💯 The Difference between CD, DVD and Blu-Ray Technologies Explained

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Question 6

What does “DVD” stand for?

(A) Digital Versatile Disc - correct answer
(B) Digital Video Disc
(C) Dense Versatile Disc
(D) Dense Video Disc

While DVDs are often used to store video data in a very dense format, they can be used to store any digital data.

Question 7

What is the sample rate of a typical music CD?

(A) 8 kHz - This is a low-quality sample rate used to send voice messages over the Internet very quickly.
(B) 44.1 kHz - This is the sampling rate for CDs, which allows them to carry 75 minutes’ worth of uncompressed music. - correct answer
(C) 48 kHz - This sampling rate is often used to make computer music recordings.
(D) 192 kHz - This sampling rate is used for high-quality recordings, like those found on DVDs.

Question 8

The installation files of a certain computer program take up 1.8 GB of data. Assume that 1 GB is 1000 MB.

(a) If the program was stored on CDs, how many CDs would be required to store it?

A CD can store 650 MB of data.
1800 / 650 = 2.77, but a computer can’t read 0.77 of a CD.
3 CDs are required.

(b) If the program was stored on DVDs, how many DVDs would be required to store it?

A DVD can store 4.7 GB of data.
1800 / 4700 = 0.38, but a computer can’t read 0.38 of a DVD.
Only 1 DVD is required.

Question 9

(a) A CD contains 650 MB of musical data, which it plays for 75 minutes. How much data per second is played by the CD?

75 minutes is 4500 s.
650 / 4500 = 0.144MB/s

(b) If music is digitised with a sampling rate of 192 kHz, what data would be needed to store 75 minutes if each sample takes up 4 bytes?

Data per second: 192000 x 4 = 768000 bytes/s = 768 kB/s
768 kB/s x 4500 s = 3456000 kB = 3.46 GB

Question 10

On a CD, it takes an area of 3 μm by 0.5 μm to store a single 1 or 0 (1 bit of data). Given that the program area of a CD is 86.05 cm2, and that 8 bits make 1 byte, estimate the maximum amount of data that can be stored on a CD.

To store 1 bit takes 3 x 10^-6 x 0.5 x 10^-6 = 1.5 x ^-12 m2
The area of a CD is 86.05 x (10^-2)^2 = 8.605 x 10^-3 m2
Maximum data: 8.605 x 10^-3 / 1.5 x 10^-12 = 5.74 x 10^9 bits = 717 MB