CBSE Class 12 Physics Solved Paper: 2017

Solved question paper of CBSE 12th Physics board exam 2017 is available here for download in PDF format. Students can also download the complete paper with the help the download link given at the end of this article. CBSE Class 12th Physics 2017 board exam was held on 15th March.

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Completely solved CBSE 12th Physics 2017 board exam paper is given below

Guidelines given in the first page of CBSE Class 12 Physics Question Paper

General guidelines given in the paper:

• Please check that this question paper contains 16 printed pages.

• Code number given on the right hand side of the question paper should be written on title page of the answer-book by the candidate.

• Please check that this question paper contains 26 questions.

• Please write down the Serial Number of the question paper before attempting it.

• 15 minutes time has been allotted to read this paper. The question paper will be distributed at 10:15 a.m. to 10:30 a.m., the student will read the question paper only and will not write any answer on the answer-book during this period.

General Instructions given CBSE Class 12 Physics Question Paper 2017:

 All questions are compulsory. There are 26 questions in all. This question paper has five sections - Section A, Section B, Section C, Section D and Section E. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each. There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions. You may use the following values of physical constants wherever necessary: c = 3 × 108 m/s h = 6.34 × 10‒34 Js e = 1.6 × 10‒19 C μo = 4π × 10‒7 T m A‒1 εo = 8.854 × 10‒12 C2 N‒1 m‒2 {1/(4π εo)} = 9 × 109 N m2 C‒2 Mass of electron = 9.1 ×10−31 kg Mass of neutron = 1.675 ×10−27 kg Mass of proton = 1.673 ×10−27 kg Avogadro's number = 6.023 ×1023 per gram mole Boltzmann constant = 1.38 ×10−23 JK−1

Some randomly selected questions from the paper are given below

Question: What is the relation between electric current (I) and drift velocity (vd) of electrons?

Sol:

Let n be the number density of free electrons in a conductor of length l and area of cross-section

A’, the drift velocity is given by, I = n e A vd where, e is the magnitude of charge on electrons.

Question: Give two examples of Paramagnetic and Ferromagnetic material?

Sol:

Ferromagnetic material: Nickel, Cobalt

Paramagnetic material: Platinum, Aluminum.

Question: Name the network formed by joining 4 to 6 computers in an office?

Sol:

Local Area Network (LAN).

Question: In a CE transistor amplifier, there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy?

Sol:

In case of CE transistor amplifier, the power gain is very high. In this type of circuit, the extra power required for amplified output is obtained from DC source. Thus, the circuit used does not violet the law of conservation.

Question: The refractive index of a medium with respect to air is √3. What is the angle of refraction, if the unpolarized light is incident on it at the polarizing angle of the medium?

Sol:

Given, refractive index of the medium = μ = √3.

Let angle of incident be ip and angle of refraction be r.

We know that, μ = tanior ip = tan-1μ,

Therefore, ip = tan-1(√3) or ip = 60o.

Therefore, r = 90o – 60o = 30o.

Question: What is attenuation? Name three different types of modulations used for a message signal using a sinusoidal continuous carrier wave.

Sol:

It is the loss of strength of a signal while propagating through a medium.

A sinusoidal carrier wave can be modulated in three ways:

(i) Amplitude modulation

(ii) Frequency modulation

(iii) Phase modulation.

Question: Two polaroids are oriented in such way that the intensity of the light transmitted by them is maximum. To what percentage of its maximum value is the intensity of transmitted light reduced when one of the polaroids is rotated through (i) 30o (ii) 60o.

Sol:

The two polaroids are parallel to each other initially. If I1 is the intensity of light from the first polaroid, then the intensity of transmitted light by the second is: I = Io cos2 θ = Io cos 0o = Io.

(i) When θ = 30o then, the transmitted intensity is I = Io cos2 30o = Io (√3 / 2) = 0.75 Io.

Which means transmitted intensity is reduced to 75 % of the maximum intensity.

(ii) When θ = 60o then, the transmitted intensity is I = Io cos2 60o = Io (1 / 2)2  = 0.25 Io.

Which means transmitted intensity is reduced to 25 % of the maximum intensity.

Question: Define the terms ‘threshold frequency’ and ‘stopping potential’ in the study of photoelectric emission. Explain briefly the reasons why wave theory of light is not able to explain the observed features in photoelectric effect.

Sol:

Threshold Frequency: The minimum frequency of incident light which is just capable of ejecting electrons from a metal. It is denoted by vo.

Stopping Potential: The minimum retarding potential applied to anode of a photoelectric tube which is just capable of stopping photoelectric current.

The observed characteristics of photoelectric effect could not be explained on the basis of wave theory of light due to the following reasons.

(i) According to wave theory, the light propagates in the form of wavefronts and the energy is distributed uniformly over the wavefronts. With increase of intensity of light, the amplitude of waves and the energy stored by waves will also increase. These waves will then, provide more energy to electrons of metal, consequently the energy of electrons will also increase.

Therefore, according to wave theory, the kinetic energy of photoelectrons must depend on the intensity of incident light; but according to experimental observations, the kinetic energy of photoelectrons does not depend on the intensity of incident light.

(ii) According to wave theory, the light of any frequency can emit electrons from metallic surface provided the intensity of light be sufficient to provide necessary energy for emission of electrons, but according to experimental observations, the light of frequency less than threshold frequency can not emit electrons irrespective of the intensity of incident light.

(iii) According to wave theory, the energy transferred by light waves will not go to a particular electron, but it will be distributed uniformly to all electrons present in the illuminated surface.

Therefore, electrons will take some time to collect the necessary energy for their emission.

The time for emission will be more for light of less intensity and vice versa. But experimental observations show that the emission of electrons take place instantaneously after the light is incident on the metal irrespective of the intensity of the light.

Question 17: Identify the type of waves which are produced by the following way and write one application for each:

(i) Radioactive decay of the nucleus,

(ii) Rapid acceleration and decelerations of electrons in aerials,

(iii) Bombarding a metal target by high energy electrons.

Sol:

 Radioactive decay of the nucleus Gamma rays Treatment of tumors Rapid acceleration and decelerations of electrons in aerials Radio waves Radio and television Communication systems Bombarding a metal target by high energy electrons X- rays Study of crystals

Question: Sunita and her friends visited an exhibition. The policeman asked them to pass through a metal detector. Sunita’s friends were initially scared of it. Sunita, however, explained to them the purpose and working of the metal detector.

Answer the following questions:

(a) On what principle does a metal detector work?

(b) Why does the detector emit sound when a person carrying any metallic object walks through it?

(c) State some qualities which Sunita displayed while explaining the purpose of walking through the detector.

Sol:

(a) Metal detector works on the principle of resonance in AC circuits.

(b) When a person walks through the gate of a metal detector, the impedance of the circuit changes, resulting in significant change in current in the circuit that causes a sound to be emitted as an alarm.

(c) Some of the qualities are: Responsible citizen, knowledgeable person.

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