Download AM – Class IV

95AM-4
Sr. No.
EXAMINATION OF MARINE ENGINEER OFFICER
Function: Marine Engineering at Operational Level
APPLIED MECHANICS
INDIA (2001)
CLASS IV
(Time allowed - 3 hours)
Morning Paper
Total Marks 100
N.B. - (1) All questions are compulsory.
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
1. 16cm3 of water flows per second through a capillary tube of radius a cm and of length l cm when
connected to a pressure head of h cm of water. If a tube of same length and radius a/2 cm is
connected to the same pressure head. The quantity of water flowing through the tube per second will
be
a)16 cm3
b)4 cm3
c)1 cm3
d)8 cm3
2. The speed of the motor boat is reduced to one third of its original value in travelling a distance S.
Then the motor boat is brought to rest in a minimum distance of
a)9S
b)(8/9) S
c)(9/8)S
d)3S
3. A pendulum beating seconds, is taken from a point on the surface of the earth to a place, deep down
to another point where both the mass & radius is half as compared to the normal earth dimensions.
The time period of pendulum will be
a)2S
b)22 S
c)2S
d)1/2 S
4. A particle starts from rest from x = 0 at t = 0 and moves on a straight line path and again comes to rest
at x = 1 and t = 1 .If the motion is simple harmonic, the maximum accrelation is nearest to
a)1
b)3
c)5
d)7
5. Two vessels are moving in the same direction with a speed of 30 km/h. They are separated from each
other by 5 km. A third vessel moving in the opposite direction meets the two vessels after an interval
of 4 minutes. What is the speed of the third vessel?
a)30 km/h
b)35 km/h
c)40 km/h
d)45 km/h
6. A beaker containing a liquid with a body floating in the liquid falls freely under gravity. The upthrust
on the body due to liquid is
a) zero
b) equal to the weight of the liquid displaced
c) equal to the weight of the immersed portion of the body.
d) equal to the weight of the body in air.
7. Two rings of equal weight W are free to move on a rough horizontal rod the coefficient of friction
being . They are connected by a smooth string of length  on which another ring of weight 2W can
move freely.
Prove that in the equilibrium position, the position of the two rings on the rod cannot be further apart
than 2x given by the expression
2μ 
2x l
.
1  4μ 2
8. A small balloon of total mass 10 kg at a height of 50 m above the ground is moving upwards with a
velocity of 20 m/sec and acceleration 1.5 m/sec2 when a mass of 1 kg becomes detached. Find the
new acceleration of the balloon assuming that the upward force acting upon it remains unaltered.
Also find how many seconds it takes the 1 kg mass to reach the ground neglecting air resistance.
9.
(a) Derive the formula for the stress on the circumferential section of a thin cylindrical vessel
subjected to internal pressure.
The shell plate of a cylindrical boiler is 36mm thick and its diameter is 4m. The tensile strength of the
plate material is 495 MN/m2. The efficiency of the longitudinal seam is 85% and that of the
circumferential seam is 66%. Find the safe working pressure allowing a factor of safety of 5.
(b) A jet 1.2cm in diameter issues from a nozzle connected to a supply of water at a pressure of 6 bar.
Find (i)
the velocity of the jet.
(ii)
the flow in cubic metres per minute, and
(iii)
the power in kilowatt available from the jet.
10. Explain what is ‘angle of friction’ and ‘coefficient of friction’. Two blocks A and B of mass 7.5kg
and 27.5kg respectively are placed together on an inclined plane, side, by side touching each other,
with block A on the downward side and block B on the upward side. The inclined plane makes an
angle of 40O with the horizontal. Coefficient of friction between A and the plane is 0.25 and that
between B and the plane is 0.10. Calculate the force between two blocks as they slide down the
incline. What is their acceleration?
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95AM-4
Sr. No.
EXAMINATION OF MARINE ENGINEER OFFICER
Function: Marine Engineering at Operational Level
APPLIED MECHANICS
CLASS IV
(Time allowed - 3 hours)
Morning Paper
INDIA (2001)
Total Marks 100
N.B. - (1) All questions are compulsory.
(2) All questions carry equal marks.
(3) Neatness in handwriting and clarity in expression carries weightage
Answers
Answer for Question No. 1
Correct Answer : c
Answer for Question No. 2
Correct Answer : c
Answer for Question No. 3
Correct Answer : c
Answer for Question No. 4
Correct Answer : c
Answer for Question No. 5
Correct Answer : d
Answer for Question No. 6
Correct Answer : a
Answer for Question No. 7
Sol: Let A and B be the rings sliding on the rod and C the ring sliding on the string. When the distance AB is maximum and =
2x, each ring on the rod is in limiting equilibrium.
Let T be the tension of the string and R the reaction between the rod and each of the rings A and B. Draw CE vertical meeting
AB at E. If ACE = , then
sin  =
cos  =
AE
x
2x


AC 1 

2
1
4x 2

2
2  4x 2

Consider the equilibrium of ring C. Resolving vertically,
2T cos  = 2 W
or T =
W
=
cos 
W
  4x 2
2
(1)
The ring A is in equilibrium under the action of the forces R, R, W and T. Resolving horizontally and vertically,
T sin  =  R and R = W + T cos 
Eliminating R,
T sin  = (W + T cos )
or T(sin  –  cos ) =  W
 2x   2  4 x 2 

or T 
 = W

 

W 
T=
2x    2  4 x 2
(2)
Equating the two values of T given (1) and (2),
W
=
2  4x 2
W 
2x    2  4 x 2
2  4x2 =   4 x 2 or 2x = 2 2  4x2
x2 = 2(  2 – 4x2), or (1 + 42)x2 = 2  2
2
2x – 
2x =
 4 2
Hence 2x 
l
2
1  4 2
.
Answer For Question No. 8
Sol: Force acting on the balloon =
Force required to produce acceleration of 1.5 m/sec 2
+ Weight of the balloon.
=
10
565
x 1.5 + 10 =
kg-wt.
9 .8
29
When 1 kg is detached, the remaining mass is 9 kg and force acting on it is the same
565
kg-wt. Hence the accelerating force
29
565
565
124
– Weight of remainder =
–0=
kg-wt.
29
29
49
124  9.8
 New acceleration =
= 2.756 m/sec2.
49  9
=
The initial velocity of the 1 kg weight is 20 m/sec upwards and we have to find the time in which it will reach the ground, i.e.
travel 40 m downwards. Taking downward direction as positive,
u = –20, s = 50,  = 9.8 m/sec2.
 50 = –20t +
[s = ut +
1 2
t
2
1
x 9.8t2
2
whence t = 5.93 sec.
Answer for Question No. 9
Answer : (i) 34.641
m/sec (ii) 0.23495 m3/min
Answer for Question No. 10
Answer :
(i) 6.65 N (ii) 5.313 m/s2
(iii) 3.673 kw