GATE CS 2024 Set 2 — Question 53

Given:

• Rotation speed = 6000 RPM
• Average seek time ($T_{seek}$) = 5 ms
• Sectors per track = 500
• Total sectors to read = 3000
• Sectors are located randomly.

Step 1: Calculate Average Rotational Latency ($T_{rot}$)$$\text{Rotations per second} = \frac{6000}{60} = 100 \text{ rps}$$$$\text{Time for one full rotation} = \frac{1}{100} \text{ sec} = 10 \text{ ms}$$$$\text{Average Rotational Latency} = \frac{1}{2} \times \text{Time for one rotation} = \frac{10}{2} = 5 \text{ ms}$$Step 2: Calculate Transfer Time per Sector ($T_{transfer}$)
Since there are 500 sectors per track, the time to read one sector is the time for one rotation divided by the number of sectors:$$T_{transfer} = \frac{10 \text{ ms}}{500} = 0.02 \text{ ms}$$Step 3: Calculate Total Time
Since the sectors are located randomly, each sector access requires independent Seek, Rotational Latency, and Transfer.$$\text{Time per sector} = T_{seek} + T_{rot} + T_{transfer}$$$$\text{Time per sector} = 5 \text{ ms} + 5 \text{ ms} + 0.02 \text{ ms} = 10.02 \text{ ms}$$Total time for 3000 sectors:$$\text{Total Time} = 3000 \times 10.02 \text{ ms}$$$$\text{Total Time} = 30060 \text{ ms}$$$$\text{Total Time in seconds} = \frac{30060}{1000} = 30.06 \text{ s}$$The value 30.06 lies within the accepted range of 29.50 to 30.50.