CBSE Delhi-Set-1-2015

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  • Q1

    Define capacitor reactance. Write its S.I. units.

    Marks:1
    Answer:

    Capacitor reactance is the resistance offered by a capacitor when connected to an electric circuit. It is denoted by Xc and is given by the expression:


    where

    ω = Angular frequency

    C = Capacitance of the capacitor

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  • Q2

    What is the electric flux through a cube of side 1 cm which encloses an electric dipole?

    Marks:1
    Answer:

    According to Gauss’s Law

    Where,

    E = Electrostatic field

    Q = total charge enclosed by the cube

    = Absolute electric permittivity of free space

    In case of a dipole, net charge is equal to zero; therefore, electric flux through the cube is also zero.

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  • Q3

    A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens?

    Marks:1
    Answer:

    If a concave lens is immersed in a medium with the refractive index greater than its own refractive index then it will behave as a diverging lens.

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  • Q4

    How are side bands produced?

    Marks:1
    Answer:

    During the process of modulation, the audio frequency modulating signal wave is superimposed on a high frequency wave called carrier wave. Any form of modulation produces frequencies that are the sum and difference of the carrier and the modulating frequencies. These frequencies are called side bands. Side bands are a band of frequencies higher than or lower than carrier frequency.

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  • Q5

    Graph showing the variation of current versus voltage for a material Ga-As is shown in the figure. Identify the region
    (i) Of negative resistance

    (ii) Where Ohm’s law is obeyed

    Marks:1
    Answer:

    (i) The slope of the curve is negative in the region DE; hence, DE represents the area of negative resistance.

    (ii) According to Ohm’s law, current varies linearly with voltage. Thus, BC represents the region where Ohm’s law is obeyed.

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  • Q6

    A proton and an α-particle have the same de-Broglie wavelength. Determine the ratio of                                               
    i. their accelerating potentials

    ii. their speeds.

    Marks:2
    Answer:

    (i) The de Broglie wavelength is expressed as:

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  • Q7

    Show that the radius of the orbit in hydrogen atom varies as n2, where n is the principle quantum number of the atom.

    Marks:2
    Answer:


    Centripetal force required by an electron of mass m and velocity v during its orbit is given by

    Also, when the charge on the nucleus is Ze, the centripetal force due to force of electrostatic attraction between the electron and nucleus is given by:

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  • Q8

    Distinguish between ‘intrinsic’ and ‘extrinsic’ semiconductors.

    Marks:2
    Answer:

    Intrinsic Semiconductors

    Extrinsic Semiconductors

    1. These types of semiconductors have no impurity atoms and hence, are pure.

    These types of semiconductors are prepared by doping a small amount of impurity atoms.

    1. The number of free electrons in conduction band is equal to the number of the holes in the valence band.

    The number of electrons and holes are never equal. The number of electrons is greater than number of holes in n-type semiconductor whereas number of holes is greater than electrons in p-type semiconductors.

     

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  • Q9

    Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

    Marks:2
    Answer:

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  • Q10

    Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum?

    Marks:2
    Answer:

    Let the two polaroids PA and PB cross each other at an angle of 90˚. A polaroid sheet PC is placed between them such that it makes an angle θ with Polaroid PA.

    Then, intensity of polarized light emerging from PC is given by

    I = I0 cos2θ

    As PA and PB are perpendicular to each other, angle between PB and PC = (90 - θ)

    Therefore, intensity of light emerging from PB is given by

    I = [I0 cos2θ] cos2 (90 - θ)

    = [I0 cos2θ] sin2θ

    So, the intensity of light is maximum when

    sin2θ = 1

    2θ = 90˚

    θ = 45˚

    The maximum intensity of light is expressed as

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