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Q1. | Let JCOKE = 20 and LPEPSI =15. Consider the statement |

A. | 25 [Wrong Answer] |

B. | 23 [Correct Answer] |

C. | 15 [Wrong Answer] |

D. | 17 [Wrong Answer] |

View Answer
Explanation:-
Answer : BDiscuss 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 :**

⇒ Leak type bias is used in a plate-modulated class C amplifier to

prevent tuned circuit damping

prevent over modulation

increase the bandwidth

prevent excessive grid current

⇒ In a thyristors are

*i*=

*kv*. The value of

^{a}*a*is

about 0.5

about 2

about 5

about 20

⇒

**Assertion (A):** A half cycle of sine wave can be expressed as sum of sine wave and another sine wave shifted by *T*/2 where T is the time period.

**Reason (R):** The function *u*(*t* - *t*_{0}) - *u*(*t* - *t*_{0} - T) is a gate function occuring at *t* = *t*_{0} and of duration T.

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 true, R is false

A is false, R is true

⇒ A magic tee is to be used as CW duplexer. Then port 1, port 2, port 3 (E arm), port 4 (H arm) respectively should be connected to

matched load, antenna, receiver and CW transmitter

CW transmitter, antenna, receiver and matched load

CW transmitter, receiver, antenna and matched load

CW transmitter, matched load, receiver and antenna

⇒ A 3 phase semiconverter when supplying a certain load has a rectification efficiency of 0.6 and transformer utilisation factor of 0.35. When a 3 phase full converter is supplying the same load

rectification efficiency will be more than 0.6 and TUF will be more than 0.35

rectification efficiency will be more than 0.6 but TUF will be less than 0.35

rectification efficiency will be less than 0.6 and TUF will be less than 0.35

rectification efficiency will be less than 0.6 but TUF will be more than 0.35

⇒ In a CE amplifier, the collector resistance is 6 kΩ and load resistance is 3 kΩ. The ac load resistance is

9 kΩ

6 kΩ

3 kΩ

2 kΩ

⇒ Determine the transistor capacitance of a diffused junction varicap diode of a reverse potential of 4.2 V if C(0) = 80 pf and V

_{T}= 0.7 V

42 pf

153.03 pf

13.33 pf

Data inadequate

⇒ The number of bits of information required to indicate the correct selection of 3 independent consecutive events out of 75 equally probable events is

6.23

18.69

10.46

24.92

⇒ The type of noise that interfere very much with high frequency transmission is

white

transit time

flicker

shot

⇒ In a Hurwitz polynomial

all coefficients are non-negative

both the odd and even parts have roots on

*j*ω axis

the continued fraction expansion is the ratio of odd to even parts of a Hurwitz polynomial yields all positive quotient terms

all of the above

⇒ A voltage v = 5 + 50 sin ω

*t*/ + 5 sin 5 &omega

*t*is applied to a pure capacitor of capacitance 1 ωF. If

*f*/= 314 rad/sec, current is

1 + 0.0157 cos 314

*t*+ 0.00785 cos 1570

*t*

0.0157 cos 314

*t*+ 0.00785 cos 1570 t

0.0157 sin 314

*t*+ 0.00785 sin 1570

*t*

0.0157 sin (314

*t*/ + 45°) + 0.00785 sin (1570

*t*+ 45°)

⇒ Which is correct for a vacuum triode?

μ =

*r*

_{p}

*g*

_{m}

*r*

_{p}= μ

*g*

_{m}

*g*

_{m}= μ

*r*

_{p}

*r*

_{p}

*g*

_{m}

⇒ A transistor has

- collector
- emitter
- base

I only

II only

II and III only

I and II only

⇒ A divide by 78 counters can be realized by using

six mod-13 counters

thirteen mod-6 counters

one mod 13 counter followed by one mod-6 counter

thirteen mod-13 counters

⇒ The hexadecimal number 64 AC is equivalent to decimal number

25727

25722

25772

25777

⇒ For most static RAM the write pulse width should be at least

10

*ns*

60

*ns*

300

*ns*

1 μ

*s*

⇒ An extrinsic semiconductor sample has 6 billion silicon atoms and 3 million pentavalent impurity atoms. The number of electrons and holes is

3 million each

6 billion each

3 million free electrons and very small number of holes

3 million holes and very small number of free electrons

⇒ In

*CE*configuration, the output characteristics of a bipolar junction transistor is drawn between

I

_{C}and V

_{CB}

I

_{E}and V

_{CB}

I

_{C}and V

_{CE}

I

_{E}and V

_{CE}

⇒

Match the following:

List I | List II | ||
---|---|---|---|

A. | Ring modulator | 1. | Clock recovery |

B. | VCO | 2. | Demodulation of FM |

C. | Foster Seeley discriminator | 3. | Frequency conversion |

D. | Mixer | 4. | Summing of two points |

5. | Generation of FM | ||

6. | Generation of DSB-SC |

A-1, B-3, C-2, D-4

A-6, B-5, C-2, D-3

A-6, B-1, C-3, D-2

A-5, B-6, C-1, D-3

⇒ The maximum theoretical efficiency for class A amplifier is

0.125

0.25

0.5

0.75

⇒ The measurement of Hall coefficient of a semiconductor with one type of charge carriers gives the information about

sign of charge carrier

density of charge carrier

both sign and density of charge carrier

none of the above

⇒ The signal defined by the equations

*f*(

*t*) = 0 for

*t*< 0,

*f*(

*t*) = E for 0 ≤

*t*≤

*a*and

*f*(

*t*) = 0 for

*t*>

*a*is

a step function

a pulse function

a shifted step function originating at

*t*=

*a*

none of the above

⇒ In a RS flip-flop no change occurs during

prohibited mode

set mode

reset mode

disable mode

⇒ For the control system in the given figure, the value of K for critical dampin

1

5.125

6.831

10

⇒

**Assertion (A):** A demultiplexer can be used as a decoder.

**Reason (R):** A demultiplexer can be built by using AND gates only.

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 true, R is false

A is false, R is true

⇒ Which of the following is the point of reference JFET?

Drain

Gate

Source

None of the above

⇒

*e*

^{-3}

*is a*

^{t}energy signal

power signal

Both (a) and (b)

none

⇒ The depletion layer in a reverse biased

*p*-

*n*junction is due to the presence of

electrons

holes

both electrons and holes

immobile ions

⇒ In 2 out of 5 code, decimal number 8 is

11000

10100

11000

1010

⇒ The sum of 1110101

_{2}and 11011

_{2}in decimal form will be

65

75

85

95