Operating-Systems
October 11, 2023
2017 November NTA UGC NET Paper 1
October 11, 2023
Operating-Systems
October 11, 2023
2017 November NTA UGC NET Paper 1
October 11, 2023

Digital-Logic-Design

Question 14

Consider the circuit given below which has a four bit binary number b3b2b1b0 as input and a five bit binary number d4d3d2d1d0 as output. The circuit implements:

A
Binary of Hex conversion
B
Binary to BCD conversion
C
Binary to grey code conversion
D
Binary to radix-12 conversion
Question 14 Explanation: 
Here ф means 0.
Whenever, b2 = b3 = 1, then only 0100, i.e., 4 is added to the given binary number. Lets write all possibilities for b.

Note that the last 4 combinations leads to b3 and b2 as 1. So, in these combinations only 0010 will be added.
1100 is 12
1101 is 13
1110 is 14
1111 is 15
in binary unsigned number system.
1100 + 0100 = 10000
1101 + 0100 = 10001, and so on.
This is conversion to radix 12.
Correct Answer: D
Question 14 Explanation: 
Here ф means 0.
Whenever, b2 = b3 = 1, then only 0100, i.e., 4 is added to the given binary number. Lets write all possibilities for b.

Note that the last 4 combinations leads to b3 and b2 as 1. So, in these combinations only 0010 will be added.
1100 is 12
1101 is 13
1110 is 14
1111 is 15
in binary unsigned number system.
1100 + 0100 = 10000
1101 + 0100 = 10001, and so on.
This is conversion to radix 12.
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