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CONTENTS

1. Sample Paper (By CBSE) …..….………………………….….……………………………….. 1-20

2. Sample Paper 1 ….….………….…………………………………….….………….………….. 21-40

3. Sample Paper 2 ….….………….…………………………………….….………….………….. 41-61

4. Sample Paper 3 ….….………….…………………………………….….………….………….. 62-79

5. Sample Paper 4 ….….………….…………………………………….….………….………….. 80-98

6. Sample Paper 5 ….….………….…………………………………….….……………………. 99-119

7. Sample Paper 6 ….….………….…………………………………….….……………..…… 120-139

8. Sample Paper 7 ….….………….…………………………………….…………………….. 140-159

9. Sample Paper 8 ….….………….…………………………………….…………………….. 160-182

10. Sample Paper 9 ….….………….…………………………………….…………………….. 183-201

11. Sample Paper 10 ….….………….…………………………….….….………………..….. 202-221

12. Sample Paper 11 ….….………….……………………………….….…………………….. 222-242

13. Sample Paper 12 ….….………….……………………………….….…………………….. 243-264

14. Sample Paper 13 ….….………….……………………………….….…………………….. 265-287

15. Sample Paper 14 ….….………….……………………………….….…………………….. 288-312

16. CBSE Syllabus and Marking Scheme 2019 …………………………………….. 313-322

CBSE Sample Paper (2018-19)

Class 12 Physics

Time: 3 Hours

Marks : 70

General Instructions

All questions are compulsory. There are 27 questions in all.

This question paper has four sections: Section A, Section B, Section C and Section D. Section A contains five questions of one mark each, Section B contains seven

questions of two marks each, Section C contains twelve questions of three marks each, and Section D contains

three questions of five marks each.

There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

You may use the following values of physical constants wherever necessary

c = 3 × 108 m/s

h = 6.63 × 10-34 Js e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

mass of neutron = 1.675 × 10-27 kg mass of proton = 1.673 × 10-27 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. State the SI unit of the electric polarization vector P

2. Define temperature coefficient of resistivity

3. Name the electromagnetic waves that are widely used as a diagnostic tool in medicine.

OR

Name the current which can flow even in the absence of electric charge.

4. A ray of light is incident on a medium with angle of incidence ‘i’ and is refracted into a second medium with angle of refraction ‘r’. The graph of sin i versus sin r is as shown. Find the ratio of the velocity of light in the first medium to the velocity of light in the second medium.

5. Two particles have equal momenta. What is the ratio of their de-Broglie wavelengths?

OR

Monochromatic light of frequency 6.0 1014 Hz is produced by a laser. What is the energy of a photon in the light beam?

Section-B

6. A network of resistors is connected to a 16 V battery with internal resistance of 1 as shown in the following figure. Compute the equivalent resistance of the network.

OR

A 9 V battery is connected in series with a resistor .The terminal voltage is found to be 8

V. Current through the circuit is measured as 5 A. What is the internal resistance of the battery?

7. The diagram below shows a potentiometer set up. On touching the jockey near to the end X of the potentiometer wire, the galvanometer pointer deflects to left. On touching the jockey near to end Y of the potentiometer, the galvanometer pointer again deflects to left but now by a larger amount. Identify the fault in the circuit and explain, using appropriate equations or otherwise, how it leads to such a one-sided deflection.

OR

Following circuit was set up in a meter bridge experiment to determine the value

X of an unknown resistance.

a. Write the formula to be used for finding X from the observations. b. If the resistance R is increased, what will happen to balancing

length?

8. The figure shows two sinusoidal curves representing oscillating supply voltage and current in an ac circuit.

Draw a phasor diagram to represent the current and supply voltage appropriately as phasors. State the phase difference between the two quantities.

9. Compare the following

i. Wavelengths of the incident solar radiation absorbed by the earth’s surface and the radiation re-radiated by the earth.

ii. Tanning effect produced on the skin by UV radiation incident directly on the skin and that coming through glass window.

10. A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength λ equals to 6000 Å and the angular width of the central maxima 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’.

11. What are universal gates? How can AND gate be realized using an appropriate combination of NOR gates?

12. A TV transmission tower antenna is at a height of 20 m. How much range can it cover if the receiving antenna is at a height of 25 m?

Section-C

13. A particle, having a charge +5 μC, is initially at rest at the point x = 30 cm on the x axis.

The particle begins to move due to the presence of a charge Q that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved 15 cm from its initial position if (a) Q =+15μC and (b) Q = -15μC

14.

a. An electric dipole is kept first to the left and then to the right of a negatively charged infinite plane sheet having a uniform surface charge density. The arrows p1 and p2 show the directions of its electric dipole moment in the two cases.

Identify for each case, whether the dipole is in stable or unstable equilibrium. Justify each answer.

b. Next, the dipole is kept in a similar way (as shown), near an infinitely long straight wire having uniform negative linear charge density.

Will the dipole be in equilibrium at these two positions? Justify your answer.

15. Two material bars A and B of equal area of cross-section, are connected in series to a DC

supply. A is made of usual resistance wire and B of an n-type semiconductor. a. In which bar is drift speed of free electrons greater?

b. If the same constant current continues to flow for a long time, how will the voltage drop across A and B be affected?

Justify each answer.

16. Derive an expression for the velocity vC of a positive ions passing undeflected through a region where crossed and uniform electric field E and magnetic field B are simultaneously present.

Draw and justify the trajectory of identical positive ions whose velocity has a magnitude less than IvcI.

OR

A particle of mass m and charge q is in motion at speed v parallel to a long straight conductor carrying current I as shown below.

Find magnitude and direction of electric field required so that the particle goes undeflected.

17. A sinusoidal voltage of peak value 10 V is applied to a series LCR circuit in which resistance, capacitance and inductance have values of , and 1H respectively. Find (i) the peak voltage across the inductor at resonance (ii) quality factor of the circuit.

18. a. What is the principle of transformer?

b. Explain how laminating the core of a transformer helps to reduce eddy current losses in it.

c. Why the primary and secondary coils of a transformer are preferably wound on the same core.

OR

Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time.

19. Draw a labelled ray diagram to show the image formation in a refracting type astronomical telescope in the normal adjustment position. Write two drawbacks of refracting type telescopes.

OR

a. Define resolving power of a telescope. Write the factors on which it depends. b. A telescope resolves whereas a microscope magnifies. Justify the statement.

20. A jar of height h is filled with a transparent liquid of refractive index μ. At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when it is placed on the top surface symmetrically about the centre, the dot is invisible

21. a. In photoelectric effect, do all the electrons that absorb a photon come out as photoelectrons irrespective of their location? Explain.

b. A source of light, of frequency greater than the threshold frequency, is placed at a distance ‘d’ from the cathode of a photocell. The stopping potential is found to be V. If the distance of the light source is reduced to d/n (where n>1), explain the changes that are likely to be observed in the (i) photoelectric current and (ii) stopping potential.

22. A monochromatic radiation of wavelength 975 Å excites the hydrogen atom from its ground state to a higher state. How many different spectral lines are possible in the resulting spectrum? Which transition corresponds to the longest wavelength amongst them?

23. Binding energy per nucleon versus mass number curve is as shown.

are four nuclei indicated on the curve.

Based on the graph:

a. Arrange X, W and S in the increasing order of stability.

b. Write the relation between the relevant A and Z values for the following nuclear reaction.

S X + W

c. Explain why binding energy for heavy nuclei is low.

OR

How are protons, which are positively charged, held together inside a nucleus? Explain the variation of potential energy of a pair of nucleons as a function of their separation. State the significance of negative potential energy in this region?

24. A sinusoidal carrier wave of amplitude Ac and angular frequency ωc is modulated in accordance with a sinusoidal information signal of amplitude Am and angular frequency ωm. Show that the amplitude modulated signal contains three frequencies centered around ωc. Draw the frequency spectrum of the resulting modulated signal.

Section-D

25. a. Write the expression for the equivalent magnetic moment of a planer current loop of area A, having N turns and carrying a current i. Use the expression to find the magnetic dipole moment of a revolving electron.

b. A circular loop of radius r, having N turns and carrying current I, is kept in the XY plane. It is then subjected to a uniform magnetic field B = Bx i + By j + Bz k. Obtain expression for the magnetic potential energy of the coil-magnetic field system.

OR

a. A long solenoid with air core has n turns per unit length and carries a current I. Using Ampere’s circuital law, derive an expression for the magnetic field B at an interior point on its axis. Write an expression for magnetic intensity H in the interior of the solenoid.

b. A (small) bar of material, having magnetic susceptibility χ, is now put along the axis and near the centre, of the solenoid which is carrying a d.c. current through its coils. After some time, the bar is taken out and suspended freely with an unspun thread. Will the bar orient itself in magnetic meridian if (i) (ii) ?

Justify your answer in each case.

26. a. There are two sets of apparatus of Young’s double slit experiment. In set A, the phase difference between the two waves emanating from the slits does not change with

time,

whereas in set B, the phase difference between the two waves from the slits changes rapidly with time. What difference will be observed in the pattern obtained on the screen in the two set ups?

b. Deduce the expression for the resultant intensity in both the above mentioned set ups (A and B), assuming that the waves emanating from the two slits have the same amplitude A and same wavelength .

OR

a. The two polaroids, in a given set up, are kept ‘crossed’

with respect to each other. A third polaroid, now put in between these two polaroids, can be rotated. Find an expression for the dependence of the intensity of light I, transmitted by the system, on the angle between the pass axis of first and the third polaroid. Draw a graph showing the dependence of I on .

b. When an unpolarized light is incident on a plane glass surface, find the expression for the angle of incidence so that the reflected and refracted light rays are perpendicular to each other. What is the state of polarisation, of reflected and refracted light, under this condition?

27. a. Draw the circuit diagram to determine the characteristics of a pnp transistor in common emitter configuration.

Explain, using I-V characteristics, how the collector current changes with the base current. How can (i) output resistance and (ii) current amplification factor be determined from the IV characteristics?

OR

a. Why are photodiodes preferably operated under reverse bias when the current in the forward bias is known to be more than that in reverse bias?

The two optoelectronic devices: – Photodiode and solar cell, have the same working principle but differ in terms of their process of operation. Explain the difference between the two devices in terms of (i) biasing, (ii) junction area and (iii) I-V characteristics.

CBSE Sample Paper (2018-19) Class 12 Physics Answer

1. C/m2

2. Fractional change in resistivity per unit change in temperature.

3. X-rays

OR

Displacement current

4. From the graph

5. P1=P2

Ratio

OR

Each photon has an energy, E=h.ν

= (6.63 10-34 J s) (6.0 1014 Hz)

= 3.98 10–19 J

6. Equivalent Resistance = R1.R2/ (R1+R2)+R3+ R4.R5)/(R4+R5)

= [(4 4)/(4 + 4)] +1+[(12 6)/(12 + 6)]

= 7

OR

=

= 0.2

7. The positive of E1 is not connected to terminal X.

In loop PGJX, E1 – VG + EXN=0

VG = E1 + EXN

VG = E1 + k

So, VG (or deflection) will be maximum when ℓ is maximum i.e. when jockey is touched near end Y. Also, VG (or deflection) will be minimum when ℓ is minimum i.e. when jockey is touched near end X.

OR

a. X = (100 – ) R/

b. Balancing length will increase on increase of resistance R.

8. Phasor diagram

Equal length of phasors current leads voltage phase difference is /4

9.

i. Radiation re-radiated by earth has greater wavelength

ii. Tanning effect is significant for direct UV radiation; it is negligible for radiation coming through the glass.

10. Angular width

Given = 6000

In Case of new λ (assumed λ’ here), angular width decreases by 30%

=

= 0. 70

2 λ’/d = 0.70 (2 λ/d)

11. Universal gates (like the NAND and the NOR gates) are gates that can be appropriately combined to realize all the three basic gates.

12. Range

d = 33.9 km

13. From energy conservation, Ui + Ki = Uf + Kf

kQq/ri + 0 = kQq/rf + Kf

Kf = kQq (1/ri – 1/rf)

When Q is +15 μC, q will move 15 cm away from it. Hence rf = 45 cm Kf = 9x 109 x 15 x 10-

6 5 10-6 [1/(30 10-2) – 1/(45 10-2)]

= 0.75 J

When Q is -15 μC, q will move 15 cm towards it. Hence rf = 15 cm Kf = 9 109 (-15 x 10-

6) 5 10-6 [ 1/(30 10-2) – 1/(15 10-2)]

= 2.25 J

14.

a. p1: stable equilibrium

p2: unstable equilibrium

The electric field, on either side, is directed towards the negatively charged sheet and its magnitude is independent of the distance of the field point from the sheet. For position p1, dipole moment and electric field are parallel. For position p2, they are

antiparallel.

b. The dipole will not be in equilibrium in any of the two positions. The electric field due to an infinite straight charged wire is non- uniform (E 1/r).

Hence there will be a net non-zero force on the dipole in each case.

15.

a. Drift speed in B (n-type semiconductor) is higher

Reason: I = neAvd is same for both

n is much lower in semiconductors.

b. Voltage drop across A will increase as the resistance of A increases with increase in temperature. Voltage drop across B will decrease as resistance of B will decrease with increase in temperature.

16. E = E j and B = B k

Force on positive ion due to electric field FE = qEj

Force due to magnetic field FB = q (vc B) For passing undeflected, FE = – FB

qEj = – q (vc Bk)

This is possible only if qvc Bk = qvcBj or vc = (E/B)i

The trajectory would be as shown.

Justification: For positive ions with speed v <vc

Force due to electric field = F’E = qE = FE

due to magnetic field F’B = qvB <FB since v <vc Now forces are unbalanced, and hence, ion will experience an acceleration along E. Since initial velocity is perpendicular to E, the trajectory would be parabolic.

OR

For the charged particle to more undeflected

Net force F’E = – Fm

electric force, magnetic force

= 1 –

qE = Bqvsin90o = Bqv

E = VB

Using (4) and (3)

Magnetic force Fm is towards wire.

Electric force and electric field should be away from the line.

17. I0 = V0/R = 10/10 = 1 A

ωr = 1/√LC = 1/√ (1 1 10-6) = 103 rad/s

V0 = I0 XL = I0 ωr L

= 1 103 1 = 103 V Q = ωr L/R

= (103 1)/10 = 100

18.

a. Principle of transformer

b. Laminations are thin, making the resistance higher. Eddy currents are confined within each thin lamination. This reduces the net eddy current.

c. For maximum sharing of magnetic flux and magnetic flux per turn to be the same in both primary and secondary.

OR

At an instant t, charge q on the capacitor and the current i are given by: Q (t) = q0 cos

i (t) = – q0

Energy stored in the capacitor at time t is

Energy stored in the capacitor at time t is

Sum of energies

This sum is constant in time as q0 and C, both are time-independent.

19. Ray diagram:

Drawbacks:

i. Large sized lenses are heavy and difficult to support

ii. large sized lenses suffer from chromatic and spherical aberration.

OR

a. Resolving power of a telescope is the reciprocal of the smallest angular separation between the two objects which can be just distinctly seen.

Factors: diameter of the objective, wavelength of the incident light

b. a telescope produces image of far objects nearer to our eye. Objects which are not

resolved at far distance, can be resolved by telescope. A microscope, on the other hand magnifies objects nearer to us and produces their large image.

20. Let d be the diameter of the disc. The spot shall be invisible if the incident rays from the dot at O, at the center of the disc, are incident at the critical angle of incidence Let i be the critical angle of incidence.

Then Sin

Now, = tan i

21. a. No, it is not necessary that if the energy supplied to an electron is more than the work function, it will come out. The electron after receiving energy, may lose energy to the metal due to collisions with the atoms of the metal. Therefore, most electrons get scattered into the metal. Only a few electrons near the surface may come out of the surface of the metal for whom the incident energy is greater than the work function

of the metal.

b. on reducing the distance, intensity increases. Photoelectric current increases with the increase in intensity. Stopping potential is independent of intensity, and therefore remains unchanged.

22. Energy corresponding to the given wavelength: E (in eV) =

The excited state:

En – E1 = 12. 71

Total no. of spectral lines emitted:

Longest wavelength will correspond to the transition

n = 4 to n = 3

23. a. S,W,X

b. Z = Z1 + Z2

A = A1 + A2

c. Reason for low binding energy:-

In heavier nuclei, the Coloumbian repulsive effects can increase considerably and can match/ offset the attractive effects of the nuclear forces. This can result in such nuclei being unstable.

OR

Nuclear force binds the protons inside the nucleus.

For Graph and explanation, refer to NCERT page no 445 Significance of negative potential energy: Force is attractive in nature

24. The modulated signal:

Frequency Spectrum :-

25. a. The equivalent magnetic moment is given by μ = NiA

The direction of μ is perpendicular to the plane of current carrying loop. It is directed along the direction of advance of a right-handed screw rotated along the direction of flow of current derivation of expression for μ of electron revolving around a nucleus

b. for the loop, μ =

Magnetic potential energy = μ.B

=

OR

c. Derivation

H = nI

The direction of H is along the axis of the solenoid, directed along the direction of advance of a right-handed screw rotated along the direction of flow of current

d. i. Not necessarily.

Reason: material is diamagnetic. After

removal of magnetising field, no magnetisation will remain in the material and hence earth’s magnetic field will not affect it.

ii. Yes Reason: The material is ferromagnetic. It will remain magnetised even after removal from the solenoid and hence align with magnetic meridian.

26. a. Set A: stable interference pattern, the positions of maxima and minima does not change with time.

Set B : positions of maxima and minima will change rapidly with time and an average uniform intensity distribution will be observed on the screen.

b. Expression for intensity of stable interference pattern in set –A If the displacement produced by slit S1 is

y1 = a cos

then, the displacement produced by S2 would be

and the resultant displacement will be given by y = y1 + y2

The amplitude of the resulant displacement is 2a cos and thereore the intensity at that point will be

In set B, the intensity will be given by the average intensity

OR

a. Refer to NCERT example 10.8 on page no. 378.

b. Expression for incident angle:

Nature of polarisation:

Reflected light: Linearly polarised

Refracted light: Partially polarised

27. a. Circuit diagram

b. output characteristics is the variation of collector current with collector -emitter voltage for the different fixed value of IB. If VBE is increased by a small amount, both the hole current and electron current in the base region increases. As a result, both IB and IC increases proportionately.

Output resistance is the ratio of change in collector-emitter voltage to the change in collector current.

Current amplification factor is ratio of change in collector current to the change in base current at constant collector- emitter voltage.

OR

a. The fractional change in majority charge carriers is very less compared to the fractional change in minority charge carriers on illumination.

b. The difference in the working of two devices:

Photodiode

Solar cell

Biasing

Used in Reverse biasing

No external biasing is given

Junction Area

Small

Large for solar radiation to be incident on it.

I-V character istics

CBSE Class 12 Physics

Sample Paper-01

Time allowed: 3 hours

Maximum Marks: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

3. Section A contains five questions of one mark each, section B contains seven questions of two marks each, Section C contains twelve questions of three marks each, and Section D contains three questions of five marks each.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

mass of neutron = 1.675 × 10-27 kg mass of proton = 1.673 × 10-27 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. Name the series of hydrogen spectrum which does not lie in the visible region.

2. Consider three charged bodies P, Q and R. If P and Q repel each other and P attracts R, what is the nature of force between Q and R?

3. Write the expression for speed of electromagnetic waves in free space.

OR

Name the electromagnetic spectrum to which the following wavelengths belong:

a. 10-2m b.

4. On inserting a dielectric between the plates of a capacitor, its capacitance is found to increase 5 times. What is the relative permittivity of the dielectric?

5. What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?

OR

What is the value of the horizontal component of the earth’s magnetic field at magnetic poles?

Section-B

6. A capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

OR

Find the capacitance of the capacitor that have a reactance of 100 when used with an a.c. source of frequency .

7. A negligibly small current is passed through a wire of length 15 m and uniform cross section , and its resistance is measured to be . What is the resistivity of the material at the temperature of the experiment?

OR

Write an expression for the resistivity of a metallic conductor showing its variation over a limited range of temperatures.

8. The figure shows a conductor of length l = 0.5 m and resistance r = 0.5 ohm sliding without friction at a velocity v = 2 m/s over two conducting parallel rods ab and cd lying in a horizontal plane. A resistance R = 2.5 connects the ends b and c. A vertical uniform magnetic field of induction B = 0.6 T exists over the region.

Determine:

i. the current in the circuit,

ii. the force in the direction of motion to be applied to the conductor for the latter to move with the velocity v and

iii. the thermal power dissipated by the circuit. Neglect the resistance of the guiding rods ab and cd

.

9. What is the cause of conduction current?

10. Explain the following terms: (i) Ground waves (ii) Spacewaves (iii) Sky waves

11. A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

12. A 50 turn coil as shown in the figure below carries of 2 A in a magnetic field B = 0.25 Wb m-2. Find the torque acting on the coil. In what direction will it rotate?

Section-C

13. An applied voltage signal consists of a superposition of a dc voltage and an a.c. voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.

14. Explain why high frequency carrier waves are needed for effective transmission of signals. A message signal of 12 kHz and peak voltage 20 V is used to modulate a carrier wave of frequency 12 MHz and peak voltage 30 V. Calculate the (i) modulation index and (ii) side band frequencies.

15. An AM transmitter records an antenna current of 10.5 A. The antenna current drops to 10

A when only carrier is transmitted. What is the percentage modulation?

16. The ground state energy of hydrogen atoms is -13.6 eV.

i. Which are the potential and kinetic energy of an electron in the third excited state?

ii. If the electron jumps to the ground state from the third excited state. Calculate the frequency of photon emitted.

OR

The ground state energy of hydrogen atom is -13.6 eV . What are the kinetic and potential energies of the electron in this state?

17. Explain how electron mobility changes for a good conductor when (i) the temperature of the conductor is decreased at constant potential difference and (ii) applied potential difference is double at constant temperature.

18. The following graph shown the variation of stopping potential V0with the frequency of the incident radiation for two photosensitive metals P and Q.

i. Explain which metal has smaller threshold wavelengths.

ii. Explain, giving reason, which metal emits photoelectrons having smaller kinetic energy.

iii. If the distance between the light source and metal P is doubled, how will the stopping potential change?

OR

Find the typical deBroglie wavelength associated with a He atom in helium gas at room temperature (27° C) and 1 atm pressure and compare it with the mean separation between two atoms under these conditions.

19. The current in a coil of self inductance L = 2 H is increasing according to the law i = 2 sin t2 . Find the amount of energy spent during the period when the current changes from 0 to 2 ampere.

OR

How is mutual inductance of a pair of coils affected when:

i. separation between the coils is increased

ii. the number of turns of each coil is increased

iii. a thin iron sheet is placed between the two coils, other factors remaining the same.

20. Explain, with the help of a circuit diagram, how the thickness of depletion layer in a p-n junction diode changes when it is forward biased. In the following circuit which one of the two diodes is forward biased and which is reverse biased?

21. Explain, with the help of a circuit diagram, how the thickness of depletion layer in a p-n junction diode changes when it is forward biased. In the following circuits which one of the two diodes is forward biased and which is reverse biased?

22. A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder in earthed and the inner cylinder is given a charge of . Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e. bending of field lines at the ends).

23. Two circular coils made of similar wires but of radii 20 cm and 40 cm are connected in parallel. What will be the ratio of the magnetic field at their centres?

OR

Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

24. Why is it necessary to slow down the neutrons, produced through the fission of

nuclei (by neutrons), to sustain a chain reaction? What type of nuclei are (preferably)

needed for slowing down fast neutrons?

Section-D

25. Define the term electric field intensity. Write its SI unit. Derive an expression for the electric field intensity at a point on the axis of an electric dipole.

OR

Find the magnitude of the resultant force on a charge of held at P due to two charges of and -10-8C at A and B respectively. Given AP = 10 cm and BP = 5 cm

26. A prism is found to give a minimum deviation of 51°. The same prism gives a deviation of

62o48′ for two values of the angles of incidence, namely, 46o6′ and 82o42′. Determine the refractive angle of the prism and the refractive index of its material.

OR

Two convex lenses A and B of focal lengths 20 cm and 10 cm are placed coaxially 10 cm apart. An object is placed on the common axis at a distance of 10 cm from lens A. Find the position and magnification of the final image.

27. In a double slit experiment, two coherent sources have slightly different intensities I and

, such that , show that resultant intensity at maxima is near 4 I, while that at minima is nearly .

OR

A double slit is illuminated by light of wavelength 6000 . The slits are 0.1 cm apart and the screen is placed 1 m away.

Calculate:

i. angular position of 10th maximum in radian ii. separation of two adjacent minima.

CBSE Class 12 Physics

Sample Paper – 01

Answers

1. The series is named after its discoverer, Theodore Lyman, who discovered the spectral lines from 1906–1914. All the wavelengths in the Lyman series are in the ultraviolet band.

2. Attractive. This is because Q and R are oppositely charged.

3. The expression for speed of electromagnetic waves in free space,

where, = absolute permeability and is the absolute permeability of the free space.

OR

a. Microwaves [ Range : 0.3 to 10-3m]

b. X -rays [Range: to ]

4.

5. Ceramics.

OR

Zero

6. Given,

Erms = 110V

f = 60 Hz

Putting the values,

= 2.49 A.

OR

or

7. Given, R = 5 W, l = 15 m

(Area of cross-section of wire)

Resistivity

OR

The resistivity of a metallic conductor is given by

Where PT be the resistivity at temperature T and be the resistivity at temperature

T0 and be the temperature coefficient.

8. The conductor ef moves with a velocity v perpendicular to a uniform magnetic induction B and hence induces an emf E = Blv.

The resistance of the circuit = (R + r)

i. Hence, the current in the circuit

= 0.2 A

ii. The power spend in the system

A force of 0.06 N is required to maintain the motion of the conductor iii. The power generated

= 0.12 W

9. The cause of conduction current is the flow of electrons in the conductor under the

effect of potential difference applied.

10. i. Ground waves: A radio wave that travels directly from one point to another following the surface of the earth is called ground wave or surface wave.

ii. Space waves: A radio wave that travels directly from a high transmitting antenna to the receiving station is called space wave.

iii. Sky waves: A radio wave transmitted towards the sky and reflected by the ionosphere towards the desired location of the earth is called a sky wave.

11. This defect is called Astigmatism. It arises because curvature of the cornea plus eye- lens refracting system is not the same in different planes. As vertical lines are seen distinctly, the curvature in the vertical plane is enough, but in the horizontal plane, curvature is insufficient.

This defect is removed by using a cylindrical lens with its axis along the vertical.

12. The sides AB and DC are along the field lines hence the force on each side is zero. The force on each vertical wire is given as

= 0.3 N-m clockwise

13. For high frequency, the inductive reactance for a.c. and capacitance of reactance . Hence, capacitor does not offer any resistance for a.c.. Thus a.c. components of voltage appears across L only.

Consequently XL for d.c. and

e

Therefore, d.c. components of voltage appears across C only.

14. For the effective transmission of signals, the high frequency carrier waves are used because these high frequency carrier waves travel through space or medium with the speed of light and they are not obstructed by earth’s atmosphere.

Numerical : Here Vm = 12kHz, Em = 20V Vc = 12 MHz = 12000 kHz, Ec = 30 V

i. Modulation index, = 0.67

ii. The side bands are

USB = vc + vm= 12000 + 12

= 12012 kHz

LSB = vc – vm = 12000 – 12

= 11988 kHz

15. Here, It = 10.5 A and Ic = 10A Since,

= 0.453 = 45.3%

16. i. In hydrogen atom in the ground state

Potential energy of electron in third excited state

= -1.7 eV

ii.

OR

Here, Ground Energy, E = -13.6 eV Kinetic energy,

and Potential energy, Total energy, E = Ek + Ep

17. Electron mobility of a conductor, and

i. When the temperature of the conductor increases, the relaxation time of free electrons increases. So mobility increases.

ii. Mobility is independent of applied potential difference.

18. i. Suppose the frequency of incident radiations of metal Q and P be and

respectively.

Therefore, metal ‘Q’ has smaller wavelength ii. As we know,

Hence, metal ‘P’ has smaller kinetic energy.

iii. Stopping, potential remains unaffected because the value of stopping potential for a given metal surface does not depend on the intensity of the incident radiation. It depends on the frequency of incident radiation.

OR

Mass of atom,

or

or

Now, PV = RT = kNT (The kinetic gas equation for one mole of a gas )

or

Mean separation,

or

The mean separation between two atoms is much less than the de-Broglie wavelength.

19. Let the current be 2 ampere at t = t2

Then

When the instantaneous current is i, the self induced emf is . If dq amount of charge is displaced in time dt then elementary work done

Let

The integral

= -L cos 2 t2

or

= 4 joules.

OR

i. When separation between the coils is increased, magnetic flux linked with secondary coil decreases. Therefore, mutual inductance (M) of pair of coils decreases.

ii. When number of turns of each coil is increased. M increases, because

iii. As therefore, mutual inductance will increase on placing a thin iron

sheet between the two coils.

20. i. In this case, the p-side is at -10V, whereas the n-side is at 0 V. VP < VN’ hence, the diode is reverse biased.

ii. In this case, the p-side is at 0V, whereas the n-side is at -10V. VP > VN’ hence the diode is forward biased.

21. When the applied voltage is such that n-side is negative and p-side is positive, the applied voltage is opposite to the barrier potential. Hence, the effective barrier potential becomes VB – V, and the energy barrier across the junction decreases. Thus, the junction width decreases.

i. p-n junction is forward biased. ii. p-n junction is reverse biased.

a. p-n junction diode under forward bias. b. Barrier potential

1. without battery

2. Low voltage battery, and

3. High voltage battery.

22. Given,

a = 1.4 cm

b = 1.5 cm

l = 15 cm

The potential of inner cylinder will be equal to the potential difference between inner and outer cylinder as outer cylinder is earthed.

Hence, potential of inner cylinder

23. Magnetic field at the centre of circular coil of radius r, turns N, and current I passing in coil is

[ R = resistance of coil]

[x is resistance per unit length]

As coils are is parallel so potential difference ‘V’ are equal in both coils or

or B1 : B2 = 4 : 1

OR

Frequency is given by

24. Since slow neutrons have a much higher intrinsic probability of inducing fission in

than fast neutrons.

Any substance which is used to slow down fast moving neutrons to thermal energies is called a moderator. Moderators are provided along with the fissionable nuclei for slowing down fast neutrons. The commonly used moderators are water, heavy water (D2D) and graphite.

25. The force experienced by a unit positive charge placed at a point is termed as the electric field intensity at that point. It is vector quantity it’s direction is in the force acting on +ve charge.

The SI unit of electric field intensity is NC-1

Electric field at an axial point of electric dipole:

Assume point P is located at distance r from the centre of an electric dipole as shown in the figure.

Electric field at point P is

or

or [ p = q (2a)]

or

or [ For r > > a ]

OR

Here, F = ? Charge at P,

Charge at A,

Charge at B, q2 = -10-8C AP = 10 cm = 0.1 m,

BP = 5 cm = 0.05 m

APB = 90°

Force at P due to q1 charge at A,

along AP produced

Force at P due to q2 charge at B,

along BP

As angle between and is 90°,

Resultant Force

26. The incident ray is deviated through when angle i = 40o6′. From the principle of reversibility of light, it is clear from the figure that the emergent ray (for which angle e = 82o42′) is also deviated through the same angle . Now,

or

or A = 60°

which is the refractive angle of the prism. For minimum deviation, i = e

Hence,

or

Which is the angle of incidence at minimum deviation? The refractive index of the material of the prism is given by

or

or

OR

From figure below, we have, for lens A

f1 = +20cm and u1 = -10cm

The image distance v1 is given by

which gives v1 = -20 cm

Thus, a virtual image is formed at I1 at a distance of 20 cm from lens A, if the lens B were absent. This image acts as a virtual object for lens B which forms the final image at I2 at a distance v2 from lens B. For lens B we have,

x = 10 cm, u2 = -(20 + 10) = -30 cm f2 = + 10 cm

The image distance v2 is given by

Which gives v2 = +15cm

Thus, a real image I2 is formed at a distance of 15 cm from lens B. Magnification due to A

Magnification due to B Magnification of the final image is

This shows that the final image is inverted and is of the same size as the object.

27. From

As , therefore, = 4I Again from

OR

i. Angular position of nth maximum

Angular position of 10 th maximum,

radian

ii. Separation between two adjacent minima i.e. fringe width,

= 0.6 mm

CBSE Class 12 Physics

Sample Paper – 02

Time allowed: 3 hours

M. M: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

3. Section A contains five questions of one mark each, section B contains seven questions of two marks each, Section C contains twelve questions of three marks each, and Section D contains three questions of five marks each.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

mass of neutron = 1.675 × 10-27 kg mass of proton = 1.673 × 10-27 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. Can a hydrogen atom absorb a photon having energy more than 13.6 eV?

2. What is the net force on an electric dipole placed in a uniform electric field?

3. If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?

OR

Name the EM waves used for studying crystal structure of solids. What is its frequency range?

4. What will be the effect on capacity of a parallel plate condenser when area of each plate is doubled and distance between them is also doubled?

5. The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetization, which piece will dissipate greater heat energy?

OR

Does the earth’s magnetic field at a point vary with time? Is this variation appreciable?

Section-B

6. A potential of is applied across a resistor of

resistance. Find

i. rms value of potential ii. frequency of a.c.

iii. initial phase

iv. rms value of current

OR

What do you mean by power factor? On what factors does it depend?

7. Six lead acid type of secondary cells each of emf 2.0 V and internal resistance

are joined in series to provide a supply to a resistance of . What are the current drawn from the supply and its terminal voltage?

OR

A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantitites is constant along the conductor : current, current density, electric field, drift speed?

8. The motion of a copper plate is damped when it is allowed to oscillate between the pole pieces of a magnet. State the cause of this damping.

9. Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?

10. State the applications of Ultraviolet radiations.

11. A converging lens of refractive index 1.5 is kept in a liquid medium having same refractive index. What would be the focal length of the lens in this medium?

12. An electron is moving along +ve x-axis in the presence of uniform magnetic field along +ve y-axis. What is the direction of force acting on it?

Section-C

13. A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an a.c. line.

14. Give reasons for the following:

i. For ground wave transmission, size of antenna (l) should be comparable to wavelength of signal i.e. .

ii. Audio signals, converted into an em wave, are not directly transmitted as such. iii. The amplitude of modulating signal is kept less than the amplitude of carrier

wave.

15. What is an optical detector? State its three essential characteristics. Name the factor which decides how good a detector is.

16. Which state of the triply ionized Be+++ has the same orbital radius as that of the ground state of hydrogen? Compare the energies of two states.

OR

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

17. Determine the equivalent resistance of networks shown in the figure (a) and (b)

below:

18. A proton and an alpha particle are accelerated through the same potential. Which one of the two has greater value of de-Broglie wavelength associated with it, and (ii) less kinetic energy?

OR

A proton and an electron have same de-Broglie wavelength which of them moves fast and which possesses more K.E.

19. The figure shows two identical rectangular loops (1) and (2) placed on a table along with a straight long current carrying conductor between them.

(i) What will be the directions of the induced current in the loops when they are pulled away from the conductor with same velocity v?

(ii) Will the emf induced in the two loops be equal?

OR

A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly rotated by 90°. The total charge flown in the coil is 7.5 mC. The resistance of the rod is . Estimate the field strength of magnetic field.

20. For a common emitter-transistor, amplifier, the audio signal voltage across the collector resistance of 2k is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1k .

21. Show that a bubbled OR gate is equivalent to a NAND gate. Hence prove the identity.

22. A parallel plate capacitor with air as dielectric is charged by ad.c. source to a potential V. Without disconnecting the capacitor from the source, air is replaced by another dielectric medium of dielectric constant K. State with reason, how does

i. potential difference

ii. electric field between the plates iii. capacity

iv. charge and

v. energy stored in the capacitor change.

23. Define current sensitivity and voltage sensitivity of a galvanometer. Increasing the current sensitivity may not necessarily increase the voltage sensitivity of galvanometer. Justify

OR

When a dielectric is inserted between the plates of a charged parallel plate capacitor, fully, occupying the intervening region, how does the polarization of the dielectric medium affect the net electric field? For linear dielectrics, show that the introduction of a dielectric increases its capacitance by a factor k, characteristic of the dielectric.

24. Two radioactive nuclei X and Y initially contain equal number of atoms. The half life is 1 hour and 2 hours respectively. Calculate the ratio of their rates of disintegration after two hours.

Section-D

25. Using Gauss’s law, derive an expression for the electric field intensity at any point outside a uniformly charged thin spherical shell of radius R and charge density

. Draw the field lines when the charge density of the sphere is:

(i) positive, (ii) negative. (b) A uniformly charged conducting sphere of 2.5 m in diameter has a surface charge density of .

Calculate the (i) Charge on the sphere (ii) Total electric flux passing through the sphere.

OR

Two identical charged bodies have and charge respectively. These bodies experience a force of 48 N at certain separation. The bodies are touched and placed at the same separation again. Find the new force between the bodies.

26. Determine the ‘effective focal length’ of the combination of the two lenses having focal lengths 30 cm and -20cm if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of

parallel light is incident? Is the notion of effective focal length of this system useful at all?

a. An object 1.5 cm in size is placed on the side of the convex lens in the arrangement above.

b. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

OR

An object is placed 40 cm in front of the curved surface of a thin plano convex lens whose plane surface is silvered. Due to refraction at the curved surface and reflection at the silvered surface, the real image of the object is 60 cm from the lens on the same side as the object. Find the focal length of the lens.

27. A ray of light incident normally on one of the faces of a right angled isosceles prism is found to be totally reflected as shown.

a. What is the minimum value of the refractive index of the material of the prism?

b. When the prism is immersed in water trace the path of the emergent ray for the same incident ray indicating the values of all the angles. ( of water = 4/3).

OR

a. What are coherent sources of light? Two slits in Young’s double slit experiment are illuminated by two different sodium lamps emitting light of the same wavelength. Why is no interference pattern observed?

b. Obtain the condition for getting dark and bright fringes in Young’s experiment.

Hence write the expression for the fringe width.

c. If s is the size of the source and its distance from the plane of the two slits, what should be the criteria for the interference fringes to be seen?

CBSE Class 12 Physics

Sample Paper – 02

Answers

1. Yes, it can absorb. But the atom would be ionized.

2. Zero.

3. The temperature of the earth would be lower because the green house effect of the atmosphere would be absent.

OR

X-rays are used for studying crystals structure of solids. Their frequency range is 1016

Hz to .

4. It will remain unaffected.

5. Carbon steel piece, because heat lost per cycle is proportional to the area of hysteresis loop.

OR

Yes. The variation may be appreciable over a very large interval of time.

6. i.

ii. Frequency, iii. Initial phase iv.

OR

The power factor is defined as the cosine of the phase angle between alternating e.m.f. and current in an a.c. circuit.

Power factor of a.c. circuit is given by

It depends upon the frequency of the a.c. source.

7. (a) Here, E = 2.0 V, n = 6,

and

Current,

Terminal voltage V = IR = 1.4 8.5 = 11.9 V

OR

Only current because it is given to be steady. The rest depends on the area of cross section inversely.

8. This damping is due to the eddy current developed in the copper plate. When copper plate oscillates between the pole pieces of the magnet, the flux linked with copper plate changes and eddy currents (induced current) are developed which opposes the cause of the production.

9. Atmosphere absorbs Xrays while visible and radiowaves can penetrate it. That is why optical and radio telescopes can work on earth’s surface but Xray astronomical telescopes must be used on satellites orbiting the earth in order to minimise the interference & maximising output.

10. Ultraviolet radiations are used :

i. to preserve the food stuff,

ii. to sterilizing the surgical instruments.

11. The lens in the liquid will act like a plane sheet of glass.

Its focal length will be infinite

By using the formula of lens maker

or

as,

Thus,

12. The direction of the force is along -ve z-axis.

13. For a steady state d.c. L has no effect even if it is increased by an iron core. For a.c. the lamp will shine dimly because of additional impedance of the choke. It will dim further when the iron core is inserted which increases the choke’s impedance.

14. i. When the size of antenna is comparable to wavelength of the signal the time variation of the signal is properly sensed by the antenna.

ii. An em wave, of audio signal frequency, would have a very high wavelength. It would, therefore, need an antenna, whose size will be practically unattainable.

iii. The amplitude of modulating signal is kept less than the amplitude of carrier wave to avoid distortion.

15. Optical detector is a device at the receiver end, which converts light into electrical signal so that the transmitted information may be decoded. Characteristics of optical detector:

– Size compatible with the fibre.

– High sensitivity at the desired optical wave length.

– High response time for fast speed data transmission / reception.

The efficiency of generatingelectron hole pairs in a photo diode decides the quality of detector.

16. Radius of nth orbit is given by

i.e.

For hydrogen, Z = 1, n =1 in ground state

For Beryleum, Z = 4, as orbital radius is same,

Hence n = 2 level of Be has same radius as n =1 level of hydrogen. Now, energy of electron in nth orbit is

OR

Energy of an electron in nth orbit of H atom

E1 = -13.6eV

Energy is 4th (n = 4) level

= -0.85 + 13.6 hv = 12.75 eV

17. Equivalent resistance of network in figure (a). The given network is a series combination of four identical units. Let us consider one such unit shown in figure. It is equivalent to a parallel combination of and its equivalent resistance is:

i.e. i.e.

So, the given electrical network is a series combination of four resistors, each equal to

,

Thus, the combined resistance is or

Equivalent resistance of network in figure (b) Suppose a battery is connected between

A and B. Same current will flow through all the resistors. So all the resistors are

connected in series.

Equivalent resistance, RS = R + R + R + R + R = 5R

18. i. The de-Broglie wavelength associated with same potential V is

or

As proton’s mass is less than the mass of alpha particle thus, K.E. = hv

ii. As,

Since,

Thus, kinetic energy of proton will be lesser than that of alpha particle.

OR

Kinetic energy of particle of mass m having momentum p is

or

de Broglie wavelength,

….(i)

And ….(ii)

If is constant, then from (i), p = a constant i.e. mpvp = meve or or vp < ve If is constant, then from (ii),

or Kp < Ke

It means the velocity of electron is greater than that of proton. Kinetic energy of electron is greater than that of proton.

19. (i) The direction of induced current will be such that it tends to maintain the original flux. So induced current flows anticlockwise in loop 1 and clockwise in loop 2.

(ii) No, the emf’s induced in the two loops will not be equal.

OR

The total charge flowing through the coil,

Where, induced EMF,

Hence, ….(i) Given, N = 25, ,

Putting these values in equations (i), we get

,

(From I)

= 0.75 T

20. Here, R0 = 2000 V0 = 2V, = 100; Vi = ?, Ib = ? Ri = 1000

As

or

21. A bubbled OR gate is the combination of two NOT gates and one OR gate i.e. the output of two NOT gates is made as input of OR gate. The Boolean expression for output of this gate is (From De-Morgan’s Theorem). This expression is for NAND gate. Hence, a bubbled OR gate is equivalent to a NAND gate.

22. i. Since the capacitor remains connected to battery, therefore, the potential difference would remain unchanged.

ii. Since neither the potential difference nor the separation between the plates is changed therefore the electric field remains unchanged.

iii. Capacity would increase by a factor of K. So C = KC0

iv. Since C is increased by a factor of K and V remains unchanged therefore Q is increased by a factor of K. Additional charge flows from the battery to the plates.

v. . So, the energy is increased by a factor of K.

23. The definition of current sensitivity and voltage sensitivity are given

Let the deflection produced in applying voltage V is then

Voltage sensitivity

The voltage sensitivity may be increased by (i) increasing, N, B, A (ii) decreasing k and

Current sensitivity can be increased by

(i) increasing NBA (ii) decreasing k.

Hence increasing the current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer.

OR

The capacitance C0 is given by

…… (i)

The capacitance C, with dielectric between the plates is

……. (ii)

The product is called the permittivity of the medium and is denoted by

Where k is called the dielectric constant of the substance. Dividing (ii) by (i) we get

Therefore, the dielectric constant of a substance is the factor > 1. By which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of a capacitor.

24. After two hours, the ratio of radioactive sample of X left, After two hours, the ratio of radioactive sample of Y left,

(R is rate of decay )

or

So

Hence, the activity of the two samples will be equal after 2 hours.

25. a. Electric field intensity at any point outside a uniformly charged spherical shell:

Consider a thin spherical shell of radius R with centre O. Let charge +q is uniformly distributed over the surface of the shell.

Let P be any point on the sphere S1 with centre O and radius r. According to Gauss’s law:

If is charge density,

Electric field lines when the charged density of the sphere:

i. Positive ii. Negative

b. Here diameter

= 2.5 m

Charge density

i.

ii. Total electric flux

OR

When two identical bodies having different magnitude of charge are touched, the redistribution of charge takes place and both the bodies acquire same charge.

Charge on each body after touching

The new force between the bodies

but

= 2 N

26. a. i. Here, f1 = 30cm, f2 = -20cm, d = 8.0 cm

Let a parallel beam be incident on the convex lens first. If second lens were absent, then

and f1 = 30 cm

As

or v1 = 30 cm

This image would now act as virtual object for second lens.

u2 = +(30 – 8) = + 22 cm f2 = -20cm

Since,

v2 = -220 cm

Parallel incident beam would appear to diverge from a point 220-4 = 216 cm from the centre of the two lens system.

ii. Assume that a parallel beam of light from the left is incident first on the

concave lens.

As

, f1 = -20 cm

v1 = -20cm

This image acts as a real object for the second lens u2 = -(20 + 8) = -28cm, f2 = 30 cm

Since,

v2 = -420 cm

The parallel beam appears to diverge from a point 420 – 4 = 416 cm, on the left of the centre of the two lens system.

We finally conclude that the answer depends on the side of the lens system where the parallel beam is incident. Therefore, the notion of effective focal length does not seem to be meaningful here.

b. For convex lens

u = -40cm, f = 30 cm, O = 1.5 cm

Using lens formula

We get or

v = 120 cm (for real object) From relation,

, we get

= + 3

The image formed by the convex lens becomes object for concave lens at a distance of 120 – 8 = 112 cm on the other side.

For concave lens, f = -20cm, u = + 112 cm (on the other side)

v = ?

Using lens formula, we get

Now,

or

(for virtual object)

From relation, , we get

Net magnification

= 0.652 (negative due to virtual image) As

= 0.98 cm (size of final image)

OR

The incident ray OA is refracted along AB due to refraction at the curved surface. This ray AB appears to come from I1, the virtual image point due to refraction at the

curved surface. For this refraction we have

Since the incident rays are from left to right, v’ = +v’, u = -u and R = +R .

Therefore, we have

….(i) Where u = + 40 cm (given).

I2 such that QI1 = QI2 = v’

The ray BC suffers refraction at the curved surface and the final image is formed at I. Thus I is the image of the virtual object I2 due to refraction at the curved surface. For this refraction, since the incident ray travels from glass to air, we have

From our sign convention, v = -v, v’ = QI2 = PI2 (since lens is thin) = -v’ and R = + R Therefore, we have

…..(ii) where v = +60 cm (given) Subtracting (ii) from (i), we get

or But

where the focal length of the lens. Hence

or f1 = 48 cm.

27. a. ABC is the section of the prism, B is a right angle. A and C are equal angles i.e. A = C = 45°.

The ray PQ is normally incident on the face AB. Hence it is normally refracted and the ray QR strikes the face AC at an angle of incidence 45°. It is given that the ray does not undergo refraction but is totally reflected at the face AC. This gives a maximum value for the critical angle as 45°.

sin C = sin 45° in the limit

Since

or,

The minimum value of refractive index .

b. When the prism is immersed in water the critical angle for the glass water interface is given by

C1 = 70.53°

The angle of incidence at R continues to be 45° and since 45° < 70.53°

There is refraction taking place now and the refracted ray is RS. The angle of refraction r is given by

The angle of refraction in water =

OR

a. Coherent sources:

Sources emitting waves of same frequency or wavelength having either a zero or a constant phase difference are said to be coherent sources of light. Two independent sources of light do not fulfill the requirement of constant phase

difference. As the case of two different sodium lamps is given here. Hence, such sources cannot be used for producing interference pattern.

b. For bright fringes (maxima), Path difference,

Where n = 0, 1, 2, 3, . . .

For dark fringes (minima). Path difference,

Where n = 0, 1, 2, 3,. . .

The separation between the centre of two consecutive bright fringes is the width of a dark fringe.

Fringe width,

c. The condition for interference fringes is to be:

CBSE Class 12 Physics

Sample Paper – 03

Time allowed: 3 hours

M. M: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

3. Section A contains five questions of one mark each, section B contains seven questions of two marks each, Section C contains twelve questions of three marks each, and Section D contains three questions of five marks each.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. Name the series of hydrogen spectrum lying in the infrared region.

2. Two point charges of each are 100 cm apart. At what point on the line joining the charges will the electric intensity be zero?

3. What is the approximate wavelength range for visible part of electromagnetic spectrum?

OR

It is necessary to use satellites for long distance TV transmission. Why?

4. What are the expressions for energy of a charged capacitor?

5. If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or

(slightly) less than when the core is empty?

OR

Give an example of magnetic dipole.

Section-B

6. In an LR circuit reactance and resistance are equal. Calculate phase by which voltage differ current?

OR

A capacitor and a resistor are connected in series with an a.c. source. If the potential difference across C, R are 120 V, 90 V respectively and if the r.m.s. current of the circuit is 3 A, calculate the (i) impedance(ii) power factor of the circuit.

7. A (i) series (ii) parallel combination of two given resistors is connected, one by one, across a cell. In which case will the terminal potential difference, across the cell, have a higher value?

OR

A 60 watt bulb carries a current of 0.5 ampere. Find the total charge passing through it in 1 hour.

8. Why does the acceleration of a magnet falling through a long solenoid decrease?

9. State two applications of Infrared radiations.

10. Name the part of electromagnetic spectrum which is suitable for;

i. radar systems used in aircraft navigation. ii. treatment of cancer tumours.

11. Two thin lenses of power +4D and -2D are in contact. What is the focal length of the combination?

12. A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Section-C

13. A 20 volts 5 watt lamp is used in a.c. main of 220 volts 50 c.p.s calculate the i. capacitance of capacitor.

ii. inductance of inductor, to be put in series to run the lamp.

iii. what pure resistance should be included in place of the above device so that the lamp can run on its voltage?

iv. which of the above arrangements will be more economical and why?

14. Distinguish between frequency modulation and amplitude modulation. Why is an FM

signal less susceptible to noise than AM signal?

15. A block diagram of a receiver is as shown:

Identify X and YState their function.

16. The ground state energy of hydrogen atom is -13.6eV

i. What is the potential energy of an electron in the 3rd excited state?

ii. If the electron jumps to the ground state from the 3rd excited state, calculate the wavelength of the photon emitted.

OR

The energy levels of hydrogen atoms are as shown in figure. Which of the shown transition will result in the emission of a photon of wavelength 275 nm? Which of the transition corresponds to emission of radiation of (i) maximum and (ii) minimum wavelength?

17. Write the mathematical relation for the resistivity of a material in terms of relaxation time, number density and mass and charge of charge carriers in it. Explain, using this relation, why the resistivity of a metal increases and that of a semiconductor decreases with rise in temperature.

18. The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de- Broglie wavelength.

OR

Calculate the :

a. Momentum

b. de-Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

19. The two rails of a railway track, insulated from each other, and the ground, are connected to a millivoltmeter. What is the reading of the voltmeter when a train

travels at a speed of 180 km h-1 along the track? Given vertical component of earth’s magnetic field and the separation between the rails = 1m.

OR

Two circular coils, one of radius r and the other of radius R are placed coaxially with their centres coinciding. For R > > r, obtain an expression for the mutual inductance of the arrangement.infinitesimally

20. An a.c. supply of 230 V is applied to a half wave rectifier circuit through a transformer of turn ratio 10 : 1. Find the output d.c. voltage. Assume the diode to be ideal.

21. If the output of a 2 input NOR gate is fed as both inputs A and B to another NOR gate, write down a truth table to find the final output, for all the combination of A, B.

22. The given graph shows the variation of charge q versus potential difference for two capacitors C1and C2. The two capacitors have same plate separation, but the plate area of C2 is double that of C1. Which of the lines in the graph corresponds to C1 and C2 why?

23. An electron of 45 eV energy is revolving in a circular path in a magnetic field of intensity .

Determine (i) the speed of the electron (ii) radius of the circular path.

OR

A charge ‘q’ moving along the X-axis with a velocity is subjected to a uniform magnetic field B acting along the Z-axis as it crosses the origin O.

i. Trace its trajectory.

ii. Does the charge again kinetic energy as it enters the magnetic field? Justify

24. A source contains two phosphorous radio nuclides and . Initially, 10% of the decays come from . How long one must wait until 90% do so?

Section-D

25. Two pieces of copper, each weighing 0.01 kg are placed at a distance of 0.1 m from each other one electron from per 1000 atoms of one piece is transferred to other piece of copper. What will be the coulomb force between two piece after the transfer of electrons? Atomic weight of copper is 63.5 g/mole. Avagadro’s number

.

OR

S1 and S2 are two hollow concentric spheres enclosing charges 2Q and 4Q respectively as shown in the figure.

i. What is the ratio of electric flux through S1 and S2 ?

ii. How will the electric flux through the sphere S1 change, if a medium of dielectric constant 6 is introduced in the space inside S1 in the place of air?

26. The direct image formed by the lens (f = 10 cm) of an object placed at O, and that formed after reflection from the spherical mirror are formed at same point O. What is the radius of curvature of the mirror?

OR

A thin equiconvex lens (radius of curvature of either face being 33 cm) is placed on a horizontal plane mirror and a pin held 20 cm vertically above the lens coincides in position with its own image. The space between the lower surface of the lens andthe mirror is filled with a liquid and then, to coincide with the image as before, the pin has to be raised to a distance of 25 cm from the lens. Find the refractive index of the liquid.

27. What is diffraction of light? Draw a graph showing the variation of intensity with angle in a single slit diffraction experiment. Write one feature which distinguish the observed pattern from the double slit interference pattern. How would the diffraction

pattern of a single slit be affected when:

i. the width of the slit is decreased?

ii. the monochromatic source of light is replaced by a source of white light?

OR

Explain how Newton’s Corpuscular theory predicts the speed of light in a medium, say water, to be greater than the speed of light in vacuum. Is the prediction confirmed by the experimental determination of speed of light in water? If not, which alternative picture of light is consistent with experiment?

CBSE Class 12 Physics

Sample Paper – 03

Answers

1. The series of lines in the hydrogen spectrum which lie in the infrared region are:

Paschen series : Near infrared region Brackett series : Infrared region Pfund series : Far infrared region

Humphrey series : Very far infrared region

2. The electric intensity will be zero at a point mid-way between the two charges.

3. 390 nm to 770 nm.

OR

It is so because television signals are not properly reflected by the ionosphere. Therefore for reflection of signals, satellites are needed.

4.

5. Slightly less, since bismuth is diamagnetic.

OR

Bar magnet.

6.

OR

Given, Irms = 3A

VR = 90V, VC = 120V

i. Impedance,

ii. Power factor,

7. Terminal potential difference of the cell is

Terminal potential difference is higher in parallel combination as compared in series combination.

OR

Charge, q = It = 1800 coulomb.

8. As it falls,its flux increases because of solenoid turns.By Lenz law the direction of induced emf in a way that it generates a field in the solenoid repelling,the falling magnet. Thereby, opposing the increasing flux. The repulsion by the solenoid decreases the acceleration of the falling magnet.

9. Infrared radiations are used for;

i. to treat muscular strain,

ii. for taking photographs during the conditions of fog, smoke etc.

10. i. Radar systems used in aircraft navigation Microwave ii. Treatment of cancer tumours Gamma rays

11. P = P1 + P2 = 4 – 2 = +2D Since focal length

= 50 cm

12. Given, I = 35 A, r = 20 cm, = 0.2 m

13. The current required by the lamp

= 0.25 amp

The resistance of the lamp

= 80 ohm

So for proper running of the lamp, the current through the lamp should be 0.25 amp. i. When the condenser C is placed in series with lamp, then

The current through the circuit

= 0.25

or = 0.25

Solving it for C, we get

ii. When inductor L henry is placed in series with the lamp, then or

or = 0.25

Solving it for L we get L = 2.53 henry.

iii. When resistance r ohm is placed in series with lamp of resistance R, then

or

r = 720 ohms

iv. It will be more economical to use inductance or capacitance in series with the lamp to run it as it consumes no power while there would be dissipation of power when resistance is inserted in series with the lamp.

14. AM:

– It is cheap and instruments are simple.

– The modulated signal’s amplitude varies as per the modulating signal. FM:

– FM are more costly and complex than AM.

– The modulated signal’s frequency varies as per the modulating signal.

– It is less noisy and gives better quality transmission and has a larger bandwidth. In FM, the message signal is in the form of frequency variations and, therefore, the

atmospheric or man made noises (which are generally amplitude changes) have very little impact. It is preferred for transmission of music.

15. X is F stage

Y is amplifier

Function of IF stage is to change carrier frequency to a (standard) lower frequency function of amplifier to amplify the signal detected because it may not be strong enough to be made use of it.

16. The energy of an electron in nth orbit is given by

i. For 3 rd excite state, n = 4

ii. Required energy to jump electron to the ground state from the 3 rd excited state

E = E4 – E1

= – 0.85 + 13.6 = 12.75 eV

Wavelength of the photon emitted as

OR

The energy of photon of wavelength 275 nm

= 6.0 eV

Thus transition is the result. (i) Transition A

(ii) Transition D.

17. Resistivity

i. The thermal speed of electrons increases as the temperature increases. Free electrons collide more frequently with the positive metal ions. The relaxation time

decreases. Consequently, the resistivity of the metal increases.

ii. The relaxation time does not change with temperature in semiconductor. But the number density (n) of free electrons increases exponentially with temperature. As a result, the resistivity of semiconductor decreases exponentially with the increase in temperature.

18.

a. For electron,

= 7.03 10-25J

b. For neutron,

= 3.81 10-28J

OR

Energy of electron accelerated through potential difference of 56V

= 56 eV = 56 1.6 10-19J

a. As

b.

or

p = 4.02 10-24kg ms-1

= 1.64 10-10 m = 0.164 10-9m

or

19. The induced emf generated is given by

where A is the area and B, the magnetic field. If l is the distance between the rails and v, the speed of the train, then

Here l = 1 m and B Thus we have

= 1 mV

Hence, the millivoltmeter will read 1 mV.

OR

Suppose a current I2flows through the outer circular coil. Magnetic field at the centre of the coil is

Field B2 may be considered constant over the cross-sectional area of the inner smaller coil.

Hence

20. Here, = 10 R.M.S. primary voltage, Vms = 230V

Maximum primary voltage, = 325.3 V Maximum secondary voltage, = 32.53 V

Half wave rectified current, Output dc voltage

21. Output of NOR gate if fed to both inputs of another NOR gate.

A

B

X

Y

0

0

1

0

0

1

0

1

1

0

0

1

1

1

0

1

22. As , and plate area of C2 is double than that of C1, therefore C2 > C1. Now

, which is greater for line A.

The line A of the graph corresponds to C2 and line B corresponds to C1.

23.

Now, or

OR

i. The trajectory of charge q moving along X-axis will be helical.

ii. The speed and kinetic energy of the particle remain constant but the velocity of the charged particle changes only in direction.

24. We know that .

So, clearly the initial ratio of the amounts of and is 1 : 9. We have to find the time after which the ratio is 9 : 1.

Initially, if the amount of is x, the amount of is 9x . Finally, if the amount of

is 9 y, the amount of is y. Using

Dividing

or 81 = 2

or, 81 = 2

or

or

25. Number of atoms in each piece of copper

Number of electrons transferred Charges on each piece after transfer

r = 0.1 m

OR

i.

ii. Using Gauss’s theorem, And

26. Here, f = 10 cm, u= -15 cm

Let v be the distance of image I formed by the lens L.

From

v = LI = 30 cm

The mirror will form the image at I itself provided I coincides with centre of curvature C of the mirror. R = PC = PI = PL – LI = 50 – 30 = 20cm

OR

In the first case, the image will coincide with the pin if the rays from the pin, after refraction through the lens, fall normally on the mirror and retrace their path, as shown in figure. This means that the focal length of the convex lens is 20 cm.

f = 20 cm

In the second case, the focal length F of the combination of the convex lens and the planoliquid lens is 25 cm [see Fig] i.e.

F = 25 cm

Let f2 be the focal length of the liquid lens, then

or f2 = -100 cm

For the liquid lens, R1 = -33 cm, the radius of curvature of the common surface and R2

= , if is refractive index of the liquid.

or

or

or

27. Diffraction of light: Phenomenon of bending of light around the corners of an obstacle or aperture is called diffraction.

The intensity distribution wave for diffraction is shown in the diagram below:

In interference, by 2 slits all bright fringes are of same intensity. In diffraction, the intensity of bright fringes decreases with the increase in distance from the central bright fringe.

i. The diffraction pattern becomes narrower if the width of the slit is decreased.

ii. A coloured diffraction pattern is obtained if monochromatic source is replaced by white light source. The central band is white therefore, the red fringe is wider than the violet fringe etc.

OR

According to Newton’s Corpuscular theory of light , when corpuscles of light strike the interface XY, figure separating a denser medium from a rarer medium, the

component of their velocity along XY remains the same: If v1 is velocity of light in rarer medium (air)

v2 is velocity of light in denser medium (water)

i is the angle of incidence, r is angle of refraction,

Then component of v1 along XY = v1 sin i

Component of v2 along XY = v2 sin r

As v1 sin i = v2 sin r

As > 1 v2 > v1

i.e. light should travel faster in water than in air. This prediction of Newton’s theory is opposite to the experiment result.

Huygens wave theory predicts that v2 < v1, which is consistent with experiment.

CBSE Class 12 Physics

Sample Paper – 04

Time allowed: 3 hours

M. M: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. Which one is unstable among neutron, proton, electron and particle.

2. Is the mass of a body affected on charging?

3. What is the effect of electromagnetic waves on charged particles?

OR

Long distance radio broadcasts use short wave bands. Why?

4. Is potential difference a scalar or a vector?

5. Which materials have permeability > 1 ?

Section-B

6. In a series LCR a.c. circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltage across the series elements of the circuit? Is the same true for r.m.s voltage?

OR

A capacitor is connected to 220 V, 50 Hz source. Find the capacitive reactance and the rms current.

7. A carbon resistor has three strips of red colour on its surface and a gold strip at one end of it. What is the value of this resistance?

OR

At room temperature (27.0°C) the resistance of a heating element is . What is the temperature of the element if the resistance is found to be given that the temperature co-efficient of the material of the resistor is .

8. A cylindrical bar magnet is kept along the axis of a circular coil. Will there be a current induced in the coil if the magnet is rotated about its axis? Give reasons.

9. Special devices like the klystron value or the magnetron value, are used for production of electromagnetic waves. Name these waves and also write one of their applications.

10. For which frequency of light, the eye is most sensitive?

11. What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

12. Two long parallel straight wires X and Y separated by a distance of 5 cm in air carry currents of 10 A and 5 A respectively in opposite directions. Calculate the magnitude and direction of the force on a 20 cm length of the wire Y.

Section-D

13. Power factor can often be improved by the use of capacitor of appropriate capacitance in the circuit. Justify

14. A signal wave of frequency 11.5 kHz is modulated with a carrier wave of frequency

3.45 MHz. What are the deviation in frequency of FM wave and bandwidth of it?

15. Is it necessary for the transmitting antenna and receiving antenna to be of the same height for line of sight communication? Find an expression for maximum line of sight distance dM between these two antenna of heights hT and hR.

16. A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

OR

(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels.

(b) Calculate the orbital period in each of these levels.

17. Three resistors of values 4 ohm, 6 ohm and 7 ohm are in series and a potential difference of 34 V is applied across the grouping.Find the potential drop across each resistor.

18. The emitter in a photoelectric tube has a threshold wavelength of 6000 . Determine the wavelength of the light incident on the tube if the stopping potential for this light is 2.5 V.

OR

Monochromatic radiation of wavelength 640.2 nm (1nm = 10-9m) from a neon lamp irradiates a photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its

427.2 nm line irradiates the same photocell. Predict the new stopping voltage.

19. An infinitesimally small bar magnet of dipole moment is pointing and moving with the speed v in the x-direction. A small closed circular conducting loop of radius a and of negligible self inductance lies in the y-z plane with its centre at x = 0, and its axis coinciding with the x-axis. Find the force opposing the motion of the magnet, if the resistance of the loop is R. Assume that the distance x of the magnet from the centre of the loop is much greater than a.

OR

Use Lenz’s law to determine the direction of induced current in the situation described by figure:

a. A wire of irregular shape turning into a circular shape.

b. A circular loop being deformed into a narrow straight wire.

20. You are given two circuits as shown in figure, which consists of NAND gates. Identify the logic operation carried out by the two circuits.

21. The following figure shows the V-I characteristics of a semiconductor diode i. Identify the semiconductor diode used.

ii. Draw the circuit diagram to obtain the given characteristics of this device. iii. Briefly explain how this diode can be used as a voltage regulator

22. A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2on the shell is.

23. An electron traveling west to east enters a chamber having a uniform electrostatic

field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

OR

A wire AB is carrying a current of 12 A and is lying on the table. Another wire CD, carrying current 5 A is arranged just above AB at a height of 1 mm. What should be the weight per unit length of the wire so that CD remains suspended at its position? Indicate the directions of current in CD and the nature of force between two wires.

24. Calculate the height of the potential barrier for a head on collision of two deuterons.

Assume that they can be taken as hard spheres of radius 2.0 fm.

Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other.

Section-D

25. A charge is placed at the distance of 10 cm from a linear charge of

uniformly distributed once the length of 10 cm as shown in figure. Find the force on

charge.

OR

An electric dipole consists of charges of separated by a distance of 2 mm. It is placed near a long line charge of density as shown in the figure below, such that the negative charge is at a distance of 2 cm from the line charge. Calculate the force acting on dipole.

26. Two concave glass refracting surfaces, each with radius of curvature R = 35 cm and refractive index , are placed facing each other in air as shown in figure. A point object O is placed at a distance of R/2 from one of the surfaces as shown. Find

the separation between the images of O formed by each refracting surface.

OR

A magnifying lens has a focal length of 10 cm.

i. Where should the object be placed if the image is to be 30 cm from the lens?

ii. What will be the magnification?

27. A beam of light consisting of two wavelengths 6500 and 5200 , is used to obtain interference fringes in a Young’s double slit experiment.

i. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 6500 .

ii. What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.

OR

What is interference of light? Write two essential conditions for sustained interference pattern to be produced on the screen. Draw a graph showing the variation of intensity versus the position on the screen in Young’s experiment when (a) both the slits are opened and (b) one of the slit is closed. What is the effect on the interference pattern in Young’s double-slit experiment when:

i. Screen is moved closer to the plane of slits?

ii. Separation between two slits is increased. Explain.

CBSE Class 12 Physics

Sample Paper – 04

Answers

1. Neutron. It decays into proton and electron.

2. Yes, very slightly. The negatively charged body gains mass also along with electrons.

3. EM waves can accelerate charged particle and produce oscillating currents.

OR

This is because ionosphere reflects waves in these bands.

4. It is a scalar

5. Para and ferromagnetic materials.

6. Yes,

But it is not true for r.m.s. voltage. In case of r.m.s voltage

OR

Given

V = 220 V, f = 50 Hz

Capacitive reactance,

or

7.

OR

Given, Rt = 117, R27 = 100

Since,

Putting values,

t = 1000 + 27 = 1027°C

8. No because, = NBA = constant

; i = 0

9. Klystron valve or magnetron value are used for production of micro waves.

Microwaves are used in Microwave Ovens.

10. The eye is most sensitive to the light of wavelength

. Therefore, its frequency

11. Here f1 = 30cm, f2 = -20cm

Since,

f = -60cm

The combination of lenses behaves as a concave lens. The system is not converging.

12. By using formula,

The direction of force is perpendicular to the length of wire Y and acts away from X.

13. Power factor . As the value of C is changed, the value of Z also changes, hence the power factor can be improved with the help of appropriate

capacitance in the circuit.

14. Here, vs = 11.5 kHz, vc = 3.45 MHz = 3450 kHz

= Deviation in frequency

= vc – vs

= 3450 – 11.5

= 3438.5 KHz

Bandwidth = 2vs =

= 23.0 KHz

15. No, it is not necessary to have the same height.

For transmitting antenna of height ht’

x = dt or

( is negligible)

dr is also called the radio horizon of the transmitting antenna.

The maximum line of sight distance dx between the two antennas is

where hR is the height of receiving antenna.

16. As,

or

Frequency,

OR

a. From , where

= 0.0073

b. Orbital period, As

As r2 = 4 r1 and

As r3 = 9r1 and

17. The current through the circuit

Potential difference across 4 ohm resistor = lR = = 8 V Potential difference across 6 ohm resistor = 12 V Potential difference across 7 ohm resistor = 14 V

18. The work function is

= 2.07 eV

The photoelectric equation then gives

or

or

OR

Here, for neon lamp,

V = 0.54 V As,

= 0.54 + 0.97

V2 = 1.51 V

19. Field due to the bar magnet at distance x (near the loop)

(axial line)

Flux linked with the loop : Emf induced in the loop :

Induced current :

Let F = force opposing the motion of the magnet.

Power due to the opposing force = Heat dissipated in the coil per second.

Fv = i2R

OR

a. When a wire of irregular shape turns into a circular loop, area of the loop tends to increase. Therefore, magnetic flux linked with the loop increases. According to Lenz’s law, the direction of induced current must oppose the magnetic field, for which induced current should flow along adcba.

b. In this case, the magnetic flux tends to decrease. Therefore, induced current must support the magnetic field for which induced current should flow along adcba.

20. The truth table for first circuit : (a)

A B Y’ Y

0

0

1

0

0

1

1

0

1

0

1

0

1

1

0

1

If we remove the intermediate output Y’, the overall result stands for the AND gate.

A

B

Y

0

0

0

0

1

0

1

0

0

1

1

1

Hence, the circuit is an equivalent of AND gate.

The truth table for second circuit (b)

A

B

Y

0

0

1

1

0

0

1

1

0

1

1

0

0

1

1

1

1

0

0

1

If we remove the intermediate output and , the overall result stands for the OR

gate.

A

B

Y

0

0

0

0

1

1

1 0 1

1

1

1

This can be represented as Y = A + B

Hence, the circuit is an equivalent of OR gate.

21. i. The semiconductor diode whose V-I characteristic is shown in figure is Zener diode.

ii. Circuit diagram to obtain the given characteristic is shown in figure.

iii. The circuit of Zener diode used as voltage regulator is shown in figure.

The voltage to be regulated is applied across Zener diode as shown in circuit. When input voltage increases the current in Zener diode circuit increases and voltage drop across series resistance RS increases and across RL remain same i.e. the voltage drop across Zener diode. Similarly when voltage decreases, the current in the Zener diode circuit decreases and voltage drop across series RS resistance decreases but across the load resistance remains same, hence the voltage is regulated.

22. Potential of inner sphere due to its one charge

Potential of inner sphere due to its presence inside the shell

Thus total potential of inner sphere = V1 + V2

Potential of shell

Potential difference between inner sphere and shell V – V’

Thus, we can conclude that potential difference is independent of the charge q2 on the shell. Potential of sphere is positive since q1is positive. From inner sphere to shell charge (if positive) will always flow no matter whatsoever charge q2 on the shell is.

23. The electrostatic field is directed towards south. Since the electron is a negatively charged particle, therefore, the electrostatic field shall exert a force directed towards north. So, if the electron is to be prevented from deflection from straight path, by the magnetic force on the electron should be directed towards south. Now

is towards south, is due east. Applying Fleming’s left

hand rule we find that magnetic field should be in the vertically downward direction.

OR

Weight of the wire will be balanced by the force of repulsion between the wires AB

and CD.

or weight per unit length

The direction of current is CD will be opposite to that of AB.

24. Suppose the two particles are fired at each other with the same kinetic energy K so that they are brought to rest by their mutual Coulomb repulsion when they are just touching each other. We can take this value of K as a representative measure of the height of Coulomb barrier.

P.E. = 2 K.E

= 179.9 keV

25. Force on charge due to an elementary part of the linear charge,

Net force on charge

OR

Electric field intensity at a distance r from line charge of density is

Field intensity on negative charge (r = 0.02 m) Force on negative charge

It is directed towards the line charge.

Similarly field intensity at positive charge (r = 0.022 m)

Force on positive charge

It is directed away from the line charge.

Net force on the dipole,

F = F1 – F2 = (7.2 – 6.54)N = 0.66 N F is towards the line charge.

26. For refraction at the first surface, we have

R = -35cm

The distance v1 of the image I1from A is given by the relation

or

or v1 = -45cm or AI1 = 45cm

Since the radius of curvature is 35 cm, this image is 10 cm to the right of P, i.e. PI1 =

10cm.

For refraction at second surface we have

The distance v2 of the image I2 from the second surface is given by

Which gives v2 = -21cm or BI2 = 21cm. Thus, the image is 14 cm to the right of P i.e. PI2 = 14cm. Hence, the separation between the two images (I, I2) = 14 – 10 = 4 cm.

OR

i. In case of magnifying lens, the lens is convergent and the image is erect, virtual and enlarged and between infinity and the object on the same side of the lens as shown in figure above diagram,

So, here

f = +10cm, v = -30cm

Let ‘x’ be the object distance using lens formula

We have,

ii.

x = 7.5 cm

Thus, the image is erect and virtual and four times of the object.

27. i. The distance of the mth bright fringe from the central maximum is given by

= 1.17 mm

ii. Let the nth bright fringe of wavelength and the mth bright fringe of wavelength coincide at a distance y from the central maximum, then

or

The least integral value of m and n which satisfy the above condition are m = 5 and n = 4

i.e. the 5 th bright fringe of wavelength 5200 coincides with the 4th bright fringe of wavelength 6500 . The smallest value of y at which this happens is:

= 1.56 mm

OR

Interference of light: Phenomenon of redistribution of light energy in a medium on account of superposition of light waves from two coherent sources is called interference of light.

Conditions for sustained interference: The two essential conditions of sustained interference are as follows:

(i) The two sources of light should emit light continuously.

(ii) The light waves should be of same wavelength. (Monochromatic).

When both the slits are open, we get interference pattern on the screen. Then the following intensity distribution curve is obtained.

When one of the slits is closed, diffraction pattern is obtained on the screen. The

following intensity curve is obtained.

Fringe width,

(i) The distance D decreases, the fringe width also decreases if screen is moved closer to the plane of the slits.

(ii) Fringe width decreases if separation d between two slits is increased.

CBSE Class 12 Physics

Sample Paper – 05

Time allowed: 3 hours

Maximum Marks: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. Write the equation of decay of the radioactive nuclei.

2. Which orientation of an electric dipole in a uniform electric field would correspond to stable equilibrium?

3. How does the resolving power of telescope change when the aperture of the objective is increased?

OR

What will be the effect on the fringes, if Young’s double slit experiment set up is immersed in water?

4. What is the work done in moving a test charge q through a distance of 1 cm along the equatorials axis of an electric dipole?

5. What is the phase difference between voltage across an inductor and a capacitor in an a.c. circuit?

OR

The number of turns in the secondary coil of a transformer is 500 times that in primary. What power is obtained from the secondary when power fed to the primary is 10 W?

Section-B

6. The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?

OR

A magnetic needle free to rotate in a vertical plane, orients itself with its axis vertical at a certain place on the earth. What are the values of (a) horizontal component of earth’s field? (b) angle of dip at this place?

7. How does the drift velocity of electrons in a metallic conductor vary with increase in temperature?

OR

Write the condition under which the potential difference between the terminals of a battery and its emf are equal.

8. An astronomical telescope consists of two thin lenses set 36 cm apart and has a magnifying power of 8 in normal adjustment. Calculate the focal lengths of lenses.

9. ‘Microwaves are used in Radar’. Why?

10. The small ozone layer on top of the stratosphere is crucial for human survival. Why?

11. Why cannot we obtain interference using two independent sources of light?

12. A beam of protons with a velocity enters a uniform magnetic field of 0.3

T at an angle 60° to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of the helix mp

Section-C

13. The power factor of an a.c. circuit is 0.5. What will be the phase difference between voltage and current in this circuit?

14. A TV tower has a height of 70 m

a. how much population is covered by the T.V. broadcast it the average population density around the tower is 1000 per mk-2. Radius of earth is .

b. by how much the height of tower be increased to double its coverage range?

15. Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.

16. (a) The energy levels of an atom are as shown below. Which of them will result in the transition of a photon of wavelength 275 nm?

(b) Which transition corresponds to emission of radiation of maximum wavelength?

OR

The wavelength of the first member of the Balmer series in hydrogen spectrum is

6563 A. What is the wavelength of the first member of Lyman series?

17. Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cable. ( ,

, Relative density of Al = 2.7, of Cu = 8.9).

18. Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy, an X-ray photon or the electron? (For quantitative comparison, take the wavelength of the

probe equal to 1 , which is of the order of interatomic spacing in the lattice)

OR

Calculate de-Broglie wavelength in nm associated with a ball of mass 66 g moving with a velocity . Given .

19. Two coherent light sources of intensity ratio 25 : 4 are employed in an interference experiment what is the ratio of the intensities of the maxima and minima in the interference pattern?

OR

Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of

(a) reflected, and (b) refracted light? Refractive index of water is 1.33.

20. Using a suitable combination from a NOR, an OR and a NOT gate. Draw circuits to obtain the truth table given below:

A

B

Y

A

B

Y

0

0

0

0

0

1

0

1

0

0

1

1

1

0

1

1

0

0

1

1

0

1

1

1

(i)

(ii)

21. The ratios of number density of free electron to holes, , for two different materials A and B, are equal to one and less than one respectively. Name the type of

semiconductors to which A and B belong. Draw energy level diagram for A and B.

22. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape ? Explain.

23. A toroid has a core(non ferromagnetic material) of inner radius 25 cm and outer radius 26 cm around which 3500 turns of wire are wound. If the current in the wire is

11 A, what is the magnetic field

a. outside the toroid

b. inside the core of the toroid

c. in the empty space surrounded by the toroid?

OR

An electron moves around the nucleus in a hydrogen atom of radius 0.51 , with a velocity of ..

Calculate the following:

i. the equivalent current due to the orbital motion of the electron, ii. the magnetic field produced at the centre of the nucleus

iii. the magnetic moment associated with the electron.

24. Two nuclei P, Q have equal number of atoms at t = 0. Their half lives are 3 hours and 9 hours respectively. Compare their rates of disintegration, after 18 hours from the

start.

Section-D

25. A thin insulating rod of length L carries a uniformly distributed charge Q. Find the electric field strength at a point along its axis at a distance ‘a’ from one end.

OR

Two equally charged particles, held apart, are released from rest. The initial acceleration of the first particle is observed to be 7.0 m/s2 and that of the second to be 9.0 m/s2 . If the mass of the first particle is , what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle?

26. A parallel beam of light traveling in water (refractive index = ) is refracted by a spherical air bubble of radius 2 mm situated in water.Assuming the light rays to be paraxial,

a. find the position of the image due to refraction at the first surface and the position of the final image and

b. draw a ray diagram showing the positions of both the images.

OR

The plane surface of a planoconvex lens of focal length 60 cm is silvered. A point object is placed at a distance 20 cm from the lens. Find the position and nature of the final image formed.

27. Two long parallel horizontal rails, distance d apart and each having a resistance per unit length, are joined at one end by a resistance R. A perfectly conducting rod MN of mass m is free to slide along the rails without friction. There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper. A variable force F is applied to the rod MN such that as the rod moves, constant current flows through R.

i. Find the velocity of the rod and the applied force F as function of the distance x of the rod from R.

ii. What fraction of the work done per second by F is converted into heat?

OR

A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius

20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm.

(a) What is the flux linking the bigger loop if a current of 2.0 A flows through the smaller loop?

(b) Obtain the mutual inductance of the two loops.

CBSE Class 12 Physics

Sample Paper – 05

Answers

1.

2. The dipole is in stable equilibrium when the direction of the electric dipole moment of an electric dipole is in the direction of electric field.

3. The resolving power of the telescope increases on increasing the aperture of an objective lens.

OR

The wavelength of light in water is less than that in the air. The fringe width is

directly proportional to the wavelength. Therefore, the fringes will become narrower.

4. As

Work done in moving a positive test charge q through a distance 1 cm is:

5. 180°

OR

If there is no loss of energy, then the output power will be 10 W.

6. Since Britain is closer to the magnetic north pole, we can expect a greater dip angle in

Britain. It is about 70°.

OR

The place is clearly the magnetic pole of earth as needle rotates in vertical plane not in horizontal plane

a. (horizontal component)

b. (angle of dip)

7. Drift velocity of electrons in a metallic conductor decreases as the temperature

increases as number of collisions increases on increasing temperature.

OR

When the internal resistance of battery is zero, the potential difference is equal to emf.

8. In the normal adjustment the final image is formed at infinity. f0 + fe = 36

and

f0 = 8fe

From equations (i) and (ii), we have

8fe + fe = 36

9fe = 36 fe = 4 cm

and = 32 cm

9. Due to their smaller wavelength, microwaves can be transmitted as beam signals in a particular direction, much better than radio waves. Microwaves do not bend around the corners of any obstacle coming in their path & It improve accuracy of radar immensely.

10. The small ozone layer on the top of the stratosphere absorbs ultraviolet radiations,

etc., from the sun. Hence these harmful radiations, which can cause genetic damage to the living cells, are prevented from reaching the earth. Thus, the small ozone layer on top of the stratosphere is crucial for human survival.

11. A sustained interference pattern cannot be obtained by using two independent sources of light. It is because of the following reasons:

Two independent sources of light can’t emit waves continuously.

The waves emitted by two independent sources of light do not have same phase or a constant phase difference

Two independent sources of light cannot be coherent, as their relative phases are changing randomly.

12.

or

13. Power factor,

The phase difference between voltage an current is 60° or radian.

14. Here, h = 70 m

a. Population covered = (a) Population covered

= 28.16 lakhs.

b. If h’ is the new height of TV tower to double the coverage range, then

or

= 280 m

Increase in height of tower = h’ – h = 280 – 70

15. The received signal is given and carrier signal as: and

The carrier is available at the receiving station. By multiplying the two

signals, we get

If this signal is passed through a low pass filter, the modulating signal, which is recorded is given by

16. (a) For element A

Ground state energy, E1 = -2eV

Excited state energy, E2 = 0eV

Energy of photon emitted, E = E2 – E1

= 0 – (-2) = 2eV

Wavelength of photon emitted,

For element B

= 621.1 nm

E1 = -4.5 eV, E2 = 0eV E = 0 – (-4.5) = 4.5eV

For element C

= 276 nm

E1 = -4.5eV, E2 = -2eV E = -2 – (4.5) = 2.5 eV

For element D

E1 = -10eV, E2 = -2eV E = -2 – (-10) = 8eV

(b) Element A has radiation of maximum wavelength 621 nm.

OR

Balmer series

17.

where V = volume of wire = Al m = mass of wire

d = density of wire material mass = volume density = Ald

For aluminium wire

For copper wire,

Since RAl = Rcu and 1Al = 1Cu

It indicates that aluminium wire is lighter than copper wire. Therefore, aluminium wires are preferred in overhead cables.

18. Energy of photon,

= 12.43 keV For the case of electron,

or

or

or E = 150.8 eV

For the same given wavelength, kinetic energy of a photon is much greater than that of electron.

OR

Given: m = 66g We know,

19. Let I1 and I2 be the intensities of the two coherent beams and A1 and A2 their respective amplitudes.

Now,

Intensity ratio, therefore Amplitude ratio

i.e. A1 = 5 units

and A2 = 2 units

At maxima : Amax = A1 + A2 = 7 units At minima : Amin = A1 – A2 = 3 units Hence,

OR

Given,

a. For reflected light

Wavelength

Speed

b. For refracted light

As frequency remains unaffected on entering another medium, therefore,

Speed,

20. Circuit diagram for truth table :

Verification:

A

B

0

0

1

1

0

0

1

1

1

0

1

0

0

0

1

1

1

0

1

0

(ii) Circuit diagram for truth table:

Verification:

A

B

0

0

1

1

0

1

1

1

1

0

0

0

1 1 0 1

21. A is intrinsic semiconductor and B is p-type semiconductor.

22. a. As a +q charge place at the centre of the shell, it will create a ‘-q’ charge on the inner surface of the shell.

The charge on the outer surface will increase by +q due to the -q charge on the outer surface will increase by +q due to the -q charge on the inner surface by induction. Therefore, there will be total (Q+q) charge on the outer surface of the shell and -q charge on the inner surface of the shell.

Now surface area of inner surface

And surface area of outer surface

Thus, charge density on the outer surface and charge density on the inner surface

b. As charge in shell reside on outer surface so, the net charge on the inner surface of the cavity is zero as per the Gaussian theorem. Although the net charge is zero yet the electric field may not be zero if the cavity is not spherical. The surface may not have equal number of positive and negative charges. We assume a loop for this reason, some portion of which is inside the cavity and rest of its part is inside the conductor. Now, consider, that there is some electric field inside the cavity. Since inside the conductor total electric field is zero and network done by the field in bringing a test charge over this loop will not be zero. But this is not possible for an electrostatic field. Therefore, we must conclude that there is no electric field inside the cavity irrespective of its shape.

23. Given

r1 = 0.25 m, r2 = 0.26 m N = 3500

I = 11 A

a. The magnetic field is zero outside the toroid. b. Magnetic field inside the core of the toroid

or ( Number of turns per unit length)

Putting the values

c. The magnetic field is zero in the empty space surrounded by the toroid.

OR Given, v = 2 105m/s

i. Equivalent current

or

ii. Magnetic field

or

iii. Magnetic moment, M = I A = I( r2)

or

24. Numerical:

Number of half lives of P in 18 hours

Number of nuclei of P left undecayed after 6 half lives

Number of half lives of Q in 18 hours

Number of nuclei Q left undecayed after 2 half lives

Hence, R1 : R2 = 3 : 16

25. Let us consider an infinitesimal element of length dx at a distance x from the point P.

The charge on this element is , where is the linear charge density.

The magnitude of the electric field at P due to this element is:

and its direction is to the right since is positive. The total electric field strength E is

given by:

OR

Given, a1 = 7.0m/s2, a2 = 9.0 m/s2

As F1 = F2

m1a1 = m2a2

As

26. Refer to figure below:

(a) For refraction at the first surface, we use

where and and R = 2 mm. Thus

which gives v1 = -6mm, the negative sign indicates that the image I1 is virtual and is on the same side as the object at a distance of 6 mm from the first surface.

For refraction at the second surface, the image I1 serves as the virtual object which is at a distance of 6 mm + 4 mm = 10 mm from the second surface. For this refraction, we use

where u = -10mm and R = -2mm.Thus

Which given v2 = -5 mm. The final image I2 is virtual and is formed at a distance of 5 mm from the second surface to the left of the second surface, i.e., the final image is formed at a distance of 1 mm from the first surface.

OR

Let f be the focal length of the equivalent spherical mirror. We have,

or

Here f1 = + 60cm

f = + 30 cm

The problem is reduced to a simple case where a point object is placed in front of a concave mirror.

Now, using mirror formula

we have

27. Let the distance from R to MN be x. Then the area of the loop between MN and R is xd and the magnetic flux linked with the loop is B x d. As the rod moves, the emf induced in the loop is given by

Where v is the velocity of MN. The total resistance of the loop between R and MN is

. The current in the loop is given by

i. Force acting on the rod,

or

On integrating both sides, we get

and Force

ii. Work done per second = Fv

Heat produced per second

= F . v

Thus, the ratio of heat produced to work done is I. The entire work done by F per second is converted into heat.

OR

We know from the considerations of symmetry that M12 = M21. Direct calculation of flux linking the bigger loop due to the field by the smaller loop will be difficult to handle. Instead, let us calculate the flux through the smaller loop due to a current in the bigger loop. The smaller loop is so small in area that one can take the simple formula for field B on the axis of the bigger loop and multiply B by the small area of

the loop to calculate flux without much error. Let 1 refer to the bigger loop and 2 the smaller loop. Field B2 at 2 due to I1 in 1 is:

Here x is distance between the centres.

But

Using the given data

CBSE Class 12 Physics

Sample Paper – 06

Time allowed: 3 hours

Maximum Marks: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. Name two radioactive elements which are not found in observable quantities?

2. Is torque on an electric dipole a vector or scalar?

3. A lens of glass is immersed in water. What will be its effect on the power of the lens?

OR

In a simple microscope why the focal length of the lens should be small?

4. What is the shape of the equipotential surfaces for a uniform electric field?

5. Is a motor starter a variable R or L or?

OR

State two factors by which the range of transmission of signals by a TV tower can be increased?

Section-B

6. Interstellar space has an extremely weak magnetic field of the order of 10-12T. Can such a weak field be of any significant consequence ? Explain.

OR

The vertical component of earth’s magnetic field at a place is times the horizontal component. What is the value of angle of dip at this place?

7. Two electric field lines never cross each other. Why?

OR

Is the electric force between two electrons greater than the gravitational force between them? If so by what factor?

8. How does the angular separation of interference fringes change, in Young’s experiment, if the distance between the slits is increased?

9. The monochromatic source of light in Young’s double slit experiment is replaced by

another monochromatic source of shorter wavelength. What will be the effect?

10. A radio transmitter operates at a frequency of 880 kHz, and a power of 10 kW. Find the number of photons per second emitted.

11. A compound microscope with an objective of 1.0 cm focal length and an eyepiece of

2.0 cm focal length has a tube length of 20 cm. Calculate the magnitude power of microscope, if the final image is formed at the near point of the eye.

12. A magnetic dipole is situated in the direction of a magnetic field. What is its potential energy? If it is rotated by 180°, then what amount of work will be done?

Section-C

13. If the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence explain statement that at very high frequency, an inductor in circuit nearly amounts to an open circuit. How does an inductor behave in a d.c. circuit after the steady state?

14. What is the meaning of the term ‘attenuation’ in communication system?

15. Why communication at frequencies above 20 MHz is generally carried out with the help of satellites?

16. Calculate the nearest distance of approach of an -particle of energy 2.5 eV being scattered by a gold nucleus (Z = 79).

OR

Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e. an atom in which a negatively charged muon of mass about 207 me orbits around a proton].

17. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is where is the unit vector in the outward normal direction and is the surface charge density near the hole.

18. The fission properties of are very similar to those of . The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms

in 1 kg of pure undergo fission?

OR

Light of wavelength 5000 falls on a sensitive plate with work function 1.90 eV. Calculate:

a. the energy of the photon in eV,

b. kinetic energy of the emitted photon-electrons and c. stopping potential

19. The ratio of the intensities at minima in the interference pattern is 9 : 25. What will be the ratio of the widths of the two slits in the Young’s double slit experiment?

OR

Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.

20. Draw a circuit diagram for use of n-p-n transistor as an amplifier in common emitter configuration. The input resistance of a transistor is 1000 . On changing its base current by , the collector current increases by 2 mA. If a load resistance of 5k

is used in the circuit, calculate: (i) the current gain,(ii) voltage gain of the amplifier.

21. 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.

22. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. a. What is the total capacitance of the combination?

b. Determine the charge on each capacitor if the combination is connected to a 100 V

supply.

23. A galvanometer of resistance gives full scale deflection for a current of 0.05 A.

Calculate the length of the shunt wire required to convert the galvanometer into an ammeter of range 0 to 5 A. The diameter of the shunt wire is 2 mm and its resistivity is

OR

Calculate the potential at P due to the charge configuration as shown in the following figure. If r >>a, then how will you modify the result?

24. It is found from an experiment that the radioactive substance emits one beta particle for each decay process. Also an average of 8.4 beta particles are emitted each second by 2.5 miligram of substance. The atomic weight of substance is 230. What is the half life?

Section-D

25. A potentiometer wire of length 100 cm has a resistance of . It is connected in series with a resistance and an accumulator of emf 2V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. What is the value of the external resistance?

OR

Determine the current in each branch of the network shown in figure.

26. With the help of ray diagram, show the formation of image of a point object by refraction of light at a spherical surface separating two media of refractive indices n1 and n2 (n2 > n1)respectively. Using this diagram, derive the relation.

Write the sign conventions used. What happens to the focal length of convex lens when it is immersed in water?

OR

If you sit in a parked car, you glance in the rear view mirror R = 2 m and notice a jogger approaching, f the jogger is running at a speed of 5ms-1, how fast is the image of the jogger moving when the jogger is (a) 39 m (b) 29 m (c) 19 m and (d) 9 m away?

27. A current of 10 A is flowing in a long straight wire situated near a rectangular coil.

The two sides, of the coil, of length 0.2 m are parallel to the wire. One of them is at a distance of 0.05 m and the other is at a distance of 0.10 m from the wire. The wire is in the plane of the coil. Calculate the magnetic flux through the rectangular coil. If the current uniformly to zero in 0.02 s, find the emf induced in the coil and indicate the direction in which the induced current flows.

OR

A wheel with 10 metallic spokes each 0.5 m long is rotated with angular speed of 120 revolutions per minute in a plane normal to the earth’s magnetic field. If the earth’s magnetic field at the given place is 0.4 gauss, find the EMF induced between the axle and the rim of the wheel.

CBSE Class 12 Physics

Sample Paper – 06

Answers

1. Tritium and Plutonium.

2. Torque is vector quantity.

3. Power increases.

OR

This is because the angular magnification is inversely proportional to the focal length.

4. The equipotential surfaces are perpendicular to the direction of electric field.

5. Variable R

OR

i. By increasing the height of the transmission tower.

ii. By increasing the height of the receiving antenna, so that it may directly intercept the signal from the transmitting antenna.

6. From the relation we find that an extremely minute field bends charged particles in a circle of very large radius. Over a small distance, the

deflection due to the circular orbit of such large R may not be noticeable, but over the

gigantic interstellar distances, the deflection can significantly affect the passage of charged particles e.g. cosmic rays.

OR

(angle of dip)

7. Two electric field lines never cross each other. If they intersect, then there will be two directions of electric field at the point of intersection which is not possible.

OR

Yes,

Electric force between 2 e- = = =

Gravitational Force between them = = Hence,

Hence by a factor of , EF is greater than GF

8. As,

When separation between slits (d) is increased, then fringe width is decreased.

9. Both the fringe width and the angular separation decrease as

10. Given:

P = 10 kW= 10x 103 W

Frequency ()=880 kHz =880 x 103 HZ Number of photons=

11. Given:

l = 20 cm D= 25 cm, f0 = 1cm

fe = 2

As per formula ,

12. P.E. of dipole = -MB cos0o = -MB

Work done = MB(cos 0o – cos 180o) = M B (1 + 1 ) = 2 MB

13. Here, L = 0.50 H, R =

Vrms = 240V, f = 10 kHz = 104Hz

Peak voltage, = 339.36 V Maximum current

(Neglecting R)

= 0.01212 A

This current is much smaller than for the low frequency showing that the inductive reactance is very large at high frequency and L nearly amounts to an open circuit. In d.c. circuit (after steady state) .

i.e. inductance L behaves like a pure inductor.

14. It is the loss of strength of a signal while propagating through a medium.

15. Frequency above 20 MHz is carried out with the help of satellites because

a. This frequency is greater than the critical frequency and ionosphere cannot reflect it.

b. The co-axial cables and wires are extremely radiating in nature at this high frequency.

16. We know that the electrostatic potential at a distance x due to nucleus is given by

where Ze is the charge on the nucleus.

The potential energy of an particle when it is a distance x, from the nucleus is given by

2 e being the charge on particle.

Since the -particle is momentarily stopped at a distance x, its initial kinetic energy is completely changed into potential energy here. Hence

(at nearest approach K.E. = P.E. )

or Now energy of -particle = 2.5 MeV

Substituting values we get

OR

The first Bohr’s radius of H-atom is given by

If r1′ is the first Bohr’s radius of muonic hydrogen atom, then

The ground state (n = 1) energy of H-atom is given by

= 13.6 eV

If is ground state energy of muonic hydrogen atom, then

= -2815.2 eV

17. Let us take a charged conductor with the hole filled up, as shown by shaded portion in the figure.

We find with the application of Gaussian theorem that field inside is zero and just outside is . This field can be viewed as the superposition of the field E2 due to the filled up hole plus the field E1 due to the rest of the charged conductor.

The two fields (E1 and E2) must be equal and opposite as the field vanishes inside the conductor.

Thus, E1 – E2 = 0

Now, the field outside the conductor is given by

or

Therefore, field in the hole (due to the rest of the conductor) is given as:

( unit vector in the outward normal direction)

18. Energy released per fission of

Quantity of fissionable material = 1 kg

In 239 gm Pu, number of fissionable atom or nuclei

In 1 g of Pu, number of fissionable atom or nuclei

In 1000 gm of Pu, number of fissionable atom or nuclei

Energy released in fission of single Pu nucleus

= 180 MeV

Energy released in fission of nucleus or in fission of 1 kg pure Pu.

OR

Given,

a. E = hv =

or,

= 2.48 eV

b. K.E. = (2.48 – 1.90)eV = 0.58 eV

c. eV0 = K.E.or = 0.58 V

19. Intensity is proportional to width of slit. So, amplitude a1 and a2 I proportional to the square root of the width of the slit.

Here and represent the widths of the two slits. Now,

or

or

or Thus,

or , = 16 : 1

OR

For diffraction or bending of waves by obstacles or apertures by a large angle, the size of the latter should be comparable to wavelength. If the size of the obstacle / aperture is much too large compared to wavelength, diffraction is by a small angle. Here the size of partition wall is of the order of a few meters. The wavelength of light is about 5

10-7m, while sound wave of say 1 kHz have wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot.

20. Circuit of n-p-n transistor as an amplifier in common emitter configuration is given below:

Numerical

Given,

i. The current gain,

ii. Voltage gain,

21. Input frequency for half wave and full wave rectifier = 50 Hz.

Input and output waveforms of half wave and full wave rectifier are shown in figure. Half wave rectifier conducts once during a cycle and full wave rectifier does so twice. Therefore, if input frequency is 50 Hz, output frequency for half wave and full wave rectifier are 50 and 100 Hz respectively.

22. Given,

a. Since the capacitors are connected in parallel then

C = C1 + C2 + C3

b. Given, V = 100 volt

23. By using the formula, Putting the values,

Now, or Putting the values,

= 3.175 m

OR

Potential at p due to the given charge configuration is the sum of the potentials due to charges -q +q and +q. These charges are at distances r +a, r and r – a respectively from the point P.

or

if r > > a, r2 – a2 = r2

24. The activity = 8.4 sec-1

Number of atoms in kilomole (i.e. 230 kg) =

Half life

25. Let AB be the potentiometer wire and R, the external resistance, as shown in the figure. Potential drop across the wire AB = current resistance

Therefore, the potential drop per cm of the wire is

The fall of potential across 40 cm of the wire is

Which must be equal to the emf of the source when the balance is achieved. Thus,

R + 100 = 800 or R = 790

OR

Applying Kirchhoff’s law to the mesh ABDA,

-101 – 5g + (I – I1)5 = 0

……. (i)

Again, applying Kirchhoff’s second law to the mesh BDCB,

– 5Ig – 10(I – I1 + Ig) + 5(I1 – Ig) = 0

3I1 – 2I – 4Ig = 0 …… (ii)

Applying Kirchhoff’s second law to the mesh ABCEA,

-10I1 – 5(I1 – Ig) – 10I + 10 = 0

3I1 + 2I – Ig = 2 ……. (iii) Adding (i) and (iii), we get

6I1 + I = 2 ……. (iv)

Multiplying (i) by 4 and adding in (ii), we get

15I1-6I=0……(v)

Solving equations (iv) and (v), we get

= 0.235 A

So, current in branch AB is 0.235 A

Putting the value of I1 in equation (v) and simplifying, we get

Total current, = 0.588 A

Putting the values of I and I1 in equation (iii) and simplifying, we get

The negative sign indicates that the direction of current is opposite to that shown in the figure above

So current in branch BD is “-0.118A” Current in branch BC is (I1 – Ig)i.e.

i.e. or 0.353 A

Current in branch AD is (I – I1)

i.e. i.e. or 0.353 A Current in branch DC is (I – I1 + Ig)

i.e. or or 0.235 A

26. AMB is a convex surface separating two media of refractive indices n1 and n2(n2 > n1).

Consider a point object O placed on the principal axis. A ray ON is incident at N and refracts along NI. The ray along ON goes straight and meets the previous ray at I. Thus I is the real image of O.

From Snell’s law,

n1sin i = n2sin r

or n1i = n2r [ is very small] From

From

or

or

But [P is close to M]

or

Using Cartesian sign convention, OM = -u, MI = +v, MC = +R

or

OR

Using mirror equation,

or

(a) For convex mirror, u = -39m, f = 1m

Since the jogger moves at a constant speed of 5ms-1 after 1s the position of the image v for u = -39 + 5 = -34m is

Difference in the position of image is

For u = -29m, -19m and -9m the speed of image is and

respectively.

The speed becomes very high as the jogger approaches the car. The change in speed can be experienced by anybody while travelling in a bus or a car.

27. Consider a strip of width dr at a distance r from the straight wire.

Magnetic field at the location of the strip due to the wire, z

Area of strip, dA = ldr

Magnetic flux linked with the strip,

Total magnetic flux linked with the coil,

Induced emf

Magnetic field, due to wire, at the location of the coil is perpendicular to the plane of the coil and directed inwards. When current is reduced to zero, this magnetic field decreases. To oppose this decrease, induced current shall flow clockwise, so that its magnetic field is also perpendicular to the plane of the coil and downward.

OR

The area covered by an angle , The induced EMF,

v = 120 rev/min = 2 rev/s

r = 0.5 m

B = 0.4 G

Each spoke will act as a parallel source of EMF. Hence, the EMF will be

.

CBSE Class 12 Physics

Sample Paper-07

Time allowed: 3 hours

Maximum Marks: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. A radioactive material has a half-life of 1 minute. If one of the nuclei decays now, when will the next one decay?

2. Is torque on an electric dipole a vector or scalar?

3. If a telescope is inverted, will it be able to work as a microscope?

OR

A ray of light falls on a mirror normally. What are the values of angle of incidence and the angle of reflection.

4. We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

5. What is the phase difference between voltage across an inductor and a capacitor in an a.c. circuit?

OR

The number of turns in the secondary coil of a transformer is 500 times that in primary. What power is obtained from the secondary when power fed to the primary is 10 W?

Section-B

6. What are S.I. units of susceptibility?

OR

Explain qualitatively on the basis of domain picture the reversibility in the magnetization curve of a ferromagnet.

7. Define the term electric dipole moment. Is it a scalar or a vector quantity?

OR

Can a charged body attract another uncharged body? Explain.

8. Two towers on top of two hills are 40 km apart. The line joining them passes 50 m

above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?

9. If the angle between the pass axis is polarizer and the analyzer is 45°, write the ratio of the intensities of original light and the transmitted light after passing through the analyzer.

10. The magnetic field of an electromagnetic wave oscillates parallel to a y-axis and is given by By = B0y sin(kz – t).

a. In what direction does the wave travel and

b. parallel to which axis does the associated electric field oscillate?

11. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT . What is the amplitude of the electric field part of the wave?

12. Give two factors by which the current sensitivity / voltage sensitivity of a moving coil galvanometer can be increased.

Section-C

13. The instantaneous current and voltage of an a.c circuit are given byi = 10 sin 314 t A

and v = 50 sin (314 t + ) V.What is the power dissipation in the circuit?

14. What does the term ‘LOS communication’ mean? Name the types of waves that are used for this communication. Give typical examples, with the help of suitable figure, of communication systems that use space wave mode propagation.

15. What is meant by deviation in frequency of FM wave?

16. The Rydberg constant for hydrogen is 10967700 m-1. Calculate the short and long wavelength limits of Lyman series.

OR

Using Rydberg formula, calculate the wavelengths of the first four spectral lines in the

Balmer series of hydrogen atom spectrum.

17. Suppose that the particle in an electron projected with velocity

. If E between the plates separated by 0.5 cm is , where will the electron strike the upper plate? .

18. Ultravoilet light of wavelength 800 and 700 , when allowed to fall on hydrogen atom in their ground state, is able to liberate electrons with kinetic energy 1.8 eV and

4.0 eV respectively. Find the value of Planck’s constant.

OR

What reasoning led de-Broglie to put forward the concept of matter wave? The wavelength , of a photon, and de-Broglie wavelength associated with a particle of mass m has the same value, say . Show that the energy of photon is times the kinetic energy of the particle

19. In Young’s double slit experiment, monochromatic light of wavelength 600 m illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 10 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes

are separated by 8 mm. Find the wavelength of light from the second source. What is the effect on the interference fringes if the monochromatic source is replaced by a source of white light?

OR

What are coherent sources? How does the width of interference fringes in Young’s experiment change when

a. the distance between the slits and screen is decreased?

b. frequency of the source is increased?

20. How many NAND gates are required to make one NOT gate?

21. State the basic electronic circuit involved in the electronic circuit of figure and find the value of the output y. Name the gate formed and set up a truth table for it.

22. The separation between the plates of a charged capacitor is to be increased. Explain when work done will be more in case battery is removed after charging the capacitor or battery remains connected.

23. A dielectric slab of thickness ‘t’ is kept in between the plates, each of area ‘A’, of a parallel plate capacitor separated by a distance ‘d’. Derive an expression for the capacitance of this capacitor for t < < d.

OR

A magnetic field of 100 G (1G = 10-4T) is required which is uniform in a region of linear dimension about 10 cm and area of cross section about 10-3m2. The maximum current carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m-1. Suggest some

appropriate design particulars of a solenoid for the required purpose. Assume the

core is not ferromagnetic.

24. A radioactive sample contains 2.2 mg of pure , which has half life period of 1224 seconds.

Calculate:

i. The number of atoms present initially.

ii. The activity when of the sample will be left.

Section-D

25. State the two rules that serve as general rules for analysis of electric circuit.Use these

rules to write the three equations that may be used to obtain the values of the three unknown currents in the branches of the circuit given below:

OR

Eight identical resistors r, each are connected along the edges of a pyramid having square base ABCD as shown in figure below. Calculate equivalent resistance between A and B. Solve the problem

i. without using Kirchhoff’s laws

ii. by using Kirchhoff’s laws.reactance

26. An angular magnification of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?

OR

A compound microscope consists of an objective lens of focal length 2.0 cm and an

eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at

a. the least distance of distinct vision (25 cm), and

b. at infinity? What is the magnifying power of the microscope in each case?

27. A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius

20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm.

(a) What is the flux linking the bigger loop if a current of 2.0 A flows through the smaller loop?

(b) Obtain the mutual inductance of the two loops.

OR

State Lenz’s law. Give one example to illustrate this law. “The Lenz’s law is a consequence of the principle of conservation of energy.” Justify this statement.

CBSE Class 12 Physics

Sample Paper-07

Answers

1. The next nucleus can decay any time.

2. Torque is vector quantity.

3. An inverted telescope will not work as a microscope.

OR

Both angle of incidence and the angle of reflection is zero.

4. No, since electric potential is a scalar quantity, it is continuous everywhere.

5. 180°

OR

If there is no loss of energy, then the output power will be 10 W.

6.

has no units.

Here I is intensity of magnetization and H is magnetizing field.

OR

The atomic dipoles are grouped together in domains in a ferromagnetic substance. All the dipoles of a domain are aligned in the same direction and have net magnetic moment. In an unmagnetised substance these domains are randomly distributed so that the resultant magnetization is zero. These domains align themselves in the direction of the field when the substance is placed in an external magnetic field. Some energy is spent in the process of alignment. These domains do not come back into

their original random positions completely when the external field is removed. Some magnetization is retained by the substance. The energy spent in the process of magnetization is not fully recovered. The balance of energy is lost as heat. This is the

basic cause for irreversibility of the magnetization curve of a ferromagnetic substance.

7. The product of either charge and separation between two charges is termed as

electric dipole moment. It is a vector quantity in the direction of the dipole axis from – q to +q.

or

OR

Yes, a charged body can attract another uncharged body. When the charged body is placed near the uncharged body, the induced charges of opposite kind are produced on the uncharged body and the uncharged body is attract by charged body.

8. Size of aperture, a = 50 m

Distance of aperture from tower

Fresnel distance,

= 12.5 cm

9. Given,

Using formula,

or

or

10. 1. The phase is changing with z and t. So, the wave is traveling along z-direction.

2. Since and the direction of propagation are mutually perpendicular

therefore should be along the x-axis.

11. Given = 510 nT For electromagnetic waves,

or E0 = C B0

= 153 NC-1

12. i. Increasing the number of turns in the galvanometer coil. ii. Decreasing the torsion constant of the suspension fibre.

13. Given, i = 10 sin 314 t A

v = 50 sin (314 t + ) V

i0 = 10, V0 = 50 and

Power dissipation in the circuit is given by

= 0

14. Mode of radio wave propagation by space waves, in which the wave travels in a straight line from transmitting antenna to the receiving antenna, is called line of sight (LOS) communication.

Two types of waves that are used for LOS communication are : space wave and

ground wave. At frequencies above 40 MHz, LOS communication is essentially limited to line-of sight paths.

15. Deviation in frequency

Where is constant frequency of carrier wave and is instantaneous frequency of the FM wave at any time t.

16. For Lyman series, the wave number is given by

For the short wavelength limit

or

For long wavelength limit

OR

The Rydberg formula is

E0 = -13.6 eV

The wavelengths of the first four lines in the Balmer series correspond to transitions from n1 = 3, 4, 5, 6, to n2 = 2.

Substituting n1= 3, 4, 5, and 6, we get

and

17. Acceleration,

Using formula We get,

Simplifying for value of t, we get

The electron covers vertical distance is shown as y = vxt

= 1.6 cm

18. The energy E is related to wavelength , velocity of light c and Planck’s constant is

( h Planck’s constant)

For radiations of wavelengths and respectively, energies E1 and E2 . and

Subtracting,

or

or h = 6.57 10-34Js

OR

de-Broglie put forward the bold hypothesis that moving particles of matter should display wave like properties under suitable conditions. He reasoned that nature is symmetrical and that the two basic physical entities, matter and energy must have symmetrical character. If radiation shows a dual nature, so should matter.

or

Also E (Photon)

or

19. Here,

Let d be the slit width and D be the distance between slit and screen, then

and

or

= 480 nm

If monochromatic source is replaced by white light. We will not fringes on screen as the white light slits will not act as coherent sources of light.

OR

Two sources which produce waves of same frequency and the phase difference of produced waves does not change with time then the sources are said to be coherent sources.

a. The fringe width decreases of the distance D between the slits and screen is decreased. Since .

b. The fringe width decreased if the frequency v of the source is increased.

Since .

20. Only one NAND gate is required to make a NOT gate. If one input of a NAND gate is made permanently high, we get a NOT gate.

Alternatively, if both the inputs of a NAND gate are tied together, we get a NOT gate.

21. The electronic circuit shown is a bubbled AND gate. It is a combination of two NOT gates and one AND gate, i.e. the output of two NOT gates is made as an input of AND gate. The Boolean expression for the output of this gate is

(From De-Morgan’s theorem)

This expression is for Nor gate. Hence, a bubbled AND gate is equivalent to a NOR

gate. Truth table of this gate is as shown in the table.

A

B

A + B

0

0

0

1

1

0

1

0

0

1

1

0

1

1

1

0

22. With increase in separation between the two plates, capacity C decreases.

When battery is removed, charge Q and electric field = would remain constant. But when battery remains connected, V is constant, Q ( = CV) decreases and hence E decreases. Clearly, more work is required to be done in the first case.

23. Let A is the area of the two plates of the parallel plate capacitor and d is the separation between them. A dielectric slab of thickness t < d and area A is kept between the two plates. The total electric field inside the dielectric slab will be:

where is the opposite field developed inside the slab due to polarization of slab. Total potential difference between the plates,

where q is the charge on each plate. Since,

or

or

OR

Given, B = 100 G = 10-2T I = 15 A, n = 1000m-1

Magnetic field inside a solenoid is

We may have I = 10 A and n = 800

The solenoid may have length 50 cm and cross section (five times given values) so as to avoid edge effects etc.

24. i. The number of atoms present initially ii. The number of atoms present in

The activity,

25. Two rules of Kirchhoff are used for analysis of the electrical circuit

In loop ABCA

-I2R1 – (I1 + I2)R3 – I1r + E1 = 0

I2R1 + (I1 + I2)R3 + I1r = E1

In loop ACDA

I1r – (I2 + I3 – I1)R4 + (I1 – I2)R2 – E1 = 0

I1r – (I2 + I3 – I1)R4 + (I1 – I2)R2 = E1 ….(ii) In loop ABCDA

-I2R1 – (I1 + I2)R3 – (I2 + I3 – I1)R4 + (I1 – I2)R2 – E1 = 0

I2R1 + (I1 + I2)R3 + (I2 + I3 – I1)R4 – (I1 – I2)R2 = E1 ….(iii)

OR

i. Without using Kirchhoff’s laws

Consider a battery connected between A and B. The circuit now has a plane of symmetry. This plane of symmetry passes through the mid-points of AB and CD and the vertex O. So, the currents are same in (i) AO and OB (ii) DO and OC. Now, OA and OB can be treated as a series combination which gives resistance 25. Also DO and OC are in series combination. This gives 2r. It is in parallel with DC. This

gives a resistance of or . This is in series with resistances AD and CB. This gives i.e. . Now, , resistor AB and combination of AO and BO i.e. 2r are in parallel. If R is the equivalent resistance, then

ii. By using Kirchhoff’s laws

Using Kirchhoff’s second law in loop DOCD, we get

-I3r – I3r + (I2 – I3) = 0 or, -3I3r + I2r = 0

or …(i)

Again, using Kirchhoff’s second law to loop AOBA, we get

-I1r – I1r + (I – I1 – I2)r = 0 or 3I1 + I2 = 1 …(ii)

Considering loop ADCBA, we get

-I2r – (I2 – I3)r – I2r + (I – I1 – I2)r = 0 or Ir – I1r – 4I2r + I3r = 0

or I = I1 + 4I2 – I3 or

Using equation (i), Using equation (ii),

From equation (ii),

or

Considering circuit ABEA, E – (I – I1 – I2)r = 0

or E = (I – I1 – I2)r

or E = 2I1r

If R is the total resistance then E = I r or

(from (iii))

or

26. In normal adjustment, image is formed at least distance of distinct vision, d = 25 cm

Angular magnification of eyepiece

Since the total magnification is 30, magnification of objective lens,

As

or

i.e. object should be held at 1.5 cm in front of objective lens. As v0 = -5u0

From

Separation between the objective lens and eyepiece

= 4.17 + 7.5 = 11.67 cm

OR

f0 = 2cm, fe = 6.25cm, u0 = ?

a. For eyepiece, ve = -25cm, fe = 6.25cm, ue = ?

or

or ue = -5cm

Now, v0 = 15-|ue| = 10cm

For objective, or

or or

Magnifying power,

b. The final image will be formed at infinity only if the image formed by the objective is in the focal plane of the eyepiece.

|v0| = (15 – 6.25)cm = 8.75 cm

=

or Magnifying power, M

27. We know from the considerations of symmetry that M12 = M21. Direct calculation of flux linking the bigger loop due to the field by the smaller loop will be difficult to handle. Instead, let us calculate the flux through the smaller loop due to a current in the bigger loop. The smaller loop is so small in area that one can take the simple formula for field B on the axis of the bigger loop and multiply B by the small area of the loop to calculate flux without much error. Let 1 refer to the bigger loop and 2 the smaller loop. Field B2 at 2 due to I1 in 1 is:

Here x is distance between the centres.

But

Using the given data

OR

According to Lenz’s law, the direction of the induced current (caused by induced emf)

is always such as to oppose the change causing it.

where k is a positive constant. The negative sign expresses Lenz’s law. It means that the induced emf is such that, if the circuit is closed, the induced current opposes the change in flux.

CBSE Class 12 Physics

Sample Paper 08

Time allowed: 3 hours

Maximum Marks: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. Name the absorbing material used to control the reaction rate of neutrons in a nuclear reactor.

2. Name the process of superimposing signal frequency (i.e. audio wave) on the carrier wave.

3. An object is placed at the focus of concave lens. Where will its image be formed?

OR

Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?

4. How does the resolving power of microscope change on

i. decreasing wavelength of light,

ii. decreasing diameter of objective lens.

5. The electric current passing through a wire in the direction from Q to P is decreasing.

What is the direction of induced current in the metallic loop kept above the wire as shown in the figure?

OR

How does the mutual inductance of a pair of coils change when the number of turns in each coil is decreased?

Section-B

6. A long solenoid of 20 turns per cm has a small loop of area 4 cm2 placed inside the solenoid normal to its axis. If the current by the solenoid changes steadily from 4 A to

6A in 0.25, what is the (average) induced emf in the loop while the current is

changing?

OR

The loops in the figure move into or out of the field which is along the inward normal to the plane of the paper. Indicate the direction of currents in loops 1, 2, 3, 4.

7. What does q1 + q2 = 0 signify in electrostatics?

OR

How does the force between two point charges change, if the dielectric constant of the medium in which they are kept decreases?

8. What is the shape of the wavefront in each of the following cases:

a. Light diverging from a point source.

b. Light emerging out of a convex lens when a point source is placed at its focus.

c. The portion of the wavefront of light from a distant star intercepted by the Earth.

9. A ray of light falls an a transparent slab of . If reflected and refracted rays are mutually perpendicular, what is the angle of incidence?

10. Why is ground wave transmission of signals restricted to a frequency of 1500 kHz?

11. Write any two applications of X-rays.

12. Magnetic lines of force are endless?

Section-C

13. In a transformer with transformation ratio 0.1, 220 volt a.c. is fed to primary. What voltage is obtained across the secondary?

14. The near point of a hypermetropic person is 50 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye?

15. A modulating signal is a square wave as shown below:

The carrier wave is given by c (t) = 2 sin volt. i. Sketch the amplitude modulated waveform:

ii. What is the modulation index?

16. The wavelength of the first member of Lyman series is 1216 . Calculate the wavelength of second member of Balmer series.

OR

Suppose you are given a chance to repeat the alpha particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K). What results do you expect?

17. Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration.

a. Show that the equilibrium of the test charge is necessarily unstable.

b. Verify this result far the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

18. What two main observations in photo electricity led Einstein to suggest the photon theory for the interaction of light with the free electron in metal? Obtain an

expression for threshold frequency for photoelectric emission in terms of the work function of the metal.

OR

Sketch the graphs showing the variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies respectively:

i. Which of the two metals, A or B has higher work function?

ii. What information do you get from the slope of the graphs?

iii. What does the value of intercept of graph ‘A’ on the potential axis represent?

19. In a Young’s double slit experiment, the two slits are kept 2 mm apart and the screen is positioned 140 cm away from the plane of the slits. The slits are illuminated with light of wavelength 600 nm. Find the distance of the third bright fringe, from the central maximum, in the interference pattern obtained on the screen.If the wavelength of the incident light were changed to 480 nm, find out the shift in the position of third bright fringe from the central maximum.

OR

Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.

20. The output of an OR gate is connected to both the inputs of a NAND gate. Draw the logic circuit of this combination of gates and write its truth table.

3.0

21. Draw the (i) symbol and (ii) the reverse I-V characteristics of a zener diode. Explain briefly, which property of the characteristics enables us to use Zener diode as voltage regulator.

22. Name the electromagnetic waves used for the following and arrange them in increasing order of their penetrating power.

a. Water purification

b. Rewrite sensing

c. Treatment of cancer

23. Two concentric coplanar semi-circular conductors form part of two current loops as shown in the figure. If their radii are 11 cm and 4 cm, calculate the magnetic induction at the centre.

OR

The area of each plate of parallel plate air capacitor is 150 cm2. The distance between the plates is 0.8 mm. It is charged to a potential difference of 1200 volt. What will be its energy? What will be the energy when it is filled with a medium of k =3 and then charged. If it is charged first as an air capacitor and then filled with this dielectric what will happen to energy?

24. A radioactive isotope X has a half life of 3 seconds. At t = 0 second, a given sample of this isotope X contains 8000 atoms. Calculate (a) its decay constant (b) the time t1, when 1000 atoms of the isotope X remains in the sample and (c) the number of decays per second in the sample at t = t1 second.

Section-D

25. A network of resistors is connected to a 16 V battery with internal resistance of as shown in figure

a. Compute the equivalent resistance of the network. b. Obtain the current in each resistor

c. Also obtain the voltage drops VAB, VBC and VCD.

OR

An infinite ladder network of resistances is constructed with 1 ohm and 2 ohm resistors as shown in figure below. The 6 volt battery between A and B has negligible internal resistance.

i. Show that effective resistance between A and B is 2 ohms

ii. What is the current that passes through the 2 ohm resistance nearest to the battery?

26. A short bar magnet is placed in a horizontal plane with its axis in the magnetic meridian. Null points are found on its equatorial line (i.e. its normal bisector ) at 12.5 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.38 gauss, and the angle of dip is zero.

a. What is the total magnetic field at points on the axis of the magnet located at the same distance (12.5 cm) as the null points from the centre?

b. Locate the null points when the bar is turned around by 180°. Assume that the length of the magnet is negligible compared to the distance of the null points from

the centre of the magnet.

OR

A mono energetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.40 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm. Given mass of electron and charge on electron .

27. Obtain the equivalent capacitance of the network shown in figure below. For a 300 V

supply, determine the charge and voltage across each capacitor.

OR

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

CBSE Class 12 Physics

Sample Paper 08

Answers

1. Heavy water

2. Modulation.

3. Image is formed at infinity.

OR

The apparent depth of oblique viewing decreases from its value for near normal viewing.

4. i. increases, ii. decreases.

5. Clockwise.

OR

It will decrease .

6. The induced emf,

= 4 volt.

OR

In 1, flux decreases and so induced current must be clockwise to increase the flux. Due to the same reason currents in 3 and 4 are clockwise but in 2 current must be anticlockwise as flux is increasing.

7. q1 = -q2. So, the charges q1 and q2 are equal and opposite.

OR

As the dielectric constant of a medium is the ratio of forces between two charges placed in vacuum and medium without no change in magnitude of charges and their distance. The force will increase.

or

or

8. a. Spherical b. Plane

c. Plane (Because a small area on the surface of a large sphere is nearly planar)

9. Using formula,

or

or

10. In ground wave propagation, loss of energy due to interaction with matter increases with the increase in frequency. Frequencies above 1500 kHz get heavily damaged. So ground wave propagation is restricted to a frequency of 1500 kHz.

11. X-rays are used

i. For the detection of explosives, opium and gold in the body of the smugglers. ii. In detecting fractures, diseased organs in human body.

12. This is because magnetic lines of force are continuous closed loops and mono pole is not possible.

13. Transformation ratio

14. Given, u = -25cm, v = -50 cm, f = ?

Using lens formula,

f = 50 cm

Hence, the corrective lens is convex.

15. i.

ii.

16.

For first member of Lyman series, n1 = 1 and n2 = 2.

or

or …(i)

For second member of Balmer series, n1 = 2, n2 = 4

or …. (ii)

Dividing equation (ii) by (i), we get

= 4 or

OR

The nucleus of a hydrogen atom is a proton (mass ) which has only about one-fourth of the mass of an alpha particle . Because the alpha particle is more massive, it won’t bounce back in even a head on collision with a proton. It is like a bowling ball colliding with a ping pong ball at rest. Thus, there would be no large angle scattering in this case. In Rutherford’s experiment, by contrast, there was large angle scattering because a gold nucleus is more massive

than an alpha particle. The analogy there is a ping pong ball hitting a bowling ball at rest.

17. a. It can be proved by contradiction. Assume that the test charge placed at null point be in stable equilibrium. The test charge displaced slightly in any direction will experience a restoring force towards the nullpoint as the stable equilibrium requires restoring force in all directions. That is, all field lines near the null point should be directed inwards towards the null point. This indicates that there is a net inward flux of electric field through a closed surface around the null point. But, according to Gauss law, the flux of electric field through a surface enclosing no charge must be zero. This contradicts our assumption. Therefore, the test charge placed at null point must be necessarily in unstable equilibrium.

b. On the mid point of the line joining the two charges, the null point lies. The test charge will experience a restoring force if it is displaced slightly on either side of the null point along this line. While the net force takes it away from the null point if it is displaced normal to this line. That is no restoring force acts in the normal direction. But restoring force in all directions is demanded by stable equilibrium, therefore, test charge placed at null point will not be in stable equilibrium.

18. The two main observations are:

i. The maximum kinetic energy of emitted photoelectron is independent of intensity of light.

ii. For each photoelectron, there must be a threshold frequency of incident light below which no emission takes place.

For the metal of work function . The kinetic energy of photoelectron emitted due to falling of photon of frequency is

= hv – hv0

where v0 is the threshold frequency. For just emission vmax = 0

hv = hv0 =

or

OR

i. Since work function,

Metal ‘A’ has higher work function as

ii. Slope of

iii. Intercept of graph A on the potential axis

19. Given, d = 2 mm

D = 140 cm = 1.40 m

= 600 nm

Position of bright fringes is given by

Distance of the third bright fringe is

= 1.26 mm

For = 480 nm

Distance of the third bright fringe is

= 1.01 mm

Shift in the position of third bright fringe

= 1.26 – 1.01 = 0.25 mm.

OR

Here, a = 4 mm

Ray optics is good approximation upto distances equal to Fresnel’s distance

20. Logic circuit:

Truth table:

A

B

Y’ = A + B

0

0

0

1

0

1

1

0

1

0

1

0

1

1

1

0

21. i. Symbol of Zener diode:

ii. I-V characteristics:

Zener diode as voltage regulator: For widely different Zener currents, the voltage across the Zener diode remains constant. On account of this fact we can use Zener diode as a d.c. voltage regulator.

For input voltage Vi > Vz’ Zener diode is in the breakdown condition. Thus, for

wide range of values of load (RL), current through the Zener diode may change but the voltage across it remains constant.

22. Following are the electromagnetic waves used for;

a. Water purification – Ultra violet waves b. Rewrite sensing – Micro waves

c. Treatment of cancer – Gamma rays

In increasing order of penetration of given waves can be written as, (b) remote sensing < (a) water purification < (c) treatment of cancer.

23. Magnetic field due to circular loop at centre

Magnetic induction at 0 due 4 quadrant of wire

weber/m2(inward)

OR

Given,

A = 150 cm2

V0 = 1200V, C0 = ? and E0 = ?

With the dielectric medium, the capacitance becomes

C = kC0 When charged to same potential,

V = 1200 V,

Energy

= k(E0)

When capacitor is charged first as air capacitor, then on filling with the dielectric, its potential becomes

= 400 volt

because capacitance becomes 3 times whereas charge remains the same.

New energy of capacitor

24. a.

b.

c.

25. a. Equivalent resistance of two resistors in parallel is i.e. . Equivalent

resistance of and resistors in parallel is i.e. or . Now

and ( equivalent of and ) are in series. So, total resistance is

, i.e. . b. = 2 A

Consider the resistors between A and B. It is a case of two equal resistors connected in parallel. So, current in each resistor is 1A. current through is clearly 2A. Let us now consider resistors between C and D. It is a parallel combination of two resistances. Current would be divided in the inverse ratio of resistances. If I1 is the current through and I2 is the current through , then

. So, current through resistor is . Similarly, current through resistor is .

c. The voltage VAB between A and B is the product of total current between A and B

and the equivalent resistance between A and B.

Similarly = 2 V = 8 V

OR

Let circuit be broken as shown in figure (a) and (b).

a.

b.

Since the circuit is infinitely long, its total resistance remains unaffected by removing one mesh from it. Let the effective resistance of the infinite network be R. The effective resistance of the remaining part of the circuit beyond CD is also R. The circuit can be recombined as shown in figure (b). The resistance R and are in parallel. Their combined resistance is

R’ is in series with remaining resistance. The total combined resistance is: Which must be equal to the total resistance of the infinite network. Therefore

or R2 + 2R = 3R + 2

R2 – R – 2 = 0

Since R can not be negative. Applying Kirchhoff’s second law to the two neighbouring meshes in figure (b) we get

2(I – I’) – 2I’ = 0

From second equation I = 2I’

4I’ = 6 or I’ = 1.5A

26. a. From figure (a) it is clear that null points are obtained on the normal bisector when the magnet’s north and south poles face magnetic north and south respectively. Magnetic field on the normal bisector at a distance r from the centre

is given by , provided r is much greater than the length of the magnet. [The above equation is strictly true only for a point dipole]. At a null point, this field is balanced by the earth’s field.

So, Be = BH

….(i)

(since the dip angle (s) is zero, therefore, Bv = 0 and the horizontal component of the earth’s field equals the field itself)

Next, magnetic field due to a magnet on its axis at point distant r from the centre is given by

….(ii)

provided r is much greater than the length of the magnet.

From figure (a) it is clear that on the axis, this field adds up to the earth’s field. Thus, the total field at a point on the axis has a magnitude equal to Ba + BH

i.e. ….(iii)

and direction along [which is parallel to the earth’s field in case (a)].

Thus, for the same distance on the axis as the distance of the null point, the total field, using equations (i) and (iii) is

i.e.

i.e.

This field is directed along .

b. When the bar is turned around by 180°, the magnet’s north and south poles face

magnetic south and north respectively i.e., in this case, is antiparallel to the earth’s field. From figure (b) it is clear that the null points now lie on the axis of the magnet at a distance r’ given by

or ….(iv)

Comparing equations (iv) and (i), we get or r’3 = 2r3 or r’ = (2)1/3r

For r = 12.5 cm, r’ = 15.7 cm

OR

Here, energy E = 18 keV B = 0.40 G

x = 30 cm = 0.3 m

As

In a magnetic field electron is deflected along a circular arc of radius r, such that

= 11.3 m

y = r – OC

By x sin g binomial

= 4 mm

27. A similar network is drawn below as given in the problem.

Here,

Assume that the series combination of C2and C3is C23

Or

Suppose that the parallel combination of C1 and C23 is C which is given as

C’ = C1 + C23 = 100 + 100 = 200 pF

Let the series combination of C4 and C’ is C which is given as

or

Total charge on all capacitors, q = CV

q1 = C1V1

V2 = V3

V2 + V3 = 100V

2V2 = 100

V2 = 50V

V2 = Vs = 50V

q2 = C2V2

OR

Suppose that two connected conducting spheres of radii a and b possess charges q1and q2respectively. On the surface of the two spheres, the potential will be

Till the potentials of two conductors become equal the flow of charges continue. V1 = V2

or

or

Now, the electric field on the two spheres is given as

or

Therefore, b : a is the ratio of the electric field of the first sphere to that of the second sphere. The surface charge densities of the two spheres are given as

(As the charges are distributed uniformly over the surfaces of conducting

spheres)

Therefore, the surface charge densities are inversely related with the radii of the sphere. The surface charge density on the sharp and pointed ends of a conductor is higher than on its flatter portion since a flat portion may be taken as a spherical surface of large radius and a pointed portion as that of small radius.

CBSE Class 12 Physics

Sample Paper – 09

Time Allowed: 3 hours

Maximum Marks: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. When a magnet falls through a vertical coil, will its acceleration be different from the

‘acceleration due to gravity’?

2. What type of modulation is employed in India for radio transmission?

3. Which three phenomena establish the wave nature of light?

OR

What is the relation of a wavefront with a ray of light?

4. Name the factors which decide the quality of reproduced document sent by Fax.

5. The south pole of a magnet is brought near a conducting loop. What is the direction of induced current as seen by a person on the other side of the loop?

OR

A ring is fixed to the wall of a room. What south pole of a magnet is brought near the rign. What shall be the direction of induced current in the ring?

Section-B

6. If the number of turns of solenoid is doubled, keeping the other factors constant, how does the self induction of the solenoid change?

OR

A jet plane is traveling west at 450 w/s. If the horizontal component of earth’s magnetic field at that place is tesla and the angle of dip is 30°, find the emf induced between the ends of wings having a span of 30 m.

7. What is the electrostatic potential due to an electric dipole at an equatorial point?

OR

How does a free electron at rest move in an electric field?

8. What is the value of refractive index of a medium of polarizing angle 60°?

9. What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5)

10. Write any two advantages of optical fibre.

11. An antenna behaves as resonant circuit only when its length is:

12. Write one condition under which an electric charge does not experience a force in a magnetic field.

Section-C

13. An alternating voltage of frequency f is applied across a series LCR circuit. Let fr be the resonance frequency for the circuit. Will the current in the circuit lag, lead or remain in phase with the applied voltage when (i) f > fr, (ii) f < fr? Explain your answer in each case.

14. The focal length of objective and eyepiece of an astronomical telescope are 25 cm and

2.5 cm respectively. The telescope is focused on an object 1.5 m from objective, the final image is at 25 cm from the eye of the observer. Calculate the length of the telescope.

15. Find the amplitude of the electric field in a parallel beam of light of intensity 7.0

W/m2.

16. Calculate the radius of the first orbit of hydrogen atom. Show that the velocity of electron in the first orbit is times the velocity of light.

OR

Prove that the ionization energy of hydrogen atom is 13.6 eV.

17. a. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is ? The radii of A and B are negligible compared to the distance of separation.

b. What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

18. Mention the significance of Davission and Germer’s experiment. An – particle and a

proton are accelerated from rest through the same potential difference V. Find the ratio of de-Broglie wavelengths associated with them.

OR

Monochromatic light of wavelength 632.8 nm is produced by a helium neon laser. The power emitted is 9.42 mW.

a. Find the energy and momentum of each photon in the light beam.

b. How many photons per second, on the average, arrive at a target irradiated by this beam? (assume the beam to have uniform cross-section which is less than the target area), and

c. How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

19. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up

to the same height, by what distance would the microscope have to be moved to focus on the needle agains?

OR

A convex lens of focal length 15 cm, and a concave mirror of radius of curvature 20 cm are placed coaxially 10 cm apart. An object is placed in front of the convex lens so that there is no parallex between the object and its image formed by the combination. Find the position of the object.

20. A transistor has a current gain of 50. In a CE amplifier circuit, the collector resistance is chosen as 5 kilo ohms and the input resistance is 1 kilo ohm. Calculate the output voltage if input voltage is 0.01 V.

21. Write the truth table for circuit given in the figure below consisting of NOR gates and identify the logic operation (OR, AND, NOT), which this circuit is performing.

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y = 1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one).

22. What is intensity of electromagnetic wave? Give its relation in terms of electric field E

and magnetic field B.

23. Two parallel wires Q and R carry currents 5 A and 10 A respectively in opposite directions. A plane lamina ABCD intersects the wires at right angles at points Q and R. Find the magnitude of the total magnetic induction at point P located in the lamina as shown in the figure.

OR

a. A circular coil of 30 turns and radius 8.0 cm, carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal to the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

b. Would your answer change if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area?

24. In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are (23.98504 u), (24.98584 u) and

(25.98259 u). The natural abundance of is 78.99% by mass. Calculate the abundance of other two isotopes.

Section-D

25. Six resistors, each of value are joined together in a circuit as shown in the figure.

Calculate equivalent resistance across the points A and B. If a cell of emf 2V is connected across AB, compute the current through the arms AB and DF of the circuit.

OR

Three piece of copper wires of lengths in the ratio 2:3:4 and with diameters in the ratio 4:5:6 are connected in parallel. Find the current in each branch if the main current is 5A.

26. If and be the angles of dip observed in two planes at right angles to each other and is the true angle of dip, then prove that

OR

If the bar magnet is turned around by 180°, where will be new null points be located?

27. A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of

. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

a. Determine the capacitance of the capacitor. b. What is the potential of the sphere?

c. Compare the capacitance of this capacitor with that of an isolated sphere of radius

12 cm. Explain why the latter is much smaller.

OR

A capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

CBSE Class 12 Physics

Sample Paper 09

Answers

1. Yes, this is because the motion of the magnet will be opposed in accordance with

Lenz’s law.

2. Amplitude modulation.

3. Interference, diffraction and polarization of light.

OR

Notice that the rays are always perpendicular to the wavefronts. In other words, the wavefront always moves in a direction at right angles to itself.

For example As the waves move farther and farther from the centre, where the rock hit the water, the wavefronts are larger and larger circles. But if you look at a small piece of the wavefront, it nearly looks flat.

4. It depends on the quality of the optical scanning of the original document.

5. Anticlockwise

OR

The induced current is clockwise as seen from the side of the magnet.

6. Coefficient of self induction

when l is made doubled, the coefficient of self induction becomes half.

OR

Here, v = 450 m/s l = 30 m B Induced e.m.f

= 5.4 V

The direction of the wing does not affect this emf.

7. Electric potential at any point in the equatorial plane of dipole is zero.

OR

When the electron is released, it will move in a direction opposite to the direction of electric field.

8.

or

9. Given,

Using formula,

or

= tan-1 (1.5) Thus, ip = 56.3o

10. i. High bandwidth

ii. High data transmission capacity.

11. or integral multiple of .

12. F = qvB sin

When the electric charge is either at rest (v = 0) or parallel ( = 0) to magnetic field, it does not experience force in magnetic field.

13. and

i. When f > fr, XL is large and XC is small. The circuit is inductive. So current lags behind the applied voltage.

ii. When f < fr, XL , is small and XC is large.

The circuit is capacitive. The current leads the voltage in phase.

14. Here, f0 = 25cm, fe = 2.5 cm

u0 = -1.5 m = -150cm, ve = 25cm

From

v0 = 30 cm

From

or

= 32.8 cm

15. The intensity of plane electromagnetic wave is given by:

or

or

= 77.6 NC-1

16. Since, Using n = 1 for 1 st orbit

Z = 1 for hydrogen, coulomb

We get,

Also,

OR

We know that

(n = 1, Z = 1)

Ionization energy is the energy required to remove an electron from ground state to

infinity.

Here, n1 = 1, n2 =

or

= 13.6 eV

17. i.

r = 50 cm = 0.50 m

F = ? Coulomb’s law

ii. Now, if each sphere is charged double, and the distance between them is halved then the force of repulsion is:

F = 0.24 N

18. The Davisson and Germer’s experiment established famour de-Broglie hypothesis of wave particle duality by confirming the wave nature of moving particle.

The energy acquired by alpha-particle,

The energy acquired by proton, Ep = eV

De Broglie wavelength

Hence, the ratio of wavelengths,

or

OR

Given,

Wavelength, Frequency,

a. E = hv

p (momentum)

b. Power emitted, P = 9.42 mW P = nE

c. Velocity of hydrogen atom

= 0.63ms-1

19. Case I: Given, real depth = 12.5 cm apparent depth = 9.4 cm

As,

or

Case II : real depth = 12.5 cm

Apparent depth

Distance through which microscope has to be moved downward.

OR

Here u = ?

As there is no parallex between the object and the image.

v = u

From convex lens formula, we get

u = -30 cm

Now the image will form at Q which is the virtual object for concave mirror. The final image will form at the optical centre. Hence the distance of the object is 30 cm from the mirror.

20. Given, Current gain,

The voltage gain,

21. In the given figure first portion represents NOR gate, second represents NOT gate.

First we calculate output of 1 which acts as input for 2.2 inverts it and we get final output Y.

A

B

A + B

0 0 0 1

0

1

1

0

1

0

1

0

1

1

1

0

A

B

0

0

1

0

0

1

0

1

1

0

0

1

1

1

0

1

Which is truth table for OR gate.

22. Intensity of electromagnetic wave is defined as the energy crossing per second per unit area perpendicular to the direction of propagation of electromagnetic waves. The intensity of electromagnetic wave at a point is;

where

and c is the velocity of electromagnetic wave,

Here E0 and B0 are maximum values of electric field and magnetic field respectively.

23. The magnetic induction at P due to the 10 A current, The magnetic induction at P due to the 5 A current,

Resultant magnetic induction,

where QR2 = PQ2 + PR2 – 2PQ.PR cos

= 0.7469

OR

given, N = 30, I = 6.0 A, B = 1.0 T, = 60°

r = 8.0 cm = Area of the coil,

a. Now,

24.

b. If the area of the loop is the same, the torque will remain unchanged as the torque on the planar loop does not depend upon the shape.

Isotope

Abundance Y

Atomic mass (Z)

12Mg24

78.99

23.98504

12Mg25

x

24.98584

12Mg26

100 – (78.99 + x) = 21.1 – x

25.98259

Mean atomic mass = 24.312

Average atomic mass

or, 2431.2 = 1894.58 + 24.98584x + 545.89 – 25.98254x

or 2431.2 = 2440.47 – 99675x

or 99675x = 2440.47 – 2431.2 = 9.27 or

21.01 – x = 21.01 – 9.30 = 11.71

Relative abundance of Relative abundance of

25. The equivalent circuits are shown as arms ratio of resistances of Wheatstone bridge are equal. So, no current flows (current = 0) in resistor Rg.

Equivalent resistance across AB is given by

or

Current through arm AB

Current through arm DF = 0 (Balanced Wheatstone Bridge).

OR

Let l1, l2, l3 be lengths of three copper wires and D1, D2, D3 be their diameters and A1, A2, A3 be there area of cross section.

Given, l1 : l2 : l3 = 2 : 3 : 4 l1 = 2l, l2 =3l and l3 = 4l

Also given, D1 : D2 : D3 = 4 : 5 : 6

A1 : A2 : A3 = (4)2 : (5)2 : (6)2= 16 : 25 : 36

A1 = 16A, A2 = 25A and A3 = 36A

If is the resistivity of copper, then

and

Or R1 : R2 : R3= 225 : 216 : 200

R1 = 225 R, R2 = 216 R and R3 = 200 R

Let I1, I2 and I3 be the currents through the wires of resistances R1, R2and R3

respectively. Then, I1 + I2 + I3 = 5 …(i)

and

or

and

Putting values in equation (i) we get

I1 + 1.04I1 + 1.125 I1 = 5

On solving I1 = 1.58A

and = 1.78 A

26. If horizontal and vertical components of earth’s magnetic field are represented by BH

and BV respectively, then

Let be the (apparent) dip in a plane which makes angle with the magnetic meridian. In this plane, the vertical component will be BV only but the effective horizontal component will be .

or

or ….(i)

Let be the (apparent) dip in the second plane. The angle made by this plane with the magnetic meridian will be

Effective horizontal component in this plane is i.e. . The

vertical component will be BV only.

Or ….(ii)

Squaring and adding equation (i) and (ii) we get

or

OR

In this case, the neutral point will be on the equatorial line. If r’ is the distance of the

neutral point from the centre of the magnet, then

or

or

27. Given,

k = 32

a. From formula,

b. Potential of inner sphere,

c. Capacitance of sphere

Total potential in case of concentric spheres is distributed over two spheres and the potential difference between the two spheres becomes smaller that is why the capacitance of an isolated sphere is much small than that of concentric spheres. Since the capacitance is inversely proportional to the potential difference

.

OR

When capacitor is charged by 200 V then charge on it is given as

Now it is connected to another uncharged capacitor of capacitance Thus, (4 + 2) =

Until both the capacitor acquire a common potential charge on the first capacitor is shared between them.

After the combination, the common potential = Q/C

V’ = 133 V

Before the combination the electrostatic potential energy of the first capacitor

After the combination electrostatic potential energy of the system

Now, lost electrostatic energy by the first capacitor in the form of heat and electromagnetic radiation

CBSE Class 12 Physics

Sample Paper – 10

TIME ALLOWED: 3 hour

Max Marks: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. Write three factors on which the self inductance of a coil depends.

2. What is meant by the transverse nature of electromagnetic waves?

3. Name the type of modulation scheme preferred for digital communication.

OR

Name one impurity each, which when added to pure Si, produces i. ntype and

ii. ptype semiconductor.

4. Name the device which can represent digital data by analog signals and vice-versa.

5. A vertical metallic pole falls down through the plane of magnetic meridian. Will any emf be induced between its ends?

OR

A coil of number of turns N, area A, is rotated at a constant angular speed , in a uniform magnetic field B, and connected to a resistor R.

Deduce expressions for: (i) maximum emf induced in the coil. (ii) power dissipation in the coil.

Section-B

6. How is the mutual inductance of a pair of coils affected when:

i. separation between the coils is increased?

ii. the number of turns of each coil is increased?

iii. A thin iron sheet is placed between the two coils, other factors remaining the same?

OR

A rectangular loop and a circular loop are moving out of a same magnetic field to a field free region with constant velocity.

It is given that the field is normal to the plane of both the loops, Draw the expected shape of the graphs. Showing the variation of flux with time in both the cases. What is the cause of the difference in the shape of the two graphs?

7. Two charged particles having charge each are joined by an insulating string of length 1 m and the system is kept on a smooth horizontal table. Find the tension in the string.

OR

Does Coulomb’s law of electric force obey Newton’s third law of motion?

8. What is the ratio of slit widths when amplitudes of light waves from them have a ratio

.

9. In the double slit experiment, the pattern on the screen is actually a superposition of single slit diffraction from each slit and the double slit interference pattern in a double slit pattern. In whatway is the diffraction from each slit related to the interference pattern in a double slit experiment? Explain.

10. It is necessary to use satellites for long distance TV transmission. Why?

11. What is space wave propagation?

12. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Section-C

13. An LC circuit contains a 20 mH inductor and a capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed

be t = 0.

a. What is the total energy stored initially? Is it conserved during LC oscillations?

b. What is the natural frequency of the circuit?

c. At what time is the energy stored

i. completely electrical (i.e. stored in the capacitor)?

ii. completely magnetic (i.e. stored in the inductor)?

d. At what times is the total energy shared equally between the inductor and the capacitor?

e. If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

14. 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?

15. Why sky wave propagation of electromagnetic waves cannot be used for TV transmission? Suggest two methods by which range of TV transmission can be increased.

16. In a Rutherford’s scattering experiment with thin gold foil, 8100 scintillations per minute are observed at an angle of 60°. What will be the number of scintillations per minute at an angle of 120°?

OR

A doubly ionized lithium atom is hydrogen like with atomic number 3:

i. Find the wavelength of the radiation required to excite the electron in Li++ from the first to the third Bohr orbit. (Ionization energy of the hydrogen atom equals

13.6 eV)

ii. How many spectral lines are observed in the emission spectrum of the above excited system?

17. Show the electric field due to line charge at any plane is same in magnitude and directed radially upward.

18. Find the maximum velocity of photoelectrons emitted by radiation of frequency 3

1015Hz from a photoelectric surface having a work function 4.0 eV.

OR

The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length scale of 10-15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV)

19. Define refractive index of a transparent medium. A ray of light phases through a triangular prism. Plot a graph showing the variation of the angle of deviation with the angle of incidence.

OR

A myopic person has been using spectacles of power -1.0 dioptre for distant vision. During old age, he also needs to use separate reading glass of power +2.0 dioptres. Explain what may have happened.

20. The black box, shown here, converts the input voltage waveform into the output voltage waveform as is shown in the figure:

Draw the circuit diagram of the circuit present in the black box and give a brief description of its working.

21. The a.c. current gain of a transistor is 120. What is the change in the collector current in the transistor whose base current changes by ?

22. Draw a diagram showing the propagation of an electromagnetic wave along the x direction indicating clearly the directions of the oscillating electric and magnetic fields associated with it.

23. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10

cm section of wire A.

OR

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

24. From the relation R = R0A1/3 where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

Section-D

25. For the network shown in figure below. Calculate the equivalent resistance between points A and B.

OR

In the given Wheatstone bridge, the current 3R is zero. Find the value of R, if carbon resistor, connected in one arm of the bridge has the colour sequence of red, red and orange. The resistance of BC and CD arm are now interchanged and another carbon resistor is connected in place of R so that the current through arm BD is again zero. Write the sequence of colour bands of the carbon resistor. Also find the current through it.

26. A short bar magnet of magnetic moment is placed with its axis perpendicular to earth’s field direction. At what distance from the centre of the magnet, is the resultant field inclined at 45° with earth’s field on (i) its normal

bisector. (ii) its axis? Magnitude of earth’s field at the place 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

OR

Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

27. An electrical technician requires a capacitance of in a circuit across a potential difference of 1kV. A large number of capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

OR

If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion . In the ground state of a , the two protons are separated by roughly

, and the electron is roughly from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

CBSE Class 12 Physics sample paper 10

Answers

1. i. Area of cross-section, ii. Number of turns

iii. Permeability of material of the core.

2. Transverse nature of electromagnetic waves means the electric and magnetic fields in an electromagnetic wave are perpendicular to each other and to the direction of propagation.

3. Pulse code modulation (PCM).

OR

i. As (Arsenic)

ii. In (Indium)

4. Modem.

5. No. When and are parallel, no emf is induced.

OR

The flux linking to the coil,

The induced EMF, (i) The maximum EMF,

(ii) The power dissipated in the coil,

6. i. Mutual inductance (M) decreases because the quantity of flux linking to a coil due to the other one will decrease.

ii. M increases because as the number of turns increase, the overall flux density also increases and hence the mutual inductance will also increase.

iii. M increases because iron is ferromagnetic in nature hence, it will increase the flux density.

OR

7. Here

r = 1 m

Tension in the string is the force of repulsion (F) between the two charges. According to Coulomb’s law.

OR

Yes. Coulomb forces are equal in magnitude, opposite in direction and act on different bodies.

8.

9. The pattern shows a broader diffraction peak in which there appear several fringes of smaller width due to the double slit interference. The number of interference fringes depends upon the ratio of the distance between the two slits to the width of a slit.

10. Television signals are not reflected back by the layer of atmosphere called ionosphere thus TV signals from air earth station are reflected back to the earth by means of an artificial satellite

11. Space wave propagation is that mode of wave propagation in which the radio waves

emitted from the transmitter antenna reach the receiving antenna through space. These radio waves are called space waves. The space waves are the radio waves of frequency range from 54 MHz to 4.2 GHz.

12. Given, I = 0.40 A, r = 8.0 cm

n = 100

13. a. Total initial energy

= 1 J

This energy shall remain conserved in the absence of resistance. b. Angular frequency,

= 103 rads-1

c.

Or , where = 6.3 ms

Energy stored is completely electrical at t = 0, T/2, 3T/2 . . .

Electrical energy is zero i.e. energy stored is completely magnetic at

d. At

Electrical energy , which is half of the total energy.

e. R damps out the LC oscillations eventually. The whole of the initial energy 1.0 J is eventually dissipated as heat.

14. Taller, when the object is in rarer medium and the observer is in denser medium, then the “apparent depth” is greater than “real depth”. In figure AB represents a fisherman standing on the bank of the lake. The rays of light from the head (A) of the fisherman suffer refraction and bend towards the normals. The refracted rays appear to come from A’. The fisherman appear as A’B (>AB) i.e. taller than what he actually is.

15. Sky waves have frequencies of 100 – 200 MHz, which penetrate ionosphere

(frequencies > 30 MHz cannot be used), hence it cannot be used for TV transmission. Two methods by which range of TV transmission can be increased:

(a) use of tall antenna

(b) use of repeaters between transmitters and receivers (line-of-sight transmission).

16. n1 (number of scintillations per minute at an angle 60°) = 8100 n2 (number of scintillations per minute at an angle 120°) = ?

The scattering in the Rutherford’s experiment is proportional to

or

Hence,

= 100

OR

i. The energy difference of electron in Li++ between the first and the third orbit

= E3 – E1

Therefore, the equivalent wavelength is given by or

ii. The following three spectral lines are observed due to the following transitions:

3 rd to 1 st orbit

3 rd to 2 nd orbit

2 nd to 1 st orbit

17. A thing long straight line L of charge having uniform linear charge density is shown by figure. By symmetry, it follows that electric field due to line charge at distance r in any plane is same in magnitude and directed radially upward.

18.

or

OR

Applying formula,

Use the relativistic formula for energy:

or

The second term (rest mass energy) being negligible. Therefore,

or =1.24 BeV

Thus, electron energies from the accelerator must have been of the order of a few

BeV.

19. Ratio of the speed of light in air (vacuum) to the speed of light in the medium is called the refractive index of the medium.

Graph: The graph of angle of deviation versus angle of incidence (i) for a triangular prism is given as follow.

OR

As the person is using spectacles of power -1.0 dioptre (i.e., focal length -100cm), the far point of the person is at 100 cm. Near point of the eye might have been normal (i.e., 25 cm). The objects at infinity produce virtual images at 100 cm (using spectacles). To see objects between 25 cm to 100 cm, the person uses the ability of accommodation of his eye lens. This ability is partially lost in old age. The near point of the eye may recede to 50 cm. He has, therefore, to use glasses of suitable power for reading.

Here u = -25cm, v = -50 cm

Since

or f = 50 cm

As dioptre.

20. Circuit diagram present in black box is drawn in figure.

It is full wave rectifier. For half cycle of the applied input a.c. one junction diode conducts and other does not conduct to work as half wave rectifier. Similarly for another half cycle the other junction diode conducts and the first one does not conduct to work is another half wave rectifier, hence the complete applied a.c. input is rectified which is drawn across the load resistance RL. The input and output waveform diagram is shown in the given figure.

21.

or

22.

23. Given, I1 = 8.0, I2 = 5A

r = 4.0 cm = 4 10-2m l = 10 cm = 0.1 m

F = ?

Force on length l,

OR

Here, l = 3.0 cm = 3 10-2m

I = 10 A, B = 0.27 T, , F = ? By the formula

24. It is found that a nucleus of mass number A has a radius

R = R0A1/3

Where,

This implies that the volume of the nucleus, which is proportional to R3 is proportional to A.

Volume of nucleus Density of nucleus

Above derived equation shows that density of nucleus is constant, independent of A, for all nuclei and density of nuclear matter is approximately which is very large as compared to ordinary matter, say water which is 103 kg m-3 .

25. The distribution of current in the circuit will be as shown in figure below following Kirchhoff’s first law. Here point F is not a true junction, hence shown separate. If R’ is the effective resistance of circuit between A and B then

E = IR’ …… (i)

In a closed circuit EABE E = (I – I1)R + (I – I1)R

= 2(I – I1)R ……. (ii)

In a closed circuit GDCG,

or …… (iii)

In a closed circuit AGCBA, we have

I1R + (I1 – I2)R + I1R – (I – I1)R – (I – I1)R = 0 or 5I1 – I2 = 2I

or [ from (iii)]

or 9I1 = 4I or

Putting this value in (ii) we get

……. (iv) Comparing (i) and (iv) we get

OR

For no current through BD, the Wheatstone bridge is balanced and the resistance of

carbon resistor is

R’ = R

= 2000

when resistance of BC and CD arms are interchanged and another carbon resistor is connected n place or R, the current through BD is again zero. Therefore, again for balanced Wheatstone bridge.

R’ = 4R

For the sequence of colour bands is gray, orange

Now net resistance of arm ADC,

= 88000 + 44000

= 132000

Current through carbon resistor,

26. Here, r = ?

Earth’s field

i. At a point P distant r on normal bisector, Figure (a) field due to the magnet is

Magnetic field B2 due to magnet at equatorial line

along PA || NS

The resultant field will be inclined at 45° to the earth’s field along PQ’, only when

Which gives, r = 0.05 m = 5 cm

ii. When the point P lies on axis of the magnet such that OP = r, field due to magnet fig. (b) is

along PO Earth’s field is along .

The resultant field will be inclined at 45° to earth’s field only when

Which gives

OR

It is along the axis of the solenoid. The direction is determined by the sense of flow of the current. Solenoid acts like a bar magnet.

27. Let possible arrangement requires N capacitors of each is n capacitors in series and m series arrangement in parallel

Total capacitors

As arrangement works on 1000 V.

P.D. across each capacitor in series arrangement is 400 V given

So

n = 2.5

As number of capacitor cannot be in fraction n = 3 equivalent capacitors in each row of series.

as the capacitors are in m rows so resultant capacitance of all capacitors equal to

lines = 2

m = 6 rows

n = = 18

OR

Suppose that the distance between two protons p1 and p2 be r12 and electron (e-) is placed at r13 and r23 distance from protons p1 and p2 respectively.

Given,

r13 = r23 = 1 = 10-10m

(protons)

(electrons)

Now, the potential energy of the system

+

= 19.2eV

CBSE Class 12 Physics

Sample Paper – 11

TIME ALLOWED: 3 hours

MAX MARKS: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. A vertical metallic pole falls down through the plane of magnetic meridian. Will any e.m.f be induced between its ends?

2. Identify the part of the electromagnetic spectrum to which the following wavelengths belong. (i) 1 mm (ii) 10-11m

3. A device X can convert one from of energy into another. Another device Y can be regarded as a combination of a transmitter and a receiver. Name the device X and Y.

OR

Name the type of biasing of a p-n junction diode so that the junction offers very high resistance.

4. What is the main cause of electron’s diffusion from ntype region to ptype region, even when there is no external supply used?

5. Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why?

OR

How does the self-inductance of a coil change when an iron rod is introduced in the coil?

Section-B

6. Show that the energy stored in an inductor i.e. the energy required to build current in the circuit from zero to I is , where L is the self inductance of the circuit.

OR

What is magnetic flux linked with a coil held in a magnetic field zero?

7. An electrostatic field line cannot be discontinuous. Why

OR

Does motion of a body affect its charge?

8. Is the speed of light in glass independent of the colour of light?

9. In a single slit diffraction experiment, the width of the slit is halved. How does it affect the size and intensity of the central maximum?

10. A photodetector is made from a semiconductor having Eg = 0.73eV. What is the maximum wavelength which it can detect?

11. A TV tower has a height of 100 m, how much population will be covered by the TV

broadcast if the average population density around the tower is 10000 per km2.

12. What is the direction of the force acting on a charged particle q, moving with a velocity in a uniform magnetic field ?

Section-C

13. Why is choke coil needed in the use of fluorescent tubes with ac mains?

14. A 5 cm long needle is placed 10 cm from a convex mirror of focal length 40 cm. Find the position, nature and size of image of the needle. What happens to the size of the image when the needle is moved further away from the mirror?

15. An electromagnetic wave travels in vacuum along z-direction.

1. What can you say about the directions of its electric and magnetic field vectors?

2. If the frequency of the wave is 30 MHz, what is its wavelength?

16. The energy of the electron, the hydrogen atom, is known to be expressible in the form

(n = 1, 2, 3, . . . )

Use this expression to show that the

i. Electron in the hydrogen atom can not have an energy of -2V.

ii. Spacing between the lines (consecutive energy levels) within the given set of the observed hydrogen atom spectrum decreases as n increases.

OR

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature.

What series of wavelengths will be emitted?

17. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consists of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3) e, and the ‘down’ quark (denoted by d) of

charge , together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter). Suggest a possible quark composition of a proton and a neutron.

18. An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (-10-2mm of Hg). A magnetic field of curves the path of the electrons in a circular orbit of radius 12.0 cm (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture this method is known as

the ‘fine beam tube’ method.) Determine e/m from the data.

OR

A nucleus of mass M initially at rest splits into two fragments of masses m/3 and 2m/3. Find the ratio of de-Broglie wavelength of the two fragments.

19. Two thin lenses, both of 10 cm focal length one convex and other concave, are placed

5 cm apart. An object is placed 20 cm in front of the convex lens. Find the nature and position of the final image.

OR

A prism of refractive index of has a refracting angle of 60°. At what angle must a ray be incident on it so that it suffers a minimum deviation?

20. Draw a circuit diagram of a common emitter amplifier using n-p-n transistor. Show the input and output voltage graphically. The current gain for common emitter amplifier is 59. If the emitter circuit is 6.0 mA, find (i) base current and (ii) collector current.

21. The characteristics curve of a diode is shown in the above figure. Determine the d.c.

and a.c. resistance around point.

22. A plane electromagnetic wave of angular frequency is propagating with velocity C along the Z-axis. Write to vector equations of oscillating electric and magnetic fields and show those fields diagrametically.

23. A uniform magnetic field of 3000 G is established along the positive zdirection. A rectangular loop of sides 10 cm and 5 cm carries a current 12 A. What is the torque on the loop in the different cases shown in the figure below. What is the force on each case? Which case corresponds to stable equilibrium?

OR

The wire shown in figure below carries a current of 60 A. Find the magnetic field B at

P.

24. Obtain approximately the ratio of the nuclear radii of the gold isotope and the silver isotope .

Section-D

25. a. Find the emf E1 and E2 in the circuit of the following diagram and the potential difference between the points a and b.

b. If in the circuit, the polarity of the battery E1, be reversed, what will be the potential difference between a and b?

OR

State the working principle of a potentiometer with the help of a circuit diagram. Describe a method to find the internal resistance of a primary cell. In a potentiometer arrangement, a cell of emf 1.20 volt gives a balance point at 30 cm length of the wire. This cell is now replaced by another cell of unknown emf. If the ratio of the length of the two cells is 1.5. Calculate the difference in the balancing length of the potentiometer wire in the two cases.

26. The magnetic moment of a short bar magnet is 1.6 m2. It is placed in the magnetic meridian with north pole pointing south. The neutral point is obtained at 20 cm from the centre of the magnet. Calculate the horizontal component H of earth’s field. If magnet be reversed i.e. north pole pointing north, find the position of the neutral point.

OR

A tangent galvanometer is set with its plane making an angle of 30° with the magnetic meridian. When a current is passed through 2 turns coil of radius 0.10 m, in the anti clockwise direction, the compass needle shows a deflection of 30°. Calculate the strength of current passed. Take H .

27. Two tiny spheres carrying charges and are located 30 cm apart. Find the potential and electric field:(a) at the mid point of the line joining the two charges, and(b) at a point 10 cm from this mid point in a plane normal to the line and passing through the mid point.

OR

Define the term electric potential due to a point charge. Calculate the electric potential at the centre of a square of side , having charge and

at the four corners of the square.

CBSE Class 12 Physics

Sample Paper 11

Answers

1. No, because the pole intercepts neither H nor V.

2.

Wavelength

Band of EM Spectrum

1 mm

Microwave

10-11m

Gamma Rays

3. X is transducer

Y is repeater

OR

Reverse biasing.

4. Because of difference in free electron density and mobility between n-type and p-type region.

5. Because the magnetic field induction outside the toroid is zero.

OR

The self-inductance shall increase, since .

6. Energy spent by the source to increase current from i to i + di in time dt in an inductor.

= Li di

Energy required to increase current from 0 to I

OR

When plane of coil is along the direction of field, the magnetic flux linked with a coil is zero.

7. An electrostatic field line cannot be a discontinuous curve i.e., it cannot have breaks.

If it is so, it will indicate the absence of electric field at the break points. But the electric field vanishes only at infinity.

OR

No, charge on a body does not change with motion of the body.

8. No, the speed of light is not independent of the colour (wavelength) of the light. The violet colour travels slower than the red light in a glass prism. The refractive index and hence the speed of light in a medium depend on the wavelength.

9. Since, size of central maximum

Thus, size becomes double and intensity reduces to

10. Since,

= 1703 nm

11.

= 3.6 104m

Area

= 4.1 1013m2

The population

= 4.1 107 km2 10000 per km2

= 4.1 1011

12.

Magnetic force is always normal to plane of and .

13. A choke coil is needed in the use of fluorescent tubes to reduce a.c. without loss of power, it we use an ordinary resistor, a.c. will reduce, but if loss of power due to heating will be there.

Power dissipated In a resistor

Power dissipated

= EvIv = max

In a choke coil,

Power dissipated = EvIv cos 90o = zero.

14. For mirror,

or Magnification,

When needle moves further away from the convex mirror, the image moves further behind the mirror towards the focus and size decreases. When it is far off, it appears almost as a point image of the object.

15. and lie in x-y plane and are mutually perpendicular, Since;

Wavelength of the wave is 10m

16. As,

Putting n = 1, 2, 3, . . ., n we get

i. Hence, it can be observed that the electron in the hydrogen atom can not have an energy of -2V.

ii. As n increases, energies of the excited states come closer and closer together.

Therefore, as n increases, En becomes less negative until at i.e. En = 0.

OR

In ground state, energy of gaseous hydrogen at room temperature . When it is bombarded with 12.5 eV electro beam, the energy becomes

So

The electron would jump from n = 1 to n = 3, where

.

On de-excitation the electron may jump from n = 3 to n = 2 giving rise to Balmer series. It may also jump from n = 3 to n = 1, giving rise to Lyman series.

So, number of spectral line

= 3 spectral lines appear.

17. Charge on ‘up’ quark Charge on ‘down’ quark Charge on a proton = e Charge on a neutron = 0

Let a proton contains x ‘up’ quarks and (3 – x) ‘down’ quarks. Then total charge on a proton is,

ux + d(3 – x) = e or,

or,

or, x = 2

and 3 – x = 3 – 2 = 1

i.e. proton contains 2 ‘up’ quarks and 1 ‘down’ quark. Its quarks composition should be uud.

Let a neutron contains y ‘up’ quarks and (3 – y) ‘down’ quarks. Then total charge on a neutron is:

ny + d(3 – y) = 0 or,

or,

or, y = 1

and 3 – y = 3 – 1 = 2

i.e. neutrons contain 1 ‘up’ quark and 2 ‘down’ quarks. Its quark composition should be udd.

18. Here, V = 100 V Magnetic field, r = 12.0 cm

When electrons are accelerated through V volt, the gain in K.E. of the electron is given by

….(i)

Since the electron moves in circular orbit under magnetic field, therefore, force on the electron due to magnetic field provides the centripetal force to the electron.

or

or, …(ii)

From equations (i) and (ii), we get

or

= 1.73 1011Ckg-1

OR

Let and

19.

But

Following the low of conservation of linear momentum, m1v1 + m2v2 = 0

Also,

Hence,

For refraction at the convex lens, we have u = -20cm; f1 = 10cm; v = v1 =?

Using lens formula, we have

The convex lens produces converging rays trying to meet at I1, 20cm from the convex lens, i.e. 15 cm behind the concave lens.

I1 will serve as a vertical object for the concave lens. For refraction at the concave lens, we have

For concave lens, u = 20 – 5 = 15 cm

f = -10cm

As per sign convention u = -15

f = -10

v = -5cm

i.e. This image is in the side of object 5 cm Right to concave and 10 cm (5 + 5) from convex

u = + 15 cm v = ? , f = -10cm

Using lens formula, we have

v = -30cm

Hence, the final image is virtual and is located at 30 cm to the left of the concave lens.

OR

For minimum deviation i = e. Therefore,

or for minimum i = r

Now,

or

sin 30°

Hence,

20. For circuit diagram,

Numerical: Given, Since,

But

or = 5.9 mA IB = IE – IC = 6.0 – 5.9 = 0.1 mA

21. d.c. resistance a.c. resistance

22. Given: Angular frequency = Velocity = C

As per vector deduction,

vector in X axis direction is

vector in Y axis direction is Where

23. a. Torque on the loop,

Where is the angle between the plane of loop and direction of magnetic field. Here, B = 3000 gauss

= 0.3 T

I = 12 A, A

The direction of torque or force on arm 5 cm, lower arm +x axis upper arm

axis by Fleming’s left hand rule.

b. Similar to (a) but torque act on side of 10 cm.

c. along -x direction of torque on lower arm of 5 cm towards -y axis.

d. This case is similar to (c). Direction of torque is 60°.

e. Zero ( angle between plane of loop and direction of magnetic field is 90°)

f. zero.

Force is zero in each case. Stable equilibrium is corresponded by case (e).

OR

The field at P arises from 3/4 th of the circular loop only because P lies on the straight wires themselves. Thus

24. As,

= 0.08827

= antilog (0.08827)

= 1.23

25. a. It is clear that 1 A current flows in the circuit from B to A.

Applying Kirchhoff’s law to the loop PAQBP,

Hence, E2 = 2V

Thus the potential difference between the points a and b is:

b. On reversing the polarity of the battery E1, the current distributions will be changed. Let the currents be I1 and I2 as shown in the following figure.

Applying Kirchhoff’s law for the loop PABP,

20 + E1 = (6 + 1)I1 – (4 + 1)I2 or 38 = 7I1 – 5I2 …… (i) Similarly for the loop ABQA,

4I2 + I2 + 18 + 2(I1 + I2) + (I1 + I2) + 7 = 0

Or 3I1 + 8I2 = – 25 …… (ii)

Solving equation (i) and (ii) for I1 and I2 we get

and

Hence,

= -20.35 + 18

= -2.35V

OR

The connections are made as shown in the figure. E is the cell whose internal resistance r is to be determined. A resistance R is connected across the cell through a key k. The key k’ is closed and k is kept open. The balance point is found out. Let the balancing length be 11. Then,

…(i)

A suitable resistance R is introduced in the resistance box R and with the key k’ closed, the balancing length l2 is found out. When the circuit is closed, the potential

difference across the cell falls to . Then,

….(ii)

Dividing equation (i) by (ii) we get

or or Given,

or = 20 cm

Difference in the balancing length,

11 – 12 = 30 – 20 = 10 cm

26. Here, M = 1.6 Am2

, H = ?

When north pole of magnet is pointing south, neutral points lie on axial line of magnet. At neutral point.

or

When magnet is reversed i.e. north pole is pointing north, neutral points lie on equatorial line of the magnet. We have to calculate .

As

= 4000 cm3

d2 = (4000)1/3= 15.87 cm

OR

Before passing current, compass needle was along H. On passing current, the needle shows a deflection of 30° (with H) under the combined effect of H and B, magnetic field due to current (in a direction perpendicular to the plane of the coil).

As compass needle is inclined equally to H and B, figure therefore. B = H

or I = 2.54 A

27. a. Potential at the mid point of the line joining the two charges is

Or

Electric field at the mid point O due to charge at A

along OB

Electric field at the mid point O due to charge at B

along OA

Thus, the total electric field at the mid point O is

(along BA)

b. Potential at the point C due to the two charges is

Electric field at C due to charge at A

Electric field at C due to charge at B

If the angle between E1 and E2 be then

Thus, magnitude of resultant field at C is

Let the field E makes angle with the field E1

Now,

= 4.2876

= 76.9°

If field E makes angle with the direction BA, then

= 90° + 56.3° – 76.9° = 69.4°

Therefore, angle of 69.4° is made by the electric field with the line joining the two charges q2 to q1.

OR

The amount of work required to carry a unit positive charge with zero acceleration from infinity to a point in electric field against the electrostatic forces is termed as electric potential at that point.

Diagonal of the square

r = Half diagonal = 1 m

The potential at the centre of the square,

CBSE Class 12 Physics

Sampler Paper – 12

TIME ALLOWED: 3 hours

MAX MARKS: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. A train is moving with uniform velocity from north to south. Will any induced emf appear across the ends of the axle?

2. The charging current for a capacitor is 0.25 A. What is the displacement current across its plates?

3. Identify the parts X and Y in the following block diagram of a generalized communication system:

OR

Give the logic symbol of NAND gate.

4. In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by m towards the slits, the change in fringe width is . If the distance between the slits is

10-3m. calculate the wavelength of light used.

5. The number of turns in secondary coil of a transformer is 100 times the number of turns in the primary coil. What is the transformation ratio?

OR

What are the values of first and second excitation potential of hydrogen atom?

Section-B

6. A coil of inductance 2 mH carrying a current 2A is given. If the current is reversed in

0.01 seconds, how much back emf is produced?

OR

When current in a coil changes with time, how is the back emf induced in the coil related to it?

7. Dielectric constant of a medium is unity. What will be its permittivity?

OR

What kind of charges are produced on each when: (i) a glass rod is rubbed with silk and

(ii) an ebonite rod is rubbed with wool?

8. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.

9. Monochromatic light from a narrow slit illuminates two narrow slits 0.3 mm apart, producing an interference pattern with bright fringes 1.5 mm apart on a screen 75 cm away. Find the wavelength of the light. How will the fringe width be altered if (a) the distance of the screen is doubled and (b) the separation between the slits is doubled?

10. What should be the length of the dipole antenna for a carrier wave of frequency

?

11. Communication in UHF / VHF regions can be established by space or tropospheric wave. Why is it restricted to or limited to line-of-sight distance of about 40 km?

12. What is the approximate distance upto which earth’s magnetic field extends?

Section-C

13. A capacitor in series with a resistance is connected to a 110 V, 60 Hz supply.

a. What is the maximum current in the circuit?

b. What is the time lag between current maximum and voltage maximum?

14. The least distance of distinct vision of a long sighted man is 50 cm. He reduces this distance to 10 cm by wearing spectacles. What is the power and type of lenses used by him?

15. Calculate the de-Broglie wavelength of a beam of electrons, accelerated through a potential difference of 10 kV.

16. A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s-1.

(a) What is the rms value of the conduction current?

(b) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.

17. Obtain an expression for the frequency of radiations emitted when a hydrogen atom de-excites from level n to level (n – 1). For large n, show that the frequency equals the classical frequency of revolution of the electron in the orbit.

OR

In a hydrogen like atom the ionization energy equals 4 times Rydberg’s constant for hydrogen. What is the wavelength of radiations emitted when a jump takes place

from the first excited state to the ground state? What is the radius of first Bohr’s orbit?

18. A point charge is a distance 5 cm directly above the centre of a square of side

10 cm, as shown in figure. What is the magnitude of the electric flux through the square?(Hint : Think of the square as one face of a cube with edge 10 cm

19. From the output characteristics shown in figure calculate the values of current amplification factor of the transistor when VCE is 2V.

OR

Explain.Given (b) Obtain the de-Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

20. When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

OR

a. At what distance should the lens be held in order to view the squares distinctly with the maximum possible magnifying power?

b. What is the magnification in this case?

c. Is the magnification equal to the magnifying power in this case? Explain.

21. Justify the output wave form (Y) of the OR gate for inputs (A) and (B) as given in the following figure:

22. Identify the gate shown in the figure. Explain with the help of a circuit diagram, how this gate is realized in practice.

23. Find the amount of work done in rotating an electric dipole of dipole moment

from its position of stable equilibrium to the position unstable equilibrium in a uniform electric field of intensities 104NC-1 .

OR

A straight horizontal conducting rod of length 0.60 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wire.

a. What magnetic field should be set up normal to the conductor in order that the tension in the wire is zero?

b. What will be total tension in the wires if the direction of current is reversed, keeping the magnetic field same as before g = 10 ms-1

24. Would the energy be released or needed for the following D-T reaction

to occur?Given: ,

= 4.002603 u, = 1.008665 u. Calculate this

energy in MeV.

25. A battery if 24 cells each of emf 1.5 V and internal resistance , is to be connected in order to send the maximum current through a resistor. How are they to be connected? Find the current in each cell and the potential difference across the external resistance.

OR

A homogeneous poorly conducting medium of resistivity fills up the space between two thin coaxial ideally conducting cylinders. The radii of the cylinders are equal to a and b with a < b, the length of each cylinder is l. Neglecting the edge effects, find the resistance of the medium between the cylinders.

26. If and be the angles of dip observed in two planes at right angles to each other and is the true angle of dip, then prove that

OR

A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius

20 cm. The centre of the small loop is on the axis of the bigger loop. The distance

between their centres is 15 cm.

(a) What is the flux linking the bigger loop if a current of 2.0 A flows through the smaller loop?

(b) Obtain the mutual inductance of the two loops.

27. Electric charge is distributed uniformly on the surface of a spherical rubber balloon.

Show how the value of electric intensity and potential vary

i. on the surface, ii. inside and

iii. outside?

OR

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

CBSE Class 12 Physics

Sample Paper 12

Answers

1. Yes. The vertical component of earth’s magnetic field shall induce emf.

2. Displacement current is 0.25 A. Since displacement current is equal to conduction current.

3. X Source of information

Y Communication channel.

OR

4.

or

5. Transformation ratio

Since Thus

OR

This is the energy required in ev to excite a ground state atom to its excited state. When electrons jumps from ground state level(n=1) to another another energy level(n=2) the corresponding energy is called 1st excitation potential energy and corresponding potential is called 1st excitation potential.

1st excitation energy = 3.4ev-(13.6ev)=10.2ev

1st excitation potential=10.2v

Similarly when electron jumps from n=1 to the another energy level n=3 the corresponding energy is called 2nd excitation potential energy and corresponding potential is called 2nd excitation potential.

2nd excitation energy =

2nd excitation potential = 12.1v

6. The current reverses in 0.01 s, this means the time period is 0.01/2 = 0.005 s

Induced EMF,

= 0.8 V

OR

The back emf,

where, I is the current and L is the self inductance.

7. We know that dielectric constant of a medium is

OR

(i) Positive charge will be produced on glass rod and negative charge will be produced on silk.

(ii) Negative charge will be produced on ebonite rod and positive charge will be produced on wool.

8. Given, D = 1 m, n = 1

Using formula,

or = 0.2 mm

9. Here, d = 0.3 mm = 0.3 10-3m

= 1.5 mm = 1.5 10-3m D = 7.5 cm = 0.75 m Wavelength is given by

a.

When D is doubled, fringe width is also doubled.

= 1.5 2= 3.0 mm

b. when d is doubled, is reduced to half i.e.

10. Wavelength,

The size of the antenna should be of the wavelength, i.e. .

11. Considering a range of 40 km, the height of the antenna must be: = 125 m

It is impractical to use an antenna of length greater than this.

12. The magnetic field of earth extends to nearly five times the radius of the earth i.e.

13. Here,

=?

v = 60 Hz, I0

In RC circuit as

In RC circuit, voltage lags behind the current by phase angle .

where

Time lag

14. For such a person, u = -10cm and v = -50cm

Using formula,

= + 12.5 cm

The type of lenses used is convex.

15. The de-Broglie wavelength of electron

(Since k=eV and K = )

Putting values of h, m and e we get

Given V = 10 kV = 10 x 103 V = 104 V

0.01227 x 10-9 m = 0.1227 x 10-10 m

16. (a) Here, a = 6.0 cm

C = 100 pF = Erms = 230V

(b) since,I = ID whether I is steady d.c. or a.c. This is shown below: Or

We know that

This formula goes through even if ID (and therefore B) oscillates in time. The formula shows that they oscillate in phase. Since ID = I , we have

If I = I0 , the maximum value of current, then amplitude of B = maximum value of B

17. The frequency of the emitted radiation when a hydrogen atom de-excites from level n to level (n – 1)is

E = hv = E2 – E1

where find structure constant

For large n, (2n – 1) = 2n, and (n – 1) = n

Putting we get

In Bohr’s atom model velocity of electron in nth orbit is and radius of nth orbit is

Frequency of revolution of electron

which is the same as (i)

Hence for large values of n, classical frequency of revolution of electron in nth orbit is the same as the frequency of radiation emitted when hydrogen atom de-excites from level (n) to level (n – 1).

OR

The ionization energy E of a hydrogen like Bohr atom of atomic number z is given by where the Rydberg constant

As ionization energy

E = Rydberg constant

= 4 R, we have

or Z = 2

i. Energy of radiation emitted E when the electron jumps from the first excited state to the ground state is given by

= 3 R

or

Wavelength of the radiation emitted,

ii. Radius of the first Bohr orbit

18. Let us assume that the charge is placed at a distance of 5 cm from the square ABCD of each side 10 cm. The square ABCD can be considered as one of the six faces of a cubic Gaussian surface of each side 10 cm.

Now the total electric flux through the faces of the cube as per Gaussian theorem

Therefore, the total electric flux through the square ABCD will be

19. As,

Consider characteristics for any two values of IB (say 10 and ). Then for VCE = 2V

from the graph, we have

Therefore, .

OR

(b) Here, T = 27 + 273 = 300 K

Boltzmann’s constant, k

We know, average K.E. of neutron at absolute temperature T is given by .

Where k is the Boltzmann’s constant.

Now,

Since this wavelength is comparable to interatomic spacing in a crystal, therefore, thermal neutrons are suitable probe for diffraction experiments: so a high energy neutron beam should be first thermalised before using it for diffraction.

20. The image of the objective lens in the eyepiece is known as the ‘eyering’. All the rays from the object refracted by the objective go through the eye ring. Therefore, it is an ideal position for our eyes for viewing. If we place our eyes too close to the eyepiece, we shall not collect much of the light and also reduce our field of view. If we position our eyes on the eye ring and the area of the pupil of our eye is greater or equal to the area of the eyering, our eyes will collect all the light refracted by the objective. The precise location of the eye ring naturally depends on the separation between the objective and the eyepiece and the focal length of the eyepiece. When we view through a microscope by placing our eyes on one end, the ideal distance between the eye and the eyepiece is usually built in the design of the instrument.

OR

a. Maximum magnifying power is obtained when the image is at the near point (25 cm). Thus,

v = -25cm, f = + 10 cm, u = ? As

or

So the lens should be held 7.14 cm away from the figure. b. Magnitude of magnification is

c. Magnifying power,

21.

Yes, the magnifying power is equal to the magnitude of magnification because the image is formed at the least distance of distinct vision.

Time interval

Input A

Input B

Output Y = A+ B

t < t1

0

0

0

t1 < t < t2

1

0

1

t2 < t < t3

1

1

1

t3 < t < t4

0

1

1

t4 < t < t5

0

0

0

t5 < t < t6

1

0

1

t > t6

0

1

1

22. OR gate

i. When A = 0, and B = 0, both diodes are reverse biased, hence Y = 0. ii. When A = 0, and B = 1, diode D2is forward biased, hence Y = 1.

iii. When A = 1, and B = 0, diode D1is forward biased, hence Y = 1.

iv. When A = 1, and B = 1, diode D1 and D2 both are forward biased, hence Y = 1.

23. Work done in rotating the dipole from an angle to of dipole moment with

electric field,

For stable equilibrium position

and for unstable equilibrium position

= 60 J

OR

Given I = 0.60 m, m = 60 g I = 5.0 A

a. Tension in the wire is zero if the force on the wire carrying current due to magnetic field is equal and opposite to the weight of wire , i.e.

B I l = mg

or

The force on the conductor due to magnetic field will be upwards if the direction of magnetic field is horizontal and normal to the conductor.

b. When direction of current is reversed, BIl and mg will act vertically downwards, the effective tension in the wires,

T = BIl + mg

= 1.2 N

24. Mass = 2.014102 + 3.016049

= 5.030151 u

= 5.011268 u

= 4.002603 + 1.008665

Energy is released in this reaction. ( as mass of reactant is larger than products) Mass defect ( m) = 5.030151 – 5.011268

= 0.018863 u

Now, energy required

= 17.561453 MeV

= 17.56 MeV

25. Let x be the number of cells in series in each row and let there be y such rows in parallel.

Total number of cells = xy = 24

Resistance of each row in series = 2 x ohms

Total internal resistance due to all xy batteries = R

times

Total internal resistance ohms (because there are y rows in parallel)

The maximum current passes through the circuit when the internal resistance of the battery of cells equal the external resistance.

Thus, Or

But xy = 24

Hence x = 12 and y = 2

i.e. there should be two rows of 12 cell in series

The current in the circuit is

= 0.75 A

Because of two rows have the same resistance, the current in each arm must be

= 0.375 A

Therefore, current through each cell = 0.375 A

The potential difference across the external resistance is

= 9 V

OR

The current will be conducted radially outwards from the inner conductor (say) to the

outer. The area of cross section for the conduction of the current is, therefore, the area of an elementary cylindrical shell and which varies with radius. The length of

the conducting shell is measured radially from radius a to radius b.

Consider an elementary cylindrical shell of radius r and thickness dr. Its area of cross section (normal to flow of current) and its length = dr.

Hence, the resistance of the elementary cylindrical shell of the medium is

The resistance of the medium is obtained by integrating for r from a to b. Hence required resistance

26. If horizontal and vertical components of earth’s magnetic field are represented by BH

and BV respectively, then

Let be the (apparent) dip in a plane which makes angle with the magnetic meridian. In this plane, the vertical component will be BV only but the effective horizontal component will be .

or

or ….(i)

Let be the (apparent) dip in the second plane. The angle made by this plane with the magnetic meridian will be

Effective horizontal component in this plane is i.e. . The

vertical component will be BV only.

Or ….(ii)

Squaring and adding equation (i) and (ii) we get

or

OR

We know from the considerations of symmetry that M12 = M21. Direct calculation of flux linking the bigger loop due to the field by the smaller loop will be difficult to handle. Instead, let us calculate the flux through the smaller loop due to a current in the bigger loop. The smaller loop is so small in area that one can take the simple formula for field B on the axis of the bigger loop and multiply B by the small area of

the loop to calculate flux without much error. Let 1 refer to the bigger loop and 2 the smaller loop. Field B2 at 2 due to I1 in 1 is:

Here x is distance between the centres.

But

Using the given data

27. On the surface E = constant, V = constant. Inside the surface, E = 0, V = constant =

potential on surface. Outside the balloon.

and

where r is the distance of the point from the centre of the balloon.

OR

Suppose that two connected conducting spheres of radii a and b possess charges q1and q2respectively. On the surface of the two spheres, the potential will be

Till the potentials of two conductors become equal the flow of charges continue. V1 = V2

or

or

Now, the electric field on the two spheres is given as

or

Therefore, b : a is the ratio of the electric field of the first sphere to that of the second sphere. The surface charge densities of the two spheres are given as

(As the charges are distributed uniformly over the surfaces of conducting

spheres)

Therefore, the surface charge densities are inversely related with the radii of the sphere. The surface charge density on the sharp and pointed ends of a conductor is higher than on its flatter portion since a flat portion may be taken as a spherical surface of large radius and a pointed portion as that of small radius.

CBSE Class 12 Physics

Sample Paper – 13

TIME ALLOWED: 3 hours

MAX MARKS: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. Name the type of modulation scheme preferred for digital communication.

2. Name the EM waves used for studying crystal structure of solids. What is its frequency range?

3. Name the experiment for which the following graph, showing the variation of intensity of scattered electrons with the angle of scattering, was obtained. Also name the important hypothesis that was confirmed by this experiment.

OR

Which three phenomena establish the wave nature of light?

4. Name the process of superimposing signal frequency (i.e. audio wave) on the carrier wave.

5. Is a motor starter a variable R or L or?

OR

What is the phase difference between voltage across an inductor and a capacitor in an a.c. circuit?

Section-B

6. A conducting U tube can slide inside another U-tube maintaining electrical contact between the tubes. The magnetic field is perpendicular to the plane of paper and is directed inward. Each tube moves towards the other at constant speed v. Find the magnitude of induced emf across the ends of the tube in terms of magnetic field B, velocity v and width of the tube?

OR

Give the direction in which the induced current flows in the coil mounted on an insulating stand when a bar magnet is quickly moved along the axis of the coil from one side to the other as shown in the figure.

7. How does a torque affect the dipole in an electric field?

OR

Two small balls having equal positive charge q coulomb are suspended by two insulating strings of equal length l metre from a hook fixed to a stand. The whole set up is taken in a satellite into space where there is no gravity. What is the angle between the two strings and the tension in each string?

8. When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?

9. How would the angular separation of interference fringes in Young’s double slit experiment change when the distance of separation between slits and screen is doubled?

10. A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of

75%?

11. A TV tower has a height of 150 m. By how much the height of tower be increased to double its coverage range?

12. Define magnetic flux. Give its SI units.

Section-C

13. Suppose the circuit has a resistance of . Obtain the average power transferred to each element of the circuit, and the total power absorbed.

14. An object 0.5 cm high is placed 30 cm from a convex mirror whose focal length is 20 cm. Find the position, size and nature of the image.

15. A plane electromagnetic wave of angular frequency is propagating with velocity C along the Z-axis. Write to vector equations of oscillating electric and magnetic fields and show those fields diagrametically.

16. Calculate the radius of the first orbit of hydrogen atom. Show that the velocity of electron in the first orbit is times the velocity of light.

OR

The ground state energy of hydrogen atom is -13.6 eV . What are the kinetic and potential energies of the electron in this state?

17. A thin conducting spherical shell of radius R has charge +q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell. Draw a graph of electric field E(r) with distance r from the centre of the shell for .

18. The wavelength of light in the visible region is about 390 nm for violet colour, about

550 nm (average wavelength) for yellow-green colour and about 760 nm for red colour.What are the energies of photons in (eV) at the

i. violet and

ii. average wavelength, yellow-green colour, and

iii. red end of the visible spectrum? (Take h and 1eV )

OR

Define the terms: (i) work function, (ii) threshold frequency and (iii) stopping

potential, with reference to photoelectric effect.Calculate the maximum kinetic energy of electrons emitted from a photosensitive surface of work function 3.2 eV, for the incident radiation of wavelength 300 nm.

19. Use the mirror equation to deduce that:

a. an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

b. a convex mirror always produces a virtual image independent of the location of the object.

c. the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

d. an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

OR

A convex mirror always produces a virtual image independent of the location of the object. Use mirror equation to prove it.

20. The output of an OR gate is connected to both the inputs of a NAND gate. Draw the logic circuit of this combination of gates and write its truth table.

21. 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.

22. Find the amplitude of the electric field in a parallel beam of light of intensity 7.0

W/m2.

23. A charge 2Q is spread uniformly over an insulated ring of radius R/2. What is the magnetic moment of the ring if it is rotated with an angular velocity with respect to normal axis?

OR

Obtain an expression for the capacitance of a parallel plate (air) capacitor. The given figure shows a network of five capacitors connected to a 100 V supply. Calculate the

total charge and energy stored in the network.

24. Draw the graph to show variation of binding energy per nucleon with mass number of different atomic nuclei. Calculate binding energy/nucleon of nucleus.Given: Mass of = 39.962589 u Mass of proton = 1.007825 uMass of neutron = 1.008665 u And 1 u = 931MeV/c2

Section-D

25. Deduce the condition for balance in a Wheatstone bridge. Using the principle of Wheatstone bridge, describe the method to determine the specific resistance of a wire in the laboratory. Draw the circuit diagram and write the formula used. Write any

two important precautions you would observe while performing the experiment.

OR

The figure shows a cube made of wires each having a resistance R. The cube is connected into a circuit across a body diagonal AB as shown. Find the equivalent resistance of the network in this case.

26. The magnetic moment of a short bar magnet is 1.6 m2. It is placed in the magnetic meridian with north pole pointing south. The neutral point is obtained at 20 cm from the centre of the magnet. Calculate the horizontal component H of earth’s field. If magnet be reversed i.e. north pole pointing north, find the position of the neutral point.

OR

A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.40 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm. Given mass of electron and charge on electron .

27. Obtain the equivalent capacitance of the network shown in figure below. For a 300 V

supply, determine the charge and voltage across each capacitor.

OR

Two charges and are placed at points A and B, 6 cm apart. a. Identify an equipotential surface of the system.

b. What is the direction of electric field at every point on this surface?

CBSE Class 12 Physics

Sample Paper 13

Answers

1. Pulse code modulation (PCM).

2. X-rays are used for studying crystals structure of solids. Their frequency range is 1016

Hz to .

3. Davisson and Germer’s experiment.

OR

Interference, diffraction and polarization of light.

4. Modulation.

5. Variable R

OR

180°

6. Relative velocity of the tube of width l = v – (-v) = 2v

induced emf,

= 2 Blv

OR

i. If seen from the right hand side, current flows clockwise when S-pole moves towards the coil.

ii. If seen from the right hand side, current flows anticlockwise when N-pole moves

away from the coil.

7. The torque tends to align the dipole in the direction of the electric field.

OR

In a satellite, there is condition of weightlessness. Therefore, mg = 0. On account of electrostatic force of repulsion between the balls, the strings would become horizontal. Therefore, angle between the strings = 180°

Also, tension in each string = force of repulsion

8. Wave diffracted from the edge of the circular obstacle interface constructively at the centre of the shadow producing a bright spot.

9. Angular separation as

It does not depend on D, the distance of separation between slits and screen.

Therefore remains unaffected.

10.

Hence, Am = 0.75 Ac

11.

or

= 600 m

Increase in height of tower

= 600 – 150

= 450 m

12. The total number of magnetic lines of force crossing the surface A in a magnetic field

is termed as magnetic flux.

Its SI unit is Weber. It is a scalar quantity.

13. Here, Impedance,

Average power transferred to Average power transferred to Average power transferred to

= 789.5 W

14. We have, u = -30cm; v = ?; f = +20cm

We know,

or = + 12 cm

The image is formed 12 cm behind the mirror. It is virtual and erect. Now,

or = + 0.2 cm

Hence, the height of the image = +0.2cm

The positive sign indicates that the image is erect.

15. Given: Angular frequency = Velocity = C

As per vector deduction,

vector in X axis direction is

vector in Y axis direction is Where

16. Since,

Using n = 1 for 1 st orbit

Z = 1 for hydrogen, coulomb

We get,

Also,

OR

Here, Ground Energy, E = -13.6 eV Kinetic energy,

and Potential energy, Total energy, E = Ek + Ep

17. Electric field intensity at any point outside a uniformly charged spherical shell: Assume a thin spherical shell of radius R with centre O. Let charge +q is uniformly distributed over the surface of the shell.

Let P be any point on the Gaussian surface sphere S1 with centre O and radius r (r > R). According to Gauss’s law.

At any point on the surface of the shell, r = R

If is charge density

Therefore,

Graph : As charge on shell reside on outer surface so there is no charge inside shell so electric field by Gauss’s law will be zero.

So inside shell r < R

q = 0 or

E = 0

The variation of the electric field intensity E(r) with distance r from the centre for shell is shown below.

18. Energy of the incident photon,

i. For violet light,

= 390 nm (lower wavelength end) Incident photon energy

ii. For yellow-green light, (average wavelength) Incident photon energy,

iii. For red light, = 760 nm (higher wavelength end) Incident photon energy,

OR

For a definition of terms

– work function

– threshold frequency

– stopping potential

Numerical:

The maximum kinetic energy of emitted photoelectron is given by :

19. a. The mirror formula is:

Now for a concave mirror, f < 0 and for an object on the left u < 0.

2 f < u < f or

or or or

This implies that v < 0 so that real image is formed on left. Also the above inequality implies that

2 f > v

or | 2 f | > | v | [ 2f and v are -ve]

i.e. real image is formed beyond 2f.

b. Now, for convex mirror, f > 0 and for an object of left, u < 0.

From mirror formula

or v > 0

This shows that whatever be the value of u, a convex mirror form a virtual image on the right.

c. For convex mirror f > 0 and for an object on left u < 0 so from mirror formula

[ v is +ve u is -ve]

or ( is a +ve quantity)

This shows that the image is located between the pole and the focus of the mirror. Also from the mirror formula

Multiply v to both side

[ v is +ve]

1 > m ( u < 0)

Magnitude of magnification

So the image is diminished in size.

d. From the mirror formula, for a concave mirror, f < 0 and for an object located between the pole and focus of a concave mirror.

f < u < 0

or

or or

i.e. a virtual image is formed on the right. Also, or v > |u|

OR

Using mirror formula,

For convex mirror,

Object distance u is negative and the focal length f is positive.

a positive quantity.

Therefore, the image distance is always positive in case of convex mirror.

The image is always formed on the other side of the object i.e. behind the convex mirror.

is +ve i.e. v is +ve but u is always (-) so m will be +ve i.e. an erect image which is virtual formed is convex mirror.

Hence, the convex mirror always form a virtual image.

20. Logic circuit:

Truth table:

A

B

Y’ = A + B

0

0

0

1

0

1

1

0

1

0

1

0

1

1

1

0

21. Input frequency for half wave and full wave rectifier = 50 Hz.

Input and output waveforms of half wave and full wave rectifier are shown in figure. Half wave rectifier conducts once during a cycle and full wave rectifier does so twice. Therefore, if input frequency is 50 Hz, output frequency for half wave and full wave rectifier are 50 and 100 Hz respectively.

22. The intensity of plane electromagnetic wave is given by:

or

or

= 77.6 NC-1

23. Charge on the element of length dl of the ring is Current due to circular motion of this charge is

Magnetic moment due to current dl

or

OR

In the given figure the capacitors C1 and C2in parallel combination

The circuit reduces to

Capacitors C4 and C5 are in parallel combinations

Now, the circuit reduces to

Capacitors C12 and C45 are in series combinations.

Now, C1245 and C3 are in parallel combinations, thus equivalent capacitance is

Total charge q = CV Energy stored

24. The graph showing the variation of binding energy per nucleon with mass number of atomic nuclei is shown on below:

Numerical: Q = [20mn + 20mp – m(Ca)]c2

= (20 1.008665 + 20 1.007825 – 39.962589)c2

= 342 MeV

Hence, the B.E. per nucleon of nucleus

= 8.55 MeV/nucleon.

25. Four resistances P, Q, R and S are connected to form quadrilateral ABCD. A galvanometer G is connected between B and D. A battery is connected between A and C. The resistances are so adjusted that no current flows in the galvanometer G. The same current I1 will flow in arms AB and BC. Similarly current I2 flows in arms AD

and DC.

Applying Kirchhoff’s second law for mesh ABCD, I1P – I2R = 0

or I1P = I2R …(i) For mesh BCDB, I1Q – I2S = 0

or I1Q = I2S …(ii)

Dividing (i) by (ii) we get

This is the balanced condition of the Wheatstone bridge.

Measurement of specific resistance: Slide wire or meter bridge is a practical form of

Wheatstone bridge.

In the figure X is unknown resistor and R.B is resistance box. After inserting the key k, jockey is moved on wire AC till galvanometer shows no deflection (point B). If k is the resistance per unit length of wire AC.

P = resistance of AB = kl

Q = resistance of BC = k(100 – l)

or

If r is the radius of wire and l be its length, then its resistivity will be

Precautions: (i) The null point should lie in the middle of the wire.

(ii) The current should not be allowed to flow in the wire for a long time.

OR

Let us search the points of the same potential. Since the three edges of the cube from

A viz. AC, AC1 and AC2 are identical in all respects the circuit points C, C1 and C2 are at the same potential. Similarly for the point B the sides BD, BD1 and BD2 are symmetrical and the points D, D1and D2 are at the same potential.

Next let us bring together the points C, C1 and C2 and also D, D1 and D2. Then the cube will look as shown above.

The resistance between A and The resistance between C and The resistance between D and

The circuit is equivalent to and in series which is equal to .

26. Here, M = 1.6 Am2

, H = ?

When north pole of magnet is pointing south, neutral points lie on axial line of magnet. At neutral point.

or

When magnet is reversed i.e. north pole is pointing north, neutral points lie on equatorial line of the magnet. We have to calculate .

As

= 4000 cm3

d2 = (4000)1/3= 15.87 cm

OR

Here, energy E = 18 keV B = 0.40 G

x = 30 cm = 0.3 m

As

In a magnetic field electron is deflected along a circular arc of radius r, such that

= 11.3 m

y = r – OC

By x sin g binomial

= 4 mm

27. A similar network is drawn below as given in the problem.

Here,

Assume that the series combination of C2and C3is C23

Or

Suppose that the parallel combination of C1 and C23 is C which is given as

C’ = C1 + C23 = 100 + 100 = 200 pF

Let the series combination of C4 and C’ is C which is given as

or

The total charge on all capacitors, q = CV

q1 = C1V1

V2 = V3

V2 + V3 = 100V

2V2 = 100

V2 = 50V

V2 = Vs = 50V

q2 = C2V2

OR

Given,

a. Potential will be zero due to both charges at equipotential surface

or

or

or x = 0.06 – x

i.e. the plane normal to AB and passing through its midpoint has zero potential everywhere.

b. The direction of the electric field is normal to the plane in the AB direction.

CBSE Class 12 Physics

Sample Paper – 14

TIME ALLOWED: 3 hours

MAX MARKS: 70

General Instructions:

1. All questions are compulsory. There are 27 questions in all.

2. This question paper has four sections: section A, Section B, Section C and Section D.

4. There is no overall choice. However, internal choices have been provided in two questions of one mark, two questions of two marks, four questions of five mark, three marks and three questions of five marks weightage. You have to attempt only one of the choices in such questions.

5. You may use the following values of physical constants wherever necessary c = 3 × 108 m/s

h = 6.63 × 10-34 Js

e = 1.6 × 10-19 C

µo = 4 × 10-7 T m A-1

= 8.854 × 10-12 C2 N-1 m-2

= 9 × 109 N m2 C-2

me = 9.1 × 10-31 kg

Avogadro’s number = 6.023 × 1023 per gram mole

Boltzmann constant = 1.38 × 10-23 JK-1

Section-A

1. What type of modulation is employed in India for radio transmission?

2. Name the electromagnetic spectrum to which the following wavelengths belong:

a. 10-2m b.

3. Name the experiment for which the following graph, showing the variation of intensity of scattered electrons with the angle of scattering, was obtained. Also name the important hypothesis that was confirmed by this experiment.

OR

How does the resolving power of telescope change when the aperture of the objective is increased?

4. Identify the parts X and Y in the following block diagram of a generalized communication system:

5. The number of turns in the secondary coil of a transformer is 500 times that in primary. What power is obtained from the secondary when power fed to the primary is 10 W?

OR

What is the phase difference between voltage across an inductor and a capacitor in an a.c. circuit?

Section-B

6. Why the coil of a dead beat galvanometer is wound on a metal frame?

OR

A coil of inductance 0.25 H is connected to 18 V battery. Calculate the rate of growth of current?

7. How is force between two charges affected when dielectric constant of the medium in which they are held increases?

OR

The electric field E due to a point charge at any point near it is defined as , where q is the test charge and F is the force acting on it. What is the physical

significance of in this expression? Draw the electric field lines of a point charge Q

when (i) Q > 0 and (ii) Q < 0.

8. In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band.

9. How will the intensity of maxima and minima, in the Young’s double experiment change, if one of the two slits is covered by a transparent paper which transmits only half of light intensity?

10. A TV tower has a height of 400 m at a given place. Calculate as coverage range, if the radius of the earth is 6400 km.

11. For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2V. Determine the modulation index, . What would be the value of if the minimum amplitude is zero volt?

12. What are the units of magnetic permeability?What is the maximum current in the circuit?

Section-C

13. A circuit is set up by connecting L = 100 mH, C = and in series. An alternating emf of volt, is applied across this combination.

Calculate the impedance of the circuit. What is the average power dissipated in (a) the resistor (b) the capacitor (c) the inductor and (d) the complete circuit?

14. A ray of light falls on one side of a prism whose refracting angle is 60°. Find the angle of incidence in order that the emergent ray may just graze the other side. .

15. An electromagnetic wave travels in vacuum along z-direction.

1. What can you say about the directions of its electric and magnetic field vectors?

2. If the frequency of the wave is 30 MHz, what is its wavelength?

16. A system has two charges and located at points A: (0, 0, -15 cm)and B : (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

OR

An electric dipole with dipole moment is aligned at 30° with the direction of a uniform electric field of magnitude . Calculate the magnitude of the torque acting on the dipole.

17. Obtain an expression for the frequency of radiations emitted when a hydrogen atom de-excites from level n to level (n – 1). For large n, show that the frequency equals the classical frequency of revolution of the electron in the orbit.

18. Obtain the de-Broglie wavelength of a neutron of kinetic energy 150 eV. An electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable?

OR

The work function for the following metals is given: Na : 2.75 eV K : 2.30 eV M0 4.17 eV Ni : 5.15 eV, Which of these metals will not give photoelectric emission for a radiation

of wavelength from a He-Cd laser placed 1m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?

19. A vessel 20 cm deep is half filled with oil of refractive index 1.37 and the other half is filled with water of refractive index 1.33. Find the apparent depth of the vessel when viewed from above.

OR

A prism is made of glass of unknown refractive index. A parallel beam of light is

incident on a face of the prism. The angle of minimum deviation is measured to be

40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water predict the new angle of minimum deviation of a parallel beam of light.

20. Draw a circuit diagram for a two input OR gate and explain its working with the help of input, output waveforms.

21. Write the truth table for the circuits given in figure. Consisting of NOR gates only.

Identify the logic operations ( OR, AND, NOT ) performed by the two circuits.

22. A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,

a. the wire intersects the axis

b. the wire is turned from N-S to northwest direction,

c. the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

OR

A galvanometer coil has a resistance of and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to

18 V?

23. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of atoms ( of mass 62.92960 u)

24. Name the electromagnetic waves used for the following and arrange them in increasing order of their penetrating power.

a. Water purification b. Rewrite sensing

c. Treatment of cancer

Section-D

25. Find the potential difference across each cell and the rate of energy dissipation in R.

OR

Determine the current in each branch of the network shown in the figure.

26. A short bar magnet is placed in a horizontal plane with its axis in the magnetic meridian. Null points are found on its equatorial line (i.e. its normal bisector ) at 12.5

cm from the centre of the magnet. The earth’s magnetic field at the place is 0.38 gauss, and the angle of dip is zero.

a. What is the total magnetic field at points on the axis of the magnet located at the same distance (12.5 cm) as the null points from the centre?

b. Locate the null points when the bar is turned around by 180°. Assume that the length of the magnet is negligible compared to the distance of the null points from the centre of the magnet.

OR

A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.40 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm. Given mass of electron and charge on electron .

27. Two tiny spheres carrying charges and are located 30 cm apart. Find the potential and electric field:(a) at the mid point of the line joining the two charges, and(b) at a point 10 cm from this mid point in a plane normal to the line and passing through the mid point.

OR

If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion . In the ground state of a , the two protons are separated by roughly

, and the electron is roughly from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

CBSE Class 12 Physics

Sample Paper 14

Answers

1. Amplitude modulation.

2. a. Microwaves [ Range : 0.3 to 10-3m]

b. X -rays [Range: to ]

3. Davisson and Germer’s experiment.

OR

The resolving power of telescope increases on increasing the aperture of objective lens.

4. X Source of information

Y Communication channel.

5. If there is no loss of energy, then the output power will be 10 W.

OR

180°

6. On switching ON the current in a galvanometer, the coil of the galvanometer does not come to rest immediately. It oscillates about its equilibrium position but the coil of a dead beat galvanometer comes to rest immediately. It is due to the reason that the eddy currents are set up in the metallic frame, over which the coil is wound and the eddy currents oppose the oscillatory motion of the coil.

OR

Given, L = 0.25 H

e = 18 V

Using formula,

7. As , therefore, force decreases, when K increases.

OR

It is indicated by the that the test charge q is so small that its presence does not disturb the distribution of source charge and therefore, its electric field. The electric

field of the point charge Q are shown below.

8. Width of central diffraction b and , so on doubling the width of the slit, the size of the central diffraction band reduces to half value. But the light amplitude becomes double, which increases the intensity four fold.

9. if one of the slit transmits only half of the light intensity, then the intensity in the interference pattern will vary fro maximum to minimum but not become zero. therefore contrast will be reduced & the interference pattern will become less sharp. So we can say that the intensity of maxima decreases and the intensity of minima increases.

10.

= 71.554 km

11. The AM wave is given by

The maximum amplitude is M1 = Ac + Am while the minimum amplitude is M2 = Ac – Am. Hence, the modulation index is:

With M2 = 0, clearly , irrespective of M1.

12. Tesla metre / ampere (TmA-1).

13.

or = 1.5 A

a. Power consumed in resistor is = W = 225 W

b. Power consumed in capacitor is zero. c. Power consumed in inductor is zero

d. Power consumed in circuits same as power consumed in resistor i.e .225 W

14. Given, A = 60°, e= 90°

r2 = C, the critical angle of the prism.

Now,

or

Again A = r1 + r2

For the refraction at the surface AB, we have

(Snell’s law)

or

= 0.46815

15. and lie in x-y plane and are mutually perpendicular, Since;

Wavelength of the wave is 10m

16. Total charge, q = qA + qB

a = AB = 15 – (-15) = 30 cm = 0.3 m

Electric dipole moment p = q . a

(along Z-axis).

OR

Given, Torque,

or

or

17. The frequency of the emitted radiation when a hydrogen atom de-excites from level n to level (n – 1)is

E = hv = E2 – E1

where find structure constant

For large n, (2n – 1) = 2n, and (n – 1) = n

Putting we get

In Bohr’s atom model velocity of electron in nth orbit is and radius of nth orbit is

Frequency of revolution of electron

which is the same as (i)

Hence for large values of n, classical frequency of revolution of electron in nth orbit is

the same as the frequency of radiation emitted when hydrogen atom de-excites from level (n) to level (n – 1).

18. Here, K.E. of neutron, E = 150 eV Mass of neutron,

We know, K.E. of neutron, or

The interatomic spacing is about a hundred times greater than this wavelength. Therefore, a neutron beam of energy 150 eV is not suitable for diffraction experiment.

OR

(i) Work function of Na is

E = 3.75 eV

It is observed that energy of incident radiation is less than Ni and Mo but larger than Na and K. So photoemission current take place from Na and K but not from Mo and Ni. Therefore, Mo and Ni will not give photoelectric emission. If the laser is brought closer the intensity of radiation increases without any change in frequency. This therefore, will not affect the result. However, photoelectric current from Na and K will increase.

19. Referring to figure given below, the apparent depth of the vessel is

where t = 20 cm, and Substituting for t, and we get

AI = 15.37 cm

OR

A = 60°,

After the prism is placed in water,

or

or

on simplification .

20. The OR gate can be made with the help of two p-n junction diodes D1 and D2 as given in the circuit follows.

Working: The negative terminal of the battery is grounded and corresponds to the 0 states and the positive (i.e. voltage 5 V in the present case) to the 1 state.

When both A and B are connected to 0, no current passes through the diode and therefore no voltage develops across R and the output is zero.

When input A is connected to zero and B to 1, the diode D2 is forward biased and the current through it is limited by a current limiting resistance. This current causes a 5 V drop across the resistance assuming the diode to be ideal and this gives an output of 5

V or 1. Interchanging A and B to 1 and 0 will still give a 5 V drop across the resistance as D1 will conduct.

When the terminals A and B are connected to 1, then both the diodes D1 and

D2conduct. However, the voltage drop across R cannot exceed 5 V and the output is 1.

21. Figure (a) represents NOT gate.

Here when A = 1, Y = 0 and when A = 0, Y = 1

Figure (b), represents two NOT gates whose outputs are given to NOR gate. Its truth table is

A

B

Y

0

0

1

1

0

0

1

1

0

0

1

0

0

1

0

1

1

0

0

1

Y is truth table of AND gate.

Logic operation performed by figure (b) is AND operation.

22. a. Diameter of cylindrical region

= 20 cm = 0.20 m

Clearly, l = 0.20 m, Also

F = BI sin

= 2.1 N

Using Fleming’s left hand rule, we find that the force is directed vertically downwards.

b. If I1 is the length of the wire is the magnetic field, then

F1 = BIl1 sin45o

But l1 sin45o = 1

The force is directed vertically downwards by Fleming’s left hand rule.

c. When the wire is lowered by 6 cm, the length of the wire in the cylindrical magnetic field is 2x.

Now x2 = 102 – 62

2x = 16 cm

= 1.68 N

The force is directed vertically downwards.

OR

Given ,

V = 18 v, R = ?

By using formula

V = Ig(R + Rg)

or

or

23. Mass of atom = 62.92960 u

Mass of 29 electrons

= 0.015892 u

Mass of nucleus = (62.9296 – 0.015892)u

= 62.913708 u

Mass of 29 protons

= 29.226925 u

Mass of (63 – 29) i.e. 34 neutrons

= 34.29461 u

Total mass of protons and neutrons

= (29.226925 + 34.29461) u

= 63.521535 u

Binding energy = Required energy

24. Following are the electromagnetic waves used for;

a. Water purification – Ultra violet waves b. Rewrite sensing – Micro waves

c. Treatment of cancer – Gamma rays

In increasing order of penetration of given waves can be written as, (b) remote sensing < (a) water purification < (c) treatment of cancer.

25. Applying Kirchhoff’s rule for loop ABCDA,

12 = 4(I1 + I2) + 2I1

Or 12 = 6I1 + 4I2 …….. (i) For loop DEFAD

6 = 4(I1 + I2) + I2

Or 6 = 4I1 + 5I2 ……. (ii)

Equation (i) and (ii) can be written as

12 = 6I1 + 4I2

and 9 = 6I1 + 7.5I2

or 3 = -3.5I2

or

Putting the value I2in equation (i)

or

or 12 (4.5) = 210 I1

or

Current in R,

Or rate of energy dissipation = 2 R

OR

Applying Kirchhoff’s law to the mesh ABDA,

-101 – 5g + (I – I1)5 = 0

……. (i)

Again, applying Kirchhoff’s second law to the mesh BDCB,

– 5Ig – 10(I – I1 + Ig) + 5(I1 – Ig) = 0

3I1 – 2I – 4Ig = 0 …… (ii)

Applying Kirchhoff’s second law to the mesh ABCEA,

-10I1 – 5(I1 – Ig) – 10I + 10 = 0

3I1 + 2I – Ig = 2 ……. (iii) Adding (i) and (iii), we get

6I1 + I = 2 ……. (iv)

Multiplying (i) by 4 and adding in (ii), we get

15I1-6I=0……(v)

Solving equations (iv) and (v), we get

= 0.235 A

So, current in branch AB is 0.235 A

Putting the value of I1 in equation (v) and simplifying, we get

Total current, = 0.588 A

Putting the values of I and I1 in equation (iii) and simplifying, we get

The negative sign indicates that the direction of current is opposite to that shown in figure above

So current in branch BD is “-0.118A” Current in branch BC is (I1 – Ig)i.e.

i.e. or 0.353 A

Current in branch AD is (I – I1)

i.e. i.e. or 0.353 A Current in branch DC is (I – I1 + Ig)

i.e. or or 0.235 A

26. a. From figure, it is clear that null points are obtained on the normal bisector when the magnet’s north and south poles face magnetic north and south respectively. Magnetic field on the normal bisector at a distance r from the centre is given by

, provided r is much greater than the length of the magnet. [The above equation is strictly true only for a point dipole]. At a null point, this field is

balanced by the earth’s field.

So, Be = BH

….(i)

(since the dip angle (s) is zero, therefore, Bv = 0 and the horizontal component of the earth’s field equals the field itself)

Next, magnetic field due to a magnet on its axis at point distant r from the centre is

given by

….(ii)

provided r is much greater than the length of the magnet.

From figure (a) it is clear that on the axis, this field adds up to the earth’s field. Thus, the total field at a point on the axis has a magnitude equal to Ba + BH

i.e. ….(iii)

and direction along [which is parallel to the earth’s field in case (a)].

Thus, for the same distance on the axis as the distance of the null point, the total field, using equations (i) and (iii) is

i.e.

i.e.

This field is directed along .

b. When the bar is turned around by 180°, the magnet’s north and south poles face

magnetic south and north respectively i.e., in this case, is anti arallel to the earth’s field. From figure (b) it is clear that the null points now lie on the axis of the magnet at a distance r’ given by

or ….(iv)

Comparing equations (iv) and (i), we get or r’3 = 2r3 or r’ = (2)1/3r

For r = 12.5 cm, r’ = 15.7 cm

OR

Here, energy E = 18 keV B = 0.40 G

x = 30 cm = 0.3 m

As

In a magnetic field electron is deflected along a circular arc of radius r, such that

= 11.3 m

y = r – OC

By x sin g binomial

= 4 mm

27. a. Potential at the mid point of the line joining the two charges is

Or

Electric field at the mid point O due to charge at A

along OB

Electric field at the mid point O due to charge at B

along OA

Thus, the total electric field at the mid point O is

(along BA)

b. Potential at the point C due to the two charges is

Electric field at C due to charge at A

Electric field at C due to charge at B

If the angle between E1 and E2 be then

Thus, magnitude of resultant field at C is

Let the field E makes angle with the field E1

Now,

= 4.2876

= 76.9°

If field E makes angle with the direction BA, then

= 90° + 56.3° – 76.9° = 69.4°

Therefore, angle of 69.4° is made by the electric field with the line joining the two charges q2 to q1.

OR

Suppose that the distance between two protons p1 and p2 be r12 and electron (e-) is

placed at r13 and r23 distance from protons p1 and p2 respectively.

Given,

r13 = r23 = 1 = 10-10m

(protons)

(electrons)

Now, the potential energy of the system

+

= 19.2eV

CLASS XII (2018-19) (THEORY)

Time: 3 hrs. Max Marks: 70

No. of Periods Marks

Unit–I Electrostatics

22

15

Chapter–1: Electric Charges and Fields

Chapter–2: Electrostatic Potential and Capacitance

Unit-II Current Electricity

20

Chapter–3: Current Electricity

Unit-III Magnetic Effects of Current and Magnetism

22

16

Chapter–4: Moving Charges and Magnetism

Chapter–5: Magnetism and Matter

Unit-IV Electromagnetic Induction and Alternating Currents

20

Chapter–6: Electromagnetic Induction

Chapter–7: Alternating Current

Unit–V Electromagnetic Waves

04

17

Chapter–8: Electromagnetic Waves

Unit–VI Optics

25

Chapter–9: Ray Optics and Optical Instruments

Chapter–10: Wave Optics

Unit–VII Dual Nature of Radiation and Matter

08

10

Chapter–11: Dual Nature of Radiation and Matter

Unit–VIII Atoms and Nuclei

14

Chapter–12: Atoms

Chapter–13: Nuclei

Unit–IX Electronic Devices

15

12

Chapter–14: Semiconductor Electronics: Materials, Devices and Simple Circuits

Unit–X Communication Systems

10

Chapter–15: Communication Systems

Total 160 70

Unit I: Electrostatics 22 Periods

Chapter–1: Electric Charges and Fields

Electric Charges; Conservation of charge, Coulomb’s law-force between two point charges, forces between multiple charges; superposition principle and continuous charge distribution.

Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in uniform electric field.

Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside).

Chapter–2: Electrostatic Potential and Capacitance

Electric potential, potential difference, electric potential due to a point charge, a dipole and system of charges; equipotential surfaces, electrical potential energy of a system of two point charges and of electric dipole in an electrostatic field.

Conductors and insulators, free charges and bound charges inside a conductor. Dielectrics and electric polarisation, capacitors and capacitance, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor.

Unit II: Current Electricity 20 Periods

Chapter–3: Current Electricity

Electric current, flow of electric charges in a metallic conductor, drift velocity, mobility and their relation with electric current; Ohm’s law, electrical resistance, V-I characteristics (linear and non-linear), electrical energy and power, electrical resistivity and conductivity, Carbon resistors, colour code for carbon resistors; series and parallel combinations of resistors; temperature dependence of resistance.

Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, Kirchhoff’s laws and simple applications, Wheatstone bridge, metre bridge.

Potentiometer – principle and its applications to measure potential difference and for comparing EMF of two cells; measurement of internal resistance of a cell.

Unit III: Magnetic Effects of Current and Magnetism 22 Periods

Chapter–4: Moving Charges and Magnetism

Concept of magnetic field, Oersted’s experiment.

Biot – Savart law and its application to current carrying circular loop.

Ampere’s law and its applications to infinitely long straight wire. Straight and toroidal solenoids (only qualitative treatment), force on a moving charge in uniform magnetic and electric fields, Cyclotron.

Force on a current-carrying conductor in a uniform magnetic field, force between two parallel current-carrying conductors-definition of ampere, torque experienced by a current loop in uniform magnetic field; moving coil galvanometer-its current sensitivity and conversion to ammeter and voltmeter.

Chapter–5: Magnetism and Matter

Current loop as a magnetic dipole and its magnetic dipole moment, magnetic dipole moment of a revolving electron, magnetic field intensity due to a magnetic dipole (bar magnet) along its axis and perpendicular to its axis, torque on a magnetic dipole (bar magnet) in a uniform magnetic field; bar magnet as an equivalent solenoid, magnetic field lines; earth’s magnetic field and magnetic elements.

Para-, dia- and ferro – magnetic substances, with examples. Electromagnets and factors affecting their strengths, permanent magnets.

Unit IV: Electromagnetic Induction and Alternating Currents 20 Periods

Chapter–6: Electromagnetic Induction

Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Eddy currents. Self and mutual induction.

Chapter–7: Alternating Current

Alternating currents, peak and RMS value of alternating current/voltage; reactance and impedance; LC oscillations (qualitative treatment only), LCR series circuit, resonance; power in AC circuits, power factor, wattless current.

AC generator and transformer.

Unit V: Electromagnetic waves 04 Periods

Chapter–8: Electromagnetic Waves

Basic idea of displacement current, Electromagnetic waves, their characteristics, their

Transverse nature (qualitative ideas only).

Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses.

Unit VI: Optics 25 Periods

Chapter–9: Ray Optics and Optical Instruments

Ray Optics: Reflection of light, spherical mirrors, mirror formula, refraction of light, total internal reflection and its applications, optical fibres, refraction at spherical surfaces, lenses, thin lens formula, lensmaker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction and dispersion of light through a prism.

Scattering of light – blue colour of sky and reddish apprearance of the sun at sunrise and sunset.

Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.

Chapter–10: Wave Optics

Wave optics: Wave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width, coherent sources and sustained interference of light, diffraction due to a single slit, width of central maximum, resolving power of microscope and astronomical telescope, polarisation, plane polarised light, Brewster’s law, uses of plane polarised light and Polaroids.

Unit VII: Dual Nature of Radiation and Matter 08 Periods

Chapter–11: Dual Nature of Radiation and Matter

Dual nature of radiation, Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation-particle nature of light.

Matter waves-wave nature of particles, de-Broglie relation, Davisson-Germer experiment

(experimental details should be omitted; only conclusion should be explained).

Unit VIII: Atoms and Nuclei 14 Periods

Chapter–12: Atoms

Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum.

Chapter–13: Nuclei

Composition and size of nucleus, Radioactivity, alpha, beta and gamma particles/rays and their properties; radioactive decay law.

Mass-energy relation, mass defect; binding energy per nucleon and its variation with mass number; nuclear fission, nuclear fusion.

Unit IX: Electronic Devices 15 Periods

Chapter–14: Semiconductor Electronics: Materials, Devices and Simple Circuits Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Semiconductor diode – I-V characteristics in forward and reverse bias, diode as a rectifier;

Special purpose p-n junction diodes: LED, photodiode, solar cell and Zener diode and their characteristics, zener diode as a voltage regulator.

Junction transistor, transistor action, characteristics of a transistor and transistor as an amplifier (common emitter configuration), basic idea of analog and digital signals, Logic gates (OR, AND, NOT, NAND and NOR).

Unit X: Communication Systems 10 Periods

Chapter–15: Communication Systems

Elements of a communication system (block diagram only); bandwidth of signals (speech, TV and digital data); bandwidth of transmission medium. Propagation of electromagnetic waves in the atmosphere, sky and space wave propagation, satellite communication. Need for modulation, amplitude modulation.

PRACTICALS (Total Periods 60)

The record to be submitted by the students at the time of their annual examination has to include:

Record of at least 15 Experiments [with a minimum of 6 from each section], to be performed by the students.

Record of at least 5 Activities [with a minimum of 2 each from section A and section B], to be demonstrated by the teachers.

The Report of the project to be carried out by the students.

Evaluation Scheme

Time Allowed: Three hours Max. Marks: 30

Two experiments one from each section 8+8 Marks

Practical record [experiments and activities] 6 Marks

Investigatory Project 3 Marks

Viva on experiments, activities and project 5 Marks

Total 30 marks

Experiments

SECTION–A

1. To determine resistance per cm of a given wire by plotting a graph for potential difference versus current.

2. To find resistance of a given wire using metre bridge and hence determine the resistivity (specific resistance) of its material.

3. To verify the laws of combination (series) of resistances using a metre bridge.

4. To verify the laws of combination (parallel) of resistances using a metre bridge.

5. To compare the EMF of two given primary cells using potentiometer.

6. To determine the internal resistance of given primary cell using potentiometer.

7. To determine resistance of a galvanometer by half-deflection method and to find its figure ofmerit.

8. To convert the given galvanometer (of known resistance and figure of merit) into a voltmeter of desired range and to verify the same.

9. To convert the given galvanometer (of known resistance and figure of merit) into an ammeter of desired range and to verify the same.

10. To find the frequency of AC mains with a sonometer.

Activities

(For the purpose of demonstration only)

1. To measure the resistance and impedance of an inductor with or without iron core.

2. To measure resistance, voltage (AC/DC), current (AC) and check continuity of a given circuit using multimeter.

3. To assemble a household circuit comprising three bulbs, three (on/off) switches, a fuse and a power source.

4. To assemble the components of a given electrical circuit.

5. To study the variation in potential drop with length of a wire for a steady current.

6. To draw the diagram of a given open circuit comprising at least a battery, resistor/rheostat, key, ammeter and voltmeter. Mark the components that are not connected in proper order and correct the circuit and also the circuit diagram.

SECTION–B

Experiments

1. To find the value of v for different values of u in case of a concave mirror and to find the focallength.

2. To find the focal length of a convex mirror, using a convex lens.

3. To find the focal length of a convex lens by plotting graphs between u and v or between 1/u and1/v.

4. To find the focal length of a concave lens, using a convex lens.

5. To determine angle of minimum deviation for a given prism by plotting a graph between angle of incidence and angle of deviation.

6. To determine refractive index of a glass slab using a travelling microscope.

7. To find refractive index of a liquid by using convex lens and plane mirror.

8. To draw the I-V characteristic curve for a p-n junction in forward bias and reverse bias.

9. To draw the characteristic curve of a zener diode and to determine its reverse break downvoltage.

10. To study the characteristic of a common – emitter npn or pnp transistor and to find out the values of current and voltage gains.

Activities

(For the purpose of demonstration only)

1. To identify a diode, an LED, a transistor, an IC, a resistor and a capacitor from a mixed collection of such items.

2. Use of multimeter to (i) identify base of transistor, (ii) distinguish between npn and pnp type transistors,

(iii) see the unidirectional flow of current in case of a diode and an LED, (iv) check whether a given electronic component (e.g., diode, transistor or IC) is in working order.

3. To study effect of intensity of light (by varying distance of the source) on an LDR.

4. To observe refraction and lateral deviation of a beam of light incident obliquely on a glass slab.

5. To observe polarization of light using two Polaroids.

6. To observe diffraction of light due to a thin slit.

7. To study the nature and size of the image formed by a (i) convex lens, (ii) concave mirror, on a screen by using a candle and a screen (for different distances of the candle from the lens/mirror).

8. To obtain a lens combination with the specified focal length by using two lenses from the given set of lenses.

Suggested Investigatory Projects

1. To study various factors on which the internal resistance/EMF of a cell depends.

2. To study the variations in current flowing in a circuit containing an LDR because of a variation in

(a) the power of the incandescent lamp, used to ‘illuminate’ the LDR (keeping all the lamps at a fixed distance).

(b) the distance of a incandescent lamp (of fixed power) used to ‘illuminate’ the LDR.

3. To find the refractive indices of (a) water (b) oil (transparent) using a plane mirror, an equi convex lens (made from a glass of known refractive index) and an adjustable object needle.

4. To design an appropriate logic gate combination for a given truth table.

5. To investigate the relation between the ratio of (i) output and input voltage and (ii) number of turns in the secondary coil and primary coil of a self designed transformer.

6. To investigate the dependence of the angle of deviation on the angle of incidence using a hollow prism filled one by one, with different transparent fluids.

7. To estimate the charge induced on each one of the two identical styrofoam (or pith) balls suspended in a vertical plane by making use of Coulomb’s law.

8. To set up a common base transistor circuit and to study its input and output characteristic and to calculate its current gain.

9. To study the factor on which the self inductance of a coil depends by observing the effect of this coil, when put in series with a resistor/(bulb) in a circuit fed up by an A.C. source of adjustable frequency.

10. To construct a switch using a transistor and to draw the graph between the input and output voltage and mark the cut-off, saturation and active regions.

11. To study the earth’s magnetic field using a tangent galvanometer.

Practical Examination for Visually Impaired Students of Classes XI and XII Evaluation Scheme

Time Allowed: Two hours Max. Marks: 30

Identification/Familiarity with the apparatus 5 marks

Written test (based on given/prescribed practicals) 10 marks

Practical Record 5 marks

Viva 10 marks

Total 30 marks

General Guidelines

The practical examination will be of two hour duration. A separate list of ten experiments is included here.

The written examination in practicals for these students will be conducted at the time of practical examination of all other students.

The written test will be of 30 minutes duration.

The question paper given to the students should be legibly typed. It should contain a total of 15 practical skill based very short answer type questions. A student would be required to answer any 10 questions.

A writer may be allowed to such students as per CBSE examination rules.

All questions included in the question papers should be related to the listed practicals. Every question should require about two minutes to be answered.

These students are also required to maintain a practical file. A student is expected to record at least five of the listed experiments as per the specific instructions for each subject. These practicals should be duly checked and signed by the internal examiner.

The format of writing any experiment in the practical file should include aim, apparatus required, simple theory, procedure, related practical skills, precautions etc.

Questions may be generated jointly by the external/internal examiners and used for assessment.

The viva questions may include questions based on basic theory/principle/concept, apparatus/ materials/chemicals required, procedure, precautions, sources of error etc.

Class XII

A. Items for Identification/ familiarity with the apparatus for assessment in practicals (All experiments)

Meter scale, general shape of the voltmeter/ammeter, battery/power supply, connecting wires, standard resistances, connecting wires, voltmeter/ammeter, meter bridge, screw gauge, jockey Galvanometer, Resistance Box, standard Resistance, connecting wires, Potentiometer, jockey, Galvanometer, Lechlanche cell, Daniell cell [simple distinction between the two vis-à-vis their outer (glass and copper) containers], rheostat connecting wires, Galvanometer, resistance box, Plug-in and tapping keys, connecting wires battery/power supply, Diode, Transistor, IC, Resistor (Wire-wound or carbon ones with two wires connected to two ends), capacitors (one or two types), Inductors, Simple electric/electronic bell, battery/power supply, Plug-in and tapping keys, Convex lens, concave lens, convex mirror, concave mirror, Core/hollow wooden cylinder, insulated wire, ferromagnetic rod, Transformer core, insulated wire.

B. List of Practicals

1. To determine the resistance per cm of a given wire by plotting a graph between voltage and current.

2. To verify the laws of combination (series/parallel combination) of resistances by Ohm’s law.

3. To find the resistance of a given wire using a meter bridge and hence determine the specific resistance

(resistivity) of its material.

4. To compare the e.m.f of two given primary cells using a potentiometer.

5. To determine the resistance of a galvanometer by half deflection method.

6. To identify a

(i) diode, transistor and IC

(ii) resistor, capacitor and inductor, from a mixed collection of such items.

7. To understand the principle of (i) a NOT gate (ii) an OR gate (iii)an AND gate and to make their equivalent circuits using a bell and cells/battery and keys /switches.

8. To observe the difference between

(i) a convex lens and a concave lens

(ii) a convex mirror and a concave mirror and to estimate the likely difference between the power of two given convex /concave lenses.

9. To design an inductor coil and to know the effect of

(i) change in the number of turns

(ii) introduction of ferromagnetic material as its core material on the inductance of the coil.

10. To design a (i) step up (ii) step down transformer on a given core and know the relation between its input and output voltages.

Note: The above practicals may be carried out in an experiential manner rather than recording observations.

Prescribed Books:

1. Physics, Class XI, Part -I and II, Published by NCERT.

2. Physics, Class XII, Part -I and II, Published by NCERT.

3. Laboratory Manual of Physics for class XII Published by NCERT.

4. The list of other related books and manuals brought out by NCERT (consider multimedia also).

PHYSICS (Code No. 042) QUESTION PAPER DESIGN CLASS – XII (2018-19)

Time 3 Hours Max. Marks: 70

S. No. Typology of

Questions Very Short Answer (VSA)

(1 mark) Short Answer-I (SA-I)

(2 marks) Short Answer –II (SA-II) (3 marks) Long Answer (LA)

(5 marks) Total

Marks % Weightage

1. Remembering – (Knowledge based Simple recall questions, to know

specific facts, terms, concepts, principles, or theories, Identify, define, or recite, information)

2

1

1

–

7

10%

2 Understanding – (Comprehension -to be familiar with meaning and

to understand conceptually,

interpret, compare, contrast, explain, paraphrase information)

–

2

4

1

21

30%

3 Application – (Use abstract information in concrete situation, to apply knowledge to new situations, Use given content to interpret a situation, provide an example, or solve a problem)

–

2

4

1

21

30%

4 High Order Thinking Skills – (Analysis & Synthesis- Classify, compare, contrast, or differentiate between different pieces of information, Organize

and/or integrate unique pieces of information from a variety of sources)

2

–

1

1

10

14%

5 Evaluation – (Appraise, judge, and/or justify the value or worth of a decision or outcome, or to predict outcomes based on values)

1

2

2

–

11

16%

TOTAL 5×1=5 7×2=14 12×3=36 3×5=15 70(27) 100%

QUESTION WISE BREAK UP

Type of Question Mark per Question Total No. of Questions Total Marks

VSA 1 5 05

SA-I 2 7 14

SA-II 3 12 36

LA 5 3 15

Total 27 70

1. Internal Choice: There is no overall choice in the paper. However, there is an internal choice in one questio n of 2 marks weightage, one question of 3 marks weightage and all the three questions of 5 marks weightage.

2. The above template is only a sample. Suitable internal variations may be made for generating similar templates keeping the overall weightage to different form of questions and typology of questions same.