CBSE Sample Papers for Class 12 Physics Term 2 Set 4 with Solutions

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 4 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.
Define the V-I characteristics of junction diode.
Answer:
The V-I characteristic of the diode is the graph drawn between the voltage V and current I in forward bias and reverse bias of a junction diode.

Question 2.
Find the relation between the three wavelengths λ₁ λ₂ and λ₃ from the energy level diagram shown.

OR
Calculate the de-Broglie wavelength of an electron.
Answer:
We know that, E_CB = \(\frac{h c}{\lambda_{1}}\)
⇒ E_BA = \(\frac{h c}{\lambda_{2}}\)
⇒ E_CA = \(\frac{h c}{\lambda_{3}}\)
Now, E_CA = E_CB + E_BA
where,
E_CB = Energy gap between level B and C,
E_BA = Energy gap between level A and B,
E_CA = Energy gap between level A and C.

\(\frac{h c}{\lambda_{3}}=\frac{h c}{\lambda_{1}}+\frac{h c}{\lambda_{2}}\)
\(\frac{h c}{\lambda_{3}}=\frac{h c}{\lambda_{1}}+\frac{h c}{\lambda_{2}}\)
λ₃ = \(\frac{\lambda_{1} \lambda_{2}}{\lambda_{2}+\lambda_{1}}\).
OR
de-Broglie Wavelength λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m \mathrm{E}}}\)
⇒ λ = \(\frac{6.66 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 10^{3}}}\)
= 0.38 × 10^-34+26
⇒ λ = 38 Å.

Question 3.
What is a photodiode ? Explain in brief. Mention some uses of photodiode.
Answer:
Photodiode is a p-n junction which is an opto-electronic device in which current carriers are generated by photons through photo-excitation, i.e., photoconduction by light.

A photodiode is a special p-n junction diode made of photosensitive semiconducting material. It is operated under reverse bias below the breakdown voltage. The conductivity of p-n junction exists at the p-n junction. The conductivity of the p-n junction photodiode increases with the increase in intensity of light falling on it.

Symbolically, a photodiode is shown in the figure (i). Figure (ii) shows an experimental arrangement for the study of V-I characteristics of a photodiode in which the photodiode is reverse biased.
When the photodiode is reverse biased with voltage less than its breakdown voltage and no light is incident on its junction, the reverse current is extremely small (almost negligible). This current is called dark current.

Section – B

Question 4.
The optical density of turpentine is higher than that of water while its mass density is lower. Figure shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in figure, the path shown in correct?

Answer:
In the figure, the path shown for the ray 2 is correct. The ray suffers two refractions: At A, ray goes from air to turpentine, bending towards normal. At B, ray goes from turpentine to water (i.e., from denser to rarer medium), bending away from normal.

Question 5.
Draw a labelled diagram of a full wave rectifier circuit. Explain briefly the functions of a its components. Sketch the input and output wave forms.
Answer:
Diagram of a full wave rectifier circuit:

In a full wave rectifier, unidirectional pulsating output current is obtained for both halves of the a.c. input voltage. Essentially, it requires two junction diodes so connected that one diode rectifies one half and second diode rectifies the second half of the input.

During the first half cycle of the a.c. input voltage, the terminal S₁ is positive relative to T and S₂ is negative. In this situation, the junction diode D₁ is forward biased and D₂ is reverse biased. Therefore D₁ conducts and D₂ does not. The conventional current flows through diode D₁, load R_L and the upper half of the secondary windings, as shown by the solid arrows. During the second half cycle of the input voltage, S₁ is negative relative to T and S₂ is positive. Now D₁ is reverse biased and does not conduct while D₂ is forward-biased and conducts. The current now flows through D₂, load R_L and the lower half of the secondary windings, as shown by the solid arrows. It may be seen that the current in the load R_Lflows in the same direction for both the half cycles of the a.c. input voltage. Thus, the output current is a continuous series of unidirectional pulses.

However, it can be made fairly steady by means of smoothing filters.

Question 6.
There are two sources of light A and B. the wavelength of light emitted from A is from 8000 Å to 11000 Å, while that from B is from 3000 Å to 6000 Å. The intensity of A is 4 times that of B. But when light of A falls on a metal, photoelectrons are not emitted, whereas light of B eject photoelctrons from the same metal. Explain its reasons.
Answer:
The photoelectrons from a metal can be emitted only when the energy of light-photons incident on the metal is not less than the work function of the metal. In other word, if the frequency of the incident light is below a certain minimum value or the wavelength is above a certain maximum value, the photoelectrons will not be emitted from the metal however high be the intensity of light. If the threshold wavelength of a metal lies between 6000 to 8000 Å, then the light source B (3000 – 6000 Å) will eject photoelectrons from that metal but not the light source A (8000 -11000 Å) though the intensity of A is 4 times that of B.

Question 7.
(a) Name the radiations of electromagnetic spectrum which are used in:
(i) Warfare to look through haze.
(ii) Radar and geostationary satellites.
(iii) Studying the structure and properties of atoms and molecules.

(b) Find the wavelength of eletromagnetic waves of frequency 5 × 10¹⁹ Hz in free space. Give its two applications.
Answer:
(a)
(i) Infrared rays
(ii) Microwaves
(iii) Gamma rays

(b) Given: Frequency (v) = 5 × 10¹⁹ Hz
Wavelength (λ) = \(\frac{c}{v}\) = \(\frac{3 \times 10^{8}}{5 \times 10^{19}}\) = 6 × 10^-12 m
This wavelength corresponds to either gamma rays or X-rays. These are used:

• for treatment of cancer, and
• for causing certain nuclear reactions.

Question 8.
(a) A ray of light incident on face AB of an equilateral glass, shows minimum deviation of 30°. Calculate
the speed of light through the prims.

(b) Find the angle of incident of face AB so that the emergent ray grazes along the face AC.
OR
(a) In a single slit diffraction pattern how is the angular width of central bright maximum changed, when:

• The slit width is decreased.
• The distance between the slit and the screen is increased.
• λ is decreased.

(b) How does the intensity of central maximum change if the width of the slit is halved in a single slit diffraction experiment?
Answer:
(i) Given δ_m = 30°
A = 60°

(ii) Let c be the critical angle, then
sin c = \(\frac{1}{\mu}\)
⇒ sin c = \(\frac{1}{\sqrt{2}}\)
⇒ sin c = sin 45°
⇒ c = 45°
∴ r₂ = c = 45°
r₁ + r₂ = A
r₁ = A – r₂
= 60° – 45°
r₁ = 15°
Now, μ = \(\frac{\sin i}{\sin r_{1}}\)
sin i = μ sin r₁
= √2 sin (15°) = 1.414 × 0.2588
sin i = 0.366
⇒ i = 21° (approx).

OR
(a) In a single slit diffraction pattern angular with of central maxima = \(\frac{2 \lambda}{d}\).

• When d is decreased, angular width will increase.
• When the distance between the slit and the screen is increased, angular width remains unaffected.
• When λ is decreased, angular width will also decrease.

(b) The width of central maxima is doubled and the intensity is reduced to one-fourth of its original value.

Question 9.
Suppose, we think of fission of a \({ }_{26}^{56} \mathrm{Fe}\) nucleus into two equal fragments, \({ }_{13}^{58} \mathrm{Al}\). Is the fission energetically possible? Argue by working out Q of the process.
Given: m(\({ }_{26}^{56} \mathrm{Fe}\)) = 55.93494u and m(\({ }_{13}^{58} \mathrm{Al}\)) = 27.981914
Answer:
The fission of \({ }_{26}^{56} \mathrm{Fe}\) can be given by,
\({ }_{26}^{56} \mathrm{Fe}\) → 2 \({ }_{13}^{28} \mathrm{Al}\)
From data given in question
Atomic mass of m(\({ }_{26}^{56} \mathrm{Fe}\)) = 55.93494 u
Atomic mass of m(\({ }_{13}^{28} \mathrm{Al}\)) = 27.98191 u
The Q-value of this nuclear reaction given by
Q = [m(\({ }_{26}^{56} \mathrm{Fe}\)) – 2m(\({ }_{13}^{28} \mathrm{Al}\))]c²
= [55.93494 – 2 × 27.98191]c² = (- 0.02888c²)u
But 1u = 931.5 MeV/c²
∴ Q = – 0.02888 × 931.5 = – 26.902 MeV
The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically possible fission reaction, the Q-value must be positive.

Question 10.
(a) Why do we use a very thin gold foil in Rutherford’s α-particle scattering experiment?
(b) What is the energy possessed by an electron in n = ∞?
Answer:
(a) A thin gold foil was taken because gold nucleus is heavy and so it can produce a large deflection in the path of a-particle and also these can be easily prepared.
(b) Energy of an electron in the n^th orbit hydrogen atom is given by
E_n = –\(\frac{13.6}{n^{2}}\)eV
Here, n = ∞
E_∞ = – \(\frac{13.6}{\infty^{2}}\) = 0

Question 11.
Define magnifying power of a telescope. Write its expression.
A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece.
OR
(a) Draw a ray diagram for the formation of image by a compound microscope.
(b) You are given the following three lenses. Which of these two lenses will you use as an eyepiece and as an objective to constructed a compound microscope?

Lenses	Power (D)	Aperture (cm)
L1	3	8
L2	6	1
L3	10	1

Answer:
When final image is formed at infinity:

When the final image is formed at least distance of distinct vision:

Magnifying power of refracting telescope (M) is defined as the ratio of the angle subtended by the image (β) at the eye to the angle subtended by the distant object at the unaided eye (α).
M = \(\frac{\beta}{\alpha}\)

We can increase the magnifying power of telescope by:

1. Increasing the focal length of the objective.
2. Decreasing the focal length of eyepiece.

Two limitations of refractive telescope are:

1. The lenses used in refractive telescope are expensive.
2. The lenses used for making refracting telescope have chromatic aberration and distortions.
   They can be minimised by using reflecting type telescope, which use concave mirror rather than a lens for the objective.

Reflecting type telescope has the following advantages:

1. They are free from chromatic aberration as mirror is used instead of lens.
2. There is no problem for mechanical support because weight of mirror is much less than the weight of the lens. It can be supported easily.

Here, f₀ = 150 cm, f_e = 5 cm
Angle subtended by 100 m tall tower at 3 km is
α = \(\frac{100}{3 \times 1000}=\frac{1}{30}\) rad
If h is the height of image formed by the objective, then
α = \(\frac{h}{f_{0}}=\frac{h}{150}\)
∴ \(\frac{h}{150}=\frac{1}{30}\) = or h = \(\frac{150}{30}\) cm = 5cm
Magnification produced by eyepiece
m_e = (1 + \(\frac{\mathrm{D}}{f_{e}}\)) = (1 + \(\frac{25}{5}\)) = 6
∴ Height of final image = h × m_e = 5 × 6 = 30 cm

(a)

(b) For constructing compound microscope, L₃ should be used as objective and L₂ as eyepiece because both the lenses of microscope have short focal lengths and the focal length of objective lens should be smaller than the eyepiece lens.

Section – C

Question 12.
Case Study:
Diffraction of light
The diffraction of light takes place when an object or obstacle comes in the path of light. In this phenomena bending of light takes place around the edges of the obstacle. The bending of light from its path is larger when the size of the obstacle is comparable to the wavelength of light.

If we place a narrow slit in the path of wavefront of light, then bending of diffracted wavefront will be more. The diffraction of sound waves also takes place. For visible light, diffraction can happen when an object of λ ≈ 10^-6 m order comes in its path.

(a) The phenomenon of bending of light around the edges is called:
(i) Interference
(ii) Refraction
(iii) Diffraction

(b) The diffraction of an incident wave will not happen when size aperture will be:
(i) Small
(ii) Very small
(iii) Very large
(iv) None of these

(c) The diffraction of light waves increases as the size of aperture of a hole:
(i) Increases
(ii) Decreases
(iii) Both (i) and (ii)
(iv) Remain same

(d) The diffraction of light can take place through:
(i) Sharp edges
(ii) Very small apertures
(iii) Very large apertures
(iv) Both (i) and (ii)

(e) In which case diffraction of light will be more for an aperture (d)?
(i) d > λ
(ii) d ≈ λ
(iii) d < λ
(iv) None of these

Answer:
(a) (iii) Diffraction
(b) (hi) Very large
(c) (ii) Decreases
(d) (iv) Both (a) and (b)
(e) (iii) d < λ