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Secondary Storage and Disk Performance PYQs for GATE CS

Solve 3+ Secondary Storage and Disk Performance previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

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Question 1

2026 Slot Set1 PYQ

Level 3: Exam Standard

Consider a hard disk with a rotational speed of 15000 rpm. The time to move the read/write head from a track to its adjacent track is 1 millisecond. Initially, the head is on track 0. The number of sectors per track is 400. The sector size is 1024 bytes. It is necessary to transfer data from 10 randomly located sectors in each of the following tracks in the order: 5, 12 and 7. The total time for the data transfer (in milliseconds) from the hard disk is _________. (rounded off to one decimal place)

Question 2

2024 Slot Set2 PYQ

Level 3: Exam Standard

Consider a disk with the following specifications: rotation speed of 6000 RPM, average seek time of 5 milliseconds, 500 sectors/track, 512-byte sectors. A file has content stored in 3000 sectors located randomly on the disk. Assuming average rotational latency, the total time (in seconds, rounded off to 2 decimal places) to read the entire file from the disk is __________

Question 3

2024 Slot Set1 PYQ

Level 3: Exam Standard

Consider a 512 GB hard disk with 32 storage surfaces. There are 4096 sectors per track and each sector holds 1024 bytes of data. The number of cylinders in the hard disk is _________

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Secondary Storage and Disk Performance PYQs for GATE CS

Solve 3+ Secondary Storage and Disk Performance previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Secondary Storage and Disk Performance

Chapter Roadmap

Welcome to Secondary Storage and Disk Performance. This chapter bridges the physical reality of magnetic disks with the mathematical performance metrics required for the exam.

The Learning Journey

Step 1: Disk Geometry, Capacity and Cylinders Current Topic

We start by looking inside the hard disk. You will learn the physical anatomy: platters, surfaces, tracks, sectors, and cylinders. By the end of this step, you will be able to calculate the total storage capacity of any disk given its physical dimensions.

Step 2: Disk Access Time, Seek and Rotational Latency

Once we know how data is stored, we measure how fast we can retrieve it. You will master the components of disk access time: seek time, rotational latency, and transfer rate, and learn to calculate the total time for complex read and write operations.

Goal: By the end of this chapter, you will seamlessly translate physical disk specifications into precise performance calculations.

The Physical Anatomy of a Hard Disk

To understand disk capacity, we must first visualize the physical hardware. Think of a hard disk as a high-speed stack of spinning records.

Core Physical Components

Platter: The physical circular disk made of aluminum or glass. A single disk drive contains a stack of multiple platters.
Surface: Each platter has two usable sides (top and bottom), coated with magnetic material. Each side is an independent storage surface.
Track: As the platter spins, the read/write head traces concentric circles on the surface. Each circular path is a track.
Sector: A track is subdivided into small arc-shaped segments. A sector is the smallest physical unit of data storage on the disk.

Secondary Storage and Disk Performance: Solved Questions with Step-by-Step Explanations (3 Problems)

Question 1 · Computer Organization and Architecture · 2026_Set1 NAT

Consider a hard disk with a rotational speed of 15000 rpm. The time to move the read/write head from a track to its adjacent track is 1 millisecond. Initially, the head is on track 0. The number of sectors per track is 400. The sector size is 1024 bytes. It is necessary to transfer data from 10 randomly located sectors in each of the following tracks in the order: 5, 12 and 7.

The total time for the data transfer (in milliseconds) from the hard disk is _________. (rounded off to one decimal place)

Correct Answer:

77.30

Step-by-Step Solution

Insight: Sectors are randomly located within their respective tracks, so we seek to each track once, then pay rotational latency for each of the 10 sectors.

Exam route: For each track, add seek time to $10 \times (T_{l a t} + T_{t r an s})$ . Sum the three tracks.

Learning route:

$T_{r o t} = \frac{60000}{15000} = 4$ ms.
$T_{l a t} = \frac{4}{2} = 2$ ms.
$T_{t r an s} = \frac{4}{400} = 0.01$ ms.
Track 5: Seek from 0 to 5 = 5 ms. Data = $10 \times (2 + 0.01) = 20.1$ ms. Total = 25.1 ms.
Track 12: Seek from 5 to 12 = 7 ms. Data = 20.1 ms. Total = 27.1 ms.
Track 7: Seek from 12 to 7 = 5 ms. Data = 20.1 ms. Total = 25.1 ms.
Grand total = $25.1 + 27.1 + 25.1 = 77.3$ ms.

Question 2 · Computer Organization and Architecture · 2024_Set2 NAT

Consider a disk with the following specifications: rotation speed of 6000 RPM, average seek time of 5 milliseconds, 500 sectors/track, 512-byte sectors.

A file has content stored in 3000 sectors located randomly on the disk. Assuming average rotational latency, the total time (in seconds, rounded off to 2 decimal places) to read the entire file from the disk is __________

Correct Answer:

30.06

Step-by-Step Solution

Insight: Randomly located sectors mean each sector incurs a full seek, rotational latency, and transfer time.

Exam route: Calculate time for 1 sector (seek + rot_lat + transfer), then multiply by 3000.

Learning route:

Rotation time $T_{r o t} = \frac{60}{6000} = 10$ ms.
Average rotational latency $T_{l a t} = \frac{10}{2} = 5$ ms.
Transfer time for 1 sector $T_{t r an s} = \frac{10}{500} = 0.02$ ms.
Access time for 1 random sector = $T_{see k} + T_{l a t} + T_{t r an s} = 5 + 5 + 0.02 = 10.02$ ms.
Total time for 3000 random sectors = $3000 \times 10.02 = 30060$ ms = 30.06 seconds.

Question 3 · Computer Organization and Architecture · 2024_Set1 NAT

Consider a 512 GB hard disk with 32 storage surfaces. There are 4096 sectors per track and each sector holds 1024 bytes of data. The number of cylinders in the hard disk is _________

Correct Answer:

4096.00

Step-by-Step Solution

Insight: Total capacity equals Surfaces $\times$ Cylinders $\times$ Sectors/track $\times$ Bytes/sector.

Exam route: Convert 512 GB to bytes, calculate capacity of one cylinder, divide total by cylinder capacity.

Learning route:

Total capacity = $512 GB = 512 \times 2^{30} bytes = 2^{39} bytes$ .
Capacity per cylinder = $Surfaces \times Sectors/track \times Bytes/sector = 32 \times 4096 \times 1024$ .
In powers of 2: $2^{5} \times 2^{12} \times 2^{10} = 2^{27} bytes$ .
Number of cylinders = $\frac{Total Capacity}{Capacity per cylinder} = \frac{2 ^{39}}{2 ^{27}} = 2^{12} = 4096$ .

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Built around you, not around a syllabus PDF

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Questions calibrated to the real exam

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Honest progress

Unlock the whole course

Secondary Storage and Disk Performance PYQs for GATE CS

Chapter Roadmap: Secondary Storage and Disk Performance

Chapter Roadmap

The Learning Journey

The Physical Anatomy of a Hard Disk

The Physical Anatomy of a Hard Disk

Core Physical Components

Secondary Storage and Disk Performance: Solved Questions with Step-by-Step Explanations (3 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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