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tion resistance between the flow and return conductors is 1000 ohms. What horse-power will be required for electrical work, and how much will be wasted ? (1 H.-P. = 746 watts.)

Ind. C. S. 1885. 156. A dynamo running at 100 volts is employed to drive a motor. The whole resistance of the circuit, including the dynamo and motor, is only half an ohm, but the whole current only 20 ampères. Account for this by elementary principles.

157. The carbon filament of an incandescent lamp being 15 cm. long and 0.025 cm. in diameter, the current in the lamp 1.5 ampère, the electromotive force between the terminals 50 volts, and the temperature of the filament 2000°, determine the emissive power of its surface for heat.

B. Sc. Honours 1884.

Ind. C. S. 1886.

ANSWERS

AND HINTS FOR SOLUTION

CHAPTER 1-DYNAMICS

7. 105 gm.

per sec.

4. ACCELERATION = 2.5; space described =

1125 cm. 5. 20 cm. per sec.

6. 6666.6 dynes. 8. 10 min. 9. 1 hr. 23 min. 20 sec. 10. 37-5 dynes. 13. 2.5 ft. per sec. 14. v/240. 16. 38,400. 17. (1) 5 oz.; (2) 2 oz. 18. 32. 19. 981. 20. 9,810,000; 1/981. 21. 4.45 X 105 dynes. 22. 785 cm. per sec.

23. 12.8; 96,000. 24. 150 ft. per sec.; 900. 25. 4 ft. per sec. per sec. 26. 4.524 x 107. 27. 1/120 of a poundal. 28. The force is to the weight of a gramme as 25,000 to 327. 29. Acceleration

= 1/70; velocity = 1/7. 31. 8.075 oz. 32. The force is equal to the weight of 5.6 lbs.; acceleration produced =0.08 ft.-sec, units. 33. 1,774,080. 34. As 80 : 7. 35. 82.5 sec. 36. Momentum = 1,471,500. 37. The force is equal to 3.2 poundals, or is one-tenth of the weight of 1 lb. 38. As 224 : 675. 39. 32 V165 ft.

40. Acceleration = 4; space described = 50 ft. 41. Acceleration = 81; tension 145,800 dynes. 42. 18 ft.

44. 72 ft. 45. Acceleration = g/4; distance 28

46. Tension 16.8 lbs. weight = 537-6 poundals. 47. 709.1 cm. per sec. 48. 5:3. 49. 242.4 ft. 51. Acceleration = 1 ft.-sec. unit;

per sec.

=

43. 3 gm.

space 6 in, 52. g = 950.

53. About 9.5 gm. 54. g=980. 55. Sin -1(1/48).

57. 26.32 poundals. 58. Tension = weight of 4024.3 gm. 59. 0.1113 ft.-sec. units. 62. g=981.05. 63. g=32.227. 64. It would have to be shortened to 3/19ths of its original length. 66. g=32191; 39.69 in. 67. The pendulum is 0.2717 in. too long, therefore 8.15 turns are required to correct it.

77. 9.6 x 106 gramme-centimetres; 9.4176 x 109 ergs. 78. 288,000 ft.-Ibs. 79. 10:27. 80. 1820 V3 ft.-Ibs. 81. 9600 ft.-lbs. 82. 1.44 x 109 ergs. 83. (1) equal; (2) inversely proportional to the masses. 84. 20:16 ft.-lbs. 85. 69.12 ft.-Ibs. 86. 67884 ft.-tons. 87. In the ratio of 400 to 88. 1634.3 ft.-tons. 89. 706.9 ft.-lbs. 90. Total work

11,000 ft.-lbs. 91. 98,175,000 ft.-Ibs.

I.

=

99. (1) 336,000 ft.-Ibs.; (2) 10,752,000 ft.-poundals. 100. 33 ft. 101. 2500 ft.-lbs. 102. 3.767 x 1010 ergs. 103. 9.81 X 107 ergs. 104. 6.25 x. 1010 ergs. 105. 4.5 x 1012 ergs. 106. Momentum 8960 V2; K.E = 143, 360 ft.-poundals, or in ft.-Ibs. = 143,360/32 =

/ 4480. 107. 1536 ft.-poundals, or 48 ft.-Ibs. 108. The initial velocity must have been 96 ft. per sec., and the final energy must = K.E. at starting = 5 * (96)2/2g=720 ft.-Ibs. 109. Force = 300 poundals; K.E. = 225,000 ft.poundals. 110. 8 x 1010 ergs. 112. 2.5 x 1010 ergs. 113, 16,940 ft.-Ibs. 114. 8.5 x 106 ergs.

116. 18 cm. 117. K.E. 1210 ft.-tons; a force equal to the weight of 1 ton. 118. 5461} ft.-poundals. 119. As 112:625. 120. 9.375 X 108 dynes. 121. 13,090 ft.-Ibs. 123. 32,000 ft.-lbs.; 320 ft. per. sec. 124, K.E. before impact 150,000. 125. Force = 716.8 poundals; work = 114,688 ft.-poundals. 126. Mass = 45; velocity = 2/3. 127. As ( 13-1): 1.

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130. (1) 4561 kilogramme-metres per min.; (2) 7.456 x 109 ergs per sec. 131. 38,016 cub. ft. 132. 911 H.P. 133. 5211 miles per hour. 134. 16.8 H.P. 135. 151.2 H.P. 136. 192 H.P. 137. 480 H.P. 138. At the rate of 3.367 H.P. 139. The unit of work would be increased ten-fold; the numerical value of the horse-power would not be changed.

EXAMINATION QUESTIONS

141, 143. See Introduction, $ 8, and Ch, I., Exs. 16 and 26. 145, 146, See SS 8 and 9 and Exs. 27 and 87. 150. The proof follows easily when the dimensions (MLT) of force are known. 156. When describing a circle of 100 ft. radius, his inclination to the ice is.84o. 162. (1) 2000 ft.-lbs. ; (2) 1562.5 ft.-Ibs. 171. See Ex. 97. 174. This example is also inserted (and solved) in the chapter on Thermodynamics (Ch. V., 40). 175. K.E. of tram-car = 431,500 ft.-poundals ; work done in a run of 3 miles = 22,065,120 ft-poundals.

CHAPTER II-HYDROSTATICS

3. The dimensions of pressure are the same as those of force, viz. MLT"; the dimensions of intensity of pressure (force per unit area) are ML-T-? 6. 124.4 lbs. per cub. ft. 7. 0.5787 oz. per cub. in. 8. 1.736 oz. per cub. in.

10. 13.824. 11. 10.98 lbs. 12. 0.7055 gm. per C.C. 13. 300.8 lbs. 14. 3.2 cub. ft. 16. Cross-section = 0.3676 sq. cm. ; diameter 0.683

17. 111.97 gm. 18. 1.4. 19. 4.5 cm. 20. As 27 : 10. 21. 19.712 gm.

23. 4.516. 25. 155/8. 26. Its density is 0.925 that of air. 27. 0.823.

31. (1) 100 gm. per sq. cm. ; (2) 98,100 dynes per sq. cm. 32. 4.273 x 104. 33. 73.53

34, 4100 gm. weight. 35. 206,991 dynes per sq.

cm.

24. 4.57.

28. 1.5.

cm.

cm.

cm.

I

36. 10:193 metres. 37. 98,100 dynes per sq.

38. 12,750 gm. weight. 39. 1/384 lb. per cub. in. 40. 18,750 lbs. per sq. ft. 41. 17:36 lbs. per sq. in. 42. 138.2 ft. 43. 337,920 lbs. per sq. ft. 44. 28.02 lbs. per sq. in. 45. 10:193 metres. 46. 0.9. 47. 49.08 ft. 48. 21.3 lbs. per sq. in. 49. Sufficient to occupy 5 in. of the tube.

50. I'cos 30°(1 – s). gm. 52. 68,360 gm. 53. 99,000 gm.

54. 18,150 gm. 55. 10,000 gm, on upper surface, 11,000 gm, on lower surface, 10,500 gm. on each of the vertical sides,

51. 150

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58. Volume = 20 c.c.; sp. gr.

3.1.

59. 22:04 gm. 60. 21.08 gm. 61. 34 c.c. 62. 425.9 oz. 63. 20.8 gm. 64. 27.5 gm. 66. sq = m.,S1/(m256 mysı+ my). 67. 20.97 gm. 68. 21:57 gm.

69. A weight equal to that of the copper. 70. Acceleration 204 ft.-sec. units; time = 1.247 sec. 71. 6.04 lbs. 72. 0.32 gm. 74. 1000 C.C. 75. 300 C.C. 76. 6437.5 cub. yds. 77. 50 c.c. 78. Sp. gr. of both solid and liquid=0.5. 79. The sphere will rest in equilibrium with 1/7th of its volume immersed in the mercury. 80. 3.3. 81. 1.625. 82. 1.4.

83. 1.204. 84. 0.8271. 85. 1.434. 86. 0.9127. 88. 11:31.

89. 1.948. 90. 0.7351. 91. 0.8. 92. 4:75. 93. 3.2 19. 94. 1.059. 99. Sp. gr. of pebble = 2.74; of spirit = 0.825. 100. 240 lbs. 101. 10,080 lbs. 102. Ratio of arms =

103. Pressure 8064 lbs.; distance

= 0.1042 in. 104. Mechanical advantage = 1440.

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:8:5

34 ft.

107. Apparent height = 20 N3 in. = 34.64 in. 108.

109. 13.6 in. 110. 25.197 ft. 111. (1) 0.0938; (2) 1.276. 112. 11.86 metres. 113. 10,546 kgm. per sq. metre. 115. 1,039,890 dynes per sq. cm. 116. 14.01 lbs. per sq. in. 117. 30,000 oz. per sq. ft. 118. 1006.I gm. per sq. cm., or 10,061 kgm. per sq. metre. 119. 4.045 x 107 dynes. 120. A change of about half a pound (0.4917 lb.) per sq. in. 124. p:p= 9:1. )

' 9

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