Kerala PSC Previous Years Question Paper & Answer

Title : FOREMAN GOVT INSTRUMENT WORKSHOP KERALA SMALL INDUSTRIES DEVELOPMENT CORPORATION LTD SIDCO
Question Code : A

Page:2


Below are the scanned copy of Kerala Public Service Commission (KPSC) Question Paper with answer keys of Exam Name 'FOREMAN GOVT INSTRUMENT WORKSHOP KERALA SMALL INDUSTRIES DEVELOPMENT CORPORATION LTD SIDCO' And exam conducted in the year 2015. And Question paper code was '154/2015'. 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.

page: 2 out of 12
Excerpt of Question Code: 154/2015



8. The process used in summer air conditioning is known as :
(^) Humidification (B) Heating and humidification
(C) De humidification (D) Cooling and dehumidification
9. Temperature of human body is 94.2°F. Its corresponding temperature in celsius scale is :
(ಹ) 73.785% (B) 34.56८
(0. 371.430 D) 110.671
10. In the cast Iron the percentage of carbon usually varies between :
ಯ) 05 10% (0) 016 0.2%
(@ 1.01 4564 D) 2.5 ७० 3.5%
11. The ratio of total emissive power of a body to the total emissive power of a black body is
called :
(ல Reflectivity (B) Absorptivity
(ಲ) Transmitivity (D) Emissivity
12. TFactor of safety is defined as the ratio of :
(ಗ) Endurance limit to yield stress രു Elastic limit to ultimate stress
)0 Yield stress to working stress (D) Breaking stress to working stress
13. A simply supported beam of span (7 carries a uniformly distributed load over the
whole span. The shear force diagram will be :
(ಸ) arectangle (03) ೩1೩೧016
(0) two equal and opposite triangles (D) two equal and opposite rectangles
‏کت‎ When a shaft of diameter d 15 subjected to torsional load T, the maximum shear
stress f5, induced in the shaft is given by the relation :
647 16T
@) fs= ட்‌
7 ಸರನೆ & ‏تیر‎
‎81" 32T
(0 ‏در[‎ 0
श्यः 0 ग्यः
15. The torsion equation is given by :
ல 7/7- ௩/0 ௨09/2. യ 7//% 278/4 ೬00/1
(0) T/R=fs/d=COIL യ (1 = £= 0 = 8
154/2015 4 A

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