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# XII Physics - Surface Tension Part II

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Summary of Surface Tension Part - II by Ednexa

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### XII Physics - Surface Tension Part II

1. 1. 9011031155 / 9011041155 Surface Tension (Part - II) Problems 1. A needle of length 5 cm can just rest on the surface of water of surface tension 0.073 N / m. Find the vertical force required to detach this floating needle from the surface of water. Sol : L = 5 cm = 5 × 10-2m T = 0.073 N/m F=? The force due to surface tension is given as, F = TL The total length of the needle in contact with water = 2 L ∴F=T×2L = 0.073 × 2 × 5 × 10-2 = 0.073 × 10-1 ∴ F = 7.3 × 10-3 N
2. 2. 9011031155 / 9011041155 Note : This force is the weight of needle. 2. A horizontal circular loop of wire of diameter 0.08 m is lowered in to a oil. The force due to surface tension required to pull the loop out of the liquid is 0.0226 N. Calculate the surface tension of the oil. Sol: d = 0.08 n ∴ r = 0.04 m F = 0.0226N, T = ? The force due to Surface tension is F = TL T F L F 2 2 r 0.0226 4 r 4 0.0226 0.16 3.14 T 0.0449 N / m 0.0226 3.14 0.04 0.0226 0.5024
3. 3. 9011031155 / 9011041155 Angle of Contact When a liquid is in contact with a solid, the angle between the surface of the liquid and the tangent drawn to the surface of the liquid, at the point of contact, on the side of the liquid is called “angle of contact” of the given solid liquid pair. Features of Angle of Contact 1. For a given solid liquid pair the angle of contact is constant. 2. If the liquid partially wets the solid, the angle of contact is acute. 3. If the liquid doesn‟t wets the solid at all, the angle of contact is obtuse.
4. 4. 9011031155 / 9011041155 4. If the liquid wets the solid completely, the angle of contact is approximately zero. 5. Any small contamination of the liquid can change the angle of contact largely. 6. The angle of contact depends on the magnitudes of adhesive forces between solid and liquid molecules and cohesive forces between liquid molecules. Explanation of Angle of Contact
5. 5. 9011031155 / 9011041155 For a liquid which completely wets the solid, the adhesive forces are so strong, as compared to the cohesive forces, that the resultant AR of these forces is along AP. So, the tangent at the point of contact is along the wall of the container. So, the liquid surface remains plane.
6. 6. 9011031155 / 9011041155 Shape of a liquid drop T1 = Force due to surface tension at the liquid - solid interface, T2 = Force due to surface tension at the air - solid interface, T3 = Force due to surface tension at the air - liquid interface, For the equilibrium of the drop T2 T1 i.e. cos T 3 cos T 2 T1 T3
7. 7. 9011031155 / 9011041155 From this equation we get following cases: 1. If T2 > T1, and T2 - T1 < T3, cos θ is positive and angle of contact θ is acute. 2. If T2 < T1, and T2 - T2 < T3, cos θ is negative and angle of contact θ is obtuse. 3. If T2 - T1 = T3, cos θ = 1 and „θ‟ is nearly equal to zero. 4. If T2 - T1 > T3 or T2 > T1 + T3, cos θ > 1 which is impossible, liquid is spread over the solid surface and drop shall not be formed.
8. 8. 9011031155 / 9011041155 Multiple choice Questions 1. A water drop of radius R is split into n smaller drops, each of radius r. If T is the surface tension of water, then the work done in this process is 4 3 1 1 R T (a) 3 r R (c) 4 R3 T 1 1 r R 3 3 1 R T (b) 4 R 1 r 1 R 1 r (d) 6 R2 T 2. One thousand small water droplets of equal size combine to form a big drop. The ratio of the final surface energy to the initial surface energy is of water drops is (a) 1:1000 (b) 10:1 (c) 1:10 (d) 1000:1 3. What is the potential energy of a soap film formed on a frame of area 5 10 3 m2 ? The surface tension of soap solution is 30 10 3 N / m . (a) 2 × 10-4J (b) 2.5 × 10-4J
9. 9. 9011031155 / 9011041155 (c) 3 × 10-4J (d) 5 × 10-4J
10. 10. 9011031155 / 9011041155 Ans: (c) 4 R3 T 1 1 r R In this process, the mass & hence the volume of the water drop, remains constant but the surface area is increased. ∴ Volume of the surface area is increased one drop = Volume of n droplets 4 3 4 3 R n. r 3 3 R3 nr 3 Change in area 4 R3 r R2 4 R2 4 (nr 2 R2 ) R2 4 dA nr 3 r n.4 r 2 Work done dW 4 R3 1 1 r R (S.T.) (Change in area) T dA 4 R3 T 1 1 r R
11. 11. 9011031155 / 9011041155 Ans: (c) 1:10 4 4 3 R3 1000 r 3 3 R3 1000 r 3 R 10 r Energy or E T A area Energy of 1000 droplets (E1 ) 1000 T (4 r 2 ) But surface tension T and Energy of the sin gle big drop E2 T 4 (R2 ) T 4 (10 r)2 4 T(100 r 2 ) E2 E1 100 r 2 1000 r 2 1 10
12. 12. 9011031155 / 9011041155 Ans: (c) 3 × 10-4J For a soap film, we have to consider double the area, as there are two free surfaces. P.E work done 2 30 10 3 10 4 J 3 T(2A) 5 10 3
13. 13. 9011031155 / 9011041155 Drops and Bubbles As it is spherical in shape, the inside pressure will be greater than that of the outside. Let the outside pressure be P0 and inside pressure be Pi, so that the excess pressure is Pi – P0. the radius of the drop increases from r to r + Δr, where Δr is very small, so that the inside pressure remains almost constant. Initial surface area (A1) = 4πr2 Final surface area (A2) = 4π(r + Δr)2
14. 14. 9011031155 / 9011041155 A2 = 4π (r2 + 2r Δr + Δr2) A2 = 4πr2 + 8πr Δr + 4πΔr2 As Δr is very small, Δr2 is neglected (i.e. 4 π Δr2 ≅ 0) Increase in surface area (dA) = A2 – A1 = 4πr2 + 8πr Δr - 4πr2 Increase in surface area (dA) = 8 πr Δr ….. (1) Work done to increase the surface area 8 πr Δr is extra surface energy. ∴ dW = TdA dW = T (8πr Δr) ….. (2) This work done is also equal to product of force and the distance Δr. Let dW = dF Δr
15. 15. 9011031155 / 9011041155 But dF = Excess pressure × area dF = (Pi – P0) 4πr2 …… (3) dW = (Pi – P0) 4πr2 Δr….. (4) By comparing equation (2) and (4) we get T (8πrΔr) = (Pi – P0) 4πr2 Δr Here P i P0 2T …… (5) r This is called Laplace‟s law of a spherical membrane. In case of soap bubble Pi P0 4T r ....... 6 The threads of rain – coat are coated with water proofing agents like resin etc. Which have very small force of adhesion with water so rain coats become water proof.