General
Easy
Physics-

The trajectory of a particle moving in vast maidan is as shown in the figure. The coordinates of a position A are open parentheses 0 , 2 close parentheses. The coordinates of another point at which the instantaneous velocity is same as the average velocity between the points are

Physics-General

  1. left parenthesis 1 comma blank 4 right parenthesis    
  2. left parenthesis 5 comma blank 3 right parenthesis    
  3. left parenthesis 3 comma blank 4 right parenthesis    
  4. left parenthesis 4 comma blank 1 right parenthesis    

    Answer:The correct answer is: left parenthesis 5 comma blank 3 right parenthesis

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    General
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    Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v blank a n d blank 2 v respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A blank,these two particles will again reach The point A ?

    A s first collision one particle having speed 2v will rotate
    240 degree open parentheses o r fraction numerator 4 pi over denominator 3 end fraction close parentheseswhile other particle having speed v blankwill rotate
    120 degree open parentheses o r fraction numerator 2 pi over denominator 3 end fraction close parentheses. At first collision they will exchange their velocities. Now as shown in figure, after two collisions they will again reach at point A.

    Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v blank a n d blank 2 v respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A blank,these two particles will again reach The point A ?

    physics-General
    A s first collision one particle having speed 2v will rotate
    240 degree open parentheses o r fraction numerator 4 pi over denominator 3 end fraction close parentheseswhile other particle having speed v blankwill rotate
    120 degree open parentheses o r fraction numerator 2 pi over denominator 3 end fraction close parentheses. At first collision they will exchange their velocities. Now as shown in figure, after two collisions they will again reach at point A.
    General
    physics-

    The potential energy of a particle varies with distance x as shown in the graph.
    The force acting on the particle is zero at

    F equals fraction numerator negative d U over denominator d x end fraction it is clear that slope of U minus x curve is zero at point B and C
    therefore F equals 0 for point B and C

    The potential energy of a particle varies with distance x as shown in the graph.
    The force acting on the particle is zero at

    physics-General
    F equals fraction numerator negative d U over denominator d x end fraction it is clear that slope of U minus x curve is zero at point B and C
    therefore F equals 0 for point B and C
    General
    physics-

    A 0.098 kg block slides down a frictionless track as shown. The vertical component of the velocity of block at A is

    fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals m g open parentheses 3 minus 1 close parentheses equals 2 m g or v equals square root of 4 g equals 2 square root of g
    Vertical component at A equals 2 square root of g sin invisible function application 30 degree equals square root of g

    A 0.098 kg block slides down a frictionless track as shown. The vertical component of the velocity of block at A is

    physics-General
    fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent equals m g open parentheses 3 minus 1 close parentheses equals 2 m g or v equals square root of 4 g equals 2 square root of g
    Vertical component at A equals 2 square root of g sin invisible function application 30 degree equals square root of g
    General
    chemistry-

    Assertion :this equilibrium favours backward direction.
    Reason :is stronger base than C H subscript 2 end subscript C O O to the power of minus end exponent

    Assertion :this equilibrium favours backward direction.
    Reason :is stronger base than C H subscript 2 end subscript C O O to the power of minus end exponent

    chemistry-General
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    maths-

    The number of rectangles in the adjoining figure is –

    The number of rectangles in the adjoining figure is –

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    The number of numbers between 1 and 1010 which contain the digit 1 is -

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    General
    maths-

    If n objects are arranged in a row, then the number of ways of selecting three of these objects so that no two of them are next to each other is -

    If n objects are arranged in a row, then the number of ways of selecting three of these objects so that no two of them are next to each other is -

    maths-General
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    maths-

    Between two junction stations A and B there are 12 intermediate stations. The number of ways in which a train can be made to stop at 4 of these stations so that no two of these halting stations are consecutive is -

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    maths-General
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    The number of non-negative integral solutions of x + y + z  n, where n  N is -

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    maths-General
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    maths-

    In a model, it is shown that an arch of abridge is semi-elliptical with major axis horizontal. If the length of the base is 9 m and the highest part of the bridge is 3 m from the horizontal, the best approximation of the height of the arch, 2 m from the centre of the base is

    In a model, it is shown that an arch of abridge is semi-elliptical with major axis horizontal. If the length of the base is 9 m and the highest part of the bridge is 3 m from the horizontal, the best approximation of the height of the arch, 2 m from the centre of the base is

    maths-General
    General
    physics-

    A 10 k g mass moves along x-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x equals 0 to x equals 8 blank c m

    Work done = Area covered in between force displacement curve and displacement axis
    = Mass cross timesArea covered in between acceleration-displacement curve and displacement axis
    equals 10 cross times fraction numerator 1 over denominator 2 end fraction open parentheses 8 cross times 10 to the power of negative 2 end exponent cross times 20 cross times 10 to the power of negative 2 end exponent close parentheses equals 8 cross times 10 to the power of negative 2 end exponent J

    A 10 k g mass moves along x-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x equals 0 to x equals 8 blank c m

    physics-General
    Work done = Area covered in between force displacement curve and displacement axis
    = Mass cross timesArea covered in between acceleration-displacement curve and displacement axis
    equals 10 cross times fraction numerator 1 over denominator 2 end fraction open parentheses 8 cross times 10 to the power of negative 2 end exponent cross times 20 cross times 10 to the power of negative 2 end exponent close parentheses equals 8 cross times 10 to the power of negative 2 end exponent J
    General
    physics-

    A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is

    By applying law of conservation of energy
    m g R equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent rightwards double arrow v equals square root of 2 R g end root

    A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is

    physics-General
    By applying law of conservation of energy
    m g R equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent rightwards double arrow v equals square root of 2 R g end root
    General
    physics-

    Three forces of magnitudes 6N, 6N and square root of 72 N at a corner of a cube along three sides as shown in figure. Resultant of these forces is

    The resultant of 5 N along O C and 5 N along O A is
    R equals square root of 6 to the power of 2 end exponent plus 6 to the power of 2 end exponent end root
    equals square root of 72N along O B
    The resultant of square root of 72 N along O B and square root of 72 N along O G is
    R to the power of ´ end exponent equals square root of 72 plus 72 end root equals 12N along O E.

    Three forces of magnitudes 6N, 6N and square root of 72 N at a corner of a cube along three sides as shown in figure. Resultant of these forces is

    physics-General
    The resultant of 5 N along O C and 5 N along O A is
    R equals square root of 6 to the power of 2 end exponent plus 6 to the power of 2 end exponent end root
    equals square root of 72N along O B
    The resultant of square root of 72 N along O B and square root of 72 N along O G is
    R to the power of ´ end exponent equals square root of 72 plus 72 end root equals 12N along O E.
    General
    physics-

    Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are vand 2 v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A

    Let initially particle x is moving in anticlockwise direction and y in clockwise direction
    As the ratio of velocities of xand y particles are fraction numerator v subscript x end subscript over denominator v subscript y end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction, therefore ratio of their distance covered will be in the ratio of 2 blank colon 1. It means they collide at point B

    After first collision at B, velocities of particles get interchanged, i. e., x will move with 2 v and particle y with v
    Second collision will take place at point C. Again at this point velocities get interchanged and third collision take place at point A
    So, after two collision these two particles will again reach the point A

    Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are vand 2 v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A

    physics-General
    Let initially particle x is moving in anticlockwise direction and y in clockwise direction
    As the ratio of velocities of xand y particles are fraction numerator v subscript x end subscript over denominator v subscript y end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction, therefore ratio of their distance covered will be in the ratio of 2 blank colon 1. It means they collide at point B

    After first collision at B, velocities of particles get interchanged, i. e., x will move with 2 v and particle y with v
    Second collision will take place at point C. Again at this point velocities get interchanged and third collision take place at point A
    So, after two collision these two particles will again reach the point A
    General
    physics-

    A man standing on a hill top projects a stone horizontally with speed v subscript 0 end subscript as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface

    Range of the projectile on an inclined plane (down the plane) is,
    R equals fraction numerator u to the power of 2 end exponent over denominator g c o s to the power of 2 end exponent beta end fraction left square bracket sin invisible function application open parentheses 2 alpha plus beta close parentheses plus sin invisible function application beta right square bracket
    Here, u equals v subscript 0 end subscript comma alpha equals 0 and beta equals theta
    therefore R equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript sin invisible function application theta over denominator g cos to the power of 2 end exponent invisible function application theta end fraction

    Now x equals R cos invisible function application theta equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan invisible function application theta over denominator g end fraction
    and y equals negative R sin invisible function application theta equals blank minus fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction

    A man standing on a hill top projects a stone horizontally with speed v subscript 0 end subscript as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface

    physics-General
    Range of the projectile on an inclined plane (down the plane) is,
    R equals fraction numerator u to the power of 2 end exponent over denominator g c o s to the power of 2 end exponent beta end fraction left square bracket sin invisible function application open parentheses 2 alpha plus beta close parentheses plus sin invisible function application beta right square bracket
    Here, u equals v subscript 0 end subscript comma alpha equals 0 and beta equals theta
    therefore R equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript sin invisible function application theta over denominator g cos to the power of 2 end exponent invisible function application theta end fraction

    Now x equals R cos invisible function application theta equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan invisible function application theta over denominator g end fraction
    and y equals negative R sin invisible function application theta equals blank minus fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction