1. electrostatics · 2016. 7. 2. · 1 1. electrostatics 1. what is quantization of electric...
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1
1. ELECTROSTATICS
1. What is quantization of electric charge?
Electric charge of any system is always an integral multiple of the least amount of charge.
q=ne
2. State coulomb law.(Jun-07,Mar-10,Jun-10)
The force of attraction or repulsion between two point charges is directly proportional to the product of the charges and
is inversely proportional to the square of the distance between them…
1 2
2
0
1
4
q qF
r
3. Define permittivity of a medium.
Permitivity of a medium is the property of the medium which determines the magnitude of the force between the charges
when placed in it.
4. Define relative permittivity of a medium.
Relative permittivity of a medium is the ratio of permittivity of the medium to the permittivity of free space.
0
r
5. Define charge of one coulomb.(Mar-06,Oct-10)
One coulomb is the quantity of charge which when placed at a distance of 1m in air or vacuum from an equal and similar
charge experiences a repulsive force of 9 x 109N
6. Define electric field and electric field intensity.
Electric field due to a charge is the space around the test charge in which it experiences a force.
Electric field intensity is defined as the force experienced by a unit positive charge kept at that point.
0
FE
q
7. Define additive nature of charge.(Oct-07)
The total electric charge of a system is equal to the algebraic sum of electric charge located in the system. For example,
if two charged bodies of charges +2q,-5q are brought in contact, the total charge of the system is -3q.
8. What is electric dipole and electric dipole moment?(Oct-09)
Two equal and opposite charges separated by a very small distance constitute an electric dipole.
The magnitude of the dipole moment is given by the product of the magnitude of one of the charges and the distance
between the two charges. P=q2d.
It is a vector quantity and it acts from –q to +q . The unit of dipole moment is Cm.
9. Define electric potential.(Mar-07,Jun-09)
Electric potential at a point in an electric field is defined as the amount of work done in moving a unit positive charge
from infinity to that point against the electric forces.
04
qV
r
10. Define electric potential difference.
Electric potential difference between two points is defined as the amount of work done in moving a unit positive charge
from one point to another point against the electric force.
11. Define one volt.
The potential difference between two points is one volt if one joule of work is done in moving a charge of one coulomb
from one point to another point against the electric force.
2
12. What are equipotential surfaces?
If all the points on a given surface have the same electric potential then the surface is called an equipotential surface.
13. Define electric flux. Give its unit.(Jun-08)
Electric flux is defined as the total number of electric lines of force crossing a given area.
.d E ds
14. What is electrostatic shielding?(Mar-08)
The process of isolating a certain region of space from electric field. It is based on the fact that electric field inside a
conductor is zero.
15. Define capacitance of one farad.
Capacitance is one farad, when a charge of 1 C is given to a conductor raises the potential of the conductor by 1 volt.
16. State Gauss law.(Jun-06,Oct-06,Mar-09,Mar-11)
The total flux of an electric field E over any closed surface is equal to
0
1
times the net charge enclosed by the surface.
0
q
17. Why it is safer to be inside a bus than to stand under a tree during a thunder storm?(Mar-06,Jun-06,Jun-09,Mar-
10)
The metal body of the bus provides electrostatic shielding. The electric discharge passes through the body of the bus and
the electric field inside the bus is zero.
18. What is electrostatic induction? Mention its uses.
It is possible to obtain charge without any contact with another charge. The phenomenon of producing charges is known
as electrostatic induction.
It is used in electrostatic machines like Van de Graff generator and capacitors.
19. Define capacitance of a conductor.(Mar-09)
The capacitor is a device to store charge.
The capacitance of a conductor is defined as the ratio of the charge given to the conductor to the potential developed by
the conductor. The unit of capacitance is farad.
q
CV
.
20. Mention the uses of capacitor.(Oct-07,Mar-11
1. They are used in the ignition system of automobile to eliminate sparking.
2. They are used to reduce voltage fluctuations in power supplies and to increase the efficiency of power
transmission.
3. They are used to generate electromagnetic oscillations and in tuning radio circuits.
21. What is dielectric? Give examples.
A dielectric is an insulating material in which all the electrons are tightly bound to the nucleus of the atom and are not
free to move under the influence of an external field. E.g Ebonite, mica and oil.
22. Define electric polarization.(Oct-06)
The alignment of the dipole moments of permanent or induced dipoles in the direction of the applied electric field is
called polarization or electric polarization.
23. What is action of points or corona discharge?(Jun-07,Oct-08)
The leakage of charges from sharp points on a charged conductor is known as action of points or corona discharge.This
principle is used in Van De Graff generator and lightning arrestors.
3
24. Differentiate polar and non polar molecule.(Mar-07,Oct-10)
No Polar molecule Non polar molecule
1 A polar molecule is one
in which the centre of
gravity of the positive
charges is separated from
the centre of gravity of
the negative charges by a
finite distance.
A non polar molecule is one in
which the centre of gravity of
the positive charges coincides
with the centre of gravity of
the negative charges.
2 They have a permanent
dipole moment.
They do not have a permanent
dipole moment
3 2 2 3, , ,H O N O HCl NH 2 2 2, ,O N H
5 MARK QUESTION.
1. Write the properties of electric lines of force.(Mar-07,08,10,11 . Oct-11)
1. Lines of force starts from positive charge and terminate at negative charge.
2. Lines of force never intersect.
3. The tangent to a line of force at any point gives the direction of the electric field at any point.
4. The number of lines per unit area, through a plane at right angles to the lines is proportional to the magnitude of E.
When the lines of force are close together , E is large and where they are far apart, E is small.
5. Each unit positive charge gives rise to
0
1
lines of force in free space. Hence number of lines of force originating
from a point charge q is
0
qN
in free space.
2. Derive an expression for the torque acting on the electric dipole in an electric field.(sep-10)
Consider a dipole AB of dipole moment P placed at an angle θ in an uniform electric field. The charge q experience of
force of Eq in the direction of the field. The charge –q experiences _Eq in opposite direction. Thus the net force on the
dipole is zero.
The two equal and unlike parallel force constitute an torque.
The magnitude of torque is
τ=one of the force x perpendicular distance between them.
2
2
Fx dSin
Eqx dSin
PESin
( q x2d = P )
In vector notation PxE
3. Derive an expression for the potential energy of an electric dipole in an electric field.
Electric potential energy of an dipole in an electrostatic field is the work done in rotating the dipole to the desired
position in the field.
A dipole placed with the electric field E, experiences a torque PESin
Work done is rotating the dipole through dθ is dw = τ. dθ
= PESin d
4
The total work done in rotating the dipole through an angle θ
W dw
W PESin d
W PECos
This work done is the potential energy (U) of the dipole
U PECos
When the dipole is aligned parallel to the field θ=00 U= -PE
4. Define electric potential and hence derive an expression for the potential due to point charge (mar-09)
Electric potential in an electric field at a point is defined as the amount of workdone in moving an unit positive charge
from infinity to that point against the electric forces.
Let q be an isolated charge situated in air at O. P is a point at a distance r from O.
Consider two points A and B at a distance x and x+dx from the point o.
The potential difference between two points is dV = - E dx
The force experienced by a unit positive charge is 2
0
1
4
qE
x
2
0
1
4
qdV dx
x
The total work done is
2
0
0
1
4
1
4
rq
V dxx
qV
r
5. Prove that the energy stored in a capacitor is q2/2C.
The capacitor is a charge storage device.
Work has to be done to store the charges in a capacitor. This work done is stored as
electrostatic potential energy in the capacitor.
Let q be the charge and V be the potential difference. The work done in storing the
additional charge dq is dw=Vdq.
q qdw dq V
c c
Total work done to charge a capacitor is
w dw
2
2
qw
c
This work done is stored as electrostatic potential energy (U) in the capacitor.
21
2
qU
C
21( )
2U CV q CV
0
w dqc
5
1.ELECTROSTATICS.
1. Derive an expression for the electric field due to an dipole along the axial line. (Mar-06,Mar-09,Sep-10,Mar-11)
-q and +q are two charges separated by a small distance constitute an electric dipole.
P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.
The electric field at the point P due to +q placed at B is
1 2
0
1
4
qE
r d
along BP
The electric field at the point P due to –q placed at A is
2 2
0
1
4
qE
r d
along PA
E1 and E2 acts in opposite direction.
The magnitude of resultant electric field acts in the direction of the vector with a greater magnitude. The resultant
electric field at P is
E = E1 – E2
2 2
0 0
1 1
4 4
q qE
r d r d
2 2
0
22 2
0
1 1
4
4
4
qE
r d r d
q rdE alongBP
r d
If the point is away from the dipole then d<<r
4
0
3
0
3
0
3
0
4
4
4
4
2 2
4
2( .2 )
4
q rdE
r
q dE
r
q dE alongBP
r
pE P q d alongBp
r
E acts in the direction of the dipole moment
2. Derive an expression of the electric field due to an dipole along the equatorial line. (Mar-07,Jun-09)
-q and +q are two charges separated by a small distance constitute an electric dipole.
P is a point along the equatorial line of the dipole at a distance r from the midpoint O of the electric dipole.
Electric field at a point P due to the charge +q of the dipole
Electric field at a point P due to the charge –q of the dipole
1 2
0
1 2 2
0
1
4
1
4
qE
BP
qE alongBP
r d
6
2 2
0
2 2 2
0
1
4
1
4
qE
AP
qE alongPA
r d
The magnitudes of E1 and E2 are equal.
E1 andE2 are resolved into two components.
The vertical component E1sinθ and E2sinθ are equal and opposite, therefore they cancel each other.
The horizontal component are added along PR.
E=E1Cosθ+E2Cosθ
E=2E1 Cosθ (E1=E2)
2 2
0
2 2
12
4
qE Cos
r d
dButCos
r d
2 2 2 20
32 20 2
32 20 2
12
4
1 2
4
12
4
q dE
r d r d
q dE
r d
PE P q d
r d
d is very small compared to r, d<<r
3
0
1
4
PE
r
The direction of E is along PR, parallel to the axis of the dipole and directed opposite to the direction of the dipole
moment.
3. Derive an expression for the electric potential due to an electric dipole .()ct-06,Mar-08,Jun-08,)
-q and +q are two charges separated by a small distance constitute an electric dipole.
P is a point at a distance r from the midpoint O of the electric dipole and θ be the angle between PO and OB. Let r1 and
r2 be the distance of the point P from +q and –q.
Potential at P due to charge +q
1
0 1
1
4
qV
r ----------(1)
Potential at P due to charge –q
2
0 2
1
4
qV
r
---------(2)
Total potential due to dipole is V= V1 +V2
0 1 2
1 1
4
qV
r r
--------(3)
Applying cosine law,
7
2 2 2
1
22 2
1 2 2
2 2
1
2
21
21
r r d rdCos
d rdCosr r
r r
d r
dCosr r
r
1
2
1
21
dCosr r
r
1
2
1
1 1 21
dCos
r r r
using binomial theorm and neglecting higher powers
1
1 11
dCos
r r r
----------(4)
2 2 2
1
2 2 2
1
22 2
1 2 2
2 2
1
2 (180 )
2
21
21
r r d rdCos
r r d rdCos
d rdCosr r
r r
d r
dCosr r
r
1
2
2
21
dCosr r
r
1
2
2
1 1 21
dCos
r r r
using binomial theorm and neglecting higher powers
2
1 11
dCos
r r r
--------(5)
Substituting (4) and (5) in eqn.(3)
0
2
0
2
0
11 1
4
2
4
4
q d dV Cos Cos
r r r
qdCosV
r
PCosV
r
2
04
PCosV
r
(P=q2d)
Special cases.
(1)when the point P lies on the axial line of the dipole on the side of +q then θ=00
2
04
PV
r
8
(2) when the point P lies on theaxial line of the dipole on the side of –q then θ=1800
2
04
PV
r
(3) when the point P lies on the equatorial line of the dipole then θ =900 V=0
4. State Gauss law. How will you find electric field (1)due to an infinitely long
straight charged wire and (2) due to an infinite charged plane sheet.(Jun-11,Mar-
12)
Law.
The total flux of the electric field E over any closed surface is equal to 1/ε0 times the
net charge enclosed by the surface.
0
q
(1)Field due to an infinite long straight charged wire.
Consider an charged wire of infinite length having linear charge density λ.
Let P be a point at a distance r from the wire.
Consider a Gaussian surface of cylinder of length l and radius r.
On the curved surface E and ds are in the same direction. θ=00 Cos 0=1.
The electric flux through curved surface
Since E and ds are at right angles to each other at plane caps, the electric flux =0.
Total flux through the Gaussian surface
2E rl .
The net charge on the Gaussian surface is
q=λl
By Gauss law.
0
0
0
2
2
q
lE rl
Er
The direction of the field E is radially outward, if the charge is positive and inward if the charge is negative.
(2)Electric field due to an infinite charged plane sheet.
Consider an infinite plane sheet of charge with surface charge density σ.
Let P be a point at a distance r from the sheet.
Consider a Gaussian surface, a cylinder of area A and length 2r.
At end cap E and ds are in the same direction.
The total flux through the closed surface is
0
.
( 0 , 1)
E ds
EdsCos
Eds Cos
9
'. . ( 0, 1)
2
P PE ds E ds Cos
EA EA
EA
If σ is the surface charge density, then the charge within the Gaussian surface is q = σA
By Gauss law
0
q
0
2A
EA
.
0
E
5. Explain the principle of capacitor. Deduce an expression for the capacitance of the parallel plate capacitor. (Jun-
10)
Principle of a capacitor.
Consider a insulated conductor with a positive charge q having potential V. The capacitance of A is C=q/V.
When another insulated metal plate B is brought near A, negative charges are induced on the side of B near A. An equal
amount of positive charge are induced on the other side of B.
The negative charge in B decreases the potential of A. The positive charge in B increases the potential of A.
But the negative charge on B is nearer to A than the positive charge on B. The net effect is that, the potential of A
decreases. Thus the potential of A increases.
If the plate B is earthed, positive charges get neutralized. Then the potential of A decreases further. Thus the capacitance
of A is considerably increased.
The capacitance depends on the geometry of the conductors and nature of the medium.
Capacitance of a parallel plate capacitor.
The parallel plate capacitor consist of parallel plates X and Y of area A, separated by distance d, having a charge density
σ.
The electric lines of force starting from plate X and ending of the plate Y are parallel to each other and perpendicular to
the plates.
By Gauss’s law, electric field at a point between the two plates is
0
E
Potential difference between the plated X and Y is
0
0
0
0
.d
d
V E dr
V dr
dV
The charge on the plate is q=σA (σ=q/A)
The capacitance of the parallel plate capacitor
The capacitance is directly proportional to the area (A)of the plates and inversely proportional to the
distance of separation(d).
0
0
/
qC
V
AC
d
AC
d
10
6. Deduce an expression for the equivalent capacitance of capacitors connected in series and parallel. (Jun-07,sep-
07)
Capacitors in series.
Consider three capacitors of capacitance C1,C2,C3 connected in series.
Each capacitor carries the same amount of charge q.
Let V1,V2,V3 be the potential difference across the capacitors C1, C2, C3 respectively.
The potential difference across each capacitor is
1 2 3
1 2 3
1 2 3
1 2 3
, ,q q q
V V VC C C
V V V V
q q qV
C C C
1 2 3
1 1 1V q
C C C
If CS is the effective capacitance of the series , then
S
qV
C
The reciprocal of the effective capacitance is equal to the sum of reciprocal of the capacitance of the individual
capacitors.
Capacitors in parallel.
Consider three capacitors of capacitance C1,C2,C3 connected in parallel.
The potential difference across the each capacitor is the same.
The charges on the capacitors are q1=C1V, q2=C2V, q3=C3V.
The total charge is
q=q1+q2+ q3
q= C1V+C2V+C3V.
But q=CPV
where CP is the effective capacitance of the system.
CPV= C1V+C2V+C3V.
CP= C1+C2+C3
Hence the effective capacitance of the capacitors is the sum of the capacitances of the individual capacitors.
7. Describe the principle, construction and working of van de graf generator.(Sep-08,Sep-09,Mar-10)
Electrostatic machine which produces large electrostatic potential difference of the order of 107 V.
Principle.
1.Electrostatic induction.
2.Action of points.
Construction.
A hollow metallic sphere A is mounted on insulating pillars.
A pulley B is mounted on the centre of the sphere and another pulley C is mounted near the bottom.
Two comb shaped conductor D and E are mounted near the pulleys. The comb D is maintained at positive potential of
104 V. The comb E is connected to the inner side of the metal sphere.
Working.
Because of high electric field, near the comb D the air gets ionized due to the action of points, the negative charges in air
moves towards the needles and the positive charges are repelled towards the belt and stick to it. The positive charges
moves up and reaches near the comb E.
Due to electrostatic induction, the comb E acquires negative charge and the sphere acquires positive charge. The
positive charge is distributed on the sphere.
The high electric field at the comb E ionizes the air. Negative charges are repelled to the belt, neutralizes the positive
charge on the belt and hence the descending belt will be left uncharged.
11
Thus continuously the positive charge are transferred to the sphere. As a result, the potential of the sphere keeps
increasing till it attains maximum valve.
After this stage it starts leaking to the surrounding due to ionization of the air.
The leakage of charge from the sphere can be reduced by enclosing it in a gas filled steel chamber at a very high
pressure.
Uses.
The high voltage is used to accelerate positive ions (protons, deuterons) for the purpose of nuclear disintegration.
12
2.CURRENT ELECTRICITY.
1. State ohm’s law.(Mar-06,Oct-07,09,Mar-10)
At constant temperature, the steady current flowing through a conductor is directly proportional to the potential
difference between the two ends of the conductor.
V=IR
2. Define specific resistance or resistivity.
The electric resistivity of a material is defined as the resistance offered to the flow of current by a conductor of unit
length, having unit area of cross section .
RA
l
3. What is super conductivity?
The ability of certain material their compounds and alloys to conduct electricity with zero resistance at very low
temperatures is called superconductivity. The material which exhibit this property is called super conductors.
4. Define critical temperature or transition temperature.
The temperature at which electrical resistivity of a material suddenly drops to zero and the material becomes a
superconductor is called the transition temperature.
5. What are the changes is observed at transition temperature?(Jun-10)
1. The electrical resistivity drops to zero.
2. The conductivity becomes infinity.
3. There magnetic flux lines are excluded from the material.
6. Define temperature co-efficient of resistance.(Jun-08,Mar-11)
The temperature coefficient of resistance α is defined as the ratio of increase in resistance per degree rise in temperature
to its resistance at 00C. Its unit is per0C
7. Differentiate EMF and potential difference.(Jun-07,Oct-08)
No. EMF Potential difference
1 It is the difference in
potential between the
two terminals of a cell
in an open circuit.
It is the difference in
potential between the two
points in closed circuit.
2 It is independent of the
external resistance
It is proportional to the
resistance between two
points.
8. Differentiate drift velocity and mobility.(Oct-06,Mar-08,Mar-09)
No Drift velocity Mobility
1 Drift velocity is defined
as the velocity with
which free electrons get
drifted towards the
positive terminal when
an electric field is
applied
Mobility is defined as the
drift velocity acquired per
unit electric field.
2 The unit is m/s The unit is m2V-1s-1
3 dv E e
m
13
9. State Kirchoff’s law (Jun-06,Mar-07,Mar-08,Mar-09)
I law.
The algebraic sum of the currents meeting at any junction in a circuit is zero.
II law.
The algebraic sum of the products of current and resistance in each part of any closed circuit is equal to the algebraic sum
of the emf in that closed circuit.
10. What is the principle of potentiometer?
The emf of the cell is directly proportional to the balancing length. E l
11. Why copper is not used as potentiometer wire?
1. Low resistivity.
2. High temperature coefficient resistance.
12. State Faraday’s law of electrolysis.(Mar-06,Jun-10,Oct-10)
I law.
The mass of substance liberated at an electrode is directly proportional to the charge passing through the electrolyte.
m q
II law.
The mass of substance liberated at an electrode by a given amount of charge is proportional to the chemical equivalent of
the substance. m E
13. Why is the internal resistance of a secondary cell is low?
The electrodes of the secondary cell are arranged in such a way so as to produce a low internal resistance, thus providing
a large amount of current to the external circuit.
14. What are the applications of secondary cells?(Oct-08)
The secondary cells are rechargeable. They have low internal resistance hence they can deliver high current.
They can be recharged for large number of times.
They are used in all automobiles like cars,trucks etc.
15. Distinguish electric power and electric energy.(Jun-08,Jun-09)
Electric power.
Electric power is defined as the rate of doing electric work. P=VI unit is watt.
Electric energy.
Electric energy is defined as the capacity to do work.
Energy = VIt unit is joule.
16. Differentiate primary cell and secondary cell.
No Primary cell Secondary cell
1 Primary cell cannot be
recharged
They can be recharged
2 They are irreversible They are irreversible
3 They have high internal
resistance.
They have low internal
resistance.
4 Daniel cell, Leclanche
cell
Lead acid accumulator,
Alkali accumulator.
14
2. CURRENT ELECTRICITY.
1. Define drift velocity and mobility. Derive an expression between the current and drift velocity.(Mar-06)
Drift velocity is defined as the velocity with which free electrons gets drifted towards the positive terminal when an
electric field is applied.
Mobility is defined as the drift velocity acquired per unit electric field.
Consider a conductor XY of length L and area of cross section A.
An electric field is applied between its ends.
Let n be the number of free electrons per unit volume.
The number of conduction electrons in the conductor = nAL.
The charge of an electron=e
The total charge passing through the conductor q=nALe
The time in which the charge pass through the conductor
d
Lt
v
The current flowing through the conductor
d
d
q nALeI
t L
v
I nAev
Current density
d
IJ nev
A
2. Write a note on superconductors.
The ability of certain metals their compounds and alloys to conduct electricity with zero resistance at very low
temperature is called superconductivity.
The material which exhibits this property is called superconductors.
The phenomenon of superconductivity was observed by H.K. Onnes. The mercury suddenly showed zero resistance at
4.2K.
The first explanation is given by BCS theory.
The temperature at which electrical resistivity of the material suddenly drops to zero and the material changes
from normal conductor to a superconductor is called transition temperature or critical temperature.
The following changes are observed at critical temperature.
1.The electrical resistivity drops to zero.
2.The conductivity becomes infinity.
3. The magnetic flux lines are excluded from the material.
3. Write the applications of superconductors.(sep-06,Mar-09,Jun-11)
1. Superconductors form the basis of energy saving power systems, namely the
superconducting generators which are smaller in size and weight, in comparison with conventional generators.
2. Superconducting magnets have been used to leviate trains above its rails. They can be driven at high speed with
minimal expenditure of energy.
3. Superconducting magnetic propulsion systems may be used to launch satellites into orbits directly from the earth
without the use of rockets.
4. It can be used for transmission lines.
5. Superconductors can be used as memory or storage elements in computers.
4. Find the effective resistance of the resistors connected in series.(Oct-06)
When the resistors are connected in series , the current flowing through each resistor is the same.
The potential difference between the resistors are V1,V2,V3and V4 respectively.
The net potential difference
V = V1 + V2 + V3 + V4,
15
By ohms law V1 = IR1 ,V2
= IR2, V3 = IR3 , V4
= IR4
IRS = IR1 +IR2+IR3+IR4
Where RS is the equivalent resistance .
RS =R1+R2+R3+R4
Thus the equivalent resistance in series is equal to the sum of
the resistance of individual resistance.
.
5. Find the effective resistance of the resistor connected in parallel.(Oct-06)
When the resistors are connected in parallel, the potential difference across each resistor is the same.
A current I entering the combination gets divided into I1,I2,I3,and I4 through R1,R2,R3, and R4 respectively.
1 2 3 4I I I I I
By ohms law 1 2 3 4
1 2 3 4
, , ,V V V V
I I I IR R R R
and
P
VI
R
Where Rp is the effective resistance of the parallel
combination.
1 2 3 4p
V V V V V
R R R R R
1 2 3 4
1 1 1 1 1
pR R R R R
When number of resistors are connected in parallel, the sum of the reciprocal of the resistance of the individual
resistors is equal to the reciprocal of the effective resistance of the combination.
6. Discuss the variation of resistance with temperature with an expression and a graph.
The resistivity of substances varies with temperature.
For conductors the resistance increases with increase with increase in temperature.
If R0 is the resistance of a conductor at 00C and Rt is the resistance of same conductor at
t0C, then
Rt = R0(1+αt)
α = temperature coefficient of resistance.
0
0
tR R
R t
The temperature coefficient of resistance is defined as the ratio of increase in resistance per degree rise in
temperature to its resistance at 00C
Metals have high positive temperature coefficient of resistance. i.e their resistance increases with increase in temperature.
Insulators and semiconductors have negative temperature coefficient of resistance. i.e their resistance decreases with
increase in temperature.
A material with a negative temperature coefficient is called a thermistor.
The temperature coefficient is low for alloys.
7. How will you determine the internal resistance of a cell using voltmeter?(Jun-08,Sep-09,Mar-11)
The circuit connections are made as in diagram. With Key open the emf of cell E is found by connecting voltmeter across
it.
A small resistance R is included in the external circuit and key K is closed.
The potential difference across R is equal to the potential difference
across cell(V).
The potential difference across R is V=IR
The potential difference is less than the emf of the cell
V = E – Ir
Ir = E – V
16
Ir E V
IR V
E Vr R
V
Since, E,V and R is known, the internal resistance of the cell can be determined.
8. State and explain Kirchoff’s law.(Jun-07)
I law.
The algebraic sum of the currents meeting at any junction in a circuit is zero.
The sum of the currents entering the junction is equal to the sum of the currents leaving the junction.
Sign convention.
The current flowing through a junction is positive and the current flowing away from a junction is negative.
1 4 5 2 3I I I I I
II law.
The algebraic sum of the product of resistance and current in each part of any
closed circuit is equal to the algebraic sum of the emf in that closed circuit.
Sign convention.
The current flowing in clockwise direction is positive and the current in the
anticlockwise direction is negative
I1R2+I2R3+I4R6+I1R1=E1-E2
9. Obtain the condition for bridge balance .(Mar-06,Jun-06,Oct-06,Mar-08,Jun-09,Mar-10)
Applying the current law at junction B
1 3 0gI I I
At junction D
2 4 0gI I I
Applying the voltage law at the closed path ABDA
1 2 0gI P I G I R
Applying the voltage law at the closed path ABDA
1 3 4 2 0I P I Q I S I R
When the galvanometer shows zero deflection then Ig = 0
1 3
2 4
1 2
I I
I I
I P I R
1 1 2 2
1 2
0I P I Q I S I R
I P Q I R S
1 2
1 2
I P Q I R S
I P I R
1 1Q S
P R
P R
Q S
17
10. Describe an experiment to find unknown resistance and temperature coefficient of resistance using meter bridge?
Metre bridge is one form of wheatstone bridge. It consist of copper strip fixed to a wooden board. There are two gaps G
and G. A uniform of magnanin wire of 1m length is fixed to the wooden board. An unknown resistance P is connected in
the gap G and a standard resistance Q is connected in the gap G A galvanometer high resistance and a jockey are
connected in series. A leclanche cell is connected between the two ends in the wire.
The jockey is adjusted for the zero deflection. The balancing length AJ and JC of the wire replace the resistances R and S
of the wheatstone bridge.
Then.
.
P R r AJ
Q S r JC
where r is the resistance per unit length.
1
2
lP Q
l
Small error will occur in the valve of l1/l2 due to end
resistance. This error can be eliminated if another set of readings are taken with P and Q are interchanged and the average
value of P is found.
Temperature coefficient of resistance.
If R1 and R2 are the resistance of a given coil of wire at the temperature t1 and t2, then the temperature coefficient of
resistance of the wire is
2 1
1 2 2 1
R R
R t R t
11. Explain the principle of potentiometer.(Oct-07)
A battery is connected between the ends A and B of a potentiometer wire. A steady current flows through the
potentiometer wire. This forms the primary circuit.
A galvanometer, high resistance and a primary cell is connected in series to the positive terminal A. This forms the
secondary circuit.
If the potential difference between A and J is equal to the emf of the cell no current flows through the galvanometer and
hence it shows zero deflection. If the balancing length is l, the potential
difference across AJ=Irl.
Where r is the resistance per unit length.
12.
12. How will you compare the emf of the two given cells using potentiometer? (Mar-07, Sep-10,Oct-11)
The potentiometer wire AB is connected in series with a battery, key, rheostat .
This forms the primary circuit.
The end A is connected to end C of DPDT switch. The end D is connected to jockey, galvanometer and high resistance.
The cell of emf E1 is connected across C1D1 and the cell of emf E2 is connected across C2D2.
Let I be the current flowing through the primary circuit and r be the resistance per unit length .
The DPDT switch is pressed towards C1D1 so that cell E1 is included in the secondary circuit. The jockey is adjusted for
zero deflection in galvanometer.
The potential difference across the balancing length l1 =Irl1.
Then by the principle of potentiometer E1=Irl1
The DPDT switch is pressed towards C2D2 so that cell E2 is included in the
secondary circuit. The jockey is adjusted
for zero deflection in galvanometer. The potential difference across the
balancing length l2 =Irl2.
Then by the principle of potentiometer E2=Irl2
18
1 1
2 2
E l
E l
13. State and verify Faraday law of electrolysis.(Jun-06,Mar-08,Ju-08,Sep-09,)
I law.
The mass of a substance liberated at an electrode is directly proportional to the charge passing through the electrolyte.
mαq or m=Zit
Z- electrochemical equivalent.
Verification.
Abattery, a rheostat and an ammeter are connected in series to an electrolyte. A current I1 is passed for a time t. The
mass of the substance deposited at the cathode is measured as m1.
A current I2 is passed for a time t. The mass of the substance deposited is m2.
It is found that
1 1
2 2
m I
m I
m I
The experiment is repeated for same current I but for different times t1 and t2.
If the masses of the deposits are m3 and m4 respectively, it is found that
1 1
2 2
m t
m t
m t
-----------(2)
From (1) and (2)
mαIt or mαq
Thus first law is verified.
II law.
The mass of a substance liberated at an electrode by a given amount of charge is proportional to the chemical equivalent
of the substance.
m α E
verification.
Two electrolytic cells containing different electrolytes CuSo4 solution and AgNo3 solution are connected in series with a
battery, rheostat and ammeter.
The current is passed for some time. Then the massesof copper and silver
deposited are found as m3 and m4.
It is found that
1 1
2 2
m E
m E
m E
14. Explain the working of Daniel cell.(Sep-08,Jun-09,Jun-10)
Daniel cell is a primary cell . It consist of a copper vessel containing a strong solution of copper sulphate. A
Zinc rod is dipped in dilute sulphuric acid contained in a porous pot. The porous pot is placed in a sulphuricacid.
The Zinc rod reacting with the sulphuric acid produced Zn++ ions and 2 electrons.
Zn++ ions pass through the pores of the porous Pot and reacts with copper
Sulphate solution, produced Cu++ ions.
The Cu++ ions deposit on the copper vessel.
When Daniel cell is connected to a circuit the two electrons on the zinc rod pass
through the external circuit and reach the copper vessel thus neutralizing the
copper ions.
This constitute an electric current from copper to zinc. Daniel cell produces
an emf of 1.08 volt.
19
15. Explain the working of leclanche cell.(Jun-07)
A leclanche cell consist of a carbon electrode packed in a porous pot containing manganese dioxide and charcoal
powder. The porous pot is immersed in asaturated solution of ammonium chloride in an glass vessel. A zinc rod is
placed in electrolytic solution.
At zinc rod due to oxidation Zn atom is converted into zn++ ions and 2 electrons.
Zn++ ions reacting with ammonium chloride produces zinc chloride and ammonia gas.
1
4 3 22 2 2 2Zn NH Cl NH ZnCl H e
The ammonia gas escapes. The hydrogen diffuses through the pores of the porous pot and react with ammonia
dioxide. In this process the positive charge of hydrogen ions transferred to carbon rod. When zinc rod and carbon rod
are connected externally, the two electrons from the zinc rod move towards carbon and neutralizes the positive
charge.
Thus current flows from carbon to zinc. The emf of the cell is 1.5 v and it can supply a current of 0.25A
16. Explain the working of lead acid accumulator.(Oct-07)
The lead consist of a container made up of a hard rubber or glass or celluloid.
The container contains dilute sulphuric acid which acts as electrolyte.
Spongy lead acts as the negative electrode.
And lead oxide acts act the positive electrode.
The electrodes are separated by suitable acid accumulator insulating material and
assembled in a way to give low internal resistance.
When the cell is connected in a circuit due to oxidation reaction at negative electrode spongy lead reaction with sulphuric
acid produces lead sulphate and two electrons.
The electrons flow in the external circuit from negative to positive electrode where the reduction reaction takes place.
At positive electrode lead oxide on reaction with sulphuric acid produces lead sulphate and the two electrons are neutralized
in the process.
The current flows from posititve electrode to negative electrode in the external circuit.
The emf of a charged cell is 2.2V and the specific gravity for the electrolyte is 1.28.
.
20
3. EFFECTS OF CURRENT
1. State joules law.
For a steady current I flowing through a conductor the amount of heat produced in time t is 1) directly proportional to the
square of the current for a given R 2)directly proportional to resistance R for a given I and 3) directly proportional to the
time of passage of current
H= I2Rt
2. Why is nichrome used as heating element in appliances?(Jun-07)
1)It has high specific resistance
2) It has high melting point
3)It does not oxidize easily
3. What is seebeck effect?
When two dissimilar metals like iron and copper are connected to form two junctions called a thermocouple an emf is
developed when the junctions are maintained at different junctions.
4. Define peltier effect.
When an electric current is passed through a circuit consisting of two dissimilar metals heat is evolved at one junction
and absorbed at the other junction. This is called peltier effect.
5. Define neutral and temperature of inversion.(Oct-08)
Keeping the temperature of the cold junction a c
constant if the temperature of the hot junction is gradually increased. The thermo-emf rises to a maximum at a
temperature (θn) called neutral temperature. And then gradually decreases and eventually becomes zero at a particular
temperature (θi) is called temperature of inversion.
6. Define peltier coefficient.(Jun-06)
The amount of heat energy absorbed or evolved at one of the junctions of a thermocouple when one ampere current flows
for one second is called peltier coefficient.
7. Define Thomson coefficient.
The amount of heat energy absorbed or evolved when one ampere current flows for one second in a metal between two
points which differ in temperature by 10C is called Thompson coefficient. Its unit is volt/0c
8. Differentiate peltier effect and joule effect.(Mar-06)
No Peltier effect Thomson effect
1 Occurs at the junction Occurs through out the body
of the material
2 It is a reversible process It is a irreversible process
3 Heat energy is absorbed or
evolved at the junctions.
Heat energy is evolved
9. State Biot Savart law.
The magnetic induction dB at P due to element of length dl is 1) directly proportional to the current(I) 2) directly
proportional to the length of t he element (dl) 3) directly proportional to the sine of the angle between dl and the line
joining element dl and the point P (sin θ ) 4) inversely proportional to the square of the distance of the point from the
element 2
1
r
0
24
IdlSindB
r
10. State Tangent law.(Mar-11)
When a magnetic needle suspended at a point where there are two crossed fields at right angles to each other will come
to rest in the direction of the resultant of the two fields.
B=BHtanθ
21
11. State Ampere circuital rule.(Mar-09)
The line integral .B dl around a closed curve is equal to μ0 times the net current I0 through the area bounded by the
curve.
0 0.B dl I
12. Mention the limitations of cyclotron.(Oct-06,Jun-10)
a. Maintaining a uniform magnetic field over a large area of the Dees is difficult.
b. At high velocities relativistic variation of mass of the particle upsets the resonance condition.
c. At high frequencies relativistic variation of mass of the electron is appreciable and hence electrons cannot be
accelerated by cyclotron.
13. State Flemming left hand rule.(Oct-10)
The forefinger, the middle finger and the thumb of the left hand are in mutually perpendicular directions. If the forefinger
points in the direction of the magnetic field , the middle finger points in the direction of the current then the thumb points
in the direction of the force on the conductor.
14. Define current sensitivity. How it is increased?(Oct-09)
Current sensitivity of a galvanometer is defined as the deflection produced when unit current passes through the
galvanometer.
It can be increased by
1. increasing the number of turns
2. increasing the magnetic induction
3. increasing the area of the coil
4. decreasing the couple per unit twist of the suspension wire.
15. Why the phosphor bronze wire is used as the suspension wire in an galvanometer?
Phosphor-bronze has small couple per unit twist. Hence current sensitivity is large.
16. How will you convert a galvanometer into voltmeter and ammeter?(Jun-09)
A galvanometer is converted into voltmeter by connecting a high resistance in series. A galvanometer is converted into
ammeter by connecting a low resistance in parallel.
17. Increasing the number of turns of the coil does not necessarily increase the voltage sensitivity of a galvanometer
why?(Mar-07)
Increasing the current sensitivity by increasing the number of turns of the coil does not necessarily increase the
voltage sensitivity of the galvanometer . This is because increasing the number of turns correspondingly increases the
resistance (G) of the coil. Hence voltage sensitivity remains unchanged.
18. What is voltage sensitivity?
The voltage sensitivity of a galvanometer is defined as the deflection per unit voltage.
nBA
V CG
19. Define one ampere.(Mar-08,Jun-08)
Ampere is defined as that constant current which when flowing through two parallel infinitely long straight conductors of
negligible cross section and placed in air or vacuum at a distance of one metre apart, experiences a force of 2 x 10-7 N per
unit length of the conductor.
22
5 marks questions.
1. Explain Thompson effect.
Thompson suggested that when a current flows through unequally heated conductors, heat energy is absorbed or evolved
throughout the body of the metal.
Positive Thompson effect.
Consider a copper bar is heated in the middle at the point C and current is passed from A to B. It is found that N shows
higher temperature than M. From A to C heat is absorbed and from C to B heat is evolved.
Negative Thompson effect.
Consider a iron bar heated in the middle at the point C.
The current is passed from A to B. M shows higher temperature compared to N. It means from A to C, heat is evolved
and from C to B heat is absorbed
Zero Thompson effect.
In the case of lead heat it at the middle point C. The point M and N equidistant from C show the same temperature when
current is flowing from A to B or from B to A. Therefore Thompson is nil. Lead is used as one of the metals to form a
thermocouple with other metals for the purpose of drawing thermo electric diagrams.
Positive Thompson effect Negative Thompson effect
2. Explain thermopile.
Thermopile is a device to detect thermal radiation.
It works on the principle of seebeck effect.
Since a single thermocouple gives a very small emf a large number of
thermocouple are connected in series. The ends are connected to a galvanometer.
One set of junctions (1,3,5) is blackened to absorb completely the thermal
radiation falling on it. The other set of junctions (2,4) called cold junction is
shielded from the radiation.
When thermal radiation falls on one set of junctions (1,3,5) a difference in
temperature between the junction is
created and a large thermo emf is produced.
The deflection in the galvanometer is proportional to the intensity of
radiation.
3. Explain magnetic Lorentz force.(Mar-07,Mar-11)
Uniform magnetic induction B acts along the Z-axis. A particle of charge q
moves with a velocity v in yz plane making an angle θ with the direction of the
field.
Under the influence of the field, the particle experiences a force F.
1.The force F on the charge is zero, if the charge is rest.
2. The force is zero, if the direction of motion of the charge is either parallel or anti parallel
to the field and the force is maximum, when the charge moves perpendicular to the field.
3. The force is proportional to the magnitude of the charge (q)
4. The force is proportional to the magnetic induction (B)
5. The force is proportional to the speed of the charge (v)
6. The direction of the force is appositively directed for charges of opposite sign
23
F q VxB
F BqVSin
4. Explain Biot savart law.(Jun-09)
Let us consider a conductor XY carrying a current I.
AB=dl is a small element of the conductor. P is a point at a distance r from a mid
point of AB.
According to Biot Savart law, the magnetic induction dB at P due to the element
of length dl is
1.directly proportional to the current(I)
2. Directly proportional to the length of the element (dl)
3. Directly proportional to the sine of the angle between dl and the line joining element dl and the point P (sinθ)
4. Inversely proportional to the square of the distance of the point from the element 2
1
r
0
24
IdlSindB
r
5. How will you convert a galvanometer into a voltmeter? (Mar-08,Mar
10,Jun-11,Oct-11)
Voltmeter is an instrument used to measure potential difference between the two ends of a current carrying
conductor.
A galvanometer can be converted to a voltmeter by connecting a high
resistance in series with it. The valve of the resistance is connected in series
decides the range of the voltmeter.
Galvanometer resistance = G
The current required to produce full deflection in the galvanometer = Ig
Range of voltmeter = V
Resistance to be measured in series = R
The current through the galvanometer
g
g
VI
R G
VR G
I
The effective resistance of the voltmeter is Rv = G+R . Rv is very large, and hence a voltmeter is connected in parallel to
a circuit.
6. How will you convert a galvanometer into an ammeter?
A galvanometer is a device used to detect the flow of current in an electrical circuit.
Galvanometer being a very sensitive instrument a large current cannot be passed through the galvanometer as it may
damage the coil.
Galvanometer is converted into an ammeter by connecting a low
resistance In parallel with it. This resistance is called shunt resistance.
Galvanometer resistance =G
Shunt resistance = S
Current in the circuit = I
Current through the shunt resistance=IS=(I-Ig)
Since the galvanometer and shunt resistance are parallel, potential is
common.
24
g g
g
g
I G I I S
I GS
I I
The effective resistance of the Ammeter Ra
1 1 1
a
a
R G S
GSR
G S
1. State and verify Joule’s law.(Jun-07)
Law
The heat produced in a conductor is (1) directly proportional to square of the current (2) directly proportional to
resistance and (3) directly proportional to the time of flow of current.
H=I2Rt
Verification.
Description.
The calorimeter consist of a coil of resistance R enclosed inside a copper vessel. The ends of the coil is connected to a
battery, key, a rheostat and an ammeter are connected in series . A voltmeter is connected in parallel to the coil.
The calorimeter is filled with water. The calorimeter is kept inside a wooden box to minimize loss of heat.
Laws of current.
The initial temperature of water is measured as θ1. A current I1 is passed for a time t. The final temperature θ2 is noted.
The quantity of heat gained by calorimeter is H1 = W(θ2-θ1). W is the heat capacity of the calorimeter.
The experiment is repeated for currents I2,I3 through the same coil for the same interval of time t and the corresponding
quantities of heat H2,H3 is calculated.
It is found that
1 2 3
2 2 2
1 2 3
2
2
.
H H H
I I I
Hconst
I
H I
Law of resistance.
Same amount of current I is passed for the same time interval t through different coil of resistance R1,R2,R3. The
corresponding quantities of heat gained H1, H2,H3 is calculated.
1 2 3
.
H H H
R R R
Hconst
R
H R
Law of time.
The same amount ofcurrent is passedthrouth the same resistance R for different intervals of time t1,t2,t3 . The
corresponding quantities of heat gained H1, H2,H3 is calculated.
25
1 2 3
.
H H H
t t t
Hconst
t
H t
2. Using Biot – savart law, determine the magnitude of the magnetic field due to an infinitely along current carrying
conductor.(Mar-06,Jun-06,Mar-08,Sep-09,Mar-10)
XY is an infinitely long straight conductor carrying a current I. P is a point at a distance a from the conductor AB is a
small element of length dl. θ is the angle between the current element I dl and the line joining the element dl and the
point P.
According to Biot-savart law,
0
24
IdlSindB
r
---------(1)
AC is drawn perpendicular to BP from A
∠OPA=υ and ∠APB=dυ
In ABC, Sinθ=AC AC
AB dl
(2)
(3)
(2) (3)
(4)
AC dlSin
APC
AC rd
from and
rd dlSin
0
2
0
0
.(4) (1)
4
(5)4
, (6)
4
subs in
IrddB
r
IddB
r
aIn OPA Cos
r
IdB Cos d
a
The total magnetic induction at P due to the conductor XY is
2
1
B dB
2
1
0
4
IB Cos d
a
01 2
4
IB Sin Sin
a
0
1 2 90
0
2
IB
a
If the conductor is placed in amedium of permeability μ,
2
IB
a
.
26
3. Determine the magnitude of the magnetic field along the axis of a circular coil carrying current.(Oct-07)
Consider a circular coil of radius a with a current I . P is a point along the axis of the coil at a distance x from the centre
O of the coil.
AB is an infinitesimally small element of length dl. C is the midpoint of AB and CP=r.
According to Biot Savart law,
0
24
IdlSindB
r
Here θ=900
0
24
IdldB
r
The direction of dB is perpendicular to the current element Idl and CP, and is along PR.
For the element A’B’ the magnitude of dB is same but it acts along PM.
dB is resolved into two components.
dBSinα along OP and
dBCosα perpendicular to OP.
dBCosα components cancel out each other whereas dBSinα are added up.
The total magnetic induction at P due to the entire coil is
0
24
B dBSin
Idl aB
r r
0
3
0
3
4
24
IaB dl
r
IaB a
r
22 2 20
32 2 2
( )
2
IaB r a x
a x
If the coil contains n turns, the magnetic induction is
2
0
32 2 22
nIaB
a x
At the centre of the coil, x=0 0
2
nIB
a
4. Explain the working of tangent galvanometer.(Jun-08)
Tangent galvanometer is a device used for measuring current.
Principle.
Tangent law.
A magnetic needle suspended at a point where there are two crossed fields at right angles to each other will come to
rest in a direction of the resultant of the two fields.
Construction.
A circular coil of wire wound over a non magnetic frame of brass or wood. The vertical frame is mounted on a
horizontal circular turn table provided with three leveling screws.
A small upright projection at the centre of the turn table supports a compass box.
The compass box consists of a small pivoted magnet to which a thin long aluminium pointer is fixed at right angles. The
scale consists of four quadrants.
The centre of the pivoted magnetic needle coincides with the centre of the coil, where there is a uniform magnetic
field. The magnetic needle used is small so that it remains in an uniform field even in deflected position.
The coil consist of 2,5 and 50 turns which are of various thickness used for measuring currents of different strength.
Theory.
27
When the plane of the coil is placed parallel to the horizontal component of Earth’s
magnetic induction (BH), there will be two magnetic fields acting perpendicular to each
other.
1. The magnetic induction (B) due to the current in the coil acting normal to the plane of
the coil, and
2. The horizontal component of Earth magnetic induction (BH).
Due to these two crossed fields, the pivoted magnetic needle is deflected through an
angle θ.
According to tangent law,
B=BH tan θ --------(1)
If a current I passes through the coil of n turns and of radius a, the magnetic induction at the centre of the coil is
0 (2)2
nIB
a
0
0
tan2
2tan
tan (3)
H
H
nIB
a
aBI
n
I K
0
2 HaBK
n is called the reduction factor of the tangent galvanometer.
The unit of reduction factor is ampere.
Since the tangent galvanometer is most sensitive at a deflection of 450. The deflection has to be adjusted to be between
300 and 600.
5. Explain the motion of a charged particle in a uniform magnetic field.(Jun-10)
Consider a magnetic induction B acting along the Z- axis. Let q be the charge of a particle moves in the XY plane with a
velocity v.
The magnetic Lorentz force on the particle
( )F q VxB
F is perpendicular to the plane containing V and B.
Since the force and velocity are perpendicular the force does not do any work. Hence, the magnitude of the velocity
remains constant and only its direction changes.
Since V and B are at right angles to each other
0
( )
90
F q VxB
F BqVSin
F BqV
This magnetic Lorentz force provides the necessary centripetal force.
2mVBqV
r
mVr
Bq
The radius of the circular path is proportional to
(1) The mass of the particle and
(2)velocity of the particle.
V Bq
r m
Bq
m
Period of rotation of the particle,
28
2
2
T
mT
Bq
The angular frequency and period of rotation of the magnetic field does not depend upon (1) the velocity of the
particle and (2) radius of the circular path.
6. Explain the working of cyclotron.(Mar-07,Sep-10,Oct-11)
Cyclotron is a device used to accelerate charged particles to high energies.
Principle.
A charged particle moving normal to a magnetic field experiences magnetic Lorentz force which makes the
particle move in a circular path.
Construction.
It consist of a hollow metallic cylinder into two separate section D1 and D2 called the Dees.
The Dees are placed between the poles of a strong electromagnet which provides a uniform magnetic field perpendicular
to the plane of the Dees.
The Dees are connected to a high frequency oscillator.
Working.
When a positive ion of charge q and mass m is emitted from the source, it is accelerated towards the Dee having a
negative potential at that instant.
Due to the magnetic field, the ion experiences magnetic Lorentz force and moves in a circular path.
By the time the ion arrives at the gap between the Dees, the polarity of the Dees gets reversed.
Hence the particle is once again accelerated and moves into the other Dee with a greater velocity and radius.
The particle moves in a spiral path of increasing radius and it is taken out with the help of a deflector plate and hits the
target T.
The magnetic Lorentz force provides the necessary centripetal force.
2
. (1)
mVBqV
r
V Bqconst
r m
The time taken to describe a semi circle
(2)
(3)
rt
V
mt
Bq
The time taken by the ion to describe a semi circle is independent of (1) the radius of the path and (2) the velocity of the
particle.
Period of rotation T=2t
2.
mT const
Bq
----------(4)
The frequency of rotation of the particle
1
2
Bq
T m
-------(5)
Cyclotron is used to accelerate protons, deuterons and α-particles.
Limitations.
1. Maintaining a uniform magnetic field over a large area of the Dees is difficult.
2. At high velocities, relativistic variation of mass of the particle upsets the resonance condition.
3. At high velocities, relativistic variation of mass of the electron is appreciable and hence electrons cannot be
accelerated by cyclotron.
29
7. Derive an expression for the force experienced by a current carrying conductor placed in a uniform magnetic
field.(Sep-08,Mar-09)
Consider a conductor PQ of length l and area of cross section A.
The conductor is placed in a uniform magnetic field of induction B making an angle θ with the field.
A current flows along PQ. Hence the electrons are drifted along QP with drift velocity Vd.
If n is the number of free electrons per unit volume in the conductor, then the current is
I=nAeVd
Multiplying both sides by the length l of the conductor
dIl nAeV l
The current element,
--------(1)
The negative sign in the equation indicates that the direction of current is opposite to the direction of drift velocity.
The magnetic Lorentz force on the moving electron
(2)df e V XB
The number of free electrons in the conductor
N=nAl ---------(3)
The magnetic Lorentz force on all the moving free electrons
F=Nf
F IlxB
F BIlSin
This total force on all the moving free electrons is the force on the current carrying conductor placed in the magnetic
field.
Magnitude of the force.
1.If the conductor is placed along the direction of the magnetic field θ=00, F=0
2. If the conductor is placed perpendicular to the magnetic field θ=900, F BIl
8. Define ampere. Derive an expression for the force between two long parallel current carrying conductors.
(Mar-11)
AB and CD are two straight very long parallel conductors placed in air at a distance a. They carry currents I1 and I2
respectively.
The magnetic induction due to current I1 in AB at a distance a is
0 11
2
IB
a
-------(1)
This magnetic field acts perpendicular to the plane of the paper and inwards.
The conductor CD with current I2 is situated in this magnetic field.
The force on a segment of length l of CD due to magnetic field B1 is
F=B1I2l
Substituting eqn. (1)
0 1 21 (2)
2
I I lB
a
By Flemming left hand rule, F acts towards left.
The magnetic induction due to current I2 flowing in CD at a distance a is
0 22 (3)
2
IB
a
dIl nAeV l
d
d
F nAl e V XB
F nAleV xB
30
This magnetic field acts perpendicular to the plane of the paper and outwards.
The conductor AB with current I1 is situated in this magnetic field.
The force on a segment of length l of AB due to magnetic field B2 is
2 1
0 1 22 (4)
2
F B I l
I I lB
a
By Flemming left hand rule, this force acts towards right.
These two forces attract each other.
Hence two parallel wires carrying currents in the same direction attract each other and if they carry currents in the
opposite direction, repel each other.
Ampere.
Ampere is defined as that constant current which when flowing through two parallel infinitely long straight
conductors of negligible cross section and placed in air or vacuum at a distance of one metre apart experience a
force of 2 x 10-7
N per unit length of the conductor.
9. Explain the principle, construction and working of moving coil galvanometer.
Moving coil galvanometer is a device used for measuring the current in a circuit.
Principle.
The current carrying coil placed in a magnetic field experiences a torque..
Construction.
It consist of a rectangular coil of a large number of turns of thin insulated copper wire wound over a metallic frame.
The coil is suspended by phosphor bronze wire between the poles of horse shoe magnet.
The lower end of the coil is connected to a hair spring(HS) .
The hemispherical magnetic poles produce a radial magnetic field in which the plane of the coil is parallel to the
magnetic field in all its positions.
A small plane mirror attached to the suspension wire is used along with a lamp and scale arrangement to measure the
deflection of the coil.
Theory.
Let PQRS be a single turn of the coil carrying a current I.
The sides QR and SP are parallel to the magnetic field. Hence they do not experience any force.
The sides PQ and RS are always perpendicular to the field.
PQ=RS=l , length of the coil
PS=QR=b, breadth of the coil
Force on PQ is, F=BI(PQ)=BIl
By Flemming left hand rule, this force is normal to the plane of the coil and acts outwards.
Force on RS is F=BI(RS)=BIl
This force is normal to the plane of the coil and acts inwards.
These two equal and opposite force constitute a couple and deflect the coil.
Moment of the deflecting couple = nBIlxb = nBIA
Now a restoring couple is set up in the wire. This couple is proportional to the twist.
Moment of the restoring couple = Cθ
At equilibrium, deflecting couple = restoring couple
nBIA=Cθ
CI
nBA
I K
Where C
KnBA
is the galvanometer constant.
I α θ
The deflection is directly proportional to the current flowing through the coil.
31
4. ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT.
1. Define magnetic flux.
The magnetic flux linked with the surface held in a magnetic field is defined as the number of magnetic lines of force
crossing a closed area.
BACos
2. Define electromagnetic induction.(Mar-08)
The phenomenon of producing an induced emf due to the changes in the magnetic flux associated with a closed circuit is
known as electromagnetic induction.
3. State Faraday law of electromagnetic induction.(Jun-06,Jun-07,Oct-07)
I law.
Whenever magnetic flux linked with the closed circuit changes an emf is induced in the coil. The induced emf lasts as
long as the change in magnetic flux continues.
II law.
The magnitude of emf induced in a closed circuit is directly proportional to the rate of change of magnetic flux linked
with the circuit.
d
edt
4. State Lenz law.(Oct-08,Jun-10)
Lenz law states that the induced current produced in a circuit always flows in such a direction that it opposes the change
or cause that produces it.
Nde
dt
5. State Flemming right hand rule.(Mar-07,Mar-09,Oct-09,Mar-10)
The forefinger, the middle finger and the thumb of the right hand are in mutually perpendicular direction.
If the forefinger points the direction of the magnetic field, the thumb points the direction of motion of the conductor then
the middle finger points in the direction of the induced current. This is also called generator rule.
6. Define coefficient of self induction.(Jun-09)
Coefficient of self induction of a coil is numerically equal to the magnetic flux linked with a coil when unit current flows
through it.
Coefficient of self induction is numerically equal to the opposing emf induced in the coil when the rate of change of
current through the coil is unity.
The unit of self inductance is henry.
7. Define 1 henry of self inductance.
One henry is defined as the self inductance of a coil in which a change in current of one ampere per second produces an
opposing emf of one volt.
8. Define the coefficient of mutual induction.
The coefficient of mutual induction of two coils is numerically equal to the magnetic flux linked with the coil when unit
current flows through the neighbouring coil.
The coefficient of mutual induction of two coils is numerically equal to the emf induced in one coil when the rate of
change of current through the other coil is unity.
9. Define quality factor.(Oct-06)
The Q factor of a series resonant circuit is defined as the ratio of the voltage across a coil or capacitor to the applied
voltage.
32
10. What is a transformer?
Transformer is a device used to convert low alternating voltage into high alternating voltage and vice versa. The principle
of transformer is electromagnetic induction.
11. What are the methods of producing of induced emf?(Ma-06,Oc-10,Mar-11)
Induced emf can be produced by changing
1. the magnetic induction
2. area enclosed by the coil
3. the orientation of the coil with respect to the magnetic field.
12. What are eddy currents?
When a mass of metal moves in a magnetic field or when the magnetic field through a stationary mass of metal is altered,
induced current is produced in the metal. This induced current is knows as eddy current. The direction of the eddy current
is given by Lenz law.
13. Can transformer with AC and DC. Explain.
The principle of transformer is mutual induction. When an DC current is applied to the applied to the primary there will
not be any magnetic flux variation in the primary and in the secondary. Hence there will not be any induced emf.
Transformer can never work with DC voltage.
14. Define rms voltage of alternating current.(Jun-07,Oct-09)
The rms valve of alternating current is defined as the valve of the steady current which when passed through a resistor for
a given time will generate the same amount of heat as generated by an alternating current when passed through the same
resistor for the same time.
15. Can a DC ammeter measure AC current.(Oct-07)
Since alternating current varies continuously with time, its average valve over one complete cycle is zero. Therefore a
DC ammeter can never measure AC current.
16. Why a capacitor blocks DC and allows AC?
The capacitive reactance 1
0CX
1 1
2CX
C C
For DC ν=0 1
0CX the capacitive offers infinite reactance to DC. Therefore it will not allows DC to pass
through it.
17. Differentiate AF choke and RF choke.(Jun-08)
No AF choke RF choke
1 Works in low frequency
AC circuits
Works in high frequency
AC circuits
2 It has a iron core It has a air core
3 Used in audio coils Used in Radio receivers.
4.ELECTROMAGNETICINDUCTION AND ALTERNATING CURRENT.
1. Derive an expression for coefficient of self induction of a solenoid.
Consider a solenoid of N turns with length l and area of a cross section a ,carrying current I. If B Is the magnetic field at
any point inside the solenoid, then
Magnetic flux per turn = B x area of each turn.
But, 0 NI
Bl
33
Magnetic flux per turn= 0NIA
l
The total magnetic flux linked with the solenoid of N turns = 0 NIAN
l
------(1)
If L is the self coefficient of self induction of the solenoid, then
LI ------(2)
From (1) and (2)
2
0
2
0
N IALI
l
N AL
l
If the core is filled with a magnetic material of permeability μ
2N AL
l
2. Derive an expression for the mutual inductance of two long solenoid (J-08)
S1 and S2 are two long solenoids of length l. If N1 and N2 are the number of turns in the solenoids . Both the solenoid
have same cross section area A. I1 isthe current flowing through the solenoid S1.
The magnetic field produced at any point inside the solenoid S1 is
1
1 0 1
NB I
l
The magnetic flux linked with each turn of S2 is B1A.
Total magnetic flux linked with solenoid S2 having N2 turns is 2 1 2B AN
0 1 12 2
0 1 2 12
2 1
N IAN
l
N N AI
l
MI
Where M is the coefficient of mutual induction
0 1 2 11
0 1 2
N N AIMI
l
N N AM
l
If the core is filled with a magnetic material of permeability μ
1 2N N AM
l
3. Explain the method of producing induced emf by changing the area of the coil.(Jun-07,Oct-07,Sep-08,Mar-09,)
PQRS is a conductor bent as in the figure. L1M1 isa sliding conductor of length l resting on the arms PQ and RS. A
uniform magnetic field B acts perpendicular to the plane.
When . L1M1 moves through a distance dx in time dt, there is a change in its area. Due to the change in area, there is a
change in the flux linked. Therefore an induced emf is produced.
Change in area dA = Area L2L1M1M2
dA=l dx.
34
change in magnetic flux dФ= B x dA
dФ = B . l dx
de
dt
Bldx dxe v
dt dt
e Blv
Where v is the velocity of the conductor.
4. Discuss the various losses of a transformer. (Jun-06,Oct-06,Sep-09,Jun-10,Sep-10,Mar-11)
1.Hysterisis loss
The repeated magnetization and demagnetization of iron core caused by an a.c. input produces loss in energy called
hysteresis loss. This loss can be minimized by using alloy like mumetal and silicon steel .
2.Copper loss.
The current flowing through the primary and secondary windings lead to joule heating effect. Hence some energy is lost
in the form of heat energy.
Thick wires with considerably low resistance are used to minimize this loss.
3.Eddy current loss.
The varying magnetic flux produces eddy current in the core. This leads to the wastage of energy In the form of heat.
This loss is minimized by using a laminated core made of stelloy.
4.Flux loss.
The flux produced in the primary is not completely linked with the secondary due to leakage. This loss can be minimized
by using a shell type of core.
5. Write a short note on choke coil.
A choke coil is an inductance coil of very small resistance used for controlling current in an a.c circuit. If a resistance is
used to control current, there is a wastage of power due to joule heating effect. If pure inductor is used there is no
dissipation of power.
Construction.
It consist of a large number of turns of insulated copper wire wound over alaminated soft iron core to minimize eddy
current loss.
Working
The inductive reactance offered by the coil is given by
XL=Lω
In the case of ideal inductor the current lags behind the emf by a phase angle π/2.
The average power consumed by the choke coil over a complete cycle is
Pav=ErmsIrmsCosπ/2
Pav=0
The average power consumed by the choke coil over a complete cycle is
Pav=ErmsIrmsCosυ
2 2 2
av rms rms
rP E I
r L
2 2 2
r
r L is the power factor.
Therefore the valve of average power dissipated is smaller than the power lost in
a resistor R.
Uses.
Choke coil are used in fluorescent tubes .
They are used in wireless receiver circuits.
6. Discuss ac connected in series with resistor.(oct-11)
Let an alternating current of emf be connected across a resistor of resistance R.
The instantaneous value of the applied emf is given by e = E0Sinωt.
35
If I is the current through the circuit at the instant, the potential drop across R is e=IR
Potential drop must be parallel to the applied emf.
Hence IR=E0Sinωt
I=I0Sinωt
I0=E0/R is the peak value of ac current in the circuit.
In a resistive circuit the applied voltage and current are in phase with each other.
4, ELECTRO MAGNETIC INDCUTION AND ALTERNATING CURRENT.
1. Explain the method of producing induced emf by changing the angle of orientation of the coil.(Jun-08,Sep-
09,Mar-10,Jun-10)
PQRS is a rectangular coil of N turns and area A placed in a uniform magnetic field B.
The coil is rotated with an angular velocity ω in the anticlockwise direction about an axis perpendicular to the direction
of the magnetic field.
If υ is the flux linked with the coil at this instant then
υ=NBA Cosθ (θ=ωt)
The induced emf is
0
( )
de
dt
d NBACos te
dt
e NBA Sin t
e E Sin t
The maximum valve of the induced emf is E0=NBAω
When ωt=0, the plane of the coil perpendicular to the
field B e=0
When ωt=π/2, the plane of the coil is parallel to B
e = E0
When ωt=π, the plane of the coil perpendicular to the
field B e=0
When ωt=3π/2, the plane of the coil is parallel to B e = -E0
When ωt=2π, the plane of the coil perpendicular to the field B e=0
36
2. Explain the working of AC generator with a neat diagram.(Mar-
07,Jun-07,oct-07,Mar-08,sep-10,Jun-11,Mar-11)
AC generator is a device used for converting mechanical energy into
electrical energy.
Principle.
Electromagnetic induction.
An emf induced in a coil when it is rotated in a uniform magnetic field.
Parts.
Armature.
It is a rectangular coil consisting of a large number of turns of insulated
copper wire wound over a laminated soft iron core.
The soft iron core not only increases the magnetic flux but also serves as
the support for the coil.
Field magnets.
The magnetic field is provided by permanent magnets in the case of low power dynamos and by electromagnet for high
power dynamos.
Slip rings.
The ends of the armature coil are connected to two metallic rings R1 and R2 called slip rings. The slip rings rotates along
with the armature.
Brushes.
B1 and B2 are two carbon brushes. They provide contact with the slip rings.
They are used to pass the current from the armature to the external power line.
Working.
Whenever there is a change in orientation of the coil. The magnetic flux linked with the coil changes, producing an
induced emf in the coil.
The direction of the induced emf is given by Flemming right hand rule.
The armature is rotated in the anticlockwise direction.
The side AB of the coil moves downwards and the side DC moves upwards.
By Flemming right hand rule, the induced current flows from B to A and from D to C.
Thus the current flows along DCBA in the coil. In the external circuit the current flows from B1 to B2.
On further rotation the arm AB moves upwards and DC moves downwards. The current flows in the direction ABCD.
In the external circuit the current flows from B2 to B1.
As the rotation of the coil continues, the induced current in the external circuit keeps changing its direction for every
half of rotation of the coil.
The induced emf at any instant is given by, e=E0 Sinωt.
The peak valve of the emf E0=NABω
N- number of turns in the coil.
B-the magnetic field
A-the area enclosed by the coil.
ω-the angular velocity of the coil.
3. Describe the construction and working of a transformer.(Mar-06)
Transformer is an electrical device used for converting low alternating voltage into high alternating voltage and vice
versa.
Principle.
Electromagnetic induction.
Construction.
A transformer consist of primary and secondary coil wound on a soft iron core and insulated from each other.
To minimize eddy current a laminated core is used.
An a.c. input is applied across the primary coil. The varying current in
the primary, produces a varying magnetic flux in the primary which in
turn produces a varying magnetic flux in the secondary. Hence an
induced emf is produced across the secondary.
37
Theory.
Let Ep and Es be the induced emf in the primary and secondary coil and let Np and Ns be the number of turns in the
primary and secondary coil.
The flux linked in the primary and secondary are the same, the emf induced per turn of the two coils must be the same.
(1)
p p
s s
s s
p p
E N
E N
E N
E N
For an ideal transformer,
input power = output power.
EpIp=EsIs where Ip and Is are the primary and secondary currents.
(2)ps
p s
IE
E I
ps s
p p s
IE Nk
E N I
where k is the transformer ratio.
In step up transformer k>1 and for step down transformer k<1.
In step up transformer Es>Ep and Is<Ip.
Thus step up transformer increases the voltage and decreases the current.
A step down transformer decreases the voltage and increases
the current.
Efficiency.
It is defined as the ratio of output power to the input power.
s s
p p
output power E I
input power E I
4. What are eddy currents? How are they minimized and
explain their applications. (Mar-07, Mar-10)
When a mass of metal moves in a magnetic field or when the
magnetic field through a stationary mass of metal is altered,
induced current is produced in the metal. This induced current is known as eddy current.
The direction of the eddy current is given by Lenz law.
If the metallic plate with holes drilled in it is made to swing inside the magnetic field, the effect of eddy current is
greatly reduced.
It is minimized by using thin laminated sheets instead of solid metal.
Applications of eddy current.
Induction furnace.
In an induction furnace high temperature is produced by generating eddy currents. The material to be melted is placed in
a varying magnetic field of high frequency. Hence a strong eddy current is developed inside the material. Due to the
heating effect of the current, the metal melts.
Induction motors.
Eddy currents are produced in a metallic cylinder called rotor, when it is placed in a rotating magnetic field.
The eddy current tries to decrease the relative motion between the cylinder and the rotating magnetic field.
As the magnetic field continues to rotate the metallic cylinder is set into rotation. These motors are used in fans.
Electro magnetic brakes.
A metallic drum is coupled to the wheels of a train. The drum rotates along with the wheel when the train is in motion.
When the brake is applied, a strong magnetic field is developed and hence, eddy currents are produced in the drum which
oppose the motion of the drum. Hence the train comes to rest.
38
5. Obtain an expression for the current flowing in a circuit containing inductance. Find the phase relationship
between voltage and current.(Mar-06,mar-08)
Let an alternating voltage is applied to a pure inductor of inductance L.
A self induced emf is generated which opposes the applied voltage. Eg. Choke coil.
The applied emf is given by
e=E0Sinωt. ---------(1)
Induced emf 'di
e Ldt
Where L is the self inductance of the coil.
In an ideal inductor induced emf is equal and opposite to the applied voltage.
e = -e’
E0Sinωt = -(di
Ldt
)
E0Sinωt = di
Ldt
0 .E
di Sin t dtL
Integrating both sides
0
0
.
.
Ei Sin t dt
L
Ei Sin t dt
L
0
0
E Cos ti
L
E Cos ti
L
0
2
Ei Sin t
L
02
i I Sin t
------------(2)
00
EI
L Here Lω is known as inductive reactance. Its unit is ohm.
From (1) and (2) an a.c. circuit containing a pure inductance the current I lags behind the voltage e by the phase
angle π/2.
Inductive reactance.
XL=Lω = 2πνL where ν is the frequency of the ac supply.
For dc ν=0, XL=0.
Thus a pure inductor offers zero resistance to dc.
For ac the reactance of the coil increases with increase in frequency.
6. Obtain an expression for the current flowing in the circuit
containing capacitance only . Find the phase relationship between
the current and voltage.
An alternating source of emf is connected across a capacitor of
capacitance C.
It is first charged in one direction and then in the other direction.
The applied emf is 0E Sin t
At any instant the potential difference across the capacitor is equal
to applied emf.
39
q
eC
Where q is the charge in the capacitor.
dq
Idt
d Ce
Idt
0d CE Sin t
Idt
0I C E Cos t
0 / 21
EI Sin t
C
0 / 2I I Sin t
0 00 1
C
E EI
X
C
1
CXC
XC is the capacitive reactance. Its unit is ohm.
Capacitive reactance.
1 1
2CX
C C
In a dc circuit ν=0 ∴XC=0
Thus a capacitor offers infinite resistance to dc. For an ac the capacitive reactance
varies inversely as the frequency of ac
7. A source of alternating emf is connected to a series combination of a resistor R, inductor and a capacitor C. Find
(1)the effective voltage (2) the impedance (3) the phase relation ship between the current and the voltage.(Jun-
06,Jun-09)
Let an ac source of emf e be connected to a series combination of a resistor R, inductor L and capacitor C.
The voltage drop across the resistor is RV IR (this is in phase with I )
The voltage drop across the inductor coil is L LV IX ( VL leads the I by π/2 )
The voltage drop across the capacitor is C CV IX ( VC lags behind I by π/2 )
VL and VC out of phase by 1800 . The resultant of VL and VC is VL-VC.
22
2 2
2 2
2 2
( ) ( )
( )
( )
R L C
L C
L C
L C
V V V V
V IR IX IX
V I R X X
VZ R X X
I
Z is the impedance Its unit is ohm.
is the phase angle between the voltage and current
40
tan
tan
tan
L C
R
L C
L C
V V
V
IX IX
IR
X X
R
I=I0Sin(ωt±υ) is the instantaneous current flowing in the circuit.
5.ELECTROMAGNETIC WAVES AND WAVE
OPTICS
1. What is an electromagnetic wave?
An electromagnetic wave is one
in which electric and magnetic
field vectors are at right angles
to each other and to the direction
of propagation.
They possess the wave
character and propagate through
free space without any material
medium.
2. What is emission and absorption spectra.(Mar-06,Oct-06)
When the light emitted directly from a source is examined with a spectrometer emission spectrum is obtained.
When the light emitted from a source is made to pass through a absorbing material and then examined with a
spectrometer, absorption spectrum is obtained.
3. What are Franhofer lines?
41
If the solar spectrum is closely examined it is found that it consist of large number of dark lines. These dark lines in the
continuous spectrum of the sun are called Franhofer lines.
Solar spectrum is an example of line absorption spectrum.
4. What is fluorescence and phosphorescence?
When atoms or molecules of a system are excited by absorption of energy and if they return to lower energy state in less
than 10-5 s, the system is found to glow brightly by emitting radiation of longer wavelength is known as Fluorescence.
When molecules of some substances are excited by the absorption of ultraviolet light and do not return immediately to
lower energy state, emission of light continues even the exciting radiation is removed. This type of delayed fluorescence
is called phosphorescence.
5. What is tyndall scattering?(Mar-07,Jun-09,Jun-10,Oct-10)
The scattering of light by the colloidal particles is called Tyndal scattering. When light passes through a colloidal
solution its path is visible.
6. State Rayleigh scattering law.
The amount of scattering is inversely proportional to the fourth power of the wavelength.
Amount of scaterrring 4
1
7. Define superposition principle.(Mar-09)
When two or more waves simultaneously pass through the same medium, each wave acts on every particle of the
medium, as if the other waves are not present. The resultant displacement of any particle is the vector addition of the
displacements due to the individual waves.
8. Differentiate photon and corpuscle.
S.No Photon Corpuscle.
1s Packets of energy It is tiny particles
which are
perfectly elastic
2 Different frequency
produce different
colours
Different sizes produce
different colours
3 Zero rest mass Massless (negligibly
small mass)
9. Why does the sky appear blue during day time?(Jun-06)
Due to Rayleigh scattering the light of lesser wavelength scatters more than the light of higher wavelength.
Since blue has lesser wavelength it scatters more than red light and the sky appears blue during day time.
10. Why does the sun appear red during sunrise and sunset?
During sunrise and sunset the rays from the sun have to travel through a larger part of the atmosphere than at noon.
Due to Rayleigh scattering law the blue is scattered more than red. Therefore the red light which is least scattered reaches
the observer. Hence sun appears red at sunrise and sunset.
11. Define wave front.
Wave front at any instant is defined as the locus of all the particles of the medium which are in the same state of vibration
and equidistant from the source.
12. What is Raman effect?
When a monochromatic light is passed through a substance, the scattered light contains additional frequencies other than
the incident frequency . This is known as Raman effect.
13. Define wave front.
42
Wave front at any instant is defined as the locus of all the particles of the medium which are in the same state of vibration
and equidistant from the source.
14. State Huygens principle.(Oct-09)
1. Each point on a given wave front can be considered as a source os secondary wavelets which spreads at the
speed of light in that medium
2. The new wave front is the forward enveloped of all the secondary wavelets at that instant.
15. Mention the conditions for the total internal reflection.(Oct-08)
1. Light must travel from a denser medium to a rarer medium.
2. The angle of incidence inside the denser medium must be greater than the critical angle.
16. Mention the conditions for sustained interference.
1. The two sources should be coherent.
2. Two sources should be very narrow.
3. The sources should lie very close to each other to form distinct and broad fringes.
17. What are the conditions for bright and broad interference bands.
1. The screen should be as far as away from the source as possible.
2. The wavelength of light used must be larger.
3. The sources should lie very close to each other to form distinct and broad fringes.
18. Why is the centre of the Newton ring dark?
The wave reflected from the denser glass plate has suffered a phase change of π while the wave reflected at the bottom of
the lens has not suffered any phase change. Hence the centre of the Newton ring appears dark.
19. State Brewster law.(Mar-11)
The tangent of the polarizing angle is numerically equal to the refractive index of the medium. tan Pi
20. Why is diffraction more pronounced in sound than in light?
Sound waves have greater wavelength, therefore diffraction effects are pronounced. Since the wavelength of light is very
small, even tiny obstacles are large is size, hence diffraction effects in light is less pronounced than in sound.
21. Define optic axis of the crystal.(Jun-07,Jun-10)
Inside the crystal there is a particular direction in which both ordinary and extra ordinary travel with same velocity. This
direction is called optic axis of the crystal.
22. Mention the factors which influence optical activity of a substance.(Jun-08)
1. Thickness of crystal.
2. Density of the crystal or concentration in the case of solutions.
3. Wavelength of light
4. Temperature of the solutions.
23. Differentiate interference and diffraction.(Oct-07)
No Interference Diffraction
1 It is due to the
superposition of
secondary wavelets from
two different wavefronts
produced by two
coherent sources
It is due to the
Superposition of secondary
wavelets emitted from
various points of the same
wave front.
2 Fringes are equally
spaced
Fringes are unequally
spaced
3
It has large number of
fringes
It has less number of
fringes
43
24. Mention the uses of ultraviolet light.
1. They are used to destroy the bacteria and for sterilizing surgical instruments.
2. They are used in detection of forged documents, finger prints in forensic lab.
3. They are used to preserve the food items.
4. They are used to find the structure of atoms.
25. Mention the uses of Infra red light(Oct-08)
1. Infra red lamp is used in physiotherapy.
2. Infra red photographs are used in weather forecasting
3. They are used to take photograph of long distance objects.
26. Differentiate Fresnel and Franhofer diffraction.(Mar-10)
No Fresnel Franhofer
1 The source and the
screen are at finite
distances from the
obstacle producing
diffraction
The source and the
screen are at infinite
distance from the
obstable producing
diffraction
2 The wavefront
Undergoing
diffraction is either
spherical or cylindrical
The wavefront
undergoing
diffraction is a plane
wavefront.
3 No lens is used to get
the diffraction
pattern
The diffracted rays are
brought to focus with the
help of convex lens.
27. Define optical activity.
When a plane polarized light is passes through certain substances the plane of polarisation of the emergent light is not the
same as that of incident light but it is rotated through some angle.
28. Define specific rotation.(Mar-08,Mar-10)
Specific rotation for a given wavelength of light at a given temperature is defined as the rotation produced by one
decimeter length of the liquid column containing 1 gram of the active material in 1cc of the solution.
29. Mention the uses of Infra red light(Oct-08)
1. Infra red lamp is used in physiotherapy.
2. Infra red photographs are used in weather forecasting
3. They are used to take photograph of long distance objects.
5.ELECTROMAGNETIC WAVES AND WAVE OPTICS.
1. List the properties of electromagnetic waves.
1.Electromagnetic waves are produced by accelerated charges.
2.They do not require any material medium for propagation.
3.In an electromagnetic wave, the electric and magnetic field vectors are at right angles to each other and to the direction of
propagation. Hence electromagnetic waves are transverse in nature.
4. Variation of maxima and minima in both E and B occur simultaneously.
5. They travel in vacuum or free space with a velocity 3 x 108 m/s.
0 0
1C
6.The electromagnetic waves are charge less, therefore they are not deflected by electric and magnetic fields.
44
2. State and prove Brewster’s law. (Jun-06,Oct-06,Jun-07,Jun-08)
Law.
The tangent of polarizing angle is numerically equal to the refractive index ofthe medium.
At polarizing angle the reflected and refracted rays are at right angles to each other.
Ip + r + 900 = 1800
r = 900 - ip
from Snell’s law
pSini
Sinr
μ- refractive index of the medium.
090
tan
p
p
p
p
p
Sini
Sin i
Sini
Cosi
i
3. Write a short note on pile of plates. (Mar-06,Jun-09)
The phenomenon of polarizing by reflection is used in the
construction of pile of plates.
It consist of a number of glass plates placed one over the other in a
tube inclined
At an angle of 32.50 to the axis of the tube.
A beam of monochromatic light falls on the pile of plates such that the
angle of
incidence is 57.50 which is the polarising angle for glass. The
vibrations
perpendicular to the plane of incidence are reflected at each surface and those
parallel to it are transmitted. The larger the number of surface the greater is the
intensity of the reflected plane polarised light.
The pile of plates is used both as polarizer and as an analyser.
4. Write a short note on Nicol prism.(Mar-07,Sep-09,Oct-11).
Nicol prism is a crystal whose length is three times the breadth. It is cut into two halves along the diagonal so that their face
angles are 720 and 108
0. The two halves are joined together by a layer of Canada balsm, a transparent cement.
For sodium light, the refractive index of ordinary light is 1.658, and for Extraordinary light is 1.486. The refractive index
for Canada balsm is 1.550 for both rays. Hence Canada balsm does not polarize light.
A monochromatic beam of unpolarised light is incident on the face of the nicol prism, due to double refraction it splits up into
two rays Ordinary ray (O)and extraordinary light(E).
The ordinary light is totally internally reflected at the layer of Canada balsm .
The extraordinary light is transmitted through the crystal and is plane polarized.
Nicol prism serves as polarizer and as an analyser.
5. Mention the uses of polariser.
1.Polaroids are used in the laboratory to produce and analyse plane polarized light.
2.Used as polarizing sun glasses.
3.They are used to eliminate the head light glare in motor cars.
4.They are used to improve colour contrasts in oil paintings.
5.They are used to produce three dimensional moving pictures.
6.They are used to study size and shape of molecule.
6. Derive an expression for the radius of the nth
dark ring in newton
45
experiment.(Jun-11)
Consider the vertical section SOP of the plano convex lens through its centre of curvature C. Let R be the radius of
curvature and O be the point of contact of lens with the plane surface. Let t be the thickness of the air film.
ST=AO=PQ=t
Let rn be the radius of the nth dark ring.
SA=AP=rn
By the law of segments
rn.rn = t(2R-t)
rn2 = 2Rt – t2
= 2Rt t2 can be neglected.
2t = rn2/R
The condition for darkness
2t = nλ
rn2/R =nλ
rn2 = nRλ
nr nR
5. ELECTROMAGNETIC WAVES AND WAVE OPTICS.
1. Explain emission and absorption spectra.(Jun-09,Mar-10)
Emission spectra.
When the light emitted directly from a source is examined with a spectrometer,
emission spectrum is obtained.
Continuous spectrum.
It consist of unbroken luminous bands of all wavelengths from violet to red.
This spectra depend only on the temperature of the source and is independent of the characteristic of the source.
Incandescent solids, liquids, carbon arc and electric filament lamps.
Line spectrum.
Line spectra are sharp lines of definite wavelengths. It is the characteristic of the emitting substance. It is used to
identify the gas.
Atoms in the gaseous state, emit line spectrum. Sodium in sodium vapour lamp and mercury in mercury vapour
lamp.
Band spectrum.
It consisits of bright bands with a sharp edge at one end fading out at the other end.
Band spectra is obtained from molecules. It is the characteristic of the molecule.
Calcium or Barium salts in a Bunsen flame, gases like carbondioxide,ammonia in the discharge tube gives band
spectra.
Using band spectra, the molecular structure of the substance can be studied.
Absorption spectra.
When the light emitted from a source is made to pass through an absorbing material and then examined with a
spectrometer, the absorption spectrum is obtained.
Continuous absorption spectra.
A pure glass plate when placed in the path of white light absorbs everything except green and gives continuous
absorption spectrum.
Line absorption spectrum.
When light from the carbon arc is made to pass through sodium vapour and then examined by a spectrometer, a
continuous spectrum of carbon arc with two dark lines in the yellow region is obtained.
Band absorption spectrum.
If white light is allowed to pass through iodine vapour or dilute solution of blood or chlorophyll dark bands on
continuous bright background are obtained.
The band absorption spectra are used for making dyes.
2. Explain Raman effect.(Mar-07,Oct-07,Mar-08,Jun-11)
nr n
46
When a monochromatic light is passed through a substance, the scattered light contains additional frequencies
other than the incident frequency. This is known as Raman effect.
The lines having frequencies lower than the incident frequency are called stoke’s line and the lines having frequencies
higher than the incident frequency are called antistoke’s line.
Stoke’s line.
If a photon strikes an atom or a molecule of the liquid, a part of the energy of the incident photon may be used to excite
the atom of the liquid and the rest is scattered. The spectral line will have lower frequency and it is called stoke’s line.
Antistoke’s line.
If a photon strikes an atom or a molecule in a liquid, which is in excited state, the scattered photon gains energy. The
spectral line will have higher frequency and it is called Antistoke’s line.
Rayleigh line.
When a light photon strikes atoms or molecules, photons may be scattered elastically. Then the photons neither gain or
lose energy. The spectral line will have unmodified frequency.
The Raman frequency ∆ν=ν0-νs
The Raman shift does not depend upon the frequency of incident light but it is the characteristic of the substance
producing Raman effect.
For stoke’s line ∆ν is positive.
For antistoke’s line ∆ν is negative.
The intensity of Stoke’s line is always greater than the Antistoke’s line.
When a system interacts with a radiation of frequency ν0, it may go to a virtual state.
Most of the molecules of the system return to the original state hence Rayleigh scattering.
A small fraction may return to states of higher and lower energy giving rise to Stoke’s line and Antistoke’s line.
3. Explain reflection of a plane wave front at a
plane surface.(Sep-08)
Let XY be a plane reflecting surface and AB be a
plane wavefront incident on the surface at A.
Let P and Q be the incident rays, and N be the
normal drawn on the surface. The wavefront and the
surface are perpendicular to the plane of the paper.
By Huygens principle each point on the wavefront acts
as the source of secondary wavelet.
By the time, the secondary wavelets from B travels a distance BD, the secondary wavelets from A would travel a
distance AC after reflection.
Taking A as centre and BD as radius an arc is drawn. From D, a tangent DC is drawn to this arc. DC is the reflected plane
wavefront and AD is the reflected ray.
Laws of reflection.
The incident wavefront AB, the reflected wavefront CD and the reflecting surface XY all lie in the same plane.
Angle of incidence
090
i PAN
i NAB
i BAD
47
Angle of reflection
090
r NAC
r CAD
r ADC
In right angled triangles BAD and CAD
090B D
BD AC
AD is common.
The two triangles are congruent.
BAC DCA
i r
Thus angle of incidence is equal to the angle of reflection.
4. Explain refraction of a plane wave front at a plane surface.
Let XY be the plane refracting surface separating two media 1 and 2 of refractive indices μ1 and μ2.
The velocities of light in two media are respectively C1 and C2.
Consider a plane wavefront AB incident on the refracting surface at A.
Let P and Q be the incident rays, and NAN1 be the normal drawn to the surface. The wavefront and the surface are
perpendicular to the plane of the paper.
By Huygens principle each point on the wave front acts as a source of secondary wavelet. By the time t, the secondary
wavelets from B, reaches D, the secondary wavelets from the point A would travel a distance AC.
BD=C1t
AC=C1t
Taking A as centre and AC as radius an arc is drawn in the second medium. From C a tangent is drawn to this arc. CD
is the refracted plane wavefront and AD is the refracted ray.
Laws of refraction.
The incident wavefront AB, the refracted wavefront CD and the refracting surface XY all lie in the same plane.
Angle of incidence
090
i PAN
i NAB
i BAD
Angle of refraction
090
r NAC
r CAD
r ADC
112
2 2
1
/
/
.
Sini BD AD BD
Sinr AC AD AC
Sini BD Cconst
CSinr CBD
C
If 1μ2 >1, the first medium is rarer and the second medium is denser.
Then 1
2
C
C >1. This means that the velocity of light in rarer medium
is greater than that in a denser medium.
5. Explain total internal reflection on the basis of Huygen’s wave theory.(Mar-06,Jun-06)
48
Let XY be a plane surface which separates a rarer medium and a denser medium. Let the velocity of the light in these
media be Ca and Cm respectively.
A plane wavefront AB passes from denser medium to rarer medium. It is incident on the surface with angle of incidence
i. Let r be the angle of refraction.
/
/
m m
a a
Sini BC AC BC C t C
Sinr AD AC AD C t C
Since m
a
C
C<1, i is less than r. This means that the refracted wave front is
deflected away from the normal.
In right angled triangle ADC, the three possibilities
1. AD<AC 2. AD=AC 3. AD>AC
When AD<AC
1
ADSinr
AC
Sinr
For valves of I, for which r < 900, a refracted wavefront is possible.
When AD=AC
Sin r = 1
r = 900
a refracted wavefront is just possible and it grazes the surface of separation of the two media.
The angle of incidence at which the angle of refraction is 900 is called the critical angle
C.
When AD > AC
Sin r > 1
No refracted wave front is possible. When the angle of incidence increases beyond the
critical angle, the incident wave front is totally reflected into the denser medium.
This is called total internal reflection.
Hence for total internal reflection
1. Light must travel from denser medium to rarer medium.
2. The angle of incidence inside the denser medium must be greater than the critical
angle. i >C
6. Obtain an expression for bandwidth in young’s double slit experiment.(Ocy-06,Jun-07,Mar-09,Jun-10,Mar-
11,Oct-11)
Let d be the distance between two coherent sources A and B. A screen is placed at a distance D from the sources. C is the
midpoint of AB. O is a point on the screen equidistant from A and B.
Waves from A and B meet at P. Draw AM perpendicular to BP.
The path difference BP AP
AP=MP
δ=BP-MP
δ=BM
ABM
BMSin
d
BM dSin
dSin
If θ is small Sinθ=θ
d
In right angled triangle COP, tanOP x
CO D
49
For small valves of θ, tan θ=θ
The path difference xd
D
Bright fringes.
( 0,1,2,...)
n
xdn n
D
n Dx
d
Dark fringes
(2 1)2
2 1 ( 0,1,2,...)2
2 12
n
xdn n
D
Dx n
d
The distance between any two consecutive bright or dark bands is called band width.
1n nx x
( 1)D n D
nd d
=
D
d
Similarly the distance between two consecutive dark
bands is D
d
condition for obtaining clear and broad interference
bands.
1.The screen should be as far away from the source as
possible.
2.The wavelength of light used must be larger.
3.The two coherent sources must be as close as possible.
7. Discuss the theory of interference in thin films. Obtain the condition for bright and dark fringes.(Jun-08,Sep-
09)
The path difference between BC and EF
( ) ( )
( )
2
m aBD DE BM
BD DE BM
BD DE
BD BM
,BM
BME SiniBE
BM BESini
50
( )
1
2,
SiniBM BESinr
Sinr
BEBL
BDL SinrBD BD
2
2
2
BE BDSinr
BM BDSin r
2
2
2 2
2 (1 )
BD BDSin r
BD Sin r
22 BDCos r
,
2
2
DL tBDL Cosr
BD BD
tBD Cosr
BD
tCosr
A ray of light travelling in air and getting reflected at the surface of a denser medium, undergoes an automatic phase
change of π or an additional path of λ/2.
The effective path difference 22
tCosr
For the constructive interference,
2
2
n
tCosr n
2 2 1
2tCosr n
For destructive interference
2 12
2 2 12 2
2
n
tCosr n
tCosr n
If light is incident normally i=0 then r=0 . Then the condition for bright fringes is 2 2 12
t n
and for dark
fringes 2 t n
8. Discuss the theory of the plane transmission grating.
The plane transmission grating is a plane sheet of transparent material on which opaque rulings are made with a fine
diamond pointer.
The combined width of a ruling and a slit is called grating element. Points separated by distance equal to the
grating element are called corresponding points.
Theory.
MN represents the section of a plane transmission grating having slits of equal width a and rulings of equal width b.
The grating element e=a+b.
According to Huygens principle, the point in the slit act as a source of secondary wavelets which spread in all
directions on the other side of the grating.
The path difference between the wavelets from one pair of corresponding points A and C is CG=(a+b)Sinθ.
The point P will be bright, when (a+b)Sinθ = mλ where m=0,1,2,3....
When m=0, θ=0; Sinθ=0.
51
In the undiffracted position (a+b)Sinθ =0 will produce maximum intensity at the centre O of the screen. This is called
zero order maximum or central maximum.
When m=1, θ=θ1; (a+b)Sinθ1=λ the first order maximum is obtained.
For second order maximum (a+b)Sinθ2=2λ.
For maximum intensity
(a+b)Sinθ=mλ
mSin
a b
Sin Nm
1
Na b
or number of lines per unit width of the grating.
When white light is used, the diffraction pattern consists of a white central maximum and on both sides coloured
images are formed.
In the undiffracted position,θ=0 and hence sin θ=0 . Therefore (a+b)Sinθ=Nmλ is satisfied for m=0 for all valves of
λ. Hence an undispersed white images is obtained.
As θ increases, (a+b)Sinθ first passes through λ/2 valves for all colours from violet to red and hence darkness results.
As θ further increases, (a+b)Sinθ passes through λ valves of all colours resulting in the formation of bright
images producing a spectrum from violet to red.
6. ATOMIC PHYSICS
1. State the principle of Milikan oil drop experiment.(Jun-06)
The study of the motion of uncharged oil drop under free fall due to gravity and charged oil drop in a uniform electric
field.
2. What are the drawbacks of Thompson atom model experiment?
1) According to electromagnetic theory, the vibrating electron should radiate energy and the frequency of the emitted
spectral line should be the same as the electron. In Thomson model gives only one spectral line of wavelength
1300A0. But the experimental observations reveal that hydrogen spectrum consist of five different series.
2) It could not account for the scattering of α particle through the angles.
3. Write the Bohr postulates .
1. An electron cannot revolve round the nucleus in all possible orbits. The electron can revolve only in those
permissible orbits in which the angular momemtum of the electron is an integral multiple of2
h
. These orbits
are called stationary orbits.
2. An atom radiates energy only when an electron jumps from a stationary orbit of higher energy to the stationary
orbits of lower energy.
2 1E E h
4. How will you detect X-rays?
1. Blackening of photographic plates.
2. Ionization produced by X-rays in a gas or vapour.
5. What is meant by energy level diagram?
The energy of the electron in the nth orbit is given by 2
13.6nE eV
n
.The energy of the first orbit is -13.6eV and the
energy of the second orbit is -3.4eV and so on.
Taking these energies on a linear scale horizontal lines are drawn which represent energy levels of the hydrogen atom.
This diagram is known as energy level diagram.
52
6. What is excitation and ionization potential?(Mar-07,Jun-09,Oct-10)
The energy required to raise the atom from its normal state into an excited state is called excitation potential energy of
the atom.
The ionization potential is that accelerating potential which makes the impinging electron acquire sufficient energy to
knock out an electron from the atom and thereby ionize the atom.
7. What are soft x-rays and hard x-rays?
Hard x-rays.
X-rays having low wavelength of the order of 1A0 have high frequency and hence high energy.
The penetrating power is high.
The are produced by high potential difference.
Soft X-rays.
X-rays having wavelength of 4A0 have lesser frequency and hence lesser energy.
The penetrating power is low.
The are produced by low potential difference.
8. How X-rays are absorbed?
If a beam of X-rays of intensity I passes through a length dx of any material, its intensity is decreased by dI. The amount
of absorbed intensity is Idx where µ is called the absorption coefficient. dI= - Idx
9. What are the factors that Laue experiment reveal?(Oct-07,Jun-08)
1. X-rays are electromagnetic waves of extremely short wavelength.
2. The atoms in a crystal are arranged in a regular three dimensional lattice
10. State Moseley law.(Mar-07,Mar-08,Mar-09,Mar-11)
The frequency of the spectral line in the characteristic X-ray spectrum is directly proportional to the square of the atomic
number of the element. 2Z
11. Define normal population.
The number of atoms in the ground state (N1) is greater than the number of atoms in the excited state(N2).
12. What are the applications of Moseley law?
1. It led to the discovery of new elements like hafnium, rhenium.
2. This law has been helpful in determine the atomic number of rare earths there by fixing their position in the
periodic table.
3. Any discrepancy in the order of the elements in the periodic table can be removed by this law.
13. Mention the characteristics of laser.(Oct-06,Jun-09,Mar-10,Jun-10)
1. Monochromatic
2. Coherent with the waves, all exactly in phase with one another.
3. Does not diverge at all
4. Extremely intense.
14. Define population inversion.
The number of atoms in the excited state (N2) is greater than the number of atoms in the ground state (N1).
15. What are the conditions to achieve laser action?(Mar-06,Jun-07)
1. There must be inverted population
2. The excited state must be a metastable state.
3. The emitted photons must stimulate further emission.
16. Mention the applications of laser in industry.(Jun-08)
1. The laser beam is used to drill extremely fine holes in diamonds, hard sheets
2. They are used for cutting thick sheets of hard metals and welding.
3. They are used to test the quality of the materials.
53
17. Mention the applications of laser in medical.(Mar-08,Oct-09)
1. In medicine, micro surgery has become possible due to narrow spread of the laser beam.
2. It can be used in the treatment of kidney stone, tumour .
3. The laser beams are used in endoscopy.
4. It can be used for the treatment of human and animal cancer.
18. Write any one drawbacks of Rutherford atom model?(Oct-08)
1. According to electromagnetic theory an accelerated electric charge must radiate energy in the form of
electromagnetic waves. If the accelerated electron lose energy by radiation, the energy of the electron
continuously decreases and it must spiral down into the nucleus. Thus the atom cannot be stable.
2. An accelerating electron must radiate energy at a frequency proportional to the angular velocity of the electron.
Therefore as the electron spiral towards the nucleus, the angular velocity tends to become infinity and hence the
frequency of the emitted energy will tend to infinity. This will result in a continuous spectrum with all possible
wavelengths. But experiments reveal only line spectra of fixed wavelength from atoms.
19. Why ordinary plane transmission grating cannot be used to produce diffraction effects in X-rays?
In order to diffract X-rays , grating with much finer rulings having distance between rulings comparable to the
wavelength of X-rays are required. But X-rays are of wavelength much shorter than that of visible light. Therefore
ordinary plane transmission cannot be used to diffract X-rays.
The regular arrangement of atoms or molecules in the cleavage plane of acrystal provide a grating element suitable to
diffract X-rays.
20. Define Bragg’s law.
The path difference 2d Sinθ is equal to integral multiple of wavelength of x-ray. i.e nλ.
2d Sinθ=nλ
21. What are the conclusions of α scattering experiment.
1.Most of the α particles either passed straight through the gold foil or scattered by only small angles of the order of a
few degrees. This observation led to the conclusion that an atom has a lot of empty space .
2.A few particles were scattered in the backward direction, which led Rutherford to conclude that the whole of the
positive charge was concentrated in a tiny space of about 10-14m. This region of the atom is nucleus.
22. What is hologram?(Oct-07)
When an object is photographed by a camera, a dimensional image of three dimensional object is obtained. A three
dimensional image of an object can be formed by holography. In holography both the phase and amplitude of the light
waves are recorded on the film. The resulting photograph is called hologram.
6. Atomic physics
1. What are the properties of cathode rays?(Sep-08,Jun-09)
1.They travel in straight lines
2. Cathode ray possess kinetic energy and momemtum.
3. Cathode rays ionize the gas through which they pass.
4. Cathode ray produce fluorescence when they strike minerals and salts.
5. Cathode ray affect photographic plates.
6. They are deflected by electric and magnetic field
7. They travel with a velocity of 1/10 of the velocity of light.
2. Derive an expression for the distance of closest approach.
As α particle directed towards the centre of the nucleus will move close upto a distance r0 where its kinetic energy will
appear as electrostatic potential energy.
After this the α particle begins to retrace its path. The kinetic energy of the particle 21
2kE mv
The electrostatic potential energy of the α particle is
54
0 0
1 2.
4p
eZeE
r
Z is the atomic number.
On reaching the distance of closest approach,
Kinetic enegy= potential energy.
EP=EK
0 0
1 2.
4
eZe
r =
21
2mv
2
0 2
0
1 4.
4
Zer
mv
3. What are the drawbacks of Rutherford atom model.
1.According to the electromagnetic theory an accelerated electric charge must radiate energy in the form of
electromagnetic waves. Therefore if the accelerated electron lose energy by radiation, the energy of the electron
continuously decreases and it must spiral down into the nucleus. Thus the atom cannot be stable.
2.According to classical mechanics, the accelerating electron must radiate energy at a frequency proportional to the
angular velocity tends to become infinity and hence the frequency of the emitted energy will tend to infinity. This will
result in a continuous spectrum. But experimental reveal only line spectra of fixed wavelength from atoms.
4. Explain the hydrogen spectral series of hydrogen atom.(Mar-06,Mar-10,Jun-10)
Lyman series.
When the electron jumps from any of the outer orbits to the first orbit, the spectral lines emitted are in the ultraviolet
region of the spectrum.
n1=1 n2=2,3,4,....
The wavenumber of Lyman series is 2
2
11R
n
Balmer series.
When the electron jumps from any of the outer orbit to the second orbit, the spectral lines emitted are in the ultraviolet
region of the spectrum.
n1=2 n2=3,4,5....
The wavenumber of Lyman series is 2
2
1 1
4R
n
Paschen series.
When the electron jumps from outer orbit to the third orbit. This series lies in the infrared region.
2 2
2
1 1
3R
n
Brackett series.
The series obtained by the transitition of the electron from n2=5,6,.. to n1=4 is called Brackett series. This lies in the
infrared region.
2 2
2
1 1
4R
n
Pfund series.
The series is obtained when the electron jumps from n2=6,7.. to n1=5. This series lies in the infrared region.
2 2
2
1 1
5R
n
55
5. What are the shortcomings of Bohr atomic theory.
1. The theory could not account for the spectra of atoms more complex than hydrogen.
2. The theory does not give any information regarding the distribution and arrangement of electrons in an atom.
3. It does not explain the intensity of the spectral lines of the element.
4. Bohr theory could not account for the fine structure of spectal lines.
5. It could not explain Zeeman effect and stark effect.
6. What are the drawbacks of sommerfield atom model.
1.It could not predict the correct number of observed fine structure of these lines.
2. It could not explain the distribution and arrangement of electrons in atoms.
3. It could not explain the spectra of alkali metals such as sodium, potassium etc
4. This model does not give any explanation for the intensities of the spectral lines.
5. It could not explain Zeeman effect and stark effect.
7. What are the properties of x-rays?(Jun-06,Mar-11)
1. They are electromagnetic waves of very short wavelength. They travel in straight lines with the velocity of light.
2. They are not deflected by electric and magnetic field.
3. They ionize the gas through which they pass.
4. They affect photographic plates.
5. x-rays penetrate through the substances which are opaque to ordinary light.
6. When x-rays fall on certain metals they liberate photoelectrons.
7. They do not pass through heavy metals such as lead and bones.
8. Derive Bragg’s law.(Jun-08,Sep-09,Oct-11)
Consider x-rays of wavelength λ incident on a crystal at a glancing angle θ. The incident rays AB and DE after
reflection from the lattice planes y and Z travel along BC and EF respectively. The path difference between
two waves ABC and DEF is equal to PE+EQ.
PBE PE
SinBE
PE = BE Sinθ
=d Sin θ
QBE EQ
SinBE
EQ = BE Sinθ
= d Sinθ
Path difference = PE + EQ
= d Sinθ + d Sinθ
= 2 d Sinθ
If the path difference2d Sinθ is equal to integral multiple of wavelength of x-ray then constructive interference will
occur between the reflected beam.
Therefore the intensity of the reflected beam is maximum.
2d Sin θ=nλ
9. How will you determine the wavelength of x-ray using Bragg spectrometer?(Mar-07,Jun-07)
Bragg spectrometer is used to determine the wavelength of the x-ay.
x-ray from an x-ray tube are made to pass through slits S1 and S2. This fine beam of x-ray is made to fall upon the
crystal C mounted on the spectrometer table.
The reflected beam after passing through the slits S3 and S4 enters the ionization chamber. The x-ray entering the
ionization chamber ionize the gas which causes a current to flow between the electrodes and the current can be
measured by galvanometer G.
The ionization current is a measure of the intensity of x-rays reflected by the crystal.
The ionization current is measured for different valves of glancing angle θ.
56
A graph is drawn between the glancing angle and ionization current. For current valves of glancing angle the ionization
current increases. This corresponds to first order, second order and so on.
From the graph, the glancing angles for different orders of reflection can be measured.
Knowing the angle θ and the spacing d for the crystal wavelength of x-rays can be determined.
10. Explain the origin of characteristic X-ray.(Mar-09,Jun-11)
Few of the fast moving electrons having velocity of about 1/10 of the velocity of light may penetrate the surface atoms
of the target material and knock out the tightly bound electrons even from the inner most shells.
When the fast moving electron knock off one electron from K-shell and the vacancy is filled by the electron from the L-
shell. This corresponds to Kα line of the series. 1K LE E h
When the electron jumps from M shell to the K shell, it gives out Kβ.
If the electron jumps from M shell to the vacant state in L shell it gives Lα
If the electron jumps from N shell to the vacant state in L shell it gives Lβ.
The frequency of radiation depends upon the target material.
The x-ray spectra consist of sharp lines and is the characteristic of target material.
57
6.ATOMIC PHYSICS.
1. Describe the J.J.Thompson method for determining the specific charge of electron.(Sep-09,Mar-10,Jun-10,Sep-
10,Oct-11)
Principle.
The cathode rays are deflected by electric and magnetic field.
Expt. Arrangement.
Cathode rays are produced by the discharge between the cathode and anodes D1 and D2. The cathode rays passes
between the metal plates P1 and P2 and strike the flat face coated with fluorescent material. A spot of light is
produced at Q. When a potential difference is applied between the plates the beam is deflected to Q1.
Theory.
1.Determination of V
With the given electric field, the magnetic field is adjusted until the beam strike the screen at the original position Q.
The downward force due to the electric field Ee is balanced by the force due to magnetic induction BeV.
Ee Bev
Ev
B
2.Determination of e/m
The magnetic induction is switched off. The deflection QQ1=Y is caused by the electric field.
Initial velocity in the downward direction U = 0
The acceleration in downward direction Ee
am
Time for which the electron moves in the electric field l
tv
The deflection produced in the downward direction
2
1
1
2
Ee ly
m v
22
1 2
1
2
Ee ly B
m Ey
2 2
1
1
2
e l By
m E
1y ky K is a constant determined by the geometry of the discharge tube.
2 21
2
e l By k
m E
2 2
2e yE
m kl B
The valve of e/m is 1.7592 x 1011 C kg-1
58
2. Describe Milikan’s oil drop experiment to determine the charge of the electron.(Jun-08,Sep-08)
Principle.
The study of the motion of uncharged oil drop under free fall due to gravity and charged oil drop in a uniform electric
field.
Expt. Arrangement.
The apparatus consist of two plates A and B.The upper plate has a hole in the middle. The plates are surrounded by a
constant temperature D. The plates are connected to a battery which provide a potential difference of 10000 V.
Theory.
A highly viscous liquid is produced by means of atomizer near the hole and enter the space between A and B.
One such droplet is viewed through the microscope as it descends under gravity. The viscous force due to air
increases and soon it attains a constant terminal velocity V.
Motion under gravity.
The gravitational force acting on the oil drop
downwards is 34
3mg a g ρ is the density of
the oil drop.
The upthrust experienced by the oil drop due to the
displaced air is 34
3a g σ is the density of air.
Net downward force acting on the oil drop= weight
of the oil drop – upthrust experienced by the oil
drop.
3 3
3
4 4
3 3
4
3
a g a g
a g
By stokes law, the viscous force on the oil drop = 6πηav
3
1
2
46
3
9
2
a g av
va
g
Motion under electric field.
The air inside the parallel plates is ionized by sending a beam of x-rays. The droplets pick up one or more
electrons from the ionized air. Let q be the charge carried by the droplet. Let E be the electric field applied between
the plate A and B. The drop moves with the terminal velocity v1.
The force due to the electric field = Eq
3
1
3
1
1
1
1
3 2
21
46
3
46
3
6 6
6
96
2
Eq a g av
Eq a g av
Eq av av
Eq a v v
vEq v v
g
59
If V is the potential difference between A and B, and the distance d between them, then v
Ed
The valve q for a large number of oil drops is determined and it is found that they are an integral multiple of a least
valve. The charge of the electron is
1.602 x 10-19 C.
3. Prove that the energy of an electron for hydrogen atom in nth
orbit is (Ma-08)
Postulates.
1.An electron cannot revolve round the nucleus in all possible orbits. The electrons can revolve the nucleus only in
those allowed orbits for which the angular momemtum of an electron is an integral multiple of h/2π. These
orbits are called stationary or non radiating orbits. 2
nhmvr
2.An atom radiates energy, only when an electron jumps from a stationary orbit of higher energy to a stationary orbit
of lower energy.
E2 – E1 = hν
Radius of nth
orbit.
Consider an atom whose nucleus has a positive charge Ze and e is the charge of the electron.
By coulomb’s law
Electrostatic force of attraction between the nucleus and the electron 2
0
1 .
4 n
Ze e
r
The centripetal force on the electron
22n
n n
n
mvmr
r ------------- (2)
Centripetal force = electrostatic force of attraction
2
0
1 .
4 n
Ze e
r=
2
n
n
mv
r
22
2
0
1
4n n
n
Zemr
r
-------------------(3)
22
3
0
1
4n
n
Ze
mr
------------(4)
The angular momemtum of an electron is 2
n n n nL mv r mr
By Bohr first postulate 2
nhL
2
2 22
2 2 4
2
(5)4
n
n
n
n
nh
mr
n h
m r
From eqn. (4) and (5)
2 2 2
3 2 2 4
0
2 2
0
2
4 4n n
n
Ze n h
mr m r
n hr
mZe
2
nr n The radii of the orbits are in the ratio 1:4:9
For hydrogen Z=1 ·
2 2
0
2n
n hr
me
Substituting the known valves
60
2 0
0
1
0.53
1
0.53
nr n x A
n
r A
This is called Bohr radius
Energy of an electron in the nth
orbit.
Potential energy of the electron in the nth orbit is
2
04p
n
ZeE
r
The kinetic energy of the electron in nth orbit is 21
2nmv
22
2
0
22
0
2 2
0 0
4
4
1
2 4 8
n
n n
n
n
K
n n
mvZe
r r
Zemv
r
Ze ZeE
r r
The total energy of the electron
2 2
0 0
2
0
4 8
8
n P K
n n
n
n
Ze ZeE E E
r r
ZeE
r
Substituting the valve of rn
2 2
0
2n
n hr
mZe
2 4
2 2 2
08n
Z meE
n h
For hydrogen atom Z=1,
4
2 2 2
0
2
8
13.6
n
n
meE
n h
E eVn
1 eV = 1.602 x 10-19
J
The energy of the electron increases as n increases.
4. Explain the working of Ruby laser with neat diagram.(Oct-06,Oct-07,Mar-09,Jun-11)
Ruby is a crystal of aluminium oxide Al2o3 in which some of aluminium ions is replaced by chromium ions. The one
end is fully silvered and the other end is partially silvered. The ruby rod is surrounded by a xenon flash tube which
provides the pumping light to raise the chromium ions to upper energy level. In xenon flash tube, each flash lasts
several milliseconds.
In normal state, most of the chromium ions are in the ground state E1.
When the ruby rod is irradiated by a flash of light, the 5500A0 radiation photons are absorbed by the chromium ions
which are pumped to the excited state E3.
The excited state gives up a part of the energy to the crystal lattice and decay to the metastable state E2 with out
giving any radiation.
The state E2 has a much longer lifetime(10-3s). Thus population inversion is achieved between E2 and E1.
61
When the excited ion from the metastable state E2 drops down spontaneously to the ground state E1 it emits a photon
on wavelength 6943A0.
This photon is reflected back and forth by the silvered ends until it stimulates other excited ion and causes it to emit a
fresh photon in phase with stimulation photon.This stimulated emission is the laser transition.
Finally a pulse of red light of wavelength 6943A0 emerges
through the partially silvered end of the crystal.
5. Explain the working of He-Ne laser with a neat diagram.(Jun-06,Mar-11)
He-Ne laser consist of quartz discharge tube containing helium and neon in the ratio 1:4 at a pressure of 10mm of Hg.
One end of the tube is fully silvered and the other end is partially reflecting mirror.
A powerful radio frequency generator is used to produce a discharge the gas.
When an electric discharge passes through the gas, the electron in the gas collide with the He and Ne atoms and excite
them into metastable states of energy 20.61eV and 20.66 eV.
Some of the excited helium atoms transfer their energy to unexcited Ne atoms by collision. Thus the He atom drops in
achieving a population inversion in Ne atoms.
Excited Ne atoms drops down spontaneously from the metastable state at 20.66eV to lower energy state 18.70eV, it
emits a 6328A0 photon in the visible region.
This photon is reflected back and forth by the reflector ends, until it stimulates an excited neon atoms and causes it to
emit a fresh 6328A0 photon in phase with the stimulating photon.
This stimulated transition from 20.66eV to 18.70eV is the laser transition.
The neon atoms drop down from the 18.70eV level to the lower state E, through spontaneous emission emitting
incoherent light.
From this level E, the Ne atoms are brought to the ground state through collision with the wall of the tube.
Hence the final transition is radiation less.
62
7. DUAL NATURE OF RADIATION AND MATTER AND RELATIVITY.
1. What is photoelectric effect?
Photoelectric emission is the phenomena by which a good number of substances chiefly metals emit electrons under the
influence of radiation such as γ rays, X-rays,ultraviolet and even visible light.
2. Define stopping potential.(Oct-09)
The minimum negative potential given to the anode for which the photo electric current becomes zero is called the cut off
potential or stopping potential.
3. Define thereshold frequency.
Thereshold frequency is defined as the minimum frequency of incident radiation below which the photoelectric emission
is not possible completely, however high the intensity of incident radiation may be.
4. Define work function.
The work function of a photo metal is defined as the minimum amount of energy required to liberate an electron from
the metal surface.
5. What is photo electric cell? What are its types?
The photoelectric cell is a device which converts light energy into electrical energy.
1. Photo emissive cell
2. Photo voltaic cell
3. Photo conductive cell.
6. What is matter wave?
The radiant energy has a dual nature. i.e. Particle and Wave nature. According to DeBroglie the moving particle should
possess wave like properties under suitable conditions. This wave is called matter waves.
7. What are the uses of electron microscope?(Mar-07)
1. It is used in the industry to study the structure of textile fibres, surface of metals.
2. In medicine and biology it is used to study virus and bacteria.
3. It is used in the investigation of atomic structure and structure under such conditions.
8. What are the limitations of electron microscope?(Mar-06,Mar-09)
An electron microscope is operated only in high vacuum. This prohibits the use of the microscope to study living
organisms which would evaporate and disintegrate under such conditions.
9. Define frame of refrence.
A system of coordinate axes which defines the position of a particle in two or three dimensional space is called a frame
of reference.
10. State the two fundamental postulates of special theory of relativity.(Oct-07,Jun-09,Mar-11)
1. The laws of physics are the same in all inertial frames of reference.
2. The velocity of light in free space is a constant in all the frame of reference.
11. Differentiate inertial and non inertial frame of reference.(Oct-0,Mar-08)
Inertial frame of reference.
A frame of reference is said to be inertial, when the bodies in this frame obey Newton’s law of inertia and other laws of
Newtonian mechanics.
In this frame, a body remains at rest or in continuous motion unless acted upon by an external force.
Non Inertial frame of reference.
A frame of reference is non inertial frame when a body not acted upon by an external force is accelerated. In this frame
Newton’s law are not valid.
63
12. If a body moves with the velocity of light what will be its mass? Comment on your result.
According to Einstein 0
2
21
mm
v
C
. If v=C then the mass becomes infinity. Therefore the body cannot move with
the speed of light.
13. Discuss the concept of time. (Jun-10)
1. The time interval between two events has the same valve for all observers irrespective of their motion.
2. If two events are simultaneous for an observer, they are simultaneous for all observers irrespective of their motion.
This means simultaneity is absolute.
14. Discuss the concept of mass.
The mass of the body is absolute and constant and independent of the motion of the body.
15. Discuss the concept of space.
1. Fixed frame of reference by which the position or motion of any object in the universe could be measured.
2. The geometrical form of an object remains the same irrespective of changes in position or state of motion of the
object of observer.
16. What are the uses of photoelectric cells.(Oct-09)
1.Photoelectric cells are used for reproducing sound in cinematography.
2.They are used for controlling the temperature of furnaces.
3.They are used for switching on and off the street lights.
7.DUAL NATURE OF RADIATION AND MATTER AND RELATIVITY.
1. Explain the variation of photoelectric current with applied voltage.
Keeping the intensity(I1) and frequency of the incident radiation as constant, the effect of potential difference between
the plates on the photoelectric current is studied.
When the positive potential of A is increased, the photoelectric current is also increased. If the positive potential is
increased then the photoelectric current reaches a maximum valve and this current is known as saturation current.
If the potential of the plate A is made negative, the photoelectric current does not immediately drop to zero but flows in
the same direction. This shows that the photo electrons move with finite velocity.
If the negative potential is further increased, the photoelectric current decreases and finally becomes zero at a particular
valve.
This minimum negative potential given to the anode for which the photo electric current becomes zero is called the cut
off or stopping potential.
The kinetic energy of the photoelectron is 21
2mv
Since at the stopping potential V0 the fastest electron is just prevented from reaching the plate A, workdone in bringing
the fastest electron to rest= kinetic energy of the fastest electron.
eV0 = 21
2mv
The experiment is repeated for different intensities. It is found that the saturation current are proportional to the
intensities of the radiation. But the stopping potential remains same for all the intensities.
The stopping potential is independent of its intensity.
64
2. State the laws of photo electric emission.(Mar-07,Mar-09Mar-11,Oct-11)
1.For a given photo sensitive material, there is a minimum frequency called the thereshold frequency, below which
emission of photoelectrons stops completely, however great the intensity may be.
2. For a given photosensitive material, the photo electric current is directly proportional to the intensity of the incident
radiation, provided the frequency is greater than thereshold frequency.
3.The photoelectric emission is an instantaneous process. There is no time lag between the incidence of radiation and the
emission of photo electrons.
4.The maximum kinetic energy of the photo electrons is directly proportional to the frequency of incident radiation, but is
independent of its intensity.
3. Derive Einstein photo electric equation.(M-06,Ju-06,Se-09,M-10,Ju-10)
According to Einstein the emission ofphoto electron is the result of the interaction between a single photon of the
incident radiation and an electron in the metal.
When a photon of energy hν is incident on a metal surface, its energy is used in two ways.
`1. A part of the energy of the photon is used in extracting the electron from the surface of metal.This energy W spent in
releasing the photo electron is known as photoelectric work function of the metal.
The work function of a photo metal is defined as the minimum amount of energy required to liberate tan electron
from its metal surface.
2. The remaining energy of the photon is used to impart kinetic energy to the liberated electron.
Energy of the photon= work function + Kinetic energy
21
2h W mv
If the electron does not lose energy by internal collisions the entire energy (hν-w) will be exhibited as the kinetic energy.
2
max
1
2h W mv
This equation is known as photoelectric equation.
When the frequency of the incident radiation is equal to the thereshold frequency (ν0) of the metal surface, kinetic energy
of the electron is zero.
hν0=W
· 2
0 max
1
2h h mv
2
0 max
1( )
2h mv
4. Mention the applications of photoelectric cell.(O-06,Ju-07,M-08,Ju-08)
1.Photo electric cells are used for reproducing sound in cinematography.
2.They are used for controlling the temperature of furnaces.
3.photoelectric cells are used for automatic switching on and off the street lights.
4.photoelectric cells are used in the study of temperature and spectra of stars.
5.These cells are used in obtaining electrical energy from sunlight during space travel.
6. These cells are used in opening and closing of door automobiles.
5. Derive an expression for debroglie wavelength.(Oct-06,Mar-07,Mar-09,Jun-10,Sep-10,Jun-11)
For a wave of frequency ν the energy associates with each photon is given by planck’s relation
E=hν where h is planck’s constant.
According to Einstein mass energy relation a mass m is equivalent to energy
E=mc2 c is the velocity of light.
hν=mc2
65
2hcmc
hmc
h
mc
For a particle moving with a velocity v, if c=v
h
p
h
p p=mv is the momemtum of the particle.
6. Derive an expression for the debroglie wavelength of an electron.
When an electron of mass m and charge e is accelerated through a potential difference V, then the energy eV is equal to
kinetic energy of the electron.
21
2
2
mv eV
eVv
m
The De Broglie wavelength is
h
mv
2
h
eVm
m
= 2
h
meV substituting the valves of known valves
012.27A
V
If V= 100 V then λ = 1.227A0 the wavelength associated with an electron accelerated by 100 volts is 1.227A0
Since E=eV is kinetic energy associated with the electron, then
2
h
mE
7. Explain the wave mechanical concept of atom.(Oct-07)
According to De Broglie an electron of mass m moving with velocity v is associated with a wave of wavelength λ is
given by
h
mv
On the basis of De Broglie hypothesis, and atom model was proposed in which the stationary orbits of Bohr model were
retained, but with the difference that electron in various orbits behaves as wave.
It was suggested that stationary orbits are those in which orbital circumference (2πr) is an integral multiple of de Broglie
wavelength λ.
2πr=nλ where n=1,2,3...
Substituting the valves of λ
2nh
rmv
2
nhmvr
66
It is seen that the total angular momentum of the moving electron is an integral multiple of h/2π. Thus de Broglie concept
confirms the Bohr postulate.
8. Explain length contraction.(Mar-06,Mar-08,Mar-10,Sep-10,Oct-11)
Consider two frames of reference S and S’ to be initially at rest. A rod is placed in the frame of reference S’ and an
observer O is in S.
The length of the rod in S’ as measured by the observer in S is l0.
Now the frame of reference S’ moves with a velocity v along the positive X- axis. Now, the length of the rod is measured
as l by the observer in S.
Then
2
21o
vl l
c
ol l .
Thus the length of the rod moving with a velocity v relative to the observer
at rest is contracted by the factor.
2
21
v
c
in the direction of motion.
This is known as Lorentz Fitzgerald contraction. Eg. A circular object will
appear as an ellipse for a fast moving observer.
9. Explain time dilation.(Jun-06,Jun-08)
Consider two frames S and S’ . Let S’ be moving with a velocity v with respect to S in the positive X-direction.gg n
Supppose a clock situated in the frame of S’ at a position gives out signals at an interval t0
If this interval is observed by an observer in frame S, then the interval t recorded by him is
0
2
21
tt
v
c
i.e t > t0.
To a stationary observer is S, the time interval appears to be lengthened by a factor .
2
2
1
1v
c
A moving clock appears to be slowed down to a stationary observer. This is known as time dilation.
67
10.Derive Einstein mass energy relation.(Jun-07)
1.
According to Newton II law of motion
( )
dF mv
dt
dv dmF m v
dt dt
The increase in kinetic energy dEK of the body is
dEK = F dx
= ( )
dv dmm v dx
dt dt
=dx dx
mdv vdmdt dt
.
=
2mvdv v dm
From Einstein theory of relativity
0
2
21
mm
v
c
2 22 0
2 2
m cm
c v
2 2 2 2 2 2
0m c m v m c
Differentiating we get,
2 2 22 2 2 0c mdm v mdm m vdv
2 2c dm v dm mvdv
2
kdE c dm
0
2
0
2
0
2 2
0
E m
k
m
k
k
dE c dm
E c m m
E mc m c
Total energy= kinetic energy of the moving body + rest mass energy
2
0
2 2 2
0 0
2
k
k
E E m c
E mc m c m c
E mc
68
8. NUCLEAR PHYSICS.
1. Define isotope.
Isotopes are atoms of the same element having the same atomic number Z but different mass number A. eg.
1 2 3
1 1 1, ,H H H
2. Define isobar.
Isobar are atoms of different elements having the same mass number A but different atomic number Z. eg. !6 16
8 7,O N
3. Define isotone.
Isotones are atoms of different elements having the same number of neutrons. Eg. !6 14
8 6,O C
4. Show that nuclear density is almost a constant for all the nuclei.
N
Nuclearmass
Nuclearvolume
3
1/3 3
0 0
4 4
3 3
N NN
Am m
r A r
Here all the valves are constant. Substituting these valves, the nuclear density is calculated as 1.816 x 1017 kg m-3 which
is almost a constant for all the nuclei irrespective of the size.
5. Define mass defect.(Oct-10)
The difference in the total mass of the nucleons and the actual mass of the nucleus is known as the mass defect.
p nZm Nm m m
6. Define Binding energy.(Oct-06)
When the protons and neutrons combine to form a nucleus, the mass that disappears is converted into an equivalent
amount of energy . This energy is called the binding energy of the nucleus.
7. Define Half life period.
The half life period of a radioactive element is defined as the time taken for one half of the radioactive element to
undergo disintegration
8. Define radioactivity.
The phenomenon of spontaneous emission of highly penetrating radiations such as α,β,γ by heavy elements having
atomic number greater than 82 is called radioactivity.
9. Compare the penetrating power and ionision power of α,β,γ rays.
Alpha rays Beta rays Gamma rays
They have less high penetrating
power
They penetrating power is greater
than that of α rays
They have a very penetrating power
The ionizing power is high The ionizing power is comparatively
less
They produce less ionization.
10. Define mean life.
The mean life of a radioactive substance is defined as the ratio of total life atom of all the radioactive atoms to the total
number of atoms in it.
11. Define activity.
The activity of a radioactive substance is defined as the rate at which the atoms decay.
69
12. Define one curie.(Oct-06,Mar-08,Mar-10,Oct-10)
Curie is defined as the quantity of a radioactive substance which gives 3.7 x 1010 disintegrations per second or 3.7 x 1010
becquerel.
13. Define artificial radioactivity.
The phenomenon by which even light elements are made radioactive by artificial or induced methods is called artificial
radioactivity.
14. How will you produce radio isotope?
Artificial isotope are produced by placing the target element in the nuclear reactor where neutrons are available.
31 ! 32
15 0 15P n P
15. Define one roentgen.(Jun-07,Oct-08)
One roentgen is defined as the quantity of radiation which produces 1.6 x 1012 pairs of ions in 1 gm of air.
16. Define nuclear fission.
The process of breaking up of the nucleus of a heavier atom into two fragments with the release of large amount of
energy is called nuclear fission.
17. Define critical mass.(Oct-08)
Critical size is defined as the minimum size in which atleast one neutron is available for further fission reaction.
18. What is breeder reactor?(Mar-09)
238 232
92 90,U Th are not fissile material. In the reactor these can be converted into fissile material 239 233
94 92,Pu U
respectively by absorption of neutrons.
19. What are the uses of reactors?
1. Nuclear reactor is used for power production.
2. Nuclear reactor is used to produce radio isotope.
3. It is used in the scientific research.
20. Define nuclear fusion.
Nuclear fusion is a process in which two or more lighter nuclei combine to form a heavier nucleus.
21. What is cosmic ray?(Jun-08,Jun-10)
The ionizing radiation many times stronger than γ rays entering the earth from all the directions from cosmic or
interstellar space is known as cosmic rays.
22. What is pair production and annihilation of matter?(Mar-06,Ju-06,Mar-07)
The conversion of a photon into an electron positron pair on its interaction with the strong electric field surrounding a
nucleus is called pair production.
The converse of pair production in which the electron and positron combine to produce a photon is known as annihilation
of matter.
23. Define Radioactive law of disintegration.(Oct-09)
The rate of disintegration at any instant is directly proportional to the number of atoms of the element present at that
instant.
24. What is α – decay?(Mar-06)
When a radioactive nucleus disintegrates by emitting an α particle the atomic number decreases by two and mass number
decreases by four.
25. What are the properties of nuclear force?(Jun-08,Mar-09)
Nuclear force is charge independent.
70
Nuclear force is the strongest known force in nature.
Nuclear force is not a gravitational force. It is about 1040 times stronger than the gravitational force
26. What are the properties of neutron?(Jun-06,Mar-08)
Neutrons are the constituent particles of all nuclei except hydrogen.
Neutrons are neutral particles with no charge and mass slightly greater than that of hydrogen.
As neutrons are neutral, they can easily penetrate any nucleus.
27. How will you classify the neutrons based on kinetic energy?(Jun-09)
Neutrons with energies from 0 to 1000 eV are called slow neutrons.
Neutrons with an average energy of about 0.025 eV in thermal equilibrium are called thermal neutrons.
Neutrons with energies 0.5MeV and 10MeV are called fast neutrons
28. What are the uses of control rods?(Oct-07)
The control rods are used to control the chain reaction. They are very good absorbers of neutron. Boron and Cadmium
rod are called control rods.
29. Write a note on leptons.(Jun-07)
Leptons are lighter particles having mass equal to or less than about 207 times the mass of an electron except neutrino
and antineutrino. This group contains particles such as electron, positron, neutrino, antineutrino, positive and negative
muons.
NUCLEAR PHYSICS.
1. Explain radio active displacement law.
α- decay.
When a radioactive nucleus disintegrated by emitting α -particle, the mass number decreases by 4 and the atomic
number decreases by two.
88Ra226 86Rn222 +2He4
β-decay.
When a radioactive nucleus disintegrated by emitting β particle, the atomic number increases by 1 and the mass number
remains the same.
90Th234 91Pa234 +-1e0
γ-decay
when a radioactive nucleus emits γ rays only the energy level of the nucleus changes and the atomic number and mass
number remains the same.
During α and β decay, the daughter nucleus is mostly in the excited state. It comes to ground state with the emission of
γ- rays
2. What are the properties of nucleons?
1.Neutrons are the constituent particles of all nuclei, except hydrogen.
2.Neutrons are neutral particles with no charge and mass slightly greater than that of protons. Hence, they are not
deflected by electric and magnetic fields.
3.Neutrons are stable, inside the nucleus. But outside the nucleus they are unstable. The free neutron decays with an
emission of proton, electron and antineutrino with half life of 13 minutes.
4.As neutrons are neutral, they can easily penetrate any nucleus.
5. Neutrons are classified according to their kinetic energy as slow neutrons and fast neutrons.
Slow neutron- 0 to 1000 eV
Fast neutron 0.5MeV to 10 MeV
Thermal neutron 0.025eV
2. Explain the biological hazards of nuclear radiations.
When γ rays or any high energy nuclear particle passes through human beings, it disrupts the entire normal functioning of
the biological system and the effect may be either pathetic or genetic.
The biological effects of nuclear radiation can be divided into three groups.
71
1.short term recoverable effects.
2.long term irrecoverable effects.
3.genetic effects.
The extent to which the human organism is damaged depends upon 1) the dose and the rate at which the radiation is
given 2) the part of the body exposed to it.
If the exposure is 100R it may cause diseases like leukemia.
When the body is exposed to 600R it causes death.
Safe limit of receiving the radiations is about 250 milli roentgen per
week.
Precautions.
1.Radioactive materials are kept in thick walled lead container.
2.Lead aprons and lead gloves are used while working in hazardous
area.
3.All radioactive samples are handled by a remote control process.
4. A small micro-film badge is always worn by the person and it is
checked periodically for the safe limit of radiation.
3. Explain Cosmic ray showers.(Mar-07)
The cascade theory of cosmic ray shower shows that the shower production involves two process. 1) radio active
radiation 2) pair production.
An energetic electron or positron present in cosmic rays loses energy when It collides with the nuclei of atoms in earth
atmosphere.
This energy loss appears as high energy photon. This photon interacts with an atomic nucleus and produce an electron
positron pair.
The electron have sufficient energies to produce more photons on interaction with nuclei, which are further capable of
bringing about pair production.
The result is the generation of a large number of photons, electrons and positrons having a common origin like a shower
and hence it is known as cosmic ray shower.
4. Write the properties of nuclear force.
1.Nuclear force is charge independent.It is the same for all the three pairs of nucleons(n-n) (p-p) and (n-p). This shows
that nuclear force is not electrostatic in nature.
2.Nuclear force is the strongest known force in nature.
3.Nuclear force is not a gravitational force. Nuclear force is about 1040 times stronger than the gravitational force.
4.Nuclear force is a short range force. It is very strong between two nucleons which are less than 10-15m apart and is
almost negligible at a distance greater than this.
5. Explain binding energy curve.(Jun-10)
1. The binding energy per nucleon increases sharply with mass number A upto 20. It increases slowly after A=20. For
A<20 there exists recurrence of peaks corresponding to those nuclei, whose mass number are multiples of four and they
contain not only equal but also even number of protons and neutrons.
2. The binding energy per nucleon reaches a maximum of 8.8 MeV at A=56, corresponding to the iron nucleus. Hence,
iron nucleus is the most stable.
3. The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number ranging from 40 and 120.
These elements are most stable and non radioactive.
4. For higher mass numbers the curve drops slowly and the BE/A is about 7.6MeV for uranium. Hence they are unstable
and radioactive fusion and nuclear fission respectively.
.
72
8.NUCLEAR PHYSICS.
1. How will you determine the isotopic mass using Bainbridge mass spectrometer.(Jun-06,Oct-06,Jun-07,Jun-
08,Mar-09,Jun-10,Sep-10)
Bain bridge mass spectrometer is an instrument used for the accurate determination of atomic masses.
Atoms with one or more electrons have a net positive charge and they become positive ions. A beam of positive ions
produced in a discharge tube is collimated into a fine beam by slits S1 and S2 and then enters into the velocity selector.
The velocity selector consist of two parallel plates P1 and P2 which produce electric field E. An electromagnet produce
uniform magnetic field B.
The deflection by the electric field is nullified by the magnetic field, so that the ions do not suffer any deflection within
the velocity selector.
The force exerted by the electric field is Eq.
The force exerted by the magnetic field is Bqv.
Eq=Bqv
V=E/B
Only those ions having this velocity pass out of thevelocity selector and then through the slit S3 enters into the evacuated
chamber D.
These ions are subjected into magnetic field B’. The ions are deflected along circular path of radius R and strike the
photographic plate.
The force due to the magnetic plate B’qv provides the necessary centripetal force.
B’qv=mv2/r
m=B’qr/v
Substituting the valve of v then,
'BB qRm
E
Ions with different mass trace semi circular path of different radii and produce dark
lines on the plate.
From this the radii R can be calculated.
Since q, B,B’ and R are known, the mass of the positive ions and hence isotopic
masses can be calculated.
2. Obtain an expression to deduce the amount of the radioactive substance present at any moment. Obtain the
relation ship a between the half life period and mean life .(Sep-08,Sep-09,Oct-11)
Rutherford and Soddy found that the rate of disintegration is independent of physical and chemical conditions.
The rate of disintegration at any instant is directly proportional to the number of atoms of the element present at
that instant.
Let N0 be the number of radioactive atoms present initially and N be the number of atoms at a given instant t. Let dN be
the number of atoms undergoing disintegration in a small interval of time dt.
The rate of disintegration is
dNN
dt
dNN
dt
λ known as decay constant.
loge
dNdt
N
N t c
0
0
0,
loge
t N N
N c
73
0log loge eN t N
0
loge
Nt
N
0
0
t
t
Ne
N
N N e
The number of atoms decreases exponentially with increase in time.
Half life period.
The half life period of a radioactive element is defined as the time taken for one half of the radioactive element to
undergo disintegration.
01
2
,2
Nt T N
1
200
2
TNN e
1
2
log 2e T
101
2
log 2 log 2 2.3026e xT
=
0.6931
The half life period is inversely proportional to the decay constant.
Mean life period.
The mean life is defined as the radio of total life time of all the radioactive atoms to the total number of atoms in it.
Mean life = Sum of life atoms of all the atoms.
Total number of atoms.
1
3. Explain the construction and working Geiger muller counter.(Mar-07,Oct-07,Jun-09,Mar-1
Geiger muller counter is used to measure the intensity of the radioactive radiation.
Principle.
When nuclear radiation passes through gas, ionization is produced.
Construction.
The GM consist of a metal tube with glass envelope(C) acting as the cathode and a fine tungsten wire ( W) along the
axis of the tube, which serves as anode.
The tube is filled with the inert gas like argon at a low pressure. One end is fitted with a thin mica sheet and this end
acts as a window through which radiations enter the tube.
A high potential difference of about 1000V is applied between the electrodes through a high resistance R of about
100 megaohm.
Operation.
When an ionizing radiation enters the counter, primary ionization takes place and a few ions are produced.
These ions are accelerated with greater energy due to the high potential difference and they cause further
ionization. Thus avalanche of electrons on reaching the anode generates a current pulse, which when passing
through R develops a potential difference.
This is amplified by electronic circuits and is used to operate an
electronic counter.
The counts in the counter is directly proportional to the intensity of the
ionizing radiation.
The ionization of the gas is independent of the type of the incident
radiation. Hence GM counter does not distinguish the type of
radiation that enters the chamber.
74
4. Describe the essential parts of the nuclear reactor.(Mar-06)
A nuclear reactor is a device in which the nuclear fission reaction takes place in a sustained manner.
Fissile material or fuel.
The fissile material or fuel generally used is U235. But this exist only in a small amount (0.7%) in natural
uranium.Natural uranium is enriched with more number of U235 (2-4%) and this low enriched uranium is
used as fuel.
U233 and Pu239 are also used as fuel.
Pressurised heavy water reactor ---- natural uranium oxide.
Pressurised light water reactor ------- low enriched uranium
Fast breeder test reactor -------------- a mixture of carbides of uranium and plutonium.
Fast breeder reactor ----- -------------- a mixture of plutonium and uranium.
Kamini ----------------------------------- U233
Moderator.
The function of the moderator is to slow down fast neutrons of about 2 MeV to thermal neutrons with an average
energy of 0.025eV.
Ordinary water and heavy water are used as moderators.
A good moderator slows down neutrons by elastic collision.
Graphite is also used as a moderator.
In Fast breeder reactors, no moderator is used.
Neutron source.
A source of neutron is required to initiate the fission chain reaction for the first time.
A mixture of beryllium with plutonium or radium or polonium is uses as a source of neutron.
Control rods.
The control rods are used to control the chain reaction. They are very good absorbers of neutrons.
Boron and Cadmium rod are uses as control rods.
By pushing them in, or pulling out, the reaction rate can be controlled.
In our country boron carbide is used as control rods.
The cooling system.
The cooling system removes the heat generated in the reactor core.
Ordinary water, heavy water and liquid sodium are used as coolants.
The coolant passes through the tube containing the fuel and carries the heat from the fuel rods to the steam generator
through the heat exchanger. The steam runs the turbines to produce electricity in power reactors.
Neutron reflector.
Neutron reflector prevent the leakage of neutrons to a large extent, by reflecting them back.
In pressurized heavy water reactors the moderator itself acts as a reflector.
Shielding.
As a protection against the harmful radiations, the reactor is surrounded by a concrete wall of thickness about 2 – 2.5
m.
5. What is cosmic rays? Explain the latitude effect and altitude effect.(Mar-08,Mar-10)
The ionizing radiation many times stronger than γ rays entering the earth from all the directions from cosmic or
interstellar space is known as cosmic rays.
Primary cosmic rays.
The primary cosmic rays are those coming from outer space and enter the outer boundary of the earth atmosphere.
The primary cosmic rays consist of 90% protons, 9% helium and remaining heavy nuclei. The energy of the
primary cosmic rays is 108 MeV.
Secondary cosmic rays.
The secondary cosmic rays are produced when primary cosmic rays interact with gases in the upper layers of the
atmosphere. They are made up of α-particles, protons, electrons, positrons, mesons, photons etc.
Latitude effect.
The intensity is maximum at the poles (θ=900) , minimum at the equator (θ=0) and constant between latitudes of 420
and 900.
The variation of cosmic ray intensity with geometric latitude is known as latitude effect.
The decrease in cosmic ray intensity is explained by earth magnetic field.
75
The charged particles approaching the earth near the poles travel almost the direction of the magnetic lines of force
experience no force and easily reach the surface of the earth and hence maximum intensity at poles.
The charged particles that approach at the equator have to travel in a perpendicular direction to The field and are
deflected away. Only those the particle with sufficient energy can reach the equator, while the slow particles are
deflected back into cosmos and hence minimum intensity at the equator.
Altitude effect.
The study of variation of cosmic ray intensity with altitude is known as altitude effect.
The intensity increases with altitude and reaches a maximum at a height of about 20 Km. Above this height there is a fall
of intensity.
9. ELECTRONICS.
1. What is forbidden energy gap?
The separation between valence band and conduction band is known as forbidden energy gap.
Insulator 10eV
Semiconductor. 0.7eV for Ge
and 1.1 eV for Si
2. What is meant by intrinsic semiconductor?(Mar-06)
A semiconductor which is pure and contains no impurity is known as an intrinsic semiconductor. In an intrinsic
semiconductor the number of free electrons and holes are equal.
3. What is meant by extrinsic crystal?(Jun-06,Jun-08,Oct-10,Mar-11)
An extrinsic semiconductor is one in which an impurity with a valency higher or lower than the valency of the pure
semiconductor is added, so as to increase the electrical conductivity of the semiconductor.
4. What is meant by doping? What are the methods of doping?(O-08,O-09)
The process of addition of a very small amount of impurity into an intrinsic semiconductor is called doping.
The methods of doping are
The impurity atoms are added to the semiconductor in its molten state.
The pure semiconductor is bombarded by ions of impurity atoms.
When the semiconductor crystal containing the impurity atoms is heated, the impurity atoms diffuse into the hot crystal.
5. What is meant by depletion layer?
A region which does not have any mobile charges very close to the junction. This region is called depletion layer.
6. What is meant by rectifier?(Mar-07,Mar-09)
The process in which alternating voltage or alternating current is converted into direct voltage of direct current is known
as rectification.
76
7. Differentiate N- type semiconductor and P-type semiconductor.(Mar-10)
No P -type
semiconductor
N-type semiconductor.
1 Trivalent impurity is added
to a pure semiconductor.
Pentavalent impurity is
added to a pure
semiconductor.
2 Impurity atoms are
known as acceptors.
Impurity atoms are
Known as donors
3 The majority carrier are
holes and the minority
carrier are electrons.
The majority carrier are
electrons and the
minority carrier are holes.
8. Explain avalanche breakdown.
When both sides of the PN junction are lightly doped and the depletion layer becomes large, avalanche breakdown takes
place. The electric field across the depletion layer is not so strong. . The minority carriers accelerated by the field collide
with the semiconductor atoms in the crystal .The covalent bonds are broken and electron holes pairs are generated. These
charge carriers in turn produce more and more carriers. This cumulative process is called avalanche multiplication and
the breakdown is called avalanche breakdown.
9. Why transistor is called current amplification device?
The input resistance of the transistor is high and the output resistance is low.Therefore when less current is given to the
input high current is produced in the output. Hence the transistor is called the current amplification device.
10. Explain zener breakdown.(Oct-08,Oct-09)
When both sides of the PN junction are heavily doped, the depletion layer is narrow. When a small reverse bias is
applied, a very strong electric field is produced across the thin depletion layer. This field breaks the covalent bonds,
extremely large number of electrons and holes are produced, which give rise to the reverse saturation current.
11. Can emitter and collector be interchanged. Explain.
In a transistor, the emitter region is heavily doped. Since emitter has to supply majority carriers. The collector region is
lightly doped. Since it has to accept majority carriers. It is physically larger in size. Hence emitter and collector cannot be
interchanged.
12. Define input impedance.(Jun-06)
The input impedance is defined as the ratio of small change in base emitter voltage to the corresponding change in base
current at a given VCE.
CE
BEi
B V
Vr
I
13. Define output impedance.
The output impedance is defined as the ratio of small change in collector emitter voltage to the corresponding change in
collector current at a constant base current.
B
CEo
C I
Vr
I
14. Why CE configuration is preferred over CB configuration for operating transistor as an amplifier?
The common emitter configuration has
1. High input impedance
2. Low output impedance
3. Higher current gain.
15. Define bandwidth of an amplifier.(Mar-07)
Band width is defined as the frequency interval between lower cut off and upper cut off frequency. BW=fU-fL
77
16. What is meant by feedback in amplifier? What are its types?
Feedback is said to exist in an amplifier circuit, when a fraction of the output signal is fed back to the input and combined
with the input signal.
The two types of feed back is
1. Negative or degenerative feedback.
2. Positive or regenerative feedback.
17. What are the advantages of negative feedback?(Jun-07,Oct-07,Jun-08,Mar-11)
1. Highly stabilized gain.
2. Reduction in the noise level.
3. Increased bandwidth.
4. Increased input impedance and decreased output impedance.
5. Less distortion.
18. What is Barkhausen condition for an oscillator?(Oct-07,Oct-08,Jun-09,Mar-10)
(1)The loop gain Aβ=1 and
(ii) the net phase shift around the loop is 00 or integral multiples of 2π.
19. What is integrated circuit?(Jun-08,Jun-09)
An integrated circuit consists of a single crystal chip of silicon containing both active (diodes and transistors) and passive
(resistance and capacitors) components and their interconnections.
20. What are the advantages of integrated circuit? (Mar-06,Oct-06,Jun-10)
1. Extremely in size
2. Low power consumption
3. Reduced cost.
4. Very small weight. 5. Easy replacement.
21. What is universal gate? Why are they called so?
The NAND and NOR gate are called universal gate. They can perform all the three basic logic functions NOT, OR and
AND gate.
22. What are the characteristics of operational amplifier?(Oct-07)
1. Very high input impedance or even infinity
2. Very low output impedance or even zero.
3. very high gain.
23. What are the uses of cathode ray oscilloscope?(Jun-09)
1.It is used to measure a.c and d.c voltage.
2.It is used to study thewaveforms of a.c voltage.
3. It is used to study the frequency of a.c voltage.
24. What is zener diode? Draw its symbol.(Oct-09,Oct-10)
Zener diode is a reverse biased heavily doped semiconductor(silicon or germanium) PN junction diode, which is operated
exclusively in the breakdown region.
25. Prove A B A B B using Boolean algebra.
0
1
1 1
A B A B
AA AB BA BB
AB BA B AA andBB B
B A A
B A A
78
26. What is light emitting diode?(Mar-07)
A light emitting diode is a forward biased PN junction diode, which emits visible light when energized.
27. Evaluate the output if A=1,B=1 and C=1
1 1
1
1
1 0
28. State Demorgan’s law.
First theorm.
The complement of a sum is equal to the product of the complements.
.A B A B
Second theorm.
The complement of a product is equal to the sum of the complements.
.A B A B
29. Prove A B A C A BC
(1 )
1 1
A B A C
AA AC BA BC
A AC BA BC AA A
A C B BC
A BC C B
30. Find the output of the logic circuit.
A A
A +B
B
B
A
B
31. Find the output of the logic circuit given below.(Mar-09)
A
A+B
B
(A+B)(A+C)=A+BC
A
C A+C
32. Give the Boolean equation for the given logic diagram.(Mar-10)
A .A B
B
C
D C.D .A B CD
79
9.ELECTRONICS.
1. Explain half wave rectifier with a neat diagram.(Jun-11,Mar-09)
A circuit which rectifies half of the a.c wave is called half wave rectifier.
The a.c voltage to be rectified is obtained across the secondary ends S1 and.
S2 of the transformer.
The P-end of the diode D is connected to S1 of the secondary coil of the
transformer and N end of the diode is connected to S2 of the transformer
through load resistance RL.
During the positive half cycleof the input ac voltage VS. S1 will be positive
and the diode is forward biased and hence it conducts. Therefore current
flows through the circuit and there is a voltage drop across RL.
During the negative half cycle of the input ac voltage S1 will be negative
and the diode D is reverse biased. Hence the diode does not conduct.
Thus corresponding to an alternating input signal, unidirectional pulsating
output is obtained.
The efficiency of half wave rectifier is 40.6%.
2. How Zener diode acts as voltage regulator.
To maintain a constant voltage across the load,even if the input voltage or load Current varies, voltage regulation is
to be made.
A zener diode working in the breakdown region can act as a voltage regulator.
The zener diode in reverse biased condition is connected in parallel with the load RL. Vdc be The unregulated dc
voltage and VZ be zener voltage. RS is the current limiting resistor Load current varies, input voltage is constant.
Let us consider that the load current increases. Zener current hence decreases, and the Current through the resistane
RS is a constant.
The output voltage is VZ=Vdc - IRS. Since the total current I remains constant,
output voltage remains constant. Input voltage varies.
Let us consider that the input voltage Vdc increases.
Now the current through Zener increases and voltage drop across Rs increases in
such a way that the load voltage remains the same.
Thus the zener diode acts a s a voltage regulator.
3. How transistor acts as a switch?(Oct-07)
Transistors are widely used in switching operations. NPN transistor is
connected in common emitter configuration and a resistor RB is connected
in series with the base.
A pulse type waveform is applied as the input to the transistor through RB.
When the input is high, base emitter junction is forward biased and current
flows through RB into the base. The base current flowing, is enough to
saturate the transistor. When the transistor is saturated, it is said to be ON.
When the input is low at 0 V, the base emitter
junction is not forward biased.
So no base current flows. Hence the transistor is said
to be OFF.
4. Explain voltage divider bias method.(M-07, J09)
This is the most widely used method of providing bias and stabilization to a
transistor.
Two resistances R1 and R2 are connected across the supply voltage VCC and
provide biasing.
The emitter resistance RE provides stabilization. The voltage drop across R2
forward bias the base emitter junction.
The stabilization is explained as follows.
β is very sensitive to temperature changes, the collector current IC increases
80
with rise in temperature. Therefore IE increases and hence the voltage drop across emitter resistance RE increases.
The voltage drop across R2 = VBE + VRE.
As voltage drop across R2 is independent of IC, VBE decreases.
This decreases IB and the reduced valve of IB tends to bring back IC to the original valve.
Hence any variation of β will have no effect in the operating point.
5. Explain the essential parts of LC oscillator.
The essential components are 1. Tank circuit 2. Amplifier and 3. Feed back
circuit.
Tank circuit.
It consist of inductance coil (L) connected in parallel with capacitor (C). The frequency of oscillations in the circuit
depends upon the valves of inductance coil and capacitance of the capacitor.
Amplifier.
The transistor amplifier receives dc power from the battery and changes it into Ac
power for supplying to the tank circuit
Feedback circuit.
It provides positive feedback. This circuit transfers a part of output energy to
LC circuit in proper phase, to maintain the oscillations.
6. Explain the OR gate using diode.
Case (i) A=0 and B=0.
When both A and B are at zero level, low the output voltage will be low, because the diodes are
non conducting.
Case(II) A=0 and B=1
When A is low and B is high, diode D2 is forward biased so that current flows through RL and
output is high.
Case (III) A=1 and B=0
When A is high and B is low, diode D1 conducts and the output is high.
Case(IV) A=1 and B=1
When A and B are high both diodes D1 and D2 are conducting and the output is high. Therefore Y is high.
7. Explain AND gate using diode.(Jun-08)
Case(I) A=0 and B=0 When A and B are zero, both diodes are in forward bias condition and they conduct and hence
the output will be zero, because the supply voltage VCC will be dropped across RL only. Therefore Y=0.
Case (II) A=0 and B=1
Diode D1 is forward biased and diode D2 is reverse biased.
The diode will now conduct due to forward biasing.
Therefore output Y =0.
Case(III) A=1 and B=0
Diode D2 will be conducting and hence the output Y=0
Case (IV) A=1 and B=1
Both the diodes are not conducting. Since D1 and D2 are in OFF condition, no current flows
through RL. The output is equal to the supply voltage. Therefore Y=1.
8. State and prove De Morgan’s theorm.(M-06,O-06,S-09,M-10,S-10,O-11)
First theorm.
The complement of a sum is equal to the product of the complements.
.A B A B
Second theorm.
The complement of a product is equal to the sum of the complements.
.A B A B
81
A B A B .A B A B A B .A B
0 0 1 1 1 1 1 1
0 1 1 0 1 1 0 0
1 0 0 1 1 1 0 0
1 1 0 0 0 0 0 0
9. Explain op-amp circuit symbol and pin out configuration.(Sep-08)
The op-amp has two input and one output. The terminal with a negative sign is called as the inverting input and with
positive sign is called as the non inverting input.
The top pin on the left side of the notch indicates pin1.
The pin number 2 is inverting input terminal.
The pin number 3 is non inverting input terminal.
The pin number 6 is the output terminal.
The pin number 7 and 4 are the power supply terminals.
Terminals 1 and 5 are used for null adjustments. Null adjustments pins are used to null the output voltage when equal
voltage are applied to the input terminals for perfect balance.
Pin 8 indicates no connection.
9.ELECTRONICS.
1. Explain the working of bridge rectifier.(Mar=06,Jun-07,Mar-10,Jun-10,Oct-11)
There are four diodes D1 ,D2 ,D3 and D4 used in the circuit, which are connected to form a network. The input ends A and
C of the network are connected to the secondary ends S1 and S2. The output ends B and D are connected to the load
resistance RL.
During positive half cycle of the ac voltage, the point A is positive with respect to C. The diodes D1 and D3 are forward
biased and conduct. The diodes D2 and D4 are reverse biased and do not conduct. Hence current flows along
S1ABCDS2 through RL.
During negative half cycle the point C is positive with respect to A. The diodes D2 and D4 are forward biased and
conduct. The diodes D1 and D3 are reverse biased and do not conduct. Hence current flows along S2CBDAS1
through RL.
It is seen that the current flows through RL in the same direction during both half cycles of the input ac signals.
The efficiency of the bridge rectifier is 81.2%.
.
2. Explain the characteristics of an NPN transistor in common emitter configuration.
82
Input characteristics.
The curve is drawn between the basecurrent (IB) and voltage between base and emitter (VBE) at constant VCE.
The input impedance of the transistor isdefined as the ratio of small change in base emitter voltage to
thecorresponding change in base current at a given VCE.
CE
BEi
B V
Vr
I
The input impedance of the transistor is very high.
Output characteristics.
The curve is drawn between IC and VCE when IB is kept constant.
Saturation region.
The initial part of the curve is called saturation region. The region in between the origin and the knee point.
Cut off region.
There is a very small collector current in the transistor even when the base current is zero. The region below the curve for
IB=0 is called cut off region. Below this region the transistor does not function.
Active region.
The central region of the curve is called active region. In this region E-B is forward biased and the C-B junction is
reverse biased.
The output impedance is defined as the ratio of variation in the collector emitter voltage to the corresponding
variation in the collector current at a constant base current.
0
B
CE
C I
Vr
I
The output impedance is very low.
Transfer characteristics.
The curve is drawn between IC and IB at constant VCE.
The current gain is defined as the ratio of a small change in the collector
current to the corresponding change in the base current at a constant VCE.
CE
C
B V
I
I
3. Explain the working of single stage CE amplifier .(Jun-08,Sep-10,Mar-11)
Biasing circuit.
The resistances R1, R2 and RE form the biasing and stabilization circuit Input capacitance. Cin.
This is used to couple the signal to the base of the transistor.
The capacitor Cin allows only ac signal to flow.
Emitter by pass capacitor. CE.
This is connected in parallel with RE to provide a low reactance path to the amplified ac signal.
If it is not used then amplified ac signal flowing through RE will cause a voltage drop across it there by shifting the
output voltage.
Coupling capacitor C.
This is used to couple the amplified signal to the output device. This capacitor C allows only ac signal to flow.
Working.
When a weak input ac signal is applied to thebase of the transistor, a small base current flows. Due to transistor action, a
much larger ac current flows through collector load RC, a large voltage appears across RC and hence at the output.
Therefore, a weak signal applied to the base appears in amplified form in the collector circuit.
Voltage gain of the amplifier is the ratio of the amplified output to the input voltage.
83
Frequency response and bandwidth.
A graph is drawn between frequency (f) along the x-axis and voltage gain along the y-axis.
It is seen that the gain decreases at low and high frequencies but it remains constant in mid frequency region.
Lower cut off frequency is defined as the frequency in the low frequency range at which the gain of the amplifier is
1
2 times the mid frequency gain.
Upper cut off frequency is defined as the frequency in the high frequency range at which the gain of the amplifier is
1
2 times the mid frequency gain.
Band width is defined as the frequency interval between cut off and upper cut off frequencies.
4. Explain the principle of feedback amplifier. (Mar-09)
Feedback is said to exist in an amplifier circuit , when a fraction of the output signal is fed back to the input and
combined with the input signal.
If the magnitude of the input signal is reduced by the feed back, the feedback is called negative or degenerative.
If the magnitude of the output signal is increased by the feed back, the feedback is called positive or regenerative.
For an ordinary amplifier without feedback , let V0 and Vi be the output and
input voltage respectively.
The voltage gain of the amplifier
0
i
VA
V
The feedback amplifier has two parts.
1. Amplifier and
2. feedback circuit.
The feedback circuit consist of passive components. A fraction of the output
voltage is fed back to the input through the feedback circuit . Let V0’ be the output
voltage with feedback.
After feedback the input voltage is
' '
0( )i iV A V V
For positive feedback β is taken as positive. For negative feedback β is negative.
For positive feedback the input voltage will be '
0( )iV V when this is amplified A
times by the amplifier the output voltage will be '
0( )iA V V
' '
0 0( )iV A V V
'
0 1 iV A AV
The voltage gain after feedback is
'
0
1f
i
V AA
V A
84
since1 1, fA A A The positive feedback increases the amplifier gain.
For negative feedback the feedback fraction is -β
1 ( ) 1
1 1,
f
f
A AA
A A
A A
The negative feedback reduces the amplifier gain.
5. Explain the working of colpitt’s amplifier.(Ju-06,Oct-06,M-08,Jun-09,J-11)
An oscillator is defined as electronic circuit which converts dc source into a periodically varying output.
Description.
The resistance of R1,R2 and R3 provide the bias for the circuit.
The frequency determining network is the parallel resonant circuits consisting of capacitors C1,C2 and the inductor L.
The function of the capacitor C4 is to block dc and provide an ac path from the collector to the tank circuit.
The voltage developed across C1 provides the positive feedback for sustained oscillations.
Working.
When the collector supply voltage is switched on, a transient current is produced in the tank circuit and damped
harmonic oscillations are produced. The oscillations across C1 are applied to the base emitter junction and appear in the
amplified form in the collector circuit.
If terminal 1 is at positive potential with respect to terminal 3 at any instant, then terminal 2 will be at negative potential
with respect to 3, since 3 is grounded.
Hence 1 and 2 are 1800 out of phase.The amplifier produces further phase shift of 1800. Thus the total phase shift is 3600.
Energy supplied to the tank circuit is in phase with the oscillations and
Aβ=1, oscillations are sustained in the circuit.
6. Explain op-amp as inverting amplifier and non inverting amplifier.(Oct-07)
Inverting amplifier.
The input voltage Vin is applied to the inverting input through the input resistor Rin. The non inverting input is grounded.
The feed back resistor Rf is connected between the output and the inverting input.
Since the input impedance of an op-amp is considered very high, no current can flow in to or out of the input terminals.
Therefore Iin must flow through Rf and is indicated by If. Since Rin and Rf are in series, then Iin = If.
The voltage between inverting and non-inverting inputs is equal to zero volt. Therefore, the inverting input is also at 0
volt. For this reason the inverting input is said to be a virtual ground.
1 2
1 2
1 2
1 2
1
2
1
2
fLC
C CC
C C
C Cf
C C
85
0
,o inf s
f S
s f
in o
S f
f
in S
f
v
S
V VI I
R R
I I
V V
R R
RV
V R
RA
R
Non inverting amplifier.
The input signal Vin is applied to the non inverting input terminal. The resistor Rin is connected From the inverting input
to ground. The feedback resistor Rf is
connected between the output and the inverting input.
Since the potential at the inverting input is same as the non inverting input
0
A in
v
A
V V
VA
V
0 1
sA o
f s
f
A s
RV V
R R
RV
V R
1f
v
s
RA
R
The input and output voltages are in phase.
7. Explain the working summing amplifier and difference amplifier.
(Mar-07,Oct-07)
Summing amplifier.
The output voltage is equal to the algebraic sum of the input voltages.
The input voltage v1 and v2 are applied through the resistors R1 and R2 to the summing junction and Rf is the feed back
resistor.
At the point P,
1 2 fi i i
1 2 0
1 2
.f
V V V
R R R
0 1 2
1 2
f fR RV V V
R R
1 2
0 1 2( )
fR R R R
V V V
Difference amplifier.
The output voltage can be obtained by using superposition principle.
86
3 401
3
21
1 2
R RV V
R
RV V
R R
3 4 201 1
3 1 2
R R RV V
R R R
402 2
3
RV V
R
0 01 02
3 4 2 40 1 2
3 1 2 3
1 2 3 4
0 1 2
V V V
R R R RV V V
R R R R
R R R R R
V V V
The voltage amplifier functions as a voltage substractor.
10. COMMUNICATION.
1. Define skip distance.(Jun-07,Mar-08,Jun-09,Oct-10,Mar-11)
For a fixed frequency, the shortest distance between the point of transmission and the point of reception along the surface
is known as skip distance.
2. Define amplitude modulation.(Mar-06,Jun-08,Oct-09,Mar-10)
When the amplitude of high frequency carrier wave is changed in accordance with the intensity of the signal, the process
is called amplitude modulation.
3. Define modulation factor.
It is defined as the ratio of the change of amplitude in carrier wave after modulation to the amplitude of the unmodulated
carrier wave.
m=signal amplitude/carrier amplitude.
4. What are the advantages of amplitude modulation? (Mar-09)
1. Easy transmission and reception.
2. Lesser bandwidth
3. Low cost.
5. What is frequency modulation?
When the frequency of carrier wave is changed in accordance with the intensity of the signal, the process is called
frequency modulation .
6. What are the advantages of frequency modulation?
1. It gives noiseless reception.
2. The operating range is quiet large.
3. The efficiency of transmission is very high.
7. What is fax of facsimile?(Oct-06)
Fax is an electronic system for transmitting graphical information by wire or radio
8. What are the pulses are used for scanning?
The saw tooth potential applied to the horizontal plates called line synchronizing pulse.
The blanking pulse which is a high negative potential is applied to the control grid.
The saw tooth potential applied to the vertical plates is called frame synchronizing pulse.
87
9. What is modem?
Modem acts as modulator and demodulator. A modem is used to convert digital signals into analog signals at
transmission end. At the receiving end of the system it converts analog signals to digital signals.
10. Define directivity.
Directivity is the ability of the antenna to concentrate the electromagnetic waves in the most desired directions(during
transmission) or to have maximum reception from most preferred directions(during reception).
11. What are the advantages of fibre optic communication?(Oct-08)
1. Transmission loss is low.
2. Fiber is lighter and less bulky than equivalent copper cable.
3. More information can be carried by each fiber than by equivalent copper cables.
12. What are the advantages of digital communication?(Mar-07,Jun-10)
The transmission quality is high and almost independent of the distance between the terminals.
The capacity of the transmission system can be increased.
The newer types of transmission media such as light beams in optical fibers and wave guides operating in the microwave
frequency extensively use digital communication.
13. What are the application of radar?(Jun-06)
1. They are used for the safety landing of air crafts.
2. The pulses can be used for discovering the position of buried metals oils and ores.
3. High flying planes and ship at sea can get detailed reports of mountains, ice bergs, lakes etc. which they can avoid.
10. COMMUNICATION.
1. Explain AM transmitter.(Oct-07,Sep-10)
AF section.
The AF section of the transmitter the modulating wave. The conversion of sound energy into electrical energy is
performed by the microphone.
The electrical energy is amplified through an amplifier. The output from the AF amplifier is fed to the AF power
amplifier. The output of the AF power amplifier is given to the modulator.
RF section.
The high frequency carrier wave is generated by a crystal controlled oscillator. The output of the crystal controlled
oscillator is power amplified by RF power amplifier. The buffer isolates the RF power amplifier from the oscillator. This
arrangement keeps the frequency of the crystal controlled oscillator as a constant.
In the modulator the RF wave and modulating AF signal are mixed to produce the amplitude modulated wave. The
output of this section is fed to the antenna for transmission.
88
2. Explain the applications of radar communication.(Mar-10)
The principle of radar is radio echoes.
1. Air and sea navigation is made entirely safe, with radar installations. High flying planes and ship at sea, can get
detailed of mountains, ice , rivers lakes which they can avoid.
2. Radar systems are used for the safe landing of air crafts.
3. Rain drops may reflect suitable radar signals and thus enable meteorologists to measure the distance of the clouds,
with great accuracy for forecasting.
4. The pulses can be used for discovering the position of buried metals, oil and ores.
3. Explain the advantages and disadvantages of digital communication.(Mar-06,Jun-06,Jun-08,Jun-09)
Advantages.
The transmission quality is high and almost independent of the distance between the terminals.
The capacity of the transmission system can be increased.
The newer types of transmission media such as light beams in optical fibres and wave guides operating in the microwave
frequency extensively use digital communication.
Disadvantages.
A digital system requires larger bandwidth.
It is very difficult to gradually change over from analog to digital transmission.
4. Explain the merits of satellite communication.
Merits.
1. Mobile communication can be easily established by satellite communication.
2. Satellite communication is economical compared with terrestrial communication
3. Compared to the optical fibre communication, satellite communication has the advantages that, quality of transmitted
signal and location of sending and receiving stations are independent of distance.
4. For thin traffic remote areas like north east regions in India, ladakh satellite communication is most economical.
5. For search, rescue and navigation satellite communication is far superior and economical compared to other systems.
Demerits.
1.Between talks there is a time gap which becomes quite annoying. This time decay also reduces the efficiency of
satellite in data transmission.
2. An imperfect impedance match may cause echo, received back after a delay.
3. Repair of satellite is almost impossible, once it has been launched.
5. Explain FM radio transmitter. (Sep-08,Sep-09,Jun-10,Jun-11)
Frequency modulated systems are operated usually at a frequency above
40 MHz.
Frequency modulated broadcasting is done in television sound, mobile radio etc. The functional block diagram of a FM
transmitter employing phase modulation.The phase modulation is essentially a frequency modulation.
It consist of a crystal oscillator, which produces the carrier wave and the output of this is fed into the phase modulator.
The buffer is a low frequency amplifier which isolates the crystal oscillator from the phase modulator.
The modulating signal is produced from a microphone. Since this AF modulating signal has uneven power, it is fed into a
network called preemphasis network, where all the frequencies in the modulating signal are made to have equal power.
The output of the preemphasis network is then amplified and sent for phase modulation. The modulated output is then
power amplifier using a phase modulation and then fed into the transmitting antenna for transmission.
89
10.COMMUNICATION SYSTEMS.
1. Explain the analysis of amplitude modulated wave.(Jun-06,Oct-06,Mar-08,Mar-09,Jun-10,Sep-10)
When the amplitude ofhigh frequency carrier wave is changed in accordance with the intensity of the signal, the process
is called amplitude modulation.
Modulation factor.
It is defined as the ratio of the change of amplitude in carrier wave after modulation to the amplitude of the
unmodulated carries wave.
Analysis
A carrier wave may be represented by c c ce E Cos t
The modulating signal is s s se E Cos t
The amplitude modulated wave is c s s ce E E Cos t Cos t
1
1
sc s c
c
c s c
c c c s c
Ee E Cos t Cos t
E
e E mCos t Cos t
e E Cos t mE Cos tCos t
22
2
cc c s c
cc c c s c s
mEe E Cos t Cos tCos t
mEe E Cos t Cos t Cos t
2 2
c cc c c s c s
mE mEe E Cos t Cos t Cos t
c cE Cos t This is carrier wave.
2
cc s
mECos t This is upper side side band.
2
cc s
mECos t this is lower side band.
The magnitude of both the upper and lower side bands is m/2 times the carrier amplitude EC.
The channel width is given bythe difference between the extreme frequencies i.e between the maximum frequency
of USB and minimum frequency of LSB.
Channel width= 2 x maximum frequency of the modulating signal. = 2 x (fs) max.
90
2. Explain the function of superheterodyne AM receiver.(Jun-07,Mar-11)
RF amplifier.
The RF amplifier uses a tuned parallel circuit. The radiowaves from various broadcasting stations are interrupted by the
receiving antenna and are coupled to this stage.
Mixer and local oscillator.
The amplified output of RF amplifier is fed to the mixer stage, where it is combined with the output of a local oscillator.
The two frequencies beat together and produce an intermediate frequency. The intermediate frequency is the difference
between oscillator frequency and radio frequency. The output of this section is always equal to the intermediate
frequency 455 KHz.
IF amplifier.
The output of the mixer circuit is fed to the tuned IF amplifier. This amplifier is tuned to one frequency (455KHz) and is
amplified.
Detector.
The output from the amplifier is coupled with input of a detector. The audio signals are extracted from the IF output.
AF amplifier.
The detected AF signal is usually weak and so it is further amplified by the AF amplifier. Then the output signal from
the amplifier is fed to the loud speaker, which converts the audio signal into sound waves corresponding to the original
sound at the broadcasting station.
3. Explain monochrome TV transmission with a block diagram.(J-08,Sep-09)
The functional block diagram can be broadly divided into two sections. 1. Amplitude modulated transmitter used for
video modulation and 2.frequency modulated transmitter used for radio modulation.
The synchronizing and scanning circuits produce sets of pulses for providing synchronizing pulses for proper
functioning of TV system.
This timing unit contains of wave generating and wave shaping circuits.
The repetition rate of its various output pulse trains is controlled by a frequency stabilized master of oscillator.
The image signals together with the synchronizing and blanking pulse are raised to a level suitable for modulating the RF
carrier wave generated in the RF channel. The picture carrier frequency is generated by the crystal controlled oscillator.
Its amplitude is made to vary in accordance with the modulating signal received from the modulating amplifier.
The microphone converts the sound associated with the picture being televised into electrical signal. The audio signal
from the microphone after amplification is frequency modulated. The output of the sound FM transmitter is finally
combined with the AM picture transmitter output, through a combining network and fed to a common antenna for
radiation of energy in the form of electromagnetic waves.
91
4. Explain the monochrome TV receiver with a block diagram.(Mar-06,Mar-07,Sep-08,Mar-10)
The receiving antenna interrupts radiated RF signals and the tuner selects the desired channel frequency band.
The antenna provides RF picture and sound signals for the RF amplifier stage. The RF amplifier stage is coupled into the
mixer stage. The mixer is connected to the local oscillator. The RF audio and video signals are heterodyned into
intermediate frequency by the mixer and local oscillator. The RF amplifier,mixer and local oscillator are combinely
called as RF tuner.
The output signal from the tuner circuit is amplified by using a common If amplifier. Then the video and audio
components are separated by a detector.
The sound signals are detected from FM waves and then fed into the loud speaker, which reproduce the sound.
The video components are first passed into a detector which separates the picture signal from the synchronizing pulses.
The line synchronizing pulses and the frame synchronizing pulses are fed into the horizontal and vertical deflector plates
of the picture tube. The blanking pulse are given to the control grid of the electron gun of the picture tube.
The picture signals are applied to the filament of the electron gun of the picture tube. According to the variations of
potential in the picture, electrons are emitted from the electron gun.
Thus the intensity of the fluorescent screen of the picture tube is in accordance with the variation of potential in the
picture and the picture is reproduced.
5. Explain the transmission and reception of radar.(Oct-07,Oct-11)
The transmitting system consists of a transmitter and a pulser. The receiving system consists of a receiver and an
indicator.
A single antenna is used for both transmission and reception and this is achieved by TR switch. This switching
arrangement iscalled as duplexer. This connects the antenna to transmitter during transmission and to the receiver during
reception.
The transmitter is a high power magnetron oscillator which generates high power pulses. This transmitter is turned on
and off with a periodic pulse from the pulser. Thus the transmitter generates periodic pulses of very short duration. These
short pulses are fed to the antenna which radiates them into the space. The antenna is highly directional.
If the transmitted pulse hits any target, a weak echo signal returns to the same antenna.
Now the TR switch is in contact with the receiver. This echo signal is amplified and demodulated by the superheterodyne
receiver.
The detected output is sent to the indicator. The indicator is a cathode ray tube. The CRT displays the original
transmitted pulse as well as the detected echo pulse along a horizontal base line.
The synchronizing pulse generated by the timer is supplied
to both transmitting and receiving systems. The indicator
records the transmitter pulse as well as the returning pulse
simultaneously.
92
3 marks PROBLEMS
ELECTROSTATICS.
1. Two charges +q and -3q are separated by a distance of 1m. At what point on its axis is the potential
zero?
Solution:
V1+V2 =0
1 2
0 0
0 0
0 0
04 4 (1 )
30
4 4 (1 )
3
4 4 (1 )
1 3
1
1 3
1 4
10.25
4
q q
x x
q q
x x
q q
x x
x x
x x
x
x m
The potential is zero at a distance of 0.25m from the charge q..
2. Find the total flux through each face of a hollow cube of side 10cm, if acharge of 8.85μC is placed at
the centre.
Solution:
By Gauss law, the electric flux is
0
6
12
6 2 1
8.854 10
8.854 10
10
q
x
x
Nm C
The electric flux through each face is
65 2 110
1.67 106
x Nm C
3. The area of each plate of a parallel plate capacitor is 4 x 10-2 m2. If the thickness of the dielectric
medium between the plates is 10-3m and the relative permittivity of the dielectric is 7. Find the
capacitance of the capacitor.
0
2 12
3
9
4 10 8.854 10 7
10
2.478 10
rAC
d
x x x xC
C x F
4. Three small identical balls have charges -3 x 10-12
C, 8 x 10-12
C and 4 x 10-12
C respectively. They are
brought in contact and then separated . Calculate (i) charge on each ball (ii) number of electrons in
excess or deficit on each ball after contact.
93
12
1 2 3
12
( 3 8 4) 10
3 3
3 10
q q q xq
q x C
Since the charge is positive, there is a shortage of electrons on each ball. 12
7
19
3 101.875 10
1.6 10
q xn x
e x
5. Two point charges +9e and +1e are kept at a distance of 16cm from each other. At what point
between these charges, should a third charge q to be placed so that it remains in equilibrium?
1 2
2
0
2 2
0 0
22
1
4
1 9 1
4 4 (16 )
9 1
16
3 1
16
48 3
4 48
4812
4
q qF
r
eq eq
x x
x x
x x
x x
x
x cm
The third charge should be placed at a distance of 12cm from the charge 9e
6. An infinite line charge produces a field of 9 x 104 N/C at a distance of 2 cm. Calculate the linear charge
density.
0
0
4 2
09 9
7 1
2
2
1 19 10 2 10 2
18 10 18 10
10
Er
Ex r
x x x xx x
Cm
7. A parallel plate capacitor with air between the plates has a capacitance of 8 PF. What will be the
capacitance, if the distance between the plates be reduced to half and the space between them is
filled with the substance of dielectric constant 6.
0
0
0
8
/ 2
2
2 6 8 96
r
r
AC PF
d
AC
d
AC
d
C x x PF PF
94
1. If 6.25 x 1018
electrons flow through a given cross section in unit time, find the current.
18 196.25 10 1.6 101
1
q neI
t t
x x xI A
2. An incandescent lamp is operated at 240V and the current is 0.5A. What is the resistance of the lamp?
V = IR
240480
0.5
VR
I
3. The resistance of a nichrome wire at 00C is 10Ω. If its temperature coefficient of resistance is 0.04/0C.
Find its resistance at boiling point of water.
Rt = R0 (1+αt)
= 10 (1+0.004x100)
= 10(1+0.4)
= 10 x 1.4
= 14 Ω
4. Two wires of same material and length have resistanes 5Ω and 10Ω respectively. Find the ratio of radii
of the two wires.
2
1 2
1
2 2
2
2
2 1
2
1 2
1 2
2 1
1 2
10 2
5 1
: 2 :1
l lR
A r
lR
r
lR
r
R r
R r
r R
r R
r r
5. The resistance of a field coil measures 50Ω at 200C and 65Ω at 700C. Find the temperature coefficient of
resistance.
2 1
1 2 2 1
0
65 50
50 70 65 20
15
3500 1300
150.0068 /
2200
R R
R t R t
x x
C
95
6. A 10Ω resistance is connected in series with a cell of emf 10V. A Voltmeter is connected in parallel to a
cell and it reads 9.9 V. Find the internal resistance of the cell.
10 9.910 0.101
9.9
E Vr R
V
r x
7. What is the drift velocity ofan electron in a copper conductor having area 10 x 10-6 m2 carrying a
current of 2 A. Assume that there are 10 x 1028 electrons/m3.
28 19 6
45
2
10 10 1.6 10 10 10
2 101.25 10 /
1.6
d
d
d
I nAeV
IV
nAe x x x x x
xV x m s
8. How much time 1028 electrons will take to flow through a point, so that the current is 200 mA?
20 19
3
2
10 1.6 10
200 10
1.6 10 16080 .
2 2
q neI
t t
ne x xt
t x
xt s
9. The resistance of a platinum wire at 00C is 4Ω. What will be the resistance of the wire at 1000C. If the
temperature coefficient of resistance of platinum is 0.0038/0C.
Rt =R0(1+αt)
=4(1+0.0038x100)
= 4(1+0.38)
=4 x 1.38
= 5.52Ω
10. Acell has a potential difference of 6 V in an open circuit, but it falls to 4 V when a current of 2 A is
drawn from it. Find the internal resistance of the cell.
6 4 4
4 2
1
E Vr R
V
E V Vr
V I
r
r
11. An electric iron of resistance 80Ω is operated at 200 V for two hours. Find the electrical energy
consumed. 2 200 200 2
80
200 5 1000 .
1 .
V t x xE
R
E x Whr
E KWhr
96
12. In a wheatstone bridge if the galvanometer shows zero deflection find the unknown resistance.
Given P=1000Ω,Q=10000Ω R=20Ω
1000 20
10000
20 10000
1000
200
P R
Q S
S
xS
S
13. A 1.5 V carbon-Zinc dry cell is connected across a load of 1000Ω. Calculate the current and power
supplied to it.
3
1.5
1000
1.5 10 1.5
VI
R
I x A mA
.
2
3
1.5 1.5
1000
2.25 10 2.25
V xP
R
P x W mW
EFFECTS OF CURRENT.
1. Calculate the resistance of the filament of a 100 W, 200 V electric bulb.
2
2 200 200
100
484
VP
R
V xR
P
R
2. A water heater is marked 1500 W, 220 V. If the voltage drops to 180 V. Calculate the power consumed
by the heater.
2 2
1 21 2
2
1 1
2
2 2
2
22 1 2
1
2
2 2
2
,
1801500
220
1004
V VP P
R R
P V
P V
VP P
V
P x
P W
3. A long straight wire carrying current produces a magnetic induction of 4 x 10-6 T at a point, 15 cm from
the wire. Calculate the current through the wire.
97
0
0
6 2
7
2
2
4 10 2 15 103
4 10
IB
a
Bx aI
x x x xI A
x
4. In a tangent galvanometer, a current of 1 A produces a deflection of 300. Find the current required to
produce a deflection of 600.
2 2
1 1
22 1
1
0
2 0
2
tan
tan
tan
tan
tan 60 31
tan30 1 / 3
3 3 3
I
I
I I x
I x
I A
5. A solenoid is 2 m and 3 cm in diameter. It has 5 layers of windings of 1000 turns each and carries a
current of 5A. Find the magnetic induction at its centre along its axis.
0 0
7
2
4 10 5000 5
2
1.57 10
NB nI I
l
x x xB
B x T
6. In a thermocouple, the temperature of the cold junction is -200C and the temperature of inversion is
6000C. If the temperature of the cold junction is 200C, Find the temperature of inversion.
0
0 0 0
2
20 600
2
580290
2
2
2 290 20 580 20 560
C in
n
n
i n C
i
T TT
T
T C
T T T
T x C
7. Find the magnetic induction at a point 10 cm from a long straight wire carrying a current of 10 A.
0
75
2
2
4 102 10
2 10 10
IB
a
xB x T
x x
8. A long solenoid of length 3 m has 4000 turns. Find the current through the solenoid if the magnetic
field produced at the centre of the solenoid along its axis is 1.8 x 10-3 T.
98
0 0
0
3
7 3
8 10 3
4 10 4 10
3 10 304.77
2 2
NB nI I
l
BlI
N
x xI
x x x
xI A
4.ALTERNATING CURRENT.
1. Magnetic field through a coil having 200 turns and cross sectional area 0.04 m2 changes from 0.1 wb m
-
2 to 0.04 wb m
-2 in 0.02 s. Find the induced emf.
2
( )
(0.04 0.1)200 4 10
0.02
24
d d NBAe
dt dt
dBe NA x x
dt
e V
2. An air craft having a wingspan of 20.48 m flies due north at a speed of 40 m/s. If the vertical
component of earth’s magnetic field at the place is 2 x 10-5 T. Calculate the emf induced between the
ends of the wings.
e = - Blv
= - 2 x 10-5 x 20.48 x 40 = - 0.0164 V
3. Calculate the mutual inductance between two coils when a current of 4 A changing to 8A in 0.5 s in one
coil, induces an emf of 50 mV in the other coil.
3
33
/
50 10
8 4
0.5
50 10 0.56.25 10
4
6.25
dIe M
dt
eM
dI dt
xM
x xM x H
M mH
4. Write the equation of a 25 cycle current sine wave having rms valve of 30 A.
0
2 2
30 2 2 25
30 1.414 157
42.42 157
rms
I I Sin t
I I Sin t
I Sin x t
I x Sin t
I Sin t
5. Two rails of a railway track insulated from each other and the ground are connected to a
millivoltmeter. The train runs at a speed of 180 km/hr. Vertical component of earth magnetic field is
0.2 x 10-4 wbm-2 and the rails are separated by 1m. Find the reading of the voltmeter.
99
Speed of the train V= 180 km/hr = 180x1000/3600=50m/s.
e = - B l v
= - 0.2 x 10-4
x 1 x 50
= - 10-3
V = -1 mV
5.ELECTRO MAGNETIC WAVES AND WAVE OPTICS.
1. In young’s experiment the width of the fringes obtained with light of wavelength 6000A0 is 2 mm.
Calculate the fringe width if the entire apparatus is immersed in a liquid of refractive index 1.33.
3 3
3
'' ( )
'
2 10 2 10 3'
4 / 3 4
' 1.5 10 1.5
D
d
D D
d d
x x x
x m mm
2. A plano convex lens of radius 3m is placed on an optically flat glass plate and is illuminated by
monochromatic light. The radius of the 8th dark ring is3.6mm. Calculate the wavelength of light used.
2
232
6 10
0
3.6 10
8 3
0.54 10 5400 10
5400
n
n
r nR
xr
nR x
x m x m
A
3. A 300 mm long tube containing 60cc of sugar solution produces a rotation of 900 when placed in a
polarimeter. If the specific rotation is 600, calculate the quantity of sugar contained in the solution.
( / )
9 603 .
3 60
SlxC lx m v
v xm gm
Sl x
4. In Newton ring experiment the diameter of certain order of dark ring is measured to be double that of
second ring. What is the order of the ring?
2
2
2
2
2
2
2
2
2
2
2
2
4 (1)
8 (2)
(1) / (2)
4
8 2
4
2
8
n
n
n
d d
d nR
d R
d nR n
d R
d n
d
n
100
5. The fringe width obtained in young’s double slit experiment while using a light of wavelength 5000A0 is
0.6cm. If the distance between the slit and the screen is halved, find the new fringe width.
D’=D/2
''
2
0.6' 0.3
2 2
D D
d d
cmcm
6. A light of wavelength 6000A0 falls normally on a thin air film, 6 dark fringes are seen between two
points. Calculate the thickness of the air film.
For dark fringe 2t=nλ
107
2
6 6000 1018 10
2
nt
x xt x m
7. The refractive index of the medium is 3 . Calculate the angle of refraction if the unpolarised light is
incident on it at the polarizing angle of the medium.
0
0
0 0
0
tan
tan 3
60
90
90 60
30
P
P
P
p
i
i
i
r i
r
r
8. A 20 cm long tube contains sugar solution of unknown strength. When observed through polarimeter,
the plane of polarization is rotated through 100. Find the strength of sugar solution in g/cc. Specific
rotation of sugar is 600/decimeter /unit concentration.
0
0
10 1
2 60 12
0.0833 /
SlC
ClC x
C g cc
6.ATOMIC PHYSICS.
1. A beam of electrons moving with a uniform speed of 4 x 107 m/s is projected normal to the uniform
magnetic field where B= 1 x 10-3 wb/m2. What is the path of the beam in magnetic field? 2
31 7
3 19
22
9.1 10 4 10
1 10 1.6 10
9 4 1022.75 10
1.6
0.2275
mvBev
r
mvr
Be
x x xr
x x x
x xr x m
r m
101
2. Calculate the longest wavelength that can be analysed by a rock salt crystal of spacing d = 2.82A0. in the
first order.
For longest wavelength Sinθ=1
2dSinθ=nλ
0
0
2
2 2.82
1
5.64
dSin
n
x A
A
3. An x-ray diffraction of a crystal gave the closest line at an angle of 6027’. If the wavelength of x-ray is
0.58A0, find the distance between the two cleavage planes.
2dSinθ = nλ
10
0
10 10
10
2
1 0.58 10
2 6 27'
0.58 10 0.58 10
2 0.1124 0.2247
2.581 10
nd
Sin
x xd
xSin
x xd
x
d x m
4. How much should be the voltage of an x-ray tube so that the electrons emitted from the cathode may
give an x-ray of wavelength 1 A0 after striking the target. 0
10
10
10
12400
12400 10
12400 10
1 10
12400 12.4
A
V
xV
xV
x
V V kV
5. Calculate the mass of an electron from the known valves of specific charge and charge of electron.
Mass of electron=charge of the electron/specific charge of the electron. 19
31
11
1.602 109.107 10 .
1.759 10
xm x kg
x
6. Find the minimum wavelength of x-rays produced by an x-ray tube at 1000kV. 0
0 10
3
10 0
12400
12400 12400 10.
1000 1000 10
0.0124 10 0.0124
A
V
A x
kV x
x A
102
7.DUAL NATURE OF RADIATION AND MATTER AND RELATIVITY.
1. The work function of Zinc is 6.8 x 10-19 J. What is the thereshold frequency for emission of
photoelectrons from zinc?
0
0
19
0 34
15
0
.
6.5 10
6.626 10
1.026 10 .
W h
W
h
x
x
x Hz
2. Calculate the deBroglie wavelength of an electron, if the speed is 105m/s.
34
31 5
88 0
6.624 10
9.1 10 10
6.626 100.7281 10 72.81
9.1
h
mv
x
x x
xx A
3. What is the deBroglie wavelength of an electron of kinetic energy 120 eV?
34
31 19
10
2
6.626 10
2 9.1 10 120 1.6 10
1.121 10
h
mE
x
x x x x x
x m
4. Find the accelerating potential of the electron, when its deBroglie wavelength is 1 A0.
0
10
10
10
2
12.27
12.27 10
12.27 10
1 10
12.27
12.27 150.55
A
V
xV
xV
x
V
V V
5. Calculate the rest energy of an electron in MeV.
E= m0C2
103
E = 9.1 x 10-31
x (3 x108)
2
= 9.1 x 10-31 x 9 x 1016
= 81.9 x 10-15
J 15
4
19
6
81.9 1051.18 10
1.6 10
0.5118 10 0.5118
xE x
x
E x eV MeV
8.NUCLEAR PHYSICS.
1. The half life of 84Po218 is 3 minute. What percentage of the sample has decayed in 15 minutes.
Sample decayed in 3 minutes = 50%
Decayed after 6 minutes = 25%
Decayed after 9 minutes = 12.5%
Decayed after 12 minutes = 6.25%
Decayed after 15 minutes= 3.125%
The sample decayed in 15 minutes = 96.875%.
2. Calculate the radius of 13Al27.
R= r0A1/3
R= 1.3 x 10-15
x(27)1/3
R= 1.3 x10-15 x 3
R= 3.9 x 10-15m
3. Tritium has a half life of 12.5 years. What fraction of the sample of will be left over after 25 years?
Fraction of the sample decayed in 12.5 years = ½
Decayed after 25 years = 1/4
Sample left after 25 years = 1/4.
4. The disintegration constant λ of a radioactive element is 0.00231/day. Calculate the half life and
mean life.
Half life T1/2=0.6931/λ
=0.6931/0.00231
= 300 days.
Mean life τ = 1/λ
τ= 1/0.00231
τ = 432.9 days.
5. A carbon specimen found in a cave contained a fraction of 1/8 of C14 to that present in a living
system. Calculate the approximate age of the specimen. Given T1/2 = 5560 years.
Fraction left after I half life = ½
Fraction left afte II half life = ¼
Fraction left after III half life = 1/8.
The age of the specimen = 3 x half life
3 x 5560 = 16681 years.
6. The half life of radon is 3.8 days. Calculate its mean life.
T1/2 = 0.6931τ
τ = 0.6931/T1/2
= 0.6931/3.8 = 5.482 days.
7. If 50% of the sample decays in 5 days, how much of the original sample will be left over after 20
days?
104
T1/2 = 5 days.
Sample left after 5 days = 50%
Sample left after 10 days = 25%
Sample left after 15 days = 12.5%
Sample left after 20 days = 6.25%
8. The isotope 84Po214 undergoes a successive disintegration of two α decays and two β decays. Find
the atomic number and mass number of the resulting isotope.
84Po214 2α 80X206 2β 82 Y
206
9. The isotope 92U238 undergoes successively undergoes three α decays and two β decays. What is the
resulting isotope?
92U238
3α 86X226
2β 88Y226
Resulting isotope is 88Ra226
.
9.ELECTRONICS.
1. When the negative feedback is appliedto an amplifier of gain 50, the gain after feedback falls to 25.
Calculate the feedback ratio.
A=50 Af =25
1
5025
1 50
501 50
25
1 50 2
50 1
10.02
50
f
AA
A
2. Prove the Boolean identity (A+B)(A+C) =A+BC
(A+B)(A+C) = AA+ AC+BA+BC
= A +AC +BA+BC ( AA = A)
=A(1+C+B)+BC
=A + BC (1+C+B =1)
3. The base current of the transistor is 50μA and collector current is 25 mA. Determine the valves of α
and β 3
6
25 10 1000500
50 10 2
500 5000.998
1 1 500 501
C
B
I x
I x
4. The gain of the amplifier is 100. If 5% of the output voltage is fed back into the input through a
negative feedback network. Find out the voltage gain after feed back.
105
1
100 10016.66
5 61 100
100
f
f
AA
A
A
x
5. Prove the following logic expression using the laws and theorm of Boolean algebra.
( )
:
( )
( 1)
( ) ( )( _)
ABC ABC ABC A B C
Solution
AC B B ABC
AC ABC B B
A C BC A C B C BC C B
6. Simplify the following logic expression using Boolean algebra.
:
( )
( 1)
(1 )
( 1 1)
Y AB AB BC CA
Solution
AB AB BC CA
A B B BC CA
A BC CA B B
A C BC
A BC C
10.COMMUNICATION.
1. An FM signal has a resting frequency of 105 MHz and highest frequency of 105.03 MHz when
modulated by a signal. Determine (i) frequency deviation (ii) carrier swing
Frequency deviation = fm – f
= 105.03-105 = 0.03 MHz.
Carrier swing = 2 x ∆f = 2 x 0.03 = 0.06 MHz.
Self evaluation
1.ELECTRO STATICS
1. A glass rod rubbed with silk acquires a charge of +8 × 10−12C. The number of electrons it has gained or lost
(a) 5 × 10-7 (gained) (b) 5 × 107 (lost)
(c) 2 × 10-8
(lost) (d) –8 × 10-12
(lost)
106
2. The electrostatic force between two point charges kept at a distance d apart, in a medium εr = 6, is 0.3 N. The force
between them at the same separation in vacuum is
(a) 20 N (b) 0.5 N (c) 1.8 N (d) 2 N
3. Electr
1 23
1 2
2 2 2
1 2 3
ic field intensity is 400 V m−1 at a distance of 2 m from a point charge. It will be 100 V m−1 at a
distance?
(a) 50 cm (b) 4 cm (c) 4 m (d) 1.5 m
4. Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining them the electric field is zero?
(a) 15 cm from the charge q (b) 7.5 cm from the charge q
(c) 20 cm from the charge 4q (d) 5 cm from the charge q
5. A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences
(a) only a net force (b) only a torque
(c) both a net force and torque (d) neither a net force nor a torque
6. If a point lies at a distance x from the midpoint of the dipole, the electric potential at this point is proportional to
(a)1/x2 (b)1/x
3 (c)1/x
4 (d)1/x
3/2
7. Four charges +q, +q, −q and –q respectively are placed at the corners A, B, C and D of a square of side a. The electric
potential at the centre O of the square is
(a)
04
q
a (b)
0
2
4
q
a (c)
0
4
4
q
a (d) zero
8. Electric potential energy (U) of two point charges is
(a) 1 2
04
q q
r (b) 1 2
2
04
q q
r (c) pE cos θ (d) pE sin θ
9. The work done in moving 500 μC charge between two points on equipotential surface is
(a) zero (b) finite positive (c) finite negative (d) infinite
10. The unit of permittivity is
(a) C2 N-1 m-2 (b) N m2 C-2 (c) H m-1 (d) N C-2 m-2
11. Which of the following quantities is scalar?
(a) dipole moment (b) electric force
(c) electric field (d) electric potential
12. The number of electric lines of force originating from a charge of 1 C is
(a) 1.129 × 1011 (b) 1.6 × 10`-19 (c) 6.25 × 1018 (d) 8.85 × 1012
13. The electric field outside the plates of two oppositely charged plane
sheets of charge density σ is
(a)
0
(b)
0
2
(c)
02
(d) zero
14. The capacitance of a parallel plate capacitor increases from 5 μf to 60 μf when a dielectric is filled between the plates.
The dielectric constant of the dielectric is
(a) 65 (b) 55 (c) 12 (d) 10
15. A hollow metal ball carrying an electric charge produces no electric field at points
(a) outside the sphere (b) on its surface
(c) inside the sphere (d) at a distance more than twice
2. CURRENT ELECTRICITY.
1. A charge of 60 C passes through an electric lamp in 2 minutes.Then the current in the lamp is
(a) 30 A (b) 1 A (c) 0.5 A (d) 5 A
2. The material through which electric charge can flow easily is
(a) quartz (b) mica (c) germanium (d) copper
3. The current flowing in a conductor is proportional to
(a) drift velocity (b) 1/ area of cross section
107
(c) 1/no of electrons (d) square of area of cross section.
4. A toaster operating at 240V has a resistance of 120Ω. The power is
(a) 400 W (b) 2 W (c) 480 W (d) 240 W
5. If the length of a copper wire has a certain resistance R, then on doubling the length its specific resistance
(a) will be doubled (b) will become 1/4th
(c) will become 4 times (d) will remain the same..
6. When two 2Ω resistances are in parallel, the effective resistance is
(a) 2 Ω (b) 4 Ω (c) 1 Ω (d) 0.5 Ω
7. In the case of insulators, as the temperature decreases, resistivity
(a) decreases (b) increases (c) remains
constant (d) becomes zero
8. If the resistance of a coil is 2 Ω at 0oc and α = 0.004 /0 C, then its
resistance at 1000 C is
(a) 1.4 Ω (b) 0 Ω (c) 4 Ω (d) 2.8 Ω
9. According to Faraday’s law of electrolysis, when a current is passed, the mass of ions deposited at the cathode is
independent of
(a) current (b) charge (c) time (d) resistance
10. When n resistors of equal resistances (R) are connected in series,the effective resistance is
(a) n/R (b) R/n (c) 1/nR (d) Nr
3. EFFECTS OF CURRENT
1. Joule’s law of heating is
(a) H =I2 Rt (b) H = V2 Rt (c) H = VIt (d) H = IR2 t
2. Nichrome wire is used as the heating element because it has
(a) low specific resistance (b) low melting point
(c) high specific resistance (d) high conductivity
3. Peltier coefficient at a junction of a thermocouple depends on
(a) the current in the thermocouple
(b) the time for which current flows
(c ) the temperature of the junction
(d) the charge that passes through the thermocouple
4. In a thermocouple, the temperature of the cold junction is 200 C, the neutral temperature is 270 C. The temperature of
inversion is
(a) 5200 C (b) 540
0 C (c) 500
0 C (d) 510
0 C
5. Which of the following equations represents Biot-savart law?
(a) 0
24
idldB
r
(b) 0
24
idlSindB
r
(c) 0
24
idlxrdB
r
(d) 0
34
idlxrdB
r
6. Magnetic induction due to an infinitely long straight conductor placed in a medium of permeability μ is
(a) 0
2
IdB
a
(b)
2
IdB
a
(c)
4
IdB
a
(d) 0
4
IdB
a
7. In a tangent galvanometer, for a constant current, the deflection is 300. The plane of the coil is rotated through 900.
Now, for the same current, the deflection will be
(a) 300 (b) 600 (c) 900 (d) 0
8. The period of revolution of a charged particle inside a cyclotron does not depend on
(a) the magnetic induction (b) the charge of the particle
(c) the velocity of the particle (d) the mass of the particle
9. The torque on a rectangular coil placed in a uniform magnetic field
is large, when
(a) the number of turns is large (b) the number of turns is less
108
(c) the plane of the coil is perpendicular to the field
(d) the area of the coil is small
10. Phosphor – bronze wire is used for suspension in a moving coil galvanometer, because it has
(a) high conductivity (b) high resistivity
(c) large couple per unit twist (d) small couple per unit twist
11. Of the following devices, which has small resistance?
(a) moving coil galvanometer (b) ammeter of range 0 – 1A
(c) ammeter of range 0–10 A (d) voltmeter
12. A galvanometer of resistance G Ω is shunted with S Ω .The effective resistance of the combination is Ra. Then, which of
the following statements is true?
(a) G is less than S (b) S is less than Ra but greater than G.
(c) Ra is less than both G and S (d) S is less than both G and Ra
13. An ideal voltmeter has
(a) zero resistance
(b) finite resistance less than G but greater than Zero
(c) resistance greater than G but less than infinity
(d) infinite resistance
.
4. ELECTROMAGNETIC INDUCTION.
1. Electromagnetic induction is not used in
(a) transformer (b) room heater (c) AC generator (d) choke coil
2. A coil of area of cross section 0.5 m2 with 10 turns is in a plane which is pendendicular to an uniform magnetic field of
0.2 Wb/m2.The flux though the coil is
(a) 100 Wb (b) 10 Wb (c) 1 Wb (d) zero
3. Lenz’s law is in accordance with the law of
(a) conservation of charges (b) conservation of flux
(c) conservation of momentum (d) conservation of energy
4. The self−inductance of a straight conductor is
(a) zero (b) infinity (c) very large (d) very small
5. The unit henry can also be written as
(a) Vs A-1 (b) Wb A-1 (c) Ω s (d) all
6. An emf of 12 V is induced when the current in the coil changes at the rate of 40 A S–1. The coefficient of self
induction of the coil is
(a) 0.3 H (b) 0.003 H (c) 30 H (d) 4.8 H
7. A DC of 5A produces the same heating effect as an AC of
(a) 50 A rms current (b) 5 A peak current
(c) 5A rms current (d) none of these
8. Transformer works on
(a) AC only (b) DC only
(c) both AC and DC (d) AC more effectively than DC
9. The part of the AC generator that passes the current from the coil to the external circuit is
(a) field magnet (b) split rings (c) slip rings (d) brushes
10. In an AC circuit the applied emf e = Eo sin (ωt + π/2) leads the
current I = Io sin (ωt – π/2) by
(a) π/2 (b) π/4 (c) π (d) 0
109
11. Which of the following cannot be stepped up in a transformer?
(a) input current (b) input voltage (c) input power (d) all
12. The power loss is less in transmission lines when
(a) voltage is less but current is more
(b) both voltage and current are more
(c) voltage is more but current is less
(d) both voltage and current are less
13. Which of the following devices does not allow d.c. to pass through?
(a) resistor (b) capacitor (c) inductor (d) all the above
14. In an ac circuit
(a) the average value of current is zero.
(b) the average value of square of current is zero.
(c) the average power dissipation is zero.
(d) the rms current is 2 time of peak current.
5.ELECTROMAGNETIC WAVES AND WAVE OPTICS.
1. In an electromagnetic wave
(a) power is equally transferred along the electric and magnetic fields
(b) power is transmitted in a direction perpendicular to both the fields
(c) power is transmitted along electric field
(d) power is transmitted along magnetic field
2. Electromagnetic waves are
(a) transverse (b) ) may be longitudinal or transverse
(c) longitudinal (d) neither longitudinal nor transverse
3. Refractive index of glass is 1.5. Time taken for light to pass through a glass plate of thickness 10 cm is
(a) 2 × 10-8 s (b) 2 × 10-10 s (c) 5 × 10-8 s (d) 5 × 10-10 s
4. In an electromagnetic wave the phase difference between electric field E and magnetic field B is
(a) π/4 (b) π/2 (c) π (d) zero
5. Atomic spectrum should be
(a) pure line spectrum (b) emission band spectrum
(c) absorption line spectrum (d) absorption band spectrum.
6. When a drop of water is introduced between the glass plate and plano convex lens in Newton’s rings system, the ring
system
(a) contracts (b) expands
(c) remains same (d) first expands, then contracts
7. A beam of monochromatic light enters from vacuum into a medium of refractive index μ. The ratio of the wavelengths
of the incident and refracted waves is
(a) μ : 1 (b) 1 : μ (c) μ2 : 1 (d) 1 : μ
2
8. If the wavelength of the light is reduced to one fourth, then the amount of scattering is
(a) increased by 16 times (b) decreased by 16 times
(c) increased by 256 times (d) decreased by 256 times
9. In Newton’s ring experiment the radii of the mth and (m + 4)th dark rings are respectively 5 mm and 7 mm. What is the
value of m?
(a) 2 (b) 4 (c) 8 (d) 10
10. The path difference between two monochromatic light waves of wavelength 4000 Å is 2 × 10-7 m. The phase difference
between them is
(a) π (b) 2π (c) 3π (d) π/2
11. In Young’s experiment, the third bright band for wavelength of light 6000 Å coincides with the fourth bright band for
another source in the same arrangement. The wave length of the another source is
(a) 4500 Å (b) 6000 Å (c) 5000 Å (d) 4000 Å
12. A light of wavelength 6000 Å is incident normally on a grating 0.005 m wide with 2500 lines. Then the maximum order is
110
(a) 3 (b) 2 (c) 1 (d) 4
13. A diffraction pattern is obtained using a beam of red light. Whathappens if the red light is replaced by blue light?
(a) bands disappear (b) no change
(c) diffraction pattern becomes narrower and crowded together
(d) diffraction pattern becomes broader and farther apart
14. The refractive index of the medium, for the polarising angle 600 is
(a) 1.732 (b) 1.414 (c) 1.5 (d) 1.468
6. ATOMIC PHYSICS
1. The cathode rays are
(a) a stream of electrons (b) a stream of positive ions
(c) a stream of uncharged particles (d) the same as canal rays
2. A narrow electron beam passes undeviated through an electric field
E = 3 × 104 V/m and an overlapping magnetic field B = 2×10-3 Wb/m2. The electron motion, electric field and
magnetic field are mutually perpendicular. The speed of the electron is
(a) 60 ms−1 (b) 10.3 × 107 ms (c) 1.5 × 10
7
ms−1 (d) 0.67 × 10-7 ms−1
3. According to Bohr’s postulates, which of the following quantities
take discrete values?
(a) kinetic energy (b) potential energy (c) angular
momentum (d) momentum
4. The ratio of the radii of the first three Bohr orbit is,
(a) 1 : 1/2 : 1/3 (b) 1 : 2 : 3 (c) 1 : 4 : 9 (d) 1 : 8 : 27
5. The first excitation potential energy or the minimum energy required to excite the atom from ground state of
hydrogen atom is,
(a) 13.6 eV (b) 10.2eV (c) 3.4 eV (d) 1.89 eV
6. According to Rutherford atom model, the spectral lines emitted by an atom
(a) line spectrum (b) continuous spectrum
(c) continuous absorption spectrum (d) band spectrum
7. Energy levels A, B, C of a certain atom correspond to increasing values of energy (i.e.,) EA < EB < EC. If λ1, λ2, λ3 are the
wavelengths of radiations corresponding to the transitions C to B, B to A and C to A respectively, which of the
following statements is correct.
(a) λ3 = λ1 + λ2 (b) 1 23
1 2
(c) 1 2 3 =0 (d)2 2 2
1 2 3
8. The elliptical orbits of electron in the atom were proposed by
(a) J.J.Thomson (b) Bohr (c) Sommerfeld (d) de Broglie
9. X−ray is
(a) phenomenon of conversion of kinetic energy into radiation.
(b) conversion of momentum
(c) conversion of energy into mass
(d) principle of conservation of charge
10. In an X-ray tube, the intensity of the emitted X−ray beam is increased by
(a) increasing the filament current
(b) decreasing the filament current
(c) increasing the target potential
(d) decreasing the target potential
11. The energy of a photon of characteristic X-ray from a Coolidge tube
comes from
111
(a) the kinetic energy of the free electrons of the target
(b) the kinetic energy of ions of the target
(c) the kinetic energy of the striking electron
(d) an atomic transition in the target.
12. A Coolidge tube operates at 24800 V. The maximum frequency of Xradiation emitted from Coolidge tube is
(a) 6 × 1018
Hz (b) 3 × 1018
Hz (c) 6 × 108 Hz (d) 3 × 10
8 Hz
13. In hydrogen atom, which of the following transitions produce a spectral line of maximum wavelength
(a) 2 → 1 (b) 4 → 1 (c) 6 → 5 (d) 5 → 2
14. In hydrogen atom, which of the following transitions produce a spectral line of maximum frequency
(a) 2 → 1 (b) 6 → 2 (c) 4 → 3 (d) 5 → 2
15. After pumping process in laser,
(a) the number of atoms in the ground state is greater than the number of atoms in the excited state.
(b) the number of atoms in the excited state is greater than the number of atoms in the ground state.
(c) the number of atoms in the ground state is equal to the number atoms in the excited state.
(d) No atoms are available in the excited state.
16. The chromium ions doped in the ruby rod
(a) absorbs red light (b) absorbs green light
(c) absorbs blue light (d) emits green light
7. DUAL NATURE OF RADIATION AND MATTER AND RELATIVITY
1. A photon of frequency ν is incident on a metal surface of threshold frequency νο. The kinetic energy of the emitted
photoelectron is
(a) h (ν – νο ) (b) hν (c) hνο (d) h (ν + νο )
2. The work function of a photoelectric material is 3.3 eV. The threshold frequency will be equal to
(a) 8 × 1014 Hz (b) 8 × 1010 Hz (c) 5 × 1020 Hz (d) 4 × 1014 Hz.
3. The stopping potential of a metal surface is independent of
(a) frequency of incident radiation (b) intensity of incident radiation
(c) the nature of the metal surface (d) velocity of the electrons emitted.
4. At the threshold frequency, the velocity of the electrons is
(a) zero (b) maximum (c) minimum (d) infinite
5. The photoelectric effect can be explained on the basis of
(a) corpuscular theory of light (b) wave theory of light
(c) electromagnetic theory of light (d) quantum theory of light
6. The wavelength of the matter wave is independent of
(a) mass (b) velocity (c) momentum (d) charge
7. If the kinetic energy of the moving particle is E, then the de Broglie
wavelength is,
(a) 2mE
(b)
2
2mE
(c)
2
mE
(d)
2
h
mE
8. The momentum of the electron having wavelength 2Å is
(a) 3.3 × 1024
kg m s−1 (b) 6.6 × 1024
kg m s−1
(c) 3.3 × 10-24 kg m s−1 (d) 6.6 × 10-24 kg m s−1
9. According to relativity, the length of a rod in motion
(a) is same as its rest length (b) is more than its rest length
(c) is less than its rest length
(d) may be more or less than or equal to rest length depending on the
speed of the rod
10. If 1 kg of a substance is fully converted into energy, then the energy
produced is
(a) 9 × 1016 J (b) 9 × 1024 J (c) 1 J (d) 3 × 10 8 J
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8. NUCLEAR PHYSICS
1. The nuclear radius of 4Be8 nucleus is
(a) 1.3 × 10-15
m (b) 2.6 × 10-15
m (c) 1.3 × 10-13
m (d) 2.6 × 10-13
m
2. The nuclei 13Al27 and 14Si28 are example of
(a) isotopes (b) isobars (c) isotones (d) isomers
3. The mass defect of a certain nucleus is found to be 0.03 amu. Its binding energy is
(a) 27.93 eV (b) 27.93 KeV (c) 27.93 MeV (d) 27.93 GeV
4. Nuclear fission can be explained by
(a) shell model (b) liquid drop model (c) quark model
(d) Bohr atom
5. The nucleons in a nucleus are attracted by
(a) gravitational force (b) electrostatic force
(c) nuclear force (d) magnetic force
6. The ionisation power is maximum for
(a) neutrons (b) α – particles
(c) γ – rays (d) β − particles
7. The half life period of a certain radioactive element with disintegration constant 0.0693 per day is
(a) 10 days (b) 14 days (c) 140 days (d) 1.4 days
8. The radio-isotope used in agriculture is
(a) 15P31 (b) 15P32 (c) 11Na23 (d) 11Na24
9. The average energy released per fission is
(a) 200 eV (b) 200 MeV (c) 200 meV (d) 200 GeV
10. The explosion of atom bomb is based on the principle of
(a) uncontrolled fission reaction (b) controlled fission reaction
(c) fusion reaction (d) thermonuclear reaction
11. Anaemia can be diagnosed by
(a) 15P31 (b)15P32 (c) 26Fe59 (d) 11Na24
12. In the nuclear reaction 80Hg198 + X → 79Au198 + 1H1, X-stands for
(a) proton (b) electron (c) neutron (d) deutron
13. In β – decay
(a) atomic number decreases by one (b) mass number decreases by one
(c) proton number remains the same (d) neutron number decreases by one
14. Isotopes have
(a) same mass number but different atomic number
(b) same proton number and neutron number
(c) same proton number but different neutron number
(d) same neutron number but different proton number
15. The time taken by the radioactive element to reduce to 1/e times is
(a) half life (b) mean life
(c) half life/2 (d) twice the mean life
16. The half life period of N13 is 10.1 minute. Its life time is
(a) 5.05 minutes (b) 20.2 minutes (c)10.1 (d) infinity
17. Positive rays of the same element produce two different traces in a Bainbridge mass spectrometer. The positive ions
have
(a) same mass with different velocity (b) same mass with same velocity
(c) different mass with same velocity (d) different mass with different velocity
18. The binding energy of 26Fe56 nucleus is
(a) 8.8 MeV (b) 88 MeV (c) 493 MeV (d) 41.3 MeV
19. The ratio of nuclear density to the density of mercury is about
(a) 1.3 × 1010 (b) 1.3 (c) 1.3 × 1013 (d) 1.3 × 104
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9.ELECTRONICS
1. The electrons in the atom of an element which determine its chemical and electrical properties are called
(a) valence electrons (b) revolving electrons
(c) excess electrons (d) active electrons
2. In an N−type semiconductor, there are
(a) immobile negative ions (b) no minority carriers
(c) immobile positive ions (d) holes as majority carriers
3. The reverse saturation current in a PN junction diode is only due to
(a) majority carriers (b) minority carriers
(c) acceptor ions (d) donor ions
4. In the forward bias characteristic curve, a diode appears as
(a) a high resistance (b) a capacitor (c) an OFF switch
d) an ON switch
5. Avalanche breakdown is primarily dependent on the phenomenon of
(a) collision (b) ionization (c) doping (d) recombination
6. The colour of light emitted by a LED depends on
(a) its reverse bias (b) the amount of forward current
(c) its forward bias (d) type of semiconductor material
7. The emitter base junction of a given transistor is forward biased and its collector−base junction is reverse biased. If the
base current is increased, then its
(a) VCE will increase (b) IC will decrease
(c) Ic will increase (d) VCC will increase.
8. Improper biasing of a transistor circuit produces
(a) heavy loading of emitter current (b) distortion in the output signal
(c) excessive heat at collector terminal (d) faulty location of load line
9. An oscillator is
(a) an amplifier with feedback (b) a convertor of ac to dc energy
(c) nothing but an amplifier (d) an amplifier without feedback
10. In a Colpitt’s oscillator circuit
(a) capacitive feedback is used (b) tapped coil is used
(c) no tuned LC circuit is used (d) no capacitor is used
11. The following arrangement performs the logic function of ____gate
(a) AND (b) OR (c) NAND (d) EXOR
12. Since the input impedance of an ideal operational amplifier is infinite,
(a) its input current is zero (b) its output resistance is high
(c) its output voltage becomes independent of load resistance
(d) it becomes a current controlled device
13. If the output (Y) of the following circuit is 1, the inputs A B C must be
(a) 0 1 0 (b) 1 0 0 (c) 1 0 1 (d) 1 1 0
14. According to the laws of Boolean algebra, the expression (A + AB) is equal to
(a) A (b) AB (c) B (d) A
15. The Boolean expression ABC can be simplified as
(a) AB + C (b) A . B . C (c) AB + BC + CA (d) A + B + C
10.COMMUNICATION
1. High frequency waves follow
(a) the ground wave propagation (b) the line of sight direction
(c) ionospheric propagation (d) the curvature of the earth
2. The main purpose of modulation is to
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(a) combine two waves of different frequencies
(b) acquire wave shaping of the carrier wave
(c) transmit low frequency information over long distances efficiently
(d) produce side bands
3. In amplitude modulation
(a) the amplitude of the carrier wave varies in accordance with the amplitude of the modulating signal.
(b) the amplitude of the carrier wave remains constant
(c) the amplitude of the carrier varies in accordance with the frequency of the modulating signal
(d) modulating frequency lies in the audio range
4. In amplitude modulation, the band width is
(a) equal to the signal frequency (b) twice the signal frequency
(c) thrice the signal frequency (d) four times the signal frequency
5. In phase modulation
(a) only the phase of the carrier wave varies
(b) only the frequency of the carrier wave varies.
(c) both the phase and the frequency of the carrier wave varies.
(d) there is no change in the frequency and phase of the carrier wave
6. The RF channel in a radio transmitter produces
(a) audio signals (b) high frequency carrier waves
(c) both audio signal and high frequency carrier waves
(d) low frequency carrier waves.
7. The purpose of dividing each frame into two fields so as to transmit 50 views of the picture per second is
(a) to avoid flicker in the picture
(b) the fact that handling of higher frequencies is easier
(c) that 50 Hz is the power line frequency in India
(d) to avoid unwanted noises in the signals
8. Printed documents to be transmitted by fax are converted into electrical signals by the process of
(a) reflection (b) scanning (c) modulation (d) light variation
EXTRA QUESTIONS.
ELECTROSTATICS.
1. If two charged bodies of charges +2q and -5q are brought in contact, the total charge of the system is
a) 3q b)-3q
c)zero d) 2q or -3q
2. The unit of electric field intensity is a) NC-1
b)N-1
C2m
-2 c) Cm d) volt.
3. Which one of the following is insulator? a)human body b) earth
c) copper d) ebonite.
4. Electric dipole moment always acts in the direction from a) q to –q b)-q
to +q c)∞ to +q d) q to∞
5. An electric dipole in a uniform field experience a a) force
b) torque c) momemtum
d)neither force nor torque
6. When a dipole is aligned with field, then potential energy is given as a)PESinθ b) 0
c) –PE d)-PE Cosθ.
7. Which one of the following is a non polar molecule? a) H2o b) CO2
c) HCl d) 02
8. Relative permittivity of vacuum is a) 3
b) 2 c) 1 d) 0
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9. The permittivity of vacuum is equal to a) 2 1 2
9
1
4 9 10C N m
x x
b)2 1 2
9
1
4 9 10C N m
x x
c)
9 2 1 29 10x C N m
d)9 2 1 29 10x C N m
10. When two capacitors are connected in series to a source of emf, then one of them will have same
a) voltage b) electric field c) both
d) charge.
11. A dipole is placed in a uniform electric field with its axis parallel to the field it experiences
a) net force only b) torque
only c) both net force and torque d) neither a net force nor a torque.
12. If the medium between two charges is replaced by air then the force between them
a) increases b) decreases
c) becomes zero d)remains constant.
13. If a Gaussian surface encloses a dipole of moment 2qd, then the total flux through the surface is
a)
0
q
b)
0
2q
c)
02
q
d) zero.
14. When air medium in capacitor is replaced by a medium of dielectric constant r the capacitance,
a) decreases by r times b) increases
by r times c)remains the same d) becomes zero.
15. Two charges 10-6
and 10-7
repel each other with a force of 400 N. The distance between the charges is
a) 0.15mm b) 1.5 mm c) 15 mm
d) 1.5m
16. The potential difference between two parallel plates is 100V and the electric field between them is 104 V/m.
Then the distance between the plates
a) 1 mm b) 1 m c) 10cm d) 1cm.
17. The intensity of the electric field that produce a force of 10N on a charge of 5 C is
a) 2 NC-1 b)50NC-1 c)5 NC-1
d) 0.5 NC-1
18. The unit of number of electric lines of force passing through a given area is
a) no unit b)NC-1
c)Nm2C
-1 d) Nm.
19. Three capacitors of capacitances 1μC, 2μC,3μC are connected in series. The effective capacitance of the
capacitors is a)11/6μF b)6/11μF c) 6 μF
d)1/6μF
20. An electric dipole of moment P is placed in a uniform electric field of intensity E at an angle θ with
respect to the field. The direction of the torque is
a) along the direction of E b) opposite to the direction of P c)
along the direction of E d) perpendicular to
the plane containing P and E.
21. When a point charge of 6μC is moved between two points in an electric field. The work done is 1.8 x 10-5 J.
The potential difference between two point is
a) 1.08 V b) 1.08μV c) 3 V d) 30 V.
22. Torque on a dipole in a uniform electric field is maximum when the angle between P and E is
a)0 b) 45 c) 90 d)
180.
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23. The equivalent capacitance of two capacitors connected in series is 1.5μF. The capacitance of one of them is
4μF.The capacitance of the other is
a) 2.4μF b)2μF c) 4μF d)6μF
24. The magnitude of the force acting on a charge of 2 x 10-10
C placed in a uniform electric field of 10 V/m is
a) 2 x 10-9 N b) 2 x 10-10 N c) 4 x 10-10 N d) 4
x 10-9
N
25. The unit for electric field intensity is a) NC
b) NC-1
c) Vm d) NC-2
26. An example for conductor is a) glass
b) human body c) dry wood d) ebonite.
27. An electric dipole is placed in a non uniform electric field with its axis at an angle θ with the field experiences
a) only a net force b) only a
torque c) both a net force and a torque d) neither a net force not torque.
28. When an electric dipole of moment P is aligned parallel to the electric field E then the potential energy of the
dipole is given as a) PE b) zero c) – PE d)
2
PE
29. A capacitor of capacitance 6μF is connected to a 100 V battery. The energy stored in the capacitor is
a) 30 J b) 3 J c) 0.03 J
d) 0.06J
30. The law that governs the force between electric charges is a) Ampere’s law. b) Faraday’s
law c) coulomb’s law d) ohm’s law.
31. The number of electric lines of force originating from a charge of 1 microcoulomb is
a) 1.129 x 105 b) 1.6 x 10-19 c) 6.25 x 1018
d) 8.85 x 10-12
32. The direction of electric field at a point on the equatorial line due to an electric dipole is
a) along the equatorial line towards the
dipole b) along the equatorial line away from the dipole
c) parallel to the axis of the dipole and opposite to the direction of dipole moment
d) parallel to the axis of the
dipole and in the direction of dipole moment.
33. When two point charges 6 C and -5 C experiences a force of 2.7 x 1011 N the distance between them is
a) 0.5 m b) 1 m c) 10 m
d) 10 cm
34. An electric dipole with dipole moment 4 x 10-9 C m is kept at 300 with the direction of uniform electric field of
magnitude 5 x 104 NC
-1 . The magnitude of torque on the dipole is
a) 10-4
Nm b) 104 Nm c) 2 X 10
-5 Nm d) 10
-5 Nm.
35. The electrostatic force between two point charges kept at a distance of d apart in air is F. The force between
them at the same separation in a medium of dielectric constant εr is
a) F.εr b) εr/F c) F/εr d) zero.
36. The energy of the capacitor is given by a) U =CV b)
21
2U CV c)
1
2U CV d)
21
2U C V ..
37. The molecule which has a permanent dipole moment is a) O2 b)
N2 c) H2 d) H2O
38. If a dielectric slab is introduced in the space between the plates of a capacitor, thecapacitance of the capacitor
a) decreases b) increases
c) remains constant d) may increase or decrease
39. Two spheres A andB of same size are charged to the same potential. A is hollow and B is solid. Which of the
two holds more charge? a) solid sphere cannot hold any charge
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b) Hollow sphere cannot hold any charge c) both have zero charge d) both
have same charge.
40. Three capacitors are connected in series. Then a) they will have equal
charges b) they will have same potential c) both d) none.
41. A charge Q is placed at the cube. The electric flux through the six faces of the cube is
a)
06
Q
b)
08
Q
c)
0
Q
d)
03
Q
2. CURRENT ELECTRICITY.
1. The unit of mobility is a)m2V-1S-1
b) m2V
-1S
-2 c) m
2V
-1S
1 d) m
2V
-1S
-1
2. Drift velocity of electron is of the order a)0.2cm/s b)
0.1 cm/s c) 0.1 m/s d) 1 cm/s
3. The conductivity of the material is obtained by a) RA
l
b)lA
R c)
l
RA d)
RA
4. At the transitition temperature the conductivity becomes a) zero
b) infinity c) minimum d) none.
5. The tolerance of silver, gold, red and brown rings in carbon resistors are respectively
a) 1%,2%,5%,10%
b)10%,2%,5%,1% c) 10%,5%,1%,2% d)10%,5%,2%,1%.
6. The temperature coefficient of resistance of insulators and semiconductors is
a) positive b) negative
c) low d) zero.
7. One kilo watt hour is equal to a) 3.6 x 105
b) 36 x 105 c)0. 36 x 105 d) 3.6 x 104
8. A charge of 600 passes through an electric lamp in 2 minutes. Then the current in the lamp is
a) 30 A b) 1 A c) 0.5 A
d) 5 A
9. If a charge of 1 C passes through an electrical equipment in 10 s, then the current flowing through it is
a) 0.5A b) 1 A c) 0.1A
d) 10A
10. A 1.15 W, 230 V water heater can drawn a current a) 0.2A
b) 2A c) 5 A d) 0.5A
11. The valve of a carbon resistor is 33kΩ. Then the colour code is a) yellow, orange, Red
b) Brown, yellow, orange c) Red, Blue, Orange d) Orange,
Orange,Orange.
12. If 6.25 x 1018 electrons flow through a given cross section in unit time, then the current is
a) 1A b) 2 A c)
0.1 A d) 0.2 A
13. The electrolyte used in Leclanche cell is a) sulphuric acid
b) copper sulphate c) ammonium chloride
d) hydrochloric acid.
14. The unit of conductivity is a) Ωm
b) Ω-1
m-1
c) Ω-1
d) m-1
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15. Resistance of a metal wire of length 10 cm is 2Ω. If the wire is stretched uniformly to 50 cm, the resistance is
a) 25Ω b) 10Ω c) 5Ω
d) 50Ω
16. The material through which electric charge can flow easily is a) quartz b) mica
c) germanium d) copper.
3. EFFECTS OFCURRENT.
1. Nichrome is an alloy of a) nickel and
manganese b)nickel, carbon and iron c)nickel and iron d)
nickel and chromium.
2. Fusewire is an alloy of a) 37%of Pb and
63% of Sn b) 63% of Pb and 37% of sn c) 73% of Pb and 27% of Sn
d) 37% of Sb and 63% of Sn.
3. In which of the following, joule heating effect is undesirable? a) electric iron
b) electric toaster c) transformer and dynamos d)
fuse wire.
4. Which of the following is not a thermo emf effect? a) peltier effect
b) Thompson effect c) Joule effect
d)Seebeck effect.
5. Complementary effect of seebeck effect a) peltier effect
b) joule effect c) positive Thompson effect d)
negative Thompson effect.
6. For small temperature difference the graph showing the variation of thermoemf with temperature of the hot
junction is a) parabola b) circle c) straight line d)
hyperbola.
7. For a given thermocouple, the neutral temperature is a) constant b) maximum c)
minimum d) zero.
8. Peltier coefficient of a junction of a thermocouple depends on a) the current in the thermocouple
b) time for which the current flows c) temperature of the
junction d) charge that passes through the junction.
9. In a thermocouple the temperature of the cold junction is 200C, the neutral temperature is 2700 C, then the
temperature of inversion is a) 5200C b) 5400C c) 5000C
d) 5100C
10. Current is flowing through a conductor of resistance10 ohm. Indicate in which of the following cases, the maximum
heat will be generated a) 5A passing for 2 minutes b) 4A passing for 3 minutes c)
3A passing for 6 minutes d) 2A passing for 5 minutes.
11. _____ is a device used to detect thermal radiation. a) thermocouple
b) thermopile c) thermometer d)
thermoscope.
12. In a tangent galvanometer, the plane of the coil should be adjusted to be i a) geographic meridian
b) magnetic meridian c) any direction d) parallel to east west
direction.
13. If the reduction factor of a tangent galvanometer is 0.9A, then the current that produces a deflection of 300 is
a) 450 mA b) 520mA c) 780 mA
d) 520 A
14. For a solenoid whose length is very large compared to its radius, the magnetic field at points outside the solenoid is
a) maximum b) minimum c) zero
d)I/μ
15. Angular frequency and period of rotation of the charged particle in a magnetic field is independent of
a) mass and radius b) velocity and radius
c) charge and velocity d) mass and charge.
119
16. Cyclotron cannot accelerate a) protons b) deuteron
c) electron d) alpha particle.
17. Work done by Lorentz force is a) positive b) negative
c) constant d) zero.
18. Two parallel straight conductors carrying currents in the same direction a) repel each other
b) attract each other c) do not experience any force d) experience a
maximum force.
19. When the number of turns in a galvanometer is doubled, then a) current sensitivity and voltage
sensitivity doubled b) current sensitivity is doubled and voltage sensitivity
remains uncharged c) current sensitivity remains uncharged d) voltage sensitivity is doubled.
20. Phosphor wire is used for suspension in moving coil galvanometer because it has
a) high conductivity b)
high resistivity c) large couple per unit twist d) small couple per unit twist.
21. If a current carrying conductor is placed parallel to the magnetic field then the force acting on the conductor is
a) zero b) maximum c) infinity
d) BIl.
22. The power of the electric heater working in 5A produces at 200 V is a) 5000 W b) 20000W
c) 100W d) 1 KW.
23. A proton enters into a uniform magnetic field of induction 2.5T with a velocity of 2.5 x 106m/s by making an angle of
300 with the direction of the field the Lorentz force on the proton is
a) 5 x 10-10
N b) 5x10-11
N c) 5x 10-12
N d) 5 x 10-13
N.
24. The unit of reduction factor of tangent galvanometer is a) no unit b) tesla c)
ampere d) ampere/degree.
25. The torque experienced by a rectangular current loop placed perpendicular to a uniform magnetic field is
a) maximum b) zero c) finite minimum d)
infinity.
26. Which of the following is used in thermopile? a) Thompson effect
b) peltier effect c) seebeck effect d) joule
effect.
27. Thermopile is used to a) measure
temperature b) measure current c) detect thermal radiation
d) measure pressure.
28. Unit of peltier coefficient is a) ohm
b) volt c) mho d) ampere.
29. An electron is moving with a velocity of 3 x 106 m/s perpendicular to a uniform magnetic field of induction 0.5T. The
force experienced by the electron is
a) 2.4 x 10-13 N b) 13.6 x 10-27 N c)13.6 x 10-11 N
d) zero.
4.ALTERNATING CURRENT .
1. A Fuse wire has a current rating of 5A. Then the peak value of the current in the fuse wire is
a)0.7A b) 1A c) 7.07A
d) 70.7A
2. The emf in an AC containing only inductance will a) lag behind the current by π/2
b) be ahead of current by π/2 c) have current in phase with the applied voltage
d) always be out of phase.
3. An aeroplane having a wingspan of 35m flies at a speed of 100m/s. If the vertical component of earth magnetic
field is 4 x 10-4 T. Then the induced emf across the wingspan is
a)28V b) 2.8V c) 14 V d) 1.4 V
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4. The coefficient of mutual induction between the pair of coil depends upon
a) size and shape of the coil b)
number of turns of the coil c) proximity of the coil d) all the above.
5. A field of induction 20 T acts at right angles to a coil of area 20 m2 with 50 turns. The flux linked with the coil is
a) 2000 wb b) 20000 wb c) 0 wb
d) 200 wb.
6. At what rate must be current change in a 65 mH coil to have a 1 volt self induced emf?
a) 25As-1
b) 17 As-1
c)
25.4 As-1 d) 15.4 As-1
7. A wire cuts across a flux of 0.2 x 10-2 wb in 0.12 second. What is the emf induced in the wire?
a)0.06 V b) 0.02 V c) 0.0167 V
d) 0.24 V
8. In a step down transformer the input voltage is 22 KV and the output voltage is 550V. The ratio of number of turns
in the primary to that in the secondary is
a) 1:40 b) 40:1 c) 1:20 d)20:1
9. In an LCR circuit if XL =XC a) current is
maximum and impedance is maximum b) current is maximum and impedance is
minimum c) current and impedance is maximum
d) current and impedance is minimum.
10. Eddy currents are used in a) nuclear
reactor b) induction motor c) dynamo
d) transformer.
11. An electric power of 11,000W is transmitted at 220V through a wire of resistance 1 ohm a then the power loss is
a) 110 W b) 2500 W c)
0.25 W d) 50W
12. The angle between the area and the area vector is a) zero b)
900 c) 450 d) 1800
13. How can you arrange the two coils so as to have the maximum coefficient of mutual induction?
a) perpendicular to each other
b) placed one after another so as to have the common axis
c) wound one over the other the have the same axis d) none
of the above.
14. An emf of 60 mV is induced in a coil when the current in the neighbouring coil changes from 11A to 5A in 0.1 s. The
coefficient of mutual induction is a) 0.003H b) 0.1H c) 0.01H d)
0.001H
15. The average power consumed over one cycle in an ac circuit is a) Erms Irms b) Erms Irms
Cosφ c) Erms Irms Sinφ d) E0 I0 Cosφ
16. If the flux associated with a coil varies at the rate of 1 wb/min then the induced emf is
a) 1V b) 1/60V
c)60V d) 0.60 V
17. In an LCR circuit when XL=XC the current a) is zero
b) is in phase with voltage c) leads the voltage
d) lags behind the voltage.
18. The generator rule is a) Flemming left
hand rule b) Flemming right hand rule c) Maxwell right hand cork screw rule d)
ampere swimming rule.
19. In an AC circuit with capacitor only, if the frequency of the signal is zero then the capacitive reactance is
a) infinity b)
zero c) finite maximum d) finite minimum.
20. The reactance offered by 300 mH inductor to an AC supply of frequency 50Hz is
a) 1046 ohm b) 94.2 ohm c) 9420 ohm d)
104.6 ohm.
121
21. In an transformer, eddy current loss is minimized by using a) laminated core made of
mumetal b)laminated core made of stelloy c) shell type of core
d) thick copper wires
22. In an ac circuit with capacitor only, if the frequency of the signal is zero then the capacitive reactance is
a) infinity b) zero c)
finite maximum d) finite minimum.
23. In RLC circuit, at resonance a) current is
minimum b) impedance is maximum c) circuit is purely inductive d) current is
in phase with the voltage.
24. In three phase AC generator the three coil are fastened rigidly together and are displaced from each other by an
angle a) 900 b) 1800 c) 1200
d) 3600
25. The core used in audio frequency chokes is a) iron b)
carbon c) lead d) steel
26. A rectangular coil is uniformly rotated in a uniform magnetic field such that the axis of rotation is perpendicular to
the direction of the magnetic field. When the plane of the coil is perpendicular to the magnetic field
a) magnetic flux is zero induced emf if zero b) magnetic flux is
maximum induced emf is maximum c) magnetic flux is maximum induced emf is zero
d) magnetic flux is zero induced emf is maximum.
27. In an AC circuit average power consumed is 200 W and the apparent power is 300 W. The power factor is
a) 1.5 b) 0.66 c) 0.33 d) 1
28. In LCR series ac circuit the phase difference between current and voltage is 300. The reactance of the circuit is
17.32Ω. The value of resistance is a) 30Ω b) 10Ω c) 17.32Ω
d) 1.732Ω
5. ELECTROMAGNETIC WAVES AND WAVE OPTICS.
1. The existence of electromagnetic waves was confirmed experimentally by a) Hertz b) Maxwell
c) Huygens d) Planck
2. When a ray of light is incident on a glass surface at polarizing angle of 57.50 the angle between the incident ray and
the reflected ray is a) 57.50 b) 32.50 c) 1150 d) 900
3. Un polarized light passes through a tourmaline crystal. The emergent light is analysed by an analyser is rotated
through 900 the intensity of light a) remains uniformly bright b) remains uniformly dark
c) varies between maximum and minimum d) varies
between maximum and zero.
4. In a plane diffraction grating the unit of grating element is a) no unit b) metre
c) m-1 d) degree
5. Which one of the following is not an electromagnetic wave ? a) X- rays b) γ-rays
c) UV rays d) β- rays.
6. If C is the velocity of light in vacuum, the velocity of light in a medium with refractive index μ
a) μC b) C/μ c)μ/C d)
1/μC
7. A ray of light passes from a denser medium into a rarer medium. For an angle of incidence of 450 the refracted ray
grazes the surface of separation of the two media. The refractive index of the denser medium a)3/2
b) 2 c) 2 d)1
2
8. Of the following which one of the following is uniaxial crystal ? a) mica b) aragonite
c) topaz d) quartz
9. The radiations used in physiotherapy are a) ultraviolet b)
infra red c) radio waves d) micro waves
10. In newton ring experiment light of wavelength 58900 is used the order of the dark ring produced where the
thickness of the air film is 0.589 nm a) 2 b) 3 c) 4 d) 5
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11. Of the following optically active material is a) sodium chloride b)
calcium chloride c) sodium d) chlorine
12. Electric filament lamp gives rise to a) line spectrum
b) continuous spectrum c) continuous absorption spectrum d) line absorption
spectrum.
13. In young slit experiment the separation between the slits is halved and the distance between the slit and the screen
is double. Then the fringe width is a) unchanged b) halved c) doubled d)
quadrupled
14. The phenomenon of light used in the formation of Newton ring is a) diffraction b)
interference c) refraction d) polarization.
15. In Raman effect the spectral line with lower frequency than the incident frequency is
a) Fraunhofer line
b) Rayleigh line c) stokes line d)
Anti stokes line.
16. The optical rotation does not depend on a) concentration of the
solution b) frequency of the light used c) the temperature of the solution d) intensity
of the light used
17. The transverse nature of light waves is demonstrated only by the phenomenon of
a) interference b) diffraction c) polarization d) reflection
18. IF the velocity of light is 2.25 x 108 m/s then the refractive index of the medium is
a) 1.5 b) 0.5 c) 1.33
d) 1.73
19. The polarizing angle for water is 5304’. If the light is incident at this angle on the surface of water. The angle of
refraction in water a) 5304’ b) 26030’ c) 3004’
d) 36056’
20. In Raman effect if the scattered photon gains energy it gives rise to a) stokes line
b) anti stokes line c) stokes and anti stokes line d)
Rayleigh line
21. In cases of Fraunhofer diffraction, the wave front undergoing diffraction is a) spherical wave front
b) cylindrical wave front c) elliptical wave front d) plane wave front.
22. Soap bubbles exhibit brilliant colours in sunlight due to a) scattering of light
b) diffraction of light c) polarization of light d)
interference of light
23. In the grating element Sinθ = Nmλ , the unit N is a) metre b)
metre-1 c) no unit d) m2
24. The ratio of the radii of the 4th
and 9th
dark ring in Newton’s ring experiment is
a) 4:9 b) 2:3 c) 16:81
d) 5 : 9
25. In Raman effect the incident photon makes collision with an excited molecule of the substance. The scattered
photon gives rise to a) stokes line b) anti stokes line c) Rayleigh line d) Zeeman
line.
26. A ray of light travelling in a rarer medium and reflected at the surface of denser medium automatically undergoes a
a) phase change of π/2 b) phase change of
2π c) phase difference of λ d) path difference of λ/2
27. The angle between the electric and magnetic component of an electromagnetic wave is
a) 0 b) π/4 c) π/2
d) π
28. If the energy of the electromagnetic wave is E then the energy associated with electric field vector
a) E b) 2E c) E/4
d) E/2
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29. Frequency of electromagnetic waves produced by Henry arrangement was about
a) 5 x 107 Hz b) 50 x 107 Hz
c) 8 x 1014
Hz d) 4 x 1014
Hz.
30. When the temperature of the solid is increased, the spectrum spread from a) red to green
b) blue to green c) red to blue d) violet to
red
31. Raman shift is a)
independent of frequency of incident light b) characteristic of the
substance c) independent of characteristic of the
substance d) both a) and b)
32. In the interference pattern, the energy is a) created at position of
maximum b) destroyed at the position of minima
c) conserved but it is redistributed d) none.
33. The distance between the two corresponding points in a grating is 2 x 10-4
cm. The number of lines per
metre width of the grating will be a) 20000 b) 2000 c) 500000
d) 50000
6. ATOMIC PHYSICS
1. The wavelength of D1 and D2 lines emitted by sodium vapour lamp is a) 589.6nm ,589 nm
b) 589nm,589.6nm c) 589.3nm,589nm d) 589 nm,
589.3nm
2. If the minimum wavelength of X- rays produced in a Coolidge tube is 0.62A0, the operating potential is
a) 20KV b) 0.2 KV c)
2KV d) 10 KV
3. Wave number is defined as the number of wave a) produced in one second
b) in a distance of 1 metre c) in a distance of 3 x 108 metre d) in a
distance of λ metre.
4. The energy of the electron in the first orbit of hydrogen atom is -13.6 eV. Its potential energy
a) -13.6 eV b) 13.6 eV c) -27.2 eV
d) 27.2 eV
5. In holography, which one of the following transitions is recorded on the photographic film?
a) frequency and amplitude b) phase and
frequency c) phase and amplitude d) frequency only
6. A crystal diffracts monochromatic x-rays. If the angle of diffraction of the second order is 900. Then that for the first
order will be a) 600 b) 45
0 c) 30
0 d) 15
0
7. If R is Rydberg constant the minimum wavelength of hydrogen spectrum is a) 1/R b) R/4
c) 4/R d) R
8. The unit of Rydberg constant is a) m
b) no unit c) m2 d) m-1
9. For first order x-ray diffraction, the wavelength of the x-ray is equal to the lattice spacing at a glancing angle is
a) 150 b) 600 c) 450 d) 300
10. If the maximum wavelength of x-ray produced from a Coolidge tube is 0.062nm then the potential difference
between the cathode and target material is
a) 2000V b)20,000V c) 2 x 105 V d) 6.2 x 103V
11. The spectral series of hydrogen atom in UV region are called a) Balmer series
b) Lyman series c) Paschen series d) Pfund series
12. Maser materials are a) diamagnetic
ions b) paramagnetic ions c) ferromagnetic ions
d)non-magnetic ions
13. Number of waves per unit length is known as a) wavelength
b) wave number c) bandwidth d) frequency.
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14. A three dimensional image of an object can be formed by a) atomic spectroscopy
b) holography c) molecular spectroscopy d) MASER
15. In a discharge tube the source of positive rays is a) cathode b)
anode c) gas atoms present in the discharge tube
d)fluorescent screen.
16. The ionization potential of hydrogen atom is a) 13.6eV b) -
13.6eV c) 13.6V d) -13.6V
17. When an electric field is applied to an atom each of the spectral lines split into several lines. This effect is known as
a) Zeeman effect b) Stark effect
c) Raman effect d) Seebeck effect.
18. The direction of viscous effect in Milikan’s oil drop experiment is a) always downward
b) always upward c) opposite to the direction of motion on
the oil drop d) either upward or downward
19. In sommerfield atom model for principal quantum number n=3, which of the following subshells represents circular
orbit? a) 3s b) 3p c) 3d d) 2s
20. If R is Rydberg constant the shortest wavelength of paschen series is a)R/6 b) 9/R c)
16/R d) 25/R
21. e/m of cathode ray particle a) depends on
the nature of the cathode b) depends upon the number of the anode
c) depends upon the nature of the gas atoms present inside the
discharge tube d) is independent of all these.
22. The ionization power is maximum for a) neutron b) α-
particles c) γ-rays d)β-particles
23. The life time of atoms for laser in excited state is a) 10-8 s b)
10-3 s c) 108 s d) 103 s
24. When an electron jumps from M shell to the vacant K shell, it contributes a) Kα line b) Kβ line
c)Lα line d) Lβ line
25. Radius of first orbit of hydrogen atom is0.53A0 then the radius of third orbit is
a)59A0 b) 4.774A0 c)
1.06A0 d) 2.12 A0
26. The charge on an oil drop is 12.82x 10-19 C, then the number of elementary charges are
a) 6 b) 2 c)
7 d) 8
7.DUAL NATURE OF RADIATION AND MATTER AND RELATIVITY
1. If the radius of third orbit in hydrogen atom is r, then debroglie wavelength of electron in this orbit is
a) r/3 b) 3r c) 2πr/3
d) 3(2πr)
2. The valve of stopping potential when the frequency of light is equal to the thereshold frequency is
a) maximum b) zero c) minimum
d) infinity
3. Two photons, each of energy 2.5eV are simultaneously incident on the metal surface. If the work function of the metal
is 4.5 eV then from the surface of the metal
a) one electron will be emitted b) two electron will be emitted c) more than
two electrons will be emitted d) not a single electron will be emitted.
4. According to relativity the only constant in all frames is a) mass b)
length c) time d) velocity of light
5. When a material particle of rest mass m0 attains the velocity of light, its mass becomes
a) 0 b) 2m2
c) 4m0 d) infinity.
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6. The particle which has zero mass but has energy is a) electron b) photon
c) proton d) neutron.
7. An electron of mass and charge acceleratedfrom rest through a potential of V volt, then its final velocity is
a) Ve
m b)
2
Ve
m
C)2Ve
m d)
2eV
m
8. Electron microscope works on the principle of a) photoelectric effect
b) particle nature of electron c) wave nature of moving electron d)
dual nature of matter
9. If C is the velocity, ν the frequency and λ the wavelength of a radiation, then its frequency is defined as
a) the number of waves in a distance of one metre
b)the number of waves in a distance of λ
c) the number of waves in a distance of C d) the
number of waves is produced in a period of T second.
10. In photo electric effect the thereshold frequency depends on a) potential difference
used b) intensity of light used c) nature of the metal surface d) none of
the above.
11. The resolving power of an electron microscope is ____ times greater than the resolving power of optical microscope.
a) 100 b) 1000 c) 2000
d) 10000
12. The maximum kinetic energy of the photoelectrons ___ a) increases with the
intensity of light b) decreases with intensity of light
c) varies linearly with the frequency of incident light
d) varies exponentially with the frequency of the incident light.
13. When the frequency of incident radiation increases the valve of stopping potential
a) will decrease b)
will increase c) remains the same d) will not increase.
14. When the intensiy of light incident on a photoelectric surface is doubled, a) the frequency of emitted
photons will be doubled b) the number of photoelectrons will be doubled
c) the number of photoelectrons will become 4 times
d) there is no effect at all.
15. If an electron is accelerated by a potential of 54 KV then its debroglie wavelength is
a) 3.34A0 b) 1.67A0 c) 16.7A0
d) 0.84A0
16. An alpha particle and a proton are accelerated through the same potential . The ratio of their debroglie wavelength is
a) 1: 1 b) 1:2 c) 1: 3
d) 1: 2 2
8.NUCLEAR PHYSICS.
1. The nuclear force is due to the continuous exchange of particle is called a) leptons b) mesons
c) hyperons d) photons.
2. Which of the following particle is a lepton? a) electron b)
proton c) neutron d) π-meson.
3. One amu is equal to a) 931 MeV
b) mass of carbon atom c) mass of oxygen atom
d) 1.66 x 10-27 kg.
4. Anaemia can be diagoned by a)32
15P
b)31
15P c)59
26Fe d)24
11Na
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5. The number of α and β particles emitted when an isotope 238
92U undergoes α and β decays to form 206
82 Pb
respectively a) 6,8 b) 4,3 c) 8,6
d) 3,4
6. Which of the following is not a moderator? a) liquid sodium
b) ordinary water c) graphite
d) heavy water
7. The fuel used in Kamini reactor is a) 92U238
b)
92U233 c) 92Pu239 d)low enriched uranium.
8. Slow neutrons are neutrons having energies between a) 1000eV to 2000eV
b) 2000 eV to 0.5 eV c) 0 eV to 1000 eV d)
0.5 eV to 10 meV
9. In β decay a) the
atomic number decreases by one b) mass number decreases by one c) proton number remains the same d)
neutron number decreases by one
10. The atom bomb is based on the principle of a) uncontrolled fission
reaction b) controlled fission reaction c) fusion reaction d)
thermonuclear reaction.
11. The mean life of radon is 5.5 days. Its half life is a) 8 days b)
2.8 days c) 0.38 days d) 3.8 days
12. The ionization power is maximum for a) neutrons b)
alpha particles c) gamma particles d) beta particles.
13. According to law of disintegration ,the number of radioactive atoms, that have been decayed during a time of t is
a) N0 b) N c)
N0-N d) N0/2
14. The coolant used in Fast breeder reactor is a) ordinary water b)
heavy water c) liquid sodium d)boron carbide.
15. Isotope are a)
element having equal number of neutrons b) identical physical properties
c) identical chemical properties d)
dissimilar chemical properties
16. Isobars are a)
different nuclei of same element b) having similar physical
and chemical properties c) identical nuclei of different elements
d) having different physical and chemical properties.
17. Nuclear density is a)
depends on atomic number b) is a constant c) depends on neutron number
d) depends on mass number.
18. In which one of the following decays the element does not change a) alpha decay b) beta decay
c) gamma decay d) neutron decay
19. Activity of one gram of radium is equal to a) 1 roentgen b) 1
curie c) 1 henry d) 1 second
20. Which one of the following is wrong? a) the sum of
initial atomic numbers is equal to sum of final atomic number b) law of conservation of charge is satisfied
c) the initial rest mass is equal to the final rest mass
d) conservation of nucleons is satisfied
21. When 5 B10 is bombarded with neutron and α particle is emitted the residual nucleus is
a) 3Li7 b) 1H2 c) 1H3
d) 2He4
22. Isotopes are the atoms a) which has
same number of neutrons but different number of protons b) which has same number of protons but different
number of neutrons c) which has equal number of protons and neutrons
d) which has same mass numbers.
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23. The ratio of radii of two nuclei is 1:2 then the ratio of their mass number is a) 1:4 b) 1:8 c)
1:9 d)1:16
24. 7/8 part of the radio active substance decays in 45 days. Then the half life period of the substance is
a) 45 days b) 25.7 days c) 141.4 days d)
282.8
25. Thehalf life period of certain radio active elements is 10 days. The disintegration constant is
a) 0.6931/day b) 0.06931/day c) 0.006931/day d) 10
26. A nucleus contains 64 nucleons. The nuclear radius is a) 2.6F b)
5.2 F c) 10.4 F d) 7.8 F
27. An elementary particle has a rest mass 500 times that of an electron. To which category does it belong?
a) photon b) lepton c)
meson d) hyperons.
9. ELECTRONICS.
1. The forbidden energy gap for germanium of the order of a) 1.1 eV b) 0.7 eV
c) 0.3 eV d) 10 eV
2. The condition for oscillator is a) Aβ=0
b) A=1/β c) Aβ=∞ d) A+β=0
3. The potential barrier of silicon PN junction diode is approximately a) 0.3 V b) 0.7 V
c) 1.1 V d) 10V
4. The Boolean expression for NAND operation is a) Y=A+B b)
Y=A.B c) Y=A d)Y AB
5. In CE single stage amplifier the voltage gain at mid frequency is 10. The voltage gain at upper cut off frequency is
a) 10 b) 14.14 c) 7.07
d) 20
6. A logic gate for which there is LOW output only when both the inputs are HIGH is
a) AND b) NAND
c) NOR d) EXOR
7. In a junction transistor the emitter region is heavily doped since emitter has to supply to the base
a) minority carrier b) majority
carrier c) acceptor ions d) donor ions
8. A logic gate which has the output 1 when the inputs are complement to each other
a) AND b) NAND c)
NOR d) EXOR
9. Of the following the donor atoms are a) silicon and
germanium b) aluminium and gallium c) bismuth and arsenic d)
boron and indium
10. In CE amplifier, the phase difference between the input and output voltage is a) 0 b) 900 c)
1800 d) 270
0
11. In CE configuration the base current of the transistor is 50μA and the collector current is 25mA. Then the current gain
is a) 50 b) 500 c) 20 d)
200
12. An example for sinusoidal oscillator is a) multivibrator
b) RC oscillator c) Colpitt oscillator d)
Crystal oscillator.
13. In pin configuration of IC 741 pin 3 represents a) inverting input
b) non inverting input c) Vcc d) output
14. What will be the input for the Boolean expression ( )( . )Y A B A B =1 ? a)0,0 b) 1,0
c) 0,1 d) 1,1
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15. Three amplifier have equal gains 10,50 and 80 respectively. When they are connected in cascade the over all gain is
a) 4000 b) 400 c) 40000
d) 140
16. In a transistor with α=0.99 then the valve of β is a) 49
b) 90 c) 99 d) 9.9
17. Common emitter configuration has a) input
impedance b) output impedance c) input admittance
d) output admittance
18. A zener diode working in the ____ region can act as voltage regulators a) normal b) saturated
c) breakdown d) constant voltage
19. Width of the potential barrier in a PN junction diode depend on a) number of electrons
b) potential difference c) number of holes d) nature of the
material.
20. In intrinsic semiconductors a) number of
free electron and is equal to number of holes. b)number of free electrons is greater than number of
holes c)number of free electons is lesser than number of holes d) number of
free electrons is zero.
10.COMMUNICATION.
1. In television blanking pulse is applied to a)horizontal plates
b) vertical plates c) control grid d) filament.
2. In an AM superheterodyne receiver, the local oscillator frequency is 1.245 MHz. The tuned station frequency is
a) 455 MHz b) 790 KHz c) 690 Khz
d) 900 KHz.
3. The radio waves after refraction from different part of ionosphere on reaching the earth are called as
a) ground waves b) sky waves c) space waves d)
micro waves.
4. The principle used for transmission of light signals through optical fibre is a) refraction
b) diffraction c) polarization d)
total internal reflection.
5. Digital signals are converted into analog signals using a) FAX b)
modem c) cable d) coaxial cable.
6. The RF channel in a radio transmitter produces a) audio signals
b) high frequency carrier wave c) both audio signals and high frequency
carrier wave d) low frequency carrier wave.
7. In interlaced scanning time taken to scan one line is a) 20ms b) 64μs
c) 50 ms d) 100s
8. Skip distance is the shortest distance between a) the point of
transmission and the reception b) the uplink and the downlink station
c) the transmitter and the target d) the receiver and the target.
9. The RF channel in a radio transmitter produces a) audio signals
b) high frequency carrier wave c) both audio signal and high frequency
carrier wave d) low frequency carrier wave.
10. An AM signal has a resting frequency of 105 MHz and highest frequency of 105.03 MHz when modulated by a signal.
Then the carrier swing is a) 0.03 MHz b) 0.06 MHz c) 0.03 KHz d) 60 MHz.
11. For FM receivers the intermediate frequency is a) 455 KHz
b)455 MHz c) 10.7KHz d) 10.7MHz.
12. A carrier wave of amplitude 10mV is modulated by a sinusoidal audio signal wave of amplitude 6 mV the modulation
factor is a) 0.6 b) 6 c) 60 d)
0.06
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13. Transmitting antenna converts the a) electrical signal into
electrical energy b) electrical signal into magnetic energy
c) electrical signal into electromagnetic energy
d) electromagnetic energy into electrical signal.
14. How many synchronizing pulses are used for scanning? a) one b)
two c) three d) four
15. In TV transmission, the picture should not be scanned during the return journey of the scanning. This is done by
a) vertical scanning pulse b) horizontal scanning pulse
c) blanking pulse d) triggering pulse.
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