COUNSELLING CODE
P.B. College of Engineering
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Campus Address :
Administration Office :
Irungattukottai, Chennai – Bangalore National Highway
Sriperumbudur Taluk, Kanchipuram Dist., Chennai – 602 117.
Tel. : 044-27198033, 9940026861, 9600112733
E-mail : pbce@pbce.in Website : www.pbce.in
3/117, Brahmin Street, Karambakkam,
Porur, Chennai – 600 116.
Tel. : 044-24766386, Telefax : 044-24766941
Cell : 9600113031, 9962206098
MODEL QUESTION PAPER – II
PHYSICS
Note : i)
Answer all the questions
ii) Choose and Write the correct answer
iii) Each question carries one mark.
PART – I
1.
The unit of electric field intensity is ____
(a) NC-1
(b) NC
(c) Vm-1
(d) Vm
Answer : (c)Vm-1
2.
A capacitor of Capacitance 6F is connected to a 100 V battery. The energy Stored
in the Capacitor is _____
(a) 30J
(b) 3J
(c) 0.03J
(d) 0.06J
Answer : (c) 0.03J
3.
The ratio of electric potential at points 10 cm and 20 cm from the centre of an
electric dipole along its axial line is
(a) 1 :2
(b) 2 :1
(e) 1 :4
(d) 4 :1
Answer : (d) 4 :1
4.
A dielectric will have
(a) Free electrons
(b) No electrons
(c) Tightly bound electron
(d) one electron in the outer orbit.
Answer : (c) Tightly bound electron
5.
A toaster operating 240V has a resistance 120. The power is
(a) 400W
(b) 2 W
(c) 480 W
(d) 240 W.
Answer : (d) 240 W.
6.
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.
Answer : (d) Small couple per unit twist.
7.
The direction of force on current carrying conductor placed in a magnetic field is
given by
(a) Fleming’s left hand rule
(c) End rule
(b) Fleming’s Right hand rule .
(d) Right hand Palm rule.
Answer : (a) Fleming’s left hand rule
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P.B. College of Engineering
8.
1222
If the flux associated with a coil varies at the rate of 1Wb/minute then the induced
e.m.f. is ___
(a) IV
(b) 1/60V
(c) 60 V
(d) 0.60 V.
Answer : (b) 1/60V
9.
A power of 11,000W is transmitted at 220V. The current through live
wires is.
(a) 50A
(b) 5A
(c) 500A
(d) 0.5A
Answer : (a) 50A
10. The Unit Henry can also be written as P1
(a) VsA-1
(b) WbA-1
(c) S
(d) all the above
Answer : (d) all the above
11. Q factor has values lying between ____for normal frequencies.
(a) 0 to 10
(b) 10 to 50
(c) 50 to 100
(d) 10 to 100.
Answer : (d) 10 to 100.
12. When a drop of water is introduced between the glass plate and plano convex lens
in Newton rings system The rings.
(a) Contracts
(b) expands
(c) remains same
(d) first expands, then contracts
Answer : (a) Contracts
13. Electric filament lamp gives rise to
(a) line spectrum
(b) Continuous Spectrum
(c) Continuous absorption spectrum
(d) line absorption spectrum.
Answer : (b) Continuous Spectrum
14. In Newton’s ring experiment, the radii of 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.
Answer : (d) 10.
15. Ratio of intensities of two waves 4 :1 then, the ratio of their amplitudes.
(a) 2:1
(b) 1:2
(c) 4:1
(d) 1:4
Answer : (a) 2:1
16. The ratio of the radii of the first three Bohr orbits is ________
(a) 1 : ½ : 1/3
(b) 1 : 2 :3
(c) 1 : 4 : 9
(d) 1 : 8 : 27
Answer : (c) 1 : 4 : 9
17. Maser Materials are
(a) diamagnetic ions (b) Paramagnetic ions. (c) Ferro magnetic ions (d) non-magnetic
ions.
Answer : (b) Paramagnetic ions.
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18. In Somerfield atom model, for principal quantum number n=3, which of the
following subshells represents circular orbit ?
(a) 2S
(b) 2P
(c) 2d
(d) None of these.
Answer : (b) 2P
19. The Value of Rydberg’s Constant is __________
(a) 1.094 x 10-7m-1
(b) 1.094 x 107m+1
(c) 1.094 x 107 m-1
(d) 1.094 x 10-7 m1
Answer : (c) 1.094 x 107 m-1
20. If 1 kg of substance is fully converted to energy, the energy produced is.
(a) 9x1016J
(b) 9x1024J
(c) 1J
(d) 3x108J
Answer : (a) 9x1016J
21. The number of debroglie waves of an electron in the nth orbit of an atom is
(a) n
(b) n-1
(c) n+1
(d) 2n
Answer : (a) n
22. The radio isotope used in agriculture is
(a)
15P31
Answer : (a)
(b)
15P32
(c)
11Na23
(d)
11Na24
15P31
23. The mean life of radon is 5.5 days. It half life is
(a) 8 days
(b) 2.8 days
(c)0.38 days
(d) 3.8 days.
Answer : (d) 3.8 days.
P2
24. Which of the following belongs to Baryon group?
(a) Photon
(b) Electron
(c) Pion
(d) Proton.
Answer : (d) Proton.
25. The half – life Period of N13 is 10.1 minute .Its life time is
(a) 5.05 minute
(b) 20.2 minute
(c)
Answer : (d) infinity
10.1 minute
0.6931
(d) infinity
26. The colour of light emitted by a LED depends on
(a) its reverse bias
(b) amount of forward current
(c) Its forward bias
(d) type of semiconductor material.
Answer : (d) type of semiconductor material.
27. The following arrangement performs the logic function of
A
Y
B
(a) AND gate
(b) OR gate .
(c) NAND gate
(d) Ex-OR gate.
Answer : (a) AND gate
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28. Bark hausen Condition for maintenance of oscillation
(a) =1/A
(b) AB =
(c) A=
(d) A = ½
Answer : (a) =1/A
29. The audio frequency range is
(a) 20HZ to 200HZ
(b) 20HZ to 2000HZ
(c) 20Hz to 200000HZ
(d) 20 HZ to 20000 HZ
Answer : (d) 20 HZ to 20000 HZ
30. Videocon Camera tube works on the Principle of
(a) Photo Conductivity (b) thermoelectric effects.
(c) themionic emission (d) seeback effect
Answer : (a) Photo Conductivity
PART – II
Note : Answer any Fifteen Questions.
15x3=45
31. Define electric flux. Give its unit :
Solution :
The electric flux is defined as the total number of electric lines of force, crossing
through the given area.
The electric flux d through the area is
d =
= Eds cos 
.
Its unit is Nm2C-1
32. Calculate the potential at a point due to a charge 4 x 10-7C located at
0.09 m away from it.
Solution :
V
=
=
1
4
q
r
9 x 109 x 4 x 10-7
9 x 10-2
= 4 x 104 V
33. Define temperature coefficient of resistance.
Solution :
The Temperature coefficient of resistance is defined as the ratio of increase in
resistance per degree rise in temperature to its resistance at 0oC
o
Its Unit is per C
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34. Two wires of same material and length have resistance 5 and 10 respectively.
Find the ratio of radii of the two wires.
Solution :
R1 = 5 ,R2 = 10
Radius of 1st wire = r1, radius of 2nd wire = r2
specific Resistance = 
Length = l

l
A
l
=
r12
(A = r 2)
R =
R1
R2
R1
=
R2
=
r2
r1 : r2
r22
r12
r22
r1

l
,R2 =
=
R1
10
5
=
2
2:1
=
35. From the following network, find the effective resistance between A and B.
R1=15
A
B
R2=15
Solution :
R1=15
A
B
R2=15
1
RP
1
=
=
=
RP
=
R1
1
15
1
+
+
R2
1
15
2
15
15
2
= 7.5 ohm.
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36. Find the magnetic induction at a point, 10 cm from a long straight wire carrying
current of 10A.
I = 10A, a = 10cm = 10 x 10-2m, B=?
Solution :
B
=
oI
2a
xx
=
2 x x
B = 2 x 10-5 T
37. State Fleming’s left hand rule.
Solution :
The fore finger, the middle finger and the thumb of the left hand are stretched
in mutually perpendicular directions.
If the fore finger 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 of the conductor.
38. What is Capacitive reactance ?
Solution :
(i) capacitive reactance is the resistance offered to the flow of current by a capacitor.
Xc =

c

= vc
its unit is ohm.
(ii) In a d.c. circuit frequency  = O, Xc = ∞It blocks D.C.
(iii) For an a.c. the capacitive reactance varies inversely as the frequency of a.c and
also inversely as the capacitance of the Capacitor.
39. Why does the sky appear blue in colour ?
Solution :
At Sunrise the blue appearance of the sky is due to scattering of sunlight by
the atmosphere. According to Rayleigh’s Scattering law, blue light to a greater
extent than red light. This Scattered radiation causes the sky to appear blue.
40. The refractive index of a medium is
3. Calculate the angle of refraction if the
unpolarised light is incident on it at the polarizing angle of the medium.
 = 3. ; r = ?
Solution :
According to Brewster’s law

= tan ip
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= 3
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or
Angle of refraction,
ip = tan-1 3 = 600
r = 90-ip
r = 90-60
r = 30o
41. Write the principle of Millikan’s oil drop experiment.
Solution :
(i)
It is based on the study of the motion of uncharged oil drop under free fall due
to gravity and charged oil drop in a uniform electric field.
(ii)
By adjusting uniform electric field suitably, a charged oil drop can be made to
move up or down or even kept balanced in the field of view for sufficiently long
time and a series of observations can be made.
42. Write any three medical applications of LASER.
Solution :
(i)
Micro surgery has become possible due to narrow angular spread of the laser
beam.
(ii)
The laser beams are used in endoscopy.
(iii) It can also be used for the treatment of human and animal cancer.
43. State the postulates of special theory of relativity.
Solution :
(i)
The laws of Physics are the same in all inertial frames of reference.
(ii)
The velocity of light in free space is a constant in all the frames of reference.
44. Define critical mass and critical size.
Solution :
(i)
Critical size of a system containing a fissile material is defined as the minimum
size in which atleast one neutron is available for further fission reaction.
(ii)
The mass of the fissile material at the critical size is called critical mass.
(iii) The chain reaction is not possible, if the size is less than the critical size.
45. Write the proton – proton cycle that takes place in sun and stars.
Solution :
1
H 1 + 1H 1 →
1
H 1 + 1H 2→
2
1
H2 + 1e0 +  (emission of positron and neutrino)
He3 + +  (emission of gamma rays)
22He3 → 2He4 + 21H1
The reaction cycle is written as
4 H 1→
1
2
He4 + 21e0 + 2 + energy (26.7 Mev)
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This four protons fuse together to form an alpha particle and two positrons with a
release of large amount of energy.
46. Define input impedance of a transistor is connected in common emitter mode.
Solution :
The Input impedance of the transistor is defined as the ratio of small change in base
emitter voltage to the corresponding change in base current at a given VCE
 Input impedance ri =
V
I
ohm
VCE
47. What is an extrinsic semi conductor ?
Solution :
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.
Types ; P-type, N-type.
48. What are the advantages of negative feedback ?
Solution :
(i) Highly established gain
(ii) Reduction in the noise level
(iii) Increased band width
(iv) Increased input impedance and decreased output impedance
(v) Less distortion
49. Draw the circuit configuration of NPN transistor in common collector (c.c.) mode.
Solution :
VBB = input voltage source
VCC = output voltage source
E = Emitter
B = Base
C = Collector
50. What is Directivity of an Antenna ?
Solution :
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.
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PART – III
Note : (i) Answer Q. No 56 Compulsorily
(ii) Answer any six of the remaining questions
(iii) Draw Diagrams wherever necessary.
51. Derive an expression for the capacitance of a parallel plate capacitor with a
dielectric medium.
Solution :
Consider a parallel plate capacitor having two conducting plates X and Y
each of area A, separated by a distance d apart. X is given a positive charge, so that
the surface charge density on it is  and Y is earthed.
Let a dielectric slab of thickness t and relative permittivity r be introduced
+σ
between the plates.
AIR
X
d
Thickness of dielectric Slab = t
Thickness of air gap = (d-t)
DIELECTRIC
t
AIR
Y
t<d
Electric field at any point in the air between the plates,
E
=


Electric field at any point, in the dielectric slab.
E΄
=

 r 
The total potential difference between the plates is the work done in crossing
unit positive charge from one plate to another in the field E over a distance (d-t) and
in the field E’ over a distance t, then.
V
= E(d-t) + E΄t
=


=


(d-t) +
[(d-t) +
t
 r 
t
r
]
The charge on the plate X, q =  A
Hence the capacitance of the capacitor is
C =
q
v
A
=


A
=
(d-t) +
t
r
(d-t) + t
r
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Effect of dielectric:
In capacitors, the region between the two plates is filled with dielectric like
mica or oil.
The Capacitance of the air filled capacitor, C =
A
d
The Capacitance of the dielectric filled capacitor , C’ =
C’
C =
r
r A
d
C’ = r C
or
52. Explain the determination of the internal resistance of a cell using voltmeter.
Solution :
The Internal resistance of a cell :
The electric current in an external circuit flows from the positive terminal to
the negative terminal of the cell, through different circuit elements.
In order to maintain continuity, the current has to flow through the electrolyte
of the cell, from its negative terminal to the positive terminal. During this process of
flow of current inside a cell, a resistance is offered to current flow by the electrolyte
of the cell.
A freshly prepared cell has low internal resistance and this increases with
aging.
Determination of internal Resistance :
The circuit connections are made as shown,
With key k open, the emf of cell E of is found by connecting a high resistance
voltmeter across it.
Since the high resistance voltmeter draws only a very feeble current for
defection, the circuit may be considered as an open circuit. Hence the voltmeter
reading gives the emf of the cell.
A small value of resistance R is included in the external circuit and key k is
closed. The potential difference R is equal to the potential difference across cell (v)
The potential drop across R
V = IR (or) IR = V
……………..
1
Due to Internal resistance r of the cell
Then V = E – Ir (or)
Ir = E-V
……………..
2
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P.B. College of Engineering
dividing 2 by 1
Ir
IR
=
r
1222
E-V
V
E-V
V
=
R.
Thus r can be calculated by knowing E, V and R.
53. In the given network, calculate the effective resistance between points A and B.
10
5
5
10
5
10
B
A
5
5
10
10
5
10
Solution :
The network has three individual units
The simplified form of one unit is
The equivalent resistance of one unit is
1
Rp
=
1
R1
+
1
R2
=
1
15 +
1
15
Rp = 7.5 
Each Unit has a resistance of 7.5 . The total network reduces to
A
The combined resistance between A and B
R = R’ + R’ = R’
= 7.5 + 7.5 + 7.5
R = 22.5 
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54. A moving Coil galvanometer of resistance 20 produces fullscale deflection for a
current of 50mA. How will you convert the galvanometer into (i) an ammeter of
range 20A and (ii) a voltmeter of range 120 volt ?
Data
G = 20
Ig = 50 x 10-3 A
I = 20A
S = ? ,V = 120V, R = ?
Solution :
(i)
S
= G
S
=
S
=
Ig
I-Ig
20 x 50 x10-3
20 – 50 x10-3
20 x 0.05
=
20 – 0.05
20 x 0.05
19.95
= 0.05 
A Shunt of 0.05  Should be connected in parallel with galvanometer.
(ii)
R
=
V
Ig
R
=
120
– 20
50 x 10-3
-G
= 2400 – 20 = 2380 
A resistance of 2380  should be connected in series with galvanometer.
55. Obtain the phase relation between current and voltage in an a.c. circuit with
inductor only.
Solution :
Let an alternating source of emf be applied to a pure inductor of inductance L.
The inductor has a negligible resistance and is wound on a laminated iron core.
Due to an alternating emf that is applied to the inductive coil. a self induced emf is
generated which opposes the applied voltage. (eg) Choke coil.
The instantaneous value of applied emf is given by,
e = Eo sin t
……….. (1)
di
Induced emf e’ = - L /
dt
where L is the self inductance of the coil. In an ideal inductor circuit induced
emf is equal and opposite to the applied voltage.
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Therefore e = -e΄
E e΄ = –
di =
–L
Eo
L
di
dt
sin t dt
Integrating both the sides
i =
=
i =
Eo
L
Eo
L
Eo
L
 sin t dt
–
cos t

=–

sin ( t –
)
2
i = Io. sin(t –
where Io =
Eo cos t
L
Eo
L

2
)
……… (2)
.
Here, L is the resistance offered by the coil. It is called inductive reactance.
Its unit is ohm.
From equations (1) and (2) it is clear that in
an a.c. circuit containing a pure inductor the current i
lags behind the voltage e by the phase angle of /2.
Conversely the voltage across L leads the current by the
phase angle of /2. This fact is presented graphically
in figure (b)
Fig. (c) represents the phasor diagram of a.c. circuit containing only L.
Inductive reactance
XL = L = 2vL, where v is the frequency of the a.c. supply.
For d.c.  = 0 ; XL = 0
Thus a pure inductor offers zero resistance to d.c. But in an a.c. circuit the
reactance of the coil increases with increase in frequency.
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56. A plane transmission grating has 5000 lines / cm. Calculate the angular separation
in second order spectrum of redline 7070Å and Blue line 5000 Å
N = 5000 lines / cm = 5000 x 102 lines /m = 5 x 105 lines/m
m = 2 ; R = 7070Å = 7.07 x10-7m
B = 5000 Å = 5 x 10-7 m
sin R = NmR = 5 x105 x 2 x 7.07 x 10-7
= 0.707
R = 450
sinB = NmB = 5 x105 x 2 x 5 x 10-7
= 0.5
B = 300
Angular Separation = 
R
-
B
= 450 - 300
= 150
(or)
A 500mm long tube containing 60 C.C. of sugar solution produces a rotation of
9° when paced in a polarimeter. If the specific rotation is 60°, Calculate the quantity
of sugar contained in the solution.
l = 500mm = 5 dm
 = 90 ;S = 600;V = 60cc; m= ?
S=

lxC
=

; m=
l x (m/v)
v
I x S=
9 x 60
5 x 60
m = 1.8 g
57. Describe laue experiment. What are the facts established by it.
Von laue suggested that a crystal can act as a three dimensional grating for an
X ray beam. X rays from the X – ray tube is collimated into a fine beam by two slits
S1, and S2. The beam is now allowed to pass through a zinc sulphide (ZnS) crystal.
The emergent rays are made to fall on a photographic plate P.
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The diffraction patterns so obtained consist of a central spot 0 and a series of
spots arranged in a definite pattern about O. The central Spot is due to the direct
beam, whereas the regularly arranged spots are due to the diffraction pattern from
the atoms of the various crystal planes.
These spots are known as laue spots. The Laue experiment has established
following two important facts.
(i)
x rays are electromagnetic waves of extremely short wave length.
(ii)
The atoms in a crystal are arranged in a regular three dimensional lattice.
58. Obtain Einstein’s photoelectric equation.
In 1905, Albert Einstein, successfully applied quantum theory of radiation to
photo electric effect.
According to Einstein, the emission of photo electron is the result of the
infraction between a single photon of the incident radiation and an electron in the
metal. When a photon of energy h is incident on a metal surface, it’s energy is
used up in two ways.
(i)
A part of the energy of the photon is used in extracting the electron from the
surface of metal, Since the electrons in the metal are bound to the nucleus.
This energy W spent in releasing the photo electron is known as photo electric
work function of the photo metal. The work function of a photo metal is
defined as the minimum amount of energy required to liberate an electron from
the metal surface.
(ii)
The remaining energy of the photon is used to import kinetic energy to the
liberated electron.
If m is the mass of an electron and v, its velocity then.
Energy of the Incident photon = Work function + kinetic energy of the electron
h = W +
1
mv2
2
If the electron does not lose energy by internal collisions, as it escapes from
the metal, the entire energy (h –w) will be exhibited as the kinetic energy of the
electron.
Thus, (h-w) represents the maximum kinetic energy of the ejected photo
electron.
If Vmax is the maximum velocity with which the photo electron can be ejected,
then
h = W+
1
2
m
2
max
…………..(2)
This equation is known as Einstein’s photo electric equation.
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When the frequency () of the incident radiation is equal to the threshold
frequency (o) of the metal surface, kinetic energy of the electron is zero.
Then equation (2) becomes,
ho = W
Substituting the value of W in equation (2)
we get,
h – ho =
1
2
mmax
2
(or) h(-o) =
1
m2max
2
This is another form of Einstein’s photo electric equation.
59. How fast would a rocket have to go relative to an observer for its length to be
corrected to 99% of its length at rest ?
Data : = 99% = 99/100 ; v = ?
Solution : l = lo
1 –
l
=
99
lo
100
l
lo
=
99
100
99
=
100
1 –
v2
c2
v2
c2
v
= 0.141c
v
= 0.141 x 3 x 108
v
= 0.423 x 108 ms-1
60. The binding energy per nucleon for 6C12 Nucleus is 7.68 Mev and that for 6C13 is
7.47 Mev, Calculate the energy required to remove a neutron from 6C13 nucleus.
Data : Binding energy per nucleon of 6C12 = 7.68 MeV Binding energy of neutron = ?
Working :
We can write the reaction as
6C13
→ 6C12 + 0n1
Total binding energy of 6C13 = 7.47 x 13
= 97.11MeV
Total binding energy of 6C12 = 7.68 x12
= 92.16 MeV
Total binding energy of the reactant = Total binding energy of the product
 97.11 MeV = 92.16 MeV + Binding energy of a neutron
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 Binding energy of a neutron = 97.11-92.16
= 4.95 MeV
61. With the circuit diagram, explain voltage divider biasing of a transistor.
Voltage divider bias :
In this method, 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 biases the base emitter junction. This causes the
base current and hence collector to flow in zero signal conditions.
The stabilization provided by RE : Since b is very Sensitive to temperature
changes, the collector current IC increases with rise in temperature, Consequently,
it can be seen that IE increases. This will cause the voltage drop across emitter
resistance RE to increase. The voltage drop across R2 = VBE + VRE. As voltage drop
across R2 is independent of IC, VBE decreases. This decreases IB and the reduced
value of IB tends to bring back IC to the original value. Hence any variation of b will
have no effect on the operating point.
62. With the help of block diagram, explain the operation of an FM super heterodyne
receiver.
An FM receiver is a super heterodyne type like a typical AM receiver. The
functional block diagram of any FM receiver is shown in the figure.
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The RF section, selects the modulated signals and is amplified. It is then fed
into the mixer and local oscillator. Here the frequency of the modulated signal is
changed to intermediate frequency. For FM receivers, this IF is 10.7 MHz. The
intermediate frequency wave is amplified using IF amplifier and then its amplitude
is maintained constant using a limiter. The output of this section is applied to the
FM detector which demodulates the modulated wave. The AF signal from the FM
detector is then passed on through a de – emphasis network, where the various
frequencies attain their original power distribution. Finally it is fed into the loud
speaker after performing AF amplification.
PART – IV
Note : (i) Answer any four question in detail
(ii) Draw diagrams wherever necessary.
63. What is an electric dipole ? Derive an expression for the electric field due to an
electric dipole at a point on its axial line.
Two Equal and opposite charges separated by a very small distance constitute
an electric dipole.
Some examples are water, ammonia, and chloroform molecules.
Derivation :
AB is an electric dipole of two point charges – q and + q separated by a small
distance 2d. 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 ,
E1 =
1
q
4
(r-d)2
(along BP)
The electric field at the point P due to – q placed at A is
E2 =
1
4
.
q
(r+d)2
(along PA)
E1 and E2 act in opposite direction.
Therefore the magnitude of resultant electric field (E) acts in the direction of the
vector with a greater magnitude. The resultant electric field at P is,
E
= E1 + (–E2)
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E
1
=
=
=
4
.
q
–
(r+d)2
q
1q
4
(r-d)2
1
4rd
4
(r2-d2)2
.
4
q
along BP
(r+d)2
1q
–
q
1222
along BP
(r+d)2
along BP
If the point P is for away from the dipole, then
d<<r
E =
E =
q
4
1
4
.
4rd
=
r4
2P
.
q
.
4
4d
r3
along BP
r3
[
Electric dipole moment p = q x 2d]
E acts in the direction of dipole moment.
64. Apply Biot – Savart law, obtain an expression for the magnetic induction at a point
due to infinitely long straight conductor carrying current.
XY is an infinitely long straight conductor carrying current I . P is a point at a
distance a from the conductor. AB is a small element of length dl.  is the angle
between the current element Idl and the line joining the element dl and the point P.
According to Biot – Savart law, the magnetic induction at the point P due to the
current element Idl is
dB =

Idl.sin
4
r2
……………….(1)
AC is drawn perpendicular to BP from A
OPA = APB =d

AC
AC
In ABC, sin 
AB
dl

AC = dl sin
……………. (2)
From  APC, AC = rd
……………. (3)
from equation (2) and (3)
rddl.sin
 
................(4)
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substituting equation (4) in equation (1)
dB =

Ird
4
r2
=
Id
4
r
………………….. (5)
a
In  OPA, cos  =
r=

r
a
cos 
substituting eqn. (6) in eqn. (5)
dB =

1
.
4
cos  d 
a
The total magnetic induction at P due to the conductor XY is

B


=
dB =

=


4a
cos  d


[ sin 1 + sin 2 ]
4a
for infinitely long conductor, 1290°

B=

2a
If the conductor is placed in a medium of permeability,
B=
l
2a .
65. Discuss with theory the method of inducing e.m.f. in a coil by changing its
orientation with respect to the direction of the magnetic field.
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  in the clockwise direction
about an axis perpendicular to the direction of the magnetic field. Suppose, initially
the coil is in vertical position, so that the angle between normal to the plane of the
coil and magnetic field is zero. After a time t, let  be the angle through which the
coil is rotated. If  is the flux linked with the coil at this instant , then
cos(since  = t)
The induced emf is
e
= –
d
dt
= – NBA
d
dt
cos (t)
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COUNSELLING CODE
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 e
1222
= NBA sint
……………. (1)
The maximum value of the induced emf is Eo = NAB



INDUCED EMF BY CHANGING THE ORIENTATION OF THE COIL
Hence, the induced emf can be represented as e – Eo sint.
The induced emf e varies sinusoidally with time t and the frequency
being  cycles per second  =
(i)

2
When t = 0, the plane of the coil is perpendicular to the field B and
hence e = 0.
(ii)
When t = /2, the plane of the coil is parallel to B and hence e = E0.
(iii) When t = , the plane of the coil is at right angle to B and hence e = 0.
(iv) When t = 3, the plane of the coil is again parallel to B and the
induced emf is e = 0.
(v)
When t = 2, the plane of the coil is again perpendicular to B and hence
e = 0.
If the ends of the coil are connected to an external circuit through a resistance R,
current flows through the circuit, which is also sinusoidal in nature.
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66. Derive an expression for band width of interference fringes in young double slit
experiment.
Expression for bandwidth : Let d be the distance between two coherent sources A
and B of wavelength . A screen XY is placed parallel to AB at a distance D from the
coherent sources. C is the midpoint of AB. O is a point on the screen equidistant
from A and B. P is a point at a distance x from O. Waves from A and B meet at P in
phase or out of phase depending upon the path difference between two waves.
Draw AM perpendicular to BP
The path difference  = BP – AP
AP = MP
  = BP – AP = BP – MP = BM
In right angled  ABM, BM = d sin 
If  is small, sin  = 
 The path difference  = .d
In right angled triangle COP, tan  =
OP
CO
=
x
D
For small values of , tan  = 
xd
 The path difference  =
D
Bright fringes : By the principle of interference, condition for constructive
interference is the path difference = n

xd
= n
D
Where n = 0,1,2…indicate the order of bright fringes.
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COUNSELLING CODE
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x =
1222
D
d
n
This equation gives the distance of the nth bright fringe from the point O.
Dark fringes : By the principle of interference, condition for destructive
interference is the path difference = (2n – 1) 
d
Where n = 1,2,3…. indicate the order of the dark fringes.
x=
D
d
(2n – 1)

d
This equation gives the distance of the nth dark fringe from the point O. Thus,
on the screen alternate dark and bright bands are seen on either side of the central
bright band .
Band width () : The distance between any two consecutive bright or dark bands is
called bandwidth. The distance between (n+1)th and nth order consecutive bright
friges from O is given by
X(n + 1) –xn = D
d
(n+1) -
D n = D 
d
d
Bandwidth,  = D 
d
Similarly, it can be proved that the distance between two consecutive dark
bands is also equal to D . Since bright and dark fringes are of same width, they
d
are equi-spaced on either side of central maximum.
67. State Bohr’s Postulates. Obtain an expression for the radius of the n th orbit of
hydrogen atom.
(i) An electron cannot revolve round the nucleus in all possible orbits. The electrons
can revolve round the nucleus only in those allowed or permissible orbits for which
h
the angular momentum of the electron is an integral multiple of
(where h is
2
Planck’s Constant = 6.626 x 10-34 Js). These orbits are called stationary orbits or
non-radiating orbits and an electron revolving in these orbits does not radiate any
energy.
If m and  are the mass and velocity of the electron in a permitted orbit of radius r
nh
then angular momentum of electron =mr =
, Where n is called principal
2
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quantum number and has the integral values 1,2,3… This is called Bohr’s
quantization condition.
(iii) An atom radiates energy, only when an electron jumps from a stationary orbit
of higher energy to an orbit of lower energy. If the electron jumps from an orbit of
energy E2 to an orbit of energy E1, a photon of energy h = E2- E1 is emitted. This
condition is called Bohr’s frequency condition.
Radius of the nth Orbit (rn) :
Consider an atom whose nucleus has a positive charge Ze, where Z is the
atomic number that gives the number of protons in the nucleus and e, the charge of
the electron which is numerically equal to that of proton. Let an electron revolve
around the nucleus in the nth orbit of radius rn
By Coulomb’s law, the electrostatic force of attraction between the nucleus
1
(Ze) (e)
and the electron =
.
………………(1)
rn2
4
Where 0 is the permittivity of the free space.
Since, the electron revolves in a circular orbit, it experiences a centripetal force,
2
mn
=
rn
mrnn2
………………(2)
where m is the mass of the electron, n and n are the linear velocity and angular
velocity of the electron in the nth orbit respectively. The necessary centripetal force
is provided by the electrostatic force of attraction. For equilibrium, from equations
(1) and (2),
1
4
1
4
From equation (4),
Ze2
.
rn2
.
Ze2
rn2
=
mn2
rn
= mrnn2
………………(3)
………………(4)
Ze2
n2 = 4mrn3
………………(5)
The angular momentum of an electron in nth orbit is,
L = mnrn = mrn2n
………………(6)
By Bohr’s first postulate, the angular momentum of the electron
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L=
1222
nh
………………(7)
2
where n is an integer and is called as the principal quantum number. From
equations, (6) and (7),
nh
mrn2n =
2
nh
(or) n =
2 mrn2
squaring both sides,
n2
=
n2h2
42 m2rn4
……………………..(8)
From equations (5) and (8),
n2h2
Ze2
4mrn3
(or)
= 42 m2r 4
n
rn
=
n2h2
mZe2
…………………….(9)
From equation (9), it is seen that the radius of the nth orbit is proportional
to the square of the principal quantum number. Therefore, the radii of the orbits
are in the ratio 1 : 4 : 9………….
For hydrogen atom, Z = 1
 From equation (9)
rn =
n2h2
me2
Substituting the known values in the above equation we get,
rn = n2 x 0.53 Å
If n = 1, r1 = 0.53 Å
This is called Bohr radius.
68. Obtain an expression for the amount of the radioactive substance present at any
moment. Obtain the relation between half –life period and decay constant.
Radioactive law of disintegration :
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.
This is known as Radioactive law of disintegration.
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Let N0 be the number of radioactive atoms present initially and N, 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. Then the rate of disintegration is
_ dN  N
dt
dN
=–N
………………… (1)
dt
where  is a constant known as decay constant or disintegration constant. The
negative sign indicate that N decreases with increase in time.
Equation (1) can be written as
dN
= -  dt
dt
Integrating,
logeN = -  t + C
………………… (2)
where C is a constant of integration,
At t = 0, N = N0
logeN0 = C
Substituting for C, equation (2) becomes,
logeN = -  t + logeN0
loge
N
N0
N
N0
=-  t

= e- t

N = N0e- t
………………… (3)
Equation (3) shows that the number of atoms of a radioactive substance decreases
exponentially with increase in time.
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Initially the disintegration takes place at a faster rate. As time increases, N
gradually decreases exponentially. Theoretically, an infinite time is required for the
complete disintegration of all the atoms.
Half life period : The half period of a radioactive element is defined as the time
taken for one half the radioactive element to undergo disintegration.
From the law of disintegration
N = N0e
-t
Let T½ be the half life period. Then, at t = T½, N =
N0
2
= N0e
N0
2
-T½
loge2 = T½
T½ =
loge2

=
log102 x 2.3026

=
0.6931

The half life period is inversely proportional to its decay constant.
For a radioactive substance, at the end of T½, 50% of the material remain
unchanged. After another T½ i.e., at the end of 2T½, 25% remain unchanged. At the
end of 3T½, 12.5% remain unchanged and so on.
69. Sketch the circuit of a colpit oscillator and explain its working.
Colpitt’s oscillater : The circuit diagram of Colpitt’s oscillator is shown in figure.
The resistance R1, R2 and RE provide the sufficient bias for the circuit.
The frequency determining network is the parallel resonant circuits consisting of
capacitors C1, C2 and the inductor L. The junction of C1 and C2 is earthed. The
function of the capacitor C4 is to block d.c. and provide an a.c 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 oscillation 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 points 1 and 2 are 180 0 out of phase.
The amplifier produce further phase shift 1800. Thus the total phase shift is 3600.
In other words, energy supplied to the tank circuit is in phase with the oscillations
and if A = 1, oscillations are sustained in the circuit.
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The frequency of oscillations is given by f =
1
where C =
2 LC
f=
1
(C1+C2)
2
LC1C2
C1C2
C1+C2
70. With the help of a block diagram, explain the function of a RADAR system.
The Block diagram of a simple radar system is shown in the figure. This block
diagram indicates that the radar system consists of both the transmitting and the
receiving system.
The transmitting system consists of a transmitter and a pulser. The receiving
system consists of a receiver and an indicator. In most of the cases, a single
antenna is used for both transmission and reception and this is achieved with the
use of TR switch (Transmitter Receiver Switch). This switching arrangement is
called as ‘duplexer’. This connects the antenna to the transmitter during
transmission and to the receiver during reception. Moreover, this switch isolates the
sensitive receiver from the damaging effects of the high power transmitter.
The transmitter is essentially 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.
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If the transmitted pulse hits any target, a weak echo signal returns to the
same antenna. But, now the TR switch puts the antenna in contact with the
receiver. This echo signal is amplified and demodulated by the superhet receiver.
The sensitivity of the receiver is very high. 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. So, the indicator records the transmitted pulse as well as the
returning pulse simultaneously. The returning echo pulse appears slightly displaced
from the transmitted pulse and this displacement is a measure of the range of the
target.
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Campus Address :
Administration Office :
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