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Q. No.
If \(X\) and \(Y\) are the inputs, the given circuit works as: 
 
1. \(\mathrm{OR}\) gate
2. \(\mathrm{AND}\) gate
3. \(\mathrm{NAND}\) gate
4. \(\mathrm{NOR}\) gate
Subtopic:  Logic gates |
Level 3: 35%-60%
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Refer to the logic circuit given below. For two inputs \((A=1, B=1) \text { and }(A=0, B=1) \text {,}\) output \((Y)\) will be:
     
1. \(1,0\) respectively
2. \(0,1\) respectively
3. \(0,0\) respectively
4. \(1,1\) respectively
Subtopic:  Logic gates |
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Level 2: 60%+
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The output \(Y\) for the given input \(A\) and \(B\) to the circuit is:
1. 2.
3. 4.
Subtopic:  Logic gates |
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Level 3: 35%-60%
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Two \(4\) bits binary numbers, \(A=1101\) and \(B=1010\) are given in the inputs of a logic circuit shown in figure below. The output (\(Y\)) will be:

1. \(Y=1101\)
2. \(Y=0010\)
3. \(Y=0111\)
4. \(Y=1000\)
Subtopic:  Logic gates |
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The given circuit works as:
         
1. \(\mathrm{AND}~\) gate 
2. \(\mathrm{NOR}~\) gate
3. \(\mathrm{NAND}~\) gate 
4. \(\mathrm{OR}~\) gate
Subtopic:  Logic gates |
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Find the correct combination of \(A, B, C~\) and \(D~\) inputs which can cause the LED to glow.
            
1. \(0100\)
2. \(0011\)
3. \(1000\)
4. \(1101\)
Subtopic:  Logic gates |
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The correct truth table for the given input data of the following logic gate is:
           
1.  
Input Output
\(A\) \(B\) \(C\) \(D\) \(Y\)
\(1\) \(1\) \(0\) \(1\) \(1\)
\(0\) \(0\) \(1\) \(1\) \(0\)
\(1\) \(0\) \(1\) \(0\) \(1\)
\(1\) \(1\) \(1\) \(1\) \(0\)

2. 
Input Output
\(A\) \(B\) \(C\) \(D\) \(Y\)
\(1\) \(1\) \(0\) \(1\) \(0\)
\(0\) \(0\) \(1\) \(1\) \(0\)
\(1\) \(0\) \(1\) \(0\) \(1\)
\(1\) \(1\) \(1\) \(1\) \(1\)

3. 
Input Output
\(A\) \(B\) \(C\) \(D\) \(Y\)
\(1\) \(1\) \(0\) \(1\) \(1\)
\(0\) \(0\) \(1\) \(1\) \(0\)
\(1\) \(0\) \(1\) \(0\) \(0\)
\(1\) \(1\) \(1\) \(1\) \(1\)

4. 
Input Output
\(A\) \(B\) \(C\) \(D\) \(Y\)
\(1\) \(1\) \(0\) \(1\) \(0\)
\(0\) \(0\) \(1\) \(1\) \(1\)
\(1\) \(0\) \(1\) \(0\) \(1\)
\(1\) \(1\) \(1\) \(1\) \(1\)
Subtopic:  Logic gates |
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Level 2: 60%+
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For the given logic gate circuit, which of the following is the correct truth table?
1. \(\begin{array}{c|c|c} {n} &{~m} & {z} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{array}\) 2. \(\begin{array}{c|c|c} {n} & {~m} & {z} \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\)
3. \(\begin{array}{c|c|c} {n} & {~m} & {z} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{array}\) 4. \(\begin{array}{c|c|c} {n} &{~m} & {z} \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{array}\)

 
Subtopic:  Logic gates |
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Level 1: 80%+
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Identify the correct truth table of the given logical circuit.
            
1.  
\(A\) \(B\) \(Y\)
\(0\) \(0\) \(0\)
\(0\) \(1\) \(1\)
\(1\) \(0\) \(1\)
\(1\) \(1\) \(0\)
2.  
\(A\) \(B\) \(Y\)
\(0\) \(0\) \(1\)
\(0\) \(1\) \(0\)
\(1\) \(0\) \(1\)
\(1\) \(1\) \(0\)
3.  
\(A\) \(B\) \(Y\)
\(0\) \(0\) \(1\)
\(0\) \(1\) \(1\)
\(1\) \(0\) \(1\)
\(1\) \(1\) \(0\)
4.  
\(A\) \(B\) \(Y\)
\(0\) \(0\) \(0\)
\(0\) \(1\) \(0\)
\(1\) \(0\) \(1\)
\(1\) \(1\) \(0\)
 
Subtopic:  Logic gates |
 82%
Level 1: 80%+
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The Boolean expression \({Y}=A \bar{B} C+\bar{A} \bar{C} \) can be realised with which of the following gate configurations.
\(\mathrm{A.}\) One \(3\text-\)input AND gate, \(3\text-\) NOT gates and one \(2\text-\)input OR gate, One \(2\text-\)input AND gate
\(\mathrm{B.}\) One \(3\text-\)input AND gate, \(1\) NOT gate, One \(2\text-\)input NOR gate and one \(2\text-\)input OR gate
\(\mathrm{C.}\) \(3\text-\)input OR gate, \(3\) NOT gates and one \(2\text-\)input AND gate
Choose the correct answer from the options given below:
1. \(\mathrm{A,C~\text{Only}}\) 2. \(\mathrm{A,B,C~\text{Only}}\)
3. \(\mathrm{B,C~\text{Only}}\) 4. \(\mathrm{A,B~\text{Only}}\)
Subtopic:  Logic gates |
Level 3: 35%-60%
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