Semiconductor Electronics Material Devices and Simple Circuits
151350
The given truth table is for : {|c|c|c|} | { Input } Output\)| |---| \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \(
1 NAND gate
2 AND gate
3 NOR gate
4 OR gate
Explanation:
B In the NAND, the output is only logical 0 when and only when all inputs of the gate are \(1 \mathrm{~s}\), and in all other cases, the output of the NAND gate is high or 1 . \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Hence, the given truth table is for NAND gate.
COMEDK-2015
Semiconductor Electronics Material Devices and Simple Circuits
151357
The given truth table is of {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \(
1 OR gate
2 AND gate
3 NOT gate
4 XOR gate
Explanation:
B Given that, The truth table {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \( Then the Boolean expression of OR gate, \(\mathrm{X}=\mathrm{A}+\mathrm{B}\) Which verifies the above truth table.
BCECE-2014
Semiconductor Electronics Material Devices and Simple Circuits
151358
The following is the truth table for {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(1 0 1\) \(0 1 1\) \(1 1 0\) \(
1 NAND
2 AND
3 XOR
4 NOT
Explanation:
B A NAND gate output as 1 when either of the input signal is low 0 and gives output as 0 only when both the input signal are high i.e. 1 The Boolean expression of NAND gate, \(\mathrm{y}=\overline{\mathrm{A} . \mathrm{B}}\)Hence, the given truth table is for NAND gate.
BCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151365
What is the equivalent expression of the decimal number 212 in binary number system?
1 11000100
2 10010100
3 11010100
4 11010110
Explanation:
C Conversion of decimal number to binary number will be {c|c|c} |2 212 |\)| |---|---| \( 2 106 0\) \( 2 53 0\) \( 2 26 1\) \( 2 13 0\) \( 2 6 1\) \( 2 3 0\) \( 2 1 1\) \( 0 1 \((212)_{10}=(11010100)_2\)
J and K CET-2012
Semiconductor Electronics Material Devices and Simple Circuits
151367
The logic gate which produces a low output only when both its inputs are high is
1 AND gate
2 OR gate
3 NAND gate
4 NOR gate
Explanation:
C A NAND gate is a logic gate which produce a low output only when both its input are high. {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(0 1 1\) \(1 0 1\) \(1 1 0\) \( Truth table of NAND gate Then its Boolean expression is, \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)
Semiconductor Electronics Material Devices and Simple Circuits
151350
The given truth table is for : {|c|c|c|} | { Input } Output\)| |---| \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \(
1 NAND gate
2 AND gate
3 NOR gate
4 OR gate
Explanation:
B In the NAND, the output is only logical 0 when and only when all inputs of the gate are \(1 \mathrm{~s}\), and in all other cases, the output of the NAND gate is high or 1 . \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Hence, the given truth table is for NAND gate.
COMEDK-2015
Semiconductor Electronics Material Devices and Simple Circuits
151357
The given truth table is of {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \(
1 OR gate
2 AND gate
3 NOT gate
4 XOR gate
Explanation:
B Given that, The truth table {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \( Then the Boolean expression of OR gate, \(\mathrm{X}=\mathrm{A}+\mathrm{B}\) Which verifies the above truth table.
BCECE-2014
Semiconductor Electronics Material Devices and Simple Circuits
151358
The following is the truth table for {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(1 0 1\) \(0 1 1\) \(1 1 0\) \(
1 NAND
2 AND
3 XOR
4 NOT
Explanation:
B A NAND gate output as 1 when either of the input signal is low 0 and gives output as 0 only when both the input signal are high i.e. 1 The Boolean expression of NAND gate, \(\mathrm{y}=\overline{\mathrm{A} . \mathrm{B}}\)Hence, the given truth table is for NAND gate.
BCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151365
What is the equivalent expression of the decimal number 212 in binary number system?
1 11000100
2 10010100
3 11010100
4 11010110
Explanation:
C Conversion of decimal number to binary number will be {c|c|c} |2 212 |\)| |---|---| \( 2 106 0\) \( 2 53 0\) \( 2 26 1\) \( 2 13 0\) \( 2 6 1\) \( 2 3 0\) \( 2 1 1\) \( 0 1 \((212)_{10}=(11010100)_2\)
J and K CET-2012
Semiconductor Electronics Material Devices and Simple Circuits
151367
The logic gate which produces a low output only when both its inputs are high is
1 AND gate
2 OR gate
3 NAND gate
4 NOR gate
Explanation:
C A NAND gate is a logic gate which produce a low output only when both its input are high. {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(0 1 1\) \(1 0 1\) \(1 1 0\) \( Truth table of NAND gate Then its Boolean expression is, \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)
Semiconductor Electronics Material Devices and Simple Circuits
151350
The given truth table is for : {|c|c|c|} | { Input } Output\)| |---| \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \(
1 NAND gate
2 AND gate
3 NOR gate
4 OR gate
Explanation:
B In the NAND, the output is only logical 0 when and only when all inputs of the gate are \(1 \mathrm{~s}\), and in all other cases, the output of the NAND gate is high or 1 . \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Hence, the given truth table is for NAND gate.
COMEDK-2015
Semiconductor Electronics Material Devices and Simple Circuits
151357
The given truth table is of {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \(
1 OR gate
2 AND gate
3 NOT gate
4 XOR gate
Explanation:
B Given that, The truth table {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \( Then the Boolean expression of OR gate, \(\mathrm{X}=\mathrm{A}+\mathrm{B}\) Which verifies the above truth table.
BCECE-2014
Semiconductor Electronics Material Devices and Simple Circuits
151358
The following is the truth table for {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(1 0 1\) \(0 1 1\) \(1 1 0\) \(
1 NAND
2 AND
3 XOR
4 NOT
Explanation:
B A NAND gate output as 1 when either of the input signal is low 0 and gives output as 0 only when both the input signal are high i.e. 1 The Boolean expression of NAND gate, \(\mathrm{y}=\overline{\mathrm{A} . \mathrm{B}}\)Hence, the given truth table is for NAND gate.
BCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151365
What is the equivalent expression of the decimal number 212 in binary number system?
1 11000100
2 10010100
3 11010100
4 11010110
Explanation:
C Conversion of decimal number to binary number will be {c|c|c} |2 212 |\)| |---|---| \( 2 106 0\) \( 2 53 0\) \( 2 26 1\) \( 2 13 0\) \( 2 6 1\) \( 2 3 0\) \( 2 1 1\) \( 0 1 \((212)_{10}=(11010100)_2\)
J and K CET-2012
Semiconductor Electronics Material Devices and Simple Circuits
151367
The logic gate which produces a low output only when both its inputs are high is
1 AND gate
2 OR gate
3 NAND gate
4 NOR gate
Explanation:
C A NAND gate is a logic gate which produce a low output only when both its input are high. {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(0 1 1\) \(1 0 1\) \(1 1 0\) \( Truth table of NAND gate Then its Boolean expression is, \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)
Semiconductor Electronics Material Devices and Simple Circuits
151350
The given truth table is for : {|c|c|c|} | { Input } Output\)| |---| \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \(
1 NAND gate
2 AND gate
3 NOR gate
4 OR gate
Explanation:
B In the NAND, the output is only logical 0 when and only when all inputs of the gate are \(1 \mathrm{~s}\), and in all other cases, the output of the NAND gate is high or 1 . \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Hence, the given truth table is for NAND gate.
COMEDK-2015
Semiconductor Electronics Material Devices and Simple Circuits
151357
The given truth table is of {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \(
1 OR gate
2 AND gate
3 NOT gate
4 XOR gate
Explanation:
B Given that, The truth table {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \( Then the Boolean expression of OR gate, \(\mathrm{X}=\mathrm{A}+\mathrm{B}\) Which verifies the above truth table.
BCECE-2014
Semiconductor Electronics Material Devices and Simple Circuits
151358
The following is the truth table for {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(1 0 1\) \(0 1 1\) \(1 1 0\) \(
1 NAND
2 AND
3 XOR
4 NOT
Explanation:
B A NAND gate output as 1 when either of the input signal is low 0 and gives output as 0 only when both the input signal are high i.e. 1 The Boolean expression of NAND gate, \(\mathrm{y}=\overline{\mathrm{A} . \mathrm{B}}\)Hence, the given truth table is for NAND gate.
BCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151365
What is the equivalent expression of the decimal number 212 in binary number system?
1 11000100
2 10010100
3 11010100
4 11010110
Explanation:
C Conversion of decimal number to binary number will be {c|c|c} |2 212 |\)| |---|---| \( 2 106 0\) \( 2 53 0\) \( 2 26 1\) \( 2 13 0\) \( 2 6 1\) \( 2 3 0\) \( 2 1 1\) \( 0 1 \((212)_{10}=(11010100)_2\)
J and K CET-2012
Semiconductor Electronics Material Devices and Simple Circuits
151367
The logic gate which produces a low output only when both its inputs are high is
1 AND gate
2 OR gate
3 NAND gate
4 NOR gate
Explanation:
C A NAND gate is a logic gate which produce a low output only when both its input are high. {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(0 1 1\) \(1 0 1\) \(1 1 0\) \( Truth table of NAND gate Then its Boolean expression is, \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)
Semiconductor Electronics Material Devices and Simple Circuits
151350
The given truth table is for : {|c|c|c|} | { Input } Output\)| |---| \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \( \(\) \(\) \(\)\) \(
1 NAND gate
2 AND gate
3 NOR gate
4 OR gate
Explanation:
B In the NAND, the output is only logical 0 when and only when all inputs of the gate are \(1 \mathrm{~s}\), and in all other cases, the output of the NAND gate is high or 1 . \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)Hence, the given truth table is for NAND gate.
COMEDK-2015
Semiconductor Electronics Material Devices and Simple Circuits
151357
The given truth table is of {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \(
1 OR gate
2 AND gate
3 NOT gate
4 XOR gate
Explanation:
B Given that, The truth table {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 0\) \(0 1 1\) \(1 0 1\) \(1 1 1\) \( Then the Boolean expression of OR gate, \(\mathrm{X}=\mathrm{A}+\mathrm{B}\) Which verifies the above truth table.
BCECE-2014
Semiconductor Electronics Material Devices and Simple Circuits
151358
The following is the truth table for {|l|l|l|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(1 0 1\) \(0 1 1\) \(1 1 0\) \(
1 NAND
2 AND
3 XOR
4 NOT
Explanation:
B A NAND gate output as 1 when either of the input signal is low 0 and gives output as 0 only when both the input signal are high i.e. 1 The Boolean expression of NAND gate, \(\mathrm{y}=\overline{\mathrm{A} . \mathrm{B}}\)Hence, the given truth table is for NAND gate.
BCECE-2013
Semiconductor Electronics Material Devices and Simple Circuits
151365
What is the equivalent expression of the decimal number 212 in binary number system?
1 11000100
2 10010100
3 11010100
4 11010110
Explanation:
C Conversion of decimal number to binary number will be {c|c|c} |2 212 |\)| |---|---| \( 2 106 0\) \( 2 53 0\) \( 2 26 1\) \( 2 13 0\) \( 2 6 1\) \( 2 3 0\) \( 2 1 1\) \( 0 1 \((212)_{10}=(11010100)_2\)
J and K CET-2012
Semiconductor Electronics Material Devices and Simple Circuits
151367
The logic gate which produces a low output only when both its inputs are high is
1 AND gate
2 OR gate
3 NAND gate
4 NOR gate
Explanation:
C A NAND gate is a logic gate which produce a low output only when both its input are high. {|c|c|c|} | \(\) \(\) \(\)\)| |---| \( 0 0 1\) \(0 1 1\) \(1 0 1\) \(1 1 0\) \( Truth table of NAND gate Then its Boolean expression is, \(\mathrm{y}=\overline{\mathrm{A} \cdot \mathrm{B}}\)
