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CBSE Class 12 Physics Term 2 Sample Paper with Answer Key

Class 12 Physics Term 2 Sample Paper

CBSE Class 12 Physics Term 2 Sample Paper with Solutions: Physics is a conceptual subject and students have to understand the concepts of physics in order to solve the numerical problems of physics. The CBSE has announced the CBSE Term 2 Exam dates. The Central Board of Secondary Education is going to conduct CBSE Term 2 Exam 2022 from 26th April 2022 for Class 10th and 12th. The students must solve sample papers to assess their preparation.

Class 12 Physics Term 2 Answer key

Class 12 Physics Term 2 Answer Key: The Central Board of Secondary Education is all set to conduct CBSE Class 12 Physics Term 2 Exam 2022 today on 20th May 2022. We have covered the CBSE Class 12 Physics Term 2 Answer Key on this page in detail. The students will get 2 hours to solve the CBSE Class 12 Physics Term 2 question paper. The Physics exam will conclude at 12:30 pm, after completion of the examination we will update the CBSE Answer Key of Physics Term 2 on this page. The students appearing in the examination can match their answers with the unofficial CBSE Class 12 Physics Term 2 Answer Key 2022 prepared by the expert facilities of Adda247.

Class 12 Physics Term 2 Answer key & Paper Solution

Class 12 Physics Term 2 Question Papers for Answer key

Paper Code: 55/4/1

Paper Code: 55/4/2

Class 12 Physics Term 2 Answer key: Paper Code 55/4/1

Q.1 What is meant by the Energy bandgap in a solid? Draw the energy band diagrams for a conductor, an insulator and a semiconductor.

Energy bands : In a solid , the energy of electrons lie within certain range. The energy levels of allowed energy are in the form of bands, these bands are separated by regions of forbidden energy called band gaps.

Q.2: Name the device which converts an ac input signal into a dc output signal. Write the principle of working of the device.

  • rectifier is a device that converts an oscillating two-directional alternating current (AC) into a single-directional direct current (DC).
  • Rectifiers can take a wide variety of physical forms, from vacuum tube diodes and crystal radio receivers to modern silicon-based diode.
  • There are two types of rectifiers: Half wave rectifier and full-wave rectifier.

1. Half-wave rectifiers work by eliminating one side of the AC, thereby only allowing one direction of current to pass through.

  • Since half of the AC power input goes unused, half-wave rectifiers produce a very inefficient conversion.

2. Full-wave rectifier eliminates AC from both the directions hence it is more efficient.

  • Rectifier mostly used in Power supply, inverters etc.

Q.3: (a) Name the spectral series for a hydrogen atom which lies in the visible region. Find the ratio of the maximum to the minimum wavelengths of this series. 

  • Hydrogen Spectrum and Spectral series: When a hydrogen atom is excited, it returns to its normal unexcited (or ground state) state by emitting the energy it had absorbed earlier.
  • This energy is given out by the atom in the form of radiations of different wavelengths as the electron jumps down from a higher to a lower orbit.
  • The transition from different orbits causes different wavelengths, these constitute spectral series which are characteristic of the atom emitting them.
  • When observed through a spectroscope, these radiations are imaged as sharp and straight vertical lines of a single color.
  • Mainly there are five series and each series is named after it’s discovered as Lyman series (n1 = 1), Balmer series (n1​ = 2), Paschen series (n1​ = 3), Bracket series (n1​ = 4), and Pfund series (n1​ = 5).
  • According to Bohr’s theory, the wavelength of the radiations emitted from the hydrogen atom is given by
    1λ=RZ2[1n12−1n22]

where

n2 = outer orbit (electron jumps from this orbit),

n1 = inner orbit (electron falls in this orbit),

Z = atomic number

R = Rydberg’s constant.

According to the equation of Balmer series
1/λ=R(1/n12−1/n22)

=λmin/λmax=(1/22−1/32)/(1/22−1/∞2)

=5/9

Hence, The ratio of minimum to maximum wavelength in the Balmer series is 5: 9.

OR

3. b) What are matter waves? A proton and an alpha particle are accelerated through the same potential difference. Find the ratio of the de Broglie wavelength associated with the proton to that with the alpha particle. 

What are matter waves: The wave associated with each moving particle is known as a matter-wave. The matter-wave has a wavelength of λ=hp, where h is the Planck’s constant and p is the moment of the moving particle.

Some of the characteristics of matter waves are as follows:

  • The faster the particle moves, the smaller is its De-Broglie wavelength.
  • Lighter the particle, greater is the De-Broglie wavelength.

By using the formula given above the ratio of De-Broglie wavelength is 2:1.
Section B
Q.4(a) Differentiate between nuclear fission and nuclear fusion.
Nuclear Fission Nuclear Fusion
When the nucleus of an atom splits into lighter nuclei through a nuclear reaction the process is termed nuclear fission. Nuclear fusion is a reaction through which two or more light nuclei collide with each other to form a heavier nucleus.
When each atom split, a tremendous amount of energy is released The energy released during nuclear fusion is several times greater than the energy released during nuclear fusion.
Fission reactions do not occur in nature naturally Fusion reactions occur in stars and the sun
Little energy is needed to split an atom in a fission reaction High energy is needed to bring fuse two or more atoms together in a fusion reaction
Atomic bomb works on the principle of nuclear fission Hydrogen bomb works on the principle of a nuclear fusion bomb.

(b) Deuterium undergoes fusion as per the reaction :

Find the duration for which an electric bulb of 500 W can be kept glowing by the fusion of 100 g of deuterium. 

The energy released per atom

Number of atoms in 100g deuterium

6.023*10^23*100/2

Total energy released- 3.27/2 MeV

Time required=total energy / power of lamp

= 500 x 60 x 60 x 24 x 365

= 1.5768 x 10^10 years

Q.6: Answer the following, giving reason: 

(a) The resistance of a p-n junction is low when it is forward biased and is high when it is reversed biased.

Forward Biased: The effective barrier potential reduces i.e. () where  = barrier potential initially, V = forward voltage applied. Therefore, the thickness of depletion layer also decreases. The junction resistance becomes very low. The holes from p region move to n-side and electrons from n-side move towards p-side. The movement of holes and electrons constitute hole current () and diffusion electron (). Total current .

Reverse Bias: When a voltage V (i.e. reverse bias voltage) is applied to a circuit. The effective barrier potential increases, it becomes (). Also the thickness of depletion layer increases. The junction resistance increases. The majority of carriers in p and n-regions respectively are attracted by negative and positive terminals of the battery. Therefore, they cannot move across the junction. There is a small saturation current due to the sweep of minority carrier in p and n-regions.

(b) Doping of intrinsic semiconductors is a necessity for making electronic devices.

Semiconductors are doped to generate either a surplus or a deficiency in valence electrons. Doping allows researchers to exploit the properties of sets of elements, referred to as dopants, in order to modulate the conductivity of a semiconductor.

(c) Photodiodes are operated in reverse bias.

If a photodiode is connected in the forward bias direction, it will conduct pretty much like a normal diode. When a photodiode is reverse biased, the width of the depletion and a small reverse current (dark current) flows through the diode.

Q.7(a): The interference pattern is not observed in Young’s double-slit experiment when the two sources S1 and S2 are far apart. Explain

In conventional light source, light comes from a large number of independent atoms, each atom emitting light for about 10^-9 sec, i.e., light emitted by an atom is essentially a pulse lasting for only 10^-9 sec. Light coming out from two slits will have a fixed phase relationship only for 10^-9 sec. Hence any interference pattern formed on the screen would last only for 10^-9 sec, and then the pattern will change. The human eye can notice intensity changes which last at least for a tenth of a second and hence we will not be able to see any interference pattern. Instead due to rapid changes in the pattern, we will only observe a uniform intensity over the screen.

Q.7(b) Mention the conditions for the two sources to be coherent.

Two sources are said to be coherent when the waves emitted from them have the same frequency and constant phase difference.

Q.9(a)(i) Draw a labelled ray diagram showing the formation of the image at infinity by an astronomical telescope. 

OR

Q.9(b)(i) Draw a labelled ray diagram showing the formation of the image at least distance of distinct vision by a compound microscope. 

Q.10(b) A plane wavefront is propagating from a rarer into a denser medium. Use Huygens principle to show the refracted wavefront and verify snell’s law.

We assume a plane wave front AB propagating in denser medium incident on the interface PP at angle i as shown in

Let T be the time taken by the wave front to travel a distance BC. If vi is the speed of the light in medium I.
So, 

In order to find the shape of the refracted wave front, we draw a sphere of radius  , where  is the speed of light in medium II (rarer medium). The tangent plane CE represents the refracted wave front
In 
and in 
       …(1)
Let c be the speed of light in vacuum
So,  and 
                                                                            …(2)
From equations (1) and (2), we have


It is known as Snell’s law.

Latest Updated: CBSE Term 2 Additional Question papers for Class 12 Physics

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Physics Term 2 Sample Paper Class 12

To help students in their preparation we have come up with the CBSE Class 12 Term 2 Sample Paper 2022. The students must solve CBSE Class 12 Term 2 Physics Sample Paper 2022 given on this page and assess their preparation. The students who are going to give CBSE Term 2 Exam 2022 must read the whole article and bookmark this page to get all the latest updates regarding CBSE Term 2 Exam 2022.

CBSE Class 12 Physics Term 2 Sample Paper-Based Important Questions

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CBSE Term 2 Sample Paper Class 12 Physics: Exam Pattern

Here we have given the CBSE Term 2 Physics pattern. Exam pattern is a prerequisite to preparing for any exam. The students will be given a log table if necessary and the use of a calculator in exams will be prohibited. CBSE Term 2 Physics question paper consists of three sections namely Section A, Section B, and Section C. Check out the other details given below:

Section A: In Section A, three questions of two marks each will be asked.

Section B: In Section B, eight questions of three marks each will be asked.

Section C:  In Section C, one case study-based question of five marks will be asked.

 

Check Out: Important Questions For Class 12 Chemistry Term 2 With Answers

Check Out: CBSE Term 2 Date sheet 2022

Class 12 Physics Term 2 Sample paper with Solution PDF

Class 12 Physics Term 2 Sample Paper with Solutions

Q.1 In a pure semiconductor crystal of Si, if antimony is added then what type of extrinsic semiconductor
is obtained. Draw the energy band diagram of this extrinsic semiconductor so formed.

As given in the statement antimony is added to pure Si crystal, then an n-type extrinsic semiconductor would be so obtained Since antimony(Sb) is a pentavalent impurity.
Energy level diagram of n-type semiconductor

Q2. Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model? Justify your answer.

No, Because according to Bohr’s model, En = -13.6/n2 and electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as

L = mvr = nh/2π

OR

Explain how does (i) photoelectric current 

The increase in the frequency of incident radiation has no effect on photoelectric current. This is because of incident photon of increased energy cannot eject more than one electron from the metal surface.

(ii) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same? Show the graphical variation in the above two cases.

The kinetic energy of the photoelectron becomes more than the double of its original energy. As the work function of the metal is fixed, so incident photon of higher frequency and hence higher energy will impart more energy to the photoelectrons.

Q3. Name the device which converts the change in intensity of illumination to a change in the electric current flowing through it. Plot I-V characteristics of this device for different intensities. State any two applications of this device.

Photodiodes are used to detect optical signals of different intensities by changing the current flowing through them.

Applications of photodiodes:
1. In detection of optical signals.
2. In demodulation of optical signals.
3. In light operated switches.
4. In speed reading of computer punched cards.
5. In electronic counters
(any two out of these or any other relevant application)

Q4. Explain with a proper diagram how an ac signal can be converted into dc ( pulsating)signal with output frequency as double than the input frequency using pn junction diode. Give its input and output waveforms.

A junction diode allows current to pass only when it is forward biased. So, if an alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased. This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier.

Working with input and output waveforms

Q5. Define wavefront. Draw the shape of refracted wavefront when the plane incident wave undergoes refraction from optically denser medium to rarer medium. Hence prove Snell’s law of refraction. 
A locus of points, which oscillate in phase is called a wavefront.
OR
A wavefront is defined as a surface of a constant phase.

Diagram
Proof n1 sin i = n2sin r (Derivation)
This is Snell’s law of refraction.

Q6. (a) Draw a ray diagram of compound microscope for the final image formed at least distance of distinct vision?

Diagram of Compound Microscope for the final image formed at D:

(a) Draw a ray diagram of the Astronomical Telescope for the final image formed at infinity.

Ray diagram of the astronomical telescope when the image is formed at infinity.

Q.7 (a) Name the e.m. waves that are suitable for radar systems used in aircraft navigation. Write the range of frequency of these waves.

Microwaves are suitable for the radar system used in aircraft navigation. The range of frequency of microwaves is 108 Hz to 1011 Hz

(b) If the Earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now? Explain.

If the Earth did not have atmosphere, then there would be absence of greenhouse effect of the atmosphere. Due to this reason, the temperature of the earth would be lower than what it is now.

(c) An e.m. wave exerts pressure on the surface on which it is incident. Justify.

An e.m. wave carries momentum with itself and given by
P = Energy of wave(U)/ Speed of the wave(c)
= U/c
when it is incident upon a surface it exerts pressure on it.

OR

(a) “If the slits in Young’s double slit experiment are identical, then intensity at any point on the screen may vary between zero and four times to the intensity due to single slit”. Justify the above statement through a relevant mathematical expression.

The total intensity at a point where the phase difference is ∅, is given by ? = ?1 + ?2 + 2√?1?2 ??? ∅. Here ?1 and ?2 are the intensities of two individual sources which are equal.
When ∅ is 0, I = 4?1.
When ∅ is 90, I = 0
Thus intensity on the screen varies between 4?1 and 0.

(b) Draw the intensity distribution as function of phase angle when diffraction of light takes place through coherently illuminated single slit.

Intensity distribution as function of phase angle, when diffraction of light takes place through coherently illuminated single slit

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CBSE Class 12 Physics Term 2 Sample Paper Solution: FAQs

Q. When will CBSE conduct CBSE Term 2 Examination?

The Central Board of Secondary Education has announced the CBSE Term 2 dates. The CBSE will conduct the Term 2 exam from 26th April 2022.

Q. Where can I get CBSE Class 12th Physics Sample Paper 2022?

On this page, you will get CBSE Class 12th Physics Sample Paper 2022.

Q. What are the passing Criteria?

To pass the exam the students have to secure 33% of the total marks in each subject.

Q. What is the duration to solve the CBSE Term 2 Physics paper?

The students will get 2 hrs to solve the CBSE Term 2 Physics paper.

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