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## CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 6 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.

Suppose a pure Si crystal has 5 × 10^{28} atoms m^{-3}. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. [Given that n_{i} = 1.5 × 10^{16} m^{-3}.]

Answer:

Note that thermally generated electrons (n_{i} ≈ 10^{16} m^{-3}) are negligibly small as compared to those produced by doping.

∴ n_{e} ≈ ND

Since n_{e}n_{h} = \(n_{i}^{2}\)

The number of holes n_{h} = \(\frac{\left(2 \cdot 25 \times 10^{32}\right)}{\left(5 \times 10^{22}\right)}\) ≈ 4.5 × 10^{9} m^{-3}

Question 2.

A beaker is filled with water to a height of 12.5 m. The apparent depth of these needle lying at the bottom of the tank as measured by a microscope is 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 upto the same height, by what distance would the microscope be moved to focus on the needle again?

OR

Describe any two characteristic features which distinguish between interference and diffraction phenomena.

Answer:

Refractive index, n = \(\frac{\text { Real depth }(\mathrm{H})}{\text { Apparent depth }(h)}\)

Given, H = 12.5 cm, h = 9.4 cm

∴ Refractive index of water, n_{w} = \(\frac{12 \cdot 5}{9 \cdot 4}\) = 1.33

Refractive index of liquid, n_{l} = 1.63

∴ Apparent height with liquid in tank, h = \(\frac{\mathrm{H}}{n_{l}}\) = \(\frac{12 \cdot 5}{1 \cdot 63}\) = 7.7 cm

∴ Displacement of microscope, x = 9.4 – 7.7 = 1.7 cm

OR

Interference | Diffraction |

1. interference is caused by superposition of two waves starting from two coherent sourcs. | 1. Diffraction is caused by superposition of a number of waves starting from the slit. |

2. All bright and dark fringes are of equal width. | 2. Width of central bright fringe is double of all other maxima. |

Question 3.

In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the * output frequency of a full-wave rectifier for the same input frequency.

Answer:

Given, Input frequency = 50 Hz

For a half-wave rectifier, the output frequency is equal to the input frequency.

∴ Output frequency = 50 Hz

For a full-wave rectifier, the output frequency is twice the input frequency.

Output frequency = 2 × 50 = 100 Hz

Section – B

Question 4.

(a) Draw a schematic labelled ray diagram of a reflecting type telescope. ;

(b) Write two important advantage justifying why reflecting type telescopes are preferred over

refracting telescopes.

(c) The objective of a telescope is of larger focal length and of larger aperture (compared to the eyepiece). Why? Give reasons.

Answer:

(a)

(b) Advantages:

- It is free from chromatic aberration.
- Its resolving power is greater than refracting telescope due to larger aperture of mirror.

(c)

- The objective of a telescope have a larger focal length to obtain large magnifying power and

greater intensity of image. - The aperture of objective lens of a telescope is taken as large because this increases the light gathering capacity of the objective from the distant object. Consequently, a brighter image is formed.

Question 5.

Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron is . hydrogen atom undergoes transition from higher energy state (quantum number ni) to the lower state ‘ (rij). When electron in hydrogen atom jumps from energy state n_{i} = 4 to n_{f} = 3, identify the spectral series to which the emission lines belong.

Answer:

According to Bohr, energy is radiated in the form of a photon when the electron of an excited hydrogen atom returns from higher energy state to the lower energy state. So, energy is radiated in the form of a photon when electron jumps from higher energy orbit (n = n_{i}) to lower energy orbit (n = n_{f}).

n_{i} > n_{f}

The energy of the emitted photon is given by

hv = E_{ni} – E_{nf},

When electron jumps from n = 4 to n = 3 state, the spectral series is called Paschen series.

Question 6.

The figure shows a ray of light passing through a prism. If the refracted ray QR is parallel to the base BC,

show that (a) r_{1} = r_{2} = \(\frac{\mathrm{A}}{2}\), (b) angle of minimum deviation, δ_{m} = 2i – A.

Answer:

(a) We know that,

r_{1} + r_{2} = A

Since QR is parallel to BC

So, r_{1} = r_{2} and i = e

∴ 2r_{1} or 2r_{2} = A

⇒ r_{1} = r_{2} = \(\frac{A}{2}\)

(b) δ_{m} = Deviation at the first face + Deviation at the second face

= (i – r_{1}) + (e – r_{2}) = (i + e) – (r_{1} + r_{2})

= 2i – A

Question 7.

Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of \(\) in MeV/c^{2}.

Answer:

Given, 1 a.m.u. = 1.6605 × 10^{-27} kg

To convert it into energy units, we multiple it by c^{2} and find that energy equivalent

= 1.6605 × 10-^{-27} × (2.9979 × 10^{8})^{2} kg m^{2}/s^{2}

= 1.4924 × 10^{-10}J

\(\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}}\) eV = 931.5 MeV

or

1 a.m.u. = 931.5 MeV/c^{2}

\(\) = Mass no. = 16

energy for 16 amu = 16 × 931.6 = 14905-6 MeV

Mass defect = \(\frac{\text { energy }}{c^{2}}\)

= \(\frac{14905 \cdot 6 \mathrm{MeV}}{c^{2}}\)

Question 8.

(a) “Two independent monochromatic sources of light cannot produce a sustained interference

pattern”. Give reason.

(b) Light wave each of amplitude V and frequency ‘ω’, emanating from two coherent light sources superpose at a point. If the displacements due to these waves is given by y_{1} = a cos ωt and y_{2} = a cos (ωt + Φ) where Φ is the phase difference between the two, obtain the expression for the resultant intensity at the point.

OR

Derive Snell’s law of refraction using Huygens’ principle.

Answer:

(a) The condition for the sustained interference is that both the sources must be coherent (i.e., they must have the same wavelength and the same frequency, and they must have the same phase or constant phase difference).

Two sources are monochromatic if they have the same frequency and wavelength. Since they are independent, i.e., they have different phases with irregular difference, they are not coherent sources.

(b) Let the displacement of the waves form the sources S_{1} and S_{2} at point P on the screen at any time t be given by:

y_{1} = a cos ω t

and y_{2} = a cos (ωt + Φ)

Where, Φ is the constant phase difference between the two waves.

By the superposition principle, the resultant displacement at point P is given by:

y = y_{1} + y_{2}

y = a cos ωt + a cos (ωt + Φ)

y = 2a[cos (\(\frac{\omega t+\omega t+\phi}{2}\)) cos (\(\frac{\omega t-\omega t-\phi}{2}\))]

y = 2a cos (ωt + \(\frac{\phi}{2}\)) cos(\(\frac{\phi}{2}\)) …………… (i)

Let 2a cos (Φ/2) = A …………….. (ii)

Then, equation (i) becomes

y = A cos (ωt + Φ/2)

Now, we have

A^{2} = 4a^{2} cos^{2}\(\frac{\phi}{2}\) ……………. (iii)

The intensity of light is directly proportional to the square of the amplitude of the wave. The intensity of light at point on the screen is given by

I = 4a^{2} cos^{2} \(\frac{\phi}{2}\)

OR

Laws of refraction: Consider a plane wavefront AB incident on a surface PQ separating two media (1) and (2). The medium (1) is rarer, having refractive index n_{1} in which the light travels with a velocity c_{1} The medium (2) is denser, having refractive index n_{2}, in which the light travels with a velocity c_{2/sub>.}

At time f = 0, the incident wavefront AB touches the boundary separating two medium at A. The secondary wavelets from point B advance forward with a velocity c_{1} and after time f seconds touches at D, thus covering a distance BD = c_{1}t. In the same time interval of f seconds, the secondary wavelets from A, advance forward in the second medium, and travels with a speed of c_{2}. With the point A as the centre and a distance AC = c_{2}t, an envelope is drawn to obtain a new refracted wave front as CD.

Consider triangles BAD and ACD,

sin i = sin (∠BAD) = \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{c_{1} t}{\mathrm{AD}}\)

sin r = sin (∠ADC) = \(\frac{\mathrm{AC}}{\mathrm{AD}}=\frac{c_{2} t}{\mathrm{AD}}\)

⇒ \(\frac{\sin i}{\sin r}\) = \(\frac{c_{1} t}{c_{2} t}=\frac{c_{1}}{c_{2}}\)

⇒ \(\frac{\sin i}{\sin r}\) = \(\frac{c_{1}}{c_{2}}\)= constant

This constant is called the refractive index of the second medium with respect to the first medium.

\(\frac{c_{1}}{c_{2}}=\frac{n_{2}}{n_{1}}\)

∴ \(\frac{\sin i}{\sin r}\) = \(\frac{c_{1}}{c_{2}}=\frac{n_{2}}{n_{1}}\) = _{1}n_{2}

This is known as the snell’s law.

Question 9.

Explain solar cell in brief. Give some uses of solar cells.

Answer:

A solar cell is a junction diode which converts light energy into electrical energy. It is based on photovoltaic effect. The surface layer of p-region is made very thin so that the incident photons may easily penetrate to reach the junction which is the active region. In an operation in the photovoltaic mode (i.e., generation of voltage due to bombardment of optical photons); the materials suitable for photocells are silicon (Si), gallium arsenide (GaAs), cadmium sulphide (CdS) and cadmium selenide (CdSe).

Working: When photons of energy greater than energy band gap of (hv > E_{g}) are made to incident on the junction, electron-hole pairs are created which moves in opposite directions due to junction field. These are collected at two sides of junction, thus producing photovoltage giving rise to photocurrent. The characteristic curve of solar cell is shown above.

Uses of solar cells:

- Solar cells are used for charging storage batteries in day time, which can supply the power during night times.
- The solar cells are also used in artificial satellite to operate the various electrical instruments kept inside the satellite.
- They are used for generating electrical energy for cooking food and pumping water.
- Solar cells are used in calculators, wrist watches and light meters (in photography).
- Solar cells are used to produce electric power in remote areas, where electric power supply is not available.
- Solar cells are used in traffic signals.
- Solar cells are used in remote radiotelephones.

Question 10.

The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is

one-third the size of the object. The wavelength of the light inside the lens is \(\frac{2}{3}\), times the wavelength in free space. Find the radius of the curved surface of the lens.

Answer:

Image is formed behind the lens.

∴ υ = + 8m

As the image is real, m = \(\frac{\mathrm{I}}{\mathrm{O}}\) = \(\frac{v}{u}\) = – \(\frac{1}{3}\)

u = – 3υ = – 3(8 m) = – 24 m

According to lens formula

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

⇒ \(\frac{1}{8}-\frac{1}{-24}\) = \(\frac{1}{f}\) or f = 6 m

Refractive index of the material of the lens is μ = Wavelength of the light in free space \(\frac{\text { Wavelength of the light in free space }}{\text { Wavelength of light inside the lens }}=\frac{\lambda_{0}}{\frac{2}{3} \lambda_{0}}=\frac{3}{2}\)

According to lens maker’s formula

\(\frac{1}{f}\) = \(\frac{(\mu-1)}{R}\) or R =f(μ – 1) ………………. (i)

Substituting the value of p and/in eqn. (i), we get

R = (6 m) (1.5 – 1) = 3 m.

Question 11.

(a) State clearly how a microwave oven works to heat up a food item containing water molecules.

(b) Why are microwaves found useful for the radar systems in aircraft navigation?

OR

(a) Two monochromatic waves emanating from two coherent sources have the displacement represented by y_{1} = a cos ωt and y_{2} = a cos (ωt + Φ) where Φ is the phase difference between the two displacements. Show that the resultant intensity at a point due to their superposition is given by I = 4I_{0} cos^{2} Φ/2, where I_{0} = a^{2}.

(b) Hence obtain the conditions for constructive and destructive interference.

Answer:

(a) In microwave oven, the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves get transferred efficiently to the kinetic energy of the molecules. This kinetic energy raises the temperature of any food containing water.

(b) Microwaves are short wavelength radio waves, with frequency of order of GHz. Due to short wavelength, they have high penetrating power with respect to atmosphere and less diffraction in the atmospheric layers. So these waves are suitable for the radar systems used in aircraft navigation.

OR

(a)

y_{1} = a cos ωt, y_{2} = a cos (ωt + Φ)

where Φ is phase difference between them. Resultant displacement at point P will be

y = y_{1}+ y_{2} = cos ωt + a cos (ωt + Φ)

= a[2cos \(\frac{(\omega t+\omega t+\phi)}{2}\) cos \(\frac{(\omega t-\omega t-\phi)}{2}\)]

y_{2} = 2a cos[ωt + \(\frac{\phi}{2}\)) cos (\(\frac{\phi}{2}\)) ……….. (i)

Let y = 2a cos (\(\frac{\phi}{2}\) = A, the equation (i) becomes

y = A cos (ωt + \(\frac{\phi}{2}\)

where A is amplitude of resultant wave,

Now,

A = 2a cos (\(\frac{\phi}{2}\))

On squaring, A^{2} = 4a^{2} cos^{2}(\(\frac{\phi}{2}\))

Hence resultant intensity, I = 4I_{0} cos^{2}(\(\frac{\phi}{2}\))

(b) Condition for constructive interference,

cos ΔΦ = + 1

2π\(\frac{\Delta x}{\lambda}\) = 0, 2π, 4π …

or Δx = nλ; n = 0,1, 2, 3,

Condition for destructive interference, cos ΔΦ = -1

2π\(\frac{\Delta x}{\lambda}\) = π, 3π, 5π …

or

Δx = (2n – 1) \(\frac{\lambda}{2}\)

Where n = 1, 2, 3 …..

Section – C

Question 12.

Photoelectric effect refers to the emission of electrons from certain metal surfaces when irradiated with high frequency light (em radiation). (UV rays falling on cesium metal is a good example).

Number of electrons emitted is proportional to the intensity of light. KE of electron emitted depends on the frequency of incident light, it does not depend on intensity of light. The minimum frequency of light required to initiate photoelectric effect is called threshold frequency (v_{0}). The minimum energy required to eject one electron from a metal surface is called its work function and is given by Φ = hv_{0}. Maximum KE of emitted electron when light of frequency v falls on the metal surface is given by KE_{max} = hv – Φ (Φ is the work function) Negative potential can be applied to the electrons to stop them first outside the metal surface.

(a) The intensity of light is plotted against number of photoelectrons . The graph is a:

(i) Straight line passing through origin

(ii) Straight line with a + y intercept

(iii) Parabola passing through the origin

(iv) None of these

(b) The work function of a metal corresponds to a frequency 2 × 10^{15} Hz of light. Which of the following frequency cannot cause photoelectric effect from the metal?

(i) 1.5 × 10^{16} Hz

(ii) 8.2 × 10^{14}HZ

(iii) 6.5 × 10^{15}Hz

(iv) 2.2 × 10^{15}Hz

(c) A particular metal can emit electrons when green light falls on it. In which of the following case, the maximum KE of emitted electrons cannot be more, compared to this case?

(i) Using blue light with the same metal

(ii) Using a metal of less work function with the same light (green)

(iii) Using red light with the first metal

(iv) Using UV light with the second metal

(d) If frequency of incident light becomes n times the initial frequency (v > v_{0}), then KE of emitted electron becomes:

(i) n times the initial KE

(ii) More than n times the initial KE

(iii) Less than n times the initial KE

(iv) No change in KE

(e) Stopping potential (negative potential required to stop the electron just outside the metal surface is related to.

(i) Mean wavelength

(ii) Shortest frequency

(iii) Maximum KE of emitted electrons

(iv) Intensity of light incident

Answer:

(a) (i) Straight line passing through origin

Explanation: n ∝ I No. electrons are emitted without suitable light ⇒ n = 0 for I = 0

(b) (ii) 8.2 × 10^{14} Hz

Explanation: Frequency of light should be more than threshold frequency for photoelectric emission.

(c) (iii) Using red light with the first metal

Explanation: KE_{max} = (hv – Φ). Increasing v and reducing Φ will make KE_{max} more

(d) (ii) More than n times the initial KE.

Explanation: KE_{1} = hv – Φ:n KE_{1} = nhv – nΦ

KE_{2} = hv – Φ

⇒ KE_{2} > n KE_{1}

(e) (ii) Shortest frequency

Explanation: Stopping potential is related to the maximum KE of emitted electrons.