Secondary Storage and Disk Performance Practice Questions for GATE CS

Solve 47+ Secondary Storage and Disk Performance practice 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 (10 Problems)

Question 1 · Computer Organization and Architecture MCQ

A disk has 8 surfaces, 500 cylinders, and 100 sectors per track. Each sector holds 512 bytes. What is the minimum number of bytes required to store data on exactly one full cylinder of this disk?

A.
409,600
B.
204,800,000
C.
819,200
D.
4,096,000

Correct Answer:

Step-by-Step Solution

Key idea: This is a contradiction question testing the Master Capacity Equation, specifically isolating the capacity of a single cylinder.

Step 1: Recall that a cylinder spans all surfaces. The capacity of one cylinder is the number of surfaces multiplied by the sectors per track multiplied by the bytes per sector.

Step 2: Identify the given values: $S = 8$ , $T_{s} = 100$ , $B = 512$ .

Step 3: Calculate the capacity of one cylinder: $8 \times 100 \times 512 = 409, 600$ bytes.

Step 4: Note that the 500 cylinders given in the problem is extra information for this specific question, as we only need the capacity of <i>one</i> cylinder.

Answer: 409,600.

Question 2 · Computer Organization and Architecture MCQ

A hard disk has 16 surfaces and 2048 tracks on each surface. What is the maximum number of cylinders that can be formed in this disk?

A.
32768
B.
2048
C.
16
D.
4096

Correct Answer:

Step-by-Step Solution

Key idea: This is an observation question testing the crucial equality between cylinders and tracks per surface.

Step 1: Recall the definition of a cylinder: it is the set of all tracks that lie at the exact same radius across all platters.

Step 2: The crucial equality states that the number of cylinders is exactly equal to the number of tracks per surface.

Step 3: The problem states there are 2048 tracks on each surface. Therefore, the number of cylinders is 2048. The number of surfaces (16) is extra information.

Answer: 2048.

Question 3 · Computer Organization and Architecture MCQ

When calculating the total time to read $N$ randomly located sectors on a disk, the access time for a single sector is multiplied by $N$ . Which of the following statements correctly bounds the components of this single-sector access time?

A.
It includes the average seek time, average rotational latency, and the transfer time for exactly one sector.
B.
It includes the total seek time for all $N$ sectors, average rotational latency, and the transfer time for one sector.
C.
It includes the average seek time, maximum rotational latency, and the transfer time for all $N$ sectors.
D.
It includes the average seek time, average rotational latency, and the transfer time for the entire track.

Correct Answer:

Step-by-Step Solution

Key idea: This is a statement truth question testing the exact components of single-sector access time for randomly located sectors.

Step 1: Recall the method for randomly located sectors. Each sector is treated as an independent read operation.

Step 2: For a single random sector, the head must seek to the track (average seek time), wait for the sector to spin under the head (average rotational latency), and read the data (transfer time for exactly one sector).

Step 3: Therefore, the single-sector access time is bounded by these three specific components.

Answer: It includes the average seek time, average rotational latency, and the transfer time for exactly one sector.

Question 4 · Computer Organization and Architecture MCQ

For a file of $N$ sectors ( $N > 1$ ), let $T_{r an d}$ be the total time to read them if they are randomly located, and $T_{co n t}$ be the total time if they are contiguous on the same track. Which of the following statements is always true?

A.
$T_{r an d} < T_{co n t}$
B.
$T_{r an d} > T_{co n t}$
C.
$T_{r an d} = T_{co n t}$
D.
$T_{r an d}$ is independent of $N$

Correct Answer:

Step-by-Step Solution

Key idea: This is a bounding question comparing the total time formulas for random versus contiguous access.

Step 1: Recall the formula for random access: $T_{r an d} = N \times (T_{see k} + T_{r o t (a v g)} + T_{t r an s f er (1)})$ .

Step 2: Recall the formula for contiguous access: $T_{co n t} = T_{see k} + T_{r o t (a v g)} + T_{t r an s f er (N)}$ .

Step 3: For $N > 1$ , $T_{r an d}$ incurs the seek and rotational latency penalties $N$ times, while $T_{co n t}$ incurs them only once.

Step 4: Therefore, $T_{r an d}$ is strictly greater than $T_{co n t}$ for any $N > 1$ .

Answer: $T_{r an d} > T_{co n t}$ .

Question 5 · Computer Organization and Architecture MCQ

A hard disk drive contains a stack of 4 platters. In a specific legacy architecture, the very top and very bottom surfaces are reserved for protection and not used for data. How many usable storage surfaces does this disk have?

A.
8
B.
6
C.
4
D.
10

Correct Answer:

Step-by-Step Solution

Key idea: This is a casework question testing the physical anatomy of platters and surfaces, including a specific legacy exception.

Step 1: Identify the number of platters, which is 4.

Step 2: The standard rule is 1 Platter = 2 Surfaces, giving $4 \times 2 = 8$ total surfaces.

Step 3: Apply the legacy exception stated in the problem: the top and bottom surfaces are unused. This means we subtract 2 surfaces.

Step 4: Calculate $8 - 2 = 6$ usable surfaces.

Answer: 6.

Question 6 · Computer Organization and Architecture MCQ

Assertion (A): When reading $N$ contiguous sectors located on the same track, the total access time is $T_{see k} + T_{r o t (a v g)} + T_{t r an s f er (N)}$ .

Reason (R): The read/write head must perform a mechanical seek and wait for rotational latency for every single sector in the contiguous block.

A.
Both A and R are true, and R is the correct explanation of A.
B.
Both A and R are true, but R is not the correct explanation of A.
C.
A is true, but R is false.
D.
A is false, but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: This is an assertion-reason question testing the physical reasoning behind contiguous sector access.

Step 1: Evaluate Assertion (A). For contiguous sectors on the same track, the head seeks once, waits for rotational latency once, and transfers all $N$ sectors continuously. The formula $T_{see k} + T_{r o t (a v g)} + T_{t r an s f er (N)}$ is correct.

Step 2: Evaluate Reason (R). The reason states that the head must seek and wait for rotational latency for <i>every single sector</i>. This is false; those penalties are paid only once for the entire contiguous block.

Step 3: Since A is true and R is false, the correct option is C.

Answer: A is true, but R is false.

Question 7 · Computer Organization and Architecture MCQ

A hard disk rotates at a constant speed of 6000 RPM. What is the maximum possible rotational latency for this disk in milliseconds?

A.
2.50
B.
5.00
C.
10.00
D.
20.00

Correct Answer:

Step-by-Step Solution

Key idea: This is an observation question testing the boundary case of rotational latency.

Step 1: Calculate the time for one full rotation: $T_{r o t a t i o n} = \frac{60 , 000 ms}{6000 RPM} = 10$ ms.

Step 2: The average rotational latency is half of this, which is 5 ms.

Step 3: The maximum possible rotational latency occurs in the worst-case boundary scenario where the desired sector is exactly one full rotation away from the head's arrival point.

Step 4: Therefore, the maximum rotational latency is equal to one full rotation time, which is 10 ms.

Answer: 10.00.

Question 8 · Computer Organization and Architecture MCQ

Assertion (A): The transfer time for $N$ contiguous sectors on the same track is exactly $N$ times the transfer time of a single sector.

Reason (R): The disk rotation speed is constant, meaning the time to read a block of sectors is directly proportional to the number of sectors in that block.

A.
A is false, but R is true.
B.
Both A and R are true, but R is not the correct explanation of A.
C.
Both A and R are true, and R is the correct explanation of A.
D.
A is true, but R is false.

Correct Answer:

Step-by-Step Solution

Key idea: This is a construction question testing the logical relationship between transfer time proportionality and constant disk rotation.

Step 1: Evaluate Assertion (A). For contiguous sectors, the transfer time is indeed $N$ times the single-sector transfer time, because the sectors are read back-to-back without interruption. (A is true).

Step 2: Evaluate Reason (R). The disk rotates at a constant RPM. Therefore, the time it takes for the disk to pass $N$ sectors under the head is directly proportional to $N$ . (R is true).

Step 3: Determine the relationship. The constant rotation speed (R) is the exact physical reason why the transfer time scales linearly with $N$ (A). Thus, R correctly explains A.

Answer: Both A and R are true, and R is the correct explanation of A.

Question 9 · Computer Organization and Architecture MCQ

According to the standard learning journey for secondary storage, calculating the total number of sectors on a disk given its physical dimensions is the primary goal of Step 1. If a disk has 2 surfaces, 100 cylinders, and 50 sectors per track, what is the total number of sectors on the disk calculated in this step?

A.
10,000
B.
5,000
C.
20,000
D.
2,500

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct substitution question based on the Step 1 capacity calculation concept.

Step 1: Identify the given physical dimensions: Surfaces ( $S$ ) = 2, Cylinders ( $C$ ) = 100, Sectors per track ( $T_{s}$ ) = 50.

Step 2: The total number of sectors is the product of these three dimensions.

Step 3: Calculate $2 \times 100 \times 50 = 10, 000$ .

Answer: 10,000.

Question 10 · Computer Organization and Architecture MCQ

A disk access operation involves moving the read/write head, waiting for the sector to rotate, and reading the data. If the seek time is 4 ms, the rotational latency is 5 ms, and the transfer time is 1 ms, what is the total access time for this single sector in milliseconds?

A.
10
B.
9
C.
5
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct substitution question testing the three pillars of disk access time.

Step 1: Identify the three components given: seek time = 4 ms, rotational latency = 5 ms, transfer time = 1 ms.

Step 2: The total access time for a single sector is the sum of these three components.

Step 3: Calculate $4 + 5 + 1 = 10$ ms.

Answer: 10.

Try a question

Login to view the complete practice questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Secondary Storage and Disk Performance Practice Questions 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 (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More practice questions in this unit