Secondary-Memory
Question 1 |
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 , 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 of the 1039th sector is
The address of the 1039th sector is
〈0, 15, 31〉
| |
〈0, 16, 30〉 | |
〈0, 16, 31〉 | |
〈0, 17, 31〉 |
Question 1 Explanation:
We know in each track there are 63 sectors. And we know there is one track per surface.
From the given options we can calculate the sector numbers as
Option A - 15*63+31 = 976
Option B - 16*63+30 = 1038
Option C - 16*63+31 = 1039
Option D - 17*63+31 = 1102
Hence Option C is the answer.
From the given options we can calculate the sector numbers as
Option A - 15*63+31 = 976
Option B - 16*63+30 = 1038
Option C - 16*63+31 = 1039
Option D - 17*63+31 = 1102
Hence Option C is the answer.
Question 2 |
Comprehension:
Question 91-95 concerns a disk with a sector size of 512 bytes, 2000 tracks per surface. 50 sectors per track, five double-sided platters, and average seek time of 10 milliseconds.
Q91: If one track of data can be transferred per revolution, then what is the data transfer rate?
2,850 KBytes/second | |
4,500 KBytes/second | |
5,700 KBytes/second | |
2,250 KBytes/second |
Question 2 Explanation:
It is given that the disk platters rotate at 5400 RPM.
In 60s - 5400 rotations Time for one rotation = 60s/5400 = (1/90) seconds In (1/90)seconds one track size can be transferred. Each track is of size 25KB. In 1/90 seconds 25KB can be transferred. In 1 seconds the amount of data that can be transferred = 25KB*90 = 2250KB So the data transfer rate = 2250 KB/s
In 60s - 5400 rotations Time for one rotation = 60s/5400 = (1/90) seconds In (1/90)seconds one track size can be transferred. Each track is of size 25KB. In 1/90 seconds 25KB can be transferred. In 1 seconds the amount of data that can be transferred = 25KB*90 = 2250KB So the data transfer rate = 2250 KB/s
Question 3 |
Comprehension:
Question 91-95 concerns a disk with a sector size of 512 bytes, 2000 tracks per surface. 50 sectors per track, five double-sided platters, and average seek time of 10 milliseconds.Q92: Given below are two statements:
Statement I: The disk has a total number of 2000 cylinders.
Statement II: 51200 bytes is not a valid block size for the disk.
In the light of the above statements, choose the correct answer from the options given below:
Both Statement I and Statement II are true | |
Both Statement I and Statement II are false | |
Statement I is correct but Statement II is false
| |
Statement I is incorrect but Statement II is true |
Question 3 Explanation:
Statement I is true because in a disk each track will be part of one cylinder. Since there are 2000 tracks there will be 2000 cylinders.
Statement II is also true , because each track has a size of 25KB. A block size can't be bigger than the track size.
Statement II is also true , because each track has a size of 25KB. A block size can't be bigger than the track size.
Question 4 |
Comprehension:
Question 91-95 concerns a disk with a sector size of 512 bytes, 2000 tracks per surface. 50 sectors per track, five double-sided platters, and average seek time of 10 milliseconds.Q93: If T is the capacity of a track in bytes, and S is the capacity of each surface in bytes, then (T,S) = ______.
(50 K, 50000 K) | |
(50 K, 50000 K) | |
(25 K, 50000 K) | |
(40 K, 36000 K) |
Question 4 Explanation:
T= size of a track in bytes = number of sectors per track * size of each sector
= 50*512 bytes = 25K bytes
S = size of each surface in bytes = Number of tracks per surface * number of sectors per track * size of each sector = 2000*50*512 bytes = 50,000 K bytes
= 50*512 bytes = 25K bytes
S = size of each surface in bytes = Number of tracks per surface * number of sectors per track * size of each sector = 2000*50*512 bytes = 50,000 K bytes
Question 5 |
Comprehension:
Question 91-95 concerns a disk with a sector size of 512 bytes, 2000 tracks per surface. 50 sectors per track, five double-sided platters, and average seek time of 10 milliseconds.Q94: What is the capacity of the disk, in bytes?
25,000 K
| |
500,000 K
| |
250,000 K | |
50,000 K |
Question 5 Explanation:
There are 5 double-sided platters. So total 10 surfaces.
There are 2000 tracks per surface and each track has 50 sectors and each sector is 512 bytes in size.
Total disk capacity = Number of surfaces * Number of tracks per surface * number of sectors per track * size of each sector = 10*2000*50*512 bytes = 500,000 K bytes
There are 2000 tracks per surface and each track has 50 sectors and each sector is 512 bytes in size.
Total disk capacity = Number of surfaces * Number of tracks per surface * number of sectors per track * size of each sector = 10*2000*50*512 bytes = 500,000 K bytes
Question 6 |
Comprehension:
Question 91-95 concerns a disk with a sector size of 512 bytes, 2000 tracks per surface. 50 sectors per track, five double-sided platters, and average seek time of 10 milliseconds.Q95: If the disk platters rotate at 5400 rpm (revolutions per minute), then approximately what is the maximum rotational delay?
0.011 seconds | |
0.11 seconds | |
0.0011 seconds | |
1.1 seconds |
Question 6 Explanation:
It is given that the disk platters rotate at 5400 RPM.
In 60s - 5400 rotations
Time for one rotation = 60s/5400 = (1/90) seconds = 0.011 seconds
The maximum rotational delay = 0.011 seconds.
In 60s - 5400 rotations
Time for one rotation = 60s/5400 = (1/90) seconds = 0.011 seconds
The maximum rotational delay = 0.011 seconds.
Question 7 |
A magnetic disk has 100 cylinders, each with 10 tracks of 10 sector. If each sector contains 128 bytes, what is the maximum capacity of the disk in kilobytes?
1,280,000 | |
1280 | |
1250 | |
128,000 |
Question 7 Explanation:
There are 100 cylinders, each with 10 tracks, and 10 sectors. Each sector is 128 bytes.
Maximum capacity of the disk = 100*10*10*128 bytes
= 10^4 * 128 bytes
= (128*10^4 / 1024) KB
= (10^4/ 8) = 1250 KB
Maximum capacity of the disk = 100*10*10*128 bytes
= 10^4 * 128 bytes
= (128*10^4 / 1024) KB
= (10^4/ 8) = 1250 KB
There are 7 questions to complete.