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Maths-

If the normal at the point P(theta) to the ellipse fraction numerator x to the power of 2 end exponent over denominator 14 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 end fraction equals 1intersects it again at the point Q(2theta) then costheta =

Maths-General

  1. –2/3    
  2. 2/3    
  3. – 6/7    
  4. 6/7    

    Answer:The correct answer is: –2/3

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    General
    maths-

    If a tangent to ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction blank+ fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1 makes an angle alpha with x- axis, then square of length of intercept of tangent cut between axes is-

    Let tangent is fraction numerator x cos invisible function application theta over denominator a end fraction+fraction numerator y sin invisible function application theta over denominator b end fraction= 1
    Slope is tan alpha = –fraction numerator b over denominator a end fraction c o t invisible function application theta S… (1)
    therefore square of length of intercept is
    = left parenthesis a s e c invisible function application theta right parenthesis to the power of 2 end exponent plus left parenthesis b c o s e c invisible function application theta right parenthesis to the power of 2 end exponent
    therefore Length is L equals a to the power of 2 end exponent plus b to the power of 2 end exponent plus a to the power of 2 end exponent t a n to the power of 2 end exponent invisible function application theta plus b to the power of 2 end exponent c o t to the power of 2 end exponent invisible function application theta … (2)
    Now use value of tan  from (1)

    If a tangent to ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction blank+ fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1 makes an angle alpha with x- axis, then square of length of intercept of tangent cut between axes is-

    maths-General
    Let tangent is fraction numerator x cos invisible function application theta over denominator a end fraction+fraction numerator y sin invisible function application theta over denominator b end fraction= 1
    Slope is tan alpha = –fraction numerator b over denominator a end fraction c o t invisible function application theta S… (1)
    therefore square of length of intercept is
    = left parenthesis a s e c invisible function application theta right parenthesis to the power of 2 end exponent plus left parenthesis b c o s e c invisible function application theta right parenthesis to the power of 2 end exponent
    therefore Length is L equals a to the power of 2 end exponent plus b to the power of 2 end exponent plus a to the power of 2 end exponent t a n to the power of 2 end exponent invisible function application theta plus b to the power of 2 end exponent c o t to the power of 2 end exponent invisible function application theta … (2)
    Now use value of tan  from (1)
    General
    physics-

    Vibrating tuning fork of frequency n is placed near the open end of a long cylindrical tube. The tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through 8.75 blank c m comma the intensity of sound changes from a maximum to minimum. If the speed of sound is 350 blank m divided by s. then n is

    When the piston is moved through a distance of 8.75 c m comma the path difference produced is 2 cross times 8.75 c m equals 17.5 c m.
    This must be equal to fraction numerator lambda over denominator 2 end fraction for maximum to change to minimum.
    therefore fraction numerator lambda over denominator 2 end fraction equals 17.5 blank c m rightwards double arrow lambda equals 35 c m equals 0.35 m
    So, v equals n lambda rightwards double arrow n equals fraction numerator v over denominator lambda end fraction equals fraction numerator 350 over denominator 0.35 end fraction equals 1000 H z

    Vibrating tuning fork of frequency n is placed near the open end of a long cylindrical tube. The tube has a side opening and is fitted with a movable reflecting piston. As the piston is moved through 8.75 blank c m comma the intensity of sound changes from a maximum to minimum. If the speed of sound is 350 blank m divided by s. then n is

    physics-General
    When the piston is moved through a distance of 8.75 c m comma the path difference produced is 2 cross times 8.75 c m equals 17.5 c m.
    This must be equal to fraction numerator lambda over denominator 2 end fraction for maximum to change to minimum.
    therefore fraction numerator lambda over denominator 2 end fraction equals 17.5 blank c m rightwards double arrow lambda equals 35 c m equals 0.35 m
    So, v equals n lambda rightwards double arrow n equals fraction numerator v over denominator lambda end fraction equals fraction numerator 350 over denominator 0.35 end fraction equals 1000 H z
    General
    maths-

    PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R and
    2 alpha, 2 beta comma blank 2 gammarespectively then tan beta tan gamma is equal to -

    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
    P (a cos 2 alpha, b sin 2 alpha), Q (a cos 2 beta , b sin 2 beta)
    R (a cos 2 gamma, b sin 2 gamma)
    chord's PQ equation
    fraction numerator x over denominator a end fractioncos (alpha + beta) + fraction numerator y over denominator b end fractionsin (alpha + beta) = cos (alphabeta)
    PQ passes through the focus (ae, 0)
    e = fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction
    PR passes through the focus (– ae, 0) the
    – e = fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction
    fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction = – fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction
    Apply componendo and dividendo, we get
    fraction numerator cos invisible function application left parenthesis alpha plus beta right parenthesis plus cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis minus cos invisible function application left parenthesis alpha minus beta right parenthesis end fraction = fraction numerator cos invisible function application left parenthesis alpha plus gamma right parenthesis minus cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis plus cos invisible function application left parenthesis alpha minus gamma right parenthesis end fraction
    fraction numerator 2 cos invisible function application alpha cos invisible function application beta over denominator 2 sin invisible function application alpha sin invisible function application beta end fraction = fraction numerator 2 sin invisible function application alpha sin invisible function application gamma over denominator 2 cos invisible function application alpha cos invisible function application gamma end fraction
    tan beta tan gamma = cot2 alpha

    PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R and
    2 alpha, 2 beta comma blank 2 gammarespectively then tan beta tan gamma is equal to -

    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 y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1
    P (a cos 2 alpha, b sin 2 alpha), Q (a cos 2 beta , b sin 2 beta)
    R (a cos 2 gamma, b sin 2 gamma)
    chord's PQ equation
    fraction numerator x over denominator a end fractioncos (alpha + beta) + fraction numerator y over denominator b end fractionsin (alpha + beta) = cos (alphabeta)
    PQ passes through the focus (ae, 0)
    e = fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction
    PR passes through the focus (– ae, 0) the
    – e = fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction
    fraction numerator cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction = – fraction numerator cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis end fraction
    Apply componendo and dividendo, we get
    fraction numerator cos invisible function application left parenthesis alpha plus beta right parenthesis plus cos invisible function application left parenthesis alpha minus beta right parenthesis over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis minus cos invisible function application left parenthesis alpha minus beta right parenthesis end fraction = fraction numerator cos invisible function application left parenthesis alpha plus gamma right parenthesis minus cos invisible function application left parenthesis alpha minus gamma right parenthesis over denominator cos invisible function application left parenthesis alpha plus gamma right parenthesis plus cos invisible function application left parenthesis alpha minus gamma right parenthesis end fraction
    fraction numerator 2 cos invisible function application alpha cos invisible function application beta over denominator 2 sin invisible function application alpha sin invisible function application beta end fraction = fraction numerator 2 sin invisible function application alpha sin invisible function application gamma over denominator 2 cos invisible function application alpha cos invisible function application gamma end fraction
    tan beta tan gamma = cot2 alpha
    General
    maths-

    The radius of the circle passing through the points of intersection of 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 = 1 and x2 – y2 = 0 is -

    Two curves are symmetrical about both axes and intersect in four points, so, the circle through their points of intersection will have centre at origin.
    Solving x to the power of 2 end exponent minus y to the power of 2 end exponent = 0 and 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= 1, we get
    x to the power of 2 end exponent equals y to the power of 2 end exponent = fraction numerator a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction
    Therefore radius of circle
    =square root of fraction numerator 2 a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction end root = fraction numerator square root of 2 a b over denominator square root of a to the power of 2 end exponent plus b to the power of 2 end exponent end root end fraction

    The radius of the circle passing through the points of intersection of 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 = 1 and x2 – y2 = 0 is -

    maths-General
    Two curves are symmetrical about both axes and intersect in four points, so, the circle through their points of intersection will have centre at origin.
    Solving x to the power of 2 end exponent minus y to the power of 2 end exponent = 0 and 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= 1, we get
    x to the power of 2 end exponent equals y to the power of 2 end exponent = fraction numerator a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction
    Therefore radius of circle
    =square root of fraction numerator 2 a to the power of 2 end exponent b to the power of 2 end exponent over denominator a to the power of 2 end exponent plus b to the power of 2 end exponent end fraction end root = fraction numerator square root of 2 a b over denominator square root of a to the power of 2 end exponent plus b to the power of 2 end exponent end root end fraction
    General
    maths-

    If  alpha comma beta are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -

    open parentheses a cos invisible function application alpha comma b sin invisible function application alpha close parentheses comma open parentheses a cos invisible function application beta comma b sin invisible function application beta close parentheses comma open parentheses a e comma 0 close parentheses are collinear.
    fraction numerator b left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis over denominator a left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis end fraction equals fraction numerator b sin invisible function application alpha – 0 over denominator a cos invisible function application alpha – a e end fraction
    therefore left parenthesis c o s invisible function application alpha minus e right parenthesis left parenthesis s i n invisible function application beta minus s i n invisible function application alpha right parenthesis equals s i n invisible function application alpha left parenthesis c o s invisible function application beta minus c o s invisible function application alpha right parenthesis
    therefore e =blank fraction numerator cos invisible function application alpha left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis – sin invisible function application alpha left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis over denominator sin invisible function application beta – sin invisible function application alpha end fraction = fraction numerator sin invisible function application left parenthesis alpha – beta right parenthesis over denominator sin invisible function application alpha – sin invisible function application beta end fraction blank= fraction numerator sin invisible function application alpha plus sin invisible function application beta over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction

    If  alpha comma beta are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -

    maths-General
    open parentheses a cos invisible function application alpha comma b sin invisible function application alpha close parentheses comma open parentheses a cos invisible function application beta comma b sin invisible function application beta close parentheses comma open parentheses a e comma 0 close parentheses are collinear.
    fraction numerator b left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis over denominator a left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis end fraction equals fraction numerator b sin invisible function application alpha – 0 over denominator a cos invisible function application alpha – a e end fraction
    therefore left parenthesis c o s invisible function application alpha minus e right parenthesis left parenthesis s i n invisible function application beta minus s i n invisible function application alpha right parenthesis equals s i n invisible function application alpha left parenthesis c o s invisible function application beta minus c o s invisible function application alpha right parenthesis
    therefore e =blank fraction numerator cos invisible function application alpha left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis – sin invisible function application alpha left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis over denominator sin invisible function application beta – sin invisible function application alpha end fraction = fraction numerator sin invisible function application left parenthesis alpha – beta right parenthesis over denominator sin invisible function application alpha – sin invisible function application beta end fraction blank= fraction numerator sin invisible function application alpha plus sin invisible function application beta over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction
    General
    maths-

    If ϕ is the angle between the diameter through any point on a standard ellipse and the normal at the point, then the greatest value of tan ϕ is–

    Any point P on ellipse is (a cos theta, b sin theta)
     Equation of the diameter CP is y = open parentheses fraction numerator b over denominator a end fraction tan invisible function application theta close parenthesesx
    The normal to ellipse at P is
    ax sec theta – by cosec theta = a2e2
    Slopes of the lines CP and the normal GP are fraction numerator b over denominator a end fractiontan theta andfraction numerator a over denominator b end fractiontan theta

     tan ϕ = fraction numerator fraction numerator a over denominator b end fraction tan invisible function application theta minus fraction numerator b over denominator a end fraction tan invisible function application theta over denominator 1 plus fraction numerator a over denominator b end fraction tan invisible function application theta. fraction numerator b over denominator a end fraction tan invisible function application theta end fraction=fraction numerator tan invisible function application theta over denominator sec to the power of 2 end exponent invisible function application theta end fraction
    =fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator a b end fractionsin  cos  = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fractionsin 2
     The greatest value of tan ϕ = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fraction.1 = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fraction .

    If ϕ is the angle between the diameter through any point on a standard ellipse and the normal at the point, then the greatest value of tan ϕ is–

    maths-General
    Any point P on ellipse is (a cos theta, b sin theta)
     Equation of the diameter CP is y = open parentheses fraction numerator b over denominator a end fraction tan invisible function application theta close parenthesesx
    The normal to ellipse at P is
    ax sec theta – by cosec theta = a2e2
    Slopes of the lines CP and the normal GP are fraction numerator b over denominator a end fractiontan theta andfraction numerator a over denominator b end fractiontan theta

     tan ϕ = fraction numerator fraction numerator a over denominator b end fraction tan invisible function application theta minus fraction numerator b over denominator a end fraction tan invisible function application theta over denominator 1 plus fraction numerator a over denominator b end fraction tan invisible function application theta. fraction numerator b over denominator a end fraction tan invisible function application theta end fraction=fraction numerator tan invisible function application theta over denominator sec to the power of 2 end exponent invisible function application theta end fraction
    =fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator a b end fractionsin  cos  = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fractionsin 2
     The greatest value of tan ϕ = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fraction.1 = fraction numerator a to the power of 2 end exponent minus b to the power of 2 end exponent over denominator 2 a b end fraction .
    General
    physics-

    A radar operates at wavelength 50.0 cm. If the beat freqency between the transmitted singal and the singal reflected from aircraft (capital delta v) is 1 kHz, then velocity of the aircraft will be

    when source is fixed and observer is moving towards it
    upsilon ´ equals fraction numerator c plus a over denominator c end fraction. upsilon
    when source is moving towards observer at rest
    upsilon " equals fraction numerator c over denominator c minus a end fraction v ´ equals fraction numerator c plus a over denominator c minus a end fraction. upsilon equals c open square brackets fraction numerator 1 plus fraction numerator a over denominator c end fraction over denominator 1 minus fraction numerator a over denominator c end fraction end fraction close square brackets upsilon
    equals   c open square brackets 1 plus fraction numerator a over denominator c end fraction close square brackets open square brackets 1 minus fraction numerator a over denominator c end fraction close square brackets to the power of negative 1 end exponent   upsilon almost equal to open square brackets 1 plus fraction numerator 2 a over denominator c end fraction close square brackets upsilon
    therefore   capital delta upsilon equals upsilon ´ minus upsilon equals fraction numerator 2 a upsilon over denominator c end fraction equals fraction numerator 2 a over denominator lambda end fraction
    therefore   a equals lambda fraction numerator capital delta upsilon over denominator 2 end fraction equals fraction numerator 0.5 x 1000 over denominator 2 end fraction equals 250   m s to the power of negative 1 end exponent
    = 900 km/hr

    A radar operates at wavelength 50.0 cm. If the beat freqency between the transmitted singal and the singal reflected from aircraft (capital delta v) is 1 kHz, then velocity of the aircraft will be

    physics-General
    when source is fixed and observer is moving towards it
    upsilon ´ equals fraction numerator c plus a over denominator c end fraction. upsilon
    when source is moving towards observer at rest
    upsilon " equals fraction numerator c over denominator c minus a end fraction v ´ equals fraction numerator c plus a over denominator c minus a end fraction. upsilon equals c open square brackets fraction numerator 1 plus fraction numerator a over denominator c end fraction over denominator 1 minus fraction numerator a over denominator c end fraction end fraction close square brackets upsilon
    equals   c open square brackets 1 plus fraction numerator a over denominator c end fraction close square brackets open square brackets 1 minus fraction numerator a over denominator c end fraction close square brackets to the power of negative 1 end exponent   upsilon almost equal to open square brackets 1 plus fraction numerator 2 a over denominator c end fraction close square brackets upsilon
    therefore   capital delta upsilon equals upsilon ´ minus upsilon equals fraction numerator 2 a upsilon over denominator c end fraction equals fraction numerator 2 a over denominator lambda end fraction
    therefore   a equals lambda fraction numerator capital delta upsilon over denominator 2 end fraction equals fraction numerator 0.5 x 1000 over denominator 2 end fraction equals 250   m s to the power of negative 1 end exponent
    = 900 km/hr
    General
    maths-

    The locus of P such that PA2 + PB2 = 10 where A = (2, 0) and B = (4, 0) is

    The locus of P such that PA2 + PB2 = 10 where A = (2, 0) and B = (4, 0) is

    maths-General
    General
    maths-

    Let L = 0 is a tangent to ellipse fraction numerator left parenthesis x minus 1 right parenthesis to the power of 2 end exponent over denominator 9 end fraction + fraction numerator left parenthesis y minus 2 right parenthesis to the power of 2 end exponent over denominator 4 end fraction = 1 and S, straight S to the power of straight prime be its foci. If length of perpendicular from S on L = 0 is 2 then length of perpendicular from straight S to the power of straight prime  on L = 0 is

    Let L = 0 is a tangent to ellipse fraction numerator left parenthesis x minus 1 right parenthesis to the power of 2 end exponent over denominator 9 end fraction + fraction numerator left parenthesis y minus 2 right parenthesis to the power of 2 end exponent over denominator 4 end fraction = 1 and S, straight S to the power of straight prime be its foci. If length of perpendicular from S on L = 0 is 2 then length of perpendicular from straight S to the power of straight prime  on L = 0 is

    maths-General
    General
    maths-

    The condition that the line ellx + my = n may be a normal to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction+ fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1 is

    The condition that the line ellx + my = n may be a normal to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction+ fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1 is

    maths-General
    General
    physics-

    Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lie on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a (m divided by s to the power of 2 end exponent ). (g is acceleration due to gravity). Normal reaction (in N) acting on block A.

    Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lie on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a (m divided by s to the power of 2 end exponent ). (g is acceleration due to gravity). Normal reaction (in N) acting on block A.

    physics-General
    General
    physics-

    Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lie on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a (m divided by s to the power of 2 end exponent ). (g is acceleration due to gravity) Normal reaction (in N) acting on block B is

    Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lie on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a (m divided by s to the power of 2 end exponent ). (g is acceleration due to gravity) Normal reaction (in N) acting on block B is

    physics-General
    General
    physics-

    A light inextensible string connects a block of mass m and top of wedge of mass M. The string is parallel to inclined surface and the inclined surface makes an angle theta with horizontal as shown in the figure. All surfaces are smooth. Now a constant horizontal force of minimum magnitude F is applied to wedge towards right such that the normal reaction on block exerted by wedge just becomes zero. The magnitude of acceleration of wedge is

    A light inextensible string connects a block of mass m and top of wedge of mass M. The string is parallel to inclined surface and the inclined surface makes an angle theta with horizontal as shown in the figure. All surfaces are smooth. Now a constant horizontal force of minimum magnitude F is applied to wedge towards right such that the normal reaction on block exerted by wedge just becomes zero. The magnitude of acceleration of wedge is

    physics-General
    General
    maths-

    The vector stack x with minus on top which is perpendicular to (2,-3,1) and (1,-2,3) and which satisfies the condition stack x with minus on top times open parentheses stack i with minus on top plus 2 stack j with minus on top minus 7 k close parentheses equals 10

    stack x with minus on top equals x stack i with minus on top plus y stack j with minus on top plus z stack k with minus on top text  and  end text stack x with minus on top times left parenthesis stack i with minus on top plus 2 stack j with minus on top minus 7 stack k with minus on top right parenthesis equals 10

    The vector stack x with minus on top which is perpendicular to (2,-3,1) and (1,-2,3) and which satisfies the condition stack x with minus on top times open parentheses stack i with minus on top plus 2 stack j with minus on top minus 7 k close parentheses equals 10

    maths-General
    stack x with minus on top equals x stack i with minus on top plus y stack j with minus on top plus z stack k with minus on top text  and  end text stack x with minus on top times left parenthesis stack i with minus on top plus 2 stack j with minus on top minus 7 stack k with minus on top right parenthesis equals 10
    General
    maths-

    Let f : R → R be a differentiable function satisfying f (x) = f (x – y) f (y) " x, y Î R and f ¢ (0) = a, f ¢ (2) = b then f ¢ (-2) is

    Let f : R → R be a differentiable function satisfying f (x) = f (x – y) f (y) " x, y Î R and f ¢ (0) = a, f ¢ (2) = b then f ¢ (-2) is

    maths-General