Answer:
d. 9 bits
Explanation:
We all know that 1 K bytes, =1024 bytes.
And we store program instructions in flash memory and of nonvolatile type. The MCUs happen to be 8 bits, and each of the instructions takes two or one 16-bit words. And the program memory size is known to us through the device name itself, and as an example, the ATmega64x line comes with 64KB of flash, and the ATmega32x line comes with 32 KB. And there is nothing like off-chip program memory, and all of the codes that are being executed by the AVR must be on the on-chip flash. Though, this limitation is never applied to the AT94 FPSLIC AVR/FPGA chips. And you should know that AVR is a family of microcontroller developed by ATEL in 1996. It is a Harvard architecture bases single-chip 8-bit RISC microcontroller. It was one of the first controllers that used flash memory to store the program as compared to one-time programmable EPROM, EEPROM that was being used by various microcontrollers of that time. They are heavily used in embedded systems.
1k byte = 1024 bytes, but if program instruction takes 16 bits.
And hence it only needs 512.
Now 512 = 10^9, and hence you require 9 bits at least.
except you can someway paginate that (like with one-bit pagination, you will be requiring only 8 bits and so on.)
Thus the answer here is 9 bits.
In this exercise we have to calculate the size of a program, so we will find:
The program is 9 bits in size.
Bit is the acronym for Binary Digit, that is, it is the smallest unit of information that can be stored or transmitted. It is commonly used in computing and information theory. A bit can only take on 2 values, such as 0 or 1.
Then performing the conversion we find that:
Now, we find:
[tex]512 = 10^9\\= 9 bits[/tex]
See more about bits at brainly.com/question/2545808
