ISC Physics 2012 Class-12 Solved Previous Year Question Paper
ISC Physics 2012 Class-12 Solved Previous Year Question Paper for practice. Step by step Solutions of Part I and II with section-A, B and C. Visit official website CISCE for detail information about ISC Class-12 Physics.
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Previous Year Question Paper ISC Physics 2012 Class-12 Solved
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Previous Year Question Paper ISC Physics 2012 Class-12 Solved
Maximum Marks: 70
Time allowed: 3 hours
- Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time
- Answer all questions in Part I and six questions from Part II, choosing two questions from each of the Sections A, B and C.
- All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer.
- The intended marks for questions or parts of questions are given in brackets [ ].
- Material to be supplied: Log tables including Trigonometric functions
- A list of useful physical constants is given at the end of this paper.
Part – I (20 Marks)
(Answer all questions)
( ISC Physics 2012 Class-12 Solved Previous Year Question Paper )
A. Choose the correct alternative A, B, C or D for each of the questions given below :  (i) A body has a positive charge of 8 × 10-19 C. It has :
(A) an excess of 5 electrons
(B) a deficiency of 5 electrons
(C) an excess of 8 electrons
(D) a deficiency of 8 electrons
(ii) Figure below shows five dc sources (cells). Their emfs are shown in the figure.
Emf of the battery AB is :
(A) 8 V
(B) 6 V
(C) 4 V
(D) 2 V
(iii) Which one of the following graphs in figure represents variation of reactance ‘Xc’ of a capacitor with frequency ‘f’ of an ac supply :
(iv) White light is passed through sodium vapors contained in a thin walled glass flask and the transmitted light is examined with the help of a spectrometer. The spectrum so obtained is :
(A) Absorption spectrum
(B) Solar spectrum
(C) Band spectrum
(D) Continuous spectrum
(v) Binding energy of a nucleus is of the order of:
(A) Electron volt (eV)
(B) Kilo electron volt (KeV)
(C) Mega electron volt (MeV)
(D) a joule (J)
B. Answer all questions briefly and to the point: 
(i) A point charge of 5 × 10-6 C experiences a force of 2 × 10-3 N when kept in a uniform electric field of intensity E. Find E.
(ii) Which conservation principle is involved in Kirchhoff’s first law of electric circuits ?
(iii) Write an expression of magnetic flux density ‘B’ at a point in end – on position or an axial position of a magnetic dipole. (Derivation not required.)
(iv) In a moving coil galvanometer, what is meant by a radial magnetic field ?
(v) Variation of alternating current T with time lt’ is shown in the graph below:
What is the rms value of this current ?
(vi) Which electromagnetic radiation has wavelength greater than that of X rays and smaller than that of visible light ?
(vii) How did Fresnel construct a biprism in order to study interference of light ?
(viii) State Brewster’s law of polarisation of light. ‘
(ix) A thin convex lens (L1) of focal length 80 cm and a thin concave lens (L2) of focal length f are kept co-axially, 20 cm apart as shown in figure below. When a narrow and parallel beam of light is incident on the convex lens, beam emerging from the concave lens is also a parallel beam Find f.
(x) What condition must be satisfied by two thin lenses kept in contact so that they form an achromatic doublet, i.e., a combination free from chromatic aberration ?
(xi) Threshold frequency of a certain metal for photoelectric emission is 5 × 1014 Hz. Calculate its work function.
(xii) What conclusion was drawn by Rutherford based on Geiger-Marsden’s experiment on scattering of alpha particles ?
(xiii) Write a balanced nuclear reaction showing emission of a β– particle by (Symbol of daughter nucleus formed in the process is Pa.)
(xiv) What is the essential difference between the working of a nuclear reactor and that of a fission bomb ?
(xv) State one important use of Zener diode.
(ii) The Kirchhoff’s first law of electric circuit is based upon the principle of conservation of electric charge which implies that at any node (junction) in an electrical circuit, the sum of current moving into that node is equal to the sum of the current flowing out of that node.
(iii) Expression for the magnetic flux density ‘B ’ at point in or an axial position of a magnetic dipole can be written as
(due to a short dipole)
where M is the moment of magnet, r is the distance of the point from mid-section of magnet.
(iv) In a moving coil galvanometer, the coil is suspended between the concave pole pieces. The magnetic field is thus directed along the radius of the concave pole pieces. This magnetic field is called radial magnetic field. The advantage is that in all positions of the coil, the magnetic field is parallel to plane of the coil.
(v) It is clear from the graph that peak current of ac is given by I0 = 5√2 A
(vi) Ultraviolet ray has a wavelength greater than X rays and smaller than that of visible light.
(vii) Biprism was constructed by the combination of two prisms of very small refracting angles, placed base to base. In practice, the prism is made from a single plate by grinding and polishing, so that it is a single prism with one of its angles about 179° and other two about 30° each.
(viii) Brewster’s states that for a given medium of refractive index μ the angle of polarisation is related as μ = tan ip, where ip is angle of polarisation. Then refractive index of a medium is equal to the tangent of the angle of polarisation.
(ix) It is clear from the figure that in the absence of the lens L2, the rays from infinity will come to focus at a distance of 80 cm from the lens L1. The image at serves as a virtual object for the concave lens and final image is formed at ∞.
∴ For the concave lens
i.e., the ratio of the dispersive power of the materials of the two lenses is the same as the ratio of their focal length. Further, one lens in the combination must be concave and the two lenses must be of different material.
(xii) The following conclusions were drawn by Rutherford from scattering of α-particles :
(1) The whole of +ve charge is concentrated in the nucleus of the atom. This is so because most of a-particles pass straight through the nucleus. Since only a very small fraction of the a-particles is scattered, nucleus must be small.
(2) The whole mass of the atom is also concentrated in a small region.
(3) Nucleus is surrounded by a cloud of electrons whose total – ve charge is equal to the total +ve charge on the nucleus.