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General
Easy

Question

For the circuit shown in figure the charge on 4muF capacitor is

  1. 40 muC    
  2. 30 muC    
  3. 24 muC    
  4. 54 muC    

The correct answer is: 24 muC


    Combined capacity of 1 blank mu F and 5 mu F blank= 1 + 5=6 blank mu F
    Now, 4 mu F blankand 6 mu F are in series.
    therefore blank fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 3 plus 2 over denominator 12 end fraction equals fraction numerator 5 over denominator 12 end fraction
    C subscript s end subscript equals fraction numerator 12 over denominator 5 end fraction mu F
    Charge in the arm containing 4 mu F blankcapacitor is
    q equals C subscript s end subscript cross times V equals fraction numerator 12 over denominator 5 end fraction cross times 10 equals 24 blank mu C

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    H subscript 1 end subscript equals H subscript 2 end subscript rightwards double arrow u subscript y end subscript subscript 1 end subscript equals u subscript y end subscript subscript 2 end subscript
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    H subscript 1 end subscript equals H subscript 2 end subscript rightwards double arrow u subscript y end subscript subscript 1 end subscript equals u subscript y end subscript subscript 2 end subscript
    Here T subscript 1 end subscript equals T subscript 2 end subscript
    Range R equals fraction numerator u to the power of 2 end exponent sin invisible function application 2 blank theta over denominator g end fraction equals fraction numerator 2 left parenthesis u sin invisible function application theta right parenthesis left parenthesis u cos invisible function application theta right parenthesis over denominator g end fraction
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    or v equals 20 blank m s to the power of negative 1 end exponent
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    open parentheses fraction numerator F r o m blank R equals fraction numerator l over denominator K A end fraction w h e r e comma over denominator K equals t h e r m a l blank c o n d u c t i v i t y end fraction close parentheses
    rightwards double arrow not stretchy integral d R equals R equals not stretchy integral subscript r subscript 1 end subscript end subscript superscript r subscript 2 end subscript end superscript fraction numerator d x over denominator 4 pi K x to the power of 2 blank end exponent end fraction
    = fraction numerator 1 over denominator 4 pi K end fraction open square brackets fraction numerator 1 over denominator r subscript 1 end subscript end fraction minus blank fraction numerator 1 over denominator r subscript 2 end subscript end fraction close square brackets equals blank fraction numerator r subscript 2 end subscript minus r subscript 1 end subscript over denominator 4 pi K left parenthesis r subscript 1 end subscript r subscript 2 end subscript right parenthesis end fraction
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    I subscript A end subscript equals fraction numerator L over denominator left parenthesis 2 r right parenthesis to the power of 2 end exponent end fraction and I subscript B end subscript equals fraction numerator L over denominator open parentheses r square root of 2 close parentheses to the power of 2 end exponent end fraction cos invisible function application theta
    equals fraction numerator L over denominator 2 r to the power of 2 end exponent end fraction. fraction numerator r over denominator r square root of 2 end fraction equals fraction numerator L over denominator 2 square root of 2   r to the power of 2 end exponent end fraction

    therefore fraction numerator I subscript A end subscript over denominator I subscript B end subscript end fraction equals fraction numerator 2 square root of 2 over denominator 4 end fraction equals fraction numerator 1 over denominator square root of 2 end fraction
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    I proportional to fraction numerator 1 over denominator r to the power of 2 end exponent end fraction rightwards double arrow fraction numerator I subscript 2 end subscript over denominator I subscript 1 end subscript end fraction equals fraction numerator r subscript 1 end subscript superscript 2 end superscript over denominator r subscript 2 end subscript superscript 2 end superscript end fraction equals fraction numerator 6 0 to the power of 2 end exponent over denominator 18 0 to the power of 2 end exponent end fraction equals fraction numerator 1 over denominator 9 end fraction
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    equals open square brackets l open parentheses 1 plus alpha subscript 2 end subscript t close parentheses close square brackets to the power of 2 end exponent minus open square brackets fraction numerator l over denominator 2 end fraction open parentheses 1 plus alpha subscript 1 end subscript t close parentheses close square brackets to the power of 2 end exponentl to the power of 2 end exponent minus fraction numerator l to the power of 2 end exponent over denominator 4 end fraction equals l to the power of 2 end exponent open parentheses 1 plus alpha subscript 2 end subscript superscript 2 end superscript t to the power of 2 end exponent plus 2 alpha subscript 2 end subscript t close parentheses minus fraction numerator l to the power of 2 end exponent over denominator 4 end fraction left parenthesis 1 plus alpha subscript 1 end subscript superscript 2 end superscript t to the power of 2 end exponent plus 2 alpha subscript 1 end subscript t right parenthesis
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    0 equals l to the power of 2 end exponent open parentheses 2 alpha subscript 2 end subscript t close parentheses minus fraction numerator l to the power of 2 end exponent over denominator 4 end fraction open parentheses 2 alpha subscript 1 end subscript t close parentheses rightwards double arrow 2 alpha subscript 2 end subscript equals fraction numerator 2 alpha subscript 1 end subscript over denominator 4 end fraction rightwards double arrow alpha subscript 1 end subscript equals 4 alpha subscript 2 end subscript

    Three rods of equal length l are joined to form an equilateral triangle P Q R. O is the mid point of P Q. Distance O R remains same for small change in temperature. Coefficient of linear expansion for P R and R Q is same, i. e. comma blank alpha subscript 2 end subscript but that for P Q is alpha subscript 1 end subscript. Then

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    open parentheses O R close parentheses to the power of 2 end exponent equals open parentheses P R close parentheses to the power of 2 end exponent minus open parentheses P O close parentheses to the power of 2 end exponent equals l to the power of 2 end exponent minus open parentheses fraction numerator l over denominator 2 end fraction close parentheses to the power of 2 end exponent
    equals open square brackets l open parentheses 1 plus alpha subscript 2 end subscript t close parentheses close square brackets to the power of 2 end exponent minus open square brackets fraction numerator l over denominator 2 end fraction open parentheses 1 plus alpha subscript 1 end subscript t close parentheses close square brackets to the power of 2 end exponentl to the power of 2 end exponent minus fraction numerator l to the power of 2 end exponent over denominator 4 end fraction equals l to the power of 2 end exponent open parentheses 1 plus alpha subscript 2 end subscript superscript 2 end superscript t to the power of 2 end exponent plus 2 alpha subscript 2 end subscript t close parentheses minus fraction numerator l to the power of 2 end exponent over denominator 4 end fraction left parenthesis 1 plus alpha subscript 1 end subscript superscript 2 end superscript t to the power of 2 end exponent plus 2 alpha subscript 1 end subscript t right parenthesis
    Neglecting alpha subscript 2 end subscript superscript 2 end superscript t to the power of 2 end exponent and alpha subscript 1 end subscript superscript 2 end superscript t to the power of 2 end exponent
    0 equals l to the power of 2 end exponent open parentheses 2 alpha subscript 2 end subscript t close parentheses minus fraction numerator l to the power of 2 end exponent over denominator 4 end fraction open parentheses 2 alpha subscript 1 end subscript t close parentheses rightwards double arrow 2 alpha subscript 2 end subscript equals fraction numerator 2 alpha subscript 1 end subscript over denominator 4 end fraction rightwards double arrow alpha subscript 1 end subscript equals 4 alpha subscript 2 end subscript
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    One of the following figures respesents the variation of particle momentum with associated de Broglie wavelength
    a)
    b)
    c)

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    a)
    b)
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    open parentheses fraction numerator negative d T over denominator d t end fraction close parentheses equals fraction numerator e A sigma over denominator m c end fraction left parenthesis T to the power of 4 end exponent minus T subscript 0 end subscript superscript 4 end superscript right parenthesis
    Where c is specific heat of material.
    or open parentheses fraction numerator negative d T over denominator d t end fraction close parentheses proportional to fraction numerator 1 over denominator c end fraction
    i e. comma blankrate of cooling varies inversely as specific heat. From the graph, for A rate of cooling is larger. Therefore, specific heat of A is smaller.

    Two circular discs A and B with equal radii are blackened. They are heated to some temperature and are cooled under identical conditions. What inference do you draw from their cooling curves?

    physics-General
    According to Newton’s law of cooling, rate of cooling is given by
    open parentheses fraction numerator negative d T over denominator d t end fraction close parentheses equals fraction numerator e A sigma over denominator m c end fraction left parenthesis T to the power of 4 end exponent minus T subscript 0 end subscript superscript 4 end superscript right parenthesis
    Where c is specific heat of material.
    or open parentheses fraction numerator negative d T over denominator d t end fraction close parentheses proportional to fraction numerator 1 over denominator c end fraction
    i e. comma blankrate of cooling varies inversely as specific heat. From the graph, for A rate of cooling is larger. Therefore, specific heat of A is smaller.