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Motion Test - 9

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Motion Test - 9
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Weekly Quiz Competition
  • Question 1
    1 / -0
    A bird flies for 4 s with a velocity of |t – 2| m/s in a straight line, where t = time in seconds. It covers a distance of
    Solution
  • Question 2
    1 / -0
    Let velocity and acceleration of a body be denoted by v and a, respectively. Which of the following relations holds true?
    Solution
    The body which is thrown upwards has zero velocity at the highest point but still has acceleration due to gravity, so option (1) is true.
  • Question 3
    1 / -0
    If a particle is moving with uniform velocity, its acceleration is
    Solution
    When the change in velocity is zero, it means that there is no force acting on the body and hence, no acceleration.
  • Question 4
    1 / -0
    A point moves with uniform acceleration and v1, v2 and v3 denote the average velocities in three successive intervals of time t1, t2 and t3. Which of the following relations is correct?
    Solution
    Let us consider the three successive intervals based on the given data:



    For first interval,



    Average velocity,



    For second interval,



    The average velocity,



    For third interval:



    The average velocity,



    Hence, equation (1) - equation (2):



    Equation (2) - Equation (3):



    Hence, equation (4)/(5):

  • Question 5
    1 / -0
    Three particles, A, B and C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. If they hit the ground with speeds VA, VB and VC, respectively, which of the following relations holds true?
    Solution
    The sum of kinetic and potential energies will always remain constant. If they have the same height and thrown with the same velocities then, when they hit the ground, their kinetic energy will be same because their potential energies are all zero.
    Therefore, if kinetic energies are all same, we can say that
    VA = VB = VC
  • Question 6
    1 / -0
    A balloon starts rising from the ground with an acceleration of 1.25 m/s2. After 8s, a stone is released from the balloon. The stone will
    Solution
    The velocity of balloon after 8 seconds is,
    V = U + at
    V = 0 + (1.25) 8
    = 10.00 m/s
    Initial velocity of stone is 10 m/s upwards
    Height reached by balloon = (Using S = ut +)
    =
    = 40 m
    Hence, option 4 is correct.
  • Question 7
    1 / -0
    An electron starting from rest has a velocity that increases linearly with time, i.e. v = Kt, where K = 2 m/s2. The distance covered in the first 4 sec will be
    Solution
    v = Kt
    v = 2 x 4 = 8 m/s

  • Question 8
    1 / -0
    The initial velocity of a particle moving along a straight line is 20 m/s and its retardation is 2 m/s2. The distance moved by the particle in the fifth second of its motion is
    Solution
    Initial velocity u = 20 m/s
    a = - 2 m/s2
    Distance moved in 5th second = Distance covered in 5 seconds - Distance covered in 4 seconds
    = S5 - S4
    S = ut -
    S5 = 20(5) - 2(5)2
    S4 = 20(4) -
    S5 - S4 = 20 -
    = 20 - 25 + 16
    = 11 m
  • Question 9
    1 / -0
    A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q with velocities 30 km/hr and 40 km/hr, respectively. The velocity of the car midway between P and Q is
    Solution
    V1 = 30 km/hr V2 = 40 km/hr
    Let acceleration = a km/hr2
    Using, V2 = V1 + at
    Now, 40 = 30 + at (Here, t is time taken to reach Q from P)
    10 = at
    a =
    S = ut +
    = 30 t +
    = 30 t + 5 t
    = 35 t km
    Half of S =
    V2 = u2 + 2 as
    V2 = (30)2 +
    V2 = 900 + 350
    V =
    Answer is 35.35 km/hr which is same as 25 km/hr
  • Question 10
    1 / -0
    A car travelling at a speed of 40 km/hr is brought to a halt in 8 m by applying brakes. If the same car is travelling at 60 km/hr, it can be brought to a halt with the same braking force in
    Solution
    As the braking force is same, acceleration is same. Let distance travelled by the car in the first case be s1 and the distance travelled in the second case be s2.

    Using
    ,

    Final velocity in both the cases is zero, so we have:



  • Question 11
    1 / -0
    Two bodies of different masses say 5 kg and 15 kg are dropped simultaneously from a tower. Which of the following statements is correct?
    Solution
    Both will reach the ground simultaneously because they will start from rest and fall under the same acceleration of 10 m/s2 (g).
  • Question 12
    1 / -0
    If a light and a heavy body are released from the same height in vacuum,
    Solution
    If a light and a heavy body are released from the same height, then they will fall at the same time if they don`t encounter air resistance.
    Hence, option 3 is correct.
  • Question 13
    1 / -0
    The relative velocity of car A with respect to car B is 20√2 m/s due north-east. The velocity of car B is 20 m/s due south. The relative velocity of car C with respect to car A is 20 m/s due north. The speed of car C and the direction (in terms of the angle it makes with the east) is:
    Solution
    Given:
    , and

    Relative velocity of A w.r.t B:





    Relative velocity of C w.r.t A:










    It makes an angle of 45° with east or directed along the northeast.
  • Question 14
    1 / -0
    Two particles are projected simultaneously in the same vertical plane from the same point, but with different speeds and at different angles to the horizontal. The path followed by one as seen by the other is:
    Solution
    A straight line making a constant angle with the horizontal is the correct option.
  • Question 15
    1 / -0
    A swimmer crosses a flowing stream of width b to and fro in time T1. The time taken to cover the same distance up and down the stream is T2. If T3 is the time that the swimmer takes to swim a distance 2b in still water, then
    Solution
    Let's consider that the speed of swimmer in still water is x and the speed of stream flow is y.

    Hence, time taken by the swimmer to cross the stream to and fro along the shortest root,

    Time taken to swim up and down the stream,



    Time taken to travel the distance in still water,

    By multiplying equations (2) and (3),


  • Question 16
    1 / -0
    A 200 m long train is moving northwards at a speed of 20 m/s. A parrot flying towards south with a speed of 5 m/s crosses the train. The time taken by the parrot to cross the train will be
    Solution
    Length of train = 200 m
    Speed of train = 20 m/s
    Parrot's speed = 5 m/s
    Relative speed = 25 m/s
    Time taken by parrot to cross the train = = 8 seconds
  • Question 17
    1 / -0
    A steam boat goes across a lake and comes back (i) on a quiet day when the water is still, and (ii) on a rough day when there is a uniform current so as to help the journey onward and to impede the journey back. If the speed of the launch, on both days, was same, the time required for the complete journey on the rough day, as compared to that on the quiet day, will be
    Solution
    Let the speed of current be = y km/hr and speed of boat in still water = x km/hr
    Let length across the lobe be = d
    Distance travelled along a journey = 2d
    (i) On a quiet day time taken across the lake =T
    Then,
    T = …….. (1)
    (ii) On a rough day time taken across the lake = t
    Then,
    t = + = = …….. (2)
    From (1) and (2),
    and
    and
    Hence, the denominator in (2) is smaller as compared to (1). Therefore, it takes longer time on a rough day.
  • Question 18
    1 / -0
    Which of the following graphs represents motion with uniform velocity?
    Solution
    The slope of s-t graph is velocity and if it`s a straight line inclined with the time axis, then its constant velocity. If it is a straight line parallel to the time axis, then body is at rest.
  • Question 19
    1 / -0
    The displacement -time graph for two particles A and B are straight lines inclined at angles of 30° and 60° with the time axis. The ratio of the speeds vA : vB is
    Solution
    The displacement-time graph for two particles A and B are straight lines inclined at angles 30° and 45°.
    The slope of S - T graph is the velocity.
    Slope of 1st graph= tan 30° = 1/
    Slope of 2nd graph= tan 60° =
    VA : VB =
    = 1 : 3
  • Question 20
    1 / -0
    A motor is making 1200 RPM. Calculate its angular velocity.
    Solution
    Q = 1200 RPM
    ù = = rad/s
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