33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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142 NEET-AIPMT Chapterwise Topicwise Solutions Physics
92. (c) : Here, Voltage gain = 50
Input resistance, R i = 100 W
Output resistance, R o = 200 W
Ro
200 Ω
Resistance gain = = = 2
Ri
100 Ω
( Voltagegain) 2 50 × 50
Power gain = = = 1250
Resistance gain 2
93. (c)
94. (c) : For common emitter, the current gain is
β= ⎛ ∆I
⎞
⎝ ⎜ C
∆I
⎠
⎟
B V CE
i.e., at a given potential difference of CE
−3 −3
( 10 × 10 − 5×
10 ) A
−3
5×
10
β=
= = 50
−6 −6
−6
( 200 × 10 − 100 × 10 ) A 100 × 10
95. (b) : One applies negative feedback, which
reduces the output but makes it very stable. For voltage
amplification amplifiers the value of output voltage
without the negative feedback could be very high. The
maximum value shown here is 100.
96. (d) : Current gain, b = DI C /DI B
( 10 − 5)
mA
−3
5×
10
=
= = 100
( 150 −100)
µA −6
50 × 10
97. (b) : A n-p-n transistor conducts when emitter-base
junction is forward biased while collector-base junction
is reverse biased.
98. (d) : The current gain of a common emitter
transistor (b) is defined as the ratio of collector current
(I C ) to the base current (I B ).
Also, I E = I B + I C ; I C /I E = 0.96 (given)
∴ β = I C IC
=
IB
IE
− IC
IE
Now, = 1
IC
096 . ∴ IE
− IC
1
= − =
IC
096 1 004 .
. 096 .
IC
096 .
∴ β = = = 24
IE
− IC
004 .
99. (a) : I C
IC
α
= α= 098 . , = β= = 49
IE
IB
1 − α
Ic
Ic
Ic
/ Ie
α
100. (b) : β = = = = .
Ib
Ie
− Ic
1 − ( Ic
/ Ie)
1 − α
V
101. (d) : Ib
= in = 001 .
Rin
10 3
−
Ic
= βIb
= 50 × 001 .
4
= 5× 10 A=
500 mA
3
10
102. (c) : In n-p-n transistor, the electrons are majority
carriers in emitter, which move from base to collector
while using n-p-n transistor as an amplifier.
103. (a) : The function of emitter is to supply the majority
carriers. So, it is heavily doped.
104. (a) : To use transistor as an amplifier the emitter
base junction is forward bias while the collector base
junction is reverse biased.
105. (a) : The phase difference between output voltage
and input signal voltage in common base transistor or
circuit is zero.
106. (d) : Radiowaves of constant amplitude can be
produced by using oscillator with proper feedback.
107. (a)
108. (a) : Distance between nearest atoms in body
centred cubic lattice (bcc), d= 3 a
37 . × 2
2
Given d = 3.7 Å, a = 43 . Å
3
109. (b) : The atomic radius in a f.c.c. crystal is
a
2 2
where a is the length of the edge of the crystal.
36 . Å
∴ Atomic radius = = 127 . Å
2 2
110. (c) : In a cubic crystal structure
a = b = c, a = b = g = 90°.
111. (b) : Lattice constant for (f.c.c.)
= a = interatomicspacing × 2 = 3. 59 Å
112. (c) : In body-centred cubic (b.c.c.) lattice there are
eight atoms at the corners of the cube and one at the
centre as shown in the figure.
Therefore number of atom per unit cell
1
= × + =
8 8 1 2
113. (b)
114. (c) : a = 4.225 Å
For bcc cubic cell, 4r
= 3×
a
3 × a 1. 732 × 4.
225
Therefore2r
= = = 366 . Å
2 2
115. (b) : Diamond is very hard due to large cohesive
energy.
116. (c) : van der Waals bonding is the weakest bonding
in solids.
vvv