General
Easy
Maths-

If m and n are order and degree of the equatioopen parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses to the power of 5 plus 4 times fraction numerator open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed over denominator fraction numerator d cubed y over denominator d x cubed end fraction end fraction plus fraction numerator d cubed y over denominator d x cubed end fraction equals x squared minus 1 then

Maths-General

  1. m = 3, n = 2
  2. m = 3, n = 3
  3. m = 3, n = 1
  4. m = 3, n = 5

    Answer:The correct answer is: m = 3, n = 2

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    Related Questions to study

    General
    physics-

    What is the potential difference between points A a n d B in the circuit shown?

    Consider the charge distribution as shown. Considering the branch on upper side, we have

    fraction numerator q over denominator V subscript x end subscript minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent
    fraction numerator q over denominator V subscript A end subscript minus V subscript y end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent
    Here, V subscript x end subscript equals 6 blank v o l t comma blank V subscript y end subscript equals 0
    therefore fraction numerator q over denominator 6 minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent…(i)
    fraction numerator q over denominator V subscript A end subscript minus 0 end fraction equals 2 cross times 10 to the power of negative 6 end exponent …(ii)
    From Eqs. (i) and (ii), we get
    fraction numerator V subscript A end subscript over denominator 6 minus V subscript A end subscript end fraction equals 2
    therefore V subscript A end subscript equals 4 v o l t
    Similarly for the lower side branch
    fraction numerator q ´ ´ over denominator 6 minus V subscript B end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent…(iii)
    fraction numerator q ´ ´ over denominator V subscript B end subscript minus 0 end fraction equals blank 4 cross times 10 to the power of negative 6 end exponent...(iv)
    From Eqs. (iii) and (iv)
    fraction numerator V subscript B end subscript over denominator 6 minus V subscript B end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction
    therefore blank V subscript B end subscript equals 2 blank v o l t
    therefore V subscript A end subscript minus V subscript B end subscript equals 4 minus 2 equals 2 blank v o l t

    What is the potential difference between points A a n d B in the circuit shown?

    physics-General
    Consider the charge distribution as shown. Considering the branch on upper side, we have

    fraction numerator q over denominator V subscript x end subscript minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent
    fraction numerator q over denominator V subscript A end subscript minus V subscript y end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent
    Here, V subscript x end subscript equals 6 blank v o l t comma blank V subscript y end subscript equals 0
    therefore fraction numerator q over denominator 6 minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent…(i)
    fraction numerator q over denominator V subscript A end subscript minus 0 end fraction equals 2 cross times 10 to the power of negative 6 end exponent …(ii)
    From Eqs. (i) and (ii), we get
    fraction numerator V subscript A end subscript over denominator 6 minus V subscript A end subscript end fraction equals 2
    therefore V subscript A end subscript equals 4 v o l t
    Similarly for the lower side branch
    fraction numerator q ´ ´ over denominator 6 minus V subscript B end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent…(iii)
    fraction numerator q ´ ´ over denominator V subscript B end subscript minus 0 end fraction equals blank 4 cross times 10 to the power of negative 6 end exponent...(iv)
    From Eqs. (iii) and (iv)
    fraction numerator V subscript B end subscript over denominator 6 minus V subscript B end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction
    therefore blank V subscript B end subscript equals 2 blank v o l t
    therefore V subscript A end subscript minus V subscript B end subscript equals 4 minus 2 equals 2 blank v o l t
    General
    physics-

    What is the potential difference across 2muF capacitor in the circuit shown?


    Net emf in the circuit here
    E equals E subscript 2 end subscript minus E subscript 1 end subscript equals 16 minus 6 equals 10 volt
    While the equivalent capacity
    C equals blank fraction numerator C subscript 1 end subscript C subscript 2 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction equals blank fraction numerator 2 cross times 3 over denominator 2 plus 3 end fraction equals fraction numerator 6 over denominator 5 end fraction mu F
    Charge on each capacitor
    q equals C V equals fraction numerator 6 over denominator 5 end fraction cross times 10 equals 12 mu C
    therefore Potential difference across 2 mu F capacitor
    V subscript 1 end subscript equals fraction numerator q over denominator C subscript 1 end subscript end fraction equals fraction numerator 12 over denominator 2 end fraction equals 6 blank v o l t

    What is the potential difference across 2muF capacitor in the circuit shown?

    physics-General

    Net emf in the circuit here
    E equals E subscript 2 end subscript minus E subscript 1 end subscript equals 16 minus 6 equals 10 volt
    While the equivalent capacity
    C equals blank fraction numerator C subscript 1 end subscript C subscript 2 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction equals blank fraction numerator 2 cross times 3 over denominator 2 plus 3 end fraction equals fraction numerator 6 over denominator 5 end fraction mu F
    Charge on each capacitor
    q equals C V equals fraction numerator 6 over denominator 5 end fraction cross times 10 equals 12 mu C
    therefore Potential difference across 2 mu F capacitor
    V subscript 1 end subscript equals fraction numerator q over denominator C subscript 1 end subscript end fraction equals fraction numerator 12 over denominator 2 end fraction equals 6 blank v o l t
    General
    physics-

    A 2mu F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch s is turned to positions 2 is

    q subscript i end subscript equals C subscript i end subscript V equals 2 V equals q (say)
    This charge will remain constant after switch is shifted from position 1 to position 2.
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 2 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 10 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 20 end fraction
    thereforeEnergy dissipatedopen parentheses fraction numerator q to the power of 2 end exponent over denominator 5 end fraction close parentheses i s blank80% of the initial stored energy open parentheses equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction close parentheses.

    A 2mu F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch s is turned to positions 2 is

    physics-General
    q subscript i end subscript equals C subscript i end subscript V equals 2 V equals q (say)
    This charge will remain constant after switch is shifted from position 1 to position 2.
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 2 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 10 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 20 end fraction
    thereforeEnergy dissipatedopen parentheses fraction numerator q to the power of 2 end exponent over denominator 5 end fraction close parentheses i s blank80% of the initial stored energy open parentheses equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction close parentheses.
    General
    physics-

    Two insulating plates are both uniformly charged in such a way that the potential difference between them is V subscript 2 end subscript minus V subscript 1 end subscript equals 20V. (i e, plate 2 is at a higher potential). The plates are separated by d equals 0.1m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e=1.6cross times 10 to the power of negative 19 end exponent C comma m subscript 0 end subscript equals 9.11 cross times 10 to the power of negative 31 end exponent k g right parenthesis

    Since V subscript 2 end subscript greater than V subscript 1 end subscript comma so electric field will point from plate 2 to plate 1.
    The electron will experience an electric force, opposite to the direction of electric field, and hence move towards the plate 2.

    Use work-energy theorem to find speed of electron when it strikes the plate 2.
    fraction numerator m subscript e end subscript v to the power of 2 end exponent over denominator 2 end fraction minus 0 equals e left parenthesis V subscript 2 end subscript minus V subscript 1 end subscript right parenthesis
    Where v is the required speed.
    therefore blank fraction numerator 9.11 cross times 10 to the power of negative 31 end exponent over denominator 2 end fraction v to the power of 2 end exponent equals 1.6 cross times 10 to the power of negative 19 end exponent cross times 20
    rightwards double arrow blank v equals blank square root of fraction numerator 1.6 cross times 10 to the power of negative 19 end exponent cross times 40 over denominator 9.11 cross times 10 to the power of negative 31 end exponent end fraction end root equals 2.65 cross times 10 to the power of 6 end exponent m s to the power of negative 1 end exponent

    Two insulating plates are both uniformly charged in such a way that the potential difference between them is V subscript 2 end subscript minus V subscript 1 end subscript equals 20V. (i e, plate 2 is at a higher potential). The plates are separated by d equals 0.1m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e=1.6cross times 10 to the power of negative 19 end exponent C comma m subscript 0 end subscript equals 9.11 cross times 10 to the power of negative 31 end exponent k g right parenthesis

    physics-General
    Since V subscript 2 end subscript greater than V subscript 1 end subscript comma so electric field will point from plate 2 to plate 1.
    The electron will experience an electric force, opposite to the direction of electric field, and hence move towards the plate 2.

    Use work-energy theorem to find speed of electron when it strikes the plate 2.
    fraction numerator m subscript e end subscript v to the power of 2 end exponent over denominator 2 end fraction minus 0 equals e left parenthesis V subscript 2 end subscript minus V subscript 1 end subscript right parenthesis
    Where v is the required speed.
    therefore blank fraction numerator 9.11 cross times 10 to the power of negative 31 end exponent over denominator 2 end fraction v to the power of 2 end exponent equals 1.6 cross times 10 to the power of negative 19 end exponent cross times 20
    rightwards double arrow blank v equals blank square root of fraction numerator 1.6 cross times 10 to the power of negative 19 end exponent cross times 40 over denominator 9.11 cross times 10 to the power of negative 31 end exponent end fraction end root equals 2.65 cross times 10 to the power of 6 end exponent m s to the power of negative 1 end exponent
    General
    physics-

    The load versus extension graph for four wires of same material is shown. The thinnest wire is represented by the line

    The load versus extension graph for four wires of same material is shown. The thinnest wire is represented by the line

    physics-General
    General
    physics-

    The graph shows the change ' ' Dl in the length of a thin uniform wire used by the application of force ‘F’ at different temperatures T1 and T2 The variation suggests that

    The graph shows the change ' ' Dl in the length of a thin uniform wire used by the application of force ‘F’ at different temperatures T1 and T2 The variation suggests that

    physics-General
    General
    physics-

    In the experiment to determine Young’s modulus of the material of a wire under tension used in the arrangement as shown. The percentage error in the measurement of length is ‘a’, in the measurement of the radius of the wire is ‘b’ and in the measurement of the change in length of the wire is ‘c’. Percentage error in the measurement of Young’s modulus for a given load is

    In the experiment to determine Young’s modulus of the material of a wire under tension used in the arrangement as shown. The percentage error in the measurement of length is ‘a’, in the measurement of the radius of the wire is ‘b’ and in the measurement of the change in length of the wire is ‘c’. Percentage error in the measurement of Young’s modulus for a given load is

    physics-General
    General
    physics-

    The stress - strain graphs for materials A and B are as shown. Choose the correct alternative

    The stress - strain graphs for materials A and B are as shown. Choose the correct alternative

    physics-General
    General
    physics-

    Calculate the force ' ' F that is applied horizontally at the axle of the wheel which is necessary to raise the wheel over the obstacle of height 4m. Radius of the wheel is 1m and its mass is open parentheses g equals 10 m s to the power of negative 2 end exponent close parentheses

    applying principle of moments
    F (0.6)=100(0.8)
    ÞF=133.3N

    Calculate the force ' ' F that is applied horizontally at the axle of the wheel which is necessary to raise the wheel over the obstacle of height 4m. Radius of the wheel is 1m and its mass is open parentheses g equals 10 m s to the power of negative 2 end exponent close parentheses

    physics-General
    applying principle of moments
    F (0.6)=100(0.8)
    ÞF=133.3N
    General
    physics-

    A wheel of radius ‘r’ and mass ‘m’ stands in front of a step of height 'h’ The least horizontal force which should be applied to the axle of the wheel to allow it to raise onto the step is

    Applying the condition of rotational equilibrium,
    F left parenthesis r minus h right parenthesis equals m g x
    But r to the power of 2 end exponent equals x to the power of 2 end exponent plus left parenthesis r minus h right parenthesis to the power of 2 end exponent rightwards double arrow x equals square root of h left parenthesis 2 r minus h right parenthesis end root
    therefore F equals fraction numerator m g square root of h left parenthesis 2 r minus h right parenthesis end root over denominator r minus h end fraction

    A wheel of radius ‘r’ and mass ‘m’ stands in front of a step of height 'h’ The least horizontal force which should be applied to the axle of the wheel to allow it to raise onto the step is

    physics-General
    Applying the condition of rotational equilibrium,
    F left parenthesis r minus h right parenthesis equals m g x
    But r to the power of 2 end exponent equals x to the power of 2 end exponent plus left parenthesis r minus h right parenthesis to the power of 2 end exponent rightwards double arrow x equals square root of h left parenthesis 2 r minus h right parenthesis end root
    therefore F equals fraction numerator m g square root of h left parenthesis 2 r minus h right parenthesis end root over denominator r minus h end fraction
    General
    maths-

    The order of differential equation open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed equals open parentheses 1 plus fraction numerator d y over denominator d x end fraction close parentheses to the power of 1 divided by 2 end exponent is

    The order of differential equation open parentheses fraction numerator d squared y over denominator d x squared end fraction close parentheses cubed equals open parentheses 1 plus fraction numerator d y over denominator d x end fraction close parentheses to the power of 1 divided by 2 end exponent is

    maths-General
    General
    maths-

    The solution of the differential equation 2 x fraction numerator d y over denominator d x end fraction minus y equals 3 represent

    The solution of the differential equation 2 x fraction numerator d y over denominator d x end fraction minus y equals 3 represent

    maths-General
    General
    physics-

    Two condensers, one of capacity C and the other of capacity fraction numerator C over denominator 2 end fraction, are connected to a V volt battery , as shown. The work done in charging fully both the condensers is

    The two capacitor the circuit are in parallel order, hence
    C to the power of ´ end exponent equals C plus fraction numerator C over denominator 2 end fraction equals fraction numerator 3 C over denominator 2 end fraction
    The work done in charging the equivalent capacitor is stored in the form of potential energy.
    Hence, W equals U equals fraction numerator 1 over denominator 2 end fraction C to the power of ´ end exponent V to the power of 2 end exponent
    equals fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator 3 C over denominator 2 end fraction close parentheses V to the power of 2 end exponent
    equals fraction numerator 3 over denominator 4 end fraction C V to the power of 2 end exponent

    Two condensers, one of capacity C and the other of capacity fraction numerator C over denominator 2 end fraction, are connected to a V volt battery , as shown. The work done in charging fully both the condensers is

    physics-General
    The two capacitor the circuit are in parallel order, hence
    C to the power of ´ end exponent equals C plus fraction numerator C over denominator 2 end fraction equals fraction numerator 3 C over denominator 2 end fraction
    The work done in charging the equivalent capacitor is stored in the form of potential energy.
    Hence, W equals U equals fraction numerator 1 over denominator 2 end fraction C to the power of ´ end exponent V to the power of 2 end exponent
    equals fraction numerator 1 over denominator 2 end fraction open parentheses fraction numerator 3 C over denominator 2 end fraction close parentheses V to the power of 2 end exponent
    equals fraction numerator 3 over denominator 4 end fraction C V to the power of 2 end exponent
    General
    physics-

    In the capacitor shown in the circuit is changed to 5 V and left in the circuit, in 12s the charge on the capacitor will become left parenthesis e equals 2.718 right parenthesis

    Final charge on capacitor is
    q equals q subscript 0 end subscript e to the power of negative t divided by R C end exponent
    where q subscript 0 end subscript= charge on capacitor at t equals 0 comma
    R C=time constant of the circuit.
    Putting q subscript 0 end subscript equals C V subscript 0 end subscript
    therefore q equals C V subscript 0 end subscript superscript negative 1 divided by R C end superscript
    Given, C equals 2 F comma V subscript 0 end subscript equals 5 blank v o l t comma R equals 6 blank capital omega comma blank t equals 12 blank s
    Hence, q equals open parentheses 2 cross times 5 close parentheses e to the power of negative left parenthesis 12 divided by 6 cross times 2 right parenthesis end exponent
    equals 10 e to the power of negative 1 end exponent equals fraction numerator 10 over denominator e end fraction C

    In the capacitor shown in the circuit is changed to 5 V and left in the circuit, in 12s the charge on the capacitor will become left parenthesis e equals 2.718 right parenthesis

    physics-General
    Final charge on capacitor is
    q equals q subscript 0 end subscript e to the power of negative t divided by R C end exponent
    where q subscript 0 end subscript= charge on capacitor at t equals 0 comma
    R C=time constant of the circuit.
    Putting q subscript 0 end subscript equals C V subscript 0 end subscript
    therefore q equals C V subscript 0 end subscript superscript negative 1 divided by R C end superscript
    Given, C equals 2 F comma V subscript 0 end subscript equals 5 blank v o l t comma R equals 6 blank capital omega comma blank t equals 12 blank s
    Hence, q equals open parentheses 2 cross times 5 close parentheses e to the power of negative left parenthesis 12 divided by 6 cross times 2 right parenthesis end exponent
    equals 10 e to the power of negative 1 end exponent equals fraction numerator 10 over denominator e end fraction C
    General
    physics-

    A cubical block of side ‘L’ rests on a rough horizontal surface with coefficient of friction ‘m’ A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

    applying the condition of rotational equilibrium
    F left parenthesis L right parenthesis equals m g open square brackets fraction numerator L over denominator 2 end fraction close square brackets semicolon rightwards double arrow F equals fraction numerator m g over denominator 2 end fraction

    A cubical block of side ‘L’ rests on a rough horizontal surface with coefficient of friction ‘m’ A horizontal force F is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is

    physics-General
    applying the condition of rotational equilibrium
    F left parenthesis L right parenthesis equals m g open square brackets fraction numerator L over denominator 2 end fraction close square brackets semicolon rightwards double arrow F equals fraction numerator m g over denominator 2 end fraction