Bukhovtsev-et-al-Problems-in-Elementary-Physics

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256 ANSWERS AND SOLUTIONS If the water covers the block entirely, then according to Archimedes' principle,Vopog=V2Plg+ (Vo- V2) P2i, where Po is the density of iron. Solving the equations, we obtain h- PI (PO-P2) Vo - P2 (Pt-P2) At If the water does not cover the block, Archimedes' principle can be Fig. 380 written as VoPog=V2Plg+hAp~, where A=V:/ 3 is the area of a block face. In this case the sought height h PoV o P2 (At +V:/ 8 ) The first solution is true when A
MECHANICS 257 Fig. 381 Fig. 382 252. A liquid moves in a syphon by the action of the forces of cohesion between the elements of the liquid. The liquid in the long elbow outweighs the liquid in the short one, and pumps it over. It could be assumed on this basis that water can be pumped over a wall of any height with the ski of a syphon. This is not so, however. At a lifting height of 10 metres the pressure inside the liquid becomes zero. The air bubbles always present in water will begin to expand, and the water column will be broken, thus stopping the action of the syphon. 253. The device will first act as a syphon and the water will flow through the narrow pipe into the reservoir. Then an air bubble will slip through A and divide the liquid in the upper part into two portions. After this the liquid will no longer flow out. . 254. The pressure of the water directly under the piston of each pump is less than atmospheric pressure by pg (H +h), where p is the density of water. Therefore, to keep the piston in equilibrium, it should be pulled upward with the force F = pg (H+h) A, where A is the area of the piston. Hence, a greater force should be applied to the pistons with a greater area. 255. The pressure on the bottom is P=pgbH+h) (Fig. 382). On the other hand, since the vessel is a cylinder, p= :R~. The height h can be determined if we equate the forces acting on the piston pghn (RI-r 2 )= G Hence, 1 ( G r 2 ) nR2p g R2~r2 H=-- m----- ~ 10 cm 256. To prevent flowing out of the water, the vessel should be given such an acceleration at which the surface of the water takes the position shown in Fig. 383. The maximum volume of the water is ~: and the mass of the entire system is M+be'; p. The required acceleration can be found from the condition that the sum of the forces acting on a small element of the water with a mass 11m near the surface is directed horizontally (Fig. 383). 17-2042
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B. B. 6yX08I4e8. B. JI. KpU8IleHIC0
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First published 1971 Second printin
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6 Chapter 5. Geometrical Optics •
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8 PROBLEMS engineer arrived at the
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10 PROBLEMS c rIII a Fig. 3 01 IIII
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18 PROBLEM~ friction if the weight
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20 PROBLEMS Fig. 28 H 11/ r ~--I---
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22 PROBLEMS Fig. 25 Fig. 26 B m; is
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a Fig. 41 Fig. 42 for the rotating
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32 PROBLEMS Fig. 48 122. In Guerick
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34 PROBLEMS (I) What work is done b
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44 PROBLEMS 178. A ball bearing sup
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a 0' Fig. 80 Fig- 8/ ces '1 and " f
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50 PROBLEMS Fig. 84 Fig. 85 205. A
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--- --r-- - Fig. 114 ~--l---I"" Fig
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CHAPTER 2 HEAT. MOLECULAR PHYSICS 2
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70 PROBLEMS What quantity of heat w
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72 PROBLEMS 315. A barometer gives
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74 PROBLEMS pressure Po = 760 mm Hg
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92 PROBLEMS Fig. 153 Fig. 154 431.
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A Fig. /66 Fig. /67 463. Two conduc
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PROBLEMS 625. Find the period of os
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138 PROBLEMS 680. Find graphically
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150 PROBLEMS 757. Two men, a far-si
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152 PROBLEMS H 767. Two point sourc
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154 PROBLEMS 773. A convergent lens
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ANSWERS AND SOLUTIONS CHAPTER 1 MEC
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166 ANSWERS AND SOLUTIONS Hence and
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o s Fig. 278 .D A B B' Fig. 277 25.
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176 ANSWERS AND SOLUTIONS ice dry s
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o • • • ~. Fig. 306 Fig. 907
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352 ANSWERS AND SOLUTIONS the ring
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366 ANSWERS AND SOLUTIONS 623. Each
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A ~-..----4-----r--------r-- Fig. 4
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CHAPTER 5 GEOMETRICAL OPTICS 5-1. P
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382 ANSWERS AND SOLUTIONS· Paper -
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384 ANSWERS AND SOLUTIONS DAB =~-=-
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386 ANSWERS AND SOLUTIONS 684. (a)
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390 ANSWERS AND SOLUTIONS Then. acc
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396 ANSWERS AND SOLUTIONS ~-"":;:'-
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8' Fig. 519 s* IIIIIIII N P H' Fig.
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414 J- ------....,.c 8- ----~--- AN
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422 S, IT ........... In --.. U' .,
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428 ANSWERS AND SOLUTIONS 779. When
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436 ANSWERS AND SOLUTIONS 805. For
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