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Test: File Systems-1

15 Questions MCQ Test Operating System | Test: File Systems-1

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This mock test of Test: File Systems-1 for Computer Science Engineering (CSE) helps you for every Computer Science Engineering (CSE) entrance exam. This contains 15 Multiple Choice Questions for Computer Science Engineering (CSE) Test: File Systems-1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: File Systems-1 quiz give you a good mix of easy questions and tough questions. Computer Science Engineering (CSE) students definitely take this Test: File Systems-1 exercise for a better result in the exam. You can find other Test: File Systems-1 extra questions, long questions & short questions for Computer Science Engineering (CSE) on EduRev as well by searching above.
QUESTION: 1

Which of the following is major part of time taken when accessing data on the disk?

Solution:

Seek time is time taken by the head to travel to the track of the disk where the data to be accessed is stored.

QUESTION: 2

Put the following disk scheduling policies results in minimum amount of head movement.

Solution:

Circular scanning works just like the elevator to some extent. It begins its scan toward the nearest end and works its way all the way to the end of the system. Once it hits the bottom or top it jumps to the other end and moves in the same direction. Circular SCAN has more head movement than SCAN (elevator) because Circular SCAN has circular jump and it does count as a head movement. SCAN (elevator) is the best choice here.

QUESTION: 3

A file system with 300 GByte disk uses a file descriptor with 8 direct block addresses, 1 indirect block address and 1 doubly indirect block address. The size of each disk block is 128 Bytes and the size of each disk block address is 8 Bytes. The maximum possible file size in this file system is

Solution:

Total number of possible addresses stored in a disk block = 128/8 = 16

Maximum number of addressable bytes due to direct address block = 8*128
Maximum number of addressable bytes due to 1 single indirect address block = 16*128
Maximum number of addressable bytes due to 1 double indirect address block = 16*16*128

The maximum possible file size = 8*128 + 16*128 + 16*16*128 = 35KB

QUESTION: 4

A computer handles several interrupt sources of which the following are relevant for this question.

. Interrupt from CPU temperature sensor (raises interrupt if CPU temperature is too high)
. Interrupt from Mouse(raises interrupt if the mouse is moved or a button is pressed)
. Interrupt from Keyboard(raises interrupt when a key is pressed or released)
. Interrupt from Hard Disk(raises interrupt when a disk read is completed)
Which one of these will be handled at the HIGHEST priority?

Solution:

Higher priority interrupt levels are assigned to requests which, if delayed or interrupted, could have serious consequences. Devices with high speed transfer such as magnetic disks are given high priority, and slow devices such as keyboard receive low priority. Interrupt from CPU temperature sensor would have serious consequences if ignored.

QUESTION: 5

A CPU generally handles an interrupt by executing an interrupt service routine

Solution:

Hardware detects interrupt immediately, but CPU acts only after its current instruction. This is followed to ensure integrity of instructions.

QUESTION: 6

Consider the data given in previous question. The address of the 1039th sector is

Solution:

(a) <0,15,31> 0th cylinder 15th surface and 31st sector So, 0 cylinders passed 0*20*63 As each cylinder has 20 surfaces and each surface has 63 sectors. + 15 surfaces passed (0-14) 15*63 As each surface has 63 sectors + We are on 31st sector So, sector no. =0*20*63+15*63+31=976 sector. Which is not equal to 1039.

(b) <0,16,30> Similarly this represents, 0*20*63 + 16*63 (0-15 sectors and each sector has 63 sectors) + 30 sectors on 16th sector Sector no = 0*20*63+16*63+30=1038 sector which is not equal to 1039.

(c) <0,16,31> Similarly this represents, 0*20*63 + 16*63 (0-15 sectors and each sector has 63 sectors) + 31 sectors on 16th sector Sector no = 0*20*63+16*63+31=1039 sector which is equal to 1039. Hence,option c is correct.

(d) <0,17,31> Similarly this represents, 0*20*63 + 17*63 (0-16 sectors and each sector has 63 sectors) + 31 sectors on 17th sector Sector no = 0*20*63+17*63+31=1102 sector which is not equal to 1039.

QUESTION: 7

A hard disk has 63 sectors per track, 10 platters each with 2 recording surfaces and 1000 cylinders. The address of a sector is given as a triple (c, h, s), where c is the cylinder number, h is the surface number and s is the sector number. Thus, the 0th sector is addressed as (0, 0, 0), the 1st sector as (0, 0, 1), and so on The address <400,16,29> corresponds to sector number:

Solution:

The data in hard disk is arranged in the shown manner. The smallest division is sector. Sectors are then combined to make a track. Cylinder is formed by combining the tracks which lie on same dimension of the platters. Read write head access the disk. Head has to reach at a particular track and then wait for the rotation of the platter so that the required sector comes under it. Here, each platter has two surfaces, which is the r/w head can access the platter from the two sides, upper and lower. So,<400,16,29> will represent 400 cylinders are passed(0-399) and thus, for each cylinder 20 surfaces (10 platters * 2 surface each) and each cylinder has 63 sectors per surface. Hence we have passed 0-399 =  400 * 20 * 63 sectors + In 400th cylinder we have passed 16 surfaces(0-15) each of which again contains 63 sectors per cylinder so 16 * 63 sectors. + Now on the 16th surface we are on 29th sector. So, sector no = 400x20x63 + 16×63 + 29 = 505037.

QUESTION: 8

For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to

Solution:

Whenever head moves from one track to other then its speed and direction changes, which is noting but change in motion or the case of inertia. So answer B

QUESTION: 9

The data blocks of a very large file in the Unix file system are allocated using

Solution:

The Unix file system uses an extension of indexed allocation. It uses direct blocks, single indirect blocks, double indirect blocks and triple indirect blocks. Following diagram shows implementation of Unix file system.

QUESTION: 10

Which of the following statements about synchronous and asynchronous I/O is NOT true?

Solution:

An interrupt service routine will be invoked after the completion of I/O operation and it will place process from block state to ready state, because process performing I/O operation was placed in blocked state till the I/O operation was completed in Synchronous I/O. However, process performing I/O will not be placed in the block state and process continues to execute the remaining instructions in Asynchronous I/O, because handler function will be registered while performing the I/O operation, when the I/O operation completed signal mechanism is used to notify the process that data is available. So, option (B) is false.

QUESTION: 11

A device with data transfer rate 10 KB/sec is connected to a CPU. Data is transferred byte-wise. Let the interrupt overhead be 4 microsec. The byte transfer time between the device interface register and CPU or memory is negligible. What is the minimum performance gain of operating the device under interrupt mode over operating it under program controlled mode?

Solution:

In programmed I/O, CPU does continuous polling, To transfer 1B CPU polls for 10^-4 sec = 10^2 micro-sec of processing In interrupt mode CPU is interrupted on completion of i\o, To transfer 1B CPU does 4 micro-sec of processing(since transfer time between other components is negligible). Gain = 10^2 / 4 = 25

QUESTION: 12

Suppose a disk has 201 cylinders, numbered from 0 to 200. At some time the disk arm is at cylinder 100, and there is a queue of disk access requests for cylinders 30, 85, 90, 100, 105, 110, 135 and 145. If Shortest-Seek Time First (SSTF) is being used for scheduling the disk access, the request for cylinder 90 is serviced after servicing ____________ number of requests.

Solution:

In Shortest-Seek-First algorithm, request closest to the current position of the disk arm and head is handled first. In this question, the arm is currently at cylinder number 100. Now the requests come in the queue order for cylinder numbers 30, 85, 90, 100, 105, 110, 135 and 145. The disk will service that request first whose cylinder number is closest to its arm. Hence 1st serviced request is for cylinder no 100 ( as the arm is itself pointing to it ), then 105, then 110, and then the arm comes to service request for cylinder 90. Hence before servicing request for cylinder 90, the disk would had serviced 3 requests. Hence option C.

QUESTION: 13

Consider an operating system capable of loading and executing a single sequential user process at a time. The disk head scheduling algorithm used is First Come First Served (FCFS). If FCFS is replaced by Shortest Seek Time First (SSTF), claimed by the vendor to give 50% better benchmark results, what is the expected improvement in the I/O performance of user programs?

Solution:

Since Operating System can execute a single sequential user process at a time, the disk is accessed in FCFS manner always. The OS never has a choice to pick an IO from multiple IOs as there is always one IO at a time

QUESTION: 14

Consider a disk drive with the following specifications: 16 surfaces, 512 tracks/surface, 512 sectors/track, 1 KB/sector, rotation speed 3000 rpm. The disk is operated in cycle stealing mode whereby whenever one byte word is ready it is sent to memory; similarly, for writing, the disk interface reads a 4 byte word from the memory in each DMA cycle. Memory cycle time is 40 nsec. The maximum percentage of time that the CPU gets blocked during DMA operation is:

Solution:

Time takes for 1 rotation = 60/3000 It reads 512*1024 Bytes in one rotation. Time taken to read 4 bytes = 153 ns 153 is approximately 4 cycles (160ns) Percentage of time CPU gets blocked = 40*100/160 = 25

QUESTION: 15

Using a larger block size in a fixed block size file system leads to :

Solution:

Using larger block size makes disk utilization poorer as more space would be wasted for small data in a block. It may make throughput better as the number of blocks would decrease. A larger block size guarantees that more data from a single file can be written or read at a time into a single block without having to move the disk ́s head to another spot on the disk. The less time you spend moving your heads across the disk, the more continuous reads/writes per second. The smaller the block size, the more frequent it is required to move before a read/write can occur. Larger block size means less number of blocks to fetch and hence better throughput. But larger block size also means space is wasted when only small size is required and hence poor utilization.