Laws of Motion

  • Question 525
    CBSEENPH11020625

    A person of mass 62 kg is inside of a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 m/s2. If g = 10 m/s2 ,the tension in the supporting cable is

    • 9680 N

    • 11000 N

    • 1200 N 

    • 8600 N

    Solution

    B.

    11000 N


    Total mass = Mass of lift + Mass of person
    = 940 +60 = 1000 kg
    T - mg = ma
    Hence, T - 1000 x 10 = 1000 x 1
     T = 11000 N
    Question 526
    CBSEENPH11020631

    A planet moving along an elliptical orbit is closer to the sun at a distance r1 and farthest away at a distance of r2. if v1 and v2 are the linear velocities at these points respectively, then the ratio v1/v2 is

    • r2/r1

    • (r2/r1)2

    • r1 /r2

    • (r1/r2)2

    Solution

    A.

    r2/r1

    From the law of conservation of angular momentum
    mr1v1 = mr2v2
    r1v1 = r2v2
    v1/v2 = r2/r1

    Question 527
    CBSEENPH11020639

    Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m is

    • 21 J

    • 26 J 

    • 13 J

    • 18 J

    Solution

    C.

    13 J

    Work done = Area under (F - x) graph
    = 2 x (7-3) + space equals space 2 space straight x space left parenthesis 7 minus 3 right parenthesis space plus 1 half space straight x space 2 space straight x space left parenthesis 12 minus 7 right parenthesis

equals space 8 space plus 1 half space straight x space 10

equals space 8 space plus space 5 space equals space 13 space straight J

    Question 528
    CBSEENPH11020640

    Uniform magnetic field acting along AB. If the magnetic force on the arm BC Is F, the force on the arm AC is

    • -F
    • F
    • square root of 2 bold F
    • negative square root of 2 space bold F

    Solution

    A.

    -F
    FAB = 0
    FAB + FBC + FCA = 0
    FBC + FCA = 0
    FBC + FCA = 0
    FCA = - FBC = - F

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