Kerala PSC Previous Years Question Paper & Answer

Title : NAVAL ARCHITECT PORT
Question Code : A

Page:11


Below are the scanned copy of Kerala Public Service Commission (KPSC) Question Paper with answer keys of Exam Name 'NAVAL ARCHITECT PORT' And exam conducted in the year 2018. And Question paper code was '050/2018/OL'. 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: 11 out of 13
Excerpt of Question Code: 050/2018/OL

Question84:-During a sway harmonic test, the hydrodynamic sway force and the ship speed are measured. Assume sway
force and sway motion as *Y_H=Y_{dotv} ™* dotv + Y v * v + Y {text{vwv}} है (v}*3" and'y" (in meters) ‘= A ¥
sin ({2pi}/{T} t) respectively. The camiage moves ahead at uniform speed *u” during the test. To determine *Y_{dotv} ,
Y v and Y_{text{vwv}}" following analysis is required:

A:-Conformal mapping

B:-Impulse response analysis

C:-Broaching

D:-Fourier analysis

Correct Answer:- Option-D
Question85:-The sway + yaw coupled oscillation at steady forward speed is given below.

“ஙா ன) 11-0௫ (ண) (60) 23 ணா "({dotv').{dotr}}
+[[(-४ ४), -11.[-4' றி ராட்டி று மா 0) 4 ங.௫ (4௧0) டல)
A:-The equation is linear
B:-The equation is nonlinear
C:-The equation is coupled
D:-(1) and (3)
Correct Answer:- Option-D
Question86:-For the ship rolling motion equation: *(I'_{text{xx}}+A' {44}) ddotphi' + b'_{44} dotphi' + C*_{44}phi=0"
: {44} :Added mass moment of inertia in the roll axis direction
B:-'b*_{44}" : roll damping
C:-*C'_{44)" : Restoring moment coefficient
0:-{1}, (2) चाध (3)
Correct Answer:- Option-D

Question87:-To transform the body fixed dynamics of a surface ship `ಟ, ೪' and 'r* (surge speed, sway speed and yawing
rate) to the NED (North East Down}) coordinate system the following parameter is required.

A:-Roll angle

B:-Pitch angle

C:-Heading angle

D:-None of the above

Correct Answer:- Option-C
Question88:-Determine the inertia tensor {matrix} about the coordinate centre for a unit which consists of two small
particles, each of mass *m* , connected by the light but rigid slender rods. Each one of the particle is attached at the
forward and aft end of the rod at a distance of *(I, 0, 0 mand (1.0. 0 m from the coordinate centre.

A:-*([0,0,01,[0,2ml~2,01,[0,0,2mI~2])"

B:-*([2ml~2,0,01,[0,0,01[0,0,2mI~2])*

C:-*([0,mI~2,0],[mI~2,0,01,[0,0,01)*

D:-*([0,0,01,[0,2mlI~2,2mI*2],[0,2miI*2,2mi~ 2]y

Correct Answer:- Option-A
Question89:-Determine the center of the circle and its radius given by the equation:

`3^2 + #^2 - 46 +6 ‏7-3و‎ 20“
A:-Centre (2, -3), Radius: 4
B:-Centre (-3, 2), Radius: 8
C:-Centre (-3, 2), Radius: 2
D:-None of the above
Correct Answer:- Option-A

Question90:-A pulley driving a belt has a diameter of 300 mm and is tuming at 2700/ pi* revolutions per minute. Find the
angular velocity of the pulley and the linear velocity of the belt assuming that no slip occurs.

A:-‘omega’ = 45 rad/sec, 'ഴ് ട 6.5 m/sec
B:-‘omega’ = 90 rad/sec, ۷۴ ‏ے‎ 13.5 m/sec
C:-"omega’ = 13.5 rad/sec, v’ = 90 m/sec
D:-None of the above

Correct Answer:- Option-B

Question91:-A towing carriage is travelling at 4.0 m/s and has wheels of diameter 1000 mm. Assume no slipping of wheel
oceurs. Determine the angular velocity of the wheels.

A:-76.4 ۳

8:-358.9 ಗಾಗಾ

C:-716.2 rpm

D:-25 ಗಾಗಾ

Correct Answer:- Option-A

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