CBSE Sample Papers for Class 12 Physics Term 2 Set 1 with Solutions

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 1 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.
In a p-n junction diode the forward bias resistance is low as compared to the reverse bias resistance. Give reason.
Answer:
When the p-n junction diode is forward biased, the diode’s depletion layer or potential barrier shrinks. This can be considered as a practical short circuit.
Hence, resistance is negligible and current flow in forward bias is maximum.
Similarly, when it is reverse bias depletion layer width (or potential barrier) increases. Hence, resistance increases and very small amount of current flow in reverse bias.

Question 2.
The wavelength of the first line of Lyman series is 1215 Å. Calculate the wavelength of first line of Balmer series.
OR
(a) For a given frequency, plot a graph between photoelectric current and intensity of light.
(b) Plot a graph between de-Broglie wavelength and the momentum of a particle.
Answer:
Here, γ_L = 1215 Å
For the first line of Lyman series,

OR

Question 3.
Why is a photodiode operated in reverse bias mode? The given figure shows reverse bias current under different illuminating intensities I₁, I₂, I₃ and I₄ for a given photodiode. Arrange the intensities lv I2,13 and I4 in decreasing order of wavelengths.

Answer:
The photodiode is used in reverse bias condition because the change in reverse current through the photodiode due to change in light flux or light intensity can be measured easily, as the reverse saturation current is directly proportional to the light flux or light intensity. But it is not so when photodiode is forward biased. As the reverse saturation current through a photodiode increases with the increase in light intensity or light flux, so, I₄ > I₃> I₂ > I₁

Section – B

Question 4.
A slit of width 0-025 mm is placed in front of a lens of focal length 50 cm. The slit is illuminated with light of wavelength 5900 Å. Calculate the distance between the centre and first dark band of diffraction pattern on a screen placed at the focal plane of the lens.
Answer:
Given, λ = 5900 Å = 5.9 × 10^-7 m and f = 50 cm = 0.5 m.
Width of the slit = a = 0.025 mm = 2.5 × 10^-5 m
According to the foomula,
sinθ = \(\frac{\lambda}{a}\)
= \(\frac{5 \cdot 9 \times 10^{-7}}{2 \cdot 5 \times 10^{-5}}\) = 0.0236
As the diffraction pattern is formed in the focal plane of the lens,
∴ sin θ = tan θ
⇒ \(\frac{x}{f}=\frac{\lambda}{a}\)
or x = f × \(\frac{\lambda}{a}\)
= 0.5 × 0.0236 = 0.0118 m

Question 5.
Consider the fusion reaction: ⁴He + ⁴He → ⁸Be
For the reaction, find (i) mass defect, (ii) Q-value, (iii) is such a fusion energetically favourable?
Atomic mass of ⁸Be is 8.0053 u and that of ⁴He is 4-0026 u.
Answer:
⁴4He + He → ⁸Be + Q
(i) Δm = 2 × 4.0026 – 8.0053
= 8.0052 – 8.0053 = – 0.0001 amu

(ii) Q = (2m_He – m_Be)c² = (2 × 4.0026 – 8.0053) × 931 MeV
= – 93.1 keV.

(iii) Since Q is negative, the fusion is not energetically favourable.

Question 6.
Find the position of the image formed by the lens combination given in the figure.

Answer:
Image formed by the first lens (given by lens formula).
\(\frac{1}{v_{1}}-\frac{1}{u_{1}}\) = \(\frac{1}{f_{1}}\)
∴ \(\frac{1}{v_{1}}-\frac{1}{-30}\) = \(\frac{1}{10}\)
or υ₁ = 15 cm.

The image formed by the first lens serves as the object for the second. This is at a distance of (15 – 5) = 10 cm to the right of the second lens. Though the image is real, it serves as a virtual object for the second lens, which means that the rays appear to come from it for the second lens.
\(\frac{1}{v_{2}}-\frac{1}{10}\) = \(\frac{1}{-10}\)
or υ₂ = ∞

The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens.
\(\frac{1}{v_{3}}-\frac{1}{u_{3}}\) = \(\frac{1}{f_{3}}\)
\(\frac{1}{v_{3}}\) = \(\frac{1}{\infty}+\frac{1}{30}\)
υ₃ = 30 cm

The final image is formed 30 cm to the right of the third lens.

Question 7.
Why are de-Broglie waves associated with a moving football not visible? The wavelength λ of a photon and the de-Broglie wavelength of an electron have the same value. Show that the energy of the photon is
\(\frac{2 \lambda m c}{h}[latex] times the kinetic energy of the electron, where m, c and h have their usual meanings.
Answer:
The magnitude of the wavelength of de-Broglie waves associated with a moving football is extremely
small (λ = [latex]\frac{h}{m v}\) < 10^-34m) much less than that of visible region and are therefore, not visible.
Wavelength of photon = λ
∴ Energy of photon, E_p = \(\frac{h c}{\lambda}\) ………… (i)
Kinetic energy of electron, E_e = \(\frac{1}{2}\) mυ²
or mυ² = 2 E_e
⇒ mυ = \(\sqrt{2 \mathrm{E}_{e} m}\)
de-Broglie wavelength of electron,
⇒ λ = \(\frac{h}{m v}=\frac{h}{\sqrt{2 \mathrm{E}_{e} m}}\)
⇒ E_e = \(\frac{h^{2}}{2 \lambda^{2} m}\) …………… (ii)
Dividing equation (i) by equation (ii), we get
\(\frac{\mathrm{E}_{p}}{\mathrm{E}_{e}}=\frac{h c}{\lambda} \cdot \frac{2 \lambda^{2} m}{h^{2}}=\frac{2 \lambda m c}{h}\)
or E_p = \(\frac{2 \lambda m c}{h}\).E_e

Question 8.
A beam of light consisting of two wavelengths 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
OR
Answer the following questions:
(a) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(b) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(c) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?
Answer:
Here, d = 0.28 mm = 0.28 × 10^-3 m, D = 1.4 m
For λ_A = 800 nm = 800 × 10^-9 m = 8 × 10^-7 m
Position of first bright fringe
X_A = λ_A.\(\frac{\mathrm{D}}{d}\)
= \(\frac{8 \times 10^{-7} \times 1 \cdot 4}{0 \cdot 28 \times 10^{-3}}\) = 40 × 10^-4 = 4 × 10^-3 m
For wavelength, λ_B = 600 nm = 600 × 10^-9 m = 6 × 10^-7 m
Position of first bright fringe is
X_B = \(\frac{6 \times 10^{-7} \times 1 \cdot 4}{0.28 \times 10^{-3}}\) = 30 × 10^-7 × 10³ = 3 × 10^-3 m
The least distance, from the central maximum, where the bright fringes coincides
X_A – X_B = 4 × 10^-3 – 3 × 10^-3
= 1 × 10^-3 m
= +1 mm.

OR
(a) The diver is in the water and the fisherman is on surface (i.e., in air). Water is a denser medium than air. As diver is viewing the fisherman, this indicates that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal resulting the fisherman will appear to be taller.

(b) Yes; Decrease
The apparent depth of a tank of water changes when viewed obliquely because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.

(c) Yes
The refractive index of diamond (2.42) is higher than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.

Question 9.
Show that in Bohr’s hydrogen atom r ∝ n², where r is the radius and n is the principal quantum number.
Answer:
According to the Bohr’s postulate, only those electronic orbits are permitted for which angular momentum of electron is an integral multiple of \(\frac{h}{2 \pi}\) .
mυr = n \(\frac{h}{2 \pi}\)
or Linear velocity of electron υ = \(\frac{n h}{2 \pi m r}\)
Also, centripetal force equals electrostatic force of attraction between electron and nucleus
\(\frac{m v^{2}}{r}\) = K\(\frac{Z e^{2}}{r^{2}}\)
Substituting the value of υ from equation (i), we get
\(\frac{m}{r} \cdot \frac{n^{2} h^{2}}{4 \pi^{2} m^{2} r^{2}}\) = K . \(\frac{\mathrm{Z} e^{2}}{r^{2}}\)
or Radius of the orbit, r = \(\frac{n^{2} h^{2}}{4 \pi^{2} m K Z e^{2}}\)
For hydrogen atom, Z = 1
∴ Radius r = \(\frac{n^{2} h^{2}}{4 \pi^{2} m K e^{2}}\)
This shows that r ∝ n².
∴ The radius of orbit is proportional to the square of the principal quantum number i.e., radii of stationary orbits are in the ratio 1² : 2² : 3² … or 1 : 4 : 9 … respectively, i.e., the orbits are not equally spaced.

Question 10.
In an intrinsic semiconductor the energy gap E_g is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that 300 K? Assume that the temperature dependence of intrinsic carrier concentration n₁ is given by
n_i = n₀ exp [-\(\frac{\mathrm{E}_{g}}{2 k_{\mathrm{B}} \mathrm{T}}\)]
Where n₀ is a constant.
Answer:
Given: Energy gap of the intrinsic semiconductor, (E_g = 1.2 eV.
The temperature dependence of the intrinsic carrier concentration is given by
n_i = n₀ exp[-\(\frac{\mathrm{E}_{g}}{2 k_{\mathrm{B}} \mathrm{T}}\)]
Here, k_B = Boltzmann constant = 8.62 x 10^-5 eV/K (T = Temperature, n₀ = Constant) Initial temperature (T₁ = 300 K.
The intrinsic carrier-concentration at this temperature can be given by
n_i1 = n₀ exp[-\(\left.\frac{\mathrm{E}_{g}}{2 k_{\mathrm{B}} \times 300}\right]\)] …………. (i)
Final temperature (T₂) = 600 K.
The intrinsic carrier-concentration at this temperature can be given as
n_i2 = n₀ exp[-\(\frac{\mathrm{E}_{g}}{2 k_{\mathrm{B}} \times 600}\)] …………. (ii)
\(\frac{n_{i_{2}}}{n_{i_{1}}}\) = \(\frac{e^{\frac{1.2 \mathrm{eV}}{2 \mathrm{~K}_{\mathrm{B}} \times 600}}}{e^{\frac{1.2 \mathrm{eV}}{2 \mathrm{~K}_{\mathrm{B}} \times 300}}}\)
= e^11.59 = 1.072 × 10⁵
The ratio between the conductive at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperature i.e., 1.072 × 10⁵.

Question 11.
The oscillating electric field of an electromagnetic wave is given by
E_y = 30 sin [2 × 10¹¹t + 300 πx] Vm^-1.
(a) Obtain the value of the wavelength of the electromagnetic wave.
(b) Write down the expression for the oscillating magnetic field.
OR
Explain the following, giving reasons:
(a) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency.
(b) When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a reduction in the energy carried by the wave?
(c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity in the photon picture of light?
Answer:
Given, E_y = 30 sin [2 × 10¹¹t + 300πx] V m^-1
Standard equation is, E_y = E₀ sin[2π(\(\frac{x}{\lambda}-\frac{t}{\mathrm{~T}}\))]
E_y = E₀ sin [latex]\frac{2 \pi x}{\lambda}-\frac{2 \pi t}{\mathrm{~T}}[/latex]
On comparing \(\frac{2 \pi}{300 \pi}=\frac{1}{150}\) = 300π and E₀ = 30 V m^-1

(a) Wavelength λ = \(\frac{2 \pi}{300 \pi}=\frac{1}{150}\) m

(b) B₀ = \(\frac{\mathrm{E}_{0}}{\mathrm{c}}\) = \(\frac{30}{3 \times 10^{8}}\) = 10^-7T
The magnetic field is perpendicular to the direction of propagation and the direction of the electric field. So, the expression for the magnetic field is
B_Z = 10^-7 sin [(2 × 10¹¹t + 300πx)] T.

OR
(a) Reflection and refraction arise through interaction of incident light with atomic constituents of matter which vibrate with the same frequency as that of the incident light. Hence frequency remains unchanged.

(b) Energy carried by a wave depends on the frequency of the wave, not on the speed of wave propagation.

(c) For a given frequency, intensity of light in the photon picture is determined by
I = \(\frac{\text { Energy of photons }}{\text { area } \times \text { time }}=\frac{n \times h v}{\mathrm{~A} \times t}\)
Where n is the number of photons incident normally on crossing area A in time t.

Section – C

Question 12.
Refraction through prism:
Prism is a glass structure which has two rectangular faced inclined at an angle. It has total 3 rectangular faces and 2 triangular faces. When light is allowed to pass through the prism, it refracts at two surfaces, due to refraction light deviates from its original path. This angle to deviation is given by,
δ = (µ – 1) A
Where
δ → Angle of deviation
µ → Refractive index of prism
A → Angle of prism.

(a) How many rectangular faced are there in a prism?
(i) 4
(ii) 3
(iii) 2
(iv) 5

(b) The number of triangular faces in prism are:
(i) 5
(ii) 3
(iii) 2
(iv) None

(c) In the above diagram, how many faces are taking part in refraction of light?
(i) 1
(ii) 2
(iii) 3
(iv) 5

(d) What will be the refractive index of prism in terms of A and 8?
(i) \(\frac{\mathrm{A}}{\delta}\) + 1
(ii) \(\frac{\mathrm{A}}{\delta}\) – 1
(iii) \(\frac{\delta}{A}\) + 1
(iv) \(\frac{\delta}{A}\) + 1

(e) The deviation made by prism of angle 5° with refractive index 1.611 be:
Answer:
(i) 2.055
(ii) 1.055
(iii) 3.065
(iv) 3.055
(a) (ii) 3
(b) (iii) 2
(c) (ii) 2
(d) (iii) \(\) + 1
∵ δ = (μ – 1)A
⇒ μ – 1 = \(\)
⇒ μ = \(\) + 1

(e) (iv) 3.055
We know, δ = (μ – 1) A
A = 5°, μ = 1.611 5
δ = (1.611 – 1) × 5 = 0.611 × 5
= 3.055