A disk has 8 equidistant tracks. The diameters of the innermost and outermost…
2005
A disk has 8 equidistant tracks. The diameters of the innermost and outermost tracks are 1 cm and 8 cm respectively. The innermost track has a storage capacity of 10 MB. If the disk has 20 sectors per track and is currently at the end of the 5th sector of the inner-most track and the head can move at a speed of 10 meters/sec and it is rotating at constant angular velocity of 6000 RPM, how much time will it take to read 1 MB contiguous data starting from the sector 4 of the outer-most track?
- A.
13.5 ms
- B.
10 ms
- C.
9.5 ms
- D.
20 ms
Attempted by 59 students.
Show answer & explanation
Correct answer: A
Key data: innermost diameter = 1 cm, outermost diameter = 8 cm; innermost track capacity = 10 MB; 20 sectors/track; disk speed = 6000 RPM; head radial speed = 10 m/s. Current head position: end of 5th sector on innermost track. Read starts at sector 4 on outermost track for 1 MB.
Compute seek (radial) distance and time: radii = 0.5 cm and 4 cm → radial distance = 3.5 cm = 0.035 m. Seek time = 0.035 m / 10 m/s = 0.0035 s = 3.5 ms.
Compute sector sizes: innermost track capacity 10 MB with 20 sectors → innermost sector = 10 MB / 20 = 0.5 MB per sector. Sector size scales with diameter (outer/inner = 8), so outer sector = 0.5 MB × 8 = 4 MB per sector.
Rotation timings: 6000 RPM = 6000/60 = 100 rev/s → period T = 1/100 = 0.01 s = 10 ms. Sector time = 10 ms / 20 = 0.5 ms per sector.
Rotational latency after seek: current angular position is at the boundary after sector 5. To reach the start of sector 4 we must wait forward (4 − 5 + 20) mod 20 = 19 sectors. Latency = 19 × 0.5 ms = 9.5 ms.
Transfer time for 1 MB on outermost track: outer sector = 4 MB, so reading 1 MB = (1/4) of a sector = 0.25 × 0.5 ms = 0.125 ms.
Total time = seek time + rotational latency + transfer time = 3.5 ms + 9.5 ms + 0.125 ms = 13.125 ms ≈ 13.13 ms.
Conclusion: The computed time is about 13.13 ms; among the given choices, the value 13.5 ms is the closest match.