10. A ROM having a total capacity of 64K bits (K = 1024 = 2^10) is given. If the ROM is known to have 8 outputs, how many address lines are there?

  1. 11
  2. 12
  3. 13
  4. 14
  5. 15

3 is CORRECT. The ROM contains 2^n x m bits of information. Thus using the capacity formula 

    C = 2^n x m, we apply values for C and m to obtain

    64K = 2^n x 8, or 8K = 2^n.
    
    Since K = 2^10 and 8 = 2^3, then

    2^3 * 2^10 = 2^n which implies that
    
    3 + 10 = n
Thus n = 13.

[ Go Back To Question 10 ]

[ Go Back To The Beginning Of Quiz ]