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Answer Keys
1

757

2.78

400

0.67

10

12.34

11

8.34

12

0.912

13

14

15

16

17

18

19

20

21

16

22

23

24

25

26

27

0.96

28

29

30

31

12

32

33

34

35

36

37

38

39

4.8

40

41

42

43

44

14.4

45

46

47

48

49

50

51

52

0.252

53

54

0.5

55

63

56

57

58

59

60

61

62

22.77

63

64

65

Explanations:1.

The constant term in the characteristic polynomial is

A = ( 2 )( 2 )( 8 ) i.e., 1 . 2 . 3 = 32

2.

z i =1
x + iy i = 1
x + (y 1)i = 1
x 2 + (y 1)2 = 1
The pointed lie on a circle centered at (0,1) with radius 1

3.

Ordering the data from least to greatest $27, 000, $29, 000, $30,000, $31,000, $32,000,
$35,500, $40,000, $43,000
th

th

n
n
term + + 1 term
2
2
median =
. observation
Here n=8,
2
4 th + 5th $31, 000 + $32,000
=
=
= $31,500
2
2

4.

1
= 1 [sin at at cos at ]
We have L1
2
( s 2 + a 2 ) 2a 3

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1 1
= [sin t t cos t ]
L1 2
( s + 1) 2

The probability that state is empty is 1 f ( E,T ) = 1

5.

1
E EF
1 + exp

KT

Where K = Boltzman constant = 8.62 105 eV / k

0.01 = 1

1
0.3
T=
= 757 K
8.62 105 x4.59
0.3
1 + exp

KT

1
1
=
= 2.777Coulomb / m3
R H 3.6 101

6.

Ch arg e density =

8.

1
L
1
= tan 1
and = tan

R
CR
1
hence circuit is under resonanace
LC
V2
So power dissipated is P =
= 400W
R
If = , then 2 =

9.

An ideal Ammeter has zero resistance.

Therefore 5 across AB is shorted. I total =

10.

10
= 1.33 Iammeter = 0.67A
7.5

R L = 3k; ( Pac )D = 0.9W; ICQ = 30mA ( without signal )

ICQ + Bo = 33mA ( with signal ) ;The increase is due to harmonic content in the signal.
B0 = 3mA; B2 = B0 = 3mA
B 2
1
( Pac )D = Pac 1 + D 22 = B12 R L 1 + 22
2
B1
2
1
1
0.9 = B12 ( 3000 ) + ( 3 103 ) 3000 B1 = 24.3mA
2
2
B2
3
D2 =
=
= 0.1234 = 12.34%
B1
24.3

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11.

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Change in Zener voltage, Vz = Vz TC T = ( 8.2 )( 0.0005 )( 60 25 ) = 144mV

VZ at 60O C = VZ + Vz = 8.2 + 144mV = 8.34V

12.

Current through R; IR = IC + 2IB =

Substitution of I B =

VCC VBE
R

IC
f VCC VBE
and solving for IC =
= 0.912 mA
F
f + 2
R

14.

In a n-bit weighted resistor DAC, if the MSB resistance value is R then the LSB resistance
value is 2n 1 R .

15.

n-bit ring counter is a MOD-n counter. So a 6-bit ring counter has 6 states.

17.

a n u ( n + 2 ) exist for n = 2, 1 ; Thus it is non-causal as it is not zero for n<0

20.

f LO = f s + f IF ;Local oscillator : 1500 to 2000 MHz 1500 : 2000 = 1:

21.

fs =

4
or 1:1.33
3

2x5x106
= 2x106 Hz
5

Hence bit rate = 8f s = 16Mbps

22.

L1 = G 3 H1 ; L 2 = G1H 2 ; L3 = G1G 2 G 3G 4 H 3 ; L 4 = G1G 5 G 4 H 3

P1 = G1G 2 G 3G 4 ; P2 = G1G 5 G 4

26.

h = 0.1, x 0 = 0, y0 = 1 and f ( x, y ) = xy 2
The Euler 's iteration formula is
yi +1 = yi + h.f ( x i , yi ) for i = 0,1,2,...
y1 = y 0 + hf ( x 0 , y 0 ) = 1 + 0 = 1
y 2 = y1 + hf ( x1 , y1 ) = 1 + ( 0.1) f ( 0.1,1) = 1.01
y3 = y 2 + hf ( x 2 , y 2 ) = 1.01 + ( 0.1) f ( 0.2,1.01) = 1.0304, is the value of y at x = 0.3

27.

1
X1 ~ B 5,
2
1
X 2 ~ B 3,
2

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1
Then X1 + X 2 ~ B 5 + 3, = B 8,
2

2
P ( X1 + X 2 2 ) =1 P ( X1 + X 2 < 2 )

= 1 {P ( X1 + X 2 = 0 ) + P ( X1 + X 2 =1)}
8

9
1
1
= 1 8 = 1 8 = 0.96
2
2
2

28.

ax xa
= 1
x a x x a a

lim

By Applying LHospital Rule

a x log a ax a 1
= 1
x a x x (1 + log x )

lim

29.

log a 1
= 1 log a = 0
1 + log a
a =1

A. dr = ( x y ) dx + ( x + y ) dy

equation of C is y = x
dy = dx
1

1



A . dr = ( x x ) dx + ( x + x ) dx
C

= 2xdx = 1
0

30.

Given

(D

+ a 2 ) y = tan ax

y p = A cos ax + Bsin ax
By method of variation of parameters
uRdx
B =
where
w
R = tan ax
u = cos ax
v = sin ax, w =
B =

u' v'

w =

cos ax

sin ax

a sin ax a cos ax

=a

cos ax tan ax
1
1
dx = sin axdx = 2 cos ax
a
a
a

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12V

31.

+
+

I1

Vx

2A

I2

2Vx

I3

Ix

Applying KCL at node Vx , we have

32.

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Vx 12
V V
2 + x x = 0 Vx = 48V I x = 12A
6
4
3

Assume any arbitrary polarity

Let

x2

Writing KVL in the loop V1-AAB-V3 = 0

VAB = V1-V3 = 3V
So VA is 3V higher than VB
or VB is 3V lower than VA
33.

VGS = 3V (seen from circuit); I DSS = 12 mA = 12 103 A

2

VP = 6 V

I D = IDSS

VGS
3
3
1
= 12 10 1 = 3mA
VP
6

VDS = VD = VDD I D R D = 35V 3 103 3.5 103 = 24.5V

Also VGS = VG = 3V

34.

V1 ( 5 )

1
2
= 2 103 ( V1 1)
1k
2
V1 = 1.55V or -2.56V; so V1 =-2.56 bcz VGS2 > VT
VGS2 = V1 ; I=

As i D1 = i D2 ; VGS1 = VGS2 ,
5 = VGS1 + V2
V2 = 5 2.56 = 2.44 Volts
35.

4 1 MUX

2 1 MUX

0
0
1
1

0
1
0
1

C
0

B'C

B + C'

F = AB + AC '+ A ' B ' C

C'
1

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36.

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Analog output Vout = KD

VFS
2N 1

K=

Full peak voltage, VFS = K(2 N 1)

= 10(212 1)
= 4095 10 (mV)

VFS = 40.95V
% resolution =

K
100
VFS

10 103
100
40.95
= 0.024%
=

37.

The BCD code for (64.23)10 is

(64.23)10 = 01100100.00100011
To get excess three code 0011 is added to each group of above pattern.

01100100.00100011
00110011.00110011

(10010111.01010110 ) BCD excess 3

38.
s4 3 5
s3 10 5
s 2 3.5 2
s1
s

1
7
2

2 right half poles

39.

1 sin 30o
1
= 0.33; 10 log dB = 4.8dB
o
1 + sin 30

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Im

40.

1 + j
135o

26.16o
3

90

Re

1 j

D = 1800 (2q + 1) + (1350 900 26.60 );q = 0,1, 2.....

with q = 0, D = 71.60 (at s = (1 + j1)and D = 71.60 [ at s = (1 j1) ]

P
10
= 7.57 PSB = 2.43kW. Hence = SB 100 = 24.3%
0.64
Pt
1+
2

43.

Pc =

44.

Average inf ormation rate H = Pi log

1
= 1.8bits / msg
pi

Information rate = 2 4 1.8kbps = 14.4 Kbps

45.

U max = 2; D =

U avg =

U max
U avg

U ( , )d =

2 sin
4

sin sin dd =
2

0 0

1
3

D=

2
1

=6

46.

47.

d =

f
2 1 c
f

377 4.3 107

6.25
2 1

= 1.298 1010 nepers / m

Array Factor (AF) = cos ( d cos + )

2

= 0; d = AF = cos ( 2 cos + 0 ) = cos ( cos )

2

3
For nulls to occur, cos ( cos ) = 0 cos = ,
2
2
1 3
2
cos = , Not possible. Hence, = ,
2 2
3 3

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R2

48.

sL

Vo

R1

V2

Vo

sL

L
V2
R
R

Vo V V V V2 2V V2
= +
=
; Vo R RV = 2sLV sLV2
sL
R
R
R
sLV2 + Vo R
Vo R + sLV2 = ( 2sL + R ) V; V=
.........(1)
2sL + R
( R + sL ) V2 ................. 2
V V2 V2
;sLV sLV2 = RV2 V =
=
( )
R
sL
sL
sLV2 + Vo R ( R + sL ) V2
V
sLR
1
1
=
2 = 2 2
=
=
2
R sL
jL jR
2sL + R
sL
Vo s L + 3sLR + R
3+
3+
+

sL R
R
L
L R
R
For oscillation to occur, imaginary part =0;
=
=
R L
L
R
V2 1
= ; Vo = 3V2 ; Vo = V2 1 + 2 = 3V2 ; R 2 = 2R 1
Vo 3
R1
49.

It is a Dead Network, that acts as some load resistor whose equivalent load resistor value can
be 0, +ve or ve. To obtain the value of the load resistor, connect an independent voltage
source and find V/I ratio.
Now V = VX
V-IR-12V = 0
-11V = IR
V R
=
I 11

51.

Ix =

52.

Neglecting time delays of emitter and collector, the base transit time is given by
B =

V2 V1
= 1A
1

1
1
W2
12
=
=
31.83

10
sec
and
also

=
B
2 f T 2 5 109
2D p

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W = 2D P B = 2 10 31.83 1012 = 0.252 m

53.

n n = P P
In case of Si,

n n
=
and n.p = n i2 P =
P P

n
P
ni n =
ni
p
n

n 1350
=
P = 1.68n i and n = 0.596 n i
P
480

54.

m 2 Pt
10.125
= 1 =
1 m = 0.5
2
Pc
9

55.

cox =

ox 3.9 8.85 1014

=
= 173nF / cm 2
t ox
2 106

High frequency capacitance in inversion, C HF =

1
1 x d,T
+
cox S

Where x d,T = depletion layer width at threshold =

=

CHF =

56.

2 s ( 2F )
qN a

2 11.9 8.85 1014 2 0.419

= 105nm
1.6 1019 1017

1
= 63nF / cm 2
1
1.05 105
+
173 109 11.9 8.85 1014

17

14
3

Bronze

Silver
7 :3

57.

Complaisance - a willingness to please others by being polite

Surliness - unfriendly and not polite

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58.

Marriage : divorce :: Incorporate : liquidate(destroy)

60.

If (D) is true, then it can directly affect the Korean economy.

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61.

Throughout the passage, particularly in the first sentence the author emphasizes the point.
Others are long drawn ones.

62.

Profit % =

Income Expenditure
100
Expenditure

Let income of company Q in 2013 = Rs. x crore

120
Then income of Q in 2014 =
x crore
100

120
x
100
x = 25
30 =

Income of Q in 2013 = Rs.25 crore

Let expenditure of Q in 2013 be Rs.E crore
25 E
Then, 10 =
100
E

25
10 = 1 100
E

10
E = 25 = 22.77
11

Expenditure of Q in 2013 = Rs.22.77 crore

63.

Let the fraction be

x
y

x+4 2
=
y3 3
3x 2y + 18 = 0 _____ (1)
x+5 1
=
y+5 2
2x y + 5 = 0 ______ (2)

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Equation ( 2 ) 2

4x 2y + 10 = 0

Equation (1)

3x 2y + 18 = 0

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____________
x 8 = 0
x =8
Substituting x in equation (2), y = 21

The fraction is

64.

8
21

Part of tank filled by A in a minute=1/60

Part of tank filled by B in a minute=1/75
Let pipe B be closed after x minutes.
1 9x
1
Part of tank filled in x minutes= x + =
60 75 300
1 40 x
Part of tank filled in (40-x) minutes= ( 40 x ) =
60
60

9x 40 x
+
=1
300
60
4x = 100 x = 25

65.

log 4

2
+ log16 0.5 = 2
x

2
1
+ log16 = 2
x
2
2
2 1
log16 = 2
x 2
2
log16 2 = 2
x
log16 2 log16 x 2 = 2

2 log16

1
2
4
7
2 log16 x =
4
7
log 4 x =
4

log16 x 2 =

log16 2 =
4

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11