Students can use CBSE Previous Year Question Papers Class 12 Physics with Solutions and CBSE Class 12 Physics Question Paper 2022 (Term-II) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Physics Question Paper 2022 (Term-II) with Solutions
Time Allowed: 2 hours
Maximum Marks: 35
General Instructions:
- This question paper contains 12 questions. All questions are compulsory.
- This question paper is divided into THREE sections, Section—A, B and C.
- Section-A Question number 1 to 3 are of 2 marks each.
- Section-B Question number 4 to 11 are of 3 marks each.
- Section-C Question number 12 is a case-study based question of 5 marks.
- There is no overall choice in the question paper. However, internal choice has been provided in some of the questions. Attempt any one of the alternatives in such questions.
- Use of log table is permitted, if necessary, but use of calculator is not permitted.
Section – A
Question 1.
With the help of a circuit diagram, explain briefly how a p-n junction diode works as a half-wave rectifier. [2]
Answer:
Semiconductor diode as a half wave Rectifier: The junction diode D
1
supplies rectified current to the band during one half of the alternating input voltage and is always in the same direction. During the first half cycle of the alternating input voltage, junction diodes D
1
will conduct each permitting current to flow during one half cycle whenever its p-terminal is positive with respect to the n-terminal.
The resulting output current is a series of unidirectional pulses with alternate gaps.
Question 2.
(a) What results do you expect if α-particle scattering experiment is repeated using a thin sheet of hydrogen in place of a gold foil? Explain.
(Hydrogen is solid at temperature below 14K).
Or
(b) Why it is the frequency and not the intensity of light source that determines whether emission of photoelectrons will occur or not? Explain. [2]
Answer:
(a) We would not be able to determine the size of nucleus if hydrogen sheet is used in place of gold foil, because hydrogen consists of a single proton and on the other hand, α-particle is bulky and less repulsion will take place between them.
Or
(b) Emission of photoelectrons has nothing to do with intensity of light. It depends on the energy of photo electrons.
Energy = \(\frac{h c}{\lambda}\) = hv ⇒ ε × v …….. where[v → frequency
∴ Ultimately emission depends on the frequency of photoelectrons.
Question 3.
Why a photo-diode is operated in reverse bias whereas current in the forward bias is much large than that in the reverse bias? Explain. Mention its two uses. [2]
Answer:
Even though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode works in reverse bias because it is easier to observe the change in the current, with change in light intensity, if a reverse bias is applied.
Uses:
- Medical instruments
- Single detection
Section – B
Question 4.
Draw a graph showing the variation of number of particles scattered (N) with scattering angle θ in Geiger-Marsden experiment. Why only a small fraction of the particles are scattered at θ > 90°? Mention two limitations of Rutherford nuclear model of an atom. [3]
Answer:
(i) Trajectory of a-particles in Geiger-Marsden Experiment:
For most of the a-particles, impact parameter is large, hence they suffer very small repulsion due to nucleus and go right through the foil.
It gives an estimate of the size of nucleus, that is relatively very very small as compared to the size of atom.
Limitations:
- It predicts that atoms are unstable because the accelerated electrons revolving around the nucleus must spiral into nucleus. This contradicts the stability of matter.
- It cannot explain the characteristic line spectra of atoms of different elements.
Question 5.
(i) Draw V-I characteristics of a p-n Junction’diode.
(ii) Differentiate between the threshold voltage and the breakdown voltage for a diode,
(iii) Write the property of a junction diode which makes it suitable for rectification of AC voltages.
Answer:
(i) V-I characteristics of p-n Junction diode:
(ii) Threshold voltage. Voltage in forward bias at which current increases exponentially with slight increase in voltage is known as threshold voltage. The magnitude of this voltage is lower than the breakdown voltage.
Breakdown voltage. Voltage in reverse bias at which current increases suddenly is known as breakdown voltage. The magnitude of the this voltage is higher than the threshold voltage.
(iii) Junction diode mainly conducts in forward biasing, hence it is suitable for rectification of AC voltages.
Question 6.
Draw a plot showing the variation of binding energy per nucleon versus the mass number (A). Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion. [3]
Answer:
(i) Nuclear fission: Binding energy per nucleon is smaller for heavier nuclei than the middle ones i.e. heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, the B.E./nucleon changes (increases) from about 7.6 MeV to 8.4 MeV. Greater binding energy of the product nuclei results in the liberation of energy. This is what happens in nuclear fission which is the basis of the atom bomb.
(ii) Nuclear fusion : The binding energy per nucleon is small for light nuclei, i.e., they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher binding energy per nucleon of the latter results in the release of energy.
Question 7.
How can you differentiate whether a pattern is produced by a single slit or double slits? Derive the expression for the angular position of
(i) bright and
(ii) dark fringes product in a single slit diffraction. [3]
Answer:
Wave diffracted from the edge of any circular obstacle undergoes constructive interference to form a bright spot at the centre of shadow.
| Young’s double slit experiment | Single slit experiment |
| 1. Light originating from two coherent sources. | Light originating from single source. |
| 2. Fringes are of equal width. | Fringe width decreases with order. |
| 3. Intensity of all the bright fringes in the brightness is the same. | Intensity falls with increasing order. The brightness of successive bright fringes goes on decreasing. |
For a slit having width ‘a’ and angle of diffraction ‘θ’.
Path difference (∆p) = a sin θ
(i) For Bright fringe : ∆p = (2n + 1)\(\frac{\lambda}{2}\)
∴ a sin θ = (2n + 1)\(\frac{\lambda}{2}\)
∴ a sin θ = \(\frac{(2 n+1) \lambda}{2 a}\) …… [n = 1,2,…….]
For small angle, sin θ ≈ θ
∴ θ = \(\frac{(2 n+1) \lambda}{2 a}\)
(ii) Dark fringe : ∆p = nλ
∴ a sin θ = nλ
∴ sin θ = \(\frac{n \lambda}{a}\)
For small angle, sin θ ≈ θ
∴ θ = \(\frac{n \lambda}{a}\)
Question 8.
(a) (i) Define SI unit of power of a lens.
(ii) A piano convex lens is made of glass of refractive index 1.5. The radius of curvature of the convex surface is 25 cm.
1. Calculate the focal length of the lens.
2. If an object is placed 50 cm in front of the lens, find the nature and position of the image formed. [3]
Answer:
Or
(b) A slit of width 0.6 mm is illuminated by a beam of light consisting of two wavelengths 600 nm and 480 nm. The diffraction pattern is observed on a screen 1.0 m from the slit.
Find:
(i) The distance of the second bright fringe from the central maximum pertaining to light of 600 nm
(ii) The least distance from the central maximum at which bright fringes due to both the wavelengths coincide. [3]
Answer:
Question 9.
(a) Calculate the energy and momentum of a photon in a monochromatic beam of wavelength 331.5 nm.
(b) How fast should a hydrogen atom travel in order to have the same momentum as that of the photon in part (a)? [3]
Answer:
(a) Energy of photon, E = \(\frac{h c}{\lambda}\)
We have, h = 6.63 × 10
-34
; c = 3 × 10
8
m/s; λ = 331.5 × 10
-9
m
E = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{331.5 \times 10^{-9}}\)
∴ E = 6 × 10
-19
J
Now, Momentum (P) = \(\frac{h}{\lambda}\) = \(\frac{6.63 \times 10^{-34}}{331.5 \times 10^{-9}}\) = 2 × 10
-27
m/s
(b) Momentum of a Hydrogen atom is equal to momentum of a photon.
P = mv ……. [P = 2 × 10
-27
From (a)]
⇒ v = \(\frac{\mathrm{P}}{m}\) = \(\frac{2 \times 10^{-27}}{1.67 \times 10^{-27}}\) ∴ v = 1.20 m/s
Question 10.
A ray of light passes through a prism of refractive index as \(\sqrt{2}\) as shown in the figure.
Find:
(i) The angle of incidence (∠r
2
) at face AC.
(ii) The angle of minimum deviation for this prism. [3]
Answer:
Question 11.
(a) (i) Arrange the following electromagnetic radiation in the ascending order of the frequencies:
X rays, Radio waves, Gamma rays, Microwaves
(ii) Write two uses of any two of these radiation.
Or
(b) With the help of a ray diagram explain the working of reflecting telescope. Mention two advantages of a reflecting telescope over a refracting telescope. [3]
Answer:
(a)
(i) Radio waves < Micro waves < X rays
(ii) Uses of radio waves:
(a) In radio broadcasting
(b) In radio telescopy
Uses of radio Microwaves:
(a) In microwave ovens
(b) In radars
Uses of Gamma rays:
(a) In cancer treatment
(b) In sterilization
Uses of X rays:
(a) In cancer treatment
(b) In diagnosis
Or
(b)
(i)
(ii) Advantages of reflecting telescope over a refracting telescope:
- Due to large aperture of the mirror used, the reflecting telescopes have high resolving power.
- Image formed by reflecting telescope is brighter than refracting telescope.
Section – C
Question 12.
A ray of light travels from a denser to a rarer medium. After refraction, it bends away from the normal. When we keep increasing the angle of incidence, the angle of refraction also increases till the refracted ray grazes along the interface of two media. The angle of incidence for which it happens is called critical angle. If the angle of incidence is increased further, the ray will not emerge and it will be reflected back in the denser medium. This phenomenon is called total internal reflection of light.
(i) A ray of light travels from a medium into water at an angle of incidence of 18°. The refractive index of the medium is more than that of water and the critical angle for the interface between the two media is 20°. Which one of the following figures best represents the correct path of the ray of light?
Answer:
(a)
When a ray of light travels from denser to rarer medium, it bends away from the normal.
(ii) A point source of light is placed at the bottom of a tank filled with water, of refractive index µ, to a depth d. The area of the surface of water through which light from the source can emerge is: [1]
(a) \(\frac{\pi d^2}{2\left(\mu^2-1\right)}\)
(b) \(\frac{\pi d^2}{\left(\mu^2-1\right)}\)
(c) \(\frac{\pi d^2}{\sqrt{2}\left(\sqrt{\mu^2-1}\right)}\)
(d) \(\frac{2 \pi d^2}{\left(\mu^2-1\right)}\)
Answer:
(b) \(\frac{\pi d^2}{\left(\mu^2-1\right)}\)
(iii) For which of the following media, with respect to air, the value of critical angle is maximum? [1]
(a) Crown glass
(b) Flint glass
(c) Water
(d) Diamond
Answer:
(c) Water
Critical angle is maximum for water.
(iv) The critical angle for a pair of two media A and B of refractive induces 2.0 and 1.0 respectively is: [1]
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Answer:
(b) 30°
As we know, θ = sin
-1
\(\left(\frac{n_2}{n_1}\right)\) = sin
-1
\(\left(\frac{1}{2}\right)\)
∴ θ sin
-1
(sin 30°) = 30°
(v) The critical angle of pair of a medium and air is 30°. The speed of light in the medium is: [1]
(a) 1 × 10
8
ms
-1
(b) 1.5 × 10
8
ms
-1
(c) 2.2 × 10
8
ms
-1
(d) 2.8 × 10
8
ms
-1
Answer:
(b) 1.5 × 10
8
ms
-1
