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Solid State Test - 10

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Solid State Test - 10
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  • Question 1
    1 / -0

    Which of the following conditions favours the existence of a substance in the solid state?

    Solution

    At sufficiently low temperature the thermal energy is low and the intrermolecular forces bring the molecules of a substance so close that they cling to one another and occupy fixed positions . They can , however , have oscillating motion about their fixed positions. Under such condition the substance exists in solid state..

     

  • Question 2
    1 / -0

    Which of the following is not a characteristic of a crystalline solid?

    Solution

    Crystalline solids exhibit anisotropic propertiies like its electrical resistance , refractive index which are found to have different values when measured along different directions in the same crystal. They are , therefore, not isotropic in nature.

     

  • Question 3
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    Which of the following is not true about the voids formed in 3 dimensional hexagonal close packed structure?

    (i)  A tetrahedral void is formed when a sphere of the second layer is present above triangular void in the first layer.

    (ii)  All the triangular voids are not covered by the spheres of the second layer.

    (iii)  Tetrahedral voids are formed when the triangular voids in the second layer lie above the triangular voids in the first layer and the triangular shapes of these voids do not overlap.

    (iv)  Octahedral voids are formed when the triangular voids in the second layer exactly overlap with similar voids in the first layer.

    Solution

    (iii)  Tetrahedral voids are formed when the triangular voids in the second layer lie above the triangular voids in the first layer and the triangular shapes of these voids do not overlap.

    (iv)  Octahedral voids are formed when the triangular voids in the second layer exactly overlap with similar voids in the first layer.

     

  • Question 4
    1 / -0

    The value of magnetic moment is zero in the case of antiferromagnetic substances because the domains .....

    (i)  get oriented in the direction of the applied magnetic field.

    (ii)  get oriented opposite to the direction of the applied magnetic field.

    (iii)  are oppositely oriented with respect to each other without the application of magnetic field.

    (iv)  cancel out each other’s magnetic moment.

    Solution

    In anti ferromagnetic substances the magnetic moments and domains are oppositely oriented and cancel out each other’s magnetic moment.(viz. MnO).

     

  • Question 5
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    Which of the following is an amorphous solid?

    Solution

    Quartz glass (SiO2) is an amorphous solid because their is no long range ordered arrangement of the constituent particles present in it .

     

  • Question 6
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    Match the defects given in Column I with the statements in given Column II.

    Column I Column II
    (i)  Simple vacancy defect (a)  shown by non-ionic solids and increases density of the solid.
    (ii)  Simple interstitial defect (b)  shown by ionic solids and decreases density of the solid.
    (iii)  Frenkel defect (c)  shown by non ionic solids and density of the solid decreases
    (iv)  Schottky defect (d)  shown by ionic solids and density of the solid remains the same.
    Solution
    Column I Column II
    (i)  Simple vacancy defect (c)  shown by non ionic solids and density of the solid decreases
    (ii)  Simple interstitial defect (a)  shown by non-ionic solids and increases density of the solid.
    (iii)  Frenkel defect (d)  shown by ionic solids and density of the solid remains the same.
    (iv)  Schottky defect (b)  shown by ionic solids and decreases density of the solid.

     

  • Question 7
    1 / -0

    Match the type of unit cell given in Column I with the features given in Column II.

    Column I Column II
    (i)  Primitive cubic unit cell (a) Each of the three perpendicular edges compulsorily have the different edge length i.e; a≠b≠c.
    (ii)  Body centred cubic unit cell  (b) Number of atoms per unit cell is one.
    (iii)  Face centred cubic unit cell (c)  Each of the three perpendicular edges compulsorily have the same edge length i.e; a = b = c
    (iv) End centred orthorhombic unit cell (d)  In addition to the contribution from the corner atoms the number of atoms present in a unit cell is one.
      (e)  In addition to the contribution from the corner atoms the number of atoms present in a unit cell is three.
    Solution

    (i) → (b), (c) (ii) → (c), (d) (iii) → (c), (e) (iv) → (a), (d)

     

  • Question 8
    1 / -0

    Assertion : The total number of atoms present in a simple cubic unit cell is one.

    Reason : Simple cubic unit cell has atoms at its corners, each of which is shared between eight adjacent unit cells.

    Solution

    In simple cubic unit cell total no. of atoms

    = 8 × 1/8 = 1

     

  • Question 9
    1 / -0

    Assertion : Graphite is a good conductor of electricity however diamond belongs to the category of insulators.

    Reason : Graphite is soft in nature on the other hand diamond is very hard and brittle.

    Solution

    The correct explanation for reason is,

    (i)  In graphite carbon atoms are arranged in different layers and each atom is covalently bonded to three of its neighbouring atoms in the same layer. The fourth valence electron of each atom is present between different layer and is free to move which makes graphite a good conductor of electricity. As different layers, can slide over each other so it is soft.

    (ii)  In case of diamond C atom is bonded covalently with their adjacent atoms through out the crystal. Thus , all the four valencies are satisfied.Covalent bonds are strong and directional in nature , therefore , atoms are held very strongly at their positions. This is why diamond is hard and has very high melting point (it decomposes before melting ) also acting as a insulator .

     

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