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Maths-

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

Maths-General

  1. 9    
  2. 10    
  3. 5    
  4. 4.5    

    Answer:The correct answer is: 4.5

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    The area bounded by y equals x squared plus 2 comma X- axis, x=1 and x=2 is

    The area bounded by y equals x squared plus 2 comma X- axis, x=1 and x=2 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 =

    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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    The x minus t graph shown in the figure represents

    Up to time t subscript 1 end subscript slope of the graph is constant and after t subscript 1 end subscriptslope is zero i. e. the body travel with constant speed up to time t subscript 1 end subscript and then stops

    The x minus t graph shown in the figure represents

    physics-General
    Up to time t subscript 1 end subscript slope of the graph is constant and after t subscript 1 end subscriptslope is zero i. e. the body travel with constant speed up to time t subscript 1 end subscript and then stops
    General
    physics-

    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

    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

    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

    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
    physics-

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

    Instantaneous velocity is given by the slope of the curve at that instant v equals fraction numerator d s over denominator d t end fraction equals tan invisible function application theta from the figure it is clear that slope of the curve is maximum at point ‘C

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

    physics-General
    Instantaneous velocity is given by the slope of the curve at that instant v equals fraction numerator d s over denominator d t end fraction equals tan invisible function application theta from the figure it is clear that slope of the curve is maximum at point ‘C
    General
    physics-

    Assertion : Owls can move freely during night.
    Reason : They have large number of rods on their retina.

    Owls can move freely during night, because they have large number of cones on their retina which help them to see in night.

    Assertion : Owls can move freely during night.
    Reason : They have large number of rods on their retina.

    physics-General
    Owls can move freely during night, because they have large number of cones on their retina which help them to see in night.
    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.
    General
    physics-

    A body is at rest at x equals 0. At t equals 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x equals 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x subscript 1 end subscript left parenthesis t right parenthesis after time ‘t’ and that of the second body by x subscript 2 end subscript left parenthesis t right parenthesis after the same time interval. Which of the following graphs correctly describes left parenthesis x subscript 1 end subscript minus x subscript 2 end subscript right parenthesis as a function of time ‘t

    x subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent and x subscript 2 end subscript equals u t blank therefore x subscript 1 end subscript minus x subscript 2 end subscript equals fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent minus u t
    y equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent minus u t.This equation is of parabola
    fraction numerator d y over denominator d t end fraction equals a t minus u and fraction numerator d to the power of 2 end exponent y over denominator d t to the power of 2 end exponent end fraction equals a
    As fraction numerator d to the power of 2 end exponent y over denominator d t to the power of 2 end exponent end fraction greater than 0 i. e. comma graph shows possess minima at t equals fraction numerator u over denominator a end fraction

    A body is at rest at x equals 0. At t equals 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x equals 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x subscript 1 end subscript left parenthesis t right parenthesis after time ‘t’ and that of the second body by x subscript 2 end subscript left parenthesis t right parenthesis after the same time interval. Which of the following graphs correctly describes left parenthesis x subscript 1 end subscript minus x subscript 2 end subscript right parenthesis as a function of time ‘t

    physics-General
    x subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent and x subscript 2 end subscript equals u t blank therefore x subscript 1 end subscript minus x subscript 2 end subscript equals fraction numerator 1 over denominator 2 end fraction blank a t to the power of 2 end exponent minus u t
    y equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent minus u t.This equation is of parabola
    fraction numerator d y over denominator d t end fraction equals a t minus u and fraction numerator d to the power of 2 end exponent y over denominator d t to the power of 2 end exponent end fraction equals a
    As fraction numerator d to the power of 2 end exponent y over denominator d t to the power of 2 end exponent end fraction greater than 0 i. e. comma graph shows possess minima at t equals fraction numerator u over denominator a end fraction
    General
    maths-

    General solution of 2 to the power of x equals n x end exponent plus 2 to the power of cos space x end exponent equals 2 to the power of 1 minus fraction numerator 1 over denominator square root of 2 end fraction end exponent is

    General solution of 2 to the power of x equals n x end exponent plus 2 to the power of cos space x end exponent equals 2 to the power of 1 minus fraction numerator 1 over denominator square root of 2 end fraction end exponent is

    maths-General
    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
    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

    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
    General
    physics-

    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 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

    Let 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

    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
    Let 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
    General
    maths-

    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

    maths-General
    General
    physics-

    In the following graph, distance travelled by the body in metres is

    Distance = Area covered between velocity and time axis
    equals fraction numerator 1 over denominator 2 end fraction open parentheses 30 plus 10 close parentheses 10 equals 200 blank m e t e r

    In the following graph, distance travelled by the body in metres is

    physics-General
    Distance = Area covered between velocity and time axis
    equals fraction numerator 1 over denominator 2 end fraction open parentheses 30 plus 10 close parentheses 10 equals 200 blank m e t e r
    General
    physics-

    Velocity-time open parentheses v minus t close parentheses graph for a moving object is shown in the figure. Total displacement of the object during the time interval when there is non-zero acceleration and retardation is

    Between time interval 20s the 4s, there is non-zero acceleration and retardation. Hence, distance travelled during this interval = Area between time interval 20 s to 40 s
    equals fraction numerator 1 over denominator 2 end fraction cross times 20 cross times 3 plus 20 cross times 1 equals 30 plus 20 equals 50 blank m

    Velocity-time open parentheses v minus t close parentheses graph for a moving object is shown in the figure. Total displacement of the object during the time interval when there is non-zero acceleration and retardation is

    physics-General
    Between time interval 20s the 4s, there is non-zero acceleration and retardation. Hence, distance travelled during this interval = Area between time interval 20 s to 40 s
    equals fraction numerator 1 over denominator 2 end fraction cross times 20 cross times 3 plus 20 cross times 1 equals 30 plus 20 equals 50 blank m