Surface Tension and Viscosity - NEET Physics Questions
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Surface Tension and Viscosity

Question 11: 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 \(T = \frac{2 \times 10^{-3} \times 800 \times 9.8 \times 10^{-2}}{4} \approx 4 \times 10^{-2}\text{ N/m}\).

Question 12: 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 \(F = 4 \times 3.14 \times 1.5 \times 10^{-2} \times 73 \times 10^{-3} \approx 1.37 \times 10^{-2}\text{ N}\).

Question 13: 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 \(r^2 : R^2 = r^2 : 100r^2 = 1 : 100\).

Question 14: 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 \(W = 8 \times \pi \times (0.2)^2 \times 0.03 \approx 3.01 \times 10^{-2}\text{ J}\).

Question 15: 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 \( \Delta P = \frac{2T}{R} = \frac{2T}{n^{1/3} r} \).

Question 16: 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 \(F_v = mg - \frac{mg}{2} = \frac{mg}{2}\).

Question 17: easy

The increase in pressure required to decrease the 400 litre volume of a liquid by 0.001% is (Bulk modulus of the liquid is \(2.1 \times 10^9 \text{ N/m}^2\))

1. 42 kPa
2. 63 kPa
3. 84 kPa
4. 21 kPa
View Answer

Bulk modulus \(B = -\frac{\Delta P}{\Delta V/V}\). Ignoring the negative sign for magnitude, \(Delta P = B \frac{\Delta V}{V} = (2.1 \times 10^9) \times \left(\frac{0.001}{100}\right) = 2.1 \times 10^4 \text{ Pa} = 21 \text{ kPa}\).

Question 18: easy

Assertion (A): A raindrop after falling through some height attains a constant velocity.


Reason (R): At constant velocity the viscous drag plus buoyant force is just equal to its weight.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Assertion (A) is true, as falling objects in a fluid reach terminal velocity when resistive forces balance gravity.


Reason (R) is true, stating the force balance condition for constant velocity: Weight = viscous drag + buoyant force. (R) correctly explains (A).

Question 19: easy

Assertion (A): Water flows faster than honey.


Reason (R): The co-efficient of viscosity of water is less than honey.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Concept: Viscosity and fluid resistance. Viscosity is a measure of a fluid's resistance to flow. A fluid with lower viscosity flows more easily and thus faster. Water has a significantly lower coefficient of viscosity than honey. Both Assertion and Reason are true, and Reason correctly explains Assertion.

Question 20: easy

Assertion (A): The angle of contact of a liquid decreases with increase in temperature.


Reason (R): With increase in temperature, the surface tension of liquid increases.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Concept: Effect of temperature on liquid properties. Assertion (A) is true; generally, the angle of contact decreases with increasing temperature as intermolecular forces weaken. Reason (R) is false; surface tension of a liquid *decreases* with an increase in temperature, not increases, because the kinetic energy of molecules increases, reducing cohesive forces.