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Class XII – Physics Practice Paper – 1 – Answers

Subject: Physics

Class XII

Time: 3 hours

Max. Marks: 70

General Instructions:

(1) There are 35 questions in all. All questions are compulsory

(2) This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections are compulsory.

(3) Section A contains eighteen MCQ of 1 mark each, Section B contains seven questions of two marks each, Section C contains five questions of three marks each, section D contains three long questions of five marks each and Section E contains two case study based questions of 4 marks each.

(4) There is no overall choice. However, an internal choice has been provided in section B, C, D and E. You have to attempt only one of the choices in such questions.

5. Use of calculators is not allowed.

SECTION A

1. According to Coulomb’s law, which is the correct relation for the following figure?

(i) q1 q2 > 0

(ii) q1 q2 < 0

(iii) q1 q2 = 0

(iv) 1 > q1/ q2 > 0

2. The electric potential on the axis of an electric dipole at a distance ‘r from its centre is V. Then the potential at a point at the same distance on its equatorial line will be

(i) 2V

(ii) -V

(iii) V/2

(iv) Zero

3. The temperature (T) dependence of resistivity of materials A and material B is represented by fig (i) and fig (ii) respectively. Identify material A and material B.

(i) material A is copper and material B is germanium

(ii) material A is germanium and material B is copper

(iii) material A is nichrome and material B is germanium

(iv) material A is copper and material B is nichrome

4. Two concentric and coplanar circular loops P and Q have their radii in the ratio 2:3. Loop Q carries a current 9 A in the anticlockwise direction. For the magnetic field to be zero at the common centre, loop P must carry

(i) 3A in clockwise direction

(ii) 9A in clockwise direction

(iii) 6 A in anti-clockwise direction

(iv) 6 A in the clockwise direction.

5. A long straight wire of circular cross section of radius a carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnitudes of magnetic field at a point distant a/2 above the surface of wire to that at a point distant a/2 below its surface is

(i) 4 :1

(ii) 1:1

(iii) 4: 3

(iv) 3 :4

6. If the magnetizing field on a ferromagnetic material is increased, its permeability

(i) decreases

(ii) increases

(iii) remains unchanged

(iv) first decreases and then increases

7. An iron cored coil is connected in series with an electric bulb with an AC source as shown in figure. When iron piece is taken out of the coil, the brightness of the bulb will

(i) decrease

(ii) increase

(iii) remain unaffected

(iv) fluctuate

8. Which of the following statement is NOT true about the properties of electromagnetic waves?

(i) These waves do not require any material medium for their propagation

(ii) Both electric and magnetic field vectors attain the maxima and minima at the same time

(iii) The energy in electromagnetic wave is divided equally between electric and magnetic fields

(iv) Both electric and magnetic field vectors are parallel to each other

9. A rectangular, a square, a circular and an elliptical loop, all in the (x-y) plane, are moving out of a uniform magnetic field with a constant velocity 𝑣⃗ = v𝚤̂. The magnetic field is directed along the negative z-axis direction. The induced emf, during the passage of these loops, out of the field region, will not remain constant for

(i) any of the four loops

(ii) the circular and elliptical loops

(iii) the rectangular, circular and elliptical loops

(iv) only the elliptical loops

10. In a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is 𝟏𝟖 th of the wavelength. The ratio of intensity at this point to that at the centre of a bright fringe is close to

(i) 0.80

(ii) 0.74

(iii) 0.94

(iv) 0.85

11. The work function for a metal surface is 4.14 eV. The threshold wavelength for this metal surface is:

(i) 4125 Å

(ii) 2062.5 Å

(iii) 3000 Å

(iv) 6000 Å

12. The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. The radius of the n =3 orbit is

(i) 1.01 x 10-10 m

(ii) 1.59 X 10-10 m

(iii) 2.12 x 10-10 m

(iv) 4.77 X 10-10 m

13. Which of the following statements about nuclear forces is not true?

(i) The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres.

(ii) The nuclear force is much weaker than the Coulomb force.

(iii) The force is attractive for distances larger than 0.8 fm and repulsive if they are separated by distances less than 0.8 fm.

(iv) The nuclear force between neutron-neutron, proton-neutron and proton-proton is approximately the same.

14. If the reading of the voltmeter V1 is 40 V, then the reading of voltmeter V2 is

(i) 30 V

(ii) 58 V

(iii) 29 V

(iv) 15 V

15. The electric potential V as a function of distance X is shown in the figure.

The graph of the magnitude of electric field intensity E as a function of X is

Ans. Option (i)

Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.

16. ASSERTION(A): The electrical conductivity of a semiconductor increases on doping.

REASON: Doping always increases the number of electrons in the semiconductor.

a) Both A and R are true and R is the correct explanation of A

b) Both A and R are true and R is NOT the correct explanation of A

c) A is true but R is false

d) A is false and R is also false

17. ASSERTION: In an interference pattern observed in Young’s double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.

REASON: Fringe width is proportional to (d/D).

a) Both A and R are true and R is the correct explanation of A

b) Both A and R are true and R is NOT the correct explanation of A

c) A is true but R is false

d) A is false and R is also false

18. ASSERTION (A): The photoelectrons produced by a monochromatic light beam incident on a metal surface have a spread in their kinetic energies.

REASON (R): The energy of electrons emitted from inside the metal surface, is lost in collision with the other atoms in the metal.

a) Both A and R are true and R is the correct explanation of A

b) Both A and R are true and R is NOT the correct explanation of A

c) A is true but R is false

d) A is false and R is also false

SECTION B

19. Electromagnetic waves with wavelength

(i) λ1 is suitable for radar systems used in aircraft navigation.

(ii) λ2 is used to kill germs in water purifiers.

(iii) λ3 is used to improve visibility in runways during fog and mist conditions.

Identify and name the part of the electromagnetic spectrum to which these radiations belong. Also arrange these wavelengths in ascending order of their magnitude.

Ans. λ1 -Microwave

λ2 – ultraviolet

λ3- infrared

Ascending order – λ2<λ3<λ1

20. A uniform magnetic field gets modified as shown in figure when two specimens A and B are placed in it.

(i) Identify the specimen A and B.

(ii) How is the magnetic susceptibility of specimen A different from that of specimen B?

Ans. A – diamagnetic

B- paramagnetic

The magnetic susceptibility of A is small negative and that of B is small positive.

21. What is the nuclear radius of 125Fe, if that of 27Al is 3.6 fermi?

Ans. From the relation R = R0A 1/3, where R0 is a constant and A is the mass number of a nucleus

R Fe/RAl = (AFe/AAl)

= (125/27)

R Fe = 5/3 RAl

=5/3 x 3.6

= 6 fermi

OR

The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 A0. Calculate the short wavelength limit for the Balmer series of the hydrogen spectrum.

Ans. Given short wavelength limit of Lyman series

22. A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Justify your answer.

Ans.

The value of (µ – 1) is negative and ‘f’ will be negative. So it will behave like diverging lens.

23. The figure shows a piece of pure semiconductor S in series with a variable resistor R and a source of constant voltage V. Should the value of R be increased or decreased to keep the reading of the ammeter constant, when semiconductor S is heated? Justify your answer

Ans. To keep the reading of ammeter constant value of R should be increased as with the increase in temperature of a semiconductor, its resistance decreases and current tends to increase.

OR

The graph of potential barrier versus width of depletion region for an unbiased diode is shown in graph A. In comparison to A, graphs B and C are obtained after biasing the diode in different ways. Identify the type of biasing in B and C and justify your answer.

Ans. B – reverse biased

In the case of reverse biased diode, the potential barrier becomes higher as the battery further raises the potential of the n side.

C -forward biased

Due to forward bias connection the potential of P side is raised and hence the height of the potential barrier decreases

24. A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength λ equal to 6000 Å and the angular width of the central maximum in the resulting diffraction pattern is measured. When the slit is next illuminated by light of wavelength λ’, the angular width decreases by 30%. Calculate the value of the wavelength λ’.

Ans. Angular width 2φ = 2λ/d

Given λ = 6000 Å

In Case of new λ (assumed λ’ here),

angular width decreases by 30%

New angular width = 0.70 (2 φ)

2 λ’/d = 0.70 X (2 λ/d)

λ’= 4200 Å

25. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.7 × 10-22 C/m2 . What is electric field intensity E:

(a) in the outer region of the first plate, and

(b) between the plates?

Ans.

Surface charge density of plate A = +17.7 × 10–22 C/m2

Surface charge density of plate B = -17.7 × 10–22 C/m2

(a) In the outer region of plate I, electric field intensity E is zero.

(b) Electric field intensity E in between the plates is given by relation

SECTION C

26. Two long straight parallel conductors carrying currents I1 and I2 are separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field produced by one exerts an attractive force on the other. Obtain the expression for this force and hence define 1 ampere.

Ans. Diagram

Derivation

The ampere is the value of that steady current which, when maintained in each of the two very long, straight, parallel conductors of negligible cross-section, and placed one metre apart in vacuum, would exert on each of these conductors a force equal to 2 × 10–7 newtons per metre of length.

27. The magnetic field through a circular loop of wire, 12cm in radius and 8.5Ω resistance, changes with time as shown in the figure. The magnetic field is perpendicular to the plane of the loop. Calculate the current induced in the loop and plot a graph showing induced current as a function of time.

Ans.

28. An a.c. source generating a voltage ε = ε0 sin ωt is connected to a capacitor of capacitance C. Find the expression for the current I flowing through it. Plot a graph of ε and I versus wt to show that the current is ahead of the voltage by π/2.

Ans. Derivation

OR

An ac voltage V = V0 sin wt is applied across a pure inductor of inductance L. Find an expression for the current i, flowing in the circuit and show mathematically that the current flowing through it lags behind the applied voltage by a phase angle of 2 p . Also draw graphs of V and i versus wt for the circuit.

Ans. Derivation

29. Radiation of frequency 1015 Hz is incident on three photosensitive surfaces A, B and C. Following observations are recorded: Surface A: no photoemission occurs Surface B: photoemission occurs but the photoelectrons have zero kinetic energy.

Surface C: photo emission occurs and photoelectrons have some kinetic energy. Using Einstein’s photo-electric equation, explain the three observations.

Ans. From the observations made (parts A and B) on the basis of Einstein’s photoelectric equation, we can draw following conclusions:

1. For surface A, the threshold frequency is more than 1015 HZ, hence no photoemission is possible.

2. For surface B the threshold frequency is equal to the frequency of given radiation. Thus, photo-emission takes place but kinetic energy of photoelectrons is zero.

3. For surface C, the threshold frequency is less than 1015 Hz. So photoemission occurs and photoelectrons have some kinetic energy

OR

The graph shows the variation of photocurrent for a photosensitive metal

(a) What does X and A on the horizontal axis represent?

(b) Draw this graph for three different values of frequencies of incident radiation ʋ1, ʋ2 and ʋ3321) for the same intensity.

(c) Draw this graph for three different values of intensities of incident radiation I1, I2 and I3 (I3> I2>I1) having the same frequency.

Ans. A – cut off or stopping potential

X – anode potential

30. The ground state energy of hydrogen atom is -13.6 eV. The photon emitted during the transition of electron from n=3 to n=1 state, is incident on a photosensitive material of unknown work function .The photoelectrons are emitted from the material with the maximum kinetic energy of 9eV.Calculate the threshold wavelength of the material used.

Ans.

SECTION D

31. (a) Draw equipotential surfaces for (i) an electric dipole and (ii) two identical positive charges placed near each other.

(b) In a parallel plate capacitor with air between the plates, each plate has an area of 6 x 10-3m2 and the separation between the plates is 3 mm.

(i) Calculate the capacitance of the capacitor.

(ii) If the capacitor is connected to 100V supply, what would be the charge on each plate?

(iii) How would charge on the plate be affected if a 3 mm thick mica sheet of k=6 is inserted between the plates while the voltage supply remains connected?

Ans.

Here, A =6 x 10-3 m2 , d =3mm=3×10-3 m

(i) Capacitance, C=0A/d = (8.85 x 10-12 x 6 x 10-3/3 x 10-3) = 17.7 x 10-12 F

(ii) Charge, Q = CV =17.7 x 10-12 x 100=17.7 x 10-10C

(iii) New charge Q’= KQ=6 x 17.7 x 10-10 =1.062 x 10-8 C

OR

(a) Three charges –q, Q and –q are placed at equal distances on a straight line. If the potential energy of the system of these charges is zero, then what is the ratio Q:q?

(b) (i) Obtain the expression for the electric field intensity due to a uniformly charged spherical shell of radius R at a point distant r from the centre of the shell outside it.

(ii) Draw a graph showing the variation of electric field intensity E with r, for r > R and r < R.

Ans. Diagram

(i) When point P lies outside the spherical shell: Suppose that we have calculate field at the point P at a distance r (r>R) from its centre. Draw Gaussian surface through point P so as to enclose the charged spherical shell. Gaussian surface is a spherical surface of radius r and centre O.

(ii)

32. (a) Explain the term drift velocity of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of drift velocity.

(b) Two cells of emfs E1 and E2 and internal resistances 𝑟1 and 𝑟2 respectively are connected in parallel as shown in the figure. Deduce the expression for the

(i) equivalent emf of the combination

(ii) equivalent internal resistance of the combination

(iii) potential difference between the points A and B.

Ans. Here, I = I1 + I2                                                                          …(i)

Let V = Potential difference between A and B.

OR

(a) State the two Kirchhoff’s rules used in the analysis of electric circuits and explain them.

(b) Derive the equation of the balanced state in a Wheatstone bridge using Kirchhoff’s laws.

Ans. (a) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction

(b) Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero Derivation

33. a) Draw the graph showing intensity distribution of fringes with phase angle due to diffraction through a single slit. What is the width of the central maximum in comparison to that of a secondary maximum?

b) A ray PQ is incident normally on the face AB of a triangular prism of refracting angle 600 as shown in figure. The prism is made of a transparent material of refractive index 2/√3. Trace the path of the ray as it passes through the prism. Calculate the angle of emergence and the angle of deviation.

Ans.

Width of central maximum is twice that of any secondary maximum

(b)

OR

a) Write two points of difference between an interference pattern and a diffraction pattern.

b) (i) A ray of light incident on face AB of an equilateral glass prism, shows minimum deviation of 30°. Calculate the speed of light through the prism.

(ii) Find the angle of incidence at face AB so that the emergent ray grazes along the face AC.

Ans. (a) (i) The interference pattern has a number of equally spaced bright and dark bands. The diffraction pattern has a central bright maximum which is twice as wide as the other maxima. The intensity falls as we go to successive maxima away from the centre, on either side.

(ii) We calculate the interference pattern by superposing two waves originating from the two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.

(b)

34. Case Study:

Read the following paragraph and answer the questions.

A number of optical devices and instruments have been designed and developed such as periscope, binoculars, microscopes and telescopes utilising the reflecting and refracting properties of mirrors, lenses and prisms. Most of them are in common use. Our knowledge about the formation of images by the mirrors and lenses is the basic requirement for understanding the working of these devices.

(i) Why the image formed at infinity is often considered most suitable for viewing. Explain

(ii) In modern microscopes multicomponent lenses are used for both the objective and the eyepiece. Why?

(iii) Write two points of difference between a compound microscope and an astronomical telescope

Ans. (i) When the image is formed at infinity, we can see it with minimum strain in the ciliary muscles of the eye.

(ii) The multi-component lenses are used for both objective and the eyepiece to improve image quality by minimising various optical aberrations in lenses.

(iii) (a)The compound microscope is used to observe minute nearby objects whereas the telescope is used to observe distant objects. (b) In compound microscope the focal length of the objective is lesser than that of the eyepiece whereas in telescope the focal length of the objective is larger than that of the eyepiece.

OR

(iii) Write two distinct advantages of a reflecting type telescope over a refracting type telescope.

Ans. (a) The image formed by reflecting type telescope is brighter than that formed by refracting telescope.

(b) The image formed by the reflecting type telescope is more magnified than that formed by the refracting type telescope.

35. Case study: Light emitting diode.

Read the following paragraph and answer the questions

LED is a heavily doped P-N junction which under forward bias emits spontaneous radiation. When it is forward biased, due to recombination of holes and electrons at the junction, energy is released in the form of photons. In the case of Si and Ge diode, the energy released in recombination lies in the infrared region. LEDs that can emit red, yellow, orange, green and blue light are commercially available. The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 eV. The compound semiconductor Gallium Arsenide – Phosphide is used for making LEDs of different colours.

LEDs of different kinds

(i). Why are LEDs made of compound semiconductor and not of elemental semiconductors?

(ii) What should be the order of bandgap of an LED, if it is required to emit light in the visible range?

(iii) A student connects the blue coloured LED as shown in the figure. The LED did not glow when switch S is closed. Explain why?

Ans. (i) LEDs are made up of compound semiconductors and not by the elemental conductor because the band gap in the elemental conductor has a value that can detect the light of a wavelength which lies in the infrared (IR) region

(ii) 1.8 eV to 3 eV

(iii) LED is reversed biased that is why it is not glowing.

OR

(iii) Draw V-I characteristic of a p-n junction diode in

(i) forward bias and (ii) reverse bias

Ans. V-I Characteristic curves of pn junction diode in forward biasing and reverse biasing.