General
Easy
Maths-

P(θ) and D space open parentheses theta plus pi over 2 close parentheses are the pts. on the ellipse b x to the power of 2 end exponent plus a to the power of 2 end exponent y to the power of 2 end exponent equals a to the power of 2 end exponent b to the power of 2 end exponent then C P to the power of 2 end exponent plus C D to the power of 2 end exponent equals

Maths-General

  1. a+b    
  2. a to the power of 2 end exponent plus b to the power of 2 end exponent    
  3. a to the power of 2 end exponent minus b to the power of 2 end exponent    
  4. ab    

    Answer:The correct answer is: a+b

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    Related Questions to study

    General
    physics-

    A ball of mass 0.2 blank k g rests on a vertical post of height 5 blank m. A bullet of mass 0.01 blank k g, travelling with a velocity V blank m divided by s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 blank mand the bullet at a distance of 100 blank mfrom the foot of the post. The initial velocity V of the bullet is

    R equals u square root of fraction numerator 2 h over denominator g end fraction end root rightwards double arrow 20 equals V subscript 1 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root and 100 equals V subscript 2 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root
    rightwards double arrow V subscript 1 end subscript equals 20 blank m divided by s comma blank V subscript 2 end subscript equals 100 blank m divided by s
    Applying momentum conservation just before and just after the collision open parentheses 0.01 close parentheses open parentheses V close parentheses equals open parentheses 0.2 close parentheses open parentheses 20 close parentheses plus left parenthesis 0.01 right parenthesis left parenthesis 100 right parenthesis
    V equals 500 blank m divided by s

    A ball of mass 0.2 blank k g rests on a vertical post of height 5 blank m. A bullet of mass 0.01 blank k g, travelling with a velocity V blank m divided by s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 blank mand the bullet at a distance of 100 blank mfrom the foot of the post. The initial velocity V of the bullet is

    physics-General
    R equals u square root of fraction numerator 2 h over denominator g end fraction end root rightwards double arrow 20 equals V subscript 1 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root and 100 equals V subscript 2 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root
    rightwards double arrow V subscript 1 end subscript equals 20 blank m divided by s comma blank V subscript 2 end subscript equals 100 blank m divided by s
    Applying momentum conservation just before and just after the collision open parentheses 0.01 close parentheses open parentheses V close parentheses equals open parentheses 0.2 close parentheses open parentheses 20 close parentheses plus left parenthesis 0.01 right parenthesis left parenthesis 100 right parenthesis
    V equals 500 blank m divided by s
    General
    physics-

    A particle P is sliding down a frictionless hemispherical bowl. It passes the point Aat t equals 0. At this instant of time, the horizontal component of its velocity v. A bead Q of the same mass as P is ejected from A to t equals 0 along the horizontal string A B (see figure) with the speed v. Friction between the bead and the string may be neglected. Let t subscript p end subscript and t subscript Q end subscript be the respective time taken by P and Q to reach the point B. Then

    For particle P, motion between A and C will be an accelerated one while between C and B a retarded one. But in any case horizontal component of it’s velocity will be greater than or equal to v on the other hand in case of particle Q, it is always equal to v. Horizontal displacement of both the particles are equal, so t subscript P end subscript less than t subscript Q end subscript

    A particle P is sliding down a frictionless hemispherical bowl. It passes the point Aat t equals 0. At this instant of time, the horizontal component of its velocity v. A bead Q of the same mass as P is ejected from A to t equals 0 along the horizontal string A B (see figure) with the speed v. Friction between the bead and the string may be neglected. Let t subscript p end subscript and t subscript Q end subscript be the respective time taken by P and Q to reach the point B. Then

    physics-General
    For particle P, motion between A and C will be an accelerated one while between C and B a retarded one. But in any case horizontal component of it’s velocity will be greater than or equal to v on the other hand in case of particle Q, it is always equal to v. Horizontal displacement of both the particles are equal, so t subscript P end subscript less than t subscript Q end subscript
    General
    physics-

    A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

    Vertical height equals h equals l cos invisible function application 30 degree
    Loss of potential energy equals m g h

    equals m g l cos invisible function application 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction m g l
    therefore Kinetic energy gained equals fraction numerator square root of 3 over denominator 2 end fraction m g l

    A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

    physics-General
    Vertical height equals h equals l cos invisible function application 30 degree
    Loss of potential energy equals m g h

    equals m g l cos invisible function application 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction m g l
    therefore Kinetic energy gained equals fraction numerator square root of 3 over denominator 2 end fraction m g l
    General
    maths-

    If the chords of contact of P space open parentheses x subscript 1 comma y subscript 1 close parentheses and Q space open parentheses x subscript 2 comma y subscript 2 close parentheses w.r.t the ellipse 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 y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 are at right angle then fraction numerator x subscript 1 end subscript x subscript 2 end subscript over denominator y subscript 1 end subscript y subscript 2 end subscript end fraction equals

    If the chords of contact of P space open parentheses x subscript 1 comma y subscript 1 close parentheses and Q space open parentheses x subscript 2 comma y subscript 2 close parentheses w.r.t the ellipse 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 y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 are at right angle then fraction numerator x subscript 1 end subscript x subscript 2 end subscript over denominator y subscript 1 end subscript y subscript 2 end subscript end fraction equals

    maths-General
    General
    maths-

    The area of an ellipse is 8π sq. units dist. between the foci is 4 square root of 3 then e=

    The area of an ellipse is 8π sq. units dist. between the foci is 4 square root of 3 then e=

    maths-General
    General
    physics-

    A particle of mass m attracted with a string of length l is just revolving on the vertical circle without slacking of the string. If v subscript A end subscript comma v subscript B end subscript and v subscript D end subscript are speed at position A comma B and D then

    At A comma blank v subscript A end subscript equals square root of g l end root
    At B comma blank v subscript B end subscript equals square root of 5 g l end root
    and at D comma blank v subscript D end subscript equals square root of 3 g l end root
    Thus, v subscript B end subscript greater than v subscript D end subscript greater than v subscript A end subscript
    Also, T equals 3 blank m g left parenthesis 1 plus cos invisible function application theta right parenthesis
    So, D comma theta equals 90 degree
    therefore blank T equals 3 blank m g open parentheses 1 plus theta close parentheses equals 3 blank m g

    A particle of mass m attracted with a string of length l is just revolving on the vertical circle without slacking of the string. If v subscript A end subscript comma v subscript B end subscript and v subscript D end subscript are speed at position A comma B and D then

    physics-General
    At A comma blank v subscript A end subscript equals square root of g l end root
    At B comma blank v subscript B end subscript equals square root of 5 g l end root
    and at D comma blank v subscript D end subscript equals square root of 3 g l end root
    Thus, v subscript B end subscript greater than v subscript D end subscript greater than v subscript A end subscript
    Also, T equals 3 blank m g left parenthesis 1 plus cos invisible function application theta right parenthesis
    So, D comma theta equals 90 degree
    therefore blank T equals 3 blank m g open parentheses 1 plus theta close parentheses equals 3 blank m g
    General
    physics-

    A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ´ ´ P ´ ´ is such that it sweeps out a length s equals t to the power of 3 end exponent plus 5, where s is in metres and t is in seconds. The radius of the path is 20 blank m. The acceleration of ´ ´ P ´ ´ when t equals 2 s is nearly

    As S equals t to the power of 3 end exponent plus 5
    fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent equals v
    therefore a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals 6 t
    at t equals 2 blank s e c
    open vertical bar stack a with rightwards arrow on top close vertical bar equals square root of a subscript c end subscript superscript 2 end superscript plus a subscript t end subscript superscript 2 end superscript end root
    equals square root of open parentheses fraction numerator v to the power of 2 end exponent over denominator R end fraction close parentheses to the power of 2 end exponent plus a subscript t end subscript superscript 2 end superscript end root equals square root of open parentheses fraction numerator 4 t to the power of 4 end exponent over denominator R end fraction close parentheses to the power of 2 end exponent plus open parentheses fraction numerator d v over denominator d t end fraction close parentheses to the power of 2 end exponent end root
    equals square root of open parentheses 7.2 close parentheses to the power of 2 end exponent plus 144 end root
    open vertical bar stack a with rightwards arrow on top close vertical bar equals 14 m divided by s to the power of 2 end exponent

    A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ´ ´ P ´ ´ is such that it sweeps out a length s equals t to the power of 3 end exponent plus 5, where s is in metres and t is in seconds. The radius of the path is 20 blank m. The acceleration of ´ ´ P ´ ´ when t equals 2 s is nearly

    physics-General
    As S equals t to the power of 3 end exponent plus 5
    fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent equals v
    therefore a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals 6 t
    at t equals 2 blank s e c
    open vertical bar stack a with rightwards arrow on top close vertical bar equals square root of a subscript c end subscript superscript 2 end superscript plus a subscript t end subscript superscript 2 end superscript end root
    equals square root of open parentheses fraction numerator v to the power of 2 end exponent over denominator R end fraction close parentheses to the power of 2 end exponent plus a subscript t end subscript superscript 2 end superscript end root equals square root of open parentheses fraction numerator 4 t to the power of 4 end exponent over denominator R end fraction close parentheses to the power of 2 end exponent plus open parentheses fraction numerator d v over denominator d t end fraction close parentheses to the power of 2 end exponent end root
    equals square root of open parentheses 7.2 close parentheses to the power of 2 end exponent plus 144 end root
    open vertical bar stack a with rightwards arrow on top close vertical bar equals 14 m divided by s to the power of 2 end exponent
    General
    physics-

    In a two dimensional motion of a particle, the particle moves from point A, position vector stack r with rightwards arrow on top subscript 1 end subscript. If the magnitudes of these vectors are respectively, r subscript 1 end subscript=3 and r subscript 2 end subscript equals 4 and the angles they make with the x-axis are theta subscript 1 end subscript equals 75 degree and 15degree, respectively, then find the magnitude of the displacement vector

    Displacement equals A B
    angle between stack r subscript 1 end subscript with rightwards arrow on top and stack r subscript 2 end subscript with rightwards arrow on top
    theta equals 75 degree minus 15 degree equals 60 degree
    From figure
    A B to the power of 2 end exponent equals r subscript 1 end subscript superscript 2 end superscript plus r subscript 2 end subscript superscript 2 end superscript minus 2 r subscript 1 end subscript r subscript 2 end subscript c o s theta
    equals 3 to the power of 2 end exponent plus 4 to the power of 2 end exponent minus 2 cross times 3 cross times 4 cos60 degree
    equals 13
    A B equals square root of 13

    In a two dimensional motion of a particle, the particle moves from point A, position vector stack r with rightwards arrow on top subscript 1 end subscript. If the magnitudes of these vectors are respectively, r subscript 1 end subscript=3 and r subscript 2 end subscript equals 4 and the angles they make with the x-axis are theta subscript 1 end subscript equals 75 degree and 15degree, respectively, then find the magnitude of the displacement vector

    physics-General
    Displacement equals A B
    angle between stack r subscript 1 end subscript with rightwards arrow on top and stack r subscript 2 end subscript with rightwards arrow on top
    theta equals 75 degree minus 15 degree equals 60 degree
    From figure
    A B to the power of 2 end exponent equals r subscript 1 end subscript superscript 2 end superscript plus r subscript 2 end subscript superscript 2 end superscript minus 2 r subscript 1 end subscript r subscript 2 end subscript c o s theta
    equals 3 to the power of 2 end exponent plus 4 to the power of 2 end exponent minus 2 cross times 3 cross times 4 cos60 degree
    equals 13
    A B equals square root of 13
    General
    physics-

    A particle of mass m moving with horizontal speed 6 blank m divided by s e c as shown in figure. If m less than less than M than for one dimensional elastic collision, the speed of lighter particle after collision will be


    v subscript 1 end subscript equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses u subscript 1 end subscript plus fraction numerator 2 m subscript 2 end subscript u subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
    Substituting m subscript 1 end subscript equals 0 comma blank v subscript 1 end subscript equals negative u subscript 1 end subscript plus 2 u subscript 2 end subscript
    rightwards double arrow v subscript 1 end subscript equals negative 6 plus 2 open parentheses 4 close parentheses equals 2 blank m divided by s
    i. e. the lighter particle will move in original direction with the speed of 2 blank m divided by s

    A particle of mass m moving with horizontal speed 6 blank m divided by s e c as shown in figure. If m less than less than M than for one dimensional elastic collision, the speed of lighter particle after collision will be

    physics-General

    v subscript 1 end subscript equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses u subscript 1 end subscript plus fraction numerator 2 m subscript 2 end subscript u subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
    Substituting m subscript 1 end subscript equals 0 comma blank v subscript 1 end subscript equals negative u subscript 1 end subscript plus 2 u subscript 2 end subscript
    rightwards double arrow v subscript 1 end subscript equals negative 6 plus 2 open parentheses 4 close parentheses equals 2 blank m divided by s
    i. e. the lighter particle will move in original direction with the speed of 2 blank m divided by s
    General
    physics-

    A block(B) is attached to two unstretched springs S subscript 1 end subscript a n d blank S subscript 2 end subscript with springs constants k blank a n d blank 4 k commarepresentively (see Fig. I) The other ends are attached to identical supports M subscript 1 end subscript and M subscript 2 end subscript not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall I by small distance x left parenthesis F i g blank I I right parenthesis and released. The block returns and moves a maximum distance y towards wall 2.Displacements x blank a n d blank y are measured with respect to the equilibrium position of the block B The ratio fraction numerator y over denominator x end fraction is

    From energy conservation,
    fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction open parentheses 4 k close parentheses y to the power of 2 end exponent
    fraction numerator y over denominator x end fraction equals fraction numerator 1 over denominator 2 end fraction

    A block(B) is attached to two unstretched springs S subscript 1 end subscript a n d blank S subscript 2 end subscript with springs constants k blank a n d blank 4 k commarepresentively (see Fig. I) The other ends are attached to identical supports M subscript 1 end subscript and M subscript 2 end subscript not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall I by small distance x left parenthesis F i g blank I I right parenthesis and released. The block returns and moves a maximum distance y towards wall 2.Displacements x blank a n d blank y are measured with respect to the equilibrium position of the block B The ratio fraction numerator y over denominator x end fraction is

    physics-General
    From energy conservation,
    fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction open parentheses 4 k close parentheses y to the power of 2 end exponent
    fraction numerator y over denominator x end fraction equals fraction numerator 1 over denominator 2 end fraction
    General
    physics-

    A small roller coaster starts at point A with a speed u blankon a curved track as shown in the figure.
    The friction between the roller coaster and the track is negligible and it always remains in contact with the track. The speed of roller coaster at point D blankon the track will be

    There is no loss of energy. Therefore, the final velocity is the same as the initial velocity.
    Hence, The speed of roller coaster at point D is u

    A small roller coaster starts at point A with a speed u blankon a curved track as shown in the figure.
    The friction between the roller coaster and the track is negligible and it always remains in contact with the track. The speed of roller coaster at point D blankon the track will be

    physics-General
    There is no loss of energy. Therefore, the final velocity is the same as the initial velocity.
    Hence, The speed of roller coaster at point D is u
    General
    maths-

    If the equation fraction numerator x to the power of 2 end exponent over denominator 9 minus k end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 minus k end fraction equals 1 represents an ellipse then

    If the equation fraction numerator x to the power of 2 end exponent over denominator 9 minus k end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 minus k end fraction equals 1 represents an ellipse then

    maths-General
    General
    physics-

    An aeroplane is flying in a horizontal direction with a velocity 600 k m h to the power of negative 1 end exponent at a height of 1960 m. when it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance A B

    From h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
    We have t subscript O B end subscript equals square root of fraction numerator 2 h subscript O A end subscript over denominator g end fraction end root
    equals square root of fraction numerator 2 cross times 1960 over denominator 9.8 end fraction end root equals 20 s
    Horizontal distance A B equals v t subscript O B end subscript
    equals open parentheses 600 cross times fraction numerator 5 over denominator 18 end fraction close parentheses left parenthesis 20 right parenthesis
    equals 3333.33 blank m equals 3.33 blank k m

    An aeroplane is flying in a horizontal direction with a velocity 600 k m h to the power of negative 1 end exponent at a height of 1960 m. when it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance A B

    physics-General
    From h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
    We have t subscript O B end subscript equals square root of fraction numerator 2 h subscript O A end subscript over denominator g end fraction end root
    equals square root of fraction numerator 2 cross times 1960 over denominator 9.8 end fraction end root equals 20 s
    Horizontal distance A B equals v t subscript O B end subscript
    equals open parentheses 600 cross times fraction numerator 5 over denominator 18 end fraction close parentheses left parenthesis 20 right parenthesis
    equals 3333.33 blank m equals 3.33 blank k m
    General
    physics-

    A body of mass m is moving with a uniform speed v along a circle of radius r, what is the average acceleration in going from A to B?

    Here, T equals fraction numerator 2 pi r over denominator 4 v end fraction equals fraction numerator pi r over denominator 2 v end fraction
    Change in velocity is going from A to B = v square root of 2
    Average acceleration equals fraction numerator v square root of 2 over denominator pi r divided by 2 v end fraction equals fraction numerator 2 square root of 2 v to the power of 2 end exponent over denominator pi r end fraction

    A body of mass m is moving with a uniform speed v along a circle of radius r, what is the average acceleration in going from A to B?

    physics-General
    Here, T equals fraction numerator 2 pi r over denominator 4 v end fraction equals fraction numerator pi r over denominator 2 v end fraction
    Change in velocity is going from A to B = v square root of 2
    Average acceleration equals fraction numerator v square root of 2 over denominator pi r divided by 2 v end fraction equals fraction numerator 2 square root of 2 v to the power of 2 end exponent over denominator pi r end fraction
    General
    physics-

    A bob of mass m accelerates uniformly from rest to v subscript 1 end subscript in time t subscript 1 end subscript. As a function of t, the instantaneous power delivered to the body is

    From v equals u plus a t comma v subscript 1 end subscript equals 0 plus a t subscript 1 end subscript open parentheses because a equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction close parentheses
    F equals m a equals m v subscript 1 end subscript divided by t subscript 1 end subscript
    Velocity acquired in t sec equals a t equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction t
    Power equals F cross times v equals fraction numerator m v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction cross times fraction numerator v subscript 1 end subscript t over denominator t subscript 1 end subscript end fraction equals fraction numerator m v subscript 1 end subscript superscript 2 end superscript t over denominator t subscript 1 end subscript superscript 2 end superscript end fraction

    A bob of mass m accelerates uniformly from rest to v subscript 1 end subscript in time t subscript 1 end subscript. As a function of t, the instantaneous power delivered to the body is

    physics-General
    From v equals u plus a t comma v subscript 1 end subscript equals 0 plus a t subscript 1 end subscript open parentheses because a equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction close parentheses
    F equals m a equals m v subscript 1 end subscript divided by t subscript 1 end subscript
    Velocity acquired in t sec equals a t equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction t
    Power equals F cross times v equals fraction numerator m v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction cross times fraction numerator v subscript 1 end subscript t over denominator t subscript 1 end subscript end fraction equals fraction numerator m v subscript 1 end subscript superscript 2 end superscript t over denominator t subscript 1 end subscript superscript 2 end superscript end fraction