(A) 0.8711 × 10⁶ N/m²

(B) 2:1

(A) 0.9 × 10^-3 J

(C) 1:16

(D) energy - is conserved

For better wettability (surface coverage), the surface tension and angle of contact of paints and lubricating oil must be low.

Let T be the surface tension of a soap solution.

The initial surface area of soap bubble = 0

The final surface area of soap bubble = 2 × 4πr²

∴ The increase in surface area = 2 × 4πr²

The work done in blowing the soap bubble is

W = surface tension x increase in surface area

=T × 2 × 4πr² = 8πr²T

Conservation of energy is the basis of the Bernoulli’s principle

Bernoulli’s equation states that the work done per unit volume of a fluid by the surrounding fluid is equal to the sum of the changes in kinetic and potential energies per unit volume that occur during the flow.

In other words, the Bernoulli’s principle is thus consistent with the principle of conservation of energy.

An open tube manometer measures the gauge pressure, p = p₀ + hρg, where p is the pressure being measured, p₀ is the atmospheric pressure, h is the difference in height between the manometric liquid of density ρ in the two arms.

For a given pressure p, the product hρ is constant.

That is, ρ should be small for h to be large.

Therefore, for noticeably large h, laboratory manometer uses a low density liquid.

A fluid which does not undergo change in volume for a large range of pressures. Thus, its density remains constant throughout the fluid for a large range of pressures. In most cases, all liquids are incompressible.

• A spherical shape has the minimum surface area-to-volume ratio of all geometric forms.
• When two drops of a liquid are brought in contact, the cohesive forces between their molecules coalesces the drops into a single larger drop.
• This is because, the volume of the liquid remaining the Same, the surface area of the resulting single drop is less than the combined surface area of the smaller drops.
• The resulting decrease in surface energy is released into the environment as heat.

Proof : Let n droplets each of radius coalesce to form a single drop of radius R. As the Volume of the liquid remains constant,

volume of the drop = volume of n droplets

\(\frac{4}{3}πR^3\) = \(n × \frac{4}{3}πr^3\)

∴ R³=nr³

∴ R = \(\sqrt[3]{n}\,r\)   ....(1)

Surface area of n droplets =n × 4πr²

Surface area of the drop = 4πR² = 4π×n^2/3r²  ..... from eq. 1

∴ Surface area of the drop = n^2/3 × 4πr²

∴ The change in the surface area

= surface area of drop — surface area of n droplets

= 4πr² (n^2/3 — n)

Since the bracketed term is negative, there is a decrease in surface area and a decrease in surface energy. .

When a tube narrows, the same volume occupies a greater length, as schematically shown in Fig.  A₁ is the cross section of the broader pipe and that of narrower pipe is A₂

By the equation of continuity, v₂, =(A₁/A₂)v₁

Fig. Speed of fluid increases as it enters a narrower pipe

Since A₁/A₂ > v₂ > v₁

For the same volume to pass points 1 and 2 in a given time, the speed must be greater at point 2.

The process is exactly reversible. If the fluid flows in the opposite direction, its speed decreases when the tube widens.

Consider a horizontal constricted tube.

Let A₁ and A₂ be the cross-sectional areas at points P₁ and P₂, respectively.

Let v₁ and v₂ be the corresponding flow speeds.

ρis the density of the fluid in the pipeline.

By the equation of continuity,

A₁v₁ = A₂v₂     ... (1)

v₂/ v₁ =A₁/A₂ > 1    (…A₁>A₂)

Therefore, the speed of the liquid increases as it passes through the constriction. Since the meter is assumed to be horizontal, from Bernoulli’s equation we get,

\(p_1+\frac{1}{2}ρv_1^2\) = \(p_2+\frac{1}{2}ρv_2^2\)

\(p_1+\frac{1}{2}ρv_1^2\) = \(p_2+\frac{1}{2}ρv_1^2(\frac{A_1}{A_2})^2\) ...... from eq.(1)

∴ \(p_1-p_2=\frac{1}{2}ρv_1^2\left[(\frac{A_1}{A_2})^2-1\right]\)

Again, since A₁>A₂ the bracketed term is positive so that p₁ > p₂. Thus, as the fluid passes through the constriction or throat, the higher speed results in lower pressure at the throat.

Consider a spherical drop as shown in Fig.

Let p_i be the pressure inside the drop and p₀ be the pressure outside it. As the drop is spherical in shape, the pressure, p_i, inside the drop is greater than p₀, the pressure outside.

Therefore, the excess pressure inside the drop is  p_i - p₀.

Let the radius of the drop increase from r to r + Δr, where Δr is very small,

so that the pressure inside the drop remains almost constant.

Let the initial surface area of the drop be

A₁ = 4πr²,

and the final surface area of the drop be

A₂ = 4π (r+Δr)²

∴ A₂ = 4π (r² + 2rΔr + Δr²)

∴ A₂ = 4π r² + 8πrΔr + 4πΔr²

As Δr is very small, Δr² can be neglected,

∴ A₂ = 4πr² + 8πrΔr

Thus, increase in the surface area of the drop is

dA = A₂ – A₁ = 4πr² + 8πrΔr - 4πr² = 8πrΔr --- (1)

Work done in increasing the surface area by dA is stored as excess surface energy.

∴ dW = TdA= T (8πrΔr) --- (2)

This work done is also equal to the product of the force F which causes increase in the area of the bubble and the displacement Δr which is the increase in the radius of the bubble.

∴ dW = FΔr --- (3)

The excess force is given by,

(Excess pressure) × (Surface area)

∴  F = (p_i – p₀) 4πr² --- (4)

Equating Eq. (2) and Eq. (3), we get,

T(8prΔr) = (p_i – p₀) 4πr² Δr

∴ (p_i – p₀) = 2T/r --- (5)

This equation gives the excess pressure inside a drop. This is called Laplace’s law of a spherical membrane or Young Laplace Equation in spherical form

Consider a fluid in steady or streamline flow, that is its density is constant. The velocity of the fluid within a flow tube, while everywhere parallel to the tube, may change its magnitude.

Suppose the velocity is v₁ at point P and v₂ at point. Q. If A₁, and A₂, are the cross-sectional areas of the tube at these two points,

the volume flux across A₁,  \(\frac{d}{dt}(V_1)\) = A₁v₁

and that across A₂, \(\frac{d}{dt}(V_2)\) =A₂v₂

By the equation of continuity of flow for a fluid,

A₁v₁ = A₂v₂ 

i.e.  \(\frac{d}{dt}(V_1)\) = \(\frac{d}{dt}(V_2)\)

If p₁ and p₂, are the densities of the fluid at P and Q, respectively, the mass flux across A₁,

\(\frac{d}{dt}(m_1)\)  = \(\frac{d}{dt}(p_1v_1)\) = A₁p₁v₁

and that across A₂,

\(\frac{d}{dt}(m_2)\)  = \(\frac{d}{dt}(p_2v_2)\) = A₂p₂v₂

Since no fluid can enter or leave through the boundary of the tube, the conservation of mass requires the mass fluxes to be equal, i.e.

\(\frac{d}{dt}(m_1)\)  = \(\frac{d}{dt}(m_2)\)

A₁p₁v₁ = A₂p₂v₂

i.e. Apv = constant

which is the required expression.

(1) When a capillary tube is partially immersed a wetting liquid, there is capillary rise and liquid meniscus inside the tube is concave, as shown in Fig. (A).

Consider four points A, B, C, D, of which point is just above the concave meniscus inside the capillary and point B is just below it. Points C and D are just above and below the free liquid surface outside.

Let P_A, P_B, P_C and P_D be the pressures at point A, B, C and D, respectively.

Now, P_A = P_C = atmospheric pressure

The pressure is the same on both sides of the free surface of a liquid, so that

P_C = P_D

∴ P_A = P_D

The pressure on the concave side of a meniscus is always greater than that on the convex side, so that

P_A > P_B

∴ P_D > P_B            (….P_A = P_D)

The excess pressure outside presses the liquid up the capillary until the pressures at B and D (at the same horizontal level) equalize, i.e., P_B becomes equal to P_D. Thus, there is a capillary rise.

(2) For a non-wetting liquid, there is capillary depression and the liquid meniscus in the capillary tube is convex, as shown in Fig. (B).

Consider again four points A, B, C and D when the meniscus in the capillary tube is at the same level as the free surface of the liquid. Points A and B are just above and below the convex meniscus.

Points C and D are just above and below the free liquid surface outside.

The pressure at B (P_B) is greater than that at A(P_A). The pressure at A is the atmospheric pressure H and at D, P_D @ H = P_A. Hence, the hydrostatic pressure at the same levels at B and D are not equal,

P_B > P_D.

Hence, the liquid flows from B to D and the level of the liquid in the capillary falls. This continues till the pressure at B’ is the same as that D’, that is till the pressures at the same level are equal.

Consider a capillary tube of radius r partially immersed into a wetting liquid of density p. Let the capillary rise be h and q be the angle of contact at the edge of contact of the concave meniscus and glass (Fig)..

If R is the radius of curvature of the meniscus then from the figure, r=R cosθ.

Surface tension T is the tangential force per unit length acting along the contact line. It is directed into the liquid making an angle θ with the capillary wall. We ignore the small volume of the liquid in the meniscus. The gauge pressure within the liquid at a depth h, i.e. at the level of the free liquid surface open to the atmosphere, is

P— P_o = ρgh ………. (1)

By Laplace’s law for a spherical membrane, this gauge pressure is

P— P_o = 2T/R  …….(2)

hρg  = 2T/R = \(\frac{2Tcosθ}{r}\)

∴ h =    \(\frac{2Tcosθ}{rρg}\)…….(3)

Thus, narrower the capillary tube, the greater is the capillary rise.

From Eq. (3),

T=   \(\frac{hrρg}{2Tcosθ}\)………(4)

Equations (3) and (4) are also valid for capillary depression h of a non-wetting liquid. In this case, the meniscus is convex and θ is obtuse. Then, cosθ is negative but so is h, indicating a fall or depression of the liquid in the capillary. T is positive in both cases.

[For capillary rise, Eq. (3) is also called the ascent formula. ]

Given : h=200 m, ρ=1060 kg/m³ p₀ = 1.013×10⁵ Pa, g = 9.8 m/s²

Absolute pressure,

P=p₀+hρg

= (1.013×10⁵ )+(200×1060×9.8)

= (1.013×10⁵ )+(20.776×10⁵)

= 21.789×10⁵ = 21.789 Mpa

Given : A₁ = 30 cm² = 3×10^-3 m² , A₂ = 1500 cm² = 0.15 m² F₁=25 N

By Pascal’s law

\(\frac{F_1}{A_1}=\frac{F_2}{A_2}\)

∴The force on the output piston,

\(F_2=F_1(\frac{A_2}{A_1})\)

= (25)\(\frac{0.15}{3×10^3}\)  =25 × 50=1250 N

Weight on output piston = 1250 N

Given: d= 1 mm, v₀ = 2 m/s, h = 1.8 × 10^-5 Ns/m²

r =d/2 =0.5 mm = 5 ×10^-4 m,

By Stoke’s law the viscous force on rain drop is

f= 6πηr v₀

= 6 ×3.142 x 1.8 × 10^-5 x 5 ×10^-4 × 2

= 3.394 ×10^-7 N

Given : F=1, A=10^-2 m^2, v₀ =2 × 10^-2 m/s, y=1.5 × 10^-3 m

Velocity gradient, dv/dy = 2 × 10^-2 / 1.5 × 10^-3 = 40/3 s^-1

Viscous Force, F= \(η\frac{dv}{dy}\)

Therefore coefficient of viscosity is

η = \(\frac{F}{A(dv/dy)} =\frac{1}{10^{-2}(40/3)}\)

= 7.5 Pa-s

Given: d=0.4 mm, r=d/2 = 0.2mm= 2×10^-4 m, , h = 0.1 Pa-s, p_L = 0.9 × 10³ kg/m³ = 900 kg/m³,  

p_air = 1.29 kg/m³ g=9.8 m/s²

Since the density of air is less than that of oil, the air bubble will rise up through the liquid. Hence, the viscous force is downward, At terminal velocity, this downward viscous force is equal in magnitude to the net upward force.

Viscous force = buoyant force - gravitational force

6πηrvt = \(\frac{4}{3}πr^3(p_L - p_{air})g\)

∴ Terminal velocity,

vt =  \(\frac{2r^2g(p_{L} - p_{air})}{9η}\)

= \(\frac{2×10^{-4}9.8(900 - 1.29)}{9×0.1}\)

= 7.829 × 10^-4 m/s = 0.7829 mm/s (upwards)

Given : d₁=10 cm, = 0.1 m, v₁ = 2 m/s, v₂ = 4 m/s

By the equation of continuity, the ratio of the speed is

v₁/v₂ = (d₂/d₁)²

∴d₂/d₁ =\(\sqrt{\frac{v_1}{v_2}}\) =\(\sqrt{\frac{2}{4}}\) =0.707

d₂ = 0.707d₁ = 0.707(0.1) = 0.0707m

Given : p-p₀ = 4 × 10⁵ N/m² , ρ = 10³ kg/m³

If a orifice at a depth h from the water surface in tank, the gauge pressure there is

p-p₀ = hρg ….. (1)

By Toricellie’s law of reflex, the velocity of efflux

v=   \(\sqrt{2gh}\)…  (2)

Substituting for h from eq. 1

v=\(\sqrt{2g(\frac{p-p_0}{ρg})}\)

=\(\sqrt{2(\frac{p-p_0}{ρ})}\)

=\(\sqrt{2(\frac{4×10^5}{10^3})}\)

=\(20\sqrt{2}\)

= 28.28 m/s

Given : p₁=3 × 10⁵ N/m² v₁=0, p₂ = 2 × 10⁵ N/m² , ρ = 1000 kg/m³

Assuming potential head to be zero, i.e. the pipe to be horizontal, the Bernoulli equation is

p₁ +(1/2) ρv₁² = p2 +(1/2) ρv₂²

∴ v₂² = \(\frac{2(p_1-p_2)}{ρ}\)  …..(..v₁=0)

∴ v₂² = \(\frac{2(3-2)10^5}{10^3}\)

v₂²=200

v₂ = \(10\sqrt{2}\)

  =14.14 m/s

Given : r=0.1 mm = 1 x 10^-4 m, θ =0°, T= 7 × 10^-2 N/m, p=10³ kg/m³,

g = 9.8 m/s^{2 ,} cosθ = cos0 =1

Capillary rise h = \(\frac{2Tcosθ}{rρg}\)

= \(\frac{2×7 × 10^{-2}×1}{ 1 × 10^{-4}×1000×9.8}\)  

= 0.143 m

Given : r=0.2 mm = 2x10^-4 m, T= 7.2 x 10^-2 N/m, ρ=10³ kg/m³

The gauge pressure inside the bubble = 2T/r

= 2(7.2 x 10^-2)/(2x10^-4) =7.2 x 10² =720 Pa

Given : r=0.1mm =1 x 10^—4 m, T = 0.072 N/m

Let R be the radius of the single drop formed due to the coalescence of 27 droplets of mercury.

Volume of 27 droplets = volume of the single drop (as the volume of the liquid remains constant.)

\(∴27×\frac{4}{3}πr^3 =\frac{4}{3}πR^3\)

27r³=R³

∴ 3r=R

Surface area of 27 droplets = 27 x 4πr²

Surface area of single drop = 4πR²

∴ Decrease in surface area = 27 x 4πr² — 4πR²

= 4π (27r²— R²)

= 4π [27r² —(3r)²]

= 4π (27r² —9r²)

= 4π x 18r²

∴The energy released

= surface tension x decrease in surface area

= T x 4π x 18r²

= 0.472 x 4 x 3.142 x 18 x (1 x 10^—4)²

= 1.628 x 10^-7 J

Let R be the radius of the drop and r be the radius of each droplet.

Given : R=0.2 cm, n =8, T = 435.5 dyn/cm

As the volume of the liquid remains constant,

volume of n droplets = volume of the drop

\(∴n×\frac{4}{3}πR^3 =\frac{4}{3}πr^3\)

\(∴r^3=\frac{R^3}{n}\)

\(∴r=\frac{R}{\sqrt[3]{8}}=\frac{R}{2}\)

Surface area of the drop = 4πR²

Surface area of n droplets = n x 4πr².

The increase in the surface area

= surface area of n droplets—surface area of drop

= 4π (nr²— R²) = 4π (8 x R²/4 — R²)

= 4π (2 — 1) R² = 4πR²

∴The work done

= surface tension x increase in surface area

= T x 4πR² = 435.5 x 4 x 3.142 x (0.2)²

= 2.19 x 10² ergs = 2.19 x 10^—5 J

Given: r= 2 cm = 2x10^-2 m, T=0.07 N/m

Initial surface area of soap bubble = 0

Final surface area of soap bubble =2 ´ 4πr²

\ Increase in surface area = 2 ´ 4πr²

The work done

= surface tension x increase in surface area

= T x 2 x 4πr²

= 0.07 x 2 x 4 x 3.142 (2 x10^-2)²

= 7.038 x 10^-4 J

Given : A₁ = 2 × 2 cm² = 4×10^-4 m² , A₂ = 3 × 3 cm² = 9×10^-4 m²

T = 3×10^-2 N/m²

As the film has two surfaces, the work done is

W=2T(A₂ – A₁)

= 2(3×10^-2)(9×10^-4−4×10^-4)

= 3.0×10^-5 J = 30μJ