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Best answer
44 votes
44 votes
Option $(C)$. $24$ bits = $3$ bytes instructions. So, PC will have multiples of $3$ in it.
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13 votes
13 votes

C) is the ans

size of instruction  =  24/8 = 3 bytes.

Program Counter can shift 3 bytes at a time to jump to next instruction.

So the given options must be divisible by 3. only 600 is satisfied.

2 votes
2 votes

As given in the question, the instruction is 24 bits, so if you convert the instruction length in terms of bytes it will be 3 Bytes because 8 bits = 1 Byte, So 24 Bits = 24/8 => 3 Bytes.

Given that the starting address of the program in the memory is 300, initially program counter (PC) will contain the value of 300.

300-----302                                                                                                                                                                                      

Starting Address Ending Address
300 302
303 305
306 308
309 311
......... ........
......... ........
......... ........
597 599
600 602

The valid value of Program counter will be 600.

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2 votes
2 votes
Here, size of instruction = 24/8 = 3 bytes. Program Counter can shift 3 bytes at a time to jump to next instruction. So the given options must be divisible by 3. only 600 is satisfied.
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