Physics-
General
Easy

Question

From an inclined plane two particles are projected with same speed at same angle theta, one up and other down the plane as shown in figure. Which of the following statements left parenthesis s right parenthesis is/are correct?

  1. The time of flight of each particle is the same.    
  2. The particles will collide the plane with same speed    
  3. Both the particles strike the plane perpendicularly    
  4. The particles will collide in mid air if projected simultaneously and time of flight of each particle is less than the time of collision    

The correct answer is: The time of flight of each particle is the same.


    Here, alpha equals 2 theta comma beta equals theta

    Time of flight of A is, T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application left parenthesis alpha minus beta right parenthesis over denominator g blank c o s blank beta end fraction
    equals fraction numerator 2 u sin invisible function application left parenthesis 2 theta minus theta right parenthesis over denominator g blank c o s blank theta end fraction equals fraction numerator 2 u over denominator g end fraction t a n invisible function application theta
    Time of flight of B is, T subscript 2 end subscript equals fraction numerator 2 u sin invisible function application theta over denominator g cos invisible function application theta end fraction equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta
    So, T subscript 1 end subscript equals T subscript 2 end subscript. The acceleration of both the particles is g downwards. Therefore, relative acceleration between the two is zero or relative motion between the two is uniform. The relative velocity of A w.r.t. B is towards A B blank, therefore collision will take place between the two in mid air.

    Related Questions to study

    General
    Physics-

    Two particles 1 and 2 are projected with same speed v as shown in figure. Particle 2 is on the ground and particle 1 is at a height h from the ground and at a horizontal distance s from particle 2. If a graph is plotted between v and s for the condition of collision of the two then (v on y-axis and s on x-axis

    Assuming particle 2 to be at rest, substituting in
    y equals x tan invisible function application theta minus fraction numerator g x to the power of 2 end exponent over denominator 2 u to the power of 2 end exponent cos to the power of 2 end exponent invisible function application theta end fraction blank left parenthesis theta equals 0 degree right parenthesis

    We have negative h equals fraction numerator negative g over denominator 2 left parenthesis 4 v to the power of 2 end exponent right parenthesis end fraction
    or v equals square root of fraction numerator g over denominator 8 h end fraction end root
    Which is a straight line passing through origin with slopesquare root of fraction numerator g over denominator 8 h end fraction end root

    Two particles 1 and 2 are projected with same speed v as shown in figure. Particle 2 is on the ground and particle 1 is at a height h from the ground and at a horizontal distance s from particle 2. If a graph is plotted between v and s for the condition of collision of the two then (v on y-axis and s on x-axis

    Physics-General
    Assuming particle 2 to be at rest, substituting in
    y equals x tan invisible function application theta minus fraction numerator g x to the power of 2 end exponent over denominator 2 u to the power of 2 end exponent cos to the power of 2 end exponent invisible function application theta end fraction blank left parenthesis theta equals 0 degree right parenthesis

    We have negative h equals fraction numerator negative g over denominator 2 left parenthesis 4 v to the power of 2 end exponent right parenthesis end fraction
    or v equals square root of fraction numerator g over denominator 8 h end fraction end root
    Which is a straight line passing through origin with slopesquare root of fraction numerator g over denominator 8 h end fraction end root
    General
    physics-

    a) Name the experiment for which the adjacent graph, showing the variation of intensity of scattered electrons with the angle of scatter ing (q) was obtained.
    b) Also name the important hypothesis that was confirmed by this experiment

    a) Name the experiment for which the adjacent graph, showing the variation of intensity of scattered electrons with the angle of scatter ing (q) was obtained.
    b) Also name the important hypothesis that was confirmed by this experiment

    physics-General
    General
    physics-

    Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v subscript 0 end subscript, then the ratio of tensions in the three sections of the string is

    Let omega is the angular speed of revolution

    T subscript 3 end subscript equals m omega to the power of 2 end exponent 3 l
    T subscript 2 end subscript minus T subscript 3 end subscript equals m omega to the power of 2 end exponent 2 l rightwards double arrow T subscript 2 end subscript equals m omega to the power of 2 end exponent 5 l
    T subscript 1 end subscript minus T subscript 2 end subscript equals m omega to the power of 2 end exponent l rightwards double arrow T subscript 1 end subscript equals m omega to the power of 2 end exponent 6 l
    T subscript 3 end subscript colon T subscript 2 end subscript colon T subscript 1 end subscript equals 3 blank colon 5 blank colon 6

    Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v subscript 0 end subscript, then the ratio of tensions in the three sections of the string is

    physics-General
    Let omega is the angular speed of revolution

    T subscript 3 end subscript equals m omega to the power of 2 end exponent 3 l
    T subscript 2 end subscript minus T subscript 3 end subscript equals m omega to the power of 2 end exponent 2 l rightwards double arrow T subscript 2 end subscript equals m omega to the power of 2 end exponent 5 l
    T subscript 1 end subscript minus T subscript 2 end subscript equals m omega to the power of 2 end exponent l rightwards double arrow T subscript 1 end subscript equals m omega to the power of 2 end exponent 6 l
    T subscript 3 end subscript colon T subscript 2 end subscript colon T subscript 1 end subscript equals 3 blank colon 5 blank colon 6
    parallel
    General
    physics-

    A ball of massopen parentheses m close parentheses 0.5 kg is attached to the end of a string having length left parenthesis L right parenthesis 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rad/s) is

    T cos invisible function application theta component will cancel m g.

    T sin invisible function application theta Component will provide necessary centripetal force the ball towards center C.
    therefore T sin invisible function application theta equals m r omega to the power of 2 end exponent equals m left parenthesis l sin invisible function application theta right parenthesis omega to the power of 2 end exponent
    o r blank T equals m l omega to the power of 2 end exponent ⟹ omega equals square root of fraction numerator T over denominator m l end fraction end root r a d divided by s
    o r blank omega subscript m a x end subscript equals square root of fraction numerator T subscript m a x end subscript over denominator m l end fraction equals square root of fraction numerator 324 over denominator 0.5 cross times 0.5 end fraction equals 36 blank r a d divided by s end root end root

    A ball of massopen parentheses m close parentheses 0.5 kg is attached to the end of a string having length left parenthesis L right parenthesis 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rad/s) is

    physics-General
    T cos invisible function application theta component will cancel m g.

    T sin invisible function application theta Component will provide necessary centripetal force the ball towards center C.
    therefore T sin invisible function application theta equals m r omega to the power of 2 end exponent equals m left parenthesis l sin invisible function application theta right parenthesis omega to the power of 2 end exponent
    o r blank T equals m l omega to the power of 2 end exponent ⟹ omega equals square root of fraction numerator T over denominator m l end fraction end root r a d divided by s
    o r blank omega subscript m a x end subscript equals square root of fraction numerator T subscript m a x end subscript over denominator m l end fraction equals square root of fraction numerator 324 over denominator 0.5 cross times 0.5 end fraction equals 36 blank r a d divided by s end root end root
    General
    maths-

    The ellipse x squared over z squared plus y squared over h squared equals 1 and the straight line y = mx + c intersect in real points only if

    fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator m to the power of 2 end exponent x to the power of 2 end exponent plus 2 m c x plus c to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1
    Þ (b2 + a2 m2) x2 + 2 mca2 x + a2 (c2 - b2) = 0
    D > 0 Þ m2c2a4 - a2 (c2 - b2)(b2 + a2m2) > 0 Þ b2 + a2m2 > c2

    The ellipse x squared over z squared plus y squared over h squared equals 1 and the straight line y = mx + c intersect in real points only if

    maths-General
    fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator m to the power of 2 end exponent x to the power of 2 end exponent plus 2 m c x plus c to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1
    Þ (b2 + a2 m2) x2 + 2 mca2 x + a2 (c2 - b2) = 0
    D > 0 Þ m2c2a4 - a2 (c2 - b2)(b2 + a2m2) > 0 Þ b2 + a2m2 > c2
    General
    Physics-

    A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

    The forces acting on the ball will be (i) in the direction opposite to its motion i e comma frictional force and(ii) weight m g.

    A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

    Physics-General
    The forces acting on the ball will be (i) in the direction opposite to its motion i e comma frictional force and(ii) weight m g.
    parallel
    General
    physics-

    Four capacitors are connected as shown in figure. The equivalent capacitance between A and B is

    fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction equals 3
    C subscript s end subscript equals fraction numerator 1 over denominator 3 end fraction
    Capacitance between A and B
    C subscript p end subscript equals fraction numerator 1 over denominator 3 end fraction plus 1
    fraction numerator 4 over denominator 3 end fraction mu F equals 1.33 mu F

    Four capacitors are connected as shown in figure. The equivalent capacitance between A and B is

    physics-General
    fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction equals 3
    C subscript s end subscript equals fraction numerator 1 over denominator 3 end fraction
    Capacitance between A and B
    C subscript p end subscript equals fraction numerator 1 over denominator 3 end fraction plus 1
    fraction numerator 4 over denominator 3 end fraction mu F equals 1.33 mu F
    General
    physics-

    A frictionless track A B C D E ends in a circular loop of radius R comma figure. A body slides down the track from point A which is at a height h equals 5 blankcm. Maximum value of R for the body to successfully complete the loop is

    For successfully completing the loop,
    h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m

    A frictionless track A B C D E ends in a circular loop of radius R comma figure. A body slides down the track from point A which is at a height h equals 5 blankcm. Maximum value of R for the body to successfully complete the loop is

    physics-General
    For successfully completing the loop,
    h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m
    General
    physics-

    In a circuit shown in figure, the potential difference across the capacitor of 2 F is

    therefore fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 1 end fraction equals fraction numerator 3 over denominator 2 end fraction
    C subscript s end subscript equals fraction numerator 2 over denominator 3 end fraction F
    Q equals C subscript s end subscript V equals fraction numerator 2 over denominator 3 end fraction cross times 12 equals 8 C
    V subscript 1 end subscript equals fraction numerator Q over denominator C subscript 1 end subscript end fraction equals fraction numerator 8 over denominator 2 end fraction equals 4 V

    In a circuit shown in figure, the potential difference across the capacitor of 2 F is

    physics-General
    therefore fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 1 end fraction equals fraction numerator 3 over denominator 2 end fraction
    C subscript s end subscript equals fraction numerator 2 over denominator 3 end fraction F
    Q equals C subscript s end subscript V equals fraction numerator 2 over denominator 3 end fraction cross times 12 equals 8 C
    V subscript 1 end subscript equals fraction numerator Q over denominator C subscript 1 end subscript end fraction equals fraction numerator 8 over denominator 2 end fraction equals 4 V
    parallel
    General
    physics-

    The effective capacitance between points X and Y shown in figure. Assuming C subscript 2 end subscript equals 10 blank muF and that outer capacitors are all 4 blank muF is

    The arrangement shows a Wheatstone bridge.
    As fraction numerator C subscript 1 end subscript over denominator C subscript 3 end subscript end fraction equals fraction numerator C subscript 4 end subscript over denominator C subscript 5 end subscript end fraction equals 1 comma blanktherefore the bridge is balanced.
    fraction numerator 1 over denominator C subscript s subscript 1 end subscript end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 2 over denominator 4 end fraction equals fraction numerator 1 over denominator 2 end fraction comma C subscript s subscript 1 end subscript end subscript equals 2 mu blank F
    Similarly, C subscript s end subscript subscript 2 end subscript equals 2 mu blank F
    therefore effective capacitance
    equals C subscript p end subscript equals C subscript s end subscript subscript 1 end subscript plus C subscript s end subscript subscript 2 end subscript equals 2 plus 2 plus equals 4 mu blank F

    The effective capacitance between points X and Y shown in figure. Assuming C subscript 2 end subscript equals 10 blank muF and that outer capacitors are all 4 blank muF is

    physics-General
    The arrangement shows a Wheatstone bridge.
    As fraction numerator C subscript 1 end subscript over denominator C subscript 3 end subscript end fraction equals fraction numerator C subscript 4 end subscript over denominator C subscript 5 end subscript end fraction equals 1 comma blanktherefore the bridge is balanced.
    fraction numerator 1 over denominator C subscript s subscript 1 end subscript end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 2 over denominator 4 end fraction equals fraction numerator 1 over denominator 2 end fraction comma C subscript s subscript 1 end subscript end subscript equals 2 mu blank F
    Similarly, C subscript s end subscript subscript 2 end subscript equals 2 mu blank F
    therefore effective capacitance
    equals C subscript p end subscript equals C subscript s end subscript subscript 1 end subscript plus C subscript s end subscript subscript 2 end subscript equals 2 plus 2 plus equals 4 mu blank F
    General
    physics-

    The four capacitors, each of 25 muF are connected as shown in figure. The DC voltmeter reads 200 V. the change on each plate of capacitor is

    Charge on each plate of each capacitor
    Q equals plus-or-minus C V equals plus-or-minus 25 cross times 10 to the power of negative 6 end exponent cross times 200
    equals plus-or-minus 5 cross times 10 to the power of negative 3 end exponent C

    The four capacitors, each of 25 muF are connected as shown in figure. The DC voltmeter reads 200 V. the change on each plate of capacitor is

    physics-General
    Charge on each plate of each capacitor
    Q equals plus-or-minus C V equals plus-or-minus 25 cross times 10 to the power of negative 6 end exponent cross times 200
    equals plus-or-minus 5 cross times 10 to the power of negative 3 end exponent C
    General
    physics-

    For the circuit shown in figure the charge on 4muF capacitor is

    Combined capacity of 1 blank mu F and 5 mu F blank= 1 + 5=6 blank mu F
    Now, 4 mu F blankand 6 mu F are in series.
    therefore blank fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 3 plus 2 over denominator 12 end fraction equals fraction numerator 5 over denominator 12 end fraction
    C subscript s end subscript equals fraction numerator 12 over denominator 5 end fraction mu F
    Charge in the arm containing 4 mu F blankcapacitor is
    q equals C subscript s end subscript cross times V equals fraction numerator 12 over denominator 5 end fraction cross times 10 equals 24 blank mu C

    For the circuit shown in figure the charge on 4muF capacitor is

    physics-General
    Combined capacity of 1 blank mu F and 5 mu F blank= 1 + 5=6 blank mu F
    Now, 4 mu F blankand 6 mu F are in series.
    therefore blank fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 3 plus 2 over denominator 12 end fraction equals fraction numerator 5 over denominator 12 end fraction
    C subscript s end subscript equals fraction numerator 12 over denominator 5 end fraction mu F
    Charge in the arm containing 4 mu F blankcapacitor is
    q equals C subscript s end subscript cross times V equals fraction numerator 12 over denominator 5 end fraction cross times 10 equals 24 blank mu C
    parallel
    General
    Physics-

    A particle originally at a rest at the highest point of a smooth circle in a vertical plane, is gently pushed and starts sliding along the circle. It will leave the circle at a vertical distance h below the highest point such that

    From law of conservation of energy, potential energy of fall gets converted to kinetic energy.

    therefore blank P E equals K E
    m g h equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
    o r blank v equals square root of 2 g h end root blank open parentheses i close parentheses
    Also, the horizontal component of force is equal centrifugal force.
    therefore blank m g cos invisible function application theta equals fraction numerator m v to the power of 2 end exponent over denominator R end fraction blank open parentheses i i close parentheses
    From Eq. (i)
    therefore blank m g cos invisible function application theta equals fraction numerator 2 m g h over denominator R end fraction blank open parentheses i i i close parentheses
    From increment A O B comma
    cos invisible function application theta equals fraction numerator left parenthesis R minus h over denominator R end fraction
    ⟹ m g open parentheses fraction numerator open parentheses R minus h close parentheses over denominator R end fraction close parentheses equals fraction numerator 2 m g h over denominator R end fraction
    ⟹ blank 3 h equals R
    ⟹ h equals fraction numerator R over denominator 3 end fraction

    A particle originally at a rest at the highest point of a smooth circle in a vertical plane, is gently pushed and starts sliding along the circle. It will leave the circle at a vertical distance h below the highest point such that

    Physics-General
    From law of conservation of energy, potential energy of fall gets converted to kinetic energy.

    therefore blank P E equals K E
    m g h equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
    o r blank v equals square root of 2 g h end root blank open parentheses i close parentheses
    Also, the horizontal component of force is equal centrifugal force.
    therefore blank m g cos invisible function application theta equals fraction numerator m v to the power of 2 end exponent over denominator R end fraction blank open parentheses i i close parentheses
    From Eq. (i)
    therefore blank m g cos invisible function application theta equals fraction numerator 2 m g h over denominator R end fraction blank open parentheses i i i close parentheses
    From increment A O B comma
    cos invisible function application theta equals fraction numerator left parenthesis R minus h over denominator R end fraction
    ⟹ m g open parentheses fraction numerator open parentheses R minus h close parentheses over denominator R end fraction close parentheses equals fraction numerator 2 m g h over denominator R end fraction
    ⟹ blank 3 h equals R
    ⟹ h equals fraction numerator R over denominator 3 end fraction
    General
    Maths-

    Polar of origin (0, 0) w.r.t. the circle x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0 touches the circle x squared plus y squared equals r squared, if

    E q u a t i o n space p o l o r space o f space t h e space o r i g i n space w. r. t. space t h e space c i r c l e space i s space
x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0
p o l a r space o f space e q u a t i o n space equals 0 comma space P space left parenthesis x subscript 1 comma y subscript 1 right parenthesis
N o w comma space x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0
rightwards double arrow x. x subscript 1 plus y. y subscript 1 plus 2. lambda open parentheses fraction numerator x plus x subscript 1 over denominator 2 end fraction close parentheses plus 2. mu open parentheses fraction numerator y plus y subscript 1 over denominator 2 end fraction close parentheses plus c equals 0
rightwards double arrow x. x subscript 1 plus y. y subscript 1 plus lambda left parenthesis x plus x subscript 1 right parenthesis plus mu left parenthesis y plus y subscript 1 right parenthesis plus c equals 0
rightwards double arrow x.0 plus y.0 plus lambda left parenthesis x plus 0 right parenthesis plus mu left parenthesis y plus 0 right parenthesis plus c equals 0
rightwards double arrow lambda x plus mu y plus c equals 0
G i v e n comma space x squared plus y squared equals r squared
rightwards double arrow left parenthesis x minus h right parenthesis squared plus left parenthesis y minus k right parenthesis squared equals r to the power of 2 end exponent p e r p e n d i c u l a r space d i s tan c e space b e t w e e n space a space p o i n t space left parenthesis x subscript 1 comma y subscript 1 right parenthesis a n d space a space l i n e space left parenthesis lambda x plus mu y plus c equals 0 right parenthesis
P equals fraction numerator open vertical bar lambda x subscript 1 plus mu y subscript 1 plus c close vertical bar over denominator square root of lambda squared plus mu squared end root end fraction rightwards double arrow r equals fraction numerator open vertical bar lambda.0 plus mu.0 plus c close vertical bar over denominator square root of lambda squared plus mu squared end root end fraction rightwards double arrow r equals fraction numerator c over denominator square root of lambda squared plus mu squared end root end fraction
s q u a r i n g space b o t h space s i d e s comma
rightwards double arrow r squared equals fraction numerator c squared over denominator lambda squared plus mu squared end fraction
rightwards double arrow r squared open parentheses lambda squared plus mu squared close parentheses equals c squared

    Polar of origin (0, 0) w.r.t. the circle x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0 touches the circle x squared plus y squared equals r squared, if

    Maths-General
    E q u a t i o n space p o l o r space o f space t h e space o r i g i n space w. r. t. space t h e space c i r c l e space i s space
x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0
p o l a r space o f space e q u a t i o n space equals 0 comma space P space left parenthesis x subscript 1 comma y subscript 1 right parenthesis
N o w comma space x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0
rightwards double arrow x. x subscript 1 plus y. y subscript 1 plus 2. lambda open parentheses fraction numerator x plus x subscript 1 over denominator 2 end fraction close parentheses plus 2. mu open parentheses fraction numerator y plus y subscript 1 over denominator 2 end fraction close parentheses plus c equals 0
rightwards double arrow x. x subscript 1 plus y. y subscript 1 plus lambda left parenthesis x plus x subscript 1 right parenthesis plus mu left parenthesis y plus y subscript 1 right parenthesis plus c equals 0
rightwards double arrow x.0 plus y.0 plus lambda left parenthesis x plus 0 right parenthesis plus mu left parenthesis y plus 0 right parenthesis plus c equals 0
rightwards double arrow lambda x plus mu y plus c equals 0
G i v e n comma space x squared plus y squared equals r squared
rightwards double arrow left parenthesis x minus h right parenthesis squared plus left parenthesis y minus k right parenthesis squared equals r to the power of 2 end exponent p e r p e n d i c u l a r space d i s tan c e space b e t w e e n space a space p o i n t space left parenthesis x subscript 1 comma y subscript 1 right parenthesis a n d space a space l i n e space left parenthesis lambda x plus mu y plus c equals 0 right parenthesis
P equals fraction numerator open vertical bar lambda x subscript 1 plus mu y subscript 1 plus c close vertical bar over denominator square root of lambda squared plus mu squared end root end fraction rightwards double arrow r equals fraction numerator open vertical bar lambda.0 plus mu.0 plus c close vertical bar over denominator square root of lambda squared plus mu squared end root end fraction rightwards double arrow r equals fraction numerator c over denominator square root of lambda squared plus mu squared end root end fraction
s q u a r i n g space b o t h space s i d e s comma
rightwards double arrow r squared equals fraction numerator c squared over denominator lambda squared plus mu squared end fraction
rightwards double arrow r squared open parentheses lambda squared plus mu squared close parentheses equals c squared
    General
    physics-

    Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 volts Photo electric effect in this metallic surface begins at a frequency 6 x 1014s -1 The frequency of the incident light in s-1 is [h=6 x 10-34J-sec;charge on the electron=1.6x10-19C]

    Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 volts Photo electric effect in this metallic surface begins at a frequency 6 x 1014s -1 The frequency of the incident light in s-1 is [h=6 x 10-34J-sec;charge on the electron=1.6x10-19C]

    physics-General
    parallel

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