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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 701. |
Assertion : p-n junction diode can be used even at ultra high frequencies. Reason : Capacitative reactance of a p-n junction diode increases as frequency increases.A. Statetment-I is True, Statement-II is True , Statement-II is a correct explanation for Statement-IB. Statement-I is True, Statement-II is True , Statement-II is NOT a correct explanation for Statement-IC. Statement-I is True, Statement-II is FalseD. Statement-I is False, Statement-II is True |
| Answer» Correct Answer - D | |
| 702. |
Statement 1: Sky wave communication is not suitable for frequencies greater than 30 MHz Statement 2: High frequency signals die out before reaching the ionosphere.A. Both statement-1 and statement-2 are true, and statement-2 is the correct explanation of statement 1.B. Both statement-1 and statement-2 are true but statement-2 is not the correct explanation of statement-2C. Statement-1 is true but statement-2 is false.D. Statement-1 is false but statement-2 true. |
| Answer» Correct Answer - C | |
| 703. |
Statement 1: Sky wave communication is not used to transmit TV signals Statement 2: The ionosphere does not reflect TV signals, it transmits them.A. Both statement-1 and statement-2 are true, and statement-2 is the correct explanation of statement 1.B. Both statement-1 and statement-2 are true but statement-2 is not the correct explanation of statement-2C. Statement-1 is true but statement-2 is false.D. Statement-1 is false but statement-2 true. |
| Answer» Correct Answer - A | |
| 704. |
Statement-1 : Electromagnetic waves with frequencies smaller than the critical frequency of ionosphere cannot be used for communications using sky wave propagation. Statement-2: The refractive index of the ionosphere becomes very high for frequencies higher than the critical frequency.A. Statetment-I is True, Statement-II is True , Statement-II is a correct explanation for Statement-IB. Statement-I is True, Statement-II is True , Statement-II is NOT a correct explanation for Statement-IC. Statement-I is True, Statement-II is FalseD. Statement-I is False, Statement-II is True |
| Answer» Correct Answer - D | |
| 705. |
Statement-I : Electromagnetic waves with frequencies smaller than the critical frequency of ionosphere cannot be used for communication using sky wave propagation Statement-II : The refractive index of the ionosphere becomes very high for frequency.A. Statetment-I is True, Statement-II is True , Statement-II is a correct explanation for Statement-IB. Statement-I is True, Statement-II is True , Statement-II is NOT a correct explanation for Statement-IC. Statement-I is True, Statement-II is FalseD. Statement-I is False, Statement-II is True |
| Answer» Correct Answer - D | |
| 706. |
For sky wave propagation of a `10 MHz` signal, what should be the minimum electron density in ionosphere?A. `~ 1.2 xx 10^(12) m^(-3)`B. `~ 10^(6) m^(-3)`C. `~ 10^(14) m^(-3)`D. `~ 10^(22) m^(-3)` |
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Answer» Correct Answer - A `F_(C)=9sqrt(N_(max))=9sqrt(10^(11))=2MHz` |
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| 707. |
Digital signals (`i`) do not provide a continuous set of values.(`ii`) represents values as descrete steps. (`iii`) can utillize binary system (`iv`) can utillize decimal as well as binary system. The true option is.A. Only (i) and (ii)B. Only (ii) and (iii)C. (i),(ii) and (iii), but not (iv)D. All the above (i) to (iv) |
| Answer» Correct Answer - C | |
| 708. |
If the source of power` 4 kW` product`10^(20)` photons //second , the radiation belongs to a part spectrum calledA. microwavesB. `gamma`-raysC. X-raysD. ultraviolet rays |
| Answer» Correct Answer - C | |
| 709. |
From the Zener diode circuit shown in figure, the current through the Zener diode is A. 10 mAB. 15 mAC. 20 mAD. 25 mA |
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Answer» Correct Answer - B `I=(150-150)/(5xx10^3)`=20 mA Load current = 5 mA `therefore` Zener current `I_Z`=15 mA |
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| 710. |
An experment is performed to find the refractive index of glass using a travelling mircroscope. In this experiment distances are measured byA. a standard laboratory scaleB. a meter scale provided on the microscopeC. a screw gauge provided on the microscopeD. a vernier scale provided on the microscope |
| Answer» Correct Answer - D | |
| 711. |
In a good conductor, what is the energy gap between the conduction band and the valence band.? |
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Answer» Correct Answer - A In a good conductor, conduction band overlaps with the valence band. Therefore, the energy gap between them is zero. |
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| 712. |
In a good conductor, what is the energy gap between the conduction band and the valence band.?A. zeroB. oneC. infiniteD. very large |
| Answer» Correct Answer - A | |
| 713. |
The energy gap separating valence band from a conduction band is 0.7 eV for a material Hence it is aA. metalB. Germanium semiconductorC. Silicon semiconductorD. Non-metal |
| Answer» Correct Answer - B | |
| 714. |
Explain concept of valence band and conduction band in solid crystal. |
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Answer» A. Valence band (V.B): 1. The topmost occupied energy level in an atom is the valence level. The energy band formed by valence energy levels of atoms in a solid is called the valence band. 2. In metallic conductors, the valence electrons are loosely attached to the nucleus. At ordinary room temperature, some valence electrons become free. They do not leave the metal surface but can move from atom to atom randomly. 3. Such free electrons are responsible for electric current through conductors. B. Conduction band (C.B): 1. The immediately next energy level that electrons from valence band can occupy is called conduction level. The band formed by conduction levels is called conduction band. 2. It is the next permitted energy band beyond valence band. 3. In conduction band, electrons move freely and conduct electric current through the solids. 4. An insulator has empty conduction band. |
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| 715. |
The depletion region of p-n junction has a thickness of the order ofA. 0.5 nm to 1 nmB. 5 nm to 10 nmC. 50 nm to 500 nmD. 500 nm to 1000 nm |
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Answer» Correct Answer - D The width of the depletion region in an unbiased p-n junction ranges from 0.5 `mu`m to 1 `mu`m. 1`mu`m = 103 nm [1 `mu`m =`10^(-6)` m and 1 nm= `10^(-9)` m]. `:.` The width varies between 500 nm to 1000 nm. |
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| 716. |
In a p-n junction, the thickness of the depletion region is `10^(-5)` m. What is the P.D. that should be applied across it, to produce an electric field of intensity `10^5` V /m ?A. 0.5 VB. 0.75 VC. 1 VD. 1.25 V |
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Answer» Correct Answer - C `E=-"dV"/"dx" therefore dV=Edx = 10^5xx10^(-5) `=1 V |
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| 717. |
In a p-n junction, the thickness of the depletion layer is `10^(-6)` m. If the potential difference across it is 0.2 V, then the electric field set up across the junction isA. `10^6` V/mB. `2xx10^5` V/mC. `10^(-6)` V/mD. `10^5` V/m |
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Answer» Correct Answer - B `E=V/d=0.2/10^(-6) = 0.2 xx10^6` `therefore E= 2 xx10^5` V/m |
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| 718. |
In a reverse biased diode, when the applied voltage changes by `1V`, the current is found to change by `0.5muA`. The reversebiase resistance of the diode isA. `2xx10^(5)Omega`B. `2xx10^(6)Omega`C. `200Omega`D. `2Omega` |
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Answer» Correct Answer - B `(DeltaV)/(DeltaI)=(1)/(0.5xx10^(-6))=2xx10^(6)Omega` |
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| 719. |
A potential barrier of 0.3 V exists across a p-n junction. If the depletion region is 1 `mu`m wide, what is the intensity of electric field in this region?A. 3 `mum`B. 5 `mum`C. 7 `mum`D. 4 `mum` |
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Answer» Correct Answer - B `E="dv"/"dx" therefore dx="dv"/E=0.5/10^5 = 5xx10^(-6) m= 5 mum` |
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| 720. |
A potential barrier of `0.50 V` exists across a `P-N` junction. If the depletion region is `5.0xx10^(-7)m`, wide the intensity of the electric field in this region isA. `1.0xx10^(6)V//m`B. `1.0xx10^(5)V//m`C. `2.0xx10^(5)V//m`D. `2.0xx10^(6)V//m` |
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Answer» Correct Answer - A `E=(V)/(d)=(0.5)/(5xx10^(-7))=10^(6)V//m` |
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| 721. |
In a reverse biased diode, when the applied voltage changes by `1V`, the current is found to change by `0.5muA`. The reversebiase resistance of the diode isA. `2xx10^5 Omega`B. `2xx10^6 Omega`C. `200 Omega`D. `2 Omega` |
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Answer» Correct Answer - B `R=(DeltaV)/(DeltaI)=1/(0.5xx1^(-6)) = 10^6/(1//2)=2xx10^6 Omega` |
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| 722. |
A square wave (-1 to +1V) is applied to a P-N junction diode as shown. Draw the output wave form across the diode which is assumed to be ideal- A. B. C. D. |
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Answer» Correct Answer - A Voltage `gt 0` will passed through diode but voltage `lt 0` will be reverse bias the junction Thus, output will be zero. |
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| 723. |
For a common emiter configuration if `a` and `beta` have their usualy meaning , the incorrrect relationship between `a and beta` is :A. `alpha=(beta^(1))/(1+beta^(2))`B. `(1)/(alpha)=(1)/(beta)+1`C. `alpha=(beta)/(1+beta)`D. `alpha=(beta)/(1+beta)` |
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Answer» Correct Answer - A::C `alpha=(I_(C))/(I_(e)),beta=(I_(C))/(I_(b))` `I_(e)=I_(b)+I_(c)rArr(I_(e))/(I_(e))=(I_(b))/(I_(c))+1rArr(I)/(alpha)=(I)/(beta)+1` `alpha = (beta)/(1+beta)` |
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| 724. |
In a common base ampifier , the phase difference between the input signal and output voltage isA. `(pi)/(4)`B. `pi`C. zeroD. `(pi)/(2)` |
| Answer» Correct Answer - C | |
| 725. |
How many minimum NAND gate are required to obtain NOR gate :-A. 3B. 2C. 1D. 4 |
| Answer» Correct Answer - D | |
| 726. |
In a transistor `10^(8)` electrons enter at the emitter in `10^(-4)` s, out of which `2%` electron go to the base. The current transfer ratio in common base configuration isA. 98B. 2C. 0.98D. 0.2 mA |
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Answer» Correct Answer - C `I_(b)=2% ` of `I_(e), I_(c)=98% ` of `I_(e), alpha=?` `alpha=(I_(c))/(I_(e))=(98% " of " I_(e))/(I_(e))=98% =0.98` |
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| 727. |
In an n-p-n transistor `10^(10)` electrons enter the emitter in `10^(-6)`s. If 2% of the electrons are lost in the base, find the current transfer ratio and the current amplification factor.A. `2xx10^(-10)` A and 49B. `1.6xx10^(-19)` A and 90C. `1.7xx10^(-11) ` A and 70D. `3.2xx10^(-9)` A and 99 |
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Answer» Correct Answer - D Let `I_E` be the emitter current . Then, `I_E=(dq)/(dt) = (200xx1.6xx10^(-19))/10^(-8)=3.2xx10^(-9)A` Since 1% electrons are lost in the base , `therefore` Base current `(I_B)=I_E/100` and the remaining `99/100 I_E` will enter the collector `therefore I_C=99/100 I_E` and the current amplification factor `beta=I_C/I_B=(0.99I_E)/(0.01 I_E)`=99 |
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| 728. |
In the common emitter configuration of n-p-n transistor `10^(10)` electrons enter the emitter in 1`mus` and 2% of the electrons are lost of the base. The current gain of the amplifier isA. 2B. 98C. 1D. 49 |
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Answer» Correct Answer - D `I_(b)=2% ` of ` I_(e), I_(c)=90% ` of `I_(e), beta=?` `beta=(I_(c))/(I_(b))=(98% " of " I_(e))/(2% " of " I_(b))=49` |
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| 729. |
In a `n-p-n` transistor `10^(10)` electrons enter the emitter in `10^(-6)s`. `2%` of the elecrons are lost in the base. The current transfer ratio and the current amplification factor will beA. `0.98,49`B. `98,0.49`C. `0.49,98`D. `49,0.49` |
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Answer» Correct Answer - A `alpha=(I_(C ))/(I_(B))=(0.98)` `beta=(I_(C ))/(I_(B))=(0.98)/(0.02)=49` |
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| 730. |
A transistor is operetaed in common-emitter configuration at `V_(c)=2V` such that a change in the base current from `100mA` to `200mA` produces a change in the collector current from `5mA` to `10mA`. The current gain isA. `75`B. `100`C. `150`D. `50` |
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Answer» Correct Answer - D `beta=(DeltaI_(C ))/(DeltaI_(B))=((10-5)xx10^(-3))/((200-100)xx10^(-6))=50` |
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| 731. |
The combination of the gates shown in the figure below produces A. NOR gateB. OR gateC. AND gateD. XOR gate |
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Answer» Correct Answer - B `C=bar(barA.barB)=bar(barA)+bar(bar(B)) =A+B=` OR gate |
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| 732. |
Which logic gate is represented by the following combination of logic gates A. ORB. NANDC. ANDD. NOR |
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Answer» Correct Answer - C `C=bar((bar(A)+bar(B)))=bar(bar(A))*bar(bar(B))=A*B =` AND gate |
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| 733. |
Identify the property which is not characteristic for a semiconductor ?A. At a very low temperatures, it behaves like an insulatorB. At higher temperatures two types of charge carriers will cause conductivityC. The charge carriers are electrons and holes in the valence band at higher temperaturesD. The semiconductor is electrically neutral |
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Answer» Correct Answer - C At the higher temperature only electrons cross the forbiden gap and reach in valence band. |
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| 734. |
If the ratio of the concentration of electron to that of holes in a semiconductor is `(7)/(5)` and the ratio of current is `(7)/(4)` then what is the ratio of their drift velocities ?A. `(5)/(8)`B. `(4)/(5)`C. `(5)/(4)`D. `(4)/(7)` |
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Answer» Correct Answer - C `I=n_(e)Av_(d)` `(I_(e))/(I_(n))=(n_(e)xx(v_(d))_(e))/(n_(h)xx(v_(d))_(h)) ` Here, `(n_(e))/(n_(h))=(7)/(5), (I_(e))/(I_(h))=(7)/(4)` `therefore (7)/(4)=(7)/(5)xx((v_(d))_(e))/((v_(d))_(h)) rArr ((v_(d))_(e))/((v_(d))_(h))=(5)/(7)xx(7)/(4)=(5)/(4)` |
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| 735. |
If the ratio of the concentration of electron to that of holes in a semiconductor is `(7)/(5)` and the ratio of current is `(7)/(4)` then what is the ratio of their drift velocities ?A. `5//8`B. `4//5`C. `5//4`D. `4//7` |
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Answer» Correct Answer - C `I=n_(e) Av_(d)` `(I_(e))/(I_(h))=(n_(e)xx(v_(d))_(e))/(n_(h)xx(v_(d))_(h))` Here `(n_(e))/(n_(h))=7/5 xx((v_(d))_(e))/((v_(d))_(h))=(7)/(4)` `(7)/(4) = (7)/(5) xx((v_(d))_(e))/((v_(d))_(h))` `rArr ((v_(d))_(e))/((v_(d))_(g))=5/7xx7/4 =5/4` |
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| 736. |
The diffusion current in a p-n junction isA. from the N-side to the P-sideB. from the P-side to the N-sideC. From the N-side to the P-side if the junction is forward biased and in the opposite direction if it is reverse biased.D. From the P-side to the N-side if the junction is forward-biased and in the opposite direction if it is reverse biased. |
| Answer» Correct Answer - B | |
| 737. |
For a reverse bias P-N junction-A. P region is positive and current is due to electronsB. P region is positive and the current is due to holesC. P region if negative and the current is due to electronsD. P region is negative and current is due to both electrons and holes |
| Answer» Correct Answer - D | |
| 738. |
A hole diffuses from the p-side to the n-side in a p-n junction.This means thatA. a bond is broken on the n-side and the electron freed from the bond jumps to the conduction bandB. a conduction electron on the p-side jumps to a broken bond to complete itC. a bond is broken on the n-side and the electron freed from the bond jumps to a broken bond on the p-side complete it.D. a bond is broken on the p-side and the electron freed from the bond jumps to a broken bond on the n-side to complete it. |
| Answer» Correct Answer - C | |
| 739. |
In a p-n junction , numbers of junction areA. 1B. 0C. 2D. 4 |
| Answer» Correct Answer - A | |
| 740. |
The drift current in a p-n junction isA. from the N-side to the P-sideB. from the P-side to the N-sideC. From the N-side to the P-side if the junction is forward biased and in the opposite direction if it is reverse biased.D. From the P-side to the N-side if the junction is forward-biased and in the opposite direction if it is reverse biased. |
| Answer» Correct Answer - A | |
| 741. |
The drift current in a p-n junction is(A) from the p region to n region.(B) from the n region to p region. (C) from n to p region if the junction is forward biased and from p to n region if the junction is reverse biased. (D) from p to n region if the junction is forward biased and from n to p region if the junction is reverse biased. |
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Answer» (B) from the n region to p region. |
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| 742. |
In a `p-n` junctionA. new holes and conduction electrons are produce continuously throughout the materialB. new holes and cinduction electrons are produced continuously throughout the material except in the depletion regionC. holes and conduction electrons recombine continuously throughout the material.D. holes and conduction electrons recombine continuously throughout the material except in the depletion region. |
| Answer» Correct Answer - A::D | |
| 743. |
Diffusion current in a p-n junction is greater than the drift current in magnitudeA. if the junction is forward-biasedB. if the junction is reverse-biasedC. if the junction is unbiasedD. in no case |
| Answer» Correct Answer - A | |
| 744. |
In a normal operation of a transistor,A. the base-emitter junction is forward-biasedB. the base-collector junction is forward-biasedC. the base-emitter juncion is reverse-baisedD. the base-coolector junction is reverse-baised |
| Answer» Correct Answer - A::D | |
| 745. |
In a transistor,A. the emiiter has the least concentration of impurityB. the collector has the least concentration of impurityC. the base has the least concentration of impurityD. all the three region have equal concentration of impurity |
| Answer» Correct Answer - C | |
| 746. |
Input resistance of transistor in comparision to output resistance is-A. LowB. HighC. Low and highD. None of these |
| Answer» Correct Answer - A | |
| 747. |
The thinest part of a transistor isA. emiiterB. baseC. collectorD. according to transistor parameters none of these |
| Answer» Correct Answer - B | |
| 748. |
Transistor can be used as :-A. amplifierB. modulatorC. oscillatorD. all of the above |
| Answer» Correct Answer - D | |
| 749. |
The length of a germanium rod is `0.58 cm` and its area of cross-section is `10cm^(2)`. If for germanium `n_(i)=2.5xx10^(19)m^(-3),mu_(h)=0.19 m^(2)//V-s,mu_(e)=0.39 m^(2)//V-s`, then the resistance of the rod will be-A. `2.5k Omega`B. `4.0kOmega`C. `5.0kOmega`D. `10.0kOmega` |
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Answer» Correct Answer - B We have, `R=(rhol)/(A)=(L)/(n_(i)e(mu_(e)+mu_(h))A)` `=(0.928xx10^(-2))/(2.5xx10^(19)xx1.6xx10^(-19)(0.19)xx10^(-6))` `=4000Omega=4.0kOmega` |
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| 750. |
In transistor symbols, the arrows shows the direction of-A. current in the emitterB. electron current in the emitterC. holes current in the emitterD. electron current in the emitter |
| Answer» Correct Answer - A | |