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Physics M.C.Q (1 Marks)

Semiconductor Devices · Maharashtra Board STD 12 (MSBSHSE - Semi English Medium). 27 questions.

Questions

M.C.Q (1 Marks)

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27 questions · auto-graded multiple-choice test.

MCQ 11 Mark
A logic gate is an electronic circuit which
  • ✓
    makes logical decisions
  • B
    allows electron flow only in one direction
  • C
    works using binary algebra
  • D
    alternates between 0 and 1 value.
Answer
Correct option: A.
makes logical decisions
makes logical decisions
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MCQ 21 Mark
Solar cell operates on the principle of
  • A
    diffusion
  • B
    recombination
  • ✓
    photovoltaic action
  • D
    carrier flow.
Answer
Correct option: C.
photovoltaic action
photovoltaic action
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MCQ 31 Mark
An LED emits visible light when its
  • A
    junction is reverse biased
  • B
    depletion region widens
  • ✓
    holes and electrons recombine
  • D
    junction becomes hot.
Answer
Correct option: C.
holes and electrons recombine
holes and electrons recombine
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MCQ 41 Mark
A series resistance is connected in the Zener diode circuit to
  • ✓
    properly reverse bias the Zener
  • B
    protect the Zener
  • C
    properly forward bias the Zener
  • D
    protect the load resistance.
Answer
Correct option: A.
properly reverse bias the Zener
properly reverse bias the Zener
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MCQ 51 Mark
In a BJT, the largest current flow occurs
  • ✓
    in the emitter
  • B
    in the collector
  • C
    in the base
  • D
    through CB junction.
Answer
Correct option: A.
in the emitter
in the emitter
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MCQ 61 Mark
In a transistor, in common emitter configuration, the ratio of power gain to voltage gain is
  • A
    $\alpha$
  • B
    $\frac{\beta}{\alpha}$
  • C
    $\beta \alpha$
  • ✓
    $\beta$
Answer
Correct option: D.
$\beta$
(d) : Voltage gain, $A_V=\beta \cdot \frac{R_L}{R_i}$
Power gain, $P_V=\beta^2 \cdot \frac{R_L}{R_i} ; \therefore \frac{\text { Power gain }}{\text { Voltage gain }}=\beta$
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MCQ 71 Mark
If the frequency of the input voltage is $50 Hz$, applied to a (a) half wave rectifier and (b) full wave rectifier. The output frequency in both cases is respectively
  • A
    $50 Hz , 50 Hz$
  • B
    $100 Hz , 100 Hz$
  • ✓
    $50 Hz , 100 Hz$
  • D
    $100 Hz , 50 Hz$
Answer
Correct option: C.
$50 Hz , 100 Hz$
(c) : In half wave rectifier, frequency remains same and for full wave rectifier, frequency is doubled.
So output frequency in both cases is $50 Hz$ and $100 Hz$ respectively.
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MCQ 81 Mark
A $5,0 V$ stabilized power supply is required to be designed using a $12 V$ DC power supply as input source. The maximum power rating of zener diode is $2.0 W$. The minimum value of resistance $R_s$ (in $\Omega$ ) connected in series with zener diode will be
  • A
    16.5
  • ✓
    17.5
  • C
    18.5
  • D
    15.5
Answer
Correct option: B.
17.5
(b) : Given, $P=2 W$
Let current is $i$.
$
2=5 i \quad(P=V i)
$
$
i=\frac{2}{5} A
$
Now, $\varepsilon=12=5+i R$
$
R=\frac{7}{i}=\frac{7 \times 5}{2}=17.5 \Omega
$
Image
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MCQ 91 Mark
To get the truth table shown, from the following logic circuit, the Gate $G$ should be
Image
  • A
    $OR$
  • ✓
    AND
  • C
    NOR
  • D
    NAND
Answer
Correct option: B.
AND
(b)
Image
So, $G$ is AND gate.
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MCQ 101 Mark
In an $n-p-n$ transistor, the collector current is 28 $mA$. If $80 \%$ of electrons reach the collector, its base current in $mA$ is
  • A
    35
  • B
    24
  • C
    14
  • ✓
    7
Answer
Correct option: D.
7
(d) : Given, $I_c=28 mA ; I_e=I_c+I_b$
$80 \%$ of $I_e=I_c=28 mA$
$
\frac{80}{100} I_e=28 mA ; \quad I_e=\frac{2800}{80} mA
$
$\begin{aligned} & I_b=I_e-I_c=\left(\frac{2000}{80}-28\right) ; I_b=28\left[\frac{10}{8}-1\right] \\ & I_b=28 \times \frac{2}{8}=7 mA \end{aligned}$
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MCQ 111 Mark
The output of following combination is same as that of
Image
  • A
    AND gate
  • B
    OR gate
  • C
    NAND gate
  • ✓
    NOR gate
Answer
Correct option: D.
NOR gate
(d) : Truth table is
begin{tabular}{ll|l}
Image
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MCQ 121 Mark
For a common emitter configuration, if ' $\alpha$ ' and ' $\beta$ ' have their usual meanings, the incorrect relation between ' $\alpha$ ' and ' $\beta$ '
  • A
    $\frac{1}{\alpha}=\frac{1}{\beta}+1$
  • ✓
    $\alpha=\frac{\beta}{1-\beta}$
  • C
    $\alpha=\frac{\beta}{1+\beta}$
  • D
    $\frac{1}{\beta}=\frac{1}{\alpha}-1$
Answer
Correct option: B.
$\alpha=\frac{\beta}{1-\beta}$
(b) : $\alpha=\frac{I_c}{I_e}, \beta=\frac{I_c}{I_b} ; I_c=I_b+I_c$
$
\begin{aligned}
& l=\frac{I_e}{I_c}=\frac{I_b}{I_c}+1 \Rightarrow \frac{1}{\alpha}=\frac{1}{\beta}+1 \\
& \frac{1}{\alpha}=\frac{1+\beta}{\beta} \Rightarrow \alpha=\frac{\beta}{1+\beta} \Rightarrow \frac{1}{\beta}=\frac{1}{\alpha}-1
\end{aligned}
$
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MCQ 131 Mark
For emission of light, a light emitting diode (LED) is
  • A
    always used in reverse biased condition.
  • B
    never used in forward or reverse biased condition.
  • C
    used both in forward and reverse biased condition depending upon its application.
  • ✓
    always used in forward biased condition.
Answer
Correct option: D.
always used in forward biased condition.
(d) : LED is used in forward biased condition.
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MCQ 141 Mark
The output of a 'NAND' gate is shown in the truthtable ( $A$ and $B$ are inputs, $Y$ is output)
ImageImageImageImage
  • A
    $P$
  • B
    R
  • C
    $Q$
  • ✓
    $S$
Answer
Correct option: D.
$S$
(d) : Trut1h-table of 'NAND' gate is:
Image
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MCQ 151 Mark
In the logic circuit diagram, $A, B$ and $C$ are the inputs, $Y$ is the output. The output $Y$ is 'HIGH'
Image
  • A
    for all the inputs 'LOW".
  • ✓
    when $A=1, B=0, C=0$
  • C
    when $A=1, B=0, C=1$
  • D
    for all the inputs "HIGH".
Answer
Correct option: B.
when $A=1, B=0, C=0$
(b) : $Y=A \cdot(\overline{B+C})$
$
Y=A \bar{B}+A \bar{C}
$
When, $A=1 ; B=0, C=0$ we will get high value of $Y$.
Image
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MCQ 161 Mark
In common emitter transistor amplifier, the output voltage and input voltage have a phase difference of
  • A
    $ \frac{2}{\pi} $
  • B
    $\frac{\pi}{2}$
  • C
    $\frac{5 \pi}{6}$
  • ✓
    $\pi$
Answer
Correct option: D.
$\pi$
(d) : The input voltage and output voltage remain in the opposite phase i.e, $180^{\circ}$.
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MCQ 171 Mark
What is the output $Y$ in the following circuit, when all the three inputs $A, B, C$ are first ' 0 ' then ' 1 '?
Image
  • A
    0,0
  • B
    0,1
  • C
    1,1
  • ✓
    1,0
Answer
Correct option: D.
1,0
(d)
Image
$
Y=\overline{(A B) C}=\overline{A B C}
$
When $A=B=C=0$, then $Y=\overline{0}=1$
When $A=B=C=1$, then $Y=\overline{1}=0$
So output is $(1,0)$
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MCQ 181 Mark
In the study of transistor as an amplifier, the ratio of collector current to emitter current is 0.98 , then the ratio of collector current to base current will be
  • A
    99
  • ✓
    49
  • C
    50
  • D
    98
Answer
Correct option: B.
49
(b) : Given, $\frac{I_C}{I_E}=\alpha=0.98$ Now, we know, $\beta=$ current gain $=\frac{I_C}{I_B}$ and $\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49$
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MCQ 191 Mark
For transistor, the current ratio ' $\beta_{d i}{ }^2$ is defined the ratio of
  • A
    collector current to emitter current
  • ✓
    collector current to base current
  • C
    base current to collector current
  • D
    emitter current to collector current
Answer
Correct option: B.
collector current to base current
(b) : For a transistor dc current gain is ratio of the collector current to the base current.
$
\therefore \beta_{ dc }=\frac{I_c}{I_b}
$
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MCQ 201 Mark
A transistor is used as a common emitter amplifier with a load resistance $2 k \Omega$. The input resistance is $150 \Omega$. Base current is changed by $20 \mu A$ which results in a change in collector current by $1.5 mA$. The voltage gain of the amplifier is
  • A
    900
  • ✓
    1000
  • C
    1100
  • D
    1200
Answer
Correct option: B.
1000
(b) : Voltage gain $=\frac{V_o}{V_i}=\frac{R_o \times I_c}{R_i \times I_b}$
$
=\frac{2000 \times 1.5 \times 10^{-3}}{150 \times 20 \times 10^{-6}}=\frac{3}{3000 \times 10^{-6}}=1000
$
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MCQ 211 Mark
For a transistor, $\alpha_{d c}$ and $\beta_{d c}$ are the current ratios, then value of $\frac{\beta_{ dc }-\alpha_{ dc }}{\alpha_{ dc } \cdot \beta_{ dc }}$ is
  • ✓
    1
  • B
    1.5
  • C
    2
  • D
    2.5
Answer
Correct option: A.
1
(a) : For a transistor, current amplification factor
$
\begin{aligned}
& \beta_{ dc }=\frac{\alpha_{ dc }}{1-\alpha_{ dc }} \\
& 1-\alpha_{ dc }=\frac{\alpha_{ dc }}{\beta_{ dc }} \\
& 1-\frac{\alpha_{ dc }}{\beta_{ dc }}=\alpha_{ dc }.....(i)
\end{aligned}
$
$
\begin{aligned}
\frac{\beta_{ dc }-\alpha_{ dc }}{\alpha_{ dc }-\beta_{ dc }} & =\frac{\beta_{ dc }\left(1-\frac{\alpha_{ dc }}{\beta_{ dc }}\right)}{\alpha_{ dc }-\beta_{ dc }}=\frac{1-\frac{\alpha_{ dc }}{\beta_{ dc }}}{\alpha_{ dc }} \\
& =\frac{\alpha_{ dc }}{\alpha_{ dc }} (From equation (i)) \\
& =1
\end{aligned}
$
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MCQ 221 Mark
The schematic symbol of light emitting diode is (LED)
ImageImageImage
  • ✓
    Image
  • C
    c
Answer
Correct option: B.
Image
(b) : Schematic symbol of light emitting diode
Image
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MCQ 231 Mark
In an oscillator, for sustained oscillations, Barkhausen criterion is $A \beta$ equal to $(A=$ voltage gain without feedback, $\beta=$ feedback factor)
  • A
    zero
  • B
    $\frac{1}{2}$
  • ✓
    1
  • D
    2
Answer
Correct option: C.
1
(c) : $A \beta=1$ is Barkhausen criterion for sustained oscillations in an oscillator.
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MCQ 241 Mark
For a transistor, the current ratio $a_{d c}=\frac{69}{70}$. The current gain $\beta_{d c}$ is
  • A
    66
  • B
    67
  • ✓
    69
  • D
    71
Answer
Correct option: C.
69
(c) : $a_{d c}=\frac{69}{70} \beta_{d c}=$ ?
We known $\beta_{d c}=\frac{\alpha_{d c}}{1-\alpha_{d c}}=\frac{\frac{69}{70}}{1-\frac{69}{70}}=69$
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MCQ 251 Mark
Which logic gate produces 'LOW' output when any of the inputs is 'HIGH'?
  • A
    AND
  • B
    $OR$
  • C
    NAND
  • ✓
    NOR
Answer
Correct option: D.
NOR
(d) : For NOR gate, $Y=\overline{A+B}$
If $A=1$ and $B=0$ or $A=0$ and $B=1$ or $A=B=1$
$
Y=\overline{1}=0
$
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MCQ 261 Mark
In common base circuit of a transistor, current amplification factor is 0.95 . Calculate the emitter current if base current is $0.2 mA$.
  • A
    $2 mA$
  • ✓
    $4 mA$
  • C
    $6 mA$
  • D
    $8 mA$
Answer
Correct option: B.
$4 mA$
(b) : Here, $\alpha=0.95$,
As $\alpha=\frac{I_C}{I_E}=0.95$
$I_B=0.2 mA$
$\therefore \quad I_C=0.95 I_E$
$
\begin{aligned}
& \text { As } I_E=I_B+I_C \\
& \therefore \quad I_E=I_B+0.95 I_E \Rightarrow 0.05 I_E=I_B \\
& I_E=\frac{I_B}{0.05}=\frac{0.2 mA }{0.05}=4 mA
\end{aligned}
$

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MCQ 271 Mark
In insulators (C.B. is conduction band and V.B. is valence band)
  • A
    V.B. is partially filled with electrons.
  • B
    C.B. is partially filled with electrons.
  • ✓
    C.B. is empty and V.B. is filled with electrons.
  • D
    C.B. is filled with electrons and V.B. is empty.
Answer
Correct option: C.
C.B. is empty and V.B. is filled with electrons.
(c) : In insulators conduction band is empty and valence band is completely filled with electrons.
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