CBSE Question Paper class 12 Physics

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CBSE Question Paper class 12 Physics conducted by Central Board of Secondary Education, New Delhi in the month of March 2018-19. CBSE previous year question papers with solution are available in myCBSEguide mobile app and cbse guide website. The Best CBSE App for students and teachers is myCBSEguide which provides complete study material and practice papers to cbse schools in India and abroad.

CBSE Question Paper 2018 Physics

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Class 12 Physics chapters wise list

Electric Charges and Fields
Electrostatic Potential and Capacitance
Current Electricity
Moving Charges and Magnetism
Magnetism and Matter
Electromagnetic Induction
Alternating Current
Electromagnetic Waves
Ray Optics and Optical Instruments
Wave Optics
Dual Nature of Radiation and Matter
Atoms
Nuclei
Semiconductor Electronic: Material, Devices and Simple Circuits
Communication Systems

CBSE Question Paper class 12 Physics

Time allowed: 3 hours
Maximum Marks: 70

General Instructions :

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 :
$c = 3 \times {10^8} m/s$
$h = 6.63 \times {10^{ - 34}} Js$
$e = 1.6 \times {10^{ - 19}}C$
${\mu _0} = 4\pi \times {10^{ - 7}}\; Tm{A^{ - 1}}$
${\varepsilon _0} = 8.854 \times {10^{ - 12}}{C^2}{N^{ - 1}} {m^{ - 2}}$
$\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9} N{m^2}{C^{ - 2}}$
Mass of electron (m_e) $= 9.1 \times {10^{ - 31}}kg$
Mass of neutron $= 1.675 \times {10^{ - 27}}kg$
Mass of proton $= 1.673 \times {10^{ - 27}}kg$
Avogadro’s number $= 6.023 \times {10^{23}}$ per gram mole
Boltzmann constant = $= 1.38 \times {10^{ - 23}}J{K^{ - 1}}$

Section A

A proton and an electron traveling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency?
Ans. Electron
Name the electromagnetic radiations used for (a) water purification, and (b) eye surgery.
Ans.(a) Ultra violet rays (b) Ultra violet rays / Laser
Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.
Ans.
Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two – the parent or the daughter nucleus – would have higher binding energy per nucleon?
Ans. Daughter nucleus
Which mode of propagation is used by short wave broadcast services?
Ans. Skywave propagation

Section B

Two electric bulbs P and Q have their resistances in the ratio of 1: 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs.
Ans.Power = I²R
The current, in the two bulbs, is the same as they are connected in series.
$\therefore \frac{{{P_1}}}{{{P_2}}} = \frac{{{I^2}{R_1}}}{{{I^2}{R_2}}} = \frac{{{R_1}}}{{{R_2}}}$
A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 $\Omega$ as shown in the figure. Find the value of current in the circuit.
ORIn a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 $\Omega$ is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.
Ans. By Kirchoff‟s law, we have, for the loop ABCD, +200 – 38i– 10 = 0
$\therefore i = \frac{{190}}{{38}}A = 5A$
Alternatively:
The two cells being in ‘opposition’,
$\therefore$ net $\varepsilon mf$ = (200 – 10)V = 190 V
Now $I = \frac{V}{R}$
$\therefore I = \frac{{190V}}{{38\Omega }} = 5A$
[Note: Some students may use the formulae, $\frac{\varepsilon }{r} = \frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}$ and $r = \frac {{r_1}{r_2}} {{r_1} + {r_2}}$
For two cells connected in parallel
They may then say that r = 0;
$\varepsilon$ is indeterminate and hence
I is also indeterminate
Award full marks(2) to students giving this line of reasoning.]ORWe have $r = \left( {\frac{{{l_1}}}{{{l_2}}} - 1} \right)R$ $= \left( {\frac{{{l_1} - {l_2}}}{{{l_2}}}} \right)R$
$CBSE Question Paper class 12 Physics$ $= \frac{{50}}{{300}} \times 9\Omega = 1.5\Omega$
1. Why are infra-red waves often called heat waves ? Explain.
2. What do you understand by the statement, ‘‘Electromagnetic waves transport momentum’’?
Ans.
1. Infrared rays are readily absorbed by the (water) molecules in most of the substances and hence increases their thermal motion.
  (If the student just writes that “infrared ray produce heating effects”, award ½ mark only)
2. Electromagnetic waves can set (and sustain) charges in motion. Hence, they are said to transport momentum.
  (Also accept the following: Electromagnetic waves are known to exert „radiation pressure‟. This pressure is due to the force associated with rate of change of momentum. Hence, EM waves transport momentum)
If light of wavelength 412·5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?
Metal Work Function
Na 1.92
K 2.15
Ca 3.20
Mo 4.17
Ans. The energy of a photon of incident radiation is given by
$E = \frac{{hc}}{\lambda }$
$\therefore E = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{(412.5 \times {{10}^{ - 9}}) \times (1.6 \times {{10}^{ - 19}})}}eV$
$\cong 3.01eV$
Hence only Na and K will show photoelectric emission
[Note: Award this ½ mark even if the student writes the name of only one of these metals] Reason: The energy of the incident photon is more than the work function of only these two metals.
A carrier wave of peak voltage 15 V is used to transmit a message signal. Find the peak voltage of the modulating signal in order to have a modulation index of 60%.
Ans. We have
$CBSE Question Paper class 12 Physics$
Here $\mu = 60\% = \frac{3}{5}$
$\therefore {A_m} = \mu {A_c} = \frac{3}{5} \times 15V$
= 9V

Metal	Work Function
Na	1.92
K	2.15
Ca	3.20
Mo	4.17

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Last Year Question Paper Class 12 Physics 2018

Download class 12 Physics question paper with solution from best CBSE App the myCBSEguide. CBSE class 12 Physics question paper 2018 in PDF format with solution will help you to understand the latest question paper pattern and marking scheme of the CBSE board examination. You will get to know the difficulty level of the question paper. CBSE question papers 2018 for class 12 Physics have 26 questions with solution

Previous Year Question Paper for class 12 in PDF

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CBSE Question Paper class 12 Physics

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Class 12 Physics chapters wise list

CBSE Question Paper class 12 Physics

Section A

Section B

Last Year Question Paper Class 12 Physics 2018

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