressible non-viscous liquid entering the cross-section `A_1` at A with a velocity `v_1` and coming a height `h_2` at B with velocity `v_2`


The P.K and KE. increase since `h_2` and `v_2` are more than `h_1` and `u_1` respectively. This is done by the pressure doing work on the liquid. If `P_1` and `P_2` are the pressure at A and B, for a small displacement at A and B, The work done on the liquid at `A = (P_1 A_I)`

`Delta x_1 = P_1 A_1 v_1 Deltat`

The work done by the liquid at B

`Deltax_2 = - (P_2 A_2)/Deltax_2 =- P_2 A_2 v_2 Deltat`

The work done on the liquid at

(Considering a small time !:!..t so that area may be same)

Net work done by pressure `= (P_1 - P_2) Av Deltat`

Since `A_1 v_1 = A_2 v_2`

From conservation of energy,

`(P_1 - P_2)` Au M = change in `(K.E. + P.E.)`

`(P_1-P_2) Av Deltat = AvpDeltatg(h_2-h_1)`

`+ 1/2 AvDeltatp (v_(2)^(2) - v_(1)^(2)`

`(i.e) P_1 + pgh_1 + p/2 v_(1)^(2) = P_2 + P_2 + pgh_2 + p/2 v_(2)^(2)`

`:. P/(pg) + h + v^2 /(2g) = Constant`

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Solution:

`"Bernoulli's Theorem."` For an incompressible, non-viscous, irrotational liquid having streamlined flow, the sum of the pressure energy, kinetic energy and potential energy permit mass is a constant, i.e.,


`P/p + v^2/2 = gh = "constant"`

`P/(pg) + (v^2)/(2g) +h= "constant"`

A liquid is said to be irrotational if the angular momentum about any point in the liquid is zero. A wheel or disc in it will not rotate .

Given : `a_1 = 0.36 pi cm^2, a_2 = 0.04 pi m^2 h = 1 m`


Since, c.s.a. at B is less velocity will be more and pressure will be less. The difference in pressure is `P_1 P_2 = hpg.` Applying Bernoulli's theorem,

`(P_1)/(pg) + (v_(1)^(2))/(2g) = (P_2)/(pg) + (v_(2)^(2))/(2g)`

`=> (P_1-p_2)/(pg) = (v_(2)^(2) - v_(1)^(2))/(2g)`

`v_(2)^(2) - v_(1)^(2) = 2gh`

for streamlined flow, `a_1 v_1 = a_2 v_2`

`v_(2)^(2) - (a_(2)^(2) v_(2)^(2))/(=a_(1)^(2)) = 2 gh`

`=> v_(2)^(2) = 2gh 1/(1-(a_(2)^(2)//a_(1)^(2))`

`v_2 = sqrt((2gh a_(1)^(2))/(a_(1)^(2) - a_(1)^(2))) = 5 m//s`

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Solution:

Pascal's law-It states that the pressure in a liquid at rest is the same at all points if they are at the same level.

`"Hydraulic lift ":`
Let, a = area of cross-section of piston in cylinder B
A = area of cross section of position in C also a < Fill the cylinders with an in compressible fluid.
Let r = downward force applied on piston B.
Then pressure exerted on liquid is

`P = fla`

This pressure is equally transmitted to piston in cylinder C (according to Pascal's law) :. upward force acting on the piston of cylinder C is

`F = Pa = f/a A`

as `A>> a, F >> f`

Therefore a small force applied on piston B
appears as a larger force on piston C.

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Solution:

(a) Archimedes' Principle. When a body is partially or completely immersed in a liquid, it loses weight due to the presence of upthrust, which is equal to the weight of liquid displaced by the submerged part of the body. Consider a liquid of density `sigma` into which a cylinder of material density p, crosssectional area A, height h is immersed such that its upper surface is at a depth `x` from `sigma` g the free surface. The pressure at the top and bottom of the cylinder are

`x sigma g` and `(x+h) sigma g`

The upthrust on the cylinder `= Deltap xx A`

`U = [h osigma] xx A = hA sigmag = V sigmag`

Weight in air,

`W_1 = Vpg`

Apparent weight

`W_a= W_1 - U`

Loss in weight

`W_1 -` Apparent weight

`= W_1 - W_a = W_1 - (W_1 - U)`

`= W_1- W_1 + u = u`
`= Vsigmag`

So, there is a loss equal to the

(b) Excess pressure `= (2sigma)/(r) = (2xx72xx10^(-3))/(0.1xx10^(-3))`

`= 1440 N//m^(2)`
`= 0.01440 xx 10^5 N//m^2`
So, net pressure `= (1.1 - 0.0144)xx 10^5`
`= 1.0856 xx 10^5 N//m^2`

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Solution:

Forces acting on balloon when held stationary are -

`U ->` the upthrust

`U = Mg + T`

Forces acting on small block when held stationary,

`T = [M/2]g`

From (i) and (ii), `U = 3/2 Mg`

When string is cut, `T = 0`

Small block will have free fall.

Balloon will have an acceleration 'a' such that
`U -Mg = Ma`


or `3/2 Mg - Mg = Ma`

`=> a = g/2` upwards

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Solution:

It is a simple device which is used for measuring the velocity of the flow at any depth in a flowing liquid. It is based on Bernoulli's theorem.

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Solution:

Consider two points A and B at two levels separated by h column of a liquid of density p surrounding the points A and B. Consider an area 'a' forming a cylinder of liquid of length h. The pressure at `A = P_A =` Atmospheric pressure `= P_0`

Weight of the liquid at centre of gravity

`W =Mg = hapg`

For equilibrium, pressure/force at B should nullity the forces acting down.

`:. P_A . a + hapg = P_B . a`
`:. P_B= P_A + hpg = P_0 + hpg`

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Solution:

Consider two points at the same height (H - h) from the ground one inside and one outside the hole. Applying Bernoulli's theorem for the points.

we have `(P_0+hpg)/(pg) + 0 (H-h)`

`= (H-h) + (V^2)/(2g) + (P_0)/(pg)`

On solving, we get `v = sqrt(2gh)` The velocity of efflux thus depends on the depth at which the hole is made from the surface of the liquid. Time taken to reach the ground,

`t = sqrt(2(H-h)/g)`

Since, initial vertical velocity is zero.

`v = sqrt(2gh)` is horizontal


and `a_x= 0 ;`

`R - vt = sqrt(2gh) sqrt(2(H-h)/g)`

`R = 2 sqrt((H-h))`

For Range to be max `(dR)/(dh) = 0`

i.e `h = H/2`

Maximum range `= R_(max)`

`=2 sqrt(H/2(H-H//2))=H`

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