seshasaiphysicsblogxii.files.wordpress.com · Web viewQ21 Show mathematically that in A.C. circuit...

BLUE PRINT OF QUESTION PAPER 1

UNIT NAME 1 MARK 2MARKS 3MARKS VBQ 4 MARKS 5MARKS TOTALELECTROSTATICS

CURRENT ELECTRICITY6(2) 4(1) 5(1) 15(4)

MAGNETISM&MAGNETIC EFFECTS OF CURRENT

AC&EMI

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

EMW&OPTICS 3(3) 6(3) 3(1) 5(1) 17(8)

DUAL NATURE OF MATTER&ATOMS &NUCLEI

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

SEMICONDUCTOR DEVICES

COMMUNICATION

12(4) 12(4)

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

*TOTAL MARKS (NO OF QUESTIONS)

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

Page 1

PHYSICS (CLASS XII) Paper-1

Time : 3 Hrs. Max. Marks : 70

General Instruction :

(i) All questions are compulsory.

(ii) There are 26 questions in total. Questions 1 to 5 are very short answer type questions and carry one mark each.

(iii) Questions 6 to 10 carry two marks each, questions 11 to 22 carry three marks each, question 23 is value based question carry 4 marks and questions 24 to 26 carry five marks each.

(iv) There is not overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all three question of five marks each weightage. You have to attempt only one of the choices in such questions.

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

(vi) 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 μ0 = 4π × 10–7 T mA–1 = 9 × 109 N m2 C–1

me = 9.1 × 10–31 kg Mass of Neutrons = 1.675 × 10–27 kg Mass of proton = 1.673 × 10–27 kg

Q1 Using the concept of force between infinite long parallel current carrying conductors, define one ampere of current.B.Sesha Sai/KV, Embassy Of India, Kathmandu/PHY/XI Science.

Page 2

Q2 A biconvex lens made of transparent material of refractive index 1.5 is immersed in water of refractive index 1.33, will the lens behave as a conversing or a diverging lens? Give reason.

Q3 To which part of the electromagnetic spectrum does a wave of frequency 5 x 1019Hz belong?

Q4 How does focal length of a lens change when red light incident on it is replaced by violet light? Give reason for your answer.

Q5 The given graph shows the variation of photo-electric current (I) versus applied voltage (V) for two difference photosensitive materials and for two different intensities of the incident radiations. Identify the pairs of curves that correspond to different materials but same intensity of incident radiation.

Q6 .State the condition which must be satisfied for two light sources to be coherent?OR

Explain two differences between telescope and a microscope.

Q7 Why is the use of AC voltage is preferred over DC voltage? Give two reasons.

Q8 A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of the combination. Will the system be converging or diverging in nature?

Q9 An ammeter of resistance 0.80 can measure current up to 1.0A.

(i) What must be the value of shunt resistance to enable the ammeter to measure current up to 5.0 A?

(ii) What is the combined resistance of the ammeter and the shunt?

Q10 (a) An em wave is travelling in a medium with a velocity. Draw a sketch showing the propagation of the em wave, indicating the direction of the oscillating electric and magnetic fields.

(b) How are the magnitudes of the electric and magnetic fields related to velocity of the em wave?


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Q11 Figure shows a block diagram of a transmitter identify the boxes ‘X’ and ‘Y’ and write their functions.

.

Q12 Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used to detect the optical signals.

OR

A Semiconductor has equal electron hole concentration of 6x108/m3 on doping with a certain impurity, electron concentration increases to 8x1012/m3.

i) Identify Semiconductor obtained after doping.ii) Calculate new hole concentration.

Q13 Write the two basic modes of communication. Show diagrammatically how an amplitude modulated wave is obtained when a modulating signal is superimposed on a carrier wave.

Q14 A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of energy stored in the combined system so that energy stored is same as initially stored in the single capacitor?

OR

A hollow cylindrical box of length 1 m and area of cross-section 25 cm2 is placed in a three dimensional coordinate

system as shown in the figure. The electric field in the region is given by , where E is NC−1 and x is in metres. Find

(i) Net flux through the cylinder. (ii) Charge enclosed by the cylinder.


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Q15 The decay constant, for a given radioactive nuclei, has a value of 1.386 day -1. After how

much time will a given sample of this radionuclide get reduced to only 6.25% of its present

number ?

Q16 (a) Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.

(b) Write the basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based.

Q17 A short bar magnet of magnetic moment 0.9 J/T is placed with its axis at 30° to a uniform magnetic field. It experiences a torque of 0.063 J.

(i) Calculate the magnitude of the magnetic field. (ii) In which orientation will the bar magnet be in stable equilibrium in the magnetic field?

Q18 Output characteristics of an n-p-n transistor in CE configuration is shown in the figure. Determine:

(i) dynamic output resistance (ii) dc current gain and (iii) ac current gain at an operating point V CE = 10 V, when IB = 30 µA.


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Q19 Draw a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. Indicate the regions in which nuclear force is (i) attractive, and (ii) repulsive.(b) Write two characteristic features of nuclear force which distinguish it from the Coulomb force.

Q20 (a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment.

(b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn to study the diffraction taking place at a single slit of aperture 2 × 10−4 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.

Q21 Show mathematically that in A.C. circuit containing only inductance, the current lags behind the voltage

by a phase of π/2.

Q22 Deduce an expression for the electric potential due to an electric dipole at any point on its axis. Mention one contrasting feature of electric field of a dipole at a point as compared to that due to single charge.

Q23 While travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm. It became very dark. He stopped driving the car and waited for thunderstorm to stop. Suddenly he noticed a child walking alone on the road. He asked the boy to come inside the car till the thunderstorm stopped. Dr. Pathak dropped the boy at his residence. The body insisted that Dr. Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak for his concern for safety of the child.

Answer the following questions based on the above information:

(a) Why is it safer to sit inside a car during a thunderstorm?

(b) Which two values are displayed by Dr. Pathak in his action?

(c) Which values are reflected in parents' response to Dr. Pathak?

(d) Give an example of similar action on your part in the past from everyday life.

Q24 a) What is the effect on the interference fringes to a Young’s double slit experiment when

(i) the separation between the two slits is decreased?

(ii) the width of a source slit is increased?

(iii) the monochromatic source is replaced by a source of white light? Justify your answer in each case.


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(b) The intensity at the central maxima in Young’s double slit experimental set-up is I0. Show that the intensity at a point where the path difference is λ/3 is I0/4.

OR

(a) Obtain the conditions for the bright and dark fringes in diffraction pattern due to a single narrow slit illuminated by a monochromatic source.

Why central maxima is more intense as compared to secondary maxima’s?

(b) When the width of the slit is made double, how would this affect the size and intensity of the central diffraction band? Justify

Q25 (a) State the working principle of a potentiometer. With the help of the circuit diagram, explain how a potentiometer is used to compare the emf's of two primary cells. Obtain the required expression used for comparing the emf’s.

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

OR

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

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

Q.26) With the help of a schematic sketch of a cyclotron explain its working principle and mention

its two limitations.

OR

With the help of a neat and labeled diagram, explain its principle, construction and working

of a moving coil galvanometer, what is the function of

i. Uniform radial magnetic fieldii. Soft iron core, in this device


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SOLUTION OF PAPER 1

1 One ampere of current can be defined as the amount of current which when flowing (in same direction) through two infinitely long parallel wires separated by one meter produces an attractive force of 2x 10-7 N/m. The wires must have negligible circular cross section and they must be placed in vacuum.2 A biconvex lens acts as a converging lens in air because the refractive index of air is less than that of the material of the lens, the refractive index of water is also less than that of refractive index of the material of the lens. So its nature will not change, it behaves as a converging lens.

3. A wave of frequency 5x1019Hz belongs to gamma rays of electromagnetic spectrum.

4 The refractive index of the material of a lens increases with the decrease in wavelength of the incident light. So, focal length will decrease with decrease in wavelength according to formula.


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Thus, when we replace red light with a violet wavelength decrease and hence the focal length of the lens also decrease.

5. Curves 1 and 2 correspond to similar materials while curves 3 and 4 represent different materials, since the value of stopping potential for the pair of curves (1 and 2) & (3 and 4) are the same. For given frequency of the incident radiation the stopping potential is independent of its intensity.

So, the pairs of curves (1 and 3) and (2 and 4) correspond to different materials but same intensity of incident radiation.

6 The two sources are said to be coherent if they emit continuous monochromatic waves of same wavelength, either in same or constant phase difference.

OR

Telescope Microscope

Construction Objective has large focal length and large aperture but eyepiece of short focal length and short aperture.

Objective has very short focal length and short aperture and eyepiece has short focal length and large aperture.( fe > fo)

Working It forms magnified image of distant object.

It forms magnified image of a small nearby object.

7 The use of AC voltage preferred over DC voltage because of the following:

i) The loss energy in transmitting the AC voltage over long distances with the help of step up transformer is negligible as compared to DC voltage.

ii) AC voltage can be stepped up and stepped down as per the requirement using a transformer.

8 We have focal length of convex lens,


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The focal length of the combination =–1 m = –100 cm.

As the focal length is in negative, the system will be diverging in nature.

9. We have, resistance of ammeter, RA = 0.80 ohm and maximum current across ammeter, IA = 1.0 A.

So, voltage across ammeter, V = .

Let the value of shunt be x.

(i) Resistance of ammeter with shunt, .

Current through ammeter, I = 5 A.

Thus, the shunt resistance is 0.2 ohm.


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

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

(b) Speed of EM wave can be given as the ratio of magnitude of electric field (E0) to the magnitude of magnetic field (B0),

Q11

Modulator: Since the frequency range of signal is quiet low and it is associated with very small amount of energy, it dies out very soon if transmitted as such. So, it is modulated by mixing with very high frequency waves called carrier waves. This is done by modulator power. Amplifier: Since the signal gets weaken after travelling through long distances it cannot be transmitted as such. Thus, we use a power amplifier to provide it necessary power before feeding the signal to the transmitting antenna.

.

Q12 Working of photo diode:

A junction diode made from light sensitive semi-conductor is called a photodiode.


Page 11

A photodiode is an electrical device used to detect and convert light into an energy signal through the use of a photo detector. It is a pn-junction whose function is controlled by the light allowed to fall on it. Suppose, the wavelength is such that the energy of a photon, hc/λ, is sufficient to break a valance bond. When such light falls on the junction, new hole-electron pairs are created. The number of charge carriers increases and hence the conductivity of the junction increases. If the junction is connected in some circuit, the current in the circuit is controlled by the intensity of the incident light.

OR

i. As electron concentration increases it is N – type semiconductor. ii. n2

2 = ne x na

na=

=(6 x 102)2 / (8 x 1012)

= 4.5 x 104/ m3

Q13 The two basic modes of communication are:

i) Point to point

ii) Broadcast:


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Q14 Let q be the charge on the charged capacitor.

Energy stored in it is given by:

U = q2/2C

When another uncharged similar capacitor is connected, then net capacitance of the system is given by :

C’ = 2C

The charge on the system remains constant. So, the energy store in the system is given by:

U’ = = q2/2C’ = = q2/4C

Thus the required ratio is given by: u’/u = (q2/4C)/ (q2/2C) = 1/2

OR


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(i)

Given,

As the electric field is only along the x-axis, so, flux will pass only through the cross-section of cylinder.

(ii) Using Gauss’s law:

Q15 T1/2 =0.693/λ = 0.693/1.386 = 0.5 day

N/ N0 = (1/2)t/T

6.25/100 =(1/2)t/T

(1/2)4 = (1/2)t/T

t/T=4 => t= 4T=4x 0.5=2 days

Q16 (a) Wave nature of radiation cannot explain the following:

(i) The instantaneous ejection of photoelectrons.

(ii) The existence of threshold frequency for a metal surface.


Page 14

(iii) The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency.

Thus, the photoelectric effect cannot be explained on the basis of wave nature of light.

(b) Photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based on particle nature of light. Its basic features are:

(i) In interaction with matter, radiation behaves as if it is made up of particles called photons.

(ii) Each photon has energy E (=hν) and momentum p (=hν/c), and speed c, the speed of light.

(iii) All photons of light of a particular frequency ν, or wavelength λ, have the same energy E (=hν=hc/λ) and momentum p (=hν/c=h/λ), whatever the intensity of radiation may be.

(iv) By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.

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

(vi) In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, number of photons may not be conserved.

Q17 (i) Magnetic moment M = 0.9 J/T τ = 0.063 J, Ɵ= 30°

We know = M × B = MB sin Ɵ

0.063 = 0.9 × B × sin 30°

(ii) Stable equilibrium is position of minimum energy.

Since

(iii) U = − M B cosƟ

Where, U is the energy stored or P.E. of the magnet inside magnetic field B.

So, when = 0, U = − MB is the minimum energy.B.Sesha Sai/KV, Embassy Of India, Kathmandu/PHY/XI Science.

Page 15

Thus, when and are parallel to each other bar magnet is in stable equilibrium

Q18 (i) Dynamic output resistance is given as:

(ii)

(iii)

Q19 (a)

(i) The potential energy is minimum at a distance of r0 = 0.8 fm. At this distance, force between nucleons is zero. For distances larger than 0.8 fm, negative P.E. goes on decreasing. The nuclear forces are attractive.

(ii) For distances less than 0.8 fm, negative P.E. decreases to zero and then becomes positive. The nuclear forces are repulsive.

(b) Properties of nuclear forces:

(i) Nuclear force is always attractive in nature.


Page 16

(ii) Nuclear force is a short range force.

Q20 (a) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of the two slits.

(b) Given that: Wavelength of the light beam, λ1 = 590 nm = 5.9 ×10-7 m

Wavelength of another light beam, λ2 = 596 nm = 5.96 ×10-7 m

Distance of the slits from the screen = D = 1.5 m

Distance between the two slits = a = 2 ×10-4 m

For the first secondary maxima,

OR

∴ Spacing between the positions of first secondary maxima of two sodium lines = 6.75 × 10-5m

Q21

As E0sinωt – L


Page 17

Q22. Let P be the axial point at a distance r from the centre of the dipole.

Electric potential at point P is V.

For a distant point r >> a,

The potential due to a dipole (at a large distance) falls off as 1/r2, but potential due to a single charge falls off as 1/r.

Q23 It is safer to be inside a car during thunderstorm because the car acts like a Faraday cage. The metal in the car will shield you from any external electric fields and thus prevent the lightning from traveling within the car.

(b) Awareness and Humanity

(c) Gratitude and obliged

(d) I once came across to a situation where a puppy was struck in the middle of a busy road during rain and was not able go cross due to heavy flow, so I quickly rushed and helped him.

Q24 a) (i) From the fringe width expression,

With the decrease in separation between two slits, ‘d’ the fringe width increases.


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(ii) For interference fringes to be seen

condition should be satisfied

Where, s = size of the source,

S = distance of the source from the plane of two slits.

As the source slit width increase, fringe pattern gets less and less sharp.

When the source slit is so wide the above condition does not satisfied and the interference pattern disappears.

(iii) The interference pattern due to different colour component of white light overlap. The central bright fringes for different colours are at the same position. Therefore central fringes are white. And on the either side of the central whit fringe coloured bands will appear.

The fringe closed on either side of central white fringe is red and the farthest will be blue. After a few fringes, no clear fringe pattern is seen.

(b) Intensity at a point is given by,

I = 4I’ cos2∅/2

Where, ∅ = phase difference

I’ = intensity produced by each one of the individual sources.

At central maxima, ∅ = 0,

I = I0 = 4I

At path difference


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Now, intensity at this point,

Hence proved.

OR

(a)

Consider a point P on the screen at which wavelets travelling in a direction making angle θ with CO are brought to focus by the lens. The wavelets from points A and B will have a path difference equal to BN.

From the right-angled ΔANB, we have

BN = AB sinθ

BN = a sinθ …(i)

Suppose BN = λ and θ = θ1

Then, the above equation gives

λ = a sin θ1

Such a point on the screen will be the position of first secondary minimum.

If BN = 2λ and θ= θ2, then

2λ = a sin θ2


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Such a point on the screen will be the position of second secondary minimum. In general, for nth minimum at point P,

If yn is the distance of the nth minimum from the centre of the screen, then from right-angled ΔCOP, we have

In case θn is small, sin θn ≈ tan θn

∴ Equations (iv) and (v) give

If and = 1′, then from equation (i), we have

Such a point on the screen will be the position of the first secondary maximum. Corresponding to path difference,

and = ′2, the second secondary maximum is produced

In general, for the nth maximum at point P,

If is the distance of nth maximum from the centre of the screen, then the angular position of the nth maximum is given by,


Page 21

In case θ’n is small,

sin θ’n ≈ tan θ’n

For n = 1

(From eq(vi), small angle approximation,

This angle is midway between two of the dark fringes. Divide the slit into three equal parts. If we take the first two third part of the slit the path difference between the two ends would be,

The first two third is divided into two halves which have path difference . The contribution due to these two halves is 180° out of phase and gets cancel. Only the remaining one third part of the slit contributes to the intensity at a point between the two minima which will be much weaker than the central maxima. Thus with increasing n the intensity of maxima gets weaker.

(b) As the number of point sources increases, their contribution towards intensity also increases. Intensity varies as square of the slit width. Thus, when the width of the slit is made double the original width, intensity will get four times of its original value.

Width of central maximum is given by,

Where, D = Distance between screen and slit,

λ = Wavelength of the light,

b = Size of slit.

So with the increase in size of slit the width of central maxima decreases. Hence, double the size of the slit would result in half the width of the central maxima


Page 22

Q25 (a) Working principle of Potentiometer:

When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.

Applications of Potentiometer for comparing emf’s of two cells:

The following Figure shows an application of the potentiometer to compare the emf of two cells of emf E1 and E2

E1, E2 are the emf of the two cells.

1, 2, 3 form a two way key.

When 1 and 3 are connected, E1 is connected to the galvanometer (G).

Jokey is moved to N1, which is at a distance l1 from A, to find the balancing length.

Applying loop rule to AN1G31A,

Φ l1 + 0 − E1 = 0 ..(1)

Where, Φ is the potential drop per unit length

Similarly, for E2 balanced against l2 (AN2),

Φ l2 + 0 −E2 = 0 ..(2)

From equations (1) and (2),

.. (3)

Thus we can compare the emf’s of any two sources. Generally, one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. The emf of the other cell is then calculated from Eq. (3).

(b) (i) The emf of the cell connected in main circuit may not be more than the emf of the primary cells whose emfs are to be compared.


Page 23

(ii) The positive ends of all cells are not connected to the same end of the wire.

OR

(a) Kirchhoff’s First Law − Junction Rule

The algebraic sum of the currents meeting at a point in an electrical circuit is always zero.

Let the currents be I1, I2I3, and I4

Convention:

Current towards the junction – positive Current away from the junction − negative

I3 + (− I1) + (− I2) + (− I4) = 0

Kirchhoff’s Second Law − Loop Rule: In a closed loop, the algebraic sum of the emfs is equal to the algebraic sum of the products of the resistances and current flowing through them.

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

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

Wheatstone Bridge:

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


Page 24

R1, R2, R3,and R4 are the four resistances.

Galvanometer (G) has a current Ig flowing through it at balanced condition, Ig = 0

Applying junction rule at B, ∴ I2 = I4

Applying junction rule at D, ∴ I1 = I3

Applying loop rule to closed loop ADBA,

Applying loop rule to closed loop CBDC,

From equations (1) and (2),

This is the required balanced condition of Wheatstone Bridge.

(b) Considering both the situations and writing them in the form of equations ,Let R' be the resistance per unit length of the potential meter wire,


Page 25

Q26 . (a)Cyclotron- Principle- A charged particle can be accelerated to very high energies by making it pass through a moderate electric field a number of times. This can be done with the help of a perpendicular magnetic field which throws the charged particle into a circular motion, the frequency of which does not depend on the speed of the particle and radius of the circular orbit. 1 mark

1 mark

Working-Suppose a positive ion, say a proton, enters the gap between the two dees and find dee D 1 to be negative. It gets accelerated towards D1. As it enters the dee D1, it does not experience any electric field due to shielding effect of the metallic dee. The perpendicular magnetic field throws it into a circular path. At the instant the proton comes out of the dee D1, it finds D1 positive and D2 negative. It now gets accelerated towards D2. It moves faster through D2

describing a semi circular path. Thus , if the frequency of the applied voltage is kept exactlythe same as the frequency of revolution of the proton, then every time the proton reaches the gap between the dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This is called cyclotron resonance condition. 2 mark

Limitations-1. It can not be used to accelerate electrons. ½ + ½

2.Neutrons, being electrically neutral, can not be accelerated in a cyclotron.

OR


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1 mark

Principle-Its principle is based on the fact that when a current carrying conductor is placed in magnetic field, it experience a torque. 1 mark

Working-As the field is radial, the plane of coil always remains parallel to the field. When a current flows through the coil, a torque acts on it. It is given as

=NIBA

This torque deflects the coil through an angle α. A restoring torque is set up in the coil due to the elasticity of the

springs such that or α, where k is the torsion constant of the spring.

In equilibrium

Restoring torque= deflecting torque => Kα=NIBA => α=NBA/K

Thus the deflection produced in the galvanometer coil is proportional to the current flowing through it. 2 mark

(i) Function of uniform radial field- It provides maximum torque irrespective of orientation(ii) Function of soft iron core:It increases the strength of the magnetic field and hence increase the sensitivity of the galvanometer. ½ + ½


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