Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations

NIMCET Previous Year Questions (PYQs)

NIMCET Computer Architecture PYQ


NIMCET PYQ
The Process when processor fetch or decode another instruction during the execution of current instruction is called





Go to Discussion

NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution

Pipelining is the process of accumulating instruction from the processor through a pipeline. It allows storing and executing instructions in an orderly process. It is also known as pipeline processing. Pipelining is a technique where multiple instructions are overlapped during execution.

NIMCET PYQ
Which of the following is used by ALU to store the intermediate results?





Go to Discussion

NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution

An accumulator is a type of register included in a CPU. It acts as a temporary storage location which holds an intermediate value in mathematical and logical calculations. Intermediate results of an operation are progressively written to the accumulator, overwriting the previous value. For example, in the operation "3 + 4 + 5," the accumulator would hold the value 3, then the value 7, then the value 12. The benefit of an accumulator is that it does not need to be explicitly referenced, which conserves data in the operation statement.

NIMCET PYQ
The Cache Memory is  more effective because of  





Go to Discussion

NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution

Locality of reference refers to a phenomenon in which a computer program tends to access same set of memory locations for a particular time period. In other words, Locality of Reference refers to the tendency of the computer program to access instructions whose addresses are near one another. 

NIMCET PYQ
To fetch data from secondary memory which one of the following register is used 





Go to Discussion

NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution

MAR register is used to access data and instructions from memory during the execution phase of instruction. MAR holds the memory location of data that needs to be accessed. When reading from memory, data addressed by MAR is fed into the MDR (memory data register) and then used by the CPU. When writing to memory, the CPU writes data from MDR to the memory location whose address is stored in MAR. MAR, which is found inside the CPU, goes either to the RAM (random-access memory) or cache.

NIMCET PYQ
Consider a computer system with speed of 106  instructions per second. A program P, having 2n2  steps is run on this system, where n is the input size. If n = 10000, what is the execution time for P?





Go to Discussion

NIMCET Previous Year PYQNIMCET NIMCET 2021 PYQ

Solution

$\text{Speed of computer} = 10^6 \text{per second} $
$\text{For n} =10000=10^4$ 
$$Time = \frac{\text{No of tasks}}{\text{Speed  of  computer}}$$ 
$$Time = \frac{2n^2}{10^6}$$ 
$$= \frac{2\times (10^4)^2}{10^6}$$
$$= \frac{2\times 10^8}{10^6}$$
$$=2 \times 10^2$$ 
$$= 200sec$$

NIMCET PYQ
Which of the following is true about Von Neumann architecture?





Go to Discussion

NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

Solution


NIMCET PYQ
Which of the following registers is used to keep track of address of the memory location where the next instruction is located?





Go to Discussion

NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

Solution


NIMCET PYQ
If a processor clock is rated as 2500 million cycles per second, then its clock period is:





Go to Discussion

NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

Solution

we know that Frequency is defined as the number of cycles in one second

Number of cycle in 1 sec = 2500 million

=> Frequency = 2500 Mhz

we know that time period is the inverse of frequency and is defined as the time taken by one cycle.

$T = \frac{1}{F}$

$T = \frac{1}{2500 \times 10^{-6}}$

$T=4 \times 10^{-10} $ sec



NIMCET PYQ
FFFF will be the last memory location in a memory of size





Go to Discussion

NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

Solution

The Hexadecimal digits are 0-9 and A-F. The Hexadecimal system represents numbers in16 symbols, zero to nine and ten to fifteen is represented by the English alphabet A-F. 

The Hexadecimal character represents 4 bits. 
The last memory location in a memory of size 64K is FFFF. 

64K is $2^{16}$ bytes, i.e. 
$16^4$ bytes = 1000 bytes in hexadecimal code. 

The last accessible address is 1000-1 = FFFF.


NIMCET


Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

NIMCET


Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

loading...