vv - gradeup 2016_paper-2_solution... · vv vv u u u 3 700 0.45 1000 0.55 865 / oil oil water water...
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1. Ans. C. ‘Lesser than’ is apt because the sentence should be in comparative degree.
2. Ans. A. ‘hanged’ means death by hanging ‘hung’ is used only with things and not with people 3. Ans. D. ‘lying prone’ means lie down flat. ‘Prone to’ means vulnerable to
4. Ans. C. Statements i and ii are not logically possible based on the given fact. 5. Ans. B.
2
3 2 6
5 6
6 / 5
6 6 36
5 5 25
3 3 6 6 3 36.
4 4 5 5 4 25
3 361 2.06
4 25
x x
x
x
Areaof square
Areaof squaretriangle
6. Ans. C.
7. Ans. B. 8. Ans. C
9. Ans. A.
10. Ans. C. 11. Ans. A. By the properties of eigen values & eigen vectors, if all
the principal minors of ‘A’ are +Ve then all the eigen values of ‘A’ are also +Ve.
2 2
10
2
1
2
A for k
sok
12. Ans. C.
The function 2
2
3 4
3 4
x xf x
x x
is not continuous at
4&1;x since f(x) does not exists at x=-4 &1.
13. Ans. A. By the L.T of standard functions 14. Ans. B.
From C-R equation; we have
&u v u v
x y y dx
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2 2
, 2
2 2
,
2 ; 2
2 2
1
u x y xy
u uky kx
x yv x y x y
u uv v
x yx yx y
ky y
k
15. Ans. A. f(x) is a linear function
16. Ans. D. Let w is the velocity after collision
17. Ans. B.
max
1 1
max
1
where, is largest principal stress
1
xy
xy
18. Ans. D.
where, p is internal pressure d is internal diameter t is thickness
19. Ans. A.
2
1
2 1
2
exp
Torque=
1 exp
31 1 1 exp 0.25 2.248
2
F
F
F F R
F R
Nm
20. Ans. B. We know that
2
2 2
n
q
m
kq m km
m
21. Ans. C.
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At location Y
At circumference
At location X
At circumference
22. Ans. A.
FBD of Lever taking moment about hinge for clockwise rotation of wheel
0
If b> c self energizing
Nb F Nc
b cFN F N
b c
So for clockwise rotation of the drum, the brake is self energizing.
23. Ans. D. Volume flow rate per unit depth between two streamlines
is given by 1 2
24. Ans. B. the variation in atmospheric pressure with height calculated from fluid statics is exponential
25. Ans. A.
For a hollow cylinder
1 2
2
1
1 2 1 2
2
1
2
ln
ln
2
th
kL T TQ
r
r
T T T T
Rr
r
kL
26. Ans. D.
1 1
1 2
12 2 1 2 1 2 2 1
2 0 0
2 2 2 2
0
0
1
2
2
1 1
i i
i
F
F
R L RAA F A F F
A R L R
RF F
R
27. Ans. D. According to Joule’s law, Internal energy of an ideal gas is a function of temperature only.
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28. Ans. B.
2 7.2. . 4
1.8R
QCOP
W
29. Ans. B. Brayton cycle:-
1
1.4 1
1.4
6
1.4
11
11
6
0.4006
p
brayton
p
r
r
Gas Turbine cycle with perfect regeneration:-
5 2 3 2
5 3
1
4
4 3
6
1.4
1
0.3
Heat supplied =
p
p p
p
r
C T T C T T
T T
T
T
C T T
4 5 2 1
4 3 2 1
T c
p p
p p
Work done W W
C T T C T T
C T T C T T
4 3 2 1
4 3
2 21
12 1 1 1
54 3 434
44
1
1
1 1
4 41
Work done
Heat supplied
1 1
1 1 1
11
11 1
11
1 0.3 6
regenerative
p p
p
p
p
p
C T T C T T
C T T
T TTTT T T T
TT T TTT
TT
rT Tr
T T
r
1.4 1
1.4 0.4994
0.40060.8021
0.4994
brayton
regenerative
30. Ans. B.
1212 / , 24 / 0.5
24hr hr
Let, expected time that a customer spend in queue is qW
0.50.5
1 0.5 6012
0.560 2.5mins
12
q sq
L LW
31. Ans. C.
For aluminium alloy solution hardening process will be used to increase strength and hardness. In this process
component will be heated to 550C above temperature so that solute particles can penetrate into the lattice easily
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32. Ans. B. Submerged arc welding uses a blanket of fusible granular flux
33. Ans. B. Final length = L/2 Initial length = L
0
1ln ln ln 0.69
2 2
f
r
L L
L L
34. Ans. A.
900
200
0.25 /
300 / min
?
/ min1000
300 / min 0.2
300478
0.2
9007.539min
0.25 478
L mm
d mm
f mm rev
v m
t
DNv m
m N
N RPM
Lt
fN
35. Ans. D. In USM,
36. Ans. D.
37. Ans. B.
Since if z=1 lies inside the closed path and z=2 lies
outside of the closed path then by cauchy’s formula.
38. Ans. D. Probability that a packet would have to be replaced i.e.,
1 ?P X [ Let ‘x’ denote the number of defective
screws]
0 5
0
5
1 1 1
1 0
1 5 0.1 0.9
1 0.9 0.40951
P X P X
P X
C
Since by the Binomial distribution when P=probability of defective screw. 39. Ans. D.
0; sin cos
3 3
03
1 1.37 1
b ah f x x x
n
x
y f x
By trapezoidal rule; we have the approximate value of the integral is
0
00
/ 3sin cos 1 1 2 1.37 0.37
2
1.822
Exact value of the integral is
sin cos cos sin 1 1 2
Error Exact value-Approximate value
=2-1.822 0.178
x x dx
x x dx x x
40. Ans. A.
2.02 / sec
1
ii
V R
Vrad
R
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2 2
2
2 2
2 2
1Loss in kinetic energy of Drum
2
1150 2 300
2
1Loss in kinetic energy of block
2
12 0
2
4000
Loss of Potential energy of block = mgh
2000 9.81 0.5
i f
drum
i f
block
J
KE Joule
m v v
KE Joule
P
9810
Total energy loss
300 4000 9810 14110
14.11
block
block block block
E Joule
KE KE PE
Joule
kJ
41. Ans. B. By definition of Torque
42. Ans. D.
Because of torsion angle of twist will be there.
2 2
4
4
2 32 4
.
T P PWhere
G J G d R G
Due to angle of twist, A will reach at A' and B will reach at B' let A'A'' be the vertical displacement
43. Ans. C.
As we know that
and slope at mid-span where moment is applied will be
U
M
(according to Costigliano’s theorem).
44. Ans. D.
F.B.D of point B
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Applying Lame's theorem
sin180 sin135 sin 45
0
Stress in AB 0
BCAB
AB
FF P
F
45. Ans. D.
5
5
200
10 /
2 10
0.2
t C
C
E MPa
mm
We know that, Axial stress gets induced in the rod when
some gap ' ' is provided is
46. Ans. A.
2Given AB
Since Rod AB is rigid, so Axial velocity of A & B should be same
cos60 cos60
2 / sec
is mid point of AB
A B
A B
V V
V V m
C
Alternate Method:
2 2 2
2
2 2cos120
2 2 2
11
2 2 2
32 2 2 3
2
3,so oc willbeperpendicular toab.2
sin 30 12
cc
x
x
x
xy
vv
47. Ans. C. Displace the block “A” & Release
Alternate method:
Energy of system remain conserved,
where,
Static elongation of spring at equilibrium which is
calculated as follows:
mgmg k
k
Differentiating Eqn.(1) w.r.t time, which will be zero because E = constant
0
0 ...(2)
dE
dt
d dvJ mV mgv k y v
dt dt
Since
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there is no slipping between rope & pulley v r
2
2
2
1 10
2
02
150010 / sec
10 5
2
n
v dv dvMr mv mgv k y v
r r dt dt
M d ym kydt
krad
Mm
48. Ans. D. Given
max min m150 , 50 , 100
50 , 200 , 300
400
a e ut
ut
MPa MPa MPa
MPa S MPa S MPa
S MPa
Equation of line 1
1 ...(1)a m
e utS S
Equation of line 2
1tan ...(2)
2
a
m
Solving above two equations to get co-ordinates of point
P ,m aS S
1200 400
2 400 ...(3)
2 ...(4)
a m
a m
m a
S S
S S
S S
from (3) & (4)
100
100. . 2
50
a
a
a
S MPa
Sf o s
49. Ans. C.
3
3
1000 /
700 /
?
0.45
0.55
water
oil
body
oil body
water body
Given kg m
kg m
V V
V V
3
700 0.45 1000 0.55
865 /
oil oil water water body body
body body body body
body
V g V g V g
V V V
kg m
50. Ans. B.
2
2
360
0.05 0.05
0.44 60 / .
26.4 /
top bottom
wall
V Vdu
dy
duJ kg m s
dy
N m
51. Ans. A
Downward force due to water = weight of water above curved surface
22
2
24
2 14
R Lg R L
gR L N
Weight of plug is neglected. 52. Ans. A.
1 2 2 1
2 2
1 /
2 /
(as both are same fluids)
1 80 50 2 10 25
h c
k c
h
c
P P
k P k k c p c c
c c
m kg s
m kg s
C C
m C t t m C t t
t t C
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11
22
1 2 1 2
1 1
2 2
80 25 5570
50 10 4025
ln ln
70 25 55 40
70 55ln ln
25 40
43.705 47.1
. .
p c
c
m m
p mp c m
CC
CC
Q U A Q U A
.
43.7050.928
47.1
cp mp c m
mpc
p mc
A A
A
A
53. Ans. C.
54. Ans. B. 3
1
1
1
3
2
21 1
1
0.4
100
80 353
0.1
Ideal gas & process is isothermal.
ln
0.1100 0.4ln
0.4
55.45
V m
P kPa
T C K
V m
VW PV
V
kJ
55. Ans. C.
Reversible cycle.
3 31 2
1 2 3
33
0
40 370 0
400 370 320
1 32064
320 5 5
Q
T
Q QQ Q
T T T
QQ kJ
56. Ans. B.
Given
1
1
2
3
2 3
236.04 /
0.9322 /
272.05 /
93.42 /
0.05 /
Heat Rejection to environment =m
0.05 272.05 93.42
8.9315
h kJ kg
s kJ kg
h kJ kg
h kJ kg
m kg s
h h
kW
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57. Ans. B.
58. Ans. A.
Ans. A
0.2 0.70.4
0.2 0.7 0.4 0.4
0.2 0.3 ...(1)
Fraction of solid =
0.3
0.2
0.20.40
0.5
L S
L S
L S L S
L S
s
S L
S
S S
59. Ans. A.
60. Ans. B.
1 2
1 1
2 2
100 40 1 & 2
140 250
180 200
10
140 2501 ...( )
180 2001 ...( )
On solving equation a and b, we have
. . 424.6
V mm mm
V V I A
V V I A
V I
CV SCC
aOCV SCC
bOCV SCC
S C C A
61. Ans. C. Cost of metal cutting = Rs 18 C/V Cost of Tooling = Rs 270 C/TV C= Constant ,V = Cutting Speed , T = tool life
1/0.250.25
4
150, 150 150 /
150 /
C VT T V
T V
4
4
3
4
Total cost =18 270
18 270
150
18 270
150
C C
V TV
C CV
V V
C CV
V
On, differentiating total cost
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2
2 4
44
18 270 3
150
18 150
3 270
57.914 / min
C C V
V
V
V m
62. Ans. A.
12 1.5GivenV V V V
1 1 1 1
1
3
33
6
3
10.02 0.02 0.2
10
7860 /
107860 /
10
55.85
7.860 /
k ohm mm ohm cmohm cm
kg m
gm cm
gm
gm cm
5 355.85: : 3.68 10 / sec
7.86 2 96600
AI IMRR Q cm
ZF
4Interelectrodegapgiven=9 m=9 10 cm
4
current destiny J=
0.2 12 1.52333.33
9 10
.
2333.33 . . / 2333.33
k V V
y
I J S A
I S A S A I
Electrode feed rate = MRR/ surface area cm/sec 53.68 10 2333.33
/ sec
0.086 10 60 / min
51.51 / min
Icm
I
mm
mm
63. Ans. D.
2 2
22
2 22
2
, ,
2
2
a R r b D R r C a b
C R r D R r
H R r C
R r R r D R r D R r
H R r D R r D
64. Ans. C.
025000 , 500 / , 20%hD kg C Rs order C of Cu
Qty (kg) Cu (Rs/kg) Ch (Rs/Kg/year)
11 750Q 70 0.2 × 70 = 14
2750 1500Q 65 0.2 × 65 = 13
3 1500Q 60 0.2 × 60 = 12
This problem belongs to inventory model with two price break.
2* o
h
DCQ
C
first checking for least unit price
*
3
2 25000 5001443.37
12Q
Now, 1443.37 < 1500 therefore, the company will not get the item at Rs 60/kg Now, checking for second minimum unit price
*
2
2 25000 5001386.75
13Q
Since, 1386.75 lies between 750 and 1500 Therefore, we need to find
65. Ans. B.
The latest finish time for node 10 is 14 days.
***