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Physics 2007 Set 1 Sol by nehalwan

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PREVIOUS YEAR PAPERS CBSE BOARD EXAM AIEEE BITSAT ISAT VITEEE IIT-JEE STUDY MATERIAL PHYSICS CLASS XI XII SAMPLE PAPERS KEY SOLUTIONS ANSWERS QUESTIONS

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									2/3/2011                                            Subjective Test Paper - Physics - Meritn…




  Physics 2007 Set 1                                                                                  Close


  Subjective Test




      (i)   All questions are compulsory.

      (ii) There are 30 questions in total.

            Questions 1 to 8 carry one mark each,

            Questions 9 to 18 carry two marks each,

            Question 19 to 27 carry three marks each and

            Question 28 to 30 carry five marks each.

      (iii) There is no overall choice. However, an internal choice has been provided.

      (iv) Wherever necessary, the diagrams drawn should be neat and properly labelled.

      (v)    Use of calculators is not permitted.


     Question 1 ( 1.0 marks)
    Is the force acting between two point electric charges q1 and q2, kept at some distance apart in air,
    attractive or repulsive, when (i) q1q2 > 0 (ii) q1q2 <0 ?

    Solution:
    (i) The force is repulsive. When q1q2 > 0, it means that charges are either both positive or both negative.
    This implies that the force between them is indeed repulsive.

    (ii) The force is attractive. When q1q2 < 0, it means that one of the charges is negative and the other is
    positive. This implies that the force between them is indeed attractive.


     Question 2 ( 1.0 marks)
    Show graphically how the stopping potential for a given photosensitive surface varies with the frequency of
    incident radiations.

    Solution:




     Question 3 ( 1.0 marks)
   A TV tower has a height of 71 m. What is the maximum distance up to which TV transmission can be
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                               1/20
                                                    Subjective Test up to - Physics - Meritn…
2/3/2011 tower has a height of 71 m. What is the maximum distancePaperwhich TV transmission can be
    A TV
    received? Given that the radius of the earth = 6.4 × 106 m.

    Solution:
    We know that:




    Where,

    dM = Maximum distance between the transmitting and receiving antenna

    hT = Height of transmitting antenna

    Therefore,




    Hence, the maximum distance up to which TV transmission can be received is 30146 m.


     Question 4 ( 1.0 marks)
    Which one of the two diodes D1 and D2 in the given figures is (i) forward biased,

    (ii) reverse biased ?




    Solution:
    (i) In the given figure, D2 is forward biased.

    (ii) In the given figure, D1 is reverse biased.


     Question 5 ( 1.0 marks)
    Suggest a possible communication channel for the transmission of a message signal which has a bandwidth
    of 5 MHz.

    Solution:
    A communication channel for the transmission of message signal, which has a bandwidth of 5 Mhz, is FM
    radio frequency.


     Question 6 ( 1.0 marks)

    Solution:
    OUT OF SYLLABUS


     Question 7 ( 1.0 marks)

    Solution:
    OUT OF SYLLABUS


     Question 8 ( 1.0 marks)
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                           2/20
2/3/2011                                            Subjective Test Paper - Physics - Meritn…

   Solution:
   OUT OF SYLLABUS


    Question 9 ( 2.0 marks)

   A spherical Gaussian surface encloses a charge of 8.85 × 10-10C.

   (i) Calculate the electric flux passing through the surface.

   (ii) How would the flux change if the radius of the Gaussian surface is doubled and why?

   Solution:

       i.   Total flux enclosed                       100 T m2

      ii.   The flux would not change if the radius of Gaussian surface is double because enclosed charge
            remains the same.


    Question 10 ( 2.0 marks)
   In a copper voltmeter, the mass of the ions deposited in 30 seconds is m grams. Thecurrent (i) vs. time (t)
   graph for the experiment is as shown in the given figure.




   Calculate the value of E.C.E. of copper in terms of the mass, m, deposited.

   Solution:
   OUT OF SYLLABUS


    Question 11 ( 2.0 marks)
   In an ammeter (consisting of a galvanometer and a shunt), 0.5% of the main current passes through the
   galvanometer. Resistance of the galvanometer coil is G. Calculate the resistance of the shunt in terms of
   galvanometer resistance, G.

   Solution:
   Let the shunt resistance be S, the Galvanometer resistance be G, and the main current be I.

   Because they are connected in parallel, their combined resistance will be        .


   Voltage drop across the combination will be I ×                .


   Current through G is 0.5% of main current (I) = 0.005I

   Current through S is 99.5% of I = 0.995I

   If we apply Ohm’s law for the shunt only, then

   V=I×R

   I×         = 0.995I × S
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                              3/20
2/3/2011                                             Subjective Test Paper - Physics - Meritn…
    I×          = 0.995I × S


   Solving, we obtain

              = 0.995


      S


   Thus, the resistance of the shunt is      .



    Question 12 ( 2.0 marks)
   An electric bulb B and a parallel plate capacitor C are connected in series to the a.c.mains as shown in the
   given figure. The bulb glows with some brightness.




   How will the glow of the bulb be affected on introducing a dielectric slab between the plates of the
   capacitor? Give reasons in support of your answer.

   Solution:
   The bulb will glow brighter.

   Reasons:

   The impedance of a capacitor is          without dielectric.


   If a dielectric is introduced inside a capacitor, then the new capacitance will be KC and the new
   impedance will be          .


   Therefore, the impedance has decreased.

   This will result in higher current through the circuit and the bulb will glow brighter.


    Question 13 ( 2.0 marks)
   What does the statement, “natural light emitted from the sun is unpolarised” mean interms of the direction
   of electric vector? Explain briefly how plane polarized light can beproduced by reflection at the interface
   separating the two media.

   Solution:
   The statement “natural light emitted from the sun is unpolarised” means that the natural light coming from
   sun is a mixture of waves, each having its electric vectors directed in random direction.

   When light falls on the interface separating two media, electrons start oscillating, which produces reflected
   ray in addition to refracted ray. As light is a transverse wave, therefore, oscillation in the transverse
   direction will produce a light wave. Parallel oscillations will not contribute to the light wave. When a light
   ray strikes an interface, the component of electric vector, which is parallel to the interface, gets reflected.
   Therefore, the reflected light wave is plane polarised light.


    Question 14 ( 2.0 marks)
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                  4/20
2/3/2011                                              Subjective Test Paper - Physics - Meritn…
    Question 14 ( 2.0 marks)
   The output of a 2-input NOR gate is fed to a NOT gate. Draw the logic circuit of this combination of
   gates and write the truth table for the output of the combination for all inputs. (i) attractive, (ii) repulsive.

   Solution:




   Truth table: for attractive and repulsive combinations.

              A B     D

              1   1   1

              1   0   1

              0   1   1

              0   0   0

   It cannot be said with certainty as to which of the two inputs is attractive or repulsive.


    Question 15 ( 2.0 marks)
   What is remote sensing? Write its two applications.

   Solution:
   Remote sensing is the science and art of acquiring information about material objects, area, or
   phenomenon, without coming into physical contact with the objects, area, or phenomenon under
   investigation.

   Its two applications are:

       i.    Meteorological observations
       ii.   Weather forecasting


    Question 16 ( 2.0 marks)
   On what principle does a metre bridge work? Draw a circuit diagram and explain how this device can be
   used for determination of an unknown resistance.

   Solution:




   Metre bridge works on the principle of Wheatstone bridge.

   Meter bridge consists of a metre scale, a wire of length 1 m and a galvanometer. Meter scale is attached
   to the wire and one end of the galvanometer is connected to the metallic strip between the resistors R and
   S and the other end is connected to a jockey, which can slide over the metallic meter scale. R is the
   unknown resistance whose value is to be determined. S is a known resistance. The jockey connected is
   moved on the metre scale such that we get zero deflection at some point. Let this point be l1 distance
   away from A.
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                    5/20
2/3/2011                                              Subjective Test Paper - Physics - Meritn…

   If Rcm is the resistance per unit length of the 1 meter wire, then the resistance of one portion of the wire
   will be Rcm l1 and the resistance of another portion will be Rcm(100 - l1).

   Using the condition for balanced bridge, we obtain




   By substituting the value of l1 in the above equation, we can find out the value of R, which is the unknown
   resistance.


    Question 17 ( 2.0 marks)
   Derive a mathematical expression for resistivity of a conductor in terms of numberdensity of charge
   carriers in the conductor and relaxation time.

   Solution:
   Average velocity of all the electrons at any instant of time under no external electric field is zero.

   Now, when the external field is present, the electrons get accelerated.

   Let the average acceleration be a. Then,

   a


   Let the initial velocity of all the electrons be denoted by v i and the final velocity be Vi. Therefore,

   Vi = v i


   The value of time (t) is the average value of the time taken by each electron during any one collision. This
   is called relaxation time.

   The average velocity of all the electrons without an external field is zero.

   Thus,

   vi = 0

   Vi


   Here, Vi is the average velocity or drift velocity v d .

   Vi = v d

   vd


   Because of the external electric field, electrons are accelerated. They move from one place to another and
   current is produced.

   For small interval dt, we have

   Idt = −q

   Where q is the total charge flowing

   Let n be the free electrons per unit area. Then, the total charge crossing area A in time dt is given by:

   Idt = neAv d dt
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                               6/20
2/3/2011                                           Subjective Test Paper - Physics - Meritn…

   Substituting the value of v d , we obtain

   Idt = neA          dt




   Where,

   J is the current density

   J = −n      Et


   From Ohm’s law,

   J = σE

   Where,

   σ is the conductivity of the material through which the current is flowing

   Thus,

   σ = −n      t


   σ=




   Substituting the value of conductivity, we obtain

   ρ


   Where,

   t is the relaxation time


    Question 18 ( 2.0 marks)
   State Ampere’s circuital law. Write the expression for the magnetic field at the centre of acircular coil of
   radius R carrying a current I. Draw the magnetic field lines due to thiscoil.

   Solution:




   Ampere’s circuital law states that the line integral of magnetic field ( ) around any closed path or circuit is
   equal to o times the total current (I) threading the closed circuit.

meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                  7/20
2/3/2011                                             Subjective Test Paper - Physics - Meritn…




    Where,

    i is the current through the loop and R is the radius of the loop


     Question 19 ( 3.0 marks)
    Write the expression for the force acting on a charged particle of charge q moving with velocity        in the
    presence of magnetic field . Show that in the presence of this force

    (i) the kinetic energy of the particle does not change

    (ii) its instantaneous power is zero

    Solution:



    (i)   is perpendicular to both           .

    If    is the instantaneous displacement of the charge,     is also perpendicular to    .

    Work done (W) by this force (      ) = Increase in kinetic energy (K.E.)



    = Fds cos 90°

    W=0

    Hence, increase in K.E. = 0

    ∴ The K.E. of the particle does not change.

    (ii) Power =

    = Fv cos 90°

    Power = 0


     Question 20 ( 3.0 marks)
    In a series LCR circuit, define the quality factor (Q) at resonance. Illustrate its significance by giving one
    example.

    Show that power dissipated at resonance in LCR circuit is the maximum.

    Solution:
    The quality factor (Q) is defined as:




    Where,      → Resonant frequency

    L → Inductance

    R → Resistance

   Another expression for quality factor is
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                  8/20
2/3/2011 expression for quality factor is
    Another                                          Subjective Test Paper - Physics - Meritn…


    Where,

    2     is the band width.

    So, larger the value of Q, the smaller the band width and sharper is the resonance.

    Power =




    Hence, power =

    ∴ Power dissipated will be the maximum at resonance.


     Question 21 ( 3.0 marks)
    A circular copper disc 10 cm in radius rotates at a speed of 20π rad/s about an axis through its centre and
    perpendicular to the disc. A uniform magnetic field of 0.2 T acts perpendicular to the disc.

    (i) Calculate the potential difference developed between the axis of the disc and the rim.

    (ii) What is the induced current if the resistance of the disc is 2 ?

                                                            OR

    A resistor of 200 and a capacitor of 40 µF are connected in series to 220 V a.c. source with angular
    frequency (ω) = 300 Hz. Calculate the voltages (rms) across the resistor and the capacitor. Why is the
    algebraic sum of these voltages more than the source voltage? How do you resolve this paradox?

    Solution:
    (i) Flux change in 1 rotation = BA = B × πr2

    = 0.2 × π × 0.12

    Rotation of 20π radians takes place in 1 second.

    Rotation of 2π radians takes place in        seconds.


    i.e., 1 rotation takes place in    seconds         second




    (ii) Induced current =


meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                               9/20
2/3/2011                                          Subjective Test Paper - Physics - Meritn…



    = 0.01 π A

    OR




    Voltage drop across resistor = IR

    = 1.02 × 200

    = 204 V

    Voltage drop across capacitor = IXC




    Total voltage drop

    = 204 + 85

    = 289 V

    Kirchoff’s voltage law will be valid only if we consider instantaneous values of voltage. 85 V and 204 V
    are rms values of voltage and hence, when they are added there value becomes more than the source
    voltage.


     Question 22 ( 3.0 marks)
    Draw a labelled diagram of Hertz’s experiment. Explain how electromagnetic radiations are produced
    using this set-up.

    Solution:
    Hertz’ experiment:




   The setup consists of two metal plates P1 and P2 held parallel to each other, which are connected to two
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                            10/20
2/3/2011                                          Subjective Test Paper - Physics - Meritn…
    The setup consists of two metal plates P1 and P2 held parallel to each other, which are connected to two
    metallic spheres S1 and S2.

    A high voltage is applied to the spheres using an induction coil so that a spark is formed between the gaps.
    This leads to the production of an electromagnetic wave.


     Question 23 ( 3.0 marks)
    How does the frequency of a beam of ultraviolet light get affected when it goes from air into glass?

    A ray of light incident on an equilateral glass prism shows minimum deviation of 30°.

    Calculate the speed of light through the glass prism.

    Solution:
    The frequency of the ultraviolet beam of light does not change when it goes from air to glass. This is
    because frequency is the property of source and does not change with medium.




    Speed of light in glass

    = 3 × 108 × sin30 ×


    = 3 × 108 ×




     Question 24 ( 3.0 marks)
    An electron, alpha particle and a proton have the same de-Broglie wavelength. Which of these particles
    has (i) minimum kinetic energy, (ii) maximum kinetic energy, and why?

    In what way has the wave nature of electron been exploited in electron microscope?

    Solution:




    Kinetic energy (K.E.) =



meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                11/20
2/3/2011                                              Subjective Test Paper - Physics - Meritn…



   Kinetic energy is inversely proportional to ‘m’.

   (i) An alpha particle has the maximum mass. Therefore, it will have minimum K.E.

   (ii) An electron has the minimum mass. Therefore, it will have maximum K.E.

   In an electron microscope, the wave nature of electrons is used. When objects of very small angular
   separation are used, it is desirable to make the diffraction pattern as small as possible Effective wavelength
   of beams of electrons is 10-5 of the wavelength of visible light. Therefore, an electron microscope permits
   the detailed examination of tiny structures.


    Question 25 ( 3.0 marks)
   State the law of radioactive decay. Establish a mathematical relation between half-life period and
   disintegration constant of a radioactive nucleus.

   Solution:
   The law of radioactive decay states that the number of nuclei undergoing the decay per unit time is
   proportional to the total number of nuclei in the given sample.

   The law of radioactive decay is given by:



    (t) → Number of nuclei left after time t

    0   → Original number of nuclei.

   λ → Decay constant

   t → Time

   Now,




   Taking ln on both sides, we obtain




   Hence, the relation between half-life period and disintegration constant of a radioactive nucleus is
   established.


    Question 26 ( 3.0 marks)
   Distinguish between nuclear fission and fusion. In a fusion reaction,

meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                 12/20
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   Calculate the amount of energy (in MeV) released. Given                = 2.014102 u;

            = 3.016049 u;             = 4.002603 u;       = 1.00867 u; 1 u = 931.5 MeV/C2.

   Solution:
   Nuclear fission is the shifting of heavy nucleus into two lighter nuclei with the liberation of high amount of
   energy whereas nuclear fusion is the joining of two lighter nuclei to form one heavy nucleus with the
   liberation of high amount of energy.



   (2.014102 u + 3.016049 u) → (4.002603 u + 1.00867 u)

   Mass lost = (2.014102 u + 3.016049 u) − (4.002603 u + 1.00867 u)

   m = 0.018878 u

   1 u = 931.5 MeV/c2

   Therefore,

   m = 0.018878 u = 0.018878 × 931.5 MeV/c2

   E = mc2

   = 0.018878 × 931.5


   E = 17.58 MeV

   Therefore, the amount of energy released is 17.58 MeV.


    Question 27 ( 3.0 marks)
   Draw a schematic diagram of a single optical fibre structure. On what principle does such a device work?
   Explain the mechanism of propagation of light signal through an optical fibre.

   Solution:




   A single optical fibre works on the principle of total internal reflection.

   An optical fibre consists of a core with higher refractive index and a cladding with a lower refractive index.
   When light enters the fibre at a suitable angle, it undergoes successive total internal reflections along the
   length of the fibre. This is how a light signal travels through the optical fibre.


    Question 28 ( 5.0 marks)
   Derive the expression for the energy stored in a parallel plate capacitor of capacitance C with air as
   medium between its plates having charges Q and − Q. Show that this energy can be expressed in terms of
   electric field as            where A is the area of each plate and d is the separation between the plates.


   How will the energy stored in a fully charged capacitor change when the separation between the plates is
   doubled and a dielectric medium of dielectric constant 4 is introduced between the plates?
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                 13/20
2/3/2011                                             Subjective Test Paper - Physics - Meritn…

   OR

   Define the term dipole moment of an electric dipole indicating its direction. Write its SI unit.

   An electric dipole is placed in a uniform electric field . Deduce the expression for the torque acting on it.
   In a particular situation, it has its dipole moment aligned with the electric field. Is the equilibrium stable or
   unstable?

   Solution:
   Consider a situation when one plate has a charge Q′ and the other plate will also have charge −Q′.

   The voltage between the plates will be       . A small charge equal to δQ′ is transferred from the negative

   plate to the positive plate. A small work δW is done in this process.

   δW = V′ δQ′ =




   Energy stored =          …(1)


   Now,

   Q = σA where σ → Surface charge density

              where A → Area of the plates and d → Separation between the plates


   Putting these values in equation (1), we obtain

   Energy stored =                                    …(2)


   Now, we know that,

   Electric field,


   Therefore, energy stored


   When distance between the plates is doubled and the dielectric constant is 4, the energy stored in the

   capacitor will be                        =


   Thus, the total energy will become half when the distance between the plates is doubled and the dielectric
   constant is 4.

   OR

   Electric dipole moment is defined as the product of any one of the charges and the length of the electric
   dipole.



   Where,

   p = Electric dipole moment
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                    14/20
2/3/2011                                             Subjective Test Paper - Physics - Meritn…
    p = Electric dipole moment

    q = One of the charges

    2a = Length of the electric dipole

    Its direction is from negative charge to positive charge.

    Its SI unit is coulomb metre.




    τ = Either force × Perpendicular distance between the two forces

    = q E (2a sin θ)

        τ = p E sin θ



        τ (θ) = p E sin θ

    For equilibrium τ (θ) = 0,

    p E sin θ = 0

        sin θ = 0

        θ = 0° or 180°

    When the dipole is aligned along the electric field,

    θ = 0°

    Now,

    U → Potential energy

    U = − p E cos θ

    = − p E cos 0°

    =−pE

    The potential energy is minimum at θ = 0.

    Hence, the dipole is in stable equilibrium.


     Question 29 ( 3.0 marks)
    Define the term ‘wavefront’. Draw the wavefront and corresponding rays in the case of a

    (i) diverging spherical wave, (ii) plane wave.

    Using Huygen’s construction of a wavefront, explain the refraction of a plane wavefront at a plane surface
    and hence verify Snell’s law.

                                                           OR

   Derive the relation between the focal length of a convex lens in terms of the radii of curvature of the two
   surfaces and refractive index of its material. Write the sign conventions and two assumptions used in the
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                              15/20
                                                Subjective Test Paper - Physics - Meritn…
2/3/2011 and refractive index of its material. Write the sign conventions and two assumptions used in the
    surfaces
    derivation of this relation.

    A convex lens of focal length 40 cm and a concave lens of focal length 25 cm are kept in contact with
    each other. What is the value of power of this combination?

    Solution:
    The locus of all the particles of the medium, which at any instant are vibrating in the same phase, is called
    the wavefront.




                                          .

    Wavefront AB strikes the surface PP′ with an angle of incidence ‘i’.

    Speed of light in medium 1 is .

    Speed of light in medium 2 is .

    Let ‘τ’ be the time taken by the wavefront to travel the distance BC.

    BC =




    Dividing (i) by (ii), we obtain




    Refractive index(n1) of medium 1 is


        v1 =


    Refractive index (n2)of medium 2 is
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                 16/20
2/3/2011                  2                        Subjective Test Paper - Physics - Meritn…



        v2 =


   Putting these values in equation (iii),




   OR




   The image I1 of the object O behaves like a virtual object for the second surface.

   For surface ABC,




   For surface ADC,




   For a thin lens, it is known that BI1 = DI1

   Adding (i) and (ii),




   Focus is the point where the image is formed when object is at infinity.

   DI = f, when OB → ∞



meritnation.com/…/C@TLOL6F7nCFj75Z…                                                            17/20
2/3/2011                                            Subjective Test Paper - Physics - Meritn…




   Using sign convention,




   We obtain:




       → Refractive index of medium 2 with respect to medium 1

   The sign conventions used here are as follows:

   (i) All distances are measured from the optical centre.

   (ii) The distances measured in the same direction as incident light are taken as positive while the distances
   measured in a direction opposite to incident light are taken as negative.

   The two assumptions used in this derivation are as follows:

   (i) The lens is thin.

   (ii) Aperture of the lens is small.

   (iii) The rays are paraxial.

   (Note: Any two of the above can be used to give the answer)

   When two thin lenses are in contact then,




   Where,

   f is the focal length of both the lenses combined.

   f1 is the focal length of the convex lens.

   f2 is the focal length of the concave lens


   Power (P) of the lens =


   It is given that

   f1 = −40 cm = −0.40 m (by sign convention)

   f2 = 25cm = 0.25 m

   Therefore,

   P


   P = 1.5 dioptre
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                18/20
2/3/2011                                               Subjective Test Paper - Physics - Meritn…
    P = 1.5 dioptre


     Question 30 ( 3.0 marks)
    Explain, with the help of a circuit diagram, the use of an n-p-n transistor as an amplifier in common-emitter
    configuration. Write the expression for voltage gain of the transistor.

    Draw frequency response curve of this amplifier.

    How will the current gain of a transistor be affected if its base region is made thicker as compared to a
    usual transistor and why?

                                                          OR

    Draw energy band diagrams for (i) an intrinsic semiconductor, (ii) p-type semiconductor.

    Draw symbolic representation of a zener diode. Draw its V-I characteristics and explain, with the help of a
    circuit diagram, its use as a voltage regulator.

    Solution:




    If a small sinusoidal voltage is applied to the input of a CE configuration, the base current and collector
    current will also have sinusoidal variations. Because the collector current drives the load, a large sinusoidal
    voltage    will be observed at the output.

    The expression for voltage gain of the transistor in CE configuration is:




    βac → ac current gain

    RL → Load resistance

    r = RB +

      → Input resistance

    RB → Base resistance


    Current gain of the transistor will decrease if the base is made thicker because current gain,


    If the base of an n-p-n transistor is made thicker, then more and more electrons will recombine with the p-
    type material of the base. This results in a decrease in collector current .

    Furthermore,      also increases.


    Hence, ac current gain (            ) decreases.


    (Draw frequency response curve of this amplifier.)

    Out of syllabus

   OR
meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                                   19/20
2/3/2011
    OR                                           Subjective Test Paper - Physics - Meritn…




   A zener diode is connected across a load such that it is always reverse biased and has sharp breakdown
   voltages. If the voltage across the zener diode increases or decreases beyond the breakdown voltage, then
   the current through the zener diode decreases or increases to keep the voltage across the diode constant.
   The figure shows a zener diode connected in reverse bias. Thus, the voltage across the zener diode
   remains constant, if it is operating beyond its breakdown voltage, even if the source voltage shows
   fluctuations.

   VZ → Breakdown voltage of zener




meritnation.com/…/C@TLOL6F7nCFj75Z…                                                                            20/20

								
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