CBSE Class 12 Physics Solved Question Paper: 2016

CBSE Class 12 Physics Solved Question Paper 2016 is available here for download in PDF format. This Solved Paper is very important for the students preparing for CBSE Class 12 Physics board exam 2018. In this article, you will also find some important links to other articles important for the preparation of CBSE board exam 2018.

Created On: Nov 3, 2017 12:12 IST CBSE Class 12 Physics Solved Question Paper: 2016

CBSE Solved Question Paper for Class 12th Physics 2016 board exam is available here. CBSE Class 12 Physics 2016 board exam was held on 05 March 2016. If you want to learn about the analysis and review of CBSE Class 12 Physics 2016 board paper then please follow the link given below

CBSE Class 12 Board Exam 2016: Physics Paper Analysis

With the help of this CBSE Class 12 Physics 2016 board exam solved question paper, students can easily understand the level of questions which are asked in CBSE Class 12th Physics board exam. Solutions provided in this papers are unique and to the point.

With these solutions, students can easily understand the correct way of writing answers in CBSE12th Physics board exam answer booklet.

CBSE Class 12 Physics Sample Paper: 2018

CBSE Class 12 Physics Solved Question Paper 2016 is given below:

Solved Question Paper

2016

Class ‒ XII

Subject ‒ Physics

All India: Set ‒ III

Time allowed: 3 hours                                                                                 Maximum Marks: 70

General Instructions:

(a) All questions are compulsory.

(b) There are 26 questions in all. Question number 1 to 5 carry one mark each, question number 6 to 10 carry two marks each, question number 11 to 22 carry three marks each, question number 23 carry four marks and question number 24 to 26 carry 5 marks each.

(c) There is no overall choice. However, internal choices have been provided in some questions.

(d) 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 CN‒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

Question1: Why does sun appear red at sunrise and sunset?

Sol:

At the time of sun rise and sun set, light from the sun has to pass maximum distance through the earth’s atmosphere. As, wavelength of red colour is maximum in comparison of other colours of visible spectrum and amount of scattering is inversely proportional wavelength of light, therefore red colour is scattered least and enter our eyes.

CBSE Class 12 Syllabus 2017-2018: All Subjects

Question2: Name the essential components of a communication system.

Sol:

Essential components of a communication system are:

Information source, transmitter, communication channel, receiver,

Question3: In what way is the behavior of a diamagnetic material different from that of a paramagnetic, when kept in an external magnetic field?

Sol:

When diamagnetic substances are placed in external magnetic field, are feebly magnetized opposite to the direction of magnetic field.

When paramagnetic substances are placed in external magnetic field, are feebly magnetized in the direction of magnetic field.

Question4: The plot of the variation of Potential difference across a combination of three identical cells in series, versus current is shown below. What is the emf and internal resistance of each cell?

Sol:

Let E and r be the emf and internal resistance of the combination of cell and E’ and r’ be the emf and internal resistance of each cell.

If V is the terminal potential difference then,

V = EI r …(i)    [where I is the current]

When, I = 0 then E = V,

From the graph, when I = 0, E = 6 V.

As, there are three cells so,

E = 3 E’E’ =2 V.

When, V = 0 then, E = I rr = E/I r =6/1 = 6 Ω.

It is given that cells are connected in series, therefore the combined resistance of the cells is given by: r = r/3 ⇒ r’ = 6/3 = 2 Ω.

Therefore, emf of each cell is 2 V and internal resistance of each cell is 2 Ω.

Question5: A charge 'q' is moved from a point A above a dipole of dipole moment ‘p’ to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.

Sol:

Net potential (ΔV) at any point on equatorial plane of a dipole is zero.

As, AB lies in the equatorial plane of the dipole, therefore the work done in moving charge (q) from A to B without acceleration is q(ΔV) = 0.

Question6: A ray PQ incident normally on the refracting face BA is refracted in the prism BAC made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray emerge? Justify your answer.

Sol:

Question7: Calculate the de-Broglie wavelength of the electron orbiting in the n = 2 state of hydrogen atom.

Sol:

Question8: Define modulation index. Why is it kept low? What is the role of a bandpass filter?

Sol:

Modulation Index indicates the depth of modulation. The modulation index (μ) of an amplitude modulated wave is defined as the ratio of the amplitude of modulating signal (Am) to the amplitude of carrier wave (Ac) i.e., μ = Am/Ac.

In order to avoid distortion modulation index is always kept low (less than 1).

A band pass filter rejects low and high frequencies and allows a band of frequencies to pass through.

Question9: A battery of emf 12V and internal resistance 2 Ω is connected to a 4 Ω resistor as shown in the figure.

(a) Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading.

(b) To record the voltage and the current in the circuit, why is voltmeter placed in parallel and ammeter in series in the circuit?

Sol:

(a) If V is the potential difference of the cell and E is the total emf of the cell, r is the internal resistance of the cell and I is the current drawn from cell then, V = EIr.

Here, E = 12 V, r = 2 Ω, ∴ V = 12 ‒ 2I … (i)

When voltmeter is connected across the cell then, V =12

On ideal voltmeter does not draw any current from the circuit,

Applying Kirchhoff’s law, we have

12 ‒ 4I ‒ 2I = 0 ⇒ I = 2 A …(ii)

When voltmeter is connected across the cell then the reading in voltmeter will be

V1 = 12 ‒ 2 (2) ⇒ V1 = 8 V.

Now when voltmeter will be connected across the resistor then,

V2 = IR = 2 × 4 = 8 V.

Clearly both the positions will give same reading.

(b) A voltmeter has very high resistance, due to which it is placed parallel to the load so that it does not affect the current flowing through the circuit.

Ammeter has very little resistance due to which it is connected in series to the circuit so that all the current flows through it and it can measures the value of current accurately.

Question10: Define ionization energy.

How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge?

Sol:

Ionization energy is the minimum amount of energy required to remove an electron from the outermost orbit of a neutral atom in its ground state.

(Ionization energy of hydrogen atom) ∝ (mass of electron), and mass of electron is constant.

But here, electron in hydrogen atom is replaced by a particle having mass 200 times of electron.

Therefore, ionization energy will be increase 200 times.

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