suitably adjusting the resistances R1 and R, a comparison of the electromotive forces of the cells may be made. Int. Sc. Honours 1885. 145. State the laws of electrolysis. [Copper is divalent.] A copper voltameter and an acidulated water voltameter are inserted in the same voltaic circuit; and it is found that 1.5 gramme of hydrogen is given off while 47.625 grammes of copper are deposited on the copper cathode. Calculate the atomic weight of copper. Oxford (Prel. Sc.) 1887. 146. Describe a series of experiments to prove Ohm's law. A current passes from A to D through a circuit composed as follows: Between A and B is a resistance of 1 ohm, between B and D an unknown small resistance; A is joined to C by a resistance of 2 ohms, and B to C by one of 99 ohms. The terminals of an electrometer (or of a high-resistance galvanometer) are joined alternately to A and C and to B and D, and the deflections are the same in the two cases. Find the value of the unknown small resistance. B. Sc. 1886. Between A and B there is a divided circuit, one branch of which (AB) has a resistance of I ohm, while the other branch (ACB) has a resistance of 101 ohms. If C is the total current, then the current through ACB is C/102, and the potential difference between the points A and C is, by Ohm's law, equal to 2 × C/102=C/51. The potential difference between the points B and D is equal to Cr, where r is the unknown small resistance. Since they are equal, it follows that the value of the unknown resistance is ohm. 147. Two points, A and B, are connected by three wires, APB, AQB, ARB, whose resistances are 1, 2, and 3 ohms respectively, and A is also joined to R, the middle point of ARB, by a wire ASR of 2 ohms resistance. How much of the total current flowing from A to B passes through each of the two branches between A and R? B. Sc. 1887. 148. A battery is connected in circuit with one coil of a differential galvanometer, shunted by a wire of resistance S, and the deflection of the needle observed. The battery is then connected with both coils of the galvanometer in series, shunted by a wire of resistance 2S, and resistance is introduced into the circuit until the galvanometer indication is the same as before. Show how the internal resistance of the battery may be obtained from these two observations. B. Sc. Honours 1885. The proportion of the total current which passes through the shunted galvanometer is the same-S -S/(G × S)--in both cases, if the resistance of the two coils is the same. Assuming further that with the same current the two coils produce the same effect on the needle, it follows that the current through the galvanometer in the first experiment is twice as strong as in the second. From this it can easily be proved that the internal resistance of the battery is equal to the additional resistance introduced into the circuit. 149. What measurements would you make to test Ohm's law? A battery of E.M.F. 10 volts is working in a circuit whose resistance is 100 ohms. How much work is done by the battery in an hour? energy required for this work produced, becomes of it? How is the and what Prel. Sc. 1888. 150. State Joule's law of the heat developed in wires by electric currents. A current is passed through a coil of fine wire of 5 ohms resistance, immersed in a vessel containing 100 grammes of water, and the same current also passes through a coil of 4 ohms resistance, immersed in a vessel containing 100 grammes of alcohol. The water rises 2° while the alcohol rises 2.5° in the same time. Find the specific heat of alcohol on the assumption that the heat absorbed by the vessels may be neglected, and the current in both cases merely passes through the wire, and not through the liquids. Vict. Int. 1885. 151. A current passes through two wires arranged in parallel arc. The first wire is of platinum, 15 cm. long and 1 mm. in diameter; the second of German-silver, 20 cm. long and 0.5 mm. in diameter. Compare the quantities of heat developed in the wires in a given time, the relative conductivities of platinum and Germansilver being 70 and 33. Prel. Sc. 1886. 152. The resistance of a copper wire, through which an electric current of unknown strength is flowing, is 3.2 ohms, and the difference of potential between its extremities is 2.5 volts. If this wire be immersed in 120 grammes of water, determine the temperature through which the water will be raised in five minutes, assuming that the water absorbs all the heat generated. [J = 41.55 × 106 ergs.] [See pp. 231-233. Int. Sc. Honours 1886. The amount of heat generated is given by the equation JH = E2t/R, where E=(2.5) × 108 and R=3.2 × 109. The rise in temperature is 1 1°.177.] 153. The resistance of an incandescent lamp is 40 ohms, and the difference of potential between its two terminals is 45 volts. Determine the heat produced in it per hour. [Mechanical equivalent of heat = 4.2 × 107ergs. I volt 108 units of electromotive = force. I ohm = 109 units of resistance.] 154. A metal disc revolves in a magnetic field about an axis through its centre and perpendicular to its plane. Determine the electromotive force between centre and circumference. Int. Sc. Honours. 1885. 155. Ninety incandescent lamps are placed in parallel circuit, and a current of 40 ampères is distributed between them, the E.M.F. between the terminals being 120 volts. The resistance of the conductors is 2 ohm, and that of the dynamo (series) is 25 ohm, and the insula 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. Ind. C. S. 1886. 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. AND HINTS FOR SOLUTION CHAPTER I-DYNAMICS 4. ACCELERATION = = 2.5; space described 1125 cm. 5. 20 cm. per sec. 6. 6666.6 dynes. 7. 105 gm. 10. 37.5 dynes. 8. 10 min. 9. 1 hr. 23 min. 20 sec. 18. 32. 19. 981. 20. 9,810,000; 17. (1) 22. 785 cm. per sec. 24. 150 ft. per sec.; 1/981. 21. 4.45 × 105 dynes. per sec. 23. 12.8; 96,000. 900. 25. 4 ft. per sec. per sec. 26. 4.524 × 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. 32 165 ft. per sec. 40. Acceleration = 4; space described = 50 ft. 41. Acceleration = 81; tension = 145,800 dynes. 42. 18 ft. 43. 3 gm. 44. 72 ft. 45. Accelera tion=g/4; distance = 2g. 46. Tension = 16.8 lbs. 48. weight = 537.6 poundals. 47. 709.1 cm. per sec. 5:3. 49. 242.4 ft. 51. Acceleration I ft.-sec. unit; = |