McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
| 51. |
Which of the following is an example of physical layer vulnerability? |
| A. | mac address spoofing |
| B. | physical theft of data |
| C. | route spoofing |
| D. | weak or non-existent authentication |
| Answer» B. physical theft of data | |
| Explanation: physical theft of data is an example of physical layer vulnerability. other such issues are unauthorized network access, damage or destruction of data & hardware and keystroke & other input logging. | |
| 52. |
Which of the following is an example of data-link layer vulnerability? |
| A. | mac address spoofing |
| B. | physical theft of data |
| C. | route spoofing |
| D. | weak or non-existent authentication |
| Answer» A. mac address spoofing | |
| Explanation: mac address spoofing is an example of data-link layer vulnerability. | |
| 53. |
Which of the following is an example of network layer vulnerability? |
| A. | mac address spoofing |
| B. | physical theft of data |
| C. | route spoofing |
| D. | weak or non-existent authentication |
| Answer» C. route spoofing | |
| Explanation: route spoofing is an example of network layer vulnerability. other examples of network layer vulnerabilities are ip address spoofing and identity & resource id vulnerability. | |
| 54. |
Which of the following is an example of physical layer vulnerability? |
| A. | mac address spoofing |
| B. | route spoofing |
| C. | weak or non-existent authentication |
| D. | keystroke & other input logging |
| Answer» D. keystroke & other input logging | |
| Explanation: keystroke & other input logging is an example of physical layer vulnerability. other such physical layer vulnerabilities are unauthorized network access, damage or destruction of data & hardware and keystroke & other input logging. | |
| 55. |
Which of the following is an example of data-link layer vulnerability? |
| A. | physical theft of data |
| B. | vlan circumvention |
| C. | route spoofing |
| D. | weak or non-existent authentication |
| Answer» B. vlan circumvention | |
| Explanation: vlan circumvention is an example of data-link layer vulnerability. mac address spoofing, as well as switches, may be forced for flooding traffic to all vlan ports are some other examples of data-link layer vulnerability. | |
| 56. |
Which transmission media provides the highest transmission speed in a network? |
| A. | coaxial cable |
| B. | twisted pair cable |
| C. | optical fiber |
| D. | electrical cable |
| Answer» C. optical fiber | |
| Explanation: fiber optics is considered to have the highest transmission speed among the all mentioned above. the fiber optics transmission runs at 1000mb/s. it is called as 1000base-lx whereas ieee standard for it is 802.3z. it is popularly used for modern day network connections due to its high transmission rate. | |
| 57. |
Bits can be sent over guided and unguided media as analog signal by |
| A. | digital modulation |
| B. | amplitude modulation |
| C. | frequency modulation |
| D. | phase modulation |
| Answer» A. digital modulation | |
| Explanation: in analog modulation, digital low frequency baseband signal (digital bit stream) is transmitted over a higher frequency. whereas in digital modulation the only difference is that the base band signal is of discrete amplitude level. the bits are represented by only two frequency levels, one for high and one for low. | |
| 58. |
The portion of physical layer that interfaces with the media access control sublayer is called |
| A. | physical signalling sublayer |
| B. | physical data sublayer |
| C. | physical address sublayer |
| D. | physical transport sublayer |
| Answer» A. physical signalling sublayer | |
| Explanation: the portion of physical layer that interfaces with the medium access control sublayer is physical signaling sublayer. the main function of this layer is character encoding, reception, decoding and performs optional isolation functions. it | |
| 59. |
The physical layer provides |
| A. | mechanical specifications of electrical connectors and cables |
| B. | electrical specification of transmission line signal level |
| C. | specification for ir over optical fiber |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: anything dealing with a network cable or the standards in use – including pins, connectors and the electric current used is dealt in the physical layer (layer 1). physical layer deals with bit to bit delivery of the data aided by the various transmission mediums. | |
| 60. |
In asynchronous serial communication the physical layer provides |
| A. | start and stop signalling |
| B. | flow control |
| C. | both start & stop signalling and flow control |
| D. | only start signalling |
| Answer» C. both start & stop signalling and flow control | |
| Explanation: in asynchronous serial communication, the communication is not synchronized by clock signal. instead of a start and stop signaling and flow control method is followed. unlike asynchronous serial communication, in synchronous serial communication a clock signal is used for communication, so the start and stop method is not really required. | |
| 61. |
The physical layer is responsible for |
| A. | line coding |
| B. | channel coding |
| C. | modulation |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the physical layer is | |
| 62. |
The physical layer translates logical communication requests from the into hardware specific operations. |
| A. | data link layer |
| B. | network layer |
| C. | trasnport layer |
| D. | application layer |
| Answer» A. data link layer | |
| Explanation: physical layer accepts data or information from the data link layer and converts it into hardware specific operations so as to transfer the message through physical cables. some examples of the cables used are optical fiber cables, twisted pair cables and co-axial cables. | |
| 63. |
A single channel is shared by multiple signals by |
| A. | analog modulation |
| B. | digital modulation |
| C. | multiplexing |
| D. | phase modulation |
| Answer» C. multiplexing | |
| Explanation: in communication and computer networks, the main goal is to share a scarce resource. this is done by multiplexing, where multiple analog or digital signals are combined into one signal over a shared medium. the multiple kinds of signals are designated by the transport layer which is the layer present on a higher level than the physical layer. | |
| 64. |
Wireless transmission of signals can be done via |
| A. | radio waves |
| B. | microwaves |
| C. | infrared |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: wireless transmission is carried out by radio waves, microwaves and ir waves. these waves range from 3 khz to above 300 ghz and are more suitable for wireless transmission. radio waves can penetrate through walls and are used in radio communications, microwaves and infrared (ir) waves cannot penetrate through walls and are used for satellite communications and device communications respectively. | |
| 65. |
In the digital communication system, signals in different frequency bands are |
| A. | orthogonal |
| B. | non orthogonal |
| C. | orthogonal & non orthogonal |
| D. | none of the mentioned |
| Answer» A. orthogonal | |
| Explanation: in digital communication system, signals from different frequency bands are orthogonal thus interference won’t occur. | |
| 66. |
Properties of impulse function are |
| A. | symmetry |
| B. | time scaling |
| C. | shifting |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: some of the properties of impulse function are symmetry, time scaling and shifting. | |
| 67. |
Properties of Fourier transform are |
| A. | duality property |
| B. | time shifting property |
| C. | modulation property |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: some of the properties of fourier transform are duality property, time scaling property, time shifting property, modulation property and many more. | |
| 68. |
A base-band signal can be up-converted using |
| A. | sine wave |
| B. | cosine wave |
| C. | filtering |
| D. | none of the mentioned |
| Answer» B. cosine wave | |
| Explanation: a base-band signal can be up- converted to band-pass filter by applying cosine wave. | |
| 69. |
A band-pass signal can be down-converted using |
| A. | sine wave |
| B. | cosine wave |
| C. | time delayed wave |
| D. | none of the mentioned |
| Answer» B. cosine wave | |
| Explanation: for down-conversion of a band-pass signal also cosine signal is used and multiplied with it. | |
| 70. |
In down-conversion multiplication with cosine wave is followed by |
| A. | low pass filter |
| B. | high pass filter |
| C. | bandpass filter |
| D. | bandstop filter |
| Answer» A. low pass filter | |
| Explanation: downconversion of bandpass signal includes multiplication with sine wave followed by low pass filtering. | |
| 71. |
Fourier transform of a signal gives the |
| A. | frequency content |
| B. | bandwidth |
| C. | frequency content & bandwidth |
| D. | none of the mentioned |
| Answer» C. frequency content & bandwidth | |
| Explanation: fourier transform of a signal give the frequency content and also determines the bandwidth of the signal. | |
| 72. |
Random things in a communication system are |
| A. | timing offset |
| B. | device frequency |
| C. | attenuation |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: some of the random things in the communication system are noise in the channel, attenuation, fading, channel filtering, device frequency, phase and timing offset. | |
| 73. |
Which can be used for periodic and non periodic? |
| A. | fourier series |
| B. | fourier transforms |
| C. | fourier series & transforms |
| D. | none of the mentioned |
| Answer» B. fourier transforms | |
| Explanation: fourier series is limited to only periodic signals where as fourier transforms and laplace transforms can be used for both periodic and non periodic signals. | |
| 74. |
A band-pass signal has a Fourier transform equal to |
| A. | one |
| B. | zero |
| C. | infinity |
| D. | cannot be determined |
| Answer» B. zero | |
| Explanation: a band-pass signal has a fourier transform equal to zero for all value in both frequency and time domain. | |
| 75. |
A band-pass signal has |
| A. | dc component |
| B. | no dc component |
| C. | no side lobes |
| D. | cannot be determined |
| Answer» B. no dc component | |
| Explanation: a band-pass signal has no dc components and has fourier transform equal to zero. outside the band it will not be exactly zero. thus this results in presence of side lobes. | |
| 76. |
Which are orthonormal signal representation? |
| A. | sine and cosine at same frequency |
| B. | wavelets |
| C. | complex sinusoids at a different frequency |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: some of the common orthonormal signal representations are sine and cosine at the same frequency, fourier serier, sinc functions centered at sampling times, wavelets etc. | |
| 77. |
Constellation diagram is plotted in |
| A. | constellation space |
| B. | signal space |
| C. | orthogonal space |
| D. | boundary space |
| Answer» B. signal space | |
| Explanation: the constellation diagram is plotted in a space called as signal space. | |
| 78. |
Cumulative distributive function is |
| A. | non negative |
| B. | non decreasing |
| C. | non negative & decreasing |
| D. | none of the mentioned |
| Answer» C. non negative & decreasing | |
| Explanation: cumulative distribution function is non negative and non decreasing function. | |
| 79. |
Which are non negative functions? |
| A. | |
| B. | pmf |
| C. | pdf & pmf |
| D. | none of the mentioned |
| Answer» C. pdf & pmf | |
| Explanation: pdf, pmf and cdf are non negative functions. | |
| 80. |
Which of the following tasks is not done by data link layer? |
| A. | framing |
| B. | error control |
| C. | flow control |
| D. | channel coding |
| Answer» D. channel coding | |
| Explanation: channel coding is the function of physical layer. data link layer mainly deals with framing, error control and flow control. data link layer is the layer where the packets are encapsulated into frames. | |
| 81. |
Which sublayer of the data link layer performs data link functions that depend upon the type of medium? |
| A. | logical link control sublayer |
| B. | media access control sublayer |
| C. | network interface control sublayer |
| D. | error control sublayer |
| Answer» B. media access control sublayer | |
| Explanation: media access control (mac) deals with transmission of data packets to and from the network-interface card, and also to and from another remotely shared channel. | |
| 82. |
Header of a frame generally contains |
| A. | synchronization bytes |
| B. | addresses |
| C. | frame identifier |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: in a frame, the header is a part of the data that contains all the required information about the transmission of the file. it contains information like synchronization bytes, addresses, frame identifier etc. it also contains error control information for reducing the errors in the transmitted frames. | |
| 83. |
When 2 or more bits in a data unit has been changed during the transmission, the error is called |
| A. | random error |
| B. | burst error |
| C. | inverted error |
| D. | double error |
| Answer» B. burst error | |
| Explanation: when a single bit error occurs in a data, it is called single bit error. when more than a single bit of data is corrupted or has error, it is called burst error. if a single bit error occurs, the bit can be simply repaired by inverting it, but in case of a burst error, the sender has to send the frame again. | |
| 84. |
CRC stands for |
| A. | cyclic redundancy check |
| B. | code repeat check |
| C. | code redundancy check |
| D. | cyclic repeat check |
| Answer» A. cyclic redundancy check | |
| Explanation: cyclic redundancy check is a code that is added to a data which helps us to identify any error that occurred during the transmission of the data. crc is only able to detect errors, not correct them. crc is inserted in the frame trailer. | |
| 85. |
Which of the following is a data link protocol? |
| A. | ethernet |
| B. | point to point protocol |
| C. | hdlc |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: there are many data link layer protocols. some of them are sdlc (synchronous data link protocol), hdlc (high level data link control), slip (serial line interface protocol), ppp (point to point protocol) etc. these protocols are used to provide the logical link control function of the data link layer. | |
| 86. |
Which of the following is the multiple access protocol for channel access control? |
| A. | csma/cd |
| B. | csma/ca |
| C. | both csma/cd & csma/ca |
| D. | hdlc |
| Answer» C. both csma/cd & csma/ca | |
| Explanation: in csma/cd, it deals with detection of collision after collision has occurred, whereas csma/ca deals with preventing collision. csma/cd is abbreviation for carrier sensing multiple access/collision detection. csma/ca is abbreviation for carrier sensing multiple access/collision avoidance. these protocols are used for efficient multiple channel access. | |
| 87. |
The technique of temporarily delaying outgoing acknowledgements so that they can be hooked onto the next outgoing data frame is called |
| A. | piggybacking |
| B. | cyclic redundancy check |
| C. | fletcher’s checksum |
| D. | parity check |
| Answer» A. piggybacking | |
| Explanation: piggybacking is a technique in which the acknowledgment is temporarily delayed so as to be hooked with the next outgoing data frame. it saves a lot of channel bandwidth as in non-piggybacking system, some bandwidth is reserved for acknowledgement. | |
| 88. |
Automatic repeat request error management mechanism is provided by |
| A. | logical link control sublayer |
| B. | media access control sublayer |
| C. | network interface control sublayer |
| D. | application access control sublayer |
| Answer» A. logical link control sublayer | |
| Explanation: the logical link control is a sublayer of data link layer whose main function is to manage traffic, flow and error control. the automatic repeat request error management mechanism is provided by the llc when an error is found in the received frame at the receiver’s end to inform the sender to re-send the frame. | |
| 89. |
In layering, n layers provide service to |
| A. | n layer |
| B. | n-1 layer |
| C. | n+1 layer |
| D. | none of the mentioned |
| Answer» C. n+1 layer | |
| Explanation: in layering n layer provides service to n+1 layer and use the service provided by n-1 layer. | |
| 90. |
Which can be used as an intermediate device in between transmitter entity and receiver entity? |
| A. | ip router |
| B. | microwave router |
| C. | telephone switch |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: ip router, microwave router and telephone switch can be used as an intermediate device between communication of two entities. | |
| 91. |
Which has comparatively high frequency component? |
| A. | sine wave |
| B. | cosine wave |
| C. | square wave |
| D. | none of the mentioned |
| Answer» C. square wave | |
| Explanation: square wave has comparatively high frequency component in them. | |
| 92. |
Which has continuous transmission? |
| A. | asynchronous |
| B. | synchronous |
| C. | asynchronous & synchronous |
| D. | none of the mentioned |
| Answer» B. synchronous | |
| Explanation: synchronous has continuous transmission where as asynchronous have sporadic transmission. | |
| 93. |
Which requires bit transitions? |
| A. | asynchronous |
| B. | synchronous |
| C. | asynchronous & synchronous |
| D. | none of the mentioned |
| Answer» B. synchronous | |
| Explanation: synchronous transmission needs bit transition. | |
| 94. |
In synchronous transmission, receiver must stay synchronous for |
| A. | 4 bits |
| B. | 8 bits |
| C. | 9 bits |
| D. | 16 bits |
| Answer» C. 9 bits | |
| Explanation: in synchronous transmission, receiver must stay synchronous for 9 bits. | |
| 95. |
How error detection and correction is done? |
| A. | by passing it through equalizer |
| B. | by passing it through filter |
| C. | by amplifying it |
| D. | by adding redundancy bits |
| Answer» D. by adding redundancy bits | |
| Explanation: error can be detected and corrected by adding additional information that is by adding redundancy bits. | |
| 96. |
Which is more efficient? |
| A. | parity check |
| B. | cyclic redundancy check |
| C. | parity & cyclic redundancy check |
| D. | none of the mentioned |
| Answer» B. cyclic redundancy check | |
| Explanation: cyclic redundancy check is more efficient than parity check. | |
| 97. |
Which can detect two bit errors? |
| A. | parity check |
| B. | cyclic redundancy check |
| C. | parity & cyclic redundancy check |
| D. | none of the mentioned |
| Answer» B. cyclic redundancy check | |
| Explanation: crc is more powerful and it can detect various kind of errors like 2 bit errors. | |
| 98. |
CRC uses |
| A. | multiplication |
| B. | binary division |
| C. | multiplication & binary division |
| D. | none of the mentioned |
| Answer» C. multiplication & binary division | |
| Explanation: crc uses more math like multiplication and binary division. | |
| 99. |
What is start frame delimeter (SFD) in ethernet frame? |
| A. | 10101010 |
| B. | 10101011 |
| C. | 00000000 |
| D. | 11111111 |
| Answer» B. 10101011 | |
| Explanation: the start frame delimiter is a 1 byte field in the ethernet frame that indicates that the preceding bits are the start of the frame. it is always set to 10101011. | |
| 100. |
MAC address is of |
| A. | 24 bits |
| B. | 36 bits |
| C. | 42 bits |
| D. | 48 bits |
| Answer» D. 48 bits | |
| Explanation: mac address is like a local address for the nic that is used to make a local ethernet (or wifi) network function. it is of 48 bits. | |
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