Kerala PSC Previous Years Question Paper & Answer
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Below are the scanned copy of Kerala Public Service Commission (KPSC) Question Paper with answer keys of Exam Name 'Overseer / Draftsman (Mechanical) Gr II (Special Recruitment From Among SC/ST) English ' And exam conducted in the year 2019. And Question paper code was '055/2019'. 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.
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The force applied on a body of mass 150 kg to produce an acceleration of 10 m/s?, 15:
(4) 156N (8) 160%
ര 1500N (D) 8000 N
Two forces act an angle of 120°. If the greater force is 50 kg and their resultant is
perpendicular to the smaller force, the smaller force is:
(4) 20kg (9) 25kg
© 30kg യ 36%
The process of finding out the resultant force is known as:
(A} Superposition of forces {B) Composition of forces
{C) Addition of forces യ) Resolution of forces
The point, at which the whole weight of the body may be considered to act, is known as:
(ಹ) Centre of mass {B) Centre of gravity
{C) Centre of curvature (D) Moment of inertia
If the resultant of forces acting on a bedy is zero, the body:
{A) isin equilibrium {B) is moving with non uniform velocity
{C) is not in equilibrium (2) none of the above
The unit of force in C.G.8. system of units, is called:
(ಉ Dyne رمق Kg
{C) Newton (D) عوط عط ال
If the angular distance, 0-25-82 (1൭ angular acceleration at t = 1sec. is:
(&) 07٥1۳9/0 (B) கல்ல
(C) 4radisec’ (D) 18186/9003
The resultant of two forces which are acting at an angle 0 18:
(५) "و- تحال +279 ००९6) ® ‰@*-९ + 7९ न)
௫ வடி ಇಗ രു னிட]
55/2019 4 A
