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This section includes 310 Mcqs, each offering curated multiple-choice questions to sharpen your Networking knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
When the Cable TV is used for data transfer then the downstream band has |
| A. | 11 channels |
| B. | 22 channels |
| C. | 33 channels |
| D. | 44 channels |
| Answer» D. 44 channels | |
| 52. |
The bandwidth is proportional to the |
| A. | baud rate |
| B. | data rate |
| C. | delay rate |
| D. | bit rate |
| Answer» B. data rate | |
| 53. |
Quadrature amplitude modulation is a combination of |
| A. | BPSK and PSK |
| B. | ASK and BPSK |
| C. | ASK and PSK |
| D. | BPSK and QPSK |
| Answer» D. BPSK and QPSK | |
| 54. |
Line coding, block coding and scrambling are three techniques of |
| A. | Digital to digital conversion |
| B. | Digital to analog conversion |
| C. | Analog to analog conversion |
| D. | Analog to digital conversion |
| Answer» B. Digital to analog conversion | |
| 55. |
MLT-3 schemes uses three levels |
| A. | 0, V,-V |
| B. | 0,1,2 |
| C. | M,L,T |
| D. | 0,-1,-2 |
| Answer» B. 0,1,2 | |
| 56. |
AM stations are allowed carrier frequencies anywhere between |
| A. | 5-10KHz |
| B. | 50-100KHz |
| C. | 250-1000KHz |
| D. | 530- 1700 kHz |
| Answer» E. | |
| 57. |
The minimum time required to download one million bytes of information for V32 modem would be |
| A. | 800 s |
| B. | 834 s |
| C. | 900 s |
| D. | 1000 s |
| Answer» C. 900 s | |
| 58. |
How many modes are currently in used for propagating light along optical channels |
| A. | one mode |
| B. | two modes |
| C. | three modes |
| D. | five modes |
| Answer» C. three modes | |
| 59. |
Cable TV networks started to distribute broadcast video signals to locations in the late |
| A. | 1940s |
| B. | 1950s |
| C. | 1960s |
| D. | 1970s |
| Answer» B. 1950s | |
| 60. |
The local loop has a bandwidth of |
| A. | 400Hz |
| B. | 4Hz |
| C. | 40Hz |
| D. | 4000Hz |
| Answer» E. | |
| 61. |
In NRZ-I the inversion or the lack of inversion determines the |
| A. | Value of the bytes |
| B. | value of the bits |
| C. | Status of Frame |
| D. | None of the above |
| Answer» C. Status of Frame | |
| 62. |
Tele is a word in |
| A. | French |
| B. | Egyptian |
| C. | German |
| D. | Greek |
| Answer» E. | |
| 63. |
In the adaptive delta modulation, for better performance, the value of delta is |
| A. | fixed |
| B. | not fixed |
| C. | infinity |
| D. | zero |
| Answer» C. infinity | |
| 64. |
The range of middle frequency is |
| A. | 3-30 kHz |
| B. | 900KHz-300 kHz |
| C. | 3khz-30 MHz |
| D. | 300 kHz-3 MHz |
| Answer» E. | |
| 65. |
Both station can transmit and receive data simultaneously in |
| A. | simplex mode |
| B. | Half duplex mode |
| C. | Full duplex mode |
| D. | None of Above |
| Answer» D. None of Above | |
| 66. |
In star topology if the central hub goes down, it effects |
| A. | One node |
| B. | No node |
| C. | whole system |
| D. | Don't know |
| Answer» D. Don't know | |
| 67. |
Downstream data are modulated using the |
| A. | 64-QAM modulation |
| B. | 68-QAM modulation |
| C. | 70-QAM modulation |
| D. | 100-QAM modulation |
| Answer» B. 68-QAM modulation | |
| 68. |
The total bandwidth required for AM can be determined from the bandwidth of the audio signal |
| A. | BAM = 2B |
| B. | BAM = 4B |
| C. | BAM = 6B |
| D. | BAM = 10B |
| Answer» B. BAM = 4B | |
| 69. |
Analog-to-analog conversion can be accomplished in |
| A. | one way |
| B. | two ways |
| C. | three ways |
| D. | four ways |
| Answer» D. four ways | |
| 70. |
The technique that is most common to change an analog signal to digital data is |
| A. | delta modulation |
| B. | pulse code modulation |
| C. | sampling |
| D. | Quantizing |
| Answer» C. sampling | |
| 71. |
The ray of light refracts and moves closer to the surface then the angle of the incidence is |
| A. | equal to the critical angle |
| B. | not equal to the critical angel |
| C. | less than the critical angle |
| D. | greater than the critical angle |
| Answer» D. greater than the critical angle | |
| 72. |
In Discrete Multitone Technique (DMT), Channel 0 is reserved for |
| A. | data communication |
| B. | audio communication |
| C. | Both a and b |
| D. | voice communication |
| Answer» E. | |
| 73. |
The Nyquist sampling rate is |
| A. | f |
| B. | 2f |
| C. | 4f |
| D. | 6f |
| Answer» C. 4f | |
| 74. |
In Binary Phase Shift Keying (BPSK), there are two values of phase i.e. 0 degree and |
| A. | 180 degree |
| B. | 90 degree |
| C. | 360 degree |
| D. | 720 degree |
| Answer» B. 90 degree | |
| 75. |
The bit rate for 1000 baud, FSK, baud rate would be |
| A. | 1000 bps |
| B. | 1500 bps |
| C. | 2000 bps |
| D. | 100 bps |
| Answer» B. 1500 bps | |
| 76. |
In Coaxial Cable, the whole cable is protected by a |
| A. | shield |
| B. | plastic cover |
| C. | insulator |
| D. | conductor |
| Answer» C. insulator | |
| 77. |
The most efficient and commonly used digital to analog conversion mechanism is |
| A. | ASK |
| B. | FSK |
| C. | PSK |
| D. | QAM |
| Answer» E. | |
| 78. |
The coaxial cable has a bandwidth that ranges from |
| A. | 5- 750MHz |
| B. | 10-300 MHz |
| C. | 5-550 MHz |
| D. | 10-3000MHz |
| Answer» B. 10-300 MHz | |
| 79. |
Infrared signals can be used for |
| A. | long-range communication |
| B. | short-range communication |
| C. | both |
| D. | none |
| Answer» C. both | |
| 80. |
Digital information in which the carrier signal is modified by one or more of its characteristics is called |
| A. | Demodulation |
| B. | Modulation |
| C. | Decoding |
| D. | Encoding |
| Answer» C. Decoding | |
| 81. |
Switched/56 service is the digital version of an |
| A. | analog leased service |
| B. | Digital data service |
| C. | analog switched service |
| D. | 900 service |
| Answer» D. 900 service | |
| 82. |
The inverse of the sampling interval is called the |
| A. | sampling rate |
| B. | delta rate |
| C. | baud rate. |
| D. | bit rate |
| Answer» B. delta rate | |
| 83. |
The services of WorldCom, Sprint and Verizon are provided by |
| A. | inter-LATA |
| B. | intra-LATA |
| C. | LATA |
| D. | local exchange carrier |
| Answer» B. intra-LATA | |
| 84. |
The V.34bis modem provides a bit rate of 28,800 with a |
| A. | 960-point constellation |
| B. | 1250-point constellation |
| C. | 1460-point constellation |
| D. | 1664-point constellation |
| Answer» B. 1250-point constellation | |
| 85. |
The value of r in analog transmission is |
| A. | log2L |
| B. | 2log L |
| C. | log L/2 |
| D. | Log L |
| Answer» B. 2log L | |
| 86. |
In BPSK, we use a polar Non Return to Zero (NRZ) signal for |
| A. | Decoding |
| B. | Encoding |
| C. | Modulation |
| D. | Demodulation |
| Answer» D. Demodulation | |
| 87. |
Which of the following statement is correct ? |
| A. | Satellite transponders contain a device that echos the radiation without change from one point on earth to another |
| B. | satellite transponders us lower frequency reception of radiation from earth stations and higher frequency for transmission to earth station |
| C. | Satellite transponders use a higher frequency for reception of radiation from earth stations and lower frequency for transmission to earth stations |
| D. | Satellite transponder contain devices that transform the message sent from one location on earth to a different code for transmission to another location |
| Answer» D. Satellite transponder contain devices that transform the message sent from one location on earth to a different code for transmission to another location | |
| 88. |
The Very high-bit-rate Digital Subscriber Line's (VDSAL) downstream rate is |
| A. | 768 kbps |
| B. | 1.5 Mbps |
| C. | 1.5-6.1 Mbps |
| D. | 25-55 Mbps |
| Answer» E. | |
| 89. |
The Asymmetric Digital Subscriber Line (ADSL), Very high-bit-rate Digital Subscriber Line (VDSL), High-bit-rate Digital Subscriber Line (HDSL) and Symmetric Digital Subscriber Line (SDSL) is referred to as |
| A. | All DSL |
| B. | yDSL |
| C. | nDSL |
| D. | xDSL |
| Answer» E. | |
| 90. |
A signal cannot be sampled with an |
| A. | infinite frequency |
| B. | finite frequency |
| C. | infinite bandwidth |
| D. | finite bandwidth |
| Answer» D. finite bandwidth | |
| 91. |
Optical fibers use reflection to guide light through a |
| A. | channel |
| B. | metal wire |
| C. | light |
| D. | plastic |
| Answer» B. metal wire | |
| 92. |
For Pulse Code Modulation (PCM), the quantization error is much greater than |
| A. | decoding |
| B. | Encoding |
| C. | delta modulation |
| D. | multiplexing |
| Answer» D. multiplexing | |
| 93. |
The designers of ADSL have divided the available bandwidth in to |
| A. | local loop evenly |
| B. | local loop unevenly |
| C. | unidirectional |
| D. | both a and c |
| Answer» D. both a and c | |
| 94. |
The term that refers to a strong material used in the fabrication of bulletproof vests is called |
| A. | Kevlar |
| B. | outer jacket |
| C. | cladding |
| D. | None of the above |
| Answer» B. outer jacket | |
| 95. |
The effectiveness of a data communications system depends on four fundamental characteristics |
| A. | delivery, accuracy |
| B. | timeliness and jitter |
| C. | jitter and delivery |
| D. | both a and b |
| Answer» E. | |
| 96. |
When the system delivers data accurately then it is called |
| A. | Accuracy |
| B. | Delivery |
| C. | Jitter |
| D. | Timelessness |
| Answer» B. Delivery | |
| 97. |
Two computers connected by an Ethernet hub are of |
| A. | LAN topology |
| B. | MAN topology |
| C. | WAN topology |
| D. | None of the above |
| Answer» B. MAN topology | |
| 98. |
Unguided communication media is |
| A. | coaxial cable |
| B. | fiber optic cable |
| C. | twisted pair cable |
| D. | satellite |
| Answer» E. | |
| 99. |
A term that refers to a code that is generated to detect some or all of errors occurred during transmission is |
| A. | Baseline Wandering |
| B. | Built in error detection |
| C. | Self-detect error code |
| D. | Bug resolver |
| Answer» C. Self-detect error code | |
| 100. |
The five components that make up a data communications system are the message, sender, receiver, medium and |
| A. | protocol |
| B. | Code |
| C. | connecting device |
| D. | both a and b |
| Answer» B. Code | |