PREVIOUS YEAR SOLVED PAPERS - December 2015 Paper 3
- Option : B
- Explanation : The priority preference for 8 X 3 priority encoder will be: 000 First 001 Second 010 Third 011 Fourth 100 Fifth 101 Sixth 110 Seventh 111 Eighth According to question second highest priority vector address will be 10100100. i.e. 1010 0100. When we convert it to hexadecimal then it will be A4. So, option (B) is correct.
- Option : B
- Explanation : In 8085 programming, the result of an operation is stored in the accumulator. So output is 82H.
3. Which of the following 8085 microprocessor hardware interrupt has the lowest priority?
- Option : D
- Explanation : 8085 microprocessor has 5 hardware interrupts. Named TRAP, RST 7.5, RST 6.5, RST 5.5 and INTR. The above order is decreasing in priority. So, option (D) is correct.
- Option : D
- Explanation : Memory cycle time = 250 nsec memory is refreshed 32 times per msec i.e. Number of refreshes in 1 memory cycle (i.e in 250 nsec) = (32 * 250 * 10-9) / 10-3 = 8 * 10-3. Time taken for each refresh = 100 nsec Time taken for 8 * 10-3 refreshes = 8 * 10-3 * 100 * 10-9. = 8 * 10-10 Percentage of the memory cycle time used for refreshing : = (Time taken to refresh in 1 memory cycle / Total time) * 100 = (8 * 10-10 / 250 * 10-9) * 100 = 0.032 * 10 = 0.32 So, option (D) is correct.
- Option : B
- Explanation : DMA controller transfers 32 bit(4 byte) words to memory(cycle stealing mode). Device transmits 4800 character per second (1 character = i byte) So, for 1 byte it will take 1 / 4800 sec. Since the controller transfers 4 byte in cycle stealing mode, it will take 4 * (1 / 4800) = 1 / 1200 sec. i.e. 1200 character will be transfered in cycle stealing mode and it is given that CPU is fetching and executing instructions at an average rate of one million instructions per second. slow down or cycle wasted % in DMA transfer = ( 1200 / 1000000) * 100 = 0.12 % So, option (B) is correct.
