Surface Tension and Viscosity

Practice Questions:

Question 11: easy

125 small droplets each of radius r, combine to form a big drop. If surface tension of liquid is T. Then loss in surface potential energy during this process will be:

1. \(T \times 16\pi r^2\)

2. \(T \times 20\pi r^2\)

3. \(T \times 100\pi r^2\)

4. \(T \times 400\pi r^2\)

View Answer

Volume conservation gives \(R = 5r\). Initial surface area is \(A_i = 125 \times 4\pi r^2 = 500\pi r^2\) and final is \(A_f = 4\pi R^2 = 100\pi r^2\). The decrease in area is \(400\pi r^2\), so loss in surface energy is \(T \times 400\pi r^2\).

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Question 12: easy

Water rises to a height of 4 cm in a capillary tube. If surface tension of water is 60 dyne/cm, then radius of capillary tube is:

1. 3 cm

2. 0.03 cm

3. 6 cm

4. 0.06 cm

View Answer

Using capillary rise formula \(h = \frac{2T \cos\theta}{r \rho g}\) with \(\theta = 0^\circ\) in CGS: \(4 = \frac{2 \times 60 \times 1}{r \times 1 \times 1000} ⇒ r = 0.03\text{ cm}\).

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Question 13: moderate

A long capillary is dipped in a beaker containing water. Water rises in capillary upto some height \(h\). Match the statements in list-I with most appropriate effects on water level mentioned in list-II:

**List-I**
(A) Soap solution is added to water
(B) Arrangement taken in a freely falling lift
(C) In a lift accelerating uniformly upward
(D) Arrangement is taken in a lift accelerating uniformly downward

**List-II**
(p) \(h\) decreases
(q) \(h\) increases
(r) \(h\) remains same
(s) water will rise upto complete height of capillary
(t) water level in capillary goes below the outside level

1. A - p, B - t, C - q, D - p

2. A - t, B - q, C - s, D - p

3. A - p, B - s, C - p, D - q

4. A - q, B - p, C - s, D - q

View Answer

Soap reduces surface tension, so \(h\) decreases (A-p). In a free fall, effective gravity \(g_{eff} = 0\), so water rises to full height (B-s). Upward acceleration increases \(g_{eff}\) hence \(h\) decreases (C-p). Downward acceleration decreases \(g_{eff}\) hence \(h\) increases (D-q).

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Question 14: easy

What is ratio of surface energy of 1 small drop and 1 large drop, if 1000 small drops combined to form 1 large drop :

1. 100 : 1

2. 1000 : 1

3. 10: 1

4. 1 : 100

View Answer

Volume conservation gives \(R = 10r\). Since surface energy \(E = T \cdot 4\pi R^2\), the ratio of surface energy of one small drop to one large drop is \(r^2 : R^2 = r^2 : 100r^2 = 1 : 100\).

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Question 15: easy

If the excess pressure inside a soap bubble is balanced by an oil column of height \(2\text{ mm}\), then the surface tension of soap solution will be : (\(r = 1\text{ cm}\) and density \(d = 0.8\text{ gm/cc}\))

1. \(4\text{ N/m}\)

2. \(4 \times 10^{-2}\text{ N/m}\)

3. \(4 \times 10^{-4}\text{ N/m}\)

4. 4 dyne/m

View Answer

Excess pressure in a soap bubble is \(\Delta P = \frac{4T}{r}\), and the pressure of the oil column is \(h d g\). Setting them equal, \(T = \frac{h d g r}{4}\). Substituting SI values gives \(T = \frac{2 \times 10^{-3} \times 800 \times 9.8 \times 10^{-2}}{4} \approx 4 \times 10^{-2}\text{ N/m}\).

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Question 16: easy

A ring of radius \(1.5\text{ cm}\) is floating horizontally on the surface of water. If this ring has to be raised up then how much additional force has to be applied to lift ring : (Surface tension of water \(73 \times 10^{-3}\text{ Newton/metre}\))

1. \(1.37 \times 10^{-2}\text{ N}\)

2. \(2.3 \times 10^{-2}\text{ N}\)

3. \(5.1 \times 10^{-3}\text{ N}\)

4. \(4 \times 10^{-2}\text{ N}\)

View Answer

The additional force required to lift the ring is \(F = 2 \times (2\pi r T) = 4\pi r T\). Substituting \(r = 1.5 \times 10^{-2}\text{ m}\) and \(T = 73 \times 10^{-3}\text{ N/m}\) gives \(F = 4 \times 3.14 \times 1.5 \times 10^{-2} \times 73 \times 10^{-3} \approx 1.37 \times 10^{-2}\text{ N}\).

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Question 17: easy

The work done in blowing a soap bubble of \(20\text{ cm}\) radius is (surface tension of soap solution is \(0.03\text{ N/m}\))

1. \(2.06 \times 10^{-2}\text{ J}\)

2. \(3.01 \times 10^{-2}\text{ J}\)

3. \(5.06 \times 10^{-2}\text{ J}\)

4. \(1.51 \times 10^{-2}\text{ J}\)

View Answer

The work done to create a soap bubble (which has two free surfaces) is given by \(W = 2T\Delta A = 8\pi R^2 T\). Substituting \(R = 0.2\text{ m}\) and \(T = 0.03\text{ N/m}\) gives \(W = 8 \times \pi \times (0.2)^2 \times 0.03 \approx 3.01 \times 10^{-2}\text{ J}\).

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Question 18: moderate

The velocity of a small ball of mass \(M\) and density \(d\), when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be

1. 2Mg

2. Mg/2

3. Mg

4. 3/2 Mg

View Answer

When the ball reaches terminal velocity, net force is zero: \(F_v + F_B = Mg\). The buoyant force is \(F_B = V \rho_{\text{glycerine}} g = V \left(\frac{d}{2}\right) g = \frac{Mg}{2}\). Thus, the viscous force is \(F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}\).

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Question 19: easy

n identical small drops of water having radius \( r \) coalesce to form a bigger drop. If surface tension of water is \( T \) then excess pressure in bigger drop will be

1. \[ \frac{n^3 4T}{r} \]

2. \[ \frac{2T}{n^{\frac{1}{3}} r} \]

3. \[ \frac{4T}{nr} \]

4. \[ \frac{2T}{n^3 r} \]

View Answer

By conserving volume, \( \frac{4}{3} \pi R^3 = n \left(\frac{4}{3} \pi r^3\right) ⇒ R = n^{1/3} r \). The excess pressure in a single-surface liquid drop of radius \( R \) is given by \( \Delta P = \frac{2T}{R} = \frac{2T}{n^{1/3} r} \).

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Question 20: easy

The velocity of a small ball of mass \(m\) and density \(d\), when dropped in a container filled with glycerine becomes constant after sometime. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be

1. \(2mg\)

2. \(\frac{mg}{2}\)

3. \(mg\)

4. \(\frac{3}{2}mg\)

View Answer

At terminal velocity, net force is zero: \(F_v + F_B = mg\). Here, buoyant force \(F_B = V\left(\frac{d}{2}\right)g = \frac{mg}{2}\). Thus, the viscous force is \(F_v = mg - \frac{mg}{2} = \frac{mg}{2}\).

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