MCQ Questions for Class 12 Medical Physics Semiconductor Electronics: Materials, Devices And Simple Circuits Quiz 2

In a

$p-n-p$ transistor the base is the n-region. Its width relative to the p-regions is

0%
smaller
0%
larger
0%
same
0%
not related

Explanation

In a

$p-n-p$ transistor the base is the n-region. Its width relative to the p-regions is smaller since in transistors the base is made very thin.

In positive logic, the logic state

$1$ corresponds to

0%
positive voltage
0%
zero voltage
0%
lower voltage level
0%
higher voltage level

Explanation

In digital logic, higher voltage is defined as logic state '

$1$ ' and lower voltage is defined as logic state '

$0$ '. The higher voltage need not necessarily be positive. For example, it is possible that state '

$0$ ' is defined as

$-10\ V$ and state '

$1$ ' is defined as

$-5\ V$ .

In Boolean algebra,

$A + B = Y$ implies that

0%
sum of $A$ and $B$ is $Y$ .
0%
$Y$ exists when $A$ exists or $B$ exists or both $A$ and $B$ exist.
0%
$Y$ exists only when $A$ and $B$ both exist.
0%
$Y$ exists when $A$ or $B$ exist but not when both $A$ and $B$ exist.

Explanation

In Boolean algebra,

$A + B = Y$ implies that

$Y$ exists when

$A$ exists or

$B$ exists or both

$A$ and

$B$ exist.

In a transistor the base is made very thin and is lightly doped with an impurity, because

0%
to enable the collector to collect about $95\%$ of the holes or electrons coming from the emitter side
0%
to enable the emitter to emit small number of holes or electrons
0%
to save the transistors from high current effects
0%
to enable the base to collect about $95\%$ of holes or electrons coming from the emitter side

Explanation

In a transistor the base is made very thin and is lightly doped with an impurity, so as to enable the collector to collect about

$95\%$ of the holes or electrons coming from the emitter side.

A p-n-p transistor is said to be in active region of operation when

0%
both emitter junction and collector junction are forward biased.
0%
both emitter junction and collector junction are reverse biased.
0%
emitter junction is forward biased and collector junction is reverse biased.
0%
emitter junction is reverse biased and collector junction is forward biased.

The part of a trasistor which is heavily doped to produce large number of majority carriers is :

0%
emitter
0%
base
0%
collector
0%
can be any of the above three

Which of the following is not the function of a NOT gate?

0%
Stop a signal.
0%
Invert an input signal.
0%
Complement a signal.
0%
Change the logic in a digital circuit.

Explanation

A NOT gate inverts the input signal which is the same as complementing a signal or changing the logic in a digital circuit. This means that when the input to the NOT gate is logic '

$0$ ', the output is logic '

$1$ '. However, it does not stop a signal.

When an input signal

$1$ is applied to a NOT gate, its output is

0%
$1$
0%
$0$
0%
either $0$ or $1$
0%
both $0$ and $1$

Explanation

NOT gate yields the reverse of the input signal in output, thus when an input signal

$1$ is applied to a NOT gate, its output is

$0$ .

The symbol represents

0%
NOT gate
0%
OR gate
0%
AND gate
0%
NOR gate

Explanation

The symbol represented above by definition is of NAND gate which gives output only when both the input signals are low i.e

$0$

A transistor has a base current of

$1\ mA$ and emitter current

$100\ mA$ . The current transfer ratio will be

0%
$0.9$
0%
$0.99$
0%
$1.1$
0%
$10.1$

Explanation

Given that:

$I_b = 1\ mA,$

$I_e = 100\ mA$

Hence,

$I_c = I_e - I_b = 100 - 1 = 99\ mA$
Current transfer ratio is

$\dfrac{I_c}{I_e} = \dfrac {99}{100} = 0.99$

An AND gate is followed by a NOT gate in series. With two inputs

$A$ &

$B$ , the Boolean expression for the out put

$Y$ will be :

0%
$A.B$
0%
$A + B$
0%
$\overline{A+B}$
0%
$\overline{A.B}$

Explanation

Expression for AND is

$A.B$ and if it is followed by a NOT gate then the Boolean expression will just be the complement of AND.

The gate that has only one input terminal

0%
NOT
0%
NOR
0%
NAN
0%
XOR

In Boolean algebra

$A.B = Y$ implies that :

0%
product of $A$ and $B$ is $Y$
0%
$Y$ exists when $A$ exists or $B$ exists
0%
$Y$ exists when both $A$ and $B$ exist but not when only $A$ or $B$ exists
0%
$Y$ exists when $A$ or $B$ exists but not both $A$ and $B$ exist.

Explanation

In Boolean algebra

$A.B = Y$ implies that

$Y$ exists when both

$A$ and

$B$ exist but not when only

$A$ or

$B$ exists.

In the Boolean algebra, the following one is wrong

0%
$1 + 0 = 1$
0%
$0 + 1 = 1$
0%
$1 + 1 = 1$
0%
$0 + 0 = 1$

Explanation

In the Boolean algebra, binary addition is given as:

$1 + 0 = 1$

$0 + 1 = 1$

$1 + 1 = 1$
Hence,

$0 + 0 = 1$ is wrong.

The output of a 2-input OR gate is zero only when its

0%
both inputs are $0$ .
0%
either input is $1$ .
0%
both inputs are $1$ .
0%
either input is $0$ .

NOR gate is the series combination of

0%
NOT gate followed by OR gate.
0%
OR gate followed by NOT gate.
0%
AND gate followed by OR gate.
0%
OR gate followed by AND gate.

The following truth table is for :

A	B	Y
1	1	0
1	0	1
0	1	1
0	0	1

0%
NAND gate
0%
AND gate
0%
XOR gate
0%
NOT gate

$A = B = 1$ , then in terms of Boolean algebra the value of

$A.B + A$ is not equal to

0%
$B.A+B$
0%
$B+A$
0%
$B$
0%
$\bar{A}.B$

Explanation

For

$A = B = 1$ ,

$\Rightarrow A.B+B = 1.1 + 1 = 1 + 1 = 1$

Therefore,

(A)

$B.A+B = 1.1 + 1 = 1 + 1 = 1$

(B)

$B+A = 1 + 1 = 1$

(C)

$B = 1$

(D)

$\bar A.B=1.1+1 = \bar 1.1 = 0.1 = 0$

Hence, option D is the correct answer.

In the Boolean algebra, of the following one which is not equal to

$A$ is

0%
$A.A$
0%
$A + A$
0%
$\bar{A}.A$
0%
$\overline{\bar A + \bar A}$

Explanation

The truth table for AND gate is given as shown in the figure.
When

$A$ is true,

$\bar{A}$ is false and vice versa.

Hence value of

$\bar{A}.A$ is always false(looking at the second and third row of the table) which is not equal to

$A$ .

Hence, option C is the answer.

In the Boolean algebra :

$A + B =$

0%
$\bar{A}$ + $\bar{B}$
0%
A.B
0%
$\bar{\bar{A}}+\bar{\bar{B}}$
0%
$\bar{\bar{A}}$ + $\bar{B}$

Explanation

Input	$\bar A+\bar B$	$A.B$	$\bar{\bar A} + \bar{\bar B}$	$\bar {\bar A} + \bar B$
$A=B=0$	1	0	0	1
$A=B=1$	0	1	1	1
$A=0, B=1$	1	0	1	1
$A=1, B=0$	1	0	1	1

A change of

$400\ mV$ in base-emitter voltage causes a change of

$200$

$\mu A$ in the base current. The input resistance of the transistor is

0%
$1K$ $\Omega$
0%
$6K$ $\Omega$
0%
$2K$ $\Omega$
0%
$8K$ $\Omega$

Explanation

Given that:

$V_{BE} = 400\ mV = 4 \times 10^{-1} V$

$I_B = 200\; \mu A = 2 \times 10^{-4} A$

The input resistance of the transistor is

$r_i = \dfrac {V_{BE}}{I_B}$

$r_i = \dfrac {4 \times 10^{-1}}{2 \times 10^{-4}}$

$r_i = 2 \times 10^{3} = 2 K\Omega$

For a p-n-p transistor in CB configuration, the emitter current

$I_{E}$ is

$1\ mA$ and

$\alpha$

$=$

$0.95$ . The base current and collector current are

0%
$0.95\ mA, 0.05\ mA$
0%
$0.05\ mA , 0.95\ mA$
0%
$9.5\ mA, 0.5\ mA$
0%
$0.5\ mA, 9.5\ mA$

Explanation

Given that:

$I_e = 1\ mA$

$\alpha = 0.95$
For transistor in CB mode,

$\alpha = \dfrac{I_c}{I_e}$

$I_c = \alpha I_e = (0.95)(1) = 0.95\ mA$
Also, for transistors

$I_e = I_b + I_c$

$I_b =I_e - I_c = 1 - 0.95 = 0.05\ mA$

$A = 1$ and

$B = 0$ , then in terms of Boolean algebra the value of

$A.A + B$ is

0%
$A$
0%
$B^2$
0%
$B$
0%
$A \cdot B$

Explanation

For

$A=1$ and

$B=0$

$A.A+B=1.1+0=1$

Hence,

$A.A+B=A$ ............(since,

$A=1$ )

$A = 1, B = 0$ then the value of

$\bar {A} + B$ in terms of Boolean algebra is

0%
$A$
0%
$B$
0%
$B + A$
0%
$A.\bar{B}$

Explanation

For

$A = 1, B = 0$

$\bar A+B = \bar 1 + 0 = 0 + 0 = 0$

Hence,

$\bar A+B = B$ .............(since,

$B= 0$ )

The following one represents logic addition is

0%
$1 + 1 = 2$
0%
$1 + 1 = 10$
0%
$1 + 1 = 1$
0%
$1 + 1 = 11$

Explanation

Logic addition is the addition of binary numbers, it is represented by

$1 + 1 = 1$

The value of

$\alpha$ of a transistor is

$0.9$ . The change in the collector current corresponding to a change of

$4\ mA$ in the base current in a common emitter arrangement is

0%
$36\ mA$
0%
$72\ mA$
0%
$18\ mA$
0%
$54\ mA$

Explanation

Given that:

$\alpha = 0.9$

$I_b = 4\ mA$
For transistors,

$\beta = \dfrac{\alpha}{1-\alpha}$

$\beta = \dfrac{0.9}{1-0.9} = \dfrac{0.9}{0.1} = 9$

Also,

$\beta = \dfrac {I_c}{I_b}$

$I_c = \beta I_b = (9)(4) = 36\ mA$

In the Boolean algebra :

$\bar{A} . \bar{B}=$ ......

0%
$\overline{A+B}$
0%
$A.B$
0%
$\overline{\bar A + \bar B}$
0%
$A + B$

Explanation

Input	$\bar {A}.\bar {B}$	$\overline {A+B}$	A.B	$\overline {\bar A+\bar B}$	A+B
$A=B=0$	1	1	0	0	0
$A=B=1$	0	0	1	1	1
$A=0, B=1$	0	0	0	1	1
$A=1, B=0$	0	0	0	1	1

The output of OR gate is 1 :-

0%
If either one or both inputs are 1
0%
Only if both inputs are 1
0%
If either input is zero
0%
If both inputs are zero

Explanation

At least one input required to activate

$or -gate$
Option A

Identify the logic gate in figure.

0%
NAND
0%
NOR
0%
NOT
0%
AND

The truth table for NOT gate is

Explanation

NOT gate gives the reverse of the input as output. Truth table for NOT gate is:

Input	Output
$1$	$0$
$0$	$1$

The logic expression which is NOT true in Boolean algebra is

0%
$[\bar{1}+\bar{1}].1=0$
0%
$[\bar{1}+0].1=0$
0%
$[\bar{1}+0].\bar{1}=0$
0%
$[1+1].1=0$

Explanation

$[\bar 1+\bar 1].1= [0+0].1 = 0$

$[\bar 1+0].1= [0+0].1 = 0$

$[\bar 1+0].\bar 1= [0+0].0 = 0$

$[1.1].1 = 1.1 = 1 \neq 0$

Which one of the following statement is false

0%
Work is a state finction
0%
temperature is a state function
0%
Change in the state is completely defined when the initial and final states are specified
0%
Work appears at the boundary of the system

Explanation

Majority carrier in a

$n-$ type semiconductor are holes.This statement is false.

Since,in

$n-$ type semiconductor, the pentavalent impurity atoms donate electrons o the host crystal and the semiconductor doped with donars (pentavalent impurity) is called

$n-$ type semiconductor.

Therefore majority carrier in a

$n-$ type semiconductor are electrons

The device that can act as a complete electronic circuit is -

0%
Junction diode
0%
Integrated circuit
0%
Junction transistor
0%
Zener diode

Boolean Expression for the gate circuit shown in the figure is

0%
$A. 1 = A$
0%
$A.\bar{A}=0$
0%
$A.A = A$
0%
$A.0 = 0$

Explanation

The input

$A$ directly goes into the AND gate and also first passes through NOT gate, giving another input of

$\bar{A}$ to the AND gate.

$A.\bar{A}=0$ since atleast one of

$A$ and

$\bar{A}$ is

$0$ .

NAND gate in the following is

The logic gate having following truth table is
A B Y
0 0 1
0 1 1
1 0 1
1 1 0

0%
XOR
0%
OR
0%
AND
0%
NAND

Why are LEDs extensively used to replace bulbs?

0%
(i) and (ii) only
0%
(ii) and (iii) only
0%
(iii) and (i) only
0%
(i), (ii) and (iii)

Semiconductors are generally made up of which substance?

0%
Silicon
0%
Carbon
0%
Phosphorus
0%
Boron

The device which produces highly coherent sources is

0%
Fresnel bi-prism
0%
Young's double slit
0%
LASER
0%
Lloyd's mirror

In an intrinsic semiconductor, the number of electrons in the conduction band is ________ the number of holes in the valence band.

0%
equal to
0%
less than
0%
greater than
0%
none of these

Explanation

In an intrinsic semiconductor, the number density of electrons is equal to the number density of holes i.e

$ne=nh$ .
Since there is no doping, no extra hole or electron is produced.

Given figure is the symbol, for which component of a circuit?

0%
LED
0%
Rheostat
0%
Transistor
0%
Resistor

Explanation

The given symbol is of

$transistor$ in electric circuit.

Answer-(C)

Which are good conductors ?

0%
silicon
0%
Silver
0%
copper
0%
glass

The energy gap in glass at room temperature is

0%
Greater than that in a semiconductor.
0%
Less than that in a good conductor.
0%
Greater than that in a good conductor.
0%
Both (1) and (3) are true.

Materials which allow only larger currents to flow through them are

0%
Insulators
0%
Semi-conductors
0%
Conductors
0%
Alloys

In an intrinsic semiconductor, if

$N_e$ is the number of electrons in the conduction band and

$N_p$ is the number of holes in the valence band then.

0%
$N_e > N_p$
0%
$N_e = N_p$
0%
$N_e < N_p$
0%
None of the above

Explanation

In an intrinsic semiconductor, the number of electrons in the conduction band is equal to the number of holes in the valence band, so

$N_{e}=N_{p}$ .

The Wavelength of a LASER beam can be used as a standard of

0%
time
0%
temperature
0%
angle
0%
length

A pure semiconductor at absolute zero has

0%
Absence of electrons in the conduction band.
0%
All the electrons occupying the valence band.
0%
Large $E_g$ value.
0%
All of the above.

Explanation

At absolute zero temperature, in a pure semiconductor, all electrons occupy the valence band and no electrons are present in the conduction band. The forbidden gap energy

$E_g$ is large.

In n-type of semiconductor, majority carries are

0%
Positron
0%
Electron
0%
Holes
0%
Impure particles

In a transistor, the base is

0%
a conductor of low resistance
0%
a conductor of high resistance
0%
an insulator
0%
an extrinsic semiconductor

A transistor has a base current of

$1\ mA$ and emitter current

$90\ mA$ . The collector current will be

0%
$90\ mA$
0%
$1\ mA$
0%
$89\ mA$
0%
$91\ mA$

Explanation

Given: Base current,

$I_b = 1\ mA$ and emittter current

$I_e = 90 \ mA$

Let collector current be

$I_c$

Using,

$I_e = I_b + I_c$

$90 = 1 + I_c \implies I_c = 89\ mA$

CBSE Questions for Class 12 Medical Physics Semiconductor Electronics: Materials, Devices And Simple Circuits Quiz 2 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Semiconductor Electronics: Materials, Devices And Simple Circuits Quiz 2 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.