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## CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 7 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.

What is the principle of a half wave rectifier?

Answer:

Principle: It is based on the principle that the junction diode offers low resistance path, when forward biased and high resistance, when reverse biased. When a.c. input is applied to a junction diode, it gets forward biased during one half cycle and reverse biased during the next opposite half cycle.

Thus, output is obtained during alternate half cycles of the a.c. input.

Question 2.

The photoelectric cut-off voltage in a certain photoelectric experiment is 2-5 V. What is. the maximum kinetic energy of photoelectrons emitted?

OR

The wavelength of some of the spectral lines obtained in hydrogen spectrum are 9,546 Ã…, 6,463 Ã… and 1,216 Ã…, which one of these wavelengths belongs to Lyman series?

Answer:

Given: Cut-off voltage, V_{0} = 2.5 V

Maximum kinetic energy of photoelectrons,

E_{k} = eV_{0} = 1.6 Ã— 10^{-19} Ã— 2.5

= 4 Ã— 10^{-19} J.

OR

Lyman series lies in ultraviolet region. Therefore, wavelength of the spectral lines in Lyman series must be less than the wavelength in visible region i.e., 3,900 A. Therefore, spectral line of wavelength 1,216 Ã… lies in Lyman series.

Question 3.

A photodiode is fabricated from a semiconductor with a band gap of 2-8 eV. Can it detect wavelength of 6000 nm? Justify.

Answer:

Wavelength, Î» = 6000 nm

Energy of incident photon, E = hv = \(\frac{h c}{\lambda}\)

= \(\frac{6 \cdot 6 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-9}}\)J

= 3.3 Ã— 10 J

\(\frac{3.3 \times 10^{-20}}{1 \cdot 6 \times 10^{-19}}\) eV

E = 0-206 eV

As the energy of the photon is less than energy band gap (E_{g} = 2-8 eV) of the semiconductor, so a wavelength of 6000 nm cannot be detected.

Section – B

Question 4.

When four hydrogen nuclei combine to form a helium nucleus estimate the amount of energy in MeV released in this process of fusion. (Neglect the masses of electrons and neutrons)

Given:

(i) Mass of \({ }_{1}^{1} \mathrm{H}\) = 1.007825 u

(ii) Mass of helium nucleus = 4.002603 u

1u = 931 MeV/c^{2}.

Answer:

Energy released = Î”m Ã— 931 MeV

Î”m = 4m(\({ }_{1}^{1} \mathrm{H}\))-m(\({ }_{2}^{4} \mathrm{He}\))

Energy released Q = [4m(latex]{ }_{1}^{1} \mathrm{H}[/latex]) – m(latex]{ }_{2}^{4} \mathrm{He}[/latex])] Ã— 931 MeV

= [4 Ã— 1.007825 – 4.002603] Ã— 931 MeV

= 26.72 MeV.

Question 5.

In the following diagram, an object ‘O’ is placed 15 cm in front of a convex lens L_{1} of focal length 20 cm and the final image is formed at I at a distance of 80 cm from the second lens L_{2}. Find the focal length of the lens L_{2}.

Answer:

As per the figure,

The image formed by the lens L is at P. Therefore

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

As per the parameters given in the question

u = -15 cm, f_{L1} = 20 cm

So, the image distance will be

\(\frac{1}{v}-\frac{1}{(-15)}=\frac{1}{20}\)

â‡’ Ï… = – 60 cm

Now, this image is acting as an object for the lens L_{2}.

We can again use the lens formula and other parameters given in the question and question figure to find the focal length of lens L_{2}.

\(\frac{1}{v_{\mathrm{L}_{2}}}-\frac{1}{u_{\mathrm{L}_{2}}}=\frac{1}{f_{\mathrm{L}_{2}}}\)

Here, U_{L2} = Ï… + (-20)

= -60-20 = -80 cm

and, Ï…_{L2} = 80cm

\(\frac{1}{80}-\frac{1}{(-80)}=\frac{1}{f_{\mathrm{L}_{2}}}\)

â‡’ f_{L2}2 = 40cm

So, the focal length of the lens L_{2} = 40 cm.

Question 6.

Write any two distinguishing features between conductors, semiconductors and insulators on the basisof energy band diagrams.

Answer:

Conductors:

(i) In case of conductors, the valence band is completely filled and the conduction band can have two cases-either it is partially filled with an extremely small energy gap between the valence and conduction bands or it is empty, with the two bands overlapping each other as shown in figure:

(ii) Even when a small current is applied, conductors can conduct electricity.

[Here, C.B. â†’ Conduction Band. V.B. â†’ Valence Band, E_{g} â†’Forbidden energy gap]

Insulators:

- En case of insulators, the energy gap between the conduction and valence bands is very large and the

conduction band is practically empty. - When an electric field is applied to such kind of material, the electrons find hard to receive such

a large amount of energy to reach the conduction band. Thus, the conduction band remains to be

empty. That is why no current flows through insulators.

Semiconductors:

- In case of semiconductor, the energy band structure of semiconductors is similar to insulators, but in this case, the size of forbidden energy gap is quite smaller than that of the insulators.
- When an electric field is applied to a semiconductor, the electrons in the valence band find it relatively easier to jump to the conduction band. So, the conductivity of semiconductors lies between the conductivity of conductors and insulators.

Question 7.

(a) Draw the intensity pattern for single slit diffraction and double slit interference.

(b) For a given single slit, the diffraction pattern is obtained on a fixed screen, first by using red light and then with blue light. In which diffraction pattern, have a larger angular width?

Answer:

(a) Single slit diffraction

Double slit interference

(b) Angular width of central maxima is given by

2Î¸ = \(\frac{2 \lambda}{a}\)

Since, Î»_{r} > Î»_{b}

Therefore, width of central maxima of red light is greater than the width of central maxima of blue light.

Question 8.

(a) Define magnifying power of a telescope. Write its expression.

(b) 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 slit of width a is illuminated by light of wavelength 6000 A. For what value of a will the:

(i) First maximum fall at an angle of diffraction of 30Â°?

(ii) First minimum fall at an angle of diffraction 30Â°?

Answer:

(a) Magnifying power of refracting telescope (M) is defined as the ratio of the angle subtended by the image (Î²) at the eye to the angle subtended by the distant object at the unaided eye (Î±).

M = \(\frac{\beta}{\alpha}\)

(b) Here, f_{0} = 150 cm,f_{e} = 5 cm

Angle subtended by 100 m tall tower at 3 km is

Î± = \(\frac{100}{3 \times 1000}=\frac{1}{30}\) rad

If h is the height of image formed by the objective, then

Î± = \(\frac{h}{f_{0}}=\frac{h}{150}\)

âˆ´ \(\frac{h}{150}=\frac{1}{30}\)

or h = \(\frac{150}{30}\) cm = 5 cm

Magnification produced by eyepiece

m_{e} = (1 + \(\frac{\mathrm{D}}{f_{e}}\)) = (1 + \(\frac{25}{5}\)) = 6

âˆ´ Height of final image = h Ã— m_{e} = 5 Ã— 6 = 30 cm.

OR

Î» = 6000 Ã… = 6000 Ã— 10^{-10} m

= 6 Ã— 10^{-7} m

Î¸_{1} = 30Â°

(i) For first maximum, sinÎ¸_{m} = \(\frac{\left(m+\frac{1}{2}\right) \lambda}{a}\)

sinÎ¸_{1} = \(\frac{3 \lambda}{2 a}\) or a = \(\frac{3 \lambda}{2 \sin \theta_{1}}\) = \(\frac{3 \times 6 \times 10^{-7}}{2 \times \sin 30^{\circ}}\)

1.8 Ã— 10^{-6}m = 1.8Î¼m

(ii) For first minimum,

sinÎ¸_{m} = \(\frac{m \lambda}{a}\)

âˆ´ sinÎ¸_{1} = \(\frac{\lambda}{a}\)

â‡’ a = \(\frac{\lambda}{\sin \theta_{1}}\) = \(\frac{6 \times 10^{-7}}{\sin 30^{\circ}}\)

= 1.2 Ã— 10^{-6} m = 1.2 Î¼m.

Question 9.

Plot a labelled graph of stopping potential of photoelectrons (Vs) versus frequency (v) of incident radiation. How will you use this graph to determine the value of Planck’s constant? Explain.

Answer:

The graph for V_{s} and v is as follows:

Determination of Planck’s constant: As from the graph it is clear that slope of the graph is given by

Slope = \(\frac{h}{e}\)

where, h = Planck’s constant

e = Charge on an electron

Now, from Einstein’s photoelectric equation

K_{max} = hv – Î¦

â‡’ eV_{0} = hv – Î¦

â‡’ V_{0} = (\(\frac{h}{e}\))v – \(\frac{\phi}{e}\)

Comparing this equation with y = mx + c.

Slope of (V_{0} – v) graph = \(\frac{h}{e}\)

So, Planck’s constant h = Slope (V_{0} – v) graph

Question 10.

In the original experiment, Geiger and Marsden calculated the distance of closest approach to the gold nucleus (Z = 79) of a 7-7 MeV a-particle before it comes momentarily to rest and reverses its direction. What is its value and how will the distance of the closest approach be affected when the kinetic energy of a-particle is doubled?

Answer:

Kinetic energy of Î±-particle,

E = \(\frac{1}{2}\) mv^{2} = 7.7 MeV

= 7.7 Ã— 1.6 Ã— 10^{-13} J

= 1.2 Ã— 10^{-12} J

Here, Z = 79, e = 1.6 Ã— 10^{-19} C

r_{0} = 3.0 Ã— 10^{-14} m

= 30 fermi.

If kinetic energy of Î±-particle is double, the distance of closest approach is halved.

Question 11.

The magnetic field in a plane electromagnetic wave is given by

B_{y} = (2 x 10^{-7}T) sin (0.5 x 10^{3} x + 1.5 >< 10^{11}t)

(i) What is the wavelength and frequency of the wave?

(ii) Write an expression for the electric field.

OR

In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.

Answer:

(i) Comparing the given equation with the standard equation of electromagnetic wave,

B = B_{0} sin (kx + Ï‰t)

We have, k = \(\frac{2 \pi}{\lambda}\) = 035 Ã— 10^{3}

âˆ´ Î» = \(\frac{2 \pi}{0.5 \times 10^{3}}\) m = 1.26 cm

Ï‰ = 2Ï€v = 1.5 Ã— 10^{11}

âˆ´ v = \(\frac{1 \cdot 5 \times 10^{11}}{2 \pi}\)Hz = 2.38 Ã— 10^{10} Hz

(ii) Given,

B_{0} = 2 Ã— 10^{-7}T

E_{0} = B_{0} c = 2 Ã— 10^{-7} Ã— 3 Ã— 10^{8}

= 60 NC^{-1}

The electric field is perpendicular to both, the magnetic field and the direction of propagation of wave. Thus,

E_{z} = (60 NC^{-1}) sin (0.5 Ã— 10^{3}x + 1.5 Ã— 10^{11}t)

OR

In Young’s double slit experiment, the wave fronts from the two illuminated slits superpose on the screen. This results in formation of alternate bright and dark fringes because of constructive and destructive interference, respectively. The intensity of light is maximum at the centre C of the screen and it is called central maxima.

Let S_{1} and S_{2} be two slits separated by a distance d. GG’ is the screen at a distance D from the slits and S_{2}. Both the slits are equidistant from point C. The intensity of light will be maximum at this point due to the path difference of the waves reaching this point will be zero.

At point P, the path difference between the rays coming from the slits S_{1} and S_{2} is S_{2}P – S_{1}P.

Now, S_{1}S_{2 } = d

EF = d_{1} and S_{2}F = D

âˆ´ In Î”S_{2}PF,

Similarly, in Î”S_{1}PE

For bright fringes (constructive interference), the path difference is an integral multiple of wavelength,

i.e., path difference is nÎ».

The separation between the centres of two consecutive dark interference fringes is the width of a bright fringes.

The separation between the centres of two consecutive dark interference fringes is the width of a bright fringes.

âˆ´ Î²_{2} = x_{n} – x_{n-1} = \(\frac{\lambda \mathrm{D}}{d}\)

Î²_{1} = Î²_{2}

All the bright and dark fringes are of equal width as Î²_{1} = Î²_{2}.

Section – C

Question 12.

Sparking Brilliance of Diamond: The total internal reflection of the light is used in polishing diamonds to create a sparking brilliance. By polishing the diamond with specific cuts, it is adjusted the most of the light rays approaching the surface are incident with an angle of incidence more than critical angle. Hence, they suffer multiple reflections and ultimately come out of diamond from the top. This gives the diamond a sparking brilliance.

(a) Light cannot easily escape a diamond without multiple internal

reflections. This is because:

(i) Its critical angle with reference to air is too large

(ii) Its critical angle with reference to air is too small

(iii) The diamond is transparent

(iv) Rays always enter at angle greater than critical angle

(b) The critical angle for a diamond is 24.4Â°. Then its refractive index is:

(i) 2.42

(ii) 0.413

(iii) 1

(iv) 1.413

(c) The basic reason for the extraordinary sparkle of suitably cut diamond is that:

(i) Low refractive index

(ii) High refractive index

(iii) High transparency

(iv) Very hard

(d) A diamond is immersed in a liquid with a refractive index greater than water. Then the critical angle for total internal reflection will:

(i) depend on the nature of the liquid

(ii) Remain the same

(iii) Decreases

(iv) Increases

(e) The following diagram shows same diamond cut in two different shapes.

The brilliance of diamond in the second diamond will be:

(i) Less than the first

(ii) Same as first

(iii) Greater than first

(iv) Will depend on the intensity of light

Answer:

(a) (ii) Its critical angle with reference to air is too small

(b) (i) 2.42

(c) (ii) High refractive index

(d) (iv) Increases

(e) (i) Less than the first