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Motion Test - 33

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Motion Test - 33
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  • Question 1
    1 / -0
    A wooden block of mass 1010 gm is dropped from the top of a cliff 100100 m high. Simultaneously a bullet of mass 1010 gm is fired from the foot of the cliff  upward with a velocity 100100 m/s. The bullet and the wooden block will meet each other after a time of:
  • Question 2
    1 / -0

    A ball is thrown upwards. It takes 4 s to reach back to the ground. Find its initial velocity.

    Solution
    Using v=u+at\text{Using}\ v=u+at
    Time of Flight =2ug= \dfrac{2u}{g}
    Where u is the initial velocity and g is acceleration due to gravity 
    4=2u104=\dfrac{2u}{10}
    u=20 ms1u = 20 \ ms^{-1}
  • Question 3
    1 / -0

    A boy standing at the top of a tower of 20 mm height drops a stone. Assuming g=10ms2g=10 ms^{-2}, the velocity with which it hits the ground is:

    Solution
    When there is a free fall, we can directly use the equation of motion :
    v2=u2+2ghv^{2} = u^{2} + 2gh
    here u=0u = 0
    so, v=2ghv = \sqrt{2gh} = 2×10×20=20 m/s\sqrt{2 \times 10 \times 20} = 20\ m/s
  • Question 4
    1 / -0
    Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms110\ ms^{-1}. It implies that the boy is :
    Solution
    • As an object moves in a merry-go-round, the speed of the object may be constant, but the direction of the object changes with the motion of the object hence its velocity will also change.
    • As velocity is changing, therefore the boy will be in an accelerated motion.

    Hence option C is correct
  • Question 5
    1 / -0
    A body is thrown vertically upwards with a velocity uu, the greatest height hh to which it will rise is:
    Solution
    The body that is thrown vertically upward with velocity uu will have final velocity  v=0v=0 at  the greatest height hh.
    Substituting the given values in the third equation of motion, v2=u2+2asv^2=u^2+2as
    we have 0=u22gh0=u^2-2gh. (taking gg in the upward direction)
    or h=u22gh=\dfrac{u^{2}}{2g}.
  • Question 6
    1 / -0

    The velocity of a bullet is reduced from 100 m/s100 \ m/s to 0 m/s0 \ m/s while travelling through a wooden block of thickness $$10\  cm$$. The retardation, assuming it to be uniform will be:

    Solution
    v2=u2+2asv^2 = u^2 +2as
    $$\text{Substituting}\ v=0, u =100 m/s, s= 10\  cm = 0.1 \ m$$
    a=10020.1×2 a = - \dfrac {100^2}{0.1\times 2 }
    a=5×104 m/s2a=- 5 \times 10^4 \ m/s^2
  • Question 7
    1 / -0
    Two bodies, A(of mass 1kg) and B(of mass 3kg),are dropped from heights of 16 m and 25 m respectively. The ratio of the time taken by them to reach the ground is:-
    Solution
    from equation of motion we know that when a body is dropped 
    h=12gt2h = \dfrac{1}{2}gt^2

    So t1t2=h112=1625=45\dfrac{t_1}{t_2} = \sqrt{\dfrac{h_1}{1_2}} = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}

    Hence (D) is correct answer
  • Question 8
    1 / -0
    A particle is moving in a circular path of radius rr. The magnitude of displacement after half a circle would be :
    Solution
    Displacement is the shortest distance traveled by an object. Here, the object goes around a circle and after half a circle, it would be 2r as it is the diameter of the circle. Thus, the displacement from A to B is 2r for half the circle.

  • Question 9
    1 / -0

    A ball is dropped from height hh and another from 2h2h. The ratio of time taken by the two balls to reach ground is:

    Solution
    Using equation of motion
    s=ut+12gt2s = ut + \dfrac{1}{2}gt^{2}
    Given s=h, u=0s = h, u = 0
    t=2hg\therefore t = \sqrt{\dfrac{2h}{g}}
    So, ratio of time taken: t1t2=h1h2=h2h\dfrac{t_1}{t_2}=\sqrt{\dfrac{h_1}{h_2}}=\sqrt{\dfrac{h}{2h}} = = 12\dfrac{1}{\sqrt 2}
  • Question 10
    1 / -0

    If a body is thrown up with the velocity of 15m/s15 m/s, then maximum height attained by the body is (g=10 m/s2)(g=10 \ m/s^2)

    Solution
    v2=u2+2asv^2=u^2+2as
    Substituting u=15 m/s,v=0 m/s,a=g=10m/s2,\text{Substituting}\ u =15\ m/s, v=0\ m/s, a=g=-10m/s^2,
    152=2×10×s15^2=2\times 10\times s
    s=225/20=11.25ms=225/20=11.25m
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