Blogger Widgets APEX INSTITUTE : Class - XII Physics Sample Paper- 4

Tuesday, September 16, 2014

Class - XII Physics Sample Paper- 4

Instructions:
1. Section-A  Q. no. 1-8 carry 1 mark each.
2. Section-B  Q. no. 9-18 carry 2 marks each.
3. Section-C  Q. no. 19-27 carry 3 marks each.
4. Section-D  Q. no. 28-30 carry 5 marks each.
Section-A
1. S1 and S2 are two hollow concentric spheres enclosing charges q and 2q respectively.



(i) What is the ratio of electric flux through S1 and S2?
(ii) How will the electric flux through sphere S1 change is a medium of dielectric constant 5 is introduced in the sphere S1 in place of air ? 
Ans. (i)
(ii)
2. For the potentiometer circuit shown in the figure, points X and Y represent the two terminals of an unknown emf E’. A student observed that when the jockey is moved from the end A to the end B, deflection in galvanometer remains in same direction. What may be two possible faults?
Ans. Two possible faults may be
(a) E < ε
(b) Polarity of terminals X and Y are not connected correctly
3. Define 1mH Inductance.
Ans. As Φ = LI ⇒ when one miliwebere flux changes due to flow of unit current through a coil then the flux linked will be one milihenery.
4. Derive an expression for capacitance of a parallel plate capacitor.
Ans. Electric field between the plates of capacitor of charge q and area of crossection A is given as E = σ/∈0
∴ P.D between the plates V = σ/0                                          
capacitance
5. How does the intensity of magnetization of paramagnetic sample vary with temperature ?
Ans.
6. A circular coil of N turn and radius R is kept normal to a magnetic field given by B=Bo cos wt. State the rule which helps to detect the direction of induced current?
Ans. Lenz’ law. As accorong to law, induced current will be such that it opposes the cause by which it is induced.
7. A bulb of resistor 10? connected to an Inductor of inductance L is in series with an a.c. source marked 100v 50Hz. If phase angle between the voltage and current is π/4 radian. Calculate the value of L.
Ans. as Φ = π/4, R = 10Ω    f = 50Hz
tanΦ = XL/R ⇒ tanπ/4 = 2πfL/R
⇒ 2πfL = R ⇒ L = R/2πf = 0.3henery
8. An electron and a proton have same De broglie wavelength associated with them. How are their Kinetic energies related to each other ?
Ans.
SECTION-B
9. Two radioactive nuclei X & Y initially contain an equal number of atoms. Their half life is 1 hour and 2 hour respectively. Calculate the ratio of their rates of disintegration after two hours.
Ans. Given Nox = Noy = No
Ty2x = IH, Ty2y = 2H.
10. A message signal of frequency 15 KHz. and peak voltage of 5 volts is used to modulation a carrier of frequency 1 MHz . and  peak voltage 20 volt .Determine the modulation index and the side bands.
Ans.
Side base band .1MHz +15 KHz and 1MHz ­- 15 KHz
i.e. 1.015 MHz and 0.985 MHz.
11. Two rectangular metal plates, each of area A are kept parallel to each other at a distance ‘d’ apart to form a parallel plate capacitor. If the area of each of the plates is doubled and their distance of separation decreases to 1/2 of its initial value. Calculate the ratio of their capacitance in the two cases?
Ans. Capacitance of parallel capacitor is given as C = 
CInitial = C1 = and

Cfinal = C2 =
         =
12. Show that energy stored in a parallel plate capacitor is 1/2CV2.
Ans. Graphically variation of stored charged on plate of capacitor & P.D between the plate is shown in figure
Area of ΔOAB =
                        
13. A conductor of length ‘l’ connected to a d.c. source of potential V. If the length of the conductor is tripled by streching it keeping ‘V’ constant. Explain how do the following factors vary in the conductor?     (2 marks)
(1) Drift speed of electrons
(2) Resistence
(3) Resistivity
Ans. (1)
(2) R:- increases 9 times as R α n2.
(3) ρ :- No effect.
14. Distinguish between paramagnetic, diamagnetic and ferromagnetic material.
Ans.
Diamagnetic
Paramagnetic
Feromagnetic
 If kept near a magnet, materials are freely repelled.
 Feebly attracted.
 Strongly attracted.
 χ : negative & small.
 χ : positive and small.
 χ: positive and large.
 μ : Slightly less than one.
 μ : Slightly more than one.
 μ : Quite larger than one.
15. A particle of mass ‘m’ with charge ‘q’ moving with a uniform speed ? normal to a uniform magnetic field B, describes a circular path of radius r. Describe an expression for the (i) time period of revolution & (ii) Kinetic energy of the particle. 
Ans. As 
It follow a  circular path 
Time to complete one circular path
and
16. Draw a schematic arrangement of Geiger-Marsden experimental setup. How does it explain the size of nucleus.  
Ans. The fact that only a small fraction of incident particles rebound back that number of α particles undergoing head on collision is small this leads the mass confined in small region.
17. Calculate the maximum kinetic energy of α particle emitted during α-decay of
92U23890Th234 + 2He4
M(92U238) = 238.05079U, M(90Th234) = 234.04363U, M(2He4) = 4.00260U
Ans. Δm = [ M(u) - M(th) - M(He) ]
      = (238.05079 – 234.04363 – 4.00560)
      = 0.00456 U
  Q = 0.00456 x 931 mev
18. State the laws of Photoelectric effect. Explain it on the basis of Einstein equation.
Ans. Laws : -
(i) It is an instantaneous process
(ii) No Photo emission takes place below threshold frequency of material, no matter how intense the incident beam.
(iii) The maximum photo current (saturation current ) does not depends upon stopping potential or frequency but depends on intensity of incident radiation.
(iv) Stopping potential is independent on intensity of incident radiation.
SECTION-C
19. Explain with the help of a labeled diagram the underlying the principle and working of a step down transformer.
Ans. Principle – It works on the principle of mutual induction i.e. when a magnetic flux linked with a coil changes an emf is induced in another coil which is placed closer to first one.
Working – when a.c. signal is applied in primary (p) coil, the flux linked with primary and secondary coil will be given as
ΦS = NSBA  and
ΦP = NPBA
emf induced in secondary will be given as
As Ns < Np ⇒ Es < Ep
20. The magnitude of magnetic field in a plane e.m. wave is given as BX = 0, BY = 2 x 10-7 sin(0.5 x 103 x + 1.5 x 1011t)T.
(a) Determine the wavelength and frequency of wave.
(b) Write an expression for the electric field.
Ans. BY = BO  sin (1.5 x 1011t + 0.5 x 103x)T
BO = 2 x 10-7 T,                          ω = 2πν = 1.5 x 1011
  ω = 1.5 x 1011                       ⇒
  K = 0.5 x 103
EX = 0 , EY = 0, EZ = CBO sin (1.5 x 1011t + 0.5 x 103x)
EZ = 60 sin (1.5x 1011t + 0.5 x 103x)
21. The energy level of an element are given below. Which corresponds to emission of spectral line of wavelength 482nm?
Ans.
For ‘B’ Transition  ΔE = 3.4 – 0.85
                                = 2.55 very closer
‘B’ is possible
23. Why do we require modulation? Explain the AM with block diagram.
Ans. As for Transmission of  signals .Size of antenna , which is too large.
Hence to reduce the size of antenna , Frequency (Other characteristics  such as amplitude, phase, should be modified, and the process is known as modulation.

24. Define coefficient of mutual inductance of two coils. A secondary coil of n2 turns is wound on a long solenoid of area of cross section A having a primary coil of n1 turns per unit length. What is the mutual inductance of the two coils ?
Ans. Mutual inductance: Mutual inductance is numerically equal to the induced emf produced in coil when the rate of change of current is unity in the neighboring coil.
Derivation:
Let    n1 = Number of turns per unit length in first and
         n2 = Number of turns per unit length in secondary coil
       Φ
21 = M21 I1
             = N2BA
             = N2(µ0n1I1) A       ----------------(1)
⇒M21 I1 = µ0 N1 N2 I1 A / L
⇒   M21 = µ0 n1 n2 AL
Similarly M12 = n1 n2 AL
25. Find the value of unknown resistance X in the following circuit, if no current flows through the section AO. Also calculate the current drawn by the circuit from the battery of emf CV.
Ans.
As
Now 2Ω & 4Ω are connected in series and 3Ω & 6Ω are connected in series and combination is connected in parallel with battery.
26. The following data was obtained for the dependence of the magnitude of electric field with distance, from a reference point o, with in the charge distribution in shaded region.
(I) Identify the charge distribution  
(II) Identify the location points A B C and A1 B1 C1 
Ans.
E Varies inversely to cube of distance Charge distribution should be Dipole , as due to dipole E α 1/r3
Hence A,B,C lies on axis & A1,B1,C1 on Bisector.
27. A short bar magnet which is placed with its axis at 30o experiences a torque of 0.016Nm in an external field  of 800G.
(i) Determine the magnetic moment of magnet.
(ii) What is work done by an external force is moving it from its most stable to most unstable position.
Ans. Given θ = 30o ,     = 0.016Nm.
B = 800 x 10-4 T
(i) = M sinθ
/ B sinθ = 0.016/800 x 10-4 x 1/2  = 16/40 =0.4Am2
(ii) Wθ=0→θ=180 = MB(cosθ1 – cosθ2)
                       = 2MB
                       = 2e = 0.032J
SECTION-D
28. (a) Derive an expression for energy stored per unit volume in parallel plate Capacitor.
(b) A 10 µF Capacitor is charged by a 30 v d.c. Supply and then connected t across an uncharged 50 µF Capacitor. Calculate the final potential difference  across the combination and loss of energy in the process?
Ans.
(a)

A capacitor of capacitance is  charged with an external source.
The graphical Variation of charge with applied potential is shown. The work done to charge upto Maximum value QO is given by
Area of OABi.e  W = 1/2 OB x AB =  VoQo
                            =  VoCVo   
                        W =  CVo2  = U       
As    ⇒ u =    V02         Put V0 = E0 x d   
U =
Energy stored per unit volume =
(b) C1 = 10μF      C2 = 50μF
     V1 = 30V       V2 = 0
 VCom = ?          ΔU = ?
As VCommon =  5 Volt
ΔU = Loss in energy = 
=
=  3.75 x 10-5 J
29. State gauss theorem. Use gauss theorem  to obtain the expression for electric field due to linearly charged wire. Sketch the graphical variation  of electric field with distance r from wire.
Ans. Statement :- It states that the net flux linked with any closed surface will be 1/∈0 times the charge enclosed i.e Φ = q/∈0
Derivation :-
According to gauss law
      Φ = q/∈0         - - - - - - - - - (i)  
30. Explain briefly, with the help of a labeled diagram the basic principle of working of a.c. generator. In an a.c. generator coil of N turns and area A is rotated at  rotation per second in a uniform magnetic field B. Write the expression of the emf produced.
Ans. Principle: when a coil is rotated in a uniform magnetic field with a uniform angular velocity w flux linked with coil changes and emf is induced.
Induced emf e = e0 sinwf
                   e = NBA w sinwf
Let θ be angle b/w and at anytime ‘t’ = wf ,  θ = wt
As , let N = Number of turns in the coil
Φ = . for ne turn
   = N (.) = NAB cosθ
Induced emf .e =  (NABcosθ)
                      = NAB(sinwt) w
                    e = NAWBsinwt
                    e = e0 sinwt

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