General
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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

Physics-General

  1. 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    
  2. 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    
  3. g    
  4. zero    

    Answer:The correct answer is: 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 gIn 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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    physics-General
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    A bullet of mass m is fired with a velocity of 50 m s to the power of negative 1 end exponent at an angle theta with the horizontal. At the highest point of its trajectory, it collides had on with a bob of massless string of length l equals 10 divided by 3m and gets embedded in the bob. After the collision, the string moves to an angle of 120degree. What is the angle theta ?

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    rightwards double arrow blank v to the power of ´ 2 end exponent equals 2 cross times 10 cross times 5 equals 100 or v to the power of ´ end exponent equals 10
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    fraction numerator 25 over denominator 2 end fraction cos invisible function application theta equals 10 rightwards double arrow cos invisible function application theta equals fraction numerator 4 over denominator 5 end fraction ortheta equals cos to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 4 over denominator 5 end fraction close parentheses

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    fraction numerator 25 over denominator 2 end fraction cos invisible function application theta equals 10 rightwards double arrow cos invisible function application theta equals fraction numerator 4 over denominator 5 end fraction ortheta equals cos to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 4 over denominator 5 end fraction close parentheses
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    A spherical hollow is made in a lead sphere of radius R such that its surface touches the outside surface of lead sphere and passes through the centre. What is the shift in the centre of lead sphere as a result of this hollowing?

    Let centre of mass of lead sphere after hollowing be at point O subscript 2 end subscript comma where O O subscript 2 end subscript equals x
    Mass of spherical hollow m equals fraction numerator fraction numerator 4 over denominator 3 end fraction pi open parentheses fraction numerator R over denominator 2 end fraction close parentheses to the power of 2 end exponent M over denominator open parentheses fraction numerator 4 over denominator 2 end fraction pi R to the power of 3 end exponent close parentheses end fraction equals fraction numerator M over denominator 8 end fraction and
    x equals O O subscript 1 end subscript equals fraction numerator R over denominator 2 end fraction

    therefore blank x equals fraction numerator M cross times 0 minus open parentheses fraction numerator M over denominator 8 end fraction close parentheses cross times fraction numerator R over denominator 2 end fraction over denominator M minus fraction numerator M over denominator 8 end fraction end fraction equals fraction numerator fraction numerator M R over denominator 16 end fraction over denominator fraction numerator 7 M over denominator 8 end fraction end fraction equals negative fraction numerator R over denominator 14 end fraction
    therefore shift equals fraction numerator R over denominator 14 end fraction

    A spherical hollow is made in a lead sphere of radius R such that its surface touches the outside surface of lead sphere and passes through the centre. What is the shift in the centre of lead sphere as a result of this hollowing?

    physics-General
    Let centre of mass of lead sphere after hollowing be at point O subscript 2 end subscript comma where O O subscript 2 end subscript equals x
    Mass of spherical hollow m equals fraction numerator fraction numerator 4 over denominator 3 end fraction pi open parentheses fraction numerator R over denominator 2 end fraction close parentheses to the power of 2 end exponent M over denominator open parentheses fraction numerator 4 over denominator 2 end fraction pi R to the power of 3 end exponent close parentheses end fraction equals fraction numerator M over denominator 8 end fraction and
    x equals O O subscript 1 end subscript equals fraction numerator R over denominator 2 end fraction

    therefore blank x equals fraction numerator M cross times 0 minus open parentheses fraction numerator M over denominator 8 end fraction close parentheses cross times fraction numerator R over denominator 2 end fraction over denominator M minus fraction numerator M over denominator 8 end fraction end fraction equals fraction numerator fraction numerator M R over denominator 16 end fraction over denominator fraction numerator 7 M over denominator 8 end fraction end fraction equals negative fraction numerator R over denominator 14 end fraction
    therefore shift equals fraction numerator R over denominator 14 end fraction
    General
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    On heating a mixture of N H subscript 4 end subscript C lans K N O subscript 2 end subscript, we get –

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    chemistry-General