CBSE 12th Term-2 Exam 2022 : Physics Analysis and Question Paper with Answer Key

CBSE 12th Term-2 Exam 2022 : Physics Analysis, Review & Question Paper with Answer Key

CBSE Class 12 Physics Exam was conducted today across the country. Students were seen entering the exam venues in proper queues and masked. They were seen adhering to the guidelines of the CBSE Board for exams.

Our reporters have reached the exam venue to collect the first response of the students who appeared in the CBSE 12th Physics Term 2 Exam 2022. The exam centres across Delhi- NCR saw students exiting with mixed reactions. Check the expert reviews question papert and student responses below in detail.

CBSE Class 12 Physics Term 2 Question Paper 2022 - Q.P. Code - 55/4/1- Set- 1

CBSE Class 12 Physics Term 2 Question Paper 2022 - Q.P. Code - 55/4/2- Set- 2

CBSE Class 12 Term 2 Study Material
CBSE Class 12 Reduced Syllabus for Term 2 Exam 2022	CBSE Class 12 Sample Paper for Term 2 Exam 2022
CBSE Class 12 Previous Year Question Paper with Solutions	CBSE Class 12 Term 2 Full Study Material 2022
CBSE Revision Notes For Class 12	CBSE Class 12 Topper Answer Sheet

CBSE Class 12 Physics Term 2 Paper Analysis: Student Reactions

The students of class 12 exited with mixed reactions. The first student that exited informed us that numerical questions were tough and theory was easy. Ratna, a student from Ghaziabad said, “ I am a little weak in numerical solving, so I found the numerical questions tricky. But the theory was very easy.”

There were too many numerical based questions as per some students in the paper. "It was not easy at all", reported a girl exiting a centre in Mayur Vihar, Delhi.

The students were also of the view that the Term 2 paper was moderately difficult. “Physics can never be very easy, we needed to think properly about the answers before writing them,” says Mayank.

Noida centres were flooded with students with happier faces than Ghaziabad. Students were happy that everything was from the syllabus. “How much can you expect from a 35 mark paper? It was easy”, said Guatami, a student from Central School.

Students are expecting better marks from the Term 1 exam this time.

CBSE Class 12 Physics Term 2 Exam: Expert Reviews

The experts are of the view that the Physics paper for class 12 CBSE Board was easier than the Term 1 exam. This may be because this time the students were used to the paper being for 35 marks and the course divided. The CBSE Class 12 Physics Exam theory part was easy, said the teacher from Apeejay School Noida. The paper was found to be balanced by many of the experts here . Many teachers were also of the view that the paper was well balanced and all the chapters were covered.

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: (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.

2. 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.

Q.3: Name the device which converts an ac input signal into a dc output signal. Write the principle of working of the device.

A 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.

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

$Class 12 Physics Term 2 Sample Paper with Answer Key PDF_60.1$

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. (Vo−V) where Vo = 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 (Ih) and diffusion electron (Ie). Total current I=Ie+Ih.

Reverse Bias: When a voltage V (i.e. reverse bias voltage) is applied to a circuit. The effective barrier potential increases, it becomes (Vo+V). 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.