Q43,p3,j14. A byte addressable computer has a memory capacity of 2^m Kbytes and can perform 2^n operations. An instruction involving 3 operands and one operator needs a maximum of

(A) 3m bits

(B) m + n bits

(C) 3m + n bits

(D) 3m + n + 30 bits

Answer D.

Memory capacity  is 2^m Kbytes= 2^m+10bytes.  Whole memory can be represented by m+10 bits.

2ⁿ operations can be represented by n bits.

Now instruction of form

Instruction op1,op2,op3. Will take  3m+30+n bits