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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.
| 501. |
Two radio antennas separated by 300 m as shown in figure simultaneously broadcast indentical singals at the same wavelength. A radio in a car traveling due north receives the signals. (a) If the car is at the position of the second maximum, what is the wavelength of the signals? (b) How much farther must the car travel to encounter the next mininum in reception? (Note: Do not use the small- angle approximation in this problem.) A. orient the receiving antenna horizontally,north-southB. orient the receiving antenna horizontally, east-westC. use a vertical receiving antennaD. Move to a town farther to the east or to the west,Use a magnetic dipole antenna instead of an electric dipole entenna. |
| Answer» Correct Answer - D | |
| 502. |
For a transistor in common emitter configuration, the voltage drop across the load of 1000 `Omega` is 0.5 V. If the value of `alpha ` for the transistor is 0. 98, then the base current will be approximately equal toA. 5 `muA`B. 8`muA`C. 10`muA`D. 15`muA` |
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Answer» Correct Answer - C `beta=0.98/(1-0.98)=98/2 = 49` `I_C=V/R = 0.5/1000 = 5xx10^(-4) A` `therefore beta=I_C/I_B` `therefore I_B=I_C/beta=(5xx10^(-4))/49 = (500xx10^(-6))/49=10 muA` |
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| 503. |
Photodiode is a deviceA. which is always operated in reverse biasB. which is always operated in forward biasC. in which photo current is independent of intensity of incident radiationD. which may be operated in forward or reverse bias |
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Answer» Correct Answer - A Photodiode is a device which is always operated in reverse bias. |
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| 504. |
The input signal given to a `CE` amplifier having a voltage gain of `150` is `V_(i)=2cos(15t+(pi)/3)`. The corresponding output signal will beA. 100 sin `(10 t + pi/3)`B. `100 sin (10 t + (4pi)/3)`C. `200 sin (10 t + (2pi)/3)`D. `100 sin (10t + pi)` |
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Answer» Correct Answer - B For output signal `V_o` = 50 x 2 and phase is `(10 t + pi/3 + pi)` `therefore ` Output signal is 100 sin `(10t + (4pi)/3)` |
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| 505. |
The energy gap of silicon is `1.14 eV`. The maximum wavelength at which silicon will begin absorbing energy isA. 1086 ÅB. 10860 ÅC. 10.86 ÅD. 108.6 Å |
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Answer» Correct Answer - B `because E=hv="hc"/lambda` `therefore lambda_"max"="hc"/lambda=(6.6xx10^(-34)xx3xx10^8)/(1.14xx1.6xx10^(-19)) ` `=(6.6xx3)/(1.14xx1.6)xx10^(-7) m` `=10.86xx10^(-7)xx10^10` Å = 10860 Å |
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| 506. |
In forward biasing of the p-n junction:A. B. C. D. |
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Answer» Correct Answer - D In p-region direction of conventional current is same as flow of holes and in n-region direction of conventional current opposite to direction of electron flow. |
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| 507. |
In a `P-N` junction diode :A. the current in the reverse biased condition is generally very smallB. the current in the reverse biased condition is small but the forward biased current is independent of the biase voltageC. the reverse biased current is strongly dependent on the applied bias voltageD. the forward biased current is very small in comparison to reverse biased current |
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Answer» Correct Answer - C The forward voltage over comes the barrier voltage. Due to which the forward current is high but depends upon the forward voltage applied. The reverse voltage supports the barrier voltage due to which the reverse current is low. Therefore, current passing through `2kOmega` resistor `=(12)/(2xx10^(3))=6xx10^(-3)A=6mA` |
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| 508. |
The temperature of germanium is decreased from room temperature to 100 K, the resistance of germaniumA. decreasesB. increasesC. remains unaffectedD. depends on external conditions |
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Answer» Correct Answer - B Germanium is a semiconductor and possesses negative temperature coefficient. Therefore, when temperature of germanium is decreased, it resistance increases. |
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| 509. |
The length of a germanium rod is 0.58 cm and its area of cross-section is `1mm^(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. `2.5kOmega`B. `5.0kOmega`C. `7.5kOmega`D. `10.0kOmega` |
| Answer» Correct Answer - A | |
| 510. |
The P-N junction is-A. an ohmic resistanceB. an non ohmic resistanceC. a positive resistanceD. a negative resistance |
| Answer» Correct Answer - B | |
| 511. |
Value of forbidden energy gap for semi conductor is:A. 1 eVB. 6 eVC. 0 eVD. 3 eV |
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Answer» Correct Answer - A Value of forbidden energy gap in semiconductor is 1 eV. |
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| 512. |
In a semiconductor it is observed that three-quarters of the current is carried by electrons and one quarters by holes. If the drift speed is three times that of the holes, what is the ratio of electrons to holes?(a) 1 : 1(b) 1 : 2(c) 2 : 1(d) 4 : 1 |
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Answer» The correct answer is (a) 1 : 1 The explanation: In a semiconductor, I = Ie + Ih Here, Ie = ^3⁄4 I and Ih = ^1⁄4 I Now ve = 3vh Ie/Ih = nve/nvh 3 = 3n/p n = p Hence the ration is, 1 : 1 |
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| 513. |
At absolute zero temperature the forbidden gap of conductor isA. zeroB. 0,67 eVC. 1.1 eVD. 6 eV |
| Answer» Correct Answer - A | |
| 514. |
The for bidden gap in the energy bands of sillcon isA. 0.5 evB. 1.1 eVC. 4 eVD. 2.6 eV |
| Answer» Correct Answer - B | |
| 515. |
A piece of copper and the other of geramnium are cooled from the room temperature to 80K . What will happen to their resistance ? |
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Answer» Correct Answer - A Copper is conductor and germanium is semiconductor . With decrease in temperature resistance of a conductor decreases and that of semiconductor increases . Therefore resistance of copper will decrease and that of semiconductor will increase. |
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| 516. |
When a transistor amplifier having current gain of `75` is given an input signal, `V_(I) = 2 sin (157 t + pi//2)`, the output signal is found to be `V_(o) = 200 sin(157 t + 3pi//2)`. The transistor is connected as :A. common base amplifierB. common emitter amplifierC. common collector amplifierD. feed back amplifier |
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Answer» Correct Answer - B There is a current gain and phase difference of `pi` between `V_i` and `V_o` |
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| 517. |
The current gain ofa transistor is 100. If the base current changes by 200 `mu`A, what is the change in collector current?A. 0.2mAB. 20 mAC. 2 mAD. 200mA |
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Answer» Correct Answer - B `I_C=beta I_B`=20 mA |
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| 518. |
The r.m.s. value of the base current ofa transistor is `10mu`A. What is the current gain (`beta`) if the peak value of the a.c. collector current is 1.414 mA ?A. 50B. 75C. 100D. 125 |
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Answer» Correct Answer - C Peak value of `I_C` = 1.414 mA `therefore` r.m.s. value of `I_C=1.414/sqrt2`= 1 mA `therefore beta=I_C/I_B=10^(-3)/(10xx10^(-6)) ` =100 |
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| 519. |
A working transitor with its three legs marked `P, Q and R` is tested using a multimeter No conduction is found between `P, Q `by connecting the common (negative) terminal of the multimeter to `R`and the other (positive) terminal to or `Q` some resistance is seen on the multimeter . Which of the following is true for the transistor ?A. it is a pnp transistor with R as collectorB. it is a pnp transistor with R as emitterC. it is an npn transistor with R as collectorD. it is an npn transistor with R as base. |
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Answer» Correct Answer - D `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),(I_(e))/(I_(h))=(7)/(4)` `(7)/(4)=(7)/(5)xx((v_(d))_(e))/((v_(d))_(h))` `implies((v_(d))_(e))/((v_(d))_(h))=(5)/(7)xx(7)/(4)=(5)/(4)`. |
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| 520. |
A working transitor with its three legs marked `P, Q and R` is tested using a multimeter No conduction is found between `P, Q `by connecting the common (negative) terminal of the multimeter to `R`and the other (positive) terminal to or `Q` some resistance is seen on the multimeter . Which of the following is true for the transistor ?A. it is a pnp transistor with R as collectorB. It is a pnp transistor with R as emitterC. it is an npn transistor with R as collectorD. It is npn transistor with R as base |
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Answer» Correct Answer - D R is base (common terminal) |
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| 521. |
In a crystal, atomic separation is around 2A to 3A. At this separation due to interatomic interaction, energies ofA. outermost electrons is changedB. innermost electrons is changedC. Both (a) and (b)D. None of these |
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Answer» Correct Answer - A Due to atomic interactions, the energy of outermost electrons changed in larger amounts. |
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| 522. |
In a P -type semi-conductor, germanium is dopped withA. BoronB. IndiumC. AluminiumD. all of these |
| Answer» Correct Answer - D | |
| 523. |
In the arrangement shown in fig. The current thorugh diode is-A. 10 mAB. 1 mAC. 20 mAD. zero |
| Answer» Correct Answer - D | |
| 524. |
Which one of the following diagrams correctly represents the energy levels in the p-type semiconductor?A. B. C. D. |
| Answer» Correct Answer - C | |
| 525. |
Consider telecommunication through optical fibres. Which of the following statements is not true?A. Optical fibres can be of graded refractive indexB. Optical fibres are subjected to electromagnetic interference from outsideC. Optical fibres have extremely low transmission lossD. Optical fibres may have homogeneous core with a suitable cladding |
| Answer» Correct Answer - B | |
| 526. |
The given truth table is for A. OR gateB. AND gateC. NOT gateD. none of these |
| Answer» Correct Answer - C | |
| 527. |
In semiconductors at a room tempretureA. the valence band is partially empty and the conduction band is partially filledB. the valence band is completely filled and the conduction band is partially filledC. `the valence band is completely filledD. the conduction band is completely empty |
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Answer» Correct Answer - A At room ttemperatre some electrons move form `VB` to `CB` |
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| 528. |
Why do holes not exist in conductor? |
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Answer» 1. In case of semiconductors, there is one missing electron from one of the covalent bonds. 2. The absence of electron leaves an empty space called as hole; each hole carries an effective positive charge. 3. In case of an conductor, number of free electrons are always available for conduction. There is no absence of electron in it. Hence holes do not exist in conductor. |
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| 529. |
The highest energy level which can be occupied by an electron in valence band at OK is known asA. Potential energyB. Ionisation energyC. Fermi energyD. Atomic energy |
| Answer» Correct Answer - C | |
| 530. |
In conduction band of solid, there is no electron at room temperature. The solid is ……………(A) semiconductors (B) insulator (C) conductor (D) metal |
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Answer» Correct option is: (B) insulator |
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| 531. |
In which of the following band an electron cannot lie in a crystal ?A. Conduction bandB. Forbidden bandC. Valence bandD. All of these |
| Answer» Correct Answer - B | |
| 532. |
Fermi energy is theA. minimum energy of electrons in a metal at 0 KB. maximum energy of electrons in a metal at 0 KC. minimum energy of electrons in a metal at 0°CD. maximum energy of electrons in a metal at 0°C |
| Answer» Correct Answer - B | |
| 533. |
In intrinsic semiconductor at room temperature, the number of electrons and holes areA. equalB. zeroC. unequalD. infinite |
| Answer» Correct Answer - A | |
| 534. |
In a semiconducting material the mobilities of electrons and holes are `mu_(e)` and `mu_(h)` respectively. Which of the following is true?A. `mu_(e)gtmu_(h)`B. `mu_(e) lt mu_(h)`C. `mu_(e)=mu_(h)`D. `mu_(e) lt 0, mu_(h) gt 0` |
| Answer» Correct Answer - A | |
| 535. |
In the following common emitter configuration an `NPN` transistor with current gain `beta=100` is used. The output voltage of the amlifier will be A. `10mV`B. `0.1V`C. `1.0V`D. `10V` |
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Answer» Correct Answer - C `DeltaI_(B)=(v_(i))/(R_(in))=(10^(-3))/(10^(3))=10^(-6)A` `DeltaI_(c )=betaDeltaI_(B)=100xx10^(-6)=10^(-4)A` `v_(0)=DeltaI-(c )R_(L)=10^(-4)xx10xx10^(3)=1V` |
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| 536. |
In the following common emitter configuration an `NPN` transistor with current gain `beta=100` is used. The output voltage of the amlifier will be A. 10 mVB. 0.1 VC. 1.0 VD. 10 V |
| Answer» Correct Answer - C | |
| 537. |
In the following common emitter configuration an `NPN` transistor with current gain `beta=100` is used. The output voltage of the amlifier will be A. 10 mVB. 0.1 VC. 1.0 VD. 10 V |
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Answer» Correct Answer - C `(V_(0))/(V_("in"))=beta(R_(i))/(R_("in"))rArrV_(0)=beta(R_(i))/(R_("in"))xxV_("in")=100xx(10kOmega)/(1kOmega)xx1mV=1V` |
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| 538. |
In the given transistor circuit, the base current is `35 muA`. The value of `R_(alpha)` is `V_(AB)` is assumed to negiligible :- A. `100 k Omega`B. `300 k Omega`C. `200 k Omega`D. `400 k Omega` |
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Answer» Correct Answer - C `R_(b)=(V_(EB))/(I_(B))=(7)/(35xx10^(-6))=200k Omega` |
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| 539. |
In frequency modulationA. The amplitude of career wave varier according to the frequency of message signalB. The frequency of career wave varies according to the amplitude of message signalC. The frequency of career wave varies according to the frequency of message signalD. The amplitude of career wave varies according to the amplitude of message signal |
| Answer» Correct Answer - B | |
| 540. |
Sound produced by a tining fork is a sort of :-A. digital signalB. analog signalC. both (1) and (2)D. neither (1) nor (2) |
| Answer» Correct Answer - B | |
| 541. |
Statement 1: In ground wave transmission the radio signals die out after travelling some distance. Statement-2: Radio signals have a very short wavelength and hence are scattered away by the dust particles and molecules of gases in the atmosphere.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 | |
| 542. |
What is the forbidden energy gap (in joule) for a Germanium crystal ?A. 0.067 evB. 6.57 evC. 2.67 eVD. 0.67 eV |
| Answer» Correct Answer - C | |
| 543. |
Given below are symbols for some logic gates :- The XOR gate and NOR gate respectively are :-A. (a) and (b)B. (b) and (c)C. (c) and (d)D. (a) and (d) |
| Answer» Correct Answer - B | |
| 544. |
An AND gateA. implements logic additionB. is equivalent to a series switching circuitC. is equivalent to a parallel switching circuitD. is a universal gate |
| Answer» Correct Answer - B | |
| 545. |
The output of a 2-input `OR` gate is zero only when itsA. either input is oneB. both inputs are oneC. either input is zeroD. both inputs are zero |
| Answer» Correct Answer - D | |
| 546. |
Which of the following pairs are universal gatesA. NAND,NOTB. NAND,ANDC. NOR,ORD. NAND,NOR |
| Answer» Correct Answer - D | |
| 547. |
The output of a two input NOR gate is in state 1 when :-A. either input terminals is at 0 stateB. either input terminals is at 1 stateC. both input terminals are at 0 stateD. both input terminals are at 1 state |
| Answer» Correct Answer - C | |
| 548. |
Modulation is not used to :-A. Reduce the bandwidth usedB. Separate the transmissions of different usersC. Ensure that intelligence may be transmitted to long distancesD. Allow the use of practical antennas |
| Answer» Correct Answer - A | |
| 549. |
Amplitude modulation is used for broad casting becauseA. It is more noise immune than other modulation systemsB. It requires less transmitting power compared with other systemsC. Its use avoids transmitter complexityD. No other modulation system can provide the necessary bandwidth faithful transmission |
| Answer» Correct Answer - C | |
| 550. |
Calculate the index of refraction of a liquid from the following into glass: (a) Reading for the bottom of an empty beaker: 11.324 cm (b) Reading for the bottom of the beaker, when partially filled with the liquid: 11.802 cm (c) Reading for the upper level of the liquid in the beaker: 12.895cmA. 1.232B. 1.389C. 1.28D. 1.437 |
| Answer» Correct Answer - D | |