ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar|Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally|Kolkata|Ahmedabad MECHANICAL ENGINEERING MOCK-A – Solutions 01. Ans: (B) Sol: By continuity equation: 2 2 1 1 V A V A Q 2 2 1 2 V . 2 d 4 V . d 4 s / m 8 2 4 V 4 V 1 2 By Bernoulli’s equation: g 2 V Z g P g 2 V Z g P 2 2 2 2 2 1 1 1 (By neglecting the losses), V 2 = 4V 1 from continuity equation. Thus, 81 . 9 2 2 0 81 . 9 1000 10 200 2 3 81 . 9 2 8 2 81 . 9 1000 P 2 2 On simplification, P 2 = 150 kN/m 2 02. Ans: (B) Sol: For ‘n’ radiation shields inserted, increase in space resistances = n Total no. of space resistances = n + 1 = 11 + 1 = 12 and increase in surface resistances = 2n Total number of surface resistances = 2n + 2 = 2 11 + 2 = 24 03. Ans: 0 (Range: 0 to 0) Sol: Since there is no external force acting on the block so the value of frictional force will be 0 N. 04. Ans: (A) Sol: The given function is odd function since f(–x) = – f(x) . For odd function 1 1 0 ) x ( f • • • d d/2 2m 2 1 2
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ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar|Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally|Kolkata|Ahmedabad
MECHANICAL ENGINEERING MOCK-A – Solutions
01. Ans: (B)
Sol:
By continuity equation:
2211 VAVAQ
2
2
1
2 V.2
d
4V.d
4
s/m824V4V 12
By Bernoulli’s equation:
g2
VZ
g
P
g2
VZ
g
P 2
2
2
2
2
1
1
1
(By neglecting the losses),
V2 = 4V1 from continuity equation.
Thus, 81.92
20
81.91000
10200 23
81.92
82
81.91000
P 2
2
On simplification,
P2 = 150 kN/m2
02. Ans: (B)
Sol: For ‘n’ radiation shields inserted,
increase in space resistances = n
Total no. of space resistances = n + 1
= 11 + 1 = 12
and increase in surface resistances = 2n
Total number of surface resistances
= 2n + 2
= 2 11 + 2
= 24
03. Ans: 0 (Range: 0 to 0)
Sol: Since there is no external force acting on
the block so the value of frictional force
will be 0 N.
04. Ans: (A)
Sol: The given function is odd function since
f(–x) = – f(x) .
For odd function
1
1
0)x(f
•
•
•
d
d/2
2m
2
1
2
: 2 : Mechanical Engineering _Solutions
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05. Ans: (C)
Sol: For a state of pure shear,
x = 0
y = 0
xy =
Principal stresses for given state are,
1 = and 2 = –
For these principal stresses, Mohr’s Circle
can be drawn as shown in the figure below.
06. Ans: (B)
Sol: If the clearance provided is greater than the
optimum clearance, no shearing action takes
place and the sheet is simply pulled into the
die.
07. Ans: (C)
Sol: Shear strain, = tan( – ) + cot
= min ,
when 2– = 90
Now, if = 0, then = 45
min = tan45 + cot45 = 2
08. Ans: (A)
Sol:
Maximum interference
= (U.L)shaft – (L.L)hole
= 0.05 – 0.03 = 0.02mm
09. Ans: (C)
Sol: In a gas turbine engines combustion is
carried out isobarically.
10. Ans: (D)
Sol: P (x = 1) = 0.5 P (x = 2)
!2
e
2
1
!1
e 2
= 4
P (x = 4) !4
e4 4
44
e3
32
24
e4
11. Ans: (B)
Sol:
(2,0) (1,0)
–
0.06
0.03 0.05
0.02
Maximum
interference
4
a d
e b
c
4
16,16 0,0
9,9
4,4
5 3 7
: 3 : GATE _ Full Length Mock Test
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag |Tirupati | Kukatpally| Kolkata
In project management, float or slack is the
amount of time that a task in a project
network can be delayed to:
Subsequent tasks (“free float”)
Project completion date (“total float”)
Here,
Slack for activity ‘b’ in months = Total float
= Lj – Ei – tij = 9 – 0 – 3 = 6
12. Ans: (A)
Sol:
Area of cross-section of threaded portion is
less than that of the shank, so stress induced
in threaded portion is higher. We know that
strain energy is,
U 2
Hence, threaded portion has higher energy
absorbing capacity.
Tensile, shear and compressive stresses are
induced on the bolt without preloading.
13. Ans: (A)
Sol: The overdamped system under free
vibration condition is non-oscillatory.
14. Ans: (C)
Sol: If runner blades of propeller turbine are
adjustable then it is called Kaplan turbine.
15. Ans: (C)
Sol:
Alligatoring Rolling
Lap Forging
Bamboo defect Extrusion
Blister Casting
16. Ans: (A)
Sol: 12
1R
TT
TCOP
T1 = –3 + 273 = 270 K
T2 = 27 + 273 = 300 K.
270300
270COP R
= 9
9P
5
P = 0.55 kW
17. Ans: (B)
Sol: The given differential equation
4y+4y+y = 0
4D3 + 4D
2 + D = 0
D(2D + 1)2 = 0
D = 0, 2
1,
2
1
yc = C1 + (C2 + C3x) e–x/2
: 4 : Mechanical Engineering _Solutions
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