Computer Network Questions and Answers

Q: In a D flip-flop, the output Q after a clock edge is equal to

Correct answer: the value of D just before the clock edgeA D flip-flop simply captures whatever is on its D input at the active clock edge and holds it until the next one. Its characteristic equation is \( Q_{next} = D \). The D stands for data or delay, and both readings are apt: the device stores one bit of data, and it delays a signal by exactly one clock period. Its virtue is that it has no forbidden or ambiguous input combination, unlike the SR flip-flop, and needs no input decoding, unlike the JK. That simplicity is why it is the workhorse of synchronous design and why registers, shift registers and pipeline stages are built from it. For the capture to be reliable the D input must be stable for the setup time before the edge and the hold time after it. Violating either can leave the output metastable - hovering between levels for an unpredictable period - which is why an asynchronous input is passed through two D flip-flops in series before being used.

Q: Which of these is a volatile memory?

Correct answer: RAMRAM is volatile: it loses its contents the moment power is removed. It is used as working memory precisely because it is fast and freely writable, and its impermanence is the price of that. The other three are non-volatile and retain data without power. ROM is written once at manufacture; EPROM can be erased by ultraviolet light and reprogrammed; flash memory is electrically erasable and is what solid-state drives and memory cards are built from. RAM itself divides in two. SRAM stores each bit in a flip-flop, needs no refreshing, and is fast but expensive, so it is used for cache. DRAM stores each bit as a charge on a tiny capacitor, which leaks and must be refreshed thousands of times a second, but it is far denser and cheaper, so it is used for main memory. The name random access distinguishes both from sequential media such as tape, where access time depends on position.

Q: How many select lines are required for an 8-to-1 multiplexer ?

Correct answer: 3A multiplexer with \(2^n\) data inputs needs \(n\) select lines to address them. Since \(2^3 = 8\), three select lines are required to choose one of the eight inputs.

Q: In Boolean algebra, the expression \( A + \bar{A} = 1 \) holds for any variable A.

Correct answer: TrueThe statement is true. This is the complement law of Boolean algebra: a variable ORed with its own complement always gives 1. The reasoning is exhaustive. If \( A = 0 \) then \( \bar{A} = 1 \), and \( 0 + 1 = 1 \). If \( A = 1 \) then \( \bar{A} = 0 \), and \( 1 + 0 = 1 \). No third case exists in a two-valued algebra, so the result is 1 whatever A may be. Its dual is equally useful: \( A \cdot \bar{A} = 0 \), since a variable and its complement can never both be 1. The complement laws are among the basic identities used to simplify expressions, alongside the identity laws \( A + 0 = A \) and \( A \cdot 1 = A \), the null laws \( A + 1 = 1 \) and \( A \cdot 0 = 0 \), and the idempotent laws \( A + A = A \) and \( A \cdot A = A \). Note that in Boolean algebra + means OR, not arithmetic addition, which is why \( 1 + 1 = 1 \) rather than 2.

Q: A JK flip-flop toggles its output when both J and K inputs are HIGH.

Correct answer: TrueThe statement is true, and it is the feature that makes the JK flip-flop the most versatile of the four types. Its behaviour on a clock edge is: J = 0, K = 0 holds the previous state; J = 0, K = 1 resets to 0; J = 1, K = 0 sets to 1; and J = 1, K = 1 toggles, so the output becomes the complement of what it was. The toggle case is precisely what the SR flip-flop cannot do. In an SR flip-flop, setting both inputs high is a forbidden condition, because it drives both outputs to the same level and leaves the next state indeterminate when the inputs are released. The JK resolves that by feeding the outputs back into the input gating, turning the invalid case into a useful one. Tying J and K together to a permanent 1 gives a T flip-flop, which toggles on every clock pulse - and that is the building block of binary counters, since each stage divides the frequency by two.

Q: Minimum number of bits required to represent B is

Correct answer: 3Three bits distinguish eight values, which is the smallest power of two large enough for the symbol set in question: \( 2^{2}=4 \) is too few and \( 2^{3}=8 \) suffices.

Q: A latch is level-triggered while a flip-flop is edge-triggered.

Correct answer: TrueA latch responds to its inputs for as long as the enable line is active, whereas a flip-flop samples only at the rising or falling edge of the clock. Edge triggering is what keeps a synchronous circuit’s stages from racing ahead of each other.

Q: What can happen at a Token Ring station

Correct answer: All of theseEvery station on a Token Ring acts as a repeater: it examines the destination address to see whether the frame is for it, regenerates the signal, and passes the frame on round the ring.

Q: Which number system is used internally by digital computers?

Correct answer: BinaryDigital computers work internally in binary, base 2, using only the digits 0 and 1. The reason is physical rather than mathematical. A circuit can distinguish two voltage levels reliably and cheaply - on or off, high or low - and that reliability degrades quickly if more levels must be told apart. Two states also map directly onto the switching behaviour of a transistor and onto the true and false of Boolean algebra, which is what makes logic gates possible. Octal and hexadecimal are used by people rather than by machines, as compact ways of writing binary: each octal digit stands for exactly three bits and each hexadecimal digit for exactly four, so conversion is done by grouping rather than by division. A byte such as 11010110 becomes D6 in hexadecimal at a glance. Decimal is what humans use, and converting between it and binary requires actual arithmetic, which is why it is not used internally.

Q: The number of distinct states in a 3-bit binary counter is

Correct answer: 8A counter with \( n \) bits passes through \( 2^{n} \) distinct states, so with three bits \[ 2^{3} = 8, \] running from 000 to 111 - decimal 0 to 7 - before returning to 000. That return is why such a device is called a modulus-8 counter, and why its highest output is the clock frequency divided by eight. Counters are therefore used as frequency dividers as often as they are used to count. The same relation works in reverse: a counter of modulus \( N \) needs the smallest \( n \) for which \( 2^{n} \ge N \). A decade counter of modulus 10 needs four flip-flops, and the six surplus states are removed by feedback that clears the counter when it reaches ten. It is the identical arithmetic that fixes the number of select lines on a multiplexer, the outputs of a decoder and the address lines of a memory - one relation worth being fluent in.

51.

In a D flip-flop, the output Q after a clock edge is equal to

A.the value of D just before the clock edge
B.the complement of D
C.the previous value of Q
D.logic 1 always

Answer: B

Correct answer: the value of D just before the clock edge

A D flip-flop simply captures whatever is on its D input at the active clock edge and holds it until the next one. Its characteristic equation is \( Q_{next} = D \). The D stands for data or delay, and both readings are apt: the device stores one bit of data, and it delays a signal by exactly one clock period. Its virtue is that it has no forbidden or ambiguous input combination, unlike the SR flip-flop, and needs no input decoding, unlike the JK. That simplicity is why it is the workhorse of synchronous design and why registers, shift registers and pipeline stages are built from it. For the capture to be reliable the D input must be stable for the setup time before the edge and the hold time after it. Violating either can leave the output metastable - hovering between levels for an unpredictable period - which is why an asynchronous input is passed through two D flip-flops in series before being used.

Discuss in Forum

52.

Which of these is a volatile memory?

A.RAM
B.ROM
C.EPROM
D.Flash memory

Answer: B

Correct answer: RAM

RAM is volatile: it loses its contents the moment power is removed. It is used as working memory precisely because it is fast and freely writable, and its impermanence is the price of that. The other three are non-volatile and retain data without power. ROM is written once at manufacture; EPROM can be erased by ultraviolet light and reprogrammed; flash memory is electrically erasable and is what solid-state drives and memory cards are built from. RAM itself divides in two. SRAM stores each bit in a flip-flop, needs no refreshing, and is fast but expensive, so it is used for cache. DRAM stores each bit as a charge on a tiny capacitor, which leaks and must be refreshed thousands of times a second, but it is far denser and cheaper, so it is used for main memory. The name random access distinguishes both from sequential media such as tape, where access time depends on position.

Discuss in Forum

53.

How many select lines are required for an 8-to-1 multiplexer ?

Answer: A

Correct answer: 3

A multiplexer with \(2^n\) data inputs needs \(n\) select lines to address them. Since \(2^3 = 8\), three select lines are required to choose one of the eight inputs.

Discuss in Forum

54.

In Boolean algebra, the expression \( A + \bar{A} = 1 \) holds for any variable A.

A.True
B.False

Answer: A

Correct answer: True

The statement is true. This is the complement law of Boolean algebra: a variable ORed with its own complement always gives 1. The reasoning is exhaustive. If \( A = 0 \) then \( \bar{A} = 1 \), and \( 0 + 1 = 1 \). If \( A = 1 \) then \( \bar{A} = 0 \), and \( 1 + 0 = 1 \). No third case exists in a two-valued algebra, so the result is 1 whatever A may be. Its dual is equally useful: \( A \cdot \bar{A} = 0 \), since a variable and its complement can never both be 1. The complement laws are among the basic identities used to simplify expressions, alongside the identity laws \( A + 0 = A \) and \( A \cdot 1 = A \), the null laws \( A + 1 = 1 \) and \( A \cdot 0 = 0 \), and the idempotent laws \( A + A = A \) and \( A \cdot A = A \). Note that in Boolean algebra + means OR, not arithmetic addition, which is why \( 1 + 1 = 1 \) rather than 2.

Discuss in Forum

55.

A JK flip-flop toggles its output when both J and K inputs are HIGH.

A.True
B.False

Answer: B

Correct answer: True

The statement is true, and it is the feature that makes the JK flip-flop the most versatile of the four types. Its behaviour on a clock edge is: J = 0, K = 0 holds the previous state; J = 0, K = 1 resets to 0; J = 1, K = 0 sets to 1; and J = 1, K = 1 toggles, so the output becomes the complement of what it was. The toggle case is precisely what the SR flip-flop cannot do. In an SR flip-flop, setting both inputs high is a forbidden condition, because it drives both outputs to the same level and leaves the next state indeterminate when the inputs are released. The JK resolves that by feeding the outputs back into the input gating, turning the invalid case into a useful one. Tying J and K together to a permanent 1 gives a T flip-flop, which toggles on every clock pulse - and that is the building block of binary counters, since each stage divides the frequency by two.

Discuss in Forum

56.

Minimum number of bits required to represent B is

Answer: C

Correct answer: 3

Three bits distinguish eight values, which is the smallest power of two large enough for the symbol set in question: \( 2^{2}=4 \) is too few and \( 2^{3}=8 \) suffices.

Discuss in Forum

57.

A latch is level-triggered while a flip-flop is edge-triggered.

A.True
B.False

Answer: A

Correct answer: True

A latch responds to its inputs for as long as the enable line is active, whereas a flip-flop samples only at the rising or falling edge of the clock. Edge triggering is what keeps a synchronous circuit’s stages from racing ahead of each other.

Discuss in Forum

58.

What can happen at a Token Ring station

A.Examination of the destination address
B.Regeneration of the frame
C.Passing of the frame to the next station
D.All of these

Answer: D

Correct answer: All of these

Every station on a Token Ring acts as a repeater: it examines the destination address to see whether the frame is for it, regenerates the signal, and passes the frame on round the ring.

Discuss in Forum

59.

Which number system is used internally by digital computers?

A.Binary
B.Decimal
C.Octal
D.Hexadecimal

Answer: D

Correct answer: Binary

Digital computers work internally in binary, base 2, using only the digits 0 and 1. The reason is physical rather than mathematical. A circuit can distinguish two voltage levels reliably and cheaply - on or off, high or low - and that reliability degrades quickly if more levels must be told apart. Two states also map directly onto the switching behaviour of a transistor and onto the true and false of Boolean algebra, which is what makes logic gates possible. Octal and hexadecimal are used by people rather than by machines, as compact ways of writing binary: each octal digit stands for exactly three bits and each hexadecimal digit for exactly four, so conversion is done by grouping rather than by division. A byte such as 11010110 becomes D6 in hexadecimal at a glance. Decimal is what humans use, and converting between it and binary requires actual arithmetic, which is why it is not used internally.

Discuss in Forum

60.

The number of distinct states in a 3-bit binary counter is

Answer: C

Correct answer: 8

A counter with \( n \) bits passes through \( 2^{n} \) distinct states, so with three bits \[ 2^{3} = 8, \] running from 000 to 111 - decimal 0 to 7 - before returning to 000. That return is why such a device is called a modulus-8 counter, and why its highest output is the clock frequency divided by eight. Counters are therefore used as frequency dividers as often as they are used to count. The same relation works in reverse: a counter of modulus \( N \) needs the smallest \( n \) for which \( 2^{n} \ge N \). A decade counter of modulus 10 needs four flip-flops, and the six surplus states are removed by feedback that clears the counter when it reaches ten. It is the identical arithmetic that fixes the number of select lines on a multiplexer, the outputs of a decoder and the address lines of a memory - one relation worth being fluent in.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Computer Network Questions and Answers || Computer Network MCQ Questions and Answers || Computer Network GK Questions and Answers || Computer Network GK MCQ Questions || Computer Network Multiple Choice Questions and Answers || Computer Network GK || GK on Computer Network || Computer Awareness Questions and Answers || Computer Awareness MCQ Questions and Answers || Computer Awareness GK Questions and Answers || GK on Computer Awareness

Computer Network Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications