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Title: Physics Study Material For Class 12th
Description: Best Physics Study Material Issued by Central Board Of Secondary Education (CBSE) For Class 12th................I Hope This Material is useful for you.........Thank you...........& All the Best for your studies....................................................
Description: Best Physics Study Material Issued by Central Board Of Secondary Education (CBSE) For Class 12th................I Hope This Material is useful for you.........Thank you...........& All the Best for your studies....................................................
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TEAM MEMBERS
Sl
...
Name
Designation
1
...
Devendra Kumar
Member
P
...
T
...
P
...
V
...
Girija Shankar
Member
P
...
T
...
S
...
V
...
1, Bhola Nath Nagar,
Shahdara, Delhi
...
K
...
Sharma
Member
P
...
T
...
P
...
V
...
042)
Time : 3 hours
Sl
...
Typology of Questions
Max
...
Remembering – (Knowledge based
Simple recall questions, to know
specific facts, terms, concepts,
principles, or theories; Identify,
define, or recite, information)
2
1
1
2
...
Application – (Use abstract
information in concrete situation, to
apply knowledge to new situations;
Use given content to interpret a
situation, provide an example, or
solve a problem)
-
4
...
Evaluation and Multi-Disciplinary
– (Appraise, judge, and/or justify the
value or worth of a decision or
outcome, or to predict outcomes
based on values)
TOTAL - 2 Projects
[Class XII : Physics]
SA-II Value based
(3M)
question
(4M)
(LA)
(5M)
Total
Marks
-
-
7
10%
4
-
1
21
30%
2
4
-
1
21
30%
2
-
1
-
1
10
14%
1
-
2
1
-
11
10%
5×1=5 5×2=10 12×3=36 1×4=4 3×5=15 70(26)
2
%
Weightage
100%
Type of Question
Mark per Question
Total No
...
Internal Choices : There is no overall choice in the paper
...
2
...
Suitable internal variations may be made for
generating similar templates keeping the overall weightage to different form of questions
and typology of questions same
...
Marks : 70
No
...
3
[Class XII : Physics]
Electric field, electric field due to a point charge electric field lines electric
dipole, electric filed due to a dipole, torque on a dipole in uniform electric field
...
Electric Potential, Potential difference, electric potential due to a point charge,
a dipole and system of charges, equipotential surfaces, electrical potential energy
of a system of two point charges and of electric dipole in an electrostatic field
...
Unit II:
Current Electricity
20 Periods
Electric current; flow of electric charges in a metllic conductor, drift velocity,
mobility and their relation with electric current
...
Internal resistance of a cell, potential difference and emf of a cell
...
Kirchhoff’s laws and simple applications,
wheatstone bridge, metre bridge
...
Unit III:
Magnetic Effects of Current and Magnetism
22 Periods
Concept of magnetic field and Oersted’s experiment
...
Ampere’s law and its applications to infinitely long straight wire, straight and
toroidal solenoids
...
Cyclotron, Force on a current carrying conductor in a uniform magnetic
field, force between two parallel current carrying conductors, definition of ampere
...
Moving coil
Galvanometer – its current sensitivity
...
Current loop as a magnetic dipole and it’s magnetic dipole moment, Magnetic
dipole moment of a revolving electron, Magnetic field intensity due to a magnetic
dipole (bar magnet) along it’s axis and perpendicular to it’s axis
...
Para-, dia- and ferro-magnetic substances with examples
...
[Class XII : Physics]
4
Unit IV:
Electromagnetic Induction and Alternating Currents
20 Periods
Electromagnetic induction; Faraday’s laws induced emf and current; Lenz’s law,
Eddy currents
...
Need for displacement current
Alternating currents, peak and rms value of altering current/voltage Reactance
and Impedance
...
LCR series circuit;
Resaonance; Power in AC circuits, wattless current
...
Unit V:
Electromagnetic Waves
04 Periods
Need for Displacement current, Electromagnetic waves and their characteristics
(qualitative ideas only)
...
Electromagnetic spectrum (radiowaves, micro-waves, infrared, visible, ultraviolet, X-rays, gamma rays) including
elementary facts about their uses
...
Refraction of
light, total internal reflection and its applications, optical fibres, refraction through
spherical surfaces, lenses thin lens formula
...
Magnification,
power of a lens, Combination of thin lenses in contact, Refraction and dispersion
of light through a prism
...
Optical Instruments; Microscopes and astronomical telescopes (reflecting and
refracting) and their magnifying powers
...
Interference, Young’s double slit experiment
and expression for fringe width coherent sources and sustained interference of
light; Diffraction due to a single slit, width of central maximum
...
Unit VII:
Dual Nature of Matter and Radiation
08 Periods
Dual nature of radiation, Photoelectric effect Hertz and Lenard’s observations;
Einstein’s photoelectrical equation, Particle nature of light
...
Unit VIII:
Atoms and Nuclei
14 Periods
Alpha-particles scattering experiment, Rutherford’s model of atom, Bohr Model,
energy levels, Hydrogen spectrum
...
Mass-energy relation, mass defect; binding energy per nucleon and its varion
with mass number; nuclear fission, nuclear fusion
...
Special purpose p-n junction diodes: LED, Photodiodes, solarcell and Zener
diode as a voltage regulator
...
Transistor as
an amplifier (common emitter configuration), basic idea and analog and digital
signals, Logic gates (OR, AND, NOT, NAND and NOR)
...
Propagation
of electromagnetic waves in the atmosphere, sky and space wave propogation,
satellite communication
...
Basic ideas about internet, mobile telephony and global positioning
system (GPS)
...
No
...
Electrostatics
9
2
...
Magnetic Effects of Current and Magnetism
61
4
...
Electromagnetic Waves
103
6
...
Dual Nature of Matter and Radiation
130
8
...
Electronic Devices
156
Communication Systems
159
10
...
Resistance
Relation b/w
current and
Drift Velocity
2
...
Electrical Conductivity
Resistances in
7
...
Series
Current density
6
...
Drift Velocity
1
...
Cells in Series
11
...
Internal Resistance
of a cell
9
...
Potentiometer
metre Bridge
16
...
Wheatstone Bridge
14
...
Cells in parallel
Equivalent Current
i = o (at a junction)
iR = E (in a closed loop)
Equivalent resistance
Equivalent e
...
f
four arms of Wheatstone Bridge
...
m
...
m = number of cells in parallel
n = no
...
[Class XII : Physics]
16
Internal Resistance
Comparison of Emf
closed circuit
...
Draw schematically an equipotential surface of a uniform electrostatic field
along x-axis
...
2
...
(i)
(ii)
Ans
...
Ans
...
Is it a scalar or a
vector quantity?
Electric field intensity
...
17
[Class XII : Physics]
4
...
5
...
6
...
Two point charges repel each other with a force F when placed in water
of dielectric constant 81
...
2 × 10–32 Cm
...
p q(2a) 2a
3
...
6 10 19
13
Net capacitance of three identical capacitors connected in parallel is 12
microfarad
...
...
Ans
...
What
will be the electric flux due to this charge through any half of the sphere
...
q
0
q
2
2 0
Draw the electric field vs distance (from the centre) graph for (i) a long
charged rod having linear charge density < 0 (ii) spherical shell of radius
R and charge Q > 0
...
r
Linear Charge
9
...
E
E
Ans
...
Ans
...
What is the
electric field and electric potential at the centre? Ans
...
Ans
...
(Additive property of charge)
Q = (n2 – n1)e
12
...
What is the total positive or negative charge present in 1 molecule of
water
...
Ans
...
Work done = p E(Cos180° – Cos0°)
= –2p E
i
...
energy decreases
...
Ans
...
Eaxial
2K p
r3
,
Eequatorial
Kp
r3
Eaxial 2 Eequatorial
...
Draw equipotential surface for a dipole
...
16
...
An uncharged conductor A placed on an insulating stand is brought near
a charged insulated conductor B
...
P
...
decreases due to induced charge on A
...
A point charge Q is placed at point O shown in Fig
...
[Class XII : Physics]
20
O
Ans
...
18
...
An electron and proton are released from rest in a uniform electrostatic
field
...
Ans
...
Calculate
the ratio of the work done by the electric field in taking the charge particle
from A to B and from B to A
...
Ans
...
If a dipole of charge 2µC is placed inside a sphere of radius 2m, what is
the net flux linked with the sphere
...
What is
the work done in bringing a test charge from to point 0
...
V0
qk
kq
kq
kq
0
AO OC OB OD
W q V00 0
22
...
Calculate number of electric field lines orgnating from one coulomb charge
...
q
lc
0
0
If the metallic conductor shown in the figure is continuously charged from
which of the points A,B,C or D does the charge leak first
...
B
A
C
D
Ans
...
Ans
...
Ans
...
What is dielectric strength? Write the value of dielectric strength of air
...
How much work is done in moving a test charge from point (b, 0, 0) to
Q (–b, 0, 0)?
W F
...
dr q Edr cos90° = 0
E along equitorial line of diple is parallel to dipole, hence perpendicular
to displacement
...
Ans
...
Gain in Energy =
[Class XII : Physics]
eV = 1
...
6 × 10–19J
22
=
27
...
6 1019
1
...
Draw schematically the equipotential surface corresponding to a field that
uniformly increases in magnitude but remains in a constant (say z) direction
...
E increases therefore, equipotential surface are closer d1 > d2
...
Figure shows five charged lumps of plastic and an electrically neutral coin
...
What is the net
electric flux through the surface?
R s
...
29
...
q
q
q1 q2
3 6
...
Explain why the capacitance
of a parallel plate capacitor reduces on increasing the separation between
the plates?
P D = V = E × d
...
23
[Class XII : Physics]
as C
30
...
...
Q2
+
Ans
...
How does the relaxation time of electron in the conductor change when
temperature of the conductor decreases
...
When temperature of the conductor decreases, ionic vibration in the
conductor decreases so relaxation time increases
...
Sketch a graph showing variation of resistivity with temperature of (i) Copper
(ii) Carbon
...
4
Ans
...
0
...
The emf of the driver cell (Auxillary battery) in the potentiometer experiment
should be greater than emf of the cell to be determined
...
If emf of a driver cell is less, then null point will not be obtained on the
potentiometer wire
...
You are required to select a carbon resistor of resistance 47k ± 10%
from a large collection
...
Yellow, Violet, Orange, Silver
...
The fig
...
What are the magnitude and
direction of the current i in the lower right-hand wire?
(5)
2A
P
4A
Ans
...
At ‘P’
+ 4 – 4 + 2 = 2A goes to B,
At ‘Q’
36
...
Which wire is thicker?
R c
lc
A
l
m m c c 1
Ac
Am
m
Am
...
37
...
You are given three constantan wires P, Q and R of length and area of
A L
cross-section (L, A), 2L, , , 2A respectively
...
2
4L
, RQ
A
A
A
4L
L
, RR
A
4A
Q has the highest resistance
...
V – I graph for a metallic wire at two different temperatures T1 and T2 is
as shown in the figure
...
Slope of T1 is large, so T1 represents higher temperature as resistance
increases with temperature for a conductor
R
39
...
l
Out of V – I graph for parallel and series combination of two metallic
resistors, which one represents parallel combination of resistors? Justify
your answer
...
The resistance for parallel combination in lesser than for series combination
for a given set of resistors
...
Hence Resistance =
is less
...
V
l
40
...
Emf measured by the potentiometer is more accurate because the cell is
in open circuit giving no current
...
Ans
...
Ans
...
How can a given 4 wires potentiometer be made more sensitive?
By connecting a resistance in series with the potentiometer wire in the
primary circuit, the potential drop across the wire is reduced
...
In the figure, what is the potential difference between A and B?
A
Ans
...
[Class XII : Physics]
26
B
12 V
44
...
45
...
What will be its new resistance?
R’ = n 2 R
Two resistances 5 and 7 are joined as shown to two batteries of emf
2V and 3V
...
What will be the current
through 5?
5
7
2V
Ans
...
I
3V
2
A
5
Calculate the equivalent resistance between points A and B in the figure
given below
...
M
R
R
B
2R
R
B
A
R
L
B
A
B
2R
R
N
47
...
What is the largest voltage that can be safely put across a resistor marked
196, 1W?
P
V2
, V2 = P R = 1 × 196 = 196
R
V = 14 Volt
...
Ans
...
A car battery is of 12V
...
5 V connected in series also
give 12V, but such a combination is not used to start a car
...
Dry cell used in series will have high resistance (=10) and hence provide
low current, while a car battery has low internal resistance (0
...
50
...
Which of the two lamps has higher resistance?
Ans
...
...
Constantan is used for making the standard resistance
...
High resistivity and low temperature Coefficient of resistance
...
Ans
...
Assuming
there is no change in density on stretching
...
R
l
l2
l2
, and V are constant
A
Al
V
R l2
l
16
1
R2
2l1
l
R1
1
R2
l2
2
2
R2 64
53
...
Ans
...
e
...
[Class XII : Physics]
28
54
...
State the Condition for maximum current to be drawn from a Cell
...
e
...
I
1
...
What is the magnitude and direction of the
electrostatic field required for this purpose? Ans : E = mg/Q, downward
2
...
854
µC
...
If q is the positive charge on each molecule of water, what is the total
positive charge in (360g) a Mug of water
...
02×10 23 C
Ans : q
18
4
...
5
...
6
...
Calculate the
electric field intensity at a point 5 cm from it
...
Ans
...
What is the ratio of electric field intensity at a point on the equatorial line
to the field at a point on axial line when the points are at the same
distance from the centre of the dipole?
Ans : 1:2
Show that the electric field intensity at a point can be given as negative
of potential gradient
...
9
...
What is the electrostatic force between them
...
An electron and a proton fall through a distance in an uniform electric field
E
...
11
...
Give the direction of electric field at points A,B,C and D
...
The electric potential V at any point in space is given V = 20x3 volt, where
x is in meter
...
Ans : 60NC–1
13
...
14
...
Ans
...
q
B
What is the electric field at O in Figures (i), (ii) and (iii)
...
A
B
q
q
A
O
q
D
r
I
A
B
O
q
C
q
D
r
II
B
O
q
C
q
D
r
III
C
q 1
2q
Ans : (i) Zero, (ii) 4 2 (iii) 4
0 r
0
[Class XII : Physics]
30
16
...
For an isolated parallel plate capacitor of capacitance C and potential
difference V, what will happen to (i) charge on the plates (ii) potential
difference across the plates (iii) field between the plates (iv) energy stored
in the capacitor, when the distance between the plates is increased?
Ans : (i) No change (ii) increases (iii) No change (iv) increases
...
Does the maximum charge given to a metallic sphere of radius R depend
on whether it is hollow or solid? Give reason for your answer
...
19
...
Under what conditions
will the electric field be zero on the line joining them (i) between the
charges (ii) outside the charge?
Ans : (i) Charge are alike (ii) Unlike charges of unequal magnitude
...
Obtain an expression for the field due to electric dipole at any point on the
equatorial line
...
21
...
1
2
1m
3
x
2m
22
...
23
...
Electric potential
of each plate is marked in Fig
...
: E 0 10 NC , E
31
4
10
1
NC
k
[Class XII : Physics]
24
...
If the capacitor is charged to 5
...
: 1
...
The figure shows the Q (charge) versus V (potential) graph for a combination
of two capacitors
...
Ans : A represents parallel combination
26
...
A
B
C
Ans : WAB = WBC = 50 x 10–8 J, WAC = 0J
27
...
Two
charges –q and +q are located at points A (0, 0, –a) and B (0, 0, +a)
respectively
...
The potential at a point A is –500V and that at another point B is +500V
...
I
...
29
...
E
...
30
...
31
...
In charging a capacitor of capacitance C by a source of emf V, energy
supplied by the sources QV and the energy stored in the capacitor is ½QV
...
33
...
If the dipole is released does it have (a)
only rotational motion (b) only translatory motion (c) both translatory and
rotatory motion?
34
...
Will the electric field intensity due to
this system also be zero
...
A point charge Q is kept at the intersection of (i) face diagonals (ii) diagonals
of a cube of side a
...
There are two large parallel metallic plates S1 and S2 carrying surface
charge dnsities 1 and 2 respectively (1 > 2) placed at a distance d
apart in vacuum
...
37
...
How does electron mobility
change when (i) temperature of conductor is decreased (ii) Applied potential
difference is doubled at constant temperature?
38
...
What are superconductors? Give one of their applications
...
Two manganin wires whose lengths are in the ratio 1 : 2 and whose
resistances are in the ratio 1 : 2 are connected in series with a battery
...
The current through a wire depends on time as i = i0 + at where i0 = 4A
and a = 2As–1
...
42
...
What will be the ratio of voltages across R1 and R2
...
(2:1)
E
R2
R1
R3
33
[Class XII : Physics]
43
...
A 100W and a 200 W domestic bulbs joined in series are connected to the
mains
...
(100W)
45
...
Which bulb will glow more brightly? Justify
...
A battery has an emf of 12V and an internal resistance of 2
...
47
...
Compute effective resistance between diametrically opposite
points A and B
...
R/4]
A
B
48
...
25V gives a balance point
at 35 cm length of the wire
...
What is the emf of the second cell?
[Ans
...
25V]
49
...
5 cm from end A
...
5
...
[Ans
...
16]
X
Y
G
A
39
...
A meterbridge is in balance condition
...
Give reason
...
Galvanometer will show no deflection
...
]
51
...
What will be effect on the
position of zero deflection in a potentiometer
...
Why should the area of cross section of the meter bridge wire be uniform?
Explain
...
Given any two limitations of Ohm’s law
...
Which one of the two, an ammeter or a milliammeter has a higher resistance
and why?
55
...
Give the
colour code for the resistor
...
If the electron drift speed is so small (~10–3 m/s) and the electron’s charge
is very small, how can we still obtain a large amount of current in a
conductor
57
...
0 volts and internal Resistance 0
...
0A
...
0 V
0
...
Why should the jockey be not rubbed against potentiometer wire?
59
...
Fi i
ve dentcalcels, each of em f E and i ernalr st
i
l
nt
esi ance r, are connected
in series to form (a) an open (b) closed circuit
...
: (a) 3E; (b) zero]
61
...
If I is the
in a circular orbit of radius
n
me2
me5
h
...
In the given circuit, with steady current, calculate the potential drop across
the capacitor in terms of V
...
A cell of e
...
f
...
Plot a graph showing the variation of terminal potential ‘V’ with
resistance ‘R’
...
1
...
Obtain expression for electrostatic
potential at a point P in the field due to a point charge
...
Calculate the electrostatic potential energy for a system of three point
charges placed at the corners of an equilateral triangle of side ‘a’
...
What is polarization of charge? With the help of a diagram show why the
electric field between the plates of capacitor reduces on introducing a
dielectric slab
...
4
...
Graphically show the variation of electric field
intensity with distance from the centre of shell
...
Three capacitors are connected first in series and then in parallel
...
6
...
Find Potential
at the common centre
...
Derive an expression for the energy density of a parallel plate capacitor
...
You are given an air filled parallel plate capacitor
...
What will be the capacitance of the capacitor
of initial area was A distance between plates d?
P
K1
P
K1
K2
K2
Q
F ig
...
2
C1 = (K1 + K2)C0
C2
9
...
The wire AB shown of length l has a liner charge
density given = kx where x is the distance measured along the wire
from end A
...
K1 K2 C0
K1 K2
S2
l
1 2
Total charge on wire AB = Q = dx k xdn = K l
2
o
o
By Gauss’s theorem
...
Explain why charge given to a hollow conductor is transferred immediately
to outer surface of the conductor
...
NCERT Vol I)
37
[Class XII : Physics]
11
...
Hence calculate the potential
energy of the dipole
...
Define electric flux
...
An electric flux of units passes
normally through a spherical Gaussian surface of radius r, due to point
charge placed at the centre
...
A conducting slab of thickness ‘t’ is introduced between the plates of a
parallel plate capacitor, separated by a distance d (t
What will be its capacitance
when t = d?
14
...
(i)
(ii)
Potential
(iii)
Capacitance
(iv)
15
...
What is an equipotential surface? Write three properties Sketch equipotential
surfaces of
(i)
Isolated point charge
(ii)
Uniform electric field
(iii)
Dipole
16
...
Give reasons to explain the above
...
Two metal spheres A and B of radius r and 2r whose centres are separated
by a di ance of 6r ar gi
st
e ven char
ge Q , ar at pot i V1 and V2
...
These spheres aqre connected to each other with the
help of a connecting wire keeping the separation unchanged, with is the
amount of charge that will flow through the wire?
[Class XII : Physics]
38
A
B
6r
18
...
Write its SI unit
...
19
...
How are the electric field E and the resistance R of the
conductor affected when (i) V is halved (ii) L is halved (iii) D is doubled
...
20
...
A conductor of length L is connected to a dc source
of emf E
...
21
...
How can potential gradient of a potentiometer be
determined experimentally
...
Which is more sensitive –A or B?
A
B
V
V o lt
l(m )
22
...
Give its SI units
...
State the principle of potentiometer
...
Write the formula used
...
The graph shows how the current I varies with applied potential difference
V across a 12 V filament lamp (A) and across one metre long nichrome
wire (B)
...
39
[Class XII : Physics]
4
I
3
am pere 2
1
0
(ii)
A
B
2 4 6 8 10 12
V (volts)
when potential difference across them is 4V
...
25
...
26
...
m
...
27
...
I
...
Draw
a graph showing the variation of resistivity with temperature for a typical
semiconductor
...
The current flowing through a conductor is 2mA at 50V and 3mA at 60V
...
29
...
Under what conditions is the heat produced in an electric circuit:
(i)
directly proportional
(ii)
inversely proportional to the resistance of the circuit
1
...
Explain its working with the
help of a neat labelled diagram
...
Derive an expression for the strength of electric field intensity at a point
on the axis of a uniformly charged circular coil of radius R carrying charge Q
...
Derive an expression for potential at any point distant r from the centre O
of dipole making an angle with the dipole
...
Suppose that three points are set at equal distance r = 90 cm from the
centre of a dipole, point A and B are on either side of the dipole on the
axis (A closer to +ve charge and B closer to B) point C which is on the
perpendicular bisector through the line joining the charges
...
6 × 10–19 Cm
at points A, B and C?
5
...
How
would the following (i) energy (ii) charge, (iii) potential be affected if dielectric
slab is introduced with battery disconnected, (b) dielectric slab is introduced
after the battery is connected
...
Derive an expression for torque experienced by dipole placed in uniform
electric field
...
7
...
Derive an expression for the electric field due to
a charged plane sheet
...
8
...
If the
dielectric slab is introduced with the battery connected, then how do the
following quantities change (i) charge (ii) potential (iii) capacitance
(iv) energy
...
Using Gauss’s theorem obtain an expression for electric field intensity
due to a plane sheet of charge
...
10
...
(See Page 68, NCERT Vol I)
...
State Kirchhoff’s rules for electrical networks
...
How is it realized in actual practice in the laboratory? State the formula
used
...
Define emf and terminal potential difference of a cell
...
41
[Class XII : Physics]
13
...
What would be the maximum current
possible if the emf of each cell is E and internal resistance is r each?
14
...
Hence
deduce ohm’s law
...
State the principle of potentiometer
...
Compare e
...
f of two cells
Measure internal resistance of a cell?
Explain how does the conductivity of a :
(i)
(ii)
Semi conductor and
(iii)
17
...
Derive expression for equivalent e
...
f and equivalent resistance of a :
(a)
Series combination
(b)
Parallel combination
of three cells with e
...
f E1, E2, E3 & internal resistances r1, r2, r3 respectively
...
Deduce the condition for balance in a Wheatstone bridge
...
Draw the circuit diagram and write
the formula used
...
1
...
Will the position change for any other value of charge q? (9 cm
from – 4 µC)
2
...
They are immersed in a medium of dielectric constant 16
...
(1/4 the original separation)
[Class XII : Physics]
42
3
...
If E is the energy stored in 20 µF capacitor what will be
the total energy supplied by the battery in terms of E
...
Two point charges 6 µC and 2 µC are separated by 3 cm in free space
...
(3
...
ABC is an equilateral triangle of side 10 cm
...
What is the force experienced by a 1 µC positive charge placed at A?
(902 × 103 N)
6
...
Another point charge of
4 µC is brought from a far point to a distance of 50 cm from origin
...
Another charge of 11 µC is brought to a point 100 cm from each of the
two charges
...
2 × 10–3J)
7
...
Calculate distance of closest approach
...
1 × 10–4 m)
8
...
(1
...
In the following fig
...
Given potential at A is 90 V
...
(Ans
...
A point charge develops an electric field of 40 N/C and a potential difference
of 10 J/C at a point
...
(2
...
Figure shows three circuits, each consisting of a switch and two capacitors
initially charged as indicated
...
12
...
3
C
A
3
4
2
3
3
B
3
3
All capacitance given in micro farad
Ans
...
A pendulum bob of mass 80 mg and carrying charge of 3 × 10–8 C is
placed in an horizontal electric field
...
Calculate the intensity of electric field
...
Eight charged water droplets each of radius 1 mm and charge 10 × 10–
10C coalesce to form a single drop
...
(3600 V)
15
...
(2
...
A 10 F capacitor can withstand a maximum voltage of 100 V across it,
whereas another 20 F capacitor can withstand a maximum voltage of
only 25 V
...
Three concentric spherical metallic shells A < B < C of radii a, b, c
(a < b < c) have surface densities , – and respectively
...
If shells A and C are at the same
potential obtain relation between a, b, c
...
Four point charges are placed at the corners of the square of edge a as
[Class XII : Physics]
44
shown in the figure
...
2 4 J
+q
–q
+q
19
...
1m F
5m F
C
Two capacitors A and B with capacitances 3 F and 2 F are charged 100
V and 180 V respectively
...
Calculate the (i) final charge on
the three capacitors (ii) amount of electrostatic energy stored in the system
before and after the completion of the circuit
...
2F
18 0V B
Two identical parallel plate capacitors connected to a battery with the
switch S closed
...
Find the ratio of the total electrostatic energy stored in both capacitors
before and after the introduction of dielectric
...
The charge passing through a conductor is a function of time and is given
as q = 2t2 – 4t + 3 milli coulomb
...
Given resistance of
conductor is 4 ohm
...
: I = 12A, V = 48 V]
23
...
00 ohm and its
resistance at steam point is 5
...
When the wire is immersed in a hot oil
bath, the resistance becomes 5
...
Calculate the temperature of the oil
bath and temperature coefficient of resistance of platinum
...
: a = 0
...
Three identical cells, each of emf 2V and internal resistance 0
...
4 ohm
...
[Ans
...
75; V = 5
...
Calculate the equivalent resistance and current shown by the ammeter in
the circuit diagram given
...
: R = 2; I = 5A]
2
4
6
4
1
6
A
2V
26
...
5 is being
charged by a 12V dc supply
...
[Ans
...
5]
[Class XII : Physics]
46
27
...
If the emf of the cell are 2V, 1V and 4
V and if their internal resistance are 4, 3 and 2 ohm respectively, find
the current through each cell
...
: I1
13
13
13
28
...
A source of emf 9 volt
is connected across one of its sides
...
[Ans
...
A length of uniform ‘heating wire’ made of nichrome has a resistance 72
...
Why is it is not advisable to use the half length of wire?
[Ans
...
400W >> 200W but since current becomes
large so it is not advisable to use half the length]
30
...
On
putting another 8 in parallel with 8 resistance in the right gap, the null
point is found to shift by 15cm
...
[Ans
...
Figure show a potentiometer circuit for comparison of two resistances
...
While that with the unknown resistance X is 134
...
Determine the
value of X
...
: 2]
E´
A
B
R
K1
G
X
E
K2
0
...
Two cells of E
...
F
...
Potentiometer is connected between points A and B
...
: 2:1]
K1
( )
A
B
E1
33
...
A graph was drawn
which was a straight line ABC
...
(ii) The internal resistance of the cell
...
: r = 5 emf = 1
...
D
...
6
1
...
2
A
B
0
...
4
...
0 8
...
1 6
...
2 4
...
Four cells each of internal resistance 0
...
4V, d are connected
(i) in series (ii) in parallel
...
Find the current through the lamp and each cell
in both the cases
...
: Is = 0
...
137A current through each cell is 0
...
In the figure an ammeter A and a resistor of resistance R = 4 have been
connected to the terminals of the source to form a complete circuit
...
Calculate
voltmeter and ammeter reading
...
: Voltmeter reading : 8V, Ammeter reading = 2A]
[Class XII : Physics]
48
V
a
a´
36
...
Calculate resistance
of voltmeter
...
: RV 150; V
3
60V
3 0 0
2 0 0
V
For the circuit given below, find the potential difference b/w points B and D
...
: 1
...
2V 2
B
1V , 1
2
37
...
(ii)
Also calculate the current through CD & ACB if a 10V d
...
B
E
R
A potentiometer wire AB of length 1m is connected to a driver cell of emf
3V as shown in figure
...
5V is used in the secondary
circuit, the balance point is found to be 60 cm
...
(ii)
Explain with reason, whether the circuit works if the driver cell is
replaced with another a cell of emf IV
...
40
...
If the current in the circuit is 0
...
A network of resistances is connected to a 16V battery with internal
resistance of 1 as shown in Fig
...
(ii)
Obtain the current in each resistor
...
[Class XII : Physics]
50
4
12
A
B
1
C
D
4
6
1
16V
42
...
5 x 1028 m–3
...
0m long to its other end? The area of cross
section of the wire is 2
...
0 A
...
A Voltmeter of resistance 400 is used to measure the potential difference
across the 100 resistor in the circuit shown in figure
...
84V
2 0 0
1 0 0
V
44
...
R
R
A
B
R
R
45
...
What would be the effective resistance of the wire, if specific
resistance of copper and nickel are c and n respectively
...
:
Pc Pe
l
r
2
;
Rn ln
51
l
2
2r – r
2
[Class XII : Physics]
R
RC Rn
3 n c
Ans
...
RC Rn
46
...
0 W
Q
–
150V
+
–
+
50V
P
2
...
: – 10V]
47
...
5 and a voltmeter of resistance 20 k
...
In each case, which of the two connection shown
should be chosen for resistance measurement?
X
Y
A
(i)
V
A
(ii)
V
[Ans
...
When resistance of 2 is connected across the terminals of a battery, the
current is 0
...
When the resistance across the terminal is 5, the current
is 0
...
(i) Determine the emf of the battery (ii) What will be current
drawn from the cell when it is short circuited
...
: E = 1
...
5A]
49
...
Calculate energy
stored in the capacitor of 4µF capacitance
...
: VAB = 20V, U = 8 × 10–4 J]
[Class XII : Physics]
52
2A
1A
3W A
3W
5W
4µ F
1W
1W
B
2W
3V 2A
1A
50
...
Calculate the net resistance between A and B
...
: 3]
A
B
51
...
What will be the percentage error in the reading of
the voltmeter
...
: 0
...
Geeta has dry hair
...
She observes that Neeta with oily hair combs her hair; the comb
could not attract small bits of paper
...
She then goes to the junior classes and shows this
phenomenon as Physics Experiment to them
...
What according to you are the values displayed Geeta?
2
...
After some
time it was raining heavily accompanied by thundering & lightening
...
Some students went inside the room
...
The other students called them to come out but they refused
...
What value was displayed by these students?
3
...
They had
learnt in school about fire caused due to electric short circult
...
They got all their friends and responsible elders
together and with the help of the electricity board, succeeded in changing
the situation
...
What value did Renu, Ritu and Kajal have?
A high tension supply of say 6 KV must have a very large internal
resistance
...
They had a 2 pin plug
...
However, Rohit
insisted that they buy a 3 pin plug before using it
...
Rohit
patiently explained the importance of using a 3 pin plug and the earthing
wire
...
If earthed, the
current would go to the earth and the potential of the metallic body would
not rise
...
(i)
What value did Rahul and Rohit have?
[Class XII : Physics]
54
(ii)
Which has greater resistance – 1 K watt electric heater or 100 watt
electric bulb, both marked 220 volts?
1
...
While the source is at
constant emf V
...
The difference between the
two goes as heat and em radiations
...
Construct a closed system such that charge is enclosed within it
...
According to Gauss’s theorem total flux through
q
the gaussian surface (both cubes) is equal to
...
0
3
...
I
5
...
V IR
ER
E
r
Rr
1
R
When R approaches infinity V becomes equal to E (or for R 00)
V
decreases position of zero deflection increases
...
7
...
m
...
Otherwise resistance per unit length of Bridge wire be different over different
length of meter Bridge
...
N
...
E
...
T page 101
10
...
To produce large deflection due to small current we need a
large number of turns we need a large number of turns in armature coil
Resistance increases
...
Temperature, Material Blue, Red, Orange Gold
10
...
the net
current can be very high even if the drift spread is low
...
V = E + ir
= 2 + 0
...
5 V
13
...
14
...
(ii)
For a small change in potential diff
...
[Class XII : Physics]
56
1
...
kq
V2
V1
V2
Vcommon
kq
r
6r
kq
kq
2r
7kq
6r
3kq kq
6r
4kq
6r
6r
7
4
2q
4 0 r 2r
2q
V´
12 0r
Charge transferred equal to
r
q´ C1 1 C1 ´
V
V
q –
2q
3
q
k
kq
r
r
k
...
3
R1
V1
50
25,000
I1 2 x10 –3
R2
3
...
V2
60
20,000
I2 3 x10 –3
As Resistance changes with I, therefore conductor is non ohmic
...
Rate of Production of heat, P = I2R, for given I, P x R, nichrome > cu
RNichrome > Rcu of same length and area of cross section
...
(i) If I in circuit is constant because H = I2 Rt
(ii) If V in circuit is Constant because H
57
V2
t
R
[Class XII : Physics]
q
q
q
VA k 1 2 3
b
c
a
1
...
2
...
V
When B is earthed VB = 0, VA = 10V and VC = – 2V
...
Before dielectric is introduced
...
1
3C V 2
2
Q
2
2C
5
CV 2
2 3C
11
2
CV ,
2
3
E´ E´A E´B
...
5A
2R 2 2
C
R
R
i1
A
B
R
i2
i3 = 0
R
i2
R
D
10V
5
...
5 2
...
5V
...
e
...
59
[Class XII : Physics]
6
...
5 (R + 3)
R = 17
V = E – Ir = 10 – 0
...
5V
7
...
I enAVd enA
t
9
...
7 104 s
I
84
84
0
...
d across Voltmeter & 100 Combination
0
...
100 400
R
B
R/2
R
A
(i)
3R
2 (ii)
B
When, l < < r,
Baxial
µ0 2M
4 r 3
Magnetic field on equatorial line of a bar magnet
[Class XII : Physics]
60
Beq
µ0
3
4
2 2 2
r l
T
61
[Class XII : Physics]
a current carrying loop
Magnetic Field at a Point on the Axis of
loop For n loops,
Magnetic field at the centre of a circular
carrying conductor
Magnetic field due to a straight current
Biot-Savart’s Law
Physical Quantity
For a << x,
When, x = 0,
Formulae
T
T
T
104 Gauss = 1T
Tesla (T);
SI Unit
[Class XII : Physics]
62
Lorentz Force (Electric and magnetic)
Or F = B qv sin
N
N
Magnetic force on a moving charge
T
m
B = 0 n I
Motion of a charged partical inside electric field
Magnetic field due to a toroidal solenoid
B = 0r nI
magnetic permeability r
If solenoid is filled with material having
T
Magnetic field due to a long straight solenoid
At the end of solenoid,
T – m
Ampere’s Circuital Law
63
[Class XII : Physics]
and
are
in a magnetic field
Force on a current carrying conduct placed
inclined to each other by an angle
The radius of helical path when
The maximum velocity
The radius corresponding to maximum velocity
Maximum energy of the positive ions
The cyclotron frequency
The period of circular motion
Radius of circular path
The Cyclotron
N
[Class XII : Physics]
64
Nm–1
J
magnetic field
The potential energy associated with
magnetic field
Period of oscillation of bar magnet if external
If coil has n turns, = n B I A sin
= n BIA sin ;
= n BIA sin
angle between notmal
drawn on the plane of
loop and magnetic field
= BIA cos
angle between loop
and magnetic field
s
Magnetic dipole moment
Torque on a rectangular current carrying loop ABCD
Am2 or JT–1
carrying conductors
Force per unit length between tow parallel current
65
[Class XII : Physics]
angle by which
Magnetic field on axial line of a bar magnet
Magnetic dipole moment
Bohr magneton
Gyromagnetic ratio
The current loop as a magnetic dipole
Voltage sensitivity
Current sensitivity
Sensitivity of a galvanometer or
the coil rotates
Current through a galvanometer
C Kg–1
T
Am2
rad V–11
JT–1 or Am2
T
rad A–1
A
[Class XII : Physics]
66
Tm2 or weber
Magnetic intensity (or Magnetic field strength)
Curie’s Law
Relative permeability ()
TmA–1
Magnetic permeability
—
C curie conctant
(or NA–2)
—
B = B0 + µ0Im T
= µ0 (H + Im)
Weber (Tm2)
Am–1
Magnetic susceptibility
Magnetic induction (or Magnetic flux
density or Magnetic field)
Magnetic flux
Intensity of magnetization
Am–1
Magnetic inclination (or Dip)
n is the no
...
M
...
AND ALTERNATING CURRENT
1
...
No, pole exists only when the source has some net magnetic moment
...
Magnetic field due to a toroid B = 0 nI
Ans
...
Ans
...
Show graphically the variation of magnetic field due to a straight conductor
of uniform cross-section of radius ‘a’ and carrying steady currently as a
function of distance r (a > r) from the axis of the conductor
...
B 1
r
r
4
...
If the current in each conductor is tripled, what would be
the value of the force per unit length between them?
67
[Class XII : Physics]
Ans
...
Ans
...
Ans
...
Ans
...
An electric current flows in a horizontal wire from East to West
...
W
E
(i) Going into the plane of the paper
(ii) Towards North
...
Ans
...
Ans
...
Ans
...
Ans
...
Ans
...
By using a ferromagnetic case
...
Where on the earth’s surface, is the vertical component of earth’s magnetic
field zero?
At equator
...
What will
be percentage increase in deflection?
1%
...
I
...
(i) Am
(ii) Am2
[Class XII : Physics]
68
13
...
When velocity v of positively charged particle is along x-axis and the
magnetic field B is along y-axis, so v B is along the z-axis (Fleming’s
s
left hand rule)
...
14
...
Bismuth is diamagnetic, hence, the overall magnetic field will be slightly
less
...
An electron beam projected along + x-axis, experiences a force due to a
magnetic field along the + y-axis
...
16
...
17
...
18
...
19
...
20
...
What is the principle of a moving coil galvanometer?
When a current carrying coil is placed in uniform magnetic field, it
experiences a torque
...
Sketch the magnetic field lines for a current carrying circular loop
...
69
[Class XII : Physics]
21
...
22
...
What is the direction of induced currents in metal rings 1 and 2 seen from
the top when current I in the wire is increasing steadily?
1
I
2
1
I
Ans
...
In which of the following cases will the mutual inductance be (i) minimum
(ii) maximum?
(a)
Ans
...
Ans
...
Ans
...
Ans
...
What is the phase difference between (i)
VL and VR (ii) VL and VC?
(i)
2
(ii)
Why can’t transformer be used to step up or step down dc voltage?
In steady current no induction phenomenon will take place
...
c
...
What is the average power
dissipated in the circuit?
As the phase difference b/w current and voltage is
Pa v Iv E v cos
27
...
vr resonant frequency
X
O
vr
v
Ans
...
2
A coil A is connected to an A
...
ammeter and another coil B to A source
of alternating e
...
f
...
A
A
B
Ans
...
71
[Class XII : Physics]
29
...
In a circuit instantaneously voltage and current are V = 150 sin 314t volt
and i = 12 cos 314 t ampere respectively
...
e
...
Hence
2
circuit is a capacitive circuit
...
Ans
...
In a series L–C–R circuit VL = VC VR
...
If the
current is now reduced to I0/2, what will be the new energy stored in the
inductor?
Ans
...
U
4
A rectangle loop a b c d of a conducting wire has been changed into a
square loop a´ b´ c´ d´ as shown in figure
...
33
...
Twelve wires of equal lengths are connected in the form of a skeleton of
a cube, which is moving with a velocity
in the direction of magnetic field
...
...
Emf in each branch will be zero since V & B are parallel for all arms
F q V B 0
34
...
R1 and R2 are resistances
of the two circuits
...
35
...
Why do we prefer carbon brushes than copper in an a
...
generator?
Ans
...
*36
...
XC
for d c = 0
XL = 0
37
...
S N
73
[Class XII : Physics]
Ans
...
The direction of current is clockwise when magnet moves towards the loop
and direction of current is anticlockwise when magnet moves away from
the loop
...
In figure, the arm PQ is moved from x = o to x = 2b with constant speed
V
...
Write
(i)
direction of induced current in rod
(ii)
polarity induced across rod
...
A wire moves with some speed perpendicular to a magnetic field
...
Lorentz force acting on the free charge carrier of conducting wire hence
polarity developed across it
...
Predict the polarity of the capacitor in the situation described in the figure
below
...
Plate a will be positive with respect to ‘b’
...
Ans
...
Ans
...
What is the average emf induced in the coil?
Zero
Define RMS Value of Current
...
[Class XII : Physics]
74
Ir m s
43
...
7
...
What are the polarities at P and Q?
P
v
Q
Ans
...
A long straight wire with current i passes (without touching) three square
wire loops with edge lengths 2L, 1
...
The loops are widely spaced
(so as to not affect one another)
...
Rank the loops according to the size of the current induced in
them if current i is (a) constant and (b) increasing greatest first
...
(a) No induced current
(b) Current will be induced only in loop 2
...
Ans
...
What is
the frequency with which magnetic energy is oscillating?
m = 2 = 8 × 106 Hz
...
Write the four measures that can be taken to increase the sensitivity of a
galvanometer
...
A galvanometer of resistance 120 gives full scale deflection for a current
of 5mA
...
3
...
Calculate the magnetic moment in each case
...
A current of 10A flows through a semicircular wire of radius 2cm as shown
in figure (a)
...
(a)
F ig
...
A proton and an alpha particle of the same enter, in turn, a region of
uniform magnetic field acting perpendicular to their direction of motion
...
6
...
Mention two properties of soft iron due to which it is preferred for making
electromagnet
...
A magnetic dipole of magnetic moment M is kept in a magnetic field B
...
[Class XII : Physics]
76
9
...
A steady current I flows along an infinitely long straight wire with circular
cross-section of radius R
...
A circular coil of n turns and radius R carries a current I
...
Calculate the ratio of magnetic moment of the new coil
and the original coil
...
A coil of N turns and radius R carries a current I
...
Calculate the ratio of the magnetic moment of the ne coil and original coil
...
At a place horizontal component of the earths magnetic field is B and
angle of dip at the place is 60°
...
(i) at Equator; (ii) at a place where dip angle is 30°
14
...
1% of the total current goes through
the coil and rest through the shunt
...
Prove that the magnetic moment of a hydrogen atom in its ground state
is eh/4m
...
16
...
Two path are indicated for the line integral
B
...
What is the value
of the integral for the path (a) and (b)
...
(b )
What is the radius of the path of an electron (mass 9 x 10–31 kg and
charge 1
...
(1 eV = 1
...
Ans
...
1 x 10–31kg x 3 x 107 ms–1/ (1
...
E = (½)mv2 = (½) 9 x 10–31 kg x 9 x 1014 m2/s2
= 40
...
5 keV
...
r EK
Radius (r)
Ans
...
Plot a graph showing the variation of the radius of the
circular path described by it with the increase in its kinetic energy, where,
other factors remain constant
...
Magnetic field arises due to charges in motion
...
Ans
...
20
...
Write the
expression for the magnetic moment when an electron revolves at a speed
‘v’, around an orbit of radius ‘r’ in hydrogen atom
...
The product of the current in the loop to the area of the loop is the
magnetic dipole moment of a current loop
...
An ac source of rms voltage V is put across a series combination of an
inductor L, capacitor C and a resistor R
...
22
...
What will be the acceleration of the magnet
(whether a > g or a < g or a = g) if (a) coil ends are not connected to each
other? (b) coil ends are connected to each other?
(a)
23
...
is in resonance state
...
The divisions marked on the scale of an a
...
ammeter are not equally
spaced
...
Circuit shown here uses an airfield parallel plate capacitor
...
Explain with reason the
effect on brightness of the bulb B
...
In the figure shown, coils P and Q are identical and moving apart with
same velocity V
...
Find I1/I2
...
A 1
...
The charging battery is then
disconnected, and a 12 mH coil is connected in series with the capacitor
so that LC Oscillations occur
...
28
...
What should be
the capacitance of the capacitor to which it should be connected in order
to impart maximum power at 50Hz?
29
...
30
...
31
...
When switch is closed and after sometime an iron rod is inserted into the
interior of inductor
...
Ans
...
Ans
...
Ans
...
Show that in the free oscillation of an LC circuit, the sum of energies
stored in the capacitor and the inductor is constant with time
...
d
...
D
...
34
...
35
...
Capacitor acts as an open key
...
Explain
...
36
...
The circuit consists of an inductor and
a capacitor in series
...
37
...
The
graph obtained for voltage produced across the coil Vs time is shown in figure
...
d
...
What is the Significance of Q-factor in a series LCR resonant circuit?
39
...
the distance between the coils is increased?
an iron rod is kept between them?
Two circular conductors are perpendicular to each other as shown in figure
...
Derive the expression for force between two infinitely long parallel straight
wires carrying current in the same direction
...
2
...
Distinguish between diamagnetic, paramagnetic and ferromagnetic
substances in terms of susceptibility and relative permeability
...
Name all the three elements of earth magnetic field and define them with
the help of relevant diagram
...
Describe the path of a charged particle moving in a uniform magnetic field
with initial velocity
(i)
parallel to (or along) the field
...
(iii)
at an arbitrary angle (0° < < 90°)
...
Obtain an expression for the magnetic moment of an electron moving with
a speed ‘v’ in a circular orbit of radius ‘r’
...
State Ampere, circuital law
...
*8
...
9
...
10
...
*11
...
*12
...
Write the necessary mathematical steps to obtain the value
of resistance required for this purpose
...
A long wire is first bent into a circular coil of one turn and then into a
circular coil of smaller radius having n turns
...
Ans
...
Obtain an expression for the self inductance of a straight solenoid of
length I and radius r (l >> r)
...
Distinguish between : (i) resistance and reactance (ii) reactance and
impedance
...
In a series L–C–R circuit XL, XC and R are the inductive reactance,
capacitive reactance and resistance respectively at a certain frequency f
...
c
...
What are eddy currents? Write their any four applications
...
In a series L–R circuit, XL = R and power factor of the circuit is P1
...
Find P1/P2
...
Instantaneous value of a
...
through an inductor L is e = e0 cos t
...
Also draw
the phasor diagram
...
In an inductor of inductance L, current passing is I0
...
In what forms is this energy stored?
21
...
(a)
(d)
X C – XL
O
frequ ency (f)
(b)
(c)
[Ans
...
A sinusoidal e
...
f
...
If the frequency of
driving source is increased
...
The figure shows, in (a) a sine curved (t) = sin t and three other
sinusoidal curves A(t), B(t) and C(t) each of the form sin (t – )
...
: (a) C, B, A;
(b) 1, A; 2, B; (c) A]
2
A (t)
1
3
4
B (t)
S (t)
C (t)
(a )
24
...
If time period of
oscillation is T them :
(i)
at what time is the energy stored completely electrical
(ii)
at what time is the energy stored completely magnetic
(iii)
at what time is the total energy shared equally between the inductor
and capacitor
...
(ii) t = T/4, 3T/4, 5T/4
...
Ans
...
8 8 8
An alternating voltage of frequency f is applied across a series LCR circuit
...
Will the current in the
circuit lag, lead or remain in phase with the applied voltage when
(i) f > fr (ii) f < fr ? Explain your answer in each case
...
VL > VC Hence VL – VC > O
(i) Current will lead, because
...
Figure (a), (b), (c) Show three alternating circuits withe equal currents
...
85
[Class XII : Physics]
R
E
(a)
Ans
...
(ii)
Current will decrease as XL increase
...
(i)
E
(c)
Current will increase as XC decrease
...
Name two main characteristics of a ferromagnetic
material which help us to decide suitability for making
...
2
...
Use it to obtain the magnetic field at an axial point,
distance d from the centre of a circular coil of radius ‘a’ and carrying
current I
...
3
...
With the help of diagram, explain the principle
and working of a cyclotron
...
*4
...
What is the importance of radial field and phosphor
bronze used in the construction of moving coil galvanometer?
5
...
c
...
Deduce the expression for emf generated
...
c
...
Explain, with the help of a neat and labelled diagram, the principle,
construction and working of a transformer
...
An L–C circuit contains inductor of inductance L and capacitor of
capacitance C with an initial charge q0
...
Let the instant the circuit is closed be t = 0
...
What is the total energy stored initially?
If a resistor is inserted in the circuit, how much energy is eventually
dissipated as heat?
An a
...
i = i0 sin t is passed through a series combination of an inductor
(L), a capacitor (C) and a resistor (R)
...
Hence show that
the current
(i)
leads the voltage when
1
LC
(ii)
is in phase with voltage when
1
...
Write two differences in each of resistance, reactance and impedance for
an ac circuit
...
1
...
Calculate the speed of electron
...
: Speed = 3
...
56 × 107 volts]
2
...
In what direction should it
be steered if declination at the place is 18° west?
[Ans
...
Calculate the magnetic field due to a circular coil of 500 turns and of mean
diameter 0
...
12m from the centre of the coil (ii) at the centre of the coil
...
: (i) 5
...
8 × 10–2 tesla]
87
[Class XII : Physics]
4
...
8 milli testa
...
[Ans
...
467 × 10 –8 s
...
If the current sensitivity of a moving coil galvanometer is increased by 20%
and its resistance also increased by 50% then how will the voltage sensitivity
of the galvanometer be affected?
[Ans
...
A uniform wire is bent into one turn circular loop and same wire is again
bent in two turn circular loop
...
[Ans
...
A horizontal electrical power line carries a current of 90A from east to west
direction
...
5m below it? [Ans
...
2 × 10–5 T south ward]
*8
...
What should be the value of resistance
to convert the galvanometer into a voltmeter of range 0V to 5V
...
: 1910 in series]
9
...
And the direction of current appear’s anticlockwise as seen from
point O which is equidistant from loop P and Q
...
I
O
R
P
x
x
Q
R
I
2
µ0IR 2
Ans
...
A cyclotron’s oscillator frequency is 10 MHz
...
6 × 10–19 C, m = 1
...
Express your answer in units
of MeV [1MeV = 1
...
[Ans
...
656T, Emax = 7
...
The coil of a galvanometer is 0
...
08 m2
...
2 tesla
...
Assuming the
magnetic field to be radial
...
1°?
[Ans
...
69 × 10–4 A; (ii) 1
...
A voltmeter reads 8V at full scale deflection and is graded according to its
resistance per volt at full scale deflection as 5000 V–1
...
: 7
...
A short bar magnet placed with its axis at 30° with an external field 1000G
experiences a torque of 0
...
(i) What is the magnetic moment of the
magnet
...
: (i) 0
...
08 J]
14
...
5 Am2
...
: BE = 7
...
625 × 10–7 T]
15
...
15T?
16
...
R1 = 10, N1 = 30, A1 = 3
...
25T
R2 = 14, N2 = 42, A2 = 1
...
50T
Given that the spring constants are the same for the two galvano meters,
89
[Class XII : Physics]
determine the ratio of (a) current sensitivity (b) voltage sensitivity of M1 &
M2
...
In the given diagram, a small magnetised needle is placed at a point O
...
The other arrows
shown different positions and orientations of the magnetic moment of
another identical magnetic needs B
B4
B5
C A
B2
B3
B6
(a)
In which configuration is the systems not in equilibrium?
(b)
In which configuration is the system
...
stable and (ii) unstable equilibrium?
Which configuration corresponds to the lowest potential energy
among all the configurations shown?
In the circuit, the current is to be measured
...
02
(c)
19
...
What is the magnetic field on the y-axis at a distance of 0
...
x = 1 cm
...
5
x
Dx
20
...
5 m carries a current of 2A
...
What is
the magnitude of the magnetic field?
21
...
0 cm
apart carrying a current of 25A
...
82 x 10–4 N towards the conductor
Hint :
F1
0 2I12
I
10–7 2 25 15 0
...
38 10 –4N attractive
4 r1
0
...
25
1
...
12
Net F = F1 – F2 = 7
...
In a chamber of a uniform magnetic field 6
...
An electron
is shot into the field with a speed of 4
...
Explain why the path of electron is a circle
...
6 x 10–19 C, me =
9
...
91
[Class XII : Physics]
Hint :
(a) r
me 9
...
8 10 6
4
...
6 10 –19 6
...
6 10–19 6
...
8 MH2
2me 2 3
...
1 10 –31
23
...
The vertical plane carrying the
needle is turned through 45° from the magnetic meridian
...
[Ans
...
Figure shows the path of an electron that passes through two regions
containing uniform magnetic fields of magnitude B1 and B2
...
(a) Which field is stronger? (b) What are the directions
of two fields? (c) Is the time spend by the electron in the B 1 , region greater
than, less than, or the same as the time spent in B 2 region?
[Ans
...
(c) Time spent in B1 < Time spent in B2]
B1
B2
25
...
What is
the (i) The peak voltage (ii) Average voltage over half cycle?
26
...
The
flux is reduced to 1 mWb in 5 s
...
27
...
Calculate :
(i)
Reactance of the circuit
...
[Class XII : Physics]
92
28
...
What should be number
of turns in its secondary to get power output at 240V?
29
...
Calculate the value of induce current at t = 15 s
...
A capacitor, a resistor and
4
henry inductor are connected in series to
2
an a
...
source of 50 Hz
...
31
...
If the input a
...
voltage is (200 V, 50 Hz), calculate (i) voltage across
capacitor and resistor
...
32
...
v
b
l
2l
The loop is moving at constant speed v and at t = 0 it just enters the field
B
...
v
(i)
Magnetic flux – time
(ii)
Induced emf – time
(iii)
Power – time
Resistance of the loop is R
...
A charged 8mF capacitor having charge 5mC is connected to a 5mH
inductor
...
the maximum current in the inductor?
the magnetic energy in the inductor at the instant when charge on
capacitor is 4mC?
A 31
...
1H inductor are connected in series to a 200V,
50Hz ac source
...
(iii)
35
...
A square loop of side 12 cm with its sides parallel to X and Y-axis is
moved with a velocity of 8 cm/s in positive x-direction
...
(i)
(ii)
Ans
...
Determine the direction and magnitude of induced current if field
changes with 10–3 Tesla/s
...
12)2 x 10–3 x 8
= 11
...
(ii) Rate of change of flux = induced emf = (0
...
44 x 10–5 Wb/s
...
Figure shows a wire of length l which can slide on a U-shaped rail of
negligible resistance
...
The wire is pulled to
the right with a constant speed v
...
Find the current in the wire
using this diagram
...
Two girls Pooja and Ritu were very good observers and performed in the
school function using their cassette player
...
But Pooja was
determined not to let down the performance so she sang the song instead
of dancing and Ritu completed the dance
...
What were the value displayed by Pooja and Ritu ?
What kind of Ferro magnetic material is using for coating magnetic
tapes used in cassette
...
Unfortunately it fell
from his hand and broke
...
Teacher
listened to him patiently and on knowing that the act was not intentional,
but just an accident, did not scold him andused the opportunity to show
the internal structure of galvanometer to the whole class
...
What are the value displayed by Tushar
...
Pooja went to the market with her mother and decided to come back home
by metro
...
Pooja passed through it and was waiting for her mother to
come
...
Pooja was confused why metal detector beeped in case of her
mother
...
Both were
satisfied with the security system
...
What values are displayed by Pooja
...
Mr
...
The dry cell was weak, giving less voltage, which was not
suffcient to give proper reading
...
(i)
(ii)
1
...
3
...
S
Ig
I
Ig
G
5 10
5 5 10
3
120 0
...
(i) – mB (ii) zero
(i)
B
10
–7
10
2 10
(ii)
4
...
B = 5p × 10–5 T (inwards)
...
2
5
...
6
...
7
...
8
...
(b)
Pole strength half; magnetic moment half
...
µ I
B 0 2r
2R
R r
B
...
= µ 0 I
B
µ0 I
r R
...
2 rN 4aN a
R
2
M2
4
M1
mnew
11
...
(a)
2
1
...
13
...
d l 0I 2 0 Tm
(b) zero
22
...
(ii) a < g because of the opposite effect caused by induced current
...
Current at resonance I
V
...
XL = I
...
V
2 L
...
C
...
97
[Class XII : Physics]
25
...
I
V
where Z
Z
XC
XC
1
2
and
1
C
1
2
Xc R
2C
, when mica sheet is introduced capacitance C increases
C
K 0 A
C
,
d
XC decreases, current increases and therefore brightness increases
...
Current I = /R =
In coil P, I1 E1 R
Bvb
R
In coil Q, I 2 E 2 R
27
...
R
em energy is conserved
µE(max) = µB(max)
12
Q
2
C
1
LI
2
2
I = 637 mA
28
...
40
...
15
...
F IL B ILBsin
Hence, force permit length, f
[Class XII : Physics]
F
IB sin30
L
98
= 8 x 0
...
6 Nm–1
16
...
25 3
...
50 1
...
5 7
N2B2 A 2
kR
2
30 0
...
6 10 –3 14
42 0
...
8 10 –3 10
1
For equilibrium, the dipole moment should be parallel or auto parallel
to B
...
(b)
(i) for stable equilibrium, the dipole moments should be parallel,
examples : AB5 and AB6 (ii) for unstable equilibrium, the dipole
moment should be anti parallel examples : AB3 and AB4
(c)
18
...
e, U = –MB Example : AB6
(a)
Tot r st
al esi ance, R G + 3 = 63
...
048A
63
Resistance of the galvanometer as ammeter is
RGrS 60 0
...
02
...
02
Total resistance R = 0
...
02
3
0
...
For the ideal ammeter, resistance is zero, the current,
I = 3/3 = 1
...
5m
o/4 = 10–7 Tm/A, = 90° so sin
10–7 10 10–2
dB
4 10 –8 T along z axis
25 10 –2
20
...
2 9
...
657 T
2 1
...
V0 = 110 volt
(ii)
26
...
22
d
dt
–25
1 – 6 10 –3
(i)
Reactance XL = L = 300 × 4 × 10–3 = 1
...
28
...
25 volt
...
5
In ideal transformer
V0 = i0
...
2 = 14
...
Pin = P0
VPIP = VsIs
VS
VP
IP
IS
NS
NP
[Class XII : Physics]
NS
240
VS
4000 = 400
N
V P
2400
P
100
29
...
= 5t2 – 4t + 2 and R = 8 I –
...
018A
When V and l in phase
1
LC
1
C
1
2
X L XC ,
1
2 2
4 L
2
4 50 50
4
2
= 2
...
31
...
32
...
(i)
Frequency of current oscillations
1
2 LC
(ii)
Frequency of electrical energy oscillation c = 2
(iii)
Maximum current in the circuit I 0
(iv)
Magnetic energy in the inductor when charge on capacitor is 4mC
...
1 q0
2 C
–
1 q
2
2
2 C
q0 – q
2
2C
q = 4mC
Current in the circuit :
V
, where Z
2
(i)
I
(ii)
XL R
2
RMS voltage across L and R
Z
VL = I
...
[Class XII : Physics]
102
UNIT V & UNIT VI
ELECTROMAGNETIC WAVES
AND OPTICS
EM waves are produced by accelerated (only by the change in speed)
charged particles
...
Properties of em waves :
(i)
Transverse nature
(ii)
Can travel though vacuum
...
(v)
In any medium v
v Speed of EM waves
...
103
av
=
[Class XII : Physics]
In an em spectrum, different waves have different frequency and
wavelengths
...
Higher, the frequency
larger the penetration power
...
A wave travelling along +x axis is represented by
Ey = Eoy cos(t – kx)
Bz = Boz cos(t – kx)
2
2
T
k
k
2
2
frequency
1
wave number
...
r
...
vacuum (or air)
Refractive index of medium ‘b’ w
...
t
...
angle up to which central
= yn – yn–1 = D/d
Path difference
Ratio of light intensity at maxima and minima
x = (2n + 1) /2
Phase difference
Destructive Interference
Phase difference
Constructive Interference
109
[Class XII : Physics]
I = I0 cos2
Law of Malus
then the number of images formed
...
r
...
I2, the resulting minima and maxima will be
If two coherent beams have different intensities I and
r + p = 90° and µ = tan p
Brewster’s Law
UNIT V & UNIT VI
1
...
2
...
3
...
4
...
5
...
6
...
7
...
Every EM wave has certain frequency
...
Electric field vector and Magnetic field vector
...
U
...
Radiation
...
Radar, T
...
Communication
...
Polarization
...
Microwave, U V radiation, -rays
Which component E or B of an em wave is responsible for visible effect?
E
[Class XII : Physics]
110
8
...
9
...
V
1
Which of the following has longest penetration power?
UV radiation, X-ray, Microwaves
...
10
...
Ans
...
Ans
...
Ans
...
Radiowaves
...
Name two physical quantities which are imparted by an em wave to a
surface on which it falls
...
Name the physical quantity with unit same as that of
d e
0
where e electric flux
...
14
...
15
...
16
...
Current
...
What is the wavelength range of em waves that were produced and
observed by J
...
Bose?
Milimeter
Name the device used for producing microwaves
...
Ans
...
What is the speed of em waves in the medium
...
1
0 r 0 r
C
9
1
0 0 r r
C
3
Identify the part of the electromagnetic spectrum to which the following
wavelengths belong :
(i) 10–1 m (ii) 10–12 m
Ans
...
Ans
...
Microwave, -ray
y
Name the part of the electromagnetic spectrum of wavelength 10–2 m and
mention its one application
...
Which of the following, if any, can act as a source of electromagnetic
waves?
(i)
(ii)
21
...
22
...
23
...
(iii)
Ans
...
A charge at rest
...
Ex and By
The charging current for a capacitor is 0
...
What is the displacement
current?
Remain same
Give the ratio of Velocities of light waves of wavelengths 4000A° and
8000A° in Vaccum
...
24
...
25
...
26
...
27
...
Zero
...
Light falls from glass to air
...
28
...
29
...
Scattering
What is the ratio of sini and sinr in terms of velocities in the given figure
...
Ans
...
µ1
i
r
Velocity = v2
Ans
...
µ2
v1
v2
What is the shape of fringes in young’s double slit experiment?
Hyperbolic
...
What will be new focal length of each half
...
32
...
33
...
34
...
35
...
30 cm
...
In which medium would the
velocity of light be minimum?
A
What is the phase difference between two points on a cylindrical wave front?
Zero
...
Zero
...
Lower half of the concave mirror is painted black
...
The intensity of the image will be reduced (in this case half) but no change
in size of the image
...
An air bubble is formed inside water
...
38
...
Diverging lens
A water tank is 4 meter deep
...
for water is 4/3
...
below the water level
...
Give one use of each of the following (i) UV ray (ii) -ray
2
...
In which electric and
magnetic fields are along y-axis and z-axis respectively
...
State the principles of production of EM waves
...
What will be the
frequency of wave in the medium?
4
...
What will be
magnetic field at this instant? (Wave is travelling in vacuum)
...
State two applications of infrared radiations
...
State two applications of ultraviolet radiations
...
State two applications of x-rays
...
Show that the average energy density of the electric field E equals the
average energy density of the magnetics fields B ?
9
...
5m
...
5 m for the remedy
of this defect
...
67D)
10
...
His
far point is infinity
...
11
...
µglass = 3/2
µwater = 4/3
A
µg
q
B
µW
C
115
[Class XII : Physics]
12
...
13
...
µ = 1
...
Define diffraction
...
15
...
16
...
5 m high are situated at distance
40m and 50m respectively from an eye
...
S1 and S2 are two sources of light separated by a distance d
...
What should be the minimum
and maximum path difference at the detector?
S 1×
d
S2×
18
...
the vehicle is stationery
the vehicle is moving with constant speed
...
Similar phenomenon is observed when vehicle is in motion
...
A person looking at a mesh of crossed wire is able to see the vertical wire
more distinctly than the horizontal wire
...
Is optical density same as mass density? Give an example
...
Optical density is the ratio of the speed of light in two media whereas
mass density e
...
mass per unit volume of a substance
...
g
...
21
...
Ans
...
22
...
Then how do we see it?
Ans
...
23
...
NCERT Pg 318
24
...
Obtain an expression for
refractive index of the medium in terms of critical angle
...
The image of a small bulb fixed on the wall of a room is to be obtained
on the opposite wall ‘s’ m away by means of a large convex lens
...
Ans
...
f = s/4
26
...
In what
sense then does magnifying glass produce angular magnification?
Ans
...
27
...
117
[Class XII : Physics]
28
...
Why?
29
...
(i)
light diverges from a point source,
(ii)
light emerges out of convex lens when a point source is placed at
its focus
...
What two main changes in diffraction pattern of single slit will you observe
when the monochromatic source of light is replaced by a source of white
light
...
You are provided with four convex lenses of focal length 1cm, 3cm, 10cm
and 100 cm
...
32
...
Sun looks reddish at sunset
clouds are generally white
Using Huygens Principle draw ray diagram for the following
(i)
(ii)
34
...
Water (refractuive index µ) is poured into a concave mirror of radius of
curvature ‘R’ up to a height h as shown in figure
...
A point source S is placed midway between two concave mirrors having
equal focal length f as shown in Figure
...
[Class XII : Physics]
118
S
d
36
...
The two halves are combined as shown in figure
...
(i)
37
...
(aµ = 4/3)?
38
...
If the angle, of prism A = 60° and µ of material of
prism is
3 then find angle
...
C
Name EM radiations used (i) in the treatment of cancer
...
(iii)
In sterilizing surgical instruments
...
H ow w oul you exper m ent l show t
d
i
aly
hat EM waves are transverse in
nature?
3
...
4
...
Give its two applications
5
...
6
...
7
...
8
...
Some oil
is poured into the beaker to a height of ‘b’ cm and it is found that microscope
has to raise through vertical distance of ‘a’ cm to bring the dot again into
focus
...
9
...
State its two conditions
...
10
...
Draw the emergent wavefront in each case
...
Explain with reason, how the resolving power of a compound microscope
will change when (i) frequency of the incident light on the objective lens
is increased
...
(iii) asperture
of objective lens is increased
...
Derive Mirror formula for a concave mirror forming real Image
...
Two narrow slits are illuminated by a single monochromatic sources
...
Draw the intensity
pattern now obtained and name the phenomenon
...
Explain briefly (i) sparkling of diamond (ii) use of optical fibre in
communication
...
Using appropriate ray diagram obtain relation for refractive index of water
in terms of real and apparent depth
...
Complete the ray diagram in the following figure where, n1, is refractive
index of medium and n2 is refractive index of material of lens
...
2
n1 = n 2
(v)
n1 > n
(iii)
2
n1 > n 2
(vi)
A converging beam of light is intercepted by a slab of thickness t and
refractive index µ
...
P
x
1
1
t
µ
t
18
...
25
...
S1
P
S
O
S2
S c reen
121
[Class XII : Physics]
1
...
Obtain the relation between critical angle and refractive indices
of two media
...
(ii)
Deviate the ray through 90°
...
2
...
Derive an expression for its magnifying power
...
Diagrammatically show the phenomenon of refraction through a prism
...
Hence for a small angle of incidence
derive the relation = (µ – 1) A
...
Name any three optical defects of eye
...
(ii)
Hypermetropic eye and corrected hypermetropic eye
...
Define diffraction
...
6
...
7
...
8
...
9
...
10
...
Deduce the relation
µ
sin A m
sin A 2
2
11
...
Derive an expression for the width of the central maximum
due to diffraction of light at a single slit
...
1
...
5
...
Find wavelength and frequency of light
in the medium
...
An EM wave is travelling in vaccum
...
Calculate amplitude of magnetic field vector
...
Suppose the electric field amplitude of an em wave is E0 = 120 NC–1 and
that its frequency is = 50
...
(a)
Determine B0, , and
(b) Find expressions for E and B
...
A radio can tune into any station of frequency band 7
...
Find the corresponding wave length range
...
The amplitude of the magnetic field vector of an electromagnetic wave
travelling in vacuum is 2
...
Frequency of the wave is 16 MHz
...
Amplitude of electric field vector and
Wavelength of the wave
...
Ey = 4 × 105 cos (3
...
57 x) N/C
...
Then find :
(i)
Wavelength
(ii)
Frequency
(iii)
Direction of propagation
(iv)
Speed of wave
(v)
Refractive index of medium
(vi)
Amplitude of magnetic
field vector
...
An object of length 2
...
5f from a concave
mirror where f is the focal length of the mirror
...
Find the size of image
...
Find the size of image formed in the situation shown in figure
...
5 c m
40 cm
O
20 cm
1
...
C
µ 2 = 1
...
A ray of light passes through an equilateral prism in such a manner that
the angle of incidence is equal to angle of emergence and each of these
angles is equal to 3/4 of angle of prism
...
[Ans
...
Critical angle for a certain wavelength of light in glass is 30°
...
[ip = tan–1 2
11
...
Find refractive
air µ
Liquid
index of liquid
...
3 2
Liqu id
At what angle with the water surface does fish in figure see the setting
sun?
µ = 1 air
Sun
µ W = 4/3
W a ter
[At critical angle, fish will see the sun
...
In the following diagram, find the focal length of lens L2
...
80 cm
A hypermetropic person whose near point is at 100 cm wants to read a
book
...
Ans
...
100
33
...
What will be the
type and power of lens which would enable him to read a book at a
distance of 60 cm?
Ans
...
So a diverging lens
16
...
3 dioptre
–30
Using the data given below, state which two of the given lenses will be
preferred to construct a (i) telescope (ii) Microscope
...
125
[Class XII : Physics]
Lenses
Power (p)
Apetune (A)
L1
6 D
1 cm
L2
3 D
8 cm
L3
10 D
1 cm
Ans
...
For microscope lens L3 is chosen as objective because of its small focal
length and lens L1, serve as eye piece because its focal length is not
larges
...
Two thin converging lens of focal lengths 15 cm and 30 cm respectively
are held in contact with each other
...
1
1
1
1
1
1
F
f1 f2
15
30
10
F = 10 cm
P = 10 D
...
Lot of people like TV program CID
...
Every member
of CID team work with full dedication
...
They use ultraviolet rays
in forensic laboratory
...
(i)
(ii)
2
...
A child is observing a thin film such as a layer of oil on water show
beautiful colours when illuminated by while light
...
His teacher explains him the reason behind it
...
(i)
(ii)
3
...
6 mm
...
Now he is not
able to observe the diffraction pattern
...
Again the teacher
replaces x-rays by yellow light and the differaction patter appears again
...
(i)
Motivation
(ii)
2
...
(ii)
1
...
10
...
Hence power
is minimum
...
sin1 8 9
12
...
sin1 3 4
14
...
Minimum path difference is zero (when p is at infinity)
Maximum path difference = d
...
Astigmatism – Cylindrical lens
29
...
A ray is a line drawn perpendicular to the wavefront in the direction of
propagation of light
...
Plane
(i)
In each diffraction order, the diffracted image of the slit gets
dispersed into component colours of white light
...
(ii)
30
...
f0 = 1 cm and fe = 3 cm for Microscope and
f0 = 100 cm and fe = 1 cm for a Telescope
33
...
C
...
R
...
Fig
...
5; Fig
...
4
...
Distance of object from p should be equal to radius of curvature
...
Distance between mirror will be 2f or 4f
...
(i) Focal length of combination is infinite
...
21 –x
21 m
21 –x
x
[Class XII : Physics]
128
...
x = 12 cm
...
deviation = 60°
...
R
...
of a compound Microscope
2µ sin
2µ sin
c
(i)
(ii)
18
...
P
...
(iii)
17
...
P
...
P
...
1
x 1 t
Path diff
...
25 + S2P – S1P)
For maxima, path diff
...
25 = (n – 0
...
2n 1
So
2
S2P – S1P = (2n + 0
...
129
[Class XII : Physics]
UNIT VII
DUAL NATURE OF MATTER
AND RADIATION
Light consists of individual photons whose energies are proportional to
their frequencies
...
Photoelectric effect : Photon of incident light energy interacts with a single
electron and if energy of photon is equal to or greater than work function,
the electron is emitted
...
Kinetic energy of emitted electron = h( – 0) Here 0 is the frequency
below which no photoelectron is emitted and is called threshold frequency
...
UNIT VIII
ATOMS AND NUCLEI
Gieger-Marsden -scattering experiment established the existence of
nucleus in an atom
...
(iii)
Electrons revolve round the nucleus in certain fixed orbits called
stationary orbits
...
The energy is emitted (or absorbed) when electrons jump from
higher to lower energy orbits
...
The frequency of the emitted radiation is given by h =Ef – Ei
...
As a result of the quantisation condition of angular momentum, the electron
orbits the nucleus in circular paths of specific radii
...
n 2 h 2
0
rn
m 2 e 2
The total energy is also quantised : E n
rn n
me
2
4
2
2 2
8n 2 0 h
13
...
In hydrogen atom, the ground state energy is – 13
...
131
[Class XII : Physics]
de Broglie’s hypothesis that electron have a wavelength = h/mv gave an
explanation for the Bohr’s quantised orbits
...
Nuclear force does not distinguish between nucleons
...
The difference in mass of a nucleus and its constituents is called the mass
defect
...
For the mass number ranging from A = 30 to 170 the binding energy per
nucleon is nearly constant at about 8MeV per nucleon
...
Mathematically :
dN
dt
N or N t N 0e
t
where is called decay constant
...
Number of radioactive atoms N in a sample at any time t can be calculated
using the formula
...
of atoms at time t = 0 and T is the half-life of the substance
...
T1 2
or
ln 2
ln 2 mean life
0
...
693
[Class XII : Physics]
132
Here
1
decay constant =
...
(Almost)
for all substances/nuclei
...
133
[Class XII : Physics]
UNIT VII & VIII
1
...
2
...
Which of the following is
correct?
(a)
Energy of each reflected photon decreases by 20%
...
of reflected photons decreases by 20%
...
Ans
...
of reflected photons decreases by 20%
...
Ans
...
Ans
...
Ans
...
?
Lower work function, sensitive to visible light
...
Photoelectric effect
...
1
2
mvmax h ho
2
Max
...
E
...
6
...
7
...
8
...
9
...
(iv) Ultraviolet (Maxi frequency)
A metal emits photoelectrons when red light falls on it
...
Photoelectric effect & Compton effect
The photoelectric cut off voltage in a certain photoelectric experiment is
1
...
What is the max kinetic energy of photoelectrons emitted?
Ans
...
5 e Joule
= 1
...
6 × 10–19 J
= 2
...
Ans
...
Ans
...
6 1034
1
...
mv
32
What factors determine the maximum velocity of the photoelectrons from
a surface?
(a) frequency of incident radiation
(b) work function of surface
...
Ans
...
Ans
...
6 eV are emitted
...
6 V
Work functions of caesium and lead are 2
...
25 eV respectively
...
14
...
15
...
Ans
...
Ans
...
Write its energy equivalent in MeV
...
m
...
2m n E k
6
1
of the mass of a carbon isotope
12
1 u = 931 MeV
What was the drawback of Rutherford’s model of atom?
Rutherford’s model of atom failed to explain the stability of atom
...
of electrons 92
No
...
18
...
Name the series of hydrogen spectrum which has least wavelength
...
Any two protons repel each other, then how is this possible for them to
remain together in a nucleus
...
Nuclear force b/w two protons is 100 times stronger than the electrostatic
force
...
Ans
...
The decay constant of radioactive substance is defined as the reciprocal
1
th
of that time in which the number of atoms of substance becomes
e
times the atoms present initially
...
Ans
...
Ans
...
Ans
...
What type of nuclear
reaction is this?
Nuclear Fusion
...
particle
...
(i) Positive charge is concentrated in the nucleus
(ii) Size of nucleus is very small in comparison to size of atom
...
Ans
...
Ans
...
P = 0, Q = 1
X is + 1e0 a positron
...
Binding energies of neutron
12 H
and -particle (2He4) are 1
...
2 MeV/nucleon respectively
...
Binding energy of 2He4 is more than deutron 1H2
...
27
...
For what angle of deviation will
the number of deflected -particles be minimum?
Ans
...
Ans
...
Ans
...
Ans
...
How many maximum
number of spectral lines can be emitted by the atom?
Max number of spectral lines =
n(n 1)
43
6
2
2
Under what conditions of electronic transition will the emitted light be
monochromatic?
Only fixed two orbits are involved and therefore single energy evolve
...
Why does only a slow neutron (
...
Slow neutron stays in the nucleus for required optimum time and disturbs
the configuration of nucleus
...
Ans
...
Ans
...
Ans
...
Write the relation for distance of closest approach
...
Name the physical quantity whose dlimensions are same as Planck’s
constant
...
Ans
...
36
...
6 V
...
37
...
6 eV
What is the energy possessed by an electron whose principal quantum
number is infinite?
[Class XII : Physics]
138
Ans
...
Ans
...
Ans
...
Ans
...
Ans
...
Ans
...
Ans
...
Ans
...
Ans
...
6
n2
eV 0
Write the value of Rydberg constant?
1
...
Lyman Series
Name the series of hydrogen spectrum lying in the infra red region
...
106 ms–1
Write a relation for Paschen series lines of hydrogen spectrum
...
Arrange radioactive radiation in the increasing order of penetrating power
...
J
1
average life
Write two units for activity of radioactive element and relate them with
number of disintegration per second
...
7×1010 decay/s
1 bequered (Bq) = 1 decaya/s
139
[Class XII : Physics]
1 Ci = 3
...
Initially, both have same number of atoms
...
Why?
Ans
...
44 TB TA > TB A < B
...
47
...
No
...
Ekmax
K
...
48
...
Compare radii of two nuclei of mass numbers 1 and 27 respectively
...
Ans
...
Ans
...
Ans
...
56Fe
26
Mention the range of mass number for which the Binding energy curve is
almost horizontal
...
[Class XII : Physics]
140
52
...
(N
...
E
...
T Page 447)
N
N0
t
53
...
Write an equation to represent decay
...
2
1
...
2
...
3
...
Which one of these
has greater de-Broglie wavelength?
4
...
75 eV, K = 2
...
14 eV, Ni = 5
...
What happens if the laser is
brought nearer and placed 50 cm away
...
Name the experiment for which the followings graph, showing the variation
of intensity of scattered electron with the angle of scattering, was obtained
...
Incide nt
electron
be am
V = 54 volt
6
...
:
141
[Class XII : Physics]
P
6
Q
V 4
(V olt)
2
4 6
v
14
v = (8 × 1 0 Hz )
–2
(i)
Which of the two metals has greater value of work function?
(ii)
Find maximum K
...
of electron emitted by light of frequency
= 8 × 1014 Hz for metal P
...
Do all the photons have same dynamic mass? If not, why?
8
...
Find the ratio of de-Broglie wavelengths associated with two electrons ‘A’
and ‘B’ which are accelerated through 8V and 64 volts respectively
...
The photoelectric current at distances r1 and r2 of light source from
I2
photoelectric cell are I1 and I2 respectively
...
1
Ans
...
Ans
...
12
...
An -particle of kinetic energy ‘K’
is bombarded on a thin gold foil
...
What will be the distance of closest approach for an -particle of double
the kinetic energy?
13
...
14
...
Some scientist have predicted that a global nuclear war on earth would be
followed by ‘Nuclear winter’
...
If the total number of neutrons and protons in a nuclear reaction is conserved
how then is the energy absorbed or evolved in the reaction?
17
...
3 × 10–11 m
...
2 × 10–11 m
...
Calculate the percentage of any radioactive substance left undecayed after
half of half life
...
Why is the density of the nucleus more than that of atom?
20
...
Why?
21
...
22
...
Show that the decay rate R of a sample of radio nuclide at some instant
is related to the number of radio active nuclei N at the same instant by the
expression R = – N
...
What is a nuclear fusion reaction? Why is nuclear fusion difficult to carry
out for peaceful purpose?
25
...
26
...
After how much time will the activity of the radioactive sample
reach the ‘safe limit’?
27
...
2
143
[Class XII : Physics]
28
...
29
...
30
...
: 820nm, (ii) 365 nm
31
...
53 Å
...
32
...
6 eV
...
Write formula of frequency to represent (i) Lyman series (ii) Balmer series
...
From the relation R = R0 A1/3 where R0 is a constant and A is the mass
number of a nucleus, show that nuclear matter density is nearly constant
...
Nuclear matter density
Mass of nucleus
Volume of nucleus
mA
mA
4
4
3
3
R
R 0 A
3
3
m
17
3
2
...
Ans
...
E = mc2
m = 1
...
6605 × 10–27 × (3 × 108)2
= 1
...
4924 × 10
1
...
9315 × 109 eV
= 931
...
Write four properties of nuclear force
...
Explain the working of a photocell? Give its two uses
...
Find the de Broglie wavelength associated with an electron accelerated
through a potential difference V
...
What is Einstein’s explanation of photo electric effect? Explain the laws of
photo electric emission on the basis of quantum nature of light
...
If kinetic energy of thermal neutron is
3
kT then show that de-Broglie
2
wavelength of waves associated with a thermal neutron of mass m at
temperature T kelvin is
h
3mkT
where k is boltz mann constant
...
Explain Davisson and Germer experiment to verify the wave nature of
electrons
...
Explain the effect of increase of (i) frequency (ii) intensity of the incident
radiation on photo electrons emitted by a metal
...
X-rays of wave length fall on a photo sensitive surface emitting electrons
...
Ans
...
2mc
A particle of mass M at rest decays into two particles of masses m1 and
m2 having velocities V1 and V2 respectively
...
1:1
9
...
Also define the Q-value of the
reaction
...
Explain how radio-active nucleus can-emit -particles even though nuclei
145
[Class XII : Physics]
do not contain these particles
...
11
...
Derive the relation
between these terms
...
State the law of radioactive decay
...
13
...
14
...
15
...
Give an example of each
...
A radioactive isotope decays in the following sequence
1
n
0
A A 1 A 2
...
Which of the three elements are
isobars?
17
...
Show that total energy is negative of K
...
and half of potential energy
...
8E 0r
18
...
19
...
4 eV
...
the kinetic energy,
Which of the answers above would change if the choice of the zero
of potential energy in changed to (i) + 0
...
5 eV
...
E
...
4 eV, using
E = – K
...
, the K
...
= + 3
...
[Class XII : Physics]
146
(b)
Since P
...
= – 2E, PE = – 6
...
(c)
If the zero of P
...
is chosen differently, K
...
does not change
...
E
...
E
...
E
...
(i)
When P
...
at is + 0
...
E
...
4 – 0,5 = – 3
...
(iii)
When P
...
at is + 0
...
E
...
4 – (–0
...
9 eV
...
What is beta decay? Write an equation to represent – and + decay
...
21
...
NCERT pg
...
State Bohr’s postulates
...
What does negative
of this energy signify?
2
...
Draw a curve between mass number
and average binding energy per nucleon
...
3
...
Hence define disintegration
constant and half life period
...
4
...
5
...
Write the conclusion made and draw the model
suggested
...
State law of radioactive decay obtain relation
(i)
N = N0 e–t
(ii)
R = R0 e–t
where
N is number of radioactive nuclei at time t and
147
[Class XII : Physics]
N0 is number of radioactive nuclei at time t0 is decay constant
R is rate of decay at any instant t
R0 is rate of decay at any time t0 (initial time)
...
Ultraviolet light of wavelength 350 nm and intensity 1W/m2 is directed at
a potassium surface having work function 2
...
(i)
Find the maximum kinetic energy of the photoelectron
...
5 percent of the incident photons produce photoelectric effect,
how many photoelectrons per second are emitted from the
potassium surface that has an area 1cm2
...
3 eV ; n 8
...
A metal surface illuminated by 8
...
52 eV the same surface is illuminated by 12
...
97eV
...
[Work Function = 3ev]
3
...
2 nm
...
(i)
(ii)
Ans
...
6
...
3
...
5 × 106 m/s; (b) 5
...
112 nm
...
A proton is accelerated through a potential difference V
...
(9
...
For what Kinetic energy of a neutron will the associated de Broglie
wavelength be 5
...
2m n K
...
h
h
K
...
6
...
6 10 –10
8
...
2
1
2m n
2
1
2 1
...
35 10
21
J
A nucleus of mass M initially at rest splits into two fragments of masses
M
2M
...
Find the ratio of de Broglie wavelength of the fragments
...
Ans
...
E
...
Ee
1
1
2
2
m and E p m p p
2 e e
2
mee
But
...
10
...
3 × 10–1m
...
[r2 = 2
...
47 × 10–10 m]
149
[Class XII : Physics]
11
...
Ans
...
43 × 10–19 J
12
...
Rydberg constant = 1
...
Ans
...
What will be the distance of closest approach of a 5 MeV proton as it
approaches a gold nucleus?
Ans
...
55 × 10–14 m
14
...
5 MeV alpha – particle approaching a gold nucleus is deflected by
180°
...
: 1
...
Determine the speed of the electron in n = 3 orbit of hydrogen atom
...
: 7
...
If half life is 20 seconds, how many nuclei will remain after 10 seconds?
Ans
...
17
...
In how much time will
15/16 of the material decay?
Ans
...
At a given instant, there are 25% undecayed radioactive nuclei in a sample
...
5%
...
Ans
...
43
19
...
37 MeV and 39
...
Which of the two nuclei is more stable? Why?
Ans
...
Find the binding energy and binding energy per nucleon of nucleus 83B209
...
0078254 u
...
008665 u
...
980388u
...
: 1639
...
84 MeV/Nucleon
21
...
934940 and
28
13Al
= 27
...
: Since Q value comes out negative, so this fission is not possible
[Class XII : Physics]
150
22
...
003576 a
...
u
...
998403 a
...
u
...
000549 a
...
u
...
Ans
...
3049 MeV
The value of wavelength in the lyman series is given as
2
93
...
Does wavelength decreases or increase
...
:
21
31
41
913
...
4 4
2
1218 A
1
913
...
3 A
o
1
41 < 31 < 21
24
...
5 × 109 years what is the
activity of 1g
...
Ans
...
5 × 109 y
= 4
...
16 × 107 s
= 1
...
025 10
238
151
23
atom
[Class XII : Physics]
= 25
...
693
T
0
...
3 10
1
...
23 × 104 Bq
...
In an experiment of photoelectric effect, Nita plotted graphs for different
observation between photo electric current and anode potential but her
friend Kamini has to help her in plotting the correct graph
...
(i)
(ii)
2
...
Draw the correct graph betweeen I and V (NCERT)
...
The auditorium has the
capacity of 400 students
...
Then science students took
responsibility at the gate
...
This
helped them to maintain discipline and counting became easy with the
help of a device used by these students
...
What value is displayed by science?
Name the device which is based on application of photoelectric
effect
...
His
nails were getting blue, he stated losing his hair
...
Doctor advised him hospitalization and suspected
he has been exposed to radiation
...
[Class XII : Physics]
152
(i)
(ii)
4
...
Name the radioactive radiations emitted from a radioactive element
...
He read that after
so many years of atomic bombing is Hiroshima or Nagasaki, Japan National
census indicated that children born even now are genetically deformed
...
He asked
his Granddaughter Medha who is studying in class XII Science
...
(i)
(ii)
X ray and Gamma rays
...
Value displayed - awareness, critical thinking, decision making
...
Name the nuclear reactions that occurred is atom bomb
...
What are the values/skills utilized by Kajal to make her grandfather
understand the reason of genetically deformity?
Nuclear-fission reactions
...
m
E
c
2
h
c
2
m depends on frequency of photon
...
Because electrons lose their energy in collision
...
9
...
It will be halved
...
Using the relation R = R0A1/3
...
14
...
(ii)
-particle
...
(iv)
-particle
...
Nuclear radioactive waste will hang like a cloud in the earth atmosphere
and will absorb sun radiations
...
The total binding energy of nuclei on two sides need not be equal
...
17
...
From relation
N
N0
N0
1
2
1
2
tT
when t T 2
12
or
N
N0
1
2
100
70
...
2
19
...
20
...
21
...
22
...
[Class XII : Physics]
154
23
...
e
...
dt
24
...
So, to carry out
fusion for peaceful purposes we need some system which can create and
bear such a high temperature
...
Nuclear forces are short range forces (within the nucleus) and do not obey
inverse equare law while coulomb forces are long range (infinite) and obey
inverse square law
...
or
or
1
2
3
1
2
3
t T1 2
t 3
t
3
t = 9 days
...
Solids are classified on the basic of
(i)
Resistivity
conductivity
Metals
(m)
(Sm–1)
10–2 – 10–8
102 – 108
Semiconductors
10–5 – 106
10–6 – 105
Insulators
(ii)
Electrical conductivity
1011 – 1019
10–19 – 10–11
Energy Bands
C
...
Metal
V
...
Band Gap
energy
Eg = 0
C
...
Eg
Semiconductor
V
...
C
...
Eg
Insulator
V
...
[Class XII : Physics]
156
Eg < 3eV
Eg > 3eV
2
...
Organic, Anthracene
Doped Pthalocyamines etc
...
In intrinsic semiconductors (Pure Si, Ge) carrier (electrons and holes) are
generated by breaking of bonds within the semiconductor itself
...
4
...
5
...
) when doped to Si or Ge give
n-type and trivalent (acceptor) atom (In, Ga, Al etc
...
6
...
7
...
8
...
9
...
10
...
11
...
12
...
Because of this property,
the diode is used as a voltage regulator
...
In a transistor current goes from low resistances (forward biasing) to high
resistance (reverse biasing)
...
Current relationship in a transistor
...
In common emitter transistor characteristic we study
Ib versus VBE at constant VCE (Input characteristic)
Ic versus VCE at constant IB (output characteristic)
Input resistance
VBE
I
B
V
CE
Output resistance
16
...
ac dc
...
17
...
(ii) as
amplifier in active region
...
In CE configuration, transistor as amplifier output differ in phase them
input by
...
Transistor as an amplifier with positive feedback works as an oscillator
...
Gates used for performing binary operations in digital electronics mainly
consist of diodes and transistors
...
NAND gates alone can be used to obtain OR gate and similarly a NOR
gates alone can be used to obtain AND gate, OR gate
...
A communication system consists of three basic elements
...
A transmitter consists of
(i)
(iii)
Transducer or Converter
(ii) Modulator
Carrier Oscillator
(iv) Transmitting Antenna
Channel : It is the medium through which the electrical signals from the
transmitter pass to reach the receiver
...
A Receiver section consists of
(i)
(iii)
Receiver Antenna
(ii) Transducer/Converter
Demodulator
Two important forms of communication system are Analog and Digital
...
In Digital communication, the information has only discrete or quantised
values
...
Need of Modulation :
(i)
To avoid interference between different base band signals
...
(iii)
To increase power radiated by antenna
...
Amplitude Modulation : In this type of modulation, the amplitude of carrier
wave is varied in accordance with the information signal, keeping the
frequency and phase of carrier wave constant
...
Space communication uses free space between transmitter and receiver
for transfer of data/information
...
They are heavily absorbed
by earth surface and not suitable for long range communication
...
The space waves are within
the troposphere region of atmosphere and have two Modes of
Transmission :
(i)
Line of sight communication
(ii)
Satellite communication
Physical Quantity
Formulae
1
SI Unit
Power radiated by an antena
Sinusoidal carrier wave
E EC cos (ct + )
V
The range of tower
d = 2Rh
R radius of earth
h Height of antena
m
[Class XII : Physics]
160
2
W
The number of channels
Bandwidth
Bandwidthper channel
The maximum rang of broadcast
d max
between transmitting and receiving
where R Radius of earth
2Rh t
2Rh r
tower ht and hr height of transmitting and
receiving towers
1
...
The Internet is a global system that makes it possible for computers worldwide to share information via a variety of languages called protocols
...
1
...
(i)
ARPANET It tends for Advance Research project Ajency network
...
In U
...
A Department of Defence sponsored a project
called ARPANET whose goal is to connect the computer of U
...
defence
...
(ii)
NSFnet It tends for national science foundation network
...
In U
...
A
...
(iii)
Internet In 1990 the internetworking of ARPANET, NSFnet and other private
network revisited into internet
...
The information of world can be
shared through internet
...
161
[Class XII : Physics]
Note : The original ARPANET was shut down in 1990, and the NSF net
was disconnected in 1995
...
1
...
1
...
2
...
3
...
1
...
Before www, internet was mainly used for
obtaining textual information only, multimedia file can’t be accessed
...
Hypertext : A hypertext is a text through which we can access or get other
information of that text by clicking on it
...
Hypermedia : A hypermedia is a file or any image or any other media
through which we can access or get other information of that media by
clicking on it
...
Hyperlink : Hyperlink refer to the link through which we can open the
other new web page by clicking on it
...
5 Hyper Text Transfer Protocol
HTTP : It tends for Hypertext transfer protocol
...
It fetch only textual informa[Class XII : Physics]
162
tion
...
Although HTTP was designed for use in the web
...
6 Web Page
A web page is a document on the internet, that can contain textual files,
multimedia files & hyper links
...
1
...
Each web site
has a unique address called URL (uniform resource location)
...
harshpublishers
...
8 Home Page
It is the top level web page of a website
...
1
...
A
web browser sends a request of user to the web server to open a desired web
page on the web browser through which user can access the information
...
10 URL (Uniform Resource Locator)
A web address is also known as an URL, which stands for Uniform Resource
Locator for example www
...
com is an URL of Harsh Publishers
Websites
...
2
...
1 The Mobile Telephone System
The traditional telephone system will still not be able to satisfy a growing
group of users: people on the go
...
Consequently, there is a tremendous amount of interest in wireless
telephony
...
It does so by connecting to a cellular
network provided by a mobile phone operator, allowing access to the public
telephone network
...
Note: The first hand-held cell phone was demonstrated by John F
...
2
pounds (1 kg)
...
2 Working of Mobile Telephone System
In this techniques a message can be divided into some fixed size of packets
that can be transferred across the network
...
The communication protocol used by mobile telephone is TCP / IP
...
TCP is responsible for dividing a
message into packets and also responsible for combining these packets into original
form at other ends
...
So
that the packets can be transferred at their proper distination
...
3 Generation of Mobile Phone
Mobile phones have gone through three distinct generations, with different
technologies:
1
...
2
...
3
...
2
...
1 First-Generation Mobile Phones : Analog Voice
This system used a single large transmitter on top of a tall building and had
a single channel, used for both sending and receiving
...
Such systems,
known as push-to-talk systems, were installed in several cities beginning in the
late 1950s
...
In the 1960s, IMTS (Improved Mobile Telephone System) was installed
...
In this case the mobile users could not hear each other (unlike
the push-to-talk system used in taxis)
...
Due to
the small number of channels, users often had to wait a long time before getting
a dial tone
...
All in all, the
limited capacity made the system impractical
...
165
[Class XII : Physics]
In all mobile phone systems, a geographic region is divided up into cells,
which is why the devices are sometimes called cell phones
...
Thus, the cellular design increases the system capacity
...
In an area where the number of users has grown to the point that the system
is overloaded, the power is reduced, and the overloaded cells are split into smaller
microcells to permit more frequency reuse,
...
(a) Frequencies are not reused in adjacent cells
...
At the center of each cell is a base station to which all the telephones in the
cell transmit
...
In a small system, all the base stations are connected
to a single device called an MTSO (Mobile Telephone Switching Office) or MSC
[Class XII : Physics]
166
(Mobile Switching Center)
...
3
...
Four systems are in use now: D-AMPS, GSM, CDMA, and PDC
...
PDC is used only in Japan and is basically D-AMPS
modified for back- ward compatibility with the first-generation Japanese analog
system
...
e
...
Originally it meant a mobile phone using the 1900 MHz band, but that distinction
is rarely made now
...
D-AMPS was
carefully designed to co-exist with AMPS so that both first- and second- generation
mobile phones could operate simultaneously in the same cell
...
Depending
on the mix of phones in a cell, the cell’s MTSO determines which channels are
analog and which are digital, and it can change channel types dynamically as the
mix of phones in a cell changes
...
S
...
Similarly GSM (Global System
for Mobile communications) is used everywhere in the world, and it is even starting
to be used in the U
...
on a limited scale
...
Both are cellular systems
...
Instead of dividing the allowed frequency range into
a few hundred narrow channels, CDMA allows each station to transmit over the
entire frequency spectrum all the time
...
CDMA also relaxes the assumption that colliding
frames are totally garbled
...
Thus, the key to CDMA is to be able to extract the desired signal while
rejecting everything else as random noise
...
3
...
This device and how to connect it is what third generation
mobile telephony is all about
...
High-quality voice transmission
...
Messaging (replacing e-mail, fax, SMS, chat, etc
...
Multimedia (playing music, viewing videos, films, television, etc
...
Internet access (Web surfing, including pages with audio an video)
2
...
IMEI (International Mobile Equipment Identity) : The IMEI number is used
by a GSM network to identify valid devices and therefore can be used for stopping
a stolen phone from accessing that network
...
This renders the phone useless on that network
and sometimes other networks too, whether or not the phone’s SIM is changed
...
4 to 2
...
Invented by telecom vendor Ericsson in 1994
Infrared–Infrared (IR) is electromagnetic radiation with longer wave length
than those of visible light, extending from the nominal red edge of the visible
spectrum at 700 nanometers (nm) to 1 mm
...
Infrared radiation
was discovered in 1800 by astronomer William Herschel
3
...
One’s exact location (longitude, latitude and height co-ordinates) accurate
to within a range of 20 m to approx
...
2
...
5ns
...
The co-ordinates and time values are determined by 28
satellites orbiting the Earth
...
GPS (the full description is: Navigation System with Timing And Ranging
Global Positioning System, NAVSTAR GPS) was developed by the U
...
Department
of Defense (DoD) and can be used both by civilians and military personnel
...
During the development of the GPS system, particular emphasis was placed
on the following three aspects:
1
...
2
...
3
...
The GPS system functions according to exactly the same principle
...
1 Generating GPS Signal Transit Time
If the time measurement is accompanied by a constant unknown error, we
will have four unknown variables in 3-D space:
longitude (X)
latitude (Y)
169
[Class XII : Physics]
height (Z)
time error (At)
It therefore follows that in three-dimensional space four satellites are needed
to determine a position
...
2 Determining a Position in 3-D Space
In order to determine these four unknown variables, four independent equations
are needed
...
1 to sat
...
The 28 GPS satellites are distributed around the globe
in such a way that at least 4 of them are always “visible” from any point on Earth
...
5-10 m
...
Ans
...
Ans
...
ne = nh
Write the value of resistance offered by an ideal diode when (i) forward
based (ii) reverse biased
...
Write any one use of (i) photodiode (ii) LED
...
(b) In demodulation of optical signal
(c) In light operated switches
(d) In electronic counters
(ii) Use of LED
(a) Infrared LEDs sare used in burglar alarm
(b) In optical communication
(c) LED’s are used as indicator lamps in radio
receivers
(d) In remote controls
171
[Class XII : Physics]
4
...
Write the truth table for a two input AND gate
...
0
1
6
...
B
0
5
...
Semiconductors do not support strong current i
...
, a semiconductor is
damaged when strong current passes through it
...
Because bonds break up, crystal breakdown takes place and crystal
becomes useless
...
Draw I–V characteristic of a solar cell
...
Vsc
V
Voc Open Circuit Voltage
Isc Short Circuit Current
I
9
...
10
...
11
...
What is the direction of diffusion current in a junction diode?
The direction of diffusion current is from P to N in a semiconductor junction
diode
...
[Class XII : Physics]
172
12
...
13
...
14
...
15
...
16
...
Name the semiconductor device that can be used to regulate an unregulated
dc power supply
...
n
...
Light emitting diode (LED)
Name any one semiconductor used to make LED
...
A current is found to pass through the circuit
...
Name the
semiconductor device
...
In the following diagram write which of the diode is forward biased and
which is reverse biased?
–10V
–12V
(i)
(ii)
Reverse biased
18
...
–5V
Forward biased
How does the energy gap in a semiconductor vary, when doped, with a
pentavalent impurity?
The energy gap decreases
...
Ans
...
Conductor - no energy gap
Semiconductor <3eV
Insulator >3eV
20
...
The ratio of the number of free electrons to holes ne/nh for two different
materials A and B are 1 and < 1 respectively
...
ne
1 ne nh Intrinsic semiconductor
nh
ne
1 ne nh
nh
21
...
22
...
23
...
p type extrinsic semiconductor
What are ground waves?
The em
...
If a radiowave from the transmitting antenna reaches to
the receiving antenna either directly or after reflection from the ground, it
is called a ground wave
...
(i) the frequency of transmitted wave
(ii) the power of the transmitter
...
What is a base band signal?
25
...
4
[Class XII : Physics]
174
26
...
Why do we use high frequencies for transmission?
To reduce the height of antena
...
Why is ionisation low near the earth and high, far away from the earth?
Ans
...
V
...
Due to this molecules get
ionised
...
At high attitude solar
intensity is high, but density of Earth’s atmosphere is low
...
On the other hand, close to the earth,
the density of Earth’s atmosphere is high but the radiation intensity is low
...
28
...
29
...
Define the modulation index
...
It is also
known as modulation factor
...
75 × 102 m
30
...
The energy loss of a ground wave increases rapidly with the increase in
frequency
...
e
...
Ans
...
Any device which converts energy from one from to another is called
transducer e
...
a microphone converts sound energy (signal) into an
electrical energy (signal)
...
A T
...
transmitting antenna is 81m tall
...
The maximum distance upto which the signal transmitted from 80m tall T
...
antena can be received
...
2km
Area = d2 m2
33
...
34
...
35
...
Why are repeaters used in communication?
Repeater is the combination of a transmitter an amplifier and a receiver
which picks up retransmits it to the receiver sometimes with a change of
carrier frequency
...
What is the significane of modulation index? What is its range?
Ans
...
High modulation index ensures better quality and better strength
...
1
...
What will be the frequency of the
pulsating output signal in case of:
(i)
2
...
Potential barrier of p
...
junction cannot be measured by connecting a
sensitive voltmeter across its terminals
...
Diode is a non linear device
...
5
...
Explain
...
The diagram shows a piece of pure semiconductor S in series with a
variable resistor R and a source of constant voltage V
...
V
S
A
R
7,
What is the field ionisation in zener diode? Write its order of magnitude
...
Power gain of a transistor is high
...
9
...
Why is a photo diode used in reverse bias?
11
...
12
...
13
...
14
...
Write the truth
table for the final output of the combination
...
15
...
y´
A
B
y
y´´
177
[Class XII : Physics]
16
...
Write its truth table and identify the gate
...
1
t2
t3
t4
t5
t6
t7 t8
In the given circuit, D is an ideal diode
...
When the applied voltage V makes the diode
...
A transistor is a current operated device
...
19
...
Here the
transistor has been represented by a circle with the emitter (e), base (b)
and collector (c) terminals marked clearly
...
(a)
What is the type of transistor pnp or npn?
(b)
Is the transistor in saturation or cutoff?
R
b
[Class XII : Physics]
c
e
178
20
...
21
...
22
...
In a transistor, base is slightly doped and is a thin layer, why?
24
...
25
...
26
...
(b)
Si junction diode
...
Which of the input and output circuits of a transistor has a higher resistance
and why?
28
...
29
...
The two
are joined end to end and connected to a battery as shown
...
Will the junction be forward biased or reversed biased?
Sketch a V–I graph for this arrangement
...
Which circuits is
it? Give reason for your answer
...
(b)
Following voltage waveform is fed into half wave rectifier that uses a silicon
diode with a threshold voltage of 0
...
Draw the output voltage
...
2v
1v
tim e
t
–2 v
32
...
33
...
Give one example of each
...
An audio signal of amplitude one fourth of the carrier wave, is used in
amplitude modulation
...
What are the essential components of a communication system? Explain
with the help of a Block diagram
...
Explain by a diagram, how space waves are used for Television broadcast
...
Long distance radio broadcasts use short wave bands
...
The short waves are the waves of wavelength less than 200m or frerquency
greater than 1
...
They are absorbed by the earth due to their high
frequency
...
These waves after
reflection from ionosphere reach the surface of earth only at a large distance
from the place of transmission
...
It is due to this reason; the short waves are used in long distance
broadscasts
...
What is modulation? Why do we need modulation? Give two reasons
...
Give two reasons for using satellite for long distance T
...
transmission
...
Explain the propagation of sky wave in ionospheric layers with the help of
a neat, labelled diagram
...
Derive an expression for maximum range of an antenna of height ‘h’ for
LOS communication
...
Plot amplitude v/s frequency for an amplitude modulated signal
...
Draw block diagram of simple modulator to obtain amplitude modulated
signal
...
It is necessary to use satellites for long distance TV transmission
...
Yes, TV signals being of high frequency are not reflected by the ionosphere
...
That is why;
satellities are used for long distance TV transmission
...
What is the basic difference between an analog communication system
and a digital communication system?
Ans
...
A digital communication system makes use of
a digital signal, which has only two valuse of votage either high or low
...
What is ground wave? Why short wave communication over long distance
is not possible via ground waves?
Ans
...
5 m to 200 m) which are travelling directly
following the surface of earth are known as ground waves
...
Moreover the ground
wave transmission becomes weaker as frequency increases
...
What is depletion region in p-n junction diode
...
2
...
3
...
181
[Class XII : Physics]
4
...
5
...
6
...
7
...
8
...
What is a transistor? Draw symbols of npn and pnp transistor
...
10
...
11
...
12
...
13
...
14
...
15
...
16
...
17
...
18
...
19
...
Write the band width of the following :
(1)
(2)
Video signal
(3)
21
...
Ground waves
Sky waves
(i) At low frequencies (v < 2MHz), radio-waves radiated by antenna travel
directly following the surface of earth and are known as ground waves
...
(iii) Frequencies between 2-20 MHz are reflected by the ionosphere and
known as sky waves (or ionospheric propagation)
22
...
Give typical examples, with the help of suitable
figure, of communication systems that use space mode propagation
...
Mode of radiowave propagation by space waves, in which the wave travells
in a straight line from transmitting antenna to the receiving antenna, is
called line-of-sight (LOS) communication
...
At
frequencies above 40 MHz, LOS communication is essentially limited to
line-of-sight paths
...
How does a transistor work as an oscillator? Explain its working with
suitable circuit diagram
...
2
...
Show these characteristics graphically
...
What is p-n junction diode? Define the term dynamic resistance for the
junction
...
4
...
5
...
6
...
Draw the output waveform in each case
...
0
0
t1 t2 t 3
t4 t5 t6 t7 t 8
t1 t2 t 3
t4 t5 t6 t7 t 8
In a p-n junction, width of depletion region is 300 nm and electric field of
7 × 105 V/m exists in it
...
(ii)
What should be the minimum kinetic energy of a conduction electron
which can diffuse from the n-side to the p-side?
2
...
If 90% of the
electrons emitted reach the collector, find the base current and emitter
current
...
An LED is constructed from a p-n junction of a certain semiconducting
material whose energy gap is 1
...
What is the wavelength of light emitted
by this LED?
4
...
(Barrier voltage for Si diode is 0
...
[Class XII : Physics]
184
D1
(S i)
I 2
...
Determine V0 and Id for the network
...
6 k
6
...
8
eV
...
7
...
Where D1 and D2 are made
of silicon
...
09 mA
Id 1 I d 2
2
I 1 0
...
Two amplifiers with voltage gain 10 and 20 are connected in series
...
01 volt
...
: 2 volt]
9
...
If the collector resistance is 6kW and
input resistance 1k
...
[Ans
...
If the current gain of a CE – Amplifier is 98 and collector current Ic = 4mA,
determine the base current
...
: Ib = 0
...
Pure Si at 300 K has equal electron (n e) and hole (nh) concentration of 1
...
Doping by indium increases nh to 4
...
Calculate ne
in the doped silicon
...
: 5× 109 m–3]
12
...
5 eV
...
(ii)
Why Cd S or CdSe (Eg ~ 2
...
185
[Class XII : Physics]
(iii)
Why we do not use materials like PbS (Eg ~ 0
...
Ans
...
Si has Eg
...
1 eV and for GaAs,
Eg
...
53 eV
...
(ii)
If we choose CdS or CdSe, we can use only the high energy
component of the solar energy for photo-conversion and a significant
part of energy will be of no use
...
is satisfied, but if we use Pbs, most of
solar radiation will be absorbed on the top-layer of solar cell and
will not reach in or near depletion region
...
A sinusoidal carrier wave of frequency 1
...
Calculate the frequency
(i) amplitude; (ii) frequencies of lower and upper side bands
...
An amplitude modulator consist of L–C circuit having a coil of inductance
8mH and capacitance of 5pF
...
[Ans
...
96 × 105 Hz; Lower side band = 786 kHz;
Upper side band = 806 kHz]
15
...
V
...
(i)
How much population is covered by the T
...
broadcast if the average
population density around the tower is 1000km–2? Radius of earth
is 6
...
(ii)
By How much should the height of the tower be increased to
double the coverage area?
[Ans
...
16 lacs;
Change in height = 70m]
16
...
If only
1% of the frequency is used as channel bandwidth for optical communication
then find the number of channels that can be accommodated for
transmission of
(i)
an Audio signal requiring a bandwidth of 8 kHz
...
V
...
5 KHz
...
Calculate the percentage increase in the range of signal reception, if the
height of TV tower is increased by 44%
...
: 20% increase]
18
...
What is the maximum distance
between them for satisfactory communication in LOS mode? Given radius
of earth 6
...
d m 2 64 105 32 2 64 10 5 50m
2
3
Sol : 64 10 10 8 10 10 m
144 10 2 10 m 45
...
A message singnal of frequency 10 kHz and peak voltage of 10 volts is
used to modulate a carrier of frequency 1 MHz and peak voltage of 20
volts
...
187
[Class XII : Physics]
Sol : (a) Modulation index = 10/20 = 0
...
20
...
What should be the peak voltage of the modulating signal in order to have
a modulation index of 75%?
Sol :
0
...
75 Ac = 0
...
A modulating signal is a square wave, as shown in figure
...
(i) Sketch the amplitude modulated waveform
(ii) What is the modulation index?
1
m( t )
(in v o lt)
Sol :
1
2 t (in s )
(i)
3
2
1
S ig n a l in
V o lts
0
t
–1
–2
–3
0
0
...
8 1 1
...
4 1
...
8 2
(ii) = 0
...
For an amplitude modulated wave, the maximum amplitude is found to be
10 V while the minimum amplitude is found to be 2 V
...
What would be the value of if the minimum amplitude is zero volt?
Sol :
The AM wave is given by (Ac + Am + Sin mt) cos ct,
The maximum amplitude is M1 = Ac + Am while the minimum
amplitude is
M2 = Ac – Am
Hence the modulation index is
Am M1 –M2 8 2
A c M1 –M2 12 3
...
A child uses a semi conductor device in listening radio & seeing pictures
on T
...
He was asked to suggest the techniques as the cost of LPG/CNG
is going up, to cope up with future situations
...
Raju was enjoying TV programme at his home with his family at night
...
Mother asked Raju
to bring candle along with matchstick from kitchen to put on TV switch off
...
Her mother was surprised and asked where from the light
was coming
...
(i)
Which values displayed by Raju?
(ii)
Which material is used in LED?
189
[Class XII : Physics]
1
...
49 & 490
...
Awareness of social problems, Generates new idea with fluency
...
(i)
NCERT (SEMI conductors)
...
He does not have much
money to spend on telephone calls
...
Deepa told her uncle that he can talk to his son with the help of computer
and told him about internet
...
He
thanked Deepa for giving useful advice
...
What according to you are the values displayed by Deepa
...
It was last date
of admission and Renu left her birth certificate at her home
...
She called her brother and he faxed the birth
certificate
...
(i)
(ii)
(i)
Caring and creating awareness
...
What value was displayed by her brother?
(ii)
3
...
ne = nh
...
(i) zero
7
...
C
...
Because bonds break up, crystal breakdown takes place and crystal
becomes useless
...
I – V characteristic of solar cell :
(ii) Infinite
1
...
Phase difference between input and output waveform is or 108°
...
Positive feedback
...
Direction of diffusion current is from P to N in a semiconductor junction
diode
...
Light emitting diode
...
GaAs, GaP
...
Constant power supply
...
The energy gap decreases
...
Conductor – no energy gap
Semi conductor – < 3 eV
Insulator – > 3 eV
...
ne/nh = 1 ne = nn intrinsic semiconductor
...
1
...
2
...
Because there is no free charge carrier in depletion region
...
On heating S, resistance of semiconductors S is decreased so to
compensate the value of resistance in the circuit R is increased
...
In this case diode is sensitive and it gives very large amount of current in
this situation
...
A
B
Y
0
0
1
0
1
1
1
0
1
1
1
1
17
...
(b) Zero
18
...
19
...
Zever diode
(ii) saturation
I (MA)
VZ
I (MA)
(i) Reverse Bias
[Class XII : Physics]
(ii) Iosmard Bias
192
22
...
24
...
C
...
R
...
pg
...
N
...
E
...
T
...
477
26
...
2V
Si ~ 0
...
27
...
28
...
(b)
30
...
(i)
V = Ed = 7 × 105 × 300 × 10–9 = 0
...
Kinetic energy = eV = 0
...
11 mA
90
193
[Class XII : Physics]
Base current
4
...
11 mA
E1 – E 2 – Vd
R
5
...
7
2
...
95 mA
V0 = E – Vsi – VGe = 12 – 0
...
1 = 12 – 1
...
2 V
Id
V0
R
[Class XII : Physics]
10
...
6 10
3
1
...
194
SOLVED QUESTION PAPER : 2014
1
...
Write its S
...
unit
...
Velocity per unit electric field with the charges is called as Mobility or
v
d S
...
unit = N1 C m s–1 or m2 V–1 s–1
[1/2]
E
2
...
Determine modulation
index
...
AM= 1, Ac = 2,
t(s)
[1/2]
Am
1/ 2 0
...
“For any charge configuration, equipotential surface through a point is
normal to the electric field
...
Ans
...
This field may move the charge along
the surface without any energy supply violating the law of conservation of
energy
...
Two spherical bobs, one metallic and the other of glass, of the same size
are allowed to fall freely from the same height above the ground
...
Due to motion, in metallic bob, eddy current will be produced which will
oppose the motion and slows the metallic bob
...
So, the glass bob will reach the ground earlier
...
Show variation of resistivity of Copper as a function of temperature in a
graph
...
6
...
T
A convex lens is placed in contact with a plane mirror
...
What is the focal length of the lens?
Since image coincides with object, it implies that ray must be falling normally
on the plane mirror
...
So, object must be at the focus of lens
...
7
...
Write the expression, in a vector form, for the Lorentz magnetic force F
due to a charge moving with velocity v in a magnetic field B
...
[1/2]
8
...
Identify the element labelled ‘X’ and write its function
...
Transmitter
User
Receiver
Message
Signal
Message
Signal
X Channel
Channel : Medium through which signal travels
...
[1/2]
9
...
Out of the two magnetic materials, ‘A’ has relative permeability slightly
greater than unity while ‘B’ has less than unity
...
Material A is paramagnetic
...
Susceptibility of A is positive and susceptibility of B is negative
...
Ans
...
What would be the flux through the same square, if the plane makes an
angle of 30° with the x-axis?
When plane of square in parallel to YZ-plane, flux of electric field
= E
...
10x 10x 10–4 50
Vm
[1/2]
[1/2]
1 50V m
When plane make 30° with x-axis
2 EA cos 5x 103
...
For a single slit of width “d”, the first minimum of the interference pattern
of a monochromatic light of wavelength occurs at an angle of
...
Explain
...
The condition for minima is single slit is
a sin = n
...
For
every secondary source in upper slit cancels the corresponding contribution
of secondary source in second slit which differ’s in path length by
...
12
...
Name the
gates used
...
S
Y
Y'
R is OR Gate
[1/2+1/2+1]
and S in AND Gate Truth table :
A
B
Y = A + B
Y = A(A+B)
0
0
0
0
0
1
1
0
1
0
1
1
1
1
1
1
[Class XII : Physics]
198
12
...
Write the
truth table for the combination
...
P in NAND Gate
Q is OR Gate
[1/2+1/2+1]
Truth Table :
A
B AB
0
1
1
0
1
1
1
1
0
1
1
1
Ans
...
B
1
0
1
State Kirchhoff s rules
...
Kirehoff s Current (Junction) Rule : At any junction, the sum of the
currents entering the junction is equal to the sum of current leaving the
junction
...
This is based on the conservation of
charge
...
This law is based on the conservation of energy
...
[1]
A capacitor ‘C’ a variable resistor ‘R’ and a bulb ‘B’ are connected in series
to the ac main; in circuit as shown
...
How will the glow of the bulb change if (i) a dielectric slab is introduced
between the plates of the capacitor, keeping resistance R to be the same;
(ii) the resistance R is increased keeping the same capacitance ?
199
[Class XII : Physics]
C
R
B
Mains
Ans
...
This increases the current
flowing in bulb and so the bulb glows brighter with eresistance unaltered
...
Glow
of bulb decreases
...
State the underlying principle of a cyclotron
...
Ans
...
Principle : A charged particle can be accelerated to very high energies
with the help of smaller values of oscillating electric field
...
[1]
Working
1
...
2
...
3
...
4
...
It
now gets accelerated towards dee D2
...
It now moves faster through D2 describing a larger semicircle than before
...
Thus, the +ve ion will go on accelerating every time it comes into the gap
between the dees and will go on describing circular path of greater and
greater radius with greater speed and finally acquire a sufficiently high
energy
...
The accelerated ion is ejected through a window by a deflecting voltage
and hits the target
...
An electric dipole of length 4 cm, when placed with its axis making an
angle of 60° with a uniform electric field, experiences a torque of 4 3 Nm
...
Ans
...
A proton and a deuteron are accelerated through the same potential
...
Give reasons to justify your
answer
...
V = Accelerating potential
(i)
Debroglie Wavelength
since
[4*1/2]
h
h
mv p
p 2 m eV
d
h
2 2m eV
h
4m eV
The de Broglie wavelength for proton is more than debroglie wavelength
of duetron
...
So momentum Pp 2meV
Kinetic energy of deutron Ka = 2eV
...
18
...
0 x 1014 Hz is produced by a laser
...
0 x 10–3 W
...
(ii)
Draw a plot showing the variation of photoelectric current versus the intensity
of incident radiation on a given photosensitive surface
...
(i) We know energy of photon = hv
= 6
...
63 x 10–20
[1+1]
So, No
...
03 1015
Energy of one photon 6
...
89
Photocurrent
(ii)
Intensity
19
...
5 eV beam of electrons is used to bombard gaseous hydrogen at
room temperature
...
Ans
...
5 eV
Energy in first shell = –13
...
6
n2
202
[1+1+1]
13
...
5
n2
13
...
6
1
...
36
n = 4 (Maximum 4th level)
(ii)
For Layman series,
1
1 1 3R
R
4
1 4
For Balmer series, n1 = 2, n2 = 3
1
1
36
1
R 2 2
5R
3
2
20
...
She tried hard to convince her parents that this move would be
a health hazard
...
m
...
(Given radius of the earth = 6400 km)
Ans
...
All
through the year it will be an action point where link will be made with
mobiles
...
Also, the signals wavering around may affect all human life in the
neighbourhood
...
203
[Class XII : Physics]
(ii)
(iii)
She has foresight of the problems that they may face at large
...
She is concerned about the welfare of the community
...
Ans
...
0 m has a resistance of 10
...
It is connected
to a 6 V battery in series with a resistance of SS1
...
At balancing point,
5V
5
A
B
G
Current flowing in the primary circuit is i
6
6
2
A
10 5 15 5
Potential difference across wire AB
V
2
10 4 volt
5
Potential gradient =
V 4
4V / m
L 1
At balance point
V
40
1 4
1
...
6 Volt
22
...
[Class XII : Physics]
204
(b)
Ans
...
The
magnification produced by the eye piece is 5
...
The distance between the objective and eyepiece is
observed to be 14 cm
...
(a) Compound Microscope :
[2+1]
Eyepiece
Objective
A
(b)
Magnification produced by eye-piece
me
ue
D
5
ue
20
4 cm
5
For eye-piece
1
1
1
v e ue fe
1
1
1
20 4 fe
1 5 1
20
fe
fe
Since
20
cm 5 cm
4
m m 0 me
v0 D v0
5
f0 ue f0
205
[Class XII : Physics]
20
v0 D v0
5
f0 ue f0
v 0 14 4 10 cm
f0
v0
10
5
5 2
...
(a) A mobile phone lies along the principal axis of a concave mirror
...
Explain why
magnification is not uniform
...
What effect this will have on the image of the
object ? Explain
...
(a)Image formation of a mobile as an object is shown
...
The magnification is not uniform along the length and height as, based on
the position of the portion of the object, magnification is formed
...
u
B
P
D
A'
F A
OD = OC
A'B' > AB
C
B'
(b)
We know intensity of image is directly proportional to area of the reflecting
surface
...
So,
intensity of image decreases
...
24
...
(b)
The electric field inside a parallel plate capacitor is E
...
[Class XII : Physics]
206
a
d
Ans
...
The small work needed to transfer a charge dq from one plate to the other:
dW Vdq
q
dq
C
q
Q2
dq
2C
0C
Q
The total work required: W
[2+1]
The energy can be considered to be stored in the electric field between
the plates
...
e
...
OR
24
...
(b)
Two charged spherical conductors of radii R, and R, when connected by
a conducting wire acquire charges q, and q
...
Find the ratio of
their surface charge densities in terms of their radii
...
(a)Let A = surface area of each plate
...
C
(b)
A
Q A
making C 0
d
V
d
0
Since, they are connected, their potentials will be equal with the charges
...
(a) State Ampere’s circuital law, expressing it in the integral form
...
A steady current “I” flow through
the inner solenoid S, to the other end B, which is connected to the outer
solenoid S, through which the same current “I” flows in the opposite direction
so as to come out at end A
...
Ans
...
dl around a closed imaginary loop is µ0
times current encircled by the loop
...
dl 0I
209
[Class XII : Physics]
B
l
A
In
1
(b)(i)
n
2
S1
S2
Magnetic field at any point inside the long solenoid is 0nI
Magnetic field due to bigger solenoid = 0n2I towards point ‘B’
Magnetic field due to smaller solenoid = 0n1I towards point ‘A’
[1]
So, net magnetic field = µ0(n1 – n2)I towards point ‘A’
(ii)
Net field due to long solenoid outside is zero, as the fields due to two
diametrically opposite elementary lengths will be zero
...
Answer the following :
(a)
Name the em waves which are suitable for radar systems used in aircraft
navigation
...
(b)
If the earth did not have atmosphere, would its average surface temperature
be higher or lower than what it is now? Explain
...
Justify
Ans
...
0 x 1011 Hz to 1 × 109 Hz
[1+1+1]
(b)
In the absence of atmosphere, the surface temperature will be less as the
higher wavelength rays reflected from the surface will not be trapped
...
So the
C
momentum transfer of smaller magnitude will exert a force and pressure
on any surface
...
(a) Deduce the expression, N N 0e l , for the law of radioactive decay
...
(b)
(i) Write symbolically the process expressing the decay of
write the basic nuclear process underlying this decay
...
Also
an isotope or
Ans
...
N = Total number of atoms left undecayed in the sample at time t
dN = A small number of atoms that disintegrate in a small interval of time
dt
Rate of disintegration of the element
R
dN
dt
[Minus sign indicates that the number
of atoms left undecayed decreases with time]
According to radioactive decay law,
R
dN
N
dt
dN
N
dt
= Disintegration constant
dN
dt
N
Integrating both side
dN
dt
N
logc N t C
C = Constant of integration
t = 0
N = N0
loge N 0 0 C
211
[Class XII : Physics]
C loge N 0
loge N t loge N0
loge N loge N 0 t
log
N
t
N0
N
t
i
...
, N e
0
N N 0e t
(b)(i)
The basic nuclear process underlying this + decay
...
28
...
Give reason
...
If the displacements due
to these waves is given by y1 = a cos t and y2 = a cos(t + ) where
is the phase difference between the two, obtain the expression for the
resultant intensity at the point
...
Find out the intensity of light at a point where path difference
is /3
...
(a)(i) In independent monochromatic sources phase difference changes at a
rate of 108 Hz
...
(ii) The phase difference between two waves arising from slits A and B is
...
Therefore, the resultant displacement will be en by
y y 1 y 2 a cos t a cos t
y 2a cos
...
When the path difference
2
, phase difference
...
I
...
(a) How does one demonstrate, using a suitable diagram, that unpolarised
light when passed through a Polaroid gets polarised ?
(b)
A beam of unpolarised light is incident on a glass-air interface
...
213
[Class XII : Physics]
Ans
...
If an electric field vector which is
perpendicular the pass axis, falls on the polariser then, it gets absorbed
...
Since, the perpendicular component gets
absorbed, the output light obtained is a polarised light whose electric field
vector is parallel to the pass axis
...
If the unpolarised light is incident at
the angles 0° or 90°, the reflected ray remains unpolarised
...
The angle of incidence in this case is called polarising
angle or Brewster’s angle (ip)
...
From the diagram,
ip + 90° + r = 180°
I p r 90 or r 90 i p
[Class XII : Physics]
214
sin i p
sinr
sini p
sin 90 i p
sini p
cosi p
tan i p
29
...
(b)
The current flowing through an inductor of self inductance L is continuously
increasing
...
dI
dt
Ans
...
To oppose the rise in magnetic flux, a
north pole is developed in the coil, this causes as induced current in the
anti-clockwise direction as seen from the side of the magnet
...
(a) Draw a schematic sketch of an ac generator describing its basic elements
...
Show a plot of variation of
(i)
(ii)
(b)
Magnetic flux and
Alternating emf versus time generated by a loop of wire rotating in
a magnetic field
...
(a)
[2+1+1]
Principle: A dynamo or generator is a device which converts mechanical
energy into electrical energy
...
Magnetic flux changes as the coils orientation varies with the
rotation BA cos BA cos t
Construction: It consists of four main parts–
[Class XII : Physics]
216
•
Field magnet: It produces the magnetic field
...
Armature coil
B
N
S
A
R1
D
B1
Field
magnet
R B2
R1 2
Carbon Brushes
Load resistance
•
Armature: It consists of a large number of turns of insulated copper wire
on a soft iron core
...
The soft iron core provides support to the coils and
increases the magnetic field through the coil
...
These rings are fixed to the shaft
which rotates the armature coil so that the rings also rotate along with the
armature
...
The output current
in external load resistance RL is taken through these brushes
...
This allows the low frequency to pass through
...
(a) State briefly the processes involved in the formation of p-n junction
explaining clearly how the depletion region is formed
...
How are these characteristics made use of in rectification ? [2+1+1+1]
Ans
...
There is concentration gradient between p
and n sides, holes diffuse from p side to n side (p n) and electrons
move from (n p) creating a layer of positive and negative charges on
n and p side respectively called depletion layer
...
V
W
p
n
Forward biased
p-n junction
(b)
p – n junction under forward bias: When p - side is connected to
positive terminal and n - side to negative terminal of external voltage, it is
said to be forward biased
...
There is minority carrier injection, hence charges
begin to flow
...
[Class XII : Physics]
218
W
metallic
contact
metallic
contact
depletion
layer
p-n junction
p – n junction under reverse bias: The direction of applied voltage is
same as direction of barrier potential, so barrier height increases to
( 0 + V)
...
Diffusion current decreases but drift of electrons and holes under
the electric field affect remains
...
p
n
W
Diode under reverse bias
Study of V – I characteristics of a diode: The circuit to study the variation
of current as a function of applied voltage is shown
...
In forward bias
we use milliammeter and reverse bias we use microammeter
...
2 0
...
6 0
...
0 V (V)
I(A)
Half wave rectifier: Junction diode allows current to pass through only if
it is forward biased, hence a pulsating voltage will appear across the load
only during positive half cycles when diode is F
...
A
P1
X
S1
Primary
R1
Secondary
Input
a
...
S2
P2
Y
B
transformer
Voltage across
Load RL
Voltage at A
(a) Diode rectification Circuit
Input a
...
time
output voltage
time
(b) Input ac and output
voltage wveforms
OR
30
...
[Class XII : Physics]
220
(b)
How is a transistor biased to be in active state?
(c)
With the help of necessary circuit diagram, describe briefly how n-p-n
transistor in CE configuration amplifies a small sinusoidal input voltage
...
Ans
...
Collector base junction is reverse biased
...
Output, VO = VCC – Ic RC
...
V1 and VO are out of
phase
...
(2)
Since VCC and RC are constant,
VO = 0 – RC IC (from (1), differentiate)
C
IC
C
RB
IB
Vi
E
II
VBB
RC
V0
VCC
Input Vi = IB RB + VBE
221
[Class XII : Physics]
Vi = RB IB + VBE
...
Rb Ib
Rb
Vce
Voltage gain, Av R I
b b
[Class XII : Physics]
222
SOLVED SAMPLE PAPER - 1
Candidate must write the Code on the title page of the answer-book
·
Please check that this question paper contains 26 questions
·
Code number given on the right hand side of the question paper should be written
on the title page of the answer -book by the candidate
...
·
15 minutes time has been allotted to read this question paper
...
15 a
...
From 10
...
m
...
30 a
...
, the students will read the
question paper only and will not write any answer on the answer script during this
period
...
What is the geometrical shape of equipotential surfaces due to a single
isolated charge?
Ans
...
(This answer can also be expressed using diagram)
...
Which of the following wave can be polarized (i) Heat waves (ii) Sound
waves ? Give reason to support your answer
...
Heat waves can be polarized because heat waves are transverse waves
whereas sound cannot be polarized as sound waves are longitudinal waves
...
Q3
...
Identify the pairs of curves
that correspond to different materials but some intensity of incident radiation
...
Curve 1 and 2 correspond to similar materials while curves 3 and 4
represent similar but different materials, since the value of stoppins potential
for the pair of curves (1 and 2) & (3 and 4) are the same
...
Q4
...
Total energy, E(ev)
En
13
...
Q
...
What are the
magnitude of conduction and displacement current, when it is fully
charged?
(ii)
Ans
...
Find the value of the current in circuit
...
Since
magnitude of conduction current is same as magnitude of displacement
current so both are zero
...
Hence, the current in the circuit is given by
I
Q7
...
R
38
State Lenz’s Law
...
Will there be an emf induced at its ends? Justify your answer
...
Lenz’s law states that the polarity of induced emf is such that it tends to
produce a current which opposes the change in magnetic flux that
produces it
...
As
metallic rod held horizontally along eastwest direction, is allowed to fall
freely under gravity then it intersect the horizontal component of earth’s
magnetic field which is along south-north direction
...
(If it is dropped exactly at pole there will be no induced emf because there
is no horizontal component of magnetic field)
...
(a) Write the necessary conditions for the phenomenon of total internal reflection
to occur
...
Ans
...
The angle of incidence should be greater than the critical angle for
the given pair of optical media
...
a and b are the rarer and denser media respectively
...
* b
Q9
...
Ans
...
For an electron moving with a
uniform speed in a circular orbit in a given radius, centrip etal force is
provided by Columb force of attraction between the electron and the nucleus
...
So,
mv 3 ko 2
2
r
r
kv 2
r
or
mv 2
where
m = mass of electron
...
Again
mvr
[Class XII : Physics]
nh
2
226
or
v
nh
2mr
From eq
...
k e e
ke 2
r
r
Explain, with the help of a circuit diagram, the working of a photo-diode
...
OR
Mention the important considerations required while fabricating a p-n junction
diode to be used as a Light Emitting Diode (LED)
...
A junction diode made from light sensitive semi-conductor is called a
photodiode
...
It is a pn-junction
whose furiction is controlled by the light allowed to fall on it
...
When such light falls on the junction, new hole-electron
pairs are created
...
If the junction is connected in some
circuit, the current in the circuit is controlled by the intensity of the incident
light
...
The reverse breakdown voltages of LEDs are very low, typically around 5V
...
2
...
Therefore, a
resistor must be used in series with the LED to avoid any damage to it
...
8 eV (spectral range of visible light is from about 0
...
7 fYm, i
...
, from about 3 eV to 1
...
Q11
...
The charge stored in it is 360 µC
...
[Class XII : Physics]
228
Calculate :
(i)
The potential V and the unknown capacitance C
(ii)
What will be the charge stored in the capacitor, if the voltage
applied had increased by 120 V?
OR
A hollow cylindrical box of length 1 m and area of cross-section 25 cmz
is placed in a three dimentional coordinate system as shown in the figure
...
Find
(i)
(ii)
Ans
...
Charge enclosed by the cylinder
...
Q1 = CV1
...
(2)
By dividing (1) by (2), we get
Q1 CV1
360
V
V 180 volts
Q2 CV2
120 V 120
C
(ii)
Q1 360 10 3
2 10 –2 F 2F
V1
180
If the voltage applied had increased by 120 V, then V3 =180 + 120 = 300
V
...
Magnitude of electric field at cross-section
A, EA = 50 × 1 = 5 N C–1
Magnitude of electric field at cross-section
B, ED 50 × 2 = 100 N C–1
The corresponding electric fluxes are:
E s 50 25 10 4 cos180 2 0
...
Using Gauss’s law:
A metallic rod of length ‘I’ is rotated with a frequency v with one end hinged
at the centre and the other end at the circumference of a circular metallic
ring of radius r, about an axis passing through the centre and perpendicular
to the plane of the ring
...
Using Lorentz force, explain how emf is induced
between the centre and the metallic ring and hence obtained the expression
for it
...
Suppose the length of the rod is greater than the radius of the
circle and rod rotates anticlockwise and suppose the direction of electrons
in the rod at any instant be along +y-direction
...
Then,
using Lorentz law, we get the following:
F a v B
F e v Bk k
j
F avBi
Thus, the direction of force on the electrons is along -x axis
...
e
...
This movement of electrons will result in current and hence it will
produce emf in the rod between the fixed end and the point touching the
ring
...
Then, area swept by the rod inside the circle
1 2
r
2
Induced emf = B
Q13
...
Write the condition under which the phenomenon of resonance
231
[Class XII : Physics]
occurs
...
Figure shows the variation of im with in a LCR series series circuit for
two values of Resistance R1 and R2 (R1 > R2)
...
wr
R2
R1
w
The condition for resonance in the LCR circuit is 0
1
LC
We see that the current amplitude is maximum at the resonant frequency
...
Quality factor or simply the Q-factor of a resonant LCR circuit is defined
as the ratio of volt-age drop across the capacitor (or inductor) to that of
applied voltage
...
Less sharp the resonance, less is the selectivity of the circuit while
higher is the Q, shai-per is the resonance curve and lesser will be the loss
in energy of the circuit
...
(a)An ammeter of resistance 0
...
0 A
...
0 A?
(ii)
What is the combined resistance of the ammeter and the shunt? What are
permanent magnets? Give one example?
[Class XII : Physics]
232
Ans
...
80 ohm and nuuciunmicurrent
across ammeter,
I A 1
...
0 0
...
8V
...
(i)
Resistance of ammeter with shunt,
R
RA x
0
...
8 X
Current through ammeter, I = 5 A
...
8x
5 0
...
8 x
0
...
8 x 4x
x
0
...
2
3
...
2 ohm
(iii)
Combined resistance of the ammeter and the shunt,
R
Ans
...
8x
0
...
2 0
...
16 ohm
0
...
8 0
...
For example : Steel, earth, Uarnaagnet etc
...
(a)In what way is diffraction from each slit related to the interference pattern
in a double slit experiment
...
The
distance between the slit and the screen is 1
...
Calculate the separation
between the positions of the first maxima of the diffraction pattern obtained
in the two cases P
...
(a) If the width of each slit is con-yarable to the wavelength of light used the
interference pattern thus obtained ui the double-slit experiment is modified
by diffraction pattern due to each slit
...
9 × 10–7 m
Wavelength of another light beam, 2 596nm 5
...
5 m
aperture = a = 2 x 10–4 m
For the first secondary maxima,
sin
or
x1
3 2 x 1
2a
D
31D
3 D
and x 2 2
2a
2a
Spacing between the positions of first sec-ondary maxima of two ocliuni
lines
x1 x 2
3D
1
2a 2
= 6
...
(i) Write the relationship between angle of incidence ‘i’, angle or prism ‘A’ and
angle of minimum deviations for a triangular prism
...
Draw a sketch
showing the propagation of the em wave, indicating the direction of the
oscillating electric and magnetic fields
...
The relation between the angle of incidence I, angle of prism, A and the
angle of minimum deviation, m for a triangular prism is given as is given
A m
...
(a) Given that velocity, V v i so electric and magnetic field should be in
X-Y plane
...
The propagation of EM wave is following:
Y
E
X
Z B
(b)
Speed of EM wave can be given as the ratio of magnitude of electric field
E0 to the magnitude of magnetic field B0
c
E0
B0
Q
...
Determine the power of the combination
...
An air bubble is formed inside water
...
25m
and focal length of concave
lens, f2 = – 20 cm = – 0
...
25
1
1
1
Power of concave lens, P2 f 0
...
25 0
...
Air bubble behave as concave lens
...
(a)In a typical nuclear reaction e
...
2
1H
2
3
1 H 2 He n 3
...
How?
Explain
...
Ans
...
So from the
0
bombarding particle
law of conservation of mass-energy, some energy (3
...
This energy is called Q-Value of the
nuclear reaction
...
66 × 10–27 kg
volume of nucleus
4
4
4
3
= nR 3 R0 A1/3 R0 A
...
66 10 2 1
...
[Class XII : Physics]
236
Using R0 1
...
97 1017 kg m 3
Q19
...
(b)
Write the basic features of photon picture of electromagnetic radiation on
which Einstein’s photoelectric equation is based
...
(a) Wave nature of radiation cannot explain the following:
(i)
(ii)
The existence of threshold frequency for a metal surface
...
The fact that kinetic energy of the emitted electrons is independent
of the intensity of light and depends upon its frequency
...
Photon picture of electromagnetic radiation on which Einstein’s photoelectric
equation is based on particle nature of light
...
However, number of photons
may not be conserved
...
Write three important factors which justify the need of modulating a message
signal
...
Ans
...
(ii)
Effective power which is radiated by antenna: Since the power radiated
from a linear antenna is inversely proportional to the square of the
transmitting wavelength
...
(iii)
The interference of signals from different transmitters: To avoid the
237
[Class XII : Physics]
interference of the signals there is need of high frequency which can be
achieved by the modulation
...
Amplitude modulated wave
Output characteristices of an n-p-n transistor in CE configuration is shown
in the figure
...
(i) Dynamic output resistance is given as:
V
R1 CI
Ic
(Note- The values 3
...
4 are arbitrary
...
e
...
c current gain,
dc
(iii)
IC 3
...
5 10 3 350
116
...
c
...
7 3
...
2 10 3
120
40 30 A
IB
10 10 6
238
in this case approximate values of Ic is used according to graph
...
Q22
...
Ans
...
Q23
...
Pathak was caught up
in a thunderstorem
...
He stoppd driving the car and
waited for thunderstorm to stop
...
He asked the boy to come inside the car till the thunderstorm
stopped
...
Pathak dropped the boy at his residence
...
Pathak should meet his parents
...
Pathak for his concern for safety of the child
...
Pathak in his action?
(c)
Which values are reflected in parents response to Dr
...
Ans
...
Electric field inside a hollow sphere is zero so it is safe
to sit in the closed car
...
Q24
...
With the help of the circuit
diagram, explain how a potentiometer is used to compare the emf’s of two
primary cells
...
(b)
Write two possible causes for one sided deflection in a potentiometer
experiment
...
Using Kirchhoff’s rules,
obtain the balance condition in terms of the resistances of four arms of
Wheatstone bridge
...
R1
R2
If a resistance of 10 is connected in series with R1, null point is, obtained
at AD = 60 cm
...
Ans
...
(ii)
The positive ends of all cells are not connected to the same end of the
wire
...
25(a)Derive the expression for the torque on a rectangular current carrying loop
suspended in a uniform magnetic field
...
Depict
their trajectories in the field
...
The moment
of inertia of the needle about the axis is ‘I’
...
Prove that it executes simple
harmonic motion
...
(b)
A compass needle, free to turn in vertical plance orient itself with its axis
vertical at a certain place on the earth
...
Ans
...
Bqv
mv 3
mv
r
r
Bq
241
[Class XII : Physics]
Let mp = mass of proton, md = mass of deutron, vp = velocity of proton and
vd = velocity of deuteron
...
Given that m pv p md v d
rp
rd
mpv p
(1)
Bq
mpv d
(2)
Bq
As (1) and (2) are equal, so rp rd r
Thus, the trajectory of both the particles will be same
...
Therefore, in equilibrium I
d 2
dt 2
mB sin
...
For small value of in radians, we approximate sin
= and get
I
[Class XII : Physics]
d 2
dt 2
mB
242
d 2
Or
dt 2
mB
I
This reprsents a simple harmonic motion
...
(a)Draw a ray diagram showing the image formation by a compound
microscoped
...
(b)
What two main changes in diffraction pattern of single slit will you observe
when the monochromatic source of light is replaced by a source of white
light
...
Using theis principle draw a diagram to show how
a plane wave front incident at the interface of the two media gets refracted
when it propagates from a rarer to a densesr medium
...
(b)
The monochromatic light travels from a rarer to a denser medium, explain
the following giving reasons :
(i)
Is the frequency of reflected and refracted light same as the
frequency of incident light?
(ii)
Does the decrease in speed imply a reduction in the energy carried
by light wave?
243
[Class XII : Physics]
Ans
...
(ii)
Since the frequency remains same, hence there is no reduction in energy
...
Give one example each of a ‘system’ that uses the following mode of
propagation
...
A concave mirror, of aperture 4 cm, has a point object placed on its
principal axis at a distance of 10 cm from the mirror
...
State the likely reason for
the same
...
Show, on a graph, the nature of variation, of the (associated) de-Broglie
wavelength (X0, with the accelerating potential (V), for an electron initially
at rest
...
Define the term ‘Transducer’ for a communication system
...
The short wavelength limits of the Lyman, Paschen and Balmer Series, in
t
he hydr
ogen spect um , ar denot
r
e
ed by L, P, B respectively
...
Q6
...
OR
A capacitor is charged with a battery and then its plate separation is
increased without disconnecting the battery
...
Draw the current versus potential difference characteristics for a cell
...
Show that a series LCR circuit connected to an a
...
source exhibits
resonance at its angular frequency equal to 1/ LC
...
An observer can see through a pin-hole the top end of a thin rod of height
h, placed as shown in the figure
...
When the beaker is filled with a liquid up to a height 2h, he can see the
lower end of the rod
...
Q10
...
to
...
The galvanometer, in each of the two given circuits, does not show any
defecton
...
Q12
...
Compare the capacitance of this combination with the
[Class XII : Physics]
246
capacitor C3, again having plates of area A each, but ‘made up’ as shown
in the figure
...
d
C2
2d
C3
A find pencil of -particles, moving with a speed v, enters a region (region
I), where a uniform electric and a uniform magnetic field are both present
...
The path of the
-particles, in the two regions, is as shown in the figure
...
(iii)
Drive the expression for the radius of the circular path of the (3particle in region II
...
S
II
q
I
Q14
...
c
...
Define the terms ‘magnetic dip’ and ‘magnetic declination’ with the help of
relevant diagrams
...
Three identical polaroid sheets Pl and P2 and P3
are oriented so that the (pass) axis of P2 and P3 are inclined at angles of
60° and 900, respectively, with respect to the (pass) axis of Pl
...
Find the intensity of this light, as observed by observers 01, O,
and 03 positioned as shown below
...
(i) The following data was recorded for values of object distance and the
corresponding values of image distance in the experiment on study of real
image formation by a convex lens of power+ 5 D
...
Identify this observation and give
reason for your choice
S
...
1
2
3
4
5
6
Object distance (cm)
25
30
35
45
5U
55
Image distance (cm)
97
61
1 37
35
2
30
(ii)
In following figure, AB is a lens
...
Give reason too
...
Write and explain three need of modulation
...
Q19
...
For a given nuclear reaction the B
...
/ nucleon of the product nucleus/
nuclei is more than that for the original nucleus/nuclei
...
OR
(a)
The number of nuclei, of a given radioactive nucleus, at times t = 0 and
t = T, are No and (NO/n) respectively
...
(b)
Identify the nature of the ‘radioactive radiations’, emitted in each step of
the ‘decay chain’ given below :
A-4
A-4 Y
A-4 W
Z
Q20
...
2 eV
...
5 mm, find the (i) energy in ev, and
(ii) value of the quantum number, nl of the electron in its initial state
...
Derive the relation between distance of object, distance of image and
radius of curvature of a convex spherical surface, when refraction takes
place from a rarer medium of refractive index µ7 to a denser medium of
refractive index µ2 and the image produced is real
...
What is the power of this lens combination in diopters?
Q22
...
What criterion is kept in mind while choosing the
semiconductor material for such a device? Write any two advantages of
Light Emitting Diode over conventional incandescent lamps
...
Q23
...
Three groups were made
...
“Why can moon be
not used as a communication satellite?” Answers were given by all the
three groups
...
Teacher told them that
reason given by each group is correct
...
(i)
(ii)
Give the correct reason for the above question
...
What values were showed by all the three groups?
Explain the term sky wave
...
A galvanometer ‘G’ is connected, as shown in the
figure and the closed circuit has a total resistance ‘R’
...
(a)
With key k open :
(i)
Find the nature of charges developed at the ends of the rod XY
...
Q25
...
Field point
A
B
C
A’
B’
C’
Magnitude of
Electric Field
E
E/8
E/27
E/2
E/16
E/54
(i)
(ii)
(b)
Identify the charge distribution and justify your answer
...
A, B and C
are three points in the field having x and y coordinates (in metre) as shown
in the figure
...
251
[Class XII : Physics]
OR
An electric dipole of dipole moment p is placed in a uniform electric field
E
...
Identify
two pairs of perpendicular vectors in the expression
...
Fig (a) and (b) show the field lines
of a single positive and negative charges respectively
...
(c)
Give the sign of the work done by the field in moving a small positive
charge from Q to P
...
Q26
...
Derive the Lens Maker’s formula for this ‘set-up’
[Class XII : Physics]
252
(ii)
A convex lens is placed over a plane mirror
...
How can this observation be used to find the focal
length of the convex lens? Give appropriate reasons in support of your
answer
...
If SS2 – SS1 = /4
...
(iii)
locate the position of the central fringe
...
·
Please write down the Serial Number of the question before attempting it
...
The question paper will
be distributed at 10
...
m
...
15 a
...
to 10
...
m
...
Maximum Marks,: 70
Time allowed : 3 hours
Q1
...
Q2
...
What is the speed of em waves in the medium
...
A metal emits photoelectrons when red light falls on it
...
Name the physical quantity whose dimensions are same as Planck’s
constant
...
Calculate number of electric field lines originating from one coulomb charge
...
The electric field component in the figure are E x 2x i , E y E z 0
...
[Class XII : Physics]
254
Q7
...
Plot a graph showing the variation of the radius of the
circular path described by it with the increase in its kinetic energy, where,
other factors remains constant
...
Water (refractive index µ) is poured into a concave mirror of radius of
curvature ‘R’ up to a height h as shown in figure
...
In a photoelectric effect experiment, the graph between the stopping potential
V and frequency of the incident radiation on two different metals P and Q
are shown in Fig
...
E
...
OR
Derive mvr
nh
using de Broglie equation
...
What is ground wave? Why short wave communication over long distance
is not possible via ground waves?
Q11
...
A conductor of length L is connected to a dc source
of emf E
...
OR
The graph shows how the current I varies with applied potential difference
V across a 12 V filament lamp (A) and across one metre long nichrome
wire (B)
...
(ii)
when potential difference across them is 4V
...
Q12
...
If the same current passes
in both the cases, find the ratio of the magnetic fields produced at the
centres in the two cases
...
In a series L-R circuit, XL = R and power factor of the circuit is P1
...
Find P1/P2
...
The figure shows, in (a) a sine curved (t) = sin wt and three other
sinusoidal curves A(t), B(t) and C(t) each of the form sin (t – )
...
List any three properties of EM waves
...
Using mirror formula show that virtual image produced by a convex mirror
is always smaller in size and is located between the focus and the pole
...
In the following diagram, find the focal length of lens L2
...
X-rays of wave length ?, fall on a photo sensitive surface emitting electrons
...
h
2mc
The total energy of an electron in the first excited state of the hydrogen
atom is about -3
...
What is
(i)
(ii)
the potential energy of the electron?
(c)
Q20
...
5 eV (ii) -0
...
Determine V0 , Id 1 and Id 2 for the given network
...
I1
Id 1 Id 2 14
...
A message signal of frequency 10 kHz and peak voltage of 10 volts is
used to modulate a carrier of frequency 1 MHz and peak voltage of 20
volts
...
257
[Class XII : Physics]
Q22
...
Explain
action of transistor
...
Rahul and Rohit bought an electric iron
...
Rahul was
keen to start using the new iron with the 2 pin plug
...
Rahul got angry
...
He said that if the metallic body of the iron came in contact with the
live wire at 220 vols, they would get an electric shock
...
The iron would then be safe to use
...
(i)
(ii)
Q24
...
How
would the following (i) energy (ii) charge, (iii) potential be affected if (a)
dielectric slab is introduced with battery disconnected, (b) dielectric slab
is introduced after the battery is connected
...
What would
be the electric potential due to the dipole of dipole moment 3
...
State Biot-Savart law
...
Also compare the magnitudes of the magnetic field of this coil
at its centre and at an axial point for which the value of d is 3a
...
R is resistence of the loop
...
(i) A thin lens, having two surfaces of radii of curvature r1 and r2, made from
a material of refractive index µ2, is kept in a medium of refractive index µ1
...
A pin is now positioned so that
there is no parallax between the pin and its image formed by this lensmirror combination
...
OR
P
s2
O
s1
The figure, drawn here, shows a modified Young’s double slit experimental
set up
...
(i)
state the condition for constructive and destructive interference
(ii)
obtain an expression for the fringe width
...
259
[Class XII : Physics]
Title: Physics Study Material For Class 12th
Description: Best Physics Study Material Issued by Central Board Of Secondary Education (CBSE) For Class 12th................I Hope This Material is useful for you.........Thank you...........& All the Best for your studies....................................................
Description: Best Physics Study Material Issued by Central Board Of Secondary Education (CBSE) For Class 12th................I Hope This Material is useful for you.........Thank you...........& All the Best for your studies....................................................