### Answers to Problems :GENERAL PHYSICS

1. 746 watts,

4. a; = ^ + - — r — ( f ~ ^ )' ^"^^'■

6 ^

8^^^ ' ^^^ t^6 distance and time
reckoned from the position where the velocity was V. Terminal
velocity, | .

8. H.p. = 38190; time=127-6 hrs. ; resistance = 236 -24 tons' weight;

coal =4353 tons. {Geographical miles are meant.)

9. 6-81 X 10-12 kgm.-weight. 10. 3762 sees.

11. Mass of sun : mass of earth = (390)» : (13)1

12. 11-2 lbs. per ton; 358-4 h.p.

13. 4400 ft.-lb.-sec. units per sec. ; 17600 ft.-poundals per sec.

14. 14015 -.g. 15- 7" • 16- *•

17. HfS^fi' ^^^- P®' ^^*'*

18. v/480/£r sees. ; ijlbgl2 radians per sec.

19. 96^7- cms. per sec. per sec.

20. If the mass M be suspended at an arm r, and the moment of inertia

of the wheel and axle be J, the acceleration = -ir-s-H^. If a Mr^ + I

length I unwrap in time t, the friction couple is
Mrg - (^Mr +-^'^j^.

21. V— •

22. When the rod has turned through an angle d, the two amounts are

\ vigl sin d and | mgl sin 6 respectively.

23. Let the added mass have a radins of gyration h, and let its centre
of gravity be x below the knife-edge ; then the positions are given

by ^< = >^ respectively. 25. (i) «.^; (") a-^-J^-

26. (i) a • -^ below the point of suspension ; (ii) — below.O

27. Becomes /^l^ times as long. 28. 52-88 cms.

29. 2.^^.

30. (i) Multiplied by 4v'2 ; (ii) divided by 2^2 ; (iii) doubled.

31. VW sec. 32. 2-8 X 10^ pounds' weight per sq. in.

33. 2-6 xlC^ dynes per sq. cm. 34. Halved.

35. s&. cm. 36. 0-1875 cm. 37. i^^i^ ? ergs per c. cm.

38. Torsional rigidity, about 2-23 x Wg dynes per sq. cm. ; simple

rigidity, 2/ir times as much.

39. Five per cent., if the errors in n and Y are in opposite senses.

40. 22001:22010-7.

41. 0000474 gm.

42. cms., assuming that the liquid wets the tube.OQort

43. 2-88x108; f^^ cms.

44. 1-2 cm.

45. 296000 dynes per sq. cm. 46. 0-0809 cm.

47. 01764 poundal per ft. 48. ^ dynes per cm. 49. -^«

50. (76s + 10) ^ + 156000, where « is the density of mercury.

51. 125960 ergs. 52. ^ gm. weight per sq. cm. 53. yj^'^"^'

KA . "*" , where P' is the external pressure and P is a normal
***• 0-2214P •atmosphere, both in dynes per sq. cm.
58. (a) 0-0192, 1-016 ; (6) 19-2, 1016 dynes per sq. cm.

60. (a) 1-152 X 10-' cm., better 1-662 x 10"' ; (6) 1-016 x lO"', better

1-466 X 10-«.

61. 000077 cm.

62. 226 cms.

63. Let the pressure outside the tyre be one atmosphere, and let

the internal pressure be n atmospheres ; then the time taken is
about 863000 (w'' - 1) log^o ( g _ ^^ _ ^ ) seconds. If n = 2, this is
infinite (as it should be); if n=3, the time is 2-75x10° sees.;
for n = 4, 3-30 xlO« sees.; for n = 10, 7-60x108; for n = 20,
15-05 X 0", When the internal pressure considerably exceeds the
external (as was probably intended), a rougher calculation is
appropriate, and leads to the result 750000 n sees. It will be
noticed that at 20 atmospheres, this is practically correct.

### COMMON COLLECTOR CONFIGURATION OF A TRANSISTOR

COMMON COLLECTOR CONNECTION

In  this  configuration  the  input  is  applied  between the  base  and  the  collector and  the  output  is  taken  from  the  collector  and  the  emitter.  Here  the  collector  is common to both the input and the output circuits as shown in Fig.

Common Collector Transistor Circuit

In  common  collector  configuration  the  input  current  is  the  base current  IB  and  the output current is the emitter current IE. The ratio of change in emitter current to the  change in the base current is called current amplification factor.

It is represented by

COMMON COLLECTOR CIRCUIT

A test  circuit  for determining the  static characteristic  of an NPN transistor is shown in Fig. In this circuit the collector is common to both the input and the output circuits.   To   measure   the   base   and   the   emitter   currents,   milli   ammeters   are connected in series with the base and the emitter circuits. Voltmeters are connected   across the input an…