SUPPORT STUDY

MATERIAL

XII Physics

Support Material, Study Notes and VBQ

UNIT-1ELECTROSTATICS

1. Where the energy of capacitor does resides?2. Do electrons tend to go to region of low or high potential?3. What is the net charge on the charged capacitor?4. A Gaussian surface encloses an electric dipole within it. What is the total flux across

sphere?5. Find the dimension of 1/2εoE2.6. In a certain l m3 of space, electric potential is found to be V Volt throughout. What is

the electric field in this Region?7. If Coulomb law involves 1/r3 instead of 1/r2 dependence, would Gauss law be still

true? 8. An electrostatic field line can’t be discontinuous, why? 9. The given graph shows that the variation of charge versus potential difference V for the two capacitors C1 & C2. The two capacitors have same plate separation but the plate area of C2 is doubled than that of C1.Which of the line in the graph corresponds to C1 &C2 and why?

C1

V C2

Q 10. Three charges, each equal to +2C are placed at the corners of an equilateral triangle. If the force between any two charges be F, then what will be the net force on either Charge? 11. A point charge q is placed at O as shown in the figure.

Is VP-VQ +ve or –ve when (i) q>0, (ii) q<0? Justify your answer.

12. An electric dipole of dipole moment 20 X 10-6 C.m is enclosed by a closed surface. What is the net flux coming out of the surface? 13. Why does the electric field inside a dielectric decrease when it is placed in an external electric field? 14. Write the magnitude and direction of electric field intensity due to an electric dipole of length 2a at the mid point of the line joining the two charges.

15. A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in fig. What is the electric field inside the emptied sphere? 16. A charged particle is free to move in an electric field. Will it always move along an electric line of

force? 17. If V (=q/4πεor) is the potential at a distance r due to a point charge q, then determine the electric field due to a point charge q, at a distance r. 18. Can electric potential at any point in space be zero while intensity of electric field at that point is not

zero?19. Devise an arrangement of three point charges separated by finite distances that has zero electric

potential energy.20. Each of the uncharged capacitor in the fig. Has a capacitance of 25μF. What charge shall flow through the meter M when the switch S is Closed?21. Charge of 2C is placed at the centre of a cube of volume 8 cm3. What is the electric flux passing through one face?22. A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It approaches Q up to a closest distance r and then returns. If q were given a speed 2v, then find the closest distance of approach.

23. Two capacitors of capacitance 6µF and 12µF are connected in series with the battery. The voltage across the 6µF capacitor is 2 volt .Compute the total battery voltage. 24. A parallel plate capacitor with air between the plates has a capacitance of 8 pF . The separation between the plates is now reduced by half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of parallel plate capacitor in second case. 25. Five identical capacitors, each of capacitance C are connected between points X and Y as shown in the figure. If the equivalent capacitance of the combination between X and Y is 5µF. Calculate the capacitance of each capacitor. 26. An uncharged capacitor is connected to a battery. Show that half of the energy supplied by the battery is lost as heat while charging the capacitor. 27. What is the angle between the electric dipole moment and electric field strength due to it on the equilateral line? 28. Find the equivalent capacitance between A & B, if capacitance of each capacitor is C. 29. Eight identically charged drops are joined to form bigger drop. By what factor the charge and

potential change?

30. A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. What is the kinetic energy of this charge at x = 4m?31. Two identical metal plates are given the charges Q1 and Q2 (Q2 < Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C then what is the potential difference between them?32. Three charges Q, +q and +q are placed at the vertices of a right angle isosceles triangle as shown. Find the magnitude of Q for which net electrostatic energy of the configuration is zero.33. An infinite number of charges each having charge ‘q’ along x-axis at x=1 ,x=2, x=4 ,x=8 an so on.

Find the electric field at x=0 due to these charges. 34. A charge Q is distributed over the two concentric hollow spheres of radii ‘r’ and ‘R’ (R>r) such that the surface densities are equal. Find the potential at the common centre. 35. An electric dipole is held in an uniform electric field. Using suitable diagram, show that it doesn’t undergo any translatory motion, and (ii)Derive an expression for torque acting on it and specify its direction. 36. The field potential inside a charged ball depends only on the distance from its centre as

V=ar2+b,where a and b are constants. Find the space charge distribution ρ(r) inside the ball.

CHAPTER-2CURRENT ELECTRICITY

1. Magnesium is used for making standard resistors, why?2. The sequence of bands marked on a carbon resistor are: Red, Red, Red, Silver. Write

the value of resistance with tolerance.3. A wire of resistively ρ is stretched to three its initial length, what will be its new resistively.4. If p.d.v applied across a conductor is increased to 2v, how will the draft velocity of

the electrons change?5. A 10Ω thick wire is stretched so that its length becomes three times. Assuming that there is no

change in its density on stretching. Calculate the resistance of new wire.6. You are given 8 Ω resistor. What length of wire of resistance 120 Ωm-1 should be

joined in parallel with it to get a value of 6 Ω ?7. Three resistance 3Ω,6Ω and 9Ω are connected to a battery. In which of them will the power

dissipation be maximum if a) They are all connected in parallel

b) They are all connected in series Give reason.8. A silver wire has a resistance of 2.1Ω at 27.5˚c and a resistance of 2.7Ω at 100˚c. determine the temperature coeff. of resistivity of silver.9. Give any two applications super conductors.10. Two wire of equal length one copper and manganin have same resistance , which wire is

thicker?.11. Why manganin is used for making standard resistor?12. A copper wire of resistivity ρ is stretched to reduce its diameter to half of its previous value .What

will be the new resistances?13. The variation of potential difference with length incase of two potentiometres A and B is given

below.Which of the two is more sensitive.

A B Potential difference

Length

14. If the length of the wire conductor is doubled by stretching it , keeping potential difference

constant by what factor the drift speed of the electron changed.15. If the temperature of the conductor increases, how does the relaxation time of electron changes.16. A heater joined in series with the 60W bulb .With change of bulb with 100 W in the circuit, the

rate heat produce by the heater will more or less or remain same.17. What will be the change in the resistance of the circular wire , when its radius is halved and length

is reduced by ¼ th of original length.18. Two 120V light bulbs , one of 25W and another of 200W are connected in series . One bulb burnt

out almost instantaneously ?.Which one was burnt and why?.19. A given copper wire is stretched to reduce its diameter is half of its original value.What will the

new resistance?.20. A student has two wire of iron and copper of equal length and diameter.He first joins two wires in

series and pass electric current through the combination which increases gradually.After that he joins two wires in parallel and repeats the process of passing current.Which wire will glow first in

each case?

21. A cylindrical metallic wire is stretched to increase its length by 5% . Calculate the percentage

change in resistances.22. A wire of resistance 4R is bend in the form of circle .What is the effective resistance between the

ends of diameter?.23. Two wires A and B have same lengths and material , have their cross sectional areas 1:4 , what

would be the ratio of heat produced in these wires when the voltage across each is constant.24. Two bulbs whose resistance are in the ratio of 1:2 are connected in parallel to a source of constant

voltage. What will be the ratio of power dissipation in these?25. Total resistance of the circuit is R/3 in which three identical resistors are connected in parallel.

Find the value of each resistance?.UNIT -3

MAGNETIC EFFECTS OF CURRENT & MAGNETISM

1. Suppose a helical spring is suspended from the roof of a room and very small weight is attached to its lower end what will happen to the spring when a current is passed through it?Give reason to support your answer?Ans Spring will contract due to the magnetic field produced by the turns of the coil and the weights will be lifted up.

2. One alpha particle and a deuteron entered perpendicularly in a uniform magnetic field with same velocity. Which one follow the greater circle?Ans: As we know for a charge particle moving in a magnetic field, the radius of circular path: r = mv/qBAs both the particles have same velocity therefore

rα/rd = mα qd/md qα

Thus both particles will follow the same

3. Out of Voltmeter and Millivoltmeter, which has the higher resistance? Ans: We know the resistance connected to galvanometer to convert it into voltmeter is

R = (V / Ig) - GSo if R is higher, range of V will also be higher, so a Voltmeter has the higher resistance.

4. Proton is moving along the axis of a solenoid carrying current of 2 A and 50 number of turns per unit length. What will be the force acting on the particle.Ans: As the magnetic field produced by solenoid is always along its axis, so direction of velocity of proton is along the direction of field, therefore

F = qvB Sin 0 = 0

5. Out of Ammeter and Milliammeter, which has the higher resistance? Ans: We know the resistance connected to galvanometer to convert it into ammeter is

i iiiiiiIIT

UYTHT

S =(Ig/( I – Ig))xGSo for higher resistance, the range of I should be small, therefore milliammeter has the higher resistance.

6. What will be the direction of magnetic field at point O

Ans: The magnetic field due to AB and EF is as the direction of length vector is along the radius vector,

Also the magnetic field due to BCE and BDE are equal opposite and equal so they cancel the effect of each other. So the net magnetic field at O is 0.

7. Can a Moving Coil Galvanometer can be used to detect an A.C. in a circuit .Give reason.Ans: As MCG detect only the average value of current and the average value of AC for a complete cycle is zero. Therefore MCG can not detect AC in a circuit.

8. Two wires of equal length are bent in the form of two loops. One loop is square whereas the other is circular. These are suspended in same magnetic field and same current is passed through them. Explain with reason which will experience greater torque?Ans: For a given length, the circle has the greatest area, as

τ = NIABi.e. torque is proportional to area, so circular current loop experiences the greater torque.

9. The pole of a magnet is brought near to a stationary charge. What will be the force experienced by pole?Ans: The force will be zero as the stationary charge particle does not produce any force.

10. A charge particle moving in a magnetic field penetrates a layer of lead and thereby losses half of its kinetic energy. How does the radius of curvature of its path change?

Ans: r = mv/qB ----------- (i) Also ------------ (ii)By equ (i) and equ (ii)

As the radius is proportional to square root of kinetic energy, so if the kinetic energy is halved the radius become √1/2 times of its initial value.

11. A Voltmeter, an ammeter and a resistance are connected in series with a battery. There is some deflection in voltmeter but the deflection of ammeter is zero. Explain why?Ans: As the resistance of V is very high so the effective resistance of circuit become very high, so the current flows in circuit is extremely low therefore the deflection is almost zero, while the V measures the potential difference between the points so it shows the reading due to battery.

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A Current ‘I’ flows along the length of an infinitely long straight thin walled pipe. What is the magnetic field at any point on the axis of pipe?Ans: Zero.

The Earth’s core contains iron but geologists do not regard this as a source of Magnetic Field, Why?Ans: Temperature in the core of earth is higher than Curie temperature of Iron.

Is the Resistance of Voltmeter larger than or smaller than the resistance of Galvanometer from which it is converted.Ans: Larger

A Magnetic Field dipole placed in a Magnetic Field experiences a net force. What can you say about the Nature of Magnetic Field? Ans: Non-uniform.

Earth’s Magnetic Field does not affect working of moving Coil Galvanometer. Why?Ans: Magnitude of Earth’s magnetic field is much smaller than magnitude of the field produced by poles of galvanometer.

Which type of Magnetism exists in all substances?Ans: Diamagnetism.

For what orientation P.E. of a Magnetic dipole placed in uniform Magnetic Field minimum?Ans: θ = 0 (Dipole is parallel to field.)

How does a ferromagnetic material change its Magnetic properties if it is heated beyond its curie temperature? Ans: Becomes Paramagnetic.

A bar magnet is cut into two pieces, along its length. How will its pole strength be affected?Ans: M1 = M/2, M=M/2

What is the work done by a magnetic force, in displacing a charged particle?Ans: Zero.

22. What is the net magnetic flux from a north (or south) pole of a magnet (dipole) ?Ans: Nil, because the number of magnetic lines entering the surface is equal to the number of lines going out of it.

23 An unmagnetised ferromagnetic substance is magnetized. Given figure shows the B-H curve. Identify the stage of saturation ,reverse region and irreversible region

stage of saturation is B to C . Reversible region O to A and A to B is irreversible region.

24. What is the magneticfield at the centre of the following circular coils carrying current I?

μoI+ μoI_ μoI _ μoI_ 2R 2П R 2R 2П R

25. Two long straight wires are set parallel to each other. Each carries a current I in the same direction and the separation between them is 2r. What is the intensity of the magnetic field midway between them?Ans: The fields of the two wires will be in the opposite directions at the midway point. B =B1 –B2 =µ0I/2πr -µ0I/2 π r =0

26. A proton is about 1840 times heavier than an electron. What will be its kinetic energy when it is accelerated by a potential difference of 1KV?Ans: Kinetic Energy gained=qv=ex1Kv=1keV

27. A circular loop of radius R carrying current I ,lies in X-Y plane with its centre at origin.What is the total magnetic flux through X-Y plane?Ans:φ =B.A = µo I Πr 2 2r = µOIRΠ 2 ie φ α R

28. A hypothetical bar magnet is cut into two equal pieces and placed as shown in the figure. What is the magnetic moment of this arrangement?

magnetic moment of the arrangement=√ M12+M2

2 +2M1M2COSθ = √M2/4+M2/4+2M/2xM/2 cos90 =√ 2M2/4 =M/√2

29. A circular current carrying coil has a radius R. What is the distance from the centre of the coil on its axis where the magnetic field is 1/8 th of its value at the centre?Ans: Baxial =1/8 Bcentre

µ0 IR 2 = 1x µ0 I 2(R2+r2)3/2 8x2R (R2+r2)3/2 =8R2

R2+r2 = 4R2

Hence , r =√ R

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A magnetic needle suspended freely in a uniform magnetic field experiences torque but no net force. A nail made up of iron kept near a bar magnet experience a force of attraction and torque .Give reason.Ans- Due to the non uniform magnetic field of bar magnet nail experience torque and translatory force.

What is the work done by a magnetic field on moving a charge? Give reason.Ans- W= FScosθ = FScos 90=0

A particle with charge q moving with velocity v in the plane of the paper enters a uniform magnetic field B acting perpendicular to the plane of the paper. Deduce an expression for the time period of the charge as it moves in a circular path in the field .Why does the kinetic energy of the charge not change while moving in the magnetic field. Ans- particle moves in circular path Bqv = mv2/r r = mv/Bq Time period T = 2 Π r/v =2 m/Bq

A solenoid of length 0.6m has a radius of 1cm and is made up of 600 turns.It carries a current of 5A.What is the magnetic field inside and at ends of solenoid.?Ans- (i)At the centre N=1000, B = µ0 ni = 4 Π x 10-7 x 1000 x 5 = 6.2 x 10-3 T

(ii) At the ends B = ½ µ0 ni = 3.1 x 10 -3 T

An element dl = dx i is placed at the origin and carries a large current I = 10A.What is the magnetic field on the y axis at a distance of 0.5m,

dB= µ0 Idl Sin θ / 4 Π r2

=10-7 x 10 x 10-2 /25 x 10-2

=4 x 10 -8 T

Direction of dB is in +Z direction

You are given a copper wire carrying current I of length L. Now the wire is turned into circular coil. Find the number of turns in the coil so that the torque at the centre of the coil is to maximum.

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Ans: Let the number of turns be = n Radius = rLength=lLength of the wire = circumference of n turns of coil L = n x 2 Π r

r = L/2 Π rMaximum torque = nIBA = nIB Π r2

= nIB Π (L/2 Π n) 2

α 1/nFor maximum torque n should be minimum i.e. n = 1.

What is the magnetic field produced at the centre of curvature of an arc of wire of radius r carrying current I subtends an angle Π /2radians at its centre. Ans: B1 = B x θ /2 Π = (µ0 I Π /2r) x (2 Π x2)

B1= µ0 I/8r

If B is the magnetic field produced at the centre of a circular coil of one turn of length L carrying current I then what is the magnetic field at the centre of the same coil which is made into 10 turns? Ans : B1 = n2 B = 102 B = 100 B.

A copper wire is bent into a square of each side 6cm.If a current of 2A is passed through a wire what is the magnetic field at the centre of the square?Ans: B1 = 4 x µ0 I/4 Π a/2 ( sin 45 + sin 45 ) = 4 x µ0 2/4 Π x3 ( 1/ 1.414 + 1/ 1.414 ) = 2 µ0 /3 Π ( 1/ 1.414 + 1/ 1.414 ) T

Find the magnetic moment of a wire of length l carrying current I bent in the form of a circle. Ans: M = IA = I x Π r2 But l = 2 Π r , i.e r = l/2 Π

M= Il2/4 Π

When current is flowing through two parallel conductors in the same direction they attract while two beams of electrons moving in the same direction repel each other. Why? Ans- Two conductors carrying current in same direction produce magnetic field and hence they attract.While two electron beams moving in the same directions repel due to its electric field (electrostatic force)

Draw diagrams to show behavior of magnetic field lines near a bar of (i) Alluminium (ii) copper and (iii) mercury cooled to a very low temperature 4.2 K Ans- (i)Alluminium --- Paramagnetic

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(ii) Copper and mercury = diamagnetic

The hysteresis loss for a sample of 6 kg is 150 J/M2/cycle. If the density of iron is 7500 kg/m3, calculate the energy loss per hour at 40cycle. Ans: Volume of sample = mass/density = 6/7500 m3

Energy loss/cycle = energy loss per volume/cycle x( volume)=150x6/7500 Energy loss/sec = 150x6x40/7500 Energy loss /hour = 150x6x40x60x60/7500 J

=1.728 x 104 J

A current carrying solenoid of 100 turns has an area of cross section 10-4 m2 .When suspended freely through its centre, it can turn in a horizontal plane .what is the magnetic moment of the solenoid for a current of 5A.Also calculate the net force and torque on solenoid if a uniform horizontal field of 10x10-2 T is set up at an angle of 30 degree with axis of solenoid when it is carrying the same current. Ans: M = nIA = 100 x 5 x 10-4 =500 x 10-4 J/TNet force = 0Torque = MB sinθ = 5x10-2 x 0.1x sin30 = 25 x 10-4 NmH = R cos ð = .4x ½ = 0.2 G

Two concentric circular coils A and B of radii 10 cm and 6 cm respectively, lie in the same vertical plane containing the north to south direction. coil A has 30 turns and carries a current of 10 A . Coil B has 40 turns and carries a current of 15 A .the sense of the current in A is anticlockwise and clockwise in B for an observer looking at the coils facing west. Give the magnitude and direction of net magnetic fieldAns: B due to the coils at the centre.

Coil A –R1 = 0.1m , n1=30,I1=10A

B1 = µ0 n1I1/2r1 = 6 Π x 10-4 T directed towards east Coil B- R2 = 0.6 m , n2=40, I2= 15 A B2 = µ0 n2I2/2r2 =2 Π x 10 – T directed towards west

Net field B = B2 - B1 = (20 Π - 6 Π ) 10-4 T towards west

The vertical component of earth’s magnetic field at a given place is 3 times its horizontal component. If the total intensity of earth’s magnetic field at a place is 0.4 G , find the value of horizontal component of earths field and angle of dip. Ans: Tan ð = V/H = 3

ð = 60

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As V = √3H andB2 = V2 + H2 = 3H2 + H2 = 4H2

(0.4)2 = 4H2 thereforeH = 0.2 G

An electron traveling west to east enters a chamber having a uniform electrostatic field in north to south direction.Specify the direction in which the uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.Ans- Due to the electrostatic field electron will be deflected towards north. To keep it neutralized the magnetic force should deflect it towards south .For this purpose the magnetic field is to be applied perpendicular to the plane of the paper inward i.e vertically downward.

A straight horizontal conducting rod of length 0.5 m and mass 50 g is suspended by two vertical wires at its ends.A current of 5A is set up in the rod sdthrough the wires.(i) What magbnetric field should be set up normal to the conductor in order that the tension in the wires is zero?(ii)What will be the tension in the wire if the direction of current is reversed keeping the magbetic field same as before?(neglect the mass if wure abd taje g=10m/s2 ) Ans: (i) Magnetic force = weight IlB sinθ = mg

IlB = mg (θ =90)B = mg/Il = 500x10-3/2.5 = 200 x 10-3 T

(ii)When the direction of fidl is reversed an additional force which was equal to weight of rod will be acting on the wires. Net tension in wires = mg + mg = 2 mg =2 x 50 x 10-3 x 10 = 1 N

A circular coil of 20 turns and radius 10cm is placed in a uniform magnetic field of 0.d10T normal to the plane of the coil.If the current in the coil is 5a,What is the (i)Total torque on the coil (ii) total force on the coil (iii) average dsforce on each electron in the coil due to the magnetic field.(coil is made of copper, A= 10-5 m2 ,free electron density in copper is 1029 /m3) Ans: N= 20, r= .1 m, B = .1T, I = 5A

(i)Total torque = nIBAsin θ (θ =0)

= 0 (ii)Total force on the coil is = 0, because force being equal and opposite and cancel eachother (iii) Average force on electron = f = evB

V = I/neA F = IB/nA = 5x 0.1/1029 x 10-5

=5 x 10 -25 N

A Rowland ring of mean radius 15 cm has 3500 turns of wore wound on a ferromagnetic core of relative permeability 800.What is the magnetic field B in the core for a magnetizing current of 1.2 A? Ans: N=3500, r= 15 x 10-2m

n = N/2 Π r = 3500/2x3.14 x .15 = 3715.5 per m

µ0 = 4 Π x 10-7 x TmA-1, µr = 800 , I = 1.2 B = µ0 µr nI = 4.48 T

50 A straight wire of mass 200g and the length 1.5m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field B. What is the magnitude of B in tesla?Ans: For equilibrium of the wire in mid – air, weight of the wire = force exerted by magnetic field Mg =IlB sin900

B = mg/Il = 200x 10 -3 x9.8 = 0.65T 2x1.5

51. A rigid circular loop of radius r and mass m lies in the x-y plane of a flat table and has a current I flowing in it. At this particular place the earth’s magnetic field is B = Bxi +Bzk. What is the value of I, so that loop starts tilting?Ans: M = IA =I r2k B = Bxi +Bzk Τ = M x B = ( Iπ r2k) x(Bxi + Bzk) = Iπ R2BXkxi = I r2BXjTorque due to the weight of the loop = mgr I r2BX = mgr Hence I = mg/ПrBX

52. In an ammeter, 10% of main current is passing through the galvanometer. If the resistance of the galvanometer is G, then what is the shunt resistance in ohms?

Ans: Ig=10% of I=0.1I

S=IgxG/I-Ig=0.1IG/I-0.1I=G/9

53. The two rails of arailway track insulated from each other and the ground is connected to a milli voltmeter. What is the reading g of the millivolmeter when the train passes at aspeed 180km/hr along the track, given that the vertical component of earth”s magnetic field is 0.2x10-4T and rails are separated by 1me = Blv = 0.2x10-4x1x180x5/18 = 10-3V = 1Mv

54 A charged particle moving in a magnetic field penetrates a layer of lead and there by looses half of its kinetic energy.How does the radius of curvature of its path changes?Radius r= mV/qBAns: If is the kinetic energy of the particle,then its momentum, p = mv√2mEk

Radius , r =√2mE /Qb rα√Ek

This shows that K.E is halved, the radius is reduced to 1/√2 times its initial value.

55 The velocities of two α particles X and Y entering in an uniform magnetic field are in the ratio 2:1.On entering the field ,they move in different circular paths .Give the ratio of the radii of their paths?Ans: qvB = mv2/r, r = mv/qB, rα v rx/ry = vx/vy = 2/1 = 2

56 In an exercise to increase current sensitivity of a galvanometer by 25 % , its resisitance is increased by 1.5 times . How does the voltage sensititvity of the galvanometer be affected. Ans: I s ‘ = I s + 25/100= 125/100 = 5/4 I s --------- 1 R’ = 1.5 R ----------- 2 V s + I s / R & Vs “ = Is’/R’ =5/4 I s/1.5 R + 5/6 V s % decrease in voltage sensitivity = (1-Vs’/V s) X 100 (1-5/6) X 100 = 16.7%

UNIT IVELECTROMAGNETIC INDUCTION & ALTERNATING CURRNT

1. Three identical coils A, B and C are placed with their planes parallel to one another. Coils A and C carry current as shown. Coil B and C are fixed. The coil A is moved towards B with uniform motion. Is there any induced current in B?ANS: - YES

2. Two coils are being moved out of magnetic field- one coil is moved rapidly and the other slowly. In which case is more work done and why?ANS: - THE ONE WHICH IS MOVED RAPIDLY

3. The figure shows two identical rectangular loops (1) and (2), placed on a table along with a straight line current carrying conductor between them. (i) What will be the directions of the induced currents in the loops when they are pulled away from the conductor with same velocity? (ii) Will the e.m.f. induced in the two loops be equal? Justify your answer.ANS: - (i) CLOCKWISE IN LOOP 1 AND ANTICLOCKWISE IN LOOP 2

(ii) EMF INDUCED IS MORE IN LOOP 2 THAN IN LOOP 1

4. 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.Ans: - ANTICLOCKWISE

5. A bar magnet M is dropped so that it falls vertically through the coil C. The graph obtained for voltage produced across the coil vs time is shown in figure (b). (i) Explain the shape of the graph. (ii) Why is the negative peak longer than the positive peak?

ANS: - (i) As magnet approaches the coil the rate of change of magnetic flux increases and the induced emf also increases. As soon as one pole of magnet enters into the coil the emf decreases due to the other pole effect and also the induced emf polarity reverses due to the same reason.

(ii) The longer peak is due to increase in the rate of change of magnetic flux as the magnet comes out of the coil.

6. A coil A is connected to a voltmeter V and the other coil B to an alternating current source D. If a large copper sheet C is placed between the two coils, how does the induced e.m.f. in the coil A change due to current in coil B?ANS: - DECREASES

7. A cylindrical bar magnet is kept along the axis of a circular coil, when the magnet is rotated (a) about its own axis, and (b) about an axis perpendicular to the length of the magnet, in which case the induced emf will be more?ANS: - IN CASE OF (b) IT IS MORE

8. How does the self inductance of an air core coil change, when (i) the number of turns in the coil is decreased, (ii) an iron rod is introduced in the coil? A copper coil L wound on a soft iron core and a lamp B are connected to a battery E through a tapping key K. When the key is suddenly opened, the lamp flashes for an instant to much greater brightness. Explain.ANS: - (i) DECREASES (ii) INCREASES. THERE IS AN OPPOSITION FOR THE CURRENT IN COIL DUE TO SELF INDUCTION SO THE BULB GETS MORE ELECTRIC CURRENT INITIALLY.

9. How is the mutual inductance of a pair of coils affected when separation between the coils is increased? The number of turns of each coil is increased? A thin iron sheet is placed between the two coils, other factors remaining the same? Explain your answer in each case.ANS: - DECREASES, INCREASES, INCREASES.

10. A rectangular wire frame, shown below, is placed in a uniform magnetic field directed upward and normal to the plane of the paper. The part AB is connected to a spring. The spring is stretched and released when the wire AB has come to the position A’B’ (t = 0). How would the induced emf vary with time? Neglect damping.ANS: - THE RATE OF CHANGE OF AREA IS MORE INITIALLY AND DECREASES WITH TIME AND SO THE INDUCED EMF.11. Why does metallic piece become very hot when it is surrounded by coil carrying high frequency alternating current?

ANS: - HIGH FREQUENCY AC PRODUCES CHANGING MAGNETIC FLUX AND THE LARGE EDDY CURRENTS PRODUCE HEAT.

12. Three students X, Y, and Z performed an experiment for studying the variation of alternating current with angular frequency in a series LCR circuit and obtained the graphs as shown. They all used a.c sources of the same r.m.s. value and inductances of the same value. What can we (qualitatively) conclude about the (i) capacitance value (ii) resistance values Used by them? In which case will the quality factor be maximum? What can we conclude about nature of the impedance of the set up at frequency wo?ANS: - (i) DECREASES FROM X TO Z (ii) DECREASES FROM X TO Z

IN CASE OF X QUALITY FACTOR IS MORE, IMPEDANCE DECREASES FROM X TO Z

13. In the circuit shown below, R represents an electric bulb. If the frequency of the supply is doubled, how should the values of C and L be changed so that glow in the bulb remains unchanged?ANS: - L SHOULD DECREASE AND C SHOULD INCREASE

14. An air cored coil L and a bulb B are connected in series to the mains as shows in the given figure: The bulb glows with some brightness. How would the glow of the bulb change if an iron rod is inserted in the coil? Give reasons in support of your answer.ANS: - DECREASES

15. When a circuit element ‘X’ is connected across an a.c. source, a current of √2A flows through it and this current is in phase with the applied voltage. When another element ‘Y’ is connected across the same a.c. source, the same current flows in the circuit but it leads the voltage by π/2 radians.(i) Name the circuit elements X and Y. (ii) Find the current that flows in the circuit when the series combination of X and Y is connected across the same a.c. voltage. ANS: - (i) X IS RESISTOR (ii) Y IS CAPACITOR16. Fig shows a light bulb (B) and iron cored inductor connected to a DC battery through a switch (S). (i) What will one observe when switch (S) is closed? (ii) How will the glow of the bulb change when the battery is replaced by an ac source of rms voltage equal to the voltage of DC battery? Justify your answer in each case.

ANS: - (i) BRIGHTNESS OF THE BULB INCREASES SLOWLY (ii) BRIGHTNESS REMAINS SAME

17. A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. (a) Obtain the current amplitude and rms values. (b) Obtain the rms values of potential drops across each element. (c) What is the average power transferred to the inductor? (d) What is the average power transferred to the capacitor? (e) What is the total average power absorbed by the circuit? ANS: - (a) 8.24A, 11.7A (b) VL=207V, VC=437V (c) zero (d) zero (e) zero.

18. A series LCR-circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply. (a) What is the source frequency for which current amplitude is maximum? Obtain this maximum value. (b) What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of this maximum power. (c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies? (d) What is the Q-factor of the given circuit? ANS: - (a) 4167 rad s-1 , 1.41A (b) 2300 W (c) 648Hz, 678Hz, I0=10A(d)21.7

19. An LC-circuit contains a 20 mH inductor and a 50 μF 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 the LC-oscillations? (b) What is the natural frequency of the circuit? (c) At what times is the energy stored (i) Completely electrical (i.e., stored in the capacitor)? (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?ANS: - (a)1 J,(b)159Hz (c)electrical at t =

20. Define self-inductance in terms of work done against the induced emf.

ANS: - L = ; Self-inductance is defined as double the work done against the induced emf in producing unit current in the coil itself

21. A circuit with a vertical copper wire bends as shown supports a small wooden piece W which floats in mercury. What do you expect when key is closed and current flows through the circuit? ANS: - The woodden block sinks when current flows through the circuit, as parallel wires carrying currents in the opposite directions repel. The given wave form shows the input current of a

transformer.

22. Draw the wave form of out put current. Substantiate your answer ANS: - 1800 phase difference due to Lenz’s law

23. An electron beam is deflected in a given field. Identify whether it is an electric field or a magnetic field in the following cases? (i) The trajectory of the beam is a parabola and its K.E changes.

(ii) The trajectory of the beam is circular and its K.E. remains the same. Justify your answer.

24. A resting electron near a stationery bar magnet does not set into motion. But a moving magnet near an electron set it into motion. Why?

25. An irregularly shaped flexible current carrying loop when placed in an external magnetic field will assume a circular shape. Give reason

26. Alpha particles (m = 6.68 X 10-27 Kg., q = +2e) accelerated through a potential difference V to 2 kV, enter a magnetic field B = 0.2 T perpendicular to their direction of motion. Calculate the radius of their path.

ANS: -

27. The above figure shows a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless track near the solenoid, the axis of the ring being along the axis of the solenoid. What will happen to the ring as the switch is closed? Justify your answer.ANS: - THE RING MOVESAWAY FROM THE SOLENOID

28. A particle with charge ‘q’ and mass ‘m’ is shot with kinetic energy K into the region between two plates as shown in the figure. If the magnetic field between the plates is B and as shown, how large must B be if the particle is to miss collision with the opposite plate? ANS: - Just to miss the opposite plate, the particle must move in a circular path with radius d so that Bqv = mv2/d, B = (2mK)1/2/(qd)

29. For the circuit shown below, find the magnitude and direction the force on wire AC, wire BC and wire AB. Also show that net force is zero.

30. A bar PQ of mass M is suspended by two wires as shown below. Assume that a uniform magnetic field B is directed into the page. Find the tension in each supporting wire when the current through the bar is I.

ANS: -According to the Flemings Right hand Rule, the magnetic force ILB is directed upward. Equilibrium in the vertical direction yields 2T + ILB = Mg, so that T = (Mg – ILB)/2

31. A bar of mass M is suspended by two springs as shown below. Assume that a magnetic field B is directed out of the page. Each spring has a spring constant K. Describe the bar’s displacement when a current I is sent through it in the direction shown.

ANS: -Due to Flemings Right hand Rule the magnetic force ILB is directed downward. This constant force shifts the equilibrium position downward by a displacement = (ILB)/2K32. An equilateral triangle is formed from a piece of uniform resistance wire. Current is fed into one corner and led out of the other as detailed in the figure below. Show that the current flowing through the sides of the triangle produces no magnetic field at its centre ‘O’ (the intersection of the medians).

ANS: - Wires A and B are in series. IA = IB = I/3, IC = 2I/3.Wire C makes a contribution to the field at O whose magnitude is twice that of A or B. By Flemings Right hand Rule, directions of field due to Wire A and B are directed down into the page. That due to wire C is upward. Net field at O is zero

33. In the following figure, the rectangular loop of wire is being pulled to the right, away from the long straight wire through which a steady current i flows upward as shown. Does the

current induced in the loop flow in the clockwise sense or in the counter clockwise sense ? Justify

ANS: -Due to Lenz’s law, the magnetic field produced by the induced current must counteract the decrease in flux and hence it must be directed into the plane of the figure (within the loop).So the induced current must be clock –wise.

34. Determine the separate effects on the induced emf of a generator if (a) the flux per pole is doubled, and (b) the speed of the armature is doubled. ANS: - In both the cases the induced emf doubles

35. An electromagnet has stored 648 J of magnetic energy when a current of 9A exists in its coils. What average emf is induced if the current is reduced to zero in 0.45 s?

ANS: - E = L = 16 H and e = 320 V

36. A 40 Ohm resistor is connected across a 15 V variable frequency electronic oscillator. Find the current through the resistor when the frequency is (a) 100 Hz and (b) 100 kHz. What is the current if the 40 Ohm resistor is replaced by a 2 mH inductor? ANS: -With resistor, current is same both for 100 Hz and 100 kHz. With inductor, the current is 11.9 A and 11.9 mA respectively

37. The axes of two magnets are collinear. One has poles of strength 80 Am separated by 125 mm, and the second has a magnetic moment of 12 A-m2 with poles of strength 160 Am. Find the attractive force between the magnets if the north pole of one is 45 mm from the south pole of the second.

ANS: - Resultant force F = 2 attractive forces + 2 repulsive forces = 520 mN (attractive)

38. A coil A is connected to voltmeter V and the other coil B to an alternating current source D. If a large copper sheet CC is placed between the two coils, how does the induced e.m.f in the coil A change due to current in coil B?

ANS: - The induced e.m.f in coil A decreases due to large copper plate introduced between the two coils as Cu is diamagnetic material

39. A magnet is moved in the direction indicated by an arrow between two coils A and B as shown below. Suggest the direction of induced current in each coil L.

ANS: - Due to Lenz’s law, end A will behave as South Pole and end B will behave as North Pole. The end face A will have clock wise direction of current and end face B will have anti clock wise direction of current when seen from the magnet side.

40. An electromagnet has stored 648 J of magnetic energy when a current of 9 A exists in its coils. What average emf is induced if the current is reduced to zero in 0.45 sec. ANS: - Calculate L = 16 H. e = L di/dt = 320 V

41. What is the magnitude of the induced current in the circular loop-A B C D of radius r, if the straight wire PQ carries a steady current of magnitude I ampere?

ANS: - Zero Induced emf.

42. Two identical loops, one of copper and another of aluminum are rotated with the same speed in the same M.F.In which case, the induced (a) e.m.f (b) current will be more and why?ANS: - Induced emf will be same in the both but Induced Current will be more in Copper loop.

43. Why is spark produced in the switch of a fan, when it is switched off?ANS: - A large Induced emf is setup across the gap in the switch.

44. Coils in the resistance boxes are made from doubled up-insulated wire. Why?ANS: - To cancel the effect of self Induced emf in the coil.

45. A galvanometer connected in an A.C. circuit does not show any deflection. Why?ANS: - A galvanometer measures mean value of a.c., which is zero over a cycle.

46. A capacitor blocks D.C. but allows A.C to pass through it. Explain. Why?

ANS: - Xc = cπ ϑ2

1 = ∞

47. Can we use transformer to step up D.C. voltage? If not, why?

ANS: - Magnetic flux linked with Primary coil does not vary with time so no Induced emf in secondary.

48. Calculate the r.m.s value of alternating current shown in the figure.ANS: - 2A.

49. The algebraic sum of potential drop across the various – elements in LCR circuit is not equal to the applied voltage. Why?ANS: - Voltages across different elements of the LCR circuit are not

in same phase.

50. A copper ring is held horizontally and bar magnet is dropped through the ring with its length along the axis of the ring. Will the acceleration of the falling magnet be equal to, greater than or less than that due to gravity?ANS: - Less than that due to gravity.

51. A magnet is moved in the direction indicated by an arrow between two coil A B and C D as shown in the figure. Suggest the direction of current in each coil.ANS: -For Coil AB: Anticlockwise.

For Coil CD: Anticlockwise.

52. Figure shows an inductor L and a resistance R connected in parallel to a battery through a switch. The resistance R Which of the bulbs lights up earlier, when K is closed?Will the bulbs be equally bright after same time?ANS: -(i) The bulb B2 will light up earlier. (ii) The bulb B1 will grow more brightly.

53. How does the self inductance of a coil change, when Number of turns in the coil is decreased? An iron rod is introduced into it. Justify your answer in each case.ANS: -i. L α n2 => L is decreased.

ii. L will Increase.

54. Figure shows two electric circuits A and B. Calculate the ratio of power factor of the circuit B to the Power factor of the circuit A.

ANS: - 2 .

55. An inductor L of reactance XL is connected in series with a bulb B to an A.C. source as shown in the figure. Briefly explain, how does the brightness of the bulb change when(a) Number of turns of the inductor is reduced and (b) A capacitor of reactance XC =XL is included in series in the same circuit.ANS: -(a) Bulb will grow more brightly.

(b) Brightness of the bulb will become maximum.

56. When a series combination of a coil of inductance L and a resistor of resistance R is connected across a 12 V-50 Hz supply, a current of 0.5.A flows through the circuit. The

current differs in phase from applied voltage by 3

Π radian. Calculate the value of L and R.

ANS: -L=0.066 H, R=12Ω57. An A.C. generator is connected to a sealed box through a pair of terminals. The box may contain R L C or the series combination of any two of the three elements. Measurements made outside the box reveal that:

E=75 Sin ωt (in volt) and

I= 1.2 Sin (ωt+ 5

Π)( in ampere)

Name the circuit elements

What is the Power factor of the circuit?What is the rate, at which energy is delivered by the generator to the circuit?ANS: -(a). Series combination of a register and a capacitor.

(b). Power factor = cosΦ = 0.81(c). Pav = EvIvCosΦ = 72.9W

58. Figure (a), (b) and (c) Show three alternating circuits with equal currents. If frequency of alternating emf be increased, what will be the effect on currents In the three cases. Explain.

ANS: -(i) No effect (ii) current will decrease (iii)Current will Increase.59. Does the current in an A.C. circuit lag, lead or remain in phase with the voltage of frequency υ applied to the circuit when(i) υ = υr (ii) υ < υr (iii) υ > υr

where υr is the resonance frequency.ANS: -(i) Current and Voltage are in the same phase.

(ii) Current leads voltage by Phase angleΦ .(iii) Current lags behind voltage by Phase angleΦ .

60. Two different coils have self inductance L1=8 mH and L2 = 2 mH. At a certain instant, the current in the two coils is increasing at the same constant rate and the power supplied to the two coils is same. Find the ratio of (a) induced voltage (b) current and (c) energy stored in the two coils at that instant?

ANS: - ⇒=dtLdIe

2

1

ee

= 4 As P= eI = const = 2

1

II

= 41

41

2

1 =∴UU

UNIT-5ELECTROMAGNETIC WAVES

1.Why is the quantity ε0 dΦE/dt called the displacement current?

2. Using a d.c. source, a capacitor has been fully charged. What are the magnitudes of

conduction and displacement currents?

3. What is the ratio of speed of infrared and ultraviolet rays in vacuum?

4. An electromagnetic wave consists of oscillating electric and magnetic fields. What is the

phase relationship between these oscillations?

5. Radio waves diffract predominately around building while light waves, which are also

electromagnetic waves, do not. Why?

6. Electromagnetic waves with wavelength

(i) λ1 are used to treat muscular strain.

(ii) λ2 are used by a FM radio station for broadcasting

(iii) λ3 are used to detect fracture in bones

(iv) λ4 are absorbed by the ozone layer of the atmosphere.

Identify and name the part of electromagnetic spectrum to which these radiations belong.

.Arrange these wavelengths in decreasing order of magnitude.

7. (a) Which of the following, if any, can act as a source of electromagnetic waves?

(i) A charge moving with a constant velocity.

(ii) A charge moving in a circular orbit.

(iii) A charge at rest.

Give reason. (b) Identify the part of the electromagnetic spectrum to which waves of

frequency (i) 1020 Hz (ii)109 Hz belong to microwaves.

8)If the area of the TV telecast is to be doubled then what will be the height of the

transmitting antenna ?.

9.Which of the physical quantity is NOT transported by the em waves?

10.What happens to the average temperature on the surface of the earth if there is no

atmosphere?.

11. Mention the law, that which asserts that the electric field lines cannot form close loops?

12.What are the characteristics properties of electromagnetic waves?

13.The energy of the electromagnetic wave is in the order of 15KV . To which part of the

spectrum does it belong?

14. Name em waves are used in telecommunication.

15.what is condition for obtaining displacement current between the plate of the capacitor?

16.Mention the pair of space and time varying E and B fields which would generate a plane

em wave travelling in the z-direction?

Ex and By

17. A plane electromagnetic wave travels, in vacuum, along the y-direction. Write (i) the ratio

of the magnitude, and (ii) the directions of its electric and magnetic field vectors.

(ii) For an electromagnetic wave traveling along y-diretion, its electric and magnetic field

vectors are along z-axis and x-axis respectively. The direction of BE

× is same as that of

direction of wave propagation and jik ˆˆˆ =× .

18. Suppose that the electric field amplitude of an electromagnetic wave is E0=120 NC-1 and

that its frequency is v=50.0 MHz.

(a) Determine, B0, w,k and λ. (b) Find expressions for E and B.

19. Answer the following questions:

(a) Long distance radio broadcasts use short-wave bands. Why?

(b) It is necessary to use satellite for long distance TV transmission. Why?

(c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only

from satellites orbiting the earth. Why?

(d) If the earth did not have an atmosphere, would its average surface temperature be higher

or lower than what it is now?

OPTICS1.When a photon collides with an electron, which of the following characteristics of the

photon increases?

2.Which of the following does not support the wave nature of light?

3.The distance travelled by the ray of light during the time octagonal mirror rotates through

90 is’L’ if the mirror rotates at N revolution per second, what is the speed of light.

4. A star appear yellow . If it starts accelerating towards earth, how will its colour appears to

change.

5.Two points A and B are situated at the same distance from the source of light, but in

opposite direction from it.What is the phase difference between the light waves passing

through A and B?

6.When the light is polarized by reflection , what is the angle between reflected and refracted

rays.

7. For double refracting crystal the refractive indices , for the ordinary and extraordinary

denoted by µo and µe. What is the relation valid along the optical axis of the crystal.

8.What is the angle between planes of electric and magnetic field oscillation in case of light

waves?

9.What is the colour of the interference fringe nearest to the white central maximum incase of

white light?

10. What happens to the fringe pattern when YDS experiment is performed in water instead

of air?

11. A man stands in front of a mirror of special shape. He finds that his image has a very

small head, a fat body and legs of normal size. What can we say about the shapes of the three

arts of the mirror?

12. In which direction relative to the normal, does a ray of light bend, when it enters

obliquely a medium in which its speed is increased?

13. For the same angle of incidence, the angles of refraction in three different media A,B and

C are 15° ,25° and 35°,respectively. In which medium will the velocity of light be minimum?

14. For what angle of incidence, the lateral shift produced by a parallel sided galss slab is

maximum?

15. If a plane glass slab is placed on letters of different colours, the red coloured letters

appear more raised up. Why?

16. Does refraction in a water tank make apparent depth same throughout?

17. The critical angle for glass-air interface is ic. Will the critical angle for glass-water

interface be greater than or less than ic?

18. An air bubble in a jar of water shines brightly. Why?

19.What happens to the shining of diamond if it is dipped in a transparent oil?

20.What type of a lens is a tumbler filled with water?

21.What type of a lens is an air bubble inside water? Give reason also.

22.A lens immersed in a transparent liquid is not visible. Under what condition can this

happen?

23. A lens whose radii of curvature are different is forming the image of an object placed on

its axis. If the lens is placed with its faces reversed, will the position of the image change?

24.What happens to focal length of a convex lens, when it is immersed in water ?

25.How does the focal length of a convex lens change if monochromatic red light is used

instead of violet light?

26.The radii of curvature of both the surfaces of a lens are equal. If one of the surfaces is

made plane by grounding, how will the focal length and power change?

27.A glass prism is held in water. How is the angle of minimum deviation affected?

28.A ray of light is normally incident on one face of an equilateral prism.Trace the course of

the ray through the prism and emerging from it.

29. What will be the colour of the sky in the absence of atmosphere?

30. Why do clouds appear white?

31. Why do sometimes we observe haloes (rings) round the sun or the moon?

32. Bees can see objects in the ultraviolet light while human beings cannot do so. Why?

33. A chicken wakes up early in the morning and goes to sleep by sunset. Why?

OR

Why is a chicken not able to see in the dim light?

35. Why is the focal length of an objective in compound microscope little shorter than the

focal length of the eyepiece?

36. You are provided with four lenses of focal length 1 cm, 3cm, 10cm and 100cm. Which

two would you prefer for a microscope and which two for a telescope?

37. Can we increase the range of a telescope by increasing the diameter of its objective?

38. A telescope has been adjusted for the relaxed eye. You are asked to adjust it for the least

distance of distinct vision, then how will you change the distance between the two lenses?

39. The distances of an object and its real image, measured from the focus of a concave

mirror, are a and b respectively. Show that f2 = ab.

40.A ray of light goes from medium 1 to medium 2. velocities of light in the two media are c1

and c2 respectively. For an angle of incidence θ in medium 1, the corresponding angle of

refraction in medium 2 is θ/2.

(i) Which of the two media is optically denser and why?

(ii) Establish the relationship between θ, c1 and c2.

41.A beam of light converges at a point on the screen. A plane parallel glass plate is introduced in the path of this converging beam. How will the point of convergence be affected? Draw the relevant ray diagram.42. A microscope is focused on a dot at the bottom of a beaker. Some oil is poured into the beaker to a height of y cm and it is found necessary to raise the microscope through a vertical distance of x cm to bring the dot again into focus. Express refractive index of oil in terms of x and y. 43. A ray of light while traveling from a denser to a rarer medium undergoes total reflection. Derive the expression for the critical angle in terms of the speed of light in the respective media.

44.Explain the twinkling of stars. Why do the planets not show twinkling effect?.

45. Only the stars near the horizon twinkle while those overhead do not twinkle. Why?46.Show that a convex lens produces an N times magnified image when the objet distances, from the lens, have magnitudes (f ± f / N). Here f is the magnitude of the focal length of the lens. Hence find the two values of object distance, for which a convex lens, of power 2.5D, will produce am image that is four times as large as the object?

47.Use the lens equation to deduce algebraically what you know otherwise from explicit ray diagrams. (a) An object placed within the focus of a convex lens produce a virtual and

enlarged image. (b) A concave lens produces a virtual and diminished image independent of the location of the object. 48. A beam of white light on passing through a hollow prism gives no spectrum. Why? 49. Give reasons for the following observations on the surface of the moon: (i) Sunrise and

sunset are abrupt. (ii) Sky appear dark. (iii) A rainbow is never formed.

.

NUMERICAL PROBLEMS

50. The bottom of a container is a 4.0 cm thick glass. (µ=1.5) slab. The container contains

two immiscible liquids And B of depths 6.0 cm and 8.0 cm respectively. What is the apparent

position of a scratch on the outer surface of the bottom of the glass slab when viewed through

the container? Refractive indices of A and B are 1.4 and 1.3 respectively.

51. The refractive index of water is 4/3. Obtain the value of the semivertical angle of the cone

within which the entire outside view would be confined for a fish under water. Draw an

appropriate ray diagram.

52. A lens forms a real image of an object. The distance of the object to the lens is 4 cm and

the distance of the image from the lens is v cm. The given graph shows the variation of v with

u. (i) What is the nature of the lens? (ii) Using this graph, find the focal length of this lens.

53. A ray of light passes through an equilateral glass prism, such that the angle of incidence is

equal to the angle of emergence. If the angle of emergence is ¾ times the angle of the prism,

Calculate the refractive index of the glass prism.

OPTICS

54. State the conditions which must be satisfied for two light sources to be coherent.

55. Two independent light sources cannot act as coherent sources. Why?

56. No interference pattern is detected when two coherent sources are infinitely close to one

another. Why?

57. If the path difference produced due to interference of light coming out of two slits for

yellow colour of light at a point on the screen be 3λ/2, what will be the colour of the fringe at

the point. Give reason also.

58. What happens to the interference pattern if the phase difference between the two sources

varies continuously?

59. Radiowaves diffract pronouncedly around the buildings, while light waves, which are

e.m. waves do not why?

60. Coloured spectrum is seen, when we look through a muslin cloth. Why.

SHORT ANSWER QUESTIONS

61. How is a wavefront different from a ray? Draw the geometrical shape of the wavefronts

when (i) light diverges from a point source, and (ii) light emerges out of convex lens when a

point source is placed at its focus.

62. In a young’s double slit experiment, the position of the first fringe coincides with S1 and

S2 respectively. What is the wavelength of light?

63. Draw the diagram showing intensity distribution of light on the screen for diffraction of

light at a single slit. How is the width of central maxima affected on increasing the

(i) Wavelength of light used (ii) width of the slit/

What happens to the width of the central maxima if the whole apparatus is immersed in water

and why?

64. What two main changes in diffraction pattern of single slit will you observe when the

monochromatic source of light is replaced by a source of white light?

65. Explain with reason, how the resolving power of a compound microscope will change

when (i) frequency of the incident light on the objective lens is increased. (ii) focal length of

the objective lens is increased, and (iii) aperture of the objective lens is increased.

66. The critical angle between a given transparent medium and air is denoted by ic, A ray of

light in air medium enters this transparent medium at an agle of incidence equal to the

polarizing angle(ip). Deduce a relation for the angle of refraction (rp) in terms of ic.

NUMERICAL QUESTIONS

67. Two Sources of Intensity I and 4I are used in an interference experiment. Find the

intensity at points where the waves from two sources superimpose with a phase difference (i)

zero (ii) π/2 (iii) π.

68. In a two slit experiment with monochromatic light, fringes are obtained on a screen

placed at some distance D from the slits. If the screen is moved 5 x 10-2 m towards the slits,

the charge in fringe width is 3 x 10 -5 m. If the distance between the slit is 10-3 m . calculate

the wavelength of the light used.

69. A narrow monochromatic beam of light of intensity I is incident a glass plate. Another

identical glass plate is kept close to the first one and parallel to it. Each plate reflects 25% of

the incident light and transmits the reaming. Calculate the ratio of minimum and maximum

intensity in the interference pattern formed by the two beams obtained after reflection from

each plate.

UNIT-VII DUAL NATURE OF MATTER AND RADIATIONS

1. If wavelength of electromagnetic waves are doubled what will happen to energy of photon?2. Alkali metals are most suitable for photoelectric emission. Why?3. Out of microwaves, UV, IR which radiation will be most effecting for emission of electrons from a metallic surface?4. Can X-rays cause photoelectric effect?5. If the intensity of incident radiation on a metal is doubled what happens to the K.E of electrons emitted?6. What is the value of stopping potential between the cathode and anode of photocell? If the max K.E of electrons emitted is 5eV?7. It is easier to remove an electron from sodium than from copper, which has a higher value of threshold wavelength? 8. What is the role of photocell in cinematography?9. An electron and photon possessing same K.E. Which one will have greater wavelength?10. In Davisson – Germer experiment if the angle of diffraction is 520 find Glancing angle?11. What is the energy associated with a photon of wavelength 6000 A0 ?12. What is the effect on the velocity photo electrons, if the wavelength of incident light

is decreased?13. Show graphically how the stopping potential for a given metal varies with a frequency of the incident radiation.14. To work functions 2ev and 5ev for two metals x and y respectively. Which metal will emit electrons, when it is irradiated with light and wave length 400nm and why?15. A photon and an electron have same de-broglie wavelength. Which has greater total energy.Explain?16. The de-broglie wave length of a photon is same as the wave length of electron. Show that K.E. of a photon is 2mc .h times K.E. of electron. Where ‘m’ is mass of electron,c is velocity of light/ג17. Derive the expression of de-broglie wave length in terms of energy of energy and temperature? 18. Light from bulb falls on a wodden table but no photon electrons are emitted why ? 19.Following table gives values of work function for a few photosensitive metal.

S.NO Metal Work function(ev)1 Na 1.922 K 2.153 Mo 4.17

If each metal is exposed to radiation of wavelength 300nm which of them will not emit photo electron.20.An electron and alpha particle and proton have same kinetic energy , which have shortest De-broglie wavelength?21. The De-broglie wave length associated with proton and neutron are equal.Which has greater kinetic energy?.

22.A stream of electron travelling with a speed at right angle to a uniform electric field E, is deflected in a circular path of radius “r” . Prove that e/m = v2/rE.23.If the potential difference used to accelerate electron is doubled , by what factor the De-broglie wave length of the electron beam changed. 24. The De-broglie wave length associated with an electron accelerated through the potential difference “V” is λ. What will be its wave length , when accelerating potential is increased to 4v?25.Visible light can not eject photo electrons from copper surface, whose work function is 4.4 ev , why? Prove mathematically.

EXTRA QUESTIONS1. Neutrons, in thermal equilibrium with matter have an average KE = 3/2 kT .

Compute de-Broglie’s wavelength associated2. A nucleus of mass M initially at rest splits in two fragments of masses M/3

& 2M/3. Find the ratio of de-Broglie’s wavelength of two fragments.3. X-rays of wavelength 0.82 A0 fall on metallic surface. Calculate de-Broglie’s

wavelength of emitted photoelectrons. Ignore ϕ of the metal.4. Wavelength of photon and de-Broglie’s wavelength of electron has same

value. Show that energy of photon is 2λmc/h times the KE of electron.5. Compare energy of electron of de-Broglie’s wavelength 1 A0 with that of an

X-ray photon of the same wavelength.6. Calculate the ratio of de-Broglie wavelength associated with deutron moving

with velocity ‘2v’ and a α-particle moving with velocity ‘v’.7. An α-particle and proton are accelerated from rest through same PD ‘V’.

Find the ratio of de-Broglie wavelength associated with them. 8. Calculate de-Broglie’s wavelength associated with an electron of energy

200eV. What will be the change in λ if accelerating potential is increased to 4 times.

9. What is the (i) momentum (ii) speed and (iii) de-Broglie wavelength of electron of kinetic energy 120 eV.

10.Show that the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength of its photon.

11. An electron, α-particle and a proton have same KE. Which of these particles have shortest wavelength?

12. The threshold frequency for a certain metal is 3.3*1014 Hz. If light of frequency 8.2*1014 Hz is incident on the metal. Predict the velocity of ejected electrons and cut-off voltage.

13.The work function of two metals A and B are respectively 1.2 eV and 2.4 eV. Light of wavelength 600nm falls on these metals. (i) Which metal / metals will give photoelectric emission? (ii) What is the maximum velocity and cut-off potential? (iii) If the source is moved away, how does it affect the stopping potential?

14. The energy flux of sun reaching the earth is 1.3888*103 Wm-2. How many photons per square m is incident on the earth per sec. Assume average wavelength of sunlight = 550nm.

15. In an experiment on photoelectric effect, the slope of cut-off voltage versus frequency of light is found to be 4.12*10-15 Vs. Calculate the value of Plank’s constant.

16.The Photoelectric cut-off voltage for certain metal is 1.5V. What is the maximum KE of photoelectrons?

17. The threshold frequency of metal is ‘f’. When the light of frequency 2f is incident on it the maximum velocity of photoelectrons is ‘v1’. When the frequency is increased to 5f, the maximum velocity of photoelectrons is ‘v2’ Find v1 /v2.

18.The work function of cesium is 2.14 eV. Find (i) threshold frequency (ii) wavelength of light if the photoelectrons are stopped with stopping potential 0.6V.

19.In a plot of photoelectric current versus anode potential, how does a. the saturation current vary with anode potential for incident

radiations of different frequencies but same intensity.b. The stopping potential varies for incident radiations of different

intensities but same frequency?c. Photoelectric current vary for different intensities but same frequency

of incident radiations?Justify your answer in each case.

20. The radiations of frequency 1015 Hz are incident on two photosensitive surfaces A and B. Following observations are recorded:

Surface A: no photo electric emission takes place.Surface B: Photoemission takes place photo electrons have zero energy.Explain the above observations on the basis of Einstein’s photoelectric equations. How will the observations with surface B change when the wavelength of incident radiations is decreased?

21. An electron, α-particle and a proton have same de-Broglie wavelength. Which of these particles has (i) minimum KE and (ii) maximum KE and why? In what way has the wave nature of electron beam exploited in electron microscope?

22.Calculate the (i) momentum and (ii) de- Broglie wavelength of electron accelerated through a potential difference of 56 V. On increasing the potential how this can improve the resolving power of a microscope.

23. Light of frequency 2.5 ν0 is incident on surface of threshold frequency ν0 and the photoelectric current is 1 mA. If frequency of light is halved and intensity is doubled, find new photoelectric current.

24. For what KE of neutron, will the associated de Broglie wavelength be 1.32*10-10 m?

25.The work function of three metals Na, K and Mo are respectively 1.92, 2.15 and 4.17 eV. If each of the metal is exposed to light of wavelength 300 nm, which of them will not emit photoelectrons and why?

26.By how much would the Stopping Potential a for given surface goes up if the frequency of the Incident radiation were to be increased from 4 × 1015Hz to 8 × 1015Hz?

27. Calculate de-Broglie wave length of (i) an electron (mass 3 × 10 –2 kg moving with speed 100 m/s. Hence

show that wave nature in hydrogen atom) moving with speed 1/100 of speed of light in vacuum and (ii) a ball of radius 5 mm and of matter is important at atomic level

but is not really relevant at the macroscopic level. 28.The de-Brouglie wavelength associated with proton and a neutron is found to be equal. Which of the two has higher value of kinetic energy?

29. Graph showing variation of VS with frequency for two M1 M2 material is given.(i) What are the values of work

function for M1 and M2 VS

(ii)The values of VS for these for frequency ν3 (ν3 > ν02) are V1 and V2. Show that slope of lines equals

V1 - V2) / (ν02 - ν01) ν01 ν02 30.Through what potential difference should an electron be accelerated so that

its de-broglie wavelength becomes 0.4 Ao.

31.Monochromatic light of frequency 6x1014 Hz is produced by a LASER. The

power emitted is 2x10-3 W. a) What is the energy of photon in the light beam. b)

How many photons per sec on the average are emitted by the source. [Ans. (a)

2.49 eV, (b) 5x1015 sec-1]

32.If 5 % of the energy supplied to an incandescent light bulb is radiated as

visible light, how many visible quantas are emitted by a 100W bulb? Assuming

the wavelength of all the visible light to be 5000 Ao.

[Ans. 1.41x1019 J]

33.If the wavelength of incident light changes from 4400 Ao to 4000 Ao, then

find the change in stopping potential. [Ans. -0.26 eV]

UNIT-VIIIATOMIC NUCLEUS

1) what conclusions were drawn from the observation in which few alpha-particle were seen rebounding from gold foil?

2) which observation led to the conclusion in the α-particle scattering exp. That atom has vast empty space?

3) Compare the radii of two nuclei with mass number 1 and 27 respectively.4) Two nuclei have mass numbers in the ratio 1:8.What is the ratio of their nuclear radii?5) which have greater ionizing power:α-particles or β-particles?6) The half life of a radioactive substance in 30 days. What is the time taken for ¾

of its original mass to disintegrate?7) Why neutrons are considered as ideal particle for nuclear reactions?8) Does the ratio of neutrons and protons in the nucleus increase, decreases or

remain the same after the emission of ά – particles?9) Why is the ionization power of ά – particle of greater than γ – rays?10) A radio isotope of silver has a half life of 20 minutes. What fraction of the

original mass would remain after one hour?11) What changes takes place in the nucleus when a γ – rays is emitted?12) Can a single nucleus emit ά – particle, β – particle and a γ – rays together?13) Two nuclei have mass no. in the ratio 1:2. What is the ratio their nuclear densities?14) Establish the relationship between half life of a radio- active substance and decay

constant.15) Explain how ά particle scattering experiment led to Rutherford to estimate the size of the

nucleus.16) The activity of a radio active material drops to 1-16th of its initial value in 30

days. Find its half life.

17) In a particular fission reaction, a U 235 nucleus captures a slow neutron. The fission products are 3 neutrons, a La 142 and fission products X y z .What is the value of Y and Z.

18) You are given two nuclides X b) Which one of the two is likely to be more stable? Give reason.19) A certain radio active substance has a half life of 30 days. What is the disintegration constant? Find its average life. 20) Find the time required to decay 3/4th of a radioactive sample whose half life is 60 days.21) Neon -23 decays in the following way 23Ne10 ---------------23Na11 + 0e-1 +γ

Find the minimum and maximum kinetic energy that the β-particle can have. The atomic masses of 23Ne10 and 23Na11 are 22.99454 and 22.98984 respectively.

22) The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. Five minutes later the rate becomes 2700 per minute. Calculate

a) Decay constantb) Half-life of the sample23) Explain with an example, whether neutron-proton ratio increases or

decreases during beta decay.24) The half life period of radioactive element A is the same as the mean half time of another radioactive element B.Initially both of them have the same number of atoms. The radioactive element B decays faster than A. Explain why?25) Obtain the binding energy of a nitrogen nucleus from the following data mh=1.007834; mn=1.00867; mN=14.03074

Give your answer in MeV.26) Write nuclear equations for a) The α-decay of 226Ra88 b) The β- -decay of 32P15

c) The β+ decay of32P15

27) A neutron is absorbed by a 6Li3 nucleus with the subsequent emission of an alpha particle.

i) Write the corresponding nuclear reactions.ii) Calculate the energy released in MeV, in this reaction. Given mass 6Li3=6.0151264; mass (neutron) =1.00966544

Mass (alpha particle)=4.00260444 and mass(triton)=3.01000004Extra Questions

1. Calculate the energy released in the following nuclear reaction. 3 Li 7 + 0 n 1 2 He 4 + 1 H 3

2. When a deutron of mass 2.0141amu is absorbed by a 3 Li 6 nucleus of

mass 6.015amu, the nucleus splits into two α-particles each of mass 4.0026amu. Calculate energy carried by each α-particles.

3. A nucleus 10 Na 23 undergoes β-decay to give 11 Na 23. Write down the

nuclear reaction. Calculate KE of electron. Given mass of 10 Na 23 = 22.994466amu, mass of 11 Na 23 = 22.989770amu.

4. A neutron is absorbed by 3 Li 6 nucleus with the subsequent emission of a α-particle. (i) Write the corresponding nuclear reaction. (ii) Calculate the energy released. M 3Li6 = 6.015126amu, m(n) = 1.0086654amu, mass of triton = 3.0100000amu.

5. The mass of the star is 5 * 1032 kg. It generates energy at the rate of 5 *

1030 W. How long does it take to convert all Helium to carbon at this rate. 3 2 He 4 6 C 12 + 7.27 MeV

6. Prove mathematically that the fraction N / N0 of a radioactive element left over after a time ‘t’ equals 1 / x where x = 2 t/T. T is half-life period.

7. The radioactive nuclei X and Y contain equal number of atoms. Their half-life periods are 1 H and 2 H respectively. Calculate the ratio of their activity after 2 Hrs.

8. How many disintegrations per sec will occur in one gm of 92 U 238, if its half-life period is 1.42 * 10 17 sec.

9. A radioactive sample contains 2.2mg of pure 6 C 11, having half-life period 1224seconds. Calculate (i) Number of active atoms (ii) Activity when 5 µgm of sample is left.

10. The half-life period of 92 U 238 against α-decay is 4.5 * 109 years. What is

the activity of 1g sample?

11. Obtain the amount of 27 Co 60 necessary to provide a radioactive source of 8 mC. The half-life period of Co60 is 5.3 years.

12. A 12.5 MeV α-particle approaching a gold nucleus is deflected back by 1800.

How close does it approach the nucleus

15. Calculate half-life period and decay constant 1.0

.75

0.5

0.25

70 140 210 280 time (days)

16. The half-life period of radioactive sample is 5500 years. Its initial activity is found to be 15 decays per min per gram. In how much time would its activity reduces to 10 decays per min per gm? (Given loge3 = 1.0986 and loge2 0.693)

17. The decay constant for a given radioactive sample is 0.3465 days-1. What % of this sample gets decayed in a period of 4 days?

18. The nucleus 92U238is unstable against α-decay with a half-life of about 4.5x 109 years. Write down the equation of this decay and estimate the KE of emitted α-particle from the following data m (92U238 = 238.05081

amu, m (2 He4 = 4.00260 amu, m (90Th234 = 234.04363 amu

(4.19 MeV)

19. How many α and β particles are emitted when 92U238 changes into 82Pb206.

20. The energy level of an atom of element X is given below. Which one of the level transitions results in the emission of photon of wavelength 620 nm. Justify your answer with proper calculation

0 -1 eV

A B C -3 eV D E

-10 eV20. Calculate the longest and shortest wavelength of Lyman series. Given R =

10967700 m-1. (911.6 A0, 1215 A0)

21. The wavelength of second line of the Balmer series in hydrogen spectrum is 4861 A0. Calculate the wavelength of first line. (6562 A0)

22. Which state of the triply ionized beryllium atom (Be +3) has the same orbital radius as that of the ground state of hydrogen atom? rn ∝ n2 /Z Ans n = 2.

23. Which level of double ionized lithium (Li +2) has same energy as the ground state energy of hydrogen atom? Compare the orbital radius of two levels. En ∝ Z2 / n2)

24. Calculate the frequency of photon, which can excite the electron to – 3.4 eV from – 13.6 eV. (2.47 × 1015 Hz)

25.Show that the shortest wavelength lines in Lyman, Balmer and Paschen series have their wavelength ratio 1: 4: 9.

26.The potential energy of the electron in ground state is –27 eV, what is its potential and kinetic energies?

27.Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its photon.

28. A radioactive sample has N0 nuclei at t = 0. Its no. of undecayed nuclei get reduced to N0 /e at t = τ. What does the term τ stand for? Write in term of τ the time interval ‘T’ in which half of original number of nuclei, of this radionucleide would have got decayed?

29. If the nucleus 26Fe56 splits into two nuclei of 13Al28. Would the energy be released or needed for this process to occur? Also calculate this energy. Given m (26Fe56) = 55.93494 amu, m(13Al28 ) = 27.98191 amu.

30.Calculate the ratio of energies of photon due to transition of electron of hydrogen atom from (i) second permitted energy level to first level (ii) highest permitted energy level to second permitted level.

31. Prove that the instantaneous rate of change of activity of a radioactive substance is inversely proportional to the square of its half-life period.

32. The nucleus of an atom 92Y235 initially at rest decays by emitting an α-

particle as per equation 92Y235 90X231 + 2He4 + energy. It is given that BE per nucleon of parent and the daughter nuclei are 7.8 MeV and 7.835 MeV and that of α-particle is 7.07 MeV / nucleon. Assuming the daughter nucleus to be formed in unexcited state and neglecting its share of in energy of the reaction, calculate the speed of emitted α-particle. Take mass of α-particle = 6.68 × 10-27 kg.

33. Four nuclei of an atom fuse together to form a heavier nucleus. If the process is accompanied by release of energy, which of the two parent or daughter nucleus have higher BE/nucleon?

34.The spectrum of a star in the visible and the ultraviolet region was

observed and wave-length of some of the lines were identified were found to be

824 A0, 970 A0, 1120 A0, 2504 A0, 5173 A0, 6100 A0

Which of these lines cannot belong to hydrogen atom spectrum. (Given R = 1.03 × 107 m-1) and 1 / R = 970 A0. Support your answer with proper calculation.

35. Why a nucleus can eject electron (B particle) though it contain no electron?

36.Why nucleuses have mass less than the sum of masses of individual nucleons in them?

37.How wills the distance of closest approach changes: a) when the kinetic energy is of the projectile is doubled? B) when the velocity of projectile is halved.

38. The second member of Lyman series in hydrogen spectrum has wavelength 5400 Ao. Find the wavelength of first number.

39.What is the effect of temperature and pressure on the radioactivity?

40. What is the value of impact parameter of alpha particle scattered through an angle of 180o.

41.Draw the graph showing the distribution of electron’s emitted during beta decay.

UNIT-9ELECTRONIC DEVICES

1.What is the order of energy gap in a conductor, semi conductor, and insulator?.2.Why does the conductivity of a semi conductor change with the rise in temperature ?3. Is the number of electrons greater than, less than (or) equal to the number of holes in an intrinsic semi conductor?4.Show in a energy band diagram the donor level for an N-type semi conductor.5.. Draw in a energy band the acceptor level for a P-type semi conductor . 6.what is knee voltage in a junction Diode?7. In transistor a current controlled (or)temperature controlled device?.8. In a given diagram ,is the diode reverse (or) forward biased?.

9.which gate is represented by the following diagram?.

10.The ratio of number of free electrons to holes ne/nh for two different materials A and B are 1 and <1 respectively. Name the type of semi conductor to which A and B belongs.11.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 rectification for the same input frequency.12. How can you relate drift velocity and mobility of an electron?13. Show by the graph how does the current vary with the voltage change for a junction diode.14. Why do semiconductors obey OHM’S law for only low fields?15. Mention the factors upon which Tranconductance of a transistor depend.16. For faster action which transistor is used and why?17. What are input and output characteristics of a transistor? Draw the graphs.Ans:

18. A germanium diode is preferred to a silicon one for rectifying small voltages. Explain why?Ans: Because the energy gap for Ge ( Eg = 0.7 ev) is smaller than the energy gap for Si (Eg = 1.1ev ) . Moreover, the germanium diode is much more open to the danger of high temperature affect than silicon at high voltage.19.Express by a truth table the output Y for all possible inputs A and B in the

circuit shown below.

A

B

Y

20. Write the Boolean equation and truth table for the circuit shown below.What is the output when all the inputs are high?

Y

AB

C

21) Construct AND gate using NAND GATE and give its truth table.23. For a common emitter amplifier, current gain = 50. If the emitter current is 6.6mA, calculate collector and base current. Also calculate current gain, when emitter is working as common base amplifier.24.The base current is 100µA and collector current is 3mA.

a) Calculate the values of β, Ie, and αb) A change of 20µA in the base current produces a change of 0.5mAin the collector

current. Calculate βa.c.

25. In NPN transistor circuit, the collector current is 5mA. If 95% of the electrons emitted reach the collector region, what is the base current?

Ans: Here Ic=95% of Ie = (95 / 100 ) Ie

Ie = (100 / 95) × 5 mA = 5.26mA Ie= Ic+ Ib Ib = 0.25 mA26. In a transistor circuit shown the figure, the emitter current is 5mA and collector current 4.75 mA. Calculate the base current and the value of Rb.

27.A circuit symbol of a logic gate and two input wave forms A and B are shown.

A

Y B A

B

a) Name the logic gate b) Give the output wave form

28. The diode shown in the figure has a constant voltage drop of 0.5V at all currents and a maximum power rating of 100mW. What should be the value of resistance R connected in series, for maximum current.?

R

29.For a transistor working as a common base amplifier, current gain is 0.96. If the emitter current is 7.2mA, then calculate the base current.

30. For a common emitter amplifier, the current gain is 70. If the emitter current is 8.8mA, calculate the collector and base current.

31.The base current of a transistor is 105 µA and collector current is 2.05 mA.

a) Deterine the value of β , Ie , and αb) A change of 27 µA in the base current produces a change of 0.65 mA in the collector

current . Find βa.c.

32. In a silicon transistor, a change of 7.89mA in the emitter current produces a change of 7.8 mA in the collector cur- rent. What change in the base current is necessary to produce an equivalent change in the collector current?

ADDITIONAL QUESTIONS(1) Out of electrons and holes, which has higher mobility?

(2) Which special type of diode can act as a voltage regulator? Give the symbol of this diode and draw the general shape of its V-I characteristics.

(3) What do you mean by rectification? (4) If a semiconductor has an intrinsic carrier concentration of when doped

with phosphorous atoms, calculate the concentration of at room temperature .

(5) In a common emitter circuit, if VCE is changed by 0.5 V, collector current Changes by 0.002 mA. What is the output resistance?

(6) Name the electrical circuits used to get smooth D.C. output from a rectified circuit. (7) How does the energy gap of an intrinsic semiconductor vary when doped with a

trivalent impurity? (8) Draw energy band diagram of n-type semiconductor.(9) A semiconductor has equal electron and hole concentration 6×108 /m3 .On doping with a

certain impurity, electron concentration increases to 8×1012 /m3 .Identify the type of semiconductor after doping.

(10) How does the dc current gain of a transistor change, if the width of the base region is increased?

(11) Why are photodiodes used preferably in reverse bias condition?

2 marks questions(1) In the working of a transistor, emitter base junction is forward biased, while the collector

base junction is reverse biased, why:To make transistor to act as an amplifier.

(2) In a transistor base is lightly doped and is a thin layer, why?To reduce the neutralization in the base emitter junction.

(3)What is the condition for the state of saturation of a transistor?

(4)Write the truth table for the following logic circuit shown in the figure below.

3 marks questions

(1) Discuss how the OR gate is realized from the NOR gate.(2) Why is the base region of a transistor usually made thin?In a common emitter mode of

transisitor, d.c. current gain is 20, the emitter current is 7 mA. Calculate (i) base current, and (ii) collector current.

(3) The input resistance of a silicon transistor is 665 Ω. Its base current is changed by 15

μA, which results in change of collector current by 2mA. This transistor is used as a common emitter amplifier with a load resistance of 5 k . Calculate (i) current gain , (ii) transconductance gm, and (iii) voltage gain Av of the amplifier.

(4) Draw the energy band diagram of a p-type semiconductor. Distinguish between p-type and n-type semiconductor.

(5)Explain briefly with the help of a circuit, how V-I characteristics of a p-n junction diode are obtained in (a) Forward bias (b) Reverse bias. Draw the shape of curves obtained.

(6)In a common emitter transistor amplifier, the input resistance of a transistor is 100 ohm. On changing its base current by 10µA, the collector current increases by 2mA.If a load resistance of 5kilo ohm is used in the circuit, calculate (a) current gain (b) voltage gain of the amplifier

(7) Two signals A, B as given below, are applied as input to (i) AND (ii) NOR and (iii) NANDgates. Draw the output wave-form in each case.

Input A

0 t t1 t2 t3 t4 t5 t6 t7 t8

Input B0 t

t1 t2 t3 t4 t5 t6 t7 t8

5 marks questions

(1) Draw a circuit diagram to obtain the characteristics of a n-p-n transistor in common emitter configuration. Describe how you will obtain input and output characteristics. Give shape of the curves.

(2) Explain the function of base region of a transistor. Draw a circuit diagram to study the input and output characteristics of NPN transistor in a common emitter (CE) configuration. Show these characteristics graphically. Explain how the current amplification factor is calculated from these characteristics.

(3) Explain the working of transistor as an oscillator with the help of a neat diagram.(4) A student has to study the input and output characteristics of a n-p-n silicon

transistor in the Common Emitter configuration. What kind of a circuit arrangement should she use for this purpose? Draw the typical shape of input characteristics likely to be obtained by her. What do we understand by the cut off, active and saturation states of the transistor? In which of these states does the transistor not remain when being used as a switch?

(5) Input signals A and B are applied to the input terminals of the ‘dotted box’ set-up shown here. Let Y be the final output signal from the box. Draw the wave forms of the signals labeled as C1 and C2 within the box, giving (in brief) the reasons for getting these wave forms. Hence draw the wave form of the final output signal Y. Give reasons for your choice. What can we state (in words) as the relation between the final output signal Y and the input signals A and B?

UNIT-10COMMUNICATIONS SYSTEMS.

1) Why are micro wave used in radars? 2) Why sky waves are not used in the transmission of television signals?3) What should be the desirable characteristic of a diode detector? 4) Give a velocity factor of a line. 5) Why is delta modulation a convenient method of digital modulation. 6) Where the two wire transmission line, Coaxial cable, Optical fiber are employed. 7).what is heterodyning?.8).What is population inversion? How is it achieved?.9).Enumerate the various types of Lasers?.10)Lists some of the applications of Lasers?.11)Name the prime elements of a telecommunications network.12) Audio signal cannot be transmitted directly in to the space why?13) What is pulse modulation?14) What is precisely meant by the term channel in a communication system “?15) Why does the electrical conductivity of earths atmosphere increase with altitude ? 16) Explain numerical aperture in fibre optical17) Differentiate between (i) PAM and (II) PPM .18)Why the transmission of signal is not possible for frequency greater than 20Mhz .19)How does the effective power radiated by the antenna vary with wavelength?20)what should be the length of the dipole antenna for a carrier wave of 5 X 10 8hz ?21)By how much should the height of the antenna be increased to double the coverage range R= 6400 Km.22) A TV. tower has a height of lOOm . How much population is covered by the TV. broadcaste if the average population density around the tower is 1000/km2

23)Ground receiver station is receiving a signal at (i) 5MH and (ii 100MHz transmitted from a round transmitter at a height of 300 m, located at a distance of 100 km from the receiver station. Identify whether the signal is coming via space wave or sky wave propagation or satellite transponder. Radius of earth = 6.4 x 106 m. Nmax of the Isosphere = 1012 m3

24)A schematic arrangement for transmitting a message signal (20 Hz to 20kHz) is given below:

Give two drawbacks from which this arrangement suffers. Describe briefly with the help of a block diagram the alternative arrangement for the transmission and reception of the message signal.25) The maximum peak-to-peak voltage of an AM wave is 16mV and the minimum peak-to peak voltage is 4mV.Calculate the modulation factor.26) An AM wave is represented by the expression:

v = 5(1+0.6cos6280t) sin 221 X 104t volts (i) What are the maximum and minimum amplitudes of the AM wave. (ii) What frequency components are contained in the modulated wave.27) An audio signal of 1 kHz is used to modulate a carrier of 500 kHz. Determine (i) Sideband frequency. (ii) Bandwidth required.28) The antenna current of an AM transmitter is 8A when only carrier is sent but it increases to 8.93A when the carrier is sinusoid ally modulated. Find the percent-age modulation index.29) A 100 MHz carrier is modulated by a 12 kHz sine wave

so as to cause a frequency swing of +50kHz. Find the modulation index.

30) The TV transmission tower at a particular place has a height of 160m. What is its coverage range? By how much should the height be increased to double its coverage range? Given that radius of earth = 6400 km. 31) A TV tower has a height of 110m. How much population is covered by the TV broadcast if the average population density around the tower is 1000 km-2? Given that radius of Earth = 6.4 X 106m.32)A microwave telephone link operating at the cenral frequency of 10 GHz has been established .If 2 % of this is available for microwave communication channel, then how many telephones channels can be simultaneously granted if each telephone is allotted a band width of 8 KHz .33) You are given three semiconductors A,B,C with respective band gaps of 3eV, 2eV and 1eV for use in a photodetector to

detect λ = 1400nm . Select the suitable semiconductor. Give reasons.34) Frequencies higher than 10MHz are found not to be reflected by the ionospere on a particular day at a place. Calculate th maximum electron density of the ionosphere.

ADDITIONAL QUESTIONS:

35) Draw a block diagram of data transmission and a data receiver. Explain them briefly.36). Explain how the optical communication system offers the possibility of millions of channels with increased band width. Give an additional advantage of optical communication system over a system employing a co axial cable.37) Give the names of the Indian satellites launched from the erstwhile USSR for remotesensing.38) Expand the following abbreviations:APPLE, INSAT, IRS, ISRO, ASLV.39) . What is the role played by launch vehicles in space exploration ? How is it differentfrom a space shuttle ? Merition any two facilities a country needs to develop to achieve self-sufficiency in space programmes.40) Explain how TV programmes are transmitted to remote areas through communicationsatellites.41) Explain (a) Modem Transmission (b) Modem Reception.42) How does Heterodyning distinguish from CW signals?.43) In a diode AM detector the output circuit consists of R =108Ω and C = 10 pF. A carriersignal of frequency 105hz is to be detected .Is it good?.44) On a particular day maximum frequency reflected from the ionosphere is 9Mhz.Onanother day it was found to increase by one Mhz.What is the ratio of the maximum electron density of the ionosphere on the two days.Hints: Nmax / nmax = (νc’/νc)2= 1.23

HINTS FOR HOTS-CLASS-XII(PHYSICS)UNIT-1

HINTS FOR ELECTROSTAICS1. In the dielectric medium between the plates.2. High potential, as electrons are negatively charged.3. Zero4. Zero5. ML-1T-2

6. Zero7. No.8. Because they are indicators of electric field, extending to infinite distance.9. C is proportional to A (area) Therefore C2 = 2C1

Since C = Q/V ,So the slope represents more capacitance. Hence P represents C2,Q represents C1

10. Each charge experiences two forces each of magnitude F inclined at an angle of 600.Theirresultant is given by [F2 + F2 +2F2cos 60]1/2 = √3.F11. (i)According to defn of P.D.,

VP>VQ .So VP-VQ is +ve for q>0.(ii)For q<0, VQ >VP .So VP-VQ is -ve

12. Flux = 0, since Qen = 013. Due to polaraisation,opposing electric field is created.14. Electric field at the midpoint of a dipole of length 2a is 2kq/a pointing towards the –vecharge or in the

direction opposite to the dipole moment. 15. Inside the cavity field at any point is uniform and non zero.16. No. If the initial velocity of the charged particle makes a certain angle with a line offorce, then the

charged particle shall not move along the line of force.17. E= -dv/dr= -d(q/4πεor)/dr = q/4πεor2

18. Yes, at the mid point of electric dipole.19. U = kq2/a- kq2/2a- kq2/2a = 0.20. C’ = C+C+C = 3C = 75μF

Therefore, charge = 75μF x 4200 V = 315 mC* 21. Total ф = q0/εo = 2/εo ,→ flux through one face = ф/6 = 1/3 εo.

22. q →------------ Q 1/2mv2 = kQq/r Or, v2 α 1/r Or, r α 1/v2 Or, r’ = r/4

23. V = V1+V2

Q =C1V = 6 X 10-6 X 2 = 12µCAs C2 is in series same amount of charge will also flow through it.

Now V2= Q/ C2 = (12 X 10-6)/ (12 X 10-6) = 1 VoltTotal battery Voltage,V = 2 + 1 = 3 Volt

24.Capacitance of parallel plate capacitor with air between the plates is C0 = ε0A/dWhen the separation between the plates reduced to half,

C1=ε0A/(d/2) = 2ε0A/dThus final capacitance is C2=10 X 8 pF=80 pF

25.The arrangement is of 5 capacitors in series.Therefore1/C’= (1/C) +(1/C) +(1/C) +(1/C) +(1/C) =(5/C)

Therefore C’=C/5 Or 5=C/5 or C=25µF. 26. The charge given to a capacitor is given by q=CV

So the remaining energy,qV- 1/2 qV =1/2 qV is lost as heat27. On equatorial line ,the direction of electric field is reversed to that of axial line. Hence theangle between

electric dipole moment& electric field strength is 1800

28. Eq. network is the Eq.capacitance= (2C series C) (2C series C)=4C/3

29. 4/3 πR3 = 8 x 4/3 πr3

30. dV = -E . dx = -2 x 103 x 4 V = -8 x 103 V E = E1 – E2 = Q1 - Q2/2A εo

Now, V1 – V2 = E . d = Q1 - Q2/2C32. U = k (q2/a + qQ/a + qQ/√2a)

But, U = 0 Therefore, Q = -2q/(2 + √2)

33. Ε= q/4пε0[1+1/4+1/16+1/64] = q/3пε0.

34. Let q & q’ be the charges on inner and outer sphere.Hence q/4πr2= q’/4πR2

q=Qr2/(R2+r2) & q’=Q-q=Q- (Qr2/R2+r2)Now potential at O is given by

35 (i) Dipole has two equal and opposite charges. In the uniform electric field they will experience equal and

opposite force. Net force is zero. So there can’t be any translatory motion.(ii) Torque, τ = r⊥F=2l sinθ qE = p X E

Torque experienced by the dipole will rotate it.So the direction of the torque will be outward from the

surface.36. We have V=ar2+b

The electric field,Er = - dV/dr=-2ar-------(1)From Gauss theorem, ∫ E.dS=q/ε0, where S is the spherical surface containing the charge qOr E.4πr2=(1/ε0 )(4π/3) r3ρ-------(2)From 1 and 2 ρ=-6πε0

UNIT-2

HINTS FOR CURRENT ELECTRICITY.1. They have high resistively and low temperature coefficient of resistance.2. 22*102 Ω ± 10%3. Resistively remains the same.4. As Vd α V.

The different velocity will be doubled.5. R=ρ(l/A)

= ρ(l2/Al) = ρl2/V)since, ρ and V are constants therefore, R α l2

(R2/R1) = (l2/l1)2=9 because R2=9R1

=9*10=90 Ω6. Now, 1/R=1/R1+1/R2

because l = 48/240 = 0.2 m7. a) in parallel, power dissipation α 1/R

Therefore 3Ω wire will dissipate more powerb) In series , power dissipation α R

Therefore 9Ω wire will dissipate more power

8. R100/R27.5 = (1+100α)/(1+27.5α)On solving, we get α = 0.0039/˚c

9. Superconductors are the materials that lose all its resistance at very low temperature=0 K

Application:Super conductor are used

a) In making very strong electromagnetsb) In producing very high speed computers .

10. Resistivity of copper is less , hence manganin wire is thicker.11. High value of resistivity and low value of temperature coefficient.12. Resistivity will be unchanged because it depends upon nature of the materials.13. B is more sensitive.14. Reduced by half.15. Relaxation time decreases with increase of temperature.16. Increase in heat.17. Reistance remains same.18. Pα 1/R. (i.e) 25Watts19. 16 times of the original reisitance.20. (i)Series - Iron

(ii)Parallel - Copper.21. R = ρ L/A.

(i.e) 10.25%22. Parallel(i.e) R.23. 1:4.24. 2:125. R.

Unit-5

HINTS FOR EM WAVE.1. Due to change in electric field.

3. Same as velocity of light.

4. 90 degree

rays, radio waves, X-rays, UV rays ץ .6

7. i)X-rays

ii)microwave.

8. doubled

9. Charge.

10..Increases.

11. Gauss’s law

13. X-rays because λ = 0.825A

14. Micro waves

15.By varying potential difference.

16. Ex and By.

UNIT-6 ANSWER KEY FOR OPTICS

1.Ans:Wavelength

2. Ans:Photo electric effect.

3. Ans:4NL.

4. Ans: It will turn gradually blue.

5. Ans:Zero.

6. Ans:90 .

7. Ans:µo= µe

8. Ans:90

9. Ans:Violet.

10. Ans: Shrinks.

11. Ans.(i) The upper part of the mirror is convex.

(ii) The middle part of the mirror is concave.

(iii) The lower part of the mirror is plane.

12. Ans. The ray of light bends away from the normal.

13. Ans.

As µ = sini/sinr = c/v or v = sinr/sini *c

For a given angle of incidence, v ∝ sinr, vA ∝ sin15° , vB ∝ sin25° , vC ∝ sin35°

But sin15° < sin25°< sin35°.

∴ vA < vB < vC .

i.e. the velocity of light is minimum in medium A.

14 Ans.For i = 90°, lateral shift is maximum and is equal to the thickness of the slab.

d = t sin( i – r )/ cos r

dmax = t sin(90°- r)/cos r = t cos r/cos r = t.

15. Ans.The apparent shift caused by a slab of thickness ‘t’ is given by

d = t( 1 -1/µ )

As the refractive index of the galass is maximum for red light, so red coloured letters are

more raised up.

16. Ans. No Apparent depth is maximum for that part of the bottom of the tank which is

observed normally. Apparent depth decreases with increasing obliquity. Due to this unequal

refraction, the flat bottom of the tank appears concave.

17. Ans. For glass-air interface, sin ic = 1/aµg

The critical angle i’c for glass water interface is given by

Sin i’c = 1/ wµg

Now wµg < aµg,

Sin i’c > Sin ic or i’c > ic

18. Ans. Light entering water is totally reflected from the air bubble. For the observer, this

light appears to come from the buble. So it shines.

19. Ans. As the critical angle for diamond-oil interface is greater than that for the diamond –

air interface, so the shining of diamond reduces when it is dipped in a transparent oil.

20. Ans. It behaves like a biconvex lens.

21. Ans. Air bubble has spherical surface and is surrounded by medium ( water) of higher

refractive index. When light passes from water to air it gets diverged. So air bubble behaves

as a concave lens.

22. Ans.When the refractive index of the liquid is same as the lens material, no light will be

reflected by the lens and hence it will not be visible. ∝

23. Ans. No, the image will be formed at the same position.From lens maker’s formula,

1/f = (µ -1) [ 1/R1 –1/R2 ] , it is clear that when we interchangeR1 and R2, the magnitude of

‘f’ remains the same.

24. Ans. focal length‘f’ of a convex lens is related to its refractive index as

f ∝ 1/(µ -1)

As wµg < aµg , so focal length of a convex lens will increase when it is immersed in water.

25. Ans.Focal length, f ∝ 1/(µ -1)

As µR < µV , so the focal length of a convex lens will increase when red light is used.

26. Ans:For the original lens: R1 = +R and R2 = -R, so we can write

1/f = (µ -1) [ 1/R +1/R ] = 2(µ -1)/R.

When one surface is made plane by grounding, we have R1 = +R and R2 = - ∞.

Therefore, 1/f ’ = (µ -1) [ 1/R +1/∞ ] =(µ -1)/R

∴ f ‘ / f = 2 or f ‘ = 2f

Thus the focal length becomes double and power becomes one –half.

27. Ans. When the prism is held in water,

wµg = Sin (A + δm/2) /SinA/2

As wµg < aµg , so the angle of minimum deviation decreases in water.

28. Ans. Total internal reflection.

29. Ans. The sunlight will not be scattered in the absence of atmosphere. So the sky will

appear dark.

30. Ans. Clouds have large particles like dust and water droplets which scatter light of all

colours almost equally, hence clouds generally appear white.

31Ans. When the sun or the moon is seen through a thin veil of high clouds, holes are seen.

These are formed due to reflection of light by the icy crystals present in the atmosphere.

32. Ans. Ultra-violet light has wavelength shorter than that of violet light. Bees have some

retinal cones that are sensitive to ultra violet light, so they can see objects in ultra-violet light.

Human eyes do not posses retinal cones sensitive to ultra-violet light, so human beings

cannot see objects in ultra-violet light. In other words, human beings are ultra-violet blind.

33. Ans. In a chicken’s eye, the retina has a large number of cones but only few rods. The

rods are sensitive to bright light only. That is why a chicken is not able to see in dim light. As

it needs bright light to see, so it wakes up early in the morning with the sunrise and goes to

sleep by sunset.

34. Ans. Magnifying power of a simple microscope ,

m = 1 + D/f

as fV < fR so the magnifying power is greater when the object is seen in violet light.

35. Ans. This is done so that the objective lens forms image within the focal length of the

eyepiece.

36. Ans. (i) We should take f0 =1 cm and fe = 3cm for a microscope.

(ii) We should take f0 = 100 cm and fe = 1 cm for a telescope.

37. Ans. Yes, because the light gathering power of objective will increase and even faint

objects will become visible.

38. Ans. For relaxed eye,

L = f0 + fe (normal adjustment)

For least distance of distinct vision,

L’ = f0 + ue , ue < fe

Therefore, L’ < L. so that distance between the two lenses should be decreased.

SHORT ANSWER TYPE QUESTIONS

39. Ans. Here u = - (f+a), v = -(f+b), f = -f

As 1/f = 1/u +1/v

F = uv/ u + v

Or -f = [-(f+a)] x [-(f+b)] / -(f+a) – f(a+b)

= f2 + af +bf +ab / -(2f+a+b)

or 2 f2 + af +bf = f2 + af +bf +ab

or f2 = ab

40.Ans. (i) Angle of refraction (θ/2) in medium 2 is less than the angle of incidence (θ) in

medoum 1 i.e. the ray bends towards the normal in medium 2. so medium 2 is optically

denser than medium 1.

(ii) From Snell’s law,

µ = sin i/sin r = sin θ/ sin θ/2 = 2sin θ/2 cos θ/2 /sin θ/2 = 2 cos θ/2

Also µ = c1 / c2

hence 2 cos θ/2 = c1 / c2 or θ = 2cos-1(c1 / 2c2) .

41. Ans. The point of convergence shifts away from the glass, as shown in the ray diagramgiven below. The screen has to be moved towards right to receive the point of convergence again.

42. Ans. Real depth = y cm

Apparent depth = y- x cm

Refractive index of oil,

µ = real depth/ apparent depth = y / y-x

43. Ans. Using Snell’s Law for refraction from glass to air,

Sin i/sin r = gµa = v / c

Where c is the speed of light in air and v is the speed of light in glass, In the consition of critical incidence, we have i = ic and r = 900

Sin ic/ sin 900 = v / c or Sin ic = v / c

Or ic = sin -1 ( v / c)

44. Ans. Twinkling of stars. The light from stars undergoes refraction continuously before itreaches earth. So the apparent position of the stars is slightly different than its actual position. Due to variation in atmosphere conditions, like change in temperature, density etc., and this apparent position keeps on changing. The amount of light entering our eyes from a particular star increases and decreases randomly with time. Sometimes, the star appears brighter and other times, it appears fainter. This gives rise to the twinkling effect of stars.

The planets do not show twinkling effect. As the planets are much closer to the earth, the greater and the fluctuations caused in the amount of light due to atmospheric refraction are negligible as compared to the amount of light received from them.

45. Ans. Light from the stars near the horizon reaches the earth obliquely through theatmosphere. Its path changes due to refraction. Frequent atmospheric disturbances change the path of light and cause twinkling of stars. Light from the stars overhead reaches the earth normally. It does not suffer refraction. There is no change in its path. Hence there is no Twinkling effect.

46. Ans. Magnification produced by any lens,

m = v/u = f / f + u

given m = ± N ±N = f / f + u

or f + u = ± f / N or u = - f ± f / N

hence magnitude of object distances,

|u| = f ± f / N

given P = 1/f = + 2.5 D

f = 1/ 2.5 = 0.4 m = 40 cm Also N = 4 |u| = 40 ± 40/4 = 40 ± 10 = 50 cm or 30 cm.

47. Ans.(a) for a convex lens, f>>0 and for an object on left, u<0. when the object is placed within the focus of a convex lens,

0 < |u| < f or 0< 1 / |u| > 1/f 1/v = 1/f+1/u=1/f-1/|u|<0

i.e. v < 0 so a virtual image is formed on left.Now as u<0 and v<0, so 1/v = 1/f + 1/u

= - 1/ |v| = 1/f – 1/|u| or 1/|u| - 1/|v| = 1/f As f>0 1/|u| - 1/|v| > 0 or 1/|u| > 1/|v| or |u|<|v|i.e. |v|>|u| |m| = |v/u| > 1

Hence image is enlarged. (b) For a concave lens f<0 and for an object on left, u<0 1/v = 1/f +1/u = 1/|f| - 1/|u|

= - [1/|f|+1/|u|] < 0 for all u. i.e. v<0 for all values of u. hence a virtual image is formed on the left.Also 1/|v| = 1/|f| + 1/|u| 1/|v| > 1/|u| Or |v|<|u| |m| = |v/u| < 1 i.e. the image is diminished in size. 48. Ans. A hollow prism contains air which does not cause dispersion. The faes AB and ACof the hollow prism behave like parallel sides of glass plates. The beam is laterally deviated at each of the two refracting faces. However, the rays of different colours emerge parallel to each other. So there is no dispersion. 49. Ans. (i) Moon has no atmosphere. There is no scattering of light. Sunlight reaches moon

straight covering shortest distance. Hence sunrise and sunset are abrupt.

(ii) Moon has no atmosphere. So there is nothing to scatter sunlight towards the moon. No

skylight reaches moon surface. Sky appears black in the day time as it does at night.

(iii) No water vapours are present at moon surface. No clouds are formed. There are no rains

on the moon. So rainbow is never observed.

NUMERICAL PROBLEMS

50. Ans. The total apparent shift in the position of the image due to all the three media is

given by

d = t1[1-1/(µ1)+ t2[1-1/(µ2)+ t3[1-1/(µ3)

Given t1 = 4.0 cm, t2=6.0 cm , t3 =8.0 cm

µ1 = 1.5 , µ2=1.4 , µ3 =1.3 cm

d= 4.0(1-1/1.5)+6.0(1-1/1.4)+8.0(1-1/1.3)

= 1.33 + 1.71 + 1.85 = 4.89 cm

51. Ans. Clearly , the fish can see the outside view of the cone with semi vertical angle,

But µ = 1 / sin ic

Or 1/3 = 1 / sin ic

Or sin ic = ¾ = 0.75

θ/2 = ic = sin-1 (0.75) = 48.60

52. Ans. (i) As the lens forms a real iamge, it must be a convex lens.

(ii) From the graph, when u=20 cm , we have v = 20 cm.

For the convex lens forming a real iamge, u is negative and v and f are positive.

U = -20 cm v = +20cm

Using this lens formula,

1/f = 1/v – 1/u = 1/20 – 1/-20 = 1/10 or f = + 10 cm

53. Ans. A=600 , δm=300

i = e = ¾ A = 450 ,

as A + δ = i + e

60 + δ = 45 + 45

or δ = 300

Refractive index,

µ = sin a+δm /2 /sin A/2 = sin 600+300/2 / sin 600/2

= sin 450/sin300 = 1/√2 / ½ = √2 = 1.414

WAVE OPTICS

54. Ans. Two light sources will be coherent if

(i) The frequency of the two light sources is same and,

(ii) The phase difference between them remains constant.

55. Two independent light sources cannot act as coherent sources. Why?

Ans. Two independent sources of light cannot be coherent. This is because light is emitted by

individual atoms, when they return to ground state. Even the smallest source of light contains

billions of atoms which obviously cannot emit light waves in the same phase.

56. Ans. Fringe width , β = λD/d

i.e. β ∝ 1/d , when d → 0, β → ∞

fringe width is very large. Even a single fringe may occupy the entire screen. The

interference pattern cannot be observed.

57. Ans. The given path difference satisfies the condition for the minimum of intensity for

yellow light, Hence if yellow light is used, a dark fringe will be formed at the given point. If

white light is used, all components of white light except the yellow one would be present at

this point.

58. Ans. The positions of bright and dark fringes will change rapidly. Such rapid changes

cannot be detected by our eyes. A uniform illumination is seen on the screen i.e. interference

pattern disappears.

59. Ans. For diffraction to take place the wave length should be of the order of the size of the

obstacle. The radio waves (particularly short radio waves) have wave length of the order of

the size of the building and other obstacles coming in their way and hence they easily get

diffracted. Since wavelength of the light waves is very small. They are not diffracted by the

buildings.

60. Ans. Muslin cloth is mde of very fine threads and as such fine slits are formed. White

light passing through these silts gets diffracted giving rise to colored spectrum. The central

maximum is white while the secondary maxima are coloured. This is because the positions of

secondary maxima (except central maximum) depend on the wavelength of light.

In a coarse cloth, the slits formed between the threads are wider and the diffraction is not so

pronounced. Hence no such spectrum is seen.

SHORT ANSWER QUESTIONS

61. Ans. A wavefront is a surface obtained by joining all points vibrating in the same phase.

A ray is a line drawn perpendicular to the wavefront in the direction of propagation of light

wave.

The wavefronts of light emerging from a point source are spherical, as shown in figure. When

a point source is placed at the focus of a convex lens, the emerging light has the plane

wavefronts, as shown in figure.

62. Ans. As shown in figure the bright fringes B1 and b2 on either side of O coincide with S1

and S2 respectively.

Clearly ,

β = d/2

As β = Dλ/d d/2 = Dλ/d or λ = d2/2D

63Ans. For intensity distribution of light in diffraction at a signal slit, see figure

Width of central maximum is given by

β0 = 2Dλ/d

(i) When wavelength of light λ used is increased, the width of central maximum increases.

(ii) When width of the slit is increased, the width of central maximum decreases.

Wavelength of light in water decreases, so width of central maximum also decreases.

64. Ans. When the monochromatic source is replaced by a source of white light, the

diffraction pattern shows following changes:

(i) In each diffraction order, the diffracted image of the slit gets dispersed into component

colours of white light. As fringe width ∝ wavelength, so the red fringe with higher

wavelength is wider than violet fringe with smaller wavelength,

(ii) In higher order spectra, the dispersion is more and it causes overlapping of different

colours.

65. Ans. R.P.of a compound microscope

= 2 µ sinθ/λ = 2 µ sinθ x v /c

(i)When the frequency v of the incident light increases, the resolving power increases (R.P. ∝

v).

(ii) Resolving power does not change with change in focal length of objective lens.

(iii) When the aperture of the objective lens increases, the semi-vertical angle θ increases and

hence the resolving power of the microscope increases.

66. Ans. According to the Brewster law, when a ray of light is incident on a transparent

refracting medium at olarising angle,

µ = tan ip

but ip + rp = 900 or ip = 900 - rp

µ = tan(900 - rp ) = Cot rp = 1/ tan rp

As ic is the critical angle for the transparent medium, so

µ = 1/sin ic

on comparing (i) and (ii) we get

tan rp = sin ic or rp =tan-1(sin ic)

NUMERICAL QUESTIONS

67. Ans. The resultant intensity at a point where phase difference is Φ is

I R = I1 +I2+2√I1I2 Cos Φ

As I1 =I and I2 = 4I therefore

I R = I +4I+2√1.4I Cos Φ = 5I +4I cos Φ

(i) when Φ =0 , I R = 5I +4I cos 0 = 9 I

(ii) when Φ =π/2 , I R = 5I +4I cos π/2 = 5 I

(iii) when Φ =π , I R = 5I +4I cos π = I

68. Ans. The fringe width in the two cases will be β = Dλ/d

And β ‘= D’λ/d

β - β’ = (D-D’)λ/d

or wavelength λ = (β - β’ )d / (D-D’)

But D-D’ = 5 x 10-2 m

And β - β’ = 3 x 10-5 m , d= 10-3m

λ = 3 x 10-5 x 10-3 / 5 x 10-2 = 6 x 10-7 m = 6000A

69. Ans. Let I be the intensity of beam I incident on first glass plate. Each plate reflects 25%

of light incident on it and transmits 75%.

Therefore,

I1 =I and I2 = 25/100I = I/4

I3 =75/100 I = 3/4I

I4 = 25/100 I3 = ¼ x ¾ I = 3/16 I

I5= 7/100 I4= ¾ x 3/16 I = 9/64 I

Amplitude ratio of beams 2 and 5 is

R = √ I2/I5 = √I/4 x 64/91 = 4/3

Imin/ Imax = [r-1/r+1]2 = [4/3-1 / 4/3+1]2 = 1/49 = 1:49

UNIT-7HINTS FOR DUAL NATURE OF MATTER

1. E = hν =hc/λ E ∝ 1/λ energy of proton reduces to half. 2. Alkali metals have too low work functions. Even visible light can eject electrons fromthem.3. UV are most effective since they have highest frequency hence more energetic.4. Yes. X-rays cause photoelectric effect in sodium, zinc & copper.5. K.E of photons remains unaffected since they do not depend6. stopping potential V0 = Kmax/e = 5ev/e =5 V7. w0 = hν0 = hc/λ0

∴λ0 ∝ 1/ w0

Since sodium has lower work functions than copper it is easier for electron ejection. As it is lower work function, higher wavelength.

8. Photocells are used for reproduction of sound.9. 1/2 mv2 = (m2 v2)/2m = p2/2m

According to De Broglie wave length λ = h/pλe/ λp = pp/pe = √(mp /me )me < mp

λe > λp

electrons have greater De broglie wavelength than proton .10. θ = 90 - φ/2

= 90 – 52/2 =64°11. Ε = hc / λ

= 3.3×10-19 J12. KE of photoelectrons is given by Einstein’s photoelectric equation.

Ek =1/2 mv2

= h w0 -ט

V α 1/ √λ As wavelength decreases velocity increases.

13. stoppingpotential

frequency

14. E= h c/ 6.6 = (ג x 10-34 x 3 x 108)/ (4 x 10-7) = 4.98 x 10-19 j E= (4.98 x 10-19)/ (1.6 x 10-19) =3ev Hence, metal x will emit electrons.

15. For a photon E1=hc/ג

For an electron ג=h/mv or m=h/גv E2/E1=c/v>1 Therefore,E2>E1.thus, electron has total energy greater than that of photon.h/mv=ג=eג=phג .16

K.E. of electrons E=1/2mv2=1/2 m [h/m 2ג]

=h2/2m 2ג

∴Eph=Ee(2mc ג/h)17 . E=1/2 mv2=m2v2/2m=p2/2m ∴p=√2Emh/p=h/√2Em=ג∴

From Kinetic theory of gases average K.E.=3/2 KTh/(√2m(3/2 KT)=h/(√3m KT)=ג

18. The energy of light obtained from the bulb is much less than work function of thewodden block.Hence no photon electrons are emitted.

19. Mo will not emit photo electron , because its work function is more than 4 ev.

20. Alpha particles due to its largest mass.

21. R α 1/q

22. Ee = Mv2/r

23. λ = λ/√2.24. λ/2.25. λ = hc/Φ = 2823 A.

UNIT-8HINTS FOR ATOMS AND NUCLEI.

1) The entire positive charge and the mass were concentrated at one place inside the atom,called the nucleus.2) A larger number of alpha particle went through undeflected.3) R=R0A1/3 R1/R2=A1

1/3/A21/3=1 1/3/27 1/3=1/3

4) R= R0A1/3 R1/R2=A11/3/A2

1/3=(1/8)1/3 =1/2R1: R2=1:2

5) α-particles have more ionizing power than β-particles.6) N/N0=(1/2)n

t=2T =2*30 =60 days7) They are neutral in nature and get absorbed by nucleus, thus distributing the neutron proton ratio.8) The ratio of neutrons to proton ratio increases, after the emission of a ά – particle.9) Owing to greater mass and charge, it is able to knock out/pull out electrons which

colliding with atoms and molecules in its path.10) T = 20 minutes t = 60 minutes

N/No = (1/2)n = (1/2)60/20 = (1/2) 3 =1/8After one hour, 1/8th of the original mass would remain.

11) The nucleus looses energy, but remains same isotope it was.12) No a nucleus either emits a ά – particle or a β – particle and if left in the exited state,it

may emit γ – ray also.

13) 1:1 (independent of A).14) At t = T ½ N = No / 2 Using N = No e- λt

No / 2 = No e- λt ½

Solving we get, T1/2 = ln2/λ = 0.693/ λ15) Size of nucleus can approximately be estimated using the concept of distance ofclosest

approach. The rebounding particle is selected and its information is substituted in the expression Ro = 1/4πEo x 2Ze2 / E for ά particle where E is its energy.

A. Using N =No (1/2)t/π/2 N =1/16No1/16 = ½(1/2)t/π/2 T1/2 = 30/4 = 7.5 days.

17) 235 = 142+Y+3Y = 90And 92 = 57+Z+0Z = 35

18) a) Yes. Since X & Y are having same atomic number.b) Y 4 3 is likely to be more stable because for its neutron to proton ratio is smaller.

19) Disintegration constant λ =0.693/T1/2=0.693/30x24x60x60

Therefore T avg =1.44xT1/2 = 1.44x30 = 43.2 days.20) Remaining amount (undecayed) =1/4N0

Using N= N0 (1/2)t/T1/2

1/4= (1/2)t/60

Solving t=2x60=120days21) Mass defect ∆m= (22.9945-22.9898) =0.00474

Energy Q= (0.00474) (931.5)=4.4MeV

Hence the energy of beta particle can range from 0 to 4.4MeV.

22) a) Using R=R0e-λt

2700=4750 e-5t λ=0.113min-1

b) Using T1/2=0.693/λ =0.693/0.113 =6.132min23) In the process of beta decay, a neutron gets converted to proton inside the nucleus .

Hence number of neutrons decreases by one whereas number of proton increases by one. Hence n/p ratio decreases. 210Bi83-------------------- 210Po84 + 0β-1 + γ Before decay =127/83After decay=126/84

24) Let λ and λ’ be the decay constant of element A and B respectively. Given is T1/2(A) = T1/2(B)

0.693/λ=1/ λ’ or λ/ λ’ =0.693Let N be the number of atoms of each of the two samples and R and R’ their

disintegration rate, thenR/R’= λN/ λ’N= λ/ λ’=0.693R’ > R

25) Find ∆m using ∆m= (7x1.00783+7x1.00867-14.003074) U

Calculate ∆Eb=∆mx931.5 MeV

26) a) 226Ra88--------------222Rn86 +4He2

b) 32P15------------------32S16 +0e-1 +γc) 32P15------------------11B5 +0e+1 + γ

27) i) 6Li3 +1n0 ----------3H1 + 4He2 +Q (energy)ii) Q=∆mx931 MeV

Where ∆m=6.01512+1.0086654-4.0026044-3.0100000UNIT-9

HINTS FOR ELECTRONIC DEVICES1. Conductor - no energy gapSemi Conductor - It is of the order of 1 ev.Insulator - 6 ev (or) more than 6 ev.2. When a semi conductor is heated more & more electrons get enough energy to jumpacross the forbidden energy gap from valence band to the conduction band, where they are free to conduct electricity. Thereby increasing the conductivity of a semi conductor.3. In an intrinsic semi conductor the number of free electrons and holes is same.4. C.B

_ _ _ _ _ _ _ _ _ _ _

V.B5.

C.B

_ _ _ _ _ _ _ _ _ _ _ _

V.B

6. About 0.3V germanium .About 0.67V for silicon.7. Transistor is a current controlled devices.8. Reverse biased.9.NAND GATE.10.If ne/nh =1 . Hence A is intrinsic semi conductor.If ne/nh <1 , ne<nh hence B is P-type.11. For half wave rectification 50 Hz. For Full wave rectification 100Hz.12. Mobility of an electron is defined as the drift velocity of electron per unit electric field,i.e. μe= Ve /EMobility of a hole is defined as the drift velocity of hole per unit electric field,i.e. μ h= Vh /E

The electrical conductivity(σ) is the reciprocal of resistivity (p), thereforeConductivity, σ =1/ ρ = e(ne µe + nh µh ) Where ne & nh are free electron density & hole density respectively.13. Diagram

Donor energy level

Acceptor energy level

14. The drift velocity of a charge carrier is proportional to electric E.Therefore V = eET/m ie. V α E

But V cannot be increased indefinitely by increasing E . At high speed relaxation time (T) begins to decrease due to increase in collision frequency. S: so drift velocity saturates at thermal velocity (lOms-1). An electric field of 106 V/m causes saturation of drift velocity. Hence semi- conduction obey ohm’s law for low electrical field and above this field ( E < 106

V/m ) current becomes independent of potential.

15. The factors upon which transconductance of a transistor depend are as follows--i) Geometry of the transistorii) Doping levels.iii) Biasing of the transistors.

16. For faster action NPN Transistor is used .In an NPN transistor, current conduction ismainly by free electron ,whereas in PNP type transistor .it is mainly holes Since electron are more mobile than holes we prefer NPN for faster action as well as high conduction current.17. graphs:

:

18. Because the energy gap for Ge ( Eg = 0.7 ev) is smaller than the energy gap for Si (Eg = 1.1ev ) . Moreover, the germanium diode is much more open to the danger of high temperature affect than silicon at high voltage.19. The output of the AND gate is Y = A.B consequently the input of the OR gate are A andA.B . Then the final Y = A + A.B

Input for AND gate Output of AND gate

Input of OR gate

output of OR gate

A B Y= A.B A Y Y = A + Y0 0 0 0 0 00 1 0 0 0 01 0 0 1 0 11 1 1 1 1

20. The output of OR gate is A+B. Consequently, the inputs of AND gate are A+B & CHence the Boolean equation for the given circuit isY=(A+B).C

21) AND Gate using NAND GATE:-

A B Y= A.B0 0 00 1 01 0 01 1 1

22. NOT gate using NAND gate:-Truth Table:-

A Y=A0 11 0

23. Here β=50Ie =6.6mAβ=Ic /Ib

NAND acts as NOT gate

Ic=βIb =50Ib

Ie=Ic + Ib

Ib=0.129mA Hence, Ic =50 × 6.6/51 = 6.47mA β = α / (1 - α) α = β / (1 + β) α = 0.98

24. Here Ib = 100µA = 0.1mA Ic = 3mA

a) β=Ic / Ib = 30 β = α / (1- α)

∴α = 0.97 α = Ic / Ie

∴Ie = 3.1 mA b) Δ Ib = 20µA 0.02 mA βa.c. = Δ Ic / Δ Ib

βa.c. = 25

25. Here Ic=95% of Ie = (95 / 100 ) Ie

Ie = (100 / 95) × 5 mA = 5.26mA Ie= Ic+ Ib Ib = 0.25 mA26. Here Ie = 5mA Ic = 4.75 mA Ie = Ic + Ib Ib = 75 × 10-5 A V = Ib Rb V = 5V Rb = V / Ib Rb = 6.67 kΩ

27. This is AND logic gate Output wave form

A

B

Y

28. R

1.5V Here e.m.f of the source, E = 1.5V Voltage drop across the diode , Vd = 0.5V Maximum power rating of the diode I = (P / Vd) I= 0.2A Potential drop across resistance R V= E – Vd

= 1V R = V / I = 1 / 0.2 = 5Ω29. α = 0.96 , Ie = 7.2mA α = Ic / Ie ∴Ic = α Ie = 6.91 mA Ib = Ie – Ic

Ib = 0.29 mA.

30. β = 70 Ie = 8.8mA Ie= Ic + Ib

∴ Ib = 8.8 / 71 Ib = 0.124 mA

31. Ib = 105 × 10-6 A Ic = 2.05 × 10-3A β = Ic / Ib = 19.5 Also, Ie = Ib + Ic = 2.155 × 10-3 A α = Ic / Ie = 0.95

Ib = 27µA = 27 × 10-6 A β = Ic / Ib = 24.132. Ie = 7.89 × 10-3 A Ic = 7.8 × 10 –3 A Now αa.c. = Ic / Ie = 0.9886

We have, βac = αac / (1-αac) = 86.72 Also, βac = Ic / Ib

∴ Ib = Ic / βac

Ib = (7.8 × 10-3 ) / (86.72) Ib = 89.94 × 10-6A

UNIT-10HINTS FOR COMMUNICATION SYSTEMS

1) In a radar, a beam signal is needed in particular direction which is possible if wavelength of signal waves is very small. Since the wavelength of microwave is a few millimeter, hence they are used in radar.2 The television signals have frequencies in 100-200 MHz range. As ionosphere cannot reflect radio waves of frequency greater than 40 M back to earth, the sky waves cannot be used in the transmission of TV signals.3) A diode detector should have the following characteristic for proper detection a) high rectification efficiency. b) negligible loading effect on previous stage c) low distortion.4) Velocity factor (VF) of a cable is the ratio of reduction speed of light in the dielectric of the cable.Velocity of light in vacuum is 3 x108 m/sec. It reduces when light passes through a medium. Velocity of light in a medium is given by v=c / √ k & where, c-velocity of light in vacuum and k- is the dielectric constant of the medium Vf. =v/c=l/√kFor a line velocity factor is generally of the order of 0.6 to 0.9.5) Delta modulation involves simple pulse coding and decoding methods. A simple delta modulation uses just one bit per sample i.e. a ‘non-zero’ sample or one per sample. Thus, this method is convenient to use.

6) Two wire transmission line and coaxial cable are employed for AF and UHF region. For optical fiber is employed for optical frequency. 7). Mixing two frequencies across nonlinear impedance.8). When light strikeds the atoms of the Laser medium, it must stimulate emission rather than be absorbed. This means that more atoms must be in an excited state than in the ground state. This is an unnatural condition and is known as population inversion.9). Solid Lasers, Semi conductor Laser, Liquid laser, Gas Laser.10) Laser surgery ,Laser printing, Optical communication .11) 1)Transmission System, 2) Switching Systems and 3) Signaling Systems. 12) 1 .The length of the antenna required is so large (L = 5000m ) that is practically impossibleto set up it. 2 The energy radiated from the antenna in audio frequency range is Practically zero3 The audio signals transmitted from the different broadcasting stations will get inseparably mixed.13) Pulse modulation is a system in which continuous wave forms are sampled at regular intervals. Information regarding the signal is transmitted only at the sampling times together with any synchronizing pulses that maybe required. Pulse modulation is the process of transmitting signals in the form of pulses ( dis continuous signals ) by using special techniques.14) The term channel is commonly used to special the frequency range allotted to a particular transmission from a broadcast station or a transmitter eg a telephone channel is also used for a link in a transmitter and receiver.

15) Atmospheric pressure decreases with in crease in altitude. The high energy particles (ie & rays and cosmic rays) coming from outer space and entering our earths atmosphere cause ionization of the atoms of the gases present there . The ionizing power of these radiation decreases rapidly as they approach the earth. due to decrease in number of collision with the gas atoms . It is due to this reason that the electrical conductivity of earths atmosphere increases with altitudereceives the signal 16) Refractive index p of core of core of optical fibre is slightly higher than glass cladding . Light propagates through and along the fibre by the series of bounces caused by internal reflection at the interface of the core and cladding. For total internal reflection the light should enter the fibre at an angle θ in accordance with core of acceptance angle θc

NA = sinθc = √(µ12 - µ2

2 ) Numerical aperature depends upon diameter of the core It decreases as the diameter of core decreases vice versa17) i) Pulse Amplitude Modulation : Amplitude of the pulse varies in accordance with the modulating signal.

(ii) Pulse Position Modulation. : Pulse position (ie) time of rise or fall of the pulse ) changes with the modulating signal.

.18)Dielectric loss increase beyond this frequency.19)How does the effective power radiated by the antenna vary with wavelength?

Power is inversely proportional to wave length20)what should be the length of the dipole antenna for a carrier wave of 5 X 10 8hz ?

L =c/2 ν.21) four times.22) d= √2hR

d= √2x 0.1x 6400 = √1280 km

Area covered by broadcaste, A = π d2 = 3.14 x 1280= 3919.2 km2

population covered = Area x population density= 3919.2 x 1000 = 3919200

23) (i) 5 MHz <fc sky wave propagation (ionospheric propagation).

(ii) 100 MHz > fc satellite mode of communication.

24)

25) Maximum voltage of AM wave,Vmax = 16 = 8mV

2 Minimum voltage of AM wave,

Vmin = 4 = 2mV 2

ma = Vmax - Vmin

Vmax + Vmin

= 8-2 = 6 = 0.6 8+2 10

26) The AM wave equation is given by ;v = 5(1+0.6cos6280t) sin 221 X 104t volts ………….(i)

(i) Maximum amplitude of AM wave = EC + maEC =5 + 0.6 X 5 = 8V

Minimum amplitude of AM wave = EC - maEC =5 - 0.6 X 5 = 2V

(ii) The AM wave will contain three frequency viz fc-fs, fc, fc+fs

336-1 336 336+1335kHz 336kHz 337kHz

27) (i) The AM wave has sideband frequency of (fc + fs) and (fc - fs). Sideband frequency = (500+1) kHz and (500-1) kHz 501 kHz and 499 kHz

(ii) Bandwidth required = 499 kHz to 501 kHz = 2 kHz

28) PS = ½ ma2 PC

1.246 = 1 + ma2

2 ma

2/2 = 0.246 ma = (2 X 0.246)1/2 = 0.701 = 70.1%

29) Modulation index , mf = Maximum frequency deviation Minimum signal frequency

30) d = (2 × 6400 X103 × 160)1/2 = 45255m Coverage range, d = (2Rh)1/2

h2 = 4h1 = 4 × 160 = 640m

31) Radius of the area covered by TV broadcast isd = (2Rh)1/2

= 37500m = 37.5 km= 4.4 × 106

32) Microwave communication channel width =

GHz10

1002 ×

=0.2 GHz band width of channel = 8 KHz

= 2.5 × 10 4

33) Energy corresponding to λ = 1400nm = 1400 × 10-9 m is

E = λhc

= eV

19

19

106.11042.1

−

−

××

= 1eV For detection E must be equal to greater then Eg. Hence only suitable semiconductor is C.

34) Critical frequency fc and maximum electron density nmax are related as

fc = 9(nmax)2/1

Squaring we get nmax = 81

2fc

Given fc = 10MHz = 61010 × =

710 Hz

ie, nmax = 81)10( 7 2

= m 3121023.1 −×

HINTS FOR HOTS-CLASS-XII(PHYSICS)UNIT-1

HINTS FOR ELECTROSTAICS1. In the dielectric medium between the plates.2. High potential, as electrons are negatively charged.3. Zero4. Zero5. ML-1T-2

6. Zero7. No.8. Because they are indicators of electric field, extending to infinite distance.9. C is proportional to A (area) Therefore C2 = 2C1

Since C = Q/V ,So the slope represents more capacitance. Hence P represents C2,Q represents C1

10. Each charge experiences two forces each of magnitude F inclined at an angle of 600.Their resultant is given by [F2 + F2 +2F2cos 60]1/2 = √3.F11. (i)According to defn of P.D.,

VP>VQ .So VP-VQ is +ve for q>0.(ii)For q<0, VQ >VP .So VP-VQ is -ve

12. Flux = 0, since Qen = 013. Due to polaraisation,opposing electric field is created.14. Electric field at the midpoint of a dipole of length 2a is 2kq/a pointing towards the –ve charge or in the

direction opposite to the dipole moment. 15. Inside the cavity field at any point is uniform and non zero.16. No. If the initial velocity of the charged particle makes a certain angle with a line of force, then the

charged particle shall not move along the line of force.17. E= -dv/dr= -d(q/4πεor)/dr = q/4πεor2

18. Yes, at the mid point of electric dipole.19. U = kq2/a- kq2/2a- kq2/2a = 0.20. C’ = C+C+C = 3C = 75μF

Therefore, charge = 75μF x 4200 V = 315 mC* 21. Total ф = q0/εo = 2/εo ,→ flux through one face = ф/6 = 1/3 εo.

22. q →------------ Q 1/2mv2 = kQq/r Or, v2 α 1/r Or, r α 1/v2 Or, r’ = r/4

23. V = V1+V2

Q =C1V = 6 X 10-6 X 2 = 12µCAs C2 is in series same amount of charge will also flow through it.

Now V2= Q/ C2 = (12 X 10-6)/ (12 X 10-6) = 1 VoltTotal battery Voltage,V = 2 + 1 = 3 Volt

24.Capacitance of parallel plate capacitor with air between the plates is C0 = ε0A/dWhen the separation between the plates reduced to half,

C1=ε0A/(d/2) = 2ε0A/dThus final capacitance is C2=10 X 8 pF=80 pF

25.The arrangement is of 5 capacitors in series.Therefore 1/C’= (1/C) +(1/C) +(1/C) +(1/C) +(1/C) =(5/C)

Therefore C’=C/5 Or 5=C/5 or C=25µF. 26. The charge given to a capacitor is given by q=CV

So the remaining energy,qV- 1/2 qV =1/2 qV is lost as heat27. On equatorial line ,the direction of electric field is reversed to that of axial line. Hence the angle between

electric dipole moment& electric field strength is 1800

28. Eq. network is the Eq.capacitance = (2C series C) (2C series C)

=4C/3 29. 4/3 πR3 = 8 x 4/3 πr3

30. dV = -E . dx = -2 x 103 x 4 V = -8 x 103 V E = E1 – E2 = Q1 - Q2/2A εo

Now, V1 – V2 = E . d = Q1 - Q2/2C 32. U = k (q2/a + qQ/a + qQ/√2a)

But, U = 0 Therefore, Q = -2q/(2 + √2)

33. Ε= q/4пε0[1+1/4+1/16+1/64] = q/3пε0.

34. Let q & q’ be the charges on inner and outer sphere.Hence q/4πr2= q’/4πR2

q=Qr2/(R2+r2) & q’=Q-q=Q- (Qr2/R2+r2)Now potential at O is given by

35 (i) Dipole has two equal and opposite charges. In the uniform electric field they will experience equal and

opposite force. Net force is zero. So there can’t be any translatory motion.(ii) Torque, τ = r⊥F=2l sinθ qE = p X E

Torque experienced by the dipole will rotate it.So the direction of the torque will be outward from the

surface.36. We have V=ar2+b

The electric field,Er = - dV/dr=-2ar-------(1)From Gauss theorem, ∫ E.dS=q/ε0, where S is the spherical surface containing the charge qOr E.4πr2=(1/ε0 )(4π/3) r3ρ-------(2)From 1 and 2 ρ=-6πε0

UNIT-2

HINTS FOR CURRENT ELECTRICITY.1. They have high resistively and low temperature coefficient of resistance.2. 22*102 Ω ± 10%3. Resistively remains the same.4. As Vd α V.

The different velocity will be doubled. 5. R=ρ(l/A) = ρ(l2/Al)

= ρl2/V) since, ρ and V are constants

therefore, R α l2

(R2/R1) = (l2/l1)2=9 because R2=9R1

=9*10=90 Ω6. Now, 1/R=1/R1+1/R2

because l = 48/240 = 0.2 m7. a) in parallel, power dissipation α 1/R Therefore 3Ω wire will dissipate more power b) In series , power dissipation α R Therefore 9Ω wire will dissipate more power

8. R100/R27.5 = (1+100α)/(1+27.5α)On solving, we get

α = 0.0039/˚c9. Superconductors are the materials that lose all its resistance at very low temperature =0 K Application: Super conductor are used

b) In making very strong electromagnets b) In producing very high speed computers .

10. Resistivity of copper is less , hence manganin wire is thicker.11. High value of resistivity and low value of temperature coefficient.12. Resistivity will be unchanged because it depends upon nature of the materials.13. B is more sensitive.14. Reduced by half.15. Relaxation time decreases with increase of temperature.16. Increase in heat.17. Reistance remains same.18. Pα 1/R. (i.e) 25Watts19. 16 times of the original reisitance.20. (i)Series - Iron

(ii)Parallel - Copper.21. R = ρ L/A.

(i.e) 10.25%22. Parallel(i.e) R.23. 1:4.24. 2:125. R.

Unit-5

HINTS FOR EM WAVE.1. Due to change in electric field.

3. Same as velocity of light.

4. 90 degree

rays, radio waves, X-rays, UV rays ץ .6

7. i)X-rays

ii)microwave.

8. doubled

9. Charge.

10..Increases.

11. Gauss’s law

13. X-rays because λ = 0.825A

14. Micro waves

15.By varying potential difference.

16. Ex and By.

UNIT-6 ANSWER KEY FOR OPTICS

1.Ans:Wavelength

2. Ans:Photo electric effect.

3. Ans:4NL.

4. Ans: It will turn gradually blue.

5. Ans:Zero.

6. Ans:90 .

7. Ans:µo= µe

8. Ans:90

9. Ans:Violet.

10. Ans: Shrinks.

11. Ans.(i) The upper part of the mirror is convex.

(ii) The middle part of the mirror is concave.

(iii) The lower part of the mirror is plane.

12. Ans. The ray of light bends away from the normal.

13. Ans.

As µ = sini/sinr = c/v or v = sinr/sini *c

For a given angle of incidence, v ∝ sinr, vA ∝ sin15° , vB ∝ sin25° , vC ∝ sin35°

But sin15° < sin25°< sin35°.

∴ vA < vB < vC .

i.e. the velocity of light is minimum in medium A.

14 Ans.For i = 90°, lateral shift is maximum and is equal to the thickness of the slab.

d = t sin( i – r )/ cos r

dmax = t sin(90°- r)/cos r = t cos r/cos r = t.

15. Ans.The apparent shift caused by a slab of thickness ‘t’ is given by

d = t( 1 -1/µ )

As the refractive index of the galass is maximum for red light, so red coloured letters are

more raised up.

16. Ans. No Apparent depth is maximum for that part of the bottom of the tank which is

observed normally. Apparent depth decreases with increasing obliquity. Due to this unequal

refraction, the flat bottom of the tank appears concave.

17. Ans. For glass-air interface, sin ic = 1/aµg

The critical angle i’c for glass water interface is given by

Sin i’c = 1/ wµg

Now wµg < aµg,

Sin i’c > Sin ic or i’c > ic

18. Ans. Light entering water is totally reflected from the air bubble. For the observer, this

light appears to come from the buble. So it shines.

19. Ans. As the critical angle for diamond-oil interface is greater than that for the diamond –

air interface, so the shining of diamond reduces when it is dipped in a transparent oil.

20. Ans. It behaves like a biconvex lens.

21. Ans. Air bubble has spherical surface and is surrounded by medium ( water) of higher

refractive index. When light passes from water to air it gets diverged. So air bubble behaves

as a concave lens.

22. Ans.When the refractive index of the liquid is same as the lens material, no light will be

reflected by the lens and hence it will not be visible. ∝

23. Ans. No, the image will be formed at the same position.From lens maker’s formula,

1/f = (µ -1) [ 1/R1 –1/R2 ] , it is clear that when we interchangeR1 and R2, the magnitude of

‘f’ remains the same.

24. Ans. focal length‘f’ of a convex lens is related to its refractive index as

f ∝ 1/(µ -1)

As wµg < aµg , so focal length of a convex lens will increase when it is immersed in water.

25. Ans.Focal length, f ∝ 1/(µ -1)

As µR < µV , so the focal length of a convex lens will increase when red light is used.

26. Ans:For the original lens: R1 = +R and R2 = -R, so we can write

1/f = (µ -1) [ 1/R +1/R ] = 2(µ -1)/R.

When one surface is made plane by grounding, we have R1 = +R and R2 = - ∞.

Therefore, 1/f ’ = (µ -1) [ 1/R +1/∞ ] =(µ -1)/R

∴ f ‘ / f = 2 or f ‘ = 2f

Thus the focal length becomes double and power becomes one –half.

27. Ans. When the prism is held in water,

wµg = Sin (A + δm/2) /SinA/2

As wµg < aµg , so the angle of minimum deviation decreases in water.

28. Ans. Total internal reflection.

29. Ans. The sunlight will not be scattered in the absence of atmosphere. So the sky will

appear dark.

30. Ans. Clouds have large particles like dust and water droplets which scatter light of all

colours almost equally, hence clouds generally appear white.

31Ans. When the sun or the moon is seen through a thin veil of high clouds, holes are seen.

These are formed due to reflection of light by the icy crystals present in the atmosphere.

32. Ans. Ultra-violet light has wavelength shorter than that of violet light. Bees have some

retinal cones that are sensitive to ultra violet light, so they can see objects in ultra-violet light.

Human eyes do not posses retinal cones sensitive to ultra-violet light, so human beings

cannot see objects in ultra-violet light. In other words, human beings are ultra-violet blind.

33. Ans. In a chicken’s eye, the retina has a large number of cones but only few rods. The

rods are sensitive to bright light only. That is why a chicken is not able to see in dim light. As

it needs bright light to see, so it wakes up early in the morning with the sunrise and goes to

sleep by sunset.

34. Ans. Magnifying power of a simple microscope ,

m = 1 + D/f

as fV < fR so the magnifying power is greater when the object is seen in violet light.

35. Ans. This is done so that the objective lens forms image within the focal length of the

eyepiece.

36. Ans. (i) We should take f0 =1 cm and fe = 3cm for a microscope.

(ii) We should take f0 = 100 cm and fe = 1 cm for a telescope.

37. Ans. Yes, because the light gathering power of objective will increase and even faint

objects will become visible.

38. Ans. For relaxed eye,

L = f0 + fe (normal adjustment)

For least distance of distinct vision,

L’ = f0 + ue , ue < fe

Therefore, L’ < L. so that distance between the two lenses should be decreased.

SHORT ANSWER TYPE QUESTIONS

39. Ans. Here u = - (f+a), v = -(f+b), f = -f

As 1/f = 1/u +1/v

F = uv/ u + v

Or -f = [-(f+a)] x [-(f+b)] / -(f+a) – f(a+b)

= f2 + af +bf +ab / -(2f+a+b)

or 2 f2 + af +bf = f2 + af +bf +ab

or f2 = ab

40.Ans. (i) Angle of refraction (θ/2) in medium 2 is less than the angle of incidence (θ) in

medoum 1 i.e. the ray bends towards the normal in medium 2. so medium 2 is optically

denser than medium 1.

(ii) From Snell’s law,

µ = sin i/sin r = sin θ/ sin θ/2 = 2sin θ/2 cos θ/2 /sin θ/2 = 2 cos θ/2

Also µ = c1 / c2

hence 2 cos θ/2 = c1 / c2 or θ = 2cos-1(c1 / 2c2) .

41. Ans. The point of convergence shifts away from the glass, as shown in the ray diagram given below. The screen has to be moved towards right to receive the point of convergence again.

42. Ans. Real depth = y cm

Apparent depth = y- x cm

Refractive index of oil,

µ = real depth/ apparent depth = y / y-x

43. Ans. Using Snell’s Law for refraction from glass to air,

Sin i/sin r = gµa = v / c

Where c is the speed of light in air and v is the speed of light in glass, In the consition of critical incidence, we have i = ic and r = 900

Sin ic/ sin 900 = v / c or Sin ic = v / c

Or ic = sin -1 ( v / c)

44. Ans. Twinkling of stars. The light from stars undergoes refraction continuously before it reaches earth. So the apparent position of the stars is slightly different than its actual position. Due to variation in atmosphere conditions, like change in temperature, density etc., and this apparent position keeps on changing. The amount of light entering our eyes from a particular star increases and decreases randomly with time. Sometimes, the star appears brighter and other times, it appears fainter. This gives rise to the twinkling effect of stars.

The planets do not show twinkling effect. As the planets are much closer to the earth, the greater and the fluctuations caused in the amount of light due to atmospheric refraction are negligible as compared to the amount of light received from them.

45. Ans. Light from the stars near the horizon reaches the earth obliquely through the atmosphere. Its path changes due to refraction. Frequent atmospheric disturbances change the path of light and cause twinkling of stars. Light from the stars overhead reaches the earth normally. It does not suffer refraction. There is no change in its path. Hence there is no Twinkling effect.

46. Ans. Magnification produced by any lens, m = v/u = f / f + u

given m = ± N ±N = f / f + u

or f + u = ± f / N or u = - f ± f / N

hence magnitude of object distances,

|u| = f ± f / N

given P = 1/f = + 2.5 D

f = 1/ 2.5 = 0.4 m = 40 cm Also N = 4 |u| = 40 ± 40/4 = 40 ± 10 = 50 cm or 30 cm.

47. Ans. (a) for a convex lens, f>>0 and for an object on left, u<0. when the object is placed within the focus of a convex lens,

0 < |u| < f or 0< 1 / |u| > 1/f 1/v = 1/f+1/u=1/f-1/|u|<0

i.e. v < 0 so a virtual image is formed on left.Now as u<0 and v<0, so 1/v = 1/f + 1/u

= - 1/ |v| = 1/f – 1/|u| or 1/|u| - 1/|v| = 1/f As f>0 1/|u| - 1/|v| > 0 or 1/|u| > 1/|v| or |u|<|v|i.e. |v|>|u| |m| = |v/u| > 1

Hence image is enlarged. (b) For a concave lens f<0 and for an object on left, u<0 1/v = 1/f +1/u = 1/|f| - 1/|u|

= - [1/|f|+1/|u|] < 0 for all u. i.e. v<0 for all values of u. hence a virtual image is formed on the left.Also 1/|v| = 1/|f| + 1/|u| 1/|v| > 1/|u| Or |v|<|u| |m| = |v/u| < 1 i.e. the image is diminished in size. 48. Ans. A hollow prism contains air which does not cause dispersion. The faes AB and AC of the hollow prism behave like parallel sides of glass plates. The beam is laterally deviated at each of the two refracting faces. However, the rays of different colours emerge parallel to each other. So there is no dispersion. 49. Ans. (i) Moon has no atmosphere. There is no scattering of light. Sunlight reaches moon

straight covering shortest distance. Hence sunrise and sunset are abrupt.

(ii) Moon has no atmosphere. So there is nothing to scatter sunlight towards the moon. No

skylight reaches moon surface. Sky appears black in the day time as it does at night.

(iii) No water vapours are present at moon surface. No clouds are formed. There are no rains

on the moon. So rainbow is never observed.

NUMERICAL PROBLEMS

50. Ans. The total apparent shift in the position of the image due to all the three media is

given by

d = t1[1-1/(µ1)+ t2[1-1/(µ2)+ t3[1-1/(µ3)

Given t1 = 4.0 cm, t2=6.0 cm , t3 =8.0 cm

µ1 = 1.5 , µ2=1.4 , µ3 =1.3 cm

d= 4.0(1-1/1.5)+6.0(1-1/1.4)+8.0(1-1/1.3)

= 1.33 + 1.71 + 1.85 = 4.89 cm

51. Ans. Clearly , the fish can see the outside view of the cone with semi vertical angle,

But µ = 1 / sin ic

Or 1/3 = 1 / sin ic

Or sin ic = ¾ = 0.75

θ/2 = ic = sin-1 (0.75) = 48.60

52. Ans. (i) As the lens forms a real iamge, it must be a convex lens.

(ii) From the graph, when u=20 cm , we have v = 20 cm.

For the convex lens forming a real iamge, u is negative and v and f are positive.

U = -20 cm v = +20cm

Using this lens formula,

1/f = 1/v – 1/u = 1/20 – 1/-20 = 1/10 or f = + 10 cm

53. Ans. A=600 , δm=300

i = e = ¾ A = 450 ,

as A + δ = i + e

60 + δ = 45 + 45

or δ = 300

Refractive index,

µ = sin a+δm /2 /sin A/2 = sin 600+300/2 / sin 600/2

= sin 450/sin300 = 1/√2 / ½ = √2 = 1.414

WAVE OPTICS

54. Ans. Two light sources will be coherent if

(i) The frequency of the two light sources is same and,

(ii) The phase difference between them remains constant.

55. Two independent light sources cannot act as coherent sources. Why?

Ans. Two independent sources of light cannot be coherent. This is because light is emitted by

individual atoms, when they return to ground state. Even the smallest source of light contains

billions of atoms which obviously cannot emit light waves in the same phase.

56. Ans. Fringe width , β = λD/d

i.e. β ∝ 1/d , when d → 0, β → ∞

fringe width is very large. Even a single fringe may occupy the entire screen. The

interference pattern cannot be observed.

57. Ans. The given path difference satisfies the condition for the minimum of intensity for

yellow light, Hence if yellow light is used, a dark fringe will be formed at the given point. If

white light is used, all components of white light except the yellow one would be present at

this point.

58. Ans. The positions of bright and dark fringes will change rapidly. Such rapid changes

cannot be detected by our eyes. A uniform illumination is seen on the screen i.e. interference

pattern disappears.

59. Ans. For diffraction to take place the wave length should be of the order of the size of the

obstacle. The radio waves (particularly short radio waves) have wave length of the order of

the size of the building and other obstacles coming in their way and hence they easily get

diffracted. Since wavelength of the light waves is very small. They are not diffracted by the

buildings.

60. Ans. Muslin cloth is mde of very fine threads and as such fine slits are formed. White

light passing through these silts gets diffracted giving rise to colored spectrum. The central

maximum is white while the secondary maxima are coloured. This is because the positions of

secondary maxima (except central maximum) depend on the wavelength of light.

In a coarse cloth, the slits formed between the threads are wider and the diffraction is not so

pronounced. Hence no such spectrum is seen.

SHORT ANSWER QUESTIONS

61. Ans. A wavefront is a surface obtained by joining all points vibrating in the same phase.

A ray is a line drawn perpendicular to the wavefront in the direction of propagation of light

wave.

The wavefronts of light emerging from a point source are spherical, as shown in figure. When

a point source is placed at the focus of a convex lens, the emerging light has the plane

wavefronts, as shown in figure.

62. Ans. As shown in figure the bright fringes B1 and b2 on either side of O coincide with S1

and S2 respectively.

Clearly ,

β = d/2

As β = Dλ/d d/2 = Dλ/d or λ = d2/2D

63Ans. For intensity distribution of light in diffraction at a signal slit, see figure

Width of central maximum is given by

β0 = 2Dλ/d

(i) When wavelength of light λ used is increased, the width of central maximum increases.

(ii) When width of the slit is increased, the width of central maximum decreases.

Wavelength of light in water decreases, so width of central maximum also decreases.

64. Ans. When the monochromatic source is replaced by a source of white light, the

diffraction pattern shows following changes:

(i) In each diffraction order, the diffracted image of the slit gets dispersed into component

colours of white light. As fringe width ∝ wavelength, so the red fringe with higher

wavelength is wider than violet fringe with smaller wavelength,

(ii) In higher order spectra, the dispersion is more and it causes overlapping of different

colours.

65. Ans. R.P.of a compound microscope

= 2 µ sinθ/λ = 2 µ sinθ x v /c

(i)When the frequency v of the incident light increases, the resolving power increases (R.P. ∝

v).

(ii) Resolving power does not change with change in focal length of objective lens.

(iii) When the aperture of the objective lens increases, the semi-vertical angle θ increases and

hence the resolving power of the microscope increases.

66. Ans. According to the Brewster law, when a ray of light is incident on a transparent

refracting medium at olarising angle,

µ = tan ip

but ip + rp = 900 or ip = 900 - rp

µ = tan(900 - rp ) = Cot rp = 1/ tan rp

As ic is the critical angle for the transparent medium, so

µ = 1/sin ic

on comparing (i) and (ii) we get

tan rp = sin ic or rp =tan-1(sin ic)

NUMERICAL QUESTIONS

67. Ans. The resultant intensity at a point where phase difference is Φ is

I R = I1 +I2+2√I1I2 Cos Φ

As I1 =I and I2 = 4I therefore

I R = I +4I+2√1.4I Cos Φ = 5I +4I cos Φ

(i) when Φ =0 , I R = 5I +4I cos 0 = 9 I

(ii) when Φ =π/2 , I R = 5I +4I cos π/2 = 5 I

(iii) when Φ =π , I R = 5I +4I cos π = I

68. Ans. The fringe width in the two cases will be β = Dλ/d

And β ‘= D’λ/d

β - β’ = (D-D’)λ/d

or wavelength λ = (β - β’ )d / (D-D’)

But D-D’ = 5 x 10-2 m

And β - β’ = 3 x 10-5 m , d= 10-3m

λ = 3 x 10-5 x 10-3 / 5 x 10-2 = 6 x 10-7 m = 6000A

69. Ans. Let I be the intensity of beam I incident on first glass plate. Each plate reflects 25%

of light incident on it and transmits 75%.

Therefore,

I1 =I and I2 = 25/100I = I/4

I3 =75/100 I = 3/4I

I4 = 25/100 I3 = ¼ x ¾ I = 3/16 I

I5= 7/100 I4= ¾ x 3/16 I = 9/64 I

Amplitude ratio of beams 2 and 5 is

R = √ I2/I5 = √I/4 x 64/91 = 4/3

Imin/ Imax = [r-1/r+1]2 = [4/3-1 / 4/3+1]2 = 1/49 = 1:49

UNIT-7HINTS FOR DUAL NATURE OF MATTER

1. E = hν =hc/λ E ∝ 1/λ energy of proton reduces to half. 2. Alkali metals have too low work functions. Even visible light can eject electrons from them.3. UV are most effective since they have highest frequency hence more energetic.4. Yes. X-rays cause photoelectric effect in sodium, zinc & copper.5. K.E of photons remains unaffected since they do not depend6. stopping potential V0 = Kmax/e = 5ev/e =5 V7. w0 = hν0 = hc/λ0

∴λ0 ∝ 1/ w0

Since sodium has lower work functions than copper it is easier for electron ejection. As it is lower work function, higher wavelength.

8. Photocells are used for reproduction of sound.9. 1/2 mv2 = (m2 v2)/2m = p2/2m According to De Broglie wave length λ = h/p λe/ λp = pp/pe = √(mp /me ) me < mp

λe > λp

electrons have greater De broglie wavelength than proton .10. θ = 90 - φ/2 = 90 – 52/2 =64° 11. Ε = hc / λ = 3.3×10-19 J12. KE of photoelectrons is given by Einstein’s photoelectric equation.

Ek =1/2 mv2

= h w0 -ט

V α 1/ √λ As wavelength decreases velocity increases.

13. stopping potential frequency

14. E= h c/ 6.6 = (ג x 10-34 x 3 x 108)/ (4 x 10-7) = 4.98 x 10-19 j E= (4.98 x 10-19)/ (1.6 x 10-19) =3ev Hence, metal x will emit electrons.

15. For a photon E1=hc/ג For an electron ג=h/mv or m=h/גv E2/E1=c/v>1 Therefore,E2>E1.thus, electron has total energy greater than that of photon.h/mv=ג=eג=phג .16 K.E. of electrons E=1/2mv2=1/2 m [h/m 2ג]

=h2/2m 2ג

∴Eph=Ee(2mc ג/h)17 . E=1/2 mv2=m2v2/2m=p2/2m ∴p=√2Emh/p=h/√2Em=ג∴ From Kinetic theory of gases average K.E.=3/2 KTh/(√2m(3/2 KT)=h/(√3m KT)=ג 18. The energy of light obtained from the bulb is much less than work function of the wodden block.Hence no photon electrons are emitted.

19. Mo will not emit photo electron , because its work function is more than 4 ev.

20. Alpha particles due to its largest mass.

21. R α 1/q

22. Ee = Mv2/r

23. λ = λ/√2.24. λ/2.25. λ = hc/Φ = 2823 A.

UNIT-8HINTS FOR ATOMS AND NUCLEI.

1) The entire positive charge and the mass were concentrated at one place inside the atom, called the nucleus.2) A larger number of alpha particle went through undeflected.3) R=R0A1/3 R1/R2=A1

1/3/A21/3=1 1/3/27 1/3=1/3

4) R= R0A1/3 R1/R2=A11/3/A2

1/3=(1/8)1/3 =1/2 R1: R2=1:25) α-particles have more ionizing power than β-particles.

6) N/N0=(1/2)n

t=2T =2*30 =60 days7) They are neutral in nature and get absorbed by nucleus, thus distributing the neutron proton ratio.8) The ratio of neutrons to proton ratio increases, after the emission of a ά – particle.9) Owing to greater mass and charge, it is able to knock out/pull out electrons which

colliding with atoms and molecules in its path.10) T = 20 minutes t = 60 minutes N/No = (1/2)n = (1/2)60/20 = (1/2) 3 =1/8

After one hour, 1/8th of the original mass would remain.11) The nucleus looses energy, but remains same isotope it was.12) No a nucleus either emits a ά – particle or a β – particle and if left in the exited state, it

may emit γ – ray also.13) 1:1 (independent of A).14) At t = T ½ N = No / 2 Using N = No e- λt

No / 2 = No e- λt ½

Solving we get, T1/2 = ln2/λ = 0.693/ λ15) Size of nucleus can approximately be estimated using the concept of distance of closest

approach. The rebounding particle is selected and its information is substituted in the expression Ro = 1/4πEo x 2Ze2 / E for ά particle where E is its energy.

B. Using N =No (1/2)t/π/2 N =1/16No1/16 = ½(1/2)t/π/2 T1/2 = 30/4 = 7.5 days.

17) 235 = 142+Y+3Y = 90And 92 = 57+Z+0Z = 35

18) a) Yes. Since X & Y are having same atomic number. b) Y 4 3 is likely to be more stable because for its neutron to proton ratio is smaller.19) Disintegration constant λ =0.693/T1/2

=0.693/30x24x60x60Therefore T avg =1.44xT1/2 = 1.44x30 = 43.2 days.

20) Remaining amount (undecayed) =1/4N0

Using N= N0 (1/2)t/T1/2

1/4= (1/2)t/60

Solving t=2x60=120days21) Mass defect ∆m= (22.9945-22.9898) =0.00474 Energy Q= (0.00474) (931.5) =4.4MeV Hence the energy of beta particle can range from 0 to 4.4MeV.

22) a) Using R=R0e-λt

2700=4750 e-5t λ=0.113min-1

b) Using T1/2=0.693/λ =0.693/0.113 =6.132min

23) In the process of beta decay, a neutron gets converted to proton inside the nucleus . Hence number of neutrons decreases by one whereas number of proton increases by one. Hence n/p ratio decreases. 210Bi83-------------------- 210Po84 + 0β-1 + γ

Before decay =127/83 After decay=126/8424) Let λ and λ’ be the decay constant of element A and B respectively. Given is T1/2(A) = T1/2(B) 0.693/λ=1/ λ’ or λ/ λ’ =0.693

Let N be the number of atoms of each of the two samples and R and R’ their disintegration rate, then

R/R’= λN/ λ’N= λ/ λ’=0.693R’ > R

25) Find ∆m using ∆m= (7x1.00783+7x1.00867-14.003074) U Calculate ∆Eb=∆mx931.5 MeV

26) a) 226Ra88--------------222Rn86 +4He2

b) 32P15------------------32S16 +0e-1 +γ c) 32P15------------------11B5 +0e+1 + γ 27) i) 6Li3 +1n0 ----------3H1 + 4He2 +Q (energy) ii) Q=∆mx931 MeV Where ∆m=6.01512+1.0086654-4.0026044-3.0100000

UNIT-9HINTS FOR ELECTRONIC DEVICES

1. Conductor - no energy gapSemi Conductor - It is of the order of 1 ev.Insulator - 6 ev (or) more than 6 ev.2. When a semi conductor is heated more & more electrons get enough energy to jump across the forbidden energy gap from valence band to the conduction band, where they are free to conduct electricity. Thereby increasing the conductivity of a semi conductor.3. In an intrinsic semi conductor the number of free electrons and holes is same.4. C.B _ _ _ _ _ _ _ _ _ _ _

V.B5.

C.B

_ _ _ _ _ _ _ _ _ _ _ _

V.B

6. About 0.3V germanium .About 0.67V for silicon.7. Transistor is a current controlled devices.8. Reverse biased.9.NAND GATE.10.If ne/nh =1 . Hence A is intrinsic semi conductor.If ne/nh <1 , ne<nh hence B is P-type.

Donor energy level

Acceptor energy level

11. For half wave rectification 50 Hz. For Full wave rectification 100Hz.12. Mobility of an electron is defined as the drift velocity of electron per unit electric field, i.e. μe= Ve /EMobility of a hole is defined as the drift velocity of hole per unit electric field,i.e. μ h= Vh /E

The electrical conductivity(σ) is the reciprocal of resistivity (p), thereforeConductivity, σ =1/ ρ = e(ne µe + nh µh ) Where ne & nh are free electron density & hole density respectively.13. Diagram

14. The drift velocity of a charge carrier is proportional to electric E. Therefore V = eET/m ie. V α E But V cannot be increased indefinitely by increasing E . At high speed relaxation time (T) begins to decrease due to increase in collision frequency. S: so drift velocity saturates at thermal velocity (lOms-1). An electric field of 106 V/m causes saturation of drift velocity. Hence semi- conduction obey ohm’s law for low electrical field and above this field ( E < 106

V/m ) current becomes independent of potential.

15. The factors upon which transconductance of a transistor depend are as follows--i) Geometry of the transistorii) Doping levels.iii) Biasing of the transistors.

16. For faster action NPN Transistor is used .In an NPN transistor, current conduction ismainly by free electron ,whereas in PNP type transistor .it is mainly holes Since electron are more mobile than holes we prefer NPN for faster action as well as high conduction current.17. graphs:

:

18. Because the energy gap for Ge ( Eg = 0.7 ev) is smaller than the energy gap for Si (Eg = 1.1ev ) . Moreover, the germanium diode is much more open to the danger of high temperature affect than silicon at high voltage.19. The output of the AND gate is Y = A.B consequently the input of the OR gate are A and A.B . Then the final Y = A + A.B

Input for AND gate Output of AND gate

Input of OR gate

output of OR gate

A B Y= A.B A Y Y = A + Y0 0 0 0 0 00 1 0 0 0 01 0 0 1 0 11 1 1 1 1

20. The output of OR gate is A+B. Consequently, the inputs of AND gate are A+B & CHence the Boolean equation for the given circuit isY=(A+B).C

21) AND Gate using NAND GATE:-

A B Y= A.B

0 0 00 1 01 0 01 1 1

22. NOT gate using NAND gate:-Truth Table:-

A Y=A0 11 0

23. Here β=50 Ie =6.6mA β=Ic /Ib

Ic=βIb =50Ib

Ie=Ic + Ib

Ib=0.129mA Hence, Ic =50 × 6.6/51 = 6.47mA β = α / (1 - α) α = β / (1 + β) α = 0.98

24. Here Ib = 100µA = 0.1mA Ic = 3mA

a) β=Ic / Ib = 30 β = α / (1- α)

∴α = 0.97 α = Ic / Ie

∴Ie = 3.1 mA b) Δ Ib = 20µA 0.02 mA βa.c. = Δ Ic / Δ Ib

βa.c. = 25

25. Here Ic=95% of Ie = (95 / 100 ) Ie

Ie = (100 / 95) × 5 mA = 5.26mA Ie= Ic+ Ib Ib = 0.25 mA26. Here Ie = 5mA Ic = 4.75 mA

NAND acts as NOT gate

Ie = Ic + Ib Ib = 75 × 10-5 A V = Ib Rb V = 5V Rb = V / Ib Rb = 6.67 kΩ

27. This is AND logic gate Output wave form

A

B

Y

28. R

1.5V Here e.m.f of the source, E = 1.5V Voltage drop across the diode , Vd = 0.5V Maximum power rating of the diode I = (P / Vd) I= 0.2A Potential drop across resistance R V= E – Vd

= 1V R = V / I = 1 / 0.2 = 5Ω29. α = 0.96 , Ie = 7.2mA α = Ic / Ie ∴Ic = α Ie = 6.91 mA Ib = Ie – Ic

Ib = 0.29 mA.

30. β = 70 Ie = 8.8mA Ie= Ic + Ib

∴ Ib = 8.8 / 71 Ib = 0.124 mA

31. Ib = 105 × 10-6 A Ic = 2.05 × 10-3A β = Ic / Ib = 19.5 Also, Ie = Ib + Ic = 2.155 × 10-3 A α = Ic / Ie = 0.95

Ib = 27µA = 27 × 10-6 A β = Ic / Ib = 24.132. Ie = 7.89 × 10-3 A Ic = 7.8 × 10 –3 A Now αa.c. = Ic / Ie = 0.9886

We have, βac = αac / (1-αac) = 86.72 Also, βac = Ic / Ib

∴ Ib = Ic / βac

Ib = (7.8 × 10-3 ) / (86.72) Ib = 89.94 × 10-6A

UNIT-10HINTS FOR COMMUNICATION SYSTEMS

1) In a radar, a beam signal is needed in particular direction which is possible if wavelength of signal waves is very small. Since the wavelength of microwave is a few millimeter, hence they are used in radar.2 The television signals have frequencies in 100-200 MHz range. As ionosphere cannot reflect radio waves of frequency greater than 40 M back to earth, the sky waves cannot be used in the transmission of TV signals.3) A diode detector should have the following characteristic for proper detection a) high rectification efficiency. b) negligible loading effect on previous stage c) low distortion.4) Velocity factor (VF) of a cable is the ratio of reduction speed of light in the dielectric of the cable.Velocity of light in vacuum is 3 x108 m/sec. It reduces when light passes through a medium. Velocity of light in a medium is given by v=c / √ k & where, c-velocity of light in vacuum and k- is the dielectric constant of the medium Vf. =v/c=l/√kFor a line velocity factor is generally of the order of 0.6 to 0.9.5) Delta modulation involves simple pulse coding and decoding methods. A simple delta modulation uses just one bit per sample i.e. a ‘non-zero’ sample or one per sample. Thus, this method is convenient to use.

6) Two wire transmission line and coaxial cable are employed for AF and UHF region. For optical fiber is employed for optical frequency. 7). Mixing two frequencies across nonlinear impedance.

8). When light strikeds the atoms of the Laser medium, it must stimulate emission rather than be absorbed. This means that more atoms must be in an excited state than in the ground state. This is an unnatural condition and is known as population inversion.9). Solid Lasers, Semi conductor Laser, Liquid laser, Gas Laser.10) Laser surgery ,Laser printing, Optical communication .11) 1)Transmission System, 2) Switching Systems and 3) Signaling Systems. 12) 1 .The length of the antenna required is so large (L = 5000m ) that is practically impossibleto set up it. 2 The energy radiated from the antenna in audio frequency range is Practically zero3 The audio signals transmitted from the different broadcasting stations will get inseparably mixed.13) Pulse modulation is a system in which continuous wave forms are sampled at regular intervals. Information regarding the signal is transmitted only at the sampling times together with any synchronizing pulses that maybe required. Pulse modulation is the process of transmitting signals in the form of pulses ( dis continuous signals ) by using special techniques.14) The term channel is commonly used to special the frequency range allotted to a particular transmission from a broadcast station or a transmitter eg a telephone channel is also used for a link in a transmitter and receiver.15) Atmospheric pressure decreases with in crease in altitude. The high energy particles (ie & rays and cosmic rays) coming from outer space and entering our earths atmosphere cause ionization of the atoms of the gases present there . The ionizing power of these radiation decreases rapidly as they approach the earth. due to decrease in number of collision with the gas atoms . It is due to this reason that the electrical conductivity of earths atmosphere increases with altitudereceives the signal 16) Refractive index p of core of core of optical fibre is slightly higher than glass cladding . Light propagates through and along the fibre by the series of bounces caused by internal reflection at the interface of the core and cladding. For total internal reflection the light should enter the fibre at an angle θ in accordance with core of acceptance angle θc

NA = sinθc = √(µ12 - µ2

2 ) Numerical aperature depends upon diameter of the core It decreases as the diameter of core decreases vice versa17) i) Pulse Amplitude Modulation : Amplitude of the pulse varies in accordance with the modulating signal.

(ii) Pulse Position Modulation. : Pulse position (ie) time of rise or fall of the pulse ) changes with the modulating signal.

.18)Dielectric loss increase beyond this frequency.19)How does the effective power radiated by the antenna vary with wavelength?

Power is inversely proportional to wave length20)what should be the length of the dipole antenna for a carrier wave of 5 X 10 8hz ?

L =c/2 ν.21) four times.22) d= √2hR

d= √2x 0.1x 6400 = √1280 km

Area covered by broadcaste, A = π d2 = 3.14 x 1280= 3919.2 km2

population covered = Area x population density= 3919.2 x 1000 = 3919200

23) (i) 5 MHz <fc sky wave propagation (ionospheric propagation).(ii) 100 MHz > fc satellite mode of communication.

24)

25) Maximum voltage of AM wave,Vmax = 16 = 8mV

2 Minimum voltage of AM wave,

Vmin = 4 = 2mV 2

ma = Vmax - Vmin

Vmax + Vmin

= 8-2 = 6 = 0.6 8+2 10

26) The AM wave equation is given by ;v = 5(1+0.6cos6280t) sin 221 X 104t volts ………….(i)

(i) Maximum amplitude of AM wave = EC + maEC =5 + 0.6 X 5 = 8V

Minimum amplitude of AM wave = EC - maEC =5 - 0.6 X 5 = 2V

(ii) The AM wave will contain three frequency viz fc-fs, fc, fc+fs

336-1 336 336+1335kHz 336kHz 337kHz

27) (i) The AM wave has sideband frequency of (fc + fs) and (fc - fs). Sideband frequency = (500+1) kHz and (500-1) kHz 501 kHz and 499 kHz

(ii) Bandwidth required = 499 kHz to 501 kHz = 2 kHz

28) PS = ½ ma2 PC

1.246 = 1 + ma2

2 ma

2/2 = 0.246 ma = (2 X 0.246)1/2 = 0.701 = 70.1%

29) Modulation index , mf = Maximum frequency deviation Minimum signal frequency

30) d = (2 × 6400 X103 × 160)1/2 = 45255m Coverage range, d = (2Rh)1/2

h2 = 4h1 = 4 × 160 = 640m

31) Radius of the area covered by TV broadcast isd = (2Rh)1/2

= 37500m = 37.5 km= 4.4 × 106

32) Microwave communication channel width =

GHz10

1002 ×

=0.2 GHz band width of channel = 8 KHz

= 2.5 × 10 4

33) Energy corresponding to λ = 1400nm = 1400 × 10-9 m is

E = λhc

= eV

19

19

106.11042.1

−

−

××

= 1eV For detection E must be equal to greater then Eg. Hence only suitable semiconductor is C.

34) Critical frequency fc and maximum electron density nmax are related as

fc = 9(nmax)2/1

Squaring we get nmax = 81

2fc

Given fc = 10MHz = 61010 × =

710 Hz

ie, nmax = 81)10( 7 2

= m 3121023.1 −×

Physics - Sample Question Papers and Answers

1

Time : 3 Hrs. Max Marks : 70General Instructions(a) All questions are compulsory.(b) There are 30 questions in total. Questions 1 to 8

carry one mark each, questions 9 to 18 carry twomarks each, questions 19 to 27 carry three markseach and questions 28 to 30 carry five marks each.

(c) There is no overall choice. However, an internalchoice has been provided in one question of twomarks, one question of three marks and all threequestions of five marks each. You have to attemptonly one of the given choices in such questions.

(d) Use of calculators is not permitted.(e) You may use the following physical constants wher-

ever necessary :c = 3 × 108ms-1

h = 6.6 × 10-34Jse = 1.6 × 10-19 Cm

0 = 4p × 10–7 TmA–1

Boltzmann constant k = 1.38 × 1023 JK-1

Avogadro’s number NA = 6.023 × 1023/moleMass of neutron m

n = 1.6 × 10-27 kg

SECTION A1. Two identical charged particles moving with same

speed enter a region of uniform magnetic field. Ifone of these enters normal to the field directionand the other enters along a direction at 300 withthe field, what would be the ratio of their angularfrequencies?

2. Why does a metallic piece become very hot whenit is surrounded by a coil carrying high frequencyalternating current?

3. How is a sample of an n-type semiconductor elec-trically neutral though it has an excess of negativecharge carriers?

4. Name the characteristics of electromagnetic wavesthat (i) increases (ii) remains constantin the electromagnetic spectrum as one moves fromradiowave region towards ultraviolet region.

5. How would the angular separation of interferencefringes in Young’s double slit experiment changewhen the distance of separation between the slitsand the screen is doubled?

6. Calculate the ratio of energies of photons produceddue to transition of electron of hydrogen atom fromits,

(i) Second permitted energy level to the first level, and(ii) Highest permitted energy level to the second per-

mitted level.7. Give expression for the average value of the a c

voltage V = V0 Sinw t over the time interval t = 0

and t =ωπ

.

8. How is the band gap, Eg, of a photodiode related to

the maximum wavelength, lm that can be detected

by it?9. Keeping the voltage of the charging source con-

stant, what would be the percentage change in theenergy stored in a parallel plate capacitor if theseparation between its plates were to be decreasedby 10%?

10. Explain how the average velocity of free electronsin a metal at constant temperature, in an electricfield, remain constant even though the electronsare being constantly accelerated by this electricfield?

11. How is the resolving power of a microscopeaffected when,

(i) the wavelength of illuminating radiations isdecreased?

(ii) the diameter of the objective lens is decreased?Justify your answer.

12. What is the basic difference between the atom ormolecule of a diamagnetic and a paramagnetic ma-terial? Why are elements with even atomic num-ber more likely to be diamagnetic?

13. Why are infrared radiations referred to as heatwaves also? Name the radiations which are next tothese radiations in electromagnetic spectrum hav-ing(i) Shorter wavelength. (ii) Longer wavelength.

14. The following data was recorded for values of ob-ject distance and the corresponding values of im-age distance in the experiment on study of real im-age formation by a convex lens of power +5D. Oneof these observations is incorrect. Identify this ob-servation and give reason for your choice:

PHYSICSSET I

Physics - Sample Question Papers and Answers

2

ORFor the circuitshown here,calculate thepotential dif-ference be-

S.No. 1 2 3 4 5 6

Object dis-tance (cm)

Image dis-tance (cm)

25 30 35 45 50 55

97 61 37 35 32 30

15. Two stu-dents X andY performan experi-ment on po-tentiometerseparatelyusing the

Keeping other things unchanged(i) X increases the value of distance R(ii) Y decreases the value of resistance S in the setup.How would these changes affect the position ofnull point in each case and why?

16. The following table gives the values of work func-tion for a few photosensitive metals

S.No.1.2.3.

MetalNaK

Mo

Work Function (eV)1.922.154.17

If each of these metals is exposed to radiations ofwavelength 300 nm, which of them will not emitphotoelectrons and why?

ORBy how much would the stopping potential for agiven photosensitive surface go up if the frequencyof the incident radiations were to be increased from4 × 1015 Hz to 8 × 1015 Hz?Given h = 6.4 × 10-34 Js, e = 1.6 × 10-19 C andc = 3 × 108 ms-1.

17. Prove that the instantaneous rate of change of theactivity of a radioactive substance is inversely pro-portional to the square of its half life.

18. What does the term LOS communication mean?Name the types of waves that are used for this com-munication. Which of the two-height of transmit-ting antenna and height of receiving antenna - canaffect the range over which this mode of commu-nication remains effective?

19. The following data was obtained for the depen-dence of the magnitude of electric field, with dis-tance, from a reference point O, within the chargedistribution in the shaded region.

Fieldpoint A B C A¢ B¢ C¢

Magnitude ofelectric field E E/8 E/27 E2 E16 E/64

(n cells in each row)

m rows

R = 1.5W.

123456123456123456123456123456

A B C

A¢

B¢

C¢

O

E –R

S

A

G

B

circuit diagram shown here.

(i) Identify thecharge distri-bution and jus-tify your an-swer.

(ii) If the potentialdue to thischarge distri-bution, has avalue V at thepoint A, what isits value at the point A¢?

20. A charge Q located at a point rr is in equilibrium

under the combined electric field of three chargesq

1,q

2, q

3. If the charges q

1,q

2 are located at points

1rr

and 2rr

respectively, find the direction of the force

on Q, due to q3 in terms of q

1,q

2,

21r.rrr

and rr .

21. 12 cells, each of emf 1.5V and internal resistance0.5W, are arranged in m rows each containing ncells connected in series, as shown. Calculate thevalues of n and m for which this combination wouldsend maximum current through an external resis-tance of 1.5W.

–2V, 2W

1V2

1W

3V, 3W

1W1V

1

A B

CD

2W

Physics - Sample Question Papers and Answers

3

23. Three students X, Y, and Z performed an ex-periment for studying the variation of alter-nating currents with angular frequency in aseries LCR circuit and obtained the graphsshown below. They all used a.c. sources ofthe same r. m. s. value and inductances of thesame value. What can we (qualitatively) con-clude about the

(i) capacitance value (ii) resistance valuesused by them? In which case will the qualityfactor be maximum?What can we conclude about nature of the im-pedance of the set up at frequency w

o?

XY

Z

I

w0 w

Liquid

25. A circular coil having 20 turns, each of radius 8 cm, isrotating about its vertical diameter with an angular speedof 50 radian s-1 in a uniform horizontal magnetic field ofmagnitude 30 mT. Obtain the maximum average andr. m. s. values of the emf induced in the coil. If the coilforms a closed loop of resistance 10W, how much poweris dissipated as heat in it? 3

26. The nucleus of an atom of Y235

92, initially at rest, decays

by emitting an a -particle as per the equation

EnergyHeXY 4

2

231

90

235

92++→ .

It is given that the binding energies per nucleon of theparent and the daughter nuclei are 7.8 MeV and 7.835MeV respectively and that of a-particle is 7.07MeV/nucleon. Assuming the daughter nucleus to be formed inthe unexcited state and neglecting its share in the energyof the reaction, calculate the speed of the emitted a-par-ticle. Take mass of a-particle to be 6.68 × 10-27 kg. 3

27. Define the term ‘modulation index’ for an AM wave. Whatwould be the modulation index for an AM wave for whichthe maximum amplitude is ‘a’ while the minimum ampli-tude is ‘b’ ? 3

28. Two circular coils X and Y having radii R and 2R re-

spectively are placed in horizontal plane with their cen-tres coinciding with each other. Coil X has a current Iflowing through it in the clockwise sense. What must bethe current in coil Y to make the total magnetic field atthe common centre of the two coils, zero? With the samecurrents flowing in the two coils, if the coil Y is now liftedvertically upwards through a distance R, what would bethe net magnetic field at the centre of coil Y?

ORA straight thick long wire of uniform cross section of ra-dius ‘a’ is carrying a steady current I. Use Ampere’s cir-cuital law to obtain a relation showing the variation of the

)A

B C

q

tween points B and D.22. A beam of light of wavelength 400 nm is in-

cident normally on a right angled prism asshown. It is observed that the light just grazesalong the surface AC after falling on it. Giventhat the refractive index of the material of theprism varies with the wavelength l as per the

relation mA= 1.2 + 2

b

λ.

Calculate the value of b and the refractive in-dex of the prism material for a wavelength l = 5000Å [(Given q = Sin-1 (0.625)]

.

24. An equiconvex lens with radii of curvature ofmagnitude r each, is put over a liquid layerpoured on top of a plane mirror. A smallneedle, with its tip on the principal axis of thelens, is moved along the axis until its inverted

real image coincideswith the needle it-self. The distance ofthe needle from thelens is measured tobe ‘a’. On removingthe liquid layer and repeating the experiment the distanceis found to be ‘b’. Given that two values of distances mea-sured represent the focal length values in the two cases,obtain a formula for the refractive index of the liquid.

Physics - Sample Question Papers and Answers

4

General instructions same as set 11. What is the angle between the directions of electric

field at any (i) axial point and (ii) equatorial pointdue to an electric dipole? 1

2. A (hypothetical) bar magnet (AB) is cut into twoequal parts. One part is now kept over the other, sothat pole C

2 is above C

1. If M is the magnetic mo-

ment of the original magnet, what would be the mag-netic moment of the combination so formed? 1

3. A rectangularwire frame,shown below,is placed in auniform mag-netic field di-rected upwardand normal tothe plane of the paper. The part AB is connected to

C1 C

2A A

B B¢

A A¢

magnetic field (Br) inside and outside the wire withdistance r, (r £ a) and (r > a) of the field point fromthe centre of its cross section. Plot a graph showingthe nature of this variation. Calculate the ratio of

magnetic field at a point 2

a above the surface of

the wire to that at a point 2

abelow its surface. What

is the maximum value of the field of this wire?5

29. State the principle which helps us to determine theshape of the wavefront at a later time from its givenshape at any time. Apply this principle to

(i) Show that a spherical/ plane wavefront continuesto propagate forward as a spherical/plane wavefront.

(ii) Derive Snell’s law of refraction by drawing the re-fracted wavefront corresponding to a planewavefront incident on the boundary separating ararer medium from a denser medium. 5

OR(a) What do we understand by ‘polarization’ of a wave?

How does this phenomenon help us to decidewhether a given wave is transverse or longitudinalin nature?

(b) Light from an ordinary source (say a sodium lamp)is passed through a polaroid sheet P

1 .The transmit-

ted light is then made to pass through a secondpolaroid sheet P

2 which can be rotated so that the angle

(q) between the two polaroid sheets varies from 00 to90o. Show graphically the variation of the intensityof light, transmitted by P

1 and P

2, as a function of the

angle q . Take the incident beam intensity as I0. Why

does the light from a clear blue portion of the sky,show a rise and fall of intensity when viewed througha polaroid which is rotated? 5

30. A student has to study the input and output charac-teristics of an n-p-n silicon transistor in the Com-mon Emitter configuration. What kind of a circuitarrangement should she use for this purpose? Drawthe typical shape of input characteristics likely tobe obtained by her. What do we understand by thecut off, active and saturation states of the transis-tor? In which of these states does the transistor notremain when being used as a switch?

OR

Input signals A and B are applied to the input termi-nals of the ‘dotted box’ set-up shown here. Let Y bethe final output signal from the box. Draw the waveforms of the signals labelled as C

1 and C

2 within the

box, giving (in brief) the rea-sons for getting these waveforms. Hence draw the waveform of the final output signalY. Give reasons for yourchoice. What can we state (inwords) as the relation betweenthe final output signal Y andthe input signals A and B ?

A

B

C1

C2

Y

SET II

Physics - Sample Question Papers and Answers

5

10. The I-V characteristicsof a resistor are ob-served to deviate froma straight line forhigher values of cur-rent as shown below.Why?

11. A charged particle moving with a uniform velocity

Vr

enters a region where uniform electric and mag-

netic fields BandErr

are present. It passes through

the region without any change in its velocity. Whatcan we conclude about the

(i) Relative directions of BandV,Errr

?

(ii) Magnitudes of BandErr

?

12. Figure shows two long coaxial solenoids, each oflength ‘l’. The outer soleniod has an area of cross-section A and number of turns/ length n

1 . The cor-

responding valuesfor the inner sole-noid are A

2 and n

2.

Write the expres-sion for self induc-tance L

1, L

2 of the

two coils and their mutual inductance M. Hence

show that M< 21

LL .

13. Two identical plane metallic surfaces A and B arekept parallel to each other in air separated by a dis-tance of 1.0 cm as shown in the figure.

–Q –Q +Q–Q

Q Q +Q –QA B

D C

A B

D C

a spring. The spring is stretched and released whenthe wire AB has come to the position A¢B¢ (t = 0).How would the induced emf vary with time? Ne-glect damping. 1

4. From the following, identify the electromagneticwaves having the (i) Maximum (ii) Minimum fre-quency. 1(i) Radio waves (ii) Gamma-rays (iii) Visible light(iv) Microwaves (v) Ultraviolet rays, and (vi) In-frared rays.

5. A partially plane polarised beam of light is passedthrough a polaroid. Show graphically the variationof the transmitted light intensity with angle of rota-tion of the polaroid.

6. The given graphsshow the variation ofphoto electric current(I) with the appliedvoltage (V) for twodifferent materialsand for two differentintensities of the in-cident radiations.Identify the pairs of curves that correspond to dif-ferent materials but same intensity of incident ra-diations.

7. Four nuclei of an element fuse together to form aheavier nucleus. If the process is accompanied byrelease of energy, which of the two - the parent orthe daughter nucleus would have a higher bindingenergy/nucleon?

8. Zener diodes have higher dopant densities as com-pared to ordinary p-n junction diodes. How does itaffect the (i) Width of the depletion layer? (ii) Junc-tion field?

9. Four point charges are placed at the four corners ofa square in the two ways (i) and (ii) as shown be-low. Will the (i) electric field (ii) Electric potential,at the centre of the square, be the same or differentin the two configurations and why? Surface A is given a positive potential of 10V and the

outer surface of B is earthed. (i) What is the magni-tude and direction of the uniform electric field be-tween points Y and Z? (ii) What is the work done inmoving a charge of 20 mC from point X and point Y?

14. In the circuit shown below, R represents an electricbulb. If the frequency u of the supply is doubled,

X

Y

B

Z1.0 cm

A

Physics - Sample Question Papers and Answers

6

22. Which two main considerations are kept in mindwhile designing the ‘objective’ of an astronomicaltelescope? Obtain an expression for the angular mag-nifying power and the length of the tube of an as-tronomical telescope in its ‘normal adjustment’ po-sition. 3

23. Calculate the de-Broglie wavelength of (i) an elec-tron (in the hydrogen atom) moving with a speed of

100

1 of the speed of light in vacuum and (ii) a ball

of radius 5mm and mass 3 × 10-2 kg moving with aspeed of 100ms-1. Hence show that the wave natureof matter is important at the atomic level but is notreally relevant at the macroscopic level. 3

A B Y0 0 00 1 01 0 11 1 0

A B Y0 0 10 1 11 0 01 1 1

(i) (ii)

ORAn air cored coil L and a bulb B are connected inseries to the ac mains as shows in the given figure:

The bulb glows with some brightness. How wouldthe glow of the bulb change if an iron rod were in-serted in the coil? Give reasons in support of youranswer. 2

15. Experimental observations have shown that X-rays(i) travel in vacuum with a speed of 3 × 108 ms-1,(ii) exhibit the phenomenon of diffraction and canbe polarized.What conclusions can be drawn about the nature ofX-rays from each of these observations? 2

16. Write the relation between the angle of incidence(i), the angle of emergence (e), the angle of prism(A) and the angle of deviation (d) for rays undergo-ing refraction through a prism. What is the relationbetween Ði and Ðe for rays undergoing minimumdeviation? Using this relation, write the expressionfor the refractive index (m) of the material of a prismin terms of ÐA and the angle of minimum devia-tion (dm). 2

17. A radioactive material is reduced to 16

1of its origi-

nal amount in 4 days. How much material shouldone begin with so that 4 × 10-3 kg of the material isleft after 6 days. 2

18. Distinguish between ‘point to point’ and ‘broad-cast’ communication modes. Give one example ofeach. 2

AC

B

Mains

how should the values of C and L be changed sothat the glow in the bulb remains unchanged?

19. In a double slit interference experiment, the twocoherent beams have slightly different intensities Iand I + dI (dI << I). Show that the resultant inten-sity at the maxima is nearly 4I while that at the

minima is nearly( )I4

I2δ

.

20. An electric dipole of dipole moment Pr is placed in

a uniform electric field Er

. Write the expression for

the torque τr experienced by the dipole. Identify two

pairs of perpendicular vectors in the expression.Show diagramatically the orientation of the dipolein the field for which the torque is (i) maximum(ii) half the maximum value (iii) zero.

ORTwo capacitors with capacity C

1 and C

2 are charged

to potential V1 and V

2 respectively and then con-

nected in parallel. Calculate the common potentialacross the combination, the charge on each capaci-tor, the electrostatic energy stored in the system andthe change in the electrostatic energy from its ini-tial value. 3

21. Using a suitable combination from a NOR, an ORand a NOT gate, draw circuits to obtain the truthtable given below:

W

R C L

K

Physics - Sample Question Papers and Answers

7

ing formula for the series combination? Two cellsof EMF 1V, 2V and internal resistances 2W and 1Wrespectively are connected in (i) series, (ii) parallel.What should be the external resistance in the circuitso that the current through the resistance be the samein the two cases? In which case more heat is gener-ated in the cells ? 5

29. (i) Describe an expression for the magnetic field ata point on the axis of a current carrying circularloop.

(ii) Two coaxial circular loops L1 and L

2 of radii 3cm

and 4cm are placed as shown. What should be themagnitude and direction of the current in the loopL

2 so that the net magnetic field at the point O be

zero?

24. Show that during the charging of a parallel platecapacitor, the rate of change of charge on each plateequals e

0 times the rate of change of electric flux

(fE) linked with it. What is the name given to the

term dt

dE

0

φε ? 3

25. The spectrum of a star in the visible and the ultra-violet region was observed and the wavelength ofsome of the lines that could be identified were foundto be :824Å, 970Å, 1120Å, 2504Å, 5173Å, 6100Å.Which of these lines cannot belong to hydrogenatom spectrum? (Given Rydberg constant

R = 1.03 × 107m-1 and R

1= 970Å. Support your

answer with suitable calculations. 326. What is space wave propagation? Which two com-

munication methods make use of this mode of propa-gation? If the sum of the heights of transmitting andreceiving antennae in line of sight of communica-tion is fixed at h, show that the range is maximum

when the two antennae have a height 2

heach.

27. Draw the transfer characteristics of a base biasedtransistor in its common emitter configuration. Ex-plain briefly the meaning of the term ‘active region’in these characteristics. For what practical use, dowe use the transistor in this ‘active region’?

28. A cell of unknown emf E and internal resistance r,two unknown resistances R

1 and R

2 (R

2 > R

1) and a

perfect ammeter are given. The current in the cir-cuit is measured in five different situations : (i) With-out any external resistance in the circuit, (ii) Withresistance R

1 only, (iii) With resistance R

2 only,

(iv) With both R1 and R

2 used in series combination

and (v) With R1 and R

2 used in parallel combina-

tion. The current obtained in the five cases are 0.42A,0.6A, 1.05A, 1.4A, and 4.2A, but not necessarily inthat order. Identify the currents in the five cases listedabove and calculate E, r,, R

1 and R

2.

ORDescribe the formula for the equivalent EMF andinternal resistance for the parallel combination oftwo cells with EMF E

1 and E

2 and internal resis-

tances r1 and r

2 respectively. What is the correspond-

I1 – IA

3 cm 4 cm

3 cm4 cm

L1

L2

OR

O 15 cm d

(i) What is the relationship between the current and themagnetic moment of a current carrying circularloop? Use the expression to derive the relation be-tween the magnetic moment of an electron movingin a circle and its related angular momentum?

(ii) A muon is a particle that has the same charge as anelectron but is 200 times heavier than it. If we hadan atom in which the muon revolves around a pro-ton instead of an electron, what would be the mag-netic moment of the muon in the ground state ofsuch an atom?

30. (i) Derive the mirror formula which gives the rela-tion between f, v and u. What is the correspondingformula for a thin lens?

(ii) Calculate the distance d, so that a real image of anobject at O, 15cm in front of a convex lens of focallength 10cm be formed at the same point O. Theradius of curvature of the mirror is 20cm. Will theimage be inverted or erect?

Physics - Sample Question Papers and Answers

8

9. The two graphs drawnbelow, show the varia-tion of electrostatic

potential (V) with r

1(r

being distance of thefield point from thepoint charge) for twopoint charges q

1 and

v

q2

q1

r

1

OR(i) Using the relation for refraction at a single spherical

refracting surface, derive the lens maker’s formula.(ii) In the accompanying

diagram, the directimage formed by thelens (f = 10cm) of anobject placed at Oand that formed afterreflection from the spherical mirror are formed atthe same point O’ . What is the radius of curvatureof the mirror?

15 cm 50 cmO

General instructions same as set 1.

1. The graph shown here, showsthe variation of the total en-ergy (E) stored in a capacitoragainst the value of the ca-pacitance (C) itself. Which ofthe two - the charge on the

capacitor or the potential used to charge it is keptconstant for this graph?

2. An a- particle and a proton are moving in the planeof the paper in a region where there is a uniform

magnetic field (Br ) directed normal to the plane of

the paper. If the two particles have equal linear mo-menta, what will be the ratio of the radii of theirtrajectories in the field? 1

3. State the condition under which a microwave ovenheats up a food item containing water moleculesmost efficiently.

4. An electrical element X, when connected to an al-ternating voltage source, has the current through it

leading the voltage by 2

πradii. Identify X and write

an expression for its reactance. 15. A double convex lens, made from a material of re-

fractive index m1, is immersed in a liquid of refrac-

tive index where m2 > m

1. What change, if any, would

occur in the nature of the lens? 16. The de Broglie wavelengths, associated with a pro-

ton and a neutron, are found to be equal. Which ofthe two has a higher value for kinetic energy? 1

7. Carbon and silicon are known to have similar lat-tice structures. However, the four bonding electronsof carbon are present in second orbit while those ofsilicon are present in its third orbit. How does thisdifference result in a difference in their electricalconductivities? 1

8. An unknown input (A) and the input (B) shown here,are used as the two inputs in a NAND gate. Theoutput Y, has the form shown below. Identify theintervals over which the input ‘A’ must be ‘low’.

q2.

(i) What are the signs of the two charges?(ii) Which of the two charges has a larger magnitude

and why? 210. Calculate the temperature at which the resistance

of a conductor becomes 20% more than its resis-tance at 270C.The value of the temperature coeffi-cient of resistance of the conductor is 2.0 × 10–4/K.

211. A student records the following data for the magni-

tudes (B) of the magnetic field at axial points at dif-ferent distances x from the centre of a circular coilof radius a carrying a current I. Verify (for any two)

SET III

Physics - Sample Question Papers and Answers

9

x® x = 0 x = a

0B225.0

x = 2a x = 3a

B® B0 0

B5039.00

B10010.0

O

a

P

I

12.An armature coil consists of 20 turns of wire, eachof area A = 0.09m2 and total resistance 15.0W. Itrotates in a magnetic field of 0.5T at a constant

frequency of .Hz150

π . Calculate the value of

(i) maximum (ii) average induced emf produced inthe coil. 2

13. Two cells of emf E1 and E

2 have internal resistance

r1 and r

2. Deduce an expression for equivalent emf

of their parallel combination.OR

A cell of emf (E) and internal resistance (r) is con-nected across a variable external resistance (R). Plotgraphs to show variation of(i) E with R ,(ii) Terminal p.d. of the cell (V) with R

14. Fig. shows a light bulb (B) and iron cored inductorconnected to a DC battery through a switch (S). 2

(i) What will one observe when switch (S) is closed?(ii) How will the glow of the bulb change when the

battery is replaced by an ac source of rms voltageequal to the voltage of DC battery? Justify youranswer in each case.

15. Electromagnetic radiations with wavelength 2(i) l

1are used to kill germs in water purifiers.

(ii) l2are used in TV communication systems.

(iii) l3play an important role in maintaining the

earth’s warmth. Name the part of electromagneticspectrum to which these radiations belong. Arrange

B

s

that these observations are in good agreement withthe expected theoretical variation of B with x.

these wavelengths in decreasing order of their mag-nitude.

16.What do the terms ‘depletion region’ and ‘barrierpotential’ mean for a p-n junction? 2

17. We do not choose to transmit an audio signal byjust directly converting it to an e.m. wave of thesame frequency. Give two reasons for the same. 2

18. Light of wavelength 550 nm. is incident as parallelbeam on a slit of width 0.1mm. Find the angularwidth and the linear width of the principal maximain the resulting diffraction pattern on a screen keptat a distance of 1.1m from the slit. Which of thesewidths would not change if the screen were movedto a distance of 2.2m from the slit? 2

19. The given figure shows the experimental set up of ametre bridge. The null point is found to be 60cmaway from the end A with X and Y in position asshown. X Y

B

A C

When a resistance of 15W is connected in series with‘Y’, the null point is found to shift by 10cm towardsthe end A of the wire. Find the position of null pointif a resistance of 30W were connected in parallelwith ‘Y’. 3

ORWhy is a potentiometer preferred over a voltmeterfor determining the emf of a cell? Two cells of emfE

1 and E

2 are connected together in two ways shown

here.

E1

E2 E

1E

2

The ‘balance points’ in a given potentiometer ex-periment for these two combinations of cells arefound to be at 351.0cm and 70.2cm respectively.Calculate the ratio of the Emfs of the two cells.

20. When a circuit element ‘X’ is connected across

an a.c. source, a current of 2 A flows through it

and this current is in phase with the applied volt-age. When another element ‘Y’ is connected acrossthe same a.c. source, the same current flows in

Physics - Sample Question Papers and Answers

10

Use these graphs to obtain the approximate valueof current amplification factor for the transistor atV

CE = 3V. 3

27. State Bohr’s postulate for the ‘permitted orbits’ forthe electron in a hydrogen atom. Use this postulateto prove that the circumference of the nth permit-ted orbit for the electron can ‘contain’ exactly nwavelengths of the deBroglie wavelength associ-ated with the electron in that orbit. 3

3mF

3mF1mF

2mF

100V

2mF

R1

B

CMA

Lv

R

+

–V

BE

+–

–

+

VCE

+V

L

–L

12.5

10

7.5

5

2.5

0 0.5 1 1.5 2 2.5 3 3.5 4

60mA50mA

40mA30mA

20mA10mA

Base current Ib

(Col

lect

or c

urre

nt I c)

in m

A

Collector to emitter voltage (VCE

) in volts

V

M1

M2

u01

u02

(i) What are the values of work functions for M1 and

M2 ?

(ii) The values of the stopping potential for M1 and M

2

for a frequency u3 (> u

02) of the incident radiations

are V1 and V

2 respectively. Show that the slope of

the lines equals 0102

21VV

υ−υ−

. 3

23. What is a wavefront? Distinguish between a planewavefront and a spherical wavefront. Explain withthe help of a diagram, the refraction of a planewavefront at a plane surface using Huygen’s con-struction. 3

24. Define the term ‘Activity’ of a radioactive sub-stance. State its SI unit. Two different radioactiveelements with half lives T

1 and T

2 have N

1 and N

2

(undecayed) atoms respectively present at a giveninstant. Determine the ratio of their activities at thisinstant. 3

the circuit but it leads the voltage by 2

πradians.

(i) Name the circuit elements X and Y.(ii) Find the current that flows in the circuit when the

series combination of X and Y is connected acrossthe same a.c. voltage.

(iii) Plot a graph showing variation of the net imped-ance of this series combination of X and Y as afunction of the angular frequency w of the appliedvoltage. 3

21. Give reasons for the following : 3(a) Astronomers prefer to use telescopes with largeobjective diameters to observe astronomical ob-jects.(b) Two identical but independent monochromaticsources of light cannot be coherent.(c) The value of the Brewster angle for a transpar-ent medium is different for lights of differentcolours.

22. The given graphsshow the variation ofthe stopping potentialV

s with the frequency

( u) of the incidentradiations for two dif-ferent photosensitivematerials M

1 and M

2.

25. (a) Draw the block diagram of a communicationsystem.(b) What is meant by ‘detection’ of a modulatedcarrier wave? Describe briefly the essential stepsfor detection. 3

26. The given circuit diagram shows a transistor con-figuration along with its output characteristics.Identify(i) the type of transistor used and(ii) the transistor configuration employed.

28. Obtain an ex-pression forthe capaci-tance of a par-allel plate(air)capacitor.The givenfigure showsa network of

Physics - Sample Question Papers and Answers

11

(a) Use Gauss’s law to ob-tain an expression forthe electric field due toan infinitely longstraight uniformlycharged wire.

(b) Electric field in the above figure is directed along+ X direction and given by Ex = 5Ax + 2B, whereE is in NC-1 and x is in metre, A and B are constantswith dimensions Taking A = 10NC–1m–1 andB = 5NC-1. calculate:(i) the electric flux through the cube.(ii) net charge enclosed within the cube. 5

29. (a) Draw the labelled diagram of moving coil gal-vanometer. Prove that in a radial magnetic field,the deflection of the coil is directly proportional tothe current flowing in the coil.(b) A galvanometer can be converted into a volt-meter to measure up to(i) ‘V’ volts by connecting a resistance R

1 in series

with coil.

(ii) 2

V volts by connecting a resistance R

2 in series

with its coil. Find the resistance (R), in terms of R1

and R2 required to convert it into a voltmeter that

can read up to ‘2V’ volts.OR

OR

M N

Z

X

Y

10 cm

five capacitors connected to a 100V supply. Calcu-late the total charge and energy stored in the net-work.

(a) Draw diagrams to depict the behaviour of mag-netic field lines near a ‘ bar’ of:(i) copper (ii)Aluminium(iii) Mercury, cooled to a very low temperature(4.2K)

(b) The vertical component of the earth’s magnetic field

at a given place is 3 times its horizontal compo-

nent. If total intensity of earth’s magnetic field atthe place is 0.4 G, find the value of :(i) angle of dip(ii) the horizontal component of earth’s magneticfield.

30. (a) Draw a ray diagram to show the refraction oflight through a glass prism. Hence obtain the rela-tion for the angle of deviation in terms of the angleof incidence, angle of emergence and the angle ofthe prism.(b) A right angled isosceles glass prism is madefrom glass of refractive index 1.5. Show that a rayof light incident normally on(i) one of the equal sides of this prism is deviatedthrough 90o

(ii) the hypotenuse of this prism is deviated through180o.

OR(a) With the help of a labelled ray diagram, showthe image formation by a compound microscope.Derive an expression for its magnifying power.(b) How does the resolving power of a compoundmicroscope get affected on(i) decreasing the diameter of its objective?(ii) increasing the focal length of its objective?

1. 1 : 12. Large induced current is produced due to electro-

magnetic induction which heats up the metallicpiece.

3. The charge of the ‘excess’ charge carriers gets bal-anced by an equal and opposite charge of the ion-ized cores in the lattice.

4. (i) Frequency (ii) Speed in free space5. No effect

(or the angular separation remains the same).6. We have

4

3.const

2

1

1

1.constE

2212=

−=−

and E•¥ ® 2

= const. 4

1.const

1

2

122

=

∞−

\ Ratio = 3 : 1

7.π

0V2

or 0V

11

7

AnswersSET I

Physics - Sample Question Papers and Answers

12

and hence a smaller length (for the same potentialgradient) would be needed for balancing it]

16. Energy of a photon of the incident radiation = λhc

= eV4eV106.110300

103104.6199

834

=××××××

−−

−

This being less than the work function of Mo, therewould be no photo-emission from Mo.

OReV

s = hu - W \ e(V

2 – V

1) = h (u

2 – u

1)

or V2 – V

1 = ( )

12e

h υ−υ

\ V2 – V

1 = ( ) 15

19

34

1048106.1

104.6 ×−××

−

−

volt = 16 volt

17. Instantaneous Activity = Ndt

dNR λ==

\ ( )dt

dNN

dt

d

dt

dR λ=λ=

l (–l N) = –l 2N = NT

log

2

21

2

e

− \ ( )2

21T

1

dt

dR ∝

18. LOS ® line of sightWaves used ® space wavesIt is both - the height of transmitting antenna as wellas the height of the receiving antenna that affectsthe range of the mode of communication.

19. We observe that the field magnitude (i) Varies asthe inverse cube of the distance of the field pointalong one line. (ii) Has a magnitude half of its mag-nitude (at an equidistant point) on the lineperperdicular to this line. These properties tell usthat the given charge distribution is a (small) elec-tric dipole centered at the reference point O. Thepoint A’ is an equatorial points for the given dipole.Hence potential of A¢ = zero.

20. We have

−

−∈π=

13

1

1

0

1rr

rr

Qq

4

1F

rr

rr

r

and ( )23

2

2

0

2rr

rr

Qq

4

1F

rr

rr

r

−−∈π

=

For equilibrium, we must have 0FFF213

=+=rrr

or ( )213FFFrrr

+−=

8. We have g

m

Ehc =λ or

g

mE

hc=λ

9. 202V

d

A

2

1CV

2

1E

∈==

\ 9

10

90

100

d

d

E

E

2

1

1

2 ===

\ %10019

10

E

EE

E

E

1

12

1

×

−=−=∆

= 11.1%

10. We have tm

EeVV

it+=

rr

\ AVAV

tAV

tt

m

EeVV +=

AVt

Vr

= zero (Random nature of motion and colli-

sions). \ τ=m

EeV

AVt

r

= constant

as t , the average time between collisions, remainscontant under constant temprature conditions.

11. (i) It increases (ii) It decreases12. The atom/molecule of a diamagnetic material has

zero net magnetic moment. For a paramagnetic ma-terial it is not so. With an even atomic number, theelectrons in an atom of an element can ‘pair off’,which can makes the net magnetic moment of eachpair as zero. This makes the element more likely tobe diamagnetic.

13. Infrared radiations get readily absorbed by watermolecules in most materials. This increases theirthermal motion and heats them up.(i) visible light (ii) Microwaves

14. Focal length of the lens = cm1005

1 × = 20 cm

Obervation at 3 is incorrect .This observation is incorrect because for an objectdistance lying between f and 2f, the image distancehas to be more than 2f.

15. For student X, the null point would shift towardsright (i.e. towards B) [Increase in R decreases thepotential gradient. Hence a greater length of wirewould be needed for balancing the same emf.For student Y, the null point would shift towardsleft (ie. toward A) [A decrease of S would decreasethe terminal p.d.V across the unknown battery

(V = E - ir and

+=

Sr

Ei increases as S decreases,

Physics - Sample Question Papers and Answers

13

Hence ( ) ( )

−−+−

−∈π=

3

2

22

13

1

1

0

3

rr

rrqrr

rr

q

4

QF rr

rrrr

rr

r

\ The direction of 3Fr

is given by the direction of

the vector ( ) ( )

−

−+−

−rr

rr

qrr

rr

q23

2

2

13

1

1rr

rr

rr

rr

21. The equivalent internal resistance of each row ofn cells in series = nr. The net equivalent internal

resistance of the combination = m

nr

Net equivalent emf of the combination = n × E(E = emf of one cell)

\ Current drawn by R =nrmR

NE

nrmR

mnE

m

nrR

nE

+=

+=

+

= mnRr2nrmR

NE2

+

−

= NRr2nrmR

NE2

+

−

For maximum current, the denominator should beminimum.

This happens when, nrmR = or m

nrR =

\ 5.1m

5.0n =× or 3m

n =

Also n × m = 12 (given)Solving, we get n = 6 and m = 2

ORWe can draw the circuit explicitly as shown. Thecurrent distribution can be taken as shown. Apply-ing Kirchhoff’s second law to loops BADB andDCBD, respectively, we get the equations:

AI

1

2V 2W I1 B

1V

1W

D 3W 3VC

1V

1W

2WI2

(1 – 1)

- 2I1 + 2 - 1 - 1× I

1 - 2I

2 = 0 or 3 I

1+ 2 I

2 = 1

and, -3(I1- I

2) + 3 - 1- 1 × (I

1- I

2) + 2 I

2 = 0

or 4 I1- 6 I

1 = 2

Solving, we get I1 = A

13

5 and I

2 = A

13

1

P.d. between B and D = V13

2V

13

12 =× = 0.154V

(Point B is at a higher potential w. r. t. point D)22. The ray must fall on the surface AC at just the criti-

cal angle, qc. The angle of incidence at the face AC

equals q.Hence q = q

c .

6.1625.0

1

sin

1

c

==θ

=µ∴ ( )27104

b2.16.1

−×+=∴

\ b = 0.4 × 16 × 10–14m2 = 6.4 × 10–14 m2

The refractive index for l =5000 Å is given by

m¢ = ( ) 25

4.62.1

105

104.62.1

27

14

+=××+

−

−

= 1.2 + 0.256 = 1.45623. (i) We have C

1 = C

2 = C

3

Resonant frequency = LC2

1

π is same for all three

and we are given that L has same value for all.(ii) We have R

1< R

2 < R

3 . Bandwidth for X< B and

width for Y< Bandwidth for Z. Max.current forX >Max. current for Y > Max.current for Z. Stu-dent X has the maximum value for the quality fac-tor because the bandwidth is least in this case. Theimpedance at the resonant frequency w

0 is purely

resistive in nature.24. The liquid layer can be regarded as forming a plane

concave lens.The first value (= a ) of the measureddistance is, therefore, the focal length of the com-bination of the given lens and the liquid lens. Thesecond value (=b ) represents the focal length ofthe lens itself. Hence, if f = 16 is the focal length ofthe liquid lens, we have

f

1

b

1

a

1 += or

−=−=ab

ab

b

1

a

1

f

1

But, ( )

−−µ=

∞+−−µ=

R

11

R

11

f

1

Physics - Sample Question Papers and Answers

14

B2

r –a

r

\ 2

R.2

I

R2

I00

′µ=µ or I¢ = 2

I

The coil Y must carry this current in the anticlockwise sense. When the coil Y is lifted through a dis-tance R, its centre becomes an axial point for coilX. Hence

( ) R24

I

RR2

IRxB 0

2322

2

oµ=

+µ=′ =

µ0 28IR

Also R2

I

2R2

21

yB 0µ=

µ=′

\ Magnitude of net field

=

−µ=′−′

4

21

R2

IxByB 0

= R

I323.0 0

µ=

This net field is in the direction of the field due tothe coil Y, i.e; perpendicular to its plane and di-rected vertically upwards.

ORConsider a closed path of radius r inside the crosssection of the wire. The current enclosed by this

path is 2

2

2

2 a

rIr

a

II =π

π=′

By Ampere’s circuital law, ∫ ′µ= Id.B0r

l

or 2

2

0ra

rIr2B µ=π \ r

a2

IB

2

0

r πµ= or B µ r (for r < a)

Outside the wire, the field of the wire is given by

B.2pr = m0 I or ( )ar

r2

IB 0 >

πµ=

The relevant graph is, therefore, as shown.

\ ( ) ( )

R

1

ab

ab

−−µ=−

, ( )

1ab

baR +−=µ

25. When the normal to the plane of the coil makes anangle with the direction of the magnetic field, theflux linked with it is f = NBA cos q = NBA coswt

\ Induced Emf = NBAdt

d =φ w sin wt

\ Max. Emf = NBA w = NB (pr2)w= 20 × 30 × 10–3 × p ( 8 × 10–2)2 × 50 volt= 0.603 volt

Average Emf = Average of Sinwt over acycle = Zero

rms value of Emf = 2

603.0

2

Emf.Max = volt

= 0.426V

Power dissipated = ( ) ( )

W018.0W10

426.0

R

E22

rms ==

26. Total B.E. of parent Nucleus = 7.8 × 235 MeV= 1833 MeV

Total B.E. of daughter nucleus= 7.835 × 231 MeV = 1809.9MeV

Total B.E. of a-particle = 7.07 × 4 MeV= 28.28 MeV

Increase in B.E. after the reaction= [(180.9 + 28.28) - (1833)]MeV = 5.18 MeVThis is the energy released in the reaction, since itassumed to be taken up totally by the a-particle,

2mv

2

1 = 5.18 × 1.6 × 10–13 J

v2 = 2214sm10

68.6

2.318.5 −×× = 17

ms1048.2−×

= 1.58 × 107 ms-1

27. The modulation index (m) for an AM wave equalsthe ratio of the peak value of the modulating signal(A

m) to the peak value of the carrier wave (A

c)

m = C

m

A

A

Given thata = Ac + A

m and b = A

c – A

m

\ Ac =

2

ba + and A

m =

2

ba − \ m =

ba

ba

+−

28. We have yx

BBrr

−=

If B1 and B

2, denote respectively, the values of the

magnetic field at points 2

a above and 2

abelow the

surface of the wire, we have

Physics - Sample Question Papers and Answers

15

Let us consider a diverging wave and let F1 F

2 rep-

resent a portion of spherical wavefront at t = 0.According to Hugygens’ principle each point ofthe wavefront is the source of a secondary distur-bance and each point on this wavefront emitswaves which travel in the medium with the samespeed as that of primary wavelets. To constructnew wavefronts draw spheres with each point onF

1 F

2 as centre and vt as radius. v is speed of light

in the medium. If we draw a common envelopetouching all these spheres, that will be the newwavefront. G

1 G

2 is the new wavefront. In a simi-

lar manner new wavefronts can be constructed fora plane wavefront also.Figure of plane wave propagation: Refer revisionpage 45

Incidentwavefront

E

A¢

C

Refractedwavefront

Medium 1

Medium 2

v1t

B

u

v2t

ir P¢P

PP¢ is the surface separating medium 1 and me-dium 2. Let v

1 and v

2 represent speed of light in

these media. AB is a plane wavefront falling ob-liquely. By the time the wave from B reaches C,the wave from A would have travelled a distancein the second medium after refraction. If t is thetime taken by the ray to travel, then BC = v

1t and

the ray would travel a distance = v2t in the second

medium in the same time.

To determine the shape of the refracted wavefront,draw a sphere of radius v

2t with the point A as cen-

tre. Let CE is the tangent drawn from the point Cto this sphere. Then AE = v

2t. CE would represent

the refracted wavefront. Consider the Dles ABCand AEC,

AC

tv

AC

BCisin 1== ;

AC

tv

AC

AErsin 2==

where i and r are angles of incidence and refrac-tion respectively.

2

1

v

v

rsin

isin = = a constant = n21

. If r < i, v2 < v

1

If c represents the speed of light in vacuum, then

n1 =

1v

c ; and n

2 =

2v

c,

1

2

n

n

rsin

isin =

n1sini

= n

2 sin r

This is Snell’s law of refraction.

OR

(a) The phenomenon of restricting the vibrations oflight (electric vector) in a particular direction per-pendicular to the direction of wave motion is calledpolarisation of light.

Polarization is possible only with transversewaves and not with longitudinal waves.

a3

I

2

a32

IB 00

1 πµ=

π

µ=and

a4

I

2

a

a2

IB 0

2

0

2 πµ==

πµ=

\ 3

4

B

B

2

1 =

The maximum value of the field is at r = a.

we have Bmax

= a2

I0

πµ

29. The scientist Huygens gave a hypothesis for geo-metrical construction of the position of commonwavefront at any instant during the propagation ofwaves. The postulates are:(a) Every point on the given wavefront called pri-mary wavefront acts as a fresh source of light calledsecondary wavelets, which travel in all directionswith the velocity of light in the medium.(b) A surface touching these secondary waveletstangentially in the forward direction at any instantgives the new wavefront at that instant. This iscalled secondary wavefront.

(ii)

Physics - Sample Question Papers and Answers

16

12345123451234512345123451234512345123451234512345

12345123451234512345123451234512345123451234512345

12345123451234512345123451234512345123451234512345

12345123451234512345123451234512345123451234512345

Polariser Analyser

T1

T2

No light

Unpolarised light Plane polarised light

I0

2

I0

90o

q

Through P2

Through P1

Typical shape of the input characteristics.

Cut off Stage : When the input voltage is less than aminimum value ( 0.6V for Si), there is no currentflow in the input or output sides of the transistor. Thetransistor is then said to be in its ‘cut-off’ stage.Active Stage : This is the stage of the transistor whenthe input is greater than about 0.6 V and there is somecurrent in the output path.Saturation stage : With increase in the input voltagebeyond a certain value, the output voltage decreasesand becomes almost constant at a near to zero value.The transistor is then said to be in the saturation state.

T1 and T

2 are two thin plates of tourmaline cut

with their faces parallel to the axis of the crystal.When unpolarised light falls on T

1, it allows only

those vibrations which are parellel to its axis.Therefore plane polarised light comes out of T

1.

When T2 is placed with its axis parallel to T

1,

light is allowed. When the crystals are in crossedposition no light is allowed.When T

2 is rotated with respect to T

1 the inten-

sity of light varies. Maximum intensity occurswhen T

1 and T

2 are parallel and zero intensity

when they are in crossed positions. This obser-vation can be explained only when we assumelight waves are transverse.Incident intensity = I

0

sphere. This scattered light is polarised. It, there-fore, shows a variation in intensity when viewedthrough a polaroid on rotation.

30. Circuit diagram for drawing the input and outputcharacteristics.

The light passingthrough P

1 re-

mains constant

i.e. 2

I0 .

The light passingthrough P

2 varies

with q as per therelation, I

2 =

I1cos2q . The

light comingfrom a clearportion of thesky is nothingbut sunlightthat has chan-ged its directiondue to scatter-ing by mol-ecules in theearth’s atmo-

Physics - Sample Question Papers and Answers

17

e

4. (i) Maximum - g-rays(ii) Minimum - Radiowaves

Looking at the shapes of A, B and Y, we can saythat : (1) The output Y is low when both A and Bare high. (2) The output Y is high when one ofthe input signals is high while the other is low.

00000 421 3

A

B

B

C1 = A B

C2 = BA

A

Y = C1 + C

2

1. 180o or antiparallel2. Nearly Zero or Zero3. Sinusoidal Variation

OR

The Transistor does not remain in the active stagewhen it is being used as a switch.

ORThe output C

1 is the output of an AND gate having

A and B as its two inputs.The output C

2 is the output of an AND gate having

A and B as its two inputs.The output Y is the output of an OR gate havingC

1 and C

2 as its two inputs. Using the truth tables

for AND and OR gates, we can or therefore get thewave forms shown for C

1, C

2 and Y.

5.

6. (1, 3), (2, 4)7. The daughter element (release of energy is accom-

panied by an increase of B.E)8. (i) ‘Depletion layer’ width decreases.

(ii) Junction field becomes very high9. (i) Potential is same (= zero) in both cases(ii) Electric field is different in the two cases.10. For higher values of current, we observe that the

current value for a given voltage is less than givenby Ohm’s law. This means that R has increased forhigher values of current. The increase of R is be-cause of the increase in temperature of the resistorat higher values of the current.

11. vE ⊥ , BE ⊥

v is not parallel or antiparallel to Bθ= sinvBE

12. L1 = l,

1

2

10A

1

2nµ l

2

2

202AnL µ=

M = m0n

1n

2 p l

2

2r , 1

r

rAA

LL

M

1

2

12

21

<==

13. (i) ( )m10

V10

dr

dVE

2−=−= = 1000 Vm–1

E∴ = 1000Vm–1

Its direction is from higher potential to lowerpotential point, i.e. from Y to Z.

(ii) The surface of a charged metal plate is an equipo-tential. \ X and Y are at the same potential.

DV = Vy – V

x = 0

\ Work done in moving a charge in an elecrticfield = qDV\ Work done in moving 20mC from X to Y

= 20 × 10–6 × 0 = 014. For same current value, the total impedance must

remain same.

\ w L C

1

ω− must remain same. Thus L and C must

SET II

Physics - Sample Question Papers and Answers

18

AB

Y

AB

Y

90o

Pr

(i)

(ii)

(iii)

Pr P

r

Pr

Er

Pr

Er

Pr

Er

Pr

Er

or

orPr

150o

30o

OR

21

2211

21CC

VCVC

CC

QV

++=

+=

Charges = Q1 = C

1V, Q

2 = C

2V

Energy stored = ( )+ VCC2

1 2

21

( )21

2

2211

CC

VCVC

2

1

++

Change in energy stored = DU

=

+−

+

+2

22

2

11

21

2

2211

VCVCCC

VCVC

2

1

= – ( )221

21

21 VVCC

CC

2

1 −+

21. Output not symmetric for A, B = (0, 1) and(1,0) NOT gate in one input.(i) has three zeros NOR gate

Thus

(ii) has three one’s Þ OR gate

Thus

22. The two main considerations(i) Large light gathering power(ii) Higher resolution (resolving power)[Both these requirements are met better when anobjective of large focal length as well as largeaperture is used]

In an astronomical telescope, an objective O oflarge focal length and large aperture is used. It

both be halved simultaneously.OR

The glow of the bulb will decrease.As the iron rod is inserted in the coil, its induc-tance increases. As inductance incresases, its reac-tance also increases resulting in an increase in theimpedance of the circuit.As a result, the current in the circuit and hence theglow of the bulb will decrease.

15. (i) X-rays are e.m. waves(ii) X-rays are transverse in nature

16.Ði + Ðe –Ðd = ÐAFor minimum deviation Ði = Ðe

Ði = 2

ma δ∠+∠

For minimum deviation, we also have

Ðr = 2

Ar

∠=′∠

\

δ+

==µ

2

Asin

2

mAsin

rsin

isin

17. Reduction factor = 42

1

16

1 = in 4 days.

Hence Half life = 1 day

For 6 days reduction factor would be64

1

2

16

=

Original amount = 4 × 10–3 × 64kg = 0.256 kg18. Point to Point : Communication over a link between

a single transmitter and receiver.Example : TelephoneBroadcast mode : Large number of receivers linkedto a single transmitter.Example : Radio

19. The two amplitudes are IIandI δ+\ Intensity at minima

( ) I.II2IIIIII 22

δ+−+δ+=−δ+

= ( ) I4I2δ

and intensity at maxima = ( )III2

+δ+

I4I.II2III 2 =δ+++δ+=

20. Eprrr

×=τ

Two Pairs : Eand,pandrrrr

ττ

Physics - Sample Question Papers and Answers

19

Let us take n2 = 1 (Lyman series of hydrogen spec-

trum)Here can take values

1

A970

1615

A970,

98

A970,

43

A9700

ooo

−−−−−

(Corresponding to n1 = 2, 3, 4, -------------- ¥ )

\ Permitted values of l are 1293.3Å, 1091Å,1034.6 Å, --------- 970ÅLet us next take n

2 = 2 (Balmer series of hydrogen

spectrum)Here l can take values

41

A970

10021

A970,

163

A970.

365

A970ooo

0

−−−

(Coresponding to n1 = 3, 4, 5----------- ¥ )Possible values of l are 6984Å, 5173. 3Å, 4619Å, -------3880ÅHence l = 824Å, 1120Å , 2504Å , 6100Å, of thegiven lines, cannot belong to the hydrogen atomspectrum.

26. Space wave : A space wave travels in a straight linefrom the transmitting antenna to the receiving an-tenna:Two ways : Line of sight communication and satel-lite communication

We have 21Rh2Rh2D +=

Let h1 = x so that h

2 = (h – x)

( )xhR2Rx2D −+=∴

( )xh2

R

x2

R

dx

dD

−−= = 0 Þ x = h/2

27.

When Vi > 0.7V, the transistor is in active state. There

will be some current Ic. From eqn.V

0 = V

cc – I

cR

c, value

of output voltage V0 decreases as the term I

cR

c increases

. With the increase of Vi, I

c increases almost linearly

and so V0 decreases linearly till V

i is less than about 1V.

Transfer charac-teristics of a basebiased transistorin its commonemiter configura-tion:

forms the image of the distant object at the focalplane of the lens.

In normal adjustment of the telescope the eye pieceis placed in such a way that the final image isformed at infinity.

m = −f

fe

0f

0 - focal length of objective

fe- focal length of eyepiece.

Eyepiece can be placed in such a way that finalimage is formed at least distance of distinct vision.

Magnification produed m = − +

f

f

f

De

e0 1

Length of the telescope tube = f0 + f

e

23.mv

h

p

h ==λ

\ 631

34

e103109

106.6

××××=λ −

−

= 2.44 × 10–10 m

100103

106.62

34

ball ×××=λ

−

−

= 2.2 × 10–34 m

le ~ size of atom, l

ball << size of ball.

24. q = CV = CEd = AEEdd

A0

0 =∈∈

= ( )EAEE0

=φφ∈ Q

dt

d

dt

dqE

0

φ∈=∴

The term dt

dE

0

φ∈ is known as displacement current.

This term has been used to modify and generalizeAmpere’s Circuital law.

25.

−=

λ=υ

2

1

2

2n

1

n

1R

1

( )2

1

2

2n1n1

R

1

−=λ∴ =

−

2

1

2

2

o

n

1

n

1

A970

Physics - Sample Question Papers and Answers

20

Beyond this the transistor goes to saturation level. Weoperate the transistor in the active region for usingit as an ‘amplifier’

28. Total resistance in the five cases are :

21

21

2121RR

RRr;RRr;Rr;Rr,r

++++++

or

++

21

21

RR

RRrr,r , r + R

1 , r + R

2, r + R

1 + R

2 in

increasing order\ The correct order of values of I are : 4.2A, 1.4A,1.05 A, 0.6 A and 0.42 A

Also ,2.4r

E = ,

,6.0Rr

E

2

=+ 42.0

RRr

E

21

=++

and 4.1

RR

RR

Er

21

21

=

+

+

Solve first four to obtain, E = 4.2V, r = 1W ,R

1 = 3W , R

2 = 6W

OR(i) (a) In series

E = E1 + E

2, r = r

1 + r

2

(ii) In parallel combination

( )rr

rErEE

21

1221

++= ,

21

121

rr

rrR

+=

(iii) Numerical

( ) ( )

Ω=⇒+×

+×+×=

+++

;4

9R

21

21211221

R21

12

More heat is generated in series.

29. (i)

According to Biot Savart law

dB Id= µπ

θ0

4l sins2 ;

B = dB∫ sin f = µ

π0

4∫ × Idl sinφ

s2

B = µ

πφ0

24Is

dsin l∫But from the symmetry of figure,

sin f = r

r x2 2+ and φ dl∫ = 2pr

B = ( ) ( )µ

ππ0

2 2 2 2 1 242I

r xr

r xr

+ + (Q s2 = r2 + x2)

The magnetic field due to a circular coil of radius rcarrying a current I on a point P at an axial distance

x from the centre is B = 2

322

20

)Rx(2

nIR

+

×µwhere n is

the number of turns.(ii) Numerical

( )( ) ( )[ ]

( )( ) ( )[ ]

×+×

×µ=

×+×

×µ=

−−

−

−−

−

232222

22

20

232222

2

0

103104

104I

104103

103IB:haveWe

Thus I2 = – A

6

9

Current is in opposite sense to that in L1 .

OR

(i) For current carrying circular loop,rr

m NIA= , whererm is the magnetic moment of the coil, I is the cur-rent.N is the number of closely wound turns in the cir-cular loop.A is the area vector of the loop

AE

1E

2

Cr

1 r2

Bx

ºA

Eeq

req

IC

ºA

E

CIB

1 B2 I

I1

I2

A

Eeq

req

C

05.1Rr

E

1

=+

dB cos f

f

f

dB sin f

P dB sin f

dB cos f

S

T

M

r

O

N

s

x

f

Physics - Sample Question Papers and Answers

21

An electron moving around the central nucleus hasa magnetic moment m

l given by

µl l= em2

. Where l is the magnitude of the angu-

lar momentum of the circulating electron aboutthe central nucleus. The smallest value of m

l is

called the Bohr magnetron. mB = 9.27 × 10–24J/T.

(ii) µπ

=

ehm4

= 4.63 × 10–26 A.m2

30. (i)

The concave mirror forms a real inverted image ofthe object between focus and centre of curvature

Right angled triangles A’B’F and MPF are similar.

\′ ′ = ′B A

PM

B F

FD or

′ ′ = ′B A

BA

B F

FD -------- (1)

Right angled triangles ′ ′A B P and ABP are similar.

\′ ′ = ′B A

BA

B P

BP ------- (2) Comparing (1) and (2)

′ = ′B F

FP

B P

BP--------- (3);

′ − = ′B P FP

FP

B P

BP ----- (4)

We have BP = u object distance

′ =B P v image distance, FP = f focal length

Substituting in (4), v f

f

v

u

− =

uv – uf = vf, Dividing by uvf,

1 1 1

f v u− = ie,

1 1 1

u v f+ =

Applying sign convention

1 1 1

−+

−=

−u v f ie,

1 1 1

u v f+ =

Thin lens formula u

1

v

1

f

1 −=

(ii) Numerical vf

1

u

1

v

1 ⇒=− = +30 cm

Distance of this image from the mirror must be20 cm [m For image to from at O,mirror must re-verse the light \ d = (30 + 20) cm = 50 cm) Thefinal image is inverted.

OR

(i)

O.c

2

n1

n2

. B DI c

1I

1

A

C

Image formation by a lens is due to refraction attwo surfaces ABC and then ADC. ABC forms thevirtual image of object at I

1 and it acts as the object

for surface ADC. Refraction at ADC results in theformation of real image I.For refraction at ABC

− + = −n

OB

n

BI

n n

BC1 2

1

2 1

1-------- (1)

For refraction at ADC,

– n

DI

n

DI

n n

DC2

1

1 1 2

2

+ = −-------- (2)

Since lens is assumed to be very thin distances mea-sured from B and D are same as measured fromopticcentreBI

1 = DI

1. Adding 1 & 2

( )− + = − −

n

OB

n

DIn n

BC DC1 1

2 11 2

1 1------ (3)

When OB = ∝ , DI = f

( )n

fn n

R R1

2 11 2

1 1= − −

------ (4)

Physics - Sample Question Papers and Answers

22

E2

E1

I1

I1

r2

r1

+ –

–

1 1 12 1

1 1 2f

n n

n R R= − −

( )11

1 121

1fn

R R= − −

------- (5)

This is lens maker’s formula.(ii) Numerical

[v = 15 cm, f = 10 cm Þ u = –30 cmDistance of O from mirror: 20 cmBut O must be at radius of curvature for rays toreverse Þ R = 20 cm

1. E = Energy stored = C

Q

2

1CV

2

1 2

2 =

(The graph is showingC

1E∝ ) ,

Hence Q the charge on capacitors is kept constant.

2. Bqvr

mv2

= , or r = Bq

p

Bq

mv =

\ r : r = qp : q

a = 1 : 2

3. The frequency of the microwaves should match theresonant frequency of the water molecules in thefood.

4. ‘X’ is a pure capacitor ; Impedance = C

1

ω

5.

−

−

µµ=

212

1

r

1

r

11

f

1, the lens would now behave

like a diverging (concave) lens.

6. mk2

h

p

h ==λ \ The proton will have a higher

K.E. (mass of proton is slightly less than that of theneutron).

7. The ionisation energy of silicon gets (considerably)reduced compared to that of carbon. Silicon (a semi-conductor), therefore, becomes a (much) better con-ductor of electricity than carbon (an insulator).

8. (0 to t1), (t

3 to t

4).

9. (i) q1 is a negative charge and q

2 is a positive charge.

(ii) r

1.

4

q

r

q

4

1V

00∈π

=∈π

=

\ Slope of the V versus graph is 0

4

q

∈π

Since the slope of the graph for q1 has a larger mag-

nitude, q1 has the larger magnitude of the two.

10. RT = R

0 [1+ a (T – T

0)]

\ 5

6

100

120 = = 1 + a ( T – T0)

\ T × 10–4 ( T – 300) = 5

1; T = 1300 K

11. ( ) ( ) 2322

0

2322

2

ax1a2

I

xa2

IaB

+µ=

+µ=

\ B = B0 ( )

µ=∴+ −

a2

IBax1 0

0

2322

\ B (at x = a) = ( )0

023

0B225.0

4

2B2B ==−

and B (at x = 2a) = ( )0

023

0B504.0

25

5B5B ==−

Thus the given values are in good agreement withthe theoretically expected values.

12. Emax

= NBAw = NBA 2pu

= 20 × 0.5 × 0.09 × 2p ×π

150 volt = 270V

Eaverage

= Zero.13.

I = I1 + I

2 =

2

2

1

1

r

VE

r

VE −+−=

+−

+

212

2

1

1

r

1

r

1v

r

E

r

E

\

+−

++=

21

21

21

1221

rr

rrI

rr

rErEV

Comparing with V = Eeq

– Ir eq

We get Eeq =

21

1221

rr

rErE

++

SET III

Physics - Sample Question Papers and Answers

23

E

R

+=−=R

r1EirEV

14. (i) The glow gradually increases till it becomesmaximumReasons : There is a back (induced) emf in the in-

ductor

−=

dt

diL when the current is growing and

this delays the growth of current to its final steadyvalue.(ii) The glow will decrease.Reasons : The impedance of circuit will increasedue to the presence of the inductive reactance ofthe circuit.

15. (i) UV rays (ii) (UHF) radio waves (iii) Infraredwaves. l

3 > l

3 > l

1

16. During the formation of pn junction two processesoccur 1) diffusion 2) drift.Due to difference in the concentration of chargecarriers in two regions of p-n junction, the electronsfrom n- region diffuse through the junction into p-region and holes from p- region diffuse into n- re-gion.When electron diffuses from n- region to p- regionof p-n junction, it leaves behind an ionised donoratom in n- region, having positive charge which isimmobile as it is bound to the surrounding atom.As diffusion continues more positively charged do-nor atoms are created in n- region and it acquirespositive charge at the junction.When holes diffuse from p- region to n-region itleaves ionised acceptor atoms in p-region havingnegative charge. This results in the formation of alayer of negative charge at the p- region. Accumu-lation of charges on p side and n side produces anelectric field across the junction. The electric field

sets a potential barrier at the junction which op-poses further diffusion of majority charges. Thepotential barrier will behave as if an imaginary bat-tery is connected with positive pole to n sectionand negative pole to p section.

17. Modulation is the process in which a low audiofrequency base band message is superimposed ona high frequency wave called carrier wave.Modulation is necessary because of the followingreasons.

1. Size of antenna or aerial:- Antenna is needed bothfor transmission and reception. Antenna should have

a minimum height of λ4

, so that time variation of

the signal is properly sensed by the antenna. Audiofrequency range is 20 Hz to 20 kHz. For an audiosignal of 15kHz frequency,

m200001015

103c

3

8

=×

×=υ

=λ The required length of

the antenna is 20000

45000= m.

It is difficult to construct and operate such a tallantenna. By modulation if transmission frequency

is raised to 1MHz, then, l = .m30010

103

6

8

=× Length

of the antenna is 300

475= m , a reasonable height.

Hence it is necessary to do modulation.2) Effective power radiated by antenna: Power radi-

ated by an antenna 2

1P

l∝

For good transmission l should be small. For smallerantenna length, λ should be small or frequencyshould be high.

18. Angular width q = 4

9

101

10550

a−

−

××=λ

radians

= 5.5 × 10–3 radiansLinear width = Dq

= 1.1 × 5.5 × 10–3 m = 6.05 mmThe angular width would not change.

19.2

3

40

60

y

x == and 11040

1060

15y

x =+−=

+

(i) OR

Physics - Sample Question Papers and Answers

24

InformationSource Transmitter Transmission

channelReceiver

MessageSignal

TransmittedSignal

ReceivedSignal

(ii) Detection is the process of recovering the modu-lating (or information) signal from the modulatedcarrier wave. The essential steps followed in theprocess of detection are (i) The AM input wave ispassed through a rectifier to obtain its rectifiedwaveform. (ii) The rectified wave is passed throughan ‘envelope detector’ which retrieves the messagesignal as the envelope of the rectified wave.

26. (i) n - p -n transistor. (ii) Common emitter(iii) Considering characterstics for

Ib = 10mA and I

b = 50mA

∆∆=

b

c

I

IB

( )( ) 150

101050

105.25.86

3

=×−×−=

−

−

27. The permitted stationary orbits for the electron ina hydrogen atom are those for which the angularmomentum of the electron is an integral multiple

of h/2p mvnr

n =

π2h

n \ mv

hnr2

n=π

But n

nmv

h λ= the associated de Broglie wavelength

R

solving, we get , x = 45W, y = 30WFor the parallel connection

Ω=Ω+×=

+=′ 15

3030

3030

y30

y30Y

\ l

l

−=

′ 100y

x ,

l

l

−=10015

45 Þ l = 75.0 cm

ORThe emf of a cell equals the p.d. between its termi-nals when it is in an open circuit i.e. not supplyingany current. A voltmeter measures p.d. (and notemf.) as it draws a (small) current for its working.The potentiometer draws no (net) current (form thecell) at the balance point. So the cell can be treatedas if it were in an open circuit.

E1 + E

2 = k (351) and E

1 – E

2 = k (70.2)

\ 1

5

2.70

351

EE

EE

21

21 ==−+

, This gives 3

2

E

E

2

1 =

20. X ® a resistor , Y ® A capacitor

2R

V = and R = Xc

\ 2R

V

XR

V

2

VI

2

c

2

=+

== , A0.12

2I ==

(iii) X ® Net Impedance

21. (a) Because such telescopes(i) have high resolving power(ii) produce brighter images

(b) Two identical but independent light sources cannotproduce light waves continuously either in the samephase or having a constant phase difference betweenthem. (c) Brewster angle (i

p) is given by tani

p = m

‘m depends upon the wavelength (l ) of the incidentlight. Hence i

p will be different for different colours

of light.22. Work functions (i)For M

1 = hu

01

For M2 = hu

02

(ii) For M1,

hu3 = hu

01 + eV

1\

e

hV

31−υ=

he

oυ 1

Similarly, For M2,

e

h

e

hV 02

32

υ−υ=

\ V1 – V

2 =

e

h (u

02 – u

01)

\ Slope of either line = 0102

21VV

e

h

υ−υ−=

23. Continuous locus of all the particles of a mediumwhich are vibrating in the same phase is called awavefront.Refer set 1, page 153 answer 29 (first part)

24. The activity of a radioactive element at any instant,equals its rate of decay at that instant. Its SI unit isBecquerel (Bq) (= 1 decay per second)

Activity NT

logN

dt

dNR

2

e=λ=−=

\ 12

21

2

2

1

1

2

1

TN

TN

T

N

T

N

R

R =÷=

25. (i)

Physics - Sample Question Papers and Answers

25

for electron in its nth orbitHence 2pr

n = nl

n

or circumference of ntn permitted orbit= n × de Broglie wavelength

associated with the electron in the nth orbit.28. Capacitance of a capacitor is the ratio of charge

given to the p. d between the plates. CQ

V= . Its

unit is farad.Capacitance of a parallel plate capacitor is

C = ε0A

d, where A is the area of the plates and d is

the separation between the plates.(ii) Net capacitance = 4mF

Energy stored (W) = 12

2CV = 0.02 J

Charge q = CV = 4 × 10–4 coulombOR

(a) Consider a long straightwire carrying a charge q.Let l be its linear chargedensity. Hence Q = l l.Let us consider aGaussian cylinder asshown in figure.In regions II and III

fII = f

III = ∫ θdsCos.E

= o

=θ⊥∴

→90,ie,dsE r

r

The flux through the curved surface [region I]

φ θI E ds= ∫ . cos

= E ds.∫ (q = 0) = E r.2π l [ ]Q ds r=∫ 2π l

The total flux through the Gaussian cylinder φ φ φ φ π= + + =I II III E r.2 l ---(1)

According to Gauss’s theorem flux φε

λε

= =Q

0 0

l ---- (2)

Comparing (1) & (2); E r.20

π λε

ll=

Er

= λπε

l

2 0

, radially outward.

(b) (i) E = 5Ax + 2 B = 50 x + 10

II

I

III

++++++++++++++++

r

E

Electric flux through the face with point M on it

ds.E1

≅φ = E ds cos 180o

f1 = –E

1ds= –(0 +10) × 0.01 = –0.1 NC–1 m2

Similarly, flux through the face having point ‘N’on itf

2 = –E

2ds cos 00 = (50 × 0.1 + 10) × 0.01

= 0.15 NC–1m2

(Flux through all other faces will be Zero)\ Total flux through the cube = f

1 + f

2

= –0.1 + 0.15 = 0.05NC–1 m2

(ii) f = 0

q

∈ \ q = φ∈0

= 44.25 × 10–14 C

29. (a) Labelled diagram of moving coil galvanometer

Deflecting torque on the coil = NI BA sinqIn radial magnetic field q = 90o

\ Deflecting torque = NIA BCounter torque provided by the spring = Kf

\ In equilibrium Kf =NIAB \ IK

NBA

=φ

The quantity in bracket is constant \ fµ I

(b) (i) GI

VR

g

1−= GR

I

V1

g

+=∴

(ii) and GI2

VR

g

2−= G2R2

I

V2

g

+=∴

On comparison G = R1 – 2R

2

GI

V2R

g

3−=∴ = 2(R

1 + G) – G = 2R

1 + G

Permanent magnetPointer

Coil

Soft-ironcore

NN

Scale

N SSp

Pivot

Physics - Sample Question Papers and Answers

26

(b) (i) tan q = 3H

H3

H

V ==

\ q = tan-1 ( )3

or 60o

(ii) Horizontal Component = BE cosq = B

E cos 60o

= 234.0 × = 0.346 G

30.(a) A

B C

PN

RT

i2

SP

i1

Q d

r1

r2

( (

The magnification produced by the eyepiece is

Me = 1+ D

fe

M A BAB

vu0

0

0

= ′ ′ =

Mvu

Dfe

= +

0

0

1

The focal length of the objective lens is very small,u

0 = f

0. The focal length of the eye piece is also

very short. So that v0 = L, where L is equal to

length of the microscope tube.

\ M

+=

fe

d1

f

L

0

(b) Resolving power of microscope RP = λβµ

22.1

sin2

(i) Decreasing diameter of objective will decreaseb Hence R P will decrease.(ii) No effect

Figure shows passage of a light ray through a prism.It undergoes two refractions and finally emergesout along RS bending towards the base.

In Quadrilateral ÐAQN + ÐARN = 180o ( normalsat Q and R)

\ ÐA + ÐN = 180o ------ (1)

In D QNR, r1 + r

2 + N = 180o ------ (2)

From (1) and (2)

r1 + r

2 = A ------- (3)

From DTQR, d = ii – r

1 + i

2 – r

2 = i

1 + i

2 –(r

1 + r

2)

d = i1 + i

2 – A --------- (4)

45o (

(45o

(45o

(i) Turning a ray by 90o (ii) Turning a ray by 180o

(b)

OR(a) Compound microscope:- It is used for observing

highly magnified images of objects. The objectivelens of small focal length and short aperture formsa real, enlarged inverted image of the object. Theeye of large focal length and large aperture is placedin such a way that the first image is within the fo-cus of the eyepiece. Eyepiece forms an enlargedvirtual image at least distance of distinct vision.

Copper

Aluminium

Mercury

III

III

= 2R1 + R

1 – 2R

2 = 3R

1 – 2R

2

OR(a) We know that (i) copper is diamagnetic (ii) Alu-

minium is paramagnetic and (iii) mercury (cooledto 4.2 K) is perfect diamagnetic. Hence thebehaviour of field lines is as shown here

Additional Support Material on Value Based

Questions

Subject: Physics

Class : XII

Central Board of Secondary Education (CBSE), whose educational process is inclusive of co-scholastic

areas of life skills, attitude and values, sports and games as well as co-curricular activities, is aiming to

strengthen its education system in the area of value education. For the same, the board will be

introducing value-based questions in the papers of final examinations in all major subjects for classes XI

and XII from the academic session 2012-13 .

The Board has decided to assess students for 5 percent weight age in classes XI and XII through

questions which will be integrated with the content of the subject and analyzed on the basis of the

values it reflects. The questions will be 3-4 marks in a question paper of 70-90 marks.

The sample value based questions deal with the life skills and values attained by students like Self

Awareness, Empathy, Critical thinking, Creative Thinking, Decision Making, Problem Solving, effective

Communication, Interpersonal Relationships, Coping with Stress and Coping with Anger, and Dealing

With Emotions.

CLASS-XII PHYSICS

VALUE BASED QUESTIONS

Q: 1. A child is observing a thin film such as a layer of oil on water show beautiful colours when

illuminated by while light. He feels happy and surprised to see this. His teacher explains him the reason

behind it .The child then gives an example of spreading of kerosene oil on water to prevent malaria and

dengue.

• What value was displayed by his teacher ?

• Name the phenomenon involved ?

Q: 2. Ravi is using yellow light in a single silt diffraction experiment with silt width of 0.6 mm. The

teacher has replaces yellow light by x-rays. Now he is not able to observe the diffraction pattern. He

feels sad. Again the teacher replaces x-rays by yellow light and the diffraction pattern appears again. The

teacher now explains the facts about the diffraction and

• Which value is displayed by the teacher ?

• Give the necessary condition for the diffraction.

Q: 3. Aditya participated in a group discussion in his school on “Human eye and its defects” in the

evening he noticed that his father is reading a book by placing it at a distance of 50 cm or more from his

eye. He advised him for his eye check-up.

• Suggest the focal length/power of the reading spectacle for him, so that he may easily read the

book placed at 25 cm from eye.

• Name the value displayed by Aditya.

Q: 4. Mohit was watching a program on the topic MOON on the Discovery channel. He came to know

from the observations recorded from the surface of Moon that the sky appears dark from there. He got

surprised and wanted to know the reason behind it. He discussed it with his friends, and they had the

reasons as 1. Phenomenon of refraction of light 2.Phenomenon of scattering of light and explained the

topic to him in detail.

(i) Name the value that was displayed by Mohit

(ii ) what values were displayed by his friends

Q: 5 A teacher has given three lenses of power 0.5 D, 4 D, 10 D to a student. He is not sure as to which

lenses would he use for constructing a good astronomical telescope. So he consults his seniors and the

teacher also and constructs a telescope. Later he shows this telescope to the junior classes and explains

about the choice of lenses.

• What values has he shown by doing these?

• Which lenses are used as objective and which one as Eyepiece?

Q: 6.A person looking at a person wearing a shirt with a pattern comprising Vertical and Horizontal lines

is able to see vertical lines more distinctly than the horizontal ones. He shares his problem with his

friend who suggests him to go to a doctor immediately.

(i) Name the value displayed by his friend

(ii ) what is this defect due to ?

(iii) How is such a defect of be corrected ?

Q: 7. Students of class XII big mirrors in their classroom for science fair. the mirrors are so arranged that

one can see six images of himself .all the students of other classes who came to see this were very

happy and Geeta of class X th was determined to know the reason behind it . She went to the library,

consulted other students and next day came up with the answer.

• What values were depicted by Geeta ?

• Give the reason for seeing six images?

Q: 8. A child uses a semi conductor device in listening radio & seeing pictures on T.V. He was asked to

suggest the techniques as the cost of LPG/CNG is going up, to cope up with future situations.

• What are the values developed by the child?

• What may be the suitable semi conductor material used for utilization of maximum solar energy

with reasons?

Q: 9. Ruchi’s uncle who was a kabadiwalah was getting weak day by day. His nails were getting blue, he

stated losing his hair. This happened immediately after he purchased a big container of heavy

mass from Delhi University Chemistry Department. Doctors advised him hospitalization and

suspected he has been exposed to radiation. His uncle didn’t know much about radiations but

Ruchi immediately convinced her uncle to get admitted and start treatment.

(i) What according to you are the values utilized by Rama to convince her uncle to get admitted in

hospital

(ii)Name the radioactive radiations emitted from a radioactive element

Q: 10. Medha’s grandfather was reading article is newspaper. He read that after so many years of

atomic bombing is Hiroshima or Nagasaki, Japan National census indicated that children born

even now are genetically deformed. His grandfather was not able to understand the reason

behind it. He asked his Granddaughter Medha who is studying in class XII science. Medha sat

with her grandfather and showed him pictures from some books and explained the harmful

effects of radiations.

(i) What are the values/ skills utilized by kajal to make her grandfather understand the

reason of genetically deformity?

(ii) Name the nuclear reactions that occurred is atom bomb.

Q: 11. Muthuswami a resident of Kundakulam was all set to leave everything and shift to another place

in view of the decision of Govt. to start nuclear thermal power plant at Kundakulam. His

granddaughter Prachi, a science student, was really upset on the ignorant decision of her

grandfather. She could finally convince him not to shift, since adequate safety measures to avoid

any nuclear mishap have already been taken by the Govt. before starting nuclear thermal plants.

• What is the value displayed by Prachi in convincing her grandfather

• What is the principle behind working of nuclear reactor

• What are the main components of nuclear reactor

• Why is heavy water used as moderator?

Q: 12. Nisha’s uncle was advised by his doctor to get an ECG for his heart,Her uncle did not Know much

about the details & significance of this Test. She told her uncle that an ECG (Electro cardio

Graph) would enable the doctors to know of the condition of his heart without causing any

harm to him.Her uncle was convinced and got the required ECGdone.The resulting information

greatly helped his doctors to treat him will

• Whatare the values displayed by Nisha

Q: 13. Geeta has dry hair. A comb ran through her dry hair attract small bits of paper. She observes that

Neeta with oily hair combs her hair; the comb could not attract small bits of paper. She consults

her teacher for this and gets the answer. She then goes to the junior classes and shows this

phenomenon as Physics Experiment to them. All the juniors feel very happy and tell her that

they will also look for such interesting things in nature and try to find the answers .she succeeds

in forming a Science Club in her school.

• What according to you are the values displayed Geeta?

Q: 14. Ranis’s mother who was illiterate was folding her synthetic saree. She saw a spark coming out of

it .She got frightened and called Rani. Rani being a science student gave the reason behind it.

After knowing the reason her mother calmed down.

• what value was displayed by Rani

Q: 15. A picnic was arranged by school for the student of XII class. After some time it was raining heavily

accompanied by thundering & lightening. The Student got afraid. Some students went inside the

room. Two students asked for the key of the car and sat inside the car folding their legs on the

seat. The other students called them to come out but they refused. They knew that charge

inside the conducting shell is zero as told by the teacher and told others not to stand near the

electric pole when it is lightening.

• What value was displayed by these students?

Q: 16. A semiconductor device is used as a rectifier that allows the EMF to flow in positive direction and

a very small value in the reverse direction. Now a days , there is a problem of supply of less voltage

that damages the house hold appliances .You are asked to give the technique to save the appliances in

use

(i) What can you think to solve the situation?

• Can a diode be fabricated in terms of Doping, and choice of material to control the input

voltage to save your appliances from damage?

Q: 17. Chitra was watching her favorite TV serial suddenly the picture started shaking on TV screen. She

asked her brother to check the dish antenna. Her brother found no problem in dish. Chitra

noticed the same problem in TV picture again after some time. At the same time she heard the

sound of low flying air craft passing over their house. She asked her brother again. He explained

the cause of shaking picture on TV screen when air craft passes over head.

• Name the values used by Chitra’sbrother ?

• Why the picture on the TV screen was shaking when air craft was passing over head?

Q: 18.Pooja went to the market with her mother and decided to come back home by metro. At Metro station they were made to pass through a gate way for security check. Pooja passed through it and was waiting for her mother to come. She heard a long beep when her mother passed through metal detector. Pooja was confused why metal detector beeped in case of her mother. She asked the duty staff, who explained her in detail. Both were satisfied with the security system.

• What values are displayed by pooja • What is cause of sound through metal detector • Write the Principle on which a Metal detector works

Q: 19. Mr. Sanjeev, a physics teacher, was doing an experiment in lab using dry cell battery. The dry cell

was weak, giving less voltage, which was not sufficient to give proper reading. One of the student asked, “Sir, can’t we step-up the voltage using a transformer?” Teacher replied, No, we cannot step up DC voltage using step-up transformer and explained the reason and working of a transformer .the student then constructed a transformer for his Physics project and studied the factors responsible for losses in a transformer.

• What values are displayed by the student • Why transformer cannot be used to step-up DC voltage

Q: 20. Raj Pal Yadav, a retired Physics Teacher was working in his field with his grandson. There was a

big high tension tower carrying thick wires in their field. Grandson wanted to know as to why can’t the tower be removed from their field, so that they may get more space for crops. Raj Pal explained him the necessity of HT tower, and said it is very high voltage AC transmission line and is a lifeline of their town.

• What values are displayed by Raj Pal Yadav? • Why Long distance AC transmission is done at very high voltage. • What is the principle of transformer? • What are the energy losses in transformer?

Q: 21. Rahul after having lived in US for 12years returned back to India. He had a discussion with his

cousin Sumit on domestic power supply in US and in India. In US domestic power supply is at 110v, 50Hz, whereas in India it is 220V, 50Hz. Rahul was stressing that US supply is better than Indian supply. Both went to Sumit’s father an electrical Engineer and asked his opinion on the issue. He explained that both the supplies have advantages as well as disadvantages.

• What values are used by Rahul and Sumit? • Write one advantage and one disadvantage of 220V supply over 110V supply

Q: 22. One Sunday Rahul and Rama were enjoying with their friends at home. Suddenly their ceiling fan stopped working. Out of shear enthusiasm Rahul first switched off the power supply of the fan and opened the cap of the fan to check the problem. Vipin tried to stop him but he did not listen. The moment he touched the interior part of the fan, he fell down because of electric shock. All friends were scared as to what has happened as the power supply was already switched off.

i )What negative trait do you think has been displayed by Rahul? ii) What could be the possible cause of electric shock? iii) Write expression of current and emf of the component used in fan with phase difference.

Q: 23.Sandeep’s mother had put lot of clothes for washing in the washing machine, but the machine did

not start and an indicator was showing that the lid did not close. Sandeep seeing his mother disturbed thought that he would close the lid by force but realized that the mechanism was different. It was a magnetic system. He went to the shop and got a small magnetic door closer and put it on the lid. The machine started working. His mother was happy that Sandeep helped her to save Rs.300/- also.

• What was the value developed by Sandeep? • What values did his mother impart to Sandeep? • Every magnetic configuration has a north pole and a south pole. What about the field due to

torpid

Q: 24.Two girls Pooja and Ritu were very good observers and performed in the school function using

their cassette player. One day when they were performing, tape got stuck up and the music

stopped. But Pooja was determined not to let down the performance so she sang the song instead

of dancing and Ritu completed the dance.

• What were the values displayed by Pooja and Ritu? • What kind of Ferro magnetic material is using for coating magnetic tapes used in cassette

players or building memories stories in modern computers?

Q: 25.Tushar was using a galvanometer in the practical class. Unfortunately it fell from his hand and broke. He was upset, some of his friends advised him not to tell the teacher but Tushar decided to tell his teacher. Teacher listened to him patiently and on knowing that the act was not intentional, but just an accident, did not scold him andused the opportunity to show the internal structure of galvanometer to the whole class. (i) What are the values displayed by Tushar. (ii) Explain the principle, Construction and working of moving coil galvanometer.

Q: 26. Group discussion was arranged in class XII on the topic atmosphere. Three groups were made.

Teacher asked the Question. “Why can moon be not used as a communication satellite?” Answers were

given by all the three groups. Each group can give only one reason. Teacher told them that reason given

by each group is correct. The groups collected all the three reasons and come to correct conclusion.

(i) What values were showed by all the three groups?

(ii) Give the correct reason for the above question.

Q: 27. Lot of people like TV program CID. In this program there is some murder mystery which has to be

solved by the team of CID people. Every member of CID team work with full dedication. They collect

information from everywhere which can lead to correct conclusion. They use ultraviolet rays in forensic

laboratory. Some people got surprised to see the advantage of ultraviolet rays because they only know

that that ultraviolet rays coming from SUN are harmful .

• What values were displayed by CID team?

• Give the use of Ultraviolet rays in forensic laboratory

Q: 28.Deepa’s uncle wants to talk to his son in USA. He does not have much money to spend on

telephone calls. He has a computer at his home. Deepa told her uncle that he can talk to his son with the

help of computer and told him about internet. Her uncle now talks to his son every day. He thanked

Deepa for giving useful advice.

• What according to you are the values displayed by Deepa.

• How does internetwork?

Q.29. In an experiment of photoelectric effect, Nita plotted graphs for different observation between

photo electric current and anode potential but her friend Kamini has to help her in plotting the correct

graph. Neeta Thanked Kamini for timely help.

a) What value was displayed were Kamini and Neeta.

b) Draw the correct graph between I and V (NCERT)

Q: 30. A function was arranged in the school auditorium. The auditorium has the capacity of 400

students. When entry started students entered in groups and counting becomes a great problem. Then

science students took responsibility at the gate. All the students entered the hall one by one. This

helped them to maintain discipline and counting became easy with the help of a device used by these

students.

(i) What value is displayed by science students?

(ii) Name the device which is based on application of photoelectric effect.

Q: 31. Monica’s mother was heating food on a gas stove. Her friend Deepti came and saw her mother

heating food on the gas stove. Deepti told Monica’s mother, “Why don’t you buy a microwave oven”?

Monica’s mother replied at once that she doesn’t like to use microwave oven. Monica and Deepti made

it clear that microwave is not harmful for cooking food. This is an easy and safe process. Monica’s

mother got convinced and ordered for a microwave oven. Monica’s mother then arranged a small party

for her friends and told them the advantages of a microwave oven.

• What value was displayed by Monica and her friend?

• What value was displayed by Monica’s mother?

• What value was displayed by Monica to her friends?

Q: 32. Renu has to take admission in some professional college. It was last date of admission and Renu

left her birth certificate at her home. College was very far from the home. She called her brother and he

faxed the birth certificate. She got the admission and thanked her brother.

• What value was displayed by Renu?

• What value was displayed by her brother?

Q: 33. Two boys were going to the market .They saw two welders using welding machine. One welder is

using the goggles and face masks with window in order to protect his face. The other one is welding with

naked eye. They went to the welder who was not using face mask and explained him the advantages of

goggles and masks. Next day the welder bought a set of goggles. He created awareness among other

welders.

• What values were displayed by two boys?

• What values were displayed by welder?

Q: 34. Mrs. Gupta family was fast asleep during Night. They had no clue that there living room has

caught fire due to a short circuit. Suddenly they heard sound of alarm and woke up. They were surprised

to see that the sound was coming from the model of fire alarm prepared by their son. They were all

happy that a small science model has saved their life

(i) Give the values displayed by the parents and son.

(ii) Name the devise use in the model.

Q: 35.Mrs Thakur left her car headlights on while parking. The car would not start when she returned.

Seeing her struggle, Mohit went to her help. Not knowing much about cars, he ran and brought a

mechanic Raju from a garage nearby. Raju realized that the battery had got discharged as the headlight

had been left on for a long time. He brought another battery and connected its terminals to the

terminals of the car battery to get the engine started. Once the engine was running, he disconnected

this second battery. This is known as “JUMP STARTING”. Mrs. Thakur thanked both Mohit and Raju for

helping her.

(i)What values did Mohit have? (ii) A storage battery of emf 8.0 volts and internal resistance 0.5 ohm is being charged by a 120 volt DC supply using a series resistor of 15.5 ohms. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Q: 36.RenuRitu and Kajal lived in a resettlement colony where they observed most houses stole power

from transmission lines using hooks. They had learnt in school about fire caused due to electric short

circuit. They decided to make people aware of the risks involved and also the importance of paying their

electricity bills. They got all their friends and responsible elders together and with the help of the

electricity board, succeeded in changing the situation.

• What values did Renu, Ritu and Kajal have? • A low voltage supply from which one needs high currents must have a very low internal

resistance, why? • A high tension supply of say 6 KV must have a very large internal resistance. Why?

Q: 37. Rahul and Rohit bought an electric iron. They had a 2 pin plug. Rahul was keen to start using the

new iron with the 2 pin plug. However, Rohit insisted that they buy a 3 pin plug before using it. Rahul

got angry. Rohit patiently explained the importance of using a 3 pin plug and the earthing wire. He said

that if the metallic body of the iron came in contact with the live wire at 220 volts, they would get an

electric shock. If earthed, the current would go to the earth and the potential of the metallic body

would not rise. The iron would then be safe to use. Hearing Rohit, Rahul calmed down and agreed.

• What values did Rahul and Rohit have? • Which has greater resistance – 1 K watt electric heater or 100 watt electric bulb, both

marked 220 volts?

Q: 38.Sachin had gone to meet his grandfather who was staying in a village. In the evening, they were

both watching TV, when suddenly the lights went off. Grandfather told Sachin that the fuse must have

blown up as all their neighbors had electricity. Luckily Sachin knew how to change a fuse. His

grandfather was happy and told him that if he had been alone, he would have had to spend the night in

the dark without a fan. Sachin felt and made up his mind to replace the fuse with a circuit breaker which

uses a solenoid with a core so that his grandfather would not have any problems in future.

• What values did Sachin have? • The main power supply of a house is through a 5 ampere fuse. How many 100 watt

bulbs can be used in the house simultaneously at 220 volts?

Q: 39.While watching Discovery Channel.Sheela was impressed that certain organisms have the ability

to sense the field lines of earth’s magnetic field. They use this ability to travel from one location to

another. Sheela wanted to find the angle of dip at her place. She got a magnetic compass , using which

she found the magnetic meridian. She then mounted the compass on a cardboard and placed it

vertically along the magnetic meridian. She was able to measure the angle of dip.

• What values did Sheela have? • Define the magnetic elements of the earth.

Q: 40. Shama, a science student, while studying, was impressed that the nervous system in animals

depends on the electrical signals to work. Neurons pass on signals from sense organs to the brain. The

passage of an electrical signal constitutes an electric current. Shama was curious to know the range of

currents in different situations. She found that current in domestic appliances is a few amperes. During

lightning, the electric current is in tens of thousands of amperes, while in the nervous system, it is only a

few microamperes. She further discussed with her teacher about the magnitude of the magnetic field

created by these currents.

• What values did Shama have? • A galvanometer coil has a resistance of 15 ohms and the meter shows full scale deflection for a

current of 4 milliampere. How will you convert the meter into an ammeter of range 0 m- 6 amperes? Q: 41. The number of electrical generators used in areas where small workshops existed created lot of

pollution. Rishab and his five friends did a survey and realized that like in multistoried apartments, a

common generator could be set up for all these small workshops so that the noise and air pollution

could be reduced considerably. They had a tough time convincing the local bodies and now they are

going to the NGOs and some financiers to help them organize funds to do the needful. It is admirable to

see their perseverance.

• What values did Rishab and his friends have? • Kamla pedals a stationary bicycle, the pedals of which are attached to a 100 turn coil of

area 0.10 sqmetres. The coil rotates at half a revolution per second and is placed in a uniform magnetic field of 0.01 Tesla perpendiculars to the axis of rotation of the coil. What is the maximum voltage generated in the coil?

Q: 42.Alka and her sister were watching a movie in which the phenomena of aurora boriolis was shown.

Alka could not believe her eyes that such a colorful display like the one during commonwealth games

could be created by nature. She went to the library, but could not find the right book. So she consulted

her teacher who guided her. Hence, Alka understood that during a solar flare, a large number of

electrons and protons are ejected from the sun. Some of these get trapped in the earth’s magnetic field

and move in a helical path along the field lines. As the density of the field lines increases near the poles,

these particles collide with atoms and molecules of the atmosphere emitting green and pink light. Alka

shared this knowledge with her class when they studied the chapter of moving charges in magnetic field.

• What values did Alka have?

• What is the radius of the path of an electron moving at a speed of 3 x 107 m / sec in a magnetic field of 6 Gauss perpendicular to it? What is its frequency? Calculate its energy in kilo electron volt.

Q: 43. Renu saw her aunt suffering from severe joint pain. Her aunt could not take any pain killer as she

was allergic to them. Renu in her quest to help her aunt found the use of magnets. She read Dr.

Philpott’s work on magnetic therapy, that most people are negative magnetic field deficient due to

electromagnetic pollution. Supplementing the body with negative field energy has shown to restore

balance and encourage healing. Renu takes her aunt to the doctor daily without fail for the treatment.

Her aunt is improving at a phenomenal speed.

(i)What values does Renu have? (ii)A short bar magnet has a magnetic moment of 0.48J/T. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10cm from the centre of the magnet on a) axial line b) the equatorial line of the magnet.

Q: 44. Raju was enjoying TV programme at his home with his family at night. Suddenly the light went off

causing darkness all over. Mother asked Raju to bring candle along with matchstick from kitchen to put

the TV switch off. Raju at once picked the mobile phone and pressed the button lighting up the

surrounding. Her mother was surprised and asked where from the light was coming. Raj proudly

showed her the mobile.

• Which valueis displayed by Raju ?

• Which material is used in LED ?

Q: 45. Garima and Gaurav want to purchase a new TV set. They visited electronic shops to look for some

branded TV. The dealer showed them LCD and LED TV, s. Now they were confused which set to buy.

Finally after discussing with friends, reading relevant literature and searching the internet, they decided

to purchase LED.

• Which value is being highlighted I by Garima and Gaurav?

• What is the difference between LED and LCD?

Q: 46. Vikas was reading semi conductor in physics. His teacher has explained that electronic

components operate at very low voltage. But at home he daily saw his father using wall socket for

charging mobile.(220V) Confused Vikas put his doubt in front of his teacher and was satisfied. Next day

he went to school and shared the information with his classmates.

• Name the values displayed byVikas?

• What is the principle of charger?

Answer key (Physics):

(1)

(i) Value –appreciation of Nature

(ii) Interference of light

(2)

(i) Motivation

(ii) As the wavelength of x-rays is much smaller than that of yellow light, so the diffraction pattern is lost

when the yellow light is replaced by x-rays

(3)

(i) As the person wants to read a book placed at 25 cm, so if u = -25 cm than its image should be formed

by spectacle lens at v=50 cm, so that the defective eye may focus it on retina.

1 1 1 1 1 1 1 1

___ = ___ - ___ = ___ - ___ = ___ - ___ = ___

F v u (-50) (-25) 25 50 50

F=+50 cms (convex lens) = 0.5 m

Power of lens p=1/f = 2 D

ii) Compassion for others

(4)

(i ) curiosity

(ii) Sharing of knowledge

(5)

(i )curiosity, sharing of knowledge

(ii)Out of these three lenses, he will use a lens of power 0.5 D for objective lens and lens of power 10D

as the Eyepiece

(6)

• Empathy

• This defect is called ASTIGMATISM ( DETAILED ANSWER IN THE CHAPER ON OPTICS )

• Can be corrected by using a cylindrical lens

(7)

(i)Determination and Critical Thinking

(ii)Two adjacent walls inclined at 90 degree will give three images and the ceiling will repeat them

(8)

(i) Awareness of social problems, Generates new idea with fluency.

(ii) See NCERT at Page No 49&490

(9)

(i) Value displayed - awareness, critical thinking, decision making

(ii) X ray and Gamma rays.

(10)

(i) Sympathy, compassion

(ii) Nuclear- fission reactions

(11)

(i) Awareness, social responsibility

(ii) Controlled chain reaction

(iii)Nuclear Fuel, Moderator, Control rods, Coolant, Shielding

(iv) Neutrons produced during fission get slowed if they collide with a nucleus of same mass. As ordinary

water contains hydrogen atoms so it can be used as a moderator. But it absorb neutron at a fast rate. To

overcome this difficulty, Heavy waters is used as a moderator which has negligible cross sections for

neutron absorption

(12)

(i) Values- Empathy, helping, caring

(13)

(i) values- Curiosity, leadership, compassion

(14)

(i) Awareness & sensitivity

(15)

(i) Sharing the knowledge, social awareness

(16)

(i) Critical thinking

(ii) Hope for improvement by looking for alternatives

(17)

(i) Critical thinking and problem solving

(ii) Low lying air crafts reflect TV signals. Due to interference between direct signal received by antenna

and reflected signal the picture on TV shakes

(18)

( i) Curiosity, Power of Observation (ii) & (iii) Resonance (NCERT)

(19)

(i) Creativeness, curiosity. (ii) DC supply does not produce changing magnetic flux in the primary hence no emf is set in the secondary coil of transformer. (20) ( i) Social awareness ( ii) To minimize power loss due to generation of heat. (iii), (iv) (NCERT, alternating current) (21) (i) Critical thinking, awareness. (ii) Advantage the power loss at 220v supply is less than at 110v Disadvantage 220v supply has peak value 311v which is much higher than peak value of 155.5V for 110v supply (22)

• Careless attitude towards life. • Capacitor • NCERT Chapter on Alternating Current

(23) i) Sympathy, responsibility, nature of helping, self-reliance ii )Appreciation, thankfulness, economical nature. iii )NCERT page 179 (24)

• Team spirit, confidence, determination and courage.

• NCERT Page 202 Ex 5.17(d)

(25) (i) Courage to tell the truth, Gratitude to his teacher for her patience and tolerance

(ii) NCERT Pg.157

(26)

• Brotherhood, team spirit, critical thinking

• NCERT chapter on Principles of Communications

(27)

• Team work, Awareness.

• In the detection of forged documents, finger prints.

(28)

(i) Caring, creating awareness.

(ii)NCRT

(29)

a) sharing and caring

b) Refer NCERT

(30)

(i) Sense of responsibility.

(ii) Photo cell.

(31)

• Sharing knowledge.

• Understanding.

• Creating awareness.

(32)

(i) Awareness

(ii) Understanding

(33)

(i) Knowledge

(ii) Awareness

(34)

(i) Knowledge, Scientific thinking

(ii) NCERT (photo cell)

(35)

• Helpful, aware of his weakness, decision making ability.

• NCERT Q p 128 Exercise 3.11 (36)

• Team spirit, critical thinking, decision making, assertive communication.

• NCERT Q p129 Exercise 3.18 c,d •

(37)

• Rahul is enthusiastic and flexible, Rohit is patient, knowledgeable, assertive,

• Compassionate and mature.

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(i) Empathy, dutifulness, determination, responsibility, compassion.

(ii) I = P / V = 100 / 220 = 5/11

No. of bulbs = 5 / (5/11) = 11

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• Nature of appreciation, curiosity, diligence, self-reliance, creative skill.

• NCERT p 186

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(i) Curiosity, nature of appreciation, critical thinking, research oriented mind and ambitious.

(ii)NCERT p 172 exercise 4.28 (41)

(i) Team spirit, patience, tolerance, magnanimity, determination, responsibility and dutifulness.

(ii) NCERT p 226 example 6.11

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• Nature of appreciation, diligence, curiosity, research mindedness, communicative skills,

• NCERT p 139 example 4.3

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• Love, sympathy, punctuality and regularity, diligent, maturity and responsibility.

• NCERT Q page201 exercise5.12

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(i) Creative thinking

(ii) NCERT (SEMI Conductors)

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(i) Interpretation skill

(ii)NCERT (SEMI Conductors)

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( i) Scientific thinking

(ii )Rectifier