The logic behind NOR gate is that which gives
A. high output when both inputs are high.
B. low output when both inputs are low.
C. high output when both inputs are low.
D. None of these.

Hint: The NOR gate is a digitally approaching language; it is the sum of the bar of the A and B. If one of the terms or both the terms input is 1 then the output will be 0.

Complete step by step solution:
The equation for the NOR gate is,
\[X = \overline {A + B} \]
Here, A and B are the inputs in the NOR gate and X represents the output of the NOR gate,
Assuming each case by substituting the digital values in the inputs A and B.
Case 1: let A is 0 and B is 0, then \[X = \overline {0 + 0} = 1\].
Case 2: let A is 1 and B is 1, then \[X = \overline {1 + 1} = 0\].
Case3: let A be 0 and B is 1, then \[X = \overline {0 + 1} = 0\].
Case4: let A be 1 and B is 0, then \[X = \overline {1 + 0} = 0\].
1 is the higher output for the inputs A and B as 0, it means that NOR gate has higher output with low input.

Therefore, the option is (C) is the correct answer, that the NOR gate has high output when both inputs are low.

Note: Be sure about the output X to the related inputs, in the NOR gate, the output will be high (it means output will be 1) only when the both inputs A and B are 0, the output of X will not be high in no other cases.