WBCHSE Class 11 Physics Viscosity And Surface Tension Short Answer Type Questions

Viscosity And Surface Tension Short Answer Type Questions

Question 1. A small spherical body of radius r made of material of density ρ is dropped into a long column of viscous liquid of density σ and coefficient of viscosity η. The graph of the terminal velocity (v) vs radius (r) of the body will be

  1. Straight line
  2. Parabola
  3. Ellipse
  4. None of these

Answer:

We know, terminal velocity, \(v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}\)

As v ∝ r² the graph v-r is a parabola.

The option 2 is correct.

Question 2. A spherical air bubble of radius r is formed in water. If T be the surface tension of water, the excess pressure inside the bubble is

  1. 4T/r
  2. 2T/r
  3. T/r
  4. None of these

Answer: The option 2 is correct.

Question 3. A During fall of water gently from a water tap the stream gets thinner at the bottom. Why?
Answer:

Velocity of the downward flow is higher. So the streamlines are crowded together. Hence the flow becomes thinner.

Question 4. According to Jurin’s law, the graph of the diameter (d) vs height of the water column (h) of the capillary tube will be

  1. Circular
  2. Parabola
  3. Hyperbola
  4. Straight line

Answer: The option 3 is correct.

Question 5. The work done in blowing a soap bubble of volume V is W. The work is to be done in blowing it of volume 2 V equals to

  1. 2W
  2. \(\sqrt[3]{4} W\)
  3. \(\sqrt[3]{2} W\)
  4. 2W

Answer:

When radius of the soap bubble is x, then excess pressure inside the bubble,

p = 4T/x [T = surface tension of soap water]

Volume, V = \(\frac{4}{3} \pi x^3\)

If the radius increases by dx, then increase in volume, dV = \(p d V=\frac{4 T}{x} \cdot 4 \pi x^2 d x=16 \pi T \cdot x d x\)

Now, work done for this increase, dW = \(p d V=\frac{4 T}{x} \cdot 4 \pi x^2 d x=16 \pi T \cdot x d x\)

If the radii of the rubbles of V and 2 V volume are R and R’ respectively,

V = \(\frac{4}{3} \pi R^3 \text { and } 2 V=\frac{4}{3} \pi R^3\)

∴ \(\frac{2 V}{V}=\left(\frac{R^{\prime}}{R}\right)^3 \text { or, } \frac{R^{\prime}}{R}=2^{1 / 3} \)

Work done to increase volume from 0 to V,

W = \(\int d W=16 \pi T \int_0^R x d x=\left.16 \pi T \frac{x^2}{2}\right|_0 ^R=8 \pi T R^2\)

Similarly, work was done to increase volume from 0 to 2 V,

W’ = \(8 \pi T R^{12}=8 \pi T R^2\left(\frac{R^{\prime}}{R}\right)^2\)

= \(W\left(2^{1 / 3}\right)^2=\sqrt[3]{4} W\)

The option 2 is correct.

Question 6. Two solid spheres of same metal having masses M and 8 M respectively fall simultaneously on a same viscous liquid and their terminal velocities are v and nv, then the value of n is

  1. 16
  2. 8
  3. 4
  4. 2

Answer:

Ratio of the masses of the two spheres =1:8

So, ratio of their volumes =1:8 and ratio of their radii = \(1: 8^{1 / 3}=1: 2\)

Thus, terminal velocity, \(v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{n} \text { i.e., } v \propto r^2\)

Hence, ratio of the terminal velocities of the two spheres
= 1:22 = 1:4 = v:4

∴ n = 4

The option 3 is correct.

Question 7. Two water drops of equal size are falling through air with constant velocity of 2 m/s. If the two drops are allowed to coalesce to form a single drop, what would be its terminal velocity?
Answer:

Ratio of volumes of the smaller and bigger water drop = 1:2

So, ratio of their radii = 1: \(2^{1 / 3}\)

Terminal velocity,

v = \(\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}, \text { i.e., } v \propto r^2\)

∴ \(\frac{v_1}{v_2}=\left(\frac{r_1}{r_2}\right)^2\)

or, \(v_2=v_1\left(\frac{r_2}{r_1}\right)^2=2 \times\left(\frac{2^{1 / 3}}{1}\right)^2=2^{5 / 3}\)

= \(\sqrt[3]{32}=3.175 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Question 8. What is the gain in potential energy of the water column in case of rise of water in a glass capillary tube? What is the work done by surface tension? Assume that the angle of contact between glass and water is 0°.
Answer:

Rise of water inside the capillary tube of radius r is,
h = \(\frac{2 T \cos \theta}{\rho g r}=\frac{2 T}{\rho g r}\) [where T = surface tension;

θ = angle of contact = 0°; ρ = density of water] Mass of water column of height h, m = nr^hp

The height of centre of mass of the water column = h/2

∴ The gain in potential energy,

U = \(m g \times \frac{h}{2}=\frac{1}{2} \pi r^2 \rho g h^2\)

= \(\frac{1}{2} \pi r^2 \rho g \times \frac{4 T^2}{\rho^2 g^2 r^2}=\frac{2 \pi T^2}{\rho g}\)

The work done by surface tension,

W = \(2 \pi r h T \cos \theta=2 \pi r \times \frac{2 T}{\rho g r} \times T \times \cos 0^{\circ}=\frac{4 \pi T^2}{\rho g}\)

Question 9. Water rises up to a height h in a certain capillary tube of a particular diameter. Another capillary tube is taken whose diameter is half that of the previous tube. The height up to which water will rise in the second tube is

  1. 3h
  2. 2h
  3. 4h
  4. h

Answer:

As, \(h \propto \frac{1}{r} \text {, so } \frac{h}{h^{\prime}}=\frac{r^{\prime}}{r}\)

Here, \(r^{\prime}=\frac{r}{2}\text { or, } r=2 r^{\prime} \)

∴ \(\frac{h}{h^{\prime}}=\frac{r^{\prime}}{2 r^{\prime}} \text { or, } h^{\prime}=2 h\)

The option 2 is correct.

Question 10. When two soap bubbles of radii r1 and r2(r2> r1) adjoin, the radius of curvature of the common surface
is

  1. \(r_2-r_1\)
  2. \(r_2+r_1\)
  3. \(\frac{\left(r_2-r_1\right)}{r_1 r_2}\)
  4. \(\frac{r_1 r_2}{r_2-r_1}\)

Answer:

If the atmospheric pressure is p0, then in case of the first bubble,

⇒ \(p_1-p_0=\frac{4 T}{r_1}\) [T = surface tension of soap solution]

In case of the second bubble,

⇒ \(p_2-p_0=\frac{4 T}{r_2}\)

⇒ \(p_1-p_2=4 T\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

Again, if the radius of curvature of the common surface is r, then, \(p_1-p_2=\frac{4 T}{r}\)

∴ \(\frac{4 T}{r}=4 T\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \text { or, } \frac{1}{r}=\frac{r_2-r_1}{r_1 r_2} \text { or, } r=\frac{r_1 r_2}{r_2-r_1}\)

The option 4 is correct.

Question 11. 27 number of droplets having same size are falling through air with the same terminal velocity of 1 m · s-1. If the small droplets merge to produce a new drop, what will be the terminal velocity of the new drop?
Answer:

The terminal velocity of a moving droplet is proportional to the square of its radius.

When 27 number of droplets having same size merge and form a new drop then the volume of the new drop becomes 27 times the volume of each small drop.

Hence, the radius of the new drop becomes 3√27 times or 3 times.

So, the terminal velocity will be (3)² or 9 times.

∴ The terminal velocity of the new drop = 9×1 = 9 m · s-1

Question 12. The dimension of surface tension is

  1. MLT-2
  2. MLT-1
  3. MT-2
  4. ML2T-2

Answer: The option 3 is correct.

Question 13. The speed of a ball of radius 2 cm in a viscous liquid medium is 20 cm · s-1. The speed of a ball of radius 1 cm in the same liquid will be

  1. 5 cm · s-1
  2. 10 cm · s-1
  3. 40 cm · s-1
  4. 80cm · s-1

Answer:

Terminal velocity, v ∝ r²

∴ \(\frac{v^{\prime}}{v}=\left(\frac{r^{\prime}}{r}\right)^2 \quad \text { or, } v=\left(\frac{1}{2}\right)^2 \times 20=5 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

The option 1 is correct.

Question 14. Write down Stokes’ law and derive it from dimensional analysis.
Answer:

[F] = MLT-2

[6πηrv] =[η]>[r] · [v] = ML-1T-1 · L · LT-1

= MLT-2

∴ F = 6πηv

Question 15. A small metal sphere of radius a is falling with a velocity through a vertical column of a viscous liquid. If the coefficient of viscosity of the liquid is η, then the sphere encounters an opposing force of

  1. \(6 \pi \eta a^2 v\)
  2. \(\frac{6 \eta \nu}{\pi a}\)
  3. \(6 \pi \eta a v\)
  4. \(\frac{\pi \eta v}{6 a^3}\)

Answer: The option 3 is correct.

Question 16. A drop of some liquid of volume 0.04 cm3 is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area 20 cm² between the surfaces of the two slides. To separate the slides, a force of 16 x 105 dyne has to be applied normally to the surfaces. The surface tension of the liquid is (in dyn · cm-1)

  1. 60
  2. 70
  3. 80
  4. 90

Answer:

Thickness of layer,

h = \(\frac{\text { volume }}{\text { area }}=\frac{0.04}{20}=0.002 \mathrm{~cm}\)

Radius of curvature of concave surface,

r = h/2 = 0.001 cm.

Force, F = T/r x area

∴ Surface tension,

T = \(\frac{F \times r}{\text { area }}=\frac{16 \times 10^5 \times 0.001}{20}=80 \mathrm{dyn} \cdot \mathrm{cm}^{-1}\)

The option 3 is correct.

Question 17. A 20 cm long capillary tube is dipped vei tically in water and the liquid rises upto 10 cm. If the entire system is kept in a freely falling platform, the length of water column in the tube will be

  1. 5 cm
  2. 10 cm
  3. 15 cm
  4. 20 cm

Answer:

Height of water in the capillary tube, h = \(\frac{2 T}{r \rho g}\)

Here, T = surface tension of water

If the entire system is kept in a freely falling platform,
g = 0

But there will be no change in T.

∴ h → ∞

So, the maximum length of water column in the tube will be 20 cm.

The option 4 is correct.

Question 18. A gas bubble of 2 cm diameter rises through a liquid of density 1.75 g · cm-3 with a fixed speed of 0.35 cm · s-1. Neglect the density of the gas. The coefficient of viscosity of the liquid is

  1. 870 poise
  2. 1120 poise
  3. 982 poise
  4. 1089 poise

Answer:

The coefficient of viscosity of the liquid,

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

Here, r = 1cm, v = -0.35 cm/s, ρ = 0, σ = 1.75 g · cm-3

∴ \(\eta=\frac{2}{9} \times \frac{(1)^2(0-1.75) \times 980}{-0.35}\)

= 1088.8 ≈ 1089 poise

The option 4 is correct.

Question 19. 1000 droplets of water having 2 mm diameter each coalesce to form a single drop. Given the surface tension of water is 0.072 N · m-1. The energy loss in the process is

  1. 8.146 x 10-4 J
  2. 4.4 x 10-4 J
  3. 2.108 x 10-5 J
  4. 4.7 x 10-1 J

Answer:

Let us consider the radius of the big drop of water = R

∴ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi(0.1)^3 \times 1000\)

or, R3 = 1 or, R = 1.0 cm

The area of the surface of the big drop to water = 4π(1.0)² – 4π cm²

The total area of the surfaces of 1000 droplets of water = 4π(0.1)² x 1000 cm²

Decrease in surface area = 47(0.1)² x 1000 -4π

= 4π(10 – 1)

= 36π cm² = 0.00367 m²

The energy loss =0.00367×0.072

= 8.143 x 10-4 J

The option 1 is correct.

Question 20. A drop of water detaches itself from the exit of a tap when [σ = surface tension of water, ρ = density of water, R =radius of the tap exit, r=radius of the drop]

  1. \(r>\left(\frac{2}{3} \frac{R \sigma}{\rho g}\right)^{\frac{1}{3}}\)
  2. \(r>\left(\frac{2}{3} \frac{\sigma}{\rho g}\right)\)
  3. \(\frac{2 \sigma}{r}>\) atmospheric pressure
  4. \(r>\left(\frac{2}{3} \frac{R \sigma}{\rho g}\right)^{\frac{2}{3}}\)

Answer:

Upward force due to surface tension

 

 

= Fsinθ = σ x 2π R sinθ

Class 11 Physics Viscosity And Surface Tension Short Answer Questions

= \(\sigma \times 2 \pi R \times \frac{R}{r}=\frac{2 \pi \sigma R^2}{r}\)

Weight of the drop of water = mg = \(\frac{4}{3} \pi r^3 \rho g\)

The drop will detach if, \(\frac{4}{3} \pi r^3 \rho g>\frac{2 \pi \sigma R^2}{r} \text { or, } r^4>\frac{3}{2} \frac{\sigma R^2}{\rho g}\)

∴ \(r>\left(\frac{3}{2} \frac{\sigma R^2}{\rho g}\right)^{\frac{1}{4}}\)

None of the options are correct.

Question 21. A uniform capillary tube of length l and inner radius r with its upper end sealed is submerged vertically into the water. The outside pressure is p0 and surface tension of water is γ. When a length x of the capillary is submerged into water, it is found that water levels inside and outside the capillary coincide. The value of x is

  1. \(\frac{l}{\left(1+\frac{p_0 r}{4 \gamma}\right)}\)
  2. \(\frac{l}{\left(1-\frac{p_0 r}{4 \gamma}\right)}\)
  3. \(\frac{l}{\left(1-\frac{p_0 r}{2 \gamma}\right)} \)
  4. \(\frac{l}{\left(1+\frac{p_0 r}{2 \gamma}\right)}\)

Answer:

When a length x of the capillary is submerged in water and the atmospheric pressure in the capillary tube is p’, then

⇒ \(p_0(l A)=p^{\prime}(l-x) A\)

or, \(p^{\prime}=\frac{p_0 l}{l-x}\) …..(1)

As the water levels inside and outside the capillary coincide, so \(p^{\prime}-p_0=\frac{2 \gamma}{r}\)…..(2)

Solving equation (1) and (2) we get,

x = \(\frac{l}{\left(1+\frac{p_0 r}{2 \gamma}\right)}\)

Question 22. What will be the approximate terminal velocity of a raindrop of diameter 1.8 x 10-3 m, when density of rain water ≈ 103 kg · m-3 and the coefficient of viscosity of air ≈ 1.8 x 10-5 N · s · m-2? (Neglect buoyancy of air)

  1. 49m · s-1
  2. 98m · s-1
  3. 392 m · s-1
  4. 980 m · s-1

Answer:

F = 6πηrv [v is the terminal velocity]

or, \(m g=6 \pi \eta r v \quad \text { or, } \frac{4}{3} \pi r^3 \times \rho \times g=6 \pi \eta r \nu\)

or, v = \(\frac{4 \pi r^3 \rho g}{3 \times 6 \pi \eta r}=\frac{2 r^2 \rho g}{9 \eta}\)

= \(\frac{2 \times 1.8 \times 1.8 \times 10^{-6} \times 10^3 \times 9.8}{4 \times 9 \times 1.8 \times 10^{-5}}=98 \mathrm{~m} / \mathrm{s}\)

The option 2 is correct.

Question 23. On heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r<<R and the surface tension of water is T, value of r just before bubbles detach is (density of water is ρ)

  1. \(R^2 \sqrt{\frac{\rho_w g}{3 T}}\)
  2. \(R^2 \sqrt{\frac{\rho_{w g} g}{6 T}}\)
  3. \(R^2 \sqrt{\frac{\rho_w g}{T}}\)
  4. \(R^2 \sqrt{\frac{3 \rho_w g}{T}}\)

None of the options are correct.

Question 24. certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then

  1. Energy = 4vt\(\left(\frac{1}{r}-\frac{1}{r}\right)\) is released
  2. Energy = 3vt\(\left(\frac{1}{r}+\frac{1}{r}\right)\) is absorbed
  3. Energy = 3vt\(\left(\frac{1}{r}-\frac{1}{r}\right)\) is released
  4. Energy is neither released nor absorbed

Answer:

⇒ \(A_f=4 \pi R^2=\frac{3}{3} 4 \pi \frac{R^3}{R}=\frac{3 V}{R}\)

⇒ \(A_i=n \times 4 \pi r^2=\frac{V}{\frac{4}{3} \pi r^3} 4 \pi r^2=\frac{3 V}{r}\)

Hence, energy released = \(\left(A_i-A_f\right) T=3 V T\left(\frac{1}{r}-\frac{1}{R}\right)\)

The option 3 is correct.

Question 25. A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m2. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be [ρair =1.2 kg/m3]

  1. 4.8 x 105 N, downwards
  2. 4.8 x 105 N, upwards
  3. 2.4 x 105 N, upwards
  4. 2.4 x 105 N, downwards

Answer:

According to Bernoulli’s theorem, \(p+\frac{1}{2} \rho v^2=\text { constant }\)

When the pressure inside the house = atmospheric pressure, P0.

Let the pressure on the roof be P.

∴ \(P+\frac{1}{2} \rho v^2=P_0+0\)

Clearly, P0 > P

So, effective pressure P0 – P will act in the upward direction.

⇒ \(P_0-P=\frac{1}{2} \rho v^2=\frac{1}{2} \times 1.2 \times(40)^2\)

Hence, effective force

= \(\frac{1}{2} \times 1.2 \times(40)^2 \times 250=2.4 \times 10^5 \mathrm{~N}\)

The option 3 is correct.

Question 26. A metal block of base area 0.2 m2 is connected to a 0.02 kg mass via a string that passes over an ideal pulley as shown in figure. A liquid film of thickness 0.6 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.17 m/s. The coefficient of viscosity of the liquid is

Class 11 Physics Unit 7 Properties Of Matter Chapter 3 Viscosity And Surface Tension A Metal Block Of Base Area

  1. 3.45 x 103 Pa · s
  2. 3.45 x 10-2 Pa · s
  3. 3.45 x 10-3 Pa · s
  4. 3.45 x 102 Pa · s

Answer:

Velocity gradient of the liquid film between the block
and the table = \(\frac{v}{x}\) [x = thickness of film]

∴ Viscous force, F = \(\eta A \frac{v}{x}\)

As the block has no acceleration, therefore the resultant force is zero.

∴ mg = F= \(\eta A \frac{v}{x}\)

or, \(\eta=\frac{m g x}{A \nu}=\frac{0.02 \times 9.8 \times\left(0.6 \times 10^{-3}\right)}{0.2 \times 0.17}\)

= 3.46 x 10-3 Pa · s

The option 3 is correct.

Question 27. A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

  1. r5
  2. r2
  3. r3
  4. r4

Answer:

The opposing force due to viscosity, F = 6πηrv

⇒ \(\frac{d Q}{d t}=P=F \cdot \nu=6 \pi \eta r v^2 \text { and } \nu \propto r^2 \text { or, } \nu=k r^2\)

∴ P = \(6 \pi \eta r\left(k r^2\right)^2=6 \pi k^2 \eta r^5\)

∴ \(P \propto r^5\)

The option 1 is correct.

Question 28. State and prove Bernoulli’s theorem.
Answer:

Let an incompressible, non-viscous liquid entering into the cross-sectional area A1 at point A with a velocity v1 and coming out at a height h2 at point B with velocity v2.

The values of the potential energy (PE) and kinetic energy (KE) are increased, since h2 > h1 and v2 > v1.

Class 11 Physics Unit 7 Properties Of Matter Chapter 3 Viscosity And Surface Tension Bernoullis Theorem

This is done by the pressure at the time of work done on (the liquid. If p1 and p2 are the pressure at A and B, for a small displacement at A and B, the work done on the liquid at A,

⇒ \(\Delta x_1=p_1 A_1 v_1 \Delta t\)

At = a small time so that the area may be same] and the work done by the liquid at B,

⇒ \(\Delta x_2=-p_2 A_2 \nu_2 \Delta t\)

So, net work done by pressure

= \(\left(p_1-p_2\right) A_0 v_0 \Delta t \text { [where } A_1 v_1=A_2 v_2=A_0 v_0 \text { ] }\)

From the conservation of energy,

⇒ \(\left(p_1-p_2\right) A_0 v_0 \Delta t=\text { change in } \mathrm{KE}+\text { change in } \mathrm{PE}\)

or, \(\left(p_1-p_2\right) A_0 \nu_0 \Delta t\)

= \(\frac{1}{2} A_0 v_0 \Delta t \rho\left(v_2^2-v_1^2\right)+A_0 v_0 \rho \Delta {tg}\left(h_2-h_1\right)\)

or, \(p_1-p_2=\rho g\left(h_2-h_1\right)+\frac{\rho}{2}\left(v_2^2-v_1^2\right)\)

or, \(p_1+\rho g h_1+\frac{\rho}{2} v_1^2=p_2+\rho g h_2+\frac{\rho}{2} v_2^2\)

∴ \(\frac{p}{\rho g}+h+\frac{v^2}{2 g}=\text { constant }\)

Question 29. Write two factors affecting viscosity. Which one is more viscous: pure water or saline water?
Amswer:

1st part: We know, viscous force F = \(\eta A \frac{d v}{d x}\)

∴ \(\eta=\frac{F}{\frac{d v}{d x}}A\)

If area A = 1, then, \(\eta=\frac{F}{\frac{d v}{d x}}\)

So the two factors which affect viscosity are force and velocity.

2nd part: Saline water is more viscous.

Question 30. The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20 °C is 6.5 cm · s-1. Compute the coefficient of viscosity of the oil at 20 °C. (Density of oil is 1.5 x 103 kg · m-3, density of copper is 8.9 x 103 kg · m-3)
Answer:

Terminal velocity of the ball, v = 6.5 cm · s-1

Radius of the ball, r = 2.0 mm = 2 x 10-3 m

Density of oil, σ = 1.5 x 103 kg · m-3

Density of copper, p = 8.9 x 103 kg · m-3.

Acceleration due to gravity, g = 9.8 m · s-2.

Hence, the coefficient of viscosity of the oil,

⇒ \(\eta=\frac{2}{9} \cdot \frac{r^2(\rho-\sigma) g}{\nu}\)

= \(\frac{2}{9} \cdot \frac{\left(2 \times 10^{-3}\right)^2(8.9-1.5) 10^3 \times 9.8}{6.5 \times 10^{-2}}\)

= 0.99 kg · m-1 · s-1

Question 31. When we try to close a water tap with our fingers fast jets of water gush through the openings between our fingers. Explain why.
Answer:

From the equation of continuity, vα = constant where v is the velocity of the fluid and α is the cross-sectional area of the tube.

When we try to close a water tap with our fingers, the cross-sectional area of the outlet of the water jet is reduced. Hence, the velocity of water increases greatly, and fast jets of water come through the openings between our fingers.

Question 32. What is the surface energy? Find the relation between surface tension and surface energy. Explain why, detergents should have small angle of contact.
Answer:

  • Surface tension of a liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. Since this force is independent of area of liquid surface, hence surface tension of a liquid is also independent of the area of the liquid surface.
  • A cloth has narrow spaces in the form of capillaries. The rise of liquid in a capillary tube is directly proportional to cosθ. If θ is small, cosθ will be large Due to this reason, there will be more capillary rise and so the detergent’s penetration in the cloth will be better. Hence, detergents should have a small angle of contact.

Question 33. Write any two limitations of Bernoulli’s theorem.
Answer:

  1. Viscous drag is not taken into account,
  2. It is assumed that there is no energy loss.

Question 33. A liquid of density ρ and surface tension S rises to a height h in a capillary tube of diameter D. What is the weight of the liquid in the capillary tube? Angle of contact is 0°.
Answer:

Rise of liquid in the capillary tube is given as,

h= \(\frac{2 S \cos \theta}{\rho r g}\)

Here, \(r=\frac{D}{2}, \theta=0^{\circ}, \cos 0^{\circ}=1\)

So, \(h=\frac{2 S \times 1}{\rho \times \frac{D}{2} \times g}=\frac{4 S}{\rho D g}\)

Now, weight of liquid in the capillary tube,

W = mass x g = (volume x density) x g

= \(\left(\pi r^2 \times h\right) \times \rho \times g\)

= \(\pi \frac{D^2}{4} \times\left(\frac{4 S}{\rho D g}\right) \times \rho \times g=\pi D S\)

This is the required result.

 

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