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Laws of Motion

    • Question 485
    CBSEENPH11020301
    Wired Faculty App

    A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 x10–2N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is

    • 0.0125 Nm-1

    • 0.1 Nm-1

    • 0.05 Nm-1

    • 0.025 Nm-1

    Solution

    D.

    0.025 Nm-1


    The force of surface tension acting on the slider balances the force due to the weight.

    ⇒F = 2Tl = w
    ⇒2T(0.3) = 1.5 x 10–2
    ⇒T = 2.5 x 10–2 N/m

    Question 486
    CBSEENPH11020304
    Wired Faculty App

    A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge (θ – θ0) and t is

    Solution

    A.

    According to Newton's law of cooling.
    dθ over dt space proportional to left parenthesis straight theta minus straight theta subscript straight o right parenthesis rightwards double arrow fraction numerator begin display style dθ end style over denominator begin display style dt end style end fraction space equals space minus straight k left parenthesis straight theta minus straight theta subscript straight o right parenthesis integral fraction numerator begin display style dθ end style over denominator begin display style straight theta minus straight theta subscript straight o end style end fraction space equals space integral negative kdt space rightwards double arrow space In space left parenthesis straight theta minus straight theta subscript straight o right parenthesis space equals space minus kt plus straight c
    Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope.

    Question 487
    CBSEENPH11020311
    Wired Faculty App

    A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

    • g

    • 2/3g

    • g/3

    • 3/2g

    Solution

    B.

    2/3g


    mg - T = ma
    TR space equals space fraction numerator mR squared straight alpha over denominator 2 end fraction straight T space equals space mRα over 2 space equals space ma over 2 mg space minus ma over 2 space equals space ma fraction numerator 3 space ma over denominator 2 end fraction space equals space mg straight a space equals space fraction numerator 2 straight g over denominator 3 end fraction
    Question 488
    CBSEENPH11020312
    Wired Faculty App

    Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly. (Surface tension of soap solution = 0.03 Nm-1)

    • 4π mJ 

    • 0.2π mJ

    • 2π mJ

    • 0.4π mJ

    Solution

    D.

    0.4π mJ

    Work done = Change in surface energy
    ⇒ W = 2T x 4π (R22-R12)
     = 2 x 0.03 x 4π [ (5)2-(3)2] x 10-4 J 
     = 0.4 π mJ

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