Answer Keys: Section-I 1 D 2 D 3 B 4 C 5 B 6 C 7 B 8 ... AWS
IN-IV Year Sample Paper
Answer Keys: Section-I 1
D
2
D
3
B
4
C
5
B
6
C
7
B
8
C
9
B
10
C
11
B
12
C
13
A
14
D
15
B
16
B
17
A
18
C
19
A
20
C
Section-II 1
C
2
A
3
A
4
A
5
D
6
A
7
C
8
D
9
B
10
B
11
A
12
C
13
B
14
A
15
A
16
A
17
B
18
A
19
D
20
A
21
C
22
A
23
C
24
A
25
C
26
C
27
D
28
B
29
A
30
A
Explanations: Section-I 1.
2.
We see that our RHS exponent is 11; therefore, we set our lowest exponent to 11. x-4 is certainly smaller than x-1 , so if we let x-4=11 we get: 214 - 211 = (23 )(211 ) - 211 = 8(211 ) - 211 = 7(211 ) 20% l 1.2l 10% b 0.9b New perimeter = 2 1.2l 0.9b
Increase in perimeter = 2 0.2l 0.1b Relation between l & b is not given We can’t find out the percentage increase/decrease 3.
If the average weight of the entire group was twice as close to the average weight of the men as it was to the average weight of the women, there must be twice as many men as women. With a 2:1 ratio of men to women of, 33 1/3% (i.e. 1/3) of the group must have been women. Consider the following rule and its proof. Alligation rule: The ratio that determines how to weight the averages of two or more subgroups in a weighted average also reflects the ratio of the distances from the weighted average to each subgroup's average.
4.
Let x be the speed of stairway 25 +15x = 13 + 24x 4 Or x 3 4 25 15 45 steps are in total 3
Email: [email protected], Website: www.engineeringolympiad.in 1
IN-IV Year Sample Paper
5.
Rate and time are always inversely related: AB's rate is 6/5, so in 1 hour, AB will do 5/6 of the job. AC's rate is 3/2, so in 1 hour, AC will do 2/3 of the job. BC's rate is 2, so in 1 hour, BC will do 1/2 of the job. Let the number of units in the total job be a number that is a multiple of 6, 3, and 2; let's say there are 18 units in the total job. Then, in one hour: AB will complete 5/6*18 = 15 units; AC will complete 2/3*18 = 12 units; BC will complete 1/2*18 = 9 units. Summing up: 2 As, 2 Bs, and 2Cs complete 36 units. So, in one hour, 2 of each of the pumps will complete two jobs. Therefore, it will take 1 of each of the pumps 1 hour to complete the job.
6.
It may be helpful to put the question in algebraic terms. The tip will be equal to a constant, c, plus an amount that is proportional to the bill: kb, where k is the fraction of the bill, and b is the amount of the bill. So the tip will be c+kb, and since we know the bill for the meal is 600/-, the tip will be c+600k. 60 = c+700k
40 = c+450k Subtract the equations, giving you the result: 20 = 250k k = 0.08 Then plug k back into one of the equations: 60 = c+700(0.08) c = 4 Therefore, tip = 4+600(0.08) = 52
7.
4 men can go in five hotels in 54 ways. Number of ways in which 4 men can go into different hotel 5! 5! = 5P4 5 4 ! Required proability
8.
5! 120 24 54 625 125
A B 40 AB A B AB 40 A 22 12 A 30
A=30 enrolled for English & included both subjects Number of students enrolled for English only = 30-12=18.
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IN-IV Year Sample Paper
9.
Let Mr. Vikas buys LCM (8, 5, 9) = 360 Apples of each variety. 360 Amount spent on the 1st variety = 45 rs. 8 360 Amount spent on the 2nd variety = 72 rs. 5 ∴ Total amount spent = 45+72 = Rs.117 Now the total (360+360) = 720 Apples are sold at 9 per rupee 720 ∴ Total revenue = 80 9 Hence the loss = 117-80 = 37 37 Loss% 100 31.62% 117
10.
Total respondents in the 21 – 30 age group = 33. Out of them 33 – 12 = 21 like any program other than singing/dancing 21 % 100 63.63 64% 33
15.
Junior/ Senior/ prior are followed by “to”
Section-II
1.
Expectation of X E x Average mean
n 1 n Xi Pi x i n 1 i 1
Where Pi is probability of occurrence of x i Let q 1 p Then probability of success in 1st, 2nd and 3rd trials are p,qp,q 2 p E x 0 p 1 qp 2 q 2 p .. qp 1 2q 3q 2 .... nq n 1 ...
2.
qp
1 q
2
qp q p2 p
As sensitivity is 1mm --------2 volts ? ---------12 volts(max) 6 mm So for 6 mm displacement, of cantilever x
Fl3 3EI
bt 3 10 103 (2 103 )3 6.67 1012 12 12 F(0.25)3 x F 1.383 N 3 180 109 6.67 1012
I
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IN-IV Year Sample Paper
3.
Porportional band
100 Kp
K p , controller gain
100 4 25
2K 2K From figure, Gain of OPAMP is 4 R1 R1 2000 R1 500 4
4.
G S
k K is negative S S T
G j 180 90 Tan 1 T 900 Tan 1 T
At 0 G j 900 and G j 00
5.
I xy x y dydx R
1
x
x y xy dydx 2
2
x 0 y x 2
x y 2 y3 2 x x dx 2 yx2 3 yx2 x 0 x
1
x4 x6 x4 x7 2 2 3 3 dx x 0 1
1
x5 x 7 x5 x8 3 10 14 15 24 x 0 56
6.
V t t for 0 t 1s it c
7.
dv t dt
2
dv t dt
2 0.8 1.6A
Capaci tan ce of cylinder, C
2x D ln 2 D1
D 2 cylinder inner diameter D1 piston outer diameter 137 1012
2 8.85 1012 x x 0.999 m 0.3 ln 0.2
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IN-IV Year Sample Paper
8.
Controllable: 1 1 A , 0 1
1 B , 0 1 1 1 0 0 1
1 AB 1
1 U B : AB 0
C 0
1
1 , U 1 0 1 0 1
Observability: 0 V CT ;A T CT ,CT , 1 1 1 0 1 A T CT 1 0 1 0 0 V 1
1 V 0 1 1 0 0 System is controllable & observable
9.
10.
10 V1 V0 V1 0 10 2V1 V0 ....... (1) 5 5 V V0 2V1 V0 V1 IL 1 0 IL 10 10 10 10 IL 1A 10 x P
2e j 1 0.25e j2
2e j 1 0.5eJ 1 0.5e j
x P
2 2 j 1 0.5e 1 0.5e j
DTFT x P x n 2 0.5 u n 2 0.5 u n pairs n
11.
0000 0000 0000 1111
FFFF
12.
n
FFFF
x n 1 x n xn
3x
2 n
FFFF FFFF
f xn ; f ' xn 2x n 1
6x n 2
3x n 2 1 6x n 2
Email: [email protected], Website: www.engineeringolympiad.in 5
IN-IV Year Sample Paper
13.
As there are no independent sources the Thevenin’s and Norton equivalent will have 0V and V 0A sources. To find RTH, a 1A source is connected as R TH x 1 Writing a nodal equation at n, Vx V V 1000 i x x x 100 3000 100
1
100
V Vx 1000 x Vx Vx 3000 1 100 3000 100 Vx i x 3000 V V 2 Vx 1 x x 100 3000 3 100 0.01Vx 0.0003Vx 0.0067 Vx 1
1000 ix
A
n ix
Vx 100
3k
Vx
1A
B
Vx 58.82 V R TH
Vx 58.82 I
k 2 103 . x (0.05) 4000m / s 2 0.025 m
14.
For Spring, F = kx = ma a
15.
Since the rate of transmission is R=105bits / second, the bit
Interval Tb is 10-5s. 2E b The probability of error in a binary PAM system is P e Q N 0 where E b A 2 Tb with P e P2 10 6 2E b 4.752 N 0 4.75 E b 0.112813 N0 2
Thus A 2 Tb 0.112813 A= 0.112813 105 106.21
16.
xe 0 y
x y2
dxdy
y 0
xe
xy
x2 / y
dx dy
x2 y x 2 2x y y y y e dx dy e dy = e y dy 2 2 2 y y 0 x y 0 0 xy
e y 1 e y 1 ye y 1 dy 0.5 2 1 0 0 1 2 1 0 2
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IN-IV Year Sample Paper N 1
17.
DFT X k x n W6kn n 0
for N=6; X(k)=4+3W6k 2W62k W63k
Y k W64k X k 4W64k 3W65k 2W66k W67k U sin g periodicity property, W66k 1; W67k W6k Y k 4W64k 3W65k 2 W61k N 1
Y k y(n)W6kn n 0
y(4) 4, y(5) 3, y(0) 2, y(1) 1 and y(2) y(3) 0 y(n) {2,1,0,0, 4,3}
18.
8 bit Johnson counter will divide frequency by 16 times 4 bit parallel counter will divide frequency by 16 times 8 bit ring counter will divide frequency by 8 times Input frequency = 16 × 6 × 8 × 10 = 20480
19.
At resonance Vc Qo Vs
20.
4x 2 dy 4x 2 dx y 2ln x c 3 x 3
22.
R 2r % strain during fibre optic bending is 1 100% Rr where R circumference bending radius r cladding radius
x 2(100) 0.25 1100 x 100 x 39900m 3.99
23. Time base Generator
y
sweep output
x(diagonal line)
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IN-IV Year Sample Paper
24.
Upper cross over voltage when, vo = +10v, At upper threshold point 5 10 20 10 2 Vth VTh 2V 5 20 10 20 10 20
Lower cross over voltage when vo = -10V 5 10 20 10 2 VTL VTL = -4V 5 20 10 20 10 20
26.
V1 Av2 B( I2 ) I1 CV2 D( I2 )
Making V2 0
Making I 2 0
V2 4 I1 0.1V2
I1 I 2 I or D 1 1 I2
V2 4I1 0.4V2 1.4V2 4I1
V1 5I1 0.3V1 0 1.3Vt 5I1 5I2 V1 5 B 3.846 1.3 I2 I 1.4 or C 1 0.35 V2 4 V1 5I1 0.3V1 V2 0 1.3V1 5 0.35V2 V2 0 1.3V1 2.75V2 V or A 1 2.11 V2 2.11 3.846 or 1 0.33 t
27.
k m d; =
d k km t f m t ,For given f M m dt 2
mean=0,S.D.=2; f rms
28.
k k 2 2
Find out the thevenin voltage across the galvanometer E.R 3 RR ER 4 R R E th ; R th 1 3 2 4 R1 R 3 R 2 R 4 R1 R 3 R 2 R1 Ig
E th R th R g
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IN-IV Year Sample Paper
29.
For the 1st stage alone, AV
0 R C ; r 0 125 25 3.125k R S r gm
A V1
125 1.2 40.3 0.6 3.125
For the 2nd stage, R S R C1 1.2 k & r is same R C2 1.2 k A V2
125 1.2 34.7 1.2 3.125
Overall voltage gain A V A V1 A V2 1400
30.
Applying Routh Hurwitz s5 s4 s3 17s2 90s 72 s5 s4 s3 s2 s1 s0
1 17
1 1 1 17 1
18
8 0 / 16 72
90 72 162 1 72 0
90 72 0
As auxiliary equation is 8s2 72 0 Differentiating we get - 16s = 0 Five roots in total, 2 on imaginary axis, one sign change on RHS, hence LHS = 5 2 1 2 .
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Answer Keys: Section-I 1
D
2
D
3
B
4
C
5
B
6
C
7
B
8
C
9
B
10
C
11
B
12
C
13
A
14
D
15
B
16
B
17
A
18
C
19
A
20
C
Section-II 1
C
2
A
3
A
4
A
5
D
6
A
7
C
8
D
9
B
10
B
11
A
12
C
13
B
14
A
15
A
16
A
17
B
18
A
19
D
20
A
21
C
22
A
23
C
24
A
25
C
26
C
27
D
28
B
29
A
30
A
Explanations: Section-I 1.
2.
We see that our RHS exponent is 11; therefore, we set our lowest exponent to 11. x-4 is certainly smaller than x-1 , so if we let x-4=11 we get: 214 - 211 = (23 )(211 ) - 211 = 8(211 ) - 211 = 7(211 ) 20% l 1.2l 10% b 0.9b New perimeter = 2 1.2l 0.9b
Increase in perimeter = 2 0.2l 0.1b Relation between l & b is not given We can’t find out the percentage increase/decrease 3.
If the average weight of the entire group was twice as close to the average weight of the men as it was to the average weight of the women, there must be twice as many men as women. With a 2:1 ratio of men to women of, 33 1/3% (i.e. 1/3) of the group must have been women. Consider the following rule and its proof. Alligation rule: The ratio that determines how to weight the averages of two or more subgroups in a weighted average also reflects the ratio of the distances from the weighted average to each subgroup's average.
4.
Let x be the speed of stairway 25 +15x = 13 + 24x 4 Or x 3 4 25 15 45 steps are in total 3
Email: [email protected], Website: www.engineeringolympiad.in 1
IN-IV Year Sample Paper
5.
Rate and time are always inversely related: AB's rate is 6/5, so in 1 hour, AB will do 5/6 of the job. AC's rate is 3/2, so in 1 hour, AC will do 2/3 of the job. BC's rate is 2, so in 1 hour, BC will do 1/2 of the job. Let the number of units in the total job be a number that is a multiple of 6, 3, and 2; let's say there are 18 units in the total job. Then, in one hour: AB will complete 5/6*18 = 15 units; AC will complete 2/3*18 = 12 units; BC will complete 1/2*18 = 9 units. Summing up: 2 As, 2 Bs, and 2Cs complete 36 units. So, in one hour, 2 of each of the pumps will complete two jobs. Therefore, it will take 1 of each of the pumps 1 hour to complete the job.
6.
It may be helpful to put the question in algebraic terms. The tip will be equal to a constant, c, plus an amount that is proportional to the bill: kb, where k is the fraction of the bill, and b is the amount of the bill. So the tip will be c+kb, and since we know the bill for the meal is 600/-, the tip will be c+600k. 60 = c+700k
40 = c+450k Subtract the equations, giving you the result: 20 = 250k k = 0.08 Then plug k back into one of the equations: 60 = c+700(0.08) c = 4 Therefore, tip = 4+600(0.08) = 52
7.
4 men can go in five hotels in 54 ways. Number of ways in which 4 men can go into different hotel 5! 5! = 5P4 5 4 ! Required proability
8.
5! 120 24 54 625 125
A B 40 AB A B AB 40 A 22 12 A 30
A=30 enrolled for English & included both subjects Number of students enrolled for English only = 30-12=18.
Email: [email protected], Website: www.engineeringolympiad.in 2
IN-IV Year Sample Paper
9.
Let Mr. Vikas buys LCM (8, 5, 9) = 360 Apples of each variety. 360 Amount spent on the 1st variety = 45 rs. 8 360 Amount spent on the 2nd variety = 72 rs. 5 ∴ Total amount spent = 45+72 = Rs.117 Now the total (360+360) = 720 Apples are sold at 9 per rupee 720 ∴ Total revenue = 80 9 Hence the loss = 117-80 = 37 37 Loss% 100 31.62% 117
10.
Total respondents in the 21 – 30 age group = 33. Out of them 33 – 12 = 21 like any program other than singing/dancing 21 % 100 63.63 64% 33
15.
Junior/ Senior/ prior are followed by “to”
Section-II
1.
Expectation of X E x Average mean
n 1 n Xi Pi x i n 1 i 1
Where Pi is probability of occurrence of x i Let q 1 p Then probability of success in 1st, 2nd and 3rd trials are p,qp,q 2 p E x 0 p 1 qp 2 q 2 p .. qp 1 2q 3q 2 .... nq n 1 ...
2.
qp
1 q
2
qp q p2 p
As sensitivity is 1mm --------2 volts ? ---------12 volts(max) 6 mm So for 6 mm displacement, of cantilever x
Fl3 3EI
bt 3 10 103 (2 103 )3 6.67 1012 12 12 F(0.25)3 x F 1.383 N 3 180 109 6.67 1012
I
Email: [email protected], Website: www.engineeringolympiad.in 3
IN-IV Year Sample Paper
3.
Porportional band
100 Kp
K p , controller gain
100 4 25
2K 2K From figure, Gain of OPAMP is 4 R1 R1 2000 R1 500 4
4.
G S
k K is negative S S T
G j 180 90 Tan 1 T 900 Tan 1 T
At 0 G j 900 and G j 00
5.
I xy x y dydx R
1
x
x y xy dydx 2
2
x 0 y x 2
x y 2 y3 2 x x dx 2 yx2 3 yx2 x 0 x
1
x4 x6 x4 x7 2 2 3 3 dx x 0 1
1
x5 x 7 x5 x8 3 10 14 15 24 x 0 56
6.
V t t for 0 t 1s it c
7.
dv t dt
2
dv t dt
2 0.8 1.6A
Capaci tan ce of cylinder, C
2x D ln 2 D1
D 2 cylinder inner diameter D1 piston outer diameter 137 1012
2 8.85 1012 x x 0.999 m 0.3 ln 0.2
Email: [email protected], Website: www.engineeringolympiad.in 4
IN-IV Year Sample Paper
8.
Controllable: 1 1 A , 0 1
1 B , 0 1 1 1 0 0 1
1 AB 1
1 U B : AB 0
C 0
1
1 , U 1 0 1 0 1
Observability: 0 V CT ;A T CT ,CT , 1 1 1 0 1 A T CT 1 0 1 0 0 V 1
1 V 0 1 1 0 0 System is controllable & observable
9.
10.
10 V1 V0 V1 0 10 2V1 V0 ....... (1) 5 5 V V0 2V1 V0 V1 IL 1 0 IL 10 10 10 10 IL 1A 10 x P
2e j 1 0.25e j2
2e j 1 0.5eJ 1 0.5e j
x P
2 2 j 1 0.5e 1 0.5e j
DTFT x P x n 2 0.5 u n 2 0.5 u n pairs n
11.
0000 0000 0000 1111
FFFF
12.
n
FFFF
x n 1 x n xn
3x
2 n
FFFF FFFF
f xn ; f ' xn 2x n 1
6x n 2
3x n 2 1 6x n 2
Email: [email protected], Website: www.engineeringolympiad.in 5
IN-IV Year Sample Paper
13.
As there are no independent sources the Thevenin’s and Norton equivalent will have 0V and V 0A sources. To find RTH, a 1A source is connected as R TH x 1 Writing a nodal equation at n, Vx V V 1000 i x x x 100 3000 100
1
100
V Vx 1000 x Vx Vx 3000 1 100 3000 100 Vx i x 3000 V V 2 Vx 1 x x 100 3000 3 100 0.01Vx 0.0003Vx 0.0067 Vx 1
1000 ix
A
n ix
Vx 100
3k
Vx
1A
B
Vx 58.82 V R TH
Vx 58.82 I
k 2 103 . x (0.05) 4000m / s 2 0.025 m
14.
For Spring, F = kx = ma a
15.
Since the rate of transmission is R=105bits / second, the bit
Interval Tb is 10-5s. 2E b The probability of error in a binary PAM system is P e Q N 0 where E b A 2 Tb with P e P2 10 6 2E b 4.752 N 0 4.75 E b 0.112813 N0 2
Thus A 2 Tb 0.112813 A= 0.112813 105 106.21
16.
xe 0 y
x y2
dxdy
y 0
xe
xy
x2 / y
dx dy
x2 y x 2 2x y y y y e dx dy e dy = e y dy 2 2 2 y y 0 x y 0 0 xy
e y 1 e y 1 ye y 1 dy 0.5 2 1 0 0 1 2 1 0 2
Email: [email protected], Website: www.engineeringolympiad.in 6
IN-IV Year Sample Paper N 1
17.
DFT X k x n W6kn n 0
for N=6; X(k)=4+3W6k 2W62k W63k
Y k W64k X k 4W64k 3W65k 2W66k W67k U sin g periodicity property, W66k 1; W67k W6k Y k 4W64k 3W65k 2 W61k N 1
Y k y(n)W6kn n 0
y(4) 4, y(5) 3, y(0) 2, y(1) 1 and y(2) y(3) 0 y(n) {2,1,0,0, 4,3}
18.
8 bit Johnson counter will divide frequency by 16 times 4 bit parallel counter will divide frequency by 16 times 8 bit ring counter will divide frequency by 8 times Input frequency = 16 × 6 × 8 × 10 = 20480
19.
At resonance Vc Qo Vs
20.
4x 2 dy 4x 2 dx y 2ln x c 3 x 3
22.
R 2r % strain during fibre optic bending is 1 100% Rr where R circumference bending radius r cladding radius
x 2(100) 0.25 1100 x 100 x 39900m 3.99
23. Time base Generator
y
sweep output
x(diagonal line)
Email: [email protected], Website: www.engineeringolympiad.in 7
IN-IV Year Sample Paper
24.
Upper cross over voltage when, vo = +10v, At upper threshold point 5 10 20 10 2 Vth VTh 2V 5 20 10 20 10 20
Lower cross over voltage when vo = -10V 5 10 20 10 2 VTL VTL = -4V 5 20 10 20 10 20
26.
V1 Av2 B( I2 ) I1 CV2 D( I2 )
Making V2 0
Making I 2 0
V2 4 I1 0.1V2
I1 I 2 I or D 1 1 I2
V2 4I1 0.4V2 1.4V2 4I1
V1 5I1 0.3V1 0 1.3Vt 5I1 5I2 V1 5 B 3.846 1.3 I2 I 1.4 or C 1 0.35 V2 4 V1 5I1 0.3V1 V2 0 1.3V1 5 0.35V2 V2 0 1.3V1 2.75V2 V or A 1 2.11 V2 2.11 3.846 or 1 0.33 t
27.
k m d; =
d k km t f m t ,For given f M m dt 2
mean=0,S.D.=2; f rms
28.
k k 2 2
Find out the thevenin voltage across the galvanometer E.R 3 RR ER 4 R R E th ; R th 1 3 2 4 R1 R 3 R 2 R 4 R1 R 3 R 2 R1 Ig
E th R th R g
Email: [email protected], Website: www.engineeringolympiad.in 8
IN-IV Year Sample Paper
29.
For the 1st stage alone, AV
0 R C ; r 0 125 25 3.125k R S r gm
A V1
125 1.2 40.3 0.6 3.125
For the 2nd stage, R S R C1 1.2 k & r is same R C2 1.2 k A V2
125 1.2 34.7 1.2 3.125
Overall voltage gain A V A V1 A V2 1400
30.
Applying Routh Hurwitz s5 s4 s3 17s2 90s 72 s5 s4 s3 s2 s1 s0
1 17
1 1 1 17 1
18
8 0 / 16 72
90 72 162 1 72 0
90 72 0
As auxiliary equation is 8s2 72 0 Differentiating we get - 16s = 0 Five roots in total, 2 on imaginary axis, one sign change on RHS, hence LHS = 5 2 1 2 .
Email: [email protected], Website: www.engineeringolympiad.in 9
