MCQ Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 8

When a forward bias is applied to p-n junction it:

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Raises the potential barrier.
0%
Reduces the majority carrier current to zero.
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Lowers the potential barrier.
0%
None of the above.

In a n - type semi conductors which of the following statement is true:

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Electrons are majority carrier and trivalent impurity are the dopants.
0%
Electrons are minority carrier and pentavalent atoms are dopants.
0%
Holes are minority carrier and pentavalent atoms are dopants.
0%
Holes are majority carrier and trivalent atoms are the dopants.

Explanation

n-type semi-conductor can be produced by doping pentavalent impurity (atoms of valence 5) such as phosphorus.

Because of extra electrons due to pentavalent impurity electrons are majority carriers and holes are minority carriers.

$\therefore$ Here holes are minority carriers and pentavalent atoms are dopants. Hence option C is correct.

The conductivity of a semi conductor is:

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Greater than the conductivity of metals.
0%
Less than the conductivity of metals.
0%
Equal to the conductivity of metals.
0%
Intermediate between the conductivity of metals and insulators.

Explanation

comparison of conductivity according to the material used is given below:-

conductivity of metal(good conductor) > conductivity of semi-conductor(conductor) > conductivity of insulators(bad conductors).

Hence the conductivity of semiconductors is less than conductivity of metals and intermediate between the conductivity of metals and insulators.

$\therefore$ option B and D are correct.

Resistance of a wire is r ohm. The wire taken is double its length, then its intial resistance in ohm will be

0%
$\dfrac{r}{2}$
0%
$4r$
0%
$2r$
0%
$\dfrac{r}{4}$

Explanation

$R=\rho\dfrac{l}{A}$

$\implies R\propto l$

Hence, on doubling length, resistance becomes twice that is

$2r$ .

Answer-(C)

The attached figure shows the characteristics of:

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Zener diode
0%
Avalanche diode
0%
Photodiode
0%
LED

Explanation

$Answer:-$ A

Zener diode is a form of semiconductor diode in which at a critical reverse voltage a large reverse current can flow.

The conductivity of a semi conductor can be increased by:

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Heating the semi conductor
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Doping the semi conductor
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Both (a) and (b)
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None

A Zener diode having break-down voltage

$5.6V$ is connected in reverse bias with a battery of emf

$10V$ and a resistance of

$100\Omega$ in series. The current flowing through the Zener is

0%
$88mA$
0%
$0.88mA$
0%
$4.4mA$
0%
$44mA$

Explanation

A Zener diode offers infinite resistance as long as the reverse voltage across it is less than the break-down value. As soon as the reverse bias voltage cross the break-down level, the residue voltage remains in the circuit with the diode offering zero resistance to it.

Hence here the voltage remaining across the Zener=

$10V-5.6V=4.4V$

Therefore the current in the circuit is

$\dfrac{4.4V}{100\Omega}$

$=44mA$

A NOR gate and a NAND gate are connected as shown in the figure. Two different sets of inputs are given to this set up. In the first case, the input to the gates are A=0, B=0, C=In the second case, the inputs are A=1, B=0, C=The output D in the first case and second case respectively are

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0 and 0
0%
0 and 1
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1 and 0
0%
1 and 1

Explanation

In the first case the output of first NOR gate is

$\overline{A+B}$

$=\overline{0+0}=1=E$

and that of NAND gate is

$\overline{E.C}=\overline{1.0}=1$

In the second case, the output of first NOT gate is

$\overline{1+0}=0$

and that of NAND gate is

$\overline{0.1}=1$

Energy band gap between valence band and conduction band for insulator is:

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Zero
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greater than $9\ eV$
0%
less than conductors
0%
approximately $1\ eV$

Energy band gap between valence band and conduction band for conductor is:

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more than semiconductors
0%
Zero
0%
more than insulators
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None of these

Conductors are materials having an electrical conductivity :

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greater than $10^3$ S/cm
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between $10^{-8}\ to \ 10^3$ S/cm
0%
equal to $10^{-8}$ S/cm
0%
Zero

Explanation

Conductors are the most conducting materials and hence have the highest conductivity or the least resistivity. Therefore, they have a conductivity of

$> 10^3$

$S/cm$ .

Insulators have the least conductivity i.e.

$< 10^{-8}$

$S/cm$ whereas semiconductors have an electrical conductivity in between

$10^{-8}$ and

$10^3$

$S/cm$ .

The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as:

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

Explanation

Hint: Single input NAND gate act as NOT gate.
Explanation:

Step 1: Draw the diagram:
Let inputs be

$A$ and

$B$ .

Then, output of OR gate is

$Y = A+B$

Step 2: Conclusion:
The above output is fed to both the inputs of NAND gate. Hence, output of NAND gate is:

$Z = \overline {Y \cdot Y} = \overline {(A+B) \cdot (A+B)} = \overline {(A+B)}$

Hence, combination serves as NOR gate.
Option D is correct.

Semiconductors are materials having an electrical conductivity :

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Less than insulators
0%
between $10^{-8}\ S/cm \ to \ 10^3\ S/cm$
0%
Equal to zero
0%
Greater than $10^3\ S/cm$

Explanation

Semiconductors are neither highly conducting nor highly insulating materials and hence have moderate conductivity. Therefore, their conductivity lies in between

$10^{-8}$ and

$10^3$ S/cm.

Insulators have the least conductivity i.e.

$< 10^{-8}$ S/cm whereas conductors have the highest conductivity which is

$> 10^3$ S/cm.

Insulators are materials having an electrical conductivity equal to:

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Between $10^{-8}$ S/cm to $10^{3}$ S/cm
0%
$<\ 10^{-8}$ S/cm
0%
Equal to $10^{3}$ S/cm
0%
$>\ 10^{3}$ S/cm

Explanation

Insulators are the least conducting materials and hence have the least conductivity or the highest resistivity. Therefore, they have a conductivity of

$< 10^{-8}$

$S/cm$ .

Conductors have the highest conductivity i.e.

$> 10^3$

$S/cm$ whereas semiconductors have an electrical conductivity in between

$10^{-8}$ and

$10^3$

$S/cm$ .

A proper combination of

$3$ NOT and

$1$ AND gates is shown. If

$A=0$ ,

$B=1$ ,

$C=1$ then the output of this combination is

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$1$
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Zero
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Not predictable
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None of these

Which of the following is a property of a Light Emitting Diode (LED)?

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Low Threshold Voltage and High Reverse Breakdown Voltage
0%
High Threshold Voltage and High Reverse Breakdown Voltage
0%
High Threshold Voltage and Low Reverse Breakdown Voltage
0%
Low Threshold Voltage and Low Reverse Breakdown Voltage

Explanation

LED or Light Emitting Diode is a basic P-N junction diode operated when forward biased. It has very high threshold voltage and low reverse breakdown voltage(say

$5$ V)

The addition of pentavalent impurities ________ of the semiconductor.

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brings no change to the conductivity
0%
decreases the conductivity
0%
increases the conductivity
0%
reduces the formation of free electrons

What change could be seen when dopant atoms are added to an intrinsic semiconductor?

0%
It will change the proton and hole carrier concentrations of the semiconductor
0%
It will change the temperature of the semiconductor
0%
It will change the electron and hole carrier concentrations of the semiconductor
0%
It will decrease the formation of free electrons in the semiconductor

Explanation

In an intrinsic semiconductor under thermal equilibrium the concentration of dopant affects the concentrations of electrons and holes.

That is,

$n=p=n=n_{i}$

where,

$n\longrightarrow$ concentration of electrons.

$p\longrightarrow$ concentration of holes.

$n_{i}\longrightarrow$ Intrinsic concentration at given temperature.

So, when dopant atoms are added to an intrinsic semiconductor, it will change the electrons and holes carrier concentration of the semiconductor.

The curve represents the

$I-V$ characteristics of which optoelectronic device?

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Solar Cell
0%
Photo-diode
0%
Light Emitting Diode (LED)
0%
None of these

Explanation

1.The span of the solar cell

$I-V$ characteristic curve ranges from short circuit current

$I_{sc}$ at zero output volts, to zero current at te full open circuit voltage

$V_{oc}$

2.With the solar cell open circuited that is not connected to any load, the current will be at its maximum(zero) and the voltage across the cell is at its maximum known as open circuit voltage

$(V_{oc})$

3.At the other extreme when the solar cell is short circuited that is the positive and negative loads connected together, the current flowing out of the cell reaches its maximum known as short circuit current

$I_{sc}$

The semiconductor diodes in which carriers are generated by photons (photo-excitation) are called devices.

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thermo-optic
0%
optical
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optoelectronic
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thermi-ionic

Energy band gap between valence band and conduction band for semiconductor is:

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approximately 1 eV
0%
less than conductors
0%
equal to conductors
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more than insulator

Match the following

a) p-type semiconductor	1) Pure semiconductor
b) Intrinsic semiconductor	2) Doped with impurity
c) Extrinsic semiconductor	3) majority carriers are electrons
d) n-type semiconductor	4) majority carriers are holes

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a-3,b-2,c-4,d-1
0%
a-1,b-2,c-4,d-3
0%
a-4,b-1,c-2,d-3
0%
a-2,b-1,c-4,d-3

What is the reason to operate the photodiodes in reverse bias?

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fractional change due to the photo-effects on the majority carrier dominated forward bias current is more easily measurable than the fractional change in the reverse bias current.
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reverse bias is more efficient than forward bias.
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fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current.
0%
reverse bias configuration result in larger amount of current (in $\mu A$ )

Explanation

1.In reverse biased

$p-n$ junction, the width of depletion region increases as we increases the applied reverse bias voltage across the diode.

2.So,by applying a larger voltage, more of the incident photons are converted to electric current, or the efficiency increases.

3.On the other hands when we apply forward bias in the

$p-n$ junction, the width of the depletion region reduces, so, only a small portion of the incident photons gets converted to electric current.

so, this means that reverse bias is more efficient than forward bias that's the reason to operate the photo diodes in reverse bias.

Examples of opto-electronic devices are

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Light Emitting Diodes (LED)
0%
Zener diode
0%
Solar cells
0%
Photodiodes

The junction area is kept much larger in a solar cell

0%
for solar radiation to be incident because we are interested in more power.
0%
for solar radiation to be incident because we are interested in less power.
0%
for solar radiation to be properly reflected from the junction.
0%
None of these.

The drift current in a

$p-n$ junction diode is

0%
motion of charge carriers due to the electric field
0%
motion of charge carriers due to the charge concentration gradient
0%
in same direction as the diffusion current
0%
opposite in direction to the diffusion current

The process that occur during formation of a

$p-n$ junction is

0%
diffusion
0%
doping
0%
drift
0%
depletion

The material selected for solar cell fabrication has

0%
band gap ( $1.0$ to $1.8\ eV$ )
0%
high optical absorption ( $10^4 \ cm^{-1}$ )
0%
electrical conductivity
0%
band gap ( $0.4$ to $1.\ eV$ )

Explanation

The materials selected for solar cell fabrication is crystalline silicon, and the band gap of crystalline silicon is around

$1.12 eV$ and band gap of amorphous silicon is

$1.7-1.8 eV$ .

They have high quality of absorption light in

$IR$ wavelength and silicon is a semiconductor so, it is electrical conductivity also.

Therefore the materials selected for solar cell fabrication has.

1.Band Gap

$(1.0 to 1.8eV)$ .

2.High optical absorption

$(10^{4}\quad cm^{-1})$

3. Electrical conductivity.

The space-charge region on either side of the junction together is known as depletion region because

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neither electrons nor holes exist in this region.
0%
the electrons and holes taking part in the initial movement across the junction freed the region of its free charges.
0%
the electrons and holes in this region are immobile.
0%
None of these

In , excess minority carriers recombine with majority carriers near the p-n junction.

0%
Photodiode
0%
Light Emitting Diode
0%
Solar Cells
0%
Zener diode

During the formation of

$p-n$ junction,

0%
holes diffuse from $p-$ side to $n-$ side ( $p \rightarrow n$ )
0%
electrons diffuse from $p-$ side to $n-$ side ( $p \rightarrow n$ )
0%
electrons diffuse from $n-$ side to $p-$ side ( $n \rightarrow p$ )
0%
holes diffuse from $n-$ side to $p-$ side ( $n \rightarrow p$ )

The I V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes because

0%
a solar cell does not draw current but supplies the same to the load.
0%
a solar cell draws current but supplies no current to the load.
0%
a solar cell draws voltage and supplies current to the load
0%
a solar cell draws current but supplies voltage to the load

Explanation

The

$V-I$ curve for solar cell is available in the fourth quadrant, i.e. where current is negative, which means that current I is supplied by the solar cell and not drawn by it.

Due to the concentration gradient across p-, and n- sides, the motion of charge carries gives rise to_______ current across the junction

0%
drift
0%
diffusion
0%
both diffusion and drift
0%
None of these

When an external voltage (

$V$ ) is applied across the diode in reverse bias with a barrier potential of

$V_o$ , the effect barrier height under reverse bias is

0%
$V_o + V$
0%
$V_o - V$
0%
$V - V_o$
0%
None of these

Explanation

When a reverse bias voltage (V) is applied externally across the diode, and the barrier potential is

$V_{o}$ . The potential difference adds to the barrier potential. The effective barrier potential increases to

$(V_{o}+V)$ .

The thickness of depletion region is of the order of

0%
$0.1 \mu m$
0%
$1 \mu m$
0%
$0.1 mm$
0%
$1 mm$

When an external voltage

$V$ is applied across a semiconductor diode such that

$p-$ side is connected to the positive terminal of the battery and

$n-$ side to the negative terminal, it is said to be

0%
reverse biased.
0%
forward biased.
0%
bias free.
0%
None of these

In the figure, the numbers 1, 2, 3 represent barrier potential (

$V_o$ ) of a p-n junction under forward bias

0%
1 - high voltage battery 2 - without battery 3 - low voltage bettery
0%
1 - high voltage battery, 2 - low voltage battery, 3 - without battery
0%
1 - without battery, 2 - low voltage battery, 3 - high voltage battery
0%
1 - low voltage battery 2 - without battery 3 - high voltage bettery

Explanation

If a forward bias voltage is applied to a p-n junction diode with a barrier potential

$V_{o}$ , the effective potential reduces to

$(V_{o}-V)$ . More is the forward bias voltage, more the potential reduces.

So, in

Barrier potential remains same, so the battery is not connected externally.
Low voltage battery is connected externally.
High voltage battery is connected externally.

In this

$I-V$ characteristic curve of a zener diode, regions 1 and 2 respectively indicate

0%
forward bias and reverse bias regions.
0%
reverse bias and forward bias regions.
0%
forward bias regions only.
0%
reverse bias regions only.

Explanation

When the zener diode is connected in forward bias it acts as a normal diode.And when the reverse bias voltage is greater than a predetermined voltage then the zener breakdown voltage occur that is denoed by

$V_z$ .

And at a certain value of reverse voltage the reverse current will increase suddenly and sharply so, it clear that in this

$I-V$ characteristic curve of a zener diode, region

$1$ is forward bias and region

$2$ is reverse bias region.

Which of the following is an example of an indirect band gap intrinsic semiconductor?

0%
Silicon
0%
Germanium
0%
Gallium Arsenide
0%
None of these

For creating a

$p-n$ junction,

0%
we take one slab of p-type semiconductor and physically join it to another n-type semiconductor
0%
continuous contact at the atomic level is necessary.
0%
two macroscopic smooth flat slabs can be used
0%
None of these

The direction of the applied voltage (

$V$ ) is opposite to the built-in potential

$V_o$ . The effective barrier height under forward bias is

0%
( $V_o + V$ ).
0%
( $V_o V$ ).
0%
( $V V_o$ ).
0%
( $V_o \times V$ ).

The logic circuit as shown below has the input wave forms

$A$ and

$B$ as shown. Pick out the correct output waveform.

By adding certain impurities to in the appropriate concentrations the conductivity can be well-controlled.

0%
semiconductors
0%
conductors
0%
insulators
0%
superconductors

Which of the following statement(s) is/are true?

0%
A hole is an electric charge carrier with a positive charge, equal in magnitude but opposite in polarity to the charge on the electron
0%
A hole is the absence of an electron in a particular place in an atom
0%
A hole is an electric charge carrier with a negative charge
0%
none of these

The logic behind NOR gate is that which gives

0%
high output when both inputs are high
0%
low output when both inputs are low
0%
high output when both inputs are low
0%
none of these

Explanation

Let the inputs of NOR gate be

$a,b$

The output of NOR gate will be

$X=\overline { a +b}$

$a=0$ and

$b=0$ then

$X=1$

$a=0$ and

$b=1$ then

$X=0$

$a=1$ and

$b=0$ then

$X=0$

$a=1$ and

$b=1$ then

$X=0$

Therefore the output will be high when both the inputs are low

So the correct option is

$C.$

The diagram of a logic circuit is given below, the output of the circuit is represented by

0%
W+(X+Y)
0%
W+(X.Y)
0%
W.(X+Y)
0%
W.(X.Y)

Explanation

Given that

$W$ and

$X$ are inputs of OR gate , so the output will be

$W+X$

Given that

$W$ and

$Y$ are inputs of OR gate , so the output will be

$W+Y$

Now

$W+X$ and

$W+Y$ are inputs for AND gate , so the out put

$F=(W+X)(W+Y)=W(1+X+Y)+XY=W+XY$

Therefore the correct option is

$B$

Electrons always moves from

0%
Valance band to conduction band
0%
Conduction to valance band
0%
Some times from valance band to conduction and sometimes from conduction band to valance band
0%
None of the above

In the following circuit, the output Y for all possible inputs A and B is expressed by the truth table

Explanation

Given that

$A$ and

$B$ are inputs for NOR gate

So we get output

$Y^{1}=\overline { A+B }$

Now

$Y^{1}$ and

$Y^{1}$ are inputs for NOR gate

So the output

$Y = A+B$

For

$A=0$ and

$B=0$ , the value of

$Y=0$

For

$A=1$ and

$B=0$ , the value of

$Y=1$

For

$A=0$ and

$B=1$ , the value of

$Y=1$

For

$A=1$ and

$B=1$ , the value of

$Y=1$

So option C is correct

When you have a P-N junction and you connect it to a DC voltage, this will lead

0%
The electrons near to the gap between the P-N junction from the N type material to move to the P side
0%
The lack of electrons in the N type side will lead to have a holes
0%
The holes near to the gap between the P-N junction from the P type material to move to the N side
0%
Both A and C

Explanation

When you have a

$p-n$ junction and we connect it to a DC voltage, this will lead to:

1. When the

$N-$ type semiconductor and

$p-$ type semiconductor materials are first joined together a very large density gradient exits between both sides of the

$P-N$ junction.

2. The results is that some of the electrons from the donar impurities atoms begin to migrate across this newly formed junction to fill up the holes in the

$P-$ type materials producing negative ions.

3. Since this electrons have moved across the

$P-N$ junction from

$n-$ type to

$p-$ type, they leave behind positively charged donar ions on the negative across the junction in the opposite direction into the region where there are large number of free electrons.

4. As a result, the charged density of the

$P-$ type along the junction is filled with negatively charged acceptors ions, and the charges density of the

$N-$ type along the junction becomes positive.

The truth table of the combination of the logic gates shown in the figure is

Explanation

When inputs

$A$ and

$B$ entered in first

$NAND$ gate , then output ,

$y_{1}=\bar {(A.B)}$

When inputs

$A$ and

$B$ entered in

$NOR$ gate , then output ,

$y_{2}=\bar {(A+B)}$

When inputs

$y_{1}$ and

$y_{2}$ entered in second

$NAND$ gate , then output ,

$y=\bar{(\bar {A.B}).(\bar{A+B})}$

This is the Boolean expression for the given diagram

Now , when

$A=1 ,B=0$

$y=\bar{(\bar {0.0}).(\bar{0+0})}=\bar{1.1}=0$

when

$A=0 ,B=0$

$y=\bar{(\bar {1.0}).(\bar{1+0})}=\bar{1.0}=1$

when

$A=0 ,B=1$

$y=\bar{(\bar {0.1}).(\bar{0+1})}=\bar{0.1}=1$

when

$A=1 ,B=1$

$y=\bar{(\bar {1.1}).(\bar{1+1})}=\bar{0.0}=1$

Option B is correct.

CBSE Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 8 - MCQExams.com

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INPUTS			OUTPUTS
A	B	C	D	E	F	Y
0	1	1	1	0	0	0

CBSE Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Physics Quiz Questions and Answers

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