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Maths-

A tea party is arranged of 16 persons along two sides of a long table with 8 chairs on each side. 4 men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated ?</span

Maths-General

  1. 8P4 × 8P2    
  2. 8P4 × 8P2 × 10!    
  3. 8P4 × 10!    
  4. None of these    

    Answer:The correct answer is: 8P4 × 8P2 × 10!

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    maths-

    If (m+n) P2 = 56 and m–nP2 = 12 then (m, n) equals-

    If (m+n) P2 = 56 and m–nP2 = 12 then (m, n) equals-

    maths-General
    General
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    A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius 3 R. The work required to take a unit mass from point P on its axis to infinity is

    W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript
    equals negative U subscript P end subscript equals negative m V subscript P end subscript
    equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses
    Potential at point P will be obtained by in integration as given below. Let d M be the mass of small rings as shown

    d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r
    equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction
    d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction
    equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r
    equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses
    therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis

    A thin uniform annular disc (see figure) of mass M has outer radius 4 R and inner radius 3 R. The work required to take a unit mass from point P on its axis to infinity is

    physics-General
    W equals increment U equals U subscript f end subscript minus U subscript i end subscript equals U subscript infinity end subscript minus U subscript P end subscript
    equals negative U subscript P end subscript equals negative m V subscript P end subscript
    equals negative V subscript P end subscript open parentheses a s blank m equals 1 close parentheses
    Potential at point P will be obtained by in integration as given below. Let d M be the mass of small rings as shown

    d M equals fraction numerator M over denominator pi left parenthesis 4 R right parenthesis to the power of 2 end exponent minus pi left parenthesis 3 R right parenthesis to the power of 2 end exponent end fraction open parentheses 2 pi r close parentheses d r
    equals fraction numerator 2 M r d r over denominator 7 R to the power of 2 end exponent end fraction
    d V subscript P end subscript equals negative fraction numerator G. d M over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction
    equals negative fraction numerator 2 G M over denominator 7 R to the power of 2 end exponent end fraction not stretchy integral subscript 3 R end subscript superscript 4 R end superscript fraction numerator r over denominator square root of 16 R to the power of 2 end exponent plus r to the power of 2 end exponent end root end fraction bullet d r
    equals negative fraction numerator 2 G M over denominator 7 R end fraction open parentheses 4 square root of 2 minus 5 close parentheses
    therefore W equals plus fraction numerator 2 G M over denominator 7 R end fraction left parenthesis 4 square root of 2 minus 5 right parenthesis
    General
    physics-

    The two bodies of mass m subscript 1 end subscript and m subscript 2 end subscript left parenthesis m subscript 1 end subscript greater than m subscript 2 end subscript right parenthesis respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

    In the pulley arrangement open vertical bar stack a with rightwards arrow on top subscript 1 end subscript close vertical bar equals open vertical bar stack a with rightwards arrow on top subscript 2 end subscript close vertical bar equals a equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g
    But stack a with rightwards arrow on top subscript 1 end subscript is in downward direction and in the upward direction i e, stack a with rightwards arrow on top subscript 2 end subscript equals negative stack a with rightwards arrow on top subscript 1 end subscript
    therefore Acceleration of centre of mass
    stack a with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack a with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack a with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator m subscript 1 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g minus m subscript 2 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g over denominator left parenthesis m subscript 1 end subscript plus m subscript 2 end subscript right parenthesis end fraction
    equals open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets to the power of 2 end exponent g

    The two bodies of mass m subscript 1 end subscript and m subscript 2 end subscript left parenthesis m subscript 1 end subscript greater than m subscript 2 end subscript right parenthesis respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

    physics-General
    In the pulley arrangement open vertical bar stack a with rightwards arrow on top subscript 1 end subscript close vertical bar equals open vertical bar stack a with rightwards arrow on top subscript 2 end subscript close vertical bar equals a equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses g
    But stack a with rightwards arrow on top subscript 1 end subscript is in downward direction and in the upward direction i e, stack a with rightwards arrow on top subscript 2 end subscript equals negative stack a with rightwards arrow on top subscript 1 end subscript
    therefore Acceleration of centre of mass
    stack a with rightwards arrow on top subscript C M end subscript equals fraction numerator m subscript 1 end subscript stack a with rightwards arrow on top subscript 1 end subscript plus m subscript 2 end subscript stack a with rightwards arrow on top subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator m subscript 1 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g minus m subscript 2 end subscript open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets g over denominator left parenthesis m subscript 1 end subscript plus m subscript 2 end subscript right parenthesis end fraction
    equals open square brackets fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close square brackets to the power of 2 end exponent g
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    maths-

    If x equals 1 plus 3 a plus 6 a squared plus 10 a cubed plus midline horizontal ellipsis. to straight infinity terms, vertical line a vertical line less than 1 comma y equals 1 plus 4 a plus 10 a squared plus 20 a cubed plus midline horizontal ellipsis straight infinity terms, vertical line a vertical line less than 1, then x colon y

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    General
    maths-

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    maths-

    open parentheses 1 plus x plus x squared plus horizontal ellipsis plus x to the power of p close parentheses to the power of n equals a subscript 0 plus a subscript 1 x plus a subscript 2 x squared plus horizontal ellipsis plus a subscript n p end subscript x to the power of n p end exponent not stretchy rightwards double arrow a subscript 1 plus 2 a subscript 2 plus 3 a subscript 3 plus horizontal ellipsis plus n p

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    physics-

    Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14m s to the power of negative 1 end exponent to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is

    At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
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    v subscript C M end subscript equals fraction numerator m subscript 1 end subscript v subscript 1 end subscript plus m subscript 2 end subscript v subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 10 cross times 14 plus 4 cross times 0 over denominator 10 plus 4 end fraction equals fraction numerator 140 over denominator 14 end fraction equals 10 m s to the power of negative 1 end exponent

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    physics-General
    At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
    Hence,
    v subscript C M end subscript equals fraction numerator m subscript 1 end subscript v subscript 1 end subscript plus m subscript 2 end subscript v subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction equals fraction numerator 10 cross times 14 plus 4 cross times 0 over denominator 10 plus 4 end fraction equals fraction numerator 140 over denominator 14 end fraction equals 10 m s to the power of negative 1 end exponent
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    maths-

    In a Δabc if b+c=3a then cot invisible function application straight B over 2 times cot invisible function application straight C over 2 has the value equal to –

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    maths-General
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    maths-

    In a capital delta A B C open parentheses fraction numerator a to the power of 2 end exponent over denominator sin invisible function application A end fraction plus fraction numerator b to the power of 2 end exponent over denominator sin invisible function application B end fraction plus fraction numerator c to the power of 2 end exponent over denominator sin invisible function application C end fraction close parentheses times s i n invisible function application fraction numerator A over denominator 2 end fraction s i n invisible function application fraction numerator B over denominator 2 end fraction s i n invisible function application fraction numerator C over denominator 2 end fraction simplifies to

    In a capital delta A B C open parentheses fraction numerator a to the power of 2 end exponent over denominator sin invisible function application A end fraction plus fraction numerator b to the power of 2 end exponent over denominator sin invisible function application B end fraction plus fraction numerator c to the power of 2 end exponent over denominator sin invisible function application C end fraction close parentheses times s i n invisible function application fraction numerator A over denominator 2 end fraction s i n invisible function application fraction numerator B over denominator 2 end fraction s i n invisible function application fraction numerator C over denominator 2 end fraction simplifies to

    maths-General
    General
    maths-

    In a triangle ABC, a: b: c = 4: 5: 6. Then 3A + B equals to :

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    maths-General