Laws of Motion

  • Question 529
    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 530
    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

    Question 531
    CBSEENPH11020674

    A mass m moving horizontally (along the x -axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y -axis) with velocity 2v. The final velocity of the combination

    • 1 half straight v space bold i with bold hat on top space plus 3 over 2 straight v space bold j with bold hat on top
    • 1 third straight v space bold i with bold hat on top space plus 2 over 3 straight v space bold j with bold hat on top
    • 2 over 3 straight v space bold i with bold hat on top space plus 1 third straight v space bold j with bold hat on top
    • 3 over 2 straight v space bold i with bold hat on top space plus 1 fourth straight v space bold j with bold hat on top

    Solution

    A.

    1 half straight v space bold i with bold hat on top space plus 3 over 2 straight v space bold j with bold hat on top

    From the law of conservation of linear momentum

    mv + (3m)(2v) = (4m)v'
    mv + 6mv j = 4 mv'

    straight v apostrophe space equals space 1 fourth space straight v space bold i space plus space 6 over 4 space straight v space bold j

straight v apostrophe space equals space 1 fourth space straight v space bold i bold space plus 3 over 2 space straight v space bold j

    Question 532
    CBSEENPH11020679

    A block of mass m is in contact with the cart C as shown in the figure


    The coefficient of static friction between the block and the cart is μ. The acceleration α of the cart that will prevent the block from falling satisfies

    • straight alpha space greater than mg over straight mu
    • straight alpha space greater than space straight g over μm
    • straight alpha space greater or equal than space straight g over straight mu
    • straight alpha space less than space straight g over straight mu

    Solution

    C.

    straight alpha space greater or equal than space straight g over straight mu

    When a cart moves with some acceleration toward right then a pseudo force (mα) acts on block towards left. The force (mα) is action force by a block on the cart. Now, block will remain static w.r.t cart if frictional force 
    μR space greater or equal than space mg

rightwards double arrow space straight mu space mα space greater or equal than space mg space space space left square bracket space as space straight R space equals space mα right square bracket

rightwards double arrow space straight alpha space greater or equal than space straight g over straight mu

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