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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.
| 551. |
A convex lens is held `45 cm` above the bottom of an empty tank. The image of a point at the bottom of a tank is formed `36 cm` above the lens. Now, a liquid is poured into the tank to a depth of `40 cm.` It is found that the distance of the image of the same point on the bottom of the tank is `48 cm` above the lens. Find the refractive index of the liquid.A. 1.28B. 1.82C. 1.47D. 3.12 |
| Answer» Correct Answer - C | |
| 552. |
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. a common base amplifierB. a common collector amplifierC. an oscillatorD. a common emitter amplifier |
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Answer» Correct Answer - D There is a phase difference of `pi` between `V_i` and `V_o` in a CE amplifier. |
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| 553. |
Consider an amplifier circuit using a transistor.The output power is several times greater than the input power.Where does the extra power come from? |
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Answer» Correct Answer - A The extra power required for amplified output is obtained from the DC souuce. |
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| 554. |
What is the resistance of and intrinsic of an intrinsic semiconductor at 0K? |
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Answer» Correct Answer - A At 0K number of holes (or number of free electrons ) in an intrinsic semiconductor become zero. Therefore, resistnace of an intrinsic semiconductor becomes infinite at 0 K. |
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| 555. |
When a potential difference is applied across, the current passing throughA. an insulator at 0 K is zeroB. a semiconductor at 0 K is zeroC. a metal at 0 K is zeroD. a reverse biased p-n junction diode at 300 K is finite |
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Answer» Correct Answer - C At OK, if the same P.D. is applied across an insulator or a semiconductor the current is zero. (The semiconductor also becomes a perfect insulator at OK.) Similarly, in the reverse biased p-n junction diode, a small current flows due to minority charge carriers. Thus (a), (b), (d) are correct. But (c) is wrong because at OK, a conductor becomes a super conductor and the current in it is very large (infinite). |
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| 556. |
n - type extrinsic semiconductors is negatively charged , while P - type extrinsic semiconductors is positively charged . Is this statement true or false? |
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Answer» Correct Answer - A False . Intrinsic as well as extrinsic semiconductors are electrically neutral. |
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| 557. |
In extrinsic semiconductors-A. The conduction band and valence band overlapB. The gap between conduction band and valence band is more than 16 eVC. The gap between conduction band and valence band is near about 1 eVD. The gap between conduction band and valence band will be 100 eV and more. |
| Answer» Correct Answer - C | |
| 558. |
if `n_(e)` and `v_(d)` be the number of electrons and drift velocity in a semiconductor. When the temperature is increased.A. `n_e` increases and `v_d` decreasesB. `n_e` decreases and `v_d` increasesC. Both `n_e` and `v_d` increasesD. Both `n_e` and `v_d` decreases |
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Answer» Correct Answer - A The conductivity of a semiconductor increases with temperature. Hence the number of free electrons (`n_e`) will l also increase. But the drift velocity `v_d=i/((n_e)eA)` Hence as `n_e` is increased, `v_d` is decreased. |
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| 559. |
if `n_(e)` and `v_(d)` be the number of electrons and drift velocity in a semiconductor. When the temperature is increased.A. `n_(e)` increases and `v_(d)`. DecreasesB. `n_(e)` decreases and `v_(d)` increasesC. Both `n_(e)` and `v_(d)` increasesD. Both `n_(e)` and `v_(d)` decreases. |
| Answer» Correct Answer - A | |
| 560. |
The circuit has two oppositively connected ideal diodes in parallel what is the current flowing in the circuit ? A. 2.31 AB. 1.33 AC. 1.71 AD. 2.00 A |
| Answer» Correct Answer - D | |
| 561. |
What is the current flowing through 1 K`Omega` restsor in the following circuit ? A. 2 mAB. 3 mAC. 4mAD. 5 mA |
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Answer» Correct Answer - D This is a zener diode circuit. Zener voltage = 5 V. Zener diode is used as a voltage regulator. `therefore` Voltage across 1 `K Omega` resistance = 5 V `therefore I=5/10^3 = 5xx10^(-3)`=5 mA |
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| 562. |
if `a,b,c,d`are inputs to a gate and `x` is its output , then as per the following time graph , the gate is : d c ba x A. NANDB. NOTC. ANDD. OR |
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Answer» Correct Answer - D Output of OR gate is 0 when all inputs are 0 & output is 1 when atleast one of the inputs is 1. Observing putput x :- It is 0 when all inputs are 0 & it is :- 1 when atleast one of the inputs is 1 `:.` OR gate. |
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| 563. |
The combination of the gates shown represents :- A. AND gateB. OR gateC. NAND gateD. NOR gate |
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Answer» Correct Answer - A `Y=(bar(A)+B).A=A.bar(A)+AB=AND` gate |
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| 564. |
Electric conduction in a semiconductor takes place due to (a) electrons only (b) holes only (c) both electrons and holes (d) neither electrons nor holes. |
| Answer» (c) both electrons and holes | |
| 565. |
In a semiconductor,(a) there are no free electrons at 0k(b) there are no free electrons at any temperature(c) there are no free electrons increases with temperature(d) there are no free electrons is less than that in a conductor. |
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Answer» (a) there are no free electrons at 0k |
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| 566. |
A semiconducting device is connected in a series circuit with a battery and a resistance.A current is found to pass through the circuit .If the polarity of the battery is reversed, the current drops to almost zero.The device may beA. an intrinsic semiconductorB. n type semiconductorC. p type semiconductorD. a p-njunction |
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Answer» Correct Answer - D The p-n junction is reversed biased in the second case. |
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| 567. |
The logic gate circuit given below acts as A. an OR gateB. a NOT gateC. an AND gateD. a NAND gate |
| Answer» Correct Answer - B | |
| 568. |
Which logic gate produces LOW output when any of the inputs in HIGHA. NANDB. NORC. ORD. AND |
| Answer» Correct Answer - A | |
| 569. |
Write down the actual logic operation carried by the following circuit. Explain your answer. |
| Answer» XOR operation `because Y = bar(A)B+A bar(B)` which is boolen expressio for XOR gate. | |
| 570. |
The correct Boolean operation represented by the circuit diagram drawn is A. ORB. ANDC. NANDD. NOT |
| Answer» Correct Answer - B | |
| 571. |
You are given two circuits as shown in following figure. The logic operation carried out by the two circuit are respectively :- A. AND,ORB. OR,ANDC. NAND,ORD. NOR,AND |
| Answer» Correct Answer - A | |
| 572. |
The circuit shown in figure (1) Contains two diodes each with a forward resistance of `50 ohm` and with infinite reverse resistance. If the battery voltage is `6 V`, the current through the `100 ohm` resistance is. .A. 0.01 AB. 0.05 AC. 0.02 AD. 0.03 A |
Answer» Correct Answer - C
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| 573. |
The circuit shown in the figure contains two diodes each with a forward resistance of 50 `Omega` and with infinite backward resistance. If the battery is 6 V, the current through the 100 `Omega` resistance (in ampere) is A. zeroB. 0.02C. 0.03D. 0.036 |
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Answer» Correct Answer - B In circuit the upper diode junction is forward biased and the lower diode junction is reverse biased. Thus, there will be no conduction across lower diode junction. Now the total resistance of circuit `=100+150+50=300Omega` Current in `100Omega=(6)/(300)=0.02A` |
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| 574. |
The emitter base junction is forward biased and the base collector junction is reverse biased, in order to use a transistor asA. ChokeB. rectifierC. amplifierD. transformer |
| Answer» Correct Answer - C | |
| 575. |
The following figure shows a logic gate circuit with two inputs `A` and `B` and the output `Y`. The voltage waveforms of `A,B` and the output `Y` are as given A. `NOR` gateB. `OR` gateC. `AND` gateD. `NAND` gate |
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Answer» Correct Answer - D `{:(A,B,Y),(1,1,0),(0,0,1),(0,1,1),(1,0,1):}` ` NAND` gate |
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| 576. |
What accounts for the flow of charge carriers in forward and reverse biasing of sillicon P-N diode-A. drift in both reverse and forward biasB. drift in forward bias and diffusion in reverse biasC. drift in reverse bias and diffusion in forward biasD. diffution in both forward and reverse bias |
| Answer» Correct Answer - C | |
| 577. |
A sinusoidal voltage of peak value `200` volts is connected to a diode and resistor `R` in the circuit shown so that half wave rectification occurs. If the forward resistance of the diode is negligible compared to `R` the `rms` voltage (in volt) across `R` is approximately A. 200B. 100C. `200/(sqrt(2))`D. 280 |
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Answer» Correct Answer - B Output of the circuit will be a half wave `V_(rms)` of half wave `=200/2=100 V` |
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| 578. |
The dominant mechanisms for motion of charge carriers in forward and reverse biased silicon P-N juntions are-A. Drift in forward bias, diffusion in reversebiasB. Diffusion in forward bias, drift in reverse biasC. Diffusion in both forward and reverse biasD. Drift in both forward and reverse bias |
| Answer» Correct Answer - B | |
| 579. |
The dominant mechanisms for motion of charge carriers in forward and reverse biased silicon `P-N` junction areA. Drift in forward bias and diffusion in reverse biasB. Diffusion in forward bias and drift in reverse biasC. Drift in both forward and reverse biasD. Diffusion in both forward and reverse bias |
| Answer» Correct Answer - B | |
| 580. |
The dominant mechanisms for motion of charge carriers in forward and reverse biased silicon `P-N` junction areA. drift in forward bias, diffusion in reverse biasB. diffusion in forward bias, drifft in reverse biasC. diffusion in both forward and reverse biasD. drift in both forward and reverse bias |
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Answer» Correct Answer - B In forward bias, diffusion current increases, drift current remains almost same, net current is due to diffusion. In reverse bias, diffusion current decrease, net current (nwgligible small)` is due to drift. |
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| 581. |
For pure 'Ge' semiconductor quantity of 'e' and hole is `10^(19) e//m^(3)` if we doped donor impurity in it with density `10^(23) e//m^(3)` then quantity of hole `(e//m^(3))` in semiconductor is :-A. `10^(15)`B. `10^(19)`C. `10^(23)`D. `10^(27)` |
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Answer» Correct Answer - A `n_(h)=(n_(i)^(2))/(n_(e))=((10^(19))^(2))/(10^(23))=10^(15)` |
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| 582. |
In a p-i-n diode solar cell, the width of the depletion region is increased by usingA. a p-type semiconductorB. an n-type semiconductorC. an intrinsic semiconductorD. an n-p-n transistor |
| Answer» Correct Answer - C | |
| 583. |
If the forward voltage in a diode is increased, the width of the depletion region-A. decreasesB. increasesC. fluctuatesD. does not change |
| Answer» Correct Answer - A | |
| 584. |
If the forward voltage in a diode is increased, the width of the depletion region-A. decreaseB. increaseC. fluctuatesD. does not change |
| Answer» Correct Answer - A | |
| 585. |
When a junction diode is reverse biased, then current called drift current is due toA. diffusion of chargeB. nature of the materialC. drift of the chargesD. both drift and diffusion of the charges |
| Answer» Correct Answer - C | |
| 586. |
A semiconductor is doped with a donor impurity.(a) the hole concentration increases.(b) the hole concentration decreases.(c) the electron concentration increases.(d) the electron concentration decreases. |
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Answer» (b) the hole concentration decreases. (c) the electron concentration increases. |
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| 587. |
A p-type semiconductor is (i) a silicon crystal doped with arsenic impurity (ii) a silicon crystal doped with aluminium impurity (iii) a germanium crystal doped with boron impurity (iv) a germanium crystal doped with phosphorus impurities.A. (i) and (ii) are correctB. (ii) and (iii) are correctC. (i) and (iv) are correctD. only (i) is correct |
| Answer» Correct Answer - B | |
| 588. |
When an impurity is doped into an intrinsic semicomiuctor, the-conductivity of the semiconductor (a) increases (b) decreases(c) remains the same (d) becomes zero. |
| Answer» (a) increases | |
| 589. |
If the forward voltage in a diode is increased, the width of the depletion region-A. decreaseB. increaseC. not changeD. increase slightly |
| Answer» Correct Answer - A | |
| 590. |
When a p-n junction diode is reverse biasedA. electrons and holes are attracted towards each other and move towards the depletion region.B. electrons and holes move away form the deplection region.C. height of the potential barrier decreasesD. no change in the current takes place |
| Answer» Correct Answer - B | |
| 591. |
In which one of the following devices the reverse biased characteristics of a p-n junction diode are used?A. AmplifierB. Zener diodeC. OscillatorD. Logic gate |
| Answer» Correct Answer - B | |
| 592. |
When arsenic is added as an impurity to silicon, the resulting material isA. an n-type semiconductorB. a p-type semiconductorC. insulatorD. good conductor |
| Answer» Correct Answer - A | |
| 593. |
Which of the following , when added as an impurity, into the silicon, produces n-type semiconductor ?A. PhosphorusB. AluminiumC. MagnesiumD. Both (b) and (c) |
| Answer» Correct Answer - A | |
| 594. |
When added an impurity into the silicon which one of the following produces `n`-type of semiconductors?A. PB. AsC. BD. Mg |
| Answer» Correct Answer - A | |
| 595. |
What are majority carriers in a semiconductor? (A) Holes in n-type and electrons in p-type.(B) Holes in n-type and p-type both. (C) Electrons in n-type and p-type both. (D) Holes in p-type and electrons in n-type. |
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Answer» (D) Holes in p-type and electrons in n-type. |
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| 596. |
When a p-n junction diode is reverse biased the flow of current across the junction is mainly due toA. drift of chargesB. diffusion of chargesC. both drift and diffusion of of chargesD. depends on the nature of the material |
| Answer» Correct Answer - B | |
| 597. |
When boron is added as an impurity to silicon,the resulting material isA. n-type conductorB. n- type semiconductorC. p-type conductorD. p-type semiconductor |
| Answer» Correct Answer - D | |
| 598. |
In `P`-type semiconductor the majority and minorty charge carriers are respectivelyA. protons and electronsB. electrons and protonsC. electrons and holesD. holes and electrons |
| Answer» Correct Answer - D | |
| 599. |
A `P-N` junction diode can withstand currents up to `10 mA`. Under forward bias, The diode has a potential drop of `0.5 V` across it which is assumed to be independent of current. The maximum voltage of the battery used to forward bias the diode when a resistance of `200 Omega` is connected in series with it is |
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Answer» Potential drop is 0.5 and current is 10 mA , so resistance of diode is `(0.5)/(10 xx 10^(-3)) = 50 Omega` Total resistance of circuit is `200 + 50 = 250 Omega` So maximum current is `(V_(max))/(250)=10 xx 10^(-3) rArr V_(max) = 2.5` volt. |
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| 600. |
An electric field us applied to a semiconductor.Let the number of charge carriers be n and the average drift speed be v.If the temperature is increased,A. Both n and v will increaseB. n will increase but v will decreaseC. v will increase but n will decreaseD. Both n and v will decrease |
| Answer» Correct Answer - B | |