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Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?
- a)1281
- b)1282
- c)1283
- d)1284
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Consider a hard disk with 16 recording surfaces (0 − 15) having ...
To 15) and 512 cylinders (0 to 511). Each cylinder contains 128 sectors (0 to 127). The disk rotates at a speed of 7200 RPM (rotations per minute). The average seek time is 8 milliseconds. The disk has an average rotational latency of 4.17 milliseconds.
To calculate the total capacity of the hard disk, we need to multiply the number of recording surfaces (16) by the number of cylinders (512) by the number of sectors per cylinder (128) and the sector size. Assuming a sector size of 512 bytes, the calculation is as follows:
Total capacity = Number of recording surfaces × Number of cylinders × Number of sectors per cylinder × Sector size
= 16 × 512 × 128 × 512 bytes
= 4,194,304 KB (kilobytes)
= 4,096 MB (megabytes)
= 4 GB (gigabytes)
Therefore, the total capacity of the hard disk is 4 GB.
To calculate the average rotational delay, we need to divide the time taken for one complete rotation (60 seconds ÷ RPM) by 2, as the average rotational latency is half the time taken for one complete rotation (assuming a uniform distribution of data). Given that the disk rotates at 7200 RPM:
Average rotational delay = (60 seconds ÷ 7200 RPM) ÷ 2
= 0.0041667 seconds
≈ 4.17 milliseconds
Therefore, the average rotational delay is approximately 4.17 milliseconds.
To calculate the average access time, we need to add the average seek time and the average rotational delay:
Average access time = Average seek time + Average rotational delay
= 8 milliseconds + 4.17 milliseconds
= 12.17 milliseconds
Therefore, the average access time is 12.17 milliseconds.
To calculate the total capacity of the hard disk, we need to multiply the number of recording surfaces (16) by the number of cylinders (512) by the number of sectors per cylinder (128) and the sector size. Assuming a sector size of 512 bytes, the calculation is as follows:
Total capacity = Number of recording surfaces × Number of cylinders × Number of sectors per cylinder × Sector size
= 16 × 512 × 128 × 512 bytes
= 4,194,304 KB (kilobytes)
= 4,096 MB (megabytes)
= 4 GB (gigabytes)
Therefore, the total capacity of the hard disk is 4 GB.
To calculate the average rotational delay, we need to divide the time taken for one complete rotation (60 seconds ÷ RPM) by 2, as the average rotational latency is half the time taken for one complete rotation (assuming a uniform distribution of data). Given that the disk rotates at 7200 RPM:
Average rotational delay = (60 seconds ÷ 7200 RPM) ÷ 2
= 0.0041667 seconds
≈ 4.17 milliseconds
Therefore, the average rotational delay is approximately 4.17 milliseconds.
To calculate the average access time, we need to add the average seek time and the average rotational delay:
Average access time = Average seek time + Average rotational delay
= 8 milliseconds + 4.17 milliseconds
= 12.17 milliseconds
Therefore, the average access time is 12.17 milliseconds.
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Community Answer
Consider a hard disk with 16 recording surfaces (0 − 15) having ...
As we know,
Since starting on disk starts from <1200,9,40> so no. of sectors left on 9
th
surface is 24.
So, on 9
th
surface total storage of "12288 B" is possible.
Now, a part from 9
th
surface, on cylinder no. 1200 only 6 surface is left.
To storage possible on these 6 surfaces are = 6 × 2
6
× 2
9
→ Storage on each sector.
No. of sectors on each surface
= 196608 B
So, total on cylinder no. 1200, storage possible is:
⇒ 196608 + 12288
= 208896 B
So, since the file size is 42797 kB and out of which 208896 B are stored on cylinder no. 1200. So, we are left only with only 43615232 B.
Since in 1 cylinder, storage possible is = 2
4
× 2
6
× 2
9
B
= 524288/B
So, we need about = 43615232 B/524288 B
= 83.189 more cylinders
So, we'll need the [1284
th
Cylinder] to completely store the file. C
02
after 1283
rd
cylinder we will left with data which will need 189.
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Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer?
Question Description
Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer?.
Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer?.
Solutions for Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer?, a detailed solution for Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of Consider a hard disk with 16 recording surfaces (0 − 15) having 16384 cylinders (0 − 16383) and each cylinder contains 64 sectors (0 − 63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no.>. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?a)1281b)1282c)1283d)1284Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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