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Q1. | The presence of some holes in an intrinsic semiconductor at room temperature is due to |

A. | valence electrons [Wrong Answer] |

B. | doping [Wrong Answer] |

C. | free electrons [Wrong Answer] |

D. | thermal energy [Correct Answer] |

View Answer
Explanation:-
Answer : DDiscuss it below :!! OOPS Login [Click here] is required to post your answer/resultHelp other students, write article, leave your comments |

**Also Read Similar Questions Below :**

⇒ Data are recorded on a 2, 400 ft reel of magnetic tape at a density of 556 characters per inch. If the record length is of 100 characters and 0.75 inch of record gap, the tape utilization factor is

0.85

0.67

0.19

0.08

⇒ Ferrimagnetic materials have anti-parallel orientation of equal moments of permanent dipoles.

TRUE

FALSE

⇒ An amplifier has open loop gain of 100, input impedance 1 kΩ and output impedance 100 Ω. If negative feedback with β = 0.99 is used, the new input and output impedances are

10 Ω and 1 Ω

10 Ω and 10 kΩ

100 kΩ and 1 Ω

100 kΩ and 10 Ω

⇒ For parallel RLC circuit, which one of the following statements is NOT correct?

The bandwidth of the circuit deceases if R is increased

The bandwidth of the circuit remains same if L is increased

At resonance, input impedance is a real quantity

At resonance, the magnitude of input impedance attains its minimum value

⇒ In Quine-Mclusky method of Boolean function reduction; __________ , are treated as __________ and therefore discrepancies may occur in the final solution, therefore this method is not applicable to the functions with __________ .

zeros, don't care conditions, zeros

one's, don't care conditions, one's

don't care conditions, zeros, don't care conditions

don't care conditions, don't care conditions, don't care conditions

⇒ In energy band diagram of

*n*type semiconductor, the donor energy level is

in valence band

in conduction band

slightly above valence band

slightly below conduction band

⇒

Match the following:

List I (Double Rectifier) | List II (Number of Diodes) | ||
---|---|---|---|

A. | Full wave Bridge rectifier | 1. | 1 |

B. | Half wave rectifier | 2. | 2 |

C. | Full wave rectifier using centre tapped transformer | 3. | 4 |

A-1, B-2, C-3

A-3, B-1, C-2

A-2, B-1, C-3

A-3, B-1, C-2

⇒ A thermal power plant has an efficiency of 0.4. If 0.6 kg of coal is burnt to generate 1 kWh, the calorific value of coal is

about 3500 kcal/kg

about 2500 kcal/kg

about 1500 kcal/kg

about 7500 kcal/kg

⇒

Match the following:

List I (Component) | List II (Dominant TE mode) | ||
---|---|---|---|

A. | Rectangular wave guide | 1. | TE_{101} |

B. | Circular waveguide | 2. | TE_{10} |

C. | Rectangular cavity | 3. | TE_{111} |

4. | TE_{11} |

A-4, B-2, C-1

A-2, B-1, C-3

A-3, B-4, C-1

A-2, B-3, C-1

⇒ A sinusoidal carrier voltage, having frequency 1000 Hz is amplitude modulated by a sinusoidal voltage of frequency 15 kHz. The frequency of upper and lower sidebands will be

1015 kHz and 985 kHz

1030 kHz and 970 kHz

13 kHz and 17 kHz

14 kHz and 16 kHz

⇒ The '

*h*' parameters of the circuit shown in the figure are

*h*

_{ib}= 25 Ω,

*h*

_{fb}= 0.999 and

*h*

_{ob}= 10

^{-6}Ω The Voltage gai

0.99

1.98

2

4

⇒ Phase velocity of waves propagating in a hollow metal waveguide is

greater than the group velocity

less than the velocity of light in free space

equal to the velocity of light in free space

equal to group velocity

⇒ Consider the following statements:

- The positiveness of coefficients of characteristic equation is necessary as well as sufficient condition for stability of first and second order systems.
- The positiveness of coefficients of characteristic equation ensures negativeness of real roots but is not sufficient condition for stability of third and higher order systems.

only 1 is correct

both 1 and 2 are correct

only 2 is correct

both are wrong

⇒ An SBS can conduct is one direction only.

TRUE

FALSE

⇒ At midband frequencies the coupling capacitor in RC coupled CE amplifier may be considered as

open circuit

short circuit

a reactance equal to load resistance

none of the above

⇒ Conversion rate, for a successive approximation ADC which uses 2 MH

*z*clock and 5 bit binary ladder containing 8 VV reference voltage is __________ .

64516.12

*conversions/sec*

400000

*conversions/sec*

400000

*bps*

2.5

*bps*

⇒ A voltage

*v*= 100 sin ω

*t*+ 10 sin 5 ω

*t*is applied to a pure capacitor having capacitance of 1 μF. If ω = 314 rad/sec, the current through the capacitor is

0.0314 cos 314

*t*+ 0.0157 cos 1570

*t*

0.0314 sin 314

*t*+ 0.0157 sin 1570

*t*

0.0314 cos 314

*t*+ 0.0314 cos 1570

*t*

0.0157 cos 314

*t*+ 0.0157 cos 1570

*t*

⇒ In monostable multivibrator

has no stable state

has one stable state

has two stable states

switches automatically from one state to other state

⇒ The initial state of MOD-16 down counter is 0110. What state will it be after 37 clock pulses?

Indeterminate

110

101

1

⇒ When the modulating frequency is doubled, the modulation index is halved and the modulating voltage remains constant. The modulation system is

FM

AM

phase modulation

none of the above

⇒ In the circuit of figure, the waveshape of load current when thyristor conducting

constant

increasing exponentially

decreasing exponentially

either (b) or (c)

⇒ For a stable system having two or more gain crossover frequencies the phase margin is measured

at lowest gain crossover frequency

at highest gain crossover frequency

at the gain crossover frequencies

none of the above

⇒ Figure shows four D type FFs are connected as a shift register using an XOR gate. The initial state and 3 subsequent states for 3 clock pulses are also g

0

1111

1001

1000

⇒ For a second order system, the position of poles is shown in the given figure. This syste

undamped

overdamped

underdamped

critically damped

⇒ A 2 μF capacitor is shunted by a 1 kΩ maintained at temperature 400 K. The rms noise voltage across the capacitor over the entire frequency band is:

6.25 x 10

^{-8}V

5.25 x 10

^{-6}V

6.25 x 10

^{-6}V

5.25 x 10

^{-8}V

⇒ For the discrete time system of the given fig

*y*- 0.5

_{k}*y*

_{k}_{ - 1}- 0.25

*y*

_{k}_{ - 2}=

*u*

_{k}*y*+ 0.5

_{k}*y*

_{k}_{ - 1}- 0.25

*y*

_{k}_{ - 2}=

*u*

_{k}*y*- 0.5

_{k}*y*

_{k}_{ - 1}+ 0.25

*y*

_{k}_{ - 2}=

*u*

_{k}*y*+ 0.5

_{k}*y*

_{k}_{ - 1}+ 0.25

*y*

_{k}_{ - 2}=

*u*

_{k}⇒ Short circuited stubs are preferred to O.C stubs because

O.C stubs are incapable of giving full range of reactances

O.C stubs are liable to radiate

O.C stubs are difficult to make and connect

none of the above

⇒

**Assertion (A):** Most PCs use static RAM for main memory.

**Reason (R):** Static RAM is faster than dynamic RAM.

Both A and R are correct and R is correct explanation of A

Both A and R are correct but R is not correct explanation of A

A is correct R is wrong

A is wrong R is correct

⇒ Which of the following uses a capacitor in the feedback path?

Active low pass filter

Active high pass filter

Both (a) or (b)

Neither (a) nor (b)

⇒ In the NMOS inverter

the driver and achieve load are enhancement type

the driver is enhancement type and load depletion type

both driver and load are depletion type

the driver and load are depletion type