## 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**

- Straight line
- Parabola
- Ellipse
- 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**

- 4T/r
- 2T/r
- T/r
- 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**

- Circular
- Parabola
- Hyperbola
- 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**

- 2W
- \(\sqrt[3]{4} W\)
- \(\sqrt[3]{2} W\)
- 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**

- 16
- 8
- 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**

- 3h
- 2h
- 4h
- 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 r _{1} and r_{2}(r_{2}> r_{1}) adjoin, the radius of curvature of the common surface**

**is**

- \(r_2-r_1\)
- \(r_2+r_1\)
- \(\frac{\left(r_2-r_1\right)}{r_1 r_2}\)
- \(\frac{r_1 r_2}{r_2-r_1}\)

**Answer:**

If the atmospheric pressure is p_{0}, 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**

- MLT
^{-2} - MLT
^{-1} - MT
^{-2} - ML
^{2}T^{-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**

- 5 cm · s
^{-1} - 10 cm · s
^{-1} - 40 cm · s
^{-1} - 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^{-1}T^{-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**

- \(6 \pi \eta a^2 v\)
- \(\frac{6 \eta \nu}{\pi a}\)
- \(6 \pi \eta a v\)
- \(\frac{\pi \eta v}{6 a^3}\)

**Answer**: The option 3 is correct.

**Question 16. A drop of some liquid of volume 0.04 cm ^{3} 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 10^{5} dyne has to be applied normally to the surfaces. The surface tension of the liquid is (in dyn · cm^{-1})**

- 60
- 70
- 80
- 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**

- 5 cm
- 10 cm
- 15 cm
- 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**

- 870 poise
- 1120 poise
- 982 poise
- 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**

- 8.146 x 10
^{-4}J - 4.4 x 10
^{-4 }J - 2.108 x 10
^{-5}J - 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, R^{3} = 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]**

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

**Answer:**

Upward force due to surface tension

= Fsinθ = σ x 2π R sinθ

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

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 p _{0} 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**

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

**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 ≈ 10^{3} kg · m^{-3} and the coefficient of viscosity of air ≈ 1.8 x 10^{-5} N · s · m^{-2}? (Neglect buoyancy of air)**

- 49m · s
^{-1} - 98m · s
^{-1} - 392 m · s
^{-1} - 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)**

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

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**

- Energy = 4vt\(\left(\frac{1}{r}-\frac{1}{r}\right)\) is released
- Energy = 3vt\(\left(\frac{1}{r}+\frac{1}{r}\right)\) is absorbed
- Energy = 3vt\(\left(\frac{1}{r}-\frac{1}{r}\right)\) is released
- 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 m ^{2}. 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 [ρ_{air} =1.2 kg/m^{3}]**

- 4.8 x 10
^{5}N, downwards - 4.8 x 10
^{5}N, upwards - 2.4 x 10
^{5}N, upwards - 2.4 x 10
^{5}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, P_{0}.

Let the pressure on the roof be P.

∴ \(P+\frac{1}{2} \rho v^2=P_0+0\)

Clearly, P_{0} > P

So, effective pressure P_{0} – 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 m ^{2} 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**

- 3.45 x 10
^{3}Pa · s - 3.45 x 10
^{-2}Pa · s - 3.45 x 10
^{-3}Pa · s - 3.45 x 10
^{2}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**

- r
^{5} - r
^{2} - r
^{3} - r
^{4}

**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 A_{1} at point A with a velocity v_{1} and coming out at a height h_{2} at point B with velocity v_{2}.

The values of the potential energy (PE) and kinetic energy (KE) are increased, since h_{2} > h_{1 }and v_{2} > v_{1}.

This is done by the pressure at the time of work done on (the liquid. If p_{1} and p_{2} are the pressure at A and B, for a small displacement at A and B, the work done on the liquid at A,

⇒ \(\Delta x_1=p_1 A_1 v_1 \Delta t\)

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 10^{3} kg · m^{-3}, density of copper is 8.9 x 10^{3} 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 10^{3} kg · m^{-3}

Density of copper, p = 8.9 x 10^{3} 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:**

- Viscous drag is not taken into account,
- 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.