Tiêu đề 10. Surface Tension Hard.pdf ✅

1. The work done in blowing a soap bubble of 10cm radius is (surface tension of the soap solution is N / m 100 3 ) (a) 4 75.36 10 −  J (b) 4 37.68 10 −  J (c) 4 150 .72 10 −  J (d) 75.36 J 2. A drop of mercury of radius 2mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 J/m2 ) (a) 23.4J (b) 18.5J (c) 26.8J (d) 16.8J 3. . The work done in increasing the size of a soap film from 10cm  6cm to 10cm  11cm is J 4 3 10 −  . The surface tension of the film is (a) 2 1 1.5 10 − −  Nm (b) 2 1 3.0 10 − −  Nm (c) 2 1 6.0 10 − −  Nm (d) 2 1 11.0 10 − −  Nm 4. A film of water is formed between two straight parallel wires of length 10cm each separated by 0.5cm. 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 10 N / m −2 =  ) (a) 6 7.22 10 −  J (b) 5 1.44 10 −  J (c) 5 2.88 10 −  J (d) 5 5.76 10 −  J 5. If the work done in blowing a bubble of volume V is W, then the work done in blowing the bubble of volume 2V from the same soap solution will be (a) W/2 (b) 2 W (c) 3 2 W (d) 3 4 W 6. Several spherical drops of a liquid of radius r coalesce to form a single drop of radius R. If T is surface tension and V is volume under consideration, then the release of energy is (a)       + r R VT 1 1 3 (b)       − r R VT 1 1 3 (c)       − r R VT 1 1 (d)       + 2 2 1 1 r R VT 7. The pressure inside a small air bubble of radius 0.1mm situated just below the surface of water will be equal to(Take surface tension of water 3 1 70 10 − −  Nm and atmospheric pressure 5 2 1.013 10 − =  Nm ) (a) Pa 3 2.054 10 (b) Pa 3 1.027 10 (c) Pa 5 1.027 10 (d) Pa 5 2.054 10 8. If the radius of a soap bubble is four times that of another, then the ratio of their excess pressures will be (a) 1 : 4 (b) 4 : 1 (c) 16 : 1 (d) 1 : 16