fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Solid State Physics, Devices and Electronics
IIT-JAM 2005
Q1.
A circuit and the signal applied at its input terminals ( Vi ) are shown in figure below.
Which one of the options correctly describes the output waveform ( V0 ). (Assume all the
devices used are ideal).
 2V
C
Vi
t
0
D
Vo
 2V
(a)  2V
0
(c)
(b)
0
t
 2V
 2V
4V
(d) 0
t
0
t
t
- 4V
Ans. : (c)
Solution: Its clamper circuit in which peak to peak remains fixed and voltage level will shift in
the direction of diode current.
Q2.
The susceptibility of a diamagnetic material is
(a) positive and proportional to temperature
(b) negative and inversely proportional to temperature
(c) negative and independent of temperature
(d) positive and inversely proportional to temperature
Ans. : (b)
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Q3.
Which of the following statements is correct for NaCl crystal structure?
(a) It is simple cubic lattice with one atom basis
(b) It is a face-centered cubic lattice with one atom basis
(c) It is a simple cubic lattice with two atom basis
(d) It is a face-centered cubic lattice with two atom basis
Ans. : (d)
IIT-JAM 2006
Q4.
In a crystalline solid, the energy band structure ( E - k relation) for an electron of mass m
is given by E 
 2 k 2k  3
. The effective mass of the electron in the crystal is
2m
(a) m
(b)
2
m
3
(c)
m
2
(d) 2m
Ans. : (c)
Solution: The expression of effective mass of electron in solid is m * 
2
d 2E 2
dk
m
dE  2
d 2 E 2
2 2
 m* 





4
k
3
4


 
2
2
dk 2m
dk
2m
m
Q5.
The truth table for the given circuit is
J
Q
K
(a)
(c)
J
0
0
1
1
K
0
1
0
1
Q
1
0
1
0
J
0
0
1
1
K
0
1
0
1
Q
0
1
0
1
(b)
(d)
J
0
0
1
1
K
0
1
0
1
Q
1
0
0
1
J
0
0
1
1
K
0
1
0
1
Q
0
1
1
0
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Ans. : (c)
Solution: Q  J .K  J .K  K
Q6.
In an intrinsic semiconductor, the free carrier
concentration n (in cm 3 ) varies with temperature
36.5
T (in Kelvin) as shown in the figure below. The
band gap of the semiconductor is (use Boltzmann
constant k B  8.625 105 eVK 1 )
35.5
lnn  34.5
33.5
(a) 1.44 eV
32.5
(b) 0.72 eV
(c) 1.38 eV
2
3 1000 / 
(d) 0.69 eV
Ans. : (d)
 Eg  1 1  
 E 
n
Solution: Since ni  n  N c N v exp   g   1  exp     
n2
 2kT 
 2k  T2 T1  
n   1 1
 Eg  2k ln  1      2  8.625 105  36.5  32.5   0.003  0.0002 
 n2   T2 T1 
 Eg  2  8.625 105  4 1103  0.69eV
IIT-JAM 2007
Q7.
Fermi energy of a certain metal M 1 is 5eV . A second metal M 2 has an electron which is
6 % higher than that of M 1 . Assuming that the free electron theory is valid for both the
metals, the Fermi energy of M 2 is closet to
(a) 5.6 eV
(b) 5.2 eV
(c) 4.8eV
(d) 4.4 eV
Ans. : (b)
Solution: The expression of Fermi energy is E F 

2
3 2 n
2m

2/3
Let us consider that n1 , E F1 is the concentration of electron and Fermi energy in metal M1
and n2 , E F2 is in metal M2.
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Given, n 2  n1  0.06n1  1.06n1
 2
3 2 n2
EF2  2m

EF1   2
3 2 n1

2
m


2/3
2/3

   n2   1.06n1   1.06 2/3  1.0396
 
2/3
2/3
2/3 
 n1 
 n1 






2/3
EF2  1.0396  EF1  1.0396  5eV  5.2 eV
Q8.
Fig. (i)
Fig. (ii)
Figure (i) and (ii) represent respectively,
(a) NOR, NOR
(b) NOR, NAND
(c) NAND, NAND
(d) OR, NAND
Ans. : (c)
IIT-JAM 2008
Q9.
The ratio of the second-neighbour distance to the nearest-neighbour distance in an fcc
lattice is
(a) 2 2
(b) 2
(c)
3
(d)
2
Ans. : (d)
Solution: In FCC lattice the first nearest is at distance  2a / 2  a / 2
Whereas the second nearest at distance of  a
The ratio of the second-neighbour distance to the nearest-neighbour distance is
a
a/ 2
Q10.
 2
Consider a doped semiconductor having the electron and the hole mobilities n and  p ,
respectively. Its intrinsic carrier density is ni . The hole concentration p for which the
conductivity is minimum at a given temperature is
(a) ni
n
p
(b) ni
p
n
(c) ni
p
n
(d) ni
n
p
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Ans. : (a)
 n2

Solution: Conductivity   e nn  p p  e  i n  p p 
 p




For minimum conductivity,
Q11.

d
 0  p  ni n /  p
dp
The logic expression for the output Y of the following circuit is
P
Y
Q
R
S
(a) P  Q  QR  S
(b) P  Q  QR  S
(c) P  Q  QR  S
(d) P  Q  QR  S
Ans. : (a)
Solution: P
PQ
P  Q  QR
P  Q  QR  S
Y
Q
R
P  Q  QR  S
QR
S
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
IIT-JAM 2009
0
Q12.
Monochromatic X -rays of wavelength 1A are incident on a simple cubic crystal. The
first order Bragg reflection from  311 plane occurs at angle of 30 0 from the plane. The
0
lattice parameter of the crystal in A is
(a) 1
(b)
3
(c)
11
2
(d) 11
Ans. : (d)
Solution: Expression of Bragg´s law is 2d sin   n ,
where, d is the inter-planar spacing defined for cubic lattice of lattice constant a in term
a
of Miller indices  h k l  as d 
h2  k 2  l 2
The lattice parameter a is
a
Q13.

2  sin 
h2  k 2  l 2 
1
1
3 2  12  12 
 11  11
0
2 1/ 2
2  sin 30
Which one of the following is an INCORRECT Boolean expression?
(a) P Q  PQ  Q
(b) P  Q P  Q   P
(c) PP  Q   Q
(d) P Q R  P Q R  PQ R  PQ R   Q
Ans. : (c)
Solution: (a) PQ  PQ   P  P  Q  1.Q  Q
(b)  P  Q   P  Q   P  QQ  P
(c) P  P  Q   P  PQ  P 1  Q   P
(d)  PQR  PQR  PQR  PQR   PQ  R  R   PQ  R  R   PQ  PQ   P  P  Q  Q
Q14.
A battery with a constant emf  and internal resistance ri provides power to an external
circuit with a load resistance made up by combining resistance RL and 2 RL in parallel.
For what value of RL will the power delivered to the load be maximum?
(a) R L 
ri
4
(b) RL 
ri
2
(c) RL 
2
ri
3
(d) RL 
3
ri
2
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Ans. : (d)
Solution: Since RL and 2 RL are in parallel so load R 
RL  2 RL 2
 RL .
RL  2 RL 3
2
  
Power through load P  I R  
 R
 ri  R 
2
For maximum power through load
 r  R   2   2 R  2  ri  R   0  r  R  2 R  0  R  r
dP
0 i
i 
i
4
dR
 ri  R 
2
Thus R  ri 
2
3
RL  ri  RL  ri
3
2
IIT-JAM 2010
Q15.
The following are the plots of the temperature dependence of the magnetic susceptibility
for three different samples.
a
 (T )
b
 (T )
0
T
0
T
c
 (T )
0
T
The plots a, b and c correspond to
(a) ferromagnet, paramagnet and diamagnet, respectively
(b) paramagnet, diamagnet and ferromagnet, respectively
(c) ferromagnet, diamagnet and paramagnet, respectively
(d) diamagnet, paramagnet and ferromagnet, respectively
Ans. : (b)
0
Q16.
The value of  at which the first-order peak in X -ray (   1.53A ) diffraction
corresponding to 111 plane of a simple cubic structure with the lattice constant,
0
  2.65 A , is approximately
(a) 150
(b) 300
(c) 450
(d) 600
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Ans. : (b)
Solution: Expression of Bragg´s law is, 2d sin   n
where, d is the inter-planar spacing defined for cubic lattice of lattice constant a in term
of Miller indices (h k l) as d 
sin  
Q17.

2d


2a
h2  k 2  l 2 
a
h2  k 2  l 2
1.53
1
1
 3     sin 1    300
2  2.65
2
2
Consider the following truth table
The logic expression for F is
(a) AB  BC  CA
(b) A B  AC  B C
(c) CA B  AB
(d) C  A  B   AB
A
B
C
F
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
0
0
0
1
1
1
0
Ans. : (d)
C
1
Solution: F  AB  C  A  B 
C
0
1
A B 00
A B 01
AC
1
AB 11
AB 10
1
1
AB
BC
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
IIT-JAM 2011
Q18.
An X - ray diffraction  XRD  experiment is carried out on a crystalline solid having
FCC structure at room temperature. The solid undergoes a phase transformation on
cooling to 200 C and shows orthorhombic structure with small decrease in its unit cell
lengths as compared to the FCC unit cell lengths. As a result the  311 line of the XRD
pattern corresponding to the FCC system
(a) will split into a doublet
(b) will split into a triplet
(c) will remain unchanged
(d) will split into two separate doublets
Ans. : (b)
Solution: In FCC the inter-planar spacing is defined as d 
a
h2  k 2  l 2
whereas in Orthorhombic the inter-planar spacing is d 
1
2
h
k2 l2


a2 b2 c2
For  311 , numbers XRD peaks in FCC is derived from Bragg’s law as
sin  

2d


2a
h2  k 2  l 2 

2a
3 2  12  12 

2a
11
Thus only one peak appears in FCC.
In Orthorhombic, sin  

2d


2
h2 k 2 l 2


a2 b2 c2
For  311 , there will be three peaks corresponding to















 ,  sin 1 
 and   sin 1 

1  sin 1 
2
3

 12 32 12 
 12 12 32 
32 12 12 
2 2  2  2 
2 2  2  2 
2 2  2  2 
 a b c 
 a b c 
 a b c 
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Q19.
Which of the following circuit does not satisfy the Boolean expression AB  A B  F
(a)


(b)
F
(c)


F


(d)
F


F
Ans. : (d)
Solution: (a) F  AB  AB


(b) F   A  B  . AB   A  B  . A  B  AB  AB


(c) F   A  B  . A  B  AB  AB
(d) F  AB  AB
IIT-JAM 2012
0
Q20.
An X -ray beam of wavelength 1.54 A is diffracted from the 110  planes of a solid with
0
a cubic lattice of lattice constant 3.08 A . The first-order Bragg diffraction occurs at
 1 
(b) sin 1 

2 2
1
(a) sin 1  
4
1
(c) sin 1  
2
 1 
(d) sin 1 

 2
Ans. : (b)
Solution: Expression of Bragg´s law is 2d sin   n
where, d is the inter-planar spacing defined for cubic lattice of lattice constant a in term
of Miller indices ( h k l ) as d 
sin  

2d


2a
h2  k 2  l 2 
a
h2  k 2  l 2
1.54
1
 1 
 2
   sin 1 

2  3.08
2 2
2 2
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fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Q21.
The Boolean expression P  P Q , where P and Q are the inputs to a circuit, represents
the following logic gate
(a) AND
(b) NAND
(c) NOT
(d) OR
Ans. : (d)
Solution:
P
Q
PQ
P  PQ
0
0
0
0
0
1
1
1
1
0
0
1
1
1
0
1
IIT-JAM 2013
Q22.
The fraction of volume unoccupied in the unit cell of the body centered cubic lattice is
(a)
8  3
8
(b)
3
8
(c)
6  2
6
(d)

3 2
Ans. : (a)
Solution: The effective number of atoms in bcc structure is
1
1
1
1
neff  nc  n f  ni   8   0  1  2
8
2
8
2
E
F


The radius of atom and lattice constant are related as
a
a 3
r
4

a
The volume occupide by the atoms in unit cell is
4
2
3
G
D
a
C
3
a 3  3 3
a

 
8
 4 
Thus amount of volume unoccupied in the unit cell is
 volume of unit cell  filled volume  a 3 
 8  3  3
a 3  
 a
8
8


 3
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11
fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Thus, the fraction of volume unoccupied in the unit cell of the body centered cubic lattice
 8  3 

is 

8


Q23.
For an ideal op-amp circuit given below, the dc gain
1 k
and the cut off frequency respectively are
1
F
2
(a) 1 and 1kHz


(b) 1 and 100 Hz
1 k
(c) 11 and 1kHz
10 k
(d) 11 and 100 Hz
Ans. : (b)
10
1
1
 11 and f H 

 1kHz
1
2 RC 2  1 1
2
A variable power supply 5V  20V is connected to a Zener diode specified by a
Solution: DC Gain  1 
Q24.
breakdown voltage of 10V (see figure). The ratio of the maximum power to the minimum
power dissipated across the load resistor is
500 
5V  20 V
1 k
Ans. : 9.02
Solution: When V  5V  open circuit voltage Vi 
 VL  Vi  3.33V  PL ,min 
1000
 5  3.33  VZ  10V
1500
Vi 2
.
RL
When V  20V  open circuit voltage Vi 
1000
 20  13.33  VZ  10V
1500
2
 VL  V z  10V  PL ,max
P
V2
V 2  10 
 Z  L ,max  Z2  
  9.02
RL
PL ,min Vi
 3.33 
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IIT-JAM 2014
Q25.
Octal equivalent of decimal number  478 10 is
(a) 736 8
(b) 6738
(c) 637 8
(d) 367 8
Ans. : (a)
Solution: 7368  7  82  3  81  6  80  448  24  6  47810
6738  6  82  7  81  3  80  384  56  3  44310
6378  6  82  3  81  7  80  384  24  7  41510
3678  3  82  6  81  7  80  192  48  7  24710
Q26.
In an ideal operational amplifier depicted below, the potential at node A is
5 k A
25 k
_
5V

IV
(a) 1V
(b) 0V
(c) 5V
(d) 25V
Ans. : (b)
Solution: Node A is virtually grounded.
Q27.
To operate a npn transistor in active region, its emitter-base and collector- base junction
respectively, should be
(a) forward biased and reversed biased
(b) forward biased and forward biased
(c) reversed biased and forward biased
(d) reversed biased and reversed biased
Ans. : (a)
Solution: In active region: emitter-base junction is F.B.
: collector-base junction is R.B.
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Q28.
Diamond lattice can be considered as a combination of two fcc lattice displaced along the
body diagonal by one quarter of its length. There are eight atoms per unit cell. The
packing fraction of the diamond structure is
(b) 0.74
(a) 0.48
Ans. : (c)
Solution: P  F 
(c) 0.34
(d) 0.68
4 3
r
3
V
neff 
Where, neff  8, V  a 3 and
3a
8r
 2r  a 
4
3
4 3
r
3
3
 PF 

 0.34 .
3
16
 8r 


 3
Thermal neutrons (energy  300 , k B  0.025 eV ) are sometimes used for structural
8
Q29.
determination of materials. The typical lattice spacing of a material for which these can
be used is
(a) 0.01nm
(b) 0.05 nm
(c) 0.1 nm
(d) 0.15 nm
Ans. : (c)
Q30.
A sine wave of 5V amplitude is applied at the input of the Vin
Vout
1 k
circuit shown in the figure. Which of the following
waveforms represents the output most closely?
5V
5V
(a)
3V
(b)
-3V
-5V
(c) 3V
(d)
3V
-3V
-5V
Ans. : (d)
Solution: If Vin  3 V , diode is OFF and V0  Vin .
If Vin  3 V , diode is ON and V0  3 V .
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Q31.
1011 binary input have been applied at X 3 X 2 X 1 X 0 input in the shown logic circuit made
of XOR gates. The binary output Y3Y2Y1Y0 of the circuit will be
X3
Y3
(a) 1101
X2
Y2
X1
Y1
X0
Y0
(c) 1111
(b) 1010
(d) 0101
Ans. : (a)
Solution:
1 X3
Y3
1
0 X2
Y2
1
1 X1
Y1
0
1
Y0
1
X0
IIT-JAM 2015
Q32.
Temperature dependence of resistivity of a metal can be described by
(a)
(b) R
R
T
T
(c)
R
(d)
T
R
T
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Ans. : (a)
Solution: Electrical resistivity of metal varies as
 T5
(For T   D )
 T
(For T   D )
where  D is the Debye temperature. Thus, correct answer is option (a)
Q33.
A Zener regulator has an input voltage in the range 15V  20V and a load current in the
range of 5 mA  20 mA . If the Zener voltage is 6.8V , the value of the series resistor
RS
should be
V0

15  20 V
(a) 390 
6 .8 V

(b) 420 
(c) 440 
(d) 460 
Ans. : Some data is missing. (No answer is possible)
Q34.
Which of the following circuits represent the Boolean expression
S  P  QR  Q P
(a) P
Q
(b) P
Q
S
(c) P
Q
R
S
(d) P
Q
S
R
S
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Ans. : (b)
 

Solution: S  P  QR  QP  P  QR QP   P  QR   QP   PQ
(a)
P
Q
(c)
P
Q
S  PQ  P  Q
(b) P
Q
(d)
PQ
S   P  Q R
R
S  PQ
P
Q
PQ
S  PQR
R
R
Q35.
R
Miller indicates of a plane in cubic structure that contains all the directions 100, 011
and 111 are
(a) 011
(b) 101
(c) 100 
(d) 110 
Ans. : (a)
y
Solution: The name of the plane containing all the directions
111
100 , 011 & 111 is  0 11 or  01 1 
The best suitable answer is option (a)
 011
x
100
z
Q36.
In an ideal Op-Amp circuit shown below R1  3k , R2  1k  and Vi  0.5sin  t
(in Volt). Which of the following statements are true? V
i
(a) The current through R1  The current through R2
(b) The potential at P is V0
R2
R1


V0
P
R2
R1
(c) The amplitude of V0 is 2V
(d) The output voltage V0 is in phase with Vi
Ans. : (a), (c) and (d)
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Solution: Voltage at P is VP 
V0 R2
.
R1  R2
Current through R1 is I1 
Vo  VP

R1
and Current through R2 is I 2 
Vo 
V0 R2
V0 R1
V0
R1  R2


R1
R1  R1  R2   R1  R2 
V0
VP
.

R2  R1  R2 
Thus I1  I 2 . Option (a) is true
The potential at P is VP 
V0 R2
. Option (b) is not true.
R1  R2
 R 
 3
V0   1  2  Vi  1   0.5sin t  2sin t  Vm  2 V . Option (c) is true
R1 
 1

Option (d) is true
Q37.
In the given circuit VCC  10V and   100 for n  p  n transistor. The collector voltage
VCC
VC (in volts) is………….
1K
100 K
VC

5V
Ans. : 5.7
Solution: I B 

5  0.7
 4.3 105 A  I C   I B  4.3 mA
3
100  10
 VC  VCC  I C RC  10  4.3  5.7 V
Q38.
A diode at room temperature kT  0.025 eV  with a current of 1A has a forward bias
voltage V F  0.4V . For V F  0.5V , the value of the diode current (in A ) is…………..
Ans. : 54.5
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Solution: I  I 0  e
V / VT
Q39.
V2 / VT
 1  e0.5/ 0.025  1  e 20  1
I2 e
 1   V / V


 54.5  I 2  54.5  A
I1  e 1 T  1  e0.4 / 0.025  1  e16  1
GaAs has a diamond structure. The number of Ga-As bonds per atom which have to be
broken to fracture the crystal in the 001 plane is………..
Ans. : 4
Solution: Diamond structure has tetrahedral bond. To fracture the diamond structure along
 0 0 1
plane, four bonds need to be broken.
IIT-JAM 2016
Q40.
The solution of the Boolean equation Y  A  B  AB is
(b) AB
(a) 1
(c) A B
(d) A  B
Ans. : (b) and (d) both are correct


Solution: Y  A  B  AB  A.B  A  B  A 1  B  B  A  B or AB
Q41.
If a constant voltage V is applied to the input of the following OPAMP circuit for a
time t , then the output voltage V0 will approach
C
R
V


Vo
(a) V exponentially
(b) V exponentially
(c) V linearly
(d) V linearly
Ans. : (d)
Solution:  I R  I C 
d  0  V0 
dV
V
V
V 0
 0 
 V0  
t c
C
dt
RC
RC
R
dt
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Q42.


A pn junction was formed with a heavily doped 1018 cm 3 p -region and lightly doped
10
14

cm 3 n - region. Which of the following statement  s  is (are) correct?
(a) The width of the depletion layer will be more in the n - side of the junction
(b) The width of the depletion layer will be more in the p - side of the junction
(c) The width of the depletion layer will be same on the both side of the junction
(d) If the pn junction is reverse biased, then the width of the depletion region increase
Ans. : (a), (d)
Solution: Since p - region is heavily doped and n - region is lightly doped, on n - side width of
depletion layer will be more and on p - side width of the depletion layer will be less.
Q43.
The addition of two binary numbers 1000.01 and 0001.11 in binary representation
is………….
Ans. : 1010
Solution:
1000.01
0001.11
1001.00
Q44.
The number of second-nearest neighbor ions to a Na + ion in NaCl crystal is__________.
Ans. : 12
Solution: The 2nd nearest neighbour is at distance 
The number of 2nd nearest neighbour 
Q45.
2a
a

2
2
Cl 
Na 
3 8
 12
2
The output voltage V0 of the OPAMP circuit given below is…………….. V
2R
1V
2V
3V
R

R

Vo
R
R
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Ans. : 6
2R
 2R 
Solution: V0  1 
 V1  3V1
R 

R/2
R/2
R/2
1 
2
3
Where, V1 
R R/2
R R/2
RR/2
1V
2V
3V
1
1
1
V1  1   2   3  2V  V0  6 V
3
3
3
Q46.
R

R

V0
R
R
In the circuit given below, the collector to emitter voltage VCE is…………… V .
(Neglect VBE , take   100 )
VCC  10 V
5k 
5k
VCE
5k
10 k 
Ans. : 2.5
Solution: VB 
V
5
5
10  5 V  VE  VB  VBE  5 V  I E  E 
 0.5 mA
55
RE 10
VCE  VCC  I C  RC  RE   10  0.5  5  10   2.5 V
Q47.
X -ray diffraction of a cubic crystal gives an intensity maximum for Bragg angle 200
corresponding to the 110  plane. The lattice parameter of the crystal is………….. nm .
(Consider wavelength of X  ray  0.15 nm )
Ans. : 0.31
Solution: According to Bragg’s law 2d sin   n

0.5 109
0.15  109
For n  1 , 2d sin     d 

d 
 0.219 nm
2sin  2  sin 200
2  0.342
a
a

Now, d 
2
h2  k 2  l 2
 a  2 d  0.31 nm  a  0.31 nm
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Q48.
Consider the following circuit with two identical Si diodes. The input ac voltage
waveform has the peak voltage VP  2V , as shown
R
P
Vin
D1
2V
D2
R
t
Q
The voltage waveform across PQ will be represented by:
VPQ
VPQ
(a)
(b)
t
t
VPQ
(c)
VPQ
(d)
t
t
Ans. : (c)
Solution:
R
P
Vin
D1
2V
D2
R
t
Q
During positive half cycle D1 is ON, when input is more than 0.7V and D1 is OFF when
input is less than 0.7V.
During negative half cycle D1 is ON, when input is more negative than   0.7  IR  and
D1 is OFF when input is less negative than   0.7  IR  .
Q49.
If the Boolean function Z  PQ  PQR  PQRS  PQRST  PQRSTU , then Z is

(a) PQ  R S  T  U
(c) P  Q

(b) P Q
(d) P  Q  R  S  T  U
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Ans. : (c)
Solution: Z  PQ  PQR  PQRS  PQRST  PQRSTU
 PQ  PQR  PQRS  PQRST 1  U   PQ  PQR  PQRS  PQRST
 PQ  PQR  PQRS 1  T   PQ  PQR  PQRS
 PQ  PQR 1  S   PQ  PQR  PQ 1  R   PQ  Z  PQ  P  Q
Q50.
Shown in the figure is a combination of logic gates. The output values at P and Q are
correctly represented by which of the following?
1
P
Q
0
(a) 0 0
(b) 1 1
(c) 0 1
(d) 1 0
Ans. : (c)
1
Solution:
P 0
1
0
Q 1
0
Q51.
A plane in a cubic lattice makes intercepts of a,
2a
a
with the three
and
3
2
crystallographic axes, respectively. The Miller indices for this plane are:
(a)  2 43
(b)  34 2 
(c)  634 
(d) 123
Ans. : (a)
Solution: Intercepts:
a,
Divide by a :
1,
Reciprocal:
1,
2,
LCM:
2,
4,
2a
3
2
3
3
2
3
Miller Indices =
2
4
3
a
,
2
1
,
2
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Q52.
Which one of the following schematic curves best represents the variation of conductivity
 of a metal with temperature T ?
(a)
(b)


0
(c)
T
(d)

0
T
0
T
0
T

Ans. : (b)
Solution: Resistivity of metal is   0   ph , where  0 is temperature independent
And,  ph  T 5 when T   D and  ph  T when T   D
The conductivity is defined as  
1




S
0
T
T
Thus correct option is (b)
Q53.
KCl has the NaCl type structure which is fcc with two-atom basis, one at  0, 0, 0  and
the other at 1/ 2,1/ 2,1/ 2  . Assume that the atomic form factors of K  and Cl  are
identical. In an X - ray diffraction experiment on KCl , which of the following  h k l 
peaks will be observed?
(a) 10 0 
(b) 110 
(c) 111
(d)  2 0 0 
Ans. : (d)
Solution In KCl the reflection from 111 layer containing K  ions is exactly out of phase with
the reflection from the Cl  close packed layers. The net effect is that the two reflections
cancel and 111 is absent. This mean the first reflection is from  200 
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Q54.
An n  p  n transistor is connected in a circuit as shown
6V
in the figure. If I C  1 mA,   50,VBE  0.7 V and the
current through R2 is 10 I B where, I B is the base current.
R1
2.5 k
Then the ratio R1 / R2 is:
VC
(a) 0.375
VCE  3V
(b) 0.25
VE  0.5V
R2
(c) 0.5
0.5k
(d) 0.275
Ans. : (b)
Solution: I1  I i  I 2  I 2  10 I B , I C  1 mA,   50,VBE  0.7 V
VB  I 2 R2  VBE  I E RE  0.7  0.5  1.2 V
6V

R1
1.2 V 1.2 V
1.2 V
 R2 


 6 k
I2
10 I B 10  1/ 50 
V R
6 6
 R1  24 k 
VB  CC 2  1.2 
R1  R2
R1  6

Q55.
6V

I1
2.5 k
C
VC
Ii
VCE  3V
B
R2
I2
VE  0.5V
E
0.5k
R2
6

 0.25
R1 24
An intrinsic semiconductor of band gap 1.25eV has an electron concentration 1010 cm 3
at 300 K . Assume that its band gap is independent of temperature and that the electron
concentration depends only exponentially on the temperature. If the electron
concentration at 200 K is Y  10 N cm 3 1  Y  10; N  integer  , then the value of N
is…………
Ans. : 4
Solution: ni  N c N v e
 Eg / 2 kT

 E / 2 kT
 Eg  1
ni1 e g 1
1010 cm 3
1 
  Eg / 2 kT2 
 exp  



ni 2 e
ni 2
 2k  300 200  
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 Eg  1
 1.25 1.6  1019  1  
1 
10
10
6
10
exp
 ni 2  1010  exp  




     10  5.7 10

23
 2  1.38  10  100  6  
 2k  300 200  
4
3
 ni 2  5.7 10 cm  N  4
Q56.
An OPAMP is connected in a circuit with a Zener diode as shown in the figure. The value
of resistance R in k  for obtaining a regulated output V0  9V is…………..
1k 
(Specify your answer to two digits after the decimal point)
R


1k 
Vin  12 V
Vo
VZ  4.7 V
Ans. : 1.09
 1
Solution: V0  1    4.7  9V  R  1.09k 
 R
IIT-JAM 2018
Q57.
Which one of the following graphs shows the correct
variation of V0 with Vi ? Here, Vd is the voltage drop across
the diode and the OP-Amp is assumed to be ideal.
V0
(a)


Vi
Vd
V0
RL
V0
(b)
Vi
0
0
Vd
Vi
V0
V0
(c)
(d)
0
Vd
Vi
0
Vi
Ans. : (a)
Solution: During positive half cycle it behaves as voltage follower i.e. v0  vi , during negative
half cycle v0  0 .
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Q58.
 
The Boolean expression AB  A  B   A  B  can be simplified to
(a) A  B
(b) AB
(c) A  B
(d) AB
Ans.: (c)
Solution: Y  AB  AB   A  B   AB  AB  AB 
 
 
  A  B  AB  AB   AB  AB  AB  A  B
Q59.
In a pn junction, dopant concentration on the p -side is higher than that on the n -side.
Which of the following statements is (are) correct, when the junction is unbiased?
(a) The width of the depletion layer is larger on the n -side.
(b) At thermal equilibrium the Fermi energy is higher on the p -side.
(c) In the depletion region, number of negative charges per unit area on the p -side is
equal to number of positive charges per unit area on the n -side
(d) The value of the built-in potential barrier depends on the dopant concentration.
Ans. : (a), (c) and (d)
Q60.
Which of the combinations of crystal structure and their coordination number is (are)
correct?
(a) body centered cubic 8
(b) face centred cubic 6
(c) diamond 4
(d) hexagonal closed packed 12
Ans. : (a), (c), (d)
Solution: Co-ordination number in different crystal structure are
(i) Body central cubic 8
(ii) Face central cubic 12
(iii) Diamond 4
(iv) Hexagonal closed packed 12
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Q61. For the given circuit, value of the base current  I b  of the npn transistor will be ____mA.
(  is the current gain and assume Op-Amp as ideal.)
(Specific your answer in mA upto two digits after the decimal point.)
10V
1k
5V

Ib

  50
1k
Ans. : 0.1
10V
Solution: v2  5V  VE
IE 
1k
5V
5
 5 mA
1
 I B  5 mA  I B 
Q62.
5
mA  0.1 mA
50
v1
 
v 
2
Ib
E   50
I E  1k
The lattice constant of unit cell of NaCl crystal is 0.563 nm . X -rays of wavelength
0.141 nm are diffracted by this crystal. The angle at which the first order maximum
occurs is _______degrees.
Ans. : 12.5
Solution: 2d sin   
 3 
  
1

  sin 
 2d 
 2a 
  sin 1 
 3  0.141 
1
0
 sin 1 
  sin  0.217   12.53
2
0.563



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Q63.
For the following circuit, the collector voltage with respect to ground will be _________
V . (Emitter diode voltage is 0.7V and  DC of the transistor is large)
(Specify your answer in volts upto one digit after the decimal point).
10V
3k
3k
1k
Ans. : 3.1
3V
Solution: VBE  0.7 V and  dc  large so I B  0
10V
From input section;
3k
C
0  0.7  I E 1  3  0
B

 I E  2.3 mA
From output section;
IB  0
10  3  2.3  VC  0
3k
 VC  10  6.9  3.1V
Q64.
E
1k
3V
In the following circuit, the time constant RC is much greater than the period of the
input signal. Assume diode as ideal and resistance R to be large. The dc output voltage
across resistance R will be ____________ V .
(Specify your answer in volts upto one digit after the decimal point)
C
24Vrms

C
R
Ans. : 68
Solution: It’s a voltage doubler circuit
VR  2Vm  2

 
2 Vrms  2

2  24  VR  68 V
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Q65.
For a metal, the electron density is 6.4  1028 m 3 . The Fermi energy is __________ eV .
( h  6.626  1034 J s, me  9.111031 kg , 1 eV  1.6 1019 J )
(Specify your answer in electron volts  eV  upto one digit after the decimal point)
Ans. : 5.84
2
Solution: EF 
3 2 n
2m


1.05 10 

34
2/3
2
2  9.111031

 3
2
 6.4 1028

2/3

 6.11039 1.53 1020  9.34 1019 J

9.34 1019
eV  5.84eV .
1.6 1019
IIT-JAM 2019
Q66.
Which one of the following crystallographic planes represent 101 Miller indices of a
cubic unit cell?
z
z
z
(a)
(b)
(c)
x
x
(d)
y
y
m
z
x
y
x
Ans. : (b)
Solution: Plane intercepts
x: y:z 
a b c a b c
: :  : :  a::c
h k l 1 0 1
 x  a , y  , z  c
Plane is parallel to y - axis and intersecting x and z - axis at a and c . Thus, option (b)
is correct.
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Q67.
For using a transistor as an amplifier, choose the correct option regarding the resistances
of base-emitter  RBE  and base-collector  RBC  junctions
(a) Both RBE and RBC are very low
(b) Very low RBE and very high RBC
(c) Very high RBE and very low RBC
(d) Both RBE and RBC are very high
Ans. : (b)
Q68.
The output of following logic circuit can be simplified to
X
Y
Z
(a) X  YZ
(b) Y  XZ
(d) X  Y  Z
(c) XYZ
Ans. : (b)
Solution: Output  XY  X Y  Z   Y Y  Z   XY  XY  XZ  Y  YZ
 XY  XZ  Y  Y 1  X   XZ  Y  XZ
Q69.
For a forward biased p-n junction diode, which one of the following energy-band
diagrams is correct (  F is the Fermi energy)
Conduction Band
p -type
n-type
(a)
Electron 
F  p
Energy
Conduction Band
p -type
n-type
(b)
 F  n
Valence Band
Conduction Band
p -type
n-type
(c)
 F  p
Valence Band
Valence Band
Conduction Band
p -type
n-type
(d)
 F  n
Electron
Energy
F
Electron
Energy
 F  p
Electron
Energy
 F  n
Valence Band
Ans. : (a)
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Q70.
The location of Cs  and Cl  ions inside the unit cell of cacl crystal is shown in the
figure. The Bravais lattice of CaCl is z
Cl 
a
y
Cs 
a
a
x
(a) Simple cubic
(b) Body centred orthorhomble
(c) Face centred cubic
(d) Base centred orthorhombic
Ans. : (a)
Solution: Cesium-Chloride is made of two interpenetrating simple cubic lattices are displaced
diagonally by half of the diagonal length. Thus, Bravais lattice of CsCl is simple cubic.
The correct option is (a).
Q71.
In the X -ray diffraction pattern recorded for a simple cubic solid (lattice) parameter
o
o
a  1 A ) using X -rays of wavelength 1 A , the first order diffraction peak(s) would
appear for the
(a) 100  planes
(b) 112  planes
(c)  210  planes
(d)  220  planes
Ans. : (a)
Solution: In simple cubic cell, planes are present. The first order diffraction peak would appear
for the first plane 10 0  .
Q72.
Sodium  Na  exhibits body-centred cubic (BCC) crystal structure with atomic radius
0.186 nm . The lattice parameter of Na unit cell is______ nm .
Ans. : 0.43
Solution: For BCC,
3a  4r
a
4r
3

4  0.186
3
 a  0.43 nm
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Q73.
For the input voltage V1  200 mV  sin  400 t  , the amplitude of the output voltage V0  of
the given OPAMP circuit is __________ V . (Round off to 2 decimal places)
35k 
35k 
Rf
10k 
R1


35k 
Rf
10k 
R2


Rf
10k 
R3


Vi
V0
Ans. : 11.03
 35 
Solution: v01   1   vi   4  5  200mV  sin  400t 
 10 
v02  
v0  
35
  4  5  200mV  sin  400t 
10
35  35 

  4  5  200mV  sin  400t 
10  10 
Vm   3.5  3.5  4.5  200  mV  11.03Volts
Q74.
The value of emitter current in the given circuit is _______  A .
(Round off to 1 decimal places)
VCC  10V
2M
B
0.3V
2k 
4k 
C
  100
E

C
 E
Ans. : 444.9
Solution: I B 
VCC  VBE
RB     1 RE
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IB 
10  0  3
9.7

A
3
2  10  101 2 10
2.202 106
6
I E     1 I B  101
The Zener current I z for the given circuit is ______ mA .

VR

IZ
VZ = 20 V
Vin = 40 V
R  10k 
RL = 20 k
Q75.
9.7
 A  444.9 A
2.202
VL
Ans. : 1
Solution: Open circuit voltage Vi 
20k
2
 40V   40  26.7 Volts
20k  10k
3
Vi  VZ , Zener “ON”
IL 
VZ 20
40  20

 1mA and I R 
 2mA
10
RL 20
I Z  I R  I L  1mA
Q76.
The decimal equivalent of the binary number 110.101 is _______.
Ans. : 6.625
Solution: 110 101  12  1 21  0  20 1 21  0  22  1 23
1
1
  4  2  0    0  
8
2
 6   0.5  0  0.125   6.625
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