# 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

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.