II Puc Preparatory Examination Physics Question paper with solutions 2023

II Puc Preparatory Examination Physics Question paper with solutions 2023


I. Pick the correct option among the four given options for all the following questions 15×1=15

1. When an insulator is charged

a. The charge remains localised

b. The charges are distributed over the surface

c. An insulator cannot be charged

d. Either (a) or (b) depending upon the shape of the insulator

Answer : (a)
Reason : In an insulator charge doesn’t redistribute on surface . It remains on the same spot where it was charged.

2. Electric potential inside a conducting sphere

a. is zero

b. increases from centre to the surface

c. decreases from centre to the surface

d. remains constant from centre to the surface

Answer : (d)

3. A medium of dielectric constant ‘k’ is introduced between the plates of parallel plate capacitor, As a result its capacitance

a. decreases k times

b. increases k times

c. decreases 2k times

d. remains unchanged

Answer : (b)

4. Ohms law holds good for

a. All conductors

b. metallic conductors

c. semiconductors

d. ionic conductors

Answer : (b)

5. A galvanometer can be converted into voltmeter by connecting

a. a small resistance in series

b. a high resistance in series

c. a small resistance in parallel

d. a high resistance in parallel

Answer : (b)

6. Substances which are strongly attracted by magnets are

a. diamagnetic

b. ferromagnetic

c. paramagnetic

d. none of these

Answer : (b)

7. The direction of induced emf is given by

a. flemings left hand rule

b. flemings right hand rule

c. biot-sarvarts law

d. lenz’s law

Answer : (b) or (d)

8. The principle behind the working of transformer

a. mutual induction

b. self induction

c. electromagnetic induction

d. Eddy currents

Answer : (a)

9. In a series RLC circuit, the frequency of ac is varied, at the resonant angular frequency ωo, the impedance of the circuit is

a. R

b. less than ωoL

c. more than ωoL

d. between ωo and 1/ωoL

Answer : (a)

10. Which of the following EM waves are used in telecommunication

a. α rays

b. UV rays

c. microwaves

d. IR rays

Answer : (c)

11. The sky appears blue because

a. red light is absorbed

b. blue light is scattered

c. blue light is absorbed

d. it is due to its natural colour

Answer : (b)

12. When light is incident at the polarising angle, which of the following is completely polarised

a. refracted ray

b. reflected ray

c. both reflected and refracted rays

d. neither refracted nor reflected rays

Answer : (b)

13. The de-broglie wavelength of an electron and proton are same. which quantity will be same for both

a. kinetic energy

b. velocity

c. momentum

d. accelerating potentials

Answer : (c)

14. Radioactive substance emits

a. α rays

b. β rays

c. γ rays

d. all of the above

Answer : (d)

15. Which of the following is a universal logic gate

a. NOT gate

b. AND gate

c. OR gate

d. NAND gate

Answer : (d)

Fill in the blanks by choosing answer given in brackets for all the questions
(Cadmium rods, Germanium, Electric field Intensity temperature, Diameter) 5×1=5

16. Force per unit charge is known as Electric Field Intensity.

17. The susceptibility of diamagnetic material is independent of Temperature.

18. The resolving power of telescope varies directly with Diameter of the objective.

19. In the nuclear reactors Cadmium Rods are used to absorb the neutrons.

20. Germanium is an intrinsic semiconductor.


Answer any five of the following questions 5×2=10

21. On what factors does the capacitance of the capacitance depend?

Answer : Factors affecting capacitance are
a. Area of the capacitor plates
b. Distance between the plates
c. Permittivity of the medium

22. Draw a neat labelled diagram of cyclotron?

Answer :

23. State and explain curie’s law in magnetism?

Answer : In a paramagnetic material the magnetization of the material is directly proportional to an applied magnetic field. However, if the material is heated, the magnetization is inversely proportional to temperature. i.e. M = C x ( B / T ) where, M is the magnetisation B is the magnetic field, measured in teslas T is absolute temperature, measured in kelvins C is a material-specific Curie constant It was first discovered experimentally by French physicist, Pierre Curie. It holds for high temperatures, or weak magnetic fields.

24. What is an AC generator? On what principle does it work?

Answer : AC generator is a machine that converts mechanical energy into electrical energy.
Principle : Electromagnetic Induction

25. Distinguish between inductive and capacitive reactance?

Answer : Inductive reactance – a. It is opposition offered to current due to an inductor
b. It increases with increase in frequency

Capacitive reactance – a. It is opposition offered to current due to an capacitor
b. It decreases with increase in frequency

26. Mention any two uses of UV rays?

Answer : killing bacteria, creating fluorescent effects, curing inks and resins, phototherapy and suntanning

27. What is the shape of wavefront from (a) Linear source (b) a point source at a large distance.

Answer : a. Cylindrical b. Plane

28. Explain briefly the observations of gieger marsden’s experiment on scattering of alpha particles by a nucleus.

Answer : a. most of the alpha particles are passing through gold foil undeflected
b. few of the alpha particles passes through gold foil with large angle deviation
c. and, very few alpha particles retraces there path.

29. Mention any two properties of nucleus.

Answer : (1) Atomic Number (Z)- It represents the total number of protons inside the nucleus.
(2) Mass Number (A)- It represents the total number of protons and neutrons inside the nucleus.
(3) Nuclear Mass- It is expressed in terms of atomic mass unit (amu)


Answer any five of the following questions 5×3=15

30. Mention nay three properties of electric field lines.

Answer : Electric field lines are outwards frm positive and inwards to negative charge
2. They never intersect
3. They never form closed loops
4. Always perpendicular to the surface of the conductor

31. Arrive at J = σE.

Answer :

32. With a circuit diagram. Explain how an galvanometer can be converted into an ammeter?

Answer :

33. Define the terms (a) magnetic intensity (b) magnetic susceptibility (c) magnetic permeability

Answer : a. Magnetic intensity is used to represent Magnetic field strength in a space. It is the force experience by a unit positive charge moving with a velocity of 1 m/s in magnetic field at an angle 90 with magnetic field
b. Magnetic susceptibility is the degree to which a material can be magnetized in an external magnetic field. If the ratio between the induced magnetization and the inducing field is expressed per unit volume
c.Magnetic permeability is defined as the ratio of the magnetic induction to the magnetic intensity. It is a scalar quantity and denoted by the symbol μ.

34. Derive an expression for motional emf induced in rod moving in magnetic field.

Answer :

35. Draw the ray diagram for the formation of image by a simple microscope. Write the expression for magnification produced by the miroscope for image formed at the least distance of distinct vision.

Answer :

36. Arrive at the expression for radius of nth orbit of H2 atom.

Answer :

37. The half life of radioactive element is 4×10^8 years. calculate its decay constant and mean life.

Answer :

38. Write any 3 three differences between n-type and p-type semiconductors.

Answer : N-type – a. An N-Type Semiconductor can be obtained by adding a pentavalent impurity such as Phosphorus
b. Four valence electron of phosphorus forms four bonds with the silicon atom and there is one valence electron which is left free;
c. Therefore, the number of electrons exceeds the number of holes which increases the conductivity
d. The majority charge carriers are electrons;

P-type semiconductors – a. An P-Type Semiconductor can be obtained by adding a trivalent impurity such as Aluminium
b. Tree valence electron of silicon forms three bonds with the Aluminium atom and there is one vacancy which is called as hole;
c. Therefore, the number of holes exceeds the number of electrons which increases the conductivity,
d. The majority carriers are holes.


Answer any three of the following questions 3×5=15

39. Obtain the expression for electric field at a point on equitorial line of an electric dipole.

Answer : Let us consider a dipole AB with charges +q and -q separated by a distance 2a , as shown in the figure. Now, let us find electric field at point P which is at a distance r from the center of the dipole along the Equatorial line.
After Resolving the Electric Field the sine components get cancelled and cos components gets added up.

40. Derive the expressions for equivalent emf and equivalent internal resistance when two cells are in parallel.

Answer :

Let us consider two cells with EMF E1,E2 and internal resistance r1,r2 connected in Parallel as shown in figure (a)
Figure (b) represents an Equivalent Circuit where the two cells are replaced with a single cell of EMF Eeq and internal resistance req

41. Deduce the expression for magnetic field at a point on the axis of a circular loop carrying current using biot-savart’s law.

Answer : Let us consider current carrying circular loop with radius ‘R’ carring Current ‘I’ as shown in the figure. Let us find magnetic field at a point P which is at a distance ‘r’ from centre of loop along the axis

42. Derive the expression for refractive index of the material of prism interms of the angle of the prism and angle of minimum deviation.

Answer : Let us consider a prism with refractive index n and angle of prism A kept in air as shown in figure

43. (i) Mention the Hallwach’s and Lenard observations on photoelectric effect.
(ii) Mention an two properties of photon.

Answer : (i) a. The photoelectric current (same as the rate of emission of electrons) is directly proportional to the intensity of light falling on the electrode. Note from the figure below that with increasing intensity, the current increases. Also, observe that as the voltage has decreased, the current also decreases. But to obtain zero current, the voltage has to be reversed to a certain V­0 known as the stopping potential. The voltage must be reversed so that the electrons cannot reach the anode. This is the maximum kinetic energy an emitted electron can achieve, Maximum Kinetic energy, KE = eVo (e is the charge of the electron).

b. The maximum kinetic energy increases with an increase in the frequency of light. With a higher frequency of light (ν), the stopping potential becomes more negative, which implies that the kinetic energy of electrons also increases.

c. All frequencies of light, however, cannot cause a photoelectric current to develop. Only light above a certain frequency (ν0) can produce a photoelectric current. This frequency is known as the threshold frequency. This varies with the electrode material. Also, the maximum kinetic energy of the electrons increases linearly with increasing light frequency. If we extend the graph below the x-axis, the intercept on the Kinetic energy axis represents the minimum energy required for the emission of the electron; this is known as the material’s work function.

d. Lastly, the electron emission occurs instantly without any time lag.

(ii) a. The rest mass of photon is zero.
b. The energy of photon is E=hc/λ
c. The momentum of photon is given by p=h/λ
d. Photon travels with speed of light

44. (i) What is rectification?
(ii) Write the circuit diagram and input and output waveforms for halfwave rectifier
(iii) Explain the working for half wave rectififer.

Answer : (i) The process of converting AC into DC is called Rectification. The electronic circuit used to convert AC to DC is called Rectifier.


(iii) a. It basically consist of an AC source connected to a transformer T Which is then connected to a diode D. Output is taken across the load resistance RL.
b. During the positive half cycle the diode D is forward biased and hence its is conducting. Therefore, there is output across the load resistance.
c. During the negetive half cycle the diode D is reversed biased and hence it is not conducting. Therefore, there is no output across the load resistance.

Answer any two of the following questions 2×5=10

45. Two capacitors of capacitances 2μF and 4μF are connected in series across a 4V battery. What is the potential difference across each capacitor. Also calculate the total energy stored in combination.

Answer :

46. Four resistance of 12, 15, 10 and 4 ohms respectively, are connected in cyclic order to form a wheatstone’s network. Is the network balanced? f not, calculate the resistance to be connected in parallel with resistance of 10 ohm to balance the network.

Answer :

47. A coil has a inductance of 0.2 H and resistance of 40 ohm. It is connected to a AC source of 220V, 50 Hz. Calculate the power dissipated in circuit.

Answer :

48. A beam of light consisting of two wavelengths 6500 A and 5200 A is used to obtain interference fringes in a youngs double slit experiment. What is the least difference from central maximum Where the bright fringes due to both the wavelengths coincide? given the distance between the slits in 2mm and the distance from slits is 1.2m.

Answer :

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