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FV(M) AM
Sr. No. 1
Examination of Engineers of Fishing Vessel (Motor)
Applied Mechanics
Time Allowed –2 ½ Hours
INDIA (2002)
Total Marks 100
N.B. -
(1) Attempt FIVE questions only .
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
1. The excess of pressure inside the first soap bubble is three times inside the second bubble. The ratio
of volume of first to second bubble is
a) 1 : 3
b) 1 : 9
c) 1 : 27
d) 3 : 1
2. Two particles A and B get 4m closer each second while travelling in opposite directions. They get
0.4m closuer every 10 seconds while travelling in the same direction. The speeds of A and B are
respectively
a) 2.2 m/sec and 0.4 m/sec
b) 2.2 m/sec and 1.8 m/sec
c) 4 m/sec and 0.4 m/sec
d) none of these
3. A metallic rod of Young's modulus 2 x 1011 N/m2 undergoes a strain of 0.05%. The energy stored
per unit volume of the rod will be
a) 0. 5 x104 Jm-3
b) 2. 5 x 104 Jm-3
c) 0. 5 x 108 Jm-3
d) 2. 5 x 108 Jm-3
4. A tube floats vertically in water having density of 1 gm /cm3 such that 4cm of the tube lies above the
surface. What will be the length of the tube if it is filled completely with a liquid of density 0.9
g/cm3.
a) 40 cm
b) 60 cm
c) 20 cm
d) 30 cm
5. A vessel , whose bottom has round holes with diameter of 0.1 mm , is filled with water . The
maximum height to which the water can be filled without leakage is (S.T. of water = 75 dyn/cm, g =
1000 cm/s2)
a) 100 cm
b) 75 cm
c) 50 cm
d) 30 cm
6. Water is flowing in streamline motion from the tube shown in figure. State whether the Pressure is
(a) More at A than that at B
(b) Equal at A and B
(c) Lesser at A than that at B
(d) Normal at A and B
7. A 13 m ladder weighing 25 kg is placed against a smooth vertical bulkhead with its lower end 5 m
from the bulkhead. The coefficient of friction between the ladder and the floor is 0.3. Show that the
ladder will remain in equilibrium in this position. What is the frictional force acting on the ladder at
the point of contact between ladder and floor? The centre of gravity of the ladder is at the mid-point
of its length.
8. A radar antenna rotates for 5 seconds with a constant angular acceleration and describes during the
time 10 radians. It then rotates with constant angular velocity and during the next 5 sec. describes 80
radians. Find the initial angular velocity and the angular acceleration.
9. A load of 4 tonnes on the deck is to be lifted by screw-jack. Mean diameter of the thread is 8 cm,
pitch of the screw is 1 cm,  = 0.1. Calculate the torque required and the work done in lifting the load
through 15 cm. What is the efficiency of the jack at this load?
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FV(M) AM
Sr. No. 1
Examination of Engineers of Fishing Vessel (Motor)
Applied Mechanics
Time Allowed –2 ½ Hours
INDIA (2002)
Total Marks 100
N.B. -
(1) Attempt FIVE questions only .
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
Answer
Answer for Question No. 1
Correct Answer : c
Answer for Question No. 2
Correct Answer : b
Answer for Question No. 3
Correct Answer : b
Answer for Question No. 4
Correct Answer : a
Answer for Question No. 5
Correct Answer : d
Answer for Question No. 6
Correct Answer : a
Answer for Question No. 7
Sol: AB is the ladder and G its middle point. The vertical through
ground at E. C is the junction of the wall and the ground.
Clearly
1
AE = EC = AC = 2.5 m.
2
G meets the
BC = AC3  AC2 = 132  52 = 12 m.
Let
R = Normal reaction at A.
F = Frictional force at A.
S = Normal reaction at B.
Resolving horizontal and vertically,
F = S,
R = 25.
Taking moments about A,
25 x 2.5 = S x 12 = F x 12 (since S = F)
F = 5.2 kg wt.
Thus for equilibrium the force of friction required is 5.2 kg wt. The maximum amount of force of friction
available is R = 0.3 x 25 kg. wt., which is more than the required amount.
Hence the ladder will remain in equilibrium and the force of friction between the ladder and the
ground is 5.2 kg.wt.
Answer for Question No. 8
Sol:
.
In the last five seconds, the radar antina describes 80 radians with constant angular velocity, say

 – 80  5 = 16 radians per second.
This must be the angular velocity at the end of the first interval.
Let 0 be the angular velocity at the commencement of the first interval of five seconds. Then for
this interval, if  be the angular acceleration,
Initial angular velocity = 0.
Final angular velocity = 16 rad/sec.
Angular displacement = 100 radians.
Time = 5 sec.
Substituting in the formulae
1
 = 0 +  t and  = 0 t +  t2,
2
we get
16 = 0 + 5
1
and
100 = 50 +  x 25.
2
Solving
 = –1.6 rad/sec2; 0 = 24 radians/sec.
Answer for Question No. 9
Sol:
Pitch p = 1 cm, diameter d = 8 cm.
1 7
p
tan  =
=
= 0.0398.
d 22  8
tan  =  = 0.1
0.0398  0.1
tan( + ) =
= 0.14.
1  0.0398  0.1
Effort required at the circumference
= W tan( + ) = 4 x 0.1 = 0.56 tonne-wt.
Torque about the axis
= 0.56 x radius of screw = 0.56 x 4 = 2.24 t-cm.
Number of turns required to lift the load through 15 cm.
= 15/1 = 15.
Angle turned through
= 15 x 2 = 30 radian.
Work done = torque x angular displacement
= 2.24 x 30 = 211.2 cm-t
Useful work done = load x distance through which load is lifted
= 4 x 15 = 60 cm-t.
0.0398
tan 
=
–
= 0.284 = 28.4%.
0.14
tan(   )