General Aptitude

4) The numbers that are multiples of 10 are

10,20,30,..190.

Average = 10 20 30 ..... 190

19

=

10 1 2 30 ..... 19 10 19

19

20

2 19 100

Sum of n natural number (1 ----n) = n n 1

2

5) x 12 12 12 ........

x 12 x

2x x 12 0

2x 4x 3x 12 0

(x 4)(x + 3) = 0

x = 4.000

8) if 2x 2x 3 11

2x 2x 3 11 or 2x 2x 3 11

2x 2x 8 0 or 2x 2x 14 0 4 4 14 < 0 2b 4ac 0

2x 4x 2x 8 0 imaginary roots

(x 4) (x + 2) = 0 (not possible as x is real)

x = 4 or x = 2

3 2x 4

x x x 64 16 4 52

233 2

x 2x x x 2 2 2 8 4 2 14

9) Let number of male students = b

Number of female student = g

2008

b 25g

No. of girls in 2009 = 2g

Boy = x

b2g

= 3 x = 6g

Initial no. of boys = b = 2.5g

Increase = 6g 2.5g = 3.5g

% increase = 3.5g

100 140%2.5g

10) When minute hand makes a complete revolution if 360 in 60 minute hour hand moves by 30

Initial angle = 180

angle after x minutes

Angle Moved Angle moved byby minute hand hour hand

360 30180 X + X

60 60

Solving, we get x = 21.8 minutes

Approximate max , time is 6.22 a.m.

Electrical Engineering

1) All eigenvalues of a real symmetric matrix are real and that of a skew symmetric matrix

are purely imaginary.

2) P n dots n

P n dots kn

dice has total of 6 faces

P total = 1

K(1) + K(2) + K(3) +K (16) = 1

K =

1 1211 2 3 4 5 6

3 1P 3 dots K 3 0.1421 7

3) For minimum

f ' x 0 & f " x 0

f(x) = 2

3x 1

1

32

f ' x x 1 03

for all x

This function is cube root of 2

x 1 which will always be positive & hence minimum

value can only be 0

minimum occurs at x = 1 and minimum value is 0 .

4)

i

i 0n 1 n 1i1 1i 0 ee e e

So values are real & non Negative.

5) Assume my x

m 1dy

mxdx

2

m 2

2

d ym m 1 x

dx

Substituting y = mx in differential equation, we get

2 m 2 m 1 mx m m 1 x x mx yx 0

mx m m 1 m 1 0

2m m m 1 0

m 1

Solutions = x and 1x

6) There are two possible connections

Leq(1)= 1 2L L 2M

Leq(2) = 1 2L L 2M

Leq(1) + Leq(2) = 2 1 2L L

= 380 + 240 = 620 H

1 2

L L = 310 H

2 M = 380 310 = 70 H

M = 35 H

7) C o1O

C o2

t2RC

C C C Ct O e

t t2RC 2RC

C 02 01 02 02 01 01t e 1 e

8) This combination can be considered as a series combination of three capacitors.

0 1 0 11

A 4AC

d d4

0 2 0 22

A 2AC

d d2

0 3 0 33

A 4AC

d d4

Voltage across 1

C & 3

C each is 2V

1 3 0 1eq1 3

C C 2AC

C C d

Voltage across eqC = 4V

0 1eq eq2A

Q C V 4d

Voltage across 2

C = 10 4 = 6 V

Q =

0 22A

6d

Q = eqQ (due to series combination)

0 2 0 1

2A 2A6 4

d d

1

2

32

9)

H1

jwjw 4 jw

A = l H w 3

jw = 2

1 1

3 16 916

A = 115

10) L(t) = 5e tu t

H(S) =

1

S 5

y(t) = 3 5te tu t e u t

y(S) =

Y S

2S 3H S

3tx t 2e u t

11) SP

S

P

NN

V

V

S

V = 2

1 sin wt = 2 sin wt

2

SPS

P

NZ Z 4 1 4

N

12) low voltage 1000

Khigh voltage 1100

KVA rating = 1

1 k

(KVA rating of 2- winding Xmer) =

150 550 KVA

10001

1100

13) No. load frequency is 50Hz

If the generator is loaded, the frequency decreases

1 2f , f 50Hz

If some load is removed, p decrease & hence frequency increases

2 1f f

14) Capacitor does not dissipate real power and hence P = 12 MW

i i

Q P tan

1i

cos 0.6

i

4Q 123

= 16 MVAR

f f

Q P tan

1f cos 0.8

f

3Q 124

= 9MVAR

Cif

Q Q Q = 7 MVAR

ve sign indicates that it is a capacitor

C

Q 7MVAR

15) Base voltage =

6Base MVA 100 10132kV

3 Base current 3 437.38

Load voltage = Pu voltage Base = 132 0.9 = 118.8 kV

16) Under light load conditions, the load side voltage becomes more than sending end due

to line charging, it is called as Ferranti effect.

So, to avoid this effect we use shunt reactors which can act as load under light loaded

condition.

17) 2

4T S

S 0.4S 4

4 S 0.4ST S41 S 0.4

S

Open loop transfer function, assuming unity feed back

G(S) =

4

S S 0.4

KP = error constant for unit step input

System is type 1 and hence error due to unit step input is 0.

18) 11E L S I A

S 1 0

S I A-1 S-1

1

2

S 1 01S I A

1 S 1S 1

=

2

1 0S 1

1 1

S 1S 1

t

t t

e 0E

te e

19) Moving iron voltmeter measures the rms value of waveform

T = 20 ms, V = 100 V

Equation of line = V

tT

rms value =

2T T22

20 0

1 Vt 1 Vdt t dt

T T T T

= 3

T2

30

V t

3T

= 2

3

V

T

3T

V

3 3

Vrms = V 100

57.73V3 3

20) Pf angle 1 2 1

2 1

P P = tan 3

P P

= 1250 100

tan 3250 100

Pf = cos = 0.8

21) The count in a BCD counter cannot go beyond the decimal value 9 and hence (D) whose

decimal equivalent is 12 cannot be a BCD counter output.

22) B BES S

V I R V

5 = BI 2k 0.7

BI = 2.15 mA

For limiting case in active region CE

V = CE

V (sat)

CC C C CEV 5V I R V sat

BC

I I = 2.15 100 = 0.215 A

5 = 0.215 C

R + 0.2

C

R = 22.32

23) The voltage across the load looks like as shown this is rectified output For limiting case,

maximum current through zener diode

LR

Vzener = 5V

Vmax = 20 V = Vzener + RSV

20 = 5 + I S

R

to avoid burn out Izener < P 1V 20

A

if LR , RSI = Izener

15 = S S

1 R R = 30020

24) output of a step-up chopper

0 i

1V V

1 D

400 =

1250

1 D

(1 D) = 58

D = 38

Off time = ( 1 D)Ts

20S

5s T 32 s8

Switching frequency = S

1T

= 31.25 kHz .

25) In low frequency region, voltage is also less and stator resistance consumes a significant

part of that voltage and hence air gap voltage reduces.

vf

and hence flux decreases from rated value.

26) y

v y 2v;2

y y

u x x = u + = u + v2 2

Jacobean =

x x 1 1u v

= 2y y 0 2

u v

f x, y dx dy f u, v J x, y u v 2u du dv

Y varies from 0 to 8 v varies from 0 to 4 y

v2

X varies from y y

12 2 u varies from 0 to 1

yu x

2

integral becomes = 4 1

0 0

2u du dv

27) The PDF looks like as shown

The shaded region is the area of interest

P (0.5 < x < 5) = Area of shaded region

= 0.5 0.2 + 0.1 3 + 0 1

= 0.1 + 0.3 = 0.4

28) 3 2f x x 3x 24x 100

For minima

f ' x 0

3f ' x 3x 6x 24 0

2x 2x 12 0

The roots of this equation are 1

x = 4.605, 2

x = 2.605

1

x xf x

= 23.52

2

x xf x

= 124.486

x 3f x = 118

x 3f x = 28

1

x does not belong to [-3, 3] & hence minimum value is at x = 3

f(x)min = 28

29) Voltage across diode branches if braches are open circuited

Th

iV

V2

The same voltage appears across both diodes branches for diode connoted to 1V source,

it will conduct iff

iV

1 02

iV 2V (waveform will be clipped)

For diode connected to 2V source, it will conduct iff.

2 iV

2 > 0

iV

2 02

i

V < 4V

So i

V must lie outside ( 4, 2) V range for output to be clipped.

30) Solving using super position theorem

Neglect current source

C

1X 0.1

10 1

;

CC

C

j XV 20

R j X

=

0.1 j20

1 0.1 j = 2 84.29

1

A 2 ; 1

84.29

Neglect voltage source

C

1X 0.20,

5 1

LX 5 1 5

SC

C

I R 10 1I 9.805 11.8

1 0.2jR j X

C C CV I jX 1.98 78.69

2

A = 1.98 , 2

D = 78.69

31) Power loss in 1 = 2

2 1 4W

Power loss in R = 1 kW 4W = 996W

Assuming no current goes into voltmeter, CI 10A

2

RP 10 R 996

R = 9.96

L

V 200V , LI 10A

L

LL

V 200Z 20

I 10

2 2LL

X Z R 17.34

32) B = 0 H

Magnetic flux density due to current along x axis =

0 04 4

k2 1 2

Density due to current along y-axis =

0 02

k2 2 2

B Total =

03

k2

totalH = 3 Ak

m2

33) Transfer function = ZI A

Characteristic equation =

Z a a -1 z a a 10

0 a 1 Z a a 1 z a

If a = 1

z 1 00

2 z 1

2

z 1 0

Z = 1 j0

34) x t 2 5sin 100 t

On sampling t = S S

s

1 1nT T

f 400

x n 2 5sin 100 2 5sin400 4

N 1N

n1

n 0

y z 1 1 z 1H z z

N NX z 1 z

h(n) =

1 0 n N 1

N

0 Otherwise

in our case, N = 400

850

samples/ cycle

y(n) = x(n) * y(n) ;

x(n) =

5 5 5 52, 2 , 2 , 2 , 2 5 , 2 ,2,............2 2 2 2

when we take convolution ,

the integration of sine term will come out to be zero and for constant term.

y(n) = N 1 7

n 0 n 0

1x k h n k 2 28

35) Due to even nature of square wave only cosine terms will be present in Fourier series.

Due to half wave symmetry only odd harmonics will be present frequency of

harmonics = 10kHz, 80kHz, 50kHz..

fundamental

The filter frequency response is centered around 30 kHz

& band width also does not allow any other harmonics.

So, output is 0 nearly perfect cosine wave at 30 kHz.

36) If it operates as a generator

L faI I I

t

ff

V 250I 2A

R 125

aI = 50 + 2 = 52 A

Eg = tV + aI aR = 250 + 52 0.6 = 281.2V

As a motor

a L fI I I

= 50 2 = 48A

at abE V I R = 250 48 0.6 = 221.2V

Since f

I is constant b

N E

bm

g g

EN 221.2

N E 281.2

g

m

N 281.21.27

N 221.2

37) If rotor is rotated anti-clockwise , stator magnetic field & rotor rotate in opposite

direction.

s rN N

SNs

f =fs + fr = f + fr

So, frequency of voltage across slip rings is (f + fr) Hz.

Frequency of voltage across commutator brushes is same as starter frequency = fHz

38) For a single phase connection

Y = 180 180

20no. of slots/pole 180 / 20

no. of slots 180

q 9no. of pole no. of phase 1 20

Distribution factor = kd =

9 20y sinsin q2 20

0.6398y 20q sinq sin 22

No. of turns = no. of slots no. of conductor

2 =

180 6

2

= 540

For a three phase connection

Y = 20

q = no. of slots 180

no.of poles no.of phase 3 20

= 3

distribution factor = kd =

20y sin 3sin q2 2

y 20q sin 3 sin2 2

= 0.9597

no. of turns = no. of slots no. of conductor

2

1

no. of phase = 180

E = 4.44 kd fN

kd = distribution factor

f = frequency

N = no. of turns ; flux

1 1

3 2

E V 4.44 0.6398 f 540

E V 4.44 0.9597 f 180 = 2

39) Hysteresis loss = 1.6h h

P k Bm f

Eddy current loss = 2 2e2P k Bm f

VBm f

Since voltage & frequency are both increased by the same amount.

mB = constant

h

P f ; 21

ff

= 1.1

1 = 1.1

2

2e2 2

e1 1

P f1.1 1.21

P f

h2 h1

P 1.1P e2 e1

P 1.21P

% change = h2 h1

h1

P P100

P

% change = e2 e1

e1

P P100

P

= h2 h1

h1

1.1P P100

P

= e2 e1

e1

1.21P P100

P

= 10% = 21%

40) tf

S

E VP sin

X

1.3 1

0.6 sin1.1

1.1 6

sin 0.50761.3

30.51

t

S

fE V 0 1.3 30.51 1 0

I 0.6098 10.305jX j 1.1

S VI* 0.6098 10.305

Q = 0.6098 sin (10.305) = 0.11 pu

41) 1 2

P P 2.5MW (i)

For generator -1

1

51.5 f1

P

f = (51.5 - 1

P ) (ii)

Similarly for generator-2

f = (51 2

P ) (iii)

from (ii) & (iii)

51.5 - 1

P = 51 - 2

P

1 2

P P = 0.5 MW .(iv)

From (i) & (iv)

1

P = 1.5 MW ; 2

P = 1MW

f = 50Hz

41) Capacitance of line of earth =

0

n

2

Dr

3

n

2

2 8.854 10C

6n

0.8 10

= 8.4 10 12 F/m

nC = 8.4 10 9 F/km

43) For a single line to ground fault

1

af

2 0

3EI

X X X

In pu values

f

3 1I 6.67pu

0.2 0.2 0.05

6

base 3

100 10I 2309.40A

3 25 10

basef fI I pu I 15,396.7A

44) Characteristic equation of system

1 + G(S) = 0

2S S 2 S 2S 2 k 0

4 3 2S 9S 6S 4S k 0

Applying Routh- Hurwitz criterion

4S 1 6

3S 4 4

2S 5 k

120 4k

S5

0S K

For marginally stable system

20 4k = 0 K = 5

45) Constant gain term

slim lim sG s =

s 0 s 0

5 s 4s

2

3.2s 0.25 s 4s 25

Corner frequency term

4 (s+4)

0.25 (S+0.25)

5 = nw 2S 4s 25

Hence maximum corner frequency is 5rad/s

46) Controllability Matrix :

C = [Q : PQ]

Q = 0

1

; 1 1 0 1

PQ 0 -3 1 3

C = 0 1

1 -3

det (C) = 3 0 1 = 1 0 hence controllable

Observability Matrix

O = [R : RP]

R = [0 1] : RP = [0 1] 0 1

1 -3

= [0 , - 3]

O = 0 1

1 -3

, det (O) = 0 not observable

47) For parallel combination

21eq1 2

R RR

R R

421R R 200 300 6 10

Tolerance of 21R R 2% (fractional errors are added in multiplication)

21R R 200 300 500

Tolerance of 21R R = 1% (absolute errors are added)

Tolerance of eqR = 2% - 1% =1% (in division fractional errors are subtracted)

48) For Ammeter X

m m shI R I Im R

0.15 1.2 = (15 0.15) sh

R

sh

R = 0.01212

For Ammeter Y

m m shI R I Im R

0.25 1.5 = (15 0.25) sh

R

sh

R = 0.02542

When both are in parallel

X

x

y

y

R 0.025I I 15 10.1A

R R 0.025 0.012

49) During t1 , positive cycle

0

V 5V

D1 gets turned ON

V = 0V 1

4

= 1.25 V

Capacitor starts getting charged

V 5V

V(0) = ?

During negative cycle, 0

V 5V , D2 is ON

V = 0V 1

2

= 0

V2

= 2.5 V

for positive cycle V(0) = 2.5 V

1t /RC1t 1.25 V V 0 V e

1t /RC3.75 1

e7.5 2

During negative cycle

2t 2.5V

V(0) = 1.25 V ; V 5V

2.5 = 5 + (1.25 (5 )) 2tRCe

2tRC2.5 2

56.25e

2

1

1 2

tRC

RCt

RC

t t

0.8e

e

e

50) F(A,B,C,D) = 0, 1, 3, 7, 11

= (2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15)

= B C A C A B C D

51) Finding the next state function for each of the option

For option (B)

T = (J+Q).(K+ Q )

nn 1Q T Q

= JQ KQ (same as JK flip flop)

52) After first instruction

A 00H, CY 0

After second instruction

B 04 H

After third instruction A 03H

Loop : 1st iteration

A 01H; CY 1 , B 03H

2nd iteration

A 80H; CY 1 , B 02H

3rd iteration

A 30H; CY 0 , B 01H

4th iteration

A 40H; CY 0 , B 00H

Hence (A) is correct

53) For a single phase full bridge converter

Distortion factor = 2 2

= 0.9

Displacement power factor = 3cos cos302

Input power factor = D.F (DPF)

= 0.9 32

= 0.78

54) Voltage across the device is zero if 0 i

V V

That is output voltage is perfectly sinusoidal & same as input voltage

It happens at a pf angle = 1wL

tanR

= 12 50 0.016

tan 455

If input & output voltages are identical , then 0 rms

V = 230 V

0rms0 rms 2 2

V 230 230I = =

Z 5 25 2 50 0.016

0rmsTi rms 2 2

V 230 230I = =

Z 5 25 2 50 0.016

ormsTi rms

II

2 (as each thyristor conduct for half cycle)

Ti rms

230I

5 2 2 = 23A

55) L

tSL

Vt 1 e

R

6

L

100 50 1050 s 1 e

500

310500

200 31023.5mA

L 50 s 23.5mA

R LlatchingI I = 40 23.5 = 16.5 mA

R = SR

Vi

= 6060.08 = 6.06 k