Question 11049 – Data-Structures
January 5, 2024Question 6 – ISRO-2018
January 5, 2024Question 11117 – Operating-Systems
A disk has 8 equidistant tracks. The diameters of the innermost and outermost tracks are 1 cm and 8 cm respectively. The innermost track has a storage capacity of 10 MB.
What is the total amount of data that can be stored on the disk if it is used with a drive that rotates it with (i) Constant Linear Velocity (ii) Constant Angular Velocity?
Correct Answer: D
Question 9 Explanation:
Constant linear velocity:
Diameter of inner track = d = 1 cm
Circumference of inner track
= 2 * 3.14 * d/2
= 3.14 cm
Storage capacity = 10 MB (given)
Circumference of all equidistant tracks
= 2 * 3.14 * (0.5+1+1.5+2+2.5+3+3.5+4)
= 113.14 cm
Here, 3.14 cm holds 10 MB
Therefore, 1 cm holds 3.18 MB.
So, 113.14 cm holds
113.14 * 3.18 = 360 MB
So, total amount of data that can be hold on the disk = 360 MB.
For constant angular velocity:
In case of CAV, the disk rotates at a constant angular speed. Same rotation time is taken by all the tracks.
Total amount of data that can be stored on the disk
= 8 * 10 = 80 MB
Diameter of inner track = d = 1 cm
Circumference of inner track
= 2 * 3.14 * d/2
= 3.14 cm
Storage capacity = 10 MB (given)
Circumference of all equidistant tracks
= 2 * 3.14 * (0.5+1+1.5+2+2.5+3+3.5+4)
= 113.14 cm
Here, 3.14 cm holds 10 MB
Therefore, 1 cm holds 3.18 MB.
So, 113.14 cm holds
113.14 * 3.18 = 360 MB
So, total amount of data that can be hold on the disk = 360 MB.
For constant angular velocity:
In case of CAV, the disk rotates at a constant angular speed. Same rotation time is taken by all the tracks.
Total amount of data that can be stored on the disk
= 8 * 10 = 80 MB
(i) 80 MB (ii) 2040 MB
(i) 2040 MB (ii) 80 MB
(i) 80 MB (ii) 360 MB
(i) 80 MB (ii) 360 MB
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