answerkeys/IN IV year 170541 IEO Exp
IV Year – Instrumentation Engineering Answer Keys 1
A
2
A
3
A
4
A
5
B
6
C
7
A
8
D
9
A
10
C
11
B
12
C
13
A
14
B
15
B
16
A
17
C
18
D
19
B
20
B
21
A
22
B
23
B
24
A
25
B
26
27
A
28
A
29
C
30
C
31
A
32
C
33
B
34
A
35
D
36
A
37
A
38
B
39
B
40
C
41
B
42
D
43
D
44
A
45
B
46
C
47
A
48
B
49
A
50
B
A
Explanations:Section-I: General Ability 1.
By using the formula, n(A ∪ B) = n(A - B) + n(A ∩ B) + n(B - A) 50 = 20 +15 + n(B - A) n(B - A) = 50 - 35 n(B - A) = 15 Now, n(B) = n(A ∩ B) + n(B - A) = 15 +15 = 30
3.
Given R = 5% per annum P = 2, 00, 000 n = from 2000 to 2002 = 3 years.
Population at the end of year 2002 = P 1+
R 100
n
= 200000 1+ = 200000×
5 100
3
21 21 21 × × 20 20 20
= 231525 So, the estimated population = 231525. 5.
First number + last number = 8 + 12 = 20. Second number + second last number = 2 + 18 = 20 and so on. So, 11 + 9 = 20 Hence, 11 will be the answer. Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Instrumentation Engineering 7.
Convert speed (km/hr) in to (m/sec). 5 So, 18 5m sec. 18
Time
Dis tan ce 200 40seconds. speed 5
8.
German that is a Nationality not a country.
9.
So, average of prime numbers between 37 to 50 will be 41 43 47 3 43.66 .... 43.67
10.
This question tests your ability to properly draw an inference from the statements that are given to you. Remember that on inference questions, your task is to evaluate each answer choice, looking for the one that has to be true if the stimulus is true. Two of the answer choices are extreme. Extreme answer choices should be avoided in inference questions because they are very unlikely to be correct. Remember that we want the choice that has to be true; extreme choices are unlikely to be the ones that have to be true. Choices (B) should be eliminated for this reason. It does not have to be true that government statisticians are the "only" source. It also does not have to be true that high incomes "never" indicate high net worth. We are only told that there is no correlation, or consistent pattern. That does not mean that there will not be some families with both high income and high net worth. Choices (A) and (D) are not extreme statements, but both do not have to follow from the statements given. Choice (A) could be true, but does not have to be true. A family could also develop high net worth by saving all their income for a short number of years. Choice (D) also could be true, but does not have to be true. Choice (C) is correct because it has to be true given the stimulus. Since we know that there is no correlation between higher income and higher net worth, it is impossible to know what the result will be of in increase in income.
11.
Let the dimension of the cuboid be 2x,3x,4x.
Then,
2 2x 3x 3x 4x 4x 2x 208 2 6x 2 12x 2 8x 2 208 2 26x 2 208 52x 2 208 x2 4 x 2cm.
Therefore volumeof cuboid 2(2) 3 2 4(2) 192cm3 . Email: [email protected], Website: www.engineeringolympiad.in IEO-2017 Partners & Sponsors:
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IV Year – Instrumentation Engineering 12.
The structure of the sentence implies that the words which fit into the two blanks are contrasting words. Option (C) is most logical in the given blank. Inspite of the squalid (filthy) surrounding in which she lived, her dress was immaculate (completely tidy). The remaining options do not make sense.
13.
Note: If a trader to sell his goods at cost price but uses false weights, then
Gain% =
error
×100 %
True value - error 200 So, here profit percentage = ×100 % 1000 - 200 200 = ×100 % = 25% 800 14.
Let „x‟ be the monthly salary of Ram.
100 31 20 % of 100 30 15 %of x 15000 49%of 55%of x 15000 49 55 x 15000 100 100 x 55658rs /
15.
Since her arrival at the native village has been trying to the best of her power spread awareness against dowry system. Replace “is” with “has been” in option B.
17.
Let the total number of mangoes be x. Then, x Mangoes sold to 1st Customer = +1 . 2 x x Remaining Mangoes = x - +1 = -1 . 2 2
1 x x 1 x 2 Mangoes sold to 2nd Customer = - 1 +1 = - +1 = + 3 2 6 3 6 3 x x 2 x x 2 x 5 Remaining Mangoes = -1 - + = - - 1+ = - 2 6 3 2 6 3 3 3 1 x 5 x 2 Mangoes sold to3rd customer = - +1 = + 5 3 3 15 3 x 5 x 2 x x 5 2 4x 7 Remaining Mangoes = - - + = - - + = - . 3 3 15 3 3 15 3 3 15 3
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IV Year – Instrumentation Engineering
4x 7 ∴ - =3 15 3 4x 16 ⇒ = 15 3 16 15 ⇒ x = × = 20. 3 4 ∴ Number of Mangoes originally the man had = 20. 18.
The error is in D. The correction is „known to stand up‟.
19.
Required Percentage =
726 + 846 + 786 + 956 + 870
980 +1050 +1020 +1240 + 940 4184 = ×100 % 5230
×100 %
= 80% Section-II: Technical x
21.
f x 2 t 2 dt f' x 1 2 x 2 2 x 2 1
The equation x 2 f ' x 0, becomes x 2 2 x 2 0 x 4 x 2 2 0 x 1 22.
Tree does not contain any loop but it must be touch the all the nodes.
23.
The diode is OFF, if V0 2.5V
If Vi 2.5V, Then the diode conducts (zero resistance) and V0 Vi V 0
The diode is open (2M resistance) when
Vi 2.5V and then V0 2.5
Vi 2.5 3
Thus for 2.5 Vi 10, the relation is V0 2.5 When Vi 10V,
24.
Tg
Vi 2.5 3
10 2.5 V0 2.5 V 5V 3
d H() d
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IV Year – Instrumentation Engineering 25.
16T 16 120 3 GD 80 109 (20 103 )3 9.55 104 radians 0.0547deg rees
26.
Active area=5 mm2=5×10-2 cm2 For 0.05 cm2, power through photodiode is 1 mW/cm2×0.05=0.05 mW As responsivity of photodiode is given as 5 A/W, current passing through
R L is 5A / W 0.05 103 W 0.25mA VL i R L 0.25 103 100 103 25V 27.
A B C
Y
Y = A B .C = A B C AB AB C 28.
29.
The second order transfer function is C 1 1 2 R s 2 4 j s 2 4 j s 4s 20 1 n 20, 5 %M p 0.2 2 e 1 Tp 2 4 n 1
3 2 2 22 22 2 2 2 2 6 32 32 6 This is true for single tone message signal only.
30.
x 2 yz 4z2 x; = 2xyz+4z2 i+x 2 zj x 2 y 8zx k a 1 1,2, 1 8i j 10k; a=2i-j-2k; a 3, a = 2i j 2k a 3 .a
31.
1 37 16 1 20 12.33 3 3
P VIcos90 0
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IV Year – Instrumentation Engineering 32.
V
gh
Vth
Vgs1 Vth 1 W/L Vgs2 Vth
W / L 2 W / L 1
L1 L2
W is fixed
Putting given values, (subscript 1 = initial, 2 = final) 5 1.5 V 1 L 2L V 1.5 3.5 2 V 6.45V gs2 gs2 Vgs 1.5 2 2 1 33.
Using convolution multiplication principle, Multiplication in time domain becomes convolution in frequency domain. Thus maximum frequency component present in x 2 (t).y(t)is
2 250 250 750 Thus sampling frequency can be 2 750 1500 rad / sec. 34.
a = ω 2x a (2 10) 2 (0.01) 39.4 m / s 2 4.02g (1g 9.8 m / s 2 )
35.
PID controller is shown as below.
36.
Inverter
f x x 2 2bx 2c2 ; g x x 2 2cx b2 f ' x 2x 2b and g' x 2x 2c f " x 2 0, g" x 2 0 f x is minimum when f' x 0 i.e. when x=-b g x is maximum when g' x 0 i.e. when x=-c b 2 2b 2 2c2 c2 2c 2 b 2 2b 2 c2 2b 2 c2 c 2b
37.
G(s) 1 1 G(s) , for a unity feedback system 1 G(s) 1 5s 5s
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IV Year – Instrumentation Engineering
K v limsG(s) s 0
38.
1 5
SSE 5 f 50kHz 0.1 f m 500kHz Bandwidth 2 1 fm
2 0.1 1 500 103 Hz 1100kHz 39.
d d2 d , x 2 2 1 , where = ; dx dx dz The given equation becomes, -1 y y y 0 2 1 y 0; We put z=logx so that x
AE is m2 1 0 m 1; y C1ez C2e z C1elog x C2e log x C1x
C2 x
Slew Rate 5 V / s 79.6kHz 2 VP P 2 10
f max
42.
In the signal flow graph there are 6 forward paths, they are abcdefg ahjmg ahdefg aifg aikjmg abmg
43.
Proximity inductance type motion transducers show the following characteristic curve.
Voltage
40.
distance
44.
Given
2 z z 4 4z 0 2 y y
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IV Year – Instrumentation Engineering
It can be put in the form, D2 4D 4 z 0, where D=
. y
A.E. is m2 4m 4 0 m 2, 2 complete solution is z= A+By e2y ................... 1 z Be2y A By 2e 2y Be 2y 2z ................. (2) y Using the given conditions in (1) and (2),we have
45.
A=0 z ye x 2y x B=e
Y A.B B.C CA The principle of duality theorem says that 1. Changing each OR sign to an AND sign 2. Changing AND sign to OR sign 3. Complementing any 0 or 1 appears in the expression
Y A.B B.C C.A Duality of Y A B B C C A
46.
y(n)=y(n-1)+y(n-2)+x(n) Apply Z-Transform to above equation Y(Z) 1 (Z2 ) X(Z) 1 Z1 For aboveTransfer function the polesare
1 5 2 z plane
j
x
1 j
1 5 2 1
x
1pole outside unit circle
Since outside area of unit circle maps RHS of s-plane. 47.
Memory capacity = 8KB 23 210 8 213 1KB 1024 210
Length of stack point er 13 48.
Only a second order type system can have that kind of polar plot.
49.
No error due to frequency variation
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IV Year – Instrumentation Engineering
Lm Lsh ..........(1) R m R sh Im 10 103 1 Zsh 3 Ish 990 10 99 Zm
2
R sh2 ( lsh ) 2 zsh R sh 2 zm Rm R m 2 Lm
L 1 sh R sh ...........(2) 2 Lm 1 Rm
(1) in (2) 1
R I R 1 sh sh R sh m 10.1 m Im R m 99 99 L 1 100 From (1) Lsh m R sh Lm H 1.01H Rm 99 99 50.
By solving using super node 10
V1
1
V2
2V 2A
2
2
7A
4
V1 V2 70 2 4 V1 V2 5 ____ 1 2 4 similarly V1 2 V2 0 2
V1 V2 2 ___ 2
solve 1 & 2 V V1 V2
22 V 3
16 V. 3
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9 of 10
A
2
A
3
A
4
A
5
B
6
C
7
A
8
D
9
A
10
C
11
B
12
C
13
A
14
B
15
B
16
A
17
C
18
D
19
B
20
B
21
A
22
B
23
B
24
A
25
B
26
27
A
28
A
29
C
30
C
31
A
32
C
33
B
34
A
35
D
36
A
37
A
38
B
39
B
40
C
41
B
42
D
43
D
44
A
45
B
46
C
47
A
48
B
49
A
50
B
A
Explanations:Section-I: General Ability 1.
By using the formula, n(A ∪ B) = n(A - B) + n(A ∩ B) + n(B - A) 50 = 20 +15 + n(B - A) n(B - A) = 50 - 35 n(B - A) = 15 Now, n(B) = n(A ∩ B) + n(B - A) = 15 +15 = 30
3.
Given R = 5% per annum P = 2, 00, 000 n = from 2000 to 2002 = 3 years.
Population at the end of year 2002 = P 1+
R 100
n
= 200000 1+ = 200000×
5 100
3
21 21 21 × × 20 20 20
= 231525 So, the estimated population = 231525. 5.
First number + last number = 8 + 12 = 20. Second number + second last number = 2 + 18 = 20 and so on. So, 11 + 9 = 20 Hence, 11 will be the answer. Email: [email protected], Website: www.engineeringolympiad.in
IEO-2017 Partners & Sponsors:
1 of 10
IV Year – Instrumentation Engineering 7.
Convert speed (km/hr) in to (m/sec). 5 So, 18 5m sec. 18
Time
Dis tan ce 200 40seconds. speed 5
8.
German that is a Nationality not a country.
9.
So, average of prime numbers between 37 to 50 will be 41 43 47 3 43.66 .... 43.67
10.
This question tests your ability to properly draw an inference from the statements that are given to you. Remember that on inference questions, your task is to evaluate each answer choice, looking for the one that has to be true if the stimulus is true. Two of the answer choices are extreme. Extreme answer choices should be avoided in inference questions because they are very unlikely to be correct. Remember that we want the choice that has to be true; extreme choices are unlikely to be the ones that have to be true. Choices (B) should be eliminated for this reason. It does not have to be true that government statisticians are the "only" source. It also does not have to be true that high incomes "never" indicate high net worth. We are only told that there is no correlation, or consistent pattern. That does not mean that there will not be some families with both high income and high net worth. Choices (A) and (D) are not extreme statements, but both do not have to follow from the statements given. Choice (A) could be true, but does not have to be true. A family could also develop high net worth by saving all their income for a short number of years. Choice (D) also could be true, but does not have to be true. Choice (C) is correct because it has to be true given the stimulus. Since we know that there is no correlation between higher income and higher net worth, it is impossible to know what the result will be of in increase in income.
11.
Let the dimension of the cuboid be 2x,3x,4x.
Then,
2 2x 3x 3x 4x 4x 2x 208 2 6x 2 12x 2 8x 2 208 2 26x 2 208 52x 2 208 x2 4 x 2cm.
Therefore volumeof cuboid 2(2) 3 2 4(2) 192cm3 . Email: [email protected], Website: www.engineeringolympiad.in IEO-2017 Partners & Sponsors:
2 of 10
IV Year – Instrumentation Engineering 12.
The structure of the sentence implies that the words which fit into the two blanks are contrasting words. Option (C) is most logical in the given blank. Inspite of the squalid (filthy) surrounding in which she lived, her dress was immaculate (completely tidy). The remaining options do not make sense.
13.
Note: If a trader to sell his goods at cost price but uses false weights, then
Gain% =
error
×100 %
True value - error 200 So, here profit percentage = ×100 % 1000 - 200 200 = ×100 % = 25% 800 14.
Let „x‟ be the monthly salary of Ram.
100 31 20 % of 100 30 15 %of x 15000 49%of 55%of x 15000 49 55 x 15000 100 100 x 55658rs /
15.
Since her arrival at the native village has been trying to the best of her power spread awareness against dowry system. Replace “is” with “has been” in option B.
17.
Let the total number of mangoes be x. Then, x Mangoes sold to 1st Customer = +1 . 2 x x Remaining Mangoes = x - +1 = -1 . 2 2
1 x x 1 x 2 Mangoes sold to 2nd Customer = - 1 +1 = - +1 = + 3 2 6 3 6 3 x x 2 x x 2 x 5 Remaining Mangoes = -1 - + = - - 1+ = - 2 6 3 2 6 3 3 3 1 x 5 x 2 Mangoes sold to3rd customer = - +1 = + 5 3 3 15 3 x 5 x 2 x x 5 2 4x 7 Remaining Mangoes = - - + = - - + = - . 3 3 15 3 3 15 3 3 15 3
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IV Year – Instrumentation Engineering
4x 7 ∴ - =3 15 3 4x 16 ⇒ = 15 3 16 15 ⇒ x = × = 20. 3 4 ∴ Number of Mangoes originally the man had = 20. 18.
The error is in D. The correction is „known to stand up‟.
19.
Required Percentage =
726 + 846 + 786 + 956 + 870
980 +1050 +1020 +1240 + 940 4184 = ×100 % 5230
×100 %
= 80% Section-II: Technical x
21.
f x 2 t 2 dt f' x 1 2 x 2 2 x 2 1
The equation x 2 f ' x 0, becomes x 2 2 x 2 0 x 4 x 2 2 0 x 1 22.
Tree does not contain any loop but it must be touch the all the nodes.
23.
The diode is OFF, if V0 2.5V
If Vi 2.5V, Then the diode conducts (zero resistance) and V0 Vi V 0
The diode is open (2M resistance) when
Vi 2.5V and then V0 2.5
Vi 2.5 3
Thus for 2.5 Vi 10, the relation is V0 2.5 When Vi 10V,
24.
Tg
Vi 2.5 3
10 2.5 V0 2.5 V 5V 3
d H() d
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4 of 10
IV Year – Instrumentation Engineering 25.
16T 16 120 3 GD 80 109 (20 103 )3 9.55 104 radians 0.0547deg rees
26.
Active area=5 mm2=5×10-2 cm2 For 0.05 cm2, power through photodiode is 1 mW/cm2×0.05=0.05 mW As responsivity of photodiode is given as 5 A/W, current passing through
R L is 5A / W 0.05 103 W 0.25mA VL i R L 0.25 103 100 103 25V 27.
A B C
Y
Y = A B .C = A B C AB AB C 28.
29.
The second order transfer function is C 1 1 2 R s 2 4 j s 2 4 j s 4s 20 1 n 20, 5 %M p 0.2 2 e 1 Tp 2 4 n 1
3 2 2 22 22 2 2 2 2 6 32 32 6 This is true for single tone message signal only.
30.
x 2 yz 4z2 x; = 2xyz+4z2 i+x 2 zj x 2 y 8zx k a 1 1,2, 1 8i j 10k; a=2i-j-2k; a 3, a = 2i j 2k a 3 .a
31.
1 37 16 1 20 12.33 3 3
P VIcos90 0
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5 of 10
IV Year – Instrumentation Engineering 32.
V
gh
Vth
Vgs1 Vth 1 W/L Vgs2 Vth
W / L 2 W / L 1
L1 L2
W is fixed
Putting given values, (subscript 1 = initial, 2 = final) 5 1.5 V 1 L 2L V 1.5 3.5 2 V 6.45V gs2 gs2 Vgs 1.5 2 2 1 33.
Using convolution multiplication principle, Multiplication in time domain becomes convolution in frequency domain. Thus maximum frequency component present in x 2 (t).y(t)is
2 250 250 750 Thus sampling frequency can be 2 750 1500 rad / sec. 34.
a = ω 2x a (2 10) 2 (0.01) 39.4 m / s 2 4.02g (1g 9.8 m / s 2 )
35.
PID controller is shown as below.
36.
Inverter
f x x 2 2bx 2c2 ; g x x 2 2cx b2 f ' x 2x 2b and g' x 2x 2c f " x 2 0, g" x 2 0 f x is minimum when f' x 0 i.e. when x=-b g x is maximum when g' x 0 i.e. when x=-c b 2 2b 2 2c2 c2 2c 2 b 2 2b 2 c2 2b 2 c2 c 2b
37.
G(s) 1 1 G(s) , for a unity feedback system 1 G(s) 1 5s 5s
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6 of 10
IV Year – Instrumentation Engineering
K v limsG(s) s 0
38.
1 5
SSE 5 f 50kHz 0.1 f m 500kHz Bandwidth 2 1 fm
2 0.1 1 500 103 Hz 1100kHz 39.
d d2 d , x 2 2 1 , where = ; dx dx dz The given equation becomes, -1 y y y 0 2 1 y 0; We put z=logx so that x
AE is m2 1 0 m 1; y C1ez C2e z C1elog x C2e log x C1x
C2 x
Slew Rate 5 V / s 79.6kHz 2 VP P 2 10
f max
42.
In the signal flow graph there are 6 forward paths, they are abcdefg ahjmg ahdefg aifg aikjmg abmg
43.
Proximity inductance type motion transducers show the following characteristic curve.
Voltage
40.
distance
44.
Given
2 z z 4 4z 0 2 y y
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7 of 10
IV Year – Instrumentation Engineering
It can be put in the form, D2 4D 4 z 0, where D=
. y
A.E. is m2 4m 4 0 m 2, 2 complete solution is z= A+By e2y ................... 1 z Be2y A By 2e 2y Be 2y 2z ................. (2) y Using the given conditions in (1) and (2),we have
45.
A=0 z ye x 2y x B=e
Y A.B B.C CA The principle of duality theorem says that 1. Changing each OR sign to an AND sign 2. Changing AND sign to OR sign 3. Complementing any 0 or 1 appears in the expression
Y A.B B.C C.A Duality of Y A B B C C A
46.
y(n)=y(n-1)+y(n-2)+x(n) Apply Z-Transform to above equation Y(Z) 1 (Z2 ) X(Z) 1 Z1 For aboveTransfer function the polesare
1 5 2 z plane
j
x
1 j
1 5 2 1
x
1pole outside unit circle
Since outside area of unit circle maps RHS of s-plane. 47.
Memory capacity = 8KB 23 210 8 213 1KB 1024 210
Length of stack point er 13 48.
Only a second order type system can have that kind of polar plot.
49.
No error due to frequency variation
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8 of 10
IV Year – Instrumentation Engineering
Lm Lsh ..........(1) R m R sh Im 10 103 1 Zsh 3 Ish 990 10 99 Zm
2
R sh2 ( lsh ) 2 zsh R sh 2 zm Rm R m 2 Lm
L 1 sh R sh ...........(2) 2 Lm 1 Rm
(1) in (2) 1
R I R 1 sh sh R sh m 10.1 m Im R m 99 99 L 1 100 From (1) Lsh m R sh Lm H 1.01H Rm 99 99 50.
By solving using super node 10
V1
1
V2
2V 2A
2
2
7A
4
V1 V2 70 2 4 V1 V2 5 ____ 1 2 4 similarly V1 2 V2 0 2
V1 V2 2 ___ 2
solve 1 & 2 V V1 V2
22 V 3
16 V. 3
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9 of 10
