Laws of Motion

  • Question 517
    CBSEENPH11020550

    A car of mass m is moving on a level circular track of radius R. If straight mu subscript straight s represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by

    • square root of straight mu subscript straight s mRg end root
    • square root of Rg divided by straight mu subscript straight s end root
    • square root of mRg divided by straight mu subscript straight s end root
    • square root of straight mu subscript straight s Rg end root

    Solution

    D.

    square root of straight mu subscript straight s Rg end root

    In this condition, centripetal force is equal to static frictional force between road and tyres, 
    so
    straight mu subscript straight s space mg space equals mv squared over straight R
straight v subscript max space equals space square root of straight mu subscript straight s Rg end root

    Question 518
    CBSEENPH11020553

    Three masses are placed on the x- axis: 300g at origin, 500g at x = 40 cm and 400 g at x = 70 cm. The distance of the centre of mass from the origin is 

    • 40 cm

    • 45 cm

    • 50 cm

    • 30 cm

    Solution

    A.

    40 cm

    straight x subscript cm space equals fraction numerator straight m subscript 1 straight x subscript 1 space plus straight m subscript 2 straight x subscript 2 space plus straight m subscript 3 straight x subscript 3 over denominator straight m subscript 1 plus straight m subscript 2 plus straight m subscript 3 end fraction
space equals space fraction numerator 300 space left parenthesis 0 right parenthesis space plus 500 space left parenthesis 40 right parenthesis space plus 400 space left parenthesis 70 right parenthesis over denominator 300 plus 500 plus 400 end fraction
equals space fraction numerator 20000 plus 28000 over denominator 1200 end fraction
space equals 48000 over 1200 space equals 40 space cm
    Question 519
    CBSEENPH11020565

    A system consists of three masses m1, m2 and m3 connected by a string passing over a pulley P. The mass m1 hangs freely and m2 and m3 are on a rough horizontal table (the coefficient of friction= straight mu). the pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is,

    • fraction numerator straight g space left parenthesis 1 minus gμ right parenthesis over denominator straight g end fraction
    • fraction numerator 2 gμ over denominator 3 end fraction
    • fraction numerator straight g space left parenthesis 1 minus 2 straight mu right parenthesis over denominator 3 end fraction
    • fraction numerator straight g left parenthesis 1 minus 2 straight mu right parenthesis over denominator 2 end fraction

    Solution

    C.

    fraction numerator straight g space left parenthesis 1 minus 2 straight mu right parenthesis over denominator 3 end fraction

    First of all consider the forces on the blocks,

    For the 1st block,
    mg - T1 = m x a        ... (i)
    Let us consider second and third block as a system,
    Then,
    T12 μmg = 2m x a
    mg (1 space minus 2 straight mu) = 3m x a
    straight a space equals space 2 over 3 left parenthesis 1 minus 2 straight mu right parenthesis 

    Question 520
    CBSEENPH11020573

    The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle straight theta without slipping and slipping down the incline without rolling is,

    • 5:7

    • 2:3

    • 2:5

    • 7:5

    Solution

    A.

    5:7

    A solid sphere rolling without slipping down an inclined plane is,

    We have in this case,
    a subscript 1 space equals space fraction numerator straight g space sin space straight theta over denominator 1 plus space begin display style straight k squared over straight R squared end style end fraction space equals space fraction numerator g space sin space theta over denominator 1 space plus open parentheses space begin display style bevelled 2 over 5 end style close parentheses end fraction
F o r space s o l i d space s p h e r e comma space K squared space equals space 2 over 5 R squared
space space space space space space equals space fraction numerator g space sin space theta over denominator begin display style bevelled 7 over 5 end style end fraction
rightwards double arrow space a subscript 1 space equals space 5 over 7 space g space sin space theta
    For a sphere slipping down an inclined plane,
    straight a subscript 2 space equals space straight g space sin space straight theta

That space is comma

space space space space space space straight a subscript 1 over straight a subscript 2 space equals space fraction numerator bevelled 5 over 7 space straight g space sin space straight theta over denominator straight g space sin space straight theta end fraction
rightwards double arrow space straight a subscript 1 over straight a subscript 2 space equals space 5 over 7

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