General
Easy
Physics

Rohit completes a semi circular path of radius R in 10 seconds. Calculate average speed and average velocity in ms-1.

PhysicsGeneral

  1. fraction numerator negative R over denominator 10 end fraction comma fraction numerator 2 R over denominator 10 end fraction
  2. πR over 10 comma R over 10
  3. fraction numerator 2 πR over denominator 10 end fraction comma fraction numerator 2 R over denominator 10 end fraction
  4. fraction numerator 2 πR over denominator 10 end fraction comma R over 10

    Answer:The correct answer is: fraction numerator negative R over denominator 10 end fraction comma fraction numerator 2 R over denominator 10 end fraction

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    If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient eta comma the Poisson’s ratio sigma will be

    Young’s modulus, Y equals fraction numerator 3 ɳ over denominator 2 end fraction
    We know that Y equals 2 ɳ open parentheses 1 plus sigma close parentheses
    therefore blank fraction numerator 3 ɳ over denominator 2 end fraction equals 2 ɳ open parentheses 1 plus sigma close parentheses
    blank sigma equals negative fraction numerator 1 over denominator 4 end fraction

    If Young’s modulus of elasticity Y for a material is one and half times its rigidity coefficient eta comma the Poisson’s ratio sigma will be

    physics-General
    Young’s modulus, Y equals fraction numerator 3 ɳ over denominator 2 end fraction
    We know that Y equals 2 ɳ open parentheses 1 plus sigma close parentheses
    therefore blank fraction numerator 3 ɳ over denominator 2 end fraction equals 2 ɳ open parentheses 1 plus sigma close parentheses
    blank sigma equals negative fraction numerator 1 over denominator 4 end fraction
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    Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

    increment L equals fraction numerator F L over denominator A Y end fraction
    Because, wires of the same material are stretched by the same load. So, F and Y will be constant.
    therefore blank increment L proportional to fraction numerator L over denominator pi r to the power of 2 end exponent end fraction
    increment L subscript 1 end subscript equals fraction numerator 100 over denominator pi cross times left parenthesis 1 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 10 to the power of negative 6 end exponent end fraction
    equals fraction numerator 100 over denominator pi end fraction cross times 10 to the power of negative 6 end exponent
    therefore increment L subscript 2 end subscript equals fraction numerator 200 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 200 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction
    equals fraction numerator 22.2 over denominator pi end fraction cross times 10 to the power of 6 end exponent
    therefore increment L subscript 3 end subscript equals fraction numerator 300 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction
    equals fraction numerator 33.3 over denominator pi end fraction cross times 10 to the power of 6 end exponent
    therefore increment L subscript 4 end subscript equals fraction numerator 400 over denominator pi cross times left parenthesis 4 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 400 over denominator pi cross times 16 cross times 10 to the power of negative 6 end exponent end fraction
    equals fraction numerator 25 over denominator pi end fraction cross times 10 to the power of 6 end exponent
    We can see that, L=100 cm and r = 1mm will elongate most.

    Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

    physics-General
    increment L equals fraction numerator F L over denominator A Y end fraction
    Because, wires of the same material are stretched by the same load. So, F and Y will be constant.
    therefore blank increment L proportional to fraction numerator L over denominator pi r to the power of 2 end exponent end fraction
    increment L subscript 1 end subscript equals fraction numerator 100 over denominator pi cross times left parenthesis 1 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 10 to the power of negative 6 end exponent end fraction
    equals fraction numerator 100 over denominator pi end fraction cross times 10 to the power of negative 6 end exponent
    therefore increment L subscript 2 end subscript equals fraction numerator 200 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 200 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction
    equals fraction numerator 22.2 over denominator pi end fraction cross times 10 to the power of 6 end exponent
    therefore increment L subscript 3 end subscript equals fraction numerator 300 over denominator pi cross times left parenthesis 3 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 100 over denominator pi cross times 9 cross times 10 to the power of negative 6 end exponent end fraction
    equals fraction numerator 33.3 over denominator pi end fraction cross times 10 to the power of 6 end exponent
    therefore increment L subscript 4 end subscript equals fraction numerator 400 over denominator pi cross times left parenthesis 4 cross times 10 to the power of negative 3 end exponent right parenthesis to the power of 2 end exponent end fraction equals fraction numerator 400 over denominator pi cross times 16 cross times 10 to the power of negative 6 end exponent end fraction
    equals fraction numerator 25 over denominator pi end fraction cross times 10 to the power of 6 end exponent
    We can see that, L=100 cm and r = 1mm will elongate most.
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    A current i is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.
    The magnetic field at the centre of the loop is fraction numerator mu subscript 0 end subscript i over denominator R end fraction times.
    left parenthesis M A equals R comma blank M B equals 2 R comma blank angle D M A equals 90 degree

    (i) Magnetic field at the centre due to the curved portion D A equals fraction numerator mu subscript 0 end subscript i over denominator 4 pi R end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction close parentheses
    According to right hand screw rule, the magnetic field will be into the plane of paper.

    (ii) Magnetic field at M due to A B is zero.
    (iii) Magnetic field at the centre due to the curved portion B C is fraction numerator mu subscript 0 end subscript i over denominator 4 pi 2 R end fraction open parentheses fraction numerator pi over denominator 2 end fraction close parentheses. According to
    right hand screw rule, the magnetic field will be into the plane of paper.
    (iv) Magnetic field at M due to D C is zero.
    Hence, the resultant magnetic field at M
    equals fraction numerator 3 mu subscript 0 end subscript i over denominator 8 R end fraction plus 0 plus fraction numerator mu subscript 0 end subscript i over denominator 16 R end fraction plus 0 equals fraction numerator 7 mu subscript 0 end subscript i over denominator 16 R end fraction

    A current i is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure.
    The magnetic field at the centre of the loop is fraction numerator mu subscript 0 end subscript i over denominator R end fraction times.
    left parenthesis M A equals R comma blank M B equals 2 R comma blank angle D M A equals 90 degree

    physics-General
    (i) Magnetic field at the centre due to the curved portion D A equals fraction numerator mu subscript 0 end subscript i over denominator 4 pi R end fraction open parentheses fraction numerator 3 pi over denominator 2 end fraction close parentheses
    According to right hand screw rule, the magnetic field will be into the plane of paper.

    (ii) Magnetic field at M due to A B is zero.
    (iii) Magnetic field at the centre due to the curved portion B C is fraction numerator mu subscript 0 end subscript i over denominator 4 pi 2 R end fraction open parentheses fraction numerator pi over denominator 2 end fraction close parentheses. According to
    right hand screw rule, the magnetic field will be into the plane of paper.
    (iv) Magnetic field at M due to D C is zero.
    Hence, the resultant magnetic field at M
    equals fraction numerator 3 mu subscript 0 end subscript i over denominator 8 R end fraction plus 0 plus fraction numerator mu subscript 0 end subscript i over denominator 16 R end fraction plus 0 equals fraction numerator 7 mu subscript 0 end subscript i over denominator 16 R end fraction
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    A part of a long wire carrying a current i is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

    The magnitude of the magnetic field at point O due to straight part of wire is
    B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction
    B subscript 1 end subscript is perpendicular to the plane of the page, directed upwards (right hand plam rule 1).
    The field at the centre O due to the current loop of radius r is
    B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction

    B subscript 2 end subscript is also perpendicular to the page, directed upwards (right hand screw rule).
    B subscript 1 end subscript plus B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction open parentheses fraction numerator 1 over denominator pi end fraction plus 1 close parentheses
    equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction open parentheses pi plus 1 close parentheses

    A part of a long wire carrying a current i is bent into a circle of radius r as shown in figure. The net magnetic field at the centre O of the circular loop is

    physics-General
    The magnitude of the magnetic field at point O due to straight part of wire is
    B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction
    B subscript 1 end subscript is perpendicular to the plane of the page, directed upwards (right hand plam rule 1).
    The field at the centre O due to the current loop of radius r is
    B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction

    B subscript 2 end subscript is also perpendicular to the page, directed upwards (right hand screw rule).
    B subscript 1 end subscript plus B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript i over denominator 2 r end fraction open parentheses fraction numerator 1 over denominator pi end fraction plus 1 close parentheses
    equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi r end fraction open parentheses pi plus 1 close parentheses
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    A wire shown in figure carries a current of 40 A. If r equals 3.14 blank c m, the magnetic field at point P will be

    Let the given circular A B C part of wire subtends an angle theta at its centre. Then, magnetic field due to this circular part is

    B to the power of ´ ´ end exponent equals B subscript c end subscript cross times fraction numerator theta over denominator 2 pi end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction cross times fraction numerator 2 pi i over denominator e end fraction cross times fraction numerator theta over denominator 2 pi end fraction
    rightwards double arrow blank B to the power of ´ ´ end exponent equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction blank. fraction numerator i over denominator r end fraction theta
    Given, i equals 40 blank A comma blank r equals 3.14 blank c m equals 3.14 cross times 10 to the power of negative 2 end exponent blank m
    theta equals 360 degree minus 90 degree equals 270 degree equals fraction numerator 3 pi over denominator 2 end fraction blank r a d.
    therefore B to the power of ´ ´ end exponent equals fraction numerator 10 to the power of negative 7 end exponent cross times 40 over denominator 3.14 cross times 10 to the power of negative 2 end exponent end fraction cross times fraction numerator 3 pi over denominator 2 end fraction
    B to the power of ´ ´ end exponent equals 6 cross times 10 to the power of negative 4 end exponent T

    A wire shown in figure carries a current of 40 A. If r equals 3.14 blank c m, the magnetic field at point P will be

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    Let the given circular A B C part of wire subtends an angle theta at its centre. Then, magnetic field due to this circular part is

    B to the power of ´ ´ end exponent equals B subscript c end subscript cross times fraction numerator theta over denominator 2 pi end fraction equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction cross times fraction numerator 2 pi i over denominator e end fraction cross times fraction numerator theta over denominator 2 pi end fraction
    rightwards double arrow blank B to the power of ´ ´ end exponent equals blank fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction blank. fraction numerator i over denominator r end fraction theta
    Given, i equals 40 blank A comma blank r equals 3.14 blank c m equals 3.14 cross times 10 to the power of negative 2 end exponent blank m
    theta equals 360 degree minus 90 degree equals 270 degree equals fraction numerator 3 pi over denominator 2 end fraction blank r a d.
    therefore B to the power of ´ ´ end exponent equals fraction numerator 10 to the power of negative 7 end exponent cross times 40 over denominator 3.14 cross times 10 to the power of negative 2 end exponent end fraction cross times fraction numerator 3 pi over denominator 2 end fraction
    B to the power of ´ ´ end exponent equals 6 cross times 10 to the power of negative 4 end exponent T
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    The figure shows the cross-section of a long cylindrical conductor of radius a carrying a uniformly distributed current i. The magnetic field due to current at P is

    The current enclosed with in the circle
    fraction numerator i over denominator pi a to the power of 2 end exponent end fraction blank comma blank pi r to the power of 2 end exponent equals fraction numerator i over denominator a to the power of 2 end exponent end fraction r to the power of 2 end exponent
    Ampere’s law not stretchy contour integral B. d l equals mu subscript 0 end subscript i ´ gives
    B blank.2 pi r equals fraction numerator mu subscript 0 end subscript i r to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction
    or B equals blank fraction numerator mu subscript 0 end subscript i r over denominator 2 pi a to the power of 2 end exponent end fraction

    The figure shows the cross-section of a long cylindrical conductor of radius a carrying a uniformly distributed current i. The magnetic field due to current at P is

    physics-General
    The current enclosed with in the circle
    fraction numerator i over denominator pi a to the power of 2 end exponent end fraction blank comma blank pi r to the power of 2 end exponent equals fraction numerator i over denominator a to the power of 2 end exponent end fraction r to the power of 2 end exponent
    Ampere’s law not stretchy contour integral B. d l equals mu subscript 0 end subscript i ´ gives
    B blank.2 pi r equals fraction numerator mu subscript 0 end subscript i r to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction
    or B equals blank fraction numerator mu subscript 0 end subscript i r over denominator 2 pi a to the power of 2 end exponent end fraction
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    A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram

    The resultant magnetic field at the centre of the coil is

    Magnetic field a B
    B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 2 pi end fraction fraction numerator i increment l over denominator r end fraction
    Magnetic field at A
    B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 2 pi end fraction fraction numerator i increment l over denominator r end fraction
    The resultant magnetic field at the centre
    open vertical bar B subscript A end subscript close vertical bar equals open vertical bar B subscript B end subscript close vertical bar
    So, magnetic field become is zero.

    A current I enters a circular coil of radius R, branches into two parts and then recombines as shown in the circuit diagram

    The resultant magnetic field at the centre of the coil is

    physics-General
    Magnetic field a B
    B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 2 pi end fraction fraction numerator i increment l over denominator r end fraction
    Magnetic field at A
    B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 2 pi end fraction fraction numerator i increment l over denominator r end fraction
    The resultant magnetic field at the centre
    open vertical bar B subscript A end subscript close vertical bar equals open vertical bar B subscript B end subscript close vertical bar
    So, magnetic field become is zero.
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    In the adjacent figure is shown a closed path P. A long straight conductor carrying a current I passes through O and perpendicular to the plane of the paper. Then which of the following holds good?

    A long straight conductor carrying current I passes through O comma then by symmetry, all points of the circular path are equivalent and hence the magnitude of magnetic field should be same at these points.
    The circulation of magnetic field along the circle is
    not stretchy contour integral B. d l equals mu subscript 0 end subscript I(using Ampere’s law)

    In the adjacent figure is shown a closed path P. A long straight conductor carrying a current I passes through O and perpendicular to the plane of the paper. Then which of the following holds good?

    physics-General
    A long straight conductor carrying current I passes through O comma then by symmetry, all points of the circular path are equivalent and hence the magnitude of magnetic field should be same at these points.
    The circulation of magnetic field along the circle is
    not stretchy contour integral B. d l equals mu subscript 0 end subscript I(using Ampere’s law)
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    A car covers one third part of its straight path with speed V1 and the rest with speed V2 . What is its average speed ?

    A car covers one third part of its straight path with speed V1 and the rest with speed V2 . What is its average speed ?

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    A particle moves 4 m in the south direction. Then it moves 3 m in the west direction. The time taken by the particle is 2 second. What is the ratio between average speed and average velocity

    A particle moves 4 m in the south direction. Then it moves 3 m in the west direction. The time taken by the particle is 2 second. What is the ratio between average speed and average velocity

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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

    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

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    A particle is projected vertically upwards with velocity .30 ms-1 Find the ratio of average speed and instantaneous velocity after 6s.[g=10 ms-1 ]

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    In angiosperms, female gametophyte is represented by

    In angiosperms, female gametophyte is represented by

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    A bus travels between two points A and B. V1 and V2 are it average speed and average velocity then

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    A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the X minus Y plane and a steady current I flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

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    If we take a small strip of d r at distance r from centre, then number of turns in this strip would be,
    d N equals open parentheses fraction numerator N over denominator b minus a end fraction close parentheses d r
    Magnetic field due to this element at the centre of the coil will be
    d B equals fraction numerator mu subscript 0 end subscript open parentheses d N close parentheses I over denominator 2 r end fraction equals fraction numerator mu subscript 0 end subscript N I over denominator 2 open parentheses b minus a close parentheses end fraction fraction numerator d r over denominator r end fraction
    B equals not stretchy integral subscript r equals a end subscript superscript r equals b end superscript d B equals fraction numerator mu subscript 0 end subscript N I over denominator 2 open parentheses b minus a close parentheses end fraction ln invisible function application open parentheses fraction numerator b over denominator a end fraction close parentheses

    A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the X minus Y plane and a steady current I flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

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    physics-General
    If we take a small strip of d r at distance r from centre, then number of turns in this strip would be,
    d N equals open parentheses fraction numerator N over denominator b minus a end fraction close parentheses d r
    Magnetic field due to this element at the centre of the coil will be
    d B equals fraction numerator mu subscript 0 end subscript open parentheses d N close parentheses I over denominator 2 r end fraction equals fraction numerator mu subscript 0 end subscript N I over denominator 2 open parentheses b minus a close parentheses end fraction fraction numerator d r over denominator r end fraction
    B equals not stretchy integral subscript r equals a end subscript superscript r equals b end superscript d B equals fraction numerator mu subscript 0 end subscript N I over denominator 2 open parentheses b minus a close parentheses end fraction ln invisible function application open parentheses fraction numerator b over denominator a end fraction close parentheses