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101
UGC NET CSE | June 2008 | Part 2 | Question: 26
The $\text{ATM}$ cells are __________ bytes long. $48$ $53$ $64$ $69$
The $\text{ATM}$ cells are __________ bytes long.$48$$53$$64$$69$
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102
UGC NET CSE | June 2008 | Part 2 | Question: 27
For slotted ALOHA, the maximum channel utilization is : $100 \%$ $50 \%$ $36 \%$ $18 \%$
For slotted ALOHA, the maximum channel utilization is :$100 \%$$50 \%$$36 \%$$18 \%$
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2 answers
103
UGC NET CSE | June 2008 | Part 2 | Question: 28
For a channel of $3 \mathrm{KHz}$ bandwidth and signal to noise ratio of $30 \mathrm{~dB}$, the maximum data rate is: $3000 \mathrm{bps}$ $6000 \mathrm{bps}$ $15000 \mathrm{bps}$ $30000 \mathrm{bps}$
For a channel of $3 \mathrm{KHz}$ bandwidth and signal to noise ratio of $30 \mathrm{~dB}$, the maximum data rate is:$3000 \mathrm{bps}$$6000 \mathrm{bps}$$15000 \mathrm{...
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1 answer
104
UGC NET CSE | June 2008 | Part 2 | Question: 29
An example of a public key encryption algorithm is : Caesar cipher algorithm DES algorithm AES algorithm Knapsack algorithm
An example of a public key encryption algorithm is :Caesar cipher algorithmDES algorithmAES algorithmKnapsack algorithm
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105
UGC NET CSE | June 2008 | Part 2 | Question: 30
With reference to hierarchical routing, the optimum number of levels for an $m$ router subnet is: $m^{2}$ $m$ $\ln m$ $\sqrt{m}$
With reference to hierarchical routing, the optimum number of levels for an $m$ router subnet is:$m^{2}$$m$$\ln m$$\sqrt{m}$
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106
UGC NET CSE | June 2008 | Part 2 | Question: 31
Assembler program is : dependent on the operating system dependent on the compiler dependent on the hardware independent of the hardware
Assembler program is :dependent on the operating systemdependent on the compilerdependent on the hardwareindependent of the hardware
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108
UGC NET CSE | June 2008 | Part 2 | Question: 33
At the end of parsing, tokens are identified. set of instructions are identified. the syntactic groups are identified. machine instructions are identified.
At the end of parsing,tokens are identified.set of instructions are identified.the syntactic groups are identified.machine instructions are identified.
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109
UGC NET CSE | June 2008 | Part 2 | Question: 34
Dead-code elimination in machine code optimization refers to : removal of all labels. removal of values that never get used. removal of function which are not involved. removal of a module after its use.
Dead-code elimination in machine code optimization refers to :removal of all labels.removal of values that never get used.removal of function which are not involved.remov...
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110
UGC NET CSE | June 2008 | Part 2 | Question: 35
A parse tree is an annotated parse tree if : it shows attribute values at each node. there are no inherited attributes. it has synthesized nodes as terminal nodes. every non-terminal nodes is an inherited attribute.
A parse tree is an annotated parse tree if :it shows attribute values at each node.there are no inherited attributes.it has synthesized nodes as terminal nodes.every non-...
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111
UGC NET CSE | June 2008 | Part 2 | Question: 36
An example of a non-preemptive $\text{CPU}$ scheduling algorithm is : Shortest job first scheduling. Round robin scheduling. Priority scheduling. Fair share scheduling.
An example of a non-preemptive $\text{CPU}$ scheduling algorithm is :Shortest job first scheduling.Round robin scheduling.Priority scheduling.Fair share scheduling.
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112
UGC NET CSE | June 2008 | Part 2 | Question: 37
There are ' $n$ ' processes in memory. A process spends a fraction ' $p$ ' of its time waiting for $\mathrm{I} / \mathrm{O}$ to complete. The CPU utilization is given by : $p^{n}$ $1-p^{n}$ $(1-p)^{n}$ $1-n p$
There are ' $n$ ' processes in memory. A process spends a fraction ' $p$ ' of its time waiting for $\mathrm{I} / \mathrm{O}$ to complete. The CPU utilization is given by ...
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113
UGC NET CSE | June 2008 | Part 2 | Question: 38
An example of a memory management system call in $\text{UNIX}$ is : fork. mmap. sigaction. execve.
An example of a memory management system call in $\text{UNIX}$ is :fork.mmap.sigaction.execve.
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114
UGC NET CSE | June 2008 | Part 2 | Question: 39
With $64$ bit virtual addresses, a $4 \mathrm{~KB}$ page and $256 \mathrm{MB}$ of RAM, an inverted page table requires : $8192$ entries. $16384$ entries. $32768$ entries. $65536$ entries.
With $64$ bit virtual addresses, a $4 \mathrm{~KB}$ page and $256 \mathrm{MB}$ of RAM, an inverted page table requires :$8192$ entries.$16384$ entries.$32768$ entries.$65...
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115
UGC NET CSE | June 2008 | Part 2 | Question: 40
A computer has $6$ tape drives with ' $n$ ' processes competing for them. Each process may need two drives. For which values of ' $n$ ' is the system deadlock free? $1$ $2$ $3$ $6$
A computer has $6$ tape drives with ' $n$ ' processes competing for them. Each process may need two drives. For which values of ' $n$ ' is the system deadlock free?$1$$2$...
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116
UGC NET CSE | June 2008 | Part 2 | Question: 41
Water fall model for software development is : a top down approach. a bottom up approach. a sequential approach. a consequential approach.
Water fall model for software development is :a top down approach.a bottom up approach.a sequential approach.a consequential approach.
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118
UGC NET CSE | June 2008 | Part 2 | Question: 43
While designing the user interface, one should : use as many short cuts as possible. use as many defaults as possible. use as many visual layouts as possible. reduce the demand on short-term memory.
While designing the user interface, one should :use as many short cuts as possible.use as many defaults as possible.use as many visual layouts as possible.reduce the dema...
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119
UGC NET CSE | June 2008 | Part 2 | Question: 44
In software cost estimation, base estimation is related to : cost of similar projects already completed. cost of the base model of the present project. cost of the project with the base minimum profit. cost of the project under ideal situations.
In software cost estimation, base estimation is related to :cost of similar projects already completed.cost of the base model of the present project.cost of the project w...
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120
UGC NET CSE | June 2008 | Part 2 | Question: 45
In clean room software engineering : only eco-friendly hardware is used. only hired facilities are used for development. correctness of the code is verified before testing. implementation is done only after ensuring correctness.
In clean room software engineering :only eco-friendly hardware is used.only hired facilities are used for development.correctness of the code is verified before testing.i...
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