CBSE Sample Papers for Class 12 Physics Term 2 Set 3 with Solutions

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 3 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

There are 12 questions in all All questions are compulsory.
This question paper has three sections: Section A, Section B and Section C.
Section A contains three questions of two marks each, Section B contains eight questions of three
marks each, Section C contains one case study-based question of five marks.
There is no overall choice However, an internal choice has been provided en one question of two marks and two questions of three marks You have to attempt only one of the choices in such questions
You may use log tables if necessary but use of calculator is not allowed.

Section – A

Question 1.
Plot a graph showing variation of current versus voltage for the material Ga.
Answer:
Current-Voltage characteristics graph of Ga.

Question 2.
State Huygens’ principle.
OR
Two sources with intensity I₀ and 4I₀ respectively, interfere at a point in a medium. Find the ratio of (i) maximum and minimum possible intensities, (ii) ratio of amplitudes.
Answer:
Huygens’ principle: Huygens’ principle is a geometrical construction which is used to determine the position of a wavefront at a later time from its given position at any instant. It states that every point on a wavefront is a source of wavelets. These wavelets spread out in forward direction, at the same speed as source wave.

It is based on the following assumptions:
(i) Each point on the given or primary wavefront acts as a source of secondary wavelets, sending out disturbance in all directions in a similar manner as the original source of light does.
(ii) The new position of the wavefront at any instant (called secondary wavefront) is the envelope of the secondary wavelets at that instant.
OR

⇒ 3r – 3 = r + 1
2r = 4
⇒ r = 2
∴ r = 2 : 1
∴ \(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\) = 2 : 1.

Question 3.
How are p-type semiconductors produced?
Answer:
When a trivalent impurity like a aluminium, indium, boron, gallium, etc., is doped with pure germanium (or silicon), then the conductivity of the crystal increases due to deficiency of electron i.e., holes and such a crystal is said to be p-type semiconductor while the impurity atoms are called acceptors.

Section – B

Question 4.
Do the frequency and wavelength change when light passes from a rarer to a denser medium?
Answer:
When the light travels from a rarer to a denser medium, its frequency remains unchanged but wavelength decreases. It is because, frequency is an inherent property of light. Since, energy of a photon of light is hv, its energy will remain the same.

Question 5.
The current in the forward bias is known to be more (~ mA) than the current in the reverse bias (~ mA). What is the reason to operate the photodiodes in reverse bias?
Answer:
Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p(i.e., n >> p). On illumination, let the excess electrons and holes generated be An and Ap, respectively.
n’ = n + Δn
p’ = p + Δp
Here n’ and p’ are the electron and hole concentrations at any particular illumination and n & p are carriers concentration when there is no illumination. Remember Δn = Δp and n >> p. Hence, the fractional change in the majority carriers (i.e., Δn/n) would be much less than that in the minority carriers (i.e., Δp/p). In general, we can state that the fractional change due to the photo-effects on the minority carriers dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity.

Question 6.
A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens?
Answer:
We know that, focal length of a concave lens is negative.
Using lens maker’s formula,
\(\frac{1}{f}\) = (\(\frac{\mu_{l}}{\mu_{m}}\) – 1) (\(\frac{1}{R_{1}}-\frac{1}{R_{2}}\))
Here μ₁ = 1.5 and μ_m = 1.65
Also, \(\frac{\mu_{l}}{\mu_{m}}\) < 1, so (\(\frac{\mu_{l}}{\mu_{m}}\) – 1) is negative
And focal length of the given lens becomes positive. Hence, it behaves like a converging lens.

Question 7.
Plot a graph showing variation of de-Broglie wavelength λ versus \(\frac{1}{\sqrt{V}}\) , where V is accelerating potential for two particles A and B carrying same charge but of masses m₁,m₂ (m₁ > m₂). Which one of the two represents a particle of smaller mass and why?
Answer:

qV = \(\frac { 1 }{ 2 }\) mv²
qV = \(\frac{p^{2}}{2 m}\)
⇒ p = \(\sqrt{2 m q V}\) = \(\frac{h}{\lambda}\)
⇒ λ = \(\frac{h}{\sqrt{2 m g V}}\)
⇒ Slope \(\frac{1}{\sqrt{m}} \cdot\)
Since, m₁ > m₂
∴ Particle with lower mass (m₂) have greater slope as shown in figure.

Question 8.
(a) How will you distinguish between a microscope and a telescope just by looking at them?
(b) A compound microscope has an objective of focal length 1 cm and an eye piece of focal length 2.5 cm. An object has to be placed at a distance of 1.2 cm away from the objective for normal adjustment. Find the angular magnification and length of the microscope tube.
OR
Explain why?
(a) A diamond glitters in a brightly lit room, but not in a dark room.
(b) A crack in a window pane appears silvery.
(c) The bubbles of air rising up in a water tank appear silvery when viewed from top.
Answer:
(a) As aperture of objective of a microscope is much smaller and that of a telescope is much larger,
therefore, front end of a microscope is narrow and front end of a telescope is much wider.
(b) Given: f₀ = 1 cm,f_e = 2.5 cm, u₀ = – 1.2 cm, m = ?, L = ?

(a) A diamond glitters in a brightly lit room because light entering the diamond from any face suffers multiple total internal reflections and does not come out. The diamond appears illuminated from inside. This would not happen in a dark room.

(b) A crack in a window pane appears silvery on a account of total internal reflection of light in the crack.
The bubbles of air rising up in a water tank appears silvery when viewed from top again on account of total internal reflection of light from the bubble.

Question 9.
A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. If slit width 2 mm and distance of screen from slits is 1.2 m.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Answer:
Here, d = 2 mm, D = 1.2 m
λ₁ = 650 nm = 650 × 10^-9 m,
λ₂ = 520 nm = 520 × 10^-9 m
(a) Distance of third bright fringe from the central maximum for the wavelength 650 nm.
y₃ = \(\frac{3 \lambda_{1} \mathrm{D}}{d}\) = \(\frac{3\left(650 \times 10^{-9}\right) 1 \cdot 2}{2 \times 10^{-3}}\) = 1.17 mm

(b) Let at linear distance ‘y’ from center of screen the bright fringes due to both wavelength coincides. Let n₁ number of bright fringe with wavelength λ₁ coincides with n₂ number of bright fringe with wavelength λ₂.
We can write
y = n₁β₁ = n₂β₂
n₁\(\frac{\lambda_{1} \mathrm{D}}{d}\) = n₂\(\frac{\mathrm{D} \lambda_{2}}{d}\) or n₁λ₁ = n₂λ₂ ……………. (i)
Also at first position of coincide, the n^th bright fringe of one will coincide with (n + 1)^thbright fringe of other.
If λ₂ < λ₁
So, then n₂ > n₁
then n₂ = n₁ + 1 …………… (ii)
Using equation (ii) in equation (i)
n₁λ₁ – (n₁ + 1)λ₂
n₁(650) × 10^-9 = (n₁ + 1) 520 × 10^-9
65n₁ – 52n₁ + 52 or 13n₁ – 52 or n₁ = 4
Thus, y = n₁\(\) = 4[latex]\frac{\left(6.5 \times 10^{-7}\right)(1.2)}{2 \times 10^{-3}}[/latex]
= 1.56 × 10^-3 m = 1.56mm
So, the fourth bright fringe of wavelength 520 nm coincides with 5^th bright fringe of wavelength 650 nm.

Question 10.
Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces.
Answer:
Plot of potential energy of a pair of nucleus as a function of their separation is given in the figure.

Conclusions:
(i) The nuclear force is much stronger than the coulomb force acting between charges or the gravitational forces between masses.
(ii) The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few fermies.
(iii) For a separation greater than r0, the force is attractive and for separation less than r₀, the force is strongly repulsive.

Question 11.
Write down properties of E.M. wave.
OR
What are electromagnetic waves? Give some characteristics of electromagnetic waves.
Answer:

Changes in electric and magnetic fields occur simultaneously. E.M. wave attain their maxima and minima at the same place and at the same time.
The electric and magnetic fields directions are mutually perpendicular to each other and as well as to the direction of propagation of wave.
The electric field vector (E) and magnetic field vector (B) are related by c = E₀/B₀, here E₀ and B₀ are the amplitudes of the respective fields and c is speed of light.
The velocity of electromagnetic waves in free space, c = 1 / \(\sqrt{\mu_{0} \varepsilon_{0}}\).
The velocity of electromagnetic waves in a material medium = 1 / \(\sqrt{\mu \varepsilon}\) Where p and e are absolute
permeability and absolute permittivity of the material medium respectively.
Electromagnetic wave follow the principle of superposition.
Electromagnetic waves transfer energy as they propagate through space. This energy is divided equally between electric and magnetic fields.
Electromagnetic waves can transfer energy as well as momentum to objects placed on their paths,
Electromagnetic waves do not require material medium to travel.

OR
Electromagnetic waves: The waves propagating in space through electric and magnetic fields, varying in space and time simultaneously are called electromagnetic waves.
Characteristics of electromagnetic waves:

The electromagnetic waves travel in free space with the speed of light (c = 3 × 10⁸ m/s) irrespective of their wavelength.
Electromagnetic waves are neutral, so they are not deflected by electric and magnetic fields.
The electromagnetic waves show properties of reflection, refraction, interference, diffraction and polarisation.
In electromagnetic waves the electric and magnetic fields are always in the same phase.
The ratio of magnitudes of electric and magnetic field vectors in free space in constant equal to c.
\(\frac{E}{B}\) = \(\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\) = c = 3 × 10⁸ m/s
The speed of electromagnetic waves in a material medium is given by
υ = \(\) , where n is the refractive index.
In an electromagnetic wave, the energy is propagated by means of electric and magnetic field vectors
in the direction of propagation of wave.
In electromagnetic wave the average values of electric energy density and magnetic energy density
are equal.(\(\frac { 1 }{ 2 }\) ε₀E²)_av = (\(\frac{\mathrm{B}^{2}}{2 \mu_{0}}\))_av
(ix) The electric vector of electromagnetic wave is responsible for optical effects and is also called the
light vector.

Section – C

Question 12.
Case Study:
Rutherford’s model of atom: This model of atom was given by Rutherford. According to him each atom have tiny core at its centre which is called nucleus of the atom. In the nucleus whole mass and positive charge of the nucleus is concentrated. The size of the nucleus is of the order of 10^-15 m and atomic size is about 10^-10 m.

The nucleus of the atom is surrounded by a number of electrons. But whole atom is electrically neutral. These electrons revolve in certain orbits around the nucleus of an atom the electrons get centripetal force from the electrostatic force between nucleus and electrons.
(a) The tiny core at the centre of atom is:
(i) Electron
(ii) Atom
(iii) Nucleus
(iv) Proton

(b) Whole mass of the atom is concentrated at:
(i) Periphery of atom
(ii) Nucleus
(iii) Proton
(iv) None of these

(d) The nucleus of the atom is surrounded by a number of:
(i) Proton
(ii) Neutron
(iii) Electron
(iv) None of these

(e) Atom is electrically:
(i) Positive
(ii) Negative
(iii) Earthed
(iv) Neutral

Answer:
(a) (iii) Nudeus
(b) (ii) Nucleus
(c) (iii) 10^-15 m
(d) (iii) Electron
(e) (iv) Neutral