Laws of Motion

  • Question 533
    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 534
    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
    Question 535
    CBSEENPH11020663

    Two stone of masses m and 2m are whirled in horizontal circles, the heavier one in a radius r/2 and the lighter one in radius r. The tangential speed of lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is

    • 2

    • 3

    • 4

    • 1

    Solution

    A.

    2

    Given, that two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a radius r/2 and lighter one in radius r as shown in a figure.


    As lighter stone is n times that of the value of heavier stone when they experience same centripetal forces, we get
    (Fc)heavier = (Fc)lighter
    fraction numerator 2 straight m left parenthesis straight v right parenthesis squared over denominator straight r divided by 2 end fraction space equals space fraction numerator straight m left parenthesis nv right parenthesis squared over denominator 2 end fraction

straight n squared space equals space 4

straight n space equals space 2

    Question 536
    CBSEENPH11020667

    Two identical piano wires kept under the same tension T have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beats/s when both the wires oscillate together would be

    • 0.02

    • 0.03

    • 0.04

    • 0.01

    Solution

    A.

    0.02

    According to law of tension, the frequency of the string varies directly as the square root of its tension

    straight n space proportional to space square root of straight T

fraction numerator increment straight n over denominator straight n end fraction space equals space 1 half. fraction numerator increment straight T over denominator straight T end fraction
or
fraction numerator increment straight T over denominator straight T end fraction space equals space 2 space straight x space fraction numerator increment straight n over denominator straight n end fraction
fraction numerator increment straight T over denominator straight T end fraction space equals space 2 space straight x space 6 over 600

equals space 0.02

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