General
Easy
Physics-

A particle of mass m is initially situated at the point P inside a hemispherical surface of radius r as shown in figure. A horizontal acceleration of magnitudea subscript 0 end subscriptis suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the sphere again is

Physics-General

  1. square root of fraction numerator 4 r blank t a n alpha over denominator a subscript 0 end subscript end fraction end root    
  2. square root of fraction numerator 4 r blank s i n alpha over denominator a subscript 0 end subscript end fraction end root    
  3. square root of fraction numerator 4 r blank c o s alpha over denominator a subscript 0 end subscript end fraction end root    
  4. None of these    

    Answer:The correct answer is: square root of fraction numerator 4 r blank c o s alpha over denominator a subscript 0 end subscript end fraction end rootLet the particle touches the sphere t the point A.
    Let P A equals 1
    therefore P B equals fraction numerator l over denominator 2 end fraction
    In increment O P B comma cos invisible function application alpha equals fraction numerator P B over denominator r end fraction

    therefore P B equals r cos invisible function application a
    or fraction numerator l over denominator 2 end fraction equals r cos invisible function application a
    therefore l equals 2 r cos invisible function application alpha
    B u t blank l equals fraction numerator 1 over denominator 2 end fraction a subscript 0 end subscript t to the power of 2 end exponent
    therefore blank t equals square root of open parentheses fraction numerator 2 l over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 2 cross times 2 r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 4 blank r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root

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    If α and β are two different solutions lying between fraction numerator negative pi over denominator 2 end fraction and  of the equation 2 Tan space theta plus Sec space theta equals 2 then Tan α + Tan β is

    If α and β are two different solutions lying between fraction numerator negative pi over denominator 2 end fraction and  of the equation 2 Tan space theta plus Sec space theta equals 2 then Tan α + Tan β is

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    A cyclist starts from the centreO of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference and returns to the point O as shown in figure. If the round trip takes 10 min, the net displacement and average speed of the cyclist (in metre and kilometer per hour) are

    Since, the initial position of cyclist coincides with final position, so his net displacement is zero.
    A v e r a g e blank s p e e d equals fraction numerator t o t a l blank d i s t a n c e blank t r a v e l l e d over denominator t o t a l blank t i m e blank t a k e n end fraction
    equals fraction numerator O P plus P Q plus Q O over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent
    equals fraction numerator 1 plus fraction numerator pi over denominator 2 end fraction cross times 1 plus 1 over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent
    equals fraction numerator pi plus 4 over denominator 20 end fraction cross times 60 blank k m h to the power of negative 1 end exponent equals 21.4 blank k m h to the power of negative 1 end exponent

    A cyclist starts from the centreO of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference and returns to the point O as shown in figure. If the round trip takes 10 min, the net displacement and average speed of the cyclist (in metre and kilometer per hour) are

    physics-General
    Since, the initial position of cyclist coincides with final position, so his net displacement is zero.
    A v e r a g e blank s p e e d equals fraction numerator t o t a l blank d i s t a n c e blank t r a v e l l e d over denominator t o t a l blank t i m e blank t a k e n end fraction
    equals fraction numerator O P plus P Q plus Q O over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent
    equals fraction numerator 1 plus fraction numerator pi over denominator 2 end fraction cross times 1 plus 1 over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent
    equals fraction numerator pi plus 4 over denominator 20 end fraction cross times 60 blank k m h to the power of negative 1 end exponent equals 21.4 blank k m h to the power of negative 1 end exponent
    General
    physics-

    A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by O A B and C D respectively Then

    Distance travelled by motor bike at t equals 18s
    s subscript b i k e end subscript equals s subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction(18)(60)=540 m
    Distance travelled by car at t equals 18s
    s subscript c a r end subscript equals s subscript 2 end subscript=(18)(60)=720 m
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    Hence, s subscript b i k e end subscript equals 540 plus 60 t and s subscript c a r end subscript equals 720 plus 40 t
    s subscript c a r end subscript minus s subscript b i k e end subscript equals 0
    rightwards double arrow blank 720 plus 40 t equals 540 plus 60 t
    rightwards double arrow blank t equals 9s beyond 18s or
    Hence, t equals open parentheses 18 plus 9 close parenthesess=27s from start and distant travelled by both is s subscript b i k e end subscript=s subscript c a r end subscript equals 1080m

    A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by O A B and C D respectively Then

    physics-General
    Distance travelled by motor bike at t equals 18s
    s subscript b i k e end subscript equals s subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction(18)(60)=540 m
    Distance travelled by car at t equals 18s
    s subscript c a r end subscript equals s subscript 2 end subscript=(18)(60)=720 m
    Therefore, separation between them at t equals 18s is 180m. Let, separation between them decreases to zero at time t beyond 18s.
    Hence, s subscript b i k e end subscript equals 540 plus 60 t and s subscript c a r end subscript equals 720 plus 40 t
    s subscript c a r end subscript minus s subscript b i k e end subscript equals 0
    rightwards double arrow blank 720 plus 40 t equals 540 plus 60 t
    rightwards double arrow blank t equals 9s beyond 18s or
    Hence, t equals open parentheses 18 plus 9 close parenthesess=27s from start and distant travelled by both is s subscript b i k e end subscript=s subscript c a r end subscript equals 1080m
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    Assertion : Owls can move freely during night.
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    Assertion : Owls can move freely during night.
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    physics-General
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    General
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    A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

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    A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

    physics-General
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    In figure, one car at rest and velocity of the light from head light is c, tehn velocity of light from head light for the moving car at velocity v, would be

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    physics-General
    Since c greater than greater than v (negligible)
    General
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    A particle starts from rest. Its acceleration open parentheses a close parentheses blank v e r s u s time open parentheses t close parentheses is as shown in the figure. The maximum speed of the particle will be

    physics-General
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    The area bounded by y=3x and y equals x to the power of 2 end exponent is

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    General
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    If Cos space 2 theta times Cos space 3 theta times cos space theta equals 1 fourth text  for  end text 0 less than theta less than pi then theta =

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    maths-General
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    The x minus t graph shown in the figure represents

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    The x minus t graph shown in the figure represents

    physics-General
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    For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds

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    physics-General
    fraction numerator open parentheses S close parentheses subscript open parentheses l a s t blank 2 s close parentheses end subscript over denominator open parentheses S close parentheses subscript 7 s end subscript end fraction equals fraction numerator table row cell fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 end cell row cell end cell end table over denominator table row cell fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 plus 2 cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times 2 cross times 10 end cell row cell end cell row cell end cell end table end fraction equals fraction numerator 1 over denominator 4 end fraction
    General
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    A particle starts from rest at t equals 0 and undergoes an acceleration a in m s to the power of negative 2 end exponent with time t in seconds which is as shown Which one of the following plot represents velocity V in m s to the power of negative 1 end exponent versus time t in seconds


    Takingthe motion from 0 to 2 blank s
    u equals 0 comma blank a equals 3 m s to the power of negative 2 end exponent comma blank t equals 2 s comma blank v equals ?
    v equals u plus a t equals 0 plus 3 cross times 2 equals 6 m s to the power of negative 1 end exponent
    Taking the motion from 2 blank s to 4 blank s
    v equals 6 plus open parentheses negative 3 close parentheses open parentheses 2 close parentheses equals 0 m s to the power of negative 1 end exponent

    A particle starts from rest at t equals 0 and undergoes an acceleration a in m s to the power of negative 2 end exponent with time t in seconds which is as shown Which one of the following plot represents velocity V in m s to the power of negative 1 end exponent versus time t in seconds

    physics-General

    Takingthe motion from 0 to 2 blank s
    u equals 0 comma blank a equals 3 m s to the power of negative 2 end exponent comma blank t equals 2 s comma blank v equals ?
    v equals u plus a t equals 0 plus 3 cross times 2 equals 6 m s to the power of negative 1 end exponent
    Taking the motion from 2 blank s to 4 blank s
    v equals 6 plus open parentheses negative 3 close parentheses open parentheses 2 close parentheses equals 0 m s to the power of negative 1 end exponent
    General
    maths-

    If gamma Sin space theta equals 3 comma gamma equals 4 left parenthesis 1 plus Sin space theta right parenthesis comma 0 less or equal than theta less or equal than 2 pi then theta = ---

    If gamma Sin space theta equals 3 comma gamma equals 4 left parenthesis 1 plus Sin space theta right parenthesis comma 0 less or equal than theta less or equal than 2 pi then theta = ---

    maths-General
    General
    physics-

    A particle starts from rest at t equals 0 and undergoes an acceleration a in m s to the power of negative 2 end exponent with time t in second which is as shownWhich one of the following plot represents velocity v in m s to the power of negative 1 end exponent blank v e r s u s time t in second?

    A particle starts from rest at t equals 0
    The equation of motion
    v equals u plus a t equals 0 plus 3 cross times 2 equals 6 blank m s to the power of negative 1 end exponent
    The velocity for next 2 s
    v to the power of ´ end exponent equals v plus a t
    blank equals 6 minus 3 cross times 2 equals 0
    v to the power of ´ end exponent equals 0

    Hence, v minus t graph will be as shown.

    A particle starts from rest at t equals 0 and undergoes an acceleration a in m s to the power of negative 2 end exponent with time t in second which is as shownWhich one of the following plot represents velocity v in m s to the power of negative 1 end exponent blank v e r s u s time t in second?

    physics-General
    A particle starts from rest at t equals 0
    The equation of motion
    v equals u plus a t equals 0 plus 3 cross times 2 equals 6 blank m s to the power of negative 1 end exponent
    The velocity for next 2 s
    v to the power of ´ end exponent equals v plus a t
    blank equals 6 minus 3 cross times 2 equals 0
    v to the power of ´ end exponent equals 0

    Hence, v minus t graph will be as shown.