Mechanical Properties of Fluids

(a) The pressure exerted by a liquid column of height h is given by P = hρg, so it increases with the vertical depth of the point. Measured from the brain, the height of the blood column standing above the feet is much greater than that above the brain. Hence, the blood pressure in human is greater at the feet than that at the brain.

(b) The density of air is maximum near the surface of the earth and decreases rapidly with height. At a height of about 6 km the air density falls to nearly half its sea-level value, and since the atmospheric pressure is produced by the weight of the air column above, the pressure at 6 km also falls to nearly half its sea-level value. Beyond 6 km the density of air decreases only slowly, so the atmosphere continues to exist up to several hundred kilometres height.

(c) Pressure is defined as the normal force exerted per unit area, $\text P = \frac{\text F}{\text A}$, where F is only the component of the force acting normal to the surface. Since the direction of this force is always perpendicular to the surface, it is already fixed by the surface itself and pressure carries no direction of its own; it gives only the magnitude of the normal force per unit area. Hence, hydrostatic pressure is a scalar quantity.

(a) For a molecule lying on the liquid surface near the wall of the tube, two forces of attraction act — the adhesive force P due to the solid and the cohesive force Q due to the neighbouring liquid molecules. The liquid surface always sets itself perpendicular to the resultant R of these two forces.

For water and glass, the adhesive force is greater than the cohesive force, so the resultant R is directed outwards from the liquid and the water surface becomes concave. Hence the angle of contact of water with glass is acute (nearly 0° for pure water and clean glass).

For mercury and glass, the cohesive force between mercury molecules is far greater than the adhesive force, so R is directed into the interior of the mercury and the surface becomes convex. Hence the angle of contact of mercury with glass is obtuse (about 135°).

(b) The adhesive force between water molecules and glass molecules is greater than the cohesive force between the water molecules themselves. Hence the water molecules cling to the glass molecules and the glass surface is wetted, so water spreads out on it. In the case of mercury the cohesive force between mercury molecules is much greater than the adhesive force between mercury and glass, so mercury molecules do not cling to the glass. Hence, water wets glass while mercury does not and mercury tends to form drops.

(c) Surface tension is defined as the force per unit length acting at right angles on either side of an imaginary line drawn in the liquid surface, T = F/l. Since it is a force per unit length and not per unit area, its value depends only upon the nature of the liquid and the medium on the other side of the surface. Hence, the surface tension of a liquid is independent of the area of the free surface.

(d) The rise of a liquid in a capillary is $\text h = \frac{2\text T\cos \theta}{\text{r}\rho \text g}$, so h is proportional to cos θ. Cloth has narrow pores which act as fine capillaries. If the angle of contact θ is small, cos θ is large and the detergent solution penetrates more deeply into the pores of the cloth and cleans it better. Hence, water with detergent dissolved in it should have a small angle of contact.

(e) When no external force acts, the shape of the drop is decided by surface tension alone. Surface tension makes the liquid surface contract to the minimum possible area, and for a given volume a sphere has the minimum surface area. This also corresponds to minimum potential energy, which is the condition of stable equilibrium. Hence, a drop of liquid under no external forces is always spherical in shape.

(a) decreases

(b) increases, decreases

(c) shear strain, rate of shear strain

(d) conservation of mass

(e) greater

Reason —

(a) The surface tension of a liquid decreases with a rise in temperature and becomes zero at the critical temperature. On heating, the molecules gain kinetic energy and the cohesive forces between them are weakened, so the surface tension falls.

(b) In a liquid, viscosity is due mainly to intermolecular cohesion. A rise in temperature weakens this cohesion and so the viscosity of liquids decreases. In a gas, viscosity is due to the transfer of momentum between layers. At a higher temperature the molecules move faster and collide more frequently, so the viscosity of gases increases.

(c) A solid can sustain a shear stress, so once the deformation is produced it remains, and the shearing force depends on the amount of shear strain. A fluid cannot sustain a shear stress and goes on deforming as long as the force acts, so by Newton's law of viscosity $\text F = -\eta \text A\frac{\text{dv}}{\text{dz}}$ the force depends on the velocity gradient, that is, on the rate of shear strain.

(d) By the equation of continuity, A × v = constant, which is derived from the conservation of mass. When the area of cross-section decreases at a constriction, the velocity must increase so that the same mass crosses every section per second.

(e) Turbulence sets in at the same value of Reynold's number, $\text R_e = \frac{\rho \text{vL}}{\eta}$. The model in the wind tunnel has a smaller characteristic length L, so to reach the same Reynold's number the speed must be greater than that for the actual plane.

(a) When we blow over the paper, the velocity of air above the upper surface becomes larger than that below it. By Bernoulli's theorem, where the velocity of a flowing fluid is large the pressure is small, so the pressure above the paper is lowered while that below it remains atmospheric. This excess pressure from below keeps the paper horizontal. If we blow under it, the pressure below would be lowered and the paper would fall.

(b) When the tap is closed with the fingers, the area of cross-section of the outlet is reduced appreciably. By the principle of continuity, A₁v₁ = A₂v₂, so the velocity of water increases enormously as the area decreases. Hence, fast jets of water gush through the openings between the fingers.

(c) By Bernoulli's theorem $\left(\text P + \frac{1}{2}\rho v^2 = \text{constant}\right)$, the total energy of the medicine flowing through the needle depends on the first power of the pressure but on the second power of the velocity. The flow therefore depends much more on velocity, which is decided by the bore of the needle. Hence, the doctor controls the flow rate by choosing a needle of proper size rather than by exerting thumb pressure.

(d) By Torricelli's theorem the liquid flows out of the hole with velocity $v = \sqrt{2\text{gh}}$, so the emerging liquid carries momentum in the outward direction. By the principle of conservation of momentum the vessel acquires an equal momentum in the opposite direction. Hence, the vessel experiences a backward thrust.

(e) When the ball spins, the layer of air near its surface is dragged round with it. On one side the velocity of this layer adds to the velocity of the air streaming past the ball and on the other side it subtracts from it. By Bernoulli's theorem the pressure is smaller on the side where the velocity is larger, so a sideways force acts on the ball. This is the Magnus effect, on account of which a spinning cricket ball follows a curved path and not a parabolic trajectory.

Given,

• Mass of the girl, m = 50 kg
• Diameter of the heel, d = 1.0 cm
• g = 9.8 N/kg

The radius of the heel is

$\text r = \frac{\text d}{2} = \frac{1.0\ \text{cm}}{2} = 0.5\ \text{cm} = 0.005\ \text m$

Pressure is the normal force exerted per unit area, and the normal force here is the weight of the girl. Therefore,

$\text p = \frac{\text F}{\text A} = \frac{\text{Mg}}{\pi \text r^2}$

Substituting the values,

$$\begin{aligned}\text p = \frac{50 \times 9.8}{3.14 \times (0.005)^2} \\[1em] = \frac{490}{3.14 \times 2.5 \times 10^{-5}}\end{aligned}$$

$= \frac{490}{7.85 \times 10^{-5}} = 6.2 \times 10^{6}\ \text{N m}^{-2}$

Hence, the pressure exerted by the heel on the horizontal floor is 6.2 × 10⁶ N m^-2.

Given,

• Density of the wine, ρ = 984 kg m^-3
• Normal atmospheric pressure, P = 1.013 × 10⁵ N m^-2
• g = 9.8 N kg^-1

In a barometer the weight of the liquid column balances the atmospheric pressure acting on the surface of the liquid in the container. If h is the height of the wine column, the pressure exerted by it is h ρ g, and this must equal the atmospheric pressure,

$\text h\rho \text g = \text P \quad \Rightarrow \quad \text h = \frac{\text P}{\rho \text g}$

Substituting the values,

$$\begin{aligned}\text h = \frac{1.013 \times 10^{5}}{984 \times 9.8} = \frac{1.013 \times 10^{5}}{9643.2} \\[1em] = 10.5\ \text m\end{aligned}$$

Hence, the height of the wine column is 10.5 m.

As wine is much lighter than mercury, the column required is very long, which is why mercury is preferred in a barometer.

Given,

• Maximum stress the structure can bear = 10⁹ Pa
• Depth of the ocean, h = 3 km = 3 × 10³ m
• Density of water, ρ = 10³ kg m^-3
• g = 9.8 N kg^-1

The structure will be suitable provided the pressure exerted on it by the ocean water is less than the maximum stress it can bear. The pressure due to a liquid column of height h is

$\text p = \text h\rho \text g$

Substituting the values,

$$\begin{aligned}\text p = (3 \times 10^{3}) \times (10^{3}) \times 9.8 \\[1em] = 2.94 \times 10^{7}\ \text{Pa}\end{aligned}$$

Since 2.94 × 10⁷ Pa is much less than the maximum stress of 10⁹ Pa,

Hence, the structure is suitable for putting up on the top of an oil well in the ocean.

Given,

• Maximum mass to be lifted, m = 3000 kg
• Area of cross-section of the larger piston, A = 425 cm² = 425 × 10^-4 m²
• g = 9.8 m s^-2

The pressure exerted on the larger piston by the load is

$\text p = \frac{\text F}{\text A} = \frac{\text{mg}}{\text A}$

Substituting the values,

$$\begin{aligned}\text p = \frac{3000 \times 9.8}{425 \times 10^{-4}} = \frac{29400}{0.0425} \\[1em] = 6.92 \times 10^{5}\ \text{N m}^{-2}\end{aligned}$$

By Pascal's law this pressure is transmitted unchanged through the liquid to the piston of smaller cross-section.

Hence, the smaller piston would have to bear a maximum pressure of 6.92 × 10⁵ Pa.

Given,

• Height of the water column, h_w = 10.0 cm
• Height of the spirit column, h_s = 12.5 cm

The mercury levels in the two arms are in the same horizontal line. In the presence of gravity, the pressure is the same at all points at the same level inside a liquid. Therefore, the pressure at A due to the water column is equal to the pressure at B due to the spirit column,

$\text h_w\rho_w\text g = \text h_s\rho_s\text g$

where the subscripts w and s refer to water and spirit respectively. Hence, the relative density of spirit is

$$\begin{aligned}\frac{\rho_s}{\rho_w} = \frac{\text h_w}{\text h_s} = \frac{10.0\ \text{cm}}{12.5\ \text{cm}} \\[1em] = 0.800\end{aligned}$$

Hence, the relative density of the spirit is 0.800.

Given,

• New height of the water column = 10.0 + 15.0 = 25.0 cm
• New height of the spirit column = 12.5 + 15.0 = 27.5 cm
• Relative density of spirit, $\frac{\rho_s}{\rho_w} = 0.800$ (from the previous question)
• Relative density of mercury, $\frac{\rho}{\rho_w} = 13.6$

On pouring the liquids, mercury falls in the arm containing water and rises in the other, because water is heavier than spirit. Let h be the difference in the mercury levels in the two arms and ρ the density of mercury. Equating the pressures at the lower common level,

pressure of the water column at A = pressure of the spirit column at B + pressure of h cm of mercury column

$(25.0)\rho_w\text g = (27.5)\rho_s\text g + \text h\rho \text g$

Dividing throughout by ρ_w g,

$25.0 = (27.5)\frac{\rho_s}{\rho_w} + \text h\frac{\rho}{\rho_w}$

Substituting the values,

$$\begin{aligned}25.0 = (27.5)(0.800) + 13.6\text h \\[1em] 25.0 = 22.0 + 13.6\text h\end{aligned}$$

$\text h = \frac{25.0 - 22.0}{13.6} = \frac{3.0}{13.6} = 0.221\ \text{cm}$

Hence, the difference in the levels of mercury in the two arms is 0.221 cm.

No, Bernoulli's equation cannot strictly be used to describe such a flow.

Bernoulli's theorem holds for an incompressible and non-viscous liquid flowing in stream-lined motion. In a river rapid the flow is highly turbulent — eddies are formed and a significant amount of energy is lost against viscous forces. Since the total energy per unit volume no longer remains constant along the flow, the conditions under which Bernoulli's theorem is derived are not satisfied.

It may, however, be applied only approximately, by considering an average flow and neglecting the turbulence.

No, it does not matter, provided the atmospheric pressure at the two points is the same.

Let P₁ and P₂ be the absolute pressures at the two points at which Bernoulli's equation is applied. For a horizontal flow,

$\text P_1 - \text P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Gauge pressure is the pressure measured relative to the atmospheric pressure, that is, P_gauge = P_absolute − P_atmospheric. If P is the atmospheric pressure at the two points, then

$$\begin{aligned}(\text P_1)_{gauge} = \text P_1 - \text P \\[1em] (\text P_2)_{gauge} = \text P_2 - \text P\end{aligned}$$

Subtracting,

$(\text P_1)_{gauge} - (\text P_2)_{gauge} = \text P_1 - \text P_2$

Since only the pressure difference enters Bernoulli's equation, and this difference is the same whether gauge or absolute pressures are used, the result is unaffected.

Given,

• Length of the horizontal tube, l = 1.5 m
• Radius of the tube, r = 1.0 cm = 0.01 m
• Mass flowing per second, m/t = 4.0 × 10^-3 kg s^-1
• Density of glycerine, ρ = 1.3 × 10³ kg m^-3
• Coefficient of viscosity of glycerine, η = 0.83 Pa s

The volume of glycerine flowing per second, that is, the rate of discharge, is

$$\begin{aligned}\text Q = \frac{\text m/\text t}{\rho} = \frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}} \\[1em] = 3.08 \times 10^{-6}\ \text{m}^3\text s^{-1}\end{aligned}$$

According to Poiseuille's formula, the rate of discharge through a pipe is

$\text Q = \frac{\pi \text{pr}^4}{8\eta \text l}$

where p is the pressure difference between the two ends. Therefore,

$\text p = \frac{8\eta \text{lQ}}{\pi \text r^4}$

Substituting the values,

$$\begin{aligned}\text p = \frac{8 \times 0.83 \times 1.5 \times 3.08 \times 10^{-6}}{3.14 \times (0.01)^4} \\[1em] = \frac{3.07 \times 10^{-5}}{3.14 \times 10^{-8}}\end{aligned}$$

$= 9.8 \times 10^{2}\ \text{Pa}$

Hence, the pressure difference between the two ends of the tube is 9.8 × 10² Pa.

Given,

• Speed of air over the upper surface, v₁ = 70 m s^-1
• Speed of air over the lower surface, v₂ = 63 m s^-1
• Area of the wing, A = 2.5 m²
• Density of air, ρ = 1.3 kg m^-3

Let P₁ and P₂ be the air pressures at the upper and the lower surfaces of the wing. By Bernoulli's theorem, for the horizontal flow of air,

$\text P_1 + \frac{1}{2}\rho v_1^2 = \text P_2 + \frac{1}{2}\rho v_2^2$

Therefore, the excess of pressure on the lower surface is

$\text P_2 - \text P_1 = \frac{1}{2}\rho (v_1^2 - v_2^2)$

Substituting the values,

$$\begin{aligned}\text P_2 - \text P_1 = \frac{1}{2} \times 1.3 \times [(70)^2 - (63)^2] \\[1em] = 0.65 \times (4900 - 3969)\end{aligned}$$

$= 0.65 \times 931 = 605\ \text{N m}^{-2}$

The force of dynamic lift on the wing is the pressure difference multiplied by the surface area of the wing,

$$\begin{aligned}\text{Lift} = (\text P_2 - \text P_1) \times \text A = 605 \times 2.5 \\[1em] = 1.5 \times 10^{3}\ \text N\end{aligned}$$

Hence, the lift on the wing is 1.5 × 10³ N.

Fig. (a) is incorrect.

At the throat of the tube the area of cross-section is the least. By the principle of continuity, A v = constant, so the velocity of the liquid is greatest at the throat.

By Bernoulli's theorem for a horizontal flow, $\text P + \frac{1}{2}\rho v^2 = \text{constant}$, wherever the velocity is greater the pressure must be smaller. Hence the height of the liquid column in the vertical tube fitted at the throat should be the shortest.

In Fig. (a) the liquid column in the manometer at the throat is shown to be the longest, which would mean a larger pressure where the velocity is largest. Hence, Fig. (a) is incorrect.

Given,

• Area of cross-section of the tube, A₁ = 8.0 cm² = 8.0 × 10^-4 m²
• Number of holes, n = 40
• Diameter of each hole = 1.0 mm, so radius r = 0.5 × 10^-3 m
• Speed of flow inside the tube, v₁ = 1.5 m min^-1

The speed of flow inside the tube in SI units is

$v_1 = \frac{1.5}{60} = 0.025\ \text{m s}^{-1}$

The total area of cross-section of all the 40 holes is

$$\begin{aligned}\text A_2 = \text n\pi \text r^2 = 40 \times 3.14 \times (0.5 \times 10^{-3})^2 \\[1em] = 40 \times 3.14 \times 2.5 \times 10^{-7} = 31.4 \times 10^{-6}\ \text m^2\end{aligned}$$

By the principle of continuity,

$\text A_1 v_1 = \text A_2 v_2 \quad \Rightarrow \quad v_2 = \frac{\text A_1 v_1}{\text A_2}$

Substituting the values,

$$\begin{aligned}v_2 = \frac{8.0 \times 10^{-4} \times 0.025}{31.4 \times 10^{-6}} = \frac{2.0 \times 10^{-5}}{31.4 \times 10^{-6}} \\[1em] = 0.637\ \text{m s}^{-1}\end{aligned}$$

Hence, the speed of ejection of the liquid through the holes is 0.637 m s^-1.

Given,

• Weight supported by the film, Mg = 1.5 × 10^-2 N
• Length of the slider, l = 30 cm = 30 × 10^-2 m

A soap film has two free surfaces, so the force acting on the slider of length l due to the surface tension of the film is

$\text F = \text T \times 2\text l$

In equilibrium this force is equal to the weight supported,

$\text T \times 2\text l = \text{Mg} \quad \Rightarrow \quad \text T = \frac{\text{Mg}}{2\text l}$

Substituting the values,

$$\begin{aligned}\text T = \frac{1.5 \times 10^{-2}}{2 \times (30 \times 10^{-2})} = \frac{1.5 \times 10^{-2}}{0.6} \\[1em] = 0.025\ \text{N m}^{-1}\end{aligned}$$

Hence, the surface tension of the film is 0.025 N m^-1.

The weight supported in Fig. (b) and Fig. (c) is also 4.5 × 10^-2 N, the same as in Fig. (a).

The weight supported by a liquid film of surface tension T on a sliding wire of length l is

$\text W = \text T \times 2\text l$

the factor 2 appearing because the film has two free surfaces.

Here the liquid is the same and the temperature is the same, so T is unchanged; and the length l of the sliding wire is 40 cm in all three figures. Since both T and l are the same, the weight supported must be the same in each case.

Hence, the weight supported in each of the figures (b) and (c) is 4.5 × 10^-2 N.

The weight supported does not depend upon the shape of the frame or the area of the film, because surface tension is a force per unit length and is independent of the surface area.

Given,

• Radius of the mercury drop, R = 3.00 mm = 3.00 × 10^-3 m
• Surface tension of mercury, T = 4.65 × 10^-1 N m^-1
• Atmospheric pressure, P = 1.01 × 10⁵ Pa

A liquid drop has only one free surface, so the excess pressure inside the drop is

$\text p = \frac{2\text T}{\text R}$

Substituting the values,

$$\begin{aligned}\text p = \frac{2 \times (4.65 \times 10^{-1})}{3.00 \times 10^{-3}} = \frac{0.93}{0.003} \\[1em] = 310\ \text{Pa}\end{aligned}$$

The total pressure inside the drop is the atmospheric pressure plus this excess pressure,

$$\begin{aligned}\text P + \text p = (1.01 \times 10^{5}) + (0.0031 \times 10^{5}) \\[1em] = 1.0131 \times 10^{5}\ \text{Pa}\end{aligned}$$

Since the data are given up to three significant figures, this is written as 1.01 × 10⁵ Pa.

Hence, the excess pressure inside the drop is 310 Pa and the total pressure inside the drop is 1.01 × 10⁵ Pa.

Given,

• Radius of the bubble, R = 5.00 mm = 5.00 × 10^-3 m
• Surface tension of soap solution, T = 2.50 × 10^-2 N m^-1
• Depth of the air bubble, h = 40.0 cm = 40.0 × 10^-2 m
• Relative density of soap solution = 1.20, so ρ = 1.20 × 10³ kg m^-3
• Atmospheric pressure, P = 1.01 × 10⁵ Pa
• g = 9.8 N kg^-1

(i) Excess pressure inside the soap bubble : A soap bubble has two free surfaces, so

$$\begin{aligned}\text p = \frac{4\text T}{\text R} = \frac{4 \times (2.50 \times 10^{-2})}{5.00 \times 10^{-3}} \\[1em] = 20.0\ \text{N m}^{-2} = 20.0\ \text{Pa}\end{aligned}$$

(ii) Pressure inside the air bubble : An air bubble formed inside the liquid has only one free surface, so its excess pressure is

$\text p' = \frac{2\text T}{\text R} = \frac{2 \times (2.50 \times 10^{-2})}{5.00 \times 10^{-3}} = 10.0\ \text{Pa}$

The pressure outside the air bubble at a depth h in the soap solution is

$$\begin{aligned}\text P' = \text P + \text h\rho \text g \\[1em] = (1.01 \times 10^{5}) + (40.0 \times 10^{-2})(1.20 \times 10^{3})(9.8)\end{aligned}$$

$= (1.01 \times 10^{5}) + (0.047 \times 10^{5}) = 1.057 \times 10^{5}\ \text{Pa}$

Therefore the total pressure inside the air bubble is

$$\begin{aligned}\text P' + \text p' = (1.057 \times 10^{5}) + 10.0 \\[1em] \simeq 1.06 \times 10^{5}\ \text{Pa}\end{aligned}$$

Hence, the excess pressure inside the soap bubble is 20.0 Pa and the pressure inside the air bubble is 1.06 × 10⁵ Pa, up to three significant figures.

$\frac{Mg}{k}\left(1 - \frac{LA\sigma}{2M}\right)$

Reason — Given,

• Length of the cylinder = L
• Mass of the cylinder = M
• Area of cross-section = A
• Density of the liquid = σ
• Spring constant = k

Three forces act on the cylinder in equilibrium — the weight Mg acting vertically downwards, the upthrust U acting vertically upwards and the spring force k x₀ acting vertically upwards.

For equilibrium,

$\text k\text x_0 = \text{Mg} - \text U \qquad \ldots(\text i)$

By Archimedes' principle, the upthrust is equal to the weight of the liquid displaced. Since the cylinder is half submerged, the volume of liquid displaced is $\text A\left(\frac{\text L}{2}\right)$, so

$\text U = \text A\left(\frac{\text L}{2}\right)\sigma \text g = \frac{\text{AL}\sigma \text g}{2}$

Substituting this in equation (i),

$\text k\text x_0 = \text{Mg} - \frac{\text{AL}\sigma \text g}{2}$

$\text x_0 = \frac{\text{Mg}}{\text k}\left(1 - \frac{\text{LA}\sigma}{2\text M}\right)$

more than half-filled if ρ_c is less than 0.5

Reason — Since ρ_c is the relative density of the material of the shell with respect to water, the relative density of water is 1.

Let V₀ be the outer volume of the shell, V_i the inner volume and V the volume of water filled inside the shell. The shell floats half-submerged, so the volume of water displaced is $\frac{\text V_0}{2}$.

By the law of floatation, the total weight of the shell together with the water inside it equals the weight of the water displaced,

$1 \times \text V \times \text g + \rho_c(\text V_0 - \text V_i)\text g = 1 \times \frac{\text V_0}{2} \times \text g$

$\text V + \rho_c(\text V_0 - \text V_i) = \frac{\text V_0}{2} \quad \Rightarrow \quad \rho_c = \frac{\frac{\text V_0}{2} - \text V}{\text V_0 - \text V_i}$

If ρ_c = 0.5 : substituting and simplifying gives $\text V = \frac{\text V_i}{2}$, that is, the shell is exactly half-filled.

If ρ_c > 0.5 : $\text V \lt \frac{\text V_i}{2}$, that is, the shell is less than half-filled.

If ρ_c < 0.5 : $\text V \gt \frac{\text V_i}{2}$, that is, the shell is more than half-filled with water.

$\frac{1 + \tan \alpha}{1 - \tan \alpha}$

Reason — Let R be the radius of the circular tube. The two liquids meet at the point A, and the radius joining the interface makes an angle α with the vertical.

From the geometry of the figure, the vertical height of the column of the liquid of density d₁ above the level of A is (R cos α + R sin α), while the vertical height of the column of the liquid of density d₂ is (R cos α − R sin α).

In equilibrium, the pressures at the point A due to the two liquid columns must be equal,

$(\text R\cos \alpha - \text R\sin \alpha)\text d_1\text g = (\text R\cos \alpha + \text R\sin \alpha)\text d_2\text g$

Therefore,

$\frac{\text d_1}{\text d_2} = \frac{\text R\cos \alpha + \text R\sin \alpha}{\text R\cos \alpha - \text R\sin \alpha}$

Dividing the numerator and the denominator by R cos α,

$\frac{\text d_1}{\text d_2} = \frac{1 + \tan \alpha}{1 - \tan \alpha}$

Given,

For sphere P :

• Radius, r_P = 0.5 cm
• Density of sphere, ρ_P = 8 gm cm^-3
• Density of liquid, σ₁ = 0.8 gm cm^-3
• Coefficient of viscosity, η₁ = 3 poiseuille

For sphere Q :

• Radius, r_Q = 0.25 cm
• Density of sphere, ρ_Q = 8 gm cm^-3
• Density of liquid, σ₂ = 1.6 gm cm^-3
• Coefficient of viscosity, η₂ = 2 poiseuille

By Stokes' law, the terminal velocity of a sphere of radius r and density ρ falling through a liquid of density σ and coefficient of viscosity η is

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta}$

Therefore, the ratio of the terminal velocities of the two spheres is

$\frac{v_P}{v_Q} = \left(\frac{\text r_P}{\text r_Q}\right)^2\left(\frac{\rho_P - \sigma_1}{\rho_Q - \sigma_2}\right)\left(\frac{\eta_2}{\eta_1}\right)$

Substituting the values,

$$\begin{aligned}\frac{v_P}{v_Q} = \left(\frac{0.5}{0.25}\right)^2\left(\frac{8 - 0.8}{8 - 1.6}\right)\left(\frac{2}{3}\right) \\[1em] = 4 \times \frac{7.2}{6.4} \times \frac{2}{3}\end{aligned}$$

$= 4 \times 1.125 \times 0.667 = 3$

Hence, the ratio of the terminal velocities of P and Q is 3 : 1.

0.025 Nm^-1

Reason — Given,

• Weight supported, mg = 1.5 × 10^-2 N
• Length of the slider, L = 30 cm = 30 × 10^-2 m

A liquid film has two free surfaces, so the upward force on the slider due to surface tension is T × 2L.

In equilibrium this force balances the weight supported,

$\text T \times 2\text L = \text{mg} \quad \Rightarrow \quad \text T = \frac{\text{mg}}{2\text L}$

Substituting the values,

$$\begin{aligned}\text T = \frac{1.5 \times 10^{-2}}{2 \times 30 \times 10^{-2}} = \frac{1}{40} \\[1em] = 0.025\ \text{N m}^{-1}\end{aligned}$$

$\frac{2T}{\rho L}$

Reason — Let the radius of the drop be x and let a layer of thickness dx evaporate from its surface.

The mass of the liquid that evaporates is

$\text m = \text{volume} \times \text{density} = 4\pi \text x^2\text{dx}\rho$

The energy required for this evaporation is

$\text Q = \text{mL} = 4\pi \text x^2\text{dx}\rho \text L \qquad \ldots(\text i)$

The change in the surface area of the drop when its radius decreases from x to (x − dx) is

$\Delta \text A = 4\pi \text x^2 - 4\pi (\text x - \text{dx})^2 = 4\pi \text x^2\left[1 - \left(1 - \frac{\text{dx}}{\text x}\right)^2\right]$

Neglecting the square of the small quantity dx,

$\Delta \text A = 4\pi \text x^2 \times \frac{2\text{dx}}{\text x} = 8\pi \text x\text{dx}$

The energy released by this decrease in surface area is

$\text Q = \text T\Delta \text A = 8\pi \text x\text{dx}\text T \qquad \ldots(\text{ii})$

Since the temperature remains unchanged, the energy released must supply the energy required. Equating (i) and (ii),

$8\pi \text x\text{dx}\text T = 4\pi \text x^2\text{dx}\rho \text L$

$\text x = \frac{2\text T}{\rho \text L}$

$\frac{2T}{b\rho g}\cos(\theta + \alpha/2)$

Reason — Since the tube is a truncated cone of apex angle α, its wall makes an angle $\frac{\alpha}{2}$ with the vertical. Hence the tangent to the liquid surface makes an angle $\left(\theta + \frac{\alpha}{2}\right)$ with the vertical, and not θ as in a uniform tube.

Let r be the radius of curvature of the meniscus at the height h, where the radius of the tube is b. From the geometry of the figure,

$\text r\cos\left(\theta + \frac{\alpha}{2}\right) = \text b \quad \Rightarrow \quad \text r = \frac{\text b}{\cos\left(\theta + \frac{\alpha}{2}\right)}$

The excess pressure on the concave side of the meniscus is

$\text p = \frac{2\text T}{\text r}$

This excess pressure supports the liquid column of height h, so

$\text h\rho \text g = \frac{2\text T}{\text r} = \frac{2\text T\cos\left(\theta + \frac{\alpha}{2}\right)}{\text b}$

$\text h = \frac{2\text T\cos\left(\theta + \frac{\alpha}{2}\right)}{\text b\rho \text g}$

2P₀Rh + Rρgh² − 2RT

Reason — Consider the vertical section ABCD, whose width is the diameter 2R of the beaker and whose height is h.

Take a thin horizontal strip of thickness dx at a depth x below the free surface. The pressure at this depth is (P₀ + ρgx) and the area of the strip is 2R dx. Hence the force on the strip is (P₀ + ρgx) × 2R dx.

The total force on the section due to the water on one side is obtained by integrating from x = 0 to x = h,

$\text F_1 = \int_0^{\text h}(\text P_0 + \rho \text{gx})2\text R\text{dx}$

Integrating,

$$\begin{aligned}\text F_1 = 2\text R\left[\text P_0\text x + \frac{\rho \text{gx}^2}{2}\right]_0^{\text h} \\[1em] = 2\text R\left(\text P_0\text h + \frac{\rho \text{gh}^2}{2}\right)\end{aligned}$$

$= 2\text P_0\text{Rh} + \text R\rho \text{gh}^2$

The surface tension acts along the line AB of length 2R at the free surface, pulling the two halves together. This force is

$\text F_2 = \text T \times 2\text R = 2\text{RT}$

Since this force is directed opposite to the pressure force, the net force on the section is

$\text F = \text F_1 - \text F_2 = 2\text P_0\text{Rh} + \text R\rho \text{gh}^2 - 2\text{RT}$

$R^2\sqrt{\frac{2\rho_w g}{3T}}$

Reason — Let θ be the angle which the radius drawn to the edge of the circular contact makes with the vertical.

From the geometry of the figure,

$\sin \theta = \frac{\text r}{\text R}$

The bubble is held down at the bottom by the vertical component of the force of surface tension acting along the circle of contact of circumference 2πr, which is

$\text F_{\text{ST}} = \text T(2\pi \text r)\sin \theta$

The bubble is pushed up by the buoyant force, which by Archimedes' principle is the weight of the water displaced,

$\text F_b = \frac{4}{3}\pi \text R^3 \rho_w\text g$

The bubble will just detach when the buoyant force becomes greater than the force of surface tension,

$$\begin{aligned}\frac{4}{3}\pi \text R^3 \rho_w\text g \geq \text T(2\pi \text r)\sin \theta \\[1em] \frac{4}{3}\pi \text R^3 \rho_w\text g \geq \text T(2\pi \text r)\frac{\text r}{\text R} \\[1em] \frac{4}{3}\pi \text R^3 \rho_w\text g \geq \frac{2\pi \text{Tr}^2}{\text R}\end{aligned}$$

Solving for r²,

$$\begin{aligned}\text r^2 \leq \frac{4\pi \text R^4\rho_w\text g}{3 \times 2\pi \text T} \\[1em] \text r^2 \leq \frac{2\text R^4\rho_w\text g}{3\text T}\end{aligned}$$

Value of r just before bubbles detatch is

$\text r = \text R^2\sqrt{\frac{2\rho_w\text g}{3\text T}}$

Given,

• Area of the hinged door, A = 20 cm² = 20 × 10^-4 m²
• Height of both the liquid columns, h = 4.0 m
• Density of water, ρ_w = 1.0 × 10³ kg m^-3
• Relative density of acid = 1.7, so ρ_a = 1.7 × 10³ kg m^-3
• g = 9.8 m s^-2

The lateral pressure exerted on the small door at the bottom by the water column is

$$\begin{aligned}\text p_w = \text h\rho_w\text g = 4.0 \times (1.0 \times 10^{3}) \times 9.8 \\[1em] = 3.92 \times 10^{4}\ \text{N m}^{-2}\end{aligned}$$

The pressure exerted by the acid column is

$$\begin{aligned}\text p_a = \text h\rho_a\text g = 4.0 \times (1.7 \times 10^{3}) \times 9.8 \\[1em] = 6.66 \times 10^{4}\ \text{N m}^{-2}\end{aligned}$$

The net pressure on the door is therefore

$$\begin{aligned}\text p_a - \text p_w = (6.66 - 3.92) \times 10^{4} \\[1em] = 2.74 \times 10^{4}\ \text{N m}^{-2}\end{aligned}$$

The net force on the door is

$$\begin{aligned}\text F = (\text p_a - \text p_w)\text A = (2.74 \times 10^{4}) \times (20 \times 10^{-4}) \\[1em] = 54.8\ \text N \simeq 55\ \text N\end{aligned}$$

Hence, a force of 55 N must be applied horizontally on the door from the compartment containing water towards the one containing acid to keep the door closed.

The base area of the tank does not affect the answer, since the pressure at a point depends only on the depth and the density of the liquid.

Given,

• Atmospheric pressure, P = 76 cm of mercury
• Difference of mercury levels in Fig. (a) = 20 cm
• Difference of mercury levels in Fig. (b) = 18 cm

(i) For Fig. (a) : The points A and B lie at the same horizontal level of mercury, so the pressures at A and B are equal. The mercury in the right limb stands 20 cm higher, therefore

$$\begin{aligned}\text{Absolute pressure} = \text P + 20 = 76 + 20 \\[1em] = 96\ \text{cm of mercury}\end{aligned}$$

The gauge pressure is the pressure measured relative to the atmospheric pressure,

$\text{Gauge pressure} = 96 - 76 = 20\ \text{cm of mercury}$

For Fig. (b) : Here the mercury level in the right limb is 18 cm lower, so

$$\begin{aligned}\text{Absolute pressure} = \text P - 18 = 76 - 18 \\[1em] = 58\ \text{cm of mercury}\end{aligned}$$

$\text{Gauge pressure} = 58 - 76 = -18\ \text{cm of mercury}$

The negative gauge pressure shows that the pressure of the gas is lower than the atmospheric pressure.

(ii) Mercury is 13.6 times heavier than water, so a water column of height 13.6 cm exerts the same pressure as a mercury column of height 1 cm.

Hence, on pouring 13.6 cm of water into the right limb, the mercury in the left limb rises by 1 cm and correspondingly the level in the right limb falls, so that

$\text{New difference in mercury levels} = 18 + 1 = 19\ \text{cm}$

Hence, the difference in the mercury levels in the two limbs becomes 19 cm.

(i) Yes, the force exerted by water on the base is the same in the two cases.

The pressure exerted by a liquid column depends only on the vertical depth of the point and the density of the liquid, p = hρg. It depends neither on the shape of the vessel nor on the amount of liquid in it. Here the height h of water and the base area A are the same for both vessels, so the force on the base, F = hρgA, is also the same. This is the familiar hydrostatic paradox.

(ii) The water also exerts force on the side walls of the vessel, and this force is always normal to the wall in contact.

For the vessel with vertical sides, this force on the walls is horizontal and has no vertical component. But for the vessel whose sides are slanting, the force normal to the wall has a non-zero vertical downward component.

Hence, the vessel with the slanting sides presses on the pan with a larger total downward force and gives a higher reading on a weighing scale, even though the force on the base is the same in both cases.

Given,

• Radius of the artery, r = 2 × 10^-3 m, so diameter D = 4 × 10^-3 m
• Coefficient of viscosity of blood, η = 2.084 × 10^-3 Pa s
• Density of blood, ρ = 1.06 × 10³ kg m^-3
• Reynold's number, R_e = 2000

(i) The largest average velocity for which the flow remains stream-lined is the critical velocity, which in terms of Reynold's number is

$v_c = \text R_e\frac{\eta}{\rho \text D}$

Substituting the values,

$$\begin{aligned}v_c = 2000 \times \frac{2.084 \times 10^{-3}}{(1.06 \times 10^{3}) \times (4 \times 10^{-3})} \\[1em] = 2000 \times \frac{2.084 \times 10^{-3}}{4.24}\end{aligned}$$

$= 0.98\ \text{m s}^{-1}$

(ii) The corresponding rate of flow of blood is

$\text Q = \text Av_c = \pi \text r^2 v_c$

Substituting the values,

$$\begin{aligned}\text Q = 3.14 \times (2 \times 10^{-3})^2 \times 0.98 \\[1em] = 3.14 \times (4 \times 10^{-6}) \times 0.98\end{aligned}$$

$= 1.23 \times 10^{-5}\ \text m^3\text s^{-1}$

Hence, the largest average velocity of blood flow is 0.98 m s^-1 and the corresponding flow rate is 1.23 × 10^-5 m³ s^-1.

Given,

• Speed of air over the upper wing surface, v₁ = 234 km/h
• Speed of air over the lower wing surface, v₂ = 180 km/h
• Area of each wing = 25 m², so total wing area A = 2 × 25 = 50 m²
• Density of air, ρ = 1.0 kg m^-3
• g = 10 m s^-2

Converting the speeds into SI units,

$$\begin{aligned}v_1 = \frac{234 \times 1000}{60 \times 60} = 65\ \text{m s}^{-1} \\[1em] v_2 = \frac{180 \times 1000}{60 \times 60} = 50\ \text{m s}^{-1}\end{aligned}$$

Let P₁ and P₂ be the air pressures at the upper and the lower surfaces of the wings. By Bernoulli's theorem,

$\text P_1 + \frac{1}{2}\rho v_1^2 = \text P_2 + \frac{1}{2}\rho v_2^2$

so the excess of pressure over the lower surface is

$\text P_2 - \text P_1 = \frac{1}{2}\rho (v_1^2 - v_2^2)$

In level flight, the weight of the plane is balanced by the upward force due to this excess pressure,

$\text{Mg} = (\text P_2 - \text P_1) \times \text A = \frac{\rho (v_1^2 - v_2^2)\text A}{2}$

$\text M = \frac{\rho (v_1^2 - v_2^2)\text A}{2\text g}$

Substituting the values,

$$\begin{aligned}\text M = \frac{1.0 \times [(65)^2 - (50)^2] \times 50}{2 \times 10} \\[1em] = \frac{1.0 \times (4225 - 2500) \times 50}{20}\end{aligned}$$

$= \frac{1725 \times 50}{20} = 4312\ \text{kg}$

Hence, the mass of the plane is 4312 kg.

Given,

• Radius of the drop, r = 2.0 × 10^-5 m
• Density of the drop, ρ = 1.2 × 10³ kg m^-3
• Coefficient of viscosity of air, η = 1.8 × 10^-5 N s m^-2
• g = 9.8 m s^-2
• Buoyancy is neglected, so σ = 0

By Stokes' law, the terminal velocity of the drop is

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta} = \frac{2}{9}\frac{\text r^2\rho \text g}{\eta}$

Substituting the values,

$$\begin{aligned}v = \frac{2}{9} \times \frac{(2.0 \times 10^{-5})^2 \times (1.2 \times 10^{3}) \times 9.8}{1.8 \times 10^{-5}} \\[1em] = \frac{2}{9} \times \frac{4.704 \times 10^{-6}}{1.8 \times 10^{-5}}\end{aligned}$$

$= 5.8 \times 10^{-2}\ \text{m s}^{-1}$

The viscous force acting on the drop at this terminal speed is given by Stokes' law,

$\text F = 6\pi \eta \text{rv}$

Substituting the values,

$$\begin{aligned}\text F = 6 \times 3.14 \times (1.8 \times 10^{-5}) \times (2.0 \times 10^{-5}) \times (5.8 \times 10^{-2}) \\[1em] = 3.93 \times 10^{-10}\ \text N\end{aligned}$$

Hence, the terminal speed of the drop is 5.8 × 10^-2 m s^-1 and the viscous force on it is 3.93 × 10^-10 N.

Given,

• Radius of the broader limb, r₁ = 3.0 mm = 3.0 × 10^-3 m
• Radius of the narrower limb, r₂ = 1.5 mm = 1.5 × 10^-3 m
• Surface tension of water, T = 7.3 × 10^-2 N m^-1
• Density of water, ρ = 1.0 × 10³ kg m^-3
• Angle of contact, θ = 0
• g = 9.8 N kg^-1

Since the angle of contact of water with glass is zero, the radii of curvature of the meniscus in the two limbs are equal to the radii r₁ and r₂ of the tube in those limbs. The pressure above the meniscus in both the limbs is atmospheric, P.

The pressure on the convex side of a curved liquid surface of radius R is greater than that on the concave side by $\frac{2\text T}{\text R}$. Hence the pressures just below the two menisci are

$\text p_1 = \text P - \frac{2\text T}{\text r_1} \quad \text{and} \quad \text p_2 = \text P - \frac{2\text T}{\text r_2}$

Therefore,

$\text p_1 - \text p_2 = 2\text T\left(\frac{1}{\text r_2} - \frac{1}{\text r_1}\right)$

But this pressure difference is also equal to hρg, where h is the difference in the levels of water in the two limbs. Thus,

$\text h\rho \text g = 2\text T\left(\frac{1}{\text r_2} - \frac{1}{\text r_1}\right) \quad \Rightarrow \quad \text h = \frac{2\text T}{\rho \text g}\left(\frac{\text r_1 - \text r_2}{\text r_1\text r_2}\right)$

Substituting the values,

$$\begin{aligned}\text h = \frac{2 \times (7.3 \times 10^{-2})}{(1.0 \times 10^{3}) \times 9.8}\left[\frac{(3.0 \times 10^{-3}) - (1.5 \times 10^{-3})}{(3.0 \times 10^{-3})(1.5 \times 10^{-3})}\right] \\[1em] = \frac{2 \times 7.3 \times 10^{-5}}{9.8} \times \frac{1}{3.0 \times 10^{-3}}\end{aligned}$$

$= 0.496 \times 10^{-2}\ \text m = 5.0\ \text{mm}$

Hence, the difference in the water levels in the two limbs of the tube is 5.0 mm.

double

Reason — The pressure exerted by a liquid column at a depth h is P = hρg. For a given liquid, ρ and g are constant, so the pressure is directly proportional to the depth. Hence, if the depth is doubled, the pressure at the bottom also becomes double.

the depth below the surface

Reason — The pressure exerted by a liquid at a point is P = hρg, so it depends only on the vertical depth of the point below the free surface and on the density of the liquid. It depends neither upon the shape of the container nor upon the amount of liquid in it. This is the reason for the hydrostatic paradox, in which vessels of different shapes filled with the same liquid up to the same height show the same pressure at their bases.

hydraulic fluid pressure

Reason — Hydraulic brakes work on Pascal's law. When the brake pedal is pressed, a piston moves into the master cylinder containing brake oil. The increase in pressure is transmitted equally and undiminished through the brake oil to the pistons of the wheel cylinder, which move outwards and press the brake-shoes against the rim of the wheel.

provide greater force multiplication

Reason — In a hydraulic system a small force applied on a piston of small area produces a pressure which is transmitted undiminished to a piston of large area, where a much larger force appears. Since the brake oil is practically incompressible, the pressure is transmitted without loss. Hence, hydraulic brakes can multiply the force far more effectively than a purely mechanical linkage.

is constant in magnitude and direction

Reason — In stream-lined flow every particle passing a given point follows exactly the same path as the particle that passed earlier, so the velocity of the fluid at any fixed point remains constant with time in both magnitude and direction. The velocity may, however, be different at different points of the tube.

increases

Reason — By the principle of continuity, A × v = constant at every section of the tube. Hence the velocity of the liquid is smaller in the wider parts of the tube and larger in the narrower parts. So, if the area of cross-section decreases, the velocity of the fluid increases.

Honey

Reason — Thicker liquids like honey, coaltar and glycerine have a larger viscosity than thinner ones like water. If water and honey are poured into separate funnels, water comes out readily from the hole while honey takes a long time, because the relative motion between the layers of honey is opposed much more strongly. The viscosity of air is very small compared with that of any of these liquids.

energy

Reason — Bernoulli's theorem states that when an incompressible and non-viscous liquid flows in stream-lined motion, the total energy per unit volume (pressure energy + kinetic energy + potential energy) is constant at every point of its path, that is,

$\text P + \frac{1}{2}\rho v^2 + \rho \text{gh} = \text{constant}$

It is therefore in one way the principle of conservation of energy applied to a flowing liquid. The conservation of mass gives the equation of continuity, not Bernoulli's theorem.

decreases

Reason — For a horizontal flow, Bernoulli's equation reduces to $\text P + \frac{1}{2}\rho v^2 = \text{constant}$. Since the sum is constant, an increase in the velocity v must be accompanied by a decrease in the pressure P. Thus, in a flowing liquid where the velocity of flow is less the pressure is larger and vice-versa.

the object's size, fluid viscosity, and density

Reason — By Stokes' law, the terminal velocity of a sphere of radius r and density ρ falling through a liquid of density σ and coefficient of viscosity η is

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta}$

Hence the terminal velocity is decided by the radius (size) of the object, the coefficient of viscosity of the fluid and the difference in the densities of the object and the fluid. It does not depend upon the shape of the container.

surface tension

Reason — Surface tension makes the free surface of a liquid contract to the minimum possible area. For a given volume, a sphere has the minimum surface area and hence the minimum potential energy, which is the condition of stable equilibrium. Hence, small water droplets, on which the effect of gravity is insignificant compared with that of surface tension, assume a spherical shape.

surface tension and adhesive forces

Reason — When a capillary tube is dipped in water, the adhesive force between water and glass being greater than the cohesive force, the meniscus becomes concave. The pressure just below a concave meniscus is less than that just above it by 2T/R. To make up this deficiency of pressure, water flows into the tube from outside and rises until hρg = 2T/R. Hence capillarity is the combined result of surface tension and the adhesive forces between the liquid and the tube.

smaller diameter

Reason — By the ascent formula, the height to which a liquid rises in a capillary tube is

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g} \quad \Rightarrow \quad \text h \propto \frac{1}{\text r}$

Thus, as r decreases h increases, that is, the narrower the tube, the greater is the height to which the liquid rises in it.

the temperature of the liquid

Reason — The coefficient of viscosity of a fluid depends only upon the nature of the fluid and not upon the area of the layer or the velocity gradient between the layers. The viscosity of liquids decreases sharply with a rise in temperature and becomes zero at the boiling temperature, whereas the viscosity of gases increases with rise in temperature.

minimize the surface area of a liquid

Reason — The molecules lying in the surface of a liquid have a greater potential energy than those in the interior, because work has to be done against the cohesive force in bringing them to the surface. A system is in stable equilibrium when its potential energy is minimum. Hence the liquid surface tends to have the minimum number of molecules in it, that is, it contracts to a minimum possible area. This tendency is exhibited as surface tension.

the area of output piston is more than the input piston

Reason — In a hydraulic lift, the pressure produced by the force F₁ on the smaller piston of area A₁ is $\text p = \frac{\text F_1}{\text A_1}$. By Pascal's law this pressure is transmitted unchanged to the larger piston of area A₂, so the force on it is

$\text F_2 = \text p \times \text A_2 = \frac{\text F_1}{\text A_1} \times \text A_2$

Since A₂ >> A₁, therefore F₂ >> F₁.

the surface tension of the liquid

Reason — The ascent formula for the rise of a liquid in a capillary tube is

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g}$

so the capillary rise is directly proportional to the surface tension T of the liquid. Viscosity does not enter this relation at all, since capillary rise is a static effect arising from surface tension.

Note: The density ρ also occurs in the ascent formula, so option 2 is not wholly wrong. The intended answer is option 4, since the capillary rise is caused by surface tension and the given options ask for the property responsible for the rise.

resistance to flow

Reason — Viscosity is the property of a liquid by virtue of which it opposes the relative motion between its adjacent layers. The coefficient of viscosity is a measure of this internal resistance to flow — the greater the viscosity, the greater is the resistance offered by the fluid to flow. For example, glycerine offers more resistance than water when it flows through a pipe.

the shape of raindrops

Reason — Surface tension causes the free surface of a liquid to contract to the minimum possible area, and for a given volume a sphere has the least surface area. Hence, raindrops assume a spherical shape. The buoyancy of ships is explained by Archimedes' principle and has nothing to do with surface tension.

is transmitted equally in all directions

Reason — Pascal's law states that if a pressure change is applied to a fluid that is completely confined, incompressible and at rest, that pressure is transmitted equally and undiminished in all directions and also acts on the walls of the container. It is this law that forms the basis of the hydraulic lift, hydraulic brakes and the hydraulic press.

random and chaotic fluid motion

Reason — Turbulent flow is a type of fluid motion in which the particles move in a highly irregular and unpredictable manner. The velocity of the fluid particles changes continuously in both magnitude and direction, the paths of the particles cross each other and swirling regions called eddies are formed. Smooth and orderly layered motion is the characteristic of laminar flow, not of turbulent flow.

conservation of mass

Reason — The equation of continuity, A₁v₁ = A₂v₂, is obtained by equating the mass of liquid entering one end of the tube per second, ρA₁v₁, to the mass of liquid leaving the other end per second, ρA₂v₂. It therefore expresses the conservation of mass in the steady flow of an incompressible fluid.

fluid resistance to flow

Reason — When one layer of a liquid slides over another layer of the same liquid, an internal frictional force acts between them which opposes their relative motion. This property of the liquid by virtue of which it opposes the relative motion between its adjacent layers is called viscosity. The coefficient of viscosity is thus a measure of the internal resistance of the fluid to flow.

streamline flow of small, spherical particles

Reason — Stokes showed that if a small sphere of radius r moves with terminal velocity v through a homogeneous medium of infinite extension, the viscous force acting on it is F = 6πηrv. The law is valid only for very small spherical bodies and assumes laminar flow; for irregular shapes or for higher speeds turbulence develops and Stokes' law fails.

Capillary rise

Reason — The rise or fall of a liquid in a capillary tube arises because the meniscus formed in the tube is curved, and the pressure on the concave side of a curved surface differs from that on the convex side by 2T/R, where T is the surface tension. Hence, capillarity is a direct consequence of the surface tension of the liquid.

random and chaotic

Reason — In turbulent flow the fluid particles move in a highly irregular and unpredictable manner, their paths crossing one another and forming eddies. This type of flow occurs when the velocity of the fluid is very high, that is, when the Reynold's number exceeds about 3000. Ordered, smooth, layered motion is characteristic of laminar flow.

non-viscous flow

Reason — Bernoulli's theorem is derived for an incompressible and non-viscous liquid flowing in stream-lined motion. Its limitations are that viscosity and friction losses are ignored, the density must be constant, the flow must be irrotational and steady, and there must be no heat exchange. Hence it cannot be applied to turbulent or viscous flows.

10,000 Pa

Reason — Given, F = 100 N and A = 0.01 m². The pressure transmitted in the fluid is

$$\begin{aligned}\text P = \frac{\text F}{\text A} = \frac{100}{0.01} \\[1em] = 10^{4}\ \text{Pa} = 10,000\ \text{Pa}\end{aligned}$$

500 N

Reason — Given, F₁ = 10 N and A₂ = 50 A₁.

The mechanical advantage of a hydraulic press is the ratio of the areas of the two pistons,

$\text{M.A.} = \frac{\text A_2}{\text A_1} = 50$

By Pascal's law, the force exerted by the larger piston is

$$\begin{aligned}\text F_2 = \text F_1 \times \text{M.A.} = 10 \times 50 \\[1em] = 500\ \text N\end{aligned}$$

the square of the sphere's radius

Reason — By Stokes' law the terminal velocity is

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta} \quad \Rightarrow \quad v \propto \text r^2$

Thus, the terminal velocity of the ball is directly proportional to the square of its radius. It is inversely proportional to the coefficient of viscosity, and it does not depend on the height from which the sphere is dropped.

cohesive forces among liquid molecules

Reason — A molecule well inside the liquid is attracted equally in all directions by the neighbouring molecules, so the resultant force on it is zero. But a molecule in the surface has its sphere of molecular activity half outside the liquid, so it experiences a resultant inward cohesive force. On account of this the surface behaves like a stretched elastic membrane and tends to contract to a minimum area. Hence, surface tension arises from the cohesive forces among the liquid molecules.

decreases

Reason — Soap and detergents are sparingly soluble impurities, and such impurities lower the surface tension of water. Because of its lower surface tension, a drop of soap solution wets a larger area of the cloth than a drop of pure water and enters the fine pores of the cloth, bringing out the dirt particles with it.

decreases

Reason — By the principle of continuity, a decrease in the area of cross-section increases the velocity of flow. By Bernoulli's theorem for a horizontal flow, $\text P + \frac{1}{2}\rho v^2 = \text{constant}$, so where the velocity is greater the pressure must be smaller. Hence the pressure decreases at the constriction.

ordered and smooth

Reason — In stream-lined flow every particle of the fluid follows a smooth, well-defined path called a streamline, and these streamlines never intersect one another. Each particle passing a given point follows exactly the same path as the particles that passed earlier, so the motion is ordered and smooth. In one line, "streamline flow shows order, laminar flow shows layered smoothness and turbulent flow shows disorder and mixing".

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The pressure exerted by a liquid column is P = hρg. For a given liquid, ρ and g are constant, so P is directly proportional to h, that is, the pressure increases linearly with depth.

Reason (R) is false: The same relation P = hρg shows that the pressure at a point in a fluid at rest depends directly on the density ρ of the fluid. Denser fluids exert more pressure at a given depth; for example, seawater exerts more pressure than freshwater at the same depth.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: A hydraulic press works on Pascal's law. The pressure produced by a small force on a piston of small cross-sectional area is transmitted through the enclosed liquid to a piston of large cross-sectional area, where a much larger force appears.

Reason (R) is false: Pascal's law states that the pressure applied to a completely confined, incompressible fluid at rest is transmitted equally and undiminished in all directions, and not that it decreases exponentially with depth.

Therefore, assertion is true but reason is false.

(A) is false, but (R) is true.

Explanation

Assertion (A) is false: In a hydraulic brake the pedal pushes the input piston of the master cylinder, of area A₁, and the pressure so produced is transmitted to the output piston of the wheel cylinder, of area A₂. The pressure produced is

$\text P = \frac{\text F_1}{\text A_1}$

and the braking force obtained at the output piston is

$\text F_2 = \text{PA}_2 = \text F_1 \times \frac{\text A_2}{\text A_1}$

If the area A₁ of the input piston is increased, then for the same force F₁ applied by the foot the pressure produced decreases, and the ratio $\frac{\text A_2}{\text A_1}$ also decreases. Hence the braking force F₂ decreases and does not increase. To increase the braking force the output piston must be made larger, or the input piston smaller, which is the opposite of what the assertion states.

Reason (R) is true: Pascal's law is correctly stated. The pressure applied to a confined fluid is transmitted equally and undiminished throughout the fluid, and it is this law which transmits the pressure from the master cylinder to the wheel cylinder.

Therefore, assertion is false and reason is true.

Note: The printed answer gives option 2, which is not correct. The foot supplies a fixed force, not a fixed pressure, so a larger input piston gives a smaller braking force. The correct answer is option 4.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Turbulent flow sets in when the velocity of the fluid becomes very high. In terms of Reynold's number, the flow becomes turbulent when R_e > 3000.

Reason (R) is false: Turbulent flow is characterized by highly irregular and chaotic motion in which the paths of the particles cross each other and eddies are formed. Smooth and orderly motion is the characteristic of stream-lined or laminar flow.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The equation of continuity is obtained from the conservation of mass. Since the liquid entering one end of the tube must leave at the other, the mass flowing per second is the same at every section, that is, the mass flow rate is constant in a steady flow.

Reason (R) is also correct: For an incompressible fluid the density ρ is constant, so equating ρA₁v₁ = ρA₂v₂ gives A₁v₁ = A₂v₂, that is, A v = constant.

The constancy of the product A v is precisely the mathematical form which the constancy of the mass flow rate takes for an incompressible fluid. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: The viscosity of gases increases with a rise in temperature. It is the viscosity of liquids which decreases sharply with rise in temperature.

Reason (R) is correct: At a higher temperature the gas molecules do move faster. However, in gases the viscosity is due to the transfer of momentum between adjacent layers, and faster molecules make more frequent collisions, so the momentum transfer and hence the viscosity increases.

Therefore, assertion is false and reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Bernoulli's theorem is not applicable to turbulent or compressible flows. Among its limitations are that it is valid only for non-viscous fluids, assumes constant density, requires irrotational and steady flow, and ignores losses due to friction.

Reason (R) is correct: Bernoulli's theorem is indeed derived for an incompressible and non-viscous liquid flowing in stream-lined motion.

It is precisely because of these restrictive assumptions that the theorem has the limitations which make the Assertion false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a small ball is dropped in a viscous liquid, its velocity goes on increasing and the viscous force also increases with the velocity. A stage is reached when the ball moves with a constant velocity called the terminal velocity.

Reason (R) is also correct: The terminal stage is reached when the upward viscous force 6πηrv together with the upthrust becomes equal to the downward weight of the ball, so that the net force on the ball is zero.

Since a body moves with constant velocity only when the net force on it is zero, the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The surface tension of a liquid decreases with rise in temperature and becomes zero at the critical temperature.

Reason (R) is also correct: On heating, the molecules of the liquid gain kinetic energy and move farther apart, which weakens the cohesive forces between them.

Since surface tension itself arises from the cohesive forces between the liquid molecules, any weakening of these forces lowers the surface tension. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a glass capillary tube is dipped vertically in water, the water rises in the tube above the level outside. This phenomenon is called capillarity.

Reason (R) is also correct: The adhesive force between water molecules and glass molecules is greater than the cohesive force between the water molecules themselves.

Because adhesion exceeds cohesion, the angle of contact is acute and the meniscus in the tube is concave. The pressure just below a concave meniscus is less than that just above it by 2T/R, and to make up this deficiency water flows into the tube and rises. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: On heating, a soap bubble expands in size.

Reason (R) is also correct: The surface tension of a liquid decreases with rise in temperature, so heating lowers the surface tension of the soap solution.

The excess pressure inside a soap bubble is $\text p = \frac{4\text T}{\text R}$, which is the pressure holding the bubble in equilibrium. When T is reduced, the inward pull of the surface is weakened and the gas inside pushes the surface outward until a new equilibrium radius is reached. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Two liquids which do not mix have different densities. Since the pressure at the common level must be the same, h₁ρ₁g = h₂ρ₂g, so the heights of the two columns must be different — the lighter liquid stands higher.

Reason (R) is correct: In the presence of gravity, the pressure is the same at all points at the same level inside a continuous fluid at rest.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Pascal's law strictly applies to incompressible and static fluids, that is, to liquids at rest. Gases are compressible, so unless they are tightly confined so that their volume cannot change, they do not follow Pascal's law exactly.

Reason (R) is also correct: An incompressible fluid does not change its volume under pressure.

Because the volume does not change, the whole of the applied pressure is passed on to every part of the liquid instead of being partly used up in compressing it. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Air bubbles reduce the effectiveness of hydraulic brakes. The fluid used in a hydraulic system must be incompressible to ensure effective transmission of pressure; any air bubbles in the system compress under pressure, making the brakes spongy and causing a loss of pressure transmission.

Reason (R) is correct in so far as it states that air is compressible. However, it is precisely this compressibility that makes the transmission of pressure worse, not better.

Therefore, assertion is false and reason is true.

Note: The Reason as worded contains two parts — that air is compressible (which is true) and that this allows better transmission of pressure (which is false). Taking the Reason as a whole, both statements would be false, but since no such option is provided, option 4 is the answer intended by the textbook.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: In a narrowing pipe the velocity of an incompressible fluid increases, so that the same mass of liquid crosses every section per second.

Reason (R) is correct: The equation of continuity, A v = constant, is a true statement.

In fact the Reason contradicts the Assertion — since A v is constant, a decrease in A must produce an increase in v.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: If water and honey are poured into separate funnels, water comes out readily from the hole while honey takes much longer, because honey is far more viscous than water.

Reason (R) is also correct: Viscosity is the property of a liquid by virtue of which it opposes the relative motion between its adjacent layers.

Since honey has a large coefficient of viscosity, the relative motion between its layers is opposed strongly and it flows slowly. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: The air pressure above the wing is lower than that below it. The wing is so shaped that the curvature of its upper surface is greater than that of the lower surface, so the air above has to travel a larger distance and its velocity is greater. It is this excess of pressure below the wing (P₂ − P₁) that provides the necessary lifting force.

Reason (R) is correct: By Bernoulli's theorem, where the velocity of a flowing fluid is larger the pressure is smaller.

Therefore, assertion is false and reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: By Stokes' law the viscous drag force is F = 6πηrv, which depends directly on the radius r of the sphere.

Reason (R) is correct: The same relation shows that the drag force is proportional to the coefficient of viscosity η of the fluid and to the radius r of the sphere.

The Reason directly contradicts the Assertion.

Therefore, assertion is false and reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Detergents and soap are sparingly soluble impurities, and such impurities decrease the surface tension of water.

Reason (R) is correct: Detergents do reduce the cohesive forces between the water molecules.

Since surface tension arises from the cohesive forces, reducing these forces lowers the surface tension, which is exactly opposite to what the Assertion states.

Therefore, assertion is false and reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: Mercury shows capillary depression, not rise, in a glass tube. Its angle of contact with glass is obtuse (about 135°), so the level of mercury inside the capillary falls below the level outside.

Reason (R) is correct: The cohesive force between mercury molecules is far greater than the adhesive force between mercury and glass.

Because cohesion exceeds adhesion, the meniscus of mercury is convex and the mercury is depressed in the capillary — again the opposite of what the Assertion states.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By Pascal's law, the pressure at a point in a static fluid is equal in all directions, that is, P₁ = P₂ = P₃ for the faces of a small prismatic element.

Reason (R) is also correct: A fluid at rest cannot sustain a tangential or shearing stress, so the force exerted by the fluid on any surface in contact with it must be purely normal to that surface.

Since the force is always normal to whichever surface is placed at the point, no direction is singled out and the pressure comes out the same for every orientation. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: In a hydraulic lift $\text F_2 = \frac{\text F_1}{\text A_1} \times \text A_2$, and since A₂ >> A₁, the force on the larger piston is much greater than that applied on the smaller piston.

Reason (R) is false: By Pascal's law the pressure is transmitted undiminished, so the pressure on both the pistons is the same. The larger force on the bigger piston arises from its larger area, not from any difference of pressure.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The pressure at the bottom of a container does not depend upon the shape or size of the container. Three vessels of different shapes filled with the same liquid up to the same height show the same pressure at their bases — the hydrostatic paradox.

Reason (R) is also correct: The pressure exerted by a liquid column is P = hρg, so for a given liquid it is directly proportional to the depth h.

Since the pressure is determined entirely by the depth and the density, the shape of the vessel cannot enter into it. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A liquid always exerts a force normal to the walls and the bottom of its container at every point, so it exerts pressure at the base.

Reason (R) is also correct: In deriving P = hρg, the weight of the imaginary liquid cylinder standing above the point, mg = A h ρ g, is balanced by the forces on its upper and lower faces. Hence the pressure at a point arises from the weight of the liquid column above it.

Since the pressure at the base is produced by the weight of the liquid standing over it, the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Under no external force, the shape of a liquid drop is decided by surface tension alone, and the drop becomes spherical.

Reason (R) is also correct: For a given volume, a sphere has the minimum surface area.

Surface tension makes the liquid surface contract to the minimum possible area so that its potential energy is minimum, and this condition is satisfied by the spherical shape. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Bernoulli's theorem states that for an incompressible and non-viscous liquid in stream-lined motion, the total energy per unit volume is constant at every point,

$\text P + \frac{1}{2}\rho v^2 + \rho \text{gh} = \text{constant}$

Reason (R) is also correct: In the proof of the theorem, the net work done on the liquid is equated to the net increase in its kinetic and potential energies, which is an application of the conservation of energy.

Since the total energy of the flowing liquid is conserved, the sum of the three terms must remain constant. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: By the ascent formula,

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g} \quad \Rightarrow \quad \text h \propto \frac{1}{\text r}$

so the height of the liquid column is inversely proportional to the radius, and hence to the diameter, of the tube.

Reason (R) is correct: Capillarity does arise from the surface tension of the liquid together with the adhesive forces between the liquid and the material of the tube, which decide the angle of contact.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Hydraulic brakes are more effective than purely mechanical brakes, since a small force on the brake pedal produces a large force on the brake-shoes.

Reason (R) is also correct: The brake oil used is practically incompressible, so by Pascal's law the pressure is transmitted equally and undiminished to the wheel cylinder with minimal loss.

Because the fluid transmits the applied pressure without loss and the wheel-cylinder pistons have a larger area, a greatly multiplied force is available at the brake-shoes. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

The pressure exerted by a column of mercury 1 mm high is called 1 torr.

$1\ \text{torr} = \frac{1}{760}\ \text{atm} \simeq 133.322\ \text{Pa}$

It is a practical unit of pressure and is popular in vacuum science, for measuring low pressures.

One atmosphere is approximately equal to the average atmospheric pressure at sea level,

$1\ \text{atm} \simeq 1.013 \times 10^{5}\ \text{N m}^{-2}\ (\text{or Pa})$

The bar is a unit of pressure equal to 100,000 Pa, that is,

$1\ \text{bar} = 10^{5}\ \text{N m}^{-2}\ (\text{or Pa})$

It is slightly less than 1 atmosphere and is commonly used in meteorology and in engineering.

Stream-line motion is the motion of a fluid in which every particle follows a smooth, well-defined path called a streamline, and these streamlines never intersect one another.

Each particle passing a given point follows exactly the same path as the particles that passed earlier, and the properties of the fluid such as velocity, pressure, density and discharge at a given point remain constant with time.

Velocity gradient is the rate of change of velocity with distance measured perpendicular to the direction of flow. If the velocity changes by Δv over a perpendicular distance Δz, the velocity gradient is $\frac{\Delta v}{\Delta \text z}$.

Unit : s^-1

Dimensions : $\frac{[\text{LT}^{-1}]}{[\text L]} = [\text T^{-1}]$

The coefficient of viscosity of a fluid is numerically equal to the viscous force per unit area which maintains a unit velocity gradient between its two parallel layers. From Newton's law of viscosity,

$\eta = \frac{\text F}{\text A\left(\frac{\text{dv}}{\text{dz}}\right)}$

Dimensions : [M L^-1 T^-1]

Unit : kg m^-1 s^-1

The SI unit of the coefficient of viscosity is kg m^-1 s^-1, that is, kg/(m-s).

Another unit is the poise, where $1\ \text{poise} = \frac{1}{10}\ \text{kg m}^{-1}\text s^{-1}$, so that 1 kg m^-1 s^-1 = 10 poise = 1 decapoise or poiseuille.

When a small ball falls through a viscous liquid, the opposing viscous force goes on increasing with its increasing velocity, until a stage is reached when this viscous force becomes equal to the effective force driving the ball. The net force on the ball then becomes zero and it moves with a constant velocity.

This constant velocity attained by the ball falling in a viscous medium is called its terminal velocity.

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta}$

where r is the radius of the ball, ρ the density of the material of the ball, σ the density of the liquid and η the coefficient of viscosity of the liquid.

The principle of continuity states that when an incompressible and non-viscous liquid flows in stream-lined motion through a tube of non-uniform cross-section, then the product of the area of cross-section and the velocity of flow is the same at every point in the tube.

$\text A \times v = \text{constant}$

Hence the velocity of the liquid is smaller in the wider parts of the tube and larger in the narrower parts.

The SI unit of surface tension is N m^-1 (newton/metre).

Since surface tension is also equal to the work done in increasing the surface area by unity, it may equally be expressed in J m^-2 (joule/metre²).

The surface tension of a liquid is affected by the following factors :

(i) Temperature : It decreases with a rise in temperature and becomes zero at the critical temperature.

(ii) Impurities (solute) : A highly soluble impurity, such as sodium chloride in water, increases the surface tension, while a sparingly soluble impurity, such as soap or phenol, decreases it.

(iii) Contamination : Dust, grease or oil on the water surface reduces its surface tension.

(iv) Nature of the liquid and the medium present on the other side of the surface.

When the surface area of a liquid is increased, molecules from the interior rise to the surface, doing work against the cohesive force. This work is stored in the newly formed surface in the form of potential energy.

This additional energy per unit area of the surface is called the surface energy of the liquid.

The surface tension of a liquid is equal to the work required to increase the surface area of the liquid film by unity at constant temperature, that is,

$\text T = \frac{\text W}{\Delta \text A}$

Numerically, therefore, the surface tension of a liquid is equal to its surface energy per unit area.

When the free surface of a liquid comes in contact with a solid, it becomes curved near the place of contact.

The angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the point of contact is called the 'angle of contact' for that pair of solid and liquid.

For pure water and clean glass it is zero, while for mercury and glass it is about 135°.

An air bubble formed inside a liquid has only one surface in contact with the liquid, so the excess pressure inside it is

$\text p = \frac{2\text T}{\text R}$

where T is the surface tension of the liquid.

Capillarity : When a glass capillary tube open at both ends is dipped vertically in a liquid, the liquid rises or falls in the capillary as compared to the liquid level outside it. This phenomenon of rising or depressing of liquid in very fine bored tubes is called capillarity.

Formula (ascent formula) :

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g}$

where T is the surface tension, θ the angle of contact, r the radius of the capillary tube and ρ the density of the liquid.

The pressure due to a liquid column of height h is P = hρg, so it increases with depth.

Hence the pressure is greatest near the bottom of a storage tank, and the walls there have to withstand the largest outward thrust. For this reason the bottom and the walls near the bottom of a storage tank are made thicker.

The apparent weight of the cork is zero.

Since the cork is floating, by the law of floatation its weight is exactly balanced by the upthrust of the water on it. Hence the apparent weight, which is the true weight minus the upthrust, becomes zero.

Any fractional volume may remain submerged.

For a body in free fall the effective value of g is zero. Hence both the weight of the ice cube and the upthrust of the water on it become zero. Since neither force acts, the ice cube can float with any fraction of its volume submerged.

The critical velocity of a liquid flowing through a tube is

$v_c = \text R_e\frac{\eta}{\rho \text D}$

Hence it depends upon :

(i) the coefficient of viscosity η of the liquid,

(ii) the density ρ of the liquid, and

(iii) the diameter D of the tube,

R_e being Reynold's number.

The layer of water in contact with the stationary bank of the river is at rest, on account of the viscosity of water. As we move away from the bank, each successive layer is dragged less by the layer below it and the velocity of the layers goes on increasing.

Hence, the velocity of the water is least at the bank and greatest in the middle of the river, which is farthest from the stationary surfaces.

Stokes showed that if a small sphere of radius r moves with a terminal velocity v through a perfectly homogeneous medium of infinite extension, the viscous force acting on the sphere is

$\text F = 6\pi \eta \text{rv}$

where,

• F = viscous force acting on the sphere,
• η = coefficient of viscosity of the medium,
• r = radius of the sphere,
• v = terminal velocity of the sphere.

The acceleration becomes zero when the net force acting on the ball becomes zero.

This happens when the upward viscous force together with the upthrust becomes equal to the downward weight of the ball. The ball then falls with a constant velocity, called its terminal velocity.

From the expression $v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta}$, the terminal velocity depends upon :

(i) the radius r of the ball, being directly proportional to its square,

(ii) the density ρ of the material of the ball,

(iii) the density σ of the liquid, and

(iv) the coefficient of viscosity η of the liquid, being inversely proportional to it.

The balls are of the same material and are falling through the same liquid, so ρ, σ and η are the same for both. Hence the terminal velocity is proportional to the square of the radius,

$v \propto \text r^2$

Therefore,

$\frac{v_1}{v_2} = \frac{\text r_1^2}{\text r_2^2}$

Hence, the ratio of their terminal velocities is r₁² : r₂².

No, the velocity does not increase continuously. As a rain drop falls, the viscous drag of the air on it increases with its velocity and finally becomes equal to the effective force of gravity. The drop then falls with a constant terminal velocity.

No, the large and the small drops do not reach the ground with the same velocity. Since the terminal velocity is directly proportional to the square of the radius of the drop, the larger drops reach the ground with a greater velocity.

Clouds are made up of extremely small water droplets. Since the terminal velocity is directly proportional to the square of the radius, these very small droplets fall with an exceedingly small terminal velocity through air.

Hence, they descend so slowly that they appear to be floating in the sky.

In the beginning the parachute is not fully opened, the viscous drag of the air is small and the man falls with nearly the acceleration due to gravity, so his velocity increases.

As the velocity increases the viscous drag of the air also increases, and as the parachute opens fully the effective acceleration is reduced greatly. A stage is soon reached when the upward viscous force becomes equal to the effective weight, so the acceleration becomes zero.

Hence, the man thereafter falls with a constant terminal speed.

For an incompressible and non-viscous liquid flowing in stream-lined motion, Bernoulli's equation is

$\text P + \frac{1}{2}\rho v^2 + \rho \text{gh} = \text{constant}$

Dividing throughout by ρ g,

$\frac{\text P}{\rho \text g} + \frac{v^2}{2\text g} + \text h = \text{constant}$

Velocity head : The term $\frac{v^2}{2\text g}$ is called the velocity head. It has the dimension of height and represents the kinetic energy of the liquid per unit weight, that is, the height through which the liquid would have to fall freely in order to acquire the velocity v.

The wing is so shaped that the curvature of its upper surface is greater than that of the lower surface, the front end being rounded and the back end flattened. During flight, the air flowing above the wing has to travel a larger distance than the air below it, so the velocity of air v₁ at the upper surface is larger than the velocity v₂ at the lower surface.

By Bernoulli's theorem the pressure P₁ at the upper surface is therefore less than the pressure P₂ at the lower surface.

Hence, because of this pressure difference (P₂ − P₁), the aeroplane receives the necessary lifting force.

When wind blows with a high velocity above a tin roof, it causes a lowering of pressure above the roof, in accordance with Bernoulli's theorem. The pressure below the roof, however, remains atmospheric.

Hence, due to this pressure-difference the tin shade is lifted up and blown off.

Bernoulli's theorem is applied here.

Statement : When an incompressible and non-viscous liquid (or gas) flows in stream-lined motion from one place to another, then at every point of its path the total energy per unit volume (pressure energy + kinetic energy + potential energy) is constant, that is,

$\text P + \frac{1}{2}\rho v^2 + \rho \text{gh} = \text{constant}$

Explanation : As the train comes in at high speed, the air between the person and the train moves with a large velocity, so by Bernoulli's theorem the pressure in that region decreases. The air behind the person is still at atmospheric pressure, which is now greater.

Hence, this pressure difference pushes the person towards the train, and for this reason people are prevented from standing near the edge of the platform. Small pieces of straw and paper are seen to fly towards the train for the same reason.

Bernoulli's theorem.

The water rising from the fountain has a very large velocity, so the air-pressure within the stream is lowered. Whenever the ball tends to move out of the stream, the outer air at the larger atmospheric pressure pushes it back into the region of low pressure. Hence the ball remains in stable equilibrium on the fountain.

Bernoulli's theorem, the effect being known as the Magnus effect.

When the ball spins, the layer of air near its surface is dragged along with it. On one side the velocity of the air relative to the ball becomes (v + u) and on the other side (v − u). By Bernoulli's theorem, the pressure is smaller on the side where the velocity is larger, and this pressure difference exerts a sideways force on the ball. Hence, the spinning ball moves along a curved path.

By the principle of continuity, A × v = constant. In deep water the area of cross-section available for the flow is much larger than in shallow water, so to maintain the same volume flow rate the velocity of flow must be smaller.

Hence, deep water flows slowly and appears calm.

The surface tension of water can be decreased :

(i) By heating it, since surface tension decreases with a rise in temperature.

(ii) By adding a sparingly soluble impurity to it, such as soap, oil or phenol.

Cause : Surface tension is caused by the intermolecular cohesive forces between the liquid molecules. A molecule well inside the liquid is attracted equally in all directions, but a molecule in the surface has its sphere of molecular activity half outside the liquid and so experiences a resultant inward pull. Hence the surface tends to contract to a minimum possible area.

Relation : If W is the work done in increasing the surface area of the liquid by ΔA at constant temperature, then

$\text T = \frac{\text W}{\Delta \text A}$

Oil on water : The adhesive force between the oil molecules and the water molecules is greater than the cohesive force between the oil molecules themselves. Hence the oil drop is pulled apart by the water and spreads out in the form of a thin film.

Water in oil : The cohesive force between the water molecules is greater than the adhesive force between the water and the oil molecules. Hence the water drop is pulled inwards by its own molecules and contracts to take the form of a globule.

A thin steel needle placed gently on water floats because the vertical components of the forces of surface tension acting on either side of it balance the weight of the needle.

When soap is dissolved in the water, the surface tension of water is lowered. The upward force of surface tension is then no longer sufficient to balance the weight of the needle.

Hence, the weight of the needle exceeds the force of surface tension acting vertically upward and the needle sinks.

When small pieces of camphor are floated on a clean surface of water, they dissolve in it. The surface tension of the camphor solution is less than that of pure water.

Since the pieces are irregular in shape, one part of the camphor dissolves more than the other, so the surface tension on one side becomes less than that on the other. The piece is therefore dragged towards the region of higher surface tension. Wherever it goes the same thing happens again.

Hence, the camphor pieces dance to and fro on the surface of water.

The surface tension of a liquid decreases with a rise in temperature. Hence the surface tension of hot soup is less than that of cold soup.

Because of its lower surface tension, the hot soup spreads over a larger area of the tongue and comes in contact with a greater number of taste buds.

Hence, the hot soup is tastier than the cold one.

When the capillary tube open at both ends is dipped in water, the water meniscus formed inside the tube is concave, since water wets glass.

The pressure just below a concave meniscus is less than the pressure just above it by 2T/R. To make up this deficiency of pressure, water flows into the tube from outside and rises until the pressure of the water column of height h becomes equal to 2T/R, that is,

$\text h\rho \text g = \frac{2\text T}{\text R}$

The fine capillaries formed in between the threads of the cloth disappear when wax is rubbed on it.

Moreover, wax increases the angle of contact between the cloth and water to an obtuse value, so that cos θ becomes negative and water is not drawn into the pores.

Hence, water cannot penetrate into the cloth and the cloth becomes water-proof.

The cotton threads of the wick have a large number of very fine pores between them, which act as fine capillaries.

By capillary action the kerosene oil rises continuously through these capillaries against gravity and reaches the top of the wick, where it burns.

The soil contains a large number of fine capillaries. If the field is not ploughed, the water of the lower layers of the soil rises up through these capillaries to the surface and evaporates.

On ploughing, the capillaries formed in the soil are broken, so the water can no longer rise to the surface.

Hence, the moisture remains in the lower layers of the soil and is available to the plants.

By the ascent formula, the height to which water rises in a capillary tube is

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g} \quad \Rightarrow \quad \text h \propto \frac{1}{\text r}$

For a given liquid and a given tube material, T, θ, ρ and g are constant, so h is inversely proportional to the radius of the tube.

Hence, the narrower the tube, the greater is the height to which the water rises in it.

Effect of temperature : The surface tension of a liquid decreases with a rise in temperature and becomes zero at the critical temperature.

Effect of contamination : If the liquid surface has dust, grease or oil on it, the surface tension is reduced. Similarly, a highly soluble impurity (such as common salt in water) increases the surface tension, while a sparingly soluble impurity (such as soap or phenol) decreases it.

By Stokes' law, the viscous force on a small sphere is F = 6πηrv, so

$\eta = \frac{\text F}{6\pi \text{rv}}$

Writing the dimensions of each quantity,

$[\eta] = \frac{[\text{MLT}^{-2}]}{[\text L][\text{LT}^{-1}]}$

Simplifying,

$[\eta] = \frac{[\text{MLT}^{-2}]}{[\text L^2\text T^{-1}]} = [\text{ML}^{-1}\text T^{-1}]$

Hence, the dimensional formula of the coefficient of viscosity is [M L^-1 T^-1].

Since the surface tension of a liquid is equal to the work required to increase its surface area by unity, the given value means that

the work required to increase the surface area of the soap solution by 1.0 m² is 3.0 × 10^-2 J.

Equivalently, a force of 3.0 × 10^-2 N acts on either side of an imaginary line of length 1 m drawn in the surface of the soap solution.

Two forces act on the floating needle :

(i) The forces of surface tension T, T acting along the water surface on either side of the needle, and

(ii) The weight W of the needle acting vertically downwards.

On resolving the tension forces into horizontal and vertical components, the horizontal components cancel each other, while the vertical components are added up and balance the weight of the needle.

Hence, the needle floats on the surface of water.

The excess pressure inside a soap bubble of radius R is

$\text p = \frac{4\text T}{\text R} \quad \Rightarrow \quad \text p \propto \frac{1}{\text R}$

Therefore,

$\frac{\text p_1}{\text p_2} = \frac{\text R_2}{\text R_1} = \frac{4}{1}$

Hence, the ratio of the excess pressures in the two bubbles is 4 : 1.

Pressure is the normal force exerted per unit area. Mathematically,

$\text P = \frac{\text F}{\text A}$

where F is the component of the force acting normal to the surface and A is the area over which the force is distributed.

Unit : The SI unit is N m^-2, which is also termed the pascal (Pa), so that 1 Pa = 1 N/m². In the CGS system it is measured in dyne/cm².

Dimensions : $[\text P] = \frac{[\text{MLT}^{-2}]}{[\text L^2]} = [\text{ML}^{-1}\text T^{-2}]$

Nature : Pressure is a scalar quantity. Although it arises from a vector force, it gives only the magnitude of the normal force per unit area and conveys no particular direction of its own, since the direction of the force is always perpendicular to the surface.

(i) Between pure water and clean glass, the angle of contact is 0°.

(ii) Between mercury and glass, the angle of contact is 135°, which is obtuse because mercury does not wet glass.

(iii) Between pure water and silver, the angle of contact is 90°. Hence, in a silver vessel the surface of water at the edges remains horizontal.

If only one hole is made in the tin, the oil does not come out freely. As the oil flows out, the pressure of the air enclosed inside the tin falls below the outside atmospheric pressure, and this excess outside pressure holds the oil back.

When two holes are made, air continuously enters the tin through the second hole. The pressure inside the tin then becomes greater than the atmospheric pressure at the lower hole.

Hence, the oil comes out steadily.

(i) Mercury is the densest common liquid, so only a column of about 0.76 m of mercury is needed to balance the atmospheric pressure. With a lighter liquid such as water the required height would be

$\text h = \frac{0.76\ \text m \times (13.6 \times 10^{3}\ \text{kg m}^{-3})}{10^{3}\ \text{kg m}^{-3}} = 10.34\ \text m$

Such a long barometer tube would be extremely inconvenient to use.

(ii) The vapour pressure of mercury is negligible, so the space above the mercury column is almost a perfect vacuum (the Torricellian vacuum) and the reading is not disturbed by the vapour of the liquid.

Initially the balloon rises because the weight of the air displaced by it, that is, the upthrust of the air on the balloon, is greater than the weight of the helium-filled balloon.

As the balloon rises up, the density of the air decreases with height, and so the upthrust on the balloon also decreases.

Hence, the balloon halts at that height where the upthrust of the air becomes equal to the weight of the helium-filled balloon.

The density of water is much greater than that of air, so by Archimedes' principle the upthrust of the water on the limbs is much greater than the upthrust of the air.

As a result, the net weight of the limbs in water is much less than that in air, and a smaller force is required to move them.

For the same reason it is easier to swim in sea water, which is heavier, than in river water.

By the principle of continuity, A v = constant, so the velocity of water in the narrower pipe is larger than that in the broader pipe.

By Bernoulli's theorem for a horizontal flow,

$\text P + \frac{1}{2}\rho v^2 = \text{constant}$

Hence, at the point where the velocity is large, the pressure is small, and so the pressure of water is reduced in the narrow pipe.

Bernoulli's theorem is derived for a non-viscous liquid, in which no energy is lost during flow.

Real water, however, is viscous. As the water flows through the pipe, work has to be done against the viscous force between its layers, and this work is taken from the pressure energy of the water.

Hence, the pressure of the water goes on decreasing along a pipe of uniform radius.

The hole is at a height h from the bottom, so the depth of the hole below the free surface is (H − h).

Velocity of efflux : By Torricelli's theorem, the velocity of efflux of a liquid from an orifice at a depth d below the free surface is $v = \sqrt{2\text{gd}}$. Here d = (H − h), so

$v = \sqrt{2\text g(\text H - \text h)}$

Horizontal range : After emerging, the water adopts a parabolic path. If it takes t second to fall through the vertical distance h, then from $\text s = \frac{1}{2}\text{at}^2$,

$\text h = \frac{1}{2}\text{gt}^2 \quad \Rightarrow \quad \text t = \sqrt{\frac{2\text h}{\text g}}$

Since there is no acceleration in the horizontal direction, the horizontal distance covered is

$$\begin{aligned}\text R = v \times \text t = \sqrt{2\text g(\text H - \text h)} \times \sqrt{\frac{2\text h}{\text g}} \\[1em] = \sqrt{4\text h(\text H - \text h)}\end{aligned}$$

Hence, the velocity of efflux is $\sqrt{2\text g(\text H - \text h)}$ and the horizontal range is $\sqrt{4\text h(\text H - \text h)}$.

The surface energy will increase.

When a big drop is broken into a large number of small drops, the total volume remains the same but the total surface area increases considerably.

Since the surface energy is the product of the surface tension and the surface area, W = T × ΔA, an increase in the surface area means an increase in the surface energy. This energy is supplied by the external work done in breaking the drop, and if no work is done, the temperature of the drops falls.

Since the density of the water-alcohol mixture is exactly equal to the density of the oil, the weight of the drop is exactly balanced by the upthrust of the mixture on it. Hence there is no effect of gravity on the drop.

The shape of the drop is then controlled by surface tension alone, which makes the surface contract to the minimum possible area. For a given volume the sphere has the minimum surface area.

Hence, the oil drop becomes perfectly spherical.

On sprinkling oil, the breeze spreads the oil over the sea-water in its own direction, so that one part of the surface is covered with oil and the other is not.

The surface tension of pure sea-water is greater than that of the oily water. Hence the water without oil pulls the oily water against the direction of the breeze.

Hence, the sea waves become calm.

Sugar : Sugar is a highly soluble impurity, so the surface tension of the sugar solution is greater than that of pure water. The surface therefore contracts near the piece of sugar, and the straw come nearer to it.

Soap : Soap is a sparingly soluble impurity, so the surface tension of the soap solution is less than that of pure water. The surface near the soap therefore spreads out, and the straw go away from it.

In both cases the straw move towards the region of higher surface tension.

Yes, the liquid will wet the solid, and it will rise in the capillary.

An acute angle of contact means that the adhesive force between the liquid and the solid is greater than the cohesive force between the liquid molecules, so the liquid clings to the solid and wets it.

Further, from the ascent formula $\text h = \frac{2\text T\cos \theta}{\text r\rho \text g}$, when θ < 90° the value of cos θ is positive, so h is positive and the liquid rises in the capillary.

The bigger bubble will expand at the cost of the smaller one.

The excess pressure inside a soap bubble is $\text p = \frac{4\text T}{\text R}$, which is inversely proportional to the radius of the bubble.

Hence the pressure inside the smaller bubble is greater than that inside the bigger bubble. Air therefore flows from the smaller bubble to the bigger one through the tube, so the smaller bubble shrinks and the bigger bubble expands.

The angle of contact for mercury and glass is obtuse (about 135°), so mercury does not wet glass and its meniscus in a glass tube is convex.

Consequently, from the ascent formula, cos θ is negative and h becomes negative, that is, mercury is depressed in a fine glass tube instead of rising in it.

Hence, when one end of the thermometer tube is dipped in mercury, the mercury descends and it becomes difficult to fill the tube.

At the critical temperature.

The surface tension of a liquid decreases with a rise in temperature and becomes zero at the critical temperature, at which the distinction between the liquid and its vapour disappears and the liquid no longer has a free surface.

Given, $\text r_A = \frac{\text r_B}{2}$, so $\frac{\text r_A}{\text r_B} = \frac{1}{2}$.

The terminal velocity of a ball falling in a viscous liquid is directly proportional to the square of its radius,

$v \propto \text r^2$

Therefore,

$\frac{v_A}{v_B} = \left(\frac{\text r_A}{\text r_B}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

Hence, the ratio of the terminal velocities of the balls A and B is 1 : 4.

Given, v_A = 4 v_B.

By the principle of continuity,

$\text A_A v_A = \text A_B v_B \quad \Rightarrow \quad \frac{v_A}{v_B} = \frac{\text A_B}{\text A_A}$

Writing the areas in terms of the diameters,

$\frac{v_A}{v_B} = \frac{\pi \text D_B^2/4}{\pi \text D_A^2/4} = \frac{\text D_B^2}{\text D_A^2}$

Substituting v_A/v_B = 4,

$4 = \frac{\text D_B^2}{\text D_A^2} \quad \Rightarrow \quad \frac{\text D_A}{\text D_B} = \frac{1}{2}$

Hence, the diameter of the pipe at A is half of that at B.

Given,

• Mass of the ball = m
• Density of the ball = ρ₁
• Density of glycerine = ρ₂

The true weight of the ball is mg, and the volume of the ball is $\frac{\text m}{\rho_1}$.

By Archimedes' principle, the buoyant force on the ball is the weight of the glycerine displaced,

$\text F_b = \frac{\text m}{\rho_1} \times \rho_2 \times \text g$

Since the velocity has become constant, the net force on the ball is zero. Hence the viscous force acting on the ball is equal to the net weight of the ball,

$\text F_v = \text{mg} - \text m\left(\frac{\rho_2}{\rho_1}\right)\text g = \text{mg}\left(1 - \frac{\rho_2}{\rho_1}\right)$

Therefore,

$\frac{\text{Viscous force on the ball}}{\text{True weight of the ball}} = \frac{\text{mg}\left(1 - \frac{\rho_2}{\rho_1}\right)}{\text{mg}} = \left(1 - \frac{\rho_2}{\rho_1}\right)$

Hence, the required ratio is $\left(1 - \frac{\rho_2}{\rho_1}\right)$.

Let r be the radius of each tiny drop. Since the total volume remains unchanged,

$\frac{4}{3}\pi \left(\frac{\text D}{2}\right)^3 = 27 \times \frac{4}{3}\pi \text r^3$

$\left(\frac{\text D}{2}\right)^3 = 27\text r^3 \quad \Rightarrow \quad \text r = \frac{\text D}{6}$

The increase in potential energy is the work done against surface tension, W = T × ΔA, where ΔA is the increase in the total surface area,

$\text W = \text T\left[27 \times 4\pi \text r^2 - 4\pi \left(\frac{\text D}{2}\right)^2\right]$

Substituting $\text r = \frac{\text D}{6}$,

$$\begin{aligned}\text W = \text T\left[27 \times 4\pi \left(\frac{\text D}{6}\right)^2 - 4\pi \left(\frac{\text D}{2}\right)^2\right] \\[1em] = 4\pi \text T\left[\frac{27\text D^2}{36} - \frac{\text D^2}{4}\right]\end{aligned}$$

$= 4\pi \text T\left[\frac{3\text D^2}{4} - \frac{\text D^2}{4}\right] = 4\pi \text T \times \frac{2\text D^2}{4}$

$\text W = 2\pi \text D^2\text T$

Hence, the increase in potential energy is 2πD²T.

Let a big drop of radius R be broken into n³ droplets each of radius r. Since the total volume remains unchanged,

$\frac{4}{3}\pi \text R^3 = \text n^3 \times \frac{4}{3}\pi \text r^3 \quad \Rightarrow \quad \text R = \text{nr}$

The work done is equal to the surface tension multiplied by the increase in the surface area,

$\text W = \text T \times [(4\pi \text r^2)\text n^3 - 4\pi \text R^2]$

Substituting R = n r,

$$\begin{aligned}\text W = \text T[4\pi \text r^2\text n^3 - 4\pi (\text{nr})^2] \\[1em] = \text T \times 4\pi \text r^2[\text n^3 - \text n^2]\end{aligned}$$

$\text W = 4\pi \text r^2\text{Tn}^3\left(1 - \frac{1}{\text n}\right)$

Hence, the work done in breaking the drop is $4\pi \text r^2\text{Tn}^3\left(1 - \frac{1}{\text n}\right)$.

For a soap bubble the excess pressure is $\text p = \frac{4\text T}{\text R}$.

For the first bubble, $\text p_1 = \frac{4\text T}{\text R_1}$, and for the second bubble, $\text p_2 = \frac{4\text T}{\text R_2}$.

Given that p₁ = 3 p₂,

$\frac{4\text T}{\text R_1} = 3 \times \frac{4\text T}{\text R_2} \quad \Rightarrow \quad \text R_2 = 3\text R_1$

The ratio of the volumes is therefore

$\frac{\text V_1}{\text V_2} = \frac{\frac{4}{3}\pi \text R_1^3}{\frac{4}{3}\pi \text R_2^3} = \left(\frac{\text R_1}{\text R_2}\right)^3 = \left(\frac{1}{3}\right)^3$

$\frac{\text V_1}{\text V_2} = \frac{1}{27}$

Hence, the ratio of the volumes of the first and the second bubble is 1 : 27.

A soap bubble has two free surfaces, so its total surface area is 2 × 4πR².

The increase in the surface area when the radius increases from R to 3R is

$$\begin{aligned}\Delta \text A = 2[4\pi (3\text R)^2 - 4\pi \text R^2] \\[1em] = 8\pi [9\text R^2 - \text R^2] = 64\pi \text R^2\end{aligned}$$

The energy spent is the work done against surface tension,

$\text W = \Delta \text A \times \text T = 64\pi \text R^2\text T$

Hence, the energy spent in blowing the bubble is 64πR²T.

No, it cannot be used.

Bernoulli's equation holds good only for the stream-lined flow of an incompressible and non-viscous liquid, in which the total energy per unit volume remains constant.

The flow of water through a rapid in a river is turbulent — eddies are formed and a considerable amount of energy is lost against viscous forces. Since the conditions assumed in deriving the theorem are not satisfied, Bernoulli's equation cannot be applied to such a flow.

Under normal conditions, water rises in a capillary tube until the upward force due to surface tension becomes equal to the weight of the water column raised, that is, until

$\text h\rho \text g = \frac{2\text T\cos \theta}{\text r}$

In a state of weightlessness the effective value of g is zero, so the weight of the water column raised in the tube becomes zero. The rising of water therefore never stops on account of its weight.

Hence, the water will rise right up to the other end of the capillary, however long the capillary may be.

A soap bubble has two free surfaces, an inner one and an outer one. Hence the total increase in the surface area in forming a bubble of radius r is

$\Delta \text A = 2 \times 4\pi \text r^2 = 8\pi \text r^2$

The work done is the surface tension multiplied by the increase in surface area,

$\text W = \text T \times \Delta \text A = 8\pi \text r^2\text T$

Hence, the work done in forming the soap bubble is 8πr²T.

(i) For a molecule on the water surface near the silver wall, P is the resultant adhesive force acting perpendicular to the silver surface and Q is the resultant cohesive force acting towards the interior of the water at 45° to the wall. Since the angle of contact is 90°, the liquid surface must be horizontal there, and the resultant R, which is always perpendicular to the liquid surface, is therefore directed vertically downward.

(ii) For the resultant R to be vertical, the horizontal components of P and Q must cancel. From the geometry of the figure, this requires

$\text P = \text Q\cos 45^\circ = \frac{\text Q}{\sqrt 2}$

$\frac{\text Q}{\text P} = \sqrt 2$

Hence, the ratio of the cohesive force to the adhesive force is $\sqrt 2 : 1$.

(iii) From the ascent formula, $\text h = \frac{2\text T\cos \theta}{\text r\rho \text g}$. Here θ = 90°, so cos 90° = 0 and hence h = 0.

Hence, the water will neither rise up nor fall down in the silver capillary.

Given,

• Radius of the first tube = 2r, radius of the second tube = 4r
• Length of each tube = l
• Pressure difference across the whole system, P − P₂ = 8.5 cm of Hg

Let P, P₁ and P₂ be the pressures at the beginning of the first tube, at the joint of the two tubes and at the end of the second tube respectively.

Since the tubes are connected in series, the rate of flow through both is the same. By Poiseuille's formula,

$\frac{\pi (\text P - \text P_1)(2\text r)^4}{8\eta \text l} = \frac{\pi (\text P_1 - \text P_2)(4\text r)^4}{8\eta \text l}$

$(\text P - \text P_1) \times 16 = (\text P_1 - \text P_2) \times 256$

$\text P - \text P_1 = 16(\text P_1 - \text P_2) \qquad \ldots(\text i)$

Also, it is given that

$\text P - \text P_2 = 8.5 \qquad \ldots(\text{ii})$

Writing equation (i) as P = 17P₁ − 16P₂ and substituting P from equation (ii), that is, P = P₂ + 8.5,

$$\begin{aligned}\text P_2 + 8.5 = 17\text P_1 - 16\text P_2 \\[1em] \text P_1 - \text P_2 = 0.5\ \text{cm of Hg}\end{aligned}$$

Therefore the pressure difference between the ends of the first tube is

$$\begin{aligned}\text P - \text P_1 = (\text P - \text P_2) - (\text P_1 - \text P_2) \\[1em] = 8.5 - 0.5 = 8.0\ \text{cm of Hg}\end{aligned}$$

Hence, the pressure difference between the ends of the first capillary tube is 8.0 cm of Hg.

(i) The student's response is correct.

A fluid at rest cannot sustain a tangential (shearing) stress. If the force were not normal to the surface, it would have a tangential component, and since a liquid begins to flow the moment a tangential force is applied to it, the liquid could not remain at rest. Hence the tangential component of the force must necessarily be zero, that is, the force exerted by a fluid at rest is always perpendicular to the surface in contact.

(ii) The student's response is correct.

The blow given to the cork exerts a pressure $\left(\text p = \frac{\text f}{\text a}\right)$ on the liquid, where f is the force and a is the cross-section area of the cork. By Pascal's law this pressure is transmitted undiminished to each part of the liquid and the bottle.

If A is the area of the surface of the bottle, the force on the bottle's surface is

$\text F = \text p \cdot \text A = \frac{\text f}{\text a}\text A$

Since A >> a, therefore F >> f, and this large force may be enough to break the bottle.

(iii) The student's response is incorrect.

The density of air is maximum near the surface of the earth and decreases rapidly with height, not the other way round. At a height of about 6 km the air density decreases to half its value at sea level, and accordingly the atmospheric pressure at 6 km decreases to nearly half its sea-level value. Beyond 6 km the density of air decreases slowly with height, so the atmosphere continues to exist up to several hundred kilometres height.

(i) Incompressible fluids only

Pascal's law applies strictly to incompressible and static fluids, that is, to liquids at rest, since they transmit pressure without any significant change of volume. Gases are compressible, so unless they are tightly confined they do not follow Pascal's law exactly.

(ii) Hydraulic lift

A hydraulic lift works on the transmission of pressure through an enclosed liquid, which is Pascal's law. The barometer and the manometer are based on the pressure exerted by a liquid column, while the Venturi meter is based on Bernoulli's theorem.

(iii) 10

The mechanical advantage of a hydraulic system is the ratio of the areas of the two pistons,

$\text{M.A.} = \frac{\text A_2}{\text A_1} = \frac{0.1}{0.01} = 10$

(iv) 1000 N

Since the area of the larger piston is 10 times that of the smaller piston, the mechanical advantage is 10. Hence the force on the larger piston is

$\text F_2 = \text F_1 \times \text{M.A.} = 100 \times 10 = 1000\ \text N$

(v) The brakes may fail to work properly

Air is compressible, so the air bubbles compress under pressure instead of transmitting it. This reduces the effective transmission of pressure to the brake-shoes, makes the brakes spongy and may cause brake failure.

(i) Incompressible and non-viscous fluids

Bernoulli's theorem is valid for an ideal fluid that is incompressible and has negligible viscosity, since the theorem assumes no energy loss due to internal friction or turbulence.

(ii) Pressure decreases

For a horizontal flow, $\text P + \frac{1}{2}\rho v^2 = \text{constant}$, so an increase in the velocity of the fluid must be accompanied by a decrease in its pressure.

(iii) Bernoulli's theorem

A Venturi meter measures the rate of flow of a fluid by creating a constriction where the velocity of the fluid increases and the pressure decreases. This inverse relation between pressure and velocity is exactly what Bernoulli's theorem expresses.

(iv) Lift on airplane wings

The velocity of air above the curved upper surface of a wing is larger than that below it, so the pressure above is smaller. This pressure difference provides the lift, which is a direct consequence of Bernoulli's theorem. Hydraulic brakes work on Pascal's law and barometers on the pressure of a liquid column.

(v) ρgh

In Bernoulli's equation the three terms represent the energy per unit volume — P is the pressure energy, $\frac{1}{2}\rho v^2$ is the kinetic energy and ρgh is the potential energy per unit volume, due to the height h of the fluid above a reference level.

(i) Fluid flow resistance

Viscosity is the property of a liquid by virtue of which it opposes the relative motion between its adjacent layers. It is therefore a measure of the internal resistance which the fluid offers to flow.

(ii) Radius

By Stokes' law, F = 6πηrv, so the drag force is directly proportional to the radius r of the sphere, to the coefficient of viscosity η and to the velocity v. It does not involve the mass, density or acceleration of the object.

(iii) Terminal velocity increases

Since $v \propto \text r^2$, a larger raindrop has a greater terminal velocity. Although the drag force also increases with size, it increases less rapidly than the gravitational force acting on the larger drop.

(iv) Honey

Honey has a much higher coefficient of viscosity than water, olive oil or air, which is why it flows slowly and resists motion more strongly.

(v) The drag force equals the gravitational force

At the terminal velocity the object moves with a constant speed, so the net force on it is zero. This happens when the sum of buoyant force and drag force equals the gravitational force.

(i) The pressure exerted by a liquid column is P = hρg, so it increases with the height of the column standing above the point. In the human body the height of the blood column above the feet is much greater than that above the brain.

Hence, the blood pressure is greater at the feet than at the brain.

(ii) Given, F = 120 N and A = 0.02 m².

The pressure on the tip of the nail is

$$\begin{aligned}\text P = \frac{\text F}{\text A} = \frac{120}{0.02} \\[1em] = 6000\ \text{N m}^{-2}\end{aligned}$$

Hence, the pressure on the tip is 6000 N m^-2.

(iii) Given, m = 3000 kg, A = 425 cm² = 425 × 10^-4 m² and g = 9.8 m s^-2.

The pressure on the piston carrying the load is

$$\begin{aligned}\text P = \frac{\text F}{\text A} = \frac{\text{mg}}{\text A} = \frac{3000 \times 9.8}{425 \times 10^{-4}} \\[1em] = 6.92 \times 10^{5}\ \text{N m}^{-2}\end{aligned}$$

By Pascal's law this pressure is transmitted undiminished to the smaller piston.

Hence, the smaller piston would have to bear a maximum pressure of 6.92 × 10⁵ N m^-2.

(i) When a ball is gently dropped in a liquid column, its velocity increases gradually. Due to the viscous force acting between the adjacent layers of the liquid, a force acts on the ball which increases with the increase in the velocity of the ball. When this force becomes equal to the effective force driving the ball, the net force on the ball becomes zero.

The constant velocity which the ball then attains is called its terminal velocity.

(ii) No, they do not reach the ground with the same velocity.

The terminal velocity is directly proportional to the square of the radius of the drop, $v \propto \text r^2$. Hence the large raindrops, having a greater terminal velocity, reach the ground with a greater velocity than the small ones.

(iii) The terminal velocity of a small ball falling freely in a viscous liquid is

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta}$

where,

• r = radius of the ball,
• ρ = density of the material of the ball,
• σ = density of the liquid,
• η = coefficient of viscosity of the liquid.

(i) Under no external force, the shape of a liquid drop is decided by surface tension alone. Surface tension makes the free surface contract so that the drop attains the minimum potential energy, and for this the liquid drop has a tendency to occupy the minimum surface area. For a given volume of liquid, the minimum surface area is that of a sphere.

Hence, a drop of liquid under no external force is always spherical in shape.

(ii) Given, R = 3.0 mm = 3.0 × 10^-3 m and T = 4.65 × 10^-1 N m^-1.

A liquid drop has only one free surface, so the excess pressure inside it is

$$\begin{aligned}\text p = \frac{2\text T}{\text R} = \frac{2 \times 4.65 \times 10^{-1}}{3.0 \times 10^{-3}} \\[1em] = 3.10 \times 10^{2}\ \text{N m}^{-2}\end{aligned}$$

Hence, the excess pressure inside the mercury drop is 3.10 × 10² N m^-2.

(iii) Given, R = 6.0 mm = 6.0 × 10^-3 m and T = 3.3 × 10^-2 N m^-1.

A soap bubble has two free surfaces, so the excess pressure inside it is

$$\begin{aligned}\text p = \frac{4\text T}{\text R} = \frac{4 \times 3.3 \times 10^{-2}}{6.0 \times 10^{-3}} \\[1em] = 22\ \text{N m}^{-2}\end{aligned}$$

Hence, the excess pressure inside the soap bubble is 22 N m^-2.

Note: In part (iii) the phrase "surface temperature of soap solution" appears to be a misprint for "surface tension of soap solution", since the value 3.3 × 10^-2 Nm^-1 is given in the unit of surface tension. The solution has been worked out accordingly.

Fluid : Liquids and gases together are called fluids. Unlike solids, a fluid cannot resist a shear stress and continuously deforms or flows under its influence. A liquid has no shape of its own and always takes the shape of its container, though it has a definite volume; a gas has neither a definite shape nor a definite volume.

A fluid exerts pressure : Consider a liquid of density ρ contained in a vessel. Let C and D be two points inside the liquid at a vertical distance h apart. Imagine a cylindrical column of the liquid of cross-sectional area A, with C and D at the centres of its upper and lower faces.

The weight of the liquid in this imaginary cylinder is

$\text{mg} = \text{volume} \times \text{density} \times \text g = \text{Ah}\rho \text g$

which acts vertically downwards. If p₁ and p₂ are the pressures at C and D, the force on the upper face is p₁A acting downwards and that on the lower face is p₂A acting upwards. The horizontal forces on the curved surface balance each other.

Since the cylinder is in equilibrium, the net force on it is zero,

$(\text p_1\text A + \text{mg}) - \text p_2\text A = 0$

$(\text p_1\text A + \text{Ah}\rho \text g) - \text p_2\text A = 0$

$\text p_2 - \text p_1 = \text h\rho \text g$

This shows that the liquid exerts a pressure which increases with the depth h. Hence, a fluid exerts pressure.

Force on a liquid in equilibrium is normal to its surface : Consider a liquid contained in a vessel in equilibrium at rest. Suppose the liquid exerts a force F on the bottom surface in an inclined direction OD. By Newton's third law, the surface exerts an equal and opposite reaction R on the liquid along OC.

This reaction R can be resolved into two components :

(i) the horizontal (tangential) component, OA = R cos θ, which acts parallel to the liquid surface, and

(ii) the vertical (normal) component, OB = R sin θ.

A liquid at rest cannot resist a tangential (shear) stress. If the tangential component R cos θ were present, the liquid particles near the point O would begin to flow and continuously change shape. But the liquid is observed to be at rest, so the tangential force must be zero,

$\text R\cos \theta = 0$

Since R ≠ 0, we must have

$\cos \theta = 0 \quad \Rightarrow \quad \theta = 90^\circ$

Hence, a liquid in equilibrium always exerts force normal to the walls or bottom of its container at every point.

Let the hole be at a depth h below the free surface of the water, so that the height of the hole above the floor is (H − h).

Velocity of efflux : By Torricelli's theorem, the velocity with which the water emerges from the hole is

$v = \sqrt{2\text{gh}}$

Time of fall : After emerging, the water adopts a parabolic path. If it takes t second to fall through the vertical distance (H − h), then from $\text s = \frac{1}{2}\text{at}^2$,

$(\text H - \text h) = \frac{1}{2}\text{gt}^2 \quad \Rightarrow \quad \text t = \sqrt{\frac{2(\text H - \text h)}{\text g}}$

Range : Since there is no acceleration in the horizontal direction, the horizontal velocity remains constant. The horizontal distance covered is

$$\begin{aligned}\text x = v \times \text t = \sqrt{2\text{gh}} \times \sqrt{\frac{2(\text H - \text h)}{\text g}} \\[1em] = 2\sqrt{\text h(\text H - \text h)}\end{aligned}$$

Condition for maximum range : The range x will be maximum when h(H − h) is maximum. Differentiating with respect to h and equating to zero,

$\frac{\text d}{\text{dh}}[\text h(\text H - \text h)] = 0$

$\frac{\text d}{\text{dh}}(\text{hH} - \text h^2) = 0 \quad \Rightarrow \quad \text H - 2\text h = 0$

$\text h = \frac{\text H}{2}$

Substituting this value of h, the maximum range is

$\text x_{max} = 2\sqrt{\left(\frac{\text H}{2}\right)\left(\text H - \frac{\text H}{2}\right)} = \text H$

Hence, the stream strikes the floor at a distance $2\sqrt{\text h(\text H - \text h)}$ from the foot of the wall, and the range is maximum, equal to H, when the hole is exactly in the middle of the wall, that is, h = H/2.

Poiseuille's concept : Experimentally it can be shown that the volume of a liquid flowing per second, that is, the rate of discharge Q through a pipe, depends upon

(i) the coefficient of viscosity η of the liquid,

(ii) the radius r of the pipe, and

(iii) the pressure gradient, that is, the change of pressure per unit length, P/l.

Derivation by the method of dimensions : We may therefore write

$\text Q = \frac{\text V}{\text t} = \text k(\eta)^a(\text r)^b\left(\frac{\text P}{\text l}\right)^c \qquad \ldots(\text i)$

where k is a constant of proportionality.

Writing the dimensions of the quantities involved,

$[\text L^3\text T^{-1}] = [\text{ML}^{-1}\text T^{-1}]^a[\text L]^b[\text{ML}^{-2}\text T^{-2}]^c$

$= [\text M^{a + c}][\text L^{-a + b - 2c}][\text T^{-a - 2c}]$

Equating the dimensions on both sides,

$$\begin{aligned}\text a + \text c = 0 \\[1em] -\text a + \text b - 2\text c = 3 \\[1em] -\text a - 2\text c = -1\end{aligned}$$

On solving these, we get a = − 1, b = 4 and c = 1.

Substituting these values in equation (i), the rate of discharge through the pipe is

$\text Q = \frac{\text{kPr}^4}{\eta \text l}$

By experiment the value of the constant k is found to be $\frac{\pi}{8}$. Hence,

$\frac{\text V}{\text t} = \frac{\pi \text{Pr}^4}{8\eta \text l}$

This relation is known as Poiseuille's relation, and with the help of this formula the viscosity of a liquid can be determined.

Statement : Pascal's law states that, "if a pressure change is applied to a fluid that is completely confined, incompressible and at rest, that pressure is transmitted equally and undiminished in all directions and also acts on the walls of the container."

Proof : Consider a tiny right-angled prismatic fluid element ABCDEF completely immersed in a liquid at rest. The element is so small that every point in it lies at essentially the same depth, so the pressure due to gravity is the same over all its faces and its own weight is negligible compared with the pressure forces.

Let the three mutually perpendicular faces be the horizontal face BEFC of area A₁, the inclined face ADFC of area A₂ and the vertical face ADEB of area A₃. If P₁, P₂ and P₃ are the fluid pressures normal to these faces, the normal forces are

$\text F_1 = \text P_1\text A_1, \quad \text F_2 = \text P_2\text A_2, \quad \text F_3 = \text P_3\text A_3$

Let θ be the angle between the normal to the inclined face and the vertical, so that the vertical component of F₂ is F₂ cos θ and its horizontal component is F₂ sin θ.

Since the element is in static equilibrium, the sum of the forces in the vertical and the horizontal directions must vanish.

Vertical equilibrium :

$\text F_1 = \text F_2\cos \theta \quad \Rightarrow \quad \text P_1\text A_1 = \text P_2\text A_2\cos \theta \qquad \ldots(\text i)$

Horizontal equilibrium :

$\text F_3 = \text F_2\sin \theta \quad \Rightarrow \quad \text P_3\text A_3 = \text P_2\text A_2\sin \theta \qquad \ldots(\text{ii})$

From the geometry of the prism, the projected areas satisfy

$\text A_1 = \text A_2\cos \theta, \qquad \text A_3 = \text A_2\sin \theta \qquad \ldots(\text{iii})$

Substituting (iii) into (i) and (ii), we get

$\text P_1 = \text P_2 \quad \text{and} \quad \text P_3 = \text P_2$

$\text P_1 = \text P_2 = \text P_3$

Hence, the pressure at a point in a static fluid is equal in all directions and acts normally to any surface, which is Pascal's law.

Examples :

(i) Hydraulic lift : A force applied on a piston of small cross-sectional area produces a pressure which is transmitted undiminished to a piston of large cross-sectional area, where a much larger force appears. This is used to lift heavy loads such as cars at a service station.

(ii) Hydraulic brakes : When the brake pedal is pressed, the increase in pressure in the brake oil of the master cylinder is transmitted undiminished to the pistons of the wheel cylinder, which press the brake-shoes against the rim of the wheel.

Stokes' law : If a small sphere of radius r is moving with a terminal velocity v through a perfectly homogeneous medium (liquid or gas) of infinite extension, then the viscous force acting on the sphere is

$\text F = 6\pi \eta \text{rv}$

where η is the coefficient of viscosity of that medium.

Derivation of terminal velocity : Consider a small ball of radius r and density ρ falling freely in a liquid of density σ and coefficient of viscosity η. When it attains the terminal velocity v, it is subjected to two forces :

(i) Effective downward force = weight of the ball − upthrust

$$\begin{aligned}= \text V\rho \text g - \text V\sigma \text g = \text V(\rho - \sigma)\text g \\[1em] = \frac{4}{3}\pi \text r^3(\rho - \sigma)\text g\end{aligned}$$

(ii) Viscous force acting upward = 6πηrv

Since the ball is moving with a constant velocity v, there is no acceleration in it, so the net force acting on it must be zero. Therefore,

$6\pi \eta \text{rv} = \frac{4}{3}\pi \text r^3(\rho - \sigma)\text g$

$v = \frac{4\pi \text r^3(\rho - \sigma)\text g}{3 \times 6\pi \eta \text r}$

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta}$

Hence, the terminal velocity of the ball is directly proportional to the square of its radius.

Forms of energy of a flowing liquid : There are three types of energies in a flowing liquid.

(i) Pressure energy : If P is the pressure on an area A of a liquid and the liquid moves through a distance l due to this pressure, then

$\text{pressure energy} = \text{force} \times \text{distance} = \text P \times \text A \times \text l$

Since the volume of the liquid is A × l,

$\text{pressure energy per unit volume} = \frac{\text P \times \text A \times \text l}{\text A \times \text l} = \text P$

(ii) Kinetic energy : If a liquid of mass m and volume V is flowing with velocity v, its kinetic energy is $\frac{1}{2}\text{mv}^2$, so

$\text{kinetic energy per unit volume} = \frac{1}{2}\left(\frac{\text m}{\text V}\right)v^2 = \frac{1}{2}\rho v^2$

(iii) Potential energy : If a liquid of mass m is at a height h from the surface of the earth, its potential energy is mgh, so

$\text{potential energy per unit volume} = \left(\frac{\text m}{\text V}\right)\text{gh} = \rho \text{gh}$

Proof that the total energy remains constant : Consider an incompressible and non-viscous liquid flowing in stream-lined motion through a tube XY of non-uniform cross-section. Let A₁ and A₂ be the areas of cross-section at the ends X and Y, at heights h₁ and h₂ from the surface of the earth. Let P₁, v₁ and P₂, v₂ be the pressures and velocities at X and Y.

The work done per second on the liquid entering at X is

$= \text P_1 \times \text A_1 \times v_1$

and the work done per second against the force at Y is P₂ × A₂ × v₂. Hence,

$\text{net work done on the liquid} = (\text P_1\text A_1v_1 - \text P_2\text A_2v_2) \qquad \ldots(\text i)$

But A₁v₁ and A₂v₂ are the volumes entering at X and leaving at Y per second, which must be equal, so

$\text A_1v_1 = \text A_2v_2 = \frac{\text m}{\rho}$

Substituting in equation (i),

$\text{net work done} = (\text P_1 - \text P_2)\frac{\text m}{\rho} \qquad \ldots(\text{ii})$

The increase in the kinetic energy of the liquid is $\frac{1}{2}\text m(v_2^2 - v_1^2)$ and the decrease in its potential energy is m g (h₁ − h₂). Hence,

$\text{net increase in energy} = \frac{1}{2}\text m(v_2^2 - v_1^2) - \text{mg}(\text h_1 - \text h_2) \qquad \ldots(\text{iii})$

This increase in energy is due to the net work done on the liquid. Equating (ii) and (iii),

$(\text P_1 - \text P_2)\frac{\text m}{\rho} = \frac{1}{2}\text m(v_2^2 - v_1^2) - \text{mg}(\text h_1 - \text h_2)$

$\text P_1 - \text P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) - \rho \text g(\text h_1 - \text h_2)$

Rearranging,

$\text P_1 + \frac{1}{2}\rho v_1^2 + \rho \text{gh}_1 = \text P_2 + \frac{1}{2}\rho v_2^2 + \rho \text{gh}_2$

$\text P + \frac{1}{2}\rho v^2 + \rho \text{gh} = \text{constant}$

Hence, in a frictionless stream-lined flow, the total energy per unit volume of the liquid remains constant at every point.

Bernoulli's theorem : When an incompressible and non-viscous liquid (or gas) flows in stream-lined motion from one place to another, then at every point of its path the total energy per unit volume (pressure energy + kinetic energy + potential energy) is constant, that is,

$\text P + \frac{1}{2}\rho v^2 + \rho \text{gh} = \text{constant}$

Velocity of efflux (Torricelli's theorem) : Let a vessel be filled with a liquid up to a height H and let there be an orifice at a depth h below the free surface of the liquid.

The pressure at the free surface of the liquid and also at the orifice is atmospheric, say P, so there is no effect of atmospheric pressure on the flow of the liquid from the orifice.

The liquid on the free surface has no kinetic energy but only potential energy, while the liquid coming out of the orifice has both kinetic and potential energies.

Applying Bernoulli's theorem, the total energy per unit volume at the free surface must equal that at the orifice,

$\text P + 0 + \rho \text{gH} = \text P + \frac{1}{2}\rho v^2 + \rho \text g(\text H - \text h)$

Cancelling P from both sides and simplifying,

$\rho \text{gH} - \rho \text g(\text H - \text h) = \frac{1}{2}\rho v^2$

$\frac{1}{2}\rho v^2 = \rho \text{gh}$

$v = \sqrt{2\text{gh}}$

Hence, the velocity of efflux of a liquid from an orifice is equal to that velocity which the liquid would acquire in falling freely from the free surface of the liquid up to the orifice. This result was first established by Torricelli in 1644 and is called Torricelli's theorem.

Surface energy : When the surface area of a liquid is increased, the molecules from the interior rise to the surface. This requires work to be done against the force of attraction of the molecules just below the surface. This work is stored in the form of potential energy in the newly formed surface.

Thus, the molecules in the surface have some additional energy due to their position, and this additional energy per unit area of the surface is called the 'surface energy of the liquid'.

Work done in blowing a soap bubble : A soap bubble has two free surfaces — one inside the bubble in contact with air and the other outside it.

The total surface area of a soap bubble of radius R is therefore

$\text A = 2 \times 4\pi \text R^2 = 8\pi \text R^2$

Since the bubble is blown from the soap solution, the initial surface area may be taken as zero. Hence the increase in the surface area is

$\Delta \text A = 8\pi \text R^2$

The work done in increasing the surface area at constant temperature is

$\text W = \text T \times \Delta \text A$

$\text W = 8\pi \text R^2\text T$

Hence, the work done in blowing the soap bubble is 8πR²T.

Angle of contact : When the free surface of a liquid comes in contact with a solid, it becomes curved near the place of contact. The angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the point of contact is called the 'angle of contact' for that pair of solid and liquid.

(i) Water and glass : The adhesive force between water and glass molecules is greater than the cohesive force between water molecules. Hence water wets glass, the meniscus is concave and the angle of contact θ is acute.

(ii) Mercury and glass : The cohesive force between mercury molecules is far greater than the adhesive force between mercury and glass. Hence mercury does not wet glass, the meniscus is convex and the angle of contact θ is obtuse, being about 135°.

Value for water and clean glass : The angle of contact between pure water and clean glass is zero. For ordinary water and glass it is about 8°.

Consider a liquid of density ρ contained in a vessel. Let C and D be two points inside the liquid at a vertical distance h apart. Imagine a cylindrical column of the liquid of cross-sectional area A such that the points C and D lie at the centres of the upper and the lower faces of the cylinder.

The weight of the liquid in this imaginary cylindrical column is

$\text{mg} = \text{volume} \times \text{density} \times \text g = \text{Ah} \times \rho \times \text g$

which acts vertically downwards. Here it is assumed that the density of the liquid is uniform and does not depend upon pressure, which is very nearly true for liquids.

Let p₁ and p₂ be the pressures at the points C and D respectively. Then,

• force on the upper face = p₁A, acting vertically downwards,
• force on the lower face = p₂A, acting vertically upwards.

The horizontal forces on the curved surface of the cylinder are mutually balanced. Since the cylinder of the liquid is in equilibrium, the net force on it must be zero,

$(\text p_1\text A + \text{mg}) - \text p_2\text A = 0$

$(\text p_1\text A + \text{Ah}\rho \text g) - \text p_2\text A = 0$

$\text p_2 - \text p_1 = \text h\rho \text g$

This is the modified Pascal's law in the presence of gravity.

If the point C is shifted to the liquid surface which is open to the atmosphere, then p₁ may be replaced by the atmospheric pressure P₀ and p₂ by P. Hence,

$\text P = \text P_0 + \text h\rho \text g$

Hence, the pressure exerted by a liquid column of height h is hρg, and it depends only upon the vertical depth of the point and the density of the liquid. It depends neither upon the shape of the container nor upon the amount of liquid in it.

Consider an ideal (incompressible and non-viscous) liquid of density ρ flowing in stream-lined motion through a horizontal tube of non-uniform cross-section.

Let P₁ and v₁ be the pressure and the velocity of flow at one end of the tube, and P₂ and v₂ the corresponding quantities at the other end.

By Bernoulli's theorem, the total energy per unit volume is constant at every point of the flow,

$\text P_1 + \frac{1}{2}\rho v_1^2 + \rho \text{gh}_1 = \text P_2 + \frac{1}{2}\rho v_2^2 + \rho \text{gh}_2$

Since the tube is horizontal, both the ends lie at the same height above the surface of the earth, that is,

$\text h_1 = \text h_2$

Hence, the potential energy terms ρgh₁ and ρgh₂ are equal and cancel out from both sides,

$\text P_1 + \frac{1}{2}\rho v_1^2 = \text P_2 + \frac{1}{2}\rho v_2^2$

Transposing the terms,

$\text P_1 - \text P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2$

$\text P_1 - \text P_2 = \frac{\rho}{2}(v_2^2 - v_1^2)$

Hence proved.

This shows that in a horizontal flow, wherever the velocity of the liquid is greater the pressure is smaller, and vice-versa.

A venturimeter is a device based on Bernoulli's theorem by which the rate of flow of water in a tube can be determined. It consists of a tube XY whose middle part Z is narrow, with two vertical tubes joined at X and Z.

Let A₁ and A₂ be the areas of cross-section of the tube at the wider and the narrower parts, v₁ and v₂ the corresponding velocities of flow, and P₁ and P₂ the pressures there.

Assuming the flow to be stream-lined, by the principle of continuity,

$\text A_1v_1 = \text A_2v_2 \quad \Rightarrow \quad \frac{\text A_1}{\text A_2} = \frac{v_2}{v_1} \qquad \ldots(\text i)$

Since the tube is horizontal, by Bernoulli's theorem,

$\text P_1 + \frac{1}{2}\rho v_1^2 = \text P_2 + \frac{1}{2}\rho v_2^2$

$\text P_1 - \text P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) = \frac{1}{2}\rho v_1^2\left(\frac{v_2^2}{v_1^2} - 1\right)$

Substituting the value of $\frac{v_2^2}{v_1^2}$ from equation (i),

$\text P_1 - \text P_2 = \frac{1}{2}\rho v_1^2\left(\frac{\text A_1^2}{\text A_2^2} - 1\right) = \frac{1}{2}\rho v_1^2\left(\frac{\text A_1^2 - \text A_2^2}{\text A_2^2}\right)$

If h is the difference in the levels of the liquid in the two vertical tubes, then P₁ − P₂ = h ρ g. Therefore,

$\text h\rho \text g = \frac{1}{2}\rho v_1^2\left(\frac{\text A_1^2 - \text A_2^2}{\text A_2^2}\right)$

$v_1^2 = \text A_2^2\frac{2\text{gh}}{\text A_1^2 - \text A_2^2} \quad \Rightarrow \quad v_1 = \text A_2\sqrt{\frac{2\text{gh}}{\text A_1^2 - \text A_2^2}}$

If Q is the volume of the liquid flowing per second through the tube, then

$\text Q = \text A_1v_1 = \text A_1\text A_2\sqrt{\frac{2\text{gh}}{\text A_1^2 - \text A_2^2}}$

Hence, by measuring h and knowing the areas of cross-section A₁ and A₂, the rate of flow Q can be determined.

(i) Excess pressure inside a liquid drop : Consider a liquid drop of radius R of a liquid of surface tension T. The molecules in the surface of the drop experience a resultant force due to surface tension acting normally inwards. Therefore, the pressure inside the drop must be greater than the pressure outside it.

Let P be the pressure outside the drop and (P + p) that inside it, so that p is the excess pressure. Let this excess pressure push the surface outwards through a small distance ΔR. The work done by the excess pressure is

$\text W = \text{force} \times \text{displacement} = (\text{excess pressure} \times \text{area}) \times \text{displacement}$

$\text W = (\text p \times 4\pi \text R^2) \times \Delta \text R \qquad \ldots(\text i)$

The increase in the surface area of the drop is

$\Delta \text A = 4\pi (\text R + \Delta \text R)^2 - 4\pi \text R^2 = 8\pi \text R(\Delta \text R)$

neglecting the smaller term (ΔR)². Hence, the increase in free surface energy is

$= \Delta \text A \times \text T = 8\pi \text R(\Delta \text R)\text T \qquad \ldots(\text{ii})$

This increase in energy is at the cost of the work done by the excess pressure. Equating (i) and (ii),

$(\text p \times 4\pi \text R^2)\Delta \text R = 8\pi \text R(\Delta \text R)\text T$

$\text p = \frac{2\text T}{\text R}$

(ii) Excess pressure inside an air bubble in a liquid : An air bubble of radius R formed in a liquid of surface tension T has, like a liquid drop, only one surface in contact with the liquid. Proceeding exactly as in the case of the liquid drop, the excess pressure inside the air bubble is

$\text p = \frac{2\text T}{\text R}$

(iii) Excess pressure inside a soap bubble : A thin soap bubble of radius R has two liquid surfaces in contact with air, one inside the bubble and the other outside it.

The work done by the excess pressure in pushing the surface outwards through ΔR is

$\text W = (\text p \times 4\pi \text R^2) \times \Delta \text R \qquad \ldots(\text i)$

The total increase in the surface area, counting both the surfaces, is

$\Delta \text A = 2[4\pi (\text R + \Delta \text R)^2 - 4\pi \text R^2] = 16\pi \text R(\Delta \text R)$

so the increase in free surface energy is

$= 16\pi \text R(\Delta \text R)\text T \qquad \ldots(\text{ii})$

Equating (i) and (ii),

$(\text p \times 4\pi \text R^2)\Delta \text R = 16\pi \text R(\Delta \text R)\text T$

$\text p = \frac{4\text T}{\text R}$

Hence, the excess pressure inside a liquid drop and an air bubble is 2T/R, while that inside a soap bubble is 4T/R.

Suppose a cleaned capillary tube of glass having a uniform bore of radius r is dipped in a liquid (say water) which rises to a height h above the level outside the tube. The liquid meniscus AEB in the tube is concave upward. Let T be the surface tension of the liquid.

The liquid meniscus in the tube is along a circle of circumference 2πr which is in contact with the glass. Due to the surface tension of the liquid, a force equal to T per unit length acts at all points of this circle. If the angle of contact for liquid-glass be θ, then this force is directed inward at an angle θ from the wall of the tube.

The wall of the tube also exerts an equal (reactionary) force T per unit length on the circumference of the liquid meniscus, directed outward. This force can be resolved into two components :

(i) T cos θ per unit length acting vertically upward, and

(ii) T sin θ per unit length acting horizontally outward.

Considering the entire circumference 2πr, for each horizontal component T sin θ there is an equal and opposite component, so the two neutralise each other. The vertical components, being in the same direction, are added up to give a total upward force

$= 2\pi \text r \times \text T\cos \theta$

It is this force which supports the weight of the liquid column so raised.

Weight of the liquid column : The curved surface AEB may be assumed hemispherical, whose radius may be taken equal to the radius r of the capillary tube. The volume of the liquid raised in the tube is

$= \pi \text r^2\text h + (\text{volume of the cylinder ABDC} - \text{volume of hemisphere AEB})$

$= \pi \text r^2\text h + \lbrace\pi \text r^2(\text r) - \frac{2}{3}\pi \text r^3 \rbrace = \pi \text r^2\left(\text h + \frac{\text r}{3}\right)$

Hence, the weight of the liquid column raised in the tube is

$= \pi \text r^2\left(\text h + \frac{\text r}{3}\right)\rho \text g$

where ρ is the density of the liquid.

In equilibrium, the upward force due to surface tension balances this weight,

$2\pi \text r \times \text T\cos \theta = \pi \text r^2\left(\text h + \frac{\text r}{3}\right)\rho \text g$

$\text T = \frac{\text r\left[\text h + \frac{\text r}{3}\right]\rho \text g}{2\cos \theta}$

Since h >> r/3, the term r/3 may be neglected in comparison to h. Then,

$\text T = \frac{\text{rh}\rho \text g}{2\cos \theta} \quad \Rightarrow \quad \text h = \frac{2\text T\cos \theta}{\text r\rho \text g}$

Hence, the height to which the liquid rises in a capillary tube is $\text h = \frac{2\text T\cos \theta}{\text r\rho \text g}$, which shows that the narrower the tube, the greater is the height to which the liquid rises in it.

For pure water and clean glass θ may be taken zero, and then $\text T = \frac{\text{rh}\rho \text g}{2}$.

(a) Stream-line flow : Stream-line flow is the motion of a fluid in which every particle follows a smooth, well-defined path called a streamline, and these streamlines never intersect one another. Each particle passing a given point follows exactly the same path as the particles that passed earlier.

Turbulent flow : Turbulent flow is a type of fluid motion in which the particles move in a highly irregular and unpredictable manner. The velocity of the fluid particles changes continuously in both magnitude and direction, and swirling regions called eddies are formed.

Difference between them :

Stream-line flow	Turbulent flow
The particles follow smooth, well-defined paths called streamlines.	The paths of the particles are irregular and cross each other.
The streamlines never intersect one another.	Eddies or whirlpools are formed.
The velocity at a given point remains constant in magnitude and direction.	The velocity at a point changes continuously in both magnitude and direction.
It occurs when the velocity of flow is below the critical velocity.	It occurs when the velocity of flow exceeds the critical velocity.
There is practically no loss of energy.	A considerable amount of energy is lost.

(b) Reynold's number : Reynold's number is a dimensionless number which predicts the nature of flow of a fluid. It is given by

$\text R_e = \frac{\rho \text{vL}}{\eta}$

where ρ is the density of the fluid, η its viscosity, L the characteristic length (for a pipe, L = D, the diameter) and v the velocity of flow.

Physically, Reynold's number gives the ratio of the inertial forces to the viscous forces in a fluid.

Critical velocity : There exists a certain maximum velocity up to which the flow of a liquid through a tube remains stream-lined. If the velocity exceeds this limit, the flow becomes turbulent. This maximum velocity is called the critical velocity (v_c). In terms of Reynold's number,

$v_c = \text R_{e_c}\left(\frac{\eta}{\rho \text D}\right)$

Characterisation of the flow : For flow in a cylindrical pipe,

(i) if R_e < 2000, the flow is laminar (stream-lined),

(ii) if 2000 < R_e < 3000, the flow is in a transition region and is unstable,

(iii) if R_e > 3000, the flow becomes turbulent.

Thus, if the viscous forces dominate (low R_e) the flow remains smooth and laminar, and if the inertial forces dominate (high R_e) the flow becomes turbulent.

Viscous force : When a liquid flows in stream-lined motion, the layer in contact with a fixed surface is at rest while the velocity of the other layers increases with distance from that surface. In between any two layers of the liquid, internal tangential forces act which try to destroy the relative motion between them. These forces are called 'viscous forces'.

The property of the liquid by virtue of which it opposes the relative motion between its adjacent layers is known as viscosity.

Factors on which it depends : According to Newton, the viscous force F acting between two layers of a liquid flowing in stream-lined motion depends upon two factors :

(i) It is directly proportional to the contact-area A of the layers (F ∝ A).

(ii) It is directly proportional to the velocity-gradient dv/dz in a direction normal to the layer (F ∝ dv/dz).

Combining these two factors,

$\text F \propto \text A\frac{\text{dv}}{\text{dz}} \quad \text{or} \quad \text F = -\eta \text A\frac{\text{dv}}{\text{dz}}$

where η is a constant called the coefficient of viscosity, the negative sign showing that the viscous force on a layer opposes its velocity relative to the adjacent layers. This is Newton's law of viscosity.

Coefficient of viscosity : The coefficient of viscosity of a fluid is numerically equal to the viscous force per unit area which maintains a unit velocity gradient between its two parallel layers. From the above formula,

$\eta = \frac{\text F}{\text A\left(\frac{\text{dv}}{\text{dz}}\right)}$

If A = 1 unit and dv/dz = 1 unit, then η = F numerically.

Dimensional formula :

$[\eta] = \frac{[\text{MLT}^{-2}]}{[\text L^2][\text{LT}^{-1}/\text L]} = \frac{[\text{MLT}^{-2}]}{[\text L^2\text T^{-1}]} = [\text{ML}^{-1}\text T^{-1}]$

MKS (SI) unit : kg m^-1 s^-1. Another unit is the poise, where 1 kg m^-1 s^-1 = 10 poise = 1 decapoise or poiseuille.

Bernoulli's theorem : When an incompressible and non-viscous liquid (or gas) flows in stream-lined motion from one place to another, then at every point of its path the total energy per unit volume (pressure energy + kinetic energy + potential energy) is constant, that is,

$\text P + \frac{1}{2}\rho v^2 + \rho \text{gh} = \text{constant}$

Pressure head, velocity head and gravitational head : Dividing the above equation throughout by ρ g, we get

$\frac{\text P}{\rho \text g} + \frac{v^2}{2\text g} + \text h = \text{constant}$

(i) Pressure head : The term $\frac{\text P}{\rho \text g}$ is called the pressure head. It represents the pressure energy of the liquid per unit weight.

(ii) Velocity head : The term $\frac{v^2}{2\text g}$ is called the velocity head. It represents the kinetic energy of the liquid per unit weight.

(iii) Gravitational head : The term h is called the gravitational head. It represents the potential energy of the liquid per unit weight.

The dimension of each of these three terms is the dimension of height, and their sum is called the total head. Hence Bernoulli's theorem may also be stated as : in the stream-lined motion of an ideal liquid, the sum of pressure head, velocity head and gravitational head at any point is always constant.

Application — Venturimeter : A venturimeter is a device based on Bernoulli's theorem by which the rate of flow of water in a tube is measured. Its middle part is made narrow, where the velocity of flow becomes larger and so the pressure becomes smaller. This pressure difference is measured by two vertical tubes, and from it the rate of flow of water is calculated.

The given equation is Bernoulli's equation for the stream-lined flow of an incompressible and non-viscous liquid. Each of its three terms represents an energy per unit volume of the liquid.

(i) P — the pressure energy per unit volume : If P is the pressure on an area A of the liquid and the liquid moves through a distance l due to this pressure, the pressure energy is P × A × l. Since the volume of the liquid is A × l, the pressure energy per unit volume is P itself. It is also called the static pressure.

(ii) $\frac{1}{2}\rho v^2$ — the kinetic energy per unit volume : If a liquid of mass m and volume V flows with velocity v, its kinetic energy is $\frac{1}{2}\text{mv}^2$, so the kinetic energy per unit volume is $\frac{1}{2}\left(\frac{\text m}{\text V}\right)v^2 = \frac{1}{2}\rho v^2$, where ρ is the density of the liquid. It is also called the dynamic pressure.

(iii) ρ g h — the potential energy per unit volume : If a liquid of mass m is at a height h above the surface of the earth, its potential energy is m g h, so the potential energy per unit volume is $\left(\frac{\text m}{\text V}\right)\text{gh} = \rho \text{gh}$.

Unit of the constant : Since the dimension of each term in this equation is the same as that of pressure, the constant also has the dimensions of pressure.

Hence, the SI unit of the constant is N m^-2 (or pascal).

Cohesive force : The force of attraction between the molecules of the same substance is called the cohesive force.

Example : The force of attraction between two water molecules. It is due to the cohesive force that two drops of a liquid, when brought in mutual contact, coalesce into one.

Adhesive force : The force of attraction between the molecules of different substances is called the adhesive force.

Example : The force of attraction between water molecules and glass molecules. It is due to the adhesive force that a glass plate becomes wet when water is poured on it.

Difference between them :

Cohesive force	Adhesive force
It acts between the molecules of the same substance.	It acts between the molecules of different substances.
Example : attraction between two water molecules.	Example : attraction between water and glass molecules.
It is responsible for surface tension and for the definite shape of solids.	It is responsible for the wetting of a solid by a liquid and for capillary rise.

Both these forces are different from the gravitational force and do not obey the inverse-square law. They act only within the molecular range c (≈ 10^-9 metre); beyond this distance the attraction between two molecules becomes negligible.

The relative magnitudes of these two forces decide whether a liquid wets a solid or not. If the adhesive force is greater than the cohesive force, the liquid wets the solid surface, as water wets glass. If the adhesive force is less than the cohesive force, the liquid does not wet the solid surface, as mercury does not wet glass.

Cause of surface tension : Laplace explained surface tension on the basis of intermolecular cohesive forces. If the distance between two molecules is less than the molecular range c (≈ 10^-9 metre), they attract each other; beyond this the attraction is negligible. Hence, if we draw a sphere of radius c with a molecule as centre, called its sphere of molecular activity, only the molecules enclosed within this sphere can attract the molecule at the centre.

A molecule D well inside the liquid is attracted equally in all directions, so the resultant force on it is zero. A molecule B a little below the surface has its sphere of molecular activity partly outside the liquid, so the number of molecules in its upper half is less than that in its lower half, and it experiences a resultant downward force. A molecule A lying in the surface has half of its sphere outside the liquid, so it experiences the maximum downward force.

When the surface area of a liquid is increased, molecules from the interior rise to the surface, doing work against this downward cohesive force. This work is stored in them as potential energy. Since a system is in stable equilibrium when its potential energy is minimum, the liquid surface tends to have the minimum number of molecules in it, that is, the surface contracts to a minimum possible area. This tendency is exhibited as surface tension.

Angle of contact : When the free surface of a liquid comes in contact with a solid, it becomes curved near the place of contact. The angle inside the liquid between the tangent to the solid surface and the tangent to the liquid surface at the point of contact is called the 'angle of contact' for that pair of solid and liquid. It is zero for pure water and clean glass, and about 135° for mercury and glass.

Units of surface tension : The SI unit of surface tension is N m^-1 (newton/metre). Since surface tension is also equal to the work done in increasing the surface area by unity, it may equally be expressed in J m^-2 (joule/metre²). Its dimensions are [M T^-2].

Expression for the work done : Let a liquid film be formed between a bent wire ABC and a straight wire PQ which can slide on the bent wire without friction.

As the film surface tends to contract, the wire PQ moves upward. To keep PQ in equilibrium, a uniform force F (which includes the weight of the wire) has to be applied in the downward direction. This force F is found to be directly proportional to the length l of the film in contact with the wire PQ. Since there are two free surfaces of the film,

$\text F \propto 2\text l \quad \text{or} \quad \text F = \text T \times 2\text l$

where T is a constant called the surface tension of the liquid.

Now, suppose the wire is moved downward through a small distance Δx. This results in an increase in the surface area of the film. The work done by the force F is

$\text W = \text F \times \Delta \text x = \text T \times 2\text l \times \Delta \text x$

But 2 l × Δx is the total increase in the area of both the surfaces of the film. Let it be ΔA. Then,

$\text W = \text T \times \Delta \text A \quad \text{or} \quad \text T = \frac{\text W}{\Delta \text A}$

Definition of surface tension on this basis : If ΔA = 1, then T = W. Hence, the surface tension of a liquid is equal to the work required to increase the surface area of the liquid film by unity at constant temperature.

Unit : Since T = W/ΔA, surface tension may be expressed in joule/metre² (J m^-2), which is the same as N m^-1.

Effect of temperature : The surface tension of a liquid decreases with an increase in temperature and becomes zero at the critical temperature. On heating, the molecules gain kinetic energy and move farther apart, which weakens the cohesive forces between them and hence lowers the surface tension.

Surface energy : When the surface area of a liquid is increased, the molecules from the interior of the liquid rise to the surface. In doing so, work has to be done against the force of attraction of the molecules just below the surface. This work is stored in the newly formed surface in the form of potential energy.

Thus, the molecules lying in the surface of a liquid possess some additional energy on account of their position. This additional energy per unit area of the surface is called the 'surface energy of the liquid'.

Relation between surface tension and surface energy : Consider a liquid film formed on a wire frame having a movable wire of length l. Since the film has two free surfaces, the force acting on the wire due to surface tension is

$\text F = \text T \times 2\text l$

If the wire is now moved through a small distance Δx, the work done is

$\text W = \text F \times \Delta \text x = \text T \times 2\text l \times \Delta \text x$

But 2 l × Δx is the total increase ΔA in the surface area of the film. Therefore,

$\text W = \text T \times \Delta \text A \quad \text{or} \quad \text T = \frac{\text W}{\Delta \text A}$

If ΔA = 1, then T = W.

Hence, the surface tension of a liquid is numerically equal to its surface energy per unit area. For this reason surface tension may be expressed either in N m^-1 or in J m^-2.

Consider water flowing in stream-lined motion through a horizontal tube of non-uniform cross-section. Let A₁ and A₂ be the areas of cross-section of the wider and the narrower parts, v₁ and v₂ the velocities of flow there, and P₁ and P₂ the corresponding pressures.

Step 1 — Applying the principle of continuity :

$\text A_1v_1 = \text A_2v_2 \quad \Rightarrow \quad \frac{v_2}{v_1} = \frac{\text A_1}{\text A_2}$

Since A₁ > A₂, therefore v₂ > v₁, that is, the velocity of water in the narrower part is greater than that in the wider part.

Step 2 — Applying Bernoulli's equation : As the tube is horizontal, h₁ = h₂ and the potential energy terms cancel out, so

$\text P_1 + \frac{1}{2}\rho v_1^2 = \text P_2 + \frac{1}{2}\rho v_2^2$

$\text P_1 - \text P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Step 3 — Conclusion : Since v₂ > v₁, the quantity $(v_2^2 - v_1^2)$ is positive, and therefore

$\text P_1 - \text P_2 \gt 0 \quad \Rightarrow \quad \text P_1 \gt \text P_2$

Hence, the pressure in the wider part of the tube is larger than that in the narrower part. In other words, in a flowing liquid, where the velocity of flow is less the pressure is larger and vice-versa.

(i) Atomizer : An atomizer is an instrument used to spray a liquid in the form of small droplets.

It consists of a vertical tube AB whose lower end A is immersed in a liquid filled in a vessel, while its upper end B opens in the narrower part of a horizontal tube. At one end of this tube there is a rubber ball, and at the other end there is a fine bore.

When the rubber ball is squeezed, it sends a stream of air which passes through the tube. The velocity of the air is very large in the constricted part of the tube, so by Bernoulli's theorem the pressure in that part is lowered.

Consequently, the liquid rises up in the tube AB due to the larger atmospheric pressure acting on the liquid surface in the vessel, and on reaching B it is mixed with air and comes out in the form of a fine spray.

The atomizer is used to spray colours and perfumes, to colour motors and to spray water on hairs and to wash nose, ear, etc.

(ii) Filter pump : A filter pump is an instrument used to produce vacuum in vessels.

It consists of a wide tube M N, in the upper part of which there is another tube A. The upper end of A is connected to a water tank, while its lower end has a fine bore through which water comes out in the form of a jet.

The vessel which is to be evacuated is connected to the tube M N. The velocity of the emerging water jet is very large, so by Bernoulli's theorem the pressure of the air near the jet becomes less than the pressure inside the vessel.

Hence, air from the vessel rushes into the tube M N and is carried out along with the water jet. Thus a partial vacuum is created in the vessel.

Whether a liquid sticks to a solid surface or not is decided by the relative magnitudes of the adhesive force between the liquid and the solid and the cohesive force between the molecules of the liquid itself.

Water and glass : The adhesive force between water molecules and glass molecules is greater than the cohesive force between the water molecules themselves. Hence, when water is poured on glass, the water molecules cling to the glass molecules and the glass surface is wetted. The meniscus formed is concave and the angle of contact is acute (zero for pure water and clean glass).

Mercury and glass : The adhesive force between mercury molecules and glass molecules is less than the cohesive force between the mercury molecules themselves. Hence the mercury molecules do not cling to the glass molecules, that is, mercury does not wet the glass. The meniscus formed is convex and the angle of contact is obtuse, being about 135°.

Hence, water sticks to the walls of a glass tube while mercury does not.

If, however, the glass surface is greasy, then water also does not wet the glass, because the adhesive force between water and grease is less than the cohesive force between the water molecules themselves.

A molecule lying in the surface of a liquid is attracted by the other molecules in the surface in all directions. The resultant force on it depends upon the shape of the surface.

(i) Plane surface : If the surface is plane, the molecule is attracted equally in all directions. Hence the resultant force on the molecule due to surface tension is zero, and there is no pressure difference between the two sides.

(ii) Convex surface : If the surface is convex, then the resultant of all the forces of attraction acting on every molecule acts normal to the surface and is directed inward.

(iii) Concave surface : If the surface is concave, then every molecule experiences a resultant force due to surface tension acting normally outward.

Obviously, for the equilibrium of a curved surface there must be a difference of pressure between its two sides, so that the excess pressure force may balance the resultant force due to surface tension.

In the case of a convex surface, the resultant force of surface tension acts inward, so the pressure on the concave side must be greater to balance it. In the case of a concave surface, the resultant force of surface tension acts outward, so again the pressure on the concave side must be greater.

Hence, in either case the pressure on the concave side of a curved liquid surface is greater than the pressure on its convex side.

This difference of pressure is equal to $\frac{2\text T}{\text R}$, where T is the surface tension and R is the radius of curvature of the surface.

Given,

• Force applied, F = 80 N
• Area of cross-section of the tip, A = 0.002 cm² = 0.002 × 10^-4 m² = 2 × 10^-7 m²

Pressure is the normal force exerted per unit area,

$\text P = \frac{\text F}{\text A}$

Substituting the values,

$$\begin{aligned}\text P = \frac{80}{2 \times 10^{-7}} \\[1em] = 4 \times 10^{8}\ \text{Pa}\end{aligned}$$

Hence, the pressure on the tip of the nail is 4 × 10⁸ Pa.

Given,

• Height of the mercury column, h = 75 cm = 0.75 m
• Density of mercury, ρ = 13.6 × 10³ kg m^-3
• g = 10 m s^-2

The pressure exerted by a liquid column of height h is

$\text P = \text h\rho \text g$

Substituting the values,

$$\begin{aligned}\text P = 0.75 \times (13.6 \times 10^{3}) \times 10 \\[1em] = 1.02 \times 10^{5}\ \text{N m}^{-2}\end{aligned}$$

Hence, the air-pressure is 1.02 × 10⁵ N m^-2.

Given,

• Depth, h = 200 m
• Average density of sea water, ρ = 1.032 × 10³ kg m^-3
• g = 9.8 m s^-2

The pressure exerted by a liquid column of height h is

$\text P = \text h\rho \text g$

Substituting the values,

$$\begin{aligned}\text P = 200 \times (1.032 \times 10^{3}) \times 9.8 \\[1em] = 2.023 \times 10^{6}\ \text{N m}^{-2}\end{aligned}$$

Hence, the pressure of sea water at a depth of 200 m is 2.023 × 10⁶ N m^-2.

Given,

• Thickness of the oil layer, dz = 1 mm = 10^-3 m
• Area of the plate, A = 10^-2 m²
• Velocity of the plate, dv = 1.5 cm/s = 1.5 × 10^-2 m s^-1
• Coefficient of viscosity of oil, η = 1 poise = 0.1 kg m^-1 s^-1

The velocity gradient between the plates is

$\frac{\text{dv}}{\text{dz}} = \frac{1.5 \times 10^{-2}}{10^{-3}} = 15\ \text s^{-1}$

By Newton's formula of viscosity, the viscous force is

$\text F = \eta \text A\frac{\text{dv}}{\text{dz}}$

Substituting the values,

$$\begin{aligned}\text F = 0.1 \times 10^{-2} \times 15 \\[1em] = 1.5 \times 10^{-2}\ \text N\end{aligned}$$

Hence, the force required to move the plate is 1.5 × 10^-2 N.

Given,

• Radius of the ball, r = 3 mm = 3 × 10^-3 m
• Density of steel, ρ = 8.0 × 10³ kg m^-3
• Density of glycerine, σ = 1.2 × 10³ kg m^-3
• Coefficient of viscosity of glycerine, η = 0.85 kg m^-1 s^-1
• g = 10 m s^-2

By Stokes' law, the terminal velocity of the ball is

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta}$

Substituting the values,

$$\begin{aligned}v = \frac{2}{9} \times \frac{(3 \times 10^{-3})^2 \times (8.0 \times 10^{3} - 1.2 \times 10^{3}) \times 10}{0.85} \\[1em] = \frac{2}{9} \times \frac{(9 \times 10^{-6}) \times (6.8 \times 10^{3}) \times 10}{0.85}\end{aligned}$$

$= \frac{2}{9} \times \frac{0.612}{0.85} = \frac{2}{9} \times 0.72$

$v = 0.16\ \text{m s}^{-1}$

Hence, the terminal velocity of the ball is 0.16 m s^-1, that is, 16 cm s^-1.

Note: The answer printed in the textbook is "16 m/s", which appears to be a misprint for 16 cm/s. The correct value in SI units is 0.16 m s^-1.

Given,

• Diameter of the shot = 2 mm, so radius r = 1 mm = 1 × 10^-3 m
• Density of the material of the shot, ρ = 8.0 × 10³ kg m^-3
• Density of the liquid, σ = 1.0 × 10³ kg m^-3
• Coefficient of viscosity of the liquid, η = 1.0 kg m^-1 s^-1
• g = 10 m s^-2

By Stokes' law, the terminal velocity is

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta}$

Substituting the values,

$$\begin{aligned}v = \frac{2}{9} \times \frac{(1 \times 10^{-3})^2 \times (8.0 \times 10^{3} - 1.0 \times 10^{3}) \times 10}{1.0} \\[1em] = \frac{2}{9} \times (1 \times 10^{-6}) \times (7.0 \times 10^{3}) \times 10\end{aligned}$$

$= \frac{2}{9} \times 7 \times 10^{-2} = 1.55 \times 10^{-2}\ \text{m s}^{-1}$

Hence, the terminal speed of the shot is 1.55 × 10^-2 m s^-1, that is, 1.55 cm s^-1.

Given,

• Radius of the drop, r = 1.0 cm = 1.0 × 10^-2 m
• Uniform (terminal) speed, v = 1.0 cm/s = 1.0 × 10^-2 m s^-1
• Coefficient of viscosity of the fluid, η = 1.0 × 10² kg m^-1 s^-1

By Stokes' law, the viscous force acting on the drop is

$\text F = 6\pi \eta \text{rv}$

Substituting the values,

$$\begin{aligned}\text F = 6\pi \times (1.0 \times 10^{2}) \times (1.0 \times 10^{-2}) \times (1.0 \times 10^{-2}) \\[1em] = 6\pi \times 10^{-2}\end{aligned}$$

$= 0.06\pi\ \text N = 0.188\ \text N$

Hence, the viscous force acting on the drop is 0.06 π N, that is, about 0.188 N.

Given,

• Velocity of flow, v = 8.0 m s^-1
• g = 10 m s^-2

On dividing Bernoulli's equation throughout by ρ g, the term $\frac{v^2}{2\text g}$ is obtained, which is called the velocity head. Thus,

$\text{velocity head} = \frac{v^2}{2\text g}$

Substituting the values,

$$\begin{aligned}\text{velocity head} = \frac{(8.0)^2}{2 \times 10} = \frac{64}{20} \\[1em] = 3.2\ \text m\end{aligned}$$

Hence, the velocity-head of the liquid is 3.2 m.

Given,

• Velocity at the first side, v₁ = 0.4 m s^-1
• Velocity at the other side, v₂ = 0.6 m s^-1
• Pressure at the first side, P₁ = 1500 N m^-2
• Density of water, ρ = 1.0 × 10³ kg m^-3

Since the pipe is horizontal, by Bernoulli's theorem,

$\text P_1 + \frac{1}{2}\rho v_1^2 = \text P_2 + \frac{1}{2}\rho v_2^2$

$\text P_2 = \text P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2)$

Substituting the values,

$$\begin{aligned}\text P_2 = 1500 + \frac{1}{2} \times (1.0 \times 10^{3}) \times [(0.4)^2 - (0.6)^2] \\[1em] = 1500 + 500 \times (0.16 - 0.36)\end{aligned}$$

$= 1500 + 500 \times (-0.20) = 1500 - 100$

$\text P_2 = 1400\ \text{N m}^{-2}$

Hence, the water pressure at the other side of the pipe is 1400 N m^-2.

Given,

• Increase in the surface area of one surface, ΔA = 1.0 cm² = 1.0 × 10^-4 m²
• Surface tension of soap solution, T = 1.8 × 10^-2 N m^-1

A soap bubble has two free surfaces, so the total increase in the surface area is 2 × ΔA.

The work done in increasing the surface area at constant temperature is

$\text W = \text T \times 2\Delta \text A$

Substituting the values,

$$\begin{aligned}\text W = (1.8 \times 10^{-2}) \times 2 \times (1.0 \times 10^{-4}) \\[1em] = 3.6 \times 10^{-6}\ \text J\end{aligned}$$

Hence, the work done is 3.6 × 10^-6 J.

Given,

• Surface tension of the soap solution, T = 2.1 × 10^-2 N m^-1
• Diameter of the bubble = 1.0 cm, so radius r = 0.5 cm = 0.5 × 10^-2 m

A soap bubble has two free surfaces, so the total surface area formed is

$\Delta \text A = 2 \times 4\pi \text r^2 = 8\pi \text r^2$

The work done is

$\text W = \text T \times \Delta \text A = \text T \times 8\pi \text r^2$

Substituting the values,

$$\begin{aligned}\text W = (2.1 \times 10^{-2}) \times 8 \times 3.14 \times (0.5 \times 10^{-2})^2 \\[1em] = (2.1 \times 10^{-2}) \times 8 \times 3.14 \times (2.5 \times 10^{-5})\end{aligned}$$

$= 1.32 \times 10^{-5}\ \text J$

Hence, the work done in forming the bubble is 1.32 × 10^-5 J.

Given,

• Initial area of one surface of the film = 50 cm²
• Final area of one surface of the film = 100 cm²
• Work done, W = 3.0 × 10^-4 J

A soap film has two free surfaces. The increase in the area of one surface is

$100 - 50 = 50\ \text{cm}^2 = 50 \times 10^{-4}\ \text m^2$

Hence, the total increase in the surface area is

$\Delta \text A = 2 \times (50 \times 10^{-4}) = 1.0 \times 10^{-2}\ \text m^2$

Since W = T × ΔA,

$$\begin{aligned}\text T = \frac{\text W}{\Delta \text A} = \frac{3.0 \times 10^{-4}}{1.0 \times 10^{-2}} \\[1em] = 3.0 \times 10^{-2}\ \text{N m}^{-1}\end{aligned}$$

Hence, the surface tension of the soap solution is 3.0 × 10^-2 N m^-1.

Given,

• Initial surface area of the bubble, A = 2.0 × 10^-3 m²
• Final surface area = 2A = 4.0 × 10^-3 m²
• Surface tension of soap solution, T = 3.0 × 10^-2 N m^-1

The increase in the area of one surface is

$(4.0 \times 10^{-3}) - (2.0 \times 10^{-3}) = 2.0 \times 10^{-3}\ \text m^2$

Since a soap bubble has two free surfaces, the total increase in the surface area is

$\Delta \text A = 2 \times (2.0 \times 10^{-3}) = 4.0 \times 10^{-3}\ \text m^2$

The work done is

$$\begin{aligned}\text W = \text T \times \Delta \text A = (3.0 \times 10^{-2}) \times (4.0 \times 10^{-3}) \\[1em] = 1.2 \times 10^{-4}\ \text J\end{aligned}$$

Hence, the work done in blowing the bubble is 1.2 × 10^-4 J.

Given,

• Initial diameter = 2 cm, so r₁ = 1 cm = 1 × 10^-2 m
• Final diameter = 5 cm, so r₂ = 2.5 cm = 2.5 × 10^-2 m
• Surface tension of soap solution, T = 3.0 × 10^-2 N m^-1

A soap bubble has two free surfaces, so the increase in the total surface area is

$\Delta \text A = 2[4\pi \text r_2^2 - 4\pi \text r_1^2] = 8\pi (\text r_2^2 - \text r_1^2)$

Substituting the values,

$$\begin{aligned}\Delta \text A = 8 \times 3.14 \times [(2.5 \times 10^{-2})^2 - (1 \times 10^{-2})^2] \\[1em] = 8 \times 3.14 \times [(6.25 \times 10^{-4}) - (1 \times 10^{-4})]\end{aligned}$$

$= 8 \times 3.14 \times (5.25 \times 10^{-4}) = 1.319 \times 10^{-2}\ \text m^2$

The work done is

$$\begin{aligned}\text W = \text T \times \Delta \text A = (3.0 \times 10^{-2}) \times (1.319 \times 10^{-2}) \\[1em] = 3.96 \times 10^{-4}\ \text J\end{aligned}$$

Hence, the work done is 3.96 × 10^-4 J.

Given,

• Radius of the water drop, R = 2.0 mm = 2.0 × 10^-3 m
• Surface tension of water, T = 0.075 N m^-1

A liquid drop has only one free surface, so the excess pressure inside it is

$\text p = \frac{2\text T}{\text R}$

Substituting the values,

$$\begin{aligned}\text p = \frac{2 \times 0.075}{2.0 \times 10^{-3}} = \frac{0.15}{0.002} \\[1em] = 75\ \text{N m}^{-2}\end{aligned}$$

Hence, the excess pressure inside the drop of water is 75 N m^-2.

Given,

• Height of the liquid column, h = 12.0 cm = 0.12 m
• Radius of the capillary tube, r = 0.02 cm = 2 × 10^-4 m
• Surface tension of the liquid, T = 9.0 × 10^-2 N m^-1
• g = 10.0 m s^-2
• Angle of contact, θ = 0 (so cos θ = 1)

By the ascent formula, the rise of a liquid in a capillary tube is

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g} \quad \Rightarrow \quad \rho = \frac{2\text T}{\text{rhg}}$

Substituting the values,

$$\begin{aligned}\rho = \frac{2 \times (9.0 \times 10^{-2})}{(2 \times 10^{-4}) \times 0.12 \times 10.0} \\[1em] = \frac{0.18}{2.4 \times 10^{-4}}\end{aligned}$$

$= 7.5 \times 10^{2}\ \text{kg m}^{-3}$

Hence, the density of the liquid is 7.5 × 10² kg m^-3.

Given,

• Height of the water column = h
• Surface tension of water, T = 9.8 × 10^-2 N m^-1
• Density of water, ρ = 10³ kg m^-3
• g = 9.8 m s^-2
• Angle of contact, θ = 0 (so cos θ = 1)

By the ascent formula,

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g} \quad \Rightarrow \quad \text r = \frac{2\text T}{\text h\rho \text g}$

Substituting the values,

$\text r = \frac{2 \times (9.8 \times 10^{-2})}{\text h \times 10^{3} \times 9.8} = \frac{0.196}{9800\text h}$

$\text r = \frac{2}{\text h} \times 10^{-5}\ \text{metre}$

Hence, the radius of the capillary is $\frac{2}{\text h} \times 10^{-5}$ metre.

Given,

• Height of the liquid column, h = 7.0 cm = 0.07 m
• Radius of the capillary tube, r = 0.1 mm = 1 × 10^-4 m
• Density of the liquid, ρ = 0.8 × 10³ kg m^-3
• Angle of contact, θ = 0, so cos θ = 1
• g = 10 m s^-2

By the ascent formula,

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g} \quad \Rightarrow \quad \text T = \frac{\text{rh}\rho \text g}{2}$

Substituting the values,

$$\begin{aligned}\text T = \frac{(1 \times 10^{-4}) \times 0.07 \times (0.8 \times 10^{3}) \times 10}{2} \\[1em] = \frac{5.6 \times 10^{-2}}{2}\end{aligned}$$

$= 2.8 \times 10^{-2}\ \text{N m}^{-1}$

Hence, the surface tension of the liquid is 2.8 × 10^-2 N m^-1.

Given,

• Radius of the capillary tube, r = 0.2 × 10^-3 m
• Density of the liquid, ρ = 0.8 × 10³ kg m^-3
• Surface tension, T = 0.08 N m^-1
• Angle of contact, θ = 0, so cos θ = 1
• g = 10 m s^-2

By the ascent formula,

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g}$

Substituting the values,

$$\begin{aligned}\text h = \frac{2 \times 0.08}{(0.2 \times 10^{-3}) \times (0.8 \times 10^{3}) \times 10} \\[1em] = \frac{0.16}{1.6}\end{aligned}$$

$= 0.10\ \text m = 10\ \text{cm}$

Hence, the liquid will rise to a height of 10 cm in the tube.

Given,

• Diameter of the capillary = 0.1 mm, so radius r = 0.05 mm = 5 × 10^-5 m
• Surface tension of water, T = 0.072 N m^-1
• Density of water, ρ = 10³ kg m^-3
• g = 9.8 m s^-2
• Angle of contact, θ = 0, so cos θ = 1

By the ascent formula,

$\text h = \frac{2\text T\cos \theta}{\text r\rho \text g}$

Substituting the values,

$$\begin{aligned}\text h = \frac{2 \times 0.072}{(5 \times 10^{-5}) \times 10^{3} \times 9.8} \\[1em] = \frac{0.144}{0.49}\end{aligned}$$

$= 0.2938\ \text m = 29.38\ \text{cm}$

Hence, the rise of the water column in the capillary is 29.38 cm.

Given,

• Velocity at the top, v₁ = 2 m s^-1
• Pressures are equal, P₁ = P₂
• Difference in height, h₁ = 3 m
• g = 10 m s^-2

Applying Bernoulli's theorem between the top and the bottom of the pipe, taking the bottom as the reference level,

$\text P_1 + \frac{1}{2}\rho v_1^2 + \rho \text{gh}_1 = \text P_2 + \frac{1}{2}\rho v_2^2 + 0$

Since P₁ = P₂, these terms cancel out, and dividing throughout by ρ,

$\frac{1}{2}v_1^2 + \text{gh}_1 = \frac{1}{2}v_2^2$

Substituting the values,

$$\begin{aligned}\frac{1}{2}(2)^2 + (10 \times 3) = \frac{1}{2}v_2^2 \\[1em] 2 + 30 = \frac{1}{2}v_2^2\end{aligned}$$

$v_2^2 = 64 \quad \Rightarrow \quad v_2 = 8\ \text{m s}^{-1}$

By the principle of continuity,

$\text A_1v_1 = \text A_2v_2 \quad \Rightarrow \quad \frac{\text A_1}{\text A_2} = \frac{v_2}{v_1} = \frac{8}{2} = 4$

Hence, the ratio A₁ : A₂ is 4 : 1.

Given,

• Weight of the person = 80 kgf, so F = 80 × 9.8 = 784 N
• Area of the body, A₁ = 1.2 m²
• Area of one foot = 60 cm²
• g = 9.8 m s^-2

(i) When he is lying : The whole area of the body is in contact with the ground, so

$$\begin{aligned}\text P_1 = \frac{\text F}{\text A_1} = \frac{784}{1.2} \\[1em] = 6.53 \times 10^{2}\ \text{Pa}\end{aligned}$$

(ii) When he is standing on both his feet : The area in contact is that of the two feet,

$\text A_2 = 2 \times 60\ \text{cm}^2 = 120\ \text{cm}^2 = 120 \times 10^{-4}\ \text m^2$

$$\begin{aligned}\text P_2 = \frac{\text F}{\text A_2} = \frac{784}{120 \times 10^{-4}} \\[1em] = 6.53 \times 10^{4}\ \text{Pa}\end{aligned}$$

Hence, the pressure exerted is 6.53 × 10² Pa when he is lying and 6.53 × 10⁴ Pa when he is standing on both his feet.

The pressure while standing is a hundred times greater, because the same force is distributed over a much smaller area.

Given,

• Radius of the air bubble, r = 0.4 mm = 4 × 10^-4 m
• Coefficient of viscosity of water, η = 1 × 10^-3 kg m^-1 s^-1
• Density of water, σ = 1.0 × 10³ kg m^-3
• Density of air is negligible, so ρ = 0
• g = 9.8 m s^-2

By Stokes' law, the terminal velocity is

$v = \frac{2}{9}\frac{\text r^2(\rho - \sigma)\text g}{\eta}$

Since ρ < σ, the terminal velocity is negative, which means that the bubble moves upward. Taking the magnitude,

$v = \frac{2}{9}\frac{\text r^2\sigma \text g}{\eta}$

Substituting the values,

$$\begin{aligned}v = \frac{2}{9} \times \frac{(4 \times 10^{-4})^2 \times (1.0 \times 10^{3}) \times 9.8}{1 \times 10^{-3}} \\[1em] = \frac{2}{9} \times \frac{1.568 \times 10^{-3}}{1 \times 10^{-3}}\end{aligned}$$

$= \frac{2}{9} \times 1.568 = 0.348\ \text{m s}^{-1}$

Hence, the terminal speed of the bubble is 0.348 m s^-1, directed vertically upward.

Given,

• Initial area of cross-section, A₁ = 0.09 π m²
• Initial speed of flow, v₁ = 1.0 m s^-1
• The diameter is halved, so $\text D_2 = \frac{\text D_1}{2}$

Since the area of cross-section is proportional to the square of the diameter,

$\frac{\text A_2}{\text A_1} = \left(\frac{\text D_2}{\text D_1}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

that is, $\text A_2 = \frac{\text A_1}{4}$.

By the principle of continuity,

$\text A_1v_1 = \text A_2v_2 \quad \Rightarrow \quad v_2 = \frac{\text A_1v_1}{\text A_2}$

Substituting,

$$\begin{aligned}v_2 = \frac{\text A_1 \times 1.0}{\text A_1/4} = 4 \times 1.0 \\[1em] = 4.0\ \text{m s}^{-1}\end{aligned}$$

Hence, the speed of flow of water through the pipe becomes 4.0 m s^-1.

Given,

• Speed at the first place, v₁ = 30 cm/s = 0.30 m s^-1
• Pressure at the first place = 10 cm of water = 0.10 m of water
• Pressure at the other place = 5 cm of water = 0.05 m of water
• Density of water, ρ = 1.0 × 10³ kg m^-3, g = 9.8 m s^-2

The pressure difference between the two places is

$$\begin{aligned}\text P_1 - \text P_2 = (0.10 - 0.05) \times \rho \text g = 0.05 \times (1.0 \times 10^{3}) \times 9.8 \\[1em] = 490\ \text{N m}^{-2}\end{aligned}$$

Since the pipe is horizontal, by Bernoulli's theorem,

$\text P_1 + \frac{1}{2}\rho v_1^2 = \text P_2 + \frac{1}{2}\rho v_2^2$

$\text P_1 - \text P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Substituting the values,

$$\begin{aligned}490 = \frac{1}{2} \times (1.0 \times 10^{3}) \times [v_2^2 - (0.30)^2] \\[1em] 0.98 = v_2^2 - 0.09\end{aligned}$$

$v_2^2 = 1.07 \quad \Rightarrow \quad v_2 = 1.034\ \text{m s}^{-1}$

Hence, the speed of water at the other place is 1.034 m s^-1, that is, 103.4 cm s^-1.

Given,

• Pressure at the first place, P₁ = 0.1 m of mercury
• Velocity at the first place, v₁ = 0.4 m s^-1
• Velocity at the other place, v₂ = 0.5 m s^-1
• Density of water, ρ = 1.0 × 10³ kg m^-3
• Density of mercury = 13.6 × 10³ kg m^-3, g = 9.8 m s^-2

Since the pipe is horizontal, by Bernoulli's theorem,

$\text P_1 - \text P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Substituting the values,

$$\begin{aligned}\text P_1 - \text P_2 = \frac{1}{2} \times (1.0 \times 10^{3}) \times [(0.5)^2 - (0.4)^2] \\[1em] = 500 \times (0.25 - 0.16)\end{aligned}$$

$= 500 \times 0.09 = 45\ \text{N m}^{-2}$

Expressing this pressure difference as a height of mercury column,

$\text{h} = \frac{45}{(13.6 \times 10^{3}) \times 9.8} = 0.000338\ \text{m of Hg}$

Therefore, the pressure at the other place is

$$\begin{aligned}\text P_2 = 0.1 - 0.000338 \\[1em] = 0.09966\ \text{m of Hg}\end{aligned}$$

Hence, the pressure at the other place is 0.09966 m of mercury column.

Given,

• Diameter at the wider place = 4 cm, so radius = 2 cm = 2 × 10^-2 m
• Diameter at the narrower place = 2 cm, so radius = 1 cm = 1 × 10^-2 m
• Pressure difference, h = 4.5 cm of water = 0.045 m of water
• g = 9.8 m s^-2

The areas of cross-section are

$$\begin{aligned}\text A_1 = \pi (2 \times 10^{-2})^2 = 1.2566 \times 10^{-3}\ \text m^2 \\[1em] \text A_2 = \pi (1 \times 10^{-2})^2 = 3.1416 \times 10^{-4}\ \text m^2\end{aligned}$$

Treating the pipe as a venturimeter, the rate of flow of water is

$\text Q = \text A_1\text A_2\sqrt{\frac{2\text{gh}}{\text A_1^2 - \text A_2^2}}$

Substituting the values,

$\text A_1^2 - \text A_2^2 = (1.5791 \times 10^{-6}) - (9.8696 \times 10^{-8}) = 1.4804 \times 10^{-6}$

$2\text{gh} = 2 \times 9.8 \times 0.045 = 0.882$

$$\begin{aligned}\text Q = (1.2566 \times 10^{-3})(3.1416 \times 10^{-4})\sqrt{\frac{0.882}{1.4804 \times 10^{-6}}} \\[1em] = (3.948 \times 10^{-7}) \times 771.8\end{aligned}$$

$= 3.047 \times 10^{-4}\ \text m^3\text s^{-1}$

Hence, the rate of flow of water in the tube is 3.047 × 10^-4 m³ s^-1, that is, about 304.7 cm³ s^-1.

Given,

• Depth of the hole below the free surface, h = 40 m
• Radius of the hole, r = 1 mm = 1 × 10^-3 m
• g = 9.8 m s^-2

By Torricelli's theorem, the velocity of efflux is

$v = \sqrt{2\text{gh}}$

Substituting the values,

$$\begin{aligned}v = \sqrt{2 \times 9.8 \times 40} = \sqrt{784} \\[1em] = 28\ \text{m s}^{-1}\end{aligned}$$

The area of cross-section of the hole is

$\text A = \pi \text r^2 = 3.14 \times (1 \times 10^{-3})^2 = 3.14 \times 10^{-6}\ \text m^2$

The rate of flow of water is

$$\begin{aligned}\text Q = \text Av = (3.14 \times 10^{-6}) \times 28 \\[1em] = 8.8 \times 10^{-5}\ \text m^3\text s^{-1}\end{aligned}$$

Hence, the velocity of efflux is 28 m s^-1 and the rate of flow of water is 8.8 × 10^-5 m³ s^-1.

Given,

• Length of each wire, l = 10 cm = 0.10 m
• Increase in the separation, Δx = 1 mm = 1 × 10^-3 m
• Surface tension of water, T = 7.2 × 10^-2 N m^-1

A liquid film has two free surfaces, so the total increase in the surface area is

$$\begin{aligned}\Delta \text A = 2 \times \text l \times \Delta \text x = 2 \times 0.10 \times (1 \times 10^{-3}) \\[1em] = 2 \times 10^{-4}\ \text m^2\end{aligned}$$

The work done is

$$\begin{aligned}\text W = \text T \times \Delta \text A = (7.2 \times 10^{-2}) \times (2 \times 10^{-4}) \\[1em] = 1.44 \times 10^{-5}\ \text J\end{aligned}$$

Hence, the work done is 1.44 × 10^-5 J.

Given,

• Diameter of the ring = 3.0 cm, so radius r = 1.5 cm = 1.5 × 10^-2 m
• Surface tension of soap solution, T = 3.0 × 10^-2 N m^-1

The film formed between the ring and the solution has two free surfaces, so the length of the film in contact with the ring is

$\text l = 2 \times (2\pi \text r) = 4\pi \text r$

The force required to break the film is

$\text F = \text T \times \text l = \text T \times 4\pi \text r$

Substituting the values,

$$\begin{aligned}\text F = (3.0 \times 10^{-2}) \times 4 \times 3.14 \times (1.5 \times 10^{-2}) \\[1em] = (3.0 \times 10^{-2}) \times (0.1884)\end{aligned}$$

$= 5.652 \times 10^{-3}\ \text N$

Hence, an upward force of 5.652 × 10^-3 N is necessary to break the film.

Given,

• Diameter of the bubble = 3.5 mm, so radius R = 1.75 mm = 1.75 × 10^-3 m
• Excess pressure inside the bubble = 8.0 mm of water column
• Density of water, ρ = 1.0 × 10³ kg m^-3, g = 9.8 m s^-2

The excess pressure expressed in N m^-2 is

$$\begin{aligned}\text p = \text h\rho \text g = (8.0 \times 10^{-3}) \times (1.0 \times 10^{3}) \times 9.8 \\[1em] = 78.4\ \text{N m}^{-2}\end{aligned}$$

A soap bubble has two free surfaces, so the excess pressure inside it is

$\text p = \frac{4\text T}{\text R} \quad \Rightarrow \quad \text T = \frac{\text{pR}}{4}$

Substituting the values,

$$\begin{aligned}\text T = \frac{78.4 \times (1.75 \times 10^{-3})}{4} = \frac{0.1372}{4} \\[1em] = 3.43 \times 10^{-2}\ \text{N m}^{-1}\end{aligned}$$

Hence, the surface tension of the soap solution is 3.43 × 10^-2 N m^-1.

Given,

• Diameter of the wider limb = 5.0 mm, so r₁ = 2.5 mm = 2.5 × 10^-3 m
• Diameter of the narrower limb = 2.0 mm, so r₂ = 1.0 mm = 1.0 × 10^-3 m
• Surface tension of water, T = 7.3 × 10^-2 N m^-1
• Density of water, ρ = 1.0 × 10³ kg m^-3, g = 9.8 m s^-2
• Angle of contact, θ = 0, so cos θ = 1

The water rises to a different height in each limb, and the difference in the two heights is

$\Delta \text h = \text h_2 - \text h_1 = \frac{2\text T\cos \theta}{\text r_2\rho \text g} - \frac{2\text T\cos \theta}{\text r_1\rho \text g}$

$\Delta \text h = \frac{2\text T}{\rho \text g}\left(\frac{1}{\text r_2} - \frac{1}{\text r_1}\right)$

Substituting the values,

$$\begin{aligned}\Delta \text h = \frac{2 \times (7.3 \times 10^{-2})}{(1.0 \times 10^{3}) \times 9.8}\left(\frac{1}{1.0 \times 10^{-3}} - \frac{1}{2.5 \times 10^{-3}}\right) \\[1em] = (1.4898 \times 10^{-5}) \times (1000 - 400)\end{aligned}$$

$= (1.4898 \times 10^{-5}) \times 600 = 8.94 \times 10^{-3}\ \text m$

Hence, the difference in the heights of the water columns is 8.94 mm.

Given,

• Height of water in the vessel, H = 35 m
• Depth of the hole below the free surface, h = 7 m
• Radius of the hole, r = 1 cm = 1 × 10^-2 m
• g = 9.8 m s^-2

(i) Velocity of efflux : By Torricelli's theorem,

$$\begin{aligned}v = \sqrt{2\text{gh}} = \sqrt{2 \times 9.8 \times 7} = \sqrt{137.2} \\[1em] = 11.7\ \text{m s}^{-1}\end{aligned}$$

(ii) Rate of flow of water :

$$\begin{aligned}\text Q = \pi \text r^2 v = 3.14 \times (1 \times 10^{-2})^2 \times 11.7 \\[1em] = 3.67 \times 10^{-3}\ \text m^3\text s^{-1}\end{aligned}$$

(iii) Range of water flow : The horizontal range is

$$\begin{aligned}\text x = 2\sqrt{\text h(\text H - \text h)} = 2\sqrt{7 \times (35 - 7)} \\[1em] = 2\sqrt{196} = 2 \times 14 = 28\ \text m\end{aligned}$$

(iv) Distance where the range is the same : From the range formula, the range remains the same whether the orifice is at a depth h or at a depth (H − h). Hence the required depth is

$\text H - \text h = 35 - 7 = 28\ \text m$

(v) Distance for maximum range : The range is maximum when h(H − h) is maximum, which on differentiating gives

$\text h = \frac{\text H}{2} = \frac{35}{2} = 17.5\ \text m$

(vi) Maximum range :

$\text x_{max} = 2\sqrt{\left(\frac{\text H}{2}\right)\left(\text H - \frac{\text H}{2}\right)} = \text H = 35\ \text m$

Hence, (i) 11.7 m s^-1, (ii) 3.67 × 10^-3 m³ s^-1, (iii) 28 m, (iv) 28 m, (v) 17.5 m and (vi) 35 m.

Given,

• Radius of the big drop, R = 1.0 mm = 1 × 10^-3 m
• Number of droplets, n = 1000
• Surface tension of mercury, T = 0.465 N m^-1

Since the total volume remains unchanged,

$\frac{4}{3}\pi \text R^3 = 1000 \times \frac{4}{3}\pi \text r^3 \quad \Rightarrow \quad \text r = \frac{\text R}{10} = 1 \times 10^{-4}\ \text m$

The work done is equal to the surface tension multiplied by the increase in the surface area,

$\text W = \text T[1000 \times 4\pi \text r^2 - 4\pi \text R^2] = 4\pi \text T[1000\text r^2 - \text R^2]$

Substituting the values,

$$\begin{aligned}\text W = 4 \times 3.14 \times 0.465 \times [1000 \times (1 \times 10^{-4})^2 - (1 \times 10^{-3})^2] \\[1em] = 5.84 \times [(1 \times 10^{-5}) - (1 \times 10^{-6})]\end{aligned}$$

$= 5.84 \times (9 \times 10^{-6}) = 5.26 \times 10^{-5}\ \text J$

Hence, the work done in the process is 5.26 × 10^-5 J.