Practice NEET Surface Tension and Viscosity Questions
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 \(d = 0.8\text{ gm/cc}\))
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 \(T = \frac{2 \times 10^{-3} \times 800 \times 9.8 \times 10^{-2}}{4} \approx 4 \times 10^{-2}\text{ N/m}\).
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 \(73 \times 10^{-3}\text{ Newton/metre}\))
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 \(F = 4 \times 3.14 \times 1.5 \times 10^{-2} \times 73 \times 10^{-3} \approx 1.37 \times 10^{-2}\text{ N}\).
What is ratio of surface energy of 1 small drop and 1 large drop, if 1000 small drops combined to form 1 large drop :
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 \(r^2 : R^2 = r^2 : 100r^2 = 1 : 100\).
The work done in blowing a soap bubble of \(20\text{ cm}\) radius is (surface tension of soap solution is \(0.03\text{ N/m}\))
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 \(W = 8 \times \pi \times (0.2)^2 \times 0.03 \approx 3.01 \times 10^{-2}\text{ J}\).
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
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 \( \Delta P = \frac{2T}{R} = \frac{2T}{n^{1/3} r} \).
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
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 \(F_v = mg - \frac{mg}{2} = \frac{mg}{2}\).
The increase in pressure required to decrease the 400 litre volume of a liquid by 0.001% is (Bulk modulus of the liquid is \(2.1 \times 10^9 \text{ N/m}^2\))
Bulk modulus \(B = -\frac{\Delta P}{\Delta V/V}\). Ignoring the negative sign for magnitude, \(Delta P = B \frac{\Delta V}{V} = (2.1 \times 10^9) \times \left(\frac{0.001}{100}\right) = 2.1 \times 10^4 \text{ Pa} = 21 \text{ kPa}\).
Reason (R): The co-efficient of viscosity of water is less than honey.
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.
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.
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.