UGC NET CS 2010 June-Paper-2
December 13, 2023Question 2750 – Data-Communication
December 13, 2023Question 10417 – Memory-Management
A certain moving arm disk storage, with one head, has the following specifications.
Number of tracks/recording surface = 200 Disk rotation speed = 2400 rpm Track storage capacity = 62,500 bits
The average latency of this device is P msec and the data transfer rate is Q bits/sec.
Write the value of P and Q.
Correct Answer: A
Question 15 Explanation:
RPM = 2400
So, in DOS, the disk rotates 2400 times.
Average latency is the time for half a rotation
= 0.5×60/2400 s
= 12.5 ms
In one full rotation, entire data in a track can be transferred. Track storage capacity = 62500 bits
So, disk transfer rate
= 62500 × 2400/60
= 2.5 × 106 bps
So,
P = 12.5, Q = 2.5×106
So, in DOS, the disk rotates 2400 times.
Average latency is the time for half a rotation
= 0.5×60/2400 s
= 12.5 ms
In one full rotation, entire data in a track can be transferred. Track storage capacity = 62500 bits
So, disk transfer rate
= 62500 × 2400/60
= 2.5 × 106 bps
So,
P = 12.5, Q = 2.5×106
P = 12.5, Q = 2.5×106
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