Kerala PSC Previous Years Question Paper & Answer
Page:9
Below are the scanned copy of Kerala Public Service Commission (KPSC) Question Paper with answer keys of Exam Name 'TECHNICIAN GR II BOILER OPERATOR PART I KCMMF LTD' And exam conducted in the year 2015. And Question paper code was '158/2015'. Medium of question paper was in Malayalam or English . Booklet Alphacode was 'A'. Answer keys are given at the bottom, but we suggest you to try answering the questions yourself and compare the key along wih to check your performance. Because we would like you to do and practice by yourself.
മു The amount of air required for complete combustion of fuel is called :
(മ) Excess air (ए) Free air
(C) Stoichiometric air (D) None of the above
64. For ideal combustion process for burning 1kg of a typical fuel containing 86% carbon, 12%
hydrogen and 2% sulphur, the theoretically required quantity of air :
ಉ ಟಟ ஐ 10kg
ര 125kg യ 14.1kg
65. Free oxygen in the products of combustion is an indication of :
(A) Stoichiometric air യ) Minimum air
ര Excess air ६ (D) None of the above
66. Orsat apparatus 15 used م determine :
(A) Products of all constituents of fuel combustion by weight
B) Products of all constituents of combustion of fuel by volume
(C) Products of only dry constituents of combustion by weight
(D) Products of only dry constituents of combustion by volume
67. Which of these devices can be used to detect leaks in boilers :
(ക Ultrasonic transmitter (8) Leak detector fluid
(©) Micromanometer (D) None of the above
68. If gravimetric analysis of products of combustion 15 Lknown, Air; fuel ratio is given by :
CxN, CxN,
A e (8) ರ್ಸ್
ക് 1860, -05800 © 2100+ +0
0 CxN,
1 ന
© 300و ) 108860, -05800
69. Volumetric analysis of sample of dry products of combustion are 695 10% 00 51% 055 8%
115 81%, ൬ വോ by weight are given by :
(^) 10:1:8:81 (B) 44:28:256:2268
(©) 22:14 :256 :2268 യ 22:14 :128 8
70. C +0, — CO, Therefore 1kg of carbon requires how much kg of oxygen?
(മ 1.67 യ) 2.67
(¢) 7 യ) 4.67
A 11 158/2015
[?.7.0.1
