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MCQOPTIONS
This section includes 12583 Mcqs, each offering curated multiple-choice questions to sharpen your Joint Entrance Exam - Main (JEE Main) knowledge and support exam preparation. Choose a topic below to get started.
| 5851. |
In a common base amplifier the phase difference between the input signal voltage and the output voltage is [CBSE PMT 1990; AIEEE 2005] |
| A. | 0 |
| B. | \[\pi /4\] |
| C. | \[\pi /2\] |
| D. | \[\pi \] |
| Answer» B. \[\pi /4\] | |
| 5852. |
The emitter-base junction of a transistor is ?? biased while the collector-base junction is ??. biased [KCET 2004] |
| A. | Reverse, forward |
| B. | Reverse, reverse |
| C. | Forward, forward |
| D. | Forward, reverse |
| Answer» E. | |
| 5853. |
Consider an NPN transistor amplifier in common-emitter configuration. The current gain of the transistor is 100. If the collector current changes by 1 mA, what will be the change in emitter current [AIIMS 2005] |
| A. | 1.1 mA |
| B. | 1.01 mA |
| C. | 0.01 mA |
| D. | 10 mA |
| Answer» C. 0.01 mA | |
| 5854. |
While a collector to emitter voltage is constant in a transistor, the collector current changes by 8.2 mA when the emitter current changes by 8.3 mA. The value of forward current ratio hfe is [KCET 2002] |
| A. | 82 |
| B. | 83 |
| C. | 8.2 |
| D. | 8.3 |
| Answer» B. 83 | |
| 5855. |
Which of the following is used to produce radio waves of constant amplitude [DCE 2004] |
| A. | Oscillator |
| B. | FET |
| C. | Rectifier |
| D. | Amplifier |
| Answer» B. FET | |
| 5856. |
In the CB mode of a transistor, when the collector voltage is changed by 0.5 volt. The collector current changes by 0.05 mA. The output resistance will be [Pb. PMT 2003] |
| A. | 10 kW |
| B. | 20 kW |
| C. | 5 kW |
| D. | 2.5 kW |
| Answer» B. 20 kW | |
| 5857. |
When NPN transistor is used as an amplifier [DCE 2002] |
| A. | Electrons move from base to emitter |
| B. | Electrons move from emitter to base |
| C. | Electrons moves from base to emitter |
| D. | Holes moves from base to emitter |
| Answer» C. Electrons moves from base to emitter | |
| 5858. |
The relation between a and b parameters of current gains for a transistors is given by [Pb. PET 2000] |
| A. | \[\alpha =\frac{\beta }{1-\beta }\] |
| B. | \[\alpha =\frac{\beta }{1+\beta }\] |
| C. | \[\alpha =\frac{1-\beta }{\beta }\] |
| D. | \[\alpha =\frac{1+\beta }{\beta }\] |
| Answer» C. \[\alpha =\frac{1-\beta }{\beta }\] | |
| 5859. |
For a transistor, in a common emitter arrangement, the alternating current gain b is given by [DPMT 2004] |
| A. | \[\beta ={{\left( \frac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}} \right)}_{{{V}_{C}}}}\] |
| B. | \[\beta ={{\left( \frac{\Delta {{I}_{B}}}{\Delta {{I}_{C}}} \right)}_{{{V}_{C}}}}\] |
| C. | \[\beta ={{\left( \frac{\Delta {{I}_{C}}}{\Delta {{I}_{E}}} \right)}_{{{V}_{C}}}}\] |
| D. | \[\beta ={{\left( \frac{\Delta {{I}_{E}}}{\Delta {{I}_{C}}} \right)}_{{{V}_{C}}}}\] |
| Answer» B. \[\beta ={{\left( \frac{\Delta {{I}_{B}}}{\Delta {{I}_{C}}} \right)}_{{{V}_{C}}}}\] | |
| 5860. |
Which of these is unipolar transistor [Pb PMT 2004] |
| A. | Point contact transistor |
| B. | Field effect transistor |
| C. | PNP transistor |
| D. | None of these |
| Answer» C. PNP transistor | |
| 5861. |
In a transistor configuration b-parameter is [Orissa PMT 2004] |
| A. | \[\frac{{{l}_{b}}}{{{l}_{c}}}\] |
| B. | \[\frac{{{l}_{c}}}{{{l}_{b}}}\] |
| C. | \[\frac{{{l}_{c}}}{{{l}_{a}}}\] |
| D. | \[\frac{{{l}_{a}}}{{{l}_{c}}}\] |
| Answer» C. \[\frac{{{l}_{c}}}{{{l}_{a}}}\] | |
| 5862. |
In a transistor, a change of 8.0 mA in the emitter current produces a change of 7.8 mA in the collector current. What change in the base current is necessary to produce the same change in the collector current |
| A. | 50 mA |
| B. | 100 mA |
| C. | 150 mA |
| D. | 200 mA |
| Answer» E. | |
| 5863. |
An oscillator is nothing but an amplifier with [MP PET 2004] |
| A. | Positive feed back |
| B. | Large gain |
| C. | No feedback |
| D. | Negative feedback |
| Answer» B. Large gain | |
| 5864. |
In a transistor circuit shown here the base current is 35 mA. The value of the resistor Rb is |
| A. | 123.5 kW |
| B. | 257 kW |
| C. | 380.05 kW |
| D. | None of these |
| Answer» C. 380.05 kW | |
| 5865. |
In case of NPN-transistors the collector current is always less than the emitter current because [AIIMS 1983] |
| A. | Collector side is reverse biased and emitter side is forward biased |
| B. | After electrons are lost in the base and only remaining ones reach the collector |
| C. | Collector side is forward biased and emitter side is reverse biased |
| D. | Collector being reverse biased attracts less electrons |
| Answer» C. Collector side is forward biased and emitter side is reverse biased | |
| 5866. |
In a common base amplifier circuit, calculate the change in base current if that in the emitter current is 2 mA and a = 0.98 [BHU 1995] |
| A. | 0.04 mA |
| B. | 1.96 mA |
| C. | 0.98 mA |
| D. | 2 mA |
| Answer» B. 1.96 mA | |
| 5867. |
For a transistor, the current amplification factor is 0.8. The transistor is connected in common emitter configuration. The change in the collector current when the base current changes by 6 mA is [Haryana CET 1991] |
| A. | 6 mA |
| B. | 4.8 mA |
| C. | 24 mA |
| D. | 8 mA |
| Answer» D. 8 mA | |
| 5868. |
The part of a transistor which is heavily doped to produce a large number of majority carriers, is [CBSE PMT 1993] |
| A. | Base |
| B. | Emitter |
| C. | Collector |
| D. | None of these |
| Answer» C. Collector | |
| 5869. |
An NPN-transistor circuit is arranged as shown in figure. It is [BHU 1994] |
| A. | A common base amplifier circuit |
| B. | A common emitter amplifier circuit |
| C. | A common collector amplifier circuit |
| D. | Neither of the above |
| Answer» C. A common collector amplifier circuit | |
| 5870. |
The most commonly used material for making transistor is [MNR 1995] |
| A. | Copper |
| B. | Silicon |
| C. | Ebonite |
| D. | Silver |
| Answer» C. Ebonite | |
| 5871. |
The symbol given in figure represents [AMU 1995, 96] |
| A. | NPN transistor |
| B. | PNP transistor |
| C. | Forward biased PN junction diode |
| D. | Reverse biased NP junction diode |
| Answer» B. PNP transistor | |
| 5872. |
A common emitter amplifier is designed with NPN transistor (a = 0.99). The input impedance is 1 KW and load is 10 KW. The voltage gain will be [CPMT 1996] |
| A. | 9.9 |
| B. | 99 |
| C. | 990 |
| D. | 9900 |
| Answer» D. 9900 | |
| 5873. |
In a PNP transistor the base is the N-region. Its width relative to the P-region is [DCE 1997] |
| A. | Smaller |
| B. | Larger |
| C. | Same |
| D. | Not related |
| Answer» B. Larger | |
| 5874. |
The phase difference between input and output voltages of a CE circuit is [MP PET 2004] |
| A. | 0o |
| B. | 90o |
| C. | 180o |
| D. | 270o |
| Answer» D. 270o | |
| 5875. |
A transistor is used in common emitter mode as an amplifier. Then [IIT-JEE 1998] |
| A. | The base-emitter junction is forward biased |
| B. | The base-emitter junction is reverse biased |
| C. | The input signal is connected in series with the voltage applied to the base-emitter junction |
| D. | The input signal is connected in series with the voltage applied to bias the base collector junction |
| Answer» D. The input signal is connected in series with the voltage applied to bias the base collector junction | |
| 5876. |
For a transistor the parameter b = 99. The value of the parameter a is [Pb CET 1998] |
| A. | 0.9 |
| B. | 0.99 |
| C. | 1 |
| D. | 9 |
| Answer» C. 1 | |
| 5877. |
The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common-emitter configuration is 1 KW. The peak value for an A.C input voltage of 0.01 V peak is [CBSE PMT 1998] |
| A. | 100 mA |
| B. | 0.01 mA |
| C. | 0.25 mA |
| D. | 500 mA |
| Answer» E. | |
| 5878. |
Least doped region in a transistor [KCET 2000] |
| A. | Either emitter or collector |
| B. | Base |
| C. | Emitter |
| D. | Collector |
| Answer» C. Emitter | |
| 5879. |
The transistors provide good power amplification when they are used in [AMU 1999] |
| A. | Common collector configuration |
| B. | Common emitter configuration |
| C. | Common base configuration |
| D. | None of these |
| Answer» C. Common base configuration | |
| 5880. |
In a common emitter transistor, the current gain is 80. What is the change in collector current, when the change in base current is 250 mA [CBSE PMT 2000] |
| A. | 80 ´ 250 mA |
| B. | (250 ? 80) mA |
| C. | (250 + 80) mA |
| D. | 250/80 mA |
| Answer» B. (250 ? 80) mA | |
| 5881. |
In an NPN transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, the emitter current (iE) and base current (iB) are given by [KCET 2001] |
| A. | iE = ? 1 mA, iB = 9 mA |
| B. | iE = 9 mA, iB = ? 1 mA |
| C. | iE = 1 mA, iB = 11 mA |
| D. | iE = 11 mA, iB = 1 mA |
| Answer» E. | |
| 5882. |
If \[{{l}_{1}},\,{{l}_{2}},\,{{l}_{3}}\] are the lengths of the emitter, base and collector of a transistor then [KCET 2002] |
| A. | \[{{l}_{1}}={{l}_{2}}={{l}_{3}}\] |
| B. | \[{{l}_{3}}<{{l}_{2}}>{{l}_{1}}\] |
| C. | \[{{l}_{3}}<{{l}_{1}}<{{l}_{2}}\] |
| D. | \[{{l}_{3}}>{{l}_{1}}>{{l}_{2}}\] |
| Answer» E. | |
| 5883. |
In a PNP transistor working as a common-base amplifier, current gain is 0.96 and emitter current is 7.2 mA. The base current is [AFMC 2002; Pb. PET 2002] |
| A. | 0.4 mA |
| B. | 0.2 mA |
| C. | 0.29 mA |
| D. | 0.35 mA |
| Answer» D. 0.35 mA | |
| 5884. |
For a common base configuration of PNP transistor \[\frac{{{l}_{C}}}{{{l}_{E}}}=0.98\] then maximum current gain in common emitter configuration will be [CBSE PMT 2002] |
| A. | 12 |
| B. | 24 |
| C. | 6 |
| D. | 5 |
| Answer» C. 6 | |
| 5885. |
When NPN transistor is used as an amplifier [AIEEE 2004] |
| A. | Electrons move from base to collector |
| B. | Holes move from emitter to base |
| C. | Electrons move from collector to base |
| D. | Holes move from base to emitter |
| Answer» B. Holes move from emitter to base | |
| 5886. |
One mole of \[{{O}_{2}}\] gas having a volume equal to 22.4 litres at \[{{0}^{o}}C\] and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is [MP PET 1993; BVP 2003] |
| A. | \[1672.5\ J\] |
| B. | 1728 J |
| C. | \[-1728J\] |
| D. | \[-1572.5\ J\] |
| Answer» E. | |
| 5887. |
If a gas is heated at constant pressure, its isothermal compressibility [MP PET 1984] |
| A. | Remains constant |
| B. | Increases linearly with temperature |
| C. | Decreases linearly with temperature |
| D. | Decreases inversely with temperature |
| Answer» B. Increases linearly with temperature | |
| 5888. |
The gas law \[\frac{PV}{T}=\] constant is true for [MNR 1974; MP PMT 1984; BHU 1995, 98, 2000] |
| A. | Isothermal changes only |
| B. | Adiabatic changes only |
| C. | Both isothermal and adiabatic changes |
| D. | Neither isothermal nor adiabatic changes |
| Answer» D. Neither isothermal nor adiabatic changes | |
| 5889. |
For an isothermal expansion of a perfect gas, the value of \[\frac{\Delta P}{P}\] is equal [CPMT 1980] |
| A. | \[-{{\gamma }^{1/2}}\frac{\Delta V}{V}\] |
| B. | \[-\frac{\Delta V}{V}\] |
| C. | \[-\gamma \frac{\Delta V}{V}\] |
| D. | \[-{{\gamma }^{2}}\frac{\Delta V}{V}\] |
| Answer» C. \[-\gamma \frac{\Delta V}{V}\] | |
| 5890. |
A vessel containing 5 litres of a gas at 0.8 m pressure is connected to an evacuated vessel of volume 3 litres. The resultant pressure inside will be (assuming whole system to be isolated) [MP PMT 1993] |
| A. | 4/3 m |
| B. | 0.5 m |
| C. | 2.0 m |
| D. | 3/4 m |
| Answer» C. 2.0 m | |
| 5891. |
During an isothermal expansion of an ideal gas [UPSEAT 2005] |
| A. | Its internal energy decreases |
| B. | Its internal energy does not change |
| C. | The work done by the gas is equal to the quantity of heat absorbed by it |
| D. | Both (b) and (c) are correct |
| Answer» E. | |
| 5892. |
In an isothermal reversible expansion, if the volume of 96 gm of oxygen at 27°C is increased from 70 litres to 140 litres, then the work done by the gas will be |
| A. | \[300\,R{{\log }_{10}}2\] |
| B. | \[81\,R{{\log }_{e}}2\] |
| C. | \[900\,R{{\log }_{10}}2\] |
| D. | \[2.3\times 900\,R{{\log }_{10}}2\] |
| Answer» E. | |
| 5893. |
The volume of an ideal gas is 1 litre and its pressure is equal to 72cm of mercury column. The volume of gas is made 900 cm3 by compressing it isothermally. The stress of the gas will be [UPSEAT 1999] |
| A. | 8 cm (mercury) |
| B. | 7 cm (mercury) |
| C. | 6 cm (mercury) |
| D. | 4 cm (mercury) |
| Answer» B. 7 cm (mercury) | |
| 5894. |
In an isothermal expansion [KCET 2000; AFMC 2001] |
| A. | Internal energy of the gas increases |
| B. | Internal energy of the gas decreases |
| C. | Internal energy remains unchanged |
| D. | Average kinetic energy of gas molecule decreases |
| Answer» D. Average kinetic energy of gas molecule decreases | |
| 5895. |
A cylinder fitted with a piston contains 0.2 moles of air at temperature 27°C. The piston is pushed so slowly that the air within the cylinder remains in thermal equilibrium with the surroundings. Find the approximate work done by the system if the final volume is twice the initial volume [BHU (Med.) 2000] |
| A. | 543 J |
| B. | 345 J |
| C. | 453 J |
| D. | 600 J |
| Answer» C. 453 J | |
| 5896. |
One mole of an ideal gas expands at a constant temperature of 300 K from an initial volume of 10 litres to a final volume of 20 litres. The work done in expanding the gas is (R = 8.31 J/mole-K) [MP PMT 1995; UPSEAT 2000] |
| A. | 750 joules |
| B. | 1728 joules |
| C. | 1500 joules |
| D. | 3456 joules |
| Answer» C. 1500 joules | |
| 5897. |
The latent heat of vaporisation of water is 2240 J/gm. If the work done in the process of expansion of 1 g is 168 J, then increase in internal energy is [Pb. PET 1998; CPMT 2000] |
| A. | 2408 J |
| B. | 2240 J |
| C. | 2072 J |
| D. | 1904 J |
| Answer» D. 1904 J | |
| 5898. |
When 1 gm of water at \[{{0}^{o}}C\] and \[1\times {{10}^{5}}\ N/{{m}^{2}}\] pressure is converted into ice of volume \[1.091\ c{{m}^{2}}\], the external work done will be |
| A. | 0.0091 joule |
| B. | 0.0182 joule |
| C. | ? 0.0091 joule |
| D. | ? 0.0182 joule |
| Answer» B. 0.0182 joule | |
| 5899. |
When heat is given to a gas in an isothermal change, the result will be [MP PET 1995; RPMT 1997] |
| A. | External work done |
| B. | Rise in temperature |
| C. | Increase in internal energy |
| D. | External work done and also rise in temp. |
| Answer» B. Rise in temperature | |
| 5900. |
When an ideal gas in a cylinder was compressed isothermally by a piston, the work done on the gas was found to be \[1.5\times {{10}^{4}}\ joules\]. During this process about [MP PMT 1987] |
| A. | \[3.6\times {{10}^{3}}\] cal of heat flowed out from the gas |
| B. | \[3.6\times {{10}^{3}}\] cal of heat flowed into the gas |
| C. | \[1.5\times {{10}^{4}}\] cal of heat flowed into the gas |
| D. | \[1.5\times {{10}^{4}}\] cal of heat flowed out from the gas |
| Answer» B. \[3.6\times {{10}^{3}}\] cal of heat flowed into the gas | |