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

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 2 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.
The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni = 1.5 × 1016 m-3. Is the material n-type or ρ-type?
Answer:
Given, Number of silicon atoms (N) = 5 × 1028 atoms/m3
Number of arsenic atoms (nAS) = 5 × 1022 atoms/ m3
Number of indium atoms (nIn) = 5 × 1020 atoms/m3
Number of thermally-generated electrons (ni) = 1.5 × 1016 electrons/m3
Number of electrons (ne) = 5 × 1022 -1.5 × 1016 ≈ 4.99 × 1022
Let, number of holes = nh
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as
nenh = ni2
or
nh = \(\frac{n_{i}^{2}}{n_{e}}\)
= \(\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.99 \times 10^{22}}\) ≈ 4.51 × 109

The number of electrons is a approximately 4.99 × 1022 and the number of holes are 4.51 × 109. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.

Question 2.
(a) What is the shortest wavelength in the Balmer series?
(b) What is the maximum number of spectral lines emitted by a hydrogen atom when it is in the third excited state?
OR
Find the de-Broglie wavelength of an electron beam accelerating through a potential difference of 60 V.
Answer:
(a) For shortest wavelength in the Balmer series: n1 = 2, n2 = ∞
OR
∴ \(\frac{1}{\lambda_{\text {min }}}\) = R\(\) = R\(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\)
= \(\frac{R}{4}\)
⇒ λmin = \(\frac{4}{R}=\frac{4}{1.097 \times 10^{7}}\) m = 364.6 nm
The shortest wavelength in the Balmer series is 364.6 nm.

(b) Number of spectral lines obtained due to transition of electron from third excited state (n = 4) to ground state,
N = \(\frac{n(n-1)}{2}=\frac{4(4-1)}{2}\) = 6
OR
Potential difference ΔV = 60 so energy E = qΔV = 1.6 × 10-19 × 60 Joule = 9.6 × 10-18 Joule
So de-Broglie wavelength λ, = \(\frac{h}{\sqrt{2 m \mathrm{E}}}\) ……………. (i)
As momentum, p = mv = \(\sqrt{2 m \mathrm{E}}\)
Putting m = 9.1 × 10-31 kg and h = 6.62 × 10-34 Js is in equation (i), we get
λ = 1.58 × 10-10 meter = 1.58 Angstrom.

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

Question 3.
Why do we get a small current (in mA) through a p-n junction diode under reverse bias condition? In which direction does this current flow through the junction?
Answer:
In the reverse biasing of a p-n junction diode, the depletion region, across the junction diode, increases, due to the flow of holes in p-region and electrons in n-region, away from the junction. But a very small current (order pA) flows across the junction due to the flow of minority carriers. This current flows through n-type to p-type semiconductor, across the junction.
CBSE Sample Papers for Class 12 Physics Term 2 Set 2 with Solutions 1

Section – B

Question 4.
Violet light is incident on a thin convex lens. If this light is replaced by red light, explain with reason, how the power of the lens would change?
Answer:
The power of a lens is given by
p = \(\frac{1}{f}\) = (n – 1)(\(\frac{1}{R_{1}}-\frac{1}{R_{2}}\))
Refractive index of lens for violet rays is more than that for red rays. So power of a lens will decrease, if violet light is replaced by red light.

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

Question 5.
State the main assumptions of Rutherford’s model of atom.
Answer:
Assumptions of Rutherford’s model of atom:

  • An atom consists of a very small central core called the nucleus.
  • The nucleus carries all the positive charge and most of the mass of the atom (99.9%).
  • The size of the nucleus is of the order of 10-15 m which is very small in comparison to the size of the atom which is of the order of 10-10 m. Thus, major portion of atom is empty space.
  • The negatively charged electrons revolve around the nucleus in circular orbits with nucleus at the centre of the orbit. The necessary centripetal force for circular motion is provided by the force of attraction between positively charged nucleus and negatively charged electron.
  • Positive charge on nucleus is exactly equal to the total negative charge of electrons so that atom is electrically neutral.

Question 6.
Distinguish between conductors, insulators and semiconductors in terms of energy band diagrams.
Draw the diagrams.
Answer:
CBSE Sample Papers for Class 12 Physics Term 2 Set 2 with Solutions 2
[Here, C.B. → Conduction Band, V.B. → Valence Band, EgForbidden Energy gap]
Conductors: Conductors are the substances which allow the flow of electric current through them. In a conductor, the valence band and the conduction band overlap. So no energy is required to move the electrons from valence band to conduction band. Free electrons are available in conduction band and thus electric current flows through it, when a potential difference is applied across it. For example: Magnesium, Silver etc.

Insulators: Insulators are the substances which do not allow the flow of electric current through them. In insulators, the valence electrons are very rigidly held in the inter atomic bands, even at high temperatures. In terms of band theory, insulator have a large energy gap between the conduction band and the valence band (greater then 5 eV). So electrons cannot jump from the valence band into the conduction band. (In the case of diamond the energy gap is 6 eV.) For example : Glass, Rubber etc.

Semiconductors: Semiconductors are the substances, whose conductivity lies in between that of a good conductors and an insulator (For example : Germanium, Silicon).

In terms of band theory, semiconductors have small energy gap between conduction band and valence band. At 0 K, the energy gap is 0.72 eV for germanium and 1.21 eV for silicon. This decreases with increase in temperature. At 0 K, the conduction band is empty and the semiconductor behaves as an insulator. With the increase in the temperature, the electrons from the valence band jump to conduction band and hence the conduction of electric current is possible.

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

Question 7.
Calculate for how many years will the fusion of 2.0 kg deuterium keep 800 W electric lamp glowing. Take the fusion reaction as:
\({ }_{1}^{2} \mathrm{H}\) + \({ }_{1}^{2} \mathrm{H}\) → \({ }_{2}^{3} \mathrm{He}\) + \({ }_{0}^{1} n\) + 3.27 MeV
Answer:
Given, m = 2 kg; P = 800 W
Here, two deuterium nuclei produces 3.27 MeV energy.
∴ E = 3.27 × 106 × 1.6 × 10-19 = 5.232 × 10-13 J
or Energy per nuclei = \(\frac{5.232 \times 10^{-13}}{2}\) = 2.616 × 10-13J
Number of deuterium atom in 2 kg = \(\frac{6.023 \times 10^{23} \times 2000}{2}\) = 6.023 × 1026 atoms
∴ Total energy = 6.023 × 1026 × 2.616 × 10-13
= 15.75 × 1013 J
Power = \(\frac{\text { Energy }}{\text { Time }}\)
∴ Time, t = \(\frac{15.75 \times 10^{13}}{800}\) 1.96 × 1011s
or t = \(\frac{1.96 \times 10^{11}}{365 \times 24 \times 60 \times 60}\)
= 6.2 × 103 years.

Question 8.
Two lenses of power 10 D and – 5 D are placed in contact.
(a) Calculate the power of combination of lens.
(b) Where should an object to be held from the lens, so as to obtain a virtual image of magnification 2?
OR
(a) A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when:
(i) the telescope is in normal adjustment.
(ii) the final image is formed at the least distance of distinct vision.

(b) Also, find the separation between the objective lens and the eye piece in normal adjustment.
Answer:

(a) Given:
p1 = 10 D, P2 = – 5 D
Power of combination, P = P1 + P2
= 10D – 5D = 5D.

(b) Focal length, F = \(\frac{1}{\mathrm{P}}\) = \(\frac{1}{5}\) = 0.20 m
= 20 cm (Convex lens)
Magnification, m = \(\frac{v}{u}\) = 2
⇒ υ = 2u
For lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
\(\frac{1}{20}=\frac{1}{2 u}-\frac{1}{u}\)
⇒ \(\frac{-1}{2 u}=\frac{1}{20}\)
⇒ u = -10cm
OR
(a) (i) Given, f0 = 140 cm and fe = 5.0 Cm
When final image is at infinity, magnifying power,
m = –\(\frac{f_{0}}{f_{\ell}}\) = –\(\frac{140}{5.0}\) = -28
Negative sign shows that, the imagais inverted.

(ii) When final image is at the least distance of distinct vision,
Magnifying power, m = – \(\frac{f_{0}}{f_{e}}\)(1 + \(\frac{f_{e}}{\mathrm{D}}\))
= –\(\frac{140}{5.0}\)(1 + \(\frac{5.0}{25}\)) = -33.6

(b) Separation between objective and eyepiece when final image is formed at infinity.
L = f0 + fe = 140 cm + 5.0 cm = 145 cm.

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

Question 9.
Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light.
Answer:
The phenomenon of bending of light round the sharp comers of an obstacle and spreading into the regions of the geometrical shadow is called diffraction.

Expression for Fringes width:
Consider a parallel light beam from a lens is incident on slit AB. As diffraction happens, the pattern is focussed on screen XY with the help of lens L2. We will get a diffraction pattern that is a central maximum at the centre O flanked by a number of dark and bright fringes known as secondary maxima and minima.
CBSE Sample Papers for Class 12 Physics Term 2 Set 2 with Solutions 3
Central Maximum: Each point on the plane wave front AB sends secondary wavelets in all directions. The waves from points equidistant from the centre C kept on the upper and lower half reach point O with zero path difference and so, reinforce each other, making maximum intensity at point O.
CBSE Sample Papers for Class 12 Physics Term 2 Set 2 with Solutions 4

Question 10.
Following observations were made during an experiment on photoelectric emission:
(i) wavelength of the incident light = 2 × 10-7 m
(ii) stopping potential = 3 V.
Using the above observations, find:
(a) kinetic energy of photoelectrons with maximum speed.
(b) work function
(c) threshold frequency (h = 6.62 × 10-34 Js)
Answer:
(a) Given, Vs = 3 V and Kmax = eVs
∴ Kmax = 3 eV

(b) Given, λ = 2 × 10-7m = 2000 Å
and hc = 12400 eVÅ
Energy of incident photon (E) = \(\frac{h c}{\lambda}\) = \(\frac{12400}{2000}\) eV
= 6.20eV
Work function, W – E – Kmax = (6.20 – 3) eV = 3.2 eV

(c) hv0 = W= 3.2 × 1.6 × 10-19J
∴ v0 = \(\frac{3.2 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}}\) 7.76 × 1014

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

Question 11.
(a) The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is B0 = 510 nT.
What is the amplitude of the electric field part of the wave?
(b) What is the frequency of electromagnetic waves produced by oscillating charge of frequency f ?
OR
In a single slit diffraction experiment, a slit of width d is illuminated by red light of wavelength 650 nm. For what value of d will (i) the first minimum fall at an angle of diffraction of 30° and (ii) the first maximum fall at an angle of diffraction of 30°?
Answer:
(a) We know that, \(\frac{\mathrm{E}_{0}}{\mathrm{~B}_{0}}\) = c
⇒ E0 – B0c
510 × 10-9 × 3 × 108
= 153 NC-1.

(b) Frequency of the electromagnet wave produced will be equal to the frequency f of the oscillating
charge.
OR
(i) For first minima, d sin θ = nλ
d sin 30° = nλ
\(\frac{d}{2}\) = nλ.

d = 2nλ
2 × 1 × 650 × 10-9
= 1,300 × 10-6m
= 1.3μm.

(ii) For first maxima, d sin θ = (n + \(\frac{1}{2}\))λ
∴ d sin30°= \(\frac{3}{2}\) × 650 nm

d = 3 × 650 × 10-9
=195 × 10-6m
= 1.96μ

Section – C

Question 12.
Case Study: Young’s double slit experiment
Young’s Double slit experiment was done by Thomas Young to demonstrate experimentally the interference of light. In this, coherent sources of monochromatic light are used. Here two sources of monochromatic light are obtained by placing two slits before a single light source. As light rays passes from the slits and interference of light takes place. The interference fringes are obtained at screen MN. The fringe obtained at the centre of screen is a bright fringe
Fringe width (β) = λD/d,
where p – Fringe width, D – Distance between screen and slit, d – Distance between slits.
CBSE Sample Papers for Class 12 Physics Term 2 Set 2 with Solutions 5
(a) What kind of light sources are used in this experiment?
(i) Monochromatic
(ii) Dichromatic
(iii) Trichromatic
(iv) None of these

(b) How many slits are used in this experiment?
(i) Two
(ii) One
(iii) Three
(iv) None of these

(c) The central fringe in the experiment is:
(i) Dark
(ii) Very dark
(iii) Bright
(iv) None of these

(d) What kind of phenomena of light is demonstrated with this experiment?
(i) Reflection
(ii) Refraction
(iii) Interference
(iv) Diffraction

(e) What will be the ratio of fringe width β1 and β2 obtained with red light of λ1 = 660 nm and ultraviolet light of λ2 = 165 nm ?
(i) 1 : 4
(ii) 4 : 1
(iii) 2 : 1
(iv) 1 : 2

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

Answer:
(a) (i) Monochromatic
(a) (i) Monochromatic
(b) (i) Two
(c) (iii) bright
(d) (iii) Interference
(e) (ii) 4 : 1
Given: λ1 = 660 nm, λ2 = 165 nm
As β = \(\frac{\lambda D}{d}\); D, d are same for lights
\(\frac{\beta_{1}}{\beta_{2}}=\frac{\lambda_{1}}{\lambda_{2}}\) = \(\frac{660}{165}\) = 4 : 1