General
Easy
Physics

A particle is thrown in upward direction with initial velocity of 60 m/s. Find average speed and average velocity after 10 seconds. [g =10 MS2]

PhysicsGeneral

  1. 26ms-1, 10ms-1
  2. 20ms-1, 10ms-1
  3. 26 m s to the power of negative 1 end exponent comma 16 m s to the power of negative 1 end exponent
  4. 15ms-1,25ms-1

    Answer:The correct answer is: 26ms-1, 10ms-1

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    Magnetic field at mid-point M in first case is B equals B subscript P Q end subscript minus B subscript R S end subscript
    left parenthesis therefore B subscript P Q end subscript a n d blank B subscript R S end subscript a r e blank i n blank o p p o s i t e blank d i r e c t i o n s right parenthesis
    equals fraction numerator 4 blank mu subscript 0 end subscript over denominator 4 pi d end fraction minus fraction numerator 2 blank mu subscript 0 end subscript over denominator 4 pi d end fraction equals fraction numerator 2 blank mu subscript 0 end subscript over denominator 4 pi d end fraction
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    PQ and RS are long parallel conductors separated by certain distance. M is the mid-point between them (see the figure). The net magnetic field at M is B. Now, the current 2A is switched off. The field at M now becomes

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    equals fraction numerator 4 blank mu subscript 0 end subscript over denominator 4 pi d end fraction minus fraction numerator 2 blank mu subscript 0 end subscript over denominator 4 pi d end fraction equals fraction numerator 2 blank mu subscript 0 end subscript over denominator 4 pi d end fraction
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    therefore B equals B subscript a end subscript plus B subscript b end subscript plus B subscript c end subscript
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    therefore B equals B subscript a end subscript plus B subscript b end subscript plus B subscript c end subscript
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    Magnetic field at mid-point due to wire C D
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    Resultant of Magnetic field B equals B subscript 1 end subscript plus B subscript 2 end subscript
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    F equals fraction numerator Y A over denominator L end fraction l
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    W o r k blank d o n e equals F cross times d l equals fraction numerator Y A over denominator L end fraction l d l
    W equals not stretchy integral subscript 0 end subscript superscript 1 end superscript fraction numerator Y A over denominator L end fraction l d l equals fraction numerator Y A over denominator L end fraction open square brackets fraction numerator l to the power of 2 end exponent over denominator 2 end fraction close square brackets subscript 0 end subscript superscript l end superscript equals fraction numerator 1 over denominator 2 end fraction Y A fraction numerator l to the power of 2 end exponent over denominator L end fraction
    equals fraction numerator 1 over denominator 2 end fraction open parentheses Y fraction numerator l over denominator L end fraction close parentheses open parentheses fraction numerator l over denominator L end fraction close parentheses open parentheses A L close parentheses
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    therefore R e s u l t a n t s comma blank f i e l d blank B subscript 1 end subscript equals B subscript 2 end subscript
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    Given, I subscript 1 end subscript equals 10 blank A comma blank r subscript 1 end subscript equals 5 comma blank r subscript 1 end subscript plus d equals 5 plus 10 equals 15 blank c m
    therefore blank I subscript 2 end subscript equals fraction numerator I subscript 1 end subscript over denominator r subscript 1 end subscript end fraction blank cross times open parentheses r subscript 1 end subscript plus d close parentheses
    I subscript 2 end subscript equals fraction numerator 10 over denominator 5 end fraction cross times 15 equals 30 blank A

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    B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction
    Magnetic field at P due to current in second conductor is
    B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
    From Fleming’s right hands rule the fields at P are directed opposite.
    therefore R e s u l t a n t s comma blank f i e l d blank B subscript 1 end subscript equals B subscript 2 end subscript
    therefore blank fraction numerator mu subscript 0 end subscript I subscript 1 end subscript over denominator 2 pi r subscript 1 end subscript end fraction equals blank fraction numerator mu subscript 0 end subscript I subscript 2 end subscript over denominator 2 pi open parentheses r subscript 1 end subscript plus d close parentheses end fraction
    Given, I subscript 1 end subscript equals 10 blank A comma blank r subscript 1 end subscript equals 5 comma blank r subscript 1 end subscript plus d equals 5 plus 10 equals 15 blank c m
    therefore blank I subscript 2 end subscript equals fraction numerator I subscript 1 end subscript over denominator r subscript 1 end subscript end fraction blank cross times open parentheses r subscript 1 end subscript plus d close parentheses
    I subscript 2 end subscript equals fraction numerator 10 over denominator 5 end fraction cross times 15 equals 30 blank A
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    T equals fraction numerator Y A l over denominator L end fraction
    Increase in length of one segment of wire
    l equals open parentheses L plus fraction numerator 1 over denominator 2 end fraction fraction numerator d to the power of 2 end exponent over denominator L end fraction close parentheses minus L equals fraction numerator 1 over denominator 2 end fraction fraction numerator d to the power of 2 end exponent over denominator L end fraction
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    A wire of length 2 L and radius r is stretched between A and B without the application of any tension. If Y is the Young’s modulus of the wire and it is stretched like A C B, then the tension in the wire will be

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    T equals fraction numerator Y A l over denominator L end fraction
    Increase in length of one segment of wire
    l equals open parentheses L plus fraction numerator 1 over denominator 2 end fraction fraction numerator d to the power of 2 end exponent over denominator L end fraction close parentheses minus L equals fraction numerator 1 over denominator 2 end fraction fraction numerator d to the power of 2 end exponent over denominator L end fraction
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    A cube of aluminium of sides 0.1 m is subjected to a sharing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

    Shearing strain = fraction numerator 0.02 blank cross times 10 to the power of negative 2 end exponent over denominator 0.1 blank end fraction equals 0.002

    A cube of aluminium of sides 0.1 m is subjected to a sharing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be

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    Shearing strain = fraction numerator 0.02 blank cross times 10 to the power of negative 2 end exponent over denominator 0.1 blank end fraction equals 0.002
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    A uniform slender rod of length L, cross-sectional area A and Young’s modulus Y is acted upon by the forces shown in the figure. The elongation of the rod is

    Net elongation of the rod is

    l equals blank fraction numerator 3 F open parentheses fraction numerator 2 L over denominator 3 end fraction close parentheses over denominator A Y end fraction plus fraction numerator 2 F open parentheses fraction numerator L over denominator 3 end fraction close parentheses over denominator A Y end fraction
    l equals fraction numerator 8 F L over denominator 3 A Y end fraction

    A uniform slender rod of length L, cross-sectional area A and Young’s modulus Y is acted upon by the forces shown in the figure. The elongation of the rod is

    physics-General
    Net elongation of the rod is

    l equals blank fraction numerator 3 F open parentheses fraction numerator 2 L over denominator 3 end fraction close parentheses over denominator A Y end fraction plus fraction numerator 2 F open parentheses fraction numerator L over denominator 3 end fraction close parentheses over denominator A Y end fraction
    l equals fraction numerator 8 F L over denominator 3 A Y end fraction
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    The graph shows the behaviour of a length of wire in the region for which the substance obeys Hooke’s law. P and Q represent

    Graph between applied force and extension will be straight line because in elastic range
    Applied force proportional to extension
    But the graph between extension and stored elastic energy will be parabolic in nature
    As U equals 1 divided by 2 blank k x to the power of 2 end exponent or U proportional to x to the power of 2 end exponent

    The graph shows the behaviour of a length of wire in the region for which the substance obeys Hooke’s law. P and Q represent

    physics-General
    Graph between applied force and extension will be straight line because in elastic range
    Applied force proportional to extension
    But the graph between extension and stored elastic energy will be parabolic in nature
    As U equals 1 divided by 2 blank k x to the power of 2 end exponent or U proportional to x to the power of 2 end exponent
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    If the shear modulus of a wire material is 5.9blank cross times 10 to the power of 11 end exponent d y n e blank c m to the power of negative 2 end exponent then the potential energy of a wire of 4 cross times 10 to the power of 3 end exponent c m in diameter and 5 cm long twisted through an angle of 10’ , is

    To twist the wire through the angle d theta comma blankit is necessary to do the work
    d W equals blank tau d theta
    And theta equals 10 to the power of ´ end exponent equals fraction numerator 10 over denominator 60 end fraction cross times fraction numerator pi over denominator 180 end fraction equals fraction numerator pi over denominator 1080 end fraction r a d
    W equals blank not stretchy integral from 0 to theta of tau blank d theta equals blank not stretchy integral from 0 to theta of fraction numerator eta pi r to the power of 4 end exponent theta d theta over denominator 2 l end fraction equals blank fraction numerator eta pi r to the power of 4 end exponent theta over denominator 4 l end fraction
    W equals blank fraction numerator 5.9 blank cross times 10 to the power of 11 end exponent cross times 10 to the power of negative 5 end exponent cross times blank pi open parentheses 2 cross times 10 to the power of negative 5 end exponent close parentheses to the power of 4 end exponent pi to the power of 2 end exponent over denominator 10 to the power of negative 4 end exponent cross times 4 cross times 5 cross times 10 to the power of negative 2 end exponent cross times open parentheses 1080 close parentheses to the power of 2 end exponent end fraction
    W equals 1.253 blank cross times 10 to the power of negative 12 end exponent blank J

    If the shear modulus of a wire material is 5.9blank cross times 10 to the power of 11 end exponent d y n e blank c m to the power of negative 2 end exponent then the potential energy of a wire of 4 cross times 10 to the power of 3 end exponent c m in diameter and 5 cm long twisted through an angle of 10’ , is

    physics-General
    To twist the wire through the angle d theta comma blankit is necessary to do the work
    d W equals blank tau d theta
    And theta equals 10 to the power of ´ end exponent equals fraction numerator 10 over denominator 60 end fraction cross times fraction numerator pi over denominator 180 end fraction equals fraction numerator pi over denominator 1080 end fraction r a d
    W equals blank not stretchy integral from 0 to theta of tau blank d theta equals blank not stretchy integral from 0 to theta of fraction numerator eta pi r to the power of 4 end exponent theta d theta over denominator 2 l end fraction equals blank fraction numerator eta pi r to the power of 4 end exponent theta over denominator 4 l end fraction
    W equals blank fraction numerator 5.9 blank cross times 10 to the power of 11 end exponent cross times 10 to the power of negative 5 end exponent cross times blank pi open parentheses 2 cross times 10 to the power of negative 5 end exponent close parentheses to the power of 4 end exponent pi to the power of 2 end exponent over denominator 10 to the power of negative 4 end exponent cross times 4 cross times 5 cross times 10 to the power of negative 2 end exponent cross times open parentheses 1080 close parentheses to the power of 2 end exponent end fraction
    W equals 1.253 blank cross times 10 to the power of negative 12 end exponent blank J
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    The Young’s modulus of the material of a wire is equal to the

    Young’s modulus of material Y equals fraction numerator L i n e a r blank s t r e s s over denominator L o n g i t u d i n a l blank s t r a i n end fraction
    If longitudinal strain is equal unity, then
    Y equals Linear stress produced

    The Young’s modulus of the material of a wire is equal to the

    physics-General
    Young’s modulus of material Y equals fraction numerator L i n e a r blank s t r e s s over denominator L o n g i t u d i n a l blank s t r a i n end fraction
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    Y equals Linear stress produced
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    If f colon R not stretchy rightwards arrow R is defined by f left parenthesis x right parenthesis equals fraction numerator x squared minus 4 over denominator x squared plus 1 end fraction then f left parenthesis x right parenthesis text  is  end text

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