Computer Science - Computer Fundamentals - Discussion
Discussion Forum : Computer Fundamentals - Section 8 (Q.No. 13)
13.
What is the byte capacity of a drum which is 5 inch high, 10 inch diameter, and which has 60 tracks per inch and bit density of 800 bits per inch?
Discussion:
6 comments Page 1 of 1.
Mahen Barfa said:
1 decade ago
How to find the result please provide a suitable formula for this question?
Neeraj pal said:
1 decade ago
Total capacity = (areal density)*(total surface area).
Areal density = 60*800=48000 bits per inch.
Total surface area = 2*pi*r*h=2*pi*5*5=50*pi.
Total capacity = (50*3.14*48000)/8=942000 bytes.
Areal density = 60*800=48000 bits per inch.
Total surface area = 2*pi*r*h=2*pi*5*5=50*pi.
Total capacity = (50*3.14*48000)/8=942000 bytes.
Ranjan said:
1 decade ago
Areal density = TPI * BPI = 60 * 800 = 48000 bits per inch square.
Total surface area of drum = 2 * pi * r * height = 50 * pi.
Total capacity = areal density * total surface area of drum = [48000 * 50 * pi]/8 = 942477 bytes.
Total surface area of drum = 2 * pi * r * height = 50 * pi.
Total capacity = areal density * total surface area of drum = [48000 * 50 * pi]/8 = 942477 bytes.
Venkatesan S said:
9 years ago
Where from 8 comes?
Bala said:
9 years ago
To convert into bytes, So that 8 comes.
Naf said:
5 years ago
Why 50*π?
Here, the Radius is 10 only.
Here, the Radius is 10 only.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers