seshasaiphysicsblogxii.files.wordpress.com€¦ · Web viewThere is not overall choice. However, an internal choice has been provided in one question of two marks, one question of

UNIT NAME 1 MARK

2 MARKS

3 MARKS

4 MARKS ( VBQ)

5 MARKS TOTAL

ELECTROSTATICS

CURRENT ELECTRICITY

6(2) 4(1) 5(1) 15(4)

MAGNETISM & MAGNETIC EFFECTS OF CURRENT

AC & EMI

1(1) 4(2) 6(2) 5(1)16(6)

EMW &

OPTICS3(3) 6(3) 3(1) 5(1)

17(8)

DUAL NATURE OF MATTER & ATOMS &NUCLEI

1(1) 9(3)10(4)

SEMICONDUCTOR DEVICES

COMMUNICATION

12(4)12(4)

TOTAL 5(5) 10(5) 36(12) 4(1) 15(3) 70(26)

BLUE PRINT

B.Sesha Sai/KV, Embassy Of India, Kathmandu/PHY/XI Science.

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MODEL QUESTION PAPER 2016-17

PHYSICS (CLASS XII)

Time : 3 Hrs. Max. Marks: 70

General Instructions :

(a) All questions are compulsory.

(b) This question paper has five sections : Section A, Section B, Section C, Section D and

Section E

(c) There are 26 questions in total. Questions 1 to 5are very short answer type questions and

carry one mark each.Questions 6 to 10 carry two marks each, questions 11 to 22 carry

three marks each, question 23 is value based question carry 4 marks and questions 24 to

26 carry five marks each.

(d) There is not overall choice. However, an internal choice has been provided in one question

of two marks, one question of three marks and all three question of five marks each

weightage. You have to attempt only one of the choices in such questions.

(e) Use of calculators is not permitted. However, you may use log tables if necessary.

(f) You may use the following values of physical constants wherever necessary.

c = 3 × 108m/s

h = 6.63 × 10–34Js

e = 1.6 × 10–19 C

μ0 = 4π × 10–7 T mA–1

14 π ∈o = 9 × 109 N m2 C–1

Mass of Electron = 9.1 × 10–31 kg

Mass of Neutron = 1.675 × 10–27 kg

Mass of proton = 1.673 × 10–27 kg

Section A

1. Draw the geometrical shape of the wavefront when light emerges out from a point source at infinity.

2. A glass lens of refractive index 1.5 is placed in a liquid. What must be the refractive index of the

liquid in order to make the lens disappear?


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3. The graph shows variation of stopping potential V0 versus frequency of incident radiation ν for two photosensitive metals A and B. What is the value of the slope of these two curves?

1

4. Name the electromagnetic waves which (i) maintain the Earth’s warmth and (ii) are used in aircraft

navigation.

5. Mention two properties of the alloy from which permanent magnets are made.

Section B

6. Write the necessary conditions for the phenomenon of total internal reflection to occur.

7. State Lenz’s Law. A metallic rod held horizontally along east-west direction, is allowed to fall

under gravity. Will there be an emf induced at its ends? Justify your answer.

8. A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is ¾ of the angle of prism. Calculate the speed of light in the prism.

9. An ammeter of resistance 0.80 can measure current up to 1.0A.(i) What must be the value of shunt resistance to enable the ammeter to measure current up to 5.0 A?(ii) What is the combined resistance of the ammeter and the shunt?

10. (a) An em wave is travelling in a medium with a velocity. Draw a sketch showing the propagation of the em wave, indicating the direction of the oscillating electric and magnetic fields.(b) How are the magnitudes of the electric and magnetic fields related to velocity of the em wave?

Section C


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11. Write down the factors on which the capacitance of a capacitor depends. A parallel plate capacitor

has a capacitance 6 F in air and 60 F when dielectric medium is introduced. What is dielectric

constant of the medium?

12. The given inputs A, B are fed to a 2-input NAND gate. Draw the output wave form of the gate.

OR

Explain with the help of a circuit diagram how a zener diode works as a DC voltage regulator. Draw its I – V characteristics.

13. A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 µC. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 µC.

Calculate: (i) The potential V and the unknown capacitance C.(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V?

14. Draw a labelled diagram of a moving coil galvanometer and explain its working. What is the function of radial magnetic field inside the coil?

15. A closely wound solenoid of 2000 turns and area of cross-section 1 .6 x 10−4 m2carrying current of

4.0A is suspended through its centre allowing it to turn in a horizontal plane.a. What is the magnetic moment associated with the solenoid?b. What is the force and torque on the solenoid if a uniform horizontal magnetic field of

7 .5 x 10−2T is set up at an angle of 300with the axis of the solenoid?

16. You are given three lenses L1, L2 and L3 each of focal length 20 cm. An object is kept at 40 cm in front of L1, as shown. The final real image is formed at the focus Iof L3. Find the separations between L1, L2 and L3.

17. Calculate the binding energy of an particle in MeV, given that:

mp= 1.007825 amu; mn = 1.008665 amu; Mass of helium nucleus

He = 4.002800 amu and 1 amu = 931 MeV

18. Block diagram of a receiver is shown in the figure:


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(a) Identify ‘X’ and ‘Y’. (b) Write their functions.

19. (a)Write two characteristic features observed in photoelectric effect which supports the photon pictures of electromagnetic radiation.

(b)Draw a graph between the frequency of incident radiation (υ) and the maximum kinetic energy of the electrons emitted from the surface of a photosensitive material state clearly how this graph can be used for finding the work function of material.

20. The energy of an electron in orbit is given by En=−13 .6

n2eV

. Calculate the energy required to take an electron from ground state to the second excited state.

21. Explain with the help of a labeled circuit diagram how a common emitter p n p transistor can be

used as an amplifier in common emitter configuration. Explain how the input & output voltages are

out of phase by 1800 for common emitter transistor amplifier. Draw input and out put waveforms.

OR

Draw the circuit diagram of a full wave rectifier using p-n junction diode. Explain its working and show the output, input waveforms.

22. Sketch a graph b/w angle of incident radiation and stopping potential for a given photosensitive material. What information can be drawn from the value of intercept on the potential axis?

Section D

23.Neeta grandmother who was illiterate was wrapping her satin saree. She found some sparks coming out of it. She frightened and called Neeta. Neeta calmed down her grandmother and explain the reason.


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What according to you are the values displaced by Neeta?Why do sparks appear when satin cloth is folded?

Section E

24. (a) Draw a ray diagram showing the image formation by a compound microscope. Hence obtained expression for total magnification shown by this microscope.

(b) A figure divided into squares, each of size 1 mm2, is being viewed at a distance of 9 cm through a magnifying lens of focal length 10 cm, held close to eye. What is the magnification produced by the lens? How much is the area of each square in the virtual image?

OR

(a) State Huygens’s principle. Using this principle draw a diagram to show how a plane wave front incident at the interface of the two media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snell's law of refraction.

(b) When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons:

(i) Is the frequency of reflected and refracted light same as the frequency of incident light?

(ii) Does the decrease in speed imply a reduction in the energy carried by light wave?

25. (a) State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf of two primary cells. Obtain the required expression used for comparing the emfs.

(b) Write two possible causes for one sided deflection in a potentiometer experiment.

OR

(a) State Kirchhoff's rules for an electric network. Using Kirchhoff's rules, obtain the balance condition in terms of the resistances of four arms of Wheatstone bridge.

(b) In the meter bridge experimental set up, the null point ‘D’ is obtained at a distance of 40 cm from end A of the meter bridge wire. If a resistance of 10Ω is connected in series with R1, null point is obtained at AD = 60 cm. Calculate the values of R1 and R2.


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26. (a) Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field.

(b) A proton and a deuteron having equal momenta enter in a region of a uniform magnetic field at right angle to the direction of a the field. Depict their trajectories in the field.

OR

(a) Using Biot-Savart’s law, derive an expression for the magnetic field at the centre of a circular coil of radius R, number of turns N, carrying currenti.

(b) at what points is the magnetic field of a current element i) zero and ii) maximum?

MARKING SCHEME

1.

(1)

2. µ= 1.5 because according to lens maker’s formula f=∞ for µ2 =µ1

(1)

3. according to Einstein’s photoelectric equation slope gives the value of h/e.

4. i) Infrared rays maintain the earth’s warmth. (ii) Microwaves are used in aircraft navigation due

to their short wavelength. (½, ½)

5. High Retaintivity and high coercivity (½, ½)

6. Necessary conditions for total internal reflection to occur are:(i) The incident ray on the interface should travel in optically denser medium. (1,1)(ii) The angle of incidence should be greater than the critical angle for the given pair of optical media.


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7. Lenz’s law states that the polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it.Yes, emf will be induced in the rod as there is change in magnetic flux. (1,1)When a metallic rod held horizontally along east-west direction, is allowed to fall freely under gravity i.e. fall from north to south , the intensity of magnetic lines of earth’s magnetic field changes through it i.e. the magnetic flux changes and hence the induced emf in it.

8. Given i=3/4 A for equilateral prism A= 60̊ ,i= 45̊Using prism formula µ =1.41 and hence µ =c/v Velocity of light in prism =2.13 x 108m/s

9. We have, resistance of ammeter, G = 0.80 ohm and maximum current across ammeter, Ig= 1.0 A. required range of current in ammeter I= 5A

Using ( shunt resistance) x= IgG/(I-Ig) x =0.2 ohm

. (1,1)(ii) Combined resistance of the ammeter and the shunt,

10. (a) Given that velocity, V = v î and electric field, E along X-axis and magnetic field, B along Z-axis.The propagation of EM wave is following: (1,1)

(b) Speed of EM wave can be given as the ratio of magnitude of electric field (E 0) to the magnitude of magnetic field (B0)

11. Since C= Kɛ0A /d) i) area of each plate ii) distance between plates iii) medium b) 60/6 = 10(1,1,1)

12. Truth table , Boolean expression and logic symbol : (1,1,1)

OR

(i). It is designed to operate in the reverse breakdown voltage region continuously without being damaged. A zener diode has unique feature that voltage drop across it , is independent of current through it.When the input d.c.voltage across zener diode increase beyond a certain limit i.e. zenervoltage , the current through the circuit rises sharply , causing a sufficient increase in voltage drop across the


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dropping resistor R. As a result of it the voltage across zener diode remain constant and hence the output voltage lower back to normal value.

(1,1,1)

13. (i) Initial voltage, V1 = V volts and charge stored, Q1 = 360 µC.Q1 = CV1 …. . . . . . . . . . .(1)

Changed potential, V2 = V − 120Q2 = 120 µCQ2 = CV2...(2)

By dividing (2) from (1), we get

(ii) If the voltage applied had increased by 120 V, then V3 = 180 + 120 = 300 VHence, charge stored in the capacitor, Q3 = CV3 = 2×10-6×300 = 600 µC (2,1)

14. Its working is based on the fact that when a current carrying coil is placed in a magnetic field, it experiences a torque.

1/2

Working: Suppose the coil PQRS is suspended freely in the magnetic field.

Let, l = Length PQ or RS of the coil b = Breadth QR or SP of the coil n = Number of turns in the coil

Area of each turn of the coil, A = l × b Let B = Strength of the magnetic field in which coil is suspended

I = Current passing through the coil in the direction PQRS

Let, at any instant, α be the angle which the normal drawn on the plane of the coil makes with the direction of magnetic field. The rectangular coil carrying current when placed in the magnetic field experiences a torque whose magnitude is given by,


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τ = nIBA sinα 1/2

Due to deflecting torque, the coil rotates and suspension wire gets twisted. A restoring torque is set up in the suspension wire.

Let θ be the twist produced in the phosphor bronze strip due to rotation of the coil and K be the restoring torque per unit twist of the phosphor bronze strip. Then,

Total restoring torque produced = kθ

In equilibrium position of the coil,

Deflecting torque = Restoring torque

∴ NIBA = kθ

(C onstant for a galvanometer)

It is known as galvanometer constant.

The uniform radial magnetic field keeps the plane of the coil always parallel to the direction of the magnetic field. That is, the angle between the plane of the coil and the magnetic field is zero in all the orientations of the coil.

(1,1,1)

15. (a) As magnetic moment M = NIA

= 2000 x 4 x 1.6x10-4Am2

= 1.28 Am2

(b) Net force F = 0

But torque T = MB sin


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= 1.28x7.5x10-2 sin 300

= 4.8 x 10-2 N-m (1½, 1½)

16. 1/fı=1/vı- 1/uı, uı= -40 cm,V3= 20 cm, u3 = ∞It shows that L2 must render the rays parallel to the common axis. It means that the image ( I1), formed by L1, must be at a distance of 20 cm from L2 ( at the focus of L2 )Therefore, distance between L1 and L2 (=40+20) = 60 cm and distance between L2 and L3 can have any value. (1,1,1)

17. An - (alpha particle) i.e., nucleus of helium atom, consists of two protons and two neutrons.

Here, mass of helium atom =4.00150a.m.u;

mp=1.00728 a.m.u. and mn= 1.00867 a.m.u.

Therefore the mass of protons and neutrons constituting an particle is = 2mp +2mn = (2mp +2mn)

– m(He) (1)

=4.0319 – 4.00150 = 0.0304 a.m.u.

Now, 1 a.m.u. =931 MeV

Therefore,binding energy of particle=0.0304 x 931= 28.32 MeV (2)18. From the given block diagram of Demodulation of a typical receiver, we can

conclude the following,

(a) X represents Intermediate Frequency (IF) stage while Y represents an Amplifier.

(b) At IF stage, the carrier frequency is changed to a lower frequency and in this process, the modulated signal is detected. While the function of amplifier is to amplify the detected signal which may not be strong enough to be made use of and hence is required.

(1,1,1)

19.The two characteristics features observed in photoelectric effect which support the photon pictures of electromagnetic radiation one: a) All photons of light of a particular frequency , or

wavelength , have the same energy and momentum whatever the intensity of radiation may be. The increase in intensity of the radiation implies an increase in the number of photons crossing a given area per second.

(b) Photons are electrically neutral and are not deflected by electric and magnetic fields.


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(i) Work function is the minimum energy required by the electron to escape out of the metal surface thus,Φ = hν.Here, ν is the threshold frequency. (1,2)

20.The energy of nth orbit of the H-atom is given by

=En=−13 .6n2

eV

¿ E2−E1=−13 .6(2 )2

−(−13 .612 )

¿−3 .4+13 .6=10 .2 eV . (1,2)

Now energy required to take an electron from ground state to second excited state.

21.

Here the zener diode is joined in reverse bias to functioning d.c. input voltage through a

resistance R of suitable value depending upon the zener voltage and power rating of zenerdiode

used. The constant output voltage is taken across a load resistance RL connected in parallel with

zener diode. When the input d.c. voltage increases beyond a certain limit, the voltage across

zener diode becomes constant equal to zener break down voltage, but the current through the


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zener diode circuit rises sharply as the dynamic resistance of zener diode becomes almost zero

after zener break down voltage. Due to which there is an increase in voltage drop across R since

RL is connected in parallel so the voltage across RL remains same as that of zener break down

voltage. (1,2)

OR

b)

(1,1,1)

22.E = h = Φo + eVo (1,2)


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23. Awareness and sensitivity.

(ii) The different portions of the cloth get charged due to friction then the flow of charge give rise to sparks.

24.(a) A compound microscope consists of two convex lenses co-axially separated by some distance. The lens nearer to the object is called the objective. The lens through which the final image is viewed is called the eyepiece.

Magnifying power, when final image is at infinity:The magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and objective.

Where, Me and M0 are the magnifying powers of the eyepiece and objective respectively.


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If u0 is the distance of the object from the objective and v0 is the distance of the image from the objective, then the magnifying power of the objective is

Where, h, h' are object and image heights respectively and f0 is the focal length of the objective.L is the tube length i.e. the distance between the second focal point of the objective and the first focal point of the eyepiece.

When the final image is at infinity,

Magnifying power of compound microscope,

(b) Area of each square (i.e., object) = 1 mm2

u = −9 cm, f = +10 cm

For a thin lens,

orv = −90 cm

Magnification,

Area of each square in the virtual image

= (10)2 × 1 mm2 = 100 mm2 (2,1,2)

OR(a)Huygens’ Principle:• Each point on the primary wave front acts as a source of secondary wavelets, sending out disturbance in all directions in a similar manner as the original source of light does.• The new position of the wave front at any instant (called secondary wave front) is the envelope of the secondary wavelets at that instant. Refraction On The Basis Of Wave Theory


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• Consider any point Q on the incident wave front.• Suppose when disturbance from point P on incident wave front reaches point on the refracted wave front, the disturbance from point Q reaches Q' on the refracting surface XY.• Since represents the refracted wave front, the time taken by light to travel from a point on incident wave front to the corresponding point on refracted wave front should always be the

same. Now, time taken by light to go from Q to Q' will beIn right-angled ΔAQK, ∠QAK = i∴ QK = AK sin i … (ii)

In right-angled ΔP'Q'K, and …. (iii)Substituting (ii) and (iii) in equation (i),

The rays from different points on the incident wave front will take the same time to reach the corresponding points on the refracted wave front i.e., t given by equation (iv) is independent of

AK. It will happen so, ifThis is the Snell’s law for refraction of light.(b) (i) The frequency of reflected and refracted light remains same as the frequency of incident light because frequency only depends on the source of light. (ii) Since the frequency remains same, hence there is no reduction in energy. (1,3,1)

25.(a) Working principle of Potentiometer:

When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.Applications of Potentiometer for comparing emf’s of two cells:The following Figure shows an application of the potentiometer to compare the emf of two cells of emf E1and E2


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E1, E2 are the emf of the two cells.1, 2, 3 form a two way key.When 1 and 3 are connected, E1 is connected to the galvanometer (G).Jokey is moved to N1, which is at a distance l1 from A, to find the balancing length.Applying loop rule to AN1G31A,Φ l1 + 0 − E1 = 0 ..(1)Where, Φ is the potential drop per unit lengthSimilarly, for E2 balanced against l2 (AN2),Φ l2 + 0 −E2 = 0 ..(2)From equations (1) and (2),

.. (3)Thus we can compare the emf’s of any two sources. Generally, one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then calculated from Eq. (3).(b) (i) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared.(ii) The positive ends of all cells are not connected to the same end of the wire. (1,2,2)OR(a) Kirchhoff’s First Law − Junction RuleThe algebraic sum of the currents meeting at a point in an electrical circuit is always zero.

Let the currents be I1, I2I3, and I4

Convention:Current towards the junction – positive Current away from the junction − negativeI3 + (− I1) + (− I2) + (− I4) = 0Kirchhoff’s Second Law − Loop Rule: In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and current flowing through them.

For closed part BACB, E1 − E2 = I1R1 + I2 R2 − I3R3


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For closed part CADC, E2 = I3R3 + I4R4 + I5R5

Wheatstone Bridge:The Wheatstone Bridge is an arrangement of four resistances as shown in the following figure.

R1, R2, R3,andR4 are the four resistances.Galvanometer (G) has a current Ig flowing through it at balanced condition, Ig = 0Applying junction rule at B, ∴I2 = I4

Applying junction rule at D, ∴ I1 = I3

Applying loop rule to closed loop ADBA,

Applying loop rule to closed loop CBDC,

From equations (1) and (2), This is the required balanced condition of Wheatstone Bridge. (3,2)(b) Considering both the situations and writing them in the form of equations ,Let R' be the resistance per unit length of the potential meter wire,

26.(a) Consider a rectangular loop ABCD carrying current I.


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Case I - The rectangular loop is placed such that the uniform magnetic field B is in the plane of loop.No force is exerted by the magnetic field on the arms AD and BC.Magnetic field exerts a force F1 on arm AB.∴F1 = IbBMagnetic field exerts a force F2 on arm CD.∴F2 = IbB = F1

Net force on the loop is zero.The torque on the loop rotates the loop in anti-clockwise direction.

Torque, τ =

= I(ab)Bτ = BIAIf there are ‘n’ such turns the torque will be nIABwhere, b → Breadth of the rectangular coila → Length of the rectangular coil A = ab → Area of the coilCase II - Plane of the loop is not along the magnetic field, but makes angle with it.

Angle between the field and the normal is θ.Forces on BC and DA are equal and opposite and they cancel each other as they are collinear.Force on AB is F1 and force on CD is F2.F1 = F2 = IbBMagnitude of torque on the loop as in the figure:


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(b) We know, Lorentz force, F = Bqvsinθwhere θ = angle between velocity of particle and magnetic field = 90o

So, Lorentz force, F = BqvThus the particles will move in circular path.

Let mp = mass of proton, md = mass of deuteron, vp= velocity of proton and vd= velocity of deuteron. The charge of proton and deuteron are equal.Given that mpvp= mdvd

Thus, the trajectory of both the particles will be same. (3,2)

OR

(a)According to Biot-Savart’s law, the magnetic field due to a current element at a point with position vector is given by the relation:

Where, μ0 is the permeability of free space

Consider a circular loop of wire having radius r and carrying a current I. Take a current element

on the loop, as shown in the following figure.

The direction of element dl is along the tangent at a point on the loop. ∴dl⊥r


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According to Biot-Savart’s law, the magnetic field at the centre O due to this current element is given as:

The magnetic field due to all such current elements will point onto the plane of paper at centre O. Hence, the total magnetic field at centre O is given by the integration:

For a coil having N turns, the magnetic field at centre,

(b)i) if Ө = 00, sin Ө = 0 so that dB = 0

ii) Ө = 900, sin Ө = 1so that dB = maximum. (3,1,1)


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