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Official PYQ · 2024Set 1 · Q44I/O SchedulingNAT2 marks

GATE CS 2024 Set 1 Q44 — I/O Scheduling

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Consider a 512512 GB hard disk with 3232 storage surfaces. There are 40964096 sectors per track and each sector holds 10241024 bytes of data. The number of cylinders in the hard disk is _________.
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Answer: 4096

To find the number of cylinders in the hard disk, we use the standard disk capacity formula:

Total Capacity=Number of Cylinders×Number of Surfaces×Sectors per Track×Sector Size\text{Total Capacity} = \text{Number of Cylinders} \times \text{Number of Surfaces} \times \text{Sectors per Track} \times \text{Sector Size}

Given data:

  • Total Capacity: 512 GB=512×230 bytes=29×230 bytes=239 bytes512 \text{ GB} = 512 \times 2^{30} \text{ bytes} = 2^9 \times 2^{30} \text{ bytes} = 2^{39} \text{ bytes}
  • Number of Surfaces (SS): 32=2532 = 2^5
  • Sectors per Track (SPTSPT): 4096=2124096 = 2^{12}
  • Sector Size (SSSS): 1024 bytes=210 bytes1024 \text{ bytes} = 2^{10} \text{ bytes}

Let the number of cylinders be CC. Substituting the values into the formula:

239=C×25×212×2102^{39} = C \times 2^5 \times 2^{12} \times 2^{10} 239=C×2272^{39} = C \times 2^{27}

Solving for CC:

C=239227=212=4096C = \frac{2^{39}}{2^{27}} = 2^{12} = 4096

Thus, the number of cylinders in the hard disk is 4096.

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