MH SET 2013: 17 February, Paper III, Electronic Science
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Electronic Science
Paper III
Time Allowed : 2½ Hours] [Maximum Marks : 150
Note : This Paper contains Seventy five (75) multiple choice questions. Each
question carries Two (2) marks. Attempt All questions.
1. JFET is a :
4. Match the following and choose the
(A) Current controlled device with
correct option from the codes given
high I/P resistance
below :
(B) Voltage controlled device with
high I/P resistance Set I Set II
(C) Current controlled current
(a) FET (i) Same crystal
source (CCCS)
(D) Voltage controlled voltage structure as that
source (VCVS)
of the substrate
2. An ideal voltage source should
have : (b) Epitaxy (ii) VVR
(A) Zero internal resistance
(c) Schmitt trigger (iii) Interconnection
(B) Infinite internal resistance
(d) Metallization (iv) Hysteresis
(C) A small internal resistance
(D) Large internal resistance Codes :
3. When by-pass capacitor across the (a) (b) (c) (d)
emitter lead in CE-amplifier is
disconnected, it causes : (A) (ii) (i) (iv) (iii)
(A) Current gain to increase
(B) (i) (iii) (ii) (iv)
(B) Voltage gain to increase
(C) (iii) (iv) (i) (ii)
(C) Current gain to decrease
(D) Voltage gain to decrease (D) (iv) (ii) (iii) (i)
1 [P.T.O.
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5. Astable multivibrator is a 8. The identical resistors are connected
multivibrator with duty cycle equal
to : in the following topology, where
(A) 50% value of each resistor is 1 Ω. The
(B) 100% effective equivalent resistance
(C) 30%
measured between Node B and Node
(D) 33%
H is .....................
6. Negative feedback in CE-amplifier
is to :
A 1Ω
(A) Increase the voltage gain B
1Ω 1Ω
(B) Increase the current gain 1Ω
1Ω D 1Ω
(C) Increase the power-band width C
(D) Decrease power band width
1Ω
F
E 1Ω
7. Important characteristics of an
1Ω
emitter follower are : 1Ω 1Ω
1Ω
(A) High I/P impedance and high H G
O/P impedance
(B) High I/P impedance and low
O/P impedance (A) 2/3 Ω
(C) Low I/P impedance and low (B) 5/6 Ω
O/P impedance
(C) 1/6 Ω
(D) Low I/P impedance and high
(D) 1/3 Ω
O/P impedance
2
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10. A series RLC circuit has a current
9. The circuit shown in the figure has which lags the applied voltage by
30°. The inductor voltage maximum
initial current iL(0 –) = 1A through is twice the capacitor voltage
maximum, and
the inductor and an initial voltage
VL = 10.0 sin 1000 t.
Vc(0–) = – 1V across the capacitor. Given that : R = 20 Ω, L and C are
respectively :
For input V(t) = u(t), the Laplace (A) L = 2.31 mH, C = 86.5 µF
(B) L = 23.1 mH, C = 86.5 µF
transform of the current i(t) for
(C) L = 23.1 mH, C = 8.65 µF
t > 0 is : (D) L = 23.1 µH, C = 86.5 nF
11. Match List I and with List II and
1Ω 1H select the correct answer from the
codes given below :
i(t) List I List II
+
V(t) 1F (a) Incidence matrix (1) Links
–
(b) Cutset matrix (2) Twigs
(c) Tie set matrix (3) Nodes
(d) Reduced incidence (4) Branch
s+2 matrix
(A) 2
s +s+1
Codes :
s−2 (A) (a)—(3), (b)—(1), (c)—(2),
(B) (d)—(4)
s2 + s + 1
(B) (a)—(3), (b)—(2), (c)—(1),
s−2 (d)—(3)
(C) 2
s + 2s + 1 (C) (a)—(1), (b)—(2), (c)—(3),
(d)—(4)
s
(D) (D) (a)—(2), (b)—(1), (c)—(3),
s2 + s + 1
(d)—(4)
3 [P.T.O.
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12. The Y-parameters of the two-port 13. The driving point impedance Z(s) of
a network has the pole-zero locations
network shown in the following
as shown in the figure. If z(0) = 3,
figure are Y 11 = Y 22 = 6 mho,
then z(s) is :
Y12 = Y21 = 4 mho. The driving
Im
point admittance at port 2-2', if the
s-plane
source voltage is 100 V with an X 1
impedance of 1 Ω is :
. .
–3 –1
Re
1I O-denotes zero
• 1
•2 X 1 X-denotes pole
1Ω I2 +
+ +
Vs V2
– V1
– –
• •2
1
3(s + 3)
1 (A)
(A) mho 2
3.714 s + 2s + 3
2(s + 3)
(B) 3.714 mho (B) 2
s + 2s + 2
3(s − 3)
(C) 0.3714 mho (C)
s2 − 2 s − 2
1 2(s − 3)
(D) mho (D) 2
0.3714 s − 2s − 3
4
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15. Match the following. Figures show
14. For the circuit shown in figure, the the conduction region for different
amplifiers for the input and choose
the correct answer from the codes
given below :
Thevenin’s equivalent impedance
(p) (1) Class B
between the output terminals
is :
(q) (2) Class C
–j4 Ω j5 Ω –j4 Ω
(r) (3) Class AB
A
3Ω (s) (4) Class A
~ 4Ω
° ∠0°
11.350
4.83∠0−°1.13
48.3 Codes :
j6 Ω
(A) (p)—(4), (q)—(2), (r)—(3), (s)—(1)
B
(B) (p)—(1), (q)—(2), (r)—(4), (s)—(3)
(C) (p)—(4), (q)—(3), (r)—(1), (s)—(2)
(A)
(D) (p)—(4), (q)—(1), (r)—(2), (s)—(3)
16. Product of gain and bandwidth is
(B)
constant when the roll-off is :
(C) (A) – 10 dB/decade
(B) – 20 dB/decade
(D) (C) – 30 dB/decade
(D) + 10 dB/decade
5 [P.T.O.
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17. In a two-stage amplifier one stage 19. A Gyrator is a circuit using OP
Amps, that simulates :
has a lower critical frequency of
(A) Inductance property with the
850 Hz and an upper critical
help of resistors and capacitors
frequency of 100 kHz. The other has
(B) Resistance property with the
correspondingly 1 kHz and 230 kHz help of inductance and
critical frequencies. The overall capacitors
(C) Capacitance property with the
bandwidth is :
help of resistance and
(A) 99 kHz inductances
(B) 229.15 kHz (D) The circuit does not use any of
the components
(C) 99.15 kHz
20. The electrothermal filters are used
(D) 229 kHz in the :
(A) less than 1 Hz to audio
18. A varactor diode exhibits :
range
(A) A variable capacitance that (B) 50 Hz to 100 kHz range
depends on reverse voltage (C) 100 kHz to 1 MHz range
(B) A variable resistance that (D) 90 MHz to 100 MHz range
21. The principle of MODEMs
depends on reverse voltage
is :
(C) A variable capacitance that
(A) FSK
depends on forward current
(B) PWM
(D) A constant capacitance over a (C) Phase adder
range of reverse voltages (D) Automatic fine tuning
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22. What is the meaning of RAM and 24. Consider the following series of
TTL :
what is its primary role ?
(i) 74
(A) Readily available memory, it is
(ii) 74H
memory that can be reached by (iii) 74L
any sub-item within a computer (iv) 74S
and at any time Write them in increasing order of
power dissipation per gate :
(B) Random access memory, it is
(A) (iii) (i) (iv) (ii)
memory used for short term
(B) (i) (ii) (iii) (iv)
temporary data storage within
(C) (iv) (i) (ii) (iii)
the computer
(D) (ii) (iii) (iv) (i)
(C) Random access memory, it is 25. Match List I with List II and select
the first level of memory used the answer using the codes given
below :
by the computer in all of its
List I List II
operations
(a) Schmitt trigger (i) Toggle switch
(D) Resettable automatic memory, (b) J-K flip-flop (ii) Delay switch
it is the memory that can be (c) D flip-flop (iii) Race around
used and then automatically condition
reset, or cleared, after being (d) T flip-flop (iv) Comparator and
read from or written to square wave
generator
23. A BCD counter is a .................
Codes :
(A) Binary counter (a) (b) (c) (d)
(B) Full-modulus counter (A) (iv) (iii) (ii) (i)
(B) (i) (ii) (iii) (iv)
(C) Decade counter
(C) (ii) (i) (iv) (iii)
(D) Johnson counter (D) (iii) (i) (iv) (ii)
7 [P.T.O.
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26. A certain system contains two Direction : Read the following paragraph
identical circuits operating in
and answer the Question Nos. 27 and
parallel. As long as both are
28 based on the same.
operating properly, the outputs of
both circuits are always the same. ‘‘Many DACs use a BCD input code
A failure in one of the circuits
where 4-bit code groups are used for
produce differing outputs. One of the
each decimal digit. In one such
following logic circuits is the
12-bit BCD digital to analog
best alternative to detect the
fault : converter the full scale output
Circuit A is 9.99 V.’’
(A) Failure
Circuit B
27. The percent resolution of above
Circuit A
referred DAC will be :
(B) Failure
Circuit B
(A) 0.001%
Circuit A
(C) Failure
(B) 0.01%
Circuit B
Circuit A (C) 0.1%
(D) Failure
Circuit B
(D) 1.0%
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28. The step size of the above referred
Direction : Read the following paragraph
DAC is :
regarding the 8085 assembly
(A) 0.001 V
language program :
(B) 0.01 V
‘‘The main program is stored at
(C) 0.1 V
(0100) H. The main program has
(D) 1.0 V
called the subroutine at (0150)H,
29. Match List I (Logic family) with
when the microprocessor is executing
List II (Typical fanout) and select
the instruction at location
the answer using the codes given
below : (0151)H. The program listing is as
List I List II follows.’’
(a) RTL (i) > 50 0100 LXI SP, 0400H
(b) TTL (ii) 5 0103 EI
(c) ECL (iii) 10
0120 CALL 0150 H
(d) CMOS (iv) 25
0130 HLT
Codes :
0150 PUSH B
(a) (b) (c) (d)
0151 LXI B, 10 FFH
(A) (ii) (iv) (iii) (i)
(B) (ii) (iii) (iv) (i) 0154 MOV C, A
(C) (i) (iv) (ii) (iii) 015E POP B
(D) (iii) (iv) (i) (ii) 015F RET
9 [P.T.O.
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Direction : Answer the Question Nos. 33. In interfacing 16 keys and 8 relays
30, 31 and 32 based on the above directly using 8255 IC with 8085
Scenario. microprocessor the minimum
30. Assuming that before CALL 0150 H, number of pins of 8255 required
the stack was not used, specify the are :
content of top two locations of the (A) 4 inputs, 4 outputs, 8 outputs
stack :
(B) 4 inputs, 8 outputs, 8 I/Os
(A) 23H, 01H
(C) 8 I/O only
(B) 23H, 00H
(D) 12 I/O only
(C) FFH, 00H
34. By using 8279 IC with 8085
(D) 00H, FFH the following functions can be
31. Specify the stack locations where the done :
contents of register pair B are (A) Relay interfacing
stored :
(B) stepper motor interfacing
(A) 03FDH and 03FCH
(C) Display interfacing only (LCD/
(B) 03FDH and 03FEH LED)
(C) 03FEH and 03FFH (D) Display (7 segment) interfacing
(D) 03FBH and 03FCH and keyboard interfacing
32. When the program was interrupted, 35. Timer ports of 8051 microcontroller
can be used for :
what is the memory address stored
on the stack ? (A) Generation of waveforms
(B) Generation of PWM
(A) FFFFH
(C) Measurement of time (ON, OFF
(B) 015FH and Freq.) of input signal with
(C) 0154H gate control
(D) As a counter and (B) and (C)
(D) 0150H
above
10
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36. Match the following and choose the
38. After execution of the following
correct answer from the codes given
below : statement :
Ports of 8051 Application X = (int *) Malloc(100);
(i) Port 0 (a) I/O + control
(assume size of int as 2 bytes)
(ii) Port 1 (b) I/O + higher order
The following amount of memory for
address bus
execution will be allotted :
(iii) Port 2 (c) Low address & data
bus (A) 2 bytes
(iv) Port 3 (d) I/O port only
(B) 100 bytes
(e) Timer port + Input
only (C) 200 bytes
Codes : (D) 4 bytes
(A) (i)—(c), (ii)—(d), (iii)—(b),
39. The output of the following
(iv)—(a)
program :
(B) (i)—(d), (ii)—(b), (iii)—(a),
(iv)—(e) void main (void)
(C) (i)—(e), (ii)—(a), (iii)—(b),
{
(iv)—(c)
(D) (i)—(a), (ii)—(b), (iii)—(c), unsigned char flag=0X00;
(iv)—(d)
flag!=(1<< 7)
37. The structure variables can be
passed to the function as a : printf("%d", flag);
(A) Copy of the structure variable (A) 0X00
(B) Pointer to a structure variable
(B) 0X70
(C) Both (A) and (B) depending on
speed of operations required (C) 0X80
(D) Both (A) and (B) (D) 0XFF
11 [P.T.O.
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40. Consider the following statement 42. What will be the output of the
fprintf(outfile, "programming in C is following program ?
fun.\n");
#include <stdio.h>
The ‘‘out file’’ in this statement is :
int abc(int u, int v)
(A) name of a variable to be
printed {
(B) file pointer previously declared int tmp;
and defined by fopen
while (v !=0) {
(C) identifier representing the file
tmp=u%v;
name
(D) pointer of the variable to be u=v;
printed v=tmp
41. (A) Assertion : Various objects can }
be drawn on the computer screen
with proper scale factors and colour return u;
selection }
because :
int main(void)
(R) Reason : Matching band rate
{
is a must for connecting a mouse to
serial port. int result;
(A) Both (A) and (R) are true result=abc(150, 35);
statements and the reason is a
correct explanation of assertion printf("result=%d", result);
(B) Both (A) and (R) are true return 0;
statements but the reason is not }
a correct explanation of
assertion (A) 35
(C) The (A) is true but (R) is (B) 4
false (C) 10
(D) The (A) is false but (R) is
(D) 5
true
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43. A uniformly charged sphere is Directions : Given the transmission line
below, obtain the following
compressed slightly on opposite quantities, for Question Nos. 45, 46
sides. It will exhibit a : and 47.
(A) Monopole
50 Ω
(B) Monpole + dipole
Vg S Z0 = 50 Ω ZL 75 Ω
(C) quadrupole
30 V
(rms)
(D) monopole + qaudrupole
2.25 λ
44. Given that the time variation of
45. The input impedance is :
electric and magnetic fields is e jωt, (A) 66.66 Ω
match the following and choose the (B) 33.33 Ω
correct option given below : : (C) 50 Ω
→ → (D) 75 Ω
(p) ∇×E (1) 0
46. The reflection coefficient
→ → →
(q) ∇×H (2) – jωµ H is :
→ → (A) 1
(r) ∇.D (3) Pν
(B) 0
→ → →
(s) ∇.B (4) (σ + jω ∈) E (C) 0.2
Options are : (D) 0.5
47. Instantaneous power delivered to
(A) (p)—(3), (q)—(1), (r)—(2), (s)—(4) the load :
(B) (p)—(4), (q)—(2), (r)—(1), (s)—(3) (A) 17.28 W
(B) 16 W
(C) (p)—(4), (q)—(3), (r)—(1), (s)—(2)
(C) 18.39 W
(D) (p)—(2), (q)—(4), (r)—(3), (s)—(1) (D) 36 W
13 [P.T.O.
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49. An incident wave has amplitude 1V.
48. A circulator with direction of flow of
1
power is shown in the figure. Corresponding reflected one is V .
3
The VSWR is :
4 (A) 3
(B) 2
(C) 1
1
3
(D)
1 2
The tuned circuit of the oscillator in
a simple AM transmitter uses a
40 µH coil and 12 nF capacitor. If
the oscillator output is modulated by
2
audio frequency of 5 kHz. Answer
questions 50 and 51 based on the
Corresponding S matrix is : above information.
50. Calculate the lower side band
0 0 0 1 frequency :
(A) 224 kHz
(A) (B)
1 0 0 0
(B) 228 kHz
0 1 0 0 (C) 442 kHz
(D) 19 kHz
0 0 1 0
51. The bandwidth needed to transmit
0 1 0 0 this amplitude modulated wave
1 0 0 0
is :
(C) 0 1 0 0 (D) 1 0 0 0
(A) 5 kHz
0 0 1 0 0 0 1 0 (B) 2 kHz
(C) 10 kHz
0 0 0 1 0 0 0 1
(D) 7 kHz
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52. Match the following and choose the 53. When input signal strength = 0.5 W,
maximum frequency deviation =
correct answer from the codes given 60 kHz, base band signal cutoff
frequency = 15 kHz, received white
below :
Gaussian noise power spectral
density = 10 –9 W/Hz and the
Column I Column II
average power of modulating signal
= 0.1 W, the SNR of an FM
(a) Amplitude (i) Frequency
limiter-discriminator demodulator
Modulator Discriminator
is ...........
(A) 36.08 dB
(b) DSBSC modulator (ii) Ring modulator
(B) 26.07 dB
(c) SSB modulator (iii) Switching (C) 46.07 dB
(D) 30 dB
modulator
54. (A) Assertion : Use of a compressor
(d) VSB modulator (iv) Phase improves the signal to quantization
noise ratio of a PCM system
discrimination because :
(R) Reason : The compressor
method provides lower gain to weak signals
and higher gain to the strong
Codes : signals.
(A) (A)—(3), (B)—(2) , (C)—(1) , (A) Both (A) and (R) are true
statements and the reason is
(D)—(4) correct explanation of the
assertion
(B) (A)—(2), (B)—(3) , (C)—(4) ,
(B) Both (A) and (R) are true
(D)—(1) statements and the reason is not
correct explanation of the
(C) (A)—(3), (B)—(1) , (C)—(2) , assertion
(D)—(4) (C) The (A) is true but (R) is
false
(D) (A)—(2), (B)—(4) , (C)—(3) , (D) The (A) is false but (R) is
(D)—(1) true
15 [P.T.O.
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55.
A3 A2 A1
Source A
Frame 3 Frame 2 Frame 1
A3 B3 C3 A2 B2 C2 A1 B1 C1
B3 B2 B1
Source B Mux
C3 C2 C1 Common medium
Source C
The above diagram 57. The correct relationship from among
represents a : the following is :
(A) FDM system (A) BW (ASK) > BW (FSK)
(B) TDM system (B) BW (BPSK) > BW (QPSK)
(C) Noise immunity of QPSK <
(C) WDM system
Noise immunity of BFSK
(D) PCM system (D) Noise immunity of ASK > =
56. For a polar NRZ signal representing Noise immunity of PSK
1 by a positive pulse X1 and 0 by 58. A thyristor of thermal resistance
a negative pulse X2 of duration T, 1.8°C/W is mounted on a neat sink
the impulse response of a matched of thermal resistance 2.0°C/W. What
is the maximum power loss of thyristor
filter is given by :
if the junction temperature is not to
(A) h(t) = X1(T – t) + X2(T – t) exceed 125°C in an ambient of 40°C.
(B) h(t) = X1(t) + X2(t) (A) 5.2 W
(C) h(t) = X1(T – t) – X2(T – t) (B) 11.4 W
(C) 18.6 W
(D) h(t) = X1(t) – X2(t)
(D) 22.4 W
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59. You need a very efficient device to 62. Match List I with List II and choose
control the speed of AC fan motor.
the correct answer from the codes
An appropriate device for the same
would be : given below :
(A) SCR (B) Triac List I List II
(C) UJT (D) PUT (a) Single mode fiber (i) Local area
network
60. The following is the circuitary for a
dimmer : (b) Step index (ii) Low cost
multimode fiber application
Lamp
(c) Graded index fiber (iii) Low bandwidth
R1
AC supply (d) PCS fibers (iv) High bandwidth
~ R2 Diac
Codes :
Triac
(a) (b) (c) (d)
C
(A) (iv) (i) (iii) (ii)
(B) (ii) (i) (iv) (iii)
What method of power control is (C) (iv) (iii) (i) (ii)
used ? (D) (i) (iii) (iv) (ii)
(A) Phase control 63. Match List I with List II and choose
(B) Feedback control the correct answer from the codes
(C) Proportional control given below :
(D) Burst firing List I List II
61. Assertion (A) : Optical communi- (a) Fusion splicing (i) Low cost
cation is called information super (b) 3 pin splicing (ii) Multimode
highway.
(c) Plastic ‘V’ grove (iii) Low insertion
Reason (R) : Optical pulses travel
splicing loss
at higher speed as compared to
copper cable. (d) Square tube (iv) Graded index
(A) Both (A) and (R) are true, and splicing fibers
(R) is the correct explanation Codes :
of (A)
(a) (b) (c) (d)
(B) Both (A) and (R) are true, but
(A) (iii) (iv) (i) (ii)
(R) is not correct explanation
of (A) (B) (ii) (i) (iv) (iii)
(C) (A) is true, but (R) is false (C) (i) (iii) (ii) (iv)
(D) (A) is false, but (R) is true (D) (iv) (ii) (iii) (i)
17 [P.T.O.
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64. The following display device is not 66. The most common type of vibration
suitable for high frame rate
sensors are :
display :
(A) Piezoelectric
(A) CRT
(B) LCD (B) Capacitive
(C) LED (C) Resistive
(D) PDP
(D) (A) and (B) both
65. Consider the following light sources.
67. The magnetic flow meter has the
Give their suitability in the order of
merit for optical communication in following disadvantages :
fibers :
(A) Suitable for high and low
(a) LED
velocities
(b) Ruby laser
(B) Suitable for conductivity
(c) Febry Perrot laser diode
greater than 20 mho/cm
(d) DFB laser
(C) Gives slow response to flow
(A) (b), (a), (c), (d)
(B) (a), (c), (b), (d) changes
(C) (c), (a), (d), (b) (D) Very low cost and require
(D) (d), (a), (b), (c) minimum maintenance
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68. Match the following and choose the 69. Match the following and choose the
correct answer from the codes given correct answer from the codes given
below : below :
Transducer Principle Measurement Measuring
(i) Photovoltaic (a) An emf is generated Quantity Unit/Symbol
across the junction
(i) Phase angle (a) Pa or N/m2
of two dissimilar
(ii) Angular velocity (b) kg/m
metals when two
(iii) Linear denstiy (c) rad/s
ends are at different
temperatures (iv) Stress (d) Radian
(ii) Moving coil (b) Motion of a coil in (e) kg/m3
generator a magnetic field Codes :
generates a voltage (A) (i)—(d), (ii)—(b) , (iii)—(a) ,
(iii) Piezoelectric (c) An emf is generated (iv)—(c)
pickup due to external force (B) (i)—(e), (ii)—(a) , (iii)—(c) ,
on quartz (iv)—(b)
(iv) Thermocouple (d) A voltage is generated
(C) (i)—(a), (ii)—(b) , (iii)—(c) ,
in a semiconductor (iv)—(e)
junction device due
(D) (i)—(d), (ii)—(c) , (iii)—(b) ,
to radiant energy (iv)—(a)
(v) Thermistor
70. The resistors in a bridge are given
Codes : by R1 = R 2 = R 3 = 120 Ω and
(A) (i)—(d), (ii)—(b) , (iii)—(c) , R4 = 121 Ω. If supply voltage is
(iv)—(a) 10.0 V dc, then the offset voltage
(B) (i)—(a), (ii)—(d) , (iii)—(b) , will be :
(iv)—(c) (A) + 20.7 mV
(C) (i)—(b), (ii)—(c) , (iii)—(d) ,
(B) – 20.7 mV
(iv)—(a)
(C) 0.0 mV
(D) (i)—(c), (ii)—(a) , (iii)—(b) ,
(iv)—(d) (D) 1.0 mV
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71. In a PID control system, the control
equation will be : 73. The system described by dx = f ( x)
dt
(A) Vout = KP * (DE) + PI * PD is :
(B) Vout = KP * (DE) + PI + PD (A) an autonomous system
(C) Vout = KP * [DE + PI + PD]
(B) a static system
(D) Vout = [KP * (DE)/PI] * PD
(C) a time dependent system
72. Match the following and select the
correct answer from the codes given (D) a non-linear system
below :
74. The characteristic equation of a
(i) Bode plot (a) Stability of linear
system is given by :
control system
s6 + 3s5 + 8s4 + 18s3 + 37s2 + 75s
(ii) Root locus plot (b) Stability in
+ 50 = 0
frequency response
(iii) Nyquist criterion (c) Analysis of
The system is :
& Nicholas chart frequency response (A) Stable
(iv) Hurwitz and (d) Method to determine (B) Unstable
Routh criterion the roots of a
(C) Marginally stable
characteristic
equation
(D) Conditionally stable
(e) Compensation and 75. The open loop transfer function of
control of close loop a feedback system is :
system k(1 + s)
Codes : G(s) H(s) = (1 − s)
(A) (i)—(d), (ii)—(b) , (iii)—(c) ,
The value of k for which the system
(iv)—(a)
is stable, is :
(B) (i)—(a), (ii)—(b) , (iii)—(c) ,
(iv)—(d) (A) k < 1
(C) (i)—(a), (ii)—(b) , (iii)—(c) , (B) k > 1
(iv)—(e) (C) k = 1
(D) (i)—(c), (ii)—(d) , (iii)—(b) ,
(D) k = 0
(iv)—(a)
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ROUGH WORK
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ROUGH WORK
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Source: Savitribai Phule Pune University. Question paper published by Savitribai Phule Pune University.