It takes one address line to choose between two modules.
20 address lines are required
The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
In the 2k*16 , the 11 address lines are required and the 16 input-output lines are required..
It takes 23 address lines to address 8 mb of memory.
How many no of address lines required in 1MB memory 11,16,22 or 24 u haven't specified correct options! 20 address lines will be required because 1 MB is 1024 KB that is 1024*1024 Byte which is equivalent to (2^10)^2 bytes if ur memory is Byte addressable then address lines required will be 20.
A 2K X 8 memory requires 11 address lines and 8 data lines
2kb=2*1024=2048 2^11=2048 therefore 11 address lines are required
17 address lines and 8 data lines. 2^17=128k
Firstly we need to convert Mb's into bits i.e 1Mb=1024x1024 = 210x210 =220 That means there are 220 memory locations and we will need 20 address lines.
If you are addressing bytes, then 512K words (16 bit words) requires 20 address lines.I gave that answer because the question was categorized 8086/8088. If you are addressing words, then the answer is 19 address lines.
10. 210 = 1024.
Let N be the number of addresses line 2 megabyte = 2*1024 =2048 N = log (size in bytes) /log 2 N= log 2048/log 2 N=11