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Physics Test - 13

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Physics Test - 13
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  • Question 1
    1 / -0

    A box of mass m kg is placed on the rear side of an open truck accelerating at 4 ms-2. The coefficient of friction between the box and the surface below it is 0.4. Then the net acceleration of the box with respect to the truck is zero. The value of m is [Given g = 10 m s-2]

    Solution

    Pseudo force on the block = m × 4 N (backward)

    Force of friction = 0.4 × m × 10 N (forward)

    Equating , m×4 = 0. 4 × m × 10 = 4m 

    Clearly the equation holds good for all values of m

    KEY CONCEPTS

    Friction#
    (i) Friction is tangential component of net contact force between two solid bodies in contact.

    (ii) This forces starts acting between them when there is relative motion ( or even the tendency of relative motion ) between them.

    (iii) Like other forces this force also makes a pair of equal and opposite forces acting on two different bodies.

    (iv) Direction of frictional force on a body is opposite to the relative motion ( or its tendency) of this body with respect to the other body.

    (v) Normal reaction N ( component of net contact force in perpendicular direction) plays very important role while deciding limiting value of static friction μsN or constant value of kinetic friction μkN.

    (vi) As long as forces are acting on a body parallel to the plane over which body is kept, normal reaction will be mg cosθ (if plane is inclined). If forces are acting at some angle with plane, normal reaction is greater than mg (or mg cos θ ) or less than this, depending upon whether the external forces are of pushing nature or pulling nature.

    (vii) static friction is of self adjusting nature with its value varying between 0 and μsN. This force acts when there is only tendency of relative motion but relative motion is not actually taking place. On the other hand, kinetic friction is constant of value μkN. This force acts when relative motion actually takes place.

    (viii) Coefficient of kinetic friction is always less than the coefficient of static friction (μs).

    (ix) If μs and μk are not separately given. But only one value of  μ is given. Then in this case limiting value of static and constant value of kinetic friction are same and equal to μN.

    When the mass is not moving, the object experiences static friction. The friction increases as the applied force increases until the block moves. After the block moves, it experiences kinetic friction, which is less than the maximum static friction.

    Static friction is friction between two or more solid objects that are not moving relative to each other. For example, static friction can prevent an object from sliding down a sloped surface. The coefficient of limiting static friction, typically denoted as μs, is usually higher than the coefficient of kinetic friction.

    The static friction force must be overcome by an applied force before an object can move. The maximum possible friction force between two surfaces before sliding begins is the product of the coefficient of static friction and the normal force: FmaxsFn. When there is no sliding occurring, the friction force can have any value from zero up to Fmax. Any force smaller than Fmax attempting to slide one surface over the other is opposed by a frictional force of equal magnitude and opposite direction. Any force larger than Fmax overcomes the force of static friction and causes sliding to occur. The instant sliding occurs, static friction is no longer applicable the friction between the two surfaces is then called kinetic friction.An example of static friction is the force that prevents a car wheel from slipping as it rolls on the ground. Even though the wheel is in motion, the patch of the tire in contact with the ground is stationary relative to the ground, so it is static rather than kinetic friction.

  • Question 2
    1 / -0

    The current from A to B is increasing in magnitude. What is the direction of induced current, if any, in the loop shown in figure.

    Solution

    Magnetic flux through the loop is upward and it is increasing due to increasing current along AB. Current induced in the loop should have magnetic flux in the downward direction so as to oppose the increase in flux. Therefore, current induced in the loop is clockwise.

    KEY CONCEPTS



  • Question 3
    1 / -0

    The alternating current I in an inductance coil varies with time according to graph given in figure.

    Which one of the following graphs, gives the variation of voltage with time?

    Solution

    As we know, that alternating volts leads the alternating current by a phase angle of 90º in an inductance coil.

    Therefore the answer is (C).

    KEY CONCEPTS

    Circular Coil

    Consider a circular coil of radius r, carrying a current I. Consider a point P, which is at a distance x from the centre of the coil. We can consider that the loop is made up of a large number of short elements, generating small magnetic fields. So the total field at P will be the sum of the contributions from all these elements. At the centre of the coil, the field will be uniform. As the location of the point increases from the centre of the coil, the field decreases.

    By Biot- Savart’s law, the field dB due to a small element dl of the circle, centered at A is given by,

    The horizontal component of the earth’s magnetic field varies greatly over the surface of the earth. For the purpose of this simulation, we will assume its magnitude to be B0 = 3.5x10-5 T. The variation of magnetic field along the axis of a circular coil is shown here.

  • Question 4
    1 / -0

    In a Rutherford scattering  experiment, when a projectile of charge z1 and mass M1 approaches a target nucleus of charge z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is

    Solution

    Energy of the projectile is equal to the potential energy at closest

    KEY CONCEPTS

    Rutherford's Nuclear Model of Atom

    Rutherford performed an alpha particle (He2+) scattering experiment on a thin gold foil and presented that :
    (i) most part of atom is empty.
    (ii) every atom possesses a highly dense, positively charged centre called "Nucleus".
    (iii) entire mass of atom is concentrated inside the nucleus.
    (iv) later Rutherford model was abandoned due to its failure to comply with classical theory of electromagnetic radiation. This theory also failed to explain the line spectrum of H-atom.

    Rutherford Experiment discovery of nucleus
    Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment in which he demonstrated that the atom has a tiny, heavy nucleus. Rutherford designed an experiment to use the alpha particles emitted by a radioactive element as probes to the unseen world of atomic structure.

    Rutherford presented his own physical model for subatomic structure, as an interpretation for the unexpected experimental results. In it, the atom is made up of a central charge (this is the modern atomic nucleus, though Rutherford did not use the term "nucleus" in his paper) surrounded by a cloud of (presumably) orbiting electrons. In this May 1911 paper, Rutherford only commits himself to a small central region of very high positive or negative charge in the atom.
    Rutherford's Theory of Alpha ScatteringRutherford Scattering

    Alpha particles from a radioactive source were allowed to strike a thin gold foil. Alpha particles produce a tiny, but visible flash of light when they strike a fluorescent screen. Surprisingly, alpha particles were found at large deflection angles and some were even found to be back-scattered.

    This experiment showed that the positive matter in atoms was concentrated in an incredibly small volume and gave birth to the idea of the nuclear atom. In so doing, it represented one of the great turning points in our understanding of nature.

    If the gold foil were 1 micrometer thick, then using the diameter of the gold atom from the periodic table suggests that the foil is about 2800 atoms thick.
     
    Rutherford's Nuclear Model of The Atom
    After α-particles scattering experiment, following conclusions were made by Rutherford as regard as atomic structure :

    (1)   Most of the mass (at least 99.95%) and all of the charge of an atom concentrated in a very small region is called atomic nucleus.

    (2)   Nucleus is positively charged and it's size is of the order of 10–15 m » 1 Fermi. The nucleus occupies only about 10–12 of the total volume of the atom or less.

    (3)   In an atom there is maximum empty space and the electrons revolve around the nucleus in the same way as the planets revolve around the sun. 

    Limitation of Rutherford Atomic Model1. An electron revolving around the nucleus gets accelerated towards the nucleus. According to the electromagnetic theory, an accelerating charged particle must emit radiation, and lose energy. Because of this loss of energy, the electron would slow down, and will not be able to withstand the attraction of the nucleus. As a result, the electron should follow a spiral path, and ultimately fall into nucleus. If it happens then the atom should collapse in about 10-8 second. But, this does not happen: atoms are stable. This indicates that there is something wrong in the Rutherford’ mass nuclear model of atom.

    2. The Rutherford’s model of atom does not say anything about the arrangement of electrons in an atom.

    3. The Rutherford model could not explain spectrum of atoms (of-H-atom which was known at that line)
    Nuclear size(approx) in Rutherford experiment

    The nucleus is tiny – about 10–15 m across. Compare this with the size of an atom – about 10–10 m across – and you can see that most of the atom is empty space. The first estimates for the size of the nucleus came from the Rutherford experiment. 

    Directly proportional to z1z2

  • Question 5
    1 / -0

    A block of mass m & charge q is released on a long smooth inclined plane. Magnetic field B is constant, uniform, horizontal and out of the plane of paper as shown. Find the time from start when block loose contact with the surface.

    Solution

    Block will loose contact with surface when force due to magnetic field will become equal to mg cosθ

    Helical Path - When the charged particle is moving at an angle to the field (other than 0, 90 or 180). Particle describes a path called helix.

    Force on a moving charge in a magnetic field

    From Lorentz law the force on the particle is given in terms of electric field E, magnetic field B and velocity of the moving particle v. Since E = 0

  • Question 6
    1 / -0

    An electrically charged particle enters into a uniform magnetic induction field in a direction perpendicular to the field with a velocity v. Then, it travels

    Solution

    By motion of charged particles in uniform magnetic field

    Helical Path - When the charged particle is moving at an angle to the field (other than 0, 90 or 180). Particle describes a path called helix.

  • Question 7
    1 / -0

    In series LCR, RMS voltage drop across them are 40 V, 40 V, 40 V. The peak value of applied ac voltage is

    Solution

    KEY CONCEPTS

    RMS value of Alternating current

    Root mean square (r.m.s.) value : Root of mean of square of voltage or current in an ac circuit for one complete cycle is called r.m.s. value. It is denoted by Vrms or irms

    Consider a series RLC circuit (one that has a resistor, an inductor and a capacitor) with a constant driving electro-motive force (emf) E. The current equation for the circuit is

  • Question 8
    1 / -0

    Which of the following is incorrect regarding Rutherford's atomic model?

    Solution

    Nucleus contains almost all the mass of atom.

    KEY CONCEPTS

    Limitation of Rutherford Atomic Model

    1. An electron revolving around the nucleus gets accelerated towards the nucleus. According to the electromagnetic theory, an accelerating charged particle must emit radiation, and lose energy. Because of this loss of energy, the electron would slow down, and will not be able to withstand the attraction of the nucleus. As a result, the electron should follow a spiral path, and ultimately fall into nucleus. If it happens then the atom should collapse in about 10-8 second. But, this does not happen: atoms are stable. This indicates that there is something wrong in the Rutherford’ mass nuclear model of atom.

    2. The Rutherford’s model of atom does not say anything about the arrangement of electrons in an atom.

    3. The Rutherford model could not explain spectrum of atoms (of-H-atom which was known at that line)

  • Question 9
    1 / -0

    What is the largest charge a metal ball of 1 mm radius can hold? Dielectric strength of air is 3 × 106 Vm-1.

    Solution

    Dielectric strength means the maximum electric field which a medium can bear. Here, if field becomes more than this, then the charge will start leaking from the metal ball. So,

    KEY CONCEPTS

    Volume Charge Distribution

    Volume charge density: Charge distributes throughout the volume of the body e.g. charge on a dielectric sphere etc.

  • Question 10
    1 / -0

    A particle is given an initial speed u inside a smooth spherical shell of radius R = 1 m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is-

    Solution

    KEY CONCEPTS

    Dynamics of Uniform Circular Motion

    For an object to be in uniform circular motion, there must be a net force acting we already know the acceleration, so can immediately write the force:

    The motion of a particle along an arbitrary curved path lying in the xy plane. If the velocity vector v (always tangent to the path) changes in direction and magnitude, the component vectors of the acceleration a are a tangential component at and a radial component ar.

    Now let us consider a particle moving along a curved path where the velocity changes both in direction and in magnitude, as shown in figure. As is always the case, the velocity vector is tangent to the path, but now the direction of the acceleration vector a changes from point to point. This vector can be resolved into two component vectors; a radial component vector ar and a tangential component vector at. Thus, a can be written as the vector sum of these component vectors;

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