MCA NIMCET Organization of a Computer Previous Year Question Paper and Solution with PDF

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Organization Of A Computer PYQ

Qus : 1MCA NIMCET PYQ

Debugger is a program that

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

Solution

A debugger is a tool used to test and debug programs. It allows programmers to pause execution at specific points (breakpoints), run code step-by-step, and inspect variables and registers to find logic or runtime errors.

Hence, the correct option is:

✅ Option 3: allows to set breakpoints, execute a segment of program, and display contents of registers.

Qus : 2MCA NIMCET PYQ

Dynamic RAM (DRAM) stores each bit of data in a separate capacitor. Due to leakage, the stored charge tends to dissipate over time and needs to be refreshed periodically. Consider the following statements:
P: DRAM requires refreshing because it uses capacitors to store bits.

Q: SRAM does not require refreshing because it uses flip-flops instead of capacitors.
R: DRAM is faster than SRAM because it needs less frequent access.
S: DRAM is more suitable for main memory than SRAM due to its density.

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

Solution

P: ✔ True — DRAM cells use capacitors that leak charge and hence must be refreshed periodically.
Q: ✔ True — SRAM uses flip-flops, so it retains data without refreshing.
R: ✗ False — DRAM is actually slower than SRAM.
S: ✔ True — DRAM is denser and cheaper, making it ideal for main memory.

✅ Correct statements: P, Q, and S

Qus : 3MCA NIMCET PYQ

In the design of a control unit of a processor, two common approaches are used: hardware control and microprogrammed control. Consider the following statements:

Hardware control units are generally faster but more difficult to modify than microprogrammed control units.
In a horizontal microprogrammed control unit, each control signal has a separate bit in the control word.
Vertical microprogramming leads to longer control words but provides greater parallelism.
Microprogrammed control units are typically easier to implement and modify than hardware control units.

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

Solution

Statement 1 – True: Hardware control units are implemented using combinational logic circuits. They generate control signals directly through hardware, making them faster but difficult to modify since any change requires redesigning the hardware.
Statement 2 – True: In a horizontal microprogrammed control unit, each control signal is represented by a dedicated bit in the control word. This allows maximum parallelism but results in a wide (long) control word.
Statement 3 – False: In vertical microprogramming, control words use encoded fields to represent control signals. This makes the word shorter, not longer, but reduces parallelism since only a few signals can be activated at once.
Statement 4 – True: Microprogrammed control units are easier to implement and modify because control logic is stored in a control memory (microprograms). Modifications can be done by changing microinstructions instead of redesigning hardware.

✅ Correct statements: 1, 2, and 4

Qus : 4MCA NIMCET PYQ

Consider a system with a CPU having 6 registers and 32-bit instructions. The maximum possible size of the main memory is 512 KB (1K = 2¹⁰). Each instruction takes two registers and one memory address as operands. Which one of the following correctly gives the maximum possible distinct instructions that can be there in the instruction set of the CPU?

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

Solution

Memory size $=512\text{ KB}=2^{19}$ bytes $\Rightarrow$ address field needs $19$ bits.

Registers $=6 \Rightarrow$ bits per register $=\lceil \log_2 6\rceil=3$ bits. For two registers $\Rightarrow 2\times3=6$ bits.

Total operand bits $=19+6=25$.

Opcode bits $=32-25=7 \Rightarrow$ maximum distinct instructions $=2^{7}= {128}$.

Qus : 5MCA NIMCET PYQ

Match all items in Group 1 with correct options from those given in Group 2.

GROUP 1	GROUP 2
P. Intermediate representation	1. Activation records
Q. Top-down parsing	2. Code generation
R. Runtime environments	3. Leftmost derivation
S. Register allocation	4. Graph colouring

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

Solution

Group 1	Group 2 (Match)	Explanation
P. Intermediate representation	→ 2. Code generation	Intermediate representation (IR) is the form of code produced by the front end of a compiler and used as input to the code generation phase.
Q. Top-down parsing	→ 3. Leftmost derivation	Top-down parsers construct a parse tree from the root using leftmost derivations.
R. Runtime environments	→ 1. Activation records	Activation records (stack frames) are part of the runtime environment used to manage function calls and local variables.
S. Register allocation	→ 4. Graph colouring	Register allocation is commonly implemented using graph-colouring algorithms to assign variables to CPU registers efficiently.

✅ Final Matching:
P – 2, Q – 3, R – 1, S – 4

Qus : 6MCA NIMCET PYQ

Consider a system running under two types of workloads:

(a) CPU-intensive jobs,

(b) I/O-intensive jobs. Which of the following statements about the relative performance of Interrupt-driven I/O and Programmed I/O is correct?

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2025 PYQ

Solution

(4) — Interrupt-driven I/O frees the CPU for compute-heavy jobs; under very I/O-intensive loads, polling (programmed I/O) can outperform by avoiding high interrupt overhead.

Qus : 7MCA NIMCET PYQ

The time required for fetching and execution of one simple machine instruction is known as

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 8MCA NIMCET PYQ

Consider the program below which uses six temporary variables a, b, c, d, e, and f.

a = 1 
b = 10
c = 20 
d = a + b
e = c + d 
f = c + e 
b = c + e 
e = b + f 
d = 5 + e
return d + f

Assuming that all the above operations take their operands from registers, the minimum number of registers needed to execute this program without spilling is

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2024 PYQ

Solution

Detailed Register Allocation Analysis

Objective: Determine the minimum number of registers needed to execute the program without spilling.

Live Range Analysis:

a: Line 1 → 4
b: Line 6 → 7
c: Line 3 → 6
d: Line 8 → 9
e: Line 7 → 8
f: Line 5 → 9

Max live variables: 3 (after lines 6 and 7)

✅ Final Answer: 3 registers are required to execute the program without spilling.

Qus : 9MCA NIMCET PYQ

Which of the following components is not a part of an instruction formation in CPU processing?

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MCA NIMCET Previous Year PYQMCA NIMCET NIMCET 2024 PYQ

Solution

Instruction Formation in CPU Processing

Question: Which of the following is not a part of instruction formation?

Options:

Opcode
Register file
Source operand
Destination operand

✅ Correct Answer: Register File

Explanation:

Opcode, Source operand, Destination operand — all are part of the instruction format.
Register File — a hardware structure that stores registers, but it is not encoded into the instruction.

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Organization Of A Computer PYQ

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Detailed Register Allocation Analysis

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Instruction Formation in CPU Processing

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