Laws of Motion

  • Question 477
    CBSEENPH11020263

    A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

    • 44%

    • 50%

    • 56%

    • 62%

    Solution

    C.

    56%

    Question 478
    CBSEENPH11020270

    A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

    • 2g/3

    • g/2

    • 5g/6

    • g

    Solution

    B.

    g/2

    For the mass m,
    mg-T = ma

    As we know, a = Rα    ... (i)
    So, mg-T = mRα
    Torque about centre of pully
    T x R = mR2α ...... (ii)
    From Eqs. (i) and (ii), we get 
    a = g/2
    Hence, the acceleration of the mass of a body fall is g/2.

    Question 479
    CBSEENPH11020301

    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 480
    CBSEENPH11020304

    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.

    NCERT Book Store

    NCERT Sample Papers

    Entrance Exams Preparation