isro 2011

Question

Answer 1

Answer is B

CS - code segment & IP - instruction pointer

since the 8086 processor uses 20 bits addressing, we can access 1MB of memory, but registers of 8086 is only 16 bits,so to access the data from the memory we are combining the values present in code segment registers and instruction pointer registers to generate a physical address, it is done by moving the value of CS by 4 bits towards left and then adding it with the value IP

value of CS is given as 1000H (hexa decimal) & value of IP is E000H.

now value of CS after moving 4 bits left is 10000H. Then after adding with IP, value becomes 1E000H which is the required memory address of the next instruction.

Comments

►  All internal registers, as well as internal and external data buses, are 16 bits wide, which firmly   established the "16-bit microprocessor" identity of the 8086.

► A 20-bit external address bus provides a 1MB physical address space 2²⁰  .