Free Mock Test : Operating System : Memory Management (Test 2)

Home

Technology

Blog

Free Mock Test - Operating System : Memory Management (Test 2)

Memory Management is one of the most important functions of an Operating System. It is responsible for managing the computer’s primary memory efficiently so that multiple processes can run smoothly without conflict. It handles tasks like allocation, deallocation, paging, segmentation, and virtual memory. Understanding memory management helps students learn how programs are loaded, executed, and optimized in a system.
Memory Management is a frequently asked topic in various competitive and academic exams such as BCA, MCA, B.Tech, and government exams. Questions related to paging, segmentation, and memory allocation strategies are commonly included. By taking this online test, students can prepare effectively and increase their chances of scoring higher marks. This online test is launched by Teachexam to provide a smart and effective learning platform for students.

Total Questions : 20

Timer : sec

Question : Suppose a 16-bit logical address uses 4 bits for segment number and 12 bits for offset.

1024 bytes

2048 bytes

4096 bytes

8192 bytes

none

Question : Which of the following is a major disadvantage of segmentation?

Internal fragmentation

External fragmentation

Fixed segment size

No logical structure for programs

none

Question : Which information is stored in each entry of the segment table?

Page number and frame size

Base address and segment limit

Block size and memory index

Process ID and segment count

none

Question : A logical address in segmentation consists of which two components?

Page number and frame number

Segment number and offset

Frame number and displacement

Block number and index

none

Question : Which of the following best describes segmentation in an operating system?

Memory is divided into equal fixed-size blocks

Memory is divided into variable-size logical units called segments

Memory is divided into frames and pages of equal size

Memory is divided only for secondary storage

none

Question : Consider a demand paging system with:
Page fault rate = 0.02
Memory access time = 100 ns
Page fault service time = 8 ms
What is the Effective Access Time (EAT)?

160,098 ns

160,000 ns

200,098 ns

160,100 ns

none

Question : Which of the following best explains why demand paging improves memory utilization?

All process pages are loaded at once

Only required pages are loaded into main memory

Pages are stored permanently in RAM

Secondary memory is not used

none

Question : Which condition best describes thrashing in a demand paging system?

When page replacement algorithms fail to execute

When CPU utilization is very high

When page fault rate becomes extremely high and CPU spends most time swapping pages

When the page table becomes full

none

Question : Which formula correctly represents the Effective Access Time (EAT) in a demand paging system?
p = page fault rate, s = page fault service time, m = memory access time

EAT = (1 − p) × s + p × m

EAT = p × s + (1 − p) × m

EAT = p × m + (1 − p) × s

EAT = s + m

none

Question : In a demand paging system, what happens when a process references a page that is not currently in main memory?

The process is immediately terminated

The OS generates a page fault and loads the page from secondary memory

The CPU ignores the instruction

The page is permanently removed from the page table

none

Question : If the physical address space = 64 KB and word size = 8 bytes, what is the number of bits required for the physical address?

10 bits

13 bits

16 bits

19 bits

none

Question : In modern operating systems, which statement about user access to physical addresses is correct?

Users can directly access physical addresses.

Physical addresses are visible only during compilation.

Users access physical memory indirectly through logical addresses.

Logical and physical addresses are always identical.

none

Question : If the physical address space = N words, the number of bits required to represent a physical address is:

2^{N}

Log2 N

N^{2}

N / 2

none

Question : Which statement correctly describes the logical address space?

It represents the size of the main memory.

It is the set of all physical addresses mapped to RAM.

It is the set of addresses generated by the CPU for a process.

It is directly accessed by hardware memory modules.

none

Question : Which component is primarily responsible for translating logical addresses into physical addresses in an operating system?

CPU Scheduler

Memory Management Unit (MMU)

Compiler

Disk Controller

none

Question : When a process tries to access a page that is not currently present in main memory, the event is called:

Page hit

Page interrupt

Page fault

Page overflow

none

Question : Which hardware component is primarily responsible for translating logical addresses into physical addresses in a paging system?

Arithmetic Logic Unit (ALU)

Memory Management Unit (MMU)

Control Unit

Cache Controller

none

Question : Which problem of memory management is significantly reduced by using paging?

Internal fragmentation

External fragmentation

Deadlock

Starvation

none

Question : In paging, logical memory is divided into pages, while physical memory is divided into:

Blocks

Frames

Segments

Clusters

none

Question : Which of the following is the main reason paging is required in an operating system?

To increase CPU clock speed

To eliminate the need for secondary storage

To allow non-contiguous allocation of physical memory to processes

To reduce the number of processes in the system

none

About us

Term & conditions

Privacy & Policy

Free Mock Test - Operating System : Memory Management (Test 2)

Cookies & Disclaimer