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Physics Test 114

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Physics Test 114
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Weekly Quiz Competition
  • Question 1
    4 / -1

    The sum & differences of two perpendicular vectors of equal magnitudes are

    Solution

     

  • Question 2
    4 / -1

    An aeroplane is moving on a circular path with a speed of 250 kmph. The change in velocity (in kmph) in half revolution is

    Solution

     

  • Question 3
    4 / -1

    If the sum of two unit vectors is also a unit vector, then the magnitude of their difference is

    Solution

     

  • Question 4
    4 / -1

    A person aims a gun at a bird from a point, at a horizontal distance of 100 m.If gun can impart a speed of 500m/s to the bullet,at what height above the bird must he aim his gun in order to hit it? (g = 10 ms-2)

    Solution

    From given the data

    Range R = 2 ⨯100 m = 200 m

    Initial velocity u = 500 m/s

    We know that Range R =
     

    Horizontal distance x = 100 m

    Velocity v = 500 m.s-1

    Time taken to travel this distance t = x/v = 100/500 = 1/5s

    Vertical distance travelled by the bullet in time 1/5s is given by,

    If the person directly aims at the bird, the bullet will hit 20 cm below

    the bird. Thus the gun must be aimed at 20 cm above the position of the bird.

     

     

  • Question 5
    4 / -1

    The hight y and distance x along the horizontal plane of a projectile on a certain planet (With no surrounding atmosphere) are given as y=y=8t-5t2 mtr. and x = 6t metres, where t is in seconds. The velocity with which the projectile is projected is 

    Solution

    Y = 8t - 5t² 

    differentiate wrt time 

    dy/dt = 8 -10t 

    we know, change in position per unit time is called , velcity .

    e.g dy/dt = Vy 

    Vy = 8 -10t 

    now at t = 0 ( initially )

    Vy =8 -10× 0 = 8 

    x = 6t 

    differentiate wrt time 

    dx/dt = 6 

    similarly , dx/dt = Vx =6 

    now, 

    V = Vx i + Vy j 

    = 8 i + 6 j

    magnitude of velocity =| V | =√(8² + 6²) =10 m/s

     

  • Question 6
    4 / -1

    A helicopter flying at 20 m/s horizontally drops a rescue bag. Ignoring air resistance & taking g = 10ms-2, the displacement of the bag 5 seconds after the drop began is nearly

    Solution

    Sx = 20 x 5

    Sy = 1/2 x w x 25

    S = √sx2 + sy2

    = 160.

     

  • Question 7
    4 / -1

    When the angular velocity of a uniformly rotating body has increased thrice, the resultant of forces applied to it increase by 60 N. Find the acceleration of the body in the two cases. The mass of the body, m = 3kg

    Solution

     

  • Question 8
    4 / -1

    In case of uniform circular motion which of the following physical quantities do not remain constant?

    Solution

    Though speed is constant but direction 0f motion and hence the velocity is variable. Therefore,the momentum is not constant 

     

  • Question 9
    4 / -1

    A parachutist of weight W strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude 3g . force exerted on him by ground during landing is

    Solution

    m = w/g

    3mg = n - w

    N = 4mg

    = 4w

     

  • Question 10
    4 / -1

    An unloaded car moving with velocity u on a road can be stopped in a distance s. If passengers add 40% to its weight and braking force remains the same, the stopping distance at velocity u is now

    Solution

     

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