General
Easy
Physics-

A body of mass m is resting on a wedge of angle theta as shown in figure. The wedge is given at acceleration alpha. What is the value of a son that the mass m just falls freely?

Physics-General

  1. g    
  2. g sin invisible function application theta    
  3. g tan invisible function application theta    
  4. g cot invisible function application theta    

    Answer:The correct answer is: g cot invisible function application thetaThe horizontal acceleration a of the wedge should be such that in time the wedge moves the horizontal distance B C. The body must fall through a vertical distance A B under gravity. Hence,
    B C equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent and A B equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
    tantheta equals fraction numerator A B over denominator B C end fraction equals fraction numerator g over denominator a end fractionor a equals fraction numerator g over denominator t a n theta end fraction equals g blank c o t theta

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    A cyclist starts from the centre O of a circular park of radius one kilometre, reaches the edge P of the park, then cycles along the circumference and returns to the centre along Q O as shown in figure. If the round trip takes ten minutes, the net displacement and average speed of the cyclist (in metre and kilometre per hour) is

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    equals 1 plus fraction numerator 2 pi cross times 1 over denominator 4 end fraction plus 1 equals fraction numerator 14.28 over denominator 4 end fraction blank k m
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    equals fraction numerator 6 cross times 14.28 over denominator 4 end fraction equals 21.42 blank k m divided by h

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    Net displacement equals 0 and total distance equals O P plus P Q plus Q O
    equals 1 plus fraction numerator 2 pi cross times 1 over denominator 4 end fraction plus 1 equals fraction numerator 14.28 over denominator 4 end fraction blank k m
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    equals fraction numerator 6 cross times 14.28 over denominator 4 end fraction equals 21.42 blank k m divided by h
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    A boy begins to walk eastward along a street infront of his house and the graph of his displacement from home is shown in the following figure. His average speed for in the whole-time interval is equal to

    D i s p l a c e m e n t blank f o r m blank 0 blank t o blank 5 blank s blank equals blank 40 blank m
    D i s p l a c e m e n t blank f r o m blank 5 blank t o blank 10 blank s blank equals blank 40 blank m
    D i s p l a c e m e n t blank f r o m blank 0 blank t o blank 15 blank s blank equals blank minus blank 20 blank m
    A n d blank d i s p l a c e m e n t blank f r o m blank 15 blank t o blank 20 blank s equals 0 blank m
    therefore N e t blank d i s p l a c e m e n t equals 40 plus 40 minus 20 plus 0 equals 60 blank m
    T o t a l blank t i m e blank t a k e n equals 5 plus 5 plus 15 plus 5 equals 30 blank m i n.
    H e n c e comma blank a v e r a g e blank s p e e d equals fraction numerator d i s p l a c e m e n t blank open parentheses m close parentheses over denominator t i m e blank open parentheses m i n close parentheses end fraction equals fraction numerator 60 over denominator 30 end fraction equals 2 m blank m i n to the power of negative 1 end exponent.

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    D i s p l a c e m e n t blank f o r m blank 0 blank t o blank 5 blank s blank equals blank 40 blank m
    D i s p l a c e m e n t blank f r o m blank 5 blank t o blank 10 blank s blank equals blank 40 blank m
    D i s p l a c e m e n t blank f r o m blank 0 blank t o blank 15 blank s blank equals blank minus blank 20 blank m
    A n d blank d i s p l a c e m e n t blank f r o m blank 15 blank t o blank 20 blank s equals 0 blank m
    therefore N e t blank d i s p l a c e m e n t equals 40 plus 40 minus 20 plus 0 equals 60 blank m
    T o t a l blank t i m e blank t a k e n equals 5 plus 5 plus 15 plus 5 equals 30 blank m i n.
    H e n c e comma blank a v e r a g e blank s p e e d equals fraction numerator d i s p l a c e m e n t blank open parentheses m close parentheses over denominator t i m e blank open parentheses m i n close parentheses end fraction equals fraction numerator 60 over denominator 30 end fraction equals 2 m blank m i n to the power of negative 1 end exponent.
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    physics-General
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    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

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    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

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    Which of the following graphs cannot possibly represent one dimensional motion of a particle

    I is not possible because total distance covered by a particle increases with time
    II is not possible because at a particular time, position cannot have two values
    III is not possible because at a particular time, velocity cannot have two values
    IV is not possible because speed can never be negative

    Which of the following graphs cannot possibly represent one dimensional motion of a particle

    physics-General
    I is not possible because total distance covered by a particle increases with time
    II is not possible because at a particular time, position cannot have two values
    III is not possible because at a particular time, velocity cannot have two values
    IV is not possible because speed can never be negative
    General
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    Which of the following graphs cannot possibly represent one dimensional motion of a particle

    I is not possible because total distance covered by a particle increases with time
    II is not possible because at a particular time, position cannot have two values
    III is not possible because at a particular time, velocity cannot have two values
    IV is not possible because speed can never be negative

    Which of the following graphs cannot possibly represent one dimensional motion of a particle

    physics-General
    I is not possible because total distance covered by a particle increases with time
    II is not possible because at a particular time, position cannot have two values
    III is not possible because at a particular time, velocity cannot have two values
    IV is not possible because speed can never be negative