General
Easy
Maths-

Solution of 2 y sin space x fraction numerator d y over denominator d x end fraction equals 2 sin space x cos space x minus y squared cos space x comma x equals pi over 2 comma y equals 1 y equals 1 is given by

Maths-General

  1. y equals sin squared space x
  2. y squared equals sin space x
  3. y squared equals cos space x plus 1
  4. None of these

    Answer:The correct answer is: y squared equals sin space x

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    Related Questions to study

    General
    maths-

    The general solution of the equation fraction numerator d y over denominator d x end fraction equals 1 plus x y is

    The general solution of the equation fraction numerator d y over denominator d x end fraction equals 1 plus x y is

    maths-General
    General
    maths-

    Equation of the curve passing through (3, 9) which satisfies the differential equation fraction numerator d y over denominator d x end fraction equals x plus 1 over x squared is

    Equation of the curve passing through (3, 9) which satisfies the differential equation fraction numerator d y over denominator d x end fraction equals x plus 1 over x squared is

    maths-General
    General
    maths-

    If f left parenthesis x right parenthesis space equals space f apostrophe left parenthesis x right parenthesis and f(1) = 2, then f(3) =

    If f left parenthesis x right parenthesis space equals space f apostrophe left parenthesis x right parenthesis and f(1) = 2, then f(3) =

    maths-General
    General
    physics-

    The figure shows electric potential V as a function of x. Rank the four regions according to the magnitude of x-component of the electric field E within them, greatest first

    Electric field
    E equals negative fraction numerator d V over denominator d x end fraction
    For I region, V subscript 1 end subscript=constant

    therefore blank fraction numerator d V subscript 1 end subscript over denominator d x end fraction equals 0 blank
    therefore blank E subscript 1 end subscript equals 0 blank
    For II region,
    V subscript 2 end subscript equals plus v e equals plus f open parentheses x close parentheses
    therefore E subscript 2 end subscript equals negative fraction numerator d V subscript 2 end subscript over denominator d x end fraction equals negative v e
    For III region.
    V subscript 3 end subscript=constant
    therefore blank fraction numerator d V subscript 3 end subscript over denominator d x end fraction equals 0 blank
    therefore blank E subscript 3 end subscript equals 0 blank
    For IV region, V subscript 1 end subscript equals negative f open parentheses x close parentheses
    therefore blank E subscript 4 end subscript equals negative fraction numerator d V subscript 4 end subscript over denominator d x end fraction equals plus v e
    From these values, we have
    E subscript 2 end subscript greater than E subscript 4 end subscript greater than E subscript 1 end subscript equals E subscript 3 end subscript

    The figure shows electric potential V as a function of x. Rank the four regions according to the magnitude of x-component of the electric field E within them, greatest first

    physics-General
    Electric field
    E equals negative fraction numerator d V over denominator d x end fraction
    For I region, V subscript 1 end subscript=constant

    therefore blank fraction numerator d V subscript 1 end subscript over denominator d x end fraction equals 0 blank
    therefore blank E subscript 1 end subscript equals 0 blank
    For II region,
    V subscript 2 end subscript equals plus v e equals plus f open parentheses x close parentheses
    therefore E subscript 2 end subscript equals negative fraction numerator d V subscript 2 end subscript over denominator d x end fraction equals negative v e
    For III region.
    V subscript 3 end subscript=constant
    therefore blank fraction numerator d V subscript 3 end subscript over denominator d x end fraction equals 0 blank
    therefore blank E subscript 3 end subscript equals 0 blank
    For IV region, V subscript 1 end subscript equals negative f open parentheses x close parentheses
    therefore blank E subscript 4 end subscript equals negative fraction numerator d V subscript 4 end subscript over denominator d x end fraction equals plus v e
    From these values, we have
    E subscript 2 end subscript greater than E subscript 4 end subscript greater than E subscript 1 end subscript equals E subscript 3 end subscript
    General
    physics-

    A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. V subscript A end subscript comma blank V subscript B end subscript comma blank V subscript c end subscriptbe the potentials at points A, B and C respectively. Then

    At each point on the surface of a conducting sphere the potential is equal.
    So, V subscript A end subscript equals V subscript B end subscript equals V subscript C end subscript

    A hollow conducting sphere is placed in an electric field produced by a point charge placed at P as shown in figure. V subscript A end subscript comma blank V subscript B end subscript comma blank V subscript c end subscriptbe the potentials at points A, B and C respectively. Then

    physics-General
    At each point on the surface of a conducting sphere the potential is equal.
    So, V subscript A end subscript equals V subscript B end subscript equals V subscript C end subscript
    General
    physics-

    Three charges Q subscript 0 end subscript comma negative q and – q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q subscript 0 end subscriptis equal to

    Here total electrostatic potential energy is zero
    i e comma blank fraction numerator negative Q subscript 0 end subscript q over denominator l end fraction minus fraction numerator q Q subscript 0 end subscript over denominator l end fraction plus fraction numerator q to the power of 2 end exponent over denominator square root of 2 l end root end fraction equals 0
    On solving,
    Q subscript 0 end subscript equals fraction numerator q over denominator 2 square root of 2 end fraction equals fraction numerator 2 q over denominator square root of 32 end fraction

    Three charges Q subscript 0 end subscript comma negative q and – q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q subscript 0 end subscriptis equal to

    physics-General
    Here total electrostatic potential energy is zero
    i e comma blank fraction numerator negative Q subscript 0 end subscript q over denominator l end fraction minus fraction numerator q Q subscript 0 end subscript over denominator l end fraction plus fraction numerator q to the power of 2 end exponent over denominator square root of 2 l end root end fraction equals 0
    On solving,
    Q subscript 0 end subscript equals fraction numerator q over denominator 2 square root of 2 end fraction equals fraction numerator 2 q over denominator square root of 32 end fraction
    General
    maths-

    The degree and order of the differential equation of all tangent lines to the parabola x2 = 4y is :

    The degree and order of the differential equation of all tangent lines to the parabola x2 = 4y is :

    maths-General
    General
    maths-

    The differential equation of all non-horizontal lines in a plane is :

    The differential equation of all non-horizontal lines in a plane is :

    maths-General
    General
    maths-

    The differential equation of all non-vertical lines in a plane is :

    The differential equation of all non-vertical lines in a plane is :

    maths-General
    General
    maths-

    The differential equation of all conics with the coordinate axes, is of order

    The differential equation of all conics with the coordinate axes, is of order

    maths-General
    General
    maths-

    If the algebraic sum of distances of points (2, 1) (3, 2) and (­-4, 7) from the line y = mx + c is zero, then this line will always pass through a fixed point whose coordinate is

     

    If the algebraic sum of distances of points (2, 1) (3, 2) and (­-4, 7) from the line y = mx + c is zero, then this line will always pass through a fixed point whose coordinate is

     

    maths-General
    General
    physics-

    In the following diagram the work done in moving a point charge from point P to point A, B and C is respectively as W subscript A end subscript comma blank W subscript B end subscript a n d W subscript C end subscript then

    When a positive charge is moved from one point to another in an electric of magnetic field, then under the influence of the field force acts on the particle and an external agent will have to do work against this force. But in the given case the charge moves under influence of no field, hence it does not experience any force therefore, no work is done.
    W subscript A end subscript equals W subscript B end subscript equals W subscript C end subscript equals 0

    In the following diagram the work done in moving a point charge from point P to point A, B and C is respectively as W subscript A end subscript comma blank W subscript B end subscript a n d W subscript C end subscript then

    physics-General
    When a positive charge is moved from one point to another in an electric of magnetic field, then under the influence of the field force acts on the particle and an external agent will have to do work against this force. But in the given case the charge moves under influence of no field, hence it does not experience any force therefore, no work is done.
    W subscript A end subscript equals W subscript B end subscript equals W subscript C end subscript equals 0
    General
    physics-

    The points resembling equal potentials are

    The points S and R are inside the uniform electric field, so these will be at equal potential.

    The points resembling equal potentials are

    physics-General
    The points S and R are inside the uniform electric field, so these will be at equal potential.
    General
    physics-

    Work required to set up the four charge configuration (as shown in the figure) is

    Work is required to set up the four charge configuration
    q subscript 1 end subscript equals plus q comma blank q subscript 2 end subscript equals negative q comma blank q subscript 3 end subscript equals plus q a n d q subscript 4 end subscript equals negative q

    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator A B end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator B C end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator C D end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator D A end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses plus q close parentheses over denominator A C end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses negative q close parentheses over denominator B D end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets negative fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator a end fraction open square brackets negative 4 plus square root of 2 close square brackets equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator a end fraction open square brackets negative 4 plus 1.414 close square brackets
    W equals negative 0.21 cross times fraction numerator q to the power of 2 end exponent over denominator epsilon subscript 0 end subscript a end fraction open parentheses a p p r o x. close parentheses

    Work required to set up the four charge configuration (as shown in the figure) is

    physics-General
    Work is required to set up the four charge configuration
    q subscript 1 end subscript equals plus q comma blank q subscript 2 end subscript equals negative q comma blank q subscript 3 end subscript equals plus q a n d q subscript 4 end subscript equals negative q

    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator A B end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator B C end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses negative q close parentheses over denominator C D end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses plus q close parentheses over denominator D A end fraction plus fraction numerator open parentheses plus q close parentheses open parentheses plus q close parentheses over denominator A C end fraction plus fraction numerator open parentheses negative q close parentheses open parentheses negative q close parentheses over denominator B D end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open square brackets negative fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction minus fraction numerator q to the power of 2 end exponent over denominator a end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction plus fraction numerator q to the power of 2 end exponent over denominator a square root of 2 end fraction close square brackets
    W equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator a end fraction open square brackets negative 4 plus square root of 2 close square brackets equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction fraction numerator q to the power of 2 end exponent over denominator a end fraction open square brackets negative 4 plus 1.414 close square brackets
    W equals negative 0.21 cross times fraction numerator q to the power of 2 end exponent over denominator epsilon subscript 0 end subscript a end fraction open parentheses a p p r o x. close parentheses
    General
    physics-

    Consider three concentric shells of metal A, B and C are having radii a, b and c respectively as shown in the figure left parenthesis a less than b less than c right parenthesis. Their surface charge densities are sigma comma negative sigma a n d sigma respectively. Calculate the electric potential on the surface of shell A

    The electric potential on the surface of shell
    V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript a end subscript over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction plus fraction numerator q subscript b end subscript over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript c end subscript over denominator c end fraction
    Or V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi a to the power of 2 end exponent sigma over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi b to the power of 2 end exponent open parentheses negative sigma close parentheses over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi c to the power of 2 end exponent sigma over denominator c end fraction
    open parentheses because q equals 4 pi r to the power of 2 end exponent sigma close parentheses
    o r blank V subscript A end subscript equals fraction numerator sigma over denominator epsilon subscript 0 end subscript end fraction left parenthesis a minus b plus c right parenthesis

    Consider three concentric shells of metal A, B and C are having radii a, b and c respectively as shown in the figure left parenthesis a less than b less than c right parenthesis. Their surface charge densities are sigma comma negative sigma a n d sigma respectively. Calculate the electric potential on the surface of shell A

    physics-General
    The electric potential on the surface of shell
    V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript a end subscript over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction plus fraction numerator q subscript b end subscript over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator q subscript c end subscript over denominator c end fraction
    Or V subscript A end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi a to the power of 2 end exponent sigma over denominator a end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi b to the power of 2 end exponent open parentheses negative sigma close parentheses over denominator b end fraction plus fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction. fraction numerator 4 pi c to the power of 2 end exponent sigma over denominator c end fraction
    open parentheses because q equals 4 pi r to the power of 2 end exponent sigma close parentheses
    o r blank V subscript A end subscript equals fraction numerator sigma over denominator epsilon subscript 0 end subscript end fraction left parenthesis a minus b plus c right parenthesis