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Magnetism and Matter Test 27

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Magnetism and Matter Test 27
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Permanent magnets are the substances having the property of
    Solution
    Permanent magnets are those substances that retain their ferromagnetic property for a long period of time at room temperature.
  • Question 2
    1 / -0
    Core of electromagnets are made of ferromagnetic material which have
    Solution
    Core of electromagnets are made of soft iron that is a ferromagnetic material with high permeability and low retentivity.
  • Question 3
    1 / -0
    The hysteresis cycle for the material of a transformer core is:
    Solution
    Transformer core is soft iron material which has small coercivity and large retentivity. Therefore its hysteresis loop is tall and narrow.
  • Question 4
    1 / -0
    The temperature of transition from ferromagnetic property to paramagnetic property is called as:
    Solution
    Curie Temperature$$(T_{c})$$
  • Question 5
    1 / -0
    Some equipotential surfaces of the magnetic scalar potential are shown in figure.Magnetic field at a point in the region is

    Solution
    As is clear from figure

    $$\triangle V = (0.2 - 0.1) 10^{-4} Tm$$
    $$\triangle x = x sin 30^0 = 0.1 \times \dfrac{1}{2}$$
    $$\therefore |B| = \dfrac{\triangle V}{\triangle x} = \dfrac{0.1 \times 10^{-4}}{0.1 /2} = 2 \times 10^{-4}T$$
  • Question 6
    1 / -0
    Assertion : When diamagnetic material is placed in a non-uniform magnetic field, it tends to move from stronger to the weaker part of the magnetic field.
    Reason : Diamagnetic materials possess strong magnetism.
    Solution
    Electrons is an atom orbiting around nucleus possess orbital angular momentum. These orbiting electrons are equivalent to current-carrying loop and thus possess orbital magnetic moment. Diamagnetic substances are the ones in which resultant magnetic moment in an atom is zero. When magnetic field is applied, those electrons having orbital magnetic moment in the same direction slow down and those in the opposite direction speed up. Thus, the substance develops a net magnetic moment in direction opposite to that of the applied field.
  • Question 7
    1 / -0
    Assertion : The product of magnetic susceptibility and absolute temperature for a paramagnetic substance is constant.
    Reason : Susceptibility is positive but very small for paramagnetic substance.
    Solution
    The susceptibility of a paramagnetic substance is inversely proportional to absolute temperature.
    $$X \propto \frac{1}{T}$$
    Hence $$XT = constant$$
    Also for paramagnetic substances the susceptibility is positive and very small.
  • Question 8
    1 / -0
    Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show the
    Solution
    Correct Option is D.
    Explanation for the correct answer:
    $$\bullet$$ Curie Temperature is that temperature above which there is sharp change in magnetic properties of all the magnetic substances all. All the Ferromagnetic Material becomes Paramagnetic Material above Curie Temperature.
    $$\bullet$$ Thus, if a temperature of Nickel which is a Ferromagnetic Material is increased above Curie Temperature, then it becomes Paramagnetic.

    This, Nickel will show Paramagnetic behavior above Curie Temperature. Option D is correct.
  • Question 9
    1 / -0
    Assertion : When a bar magnet is freely suspended it points in the north - south direction.
    Reason : The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to north
    Solution
    The magnet align itself in north-south direction, when it is freely suspended because earth behaves as a magnet.
  • Question 10
    1 / -0
    A domain in ferromagnetic iron is in the form of a cube of side length 2$$\mu $$ m then the number of iron atoms in the domain are (Molecular mass of iron = $$55 g mol^{-1}$$ and density = $$7.92 g cm^{-3}$$), maximum value of magnetisation of the given domain is (Dipole moment of an iton atom $$9.27 \times 10^{-24}A m^2)$$
    Solution
    Net dipole moment, $$M_N = N \times m = 6.92 \times 10^{11} \times 9.27 \times 10^{-24} = 6.4 \times 10^{-12} Am^2$$
    Net magnetisation =$$ \dfrac{6.4 \times 10^{-12}}{8 \times 10^{-18}} = 8 \times 10^5 A m^{-1}$$
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