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  • Question 1
    1 / -0
    Water of mass m2 = 1 kg is contained in a copper calorimeter of mass m1 = 1 kg. Their common temperature t is 10°C. Now, a piece of ice of mass m3 = 2 kg and temperature -11°C is dropped into the calorimeter. Neglecting any heat loss, the final temperature of the system is [Specific heat of copper = 0.1 Kcal/ kg°C, Specific heat of water = 1 Kcal/kg°C, Specific heat of ice = 0.5 Kcal/kg°C, Latent heat of fusion of ice = 78.7 Kcal/kg]
    Solution
    Available heat from: calorimeter + water, as temperature changes from 10°C to 0°C (water)
    = m1C110 + m2C210
    = 1 × 1 × 10 + 1 × 0.1 × 10
    = 11 kcal

    Required heat for ice as its temperature changes from -11°C to 0°C (ice)
    = m3C3 × 11
    = 2 × 0.5 × 11
    = 11 kcal

    Now, available heat to reach 0oC temperature = Required heat to reach 0oC temperature. So, the final temperature is 0oC.
  • Question 2
    1 / -0
    A particle is moving along the path given by (where C is a positive constant). The relation between the acceleration (a) and the velocity (v) of the particle, at t = 5 sec, is
    Solution
    The path of the particle is given by y. As we know that velocity is given by the derivative of path, so



    Acceleration is given by double derivative of y.



    Now, at t = 5 sec



    So at t = 5 sec, the relation between a and v is given by:

    a = v
  • Question 3
    1 / -0
    'n' number of water droplets, each of radius r, coalesce to form a single drop of radius R. The rise in temperature d is (T is the surface tension, J is the energy equivalent)
    Solution
    Let 'n' drops coalesce to from a big drop. Then,
    Work done = Decrease in surface area Surface tension =

    And we also know that work done =

    Now, from the above equations and putting density as 1 for water, we get

  • Question 4
    1 / -0
    Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle with AB. The moment of inertia of the plate about the axis CD is then equal to
    Solution
    According to the perpendicular axis theorem, .



    Since the plate is quite symmetrical about x and y, Ix, = Iy, .

    Similarly, Ix = Iy = Iz/2 = I (Moment of inertia along AB)

    So, M. I. about any axis on the surface of square plate is I.

     The required moment of inertia = I
  • Question 5
    1 / -0
    Which of the following graphs represents current-voltage characteristics for a series combination?
    Solution
    In element A, the resistance remains constant up to the potential drop of 10 V. Further increase in the voltage does not increase this current (which is constant at 1 A). This means that the ratio V/RA = constant, and this resistance RA increases linearly with voltage.
    In element B, the resistance decreases gradually up to 15 V and afterwards, the resistance RB increases linearly with voltage.
    When both A and B are in series, the current in the circuit will increase non-linearly up to 1 A, when the total voltage drop across A and B becomes 10 V + 15 V = 25 V.
    Further increase in this voltage does not bring about any change in the current as shown in solution (3). The voltage drop across A will go on increasing while that across B will remain fixed at 15 V.
  • Question 6
    1 / -0
    An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is



    (A)
    (B)
    (C)
    (D)
    Solution
    Snell's law:

    , where i = 45°

    and tan r = 1/2



  • Question 7
    1 / -0
    A sound source S and an observer O are placed as shown in the figure. Sound can reach the observer only by getting reflected by a reflector surface M, which is moving towards S along PS with a speed of 300 m/s. The angle OMS at any point is referred to as . What frequency will O perceive when = 127°? (Assume that the reflector is always able to reflect the sound towards the observer. Speed of sound = 400 m/s. Natural frequency of sound source = 400 Hz)

    Solution


    If one notices carefully from the figure, it's clear that when is 127o. O would perceive a sound which was reflected when was 90°.

    So, f¢ = f = 400 Hz
  • Question 8
    1 / -0
    In the assembly shown in the figure below, both the pulleys are massless and frictionless on their axes. The positions of the strings are adjusted in such a way that, in equilibrium position, k1 and k2 are unstretched. Now, m is given a very slight displacement in vertical direction, what will be its time period for small vertical oscillations. k1 = k2 = k3 = 62 N/m. The string that connects the left ends of k1 and k2 is inextensible, and is wrapped around pulley P1 to ensure no slipping. String that connects the mass and k3 passes over both the pulleys and does not slip over them.

    Solution
    During motion of system, k3 and either one of k1 and k2 will be associated with the motion.

    Time period =

    =

    =
  • Question 9
    1 / -0
    A particle of mass M at rest decays into two particles of masses m1 and m2 having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles l1 and l2 is
    Solution
    From the law of momentum conservation, P1 = P2 (in opposite directions).

    Now, de-Broglie wavelength is given byl = ; where h = Planck`s constant.

    Since momentum (P) of both the particles is equal, therefore l1 = l2 or l1/l2 = 1.
  • Question 10
    1 / -0
    From point A located on a highway, one has to get by car as soon as possible to point B, located in the field at a distance from the highway. It is known that the car moves in the field h times slower than on the highway. The distance from the point D one must turn off the highway i.e. distance CD is:

    Solution
    Suppose distance AD = d and CD = x and speed of car is v.






    For minimum time and on solving this equation, we will get.
  • Question 11
    1 / -0
    For a particle executing SHM, the displacement x given by x = A cos Wt. Identify the graph which represents the variation of potential energy (PE) as a function of time t and displacement x.

    Solution
    From energy conservation, P. E. + K. E. = Constant

    From the given graphs, we find that in graph III, potential energy is minimum (in this case, zero) at mean position (x = 0) and maximum at extreme positions (x = A).
    Similarly from graph I, we find that at time t = 0, x = A. Hence, PE should be maximum.
    Therefore, graph I is correct. Further in graph III, PE is minimum at x = 0. Hence, this is also correct.
  • Question 12
    1 / -0
    The half-life of 215At is 100 ms. The time taken for the radioactivity of a sample of 215At to decay to of its initial value is
    Solution
    Decay is given by:

    ........(1)

    and half-life, = 100 ms (Given) .......(2)

    Here, N = Activity of radioactive substance after time t = (given)

    Substituting in Eq. (1), we get t = .

    Now, from equation (2):

    t =
  • Question 13
    1 / -0
    If the area of each plate is S and the successive separations are d, 2d and 3d, then the equivalent capacitance across A and B is

    Solution
    Plates with distance 2d are not a capacitor because these plates are connected with a wire.

    A and B, two capacitors (with separation d and 3d), are in series.

    Hence,



    C =
  • Question 14
    1 / -0
    A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, R is equal to
    Solution
    Velocity of efflux at a depth h is given by

    Volumes of water flowing out per second from both the holes are equal.



    or
  • Question 15
    1 / -0
    A cube of mass m and density D is suspended from a point P by a spring of stiffness k. The system is kept inside a beaker filled with a liquid of density d. The elongation in the spring, assuming D > d, is

    Solution
    The cube is in equilibrium under the following three forces:

    (A) Spring force kx, where x = elongation of the spring
    (B) Gravitational force w, weight of the cube = mg
    (C) Buoyant force Fb (or upward thrust) imparted by the liquid on the cube, given as Fb = Vdg
    where
    V = Volume of the immersed portion of the cube.
    For complete immersion, V = Volume of the cube



    For equilibrium of the cube, kx + Fb = mg

    ; where V = (m/D)

  • Question 16
    1 / -0
    In the shown circuit, all three capacitors are identical and have capacitance C each. Each resistor has a resistance R. An ideal cell of emf V is connected as shown. Then, the magnitude of potential difference across capacitor C3 in steady state is

    Solution
    No current passes through capacitors in steady state. Assume potential at point '4' to be zero.

    Then, points '1' and '2' are at same potential .
    Hence, C1 and C2 can be taken in parallel.

    The potential at point 3 is .
    Equivalent circuit of all three capacitors is shown below:
    Hence, potential difference across capacitor C3
    = =

  • Question 17
    1 / -0
    In a certain organ pipe, three successive resonance frequencies are observed at 425 Hz, 595 Hz and 765 Hz, respectively. If the speed of sound in air is 340 m/s, then the length of the pipe is
    Solution
    425 : 595 : 765 5 : 7 : 9

    Only odd harmonics are present. Hence, it is a closed organ pipe.



    L = 1 m

    = 2 m
  • Question 18
    1 / -0
    The length of a metal wire is when the tension in it is T1, and when the tension is T2. The natural length of the wire is
    Solution
    and



    ................(1)

    ...............(2)



    L =
  • Question 19
    1 / -0


    ABC is an equilateral triangle. From point A, a projectile is thrown such that it just grazes over B. Given, length of AB = . The angle of projection `` is
    Solution


    sin 60 =

    H =





    Range =





  • Question 20
    1 / -0
    A simple pendulum of length and with a bob of mass M is at equilibrium under action of constant forces F1 and F2. The string makes an angle `` with the vertical and is in equilibrium. The angular frequency of oscillations of the pendulum about equilibrium position is

    Solution


    where T0 is the tension in the string in equilibrium.


    .....(1)

    ..................(2)

    From eq.(1):









  • Question 21
    1 / -0
    A weight of mass 1 kg, attached to a spring with force constant 20 N/m, is able to oscillate on a horizontal steel rod. The initial displacement from the position of equilibrium is 30 cm. If one swing is considered as the movement from maximum displacement to the equilibrium position (or) back, find how many swings will the weight make before stopping completely. Coefficient of friction between the weight and the rod is The rod is long enough. (Assume there is no surface below the weight and g = 10 m/s2)

    Solution
    Let initial amplitude be A0, next amplitude be A1.

    Then,

    Similarly,

    When An = 0, n =

    The block completes two swings for all the amplitudes, except for the initial one.

    The number of swings is N = 2n - 1 =
  • Question 22
    1 / -0
    A certain mass of an ideal diatomic gas contained in a closed vessel is heated. It is observed that the temperature remains constant, however, half the amount of gas gets dissociated. The ratio of the heat supplied to the gas to the initial internal energy of the gas will be
    Solution
    Initially,

    Volume is constant, so .

    Internal energy =







  • Question 23
    1 / -0
    In the figure shown below, find out the value of . [Assume string to be tight]

    Solution
    Total length of the string is always constant.






    On differentiating both sides with respect to time,
  • Question 24
    1 / -0
    The ratio of translational and rotational kinetic energies at 100 K temperature is 3 : 2. Then, the internal energy of one mole of gas at that temperature is [R = 8.3 J/mol-K]
    Solution
    According to law of equipartition of energy, energy is equally distributed among its degrees of freedom.
    Let translational and rotational degrees of freedom be f1 and f2, respectively.

    Hence, the ratio of translational degree to rotational degree of freedom is 3 : 2.

    Since translational degree of freedom is 3, the rotational degree of freedom must be 2.
    Internal energy (U)
  • Question 25
    1 / -0
    Wires AB and PQ lie in the same vertical plane and AB remains in equilibrium due to magnetic repulsive force. Then, the time period of oscillation of the rod AB, if it is displaced by a small distance in vertical plane, will be (where PQ is fixed on a horizontal surface)

    Solution
    mg = …(i)
    By displacing a small distance x in vertical plane,




    From here,

  • Question 26
    1 / -0
    Particle A starts from rest at t = 0 from x = 0 with constant acceleration to reach x = 1 m at t = 1 s. Particle B starts with uniform velocity at t = 0 from x = 1 to reach x = 2m at t = 1 s. What will be the distance covered by particle B in the frame of reference attached to particle A from t = 0 to t = 1 s. (upto two decimal places in meter)
    Solution
    For particle A



    now using acceleration value



    For particle B


  • Question 27
    1 / -0
    Two concentric shells, A and B, having radii R and 2R carry changes qA and qB and potential 2V and , respectively. Now, shell B is earthed and as a result, final charges on shell A and B are , respectively. Then,

    Solution
    Before earthing:



    After earthing:


    Potential difference between A and B = .
  • Question 28
    1 / -0
    A plank of mass 4m and length l is kept on a frictionless surface. A small block of mass m is kept on upper surface of the plank at point A, as shown in the figure. Coefficient of friction between the block and the plank is . What minimum velocity must be given to the block in horizontal direction so that it just comes to the other end i.e. at point B of the plank?

    Solution
    Let v be the minimum velocity given to the upper block.

    The (block + plank) system has tendency to move with common velocity.

    For minimum value of v so that block reaches point B, this common velocity must be achieved at point B.

    Let that velocity be v.

    From momentum conservation:





    From work energy theorem,

  • Question 29
    1 / -0
    A rod of length L, with sides fully insulated, is made of a material whose thermal conductivity varies with temperature as K = , where is a constant. The ends of the rod are kept at temperatures T1 and T2. The temperature T at x, where x is the distance from the end whose temperature is T1, is
    Solution


    At x = L, T = T2



  • Question 30
    1 / -0
    A small water drop falls in air with terminal velocity. If R1 and R2 are the radii of curvature at the upper and lower points of the surface of drop, respectively and h is the distance between the two points, then (R1 - R2) is (approximately) [ = Density of water, T = Surface tension)

    Solution
    Pressure just inside the upper surface: P1 = Po + 2T/R1

    Pressure just above the lower surface: P2 = Po + 2T/R2

    Net force on drop = Zero = P1 + gh = P2 P2 - P1 = hpg



    Since the water drop is small,
    R1 - R2 =
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