Question

To store a 3 band, 4-bit, $512\times512$ size image (without header) the number of storage bits required are ..................

• A. 31,45,728
• B. 6,144
• C. 10,48,576
• D. 1,25,82,912

Hint:

Storage (no header, no compression) $=$ \((\text{rows}\times\text{cols}) \times (\text{bands}) \times (\text{bits per band})\).

Answer 1

The Correct Option is A

Pixels $=512\times512=262{,}144$. Each pixel has 3 bands $\times$ 4 bits/band $=12$ bits. Total bits $=262{,}144\times12=3{,}145{,}728$ bits $=$ \(\boxed{31{,}45{,}728}\) (as written with Indian commas). (= \(3{,}145{,}728/8=393{,}216\) bytes $\approx 384$ KB.)