{"id":12385,"date":"2019-12-17T11:56:41","date_gmt":"2019-12-17T06:26:41","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=12385"},"modified":"2021-07-02T15:36:32","modified_gmt":"2021-07-02T10:06:32","slug":"2nd-puc-computer-science-question-bank-chapter-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-computer-science-question-bank-chapter-2\/","title":{"rendered":"2nd PUC Computer Science Question Bank Chapter 2 Boolean Algebra"},"content":{"rendered":"

You can Download Chapter 2 Boolean Algebra Questions and Answers, Notes, 2nd PUC Computer Science Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Computer Science Question Bank Chapter 2 Boolean Algebra<\/h2>\n

2nd PUC Computer Science Boolean Algebra One Mark Questions and Answers<\/h3>\n

Question 1.
\nWhat is another name of Boolean algebra?
\nAnswer:
\nAnother name of Boolean algebra is ‘Switching Algebra’.<\/p>\n

Question 2.
\nWhat do you understand by the term truth value?
\nAnswer:
\nThe result TRUE or FALSE of logical statement are called truth values.<\/p>\n

Question 3.
\nWhat do you understand by the term truth function?
\nAnswer:
\nThe sentences which can be determined to be true or false are called truth function.<\/p>\n

Question 4.
\nWhat do you meant by binary-valued variables?
\nAnswer:
\nThe variables that can store TRUE (1) or FALSE (0) truth values are called binary-valued variables.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nWhat do you understand by logic function?
\nAnswer:
\nThe logic function is a compound statement that consists of a logic statement with logical operators like AND, OR and NOT.<\/p>\n

Question 6.
\nGive examples for logic function.
\nAnswer:
\nThe examples for logic functions are
\nX NOT Y OR Z
\nY AND X OR Z<\/p>\n

Question 7.
\nWhat is meant by tautology and fallacy?
\nAnswer:
\nIf the result of any logical statement or expression is always TRUE is called Tautology and the result of any logical statement is always FALSE is called fallacy.<\/p>\n

Question8.
\nProve the 1+Y is a tautology and 0.Y is a fallacy.
\nAnswer:<\/p>\n\n\n\n\n\n
tautology<\/td>\nfallacy<\/td>\n<\/tr>\n
If Y=0, then 1 + 0=1<\/td>\nIf Y = 1, then 0.1 = 0<\/td>\n<\/tr>\n
If Y = 1, then 1+1 = 1<\/td>\nIf Y = 0, then 0.0=0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 9.
\nName the three logical operators.
\nAnswer:
\nThe three logical operators are AND, OR and NOT.<\/p>\n

Question 10.
\nWhat is a truth table? What is its significance?
\nAnswer:
\nTruth tables are a means of representing the results of a logic function using a table. They are constructed by defining all possible combinations of the inputs to a function and then calculating the output for each combination in turn.<\/p>\n

Question 11.
\nWhat is NOT operator?
\nAnswer:
\nThe “NOT” is simply the opposite or complement of its original value.<\/p>\n

Question 12.
\nWrite Venn diagram for NOT operator.
\nAnswer:
\nNOT – The first term but not the second is present.
\nfruit NOT apples
\n\"2nd<\/p>\n

Question 13.
\nWrite the truth table for NOT operation.
\nAnswer:
\nThe truth table for NOT operation.<\/p>\n\n\n\n\n\n
X<\/td>\nNOT X<\/td>\n<\/tr>\n
0<\/td>\n1<\/td>\n<\/tr>\n
1<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 14.
\nWhat is OR operator?
\nAnswer:
\nThe OR is a disjunction operator and denotes logical addition.<\/p>\n

Question 15.
\nWrite Venn diagram for OR operator?
\nAnswer:
\nOR – Any one of the terms are present (more than one term may be present.
\nfruit OR vegetables
\n\"2nd
\nfruit OR vegetables OR cereal
\n\"2nd<\/p>\n

Question 16.
\nWrite the truth table for OR operator.
\nAnswer:<\/p>\n\n\n\n\n\n\n\n
X<\/td>\nY<\/td>\nOUTPUT X + Y<\/td>\n<\/tr>\n
0<\/td>\n0<\/td>\n0<\/td>\n<\/tr>\n
0<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
1<\/td>\n0<\/td>\n1<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 17.
\nWhat is AND operator?
\nAnswer:
\nThe Logical AND operator is a conjunction operator and denotes logical multiplication.<\/p>\n

Question 18.
\nWrite Venn diagram for AND operator.
\nAnswer:
\nrivers AND salinity.
\n\"2nd<\/p>\n

Question 19.
\nWrite the truth table for AND operation.
\nAnswer:<\/p>\n\n\n\n\n\n\n\n
X<\/td>\nY<\/td>\nOUTPUT
\nX.Y<\/td>\n<\/tr>\n
0<\/td>\n0<\/td>\n0<\/td>\n<\/tr>\n
0<\/td>\n1<\/td>\n0<\/td>\n<\/tr>\n
1<\/td>\n0<\/td>\n0<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 20.
\nState idempotent law.
\nAnswer:
\nWhen a variable combines with itself using OR or AND operator and produce the same variable as output is called idempotent law.
\nFor example, X + X = X, X.X=X<\/p>\n

Question 21.
\nProve idempotent law using truth table.
\nAnswer:
\nIdempotent law using truth table<\/p>\n\n\n\n\n\n
X<\/td>\nX<\/td>\nOUTPUT
\nX. X<\/td>\n		X + X<\/td>\n<\/tr>\n
0<\/td>\n0<\/td>\n0<\/td>\n0<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 22.
\nDraw logic diagram to represent idempotent law.
\nAnswer:
\n\"2nd<\/p>\n

Question 23.
\nState Involution law.
\nAnswer:
\nThis law states that the double complement of a variable gives the same variable.<\/p>\n

Question 24.
\nProve Involution law using truth table.
\nAnswer:<\/p>\n\n\n\n\n\n
Y<\/td>\nY<\/td>\nOUTPUT
\n!(!Y)<\/td>\n<\/tr>\n
0<\/td>\n1<\/td>\n0<\/td>\n<\/tr>\n
1<\/td>\nO<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 25.
\nDraw logic diagram to represent Involution law.
\nAnswer:
\n\"2nd<\/p>\n

Question 26.
\nState Complementarity law.
\nAnswer:
\nThe complementarity law states that a term ANDed with its complement equals 0, and a term ORed with its complement equals 1 (AA’ = 0, A+A’ = 1).<\/p>\n

\"KSEEB<\/p>\n

Question 27.
\nProve Complementarity law using truth table.
\nAnswer:
\n\"2nd<\/p>\n

Question 28.
\nDraw logic diagram to represent Complementarity law.
\nAnswer:
\n\"2nd<\/p>\n

Question 29.
\nState Commutative law.
\nAnswer:
\nThe commutative law states that the order in which terms are written does not affect their value. For example, (AB = BA, A+B = B+A).<\/p>\n

Question 30.
\nProve Communicative law using truth table.
\nAnswer:
\n\"2nd<\/p>\n

Question 31.
\nDraw logic diagram to represent Commutative law.
\nAnswer:
\n\"2nd<\/p>\n

Question 32.
\nState Associative law.
\nAnswer:
\nIt is a simple equality statement i.e., A(BC) = ABC or A+(B+C) = A+B+C.<\/p>\n

Question 33.
\nProve Associative law using truth table.
\nAnswer:
\n\"2nd<\/p>\n

Question 34.
\nDraw logic diagram to represent Associative
\nAnswer:
\n\"2nd<\/p>\n

Question 35.
\nState Distributive law.
\nAnswer:
\n1. a term (A) ANDed with an parenthetical expression (B+C) equals that term ANDed with each term within the parenthesis: A . (B+C) = AB+AC;<\/p>\n

2. a term (A) ORed with a parenthetical expression (B . C) equals that term ORed with each term within the parenthesis: A+(BC) = (A+B) . (A+C).<\/p>\n

Question 36.
\nProve Distributive law using truth table.
\nAnswer:
\n\"2nd<\/p>\n

Question 37.
\nDraw logic diagram to represent Distributive law.
\nAnswer:
\n\"2nd
\nQuestion 38.
\nProve that X+XY = X (Absorption law)
\nAnswer:<\/p>\n\n\n\n\n\n\n\n
X<\/td>\nY<\/td>\nX.Y<\/td>\nX+XY<\/td>\n<\/tr>\n
0<\/td>\n0<\/td>\n0<\/td>\n0<\/td>\n<\/tr>\n
0<\/td>\n1<\/td>\n0<\/td>\n0<\/td>\n<\/tr>\n
1<\/td>\n0<\/td>\n0<\/td>\n1<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 39.
\nProve that X(X+Y) = X (Absorption law).
\nAnswer:<\/p>\n\n\n\n\n\n\n\n
X<\/td>\nY<\/td>\nX + Y<\/td>\nX(X+Y)<\/td>\n<\/tr>\n
0<\/td>\n0<\/td>\n0<\/td>\n0<\/td>\n<\/tr>\n
0<\/td>\n1<\/td>\n1<\/td>\n0<\/td>\n<\/tr>\n
1<\/td>\n0<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 40.
\nDraw logic diagram to represent Absorption law.
\nAnswer:
\n\"2nd<\/p>\n

Question 41.
\nProve that XY + X\\(\\hat{\\mathbf{Y}}\\) =X.
\nAnswer:
\n\"2nd<\/p>\n

Question 42.
\nProve that (X+Y) (X+ \\(\\hat{\\mathbf{Y}}\\)) = X
\nAnswer:
\n\"2nd<\/p>\n

Question 43.
\nProve that A + \\(\\hat{\\mathbf{A}}\\)B = A + B
\nAnswer:
\n\"2nd<\/p>\n

Question 44.
\nWhat is minterm?
\nAnswer:
\nA minterm is a special product of literals, in which each input variable appears exactly once.<\/p>\n

Question 45.
\nFind the minterm of AB + C.
\nAnswer:
\n\"2nd<\/p>\n

Question 46.
\nWhat is a maxterm?
\nAnswer:
\nA maxterm is a sum of literals, in which each input variable appears exactly once.<\/p>\n

Question 47.
\nFind the maxterm for X + \\(\\hat{\\mathbf{Y}}\\) + Z.
\nAnswer:
\n\"2nd<\/p>\n

Question 48.
\nWhat is the canonical form of Boolean expression?
\nAnswer:
\nBoolean expression expressed as sum of minterms or product of max terms are called canonical form.<\/p>\n

\"KSEEB<\/p>\n

Question 49.
\nGive an example for a Boolean expression in the sum of minterms form.
\nAnswer:
\nThe example for boolean expression is the sum of minterms form is
\n\"2nd<\/p>\n

Question 50.
\nGive an example for a Boolean expression in the product of max terms form.
\nAnswer:
\nThe example for boolean expression in the product of max terms form is
\n\"2nd<\/p>\n

2nd PUC Computer Science Boolean Algebra Two Marks Questions and Answers<\/h3>\n

Question 1.
\nProve algebraically that (X+Y) (X+Z) = X + YZ
\nAnswer:
\nLHS: = (X+Y) (X+Z)
\n= XX + XZ + XY + YZ
\n= X + XZ + XY + YZ
\n= X + (1 + Z +Y) YZ
\n= X + YZ
\n= RHS<\/p>\n

Question 2.
\nProve algebraically that X+xY = X + Y
\nAnswer:
\nLHS : = X + \\(\\bar{x}\\) . X + Y
\n= 1 . X + Y
\n= X + Y<\/p>\n

Question 3.
\nUse duality theorem to derive another Boolean relation from: A + \\(\\bar{A}\\) B = A+B
\nAnswer:
\n\"2nd
\nQuestion 4.
\nWhat would be complement of the following:
\nAnswer:
\n\"2nd<\/p>\n

Question 5.
\nWhat are the fundamental products for each of the input words?
\nAnswer:
\nABCD = 0010, ABCD =0110, ABCD = 1110. Write SOP expression.
\nThe SOP expression for ABCD = 0010
\n\\(\\bar{A}\\)\\(\\bar{B}\\) C\\(\\bar{D}\\) = m2<\/sub>
\nThe SOP expression for ABCD =0110,
\n\\(\\bar{A}\\) BC\\(\\bar{D}\\) = m6<\/sub>
\nThe SOP expression for ABCD = 1110
\nABC\\(\\bar{D}\\) = M14<\/sub><\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nA truth table has output 1 for each of these input;
\nABCD = 0011, ABCD =0101, ABCD = 10000, what are the fundamental products and write minterm expression.
\nAnswer:
\nFundamental product of
\nABCD = 0011 is \\(\\bar{A}\\) \\(\\bar{B}\\) C D = m3<\/sub>
\nABCD = 0101 \\(\\bar{A}\\) B \\(\\bar{C}\\) D = m5<\/sub>
\nABCD = 1000 is A \\(\\bar{B}\\) \\(\\bar{C}\\) \\(\\bar{D}\\) = m8<\/sub>
\nminterm expression is
\nf(A,B,C,D) = \u03a3(3, 5, 8)
\nor
\n\"2nd<\/p>\n

Question 7.
\nConstruct a Boolean function of three variables X, Y and Z that has an output 1 when exactly two of X, Y and Z are having values 0, and an output 0 in all other cases.
\nAnswer:
\n\"2nd<\/p>\n

Question 8.
\nConstruct a truth table for three variables A, B and C that will have an output 1 when XYZ = 100, XYZ = 101, XYZ = 110 and XYZ = 111. Write the Boolean expression for logic network in the SOP form.
\nAnswer:
\n\"2nd<\/p>\n

Question 9.
\nConvert the following expressions to canonical Product-of-Sum form:
\na) (A + C) (C + D)
\nb) A(B + C) (\\(\\bar{C}\\) + \\(\\bar{D}\\))
\nc) (X + Y) (Y + Z) (X + Z)
\nAnswer:
\nCanonical product of sum of
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

Question 10.
\nConvert the following expressions to canonical Sum-of-Product form:
\n\"2nd
\nAnswer:
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

Question 11.
\nDraw Karnaugh maps for the following expressions:
\n\"2nd
\n\"2nd
\nAnswer:
\nKarnaugh maps for \\(\\bar{x} \\bar{y}+\\bar{x} y\\)
\n\"2nd
\n\"2nd<\/p>\n

Question 12.
\nDraw a general K-map for four variables A, B, C, and D.
\nAnswer:
\n\"2nd<\/p>\n

Question 13.
\nGiven the expression in four variables, draw the k-map for the function:<\/p>\n

    \n
  1. m2<\/sub> + m3<\/sub> + m7<\/sub> + m9<\/sub> + m11<\/sub> + m13<\/sub><\/li>\n
  2. m0<\/sub> + m2<\/sub> + m4<\/sub> + m8<\/sub> + m9<\/sub> + m10<\/sub> +m11<\/sub>+ m12<\/sub> + m13<\/sub><\/li>\n<\/ol>\n

    Answer:
    \n1. m2<\/sub> + m3<\/sub> + m7<\/sub> + m9<\/sub> + m11<\/sub> + m13<\/sub>
    \n\"2nd
    \n2. m0<\/sub> + m2<\/sub> + m4<\/sub> + m8<\/sub> + m9<\/sub> + m10<\/sub> + m11<\/sub> + m12<\/sub> + m13<\/sub>
    \n\"2nd<\/p>\n

    Question 14.
    \nDraw the k-map for the function in three variables given below.<\/p>\n

      \n
    1. m0<\/sub> + m2<\/sub> + m4<\/sub> + m6<\/sub> + m7<\/sub><\/li>\n
    2. m1<\/sub> + m2<\/sub> + m3<\/sub> + m5 <\/sub> + m7<\/sub><\/li>\n<\/ol>\n

      Answer:
      \n1. m0<\/sub> + m2<\/sub> + m4<\/sub> + m6<\/sub> + m7<\/sub>
      \n\"2nd
      \n2. m1<\/sub> + m2<\/sub> + m3<\/sub> + m5 <\/sub> + m7<\/sub>
      \n\"2nd<\/p>\n

      Question 15.
      \nWrite SOP expression corresponding to the function F in the following truth table and draw the logic diagram (use OR and AND gates)
      \nAnswer:
      \n\"2nd<\/p>\n

      SOP Expression.
      \n\"2nd
      \n\"2nd
      \n\"2nd<\/p>\n

      2nd PUC Computer Science Boolean Algebra Three Marks Questions and Answers<\/h3>\n

      Question 1.
      \nState and prove any three theorems of boolean algebra.
      \nAnswer:
      \n1. Idempotance law:
      \nThis law states that when a variable is combined with itself using OR or AND operator, the output is the same variable.
      \n\"2nd<\/p>\n

      2. Involution Law:
      \nThe complement of a variable is complemented again then we get the same variable.
      \nX” = X
      \nIf X =0, then O’ = 1′ = 0
      \nIf X=1, then 1′ = 0′ = 1.<\/p>\n

      \"KSEEB<\/p>\n

      3. Complementary Law:
      \nThe variable is combined with its complement and ORed and ANDed. If ORed, it always gives True (Tautology). If ANDed, then it gives False as the result always.
      \n\"2nd<\/p>\n

      Question 2.
      \nState and prove associative law of addition and multiplication. This law states that<\/p>\n

        \n
      1. A + (B + C) = (A + B) + C<\/li>\n
      2. A. (B .C) = (A .B)<\/li>\n<\/ol>\n

        Proof:
        \n\"2nd
        \n\"2nd<\/p>\n

        Question 3.
        \nState and prove De Morgan’s theorems by the method of perfect induction.
        \nAnswer:
        \n1. De Morgan’s First Theorem:
        \nWhen the OR sum of two variables is inverted, this is the same as inverting each variable individually and then ANDing these inverted variables.
        \n\"2nd<\/p>\n

        2. De Morgan’s Second Theorem:
        \nWhen the AND product of two variables is inverted, this is the same as inverting each variable individually and then ORing them.
        \n\"2nd
        \n\"2nd<\/p>\n

        Question 4.
        \nObtain the minterm expression for the Boolean function F = A+B’C. the minterm expression for the Boolean function F = A+B’C is
        \nAnswer:
        \n= A + B’C …….(1)
        \nThe first term A is missing two variables B and C, therefore
        \nA = A (B + B’) = AB + AB’ = AB (C+C’) + AB’ (C+C’)
        \n= ABC + ABC’ + AB’C + AB’C’ …….(2)
        \nThe second term B’C is missing variable A, therefore
        \nB’C = B’C (A + A’)
        \n= AB’C + A’B’C …….(3)
        \nNow, Substitute (2) and (3) in (1) we get
        \nABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C.<\/p>\n

        Question 5.
        \nExplain with an example how to express a Boolean function in its sum of product form.
        \nAnswer:
        \nThe logical sum of two or more logical product terms is known as sum of products expression. SOP is an ORing of ofANDedvariables. The boolean expression containing all the input variables either in complemented or un complemented form in each of the product term is known as canonical SOP expression and each term is called minterm.<\/p>\n

        For example, express the product of sum from the boolean function f(x,y). For example, express the product of sum from the boolean function F(X, Y) and the truth table for which is given below:
        \n\"2nd
        \nNow by adding minterm for the output 1’s, we get the desired sum of product expression which is X’Y’ + XY.<\/p>\n

        \"KSEEB<\/p>\n

        Question 6.
        \nExplain with an example how to express a Boolean function in its product of sum form.
        \nAnswer:
        \nThe logical product of two or more logical sum terms is known as a product of sums expression. POS is an ANDing of ORedvariables. The boolean expression containing all the input variables either in complemented or un complemented form in each of the sum term is known as a canonical POS expression and each term is called maxterm. For example, express the product of sum from the boolean function F(X, Y) and the truth table for which is given below:<\/p>\n\n\n\n\n\n\n\n
        X<\/td>\nY<\/td>\nF<\/td>\nMaxterm<\/td>\n<\/tr>\n
        0<\/td>\n0<\/td>\n1<\/td>\nX+Y<\/td>\n<\/tr>\n
        0<\/td>\n1<\/td>\n0<\/td>\nX+Y’<\/td>\n<\/tr>\n
        1<\/td>\n0<\/td>\n0<\/td>\nX’+Y<\/td>\n<\/tr>\n
        1<\/td>\n1<\/td>\n1<\/td>\nX’+Y’<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Now by multiplying max terms for the output 0’s, we get the desired product of sums expression which is (X+Y1) (X’+Y).<\/p>\n

        Question 7.
        \nConstruct a truth table f r minterms and max terms for three variables and designate the terms.
        \nAnswer:
        \nThe truth table for minterms for three variables
        \n\"2nd
        \nThe truth table for max terms for three variables
        \n\"2nd<\/p>\n

        Question 8.
        \nUsing basic gates, construct a logic circuit for the Boolean expression (X’+Y) (X+Z) (Y+Z)
        \nAnswer:
        \nLogic circuit for the Boolean expression (\\(\\bar{x} \\)+Y) (X + Z) (Y + Z)
        \n\"2nd<\/p>\n

        Question 9.
        \nSimplify the following Boolean expressions and draw logic circuit diagrams of the simplified expressions using only NAND gates.
        \nAnswer:
        \nLogic circuit diagram of the simplified expressions using only NAND gates.
        \n\"2nd
        \n\"2nd
        \n\"2nd
        \n\"2nd<\/p>\n

        Question 10.
        \nFor a four-variable map in w,x,y, and z draw the sub cubes for<\/p>\n

          \n
        1. w x\\(\\bar{Y} \\)<\/li>\n
        2. w x<\/li>\n
        3. x y \\(\\bar{Z} \\) and<\/li>\n
        4. y<\/li>\n<\/ol>\n

          Answer:
          \nFor a four-variable mao in w,x,y and z, the sub cubes for<\/p>\n

            \n
          1. wx\\(\\bar{Y} \\)<\/li>\n
          2. w x<\/li>\n
          3. x y\\(\\bar{Z} \\) and<\/li>\n
          4. y are as follows<\/li>\n<\/ol>\n

            \"2nd<\/p>\n

            Question 11.
            \nConvert the following product-of-sums form into its corresponding sum-of-products form using truth table. F(x,y,z) = \u03c0(2,4,6,7)
            \nAnswer:
            \nPOS form conversion into its SOP form using truth table is F(x, y, z) = \u03a3(2, 4, 6, 7)
            \n\"2nd
            \n\"2nd<\/p>\n

            Question 12.<\/p>\n

              \n
            1. Reduce the following Boolean expression to the simplest form: A. [B+C.(AB + AC)<\/li>\n
            2. Given : F(X,Y,Z) – E(l,3,7) then prove that F'(x,y,z) = \u03c0(0,2,4,5,6)<\/li>\n<\/ol>\n

              Answer:
              \n1.
              \n\"2nd<\/p>\n

              2. Given F(x, y, z) = \u03a3(1, 3, 7)
              \n\"2ndFrom the above, the minterm list is
              \nF(x, y, z) = \u03a3(1, 3, 7)
              \nand the maxterm list is F'(x, y, z) = \u03c0(0, 2, 4, 5, 6)<\/p>\n

              2nd PUC Computer Science Boolean Algebra Five Marks Questions and Answers<\/h3>\n

              Question 1.
              \nUsing maps, simplify the following expressions in four variables W, X, Y, and Z.<\/p>\n

                \n
              1. m1<\/sub> + m3<\/sub> + m5<\/sub> + m6<\/sub> + m7<\/sub> + m9<\/sub> + m11<\/sub> + m13<\/sub><\/li>\n
              2. m0<\/sub> + m2<\/sub> + m4<\/sub> + m8<\/sub> + m9<\/sub> + m10<\/sub> + m11<\/sub> + m12<\/sub> + m13<\/sub>.<\/li>\n<\/ol>\n

                Answer:
                \n1. m1<\/sub> + m3<\/sub> + m5<\/sub> + m6<\/sub> + m7<\/sub> + m9<\/sub> + m11<\/sub> + m13<\/sub>
                \n\"2nd
                \n\"2nd
                \n2. m0<\/sub> +m2<\/sub> +m4<\/sub> +m8<\/sub> +m9<\/sub> +m10<\/sub> + m11<\/sub> + m12<\/sub> +m13<\/sub>
                \n\"2nd<\/p>\n

                Question 2.
                \nUsing maps, simplify the following expressions in four variables W, X, Y, and Z.<\/p>\n

                  \n
                1. m1<\/sub> + m3<\/sub> + m5<\/sub> + m6<\/sub> + m7<\/sub> + m9<\/sub> + m11<\/sub> + m13<\/sub><\/li>\n
                2. m0<\/sub> + m2<\/sub> + m4<\/sub> + m8<\/sub> + m9<\/sub> + m10<\/sub> + m11<\/sub> + m12<\/sub> + m13<\/sub>.<\/li>\n<\/ol>\n

                  Answer:
                  \n1. m1<\/sub> + m3<\/sub> + m5<\/sub> + m6<\/sub> + m7<\/sub> + m9<\/sub> + m11<\/sub> + m13<\/sub>
                  \n\"2nd
                  \n2. m0<\/sub> + m2<\/sub> + m4<\/sub> + m8<\/sub> + m9<\/sub> + m10<\/sub> + m11<\/sub> + m12<\/sub> + m13<\/sub>.
                  \n\"2nd
                  \nQuestion 3.
                  \nFor the Boolean function F and F’ in the truth table, find the following:
                  \n(a) List the minterms of the functions F and F’.
                  \n(b) Express F and F’ in the sum of minterms in algebraic form.
                  \n(c) Simplify the functions to an expression with a minimum number of literals.
                  \nAnswer:
                  \n3. a) Minterms of the functions F
                  \n\"2nd<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 4.
                  \nState and prove De Morgan’s theorems algebraically.
                  \nAnswer:
                  \n1. De Morgan’s First Theorem:
                  \nWhen the OR sum of two variables is inverted, this is the same as inverting each variable individually and then ANDing these inverted variables. This is written in the born of
                  \nBoolean expression as \\(\\overline{x+y}=\\bar{x} \\cdot y\\)
                  \nProof:
                  \n\"2nd<\/p>\n

                  2. De Morgan’s 2nd Theorem
                  \nWhen the AND product of two variables is inverted, this is the same as inverting each variable individually and then ORing them \\(\\overline{x. y}=\\bar{x}+\\bar{y}\\)
                  \nTo prove, we again use complementarity law
                  \n\"2nd<\/p>\n

                  Question 5.
                  \nFind the complement of F = X + YZ, then show that F.F’ = 0 and F + F’ = 1.
                  \nAnswer:
                  \nF = X + YZ
                  \n\"2nd<\/p>\n

                  Question 6.<\/p>\n

                    \n
                  1. State the two Absorption laws of Boolean algebra. Verify using the truth table.<\/li>\n
                  2. Simplify using the laws of Boolean algebra. At each step state clearly the law used for simplification. F = x.y + x.z + x.y.z.<\/li>\n<\/ol>\n

                    Answer:
                    \n1. The two Absorption law of Boolean algebra are<\/p>\n

                      \n
                    • A +AB = A<\/li>\n
                    • A (A + B) = A<\/li>\n<\/ul>\n

                      Verification using truth table
                      \n\"2nd<\/p>\n

                      2. Simplification using Boolean algebra.
                      \nF = x.y+ x.z+ x.y.z
                      \n= x(1 + y + z + yz)
                      \n= x(1) – using property of additon 1 + x = 1
                      \n= x – property of multiplication x.1 = x<\/p>\n

                      Question 7.
                      \nGiven the Boolean function F(x, y, z) = \u03a3 (0, 2, 4, 5, 6). Reduce it using the Karnuagh map method.
                      \nAnswer:
                      \nF (x, y, z) = \u03a3 (0, 2, 4, 5, 6)
                      \n\"2nd<\/p>\n

                      Question 8.
                      \n(a) State the two complement properties of Boolean algebra. Verify using the truth tables.
                      \n(b) \\(x \\cdot(\\overline{y z}+y z)\\)
                      \nAnswer:
                      \na) The two complement properties of Boolean algebra are
                      \n\"2nd
                      \n\"2nd<\/p>\n

                      Question 9.
                      \nGiven the Boolean function F{A, B, C, D) = \u03a3(5, 6, 7, 8, 9,10,14). Use Karnaugh’s map to reduce the function F using SOP form. Write a logic gate diagram for the reduced SOP expression.
                      \nAnswer:
                      \nF(A, B, C, D) = \u03a3(5, 6, 7, 8, 9,10,14)
                      \n\"2nd
                      \n\"2nd<\/p>\n

                      Question 10.
                      \nGiven: F(A, B, C, D) = (0, 2,4, 6,8,10,14). Use Karnaugh map to reduce the function F using POS form. Write a logic gate diagram for the reduced POS expression.
                      \nAnswer:
                      \nGiven F(A, B, C, D) = (0, 2, 4, 6, 8,10,14)
                      \nk-map using POS form
                      \n\"2nd
                      \n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

                      You can Download Chapter 2 Boolean Algebra Questions and Answers, Notes, 2nd PUC Computer Science Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Computer Science Question Bank Chapter 2 Boolean Algebra 2nd PUC Computer Science Boolean Algebra One Mark …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12385"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=12385"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12385\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=12385"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=12385"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=12385"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}