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{vvv}} ** (v)*3° and ‘y* (in meters) “= A “**
sin ({2pi}/{T} t)’ respectively. The carriage moves ahead at uniform speed ‘u* during the test. To determine *Y_{dotv} ,
Yv and *Y_{text{vvv}}* 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.

‘{L(m'-Y'_{dotv}),m'].[-(N'_{dotv}-m'x'_{CG}),(I_Z-N'_{dotrP]IP *((dotv’), (dotr’))
+[(-Y'_v),-YL[-(N'_v),-(N"_r-m'x'_{CG})]} *((v'),(r'))’ = *((¥'_{delta} delta_R),(N'_{delta} delta_R))*

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
' {44}° : roll damping

C:-*C'_{44}° : Restoring moment coefficient

D:-(1), (2) and (3)

Correct Answer:- Option-D
Question87:-To transform the body fixed dynamics of a surface ship ‘u, v’ 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 *(-I, 0, 0)° m from the coordinate centre.

A:-*([0,0,0],[0,2m1*2,0],[0,0,2ml*2])*
([2ml*2,0,0],[0,0,0],[0,0,2mI*2])*

C:-*([0,ml*2,0],[mI*2,0,0],[0,0,0])°

D:-* ((0,0,0],[0,2mI*2,2mI*2],[0,2mI*2,2m1*2])

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

“x42 + y*2-4x +6 y-3=0°

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 turning 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, "ഗ്‌ - 6.5 m/sec

B:- ‘omega* = 90 rad/sec, ‘v’ = 13.5 m/sec

C:- ‘omega* = 13.5 rad/sec, ` ४ = 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
occurs. Determine the angular velocity of the wheels.

A:-76.4 rpm

B:-358.9 rpm

C:-716.2 rpm

D:-25 rpm

Correct Answer:- Option-A