Physics-
General
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Question

To double the volume of a given mass of an ideal gas at 27 ℃ keeping the pressure constant, one must raise the temperature in degree centigrade to

  1. 54 to the power of o end exponent    
  2. 270 to the power of o end exponent    
  3. 327 to the power of o end exponent    
  4. 600 to the power of o end exponent    

The correct answer is: 327 to the power of o end exponent


    V proportional to T rightwards double arrow fraction numerator V subscript 1 end subscript over denominator V subscript 2 end subscript end fraction equals fraction numerator T subscript 1 end subscript over denominator T subscript 2 end subscript end fraction rightwards double arrow fraction numerator V over denominator 2 V end fraction equals fraction numerator open parentheses 273 plus 27 close parentheses over denominator T subscript 2 end subscript end fraction equals fraction numerator 300 over denominator T subscript 2 end subscript end fraction
    rightwards double arrow T subscript 2 end subscript equals 600 K equals 327 ℃

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    The function f left parenthesis x right parenthesis equals sin invisible function application fraction numerator pi x over denominator 2 end fraction plus 2 cos invisible function application fraction numerator pi x over denominator 3 end fraction minus tan invisible function application fraction numerator pi x over denominator 4 end fraction is period with period

    Period of sin invisible function application fraction numerator pi x over denominator 2 end fraction equals fraction numerator 2 pi over denominator pi divided by 2 end fraction equals 4
    Period of cos invisible function application fraction numerator pi x over denominator 3 end fraction equals fraction numerator 2 pi over denominator pi divided by 3 end fraction equals 6
    Period of tan invisible function application fraction numerator pi x over denominator 4 end fraction equals fraction numerator pi over denominator pi divided by 4 end fraction equals 4
    thereforePeriod of f left parenthesis x right parenthesis equals L.C.M. of (4, 6, 4) = 12

    The function f left parenthesis x right parenthesis equals sin invisible function application fraction numerator pi x over denominator 2 end fraction plus 2 cos invisible function application fraction numerator pi x over denominator 3 end fraction minus tan invisible function application fraction numerator pi x over denominator 4 end fraction is period with period

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    Period of sin invisible function application fraction numerator pi x over denominator 2 end fraction equals fraction numerator 2 pi over denominator pi divided by 2 end fraction equals 4
    Period of cos invisible function application fraction numerator pi x over denominator 3 end fraction equals fraction numerator 2 pi over denominator pi divided by 3 end fraction equals 6
    Period of tan invisible function application fraction numerator pi x over denominator 4 end fraction equals fraction numerator pi over denominator pi divided by 4 end fraction equals 4
    thereforePeriod of f left parenthesis x right parenthesis equals L.C.M. of (4, 6, 4) = 12
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    The value of k, for which left parenthesis cos invisible function application x plus sin invisible function application x right parenthesis to the power of 2 end exponent plus k sin invisible function application x cos invisible function application x minus 1 equals 0 is an identity, is

    Given, left parenthesis cos invisible function application x plus sin invisible function application x right parenthesis to the power of 2 end exponent plus k sin invisible function application x cos invisible function application x minus 1 equals 0 comma for all x
    Þ cos to the power of 2 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application x plus 2 cos invisible function application x sin invisible function application x plus k sin invisible function application x cos invisible function application x minus 1 equals 0 comma for all x
    Þ left parenthesis k plus 2 right parenthesis cos invisible function application x sin invisible function application x equals 0 comma for all x Þ k plus 2 equals 0 Þ k equals negative 2

    The value of k, for which left parenthesis cos invisible function application x plus sin invisible function application x right parenthesis to the power of 2 end exponent plus k sin invisible function application x cos invisible function application x minus 1 equals 0 is an identity, is

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    Þ cos to the power of 2 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application x plus 2 cos invisible function application x sin invisible function application x plus k sin invisible function application x cos invisible function application x minus 1 equals 0 comma for all x
    Þ left parenthesis k plus 2 right parenthesis cos invisible function application x sin invisible function application x equals 0 comma for all x Þ k plus 2 equals 0 Þ k equals negative 2
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    If A plus B plus C equals 27 0 to the power of o end exponent comma then cos invisible function application 2 A plus cos invisible function application 2 B plus cos invisible function application 2 C plus 4 space sin invisible function application A space sin invisible function application B space sin invisible function application C equals

    A plus B plus C equals 27 0 to the power of o end exponent rightwards double arrow A equals B equals C equals 9 0 to the power of o end exponent comma then
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    equals negative 1 minus 1 minus 1 plus 4.1.1.1 equals negative 3 plus 4 equals 1.

    If A plus B plus C equals 27 0 to the power of o end exponent comma then cos invisible function application 2 A plus cos invisible function application 2 B plus cos invisible function application 2 C plus 4 space sin invisible function application A space sin invisible function application B space sin invisible function application C equals

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    A plus B plus C equals 27 0 to the power of o end exponent rightwards double arrow A equals B equals C equals 9 0 to the power of o end exponent comma then
    cos invisible function application 2 A plus cos invisible function application 2 B plus cos invisible function application 2 C plus 4 sin invisible function application A sin invisible function application B sin invisible function application C
    equals cos invisible function application 1 8 0 to the power of o end exponent plus cos invisible function application 1 8 0 to the power of o end exponent plus cos invisible function application 1 8 0 to the power of o end exponent plus 4 sin invisible function application 9 0 to the power of o end exponent sin invisible function application 9 0 to the power of o end exponent sin invisible function application 9 0 to the power of o end exponent
    equals negative 1 minus 1 minus 1 plus 4.1.1.1 equals negative 3 plus 4 equals 1.
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    Which one of the following is not true

    It is a property of matrix multiplication.

    Which one of the following is not true

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    It is a property of matrix multiplication.
    General
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    If A equals open square brackets table row 0 1 row 1 0 end table close square brackets comma B equals open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets then left parenthesis A plus B right parenthesis to the power of 2 end exponentequals

    A B equals open square brackets table row 0 1 row 1 0 end table close square brackets open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets equals open square brackets table row i 0 row 0 cell negative i end cell end table close square brackets
    and B A equals open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets open square brackets table row 0 1 row 1 0 end table close square brackets equals open square brackets table row cell negative i end cell 0 row 0 i end table close square brackets equals negative A B
    therefore A B plus B A equals O
    Hence, left parenthesis A plus B right parenthesis to the power of 2 end exponent equals A to the power of 2 end exponent plus B to the power of 2 end exponent.

    If A equals open square brackets table row 0 1 row 1 0 end table close square brackets comma B equals open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets then left parenthesis A plus B right parenthesis to the power of 2 end exponentequals

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    A B equals open square brackets table row 0 1 row 1 0 end table close square brackets open square brackets table row 0 cell negative i end cell row i 0 end table close square brackets equals open square brackets table row i 0 row 0 cell negative i end cell end table close square brackets
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    therefore A B plus B A equals O
    Hence, left parenthesis A plus B right parenthesis to the power of 2 end exponent equals A to the power of 2 end exponent plus B to the power of 2 end exponent.
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    A to the power of 2 end exponent equals open square brackets table row 8 5 row cell negative 5 end cell 3 end table close square brackets.

    If A equals open square brackets table row 3 1 row cell negative 1 end cell 2 end table close square brackets, then A to the power of 2 end exponent equals

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    A equals open square brackets table row 3 1 row cell negative 1 end cell 2 end table close square brackets
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    If A equals open square brackets table row 0 1 row 0 0 end table close square bracketsand A B equals O, then B =

    Since open square brackets table row 0 1 row 0 0 end table close square brackets open square brackets table row cell negative 1 end cell 0 row 0 0 end table close square brackets equals open square brackets table row 0 0 row 0 0 end table close square brackets equals O equals A B
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    If A equals open square brackets table row 0 1 row 0 0 end table close square bracketsand A B equals O, then B =

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    If A equals open square brackets table row cell 1 divided by 3 end cell 2 row 0 cell 2 x minus 3 end cell end table close square brackets comma B equals open square brackets table row 3 6 row 0 cell negative 1 end cell end table close square bracketsand A B equals I, then x =

    vertical line a d j A vertical line
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    If A equals open square brackets table row cell 1 divided by 3 end cell 2 row 0 cell 2 x minus 3 end cell end table close square brackets comma B equals open square brackets table row 3 6 row 0 cell negative 1 end cell end table close square bracketsand A B equals I, then x =

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    vertical line a d j A vertical line
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    If A equals open square brackets table row 1 0 1 row 0 1 1 row 1 0 0 end table close square brackets, then A is

    capital delta equals open square brackets table row 1 0 1 row 0 1 1 row 1 0 0 end table close square brackets equals negative 1 not equal to 0, hence matrix is non-singular.

    If A equals open square brackets table row 1 0 1 row 0 1 1 row 1 0 0 end table close square brackets, then A is

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    capital delta equals open square brackets table row 1 0 1 row 0 1 1 row 1 0 0 end table close square brackets equals negative 1 not equal to 0, hence matrix is non-singular.
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    The values of x comma y comma zin order of the system of equations 3 x plus y plus 2 z equals 3 comma 2 x minus 3 y minus z equals negative 3, x plus 2 y plus z equals 4 comma are

    Put the value left parenthesis x comma y comma z right parenthesis equals left parenthesis 1 , 2 comma negative 1 right parenthesis comma which satisfies the equation. Hence, (d) is correct.

    The values of x comma y comma zin order of the system of equations 3 x plus y plus 2 z equals 3 comma 2 x minus 3 y minus z equals negative 3, x plus 2 y plus z equals 4 comma are

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    Put the value left parenthesis x comma y comma z right parenthesis equals left parenthesis 1 , 2 comma negative 1 right parenthesis comma which satisfies the equation. Hence, (d) is correct.
    General
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    If x is a positive integer, then capital delta equals open vertical bar table row cell x factorial end cell cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell row cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell row cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell cell left parenthesis x plus 4 right parenthesis factorial end cell end table close vertical bar is equal to

    capital delta equals open vertical bar table row cell x factorial end cell cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell row cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell row cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell cell left parenthesis x plus 4 right parenthesis factorial end cell end table close vertical bar
    = x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial open vertical bar table row 1 cell left parenthesis x plus 1 right parenthesis end cell cell left parenthesis x plus 2 right parenthesis left parenthesis x plus 1 right parenthesis end cell row 1 cell left parenthesis x plus 2 right parenthesis end cell cell left parenthesis x plus 3 right parenthesis left parenthesis x plus 2 right parenthesis end cell row 1 cell left parenthesis x plus 3 right parenthesis end cell cell left parenthesis x plus 4 right parenthesis left parenthesis x plus 3 right parenthesis end cell end table close vertical bar
    Applying R subscript 1 end subscript rightwards arrow R subscript 2 end subscript minus R subscript 1 end subscript comma R subscript 2 end subscript rightwards arrow left parenthesis R subscript 3 end subscript minus R subscript 2 end subscript right parenthesis we get
    equals x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial open vertical bar table row 0 1 cell 2 left parenthesis x plus 2 right parenthesis end cell row 0 1 cell 2 left parenthesis x plus 3 right parenthesis end cell row 1 cell left parenthesis x plus 3 right parenthesis end cell cell left parenthesis x plus 4 right parenthesis left parenthesis x plus 3 right parenthesis end cell end table close vertical bar
    equals 2 x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial,(on simplification).
    Trick: Put x equals 1and then match the alternate.

    If x is a positive integer, then capital delta equals open vertical bar table row cell x factorial end cell cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell row cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell row cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell cell left parenthesis x plus 4 right parenthesis factorial end cell end table close vertical bar is equal to

    Maths-General
    capital delta equals open vertical bar table row cell x factorial end cell cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell row cell left parenthesis x plus 1 right parenthesis factorial end cell cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell row cell left parenthesis x plus 2 right parenthesis factorial end cell cell left parenthesis x plus 3 right parenthesis factorial end cell cell left parenthesis x plus 4 right parenthesis factorial end cell end table close vertical bar
    = x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial open vertical bar table row 1 cell left parenthesis x plus 1 right parenthesis end cell cell left parenthesis x plus 2 right parenthesis left parenthesis x plus 1 right parenthesis end cell row 1 cell left parenthesis x plus 2 right parenthesis end cell cell left parenthesis x plus 3 right parenthesis left parenthesis x plus 2 right parenthesis end cell row 1 cell left parenthesis x plus 3 right parenthesis end cell cell left parenthesis x plus 4 right parenthesis left parenthesis x plus 3 right parenthesis end cell end table close vertical bar
    Applying R subscript 1 end subscript rightwards arrow R subscript 2 end subscript minus R subscript 1 end subscript comma R subscript 2 end subscript rightwards arrow left parenthesis R subscript 3 end subscript minus R subscript 2 end subscript right parenthesis we get
    equals x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial open vertical bar table row 0 1 cell 2 left parenthesis x plus 2 right parenthesis end cell row 0 1 cell 2 left parenthesis x plus 3 right parenthesis end cell row 1 cell left parenthesis x plus 3 right parenthesis end cell cell left parenthesis x plus 4 right parenthesis left parenthesis x plus 3 right parenthesis end cell end table close vertical bar
    equals 2 x factorial left parenthesis x plus 1 right parenthesis factorial left parenthesis x plus 2 right parenthesis factorial,(on simplification).
    Trick: Put x equals 1and then match the alternate.
    General
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    The system of equations lambda x plus y plus z equals 0 comma negative x plus lambda y plus z equals 0 comma negative x minus y plus lambda z equals 0, will have a non zero solution if real values of lambdaare given by

    Accordingly, open vertical bar table row lambda 1 1 row cell negative 1 end cell lambda 1 row cell negative 1 end cell cell negative 1 end cell lambda end table close vertical bar equals 0 rightwards double arrow lambda to the power of 3 end exponent plus 3 lambda equals 0
    Therefore lambda equals 0, since lambda equals i square root of 3does not exist.

    The system of equations lambda x plus y plus z equals 0 comma negative x plus lambda y plus z equals 0 comma negative x minus y plus lambda z equals 0, will have a non zero solution if real values of lambdaare given by

    Maths-General
    Accordingly, open vertical bar table row lambda 1 1 row cell negative 1 end cell lambda 1 row cell negative 1 end cell cell negative 1 end cell lambda end table close vertical bar equals 0 rightwards double arrow lambda to the power of 3 end exponent plus 3 lambda equals 0
    Therefore lambda equals 0, since lambda equals i square root of 3does not exist.