Trending ▼   ResFinder  

JAM 2016-GG-Q31-60-explanation-upload

1 pages, 2 questions, 0 questions with responses, 0 total responses,    0    0
Kumar Ayyasami
Presidency College, Chennai
+Fave Message
ProfileTimelineUploads
 Home > ayyasami >

Formatting page ...

Q 31. A; C; D Cataclasite is a type of cataclastic rock that is formed by fracturing and comminution during faulting. It is normally cohesive and non-foliated, consisting of angular clasts in a finer-grained matrix. Pseudotachylite is a cohesive glassy or very fine-grained rock that occurs as veins and often contains inclusions of wall-rock fragments. It is found either along fault surfaces (fault vein type), as the matrix to a fault breccia (pseudotachylite breccia), or as veins injected into the walls of the fault. Fault gouge is a tectonite (a rock formed by tectonic forces) with a very small grain size. Mylonite in Ductile shear zone. 32. B;C;D Increase in the length of tributary is not related to the main river. 33. A; D epsilon is greater than omega for positive. 1.55 is low to medium 34. D. 35. 35. A;B;C;D 36. A;D. Teredo is wood boring bivalve. Pecten is free living bivalve. 37. B;C 38. A;B 39. A;C 40. A;B;C;D 41. 25% 42. 2.0 43. 1 44. 2.8 10x20/100=2ccm. 26g+2g=28g for 10ccm. Density (for one ccm) =2.8 45. 140 If a ray stands on a line, then the sum of two adjacent angles so formed is 180 . 46. .46 47. 115 . (90 +25 ) 48. 3 (1cm=.25km) 49. 100 my 50. 21.32 51. 0.125 52. 198. convert in to centimetre & multiply. Go for percentage. Convert to metric tonne. 53. 625 [25+(25*15)+(15*15)] 54. .24 to .26 Weight of MgO in melt is 10.08g. 1 mole of MgO is 40.3. So, 10.08x1/40.3 = 0.25 mole. 3.395 (3.42 x ) + (3.28 x ) + (3.46 x ) 47.61 to 47.63 (45.58/95.72*100) 105 (sin 30 =thickness/bore length(hypotenuse)) = thickness/210; thickness=210x = 105. 400 The contact of sandstone&shale is 600m from the 400 m contour to the 200m in other block. The contour interval is 400m 200m = 200. So, 600m 200m gives 400m. 59. 80.62 square on hypotenuse = 802 x 102= 6400+100= 6500. Distance = 6500 = 80.62 60. 2.678 Pressure point =4.5Kbar-18.22bar per degree =4500-1822 (for 100 degrees) = 2678bar =2.678Kbar. 55. 56. 57. 58.

 

  Print intermediate debugging step

Show debugging info


 

 

© 2010 - 2024 ResPaper. Terms of ServiceContact Us Advertise with us

 

ayyasami chat