Computer Science - Computer Fundamentals - Discussion

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?

[A]. 942000 bytes
[B]. 9712478 bytes
[C]. 192300 bytes
[D]. 14384 bytes
[E]. None of the above

Answer: Option A

Explanation:

No answer description available for this question.

Mahen Barfa said: (Jun 28, 2013)  
How to find the result please provide a suitable formula for this question?

Neeraj Pal said: (Aug 14, 2013)  
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.

Ranjan said: (Jun 2, 2014)  
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.

Venkatesan S said: (Sep 11, 2016)  
Where from 8 comes?

Bala said: (Sep 16, 2016)  
To convert into bytes, So that 8 comes.

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