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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Computer Architecture PYQ


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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2021 PYQ

Solution

✅ Concept:

In modern processors, to improve performance, the CPU often overlaps the fetch, decode, and execute stages of multiple instructions. This overlapping is known as:

Instruction Pipelining

✅ Example:

  • Instruction 1: Being executed
  • Instruction 2: Being decoded
  • Instruction 3: Being fetched

This overlapping allows multiple instructions to be processed simultaneously at different stages of the pipeline, improving throughput.

✅ Final Answer: $\boxed{\text{Instruction Pipelining}}$


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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2021 PYQ

Solution

✅ Concept:

The Accumulator is a special-purpose register used by the ALU to store intermediate results during arithmetic and logic operations.

✅ Explanation:

  • It simplifies CPU design by reducing the number of memory accesses.
  • Results of one operation are stored in the accumulator and used in the next.
  • Widely used in simple or older CPU architectures.

✅ Final Answer: $\boxed{\text{Accumulator}}$


MCA NIMCET PYQ
The Cache Memory is  more effective because of  





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2021 PYQ

Solution

✅ Concept:

Cache memory is highly effective due to its ability to take advantage of the locality of reference. This refers to the tendency of programs to access a relatively small portion of memory repeatedly during execution.

✅ Explanation of Options:

  • Memory Localization: Not the correct answer, as it's a general concept related to memory organization.
  • Locality of Reference: The correct answer! This refers to the tendency of a program to repeatedly access the same memory locations, which cache memory leverages to speed up data access.
  • Memory Size: Cache memory is small, and its size is actually one of the factors that makes it faster, but size alone does not determine its effectiveness.
  • None of the Mentioned: This option is incorrect because "locality of reference" is the main reason cache memory is effective.

✅ Final Answer:

The most effective reason for cache memory is the $\boxed{\text{Locality of Reference}}$.


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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2021 PYQ

Solution

✅ Concept:

To fetch data from secondary memory, certain registers are used to handle the addresses and manage the data flow between secondary memory and the CPU.

✅ Explanation of Options:

  • Program Counter (PC): The Program Counter holds the address of the next instruction to execute, not used for fetching data from secondary memory.
  • Memory Address Register (MAR): This is the correct answer! The MAR holds the address in memory from which data will be fetched or written, including from secondary storage. It is responsible for managing the memory access.
  • Memory Buffer Register (MBR): The MBR temporarily holds data being transferred to or from memory but does not directly fetch data from secondary memory.
  • Accumulator: The Accumulator holds intermediate results of arithmetic or logical operations, not used for fetching data from secondary memory.

✅ Final Answer:

The register used to fetch data from secondary memory is the $\boxed{\text{Memory Address Register (MAR)}}$.


MCA 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?





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Solution

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

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





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Solution


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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2023 PYQ

Solution

Register Responsible for Next Instruction

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

✅ Correct Answer: Program Counter (PC)

Explanation:

The Program Counter (PC) holds the address of the next instruction to be fetched from memory and executed by the CPU. After fetching the current instruction, it automatically updates to point to the next one.

Final Answer: Program Counter


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





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MCA NIMCET Previous Year PYQMCA NIMCET 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



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





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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2022 PYQ

Solution

✅ Finding the Memory Size:

The last memory location "FFFF" is a hexadecimal number.

Each hexadecimal digit represents 4 bits, so:

  • "FFFF" is a 4-digit hexadecimal number, meaning it represents 16 bits or 2 bytes.
  • "FFFF" in decimal is equivalent to 65535.
  • Therefore, the total memory size is 65536 bytes or 64 KB.

✅ Final Answer:

The memory size is: 64 KB



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