Physics-
General
Easy

Question

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

  1. 1.253 blank cross times 10 to the power of negative 12 end exponent blank J    
  2. 2.00 blank cross times 10 to the power of negative 12 end exponent blank J    
  3. 1.00 blank cross times 10 to the power of negative 12 end exponent blank J    
  4. 0.8 blank cross times 10 to the power of negative 12 end exponent blank J    

The correct answer is: 1.253 blank cross times 10 to the power of negative 12 end exponent blank J


    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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    Related Questions to study

    General
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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

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    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
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    y equals f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus vertical line x vertical line end fraction comma x element of R comma y element of R text  is  end text

    w e space h a v e comma space
f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator x over denominator 1 plus x end fraction comma space i f space x greater or equal than 0 end cell row cell fraction numerator x over denominator 1 minus x end fraction comma space i f space x less or equal than 0 end cell end table close
C a s e space i comma space w h e n space x greater or equal than 0
f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus x end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 over denominator 1 plus x subscript 1 end fraction equals fraction numerator x subscript 2 over denominator 1 plus x subscript 2 end fraction rightwards double arrow x subscript 1 plus x subscript 1 x subscript 2 equals x subscript 2 plus x subscript 1 x subscript 2 rightwards double arrow x subscript 1 equals x subscript 2
S o comma space f space i s space o n e minus space o n e.
L e t space y element of R
a n d space y equals fraction numerator x over denominator 1 plus x end fraction rightwards double arrow y plus x y equals x rightwards double arrow x left parenthesis 1 minus y right parenthesis equals y rightwards double arrow x equals fraction numerator y over denominator 1 minus y end fraction
N o w comma space f open parentheses fraction numerator y over denominator 1 minus y end fraction close parentheses equals fraction numerator fraction numerator y over denominator 1 minus y end fraction over denominator 1 plus fraction numerator y over denominator 1 minus y end fraction end fraction equals fraction numerator y over denominator 1 minus y plus y end fraction equals y
S o comma space f left parenthesis x right parenthesis space i s space o n t o.

C a s e space i comma space w h e n space x less or equal than 0
f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 minus x end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 over denominator 1 minus x subscript 1 end fraction equals fraction numerator x subscript 2 over denominator 1 minus x subscript 2 end fraction rightwards double arrow x subscript 1 minus x subscript 1 x subscript 2 equals x subscript 2 minus x subscript 1 x subscript 2 rightwards double arrow x subscript 1 equals x subscript 2
S o comma space f space i s space o n e minus space o n e.
L e t space y element of R
a n d space y equals fraction numerator x over denominator 1 minus x end fraction rightwards double arrow y minus x y equals x rightwards double arrow x left parenthesis 1 plus y right parenthesis equals y rightwards double arrow x equals fraction numerator y over denominator 1 plus y end fraction
N o w comma space f open parentheses fraction numerator y over denominator 1 plus y end fraction close parentheses equals fraction numerator fraction numerator y over denominator 1 plus y end fraction over denominator 1 minus fraction numerator y over denominator 1 plus y end fraction end fraction equals fraction numerator y over denominator 1 plus y minus y end fraction equals y
S o comma space f left parenthesis x right parenthesis space i s space o n t o.

    y equals f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus vertical line x vertical line end fraction comma x element of R comma y element of R text  is  end text

    Maths-General
    w e space h a v e comma space
f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator x over denominator 1 plus x end fraction comma space i f space x greater or equal than 0 end cell row cell fraction numerator x over denominator 1 minus x end fraction comma space i f space x less or equal than 0 end cell end table close
C a s e space i comma space w h e n space x greater or equal than 0
f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 plus x end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 over denominator 1 plus x subscript 1 end fraction equals fraction numerator x subscript 2 over denominator 1 plus x subscript 2 end fraction rightwards double arrow x subscript 1 plus x subscript 1 x subscript 2 equals x subscript 2 plus x subscript 1 x subscript 2 rightwards double arrow x subscript 1 equals x subscript 2
S o comma space f space i s space o n e minus space o n e.
L e t space y element of R
a n d space y equals fraction numerator x over denominator 1 plus x end fraction rightwards double arrow y plus x y equals x rightwards double arrow x left parenthesis 1 minus y right parenthesis equals y rightwards double arrow x equals fraction numerator y over denominator 1 minus y end fraction
N o w comma space f open parentheses fraction numerator y over denominator 1 minus y end fraction close parentheses equals fraction numerator fraction numerator y over denominator 1 minus y end fraction over denominator 1 plus fraction numerator y over denominator 1 minus y end fraction end fraction equals fraction numerator y over denominator 1 minus y plus y end fraction equals y
S o comma space f left parenthesis x right parenthesis space i s space o n t o.

C a s e space i comma space w h e n space x less or equal than 0
f left parenthesis x right parenthesis equals fraction numerator x over denominator 1 minus x end fraction
t h e n comma space f left parenthesis x subscript 1 right parenthesis equals f left parenthesis x subscript 2 right parenthesis rightwards double arrow fraction numerator x subscript 1 over denominator 1 minus x subscript 1 end fraction equals fraction numerator x subscript 2 over denominator 1 minus x subscript 2 end fraction rightwards double arrow x subscript 1 minus x subscript 1 x subscript 2 equals x subscript 2 minus x subscript 1 x subscript 2 rightwards double arrow x subscript 1 equals x subscript 2
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L e t space y element of R
a n d space y equals fraction numerator x over denominator 1 minus x end fraction rightwards double arrow y minus x y equals x rightwards double arrow x left parenthesis 1 plus y right parenthesis equals y rightwards double arrow x equals fraction numerator y over denominator 1 plus y end fraction
N o w comma space f open parentheses fraction numerator y over denominator 1 plus y end fraction close parentheses equals fraction numerator fraction numerator y over denominator 1 plus y end fraction over denominator 1 minus fraction numerator y over denominator 1 plus y end fraction end fraction equals fraction numerator y over denominator 1 plus y minus y end fraction equals y
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    General
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    Two short bar magnets of equal dipole moment M are fastened perpendicularly at their centers, figure. The magnitude of resultant of two magnetic field at a distance d from the center on the bisector of the right angle is

    Resolving the magnetic moments along O P and perpendicular toO P comma figure we find that component O P perpendicular O P cancel out. Resultant magnetic moment along O P i s equals M cos invisible function application 45 degree plus M c o s 45 degree

    equals 2 blank M c o s blank 45 degree equals fraction numerator 2 M over denominator square root of 2 end fraction equals square root of 2 M
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    Two short bar magnets of equal dipole moment M are fastened perpendicularly at their centers, figure. The magnitude of resultant of two magnetic field at a distance d from the center on the bisector of the right angle is

    physics-General
    Resolving the magnetic moments along O P and perpendicular toO P comma figure we find that component O P perpendicular O P cancel out. Resultant magnetic moment along O P i s equals M cos invisible function application 45 degree plus M c o s 45 degree

    equals 2 blank M c o s blank 45 degree equals fraction numerator 2 M over denominator square root of 2 end fraction equals square root of 2 M
    The point P lies on axial line of magnet of moment
    equals square root of 2 M
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    Two magnets of equal mass are joined at 90degree each other as shown in figure. Magnet N subscript 1 end subscript S subscript 1 end subscript has a magnetic moment square root of 3 times that of N subscript 2 end subscript S subscript 2 end subscript. The arrangement is pivoted so that it is free to rotate in horizontal plane. When in equilibrium, what angle should N subscript 1 end subscript S subscript 1 end subscript make with magnetic meridian?

    physics-General
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    t a n theta equals fraction numerator M subscript 2 end subscript over denominator M subscript 1 end subscript end fraction equals fraction numerator M over denominator square root of 3 M end fraction equals fraction numerator 1 over denominator square root of 3 end fraction blank therefore theta equals 30 degree
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    The pressure applied from all directions on a cube is P. How much its temperature should be raised to maintain the original volume? The volume elasticity of the cube is beta and the coefficient of volume expansion is alpha

    physics-General
    If coefficient of volume expansion is alpha and rise in temperature is increment theta then increment V equals V alpha increment theta rightwards double arrow fraction numerator increment V over denominator V end fraction equals alpha increment theta
    Volume elasticity beta equals fraction numerator P over denominator increment V divided by V end fraction equals fraction numerator P over denominator alpha increment theta end fraction rightwards double arrow increment theta equals fraction numerator P over denominator alpha beta end fraction
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    A wire left parenthesis Y equals 2 cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent right parenthesis has length 1 m and cross-sectional area 1 m m to the power of negative 2 end exponent. The work required to increase the length by 2 mm is

    Work done = fraction numerator 1 over denominator 2 end fraction blank F increment l
    fraction numerator equals blank fraction numerator 1 over denominator 2 end fraction fraction numerator Y A increment l to the power of 2 end exponent over denominator l end fraction over denominator fraction numerator 2 cross times 10 to the power of 11 end exponent cross times 10 to the power of negative 6 end exponent open parentheses 2 cross times 10 to the power of negative 3 end exponent close parentheses to the power of 2 end exponent over denominator 2 cross times 1 end fraction end fraction open vertical bar table row cell Y equals fraction numerator F l over denominator A increment l end fraction end cell row cell o r blank F equals fraction numerator Y A increment l over denominator l end fraction end cell end table close
    = 4cross times 10 to the power of negative 1 end exponent blank J equals 0.4 blank J

    A wire left parenthesis Y equals 2 cross times 10 to the power of 11 end exponent N m to the power of negative 2 end exponent right parenthesis has length 1 m and cross-sectional area 1 m m to the power of negative 2 end exponent. The work required to increase the length by 2 mm is

    physics-General
    Work done = fraction numerator 1 over denominator 2 end fraction blank F increment l
    fraction numerator equals blank fraction numerator 1 over denominator 2 end fraction fraction numerator Y A increment l to the power of 2 end exponent over denominator l end fraction over denominator fraction numerator 2 cross times 10 to the power of 11 end exponent cross times 10 to the power of negative 6 end exponent open parentheses 2 cross times 10 to the power of negative 3 end exponent close parentheses to the power of 2 end exponent over denominator 2 cross times 1 end fraction end fraction open vertical bar table row cell Y equals fraction numerator F l over denominator A increment l end fraction end cell row cell o r blank F equals fraction numerator Y A increment l over denominator l end fraction end cell end table close
    = 4cross times 10 to the power of negative 1 end exponent blank J equals 0.4 blank J
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    The diagram shows a force-extension graph for a rubber band. Consider the following statements
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    physics-General
    Area of hysterisis loop gives the energy loss in the process of stretching and unstretching of rubber band and this loss will appear in the form of heating
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    A wire of length Land radius a rigidly fixed at one end. On stretching the other end of the wire with a force F, the increase in its length is l. If another wire of same material but of length 2L and radius 2 a is stretched with a force 2F, the increase in its length will be

    Young’s modulus Y equals fraction numerator F L over denominator A l end fraction
    equals fraction numerator F L over denominator pi a to the power of 2 end exponent l end fraction
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    Y subscript 1 end subscript equals Y subscript 2 end subscript
    orfraction numerator F L over denominator pi a to the power of 2 end exponent l end fraction equals fraction numerator open parentheses 2 F close parentheses left parenthesis 2 L right parenthesis over denominator pi open parentheses 2 a close parentheses to the power of 2 end exponent l ʹ end fraction
    orl to the power of ´ end exponent equals l

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    Young’s modulus Y equals fraction numerator F L over denominator A l end fraction
    equals fraction numerator F L over denominator pi a to the power of 2 end exponent l end fraction
    Since for same material Young’s modulus is same, i e comma
    Y subscript 1 end subscript equals Y subscript 2 end subscript
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    orl to the power of ´ end exponent equals l
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    For a dia-magnetic substance, I is negative and negative I proportional to H. Therefore, the variation is represented by O C o r O D. As magnetisation is small, O C is better choice.

    The variation of intensity of magnetization left parenthesis I right parenthesis with respect to the magnetizing field (H) in a diamagnetic substance is described by the graph in figure.

    physics-General
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    Maths-General
    n factorial
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    maths-General
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