(i) The molecules of liquid are in random motion due to their thermal velocity they collide with surface and rebound that result in the change in momentum which is transferred to surface in contact.
(ii) It is thrust exerted by 1iquid at rest per unit area of the surface in contact. The S.I unit of pressure is Nm^-2 or Pascal.
(iii) Hydrostatic pressure is a scalar quantity, because it is exerted in all direction equally. It has arbitrary direction.
(iv) Relative density of a substance is defined as the ratio of its density to the density of water at 4° C i.e..
\(\text { Relative density }=\frac{\text { density of substance }}{\text { density of water at } 4^{\circ} \mathrm{C}}\)
(v) It is equal to P₀ = Po + h\(\rho\)g
where P₀ = atmospheric Pressure
g = acceleration due to gravity.
\(\rho\) = density of liquid
(vi) Gauge pressure at a point in a liquid is the difference of total pressure at that point and atmospheric pressure. It is independent of area of cross-section A but depends upon the height h of the liquid column and density of the liquid.
(vii) The liquid pressure at a point is independent of the quantity of liquid but depends upon the depth of point below the liquid surface. This is known as hydrostatic paradox.

(i) Pascal's law states that the increase in pressure at one point of the enclosed liquid in equilibrium of rest is transmitted equally to all other points of the liquid and also to the walls of the container provided the effect of gravity is neglected.
(ii) It is because the increase in pressure at one part of liquid is communicated equally at the other parts of liquid that result in large force on walls of bottle to break the bottle.
(iii) Here \(A_{1}=\pi\left(\frac{2.5}{2}\right)^{2} \mathrm{~cm}^{2}, A_{2}=\pi\left(\frac{30}{2}\right)^{2} \mathrm{~cm}^{2}\)
F₁ = 50 kg wt
Now \(F_{2}=\frac{F_{1}}{A_{1}} \times A_{2}=\frac{50 \times \pi \times\left(\frac{30}{2}\right)^{2}}{\pi\left(\frac{2.5}{2}\right)^{2}}\)
= 7200 kg wt
(iv) In one stroke, Input workdone = Output workdone
So F₁l₁ = F₂l₂
or \(l_{2}=\frac{F_{1} l_{1}}{F_{2}}=\frac{50 \times 4}{7200}=0.028 \mathrm{~cm}\)
Distance covered after 10 strokes = 0.028 x 10= 0.28 cm
(v) Here, \(F_{1}=?, r_{1}=\frac{5}{100} \mathrm{~m}\)
F₂ = 1300 Kg wt = 1300 x 9.8 N
\(r_{2}=\frac{10}{100} \mathrm{~m}\)
As \(\frac{F_{1}}{a_{1}}=\frac{F_{2}}{a_{2}} \text { or } F_{1}=\frac{a_{1}}{a_{2}} F_{2}=\frac{\pi r_{1}^{2}}{\pi r_{2}^{2}} F_{2}\)
or \(F_{1}=\frac{r_{1}^{2}}{r_{2}^{2}} F_{2}=\frac{\left(\frac{5}{100}\right)^{2}}{\left(\frac{10}{100}\right)^{2}} \times 1300 \times 9.8\)
= 3185 N
Pressure \(P=\frac{F_{1}}{a_{1}}=\frac{3185}{\frac{22}{7} \times\left(\frac{5}{100}\right)^{2}}=4.05 \times 10^{5} \mathrm{~Pa}\) 

(i) Surface tension is the property of liquid by virtue of which the free surface of liquid at rest tends to have minimum surface area and as such it behaves as if covered with stretched membrane. Surface tension is due to force of cohesion.
(ii) Surface tension of oil is less than that of water. When oil is dropped on surface of water, water stretches the oil drops on all sides. Hence the oil spreads over the surface of water.
(iii) At critical temperature.
(iv) In order to have low value of surface tension so that it can spread over larger area.
(v) A highly soluble substance like sodium chloride when dissolved in water, increases the surface tension of water (liquid). But the sparingly substance like phenol when dissolved in water, reduces the surface tension of water.
(vi) Work done = Surface tension x area of soap bubble
\(=S \times\left(4 \pi r^{2}\right) \times 2=8 \pi S r^{2}\)
(vii) Surface energy of a given liquid surface is the amount of work done against the force of surface tension, in forming the liquid surface of given area at a constant temperature.

(i) Viscosity is the property of the fluid by virtue of which an internal frictional force comes into play when the fluid is in motion in the form of layers having relative motion of the different layers.
(ii) Force \(F=\frac{\eta A d v}{d x}\)
\(\hat{\eta}=\frac{N \times m}{m^{2} \times m s^{-1}}=\mathrm{Nm}^{-2} \mathrm{~s}\)
The SI unit of  \(\eta=\mathrm{Nm}^{-2} \mathrm{~s}\)
(iii) Forces acting on the small spherical body falling freely through a viscous medium are the following
(i) Weight of the body acting vertically downwards.
(ii) Upward thrust due to buoyancy equal to weight of body.
(iii) Viscous drag acting in the direction opposite to the direction of motion of body.
(iv) \(\text { Here } \eta=\frac{F}{6 \pi r v}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]\left[\mathrm{LT}^{-1}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]\)
(v) When upthrust acting on the body is negligible then terminal velocity of the body
\(v_{t}=\frac{m g}{6 \pi \eta r}\) 
(vi) \(v_{\mathbf{T}}=\frac{2 r^{2}(\rho-\sigma) g}{9 \eta}\)
(viii) Viscosity of liquid decreases with temperature due to decrease in intermolecular force of attraction. \(\eta \alpha \sqrt{\mathrm{T}}\) for gases viscosity increases with temperature. Viscosity of liquid increases with pressure whereas that of gases remain independent.

(i) Volume of water flowing per second
V = 31.4 litre/min \(=\frac{31.4 \times 10^{-3}}{60} \mathrm{~m}^{3} / \mathrm{s}\)
Area of cross-section \(a=\pi r^{2}\) 
= 3.14 x 10^{- 2} m² ,
since \(r=\frac{d}{2}=\frac{2}{2}=1 \mathrm{~cm}=10^{-2} \mathrm{~m}\) 
velocity of flow \(=\frac{V}{a}=\frac{5}{3} \mathrm{~ms}^{-1}\) 
(ii) a₁v₁ = a₂v₂
therefore \(v_{2}=\left(\frac{a_{1}}{a_{2}}\right) v_{1}=\left(\frac{r_{1}}{r_{2}}\right)^{2} v_{1}\)
\(=\left(\frac{1}{0.25}\right)^{2} \times 3=48 \mathrm{~cm} \mathrm{~s}^{-1}\) 
(iii) It is that flow in which every particle of the liquid follows exactly the path of its preceding particle and has the same velocity in magnitude and direction as that of its preceding particle while crossing through that point.
(iv) In a laminar flow the liquid moves in layers and one layer slides over the other layer of liquid. While in turbulent flow liquid moves with a velocity greater than its critical velocity, the motion of the particles of liquid becomes disorderly.
(v) According to equation of continuity av = constant
At high depth area of cross-section becomes very large therefore velocity of flow becomes negligible, this is reason behind the still water run deep.