General
Easy
Maths-

If y equals e to the power of negative x end exponent left parenthesis A cos space x plus B sin space x right parenthesis then y satisfies

Maths-General

  1. fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction equals 0
  2. fraction numerator d squared y over denominator d x squared end fraction minus 2 fraction numerator d y over denominator d x end fraction plus 2 y equals 0
  3. fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction plus 2 y equals 0
  4. fraction numerator d squared y over denominator d x squared end fraction plus 2 y equals 0

    Answer:The correct answer is: fraction numerator d squared y over denominator d x squared end fraction plus 2 fraction numerator d y over denominator d x end fraction plus 2 y equals 0

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    Related Questions to study

    General
    physics-

    The charge deposited on 4 mu F capacitor the circuit is


    As the capacitors 4 mu F and 2 mu F are connected in parallel and are in series with 6blank mu F capacitor, their equivalent capacitance is
    fraction numerator left parenthesis 2 plus 4 right parenthesis cross times 6 over denominator 2 plus 4 plus 6 end fraction equals 3 blank mu F
    Charge in the circuit
    Q equals blank 3 blank mu F cross times 12 V equals 36 mu C

    Since, the capacitors 4 mu F and 2mu F are connected in parallel, therefore potential difference across them is same.
    rightwards double arrow fraction numerator Q subscript 1 end subscript over denominator Q subscript 2 end subscript end fraction equals fraction numerator C subscript 1 end subscript over denominator C subscript 2 end subscript end fraction equals fraction numerator 4 over denominator 2 end fraction o r Q subscript 1 end subscript equals 2 Q subscript 2 end subscript
    Also Q equals Q subscript 1 end subscript plus Q subscript 2 end subscript
    therefore blank 36 mu C equals 2 Q subscript 2 end subscript plus Q subscript 2 end subscript or Q subscript 2 end subscript equals fraction numerator 36 mu C over denominator 3 end fraction equals 12 mu C
    Q subscript 1 end subscript equals Q minus Q subscript 2 end subscript equals 36 mu C minus 12 mu C
    equals 24 blank mu C equals 24 cross times 10 to the power of negative 6 end exponent C

    The charge deposited on 4 mu F capacitor the circuit is

    physics-General

    As the capacitors 4 mu F and 2 mu F are connected in parallel and are in series with 6blank mu F capacitor, their equivalent capacitance is
    fraction numerator left parenthesis 2 plus 4 right parenthesis cross times 6 over denominator 2 plus 4 plus 6 end fraction equals 3 blank mu F
    Charge in the circuit
    Q equals blank 3 blank mu F cross times 12 V equals 36 mu C

    Since, the capacitors 4 mu F and 2mu F are connected in parallel, therefore potential difference across them is same.
    rightwards double arrow fraction numerator Q subscript 1 end subscript over denominator Q subscript 2 end subscript end fraction equals fraction numerator C subscript 1 end subscript over denominator C subscript 2 end subscript end fraction equals fraction numerator 4 over denominator 2 end fraction o r Q subscript 1 end subscript equals 2 Q subscript 2 end subscript
    Also Q equals Q subscript 1 end subscript plus Q subscript 2 end subscript
    therefore blank 36 mu C equals 2 Q subscript 2 end subscript plus Q subscript 2 end subscript or Q subscript 2 end subscript equals fraction numerator 36 mu C over denominator 3 end fraction equals 12 mu C
    Q subscript 1 end subscript equals Q minus Q subscript 2 end subscript equals 36 mu C minus 12 mu C
    equals 24 blank mu C equals 24 cross times 10 to the power of negative 6 end exponent C
    General
    physics-

    Three capacitors C subscript 1 end subscript comma C subscript 2 end subscript a n d C subscript 3 end subscript are connected as shown in the figure to a battery of V volt. If the capacitor C subscript 3 end subscriptbreaks down electrically the change in total charge on the combination of capacitors is

    Since,C subscript 1 end subscript a n d blank C subscript 2 end subscript are parallel to their equivalent capacitance will be left parenthesis C subscript 1 end subscript plus C subscript 2 end subscript right parenthesis. Now, open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses a n d blank C subscript 3 end subscript are in series, so the net equivalent capacitances of circuit.
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C to the power of 3 end exponent end fraction plus fraction numerator 1 over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction
    equals blank fraction numerator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript over denominator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript end fraction
    C equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    Since, V is the voltage of the battery, so charge on this system
    q equals C V
    q equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    If the capacitor C subscript 3 end subscript breaks down then total equivalent capacitance
    C to the power of ´ end exponent equals blank C subscript 1 end subscript plus C subscript 2 end subscript
    therefore New charge stored
    q to the power of ´ end exponent equals C ´ V
    q ´ equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V
    Change in total charge
    increment q equals q to the power of ´ end exponent minus q left parenthesis because q to the power of ´ end exponent greater than q right parenthesis
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V minus fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets
    increment q equals blank open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets

    Three capacitors C subscript 1 end subscript comma C subscript 2 end subscript a n d C subscript 3 end subscript are connected as shown in the figure to a battery of V volt. If the capacitor C subscript 3 end subscriptbreaks down electrically the change in total charge on the combination of capacitors is

    physics-General
    Since,C subscript 1 end subscript a n d blank C subscript 2 end subscript are parallel to their equivalent capacitance will be left parenthesis C subscript 1 end subscript plus C subscript 2 end subscript right parenthesis. Now, open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses a n d blank C subscript 3 end subscript are in series, so the net equivalent capacitances of circuit.
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator C to the power of 3 end exponent end fraction plus fraction numerator 1 over denominator C subscript 1 end subscript plus C subscript 2 end subscript end fraction
    equals blank fraction numerator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript over denominator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript end fraction
    C equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    Since, V is the voltage of the battery, so charge on this system
    q equals C V
    q equals blank fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    If the capacitor C subscript 3 end subscript breaks down then total equivalent capacitance
    C to the power of ´ end exponent equals blank C subscript 1 end subscript plus C subscript 2 end subscript
    therefore New charge stored
    q to the power of ´ end exponent equals C ´ V
    q ´ equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V
    Change in total charge
    increment q equals q to the power of ´ end exponent minus q left parenthesis because q to the power of ´ end exponent greater than q right parenthesis
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V minus fraction numerator open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses C subscript 3 end subscript V over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction
    equals open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets
    increment q equals blank open parentheses C subscript 1 end subscript plus C subscript 2 end subscript close parentheses V open square brackets 1 minus fraction numerator C subscript 3 end subscript over denominator C subscript 1 end subscript plus C subscript 2 end subscript plus C subscript 3 end subscript end fraction close square brackets
    General
    physics-

    For the given figure, calculate zero correction.

    For the given figure, calculate zero correction.

    physics-General
    General
    physics-

    In the figure, a proton moves a distance d in a uniform electric field E as shown in the figure. Does the electric field do a positive or negative work on the proton? Does the electric potential energy of the proton increase or decrease?

    Since, the proton is moving against the direction of electric field so, work is done by the proton against electric field. It implies that electric field does negative work on the proton.
    Again, proton is moving in electric field from low potential region to high potential region hence, its potential energy increases.

    In the figure, a proton moves a distance d in a uniform electric field E as shown in the figure. Does the electric field do a positive or negative work on the proton? Does the electric potential energy of the proton increase or decrease?

    physics-General
    Since, the proton is moving against the direction of electric field so, work is done by the proton against electric field. It implies that electric field does negative work on the proton.
    Again, proton is moving in electric field from low potential region to high potential region hence, its potential energy increases.
    General
    maths-

    Solution of fraction numerator d y over denominator d x end fraction equals e to the power of y plus x end exponent plus e to the power of y minus x end exponent is

    Solution of fraction numerator d y over denominator d x end fraction equals e to the power of y plus x end exponent plus e to the power of y minus x end exponent is

    maths-General
    General
    maths-

    The solution of x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x is

    The solution of x squared fraction numerator d y over denominator d x end fraction minus x y equals 1 plus cos space y over x is

    maths-General
    General
    maths-

    Order and degree of differential equation fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent are

    Order and degree of differential equation fraction numerator d squared y over denominator d x squared end fraction equals open square brackets y plus open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared close square brackets to the power of 1 divided by 4 end exponent are

    maths-General
    General
    physics-

    Two charges q subscript 1 end subscriptand q subscript 2 end subscript are placed 30 cm apart, as shown in the figure. A third charge q subscript 3 end subscriptis moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k comma where k is

    When charge q subscript 3 end subscriptis at C, then its potential energy is
    U subscript C end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    Where charge q subscript 3 end subscriptis at D, then
    U subscript D end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction close parentheses
    Hence, change in potential energy
    increment U equals U subscript D end subscript minus U subscript C end subscript
    equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    b u t blank increment U equals fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k
    therefore fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    ⟹ k equals q subscript 2 end subscript open parentheses 10 minus 2 close parentheses equals 8 q subscript 2 end subscript

    Two charges q subscript 1 end subscriptand q subscript 2 end subscript are placed 30 cm apart, as shown in the figure. A third charge q subscript 3 end subscriptis moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k comma where k is

    physics-General
    When charge q subscript 3 end subscriptis at C, then its potential energy is
    U subscript C end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    Where charge q subscript 3 end subscriptis at D, then
    U subscript D end subscript equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 1 end subscript q subscript 3 end subscript over denominator 0.4 end fraction plus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction close parentheses
    Hence, change in potential energy
    increment U equals U subscript D end subscript minus U subscript C end subscript
    equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    b u t blank increment U equals fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k
    therefore fraction numerator q subscript 3 end subscript over denominator 4 pi epsilon subscript 0 end subscript end fraction k equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end subscript end fraction open parentheses fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.1 end fraction minus fraction numerator q subscript 2 end subscript q subscript 3 end subscript over denominator 0.5 end fraction close parentheses
    ⟹ k equals q subscript 2 end subscript open parentheses 10 minus 2 close parentheses equals 8 q subscript 2 end subscript
    General
    physics-

    If the time period of a pendulum is 1 sec, then what is the length of the pendulum at point of intersection of l-T and l minus T to the power of 2 end exponent graph

    T equals 2 pi square root of fraction numerator l over denominator g end fraction end root semicolon T to the power of 2 end exponent equals 4 pi to the power of 2 end exponent fraction numerator l over denominator g end fraction
    l equals fraction numerator g over denominator 4 pi to the power of 2 end exponent end fraction equals fraction numerator 9.8 over denominator 4 cross times 9.8 end fraction equals 0.25 equals 25 c m

    If the time period of a pendulum is 1 sec, then what is the length of the pendulum at point of intersection of l-T and l minus T to the power of 2 end exponent graph

    physics-General
    T equals 2 pi square root of fraction numerator l over denominator g end fraction end root semicolon T to the power of 2 end exponent equals 4 pi to the power of 2 end exponent fraction numerator l over denominator g end fraction
    l equals fraction numerator g over denominator 4 pi to the power of 2 end exponent end fraction equals fraction numerator 9.8 over denominator 4 cross times 9.8 end fraction equals 0.25 equals 25 c m
    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
    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
    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
    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
    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-

    Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential differenceV subscript A end subscript minus V subscript B end subscript equals V subscript B end subscript minus V subscript C end subscript. If t subscript 1 end subscriptand t subscript 2 end subscriptbe the distance between them. Then

    Potential difference between two equipotential surfaces A and B.
    V subscript A end subscript minus V subscript B end subscript equals k q open parentheses fraction numerator 1 over denominator r subscript A end subscript end fraction minus fraction numerator 1 over denominator r subscript B end subscript end fraction close parentheses
    equals k q open parentheses fraction numerator r subscript B end subscript minus r subscript A end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction close parentheses
    equals fraction numerator k q t subscript 1 end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction
    Or
    t subscript 1 end subscript equals fraction numerator open parentheses V subscript A end subscript minus V subscript B end subscript close parentheses r subscript A end subscript r subscript B end subscript over denominator k q end fraction
    Or t subscript 1 end subscript proportional to r subscript A end subscript r subscript B end subscript
    Similarly, t subscript 2 end subscript proportional to r subscript B end subscript r subscript C end subscript
    Since,
    r subscript A end subscript less than r subscript B end subscript less than r subscript C end subscript comma therefore r subscript A end subscript r subscript B end subscript less than r subscript B end subscript r subscript C end subscript
    therefore blank t subscript 1 end subscript less than t subscript 2 end subscript

    Figure shows three spherical and equipotential surfaces A, B and C round a point charge q. The potential differenceV subscript A end subscript minus V subscript B end subscript equals V subscript B end subscript minus V subscript C end subscript. If t subscript 1 end subscriptand t subscript 2 end subscriptbe the distance between them. Then

    physics-General
    Potential difference between two equipotential surfaces A and B.
    V subscript A end subscript minus V subscript B end subscript equals k q open parentheses fraction numerator 1 over denominator r subscript A end subscript end fraction minus fraction numerator 1 over denominator r subscript B end subscript end fraction close parentheses
    equals k q open parentheses fraction numerator r subscript B end subscript minus r subscript A end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction close parentheses
    equals fraction numerator k q t subscript 1 end subscript over denominator r subscript A end subscript r subscript B end subscript end fraction
    Or
    t subscript 1 end subscript equals fraction numerator open parentheses V subscript A end subscript minus V subscript B end subscript close parentheses r subscript A end subscript r subscript B end subscript over denominator k q end fraction
    Or t subscript 1 end subscript proportional to r subscript A end subscript r subscript B end subscript
    Similarly, t subscript 2 end subscript proportional to r subscript B end subscript r subscript C end subscript
    Since,
    r subscript A end subscript less than r subscript B end subscript less than r subscript C end subscript comma therefore r subscript A end subscript r subscript B end subscript less than r subscript B end subscript r subscript C end subscript
    therefore blank t subscript 1 end subscript less than t subscript 2 end subscript