Question 11:
easy125 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:
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 \(T \times 400\pi r^2\).
Question 12:
easyWater 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:
Using capillary rise formula \(h = \frac{2T \cos\theta}{r \rho g}\) with \(\theta = 0^\circ\) in CGS: \(4 = \frac{2 \times 60 \times 1}{r \times 1 \times 1000} ⇒ r = 0.03\text{ cm}\).
Question 13:
moderateA 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
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).
Question 14:
easyWhat 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 \(r^2 : R^2 = r^2 : 100r^2 = 1 : 100\).
Question 15:
easyIf 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 \(T = \frac{2 \times 10^{-3} \times 800 \times 9.8 \times 10^{-2}}{4} \approx 4 \times 10^{-2}\text{ N/m}\).
Question 16:
easyA 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 \(F = 4 \times 3.14 \times 1.5 \times 10^{-2} \times 73 \times 10^{-3} \approx 1.37 \times 10^{-2}\text{ N}\).
Question 17:
easyThe 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 \(W = 8 \times \pi \times (0.2)^2 \times 0.03 \approx 3.01 \times 10^{-2}\text{ J}\).
Question 18:
moderateThe 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
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 \(F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}\).
Question 19:
easyn 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 \( \Delta P = \frac{2T}{R} = \frac{2T}{n^{1/3} r} \).
Question 20:
easyThe 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 \(F_v = mg - \frac{mg}{2} = \frac{mg}{2}\).