downloads/papers/EE IV Year IEO Paper I EXP

EE-IV Year Sample Paper

|IEO-2016|

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

C

3

A

4

A

5

D

6

D

7

D

8

D

9

B

10

A

11

A

12

C

13

B

14

C

15

C

16

A

17

B

18

A

19

A

20

A

21

C

22

A

23

D

24

A

25

D

26

C

27

B

28

D

29

A

30

D

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

EE-IV Year Sample Paper

5.

|IEO-2016|

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 AB  A  B  AB 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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EE-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

|IEO-2016|

 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 

kVA 

2



qp q  p2 p

992  1240 kVA 0.8

1240  108.47 A 3  6.6 6.6  103 Phase Voltage   3810.51V 3

Current I 

E ph  (Vph cos   Ia R a ) 2  (Vph sin   I a X s ) 2  3212.28V  Regulation 

3212.28  3810.5  15.7% 3810.5

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EE-IV Year Sample Paper

3.

|IEO-2016|

At A, C1 1  C 6 I 2  i1  I1 K

C KC

V2 c  V1c  V1(kc)

I1

i1

V2  V1  V1k  V1.k  V1 (1  k)

V1

A

I2

V2  V1 (1  k)

At B i3  i 2  I2

V2

C

KC

B

V3c   V1  V2  kc  V2 c

i2

V3   V1  V2  k  V2

V3

V3  V1.k  V2 (1  k)  1 7 V2  V1 1    v1  1.16 V1  6 6 V 49 1  1 V 7 7  1 49  V3  V1    V2 1    1  .v1  1  V1  v1    6 36 6  6 6 6 6  6 36   55  V3  V1    36  Given V3  20kv  55   20  V1    V1  13.09 KV  36   1 V2  V1 1  k   13.09 1    6 V3  20kv Total voltage across string  V1  V2  V3  48.36KV

4.

G  S 

k  K is negative S S  T 

    G  j  180  90  Tan 1    900  Tan 1   T T

At   0  G  j  900 and     G  j  00

5.

I   xy  x  y  dydx  R

1

x

   x y  xy  dydx 2

x 0 y  x

2

2



x   y 2  x   y3  2 x  x x 0   2  2  3  2  dx  yx yx  



 x4 x6 x4 x7  x5 x 7 x5 x8 3    dx  x 0  2 2 3 3  10  14  15  24  56 x 0

1

1

1

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EE-IV Year Sample Paper

6.

V  t   t for 0  t  1s it  c

7.

dv  t  dt

 2

dv  t  dt

 2  0.8  1.6A

Time period T=ton+toff= (1+1.5)=2.5msec, T 1 Duty cycle   on   0.4 T 2.5 Form-factor (FF)  

 .E dc RMS Value  Average Value .E dc 1 



1 0.4

 1.58

Ripple factor (RF)  (FF)2  1  

8.

|IEO-2016|

Controllable: 1 1 A , 1 0  1 AB   1

1 1 

1  1  0.4   1.23  0.4

1  B   , 0 

C  0

1

1  1  1      0  0  1

1 U   B : AB   0 Observability:

1 , U 1 0 1  0 1

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  V1 V0  V1  0 5 5  10  2V1  V0 ....... (1) V  V0 V1  IL  1 0 10 10  10  2V1  V0  IL   IL   1A 10 10

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EE-IV Year Sample Paper

10.

|IEO-2016|

The turn on loss  1  vdc  Io  Ton 6  Turn on energy loss  1  v dc  I o  Ton 6 Turnoff energy loss  1  v dc  I o  Toff 6  Swtching loss  1  100  20  0.1  106  1  100  20  0.1  10 6  66.67  10 6 J 6 6





conduction energy loss  v o  I o   I o   R d  0.1  10 3  [1x20  (20) 2 xR d ]x104

. Avergage power loss 

2

 66.67 10

6

  20  400R d   104  0.2  10

3



T JC

 66.674  106  20  104  400R d  104  100  0.2  103  R d  0.448

11.

0000 0000 0000 1111

 FFFF

FFFF

12.

x n 1  x n 

f  xn  ; f ' xn 

3x 

 2x n  1

xn

13.

FFFF FFFF

2 n

6x n  2



3x n 2  1 6x n  2

As there are no independent sources the Thevenin’s and Norton equivalent will have 0V V and 0A sources. To find RTH, a 1A source is connected as R TH  x 1 Writing a nodal equation at n, 100 A

n

V V V  1000 i x 1 x  x  x 100 3000 100 1

Vx V  x  100 3000

Vx  1000



1000 ix 

ix

Vx 100

3k

Vx 3000

 Vx

1A



B

100

Vx    i x   3000   1

Vx V 2 Vx  x  100 3000 3 100

 0.01Vx  0.0003Vx  0.0067 Vx  1  Vx  58.82 V  R TH 

Vx  58.82 I Email: [email protected], Website: www.engineeringolympiad.in 6

EE-IV Year Sample Paper

14.

15.

y

sweep output

Time  base Generator

|IEO-2016|

 

x(diagonal line)



  30 3vms 3 2 .440 cos   cos30  514.59V   V0  E  IO R V0 

V0  E 514.59  300   10.72A R 20 Power delivered to load  V0 I 0   514.59 10.72   5521.54 W

 I0 



16.

 xe





x y2

dxdy 



y 0

0 y



xe

 x2 / y



dx dy

xy

 2       y  x 2  2x      x    y y      e y  dx   dy    e y  dy =  e  y dy 2 2  2    y    y 0  x  y  0 0   xy    

    e  y   1  e  y  1   ye  y  1    dy          0.5 2   1  0 0  1   2  1  0 2  

17.

 2Z2   2  50  V"  V    100    22.22kV  400  50   Z1  Z2 

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.

y  0.02  j0.08

yAB  y BA  y

y BC  yCB  y

yCA  yAC  y

y AA  y  y  2y

 y AA y  BA  yCA

y AB y BB yCB

A

B

y AC   2y  y  y  yBC     y 2y  y  yCC    y  y 2y  C

YBus

 0.04  j0.16 0.02  j0.08 0.02  j0.08    0.02  j0.08 0.04  j0.16 0.02  j0.08   0.02  j0.08 0.02  j0.08 0.04  j0.16  Email: [email protected], Website: www.engineeringolympiad.in 7

EE-IV Year Sample Paper

|IEO-2016|

4x 2  dy    4x 2  dx  y  2ln x  c 3 x  3

20.

21.

To avoid phase displacement, the connection of power Transformer and CT’s as shown below: PT’s CT’s  If  If

22.

Rotational loss = 500W Field circuit loss = 200  2w  400w

2A

69A

Constant loss =  500  400 w  900w

200V

For rated load

E b Ia  10  103 

  200  0.8Ia  Ia  10000  Ia  69A

[ Ia  181A is not feasible as it will result in huge losses ] So input current = 71 A  71  200w  14.2kW  10  Efficiency =   100  %  70.4%  14.2 

23.

dt  d1 d 2

24.

P 895  5 Nm m 1710  2n 60



Va1 2 Va 2 2

 Va  d 2   2  Va  1

2

 2  d1   0.9   5  4.05 Nm  

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 

C1

25.

A2

A2

A1

C1 C2

B1

B2

A1 B1

B2

Y11

C2

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EE-IV Year Sample Paper

26.

|IEO-2016|

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 or  1.3  I 2  V1  5I1  0.3V1  V2  0

  I1 1.4   0.35 C  V2 4  

1.3V1  5  0.35V2  V2  0 1.3V1  2.75V2    2.11 3.846  V or  A  1  2.11 or  1  V2 0.33  

27.

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 

28.

E th R th  R g

First multiplier output=10cos4000 t cos 4000t is shifted by 90o  cos  4000  90  t  sin 4000 t  s econd multiplier output  10cos 4000t sin 4000t y1 (t)  5sin 8000 t 5     8000       8000  j  y1(t) is passed through an LPF of cut off frequency 3kHz. y1 (i ) 

y1(j ) 4

f(kHz)

4

LPF

3

3

f(kHz)

On multiplying above two signals the output y(t) is zero.

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EE-IV Year Sample Paper

29.

|IEO-2016|

For the 1st stage alone,  R  A V  0 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.

Assume H for r>b be Hex 2r H ex  current enclose  k.(b 2  a 2 ) H ex 

k(b 2  a 2 ) 2r

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