AjneeshFlame Institute

5. The temperature of an ideal gas is kept constant as it expands. The gas does external work. During this process, the internal energy of the gas

(a) Decreases (b) Increases (c) Remains constant (d) Depends on the molecular motion Ans. c (16)

4. Heat given to a system is 35 joules and work done by the system is 15 joules. The change in the internal energy of the system will be

(a) – 50 J (b) 20 J (c) 30 J (d) 50 J Ans. b (6)

3. In changing the state of thermodynamics from A to B state, the heat required is Q and the work done by the system is W. The change in its internal energy is

(a) Q + W (b) Q – W (c) Q (d) (Q – W)/2 Ans. b (2)

43. A liquid film is formed in a loop of

area 0.05 m2. Increase in its potential energy will be (T = 0.2 N/m) (a) 5 x 10-2J (b) 2 x 10-2j (c) 3 …

41. Radius of a soap bubble is increased from R to 2R work done in this process in terms of surface tension is

(a) 24πR2S (b) 48 πR2S (c) 12πR2S (d) 36πR2S Ans. a (5)

1. First law of thermnodynamics is given by

(a) dQ = dU + PdV (b) dQ = dU x Pdv (c) dQ = (dU + dv)P (d) dQ = PdU + dV Ans. a (10)

45. If two soap bubbles of equal radii r coalesce then the radius of curvature of interface between two bubbles will be

(a) r (b) 0 (c) Infinity (d) 1/2r Ans. c (13)

44. In order to float a ring of area 0.04 m2 in a liquid of surface tension 75 N/m, the required surface energy will be

(a) 3 J (b) 6.5 J (c) 1.5 J (d) 4 J Ans. a (7)

42. The work done in blowing a soap bubble of radius 0.2 m is (the surface tension of soap solution being 0.06 N/m)

(a) 192π x 10-4 J (b) 280π x 10-4 J (c) 200π x 10-3 J (d) None of these Ans. a (12)

40. Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is

(a) 1:21/3 (b) 21/3:1 (c) 2 : 1 (d) 1 : 2 Ans. b (10)

39. A drop of mercury of radius 2 mm is split into 8 identical droplets. Find

the increase in surface energy. (Surface tension of mercury is 0.465 J/m2 ) (a) 23.4 μJ (b) 18.5 μJ (c) 26.8 μJ …

38. A film of water is formed between two straight parallel wires of length 10cm each separated by 0.5 cm. If their separation is increased by 1 mm while still maintaining their parallelism,

how much work will have to be done (Surface tension of water =7.2 x10-2 N/m) (a) 7.2 x 10-6 Joule (b) 1.44 x 10-5 Joule (c) 2.88 x 10-5 Joule (d) 5.76 x 10-5 Joule …

37. When two small bubbles join to form a bigger one, energy is

(a) Released (b) Absorbed (c) Both (a) and (b) (d) None of these Ans. a (12)

36. A mercury drop of radius 1cm is sprayed into

108 drops of equal size. The energy expended in joules is (surface tension of Mercury is 460 x 10-3N/m (a) 0.057 …

35. If work done in increasing the size of a soap film from to is then the surface tension is

(a) 2 x 10-2Nm-1 (b) 2 x 10-4Nm-1 (c) 2 x 10-6Nm-1 (d) 2 x 10-8Nm-1 …

34. 8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of

(a) 1 (b) 2 (c) 4 (d) 6 Ans. c (10)

33. The surface energy of liquid film on

a ring of area is (Surface tension of liquid = 5Nm-1) (a) 0.75 J (b) 1.5 J (c) 2.25 J (d) 3.0 J Ans. b (10)

32. 8000 identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is

(a) 1 : 10 (b) 1 : 15 (c) 1 : 20 (d) 1 : 25 Ans. c (5)

31. Which of the following statements are true in case when two water drops coalesce and make a bigger drop

(a) Energy is released (b) Energy is absorbed (c) The surface area of the bigger drop is greater than the sum of the surface areas of both the drops (d) The surface area of the bigger drop is smaller than …

24. The work done in blowing a soap bubble of 10 cm

radius is (Surface tension of the soap solution is (3/100) N/m) (a) 75.36 x 10-4 joule (b) 37.68 x 10-4 joule (c) 150.72 x 10-4 joule (d) 75.36 joule Ans. a (10)

23. The amount of work done in forming a soap film of

size 10 cm x 10 cm is (Surface tension T= 3 x 10-2 Nm (a)6 x 10-4 j (b) 3 x 10-4 j (c) 6 x 10-3 j d) 3 x 10-4 j Ans. a (10)

18. A mercury drop of 1 cm radius is broken into

106 small drops. The energy used will be (surface tension of mercury is 35 x10-3N/cm (a) 4.4 x 10-3 j (b) 2.2 x 10-4 j (c) 8.8 x 10-4 j (d) 104 j Ans. a (12)

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

(a) 100 : 1 (b) 1000 : 1 (c) 10 : 1 (d) 1 : 100 Ans. d (5)

21. The surface tension of liquid is 0.5 N/m. If a film is held on a

The surface tension of liquid is 0.5 N/m. If a film is held on a ring of area 0.02 m2, its surface energy is (a) 5 × 10-2 joule (b) 2.0 × 10-2 joule (c) 4 × 10-4 joule (d) …

19. The surface tension of a liquid at its boiling point

(a) Becomes zero (b) Becomes infinite (c) is equal to the value at room temperature (d) is half to the value at the room temperature Ans. a (9)

26. One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

(a) 1000 : 1 (b) 1 : 1000 (c) 10 : 1 (d) 1 : 10 Ans. d (3)

17. The surface tension of a soap solution

is 2 x 10-2 N/m. To blow a bubble of radius 1 cm, the work done is (a) 4 π x 10-6 J (b) 8 π x 10-6 J (c) 12 π x 10-6 J d) 16 π …

30. When

106 small drops coalesce to make a new larger drop then the drop (a) Density increases (b) Density decreases (c) Temperature increases (d) Temperature decreases Ans. c (3)

29. A big drop of radius R is formed by 1000 small droplets of water, then the radius of small drop is

(a) R/2 (b) R/5 (c) R/6 (d) R/10 Ans. d (9)

28. If σ be the surface tension, the work done in breaking a big drop of radius R in n drops of equal radius is

(a) Rn(2/3) σ (b) (n(2/3) – 1) Rσ (c) (n(1/3) – 1) Rσ (d) 4πR2(n(1/3) – 1)σ (e) [1/( n(1/3) – 1)] σR Ans. d (4)

25. A liquid drop of diameter D breaks upto into 27 small drops of equal size. If the surface tension of the liquid is σ, then change in surface energy is

(a) π D2 σ (b) 2 π D2 σ (c) 3 π D2 σ (d) 4 π D2 σ Ans. b (4)

20. Surface tension of a soap solution

is 1.9 x 10-2N/m . Work done in blowing a bubble of 2.0 cm diameter will be (a) 7.6 x 10-6 π joule (b) 15.2 x 10-6 π joule (c) 1.9 x 10-6 π joule (d) 1 x 10-4 π joule …

27. The work done in increasing the size of a soap film from

10 cm× 6 cm to 10 cm × 11 cm is 3 ×10-4 joule. The surface tension of the film is (a) 1.5 x 10-2N/m (b) 3.0 x 10-2N/m (c) 6.0 x 10-2N/m (d) 11 x 10-2N/m Ans. b …

6. A liquid is coming out from a vertical tube. The relation between the weight of the drop W, surface tension of the liquid T and radius of the tube r is given by, if the angle of contact is zero

(a) w= πr2T (b) w= 2πrT (c)w= 2πr2T (d) w= (3/4)πr3 T Ans. b (4)

7. The parts of motor cars are polished by chromium because the angle of contact between water and chromium is

(a) 0o (b) 90o (c) Less than 90o (d) Greater than 90o Ans. d (5)

10. The radius of a soap bubble is increased

from (1/√π) cm to (2/√π) cm. If the surface tension of water is 30 dynes per cm, then the work done will be (a) 180 ergs (b) 360 ergs (c) 720 ergs (d) 960 ergs Ans. c (11)