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Physics 2008 Set 2 Sol

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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 2008 Set 2                                                                           Close


 Subjective Test




     These are additional set 2 questions.



    Question 3 ( 1.0 marks)
   A converging lens of refractive index 1.5 is kept in a liquid medium having same refractive index.
   What would be the focal length of the lens in this medium?

   Solution:
   The lens in the liquid will act like a plane sheet of glass.

   ∴ Its focal length will be infinite (∞).


                               [By lens maker’s formula]




   Here,




    Question 6 ( 1.0 marks)
   How does the angular separation of interference fringes change, in Young’s experiment, if the
   distance between the slits is increased?

   Solution:

               , when separation between slits (d) is increased, fringe width β decreases.




    Question 11 ( 2.0 marks)
   Draw a ray diagram of an astronomical telescope in the normal adjustment position, state two
   drawbacks of this type of telescope.

   Solution:
   Ray diagram of astronomical telescope (Normal adjustment)




   Drawbacks:

   (i) In normal setting of telescope, final image is at infinity.

   (ii) Magnifying power is the minimum and its field of view is small.

meritnation.com/…/P8BiPvkOl8xGyYhCz…                                                                    1/5
2/3/2011                                         Subjective Test Paper - Physics - Meritn…

    Question 12 ( 2.0 marks)
   Calculate the distance of an object of height h from a concave mirror of focal length 10 cm, so as to
   obtain a real image of magnification 2.

   Solution:
   Here, f = −10 cm, m = 2, u = ?

   As the image is real,




   Applying the mirror formula,




   ∴Object distance = 15 cm



    Question 13 ( 2.0 marks)
   Draw the output wave form at X, using the given inputs A, B for logic circuit shown below. Also
   identify the gate.




   Solution:
   For the given gate, the Boolean expression is

   ∴The logic gate is AND gate.




    Question 15 ( 2.0 marks)
   Derive an expression for the potential energy of an electric dipole.

   Solution:
   Potential energy of a dipole in an external field:

   Consider a dipole with charges q1 = + q and q2 = − q placed in a uniform electric field   such that AB
   = 2a as shown in figure.




meritnation.com/…/P8BiPvkOl8xGyYhCz…                                                                        2/5
2/3/2011                                        Subjective Test Paper - Physics - Meritn…




   Since the dipole experiences no net force in a uniform electric field but experiences a torque,

   τ is given by:




   Suppose an external torque       is applied in such a way that it neutralises this torque and rotates in
   the plane of paper from angle          at an infinitesimal angular speed.

   The amount of work done by the external torque is:




   This work is stored as the potential energy of the system.



       Question 18 ( 2.0 marks)
   Prove that an ideal inductor does not dissipate power in an a.c circuit.

   OR

   Derive an expression for the self-inductance of a long air-cored solenoid of length l and number of
   turns N.

   Solution:

   In a circuit containing inductor L, current I lags behind the voltage E by a phase angle of 90° or

   .




   Work done in one complete cycle is




meritnation.com/…/P8BiPvkOl8xGyYhCz…                                                                          3/5
2/3/2011                                            Subjective Test Paper - Physics - Meritn…
   ∴Average power =


   Hence, an ideal inductor does not dissipate power in an ac circuit.

   OR

   Self inductance of a long solenoid:

   Consider a long solenoid of length 1, number of turns N, and radius r. Suppose current I flows
   through it. Then, magnetic field set up in the coil is:




   Flux through each turn =


   Where,

   A is the cross-sectional area


   Flux through N turns =


   However, Φ = LI


   ∴Self inductance,




    Question 24 ( 3.0 marks)
   How is a wave front defined? Using Huygen’s construction draw a figure showing the propagation of
   a plane wave reflecting at the interface of the two media. Show that the angle of incidence is equal
   to the angle of reflection.

   Solution:
   Waves spread out from the point of impact. At any instant after the impact, locus of all points, which
   oscillate in phase and have the maximum disturbance on them, is called a wavefront.

   Huygens principle:

   (i) Every point on a given wavefront acts as a fresh source of secondary wavelets, which travel in all
   directions with the speed of light.

   (ii) The forward envelope of these secondary wavelets gives a new wavefront at every instant.

   Laws of reflection by Huygens principle:




   By the Huygens principle, secondary wavelets start from B and cover a distance ct in time t’ and
   reaches B’.

   To obtain new wavefront, draw circles w ith point B as centre and ct             as radius. Draw a
   tangent       from the point        .

   Then,     represents the reflected wavefront.

   In                       ,

                     [                     = Radii of same circle]

               [Common]

                         [Each 90°]

                          [By R.H.S]
meritnation.com/…/P8BiPvkOl8xGyYhCz…                                                                        4/5
2/3/2011                                         Subjective Test Paper - Physics - Meritn…

                        [C.P.C.T]

   ∴Incident angle i = Reflected angle r

   ∠i = ∠r

   Therefore, angle of incidence is equal to angle of reflection.



    Question 25 ( 3.0 marks)
   A coil of number of turns N, area A, is rotated at a constant angular speed ω , in a uniform magnetic
   field B, and connected to a resistor R. Deduce expressions for:

   (i) Maximum emf induced in the coil

   (ii) Power dissipation in the coil

   Solution:
   (i) The magnetic flux linked with the coil at any instant is Φ = NBA cosωt

   ∴Induced emf will be




   = NBA ω

   emf is maximum when sin ωt = 1




   Hence,

   (ii)

   All calculations have been done considering that the coil has zero resistance.

   Therefore, power dissipated in the coil is equal to zero because the resistance of an ideal coil is
   zero.

   Power dissipated inside the coil = Current2 × Resistance of the coil

   = Current2 × Zero

   =0




meritnation.com/…/P8BiPvkOl8xGyYhCz…                                                                       5/5

								
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