General
Easy
Physics-

A fighter plane enters inside the enemy territory, at time t equals 0 with velocity v subscript 0 end subscript equals 250 blank m s to the power of negative 1 end exponent and moves horizontally with constant acceleration a equals 20 m s to the power of negative 2 end exponent (see figure). An enemy tank at the border, spot the plane and fire shots at an angle theta equals 60 degree with the horizontal and with velocity u equals 600 blank m s to the power of negative 1 end exponent. At what altitude H of the plane it can be hit by the shot?

Physics-General

  1. 1500 square root of 3 m    
  2. 2473 blank m    
  3. 125 blank m    
  4. 1400 blank m    

    Answer:The correct answer is: 1500 square root of 3 mIf it is being hit then
    d equals v subscript 0 end subscript t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent equals left parenthesis u cos invisible function application theta right parenthesis t
    or t equals fraction numerator u cos invisible function application theta minus v subscript 0 end subscript over denominator a divided by 2 end fraction

    therefore blank t equals fraction numerator 600 cross times fraction numerator 1 over denominator 2 end fraction minus 250 over denominator 10 end fraction equals 5 blank s
    H equals open parentheses u sin invisible function application theta close parentheses t minus fraction numerator 1 over denominator 2 end fraction cross times g t to the power of 2 end exponent
    equals 600 cross times fraction numerator square root of 3 over denominator 2 end fraction cross times 5 minus fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 25
    H equals 2473 blank m

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    T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application open parentheses alpha minus beta close parentheses over denominator g cos invisible function application beta end fraction
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    General
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    A particle is moving on a circular path of radius r with uniform velocity v. The change in velocity when the particle moves from P blanktoblank Q blankisblank left parenthesis angle P O Q equals 40 degree right parenthesis

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    equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L
    equals M blank 4 pi to the power of 2 end exponent open parentheses fraction numerator 2 over denominator pi end fraction close parentheses to the power of 2 end exponent L equals 16 blank M L

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    A piece of wire is bent in the shape of a parabola y equals k x to the power of 2 end exponent blank left parenthesis y-axis vertical) with a bead of mass m on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

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    physics-General
    m a cos invisible function application theta equals m g cos invisible function application left parenthesis 90 minus theta right parenthesis
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    The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through R, where P R equals

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    6 cross times P R equals 4 cross times R Q
    P R equals fraction numerator 4 over denominator 6 end fraction R Q equals fraction numerator 2 over denominator 3 end fraction R Q

    The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through R, where P R equals

    physics-General
    Equating the moments about R
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    In a two dimensional motion of a particle, the particle moves from point A, position vector stack r with rightwards arrow on top subscript 1 end subscript. If the magnitudes of these vectors are respectively, r subscript 1 end subscript=3 and r subscript 2 end subscript equals 4 and the angles they make with the x-axis are theta subscript 1 end subscript equals 75 degree and 15degree, respectively, then find the magnitude of the displacement vector

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    theta equals 75 degree minus 15 degree equals 60 degree
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    The time taken by the projectile to reach from A to B is t comma then the distance A B is equal to

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    The time taken by the projectile to reach from A to B is t comma then the distance A B is equal to

    physics-General
    Horizontal component of velocity at A

    v subscript H end subscript equals u cos invisible function application 60 degree equals fraction numerator u over denominator 2 end fraction blank therefore A C equals u subscript H end subscript cross times t equals fraction numerator u t over denominator 2 end fraction
    A B equals A C sec invisible function application 30 degree equals fraction numerator u t over denominator 2 end fraction cross times fraction numerator 2 over denominator square root of 3 end fraction equals fraction numerator u t over denominator 2 end fraction
    General
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    A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

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    A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

    physics-General
    The forces acting on the ball will be (i) in the direction opposite to its motion ie, frictional force and(ii) weight mg.    
    General
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    Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

    R=u2sin2θg=2uxvyg Range horizontal initial velocity (ux) In path 4 range is maximum so football possess maximum horizontal velocity in the path    

    Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

    physics-General
    R=u2sin2θg=2uxvyg Range horizontal initial velocity (ux) In path 4 range is maximum so football possess maximum horizontal velocity in the path    
    General
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    A body of mass m is moving with a uniform speed v along a circle of radius r, what is the average acceleration in going from A to B?

    Here, T equals fraction numerator 2 pi r over denominator 4 v end fraction equals fraction numerator pi r over denominator 2 v end fraction
    Change in velocity is going from A to B = v square root of 2
    Average acceleration equals fraction numerator v square root of 2 over denominator pi r divided by 2 v end fraction equals fraction numerator 2 square root of 2 v to the power of 2 end exponent over denominator pi r end fraction

    A body of mass m is moving with a uniform speed v along a circle of radius r, what is the average acceleration in going from A to B?

    physics-General
    Here, T equals fraction numerator 2 pi r over denominator 4 v end fraction equals fraction numerator pi r over denominator 2 v end fraction
    Change in velocity is going from A to B = v square root of 2
    Average acceleration equals fraction numerator v square root of 2 over denominator pi r divided by 2 v end fraction equals fraction numerator 2 square root of 2 v to the power of 2 end exponent over denominator pi r end fraction
    General
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    Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed omega. The discs are in the same horizontal plane. At time t equals 0, the points P and Q are facing each other as shown in figure. The relative speed between the two points P and Q is V subscript r end subscript as function of times best represented by


    So, V subscript r end subscript equals 2 omega R sin invisible function application left parenthesis omega t right parenthesis
    At t equals T divided by 2 comma blank V subscript r end subscript equals 0
    So two half cycles will take place

    Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed omega. The discs are in the same horizontal plane. At time t equals 0, the points P and Q are facing each other as shown in figure. The relative speed between the two points P and Q is V subscript r end subscript as function of times best represented by

    physics-General

    So, V subscript r end subscript equals 2 omega R sin invisible function application left parenthesis omega t right parenthesis
    At t equals T divided by 2 comma blank V subscript r end subscript equals 0
    So two half cycles will take place
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    physics-

    A small body of mass m slides down from the top of a hemisphere of radius r. The surface of block and hemisphere are frictionless. The height at which the body lose contact with the surface of the sphere is

    A small body of mass m slides down from the top of a hemisphere of radius r. The surface of block and hemisphere are frictionless. The height at which the body lose contact with the surface of the sphere is

    physics-General
    General
    physics-

    The path of a projectile in the absence of air drag is shown in the figure by dotted line. If the air resistance is not ignored then which one of the path is shown in the figure is appropriate for the projectile

    If air resistance is taken into consideration then range and maximum height, both will decrease

    The path of a projectile in the absence of air drag is shown in the figure by dotted line. If the air resistance is not ignored then which one of the path is shown in the figure is appropriate for the projectile

    physics-General
    If air resistance is taken into consideration then range and maximum height, both will decrease