ll ne - Arihant Book

Transcription

CHAPTER
ONE
Electric Charges
and Fields
TOPIC
1 Electric Charges
Introduction
When a glass rod is rubbed with silk, it acquires a power to attract light bodies such as,
small pieces of paper. The agencies which acquire the attracting power are said to be
electrified or charged.
The electricity produced by friction is called frictional electricity.
If the charge in a body does not move, then the frictional electricity is known as static
electricity. The branch of Physics which deals with static electricity is called electrostatics.
Electric Charge
The property of protons and electrons, which gives rise to electric force between them is
called electric charge. Electric charge is a characteristic that accompanies fundamental
particles, wherever they exist.
According to William Gilbert, charge is something possessed by material objects that
makes it possible for them to exert electrical force and to respond to electrical force.
Electric charge is a scalar quantity. A proton possesses positive charge while an electron
possesses an equal negative charge (where the value of the electric charge, e is 1.6 ´ 10 -19 C) .
Unlike charges attract each other, whereas like charges repel each other. A simple
apparatus used to detect charge on a body is the gold leaf electroscope. There are two kinds
of charges such as positive charge and negative charge.
An object can attain positive charge by loosing electrons while other can attain negative
charge by gaining electrons. Charges with same sign, i.e. like charges repel each other while
charges with opposite sign, i.e. unlike charges attract each other.
Charges always reside on the surface of the charged conducting object. An object can be
charged by different methods like friction, conduction and induction.
Charge is a scalar quantity, it can be added and subtracted as a number. Its SI unit is
coulomb. Dimensional formula for electric charge is [M 0L0 TA].
Chapter Checklist
Electric Charges
Coulomb's Law and
Electrostatic Field
Electric Dipole
Electric Flux
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Conductors and Insulators
Conductors are those substances which can be used to
carry or conduct electric charge/electron from one point
to other. They have electric charges that are comparatively
free to move inside the material. They allow electricity to
pass through them easily, e.g. silver, copper, iron,
aluminium, etc.
Insulators are those substances which cannot
conduct electricity. They are also called dielectrics. They
offer high resistance to the passage of electricity through
them, e.g. glass, rubber, plastic, ebonite, mica, etc.
Difference between Dielectrics
and Conductors
Dielectrics are non-conductors and do not have free
electrons at all, while conductors have free electrons in
their any volume which makes them able to pass the
electricity through them.
Charging by Induction
The process of charging a neutral body by bringing a
charged body nearby it without making contact between
the two bodies is known as charging by induction.
Using the process of charging by induction, a
conductor may be charged permanently. For charging an
uncharged conducting sphere positively by induction, a
negatively charged rod is brought close to it. The near end
of the sphere becomes positively charged, while its far end
becomes negatively charged. Now, the sphere is connected
to the earth, keeping the negatively charged rod in its
position.
The electrons will flow to the ground, while positive
charges will remain held at near end because of attractive
force of negative charges on rod. Now, disconnect the
sphere from the ground. The positive charges still
continues to be held at near end. When the negatively
charged rod is removed, the positive charge will spread
uniformly over the sphere.
Example 1. Attraction by a Comb
A comb run through one’s hair attracts small bits
of paper. What happens, if the hair are wet or if it
is a rainy day?
Sol. If the hair are wet or it is a rainy day, then the friction
between the hair and the comb reduces. The comb does
not get charged and it will not attract small bits of paper.
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Physics Class 12th
Basic Properties of
Electric Charge
As discussed above, we know that there are two types of
charges, namely positive and negative and their effects tend to
cancel each other. Now, we shall discuss some other properties
of the electric charge.
Additive Nature of Electric Charge
Electric charge is additive in nature. It means if a system
consists of two charges q1 and q 2 , then the total charge of the
system will be q1 + q 2 .
In general, if a system consists of n charges q1 , q 2 , q3 , ¼,
qn , then the total charge of the system will be
q1 + q 2 + q3 + ¼+ qn .
In order to calculate the net charge on a system, we have to
just add algebraically, all the charges present in the system.
This is known as the principle of superposition of charge.
If the sizes of charged bodies are very small as compared to
distance between them, they are considered as point charges.
Conservation of Electric Charge
During any process, the net electric charge of an isolated
system remains constant (i.e. conserved). In simple words,
charge can neither be created nor be destroyed.
In any physical process, the charge may get transferred
from one part of the system to another, but the net charge will
always remain the same. It is impossible to create or destroy
net charge carried by an isolated system although charge
carrying particles may be created or destroyed in a process.
Nature creates charged particles, i.e. a neutron turns into an
electron and proton. Thus, the electron and proton have
created equal and opposite charges and total charge is zero,
before and after their creation.
Quantisation of Electric Charge
The charge on any body can be expressed as the integral
multiple of basic unit of charge, i.e. charge on one electron.
This phenomena is called quantisation of electric charge.
It can be written as q = ± ne .
where, n = 1, 2, 3, ××× is any integer, positive or negative
and e is the basic unit of charge.
Charge is said to be quantised because it can have only
discrete values rather than any arbitrary value, i.e. free particle
has no charge at all or a charge of +10 e or - 6 e but not a free
particle with a charge of, say 3.57 e.
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Electric Charges and Fields
Origin of Electric Charge:
Electron Theory of Electrification
Experiment shows that an object becomes charged,
when a very tiny fraction of the mobile electrons with their
negative charge are transferred from one object to another.
This is why we must rub, stroke or make significant contact
between two objects for the objects to become charged.
In ordinary matter, a positive charge is much less mobile
than negative charge. For this reason, an object becomes
positively charged through the removal of negatively charged
electrons rather than through the addition of positively
charged protons.
Example 2. Discrete Nature of Charge
Is a charge of 5. 8 ´ 10 -18 C possible?
Sol. From q = ne Þ n =
q
e
=
5.8 ´ 10 -18
1.6 ´ 10 -19
Worked Out Problem
A paisa coin is made up of Al-Mg alloys and weighs
0.75 g. It has a square shape and its diagonal measures
17 mm. It is electrically neutral and contains equal
amounts of positive and negative charges.
Treating the paisa coins made up of only Al, find the
magnitude of equal number of positive and negative
charges. What conclusion do you draw from this
magnitude?
NCERT Exemplar
Step I Write the given quantities in the question.
Mass of a paisa coin, m = 0.75 g
Atomic mass of aluminium, M = 26.9815 g
Length of the diagonal of square shaped paisa
coin
=17 mm
= 36.25
Avogadro’s number, N A = 6.023 ´ 10 23
As, n is not an integer, this value of charge is not possible.
Difference between Charge and Mass
Step II Calculate the number of aluminium atoms in
one paisa coin.
NA
´m
M
6.023 ´ 10 23
=
´ 0.75g
26.9815 g
n=
The differences between mass and charge is given in the
following table:
Charge
Mass
Mass of a body is a positive quantity.
Electric charge on a
body may be positive,
negative or zero.
Charge carried by a
body does not
depend upon velocity
of the body.
Mass of a body increases with its velocity
m0
as m =
where, c is velocity of
v2
1- 2
c
light in vacuum, m is the mass of the
body moving with velocity v and m 0 is
rest mass of the body.
Charge is quantised.
The quantisation of mass is yet to be
established.
Electric charge is
always conserved.
Mass is not conserved as it can be
changed into energy and vice-versa.
The gravitational force between two
Force between
charges can be either masses is always attractive.
attractive or repulsive,
as charges are unlike
or like charges.
Þ
n = 1.6742 ´ 10 22
Step III Since, atomic number ( Z ) of Al is 13,
therefore, each atom of Al contains 13 protons
and 13 electrons. Now, find out the magnitude
of positive and negative charges present in one
paisa coin.
nZe = 1.6742 ´ 10 22 ´ 13 ´ 1.6 ´ 10 -19 C
= 3.48 ´ 10 4 C
= 34.8 kC
Step VI Now, write the conclusion drawn from this
magnitude of charge.
34.8 kC is a very large amount of charge. This
concludes that ordinary neutral matter
contains an enormously large amount of
positive and negative charges.
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EXAM Practice
Very Short Answer Type Questions
1.
Which is bigger, a coulomb or a charge on an
electron?
Sol. A coulomb of charge is bigger than the charge on an
electron.
2.
An ebonite rod held in hand can be charged by
rubbing with flannel but a copper rod cannot be
charged like this, why?
Sol. Both the human body and the copper rod conduct
electricity. When it is attempted to charge a copper
rod by rubbing, the charge flows from the rod to the
earth through the hand. However, when ebonite rod
is charged by rubbing, the charges so produced stay
on the ebonite rod as it is bad conductor of electricity.
3.
What does q1 + q2 = 0 signify in electrostatics?
Sol. The charges q1 and q2 are equal and opposite.
4.
Can a body has charge 1.5 e, where e is the
electronic charge?
Sol. No, because the physically existing charge is always
an integral multiple of e = 1.6 ´ 10 -19 C.
5.
Consider three charged bodies A, B and C. If A and B
repel each other and A attracts C, what is nature of
the force between B and C?
Sol. It is also attractive.
6.
A metal sphere has a charge of – 6 mC. When 5 ´ 10 12
electrons are removed from the sphere, what would
be net charge on it?
Sol. Here, q1 =- 6 mC and q2 = ne = 5´1012 (1.6 ´10 -19) C
= 8 .0 ´10 -7 C
= 0 .8 ´10 -6 C = 0 .8 mC
Since, electrons are removed from the sphere, q2 is
positive.
Therefore, net charge on the sphere,
q = q1 + q2 = (- 6 .0 + 0 .8) mC =-5 .2 mC
7.
A glass rod when rubbed with silk cloth acquires a
charge 1.6 ´10-13 C. What is the charge on the silk
cloth?
Sol. Silk cloth will also acquire a charge 1.6 ´10 -13 C.
However, it will be negative in nature.
[1 Mark]
8.
What is the basic cause of quantisation of charge?
Sol. The basic cause of quantisation of charge is only the
integral number of electrons which is transferred
from one body to another, i.e. ± ne.
9.
Quarks are the building blocks of nucleons and
possess fractional electronic charge. Does this
discovery violate the principle of quantisation of
charge?
Sol. If the quarks are detected in any experiment with
concrete practical evidence, then the minimum value
1
2
of quantum of charge will be either
e or
e.
3
3
However, the law of quantisation will hold good.
10.
A glass object is charged to +3 nC by rubbing, it with
a silk cloth. In this rubbing process, have protons
been added to the object or have electrons been
removed from it?
Sol. Electrons have been removed from the object.
11.
Two insulated charged copper spheres A and B of
identical size have charges qA and qB respectively.
A third sphere C of the same size but uncharged is
brought in contact with the first and then in contact
with the second and finally removed from both.
What are the new charges on A and B? Foreign 2011
Sol. When sphere C is brought in contact with A :
Charge on sphere C,
q + 0 qA
qC = A
=
2
2
and new charge on sphere A,
q
q¢A = A
2
When sphere C is brought in contact with B :
Charge on sphere C,
q + qB
q¢C = C
2
q A / 2 + qB
=
2
q A + 2 qB
=
4
q + 2 qB
\ New charge on sphere B, q¢B = A
4
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Short Answer Type Questions
[2 Marks]
1.
Ordinary rubber is an insulator. But the special
rubber tyres of aircrafts are made slightly
conducting. Why is this necessary?
Sol. During landing or take off, the tyres of aircrafts get
charged due to the friction between tyres and ground.
In case, the tyres are slightly conducting, the charge
developed on the tyres will not stay on them and it
finds its way to the earth.
[1+1]
7.
A copper slab of mass 2 g contains 2 ´ 1022 atoms.
The charge on the nucleus of each atom is 29 e.
What fraction of the electrons must be removed
from the sphere to give it a charge of +2 mC?
Sol. Total number of electrons in the slab
= 29 ´ 2 ´ 10 22
Number of electrons removed
q
2 ´10 -6
= 1.25 ´ 10 13
= =
e 1.6 ´ 10 -19
Fraction of electrons removed
1.25 ´ 10 13
= 2 .16 ´ 10 -11
=
29 ´ 2 ´ 10 22
2.
Automobile ignition failure occurs in damp weather.
Explain, why?
Sol. The insulating porcelain of the spark plugs
accumulates a film of dirt.
The surface dirt is hygroscopic and picks up moisture
from the air. Therefore, in humid weather, the
insulating porcelain of the plugs becomes
quasi-conductor. This allows an appreciable proportion
of the spark to leak across the surface of the plug
instead of discharging across the gap.
[1+1]
[1]
[1]
8.
Why does a charged glass rod attract a piece of paper?
Sol. Paper is a dielectric, so when a positively charged
glass rod is brought near it, atoms of paper get
polarised, with centre of negative charge of atoms
coming closer to the glass rod.
[1/2]
3.
Can a charged body attract another uncharged
body? Explain.
Sol. Yes, because when a charged body is brought infront
of the uncharged body, opposite kind of induced
charge is produced on the uncharged body. Therefore,
the charged body attracts the uncharged body. [1+1]
Fr
4.
A and B have identical size and same mass.
++
and B becomes B - - . Will A + + and
HOTS A becomes A
-B still have the same mass? Why?
Sol. No, they will not have same mass. B- - has more mass
as, it has gained two electrons, whereas A + + has lost
[1+1]
two electrons.
5.
Can two balls having same kind of charge on them
attract each other? Explain.
Sol. Yes, two balls having same kind of charge also attract
each other. If any one of them has more charge as
compared to the other. Then, due to the induction,
they induce opposite kind of charges on the faces of
each other when they are brought nearer. Therefore,
they behave as oppositely charged balls and hence
they attract each other.
[1+1]
6.
Can ever the whole excess charge of a body be
transferred to the other? If yes, how and if not, why?
All India 2011
Sol. Yes, the whole charge of a body P can be transferred
to a conducting body Q, when P is enclosed by Q and is
connected to it. This is because the charge always
[1+1]
resides on the outer surface of the conductor.
+
+
+ +
– ––
–
Paper
++
++
Fa
Glass rod
[1/2]
Therefore, force of attraction Fa between glass rod
and piece of paper becomes greater than the force of
repulsion Fr between the glass rod and the piece of
paper. This results in attraction of the piece of paper
toward the glass rod.
[1]
9.
In filling the gasoline tank of an aeroplane, the
metal nozzle of hose from the gasoline truck is
always carefully connected to the metal of the
aeroplane by a wire, before the nozzle is inserted in
the tank. Explain, why?
Sol. Since, the aeroplane and the gasoline truck usually
have wheels with rubber tyres, they are insulated
from the ground. Further, the service ramps are
usually made of concrete and are not necessary good
conductors to the earth. Therefore, inspite of
grounding metallic ropes, the aeroplane and the
truck could remain charged.
A spark may jump and ignite the explosive gasoline,
when the metal nozzle is brought near the aeroplane.
The connection of metal of the aeroplane and the
nozzle of the hose with a wire avoids any unbalance
of charge and hence the risk of gasoline.
[1+1]
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For the neutron, the charge on neutron is 0.
Let the number of up quarks are b and the number
of down quarks are 3 - b.
[1]
So, b ´ up quark charge + (3 - b) down quark
charge = 0
æ 2e ö
æ eö
bç ÷ + (3 - b) ç - ÷ = 0
è3ø
è 3ø
16.
It is now believed that protons and neutrons (which
constitute nuclei of ordinary matter) are themselves
built out of more elementary units called quarks.
A proton and a neutron consist of three quarks each.
Two types of quarks, the so called ‘up’ quark (denoted
2
by u) of charge + æç ö÷ e and the ‘down’ quark (denoted
è3ø
1
æ
by d) of charge ç - ö÷ e, together with electrons build
è 3ø
up ordinary matter. (Quarks of other types have also
been found which give rise to different unusual
varieties of matter). Suggest a possible quark
composition of a proton and neutron.
NCERT
Sol. For the protons, the charge on it is +e.
Let the number of up quarks are a, then the number of
down quarks are (3 - a) as the total number of quarks
are 3. So, (a ´ up quark charge + (3 - a) down quark
charge) = + e
æ2 ö
æ eö
a ´ ç e ÷ + (3 - a) ç - ÷ = e
è3 ø
è 3ø
2 ae (3 - a) e
=e
Þ
3
3
Þ
2a - 3 + a = 3
Þ
3a = 6
Þ
a =2
Thus, in the proton, there are two up quarks and one
down quark.
\Possible quark composition for proton = uud
Value Based Questions
1.
Þ
2b - 3 + b = 0
Þ
3b = 3
Þ
b =1
Thus, in neutron, there are one up quark and two
down quarks.
\ Possible quark composition for neutrons = udd.
[1]
17.
A balloon gets negatively charged by rubbing
HOTS ceilings of a wall. Does this mean that the wall is
positively charged? Why does the balloon
eventually fall?
Sol. No, this does not imply that the wall is positively
charged. The balloon induces a charge of opposite
sign in the ceiling of the wall, causing the balloon
and the ceiling to be attracted to each other. The
balloon eventually falls because its charge slowly
diminishes as it leaks to ground. Some of the
charge on the balloon could also be lost due to the
presence of positive ions in the surrounding
atmosphere, which would tend to neutralize the
negative charges on the balloon.
[1+1]
[4 Marks]
Geeta has dry hair. A comb ran through her dry hair attracts small bits of paper. She observes that Neeta with oily hair
combs her hair, the comb could not attract small bits of paper. She consults her teacher for this and gets the answer. She
then goes to the junior classes and shows this phenomenon as Physics experiment to them.
All the juniors feel very happy and tell her that they will also look for such interesting things in nature and try to find
the answers. She succeeds in forming a Science club in her school.
Read the above passage and answer the following questions.
(i) What according to you are the values displayed by Geeta?
(ii) Explain the phenomenon involved.
Sol. (i) The values displayed by Geeta are curiosity, leadership and compassion.
[2]
(ii) Frictional electricity.
Frictional electricity is the electricity produced by rubbing two suitable bodies and transfer of electrons from
one body to other.
[1+1]
Further, if the hair are oily or wet, then the friction between the hair and comb reduces and the comb will not
attract small bits of paper.
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2.
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Physics Class 12th
As it is known, that all matter is made up of atoms/ molecules. Every atom consists of a central core, called the
atomic nucleus, around which negatively charged electrons revolve in circular orbits. Every atom is electrically
neutral containing as many electrons as the number of protons in the nucleus. All the materials are electrically
neutral, they contain charges, but their charges are exactly balanced.
An object contains equal amounts of positive charge and negative charge with such an equality or balance of
charges, the object is said to be electrically neutral or uncharged. To electrify or charge a neutral body, we need to
add or remove one kind of charge. The body which gains electrons becomes negatively charged and the body which
loses electrons becomes positively charged. Further, like charges repel and unlike charges attract.
Read the above passage and answer the following questions.
(i) Every body whether a conductor or an insulator is electrically neutral. Is it true?
(ii) Charging lies in charge imbalance, i.e. excess charge or deficit charge, comment.
(iii) How do you visualize this principle being applied in our daily life?
Sol. (i) Yes, it is true. Every conductor/insulator is electrically neutral, as it contains equal amounts of positive charge
and negative charge.
[1]
(ii) This statement is true. Charging lies in charge imbalance. When a body loses some electrons, it becomes positively
charged because it has excess of protons over electrons. The reverse is also true.
[1]
(iii) Nature/God has created the universe. In original, all bodies are neutral with no forces of attraction/repulsion.
When interests of any two persons clash (i.e. two bodies are rubbed against each other), they become
charged. From the charging, arises the forces of attraction/repulsion, i.e. pulls and pressures of life.
[1]
Nature/God wants us to live in peace without stress and tensions in life. We get charged over petty things in
life and invite all sorts of pulls, pressures and tensions.
[1]
Q.1
Is the mass of an amber rod after charging with fur (a) less than, (b) greater than or (c) same as its mass
before charging? Explain.
Q.2
Predict and explain:
An electrically neutral sphere is given a negative charge.
(i) In principle, does the object’s mass increase, decrease or remain the same after being charged?
(ii) Choose the best appropriate explanation from the following:
(a) To give the sphere a negative charge, we must give it more electrons and this will increase its mass.
(b) A positive charge increases the sphere’s mass, a negative charge decreases its mass.
(c) Charge is conserved therefore the mass of the object will remain the same.
Q.3
Two objects P and Q carry charges -3.00 mC and -0.44 mC. What is the number of electrons transferred from P
to Q, so that they acquire equal charges?
Q.4
Explain the following:
(i) Would life be different, if the electron were positively charged and the proton were negatively charged?
(ii) Does the choice of signs have any bearing on physical and chemical interactions?
(iii) When a person chews a winter green life saver in a dark room, a faint flash of blue light is seen from his mouth. How?
Q.5
A copper sphere of mass 2 g contains nearly 2 ´ 1022 atoms. If charge on the nucleus of each atom is 29 e,
what will be the fraction of electrons that must be removed from the sphere to give it a charge of +2 mC ?
[Ans. 2.16 ´ 10- 11]
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TOPIC
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2 Coulomb’s Law and Electrostatic Field
Coulomb’s Law
The force of interaction (attraction or repulsion)
between two stationary point charges in vacuum is directly
proportional to the product of the charges and inversely
proportional to the square of distance between them.
Mathematically, electrostatic force between two stationary
charges is given by
q2
q1
r
F =
2
Also, r21
\
where, e 0 = 8.85 ´ 10 -12 C 2 N -1m -2 and is called the
permittivity of free space.
q 1q 2
i.e.
F = 9 ´ 10 9
r2
The coulomb force acts along the straight line
connecting the points of location of the charges. It is central
and spherically symmetric.
If q1 = q 2 = 1 C and r = 1m
1´ 1
Then,
F = 9 ´ 10 9 2
(1)
F = 9 ´ 10 9 N
i.e. one coulomb is the charge that when placed at distance
of 1m from another charge of same magnitude in vacuum,
experiences an electric force of repulsion of magnitude
9 ´ 10 9 N. Coulomb is a big unit, in practice we use smaller
units like mC or mC.
Coulomb’s Law in Vector Form
Consider two like charges q1 and q 2 present at points
A and B respectively in vacuum at a distance r apart.
r
r12
q1
[Q F12 is along the direction of unit vector r21 ]
be the unit vector pointing from charge q 2 to q1 .
1 q 1q 2
…(iii)
×
F21 =
r$12
4pe 0 r 2
[Q F21 is along the direction of unit vector r21 ]
= 9 ´ 10 9 Nm 2C -2
F12
Let r12 be the unit vector pointing from charge q1 to q 2 .
1 q 1q 2
…(ii)
F12 =
r$21
×
4pe 0 r 2
k | q 1q 2 |
r
where, k is a proportionality constant.
In SI unit, k is given by
1
k=
4pe 0
A
According to Coulomb’s law, the magnitude of force on
charge q1 due to q 2 (or on charge q 2 due to q1 ) is given by
1 q 1q 2
…(i)
×
| F12 | = | F21 | =
4pe 0 r 2
B
r21
F21
q2
Coulomb force between two charges
$r21 = - r$12
\
Eq. (ii) becomes
- 1 q 1q 2
F12 =
r$12
×
4pe 0 r 2
…(iv)
On comparing Eq. (iii) with Eq. (iv), we get
F12 = - F21
i.e., Coulomb’s law agrees with Newton’s third law.
Comparison of Coulomb’s Law with
Gravitational Law
Both the Coulomb’s and Newton’s laws are inverse
square law.
Gm1m 2
[Newton’s gravitational law]
F =
r2
kq q
[Coulomb’s law]
F = 12 2
r
The electric force is much stronger than the
gravitational force between two electrons,
F E = 10 39 FG
Electric force can be attractive or repulsive. But the
gravitational force is always attractive.
Electric force depends on the nature of the medium
between the charges but gravitational force does not.
Coulomb force and gravitational force, both are central
forces.
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EXAM Practice
Very Short Answer Type Questions
1.
Two equal balls having equal positive charge q
coulombs are suspended by two insulating strings
of equal length. What would be the effect on the
force when a plastic sheet is inserted between the
two?
All India 2014
Sol. From Coulomb’s law, electric force between the two
charged bodies, in a medium,
1 |q1 q2|
F=
4 pe 0 K r 2
where, K = dielectric constant of the medium
For vacuum
K =1
For plastic,
K >1
Therefore, after insertion of plastic sheet, the force
[1]
between the two balls will reduce.
2.
The test charge used to measure electric field at a
point should be vanishingly small. Why?
Sol. In case, test charge is not vanishingly small, it will
produce its own electric field and the measured value
of electric field will be different from the actual value
of an electric field at that point.
[1 Mark]
7.
Force experienced by an electron in an electric field
is F newton. What will be the force experienced by a
proton in the same field? Take mass of proton
1836 times the mass of an electron.
Sol. The proton will experience the same force, F newton,
but in the opposite direction.
8.
Two point charges of + 3 mC each are 100 cm apart.
At what point on the line joining the charges will the
electric field intensity be zero?
Sol. At the centre, since the electric field due to two
charges is equal and opposite at this point.
9.
A proton is placed in a uniform electric field
directed along a positive x-axis. In which direction
will it tend to move?
Delhi 2011C
Sol. Proton will tend to move along the x-axis in the
direction of uniform electric field.
10.
3.
A point charge q is placed at the origin. How does
the electric field due to the charge vary with the
distance r from the origin?
Sol. The electric field varies inversely as the square of the
distance from the point charge.
4.
Why electrostatic field be normal to the surface at
every point of a charged conductor?
Delhi 2012
Sol. For the condition of electrostatics, the electric field
lines must be normal to the surface of the conductor,
otherwise there would be a non-zero component of
electric field along the surface of conductor and
charges could not be at rest.
5.
An electrostatic field line is continuous curve. That
is, a field line cannot have sudden breaks. Why not?
Sol. An electrostatic field line cannot be a discontinuous
curve, i.e. it cannot have breaks.If it has breaks, then
it will indicate absence of electric field at the break
points. But the electric field vanishes only at infinity.
6.
Does the Coulomb force that one charge exerts on
another charge, if other charge is brought nearby?
Sol. Yes, it changes as the distance becomes less.
Two point charges having equal charges separated
by 1 m distance experience a force of 8 N. What will
be the force experienced by them, if they are held in
water, at the same distance?
(Given, K water = 80 )
All India 2011C
Two point charges system is taken from air to water
keeping other variable (e.g. distance, magnitude of
charge) uncharged. So, the only factor which may affect
the interacting force is dielectric constant of medium.
Sol. Force acting between two point charges,
F=
Þ
Fair
1
1
q1 q2
or F µ
Þ
=K
2
Fmedium
4 pe 0 K r
K
8
1
8
N
= 80 Þ Fwater =
=
80 10
Fwater
11.
Why should electrostatic field be zero inside a
conductor?
Delhi 2012
Sol. Electric field lines do not pass through a conductor.
Hence, the interior of the conductor is free from the
influence of the electric field.
12.
Why the electric field lines do not form closed loops?
All India 2014
Sol. Because it is always directed from positive charge to
negative charge.
ll
16
13.
The dimensions of an atom are of the
order of an Angstrom. Thus there must be
large electric fields between the protons
and electrons. Why, then is the
electrostatic field inside a conductor zero?
NCERT Exemplar
Sol. The electric fields find the atoms to neutral
entity. As it is known that, electrostatic
fields are caused by excess charges.
However, there is no excess charge on the
inner surface of an isolated conductor.
Therefore, electrostatic field inside a
conductor is zero.
14. A metallic spherical shell has an inner
radius R1 and outer radius R2 . A charge Q
is placed at the centre of the spherical
cavity. What will be surface charge density
on (i) the inner surface and (ii) the outer
surface?
NCERT Exemplar
An uncharged, metallic ball is suspended
in the region between two vertical metal
plates. If the two plates are charged, one
positively and one negatively, describe the
motion of the ball after it is brought into
contact with one of the plates.
2.
The sum of two point charges is 7 mC. They
repel each other with a force of 1 N when
kept 30 cm apart in free space. Calculate
the value of each charge.
Foreign 2009
+Q
of spherical cavity, as shown in the
figure. Then, charge induced on the
inner surface of a shell is - Q and charge
induced on the outer surface of a shell is
+ Q.
Therefore, surface charge density on
-Q
inner surface of shell is
and on
4 pR12
+Q
outer surface of shell is =
.
4 pR22
R2
–Q
+Q
R1
16.
In a fair weather condition, there is an electic field at the
HOTS surface of the earth, pointing down into the ground. What is
the electric charge on the ground in the situation?
Sol. Electric field lines start from positive charges and end on
negative charges. Thus, if the fair weather field is directed
down into the ground, the ground must possess a negative
charge.
[2 Marks]
Sol. Let one of two charges be x mC. Therefore, other charge be will
be (7 - x)mC.
By Coulomb’s law, F =
1
qq
× 1 22
4 pe 0 r
1 = 9 ´ 10 9 ´
Sol. The two charged plates create a region with
a uniform electric field between them,
directed from the positive toward the
negative plate.
Once the ball is disturbed so as to touch one
plate (say, the negative one), some negative
charge will be transferred to the ball and an
electric force will act on the ball, that will
accelerate it to the positive plate.
Once the ball touches the positive plate, it
will release its negative charge, acquire a
positive charge and accelerate back to the
negative plate. The metallic ball will
continue to move back and forth between
the plates until it has transferred all their
net charge, thereby making both the plates
neutral.
[2]
Physics Class 12th
Sol. When a charge +Q is placed at the centre
Short Answer Type I Questions
1.
ne
(x ´ 10 -6) ( 7 - x) ´ 10 -6
(0.3)2
9 ´ 10 -2 = 9 ´ 10 9 - 12 x (7 - x) Þ
[1]
10 = x (7 - x)
\ x 2 - 7 x + 10 = 0 Þ (x - 2) (x - 5) = 0
x = 2 mC or 5mC
Therefore, charges are 2mC and 5mC.
[1]
3.
A conducting sphere of radius 10 cm has an unknown
charge. If the electric field 20 cm from the centre of the
sphere is 1.5 ´ 103 N/C and points radially inwards, what is
the net charge on the sphere?
NCERT
Sol. Let the value of unknown charge be q.
Electric field at 20 cm away, E = 1 .5 ´ 10 3 N/C
From the formula, electric field
9 ´ 10 9 ´ q
1
q
. 2 Þ 1.5 ´ 103 =
E=
4 pe 0 r
(20 ´ 10 - 2)2
20 cm
q
P
E
1.5 ´ 10 3 ´ 20 ´ 20 ´ 10 -4
= 6 .67 ´ 10 -9 C
9 ´ 10 9
As, the electric field is radially inwards which shows that the
nature of unknown charge q is negative.
[2]
q=
ll
ne
45
Miscellaneous Questions
A set of questions based on the concepts of the chapter mingled with some other concepts of Physics,
to assess student’s approach to Multi-disciplinary Questions
1.
There are great similarities between electric and
gravitational fields. A room can be electrically
shielded, so that there are no electric fields in the
room by surrounding it with a conductor. Can a
room be gravitationally shielded? Why or why not?
æ 1 ö
÷÷ = 9 ´ 10 9 N- m 2 /C2
The value of k = çç
4
pe
0 ø
è
The value of e (charge of an electron) = 1.6 ´ 10 -19 C
The value of G (universal gravitational constant)
= 6.67 ´10 -11 N-m 2 / kg 2
There are two kinds of charge in nature, but only one
kind of mass.
The value of me (mass of electron) = 9.1 ´ 10 -31 kg
The value of mp (mass proton) = 1.67 ´ 10 -27 kg
Sol. No, the electric shielding effect of conductors depends
on the fact that there are two kind of charges : positive
and negative. Therefore, charges can move within the
conductor, so that the combination of positive and
negative charges establishes an electric field which
exactly cancel out the external field within the
conductor and any cavities inside the conductor. There
is only one type of gravitational charge, however,
because there is no negative mass. As a result,
gravitational shielding is not possible.
The value of
=
where, G =gravitational constant
me = mass of an electron, m = mass of a proton
qq
From Coulomb’s law, F = k 1 2 2
r
2
Fr
Fr 2
or k = 2
Þ
k=
q1 q2
q
æ 1 ö [MLT -2 ][L2 ]
÷÷ =
The dimensions of k = çç
[AT] [AT]
è 4 pe 0 ø
= [ML3 T -4 A -2 ]
The dimensions of e (electronic charge) =[ AT ]
The dimensions of G (universal gravitational
[MLT -2 ] [L2 ]
constant) =
= [M -1L3 T -2 ]
[M 2 ]
The dimensions of me or mp (mass of electron or mass
of proton) =[M]
ke 2
[ML3 T -4 A -2 ] [A 2 T 2 ]
The dimensions of
=
G me mp
[M -1L3 T -2 ] [M 2 ]
=[M 0L0 T 0 A 0 ]
Thus, the given ratio is dimensionless.
9 ´ 10 9 ´ (1.6 ´ 10 -19)2
6.67 ´ 10 -11 ´ 9.1 ´ 10 -31 ´ 1.67 ´ 10 -27
= 2.29 ´ 10 39
The ratio signifies that the ratio of electrostatic force
to the gravitational force is 2.29 ´ 10 39 . This means
the electrostatic force between an electron and a
proton is 2.29 ´ 10 39 times the gravitational force
between an electron and a proton.
2.
Check that the ratio ke 2 / G m e m p is dimensionless.
Look up a table of physical constants and
determine the value of this ratio. What does the
ratio signify?
NCERT
2
ke
Sol. In the ratio
, k = 4 pe 0 (constant)
Gme mp
ke 2
G me mp
3.
An oil drop of 12 excess electrons is held stationary
under a constant electric field of 2.55 ´ 104 N/C in
Millikan’s oil drop experiment. The density of the oil
is 1.26 g/cm 3. Estimate the radius of the drop.
NCERT
(g = 9.81m/s 2 ; e = 1.60 ´ 10- 19 C).
Here, oil drop is held stationary under electric field that
means the weight of the drop is balanced by the
electrostatic force applied on it.
Sol. Given, the number of excess electrons, n =12
E = 2.55 ´ 10 4 N/C
r =1.26 g/cm 3
= 1.26 ´ 10 3 kg/m 3
Electronic charge, e = 1.6 ´ 10 -19 C
Electric field,
Density of oil,
g = 9.81m/s 2
E
mg
Let the radius of drop be r.
The electrostatic force on drop = qE = neE
[Q q = ne]
The gravitational force on the drop = mg
[where, m = mass of the drop]
= volume × density × g
[QMass = Volume × Density]
4 3
= pr ´ r ´ g
3
As the drop is held stationary. So, the net force on the
drop is zero.
REVISION MAP
Coulomb’s Law and Electrostatic Field
Electric Charge
The property of protons and
electrons, which gives rise
to electric force between
them is called electric
charge.
Conductors and Insulators
Conductors are those
substances which conduct
the electricity, whereas
insulators are those
substances which cannot
conduct the electricity.
Charging by Induction
The process of charging a
neutral body by bringing a
charged body nearby it
without making contact
between the two bodies is
called charging by induction.
Quantisation of Electric
Charge
The charge on any body can
be expressed as an integral
multiple of basic unit of
charge, i.e. charge on one
electron.
It can be written as
q = ± ne, where n = 1, 2,...
Properties of Electric
Field Lines
ŸElectric lines of forces
start from positive charges
and end at negative
charges.
ŸTwo field lines never
intersect each other.
ŸThese are perpendicular
to the surface of charged
conductor.
ŸThese do not pass through
a conductor.
Coulomb’s Law
The force of interaction
(attraction or repulsion)
between two stationary point
charges in vacuum is directly
proportional to the product of
the charges and inversely
proportional to the square of
distances between them.
kq q
i.e. F = 1 2
r
2
Electric Flux
Electric Dipole
It is a pair of point
charges with equal
magnitude and opposite
in sign separated by a
very small distance.
Dipole Moment
It is the product of the
charge and separation
between the charges.
i.e. p = 2aq
Superposition Principle
Force on any charge due to
number of charges is the
vector sum of all the forces
on that charge due to other
charges, taken one at a time.
Electric Field
Intensity due to
an Electric Dipole
[
At a point on axial
line.
2kpx
E = ———
(x2–l 2)2
Electrostatic Force due to
Continuous Charge
Distribution
q
l
q
s=
A
q
=
r
v
l=
; where l is a linear
charge density.
; where s is a surface
charge density.
; where r is a volume
charge density.
Electric Field
It is the space around the
given charge in which
another charge experience
an electrostatic force of
repulsion or attraction.
Electric Field Intensity
It is the force experienced
per unit positive test charge
placed at that point without
disturbing source charge.
i.e. E = (F/q0)
Electric Field Lines
It is a path traversed by a test
charge around the given
charge.
[
Gauss Theorem
The surface integral of the
electric field intensity over any
closed surface in free space is
1 times the net charge
equal to —
e0
enclosed with in the surface.
q
fE = Šs E×d s = —
e
Application of Gauss
Theorem
ŸField due to an infinitely long
[
[
kp
E = ———
(x 2+l 2)3/2
straight charged wire:
l
E = ———
2pe0r
ŸField due to thin infinite sheet
of charge:
s
E = ———
2e0
[
Electric Field Intensity
at Any Point due to
Short Electric Dipole
Electric field intensity at
any point due to short
electric dipole is
[
Area Vector
The vector associated with area
element of a closed surface is
taken to be in the direction of
outward normal.
0
At a point on
equitorial line.
[
It is defined as the total number
of electric lines of force passing
normally through the surface.
[
P Ö3cos2q+1
|E | = ——————
4pe0r 3
Dipole in a Uniform
Electric Field
Work done on Dipole in a
Uniform Electric Field
Total work done in rotating the
dipole from orientation
q 1 to q2 is
W = PE (cos q1 – cos q2)
[
[
ŸField due to a uniformly
charged thin spherical shell:
sR 2
Outside the shell, E = ——
e0r 2
s
ŸOn the surface of shell,E = —
e0
Inside the shell, E = 0
Torque on an Electric Dipole
in a Uniform Electric Field
It is given by, t = pE sin q
ŸIf q = 0°, then t = 0.
The dipole is in stable
equilibrium.
ŸIf q = 90°, then t = PE.
The torque will be maximum.
ŸIf q = 180°, then t = 0.
The dipole is in unstable
equilibrium.
Self Assessment Sheet
8 A point charge causes an electric flux
Very Short Answer Type
1 The electric field induced in a dielectric when
placed in an external field is 1/10 times the
electric field. Calculate relative permittivity of
the dielectric.
[Ans. 10 times]
2 “Electrostatic forces are much stronger than
the gravitational forces”. Give one example to
justify this statement?
3 Give one difference between the conductors
and insulators?
4 Electric charge is additive in nature. Explain?
5 A metallic solid sphere is placed in a uniform
electric field as shown below.
10 Two identical metallic spherical shells
1
1
2
2
3
3
4
4
(a)
+
+
+
+
+
–
–
–
–
–
of - 3 ´ 10 4 N-m 2 /C to pass through a
spherical Gaussian surface.
(i) Find the value of the point
charge.
[Ans. 2 .6 ´ 10 -7 C]
(ii) If the radius of the Gaussian
surface is doubled, how much
flux would pass through the
surface?
9 Two charged conducting spheres of
radii r1 and r2 connected to each other
by a wire. Find the ratio of electric
fields at the surfaces of the two
spheres.
E
A and B having charges + 4 Q and
- 10Q are kept a certain distance apart.
A third identical uncharged sphere C is
first placed in contact with sphere A
and then with sphere B, then spheres A
and B are brought in contact and then
separated. Find the charge on the
spheres A and B.
11 What is the use of Gaussian surface?
4
4
Also, mention the importance of Gauss
theorem
12 Deduce the expression for the electric
(b)
Which path is followed by electric field lines?
6 Two identical metallic spheres of exactly equal
masses are taken. One is given a positive
charge (q) and the other an equal negative
charge by friction. Is three masses after
charging equal?
field E due to a system of two charges q1
and q 2 with position vectors r1 and r2 at
a point r with respect to a common
origin.
13 S1 and S 2 are two parallel concentric
spheres enclosing charges Q and 2Q as
shown.
Short Answer Type I
S2
7 Two small identical dipoles AB and CD, each of
dipole moment p are kept at an
2Q
S1
Q
D
P
120°
P
A
B
C
angle of 120° as shown in the figure.
What is the resultant dipole moment of this
combination? If this system is subjected to
electric field (E) directed along + X-direction,
what will be the magnitude and direction of
the torque acting on this?
(i) What is the ratio of the electric
flux through S1 and S 2 ?
[Ans. 1/3]
(ii) How will the electric flux through
the sphere S1 change, if a
medium of dielectric constant 5E0
is introduced in the space inside
S1 in place of air?
[Ans. Q /5e ]
0
18 An early model for an atom considered it to
Short Answer Type II
have a positively charged point nucleus of
charge Ze, surrounding by a uniform density
of negative charge upto a radius R. The
atom as a whole is neutral. For this model,
what is the electric field at a distance r
from the nucleus? Assuming, r > R.
14 A spherical conducting shell of inner
radius R1 and outer radius R2 has a
charge Q. A charge q is placed at the
centre of the shell.
(i) What is the surface charge
density on the
(a) inner surface,
(b) outer surface of the shell?
(ii) Write the expression for the
electric field at a point X > R2
from the centre of the shell.
é
Ze æ 1
1 öù
ç 2 - 2 ÷ú
ê Ans.
4pe 0 è r
R øû
ë
19 Total charge – Q is uniformly spread along
length of a ring of radius R. A small test
charge + q of mass m is kept at the centre of
the ring and is given a gentle push along
the axis of the ring.
(a) Show that the particle executes a
simple harmonic oscillation.
(b) Obtain its time period.
15 A large plane sheet of charge having
surface charge density 5 ´ 10 -6 C/m 2 lies
in XY -plane.
Find the electric flux through a circular
area of radius 0.1 m, if the normal to
the circular area makes an angle of 60°
with Z-axis.
(Take, e 0 = 8.85 ´ 10 -12 C 2 /Nm 2 )
[Ans. 4.44 ´ 103 Nm 2 /C]
20 The electric field components in figure are
EX = ax1/ 2 , EY = EZ = 0, in which a = 800
N/C m 1/ 2 . Calculate (a) the flux through the
cube and (b) the charge within the cube.
Assume that a = 01
. m.
16 Consider a region bounded by conical
Y
surface as shown in the figure.
a
E
Ù
nR
Ù
nL
a
a
X
q
a
l
Z
Long Answer Type
r
In this region, E is in vertical upward
direction, then find the electric field
when electric flux is passing through
curved surface?
[Ans. E ´ pr 2 ]
17 What is meant by continuous charge
distribution? How do you apply
superposition principle to obtain total
force on a point charge due to n discrete
charges and three continuous charge
distribution?
21
(i) Define electric flux and write its SI
unit.
(ii) Using Gauss’ law, prove that electric
current at a point due to a uniformly
charged infinite plane sheet is
independent of the distance from it.
(iii) How is the field directed if,
(a) the sheet is positively charged
(b) negatively charged?
22 Find the electric field intensity at a point
due to thin uniformly charged sheet having
charge density s using Gauss’ theorem.
CBSE Examinations Archive
Collection of Questions Asked in CBSE Examinations held in Last 3 Years from 2014-2012
Very Short Answer Type Questions [1 Mark]
Short Answer Type II Questions [3 Marks]
1 Two equal balls having equal positive charge q coulombs are
12 A hollow cylindrical box of length 1 m and area of
suspended by two insulating strings of equal length. What
would be the effect on the force when a plastic sheet is
inserted between the two?
All India 2014
cross-section 25 cm 2 is placed in a three
dimensional coordinate system as shown in the
figure. The electric field in the region is given by
E = 50x $i , where E is in NC -1 and x is in metre. Find
Refer to Q.no. 1 on page 15.
2 Why do the electrostatic field lines not form closed loop?
Refer to text on page 13.
All India 2014
(i) Net flux through the cylinder.
(ii) Charge enclosed by the cylinder.
All India 2014
4 Two charges of magnitudes -2 Q and +Q are located at
points (a, 0) and (4 a, 0) respectively. What is the electric flux
due to these charges through a sphere of radius 3a with its
centre at the origin?
5 Distinguish between a dielectric and a conductor.
Delhi 2012
6 Define dipole moment of an electric dipole. Is it a scalar or
vector?
Foreign 2012
7 A charge q is placed at the centre of a cube. What is the
electric flux passing through a single face to the cube?
Refer to Q.no. 11 on page. 37.
All India 2012
8 Why most electrostatic field be normal to the surface at every
point of a charged conductor?
Foreign 2012
9 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. Calculate the potential
energy of the dipole, if it has charge ± 8 nC.
Delhi 2014
10 An electric dipole is held in a uniform electric field.
(i) Show that the net force facing on it is zero.
(ii) The dipole is aligned parallel to the field.
Find the work done in rotating it through the angle of 180°.
All India 2012
11 Calculate the amount of work done in rotating a dipole, of
dipole moment 3 ´ 10-8 cm, from its position of stable
equilibrium to the position of unstable equilibrium, in a
uniform electric field of intensity 103 N / C.
Foreign 2011
O
X
1m
Z
13 State Gauss’ law in
Y
electrostatics. A cube
with each side A is kept
in an electric field given
X
by E = Cx $i , (as in
a
a
shown in the figure)
where C is a positive Z
dimensional constant.
Find out
(i) the electric flux through the cube and
(ii) the net charge inside the cube.
Foreign 2012
Long Answer Type Questions [5 Marks]
Short Answer Type Questions [2 Marks]
Refer to text on page 26 and 27.
Delhi 2014
Y
3 Why do the electric field lines never cross each other?
14 Using Gauss’ law deduce the expression for the
electric field due to a uniformly charged spherical
conducting shell of radius R at a point (i) outside
and (ii) inside the shell. Plot a graph showing
variation of electric field as a function of r > R and
r < R. (r being the distance from the centre of the
shell).
All India 2013
15 (i) Define electric flux. Write its SI units.
(ii) Using Gauss’ law, prove that the electric field at
a point due to a uniformly charged infinite plane
sheet is independent of the distance from it.
(iii) How is the field directed if (a) the sheet is
positively charged, (b) negatively charged?
(i) Refer to text on page 32.
(ii) and (iii) Refer to text on page 34.
Delhi 2012
Sample
Question Paper
1
(Fully Solved)
Physics
A Sample Question Paper for CBSE Class XII Summative Examination
Time : 3 hrs
Max. Marks : 70
General Instructions
1. All questions are compulsory. There are 26 questions in all.
2. This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
3. Section A contains five questions of one mark each, Section B contains five questions of two marks each,
Section C contains twelve questions of three marks each, Section D contains one value based question of four
marks and Section E contains three questions of five marks each.
4. There is no overall choice. However, an internal choice has been provided in one question of two marks, one
question of three marks and all the three questions of five marks weightage. You have to attempt only one of the
choices in such questions.
5. You may use the following values of physical constants wherever necessary.
c = 3 ´ 108 m/s; h = 6 .63 ´ 10-34 Js; e = 1.6 ´ 10-19 C; m0 = 4p ´ 10-7 TmA-1; e 0 = 8 .854 ´ 10-12 C 2N-1m-2 ;
1
= 9 ´ 109 Nm2C-2 ; me = 9 .1 ´ 10-31 kg; mass of neutron = 1.675 ´ 10-27 kg;
4pe0
mass of proton = 1.673 ´ 10-27 kg; Avogadro’s number = 6023
.
´ 1023 per gram mole;
Boltzmann constant = 1.38 ´ 10-23 JK -1
1. If an oil drop of weight 3.2 ´ 10 -13N is stationary in an electric field of 5 ´ 105 V/m, then find the charge on oil
drop.
2. A carbon resistor is marked in coloured bands of red, black, orange and silver. What is the resistance and the
tolerance value of the resistor?
3. What is the resistance offered by the capacitance to DC?
4. A concave mirror is held in water. What should be the change in focal length of the mirror?
5. Identify the parts X and Y in the following block diagram of a generalised communication system
X
Transmitter
Y
Receiver
Sample Question Papers
Section A
ll
542
ne
Physics Class 12th
Section C
Section B
6. Apply Kirchhoff’s laws to the loops ACBPA and
11. (i) An electrostatic field line is a continuous curve,
ACBQA to write the expressions for the currents
I1, I 2 and I 3 in the network.
i.e. a field line cannot have sudden breaks. Why
not?
+
(ii) Explain why two field lines never cross each
other at any point?
6V
–
P
(iii) A proton is placed in a uniform electric field
directed along the positive X-axis. In which
direction will it tend to move?
0.5 W
I1
A
B
12. (i) A current is set up in a long copper pipe. Is there
I2
I3
Q
+ –
10 V
a magnetic field (a) inside, (b) outside the pipe?
1W
(ii) Figure shown below shows a bar magnet
M falling under the gravity through an air cored
coil C.
C
12 W
M
Circuit diagram of loops
S
N
7. The frequency of oscillation of the electric field
vector of a certain electromagnetic wave is
5 ´ 10 4 Hz. What is the frequency of oscillation of
the corresponding magnetic field vector and to
which part of the electromagnetic spectrum does
it belong?
or
Show that linear magnification of an image
formed by a curved mirror may be expressed as,
m=
f
f-v
=
f-u
f
where, letters have their usual meanings.
8. Draw suitable graphs to show the variation of
photoelectric
potential for
current
with
collector
plate
(i) a fixed frequency but different intensities
I1 > I 2 > I3 of radiation.
(ii) a fixed intensity but different frequencies
f1 > f2 > f3 of radiation.
9. For what kinetic energy of a neutron, will be
associated by the de-Broglie wavelength be
1.32 ´ 10 -10 m? Given that mass of a neutron
= 1675
.
´ 10 -27 kg.
C
V
(a) Plot a graph showing variation of induced emf (E )
with time ( t ).
(b) What does the area enclosed by the E-t curve
depict?
or
On a smooth plane inclined at 30°° with the
horizontal, a thin current carrying metallic rod is
placed parallel to the horizontal ground. The plane
is located in a uniform magnetic field of 0.15 T in
the vertical direction. For what value of current
can the rod stationary? The mass per unit length
of the rod is 0.03 kg m -1.
13. Does the current in an AC circuit lag, lead or
remain in phase with the voltage of frequency ( f )
applied to the circuit, when
(ii ) f < fr (iii) f > fr ,
(i ) f = fr
where fr is the resonant frequency?
14. (i ) Two slits are made 1 mm apart and the screen is
10. A carrier wave of peak voltage 12 V is used to
placed 1 m away. What is the fringe separation,
when blue-green light of wavelength 500 nm is
used.
transmit a message signal. What should be the
peak voltage of the modulating signal, in order to
have a modulation index 75%?
(ii ) What should the width of each slit be to obtain 10
maxima of the double slit pattern within the
central maximum of the single slit pattern?
ll
ne
Sample Question Paper 1
15. Three light rays,
A
red (R), green (G ) and B
blue (B ) are incident G
R
on a right angled
45°
prism ABC at face AB.
B
C
The refractive indices
Light rays are
of the material of the
incident on a
prism for red, green
right angled
and blue wavelengths
prism ABC
are 1.39, 1.44 and
1.47 respectively. Out of the three, which
colour of ray will emerge out of face AC?
Justify your answer. Trace the path of
these rays after passing through face AB.
16. A beam of light, consisting of two
wavelengths 560 nm and 420 nm, is used
to obtain interference fringes in a
Young’s double-slit experiment. Find the
least distance from the central
maximum, where the bright fringes due
to both the wavelengths coincide. The
distance between the two slits is 4.0 mm
and the screen is at a distance of 1.0 m
from the slits.
17. Calculate the binding energy per nucleon
543
20. Sketch the graph, showing the variation of stopping
potential with frequency of incident radiation for two
photosensitive material A and B having threshold
frequencies v 0 ¢ and n 0 respectively ( v 0 ¢ > v 0 ).
(i) Which of the two metals, A or B has higher work
function?
(ii) What information do you get from the slope of the
graphs?
(iii) What does the value of the intercept of graph it on the
potential axis represent?
21. The diagram given below represents the block diagram of
a generalised communication system. Identify the
elements labelled as X, Y and Z in this diagram. Explain
the function of each of these elements.
Information Message
source
signal
X
Y
Z
Message
User
signal
22. Give the circuit diagram of a common emitter amplifier,
using an n-p-n transistor. Draw the input and output
waveforms of the signal. Write the expression for its
voltage gain?
Section D
40
nucleus given
20 Ca
40
m( 20 Ca ) = 39.962589 u
23. Physics teacher Mr. Sharma conducts viva-voice for
mn = 1.008665 u, mp = 1.007825 u
(take, 1 amu = 931 MeV).
18. The electron in a given Bohr orbit has a
total energy of -1.5 eV. Find
(i) its kinetic energy
(ii) potential energy
(iii) wavelength of light emitted, when the
electron makes a transition to the ground
state. (take, E g = - 13.6 eV)
board practical and asks the following two questions
from every student.
(i) Why a potentiometer be preferred over a voltmeter for
measurement of emf of a cell?
(ii) Why should a six wire potentiometer be preferred over
a three wire potentiometer?
The student A who could not answer many questions was
teacher ward. However another student B answered
following questions correctly do not belong to teacher
ward. Mr. Sharma award full marks to student B.
19. In the figure given below, circuit symbol of
Answer the following questions on the basis of given
informations.
a logic gate and two input waveforms A and
B are shown.
(ii) Write the answer of the questions asked by Mr. Sharma.
(i) Which values are displayed by Mr. Sharma?
A
Y
B
Circuit symbol
Section E
24. Find an expression for the torque acting on an electric
A
B
Input waveform
(i) Name the logic gate.
(ii) Write its truth table.
(iii) Give the output waveform.
dipole placed in uniform electric field. A system of two
charges, qA = 2.5 ´ 10 -7 C and qB = 2.5 ´ 10 -7 C located
at points A(0, 0, - 15 cm) and B(0, 0, + 15 cm) respectively.
Find the electric dipole moment of the system and the
magnitude of the torque acting on it, when it is placed in a
uniform electric field 5 ´ 10 4 NC -1, making an angle 30°.
or
of
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544
A capacitor of capacitance C is charged fully by
connecting it to a battery of emf E. It is then
disconnected from the battery. If the separation
between the plates of the capacitor is now doubled,
what will happen to
(i)
(ii)
(iii)
(iv )
(v)
(ii) Obtain the impedance of the circuit and the
amplitude of current at the resonating
frequency.
(iii) Determine the rms potential drop across the
three elements of the circuit.
26. Show that the refractive index of the material
25. Explain with the help of a neat and labelled diagram,
the principle, construction and working of a
transformer.
or
( A + dm )
2
æAö
sinç ÷
è2ø
sin
of a prism is given by m =
where the symbols have their usual meanings.
The given circuit diagram shows a series L-C-R
circuit connected to a variable frequency 230 V
source.
80 mF
Physics Class 12th
(iv ) How do you explain the observation that the
algebraic sum of the voltage across the three
elements in (C ) is greater than the supplied
voltage?
charge stored by the capacitor?
potential difference across it?
field strength between the plates?
energy stored by the capacitor?
capacitance of the capacitor?
5.0 H
ne
40 W
or
Define the term resolving power of an
astronomical telescope. How does, it get
affected on
(i) increasing the aperture of the objective lens?
(ii) increasing the wavelength of light used?
230 V
(i) Determine the source frequency which derives the
circuit in resonance.
(iii) increasing the focal length of the objective
lens?
SOLUTIONS
1. Weight of drop, W = 3.2 ´ 10 -13 N
Electric field,
5
E = 5 ´ 10 V/m
W 3.2 ´ 10 -13
=
E
5 ´ 10 5
= 6.4 ´ 10 -19 C
\ Charge on oil drop, q =
2. According to colour code of carbon resistor, a carbon
resistance of bands red and black having figures 2 and 0.
The third band of orange having multiplier 10 3 .
\ The value of resistance is given by
R = 20 ´ 103 W
But the fourth band having silver colour, which represents a
tolerance of ±10% .
Hence, the value of carbon resistor
R = 20 ´ 103 W ´ 10%
3. We know that, capacitive reactance is
XC =
1
2pf C
For DC, frequency, f C = 0
i.e.
XC = ¥
Therefore, resistance offered by capacitance to DC is
infinite.
4. No change in focal length as focal length of mirror ( f )
is independent of medium and depends only on
radius of curvature (R).
R
f=
Q
2
5. X part is information and Y part is medium or
channel.
6. Apply Kirchhoff’s Ist law, I3 = I1 + I 2
Applying Kirchhoff’s IInd law to loop ACBPA
-12 I3 - 0 .5 I1 + 6 = 0
0 .5 I1 + 12 I3 = 6
Applying Kirchhoff’s IInd law to loop ACBQA
-12 I3 - 1I 2 + 10 = 0
I 2 + 12 I3 = 10
7. For electromagnetic waves, frequency of electric
field vector and magnetic field vector is same.
\ Frequency, v = 5 ´ 104 Hz
CBSE
Examination Paper
2015 (Delhi)
Physics
Max. Marks : 70
Time : 3 hrs
General Instructions
1. All questions are compulsory. There are 26 questions in all.
2. This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
3. Section A contains five questions of one mark each, Section B contains five questions of two marks each,
Section C contains twelve questions of three marks each, Section D contains one value based question of four
marks and Section E contains three questions of five marks each.
4. There is no overall choice. However, an internal choice has been provided in one question of two marks, one
question of three marks and all the three questions of five marks weightage. You have to attempt only one of the
choices in such questions.
5. You may use the following values of physical constants wherever necessary.
c = 3 ´ 108 m/s; h = 6 .63 ´ 10-34 Js; e = 1.6 ´ 10-19 C; m 0 = 4p ´ 10-7 TmA-1; e 0 = 8 .854 ´ 10-12 C 2N-1m-2 ;
1
= 9 ´ 109 Nm2C-2 ; me = 9 .1 ´ 10-31 kg; mass of neutron = 1.675 ´ 10-27 kg;
4pe0
mass of proton = 1.673 ´ 10-27 kg; Avogadro’s number = 6023
.
´ 1023 per gram mole;
-23
-1
Boltzmann constant = 1.38 ´ 10
JK
Section A
1. Define the term ‘Self-inductance’ of a coil. Write
its SI unit?
Sol. Self Inductance is the property of a coil by virtue of
which, the coil opposes any change in the strength of
the current flowing through it by inducing an emf in
itself.
(1/2)
The SI unit of self-inductance is henry.
(1/2)
Since, the blue colour has shorter wavelength than red.
So, it is scattered much more strongly. Hence, sky looks
(1)
blue.
3. I-V graph for a metallic wire at two different
temperatures, T1 and T2 is as shown in the figure.
Which of the two temperature is lower and why?
T1
2. Why does bluish colour predominate in a clear
sky?
T2
I
Sol. Blue colour of the sky is due to scattering of light from
atmosphere’s particles. Light of shorter wavelength is
scattered more than the light of longer wavelength.
V
ll
592
Sol. Consider the figure,
ne
Physics Class 12th
Sol. (i) Size of the antenna
T1
1
T2
2
I
V
Since, slope of 1>slope of 2.
R1 < R2
(1/2)
Þ Also, we know that resistance is directly proportional
(1/2)
to the temperature. therefore, T2 >T1
4. Which basic mode of communication is used for
telephonic communication?
Sol. Point to point is a basic mode of communication.
Which is used for telephonic conversation. In this mode
of communication, communication takes place over a
link between a single transmitter and a receiver.
(1)
5. Why do the electrostatic field lines not form the
closed loops?
Sol. Electrostatic field lines are discontinuous curves. They
start from a positively charged body and end at a
negatively charged body. No electric lines of force exist
inside the charged body. Thus, electrostatic field lines do
(1)
not form any closed loops.
8. Use Kirchhoff’s rules to determine the potential
difference between the points A and D. When no
current flows in the arm BE of the electric network
shown in the figure.
3W
F
E
1V
Section B
D
R1
R
6. Which an electron in hydrogen atom jumps from
the third excited state to the ground state, how
would the de-Broglie wavelength associated with
the electron change? Justify your answer.
h
h
=
p mv
h
hr nh
or mvr = =
or,
mv =
l
l 2l
2p
2pr
or,
l=
´ hr =
n
nh
2
As,
r µn
1
Þ
l µ (n 2 ) = n
n
l
l
3
Thus we can say that 3 =
or l1 = 3
3
l1 1
Since, an antenna is needed both for transmission
and reception. Each antenna should have a size
comparable to the wavelength of the signal.
for an EM wave of frequency 20 kHz, wavelength is
15 km.
15
\ length of antenna = km = 3.75 km.
4
It is an obvious that such a long antenna is not
possible. Hence, antenna length can be made
reasonable, if the frequency is high. So, there is a
need to convert low frequency signal in to high
(1/2)
frequency signal before transmission.
(ii) Effective power radiated by an Antenna
Power P radiated from linear antenna of length l is
2
1
proportional to æç ö÷ .
èlø
Hence, antenna length can be made reasonable, if
the frequency is high.
So, there is a need to convert low frequency signal
in to high frequency signal before transmission. (1)
2W
3V
A
6V
C
B
4V
Sol. Consider the given figure,
Sol. We know that l =
F
I
3W
1V
I
E
D
R1
R
2W
(1/2)
Thus wavelength decreases 3 times as an electron
jumps from third excited state to the ground state. (1/2)
7. Write two factors which justify the need of
modulating a low frequency signal in to high
frequencies before transmission?
I
3V
I
A
6V
B I
C
4V
Applying Kirchhoff’s IInd law in AFEBA
2 I -1+3 I - 6 = 0
[Since, no current flows in the arm BE of the circuit]
7
…(i)
5I =7 Þ I = A
5
Applying Kirchhoff’s IInd law in mesh AFDCA
3 I +RI - 4 - 6 +2 I -1= 0
…(ii)
5 I + RI = 11
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ne
CBSE Examination Paper 2015 (Delhi)
Now, substitute the vales of I from equation (i) to (ii),
we get
7R
7
7
= 11
5 ´ + R ´ = 11 Þ 7 +
5
5
5
7R
20
Þ
= 4 Þ R = W.
(1)
5
7
For PD across A and D, along AFD
7
7
VA - ´ 2 + 1 - 3 ´ = VD
5
5
21
14
VA - - 1 - = VD
5
5
14 21
Þ
VA - VD = + - 1
5
5
(1)
(VA - VD ) = 7 - 1 = 6 V
9. You are given two converging lenses of focal
length 1.25 cm and 5 cm to design a compound
microscope. If it is desired to have a magnification
of 30, find out the separation between the
objective and eyepiece.
or
A small telescope has on objective lens of focal
length 150 cm and eyepiece of focal length 5 cm.
What is the magnifying power of the telescope for
viewing distance objects in normal adjustments.
If this telescope is used to view a 100 m tall tower
3 km away, what is the height of the tower formed
by the objective lens.
Sol. Given, F 0 = 1 . 25 cm, Fe = - 5 cm
Magnification, M = 30, D= 25 cm
Q If the objective is very close to the principal focus of
the objective and the image formed by the objective is
very close to eyepiece, then
Magnifying power of a microscope is given by,
1 D
1 25
(1)
Þ 30 =
M= - ×
×
f 0 fe
1 .25 5
1.25 ´ 30 ´ 5
25 ´ 30
Þ
L=
Þ L=
25 ´ 100
100
30
Þ
L=
Þ L= 7.5 cm
4
This is a required separation between the objective and
(1)
the eye piece.
or
When final image is at D,
f æ
f ö
then magnifying power, M = 0 ç1 + e ÷
fe è
Dø
In normal adjustment, M =
- f0
fe
593
For telescope,
Focal length of objective lens, f 0 = 150 cm
focal length of eye lens, f e = 5 cm
when final image forms at D = 25 cm
-f æ
f ö
\ Magnification, M = 0 ç1 + e ÷
fe è
Dø
5ö
-150 æ
ç1 + ÷
5 è 25 ø
-150 6
=
´
5
5
Þ
M = - 36
Let height of final image be h cm.
h
tanb =
Þ
25
b = Visual angle formed by final image at eye.
a = Visual angle subtended by object at objective.
100m
1
tana =
=
3000m 30
tanb
But,
M=
tana
æhö
ç ÷
è 25 ø
h
Þ
-36 =
Þ -36 = ´ 30
1
25
æ ö
ç ÷
è 30 ø
6h
-36 ´ 5
-36 =
Þ h=
Þ
5
6
h = - 30 cm
=
(1)
(1)
10. Calculate the shortest wavelength in the Balmer
series of hydrogen atom. In which region
(infra-red, visible, ultraviolet) of hydrogen
spectrum does this wavelength lie?
Sol. Since, we know that for Balmer series,
æ1
1
1ö
= R ç 2 - 2 ÷ , n 2 = 3, 4, 5, .....
l
è2
n2 ø
(1)
for shortest wavelength in Balmer series, the spectral
series is given by,
n1 = 2 ,n 2 = ¥
1
1 ö
æ1
Þ
=Rç 2 - 2 ÷
è2
l
¥ ø
1
1
1 R
4
=R´
Þ
=
Þ l=
Þ
l
4
l 4
R
4
[Q R = 1. 097 ´ 10 -7 m -1 ]
l=
7
1097
.
´ 10
Þ
l = 3.64 ´ 10 -7 m.
The lines of Balmer series are found in the visible part of
the spectrum.
(1)
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594
relaxation time of charge carriers in a conductor.
A conductor of length L is connected to a d.c.
source of emf ‘E’. If the length of the conductor is
tripled by stretching it, keeping ‘E’ constant,
explain how its drift velocity would be affected.
11. Calculate the potential difference and the energy
stored in the capacitor C 2 in the circuit shown in
the figure. Given potential at A is 90 V. C1 = 20 mF,
C 2 = 30 mF, C 3 = 15 mF.
A
C2
Physics Class 12th
12. Find the relation between drift velocity and
Section C
C1
ne
Sol.
+
C2
E
F = –eE
h
–
20 mF 30 mF
15 mF
V
A
C1
C2
C3
Let V be the potential difference applied across the ends
of a conductor of length (l), then the magnitude of
electric field is
V
E=
(1)
l
Given, C1 = 20 mF, C 2 = 30 mF, C3 = 15 mF
Potential at A = 90 V
As, we can see that capacitor C3 is earthed, therefore,
potential across C3 will be zero.
Since, capacitor C1 , C 2 and C3 are connected in series,
1
1
1
1
therefore,
= +
+
C eq C1 C 2 C3
1
1
1
1
= + +
C eq 20 30 15
3+2+ 4
1
=
Þ
C eq
60
The direction of electric field is from positive to negative
end of conductor as shown in the above figure. Since,
the charge on an electron is -e and each free electron in
the conductor experiences a force (F).
F = - eE
Acceleration of each electron is given by,
- eE
F
[Qfrom Newton’s second law, a = ]
a=
m
m
60
1
9
=
Þ C eq =
9
C eq 60
20
mF
3
Since, charge remains same in series combination,
20
So,
Q = C eq V Þ Q = ´ 90
3
Þ
Q = 600 mC Þ Q = 600 ´ 10 -6 C
Þ
C eq =
Þ
Q = 6 ´ 10 -4 C
\Potential difference across C 2 =
Q
C2
Þ
V2 =
Þ V2 =
Þ
V2 = 0 .2 ´ 10 2
Q
V2
6 ´ 10 -4
30 ´ 10 -6
Þ
V2 = 20 V
Also, energy stored in capacitor l 2 is given by
1
E = C 2V22
2
1
E = ´ 30 ´ (20) 2 ´ 10 -6
Þ
2
1
Þ
E = ´ 30 ´ 400 ´ 10 -6
2
Þ
E = 6000 ´ 10 -6
Þ
E = 6 ´ 10 -3 J
(1)
(1)
At any instant of time, the velocity acquired by electron
having thermal velocity (V1 ) is givne by
V1 = u1 + at1
where, t1 is the time elapsed.
Similarly,
V2 = u 2 + at 2 , ...Vn = u n + at n
\ Average velocity, Vd = V1 + V2 + ... + Vn
(u + at1 ) + (u 2 + at 2 ) + ... + (u n + at n )
= 1
n
(u1 + u 2 + ... + u n ) a (t1 + t 2 + ... + t n )
=
+
n
n
(u1 + u 2 + ... + u n )
We know that,
=0
n
t + t 2 + ... + t n
is called the average relaxation
and 1
n
time. It’s value is of the order of 10 -14 second.
V d = 0 + at
eE
V d = at Þ V d = - t
m
Negative sign shows that Vd is opposite to the direction
of E .
eE
\ Average drift velocity, V d = t
m
(1)
This is a required relation between drift velocity and
relaxation time of the charge carriers in a conductor. (1)
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ne
The conductor connected to d.c. source of emf E is
shown in the figure alongside.
Conductor
E
Suppose initial length of the conductor, l 2 = l 0 .
New length l f = 3l 0
We know that drift velocity
Vd µ E 0 (Electric field)
(V d ) f (E 0 ) f E/ lf
Thus,
=
=
(V d )i (E 0 ) i
E/ li
=
Thus
(V d ) f =
li l 0
1
=
=
lf 3l 0 3
(V d )i
3
Thus drift velocity decreases three times.
(1)
13. State clearly how an unpolarised light get linearly
polarised, when passed through a polaroid.
(i) Unpolarised light of intensity I 0 is incident on
a polaroid P1 which is kept near polaroid P2
whose pass axis is parallel to that of P1. How
will the intensities of light, I1 and I 2
transmitted by the polaroids P1 and P2
respectively, change on rotating P1 without
disturbing P2?
(ii) Write the relation between the intensities I1
and I 2.
Sol. (i)
Angle with planes of
transmission = q
Unpolarised
light
Polarised
light
595
When polarizer and analyzer are perpendicular to
each other
other,
q = 90°
Þ
cos q = cos 90° = 0
(1)
Þ
I=0
In unpolarised light, vibrations are probable in all
the direction in a plane perpendicular to the
direction of propagation.
Therefore, q can have any value from 0 and 2p.
1 2p
\ [cos 2 q] av = ò cos 2 q dq
2p 0
1 2p (1 + cos 2q) dq
= ò
2p 0
2
2p
sin2 q ù
1 é
=
+
0
2 úû 0
2p ´ 2 êë
1
Þ [cos 2 q] av =
2
using law of Malus, I = I 0 cos 2 q
1 1
I = I0 ´ = I0
Þ
(1)
2 2
As, per the question, I 0 is the intensity of incident
unpolarised light and I1 and I 2 are the intensities of
polaroids P1 , and P2 respectively, then we can say
that when unpolarised light of intensity I o get
polarised on passing through a polaroid P1 its
I
(1)
intensity become half i.e. I1 = 0
2
(ii) and when this polarised light of intensity I1 passes
through polaroid P2 , then its intensity will be given
by,
I 2 = I1 cos 2 q
This is a required relation between intensities I1 and
(1)
I 2.
14. Define modulation index. Why is its value kept, in
Intensity
= I0
Polariser P1
Intensity
I0
I1= —
2
Polorised E
cos2 q
Intensity = I0 —
P2
2
I2= I1 cos2 q
According to law of Malus, when a beam of
completely plane polarised light is incident on an
analyser, the resultant intensity of light (I )
transmitted from the analyzer varies directly as the
square of the cosine of angle q between the plane of
transmission of analyzer and polarizer.
i.e.
I µ cos 2 q Þ I = I 0 cos 2 q
When polarizer and analyzer are parallel, q = 0° or
180°.
So that
cosq = ± 1
Þ
I = I0
practice, less than one?
A carrier wave of frequency 1.5 mHz and
amplitude 50 V is modulated by a sinusoidal wave
of frequency 10 kHz producing 50% amplitude
modulation. Calculate the amplitude of the AM
wave and frequencies of side bands produced.
Sol. Amplitude modulation index is the ratio of the
modulating signal to the maximum amplitude of carrier
wave.
Am
It is given by, m =
Ac
Since, the amplitude modulation index (m) determines
the quality of the transmitted signal. When modulation
index is small, variation in carrier amplitude will be small.
Therefore, audio signal being transmitted will be weak.
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596
As the modulation index increases, the audio signal on
reception becomes clearer.
(1)
Given, frequency of carrier wave,
n c = 1.5 mHz = 1500 kHz
frequency of sinusoidal (modulated) wave, n m = 10 kHz
Amplitude of carrier wave, Ac = 50 V.
50 1
=
100 2
Modulation Index (m) = 50% Þ
QModulation index, m =
Am
Ac
1 Am
=
2 50
A m = 25 V
Þ
Þ
(1)
So, the amplitude of AMwave, A m = 25 V.
As, we know, the side bands are
USB = n c + n m = 1500 + 10 = 1510 kHz
LSB = n c - n m = 1500 - 10 = 1490 kHz
Physics Class 12th
mV^2
= qV^ B
r
from equation (i) and (ii), we get
V^2 m
V m
= qV^ B Þ ^ = r
r
qB
2pV^ m
T=
Þ
qB × V^
2pm
T=
Bq
…(ii)
and
(1)
Distance moved by the particle along the magnetic field
in one rotation (pitch of the helix path)
= VII ´ T
[QVII = Vparallel ]
2pm
= Vcos q ´
Bq
2pmV cos q
(1)
P=
qB
16. (i) Determine the value of phase difference
between the current and the voltage in the
given series L-C-R circuit
These are the required frequencies of the side bands
(1)
produced.
15. A Uniform magnetic field B is set up along the
positive x-axis. A particle of charge ‘q’ and mass ‘m’
moving with a velocity v enters the field at the origin
in X-Y plane such that it has velocity components
both along and perpendicular to the magnetic field
B. Trace, giving reason, the trajectory followed by
the particle. Find out the expression for the distance
moved by the particle along the magnetic field in one
rotation.
Sol.
ne
R=400 W
C=2 mF
V=V sin(1000t+f)
°
L=100 mH
(ii) Calculate the value of additional capacitor
which may be joined suitably to the capacitor
C that would make the power factor of the
circuit unity.
Sol. (i) Consider the given figure,
R=400 W
y
B
V^=V sin q
L=100 mH
V
x
VII=Vcosq
z
The path of the charged particle will be helix. As, the
charge moves linearly in the direction of the magnetic
field with velocity V cosq and also describe the circular
path due to velocity V sinq.
Time taken by the charge to complete one circular
2pr
rotation,
T=
(1)
V^
f = qV^ B
C=2 mF
V=V sin(1000t+f)
°
…(i)
Since, the alternating emf in the above L-C-R series
circuit would be represented by
V = V0 sin(1000 t + f)
Þ
w = 1000 Hz
Given, R = 400 W, C = 2mF, L= 100 mH
1
Q Capacitive reactance, X C =
wC
1
XC =
Þ
1000 ´ 2 ´ 10 -6
103
2
Þ
X C = 500 W
Q Inductive reactance, X L = wL
Þ
XC =
(1)
ll
ne
Þ
X L = 1000 ´ 100 ´ 10 -3
Þ X L = 100 W
So, we can see that X C > X L
Þ tan f is negative
Hence, the voltage lags behind the current by a
phase angle f. The A.C. circuit is capacitance
dominated circuit.
X - XC
Q phase difference, tan f = L
R
100 - 500
-400
Þ tan f =
tan f =
400
400
æpö
tan f = - 1 Þ tan f = - tanç ÷
è 4ø
-p
Þ
f=
4
This is the required value of the phase difference
between the current and the voltage in the given
(1)
series L-C-R circuit.
As, cos f < 1
(ii) Suppose new capacitance of the circuits is C¢ .
Thus, to have power factor unity
R
cos f¢ = 1 =
2
R + (X L - X C¢ ) 2
R2 = R2 + (X L - X C¢ ) 2
1
1
or,
or, w L =
X L = X C¢ =
wC ¢
w C¢
1
1
or, (1000) 2 =
or,
w2 =
[Q w = 1000]
LC¢
LC¢
1
1
or,
C¢ =
=
6
L ´ 10
100 ´ 10 -3 ´ 10 6
10
1
= 6 = 5 = 10 -5
10
10
or,
C¢ = 10 -5 F = 10 ´ 10 -6 F
= 10mF
As, C¢ > C. Hence we have to add an additional
capacitor of capacitance 8 mF (10mF- 2mF) in parallel
with previous capacitor.
(1)
or,
597
While dealing with the charging of a parallel plate
capacitor with varying current. It was found that
ampere’s circuital law is not logically consistent, because
òB × d l has not the same value on the two sides of a
plate of charged capacitor. The inconsistency of
ampere’s circuital law was removed by Maxwell by
predicting the presence of displacement current in the
region between the plates of capacitor, when the charge
on the capacitor is changing with time. So, displacement
current is the missing term (which is related with the
changing electic field which passes through the surface
(1)
between the plates of capacitor).
In this way, Maxwell pointed out that for consistency of
ampere’s circuital law, there must be displacement
current I D along with the conduction current in the
closed loop as (I + I D ) has the property of continuity.
(1)
18. Use Huygen’s principle to show how a plane wave
front propagates from a denser to rarer medium.
Hence, verify Snell's law of refraction.
Sol. According to Huygen’s principle
(i) Each point on the given wave front (called primary
wave front) is the source of a secondary disturbance
(called secondary wavelets) and the wavelets
emanating from these point spread out in all the
(1)
directions with the speed of the wave.
(ii) A surface touching these secondary wavelets,
tangentially in the forward direction at any instant
gives the new wavefront at that instant. This is
called secondary wave front.
A1
A
A2
A2 A A1
1
1
S
æ
d fE ö
÷
dt ø
(1)
Propagation of
3
light wave
New
4
wavefront
New wavefront
(Spherical) B B B plane
2
1
(a)
(b)
N¢
N
3
in
wa cid
ve en
fro t
nt
2
1
X
Medium 1
Medium 2
B
Rarer Medium
V1, m1
E
i
i
F
Y
r Denser Medium
V2, m2
r
G
ed
ct nt
fra fro
Re ave
w
òB × d l = m 0 (I + I D ) = m 0 çè1 + e 0
3
Propogation of wavefront
ampere’s circuital law. Discuss its significance
and describe briefly how the concept of
displacement current is explained through
charging/discharging of a capacitor in an electric
circuit.
by Maxwell states that
2
B2 4
B
B1
17. Write the expression for the generalised for of
Sol. The generalised form of ampere’s circuital law, modified
2
1¢
3¢
2¢
ll
598
If V1 , V2 are the speed of light in to two mediums and t is
the time taken by light to go from B to C or A to D or E to
EF FG
G through F, then t = +
V1 V2
EF
In DAFE, sini =
AF
AF sini FC sinr
FG
In DFGC, sinr =
+
Þ t=
V1
V2
FC
amplifier. Briefly explain its working and write the
expression for (i) current gain, (ii) voltage gain of
the amplifier.
Sol. Circuit diagram of a C.E. transistor amplifier
IC
C
IB
p-n-p
VCE
IE
So, t should not depend on F. This is possible only, if
sini sinr
sini V1
= 0 or
=
=m
V1
V2
sinr V2
Now, if C represents the speed of light in vacuum, then
C
C
are known as the refractive index
m1 = and m 2 =
V1
V2
of medium and medium 2 respectively.
m1 sini = m 2 sinr Þ m =
sini
sinr
This is known as Snell’s law of refraction.
(1)
19. Identify the gates P and Q shown in the figure.
Write the truth table for the combination of gates
shown.
A
P
B
Q
X
Y
Name the equivalent gate representing this circuit
and write its logic symbol.
A
P
B
Q
X
Y
P is AND gate and Q is a NOT gate.
Truth Table
B
X
X=Y
0
0
0
1
0
1
0
1
1
0
0
1
1
1
1
0
B
VCC
+
– +
VBE
Output AC
signal
IB
IC
(1)
Working
The emitter-base circuit is forward biased by a low
voltage battery VBE , that means the resistance of input
circuit is small. The collector-emitter circuit is reversed
biased by a high voltage battery VCC , that means the
resistance of the output circuit is high.
RL is a load resistance connected in collector-emitter
circuit. The weak input AC signal is applied across the
base-emitter circuit and the amplified output is
obtained across the collector emitter circuit. When no
AC voltage is supplied to the input circuit, we have
…(i)
IE = IB + IC
Due to collector current I C , the voltage drop across the
load resistance (RL) is I C RL. Therefore, the collector
emitter voltage VCE is given by,
…(ii)
VCE = VCC - I C RL
when the input AC voltage signal is applied across the
base-emitter circuit, it changes base-emitter voltage and
hence, emitter-current I E changes which in turn
changes the collector current I C . So, the
collector-emitter voltage VCE varies in the accordance
with equation (ii).
(1)
Current gain and Voltage gain of amplifier
I
(i) D.C. Current gain, b DC = C
IB
æDI ö
AC current gain, b AC = ç C ÷VCE
è DIB ø
(1)
The equivalent gate representing this circuit is NAND
gate. Its logic symbol is
A
Input AC
signal
IC
–
This variation in VCE appears as an amplified output.
(1)
A
R2
E
For rays of light from the different parts on the incident
wave front, the values of AF are different. But light from
different points of the incident wave front should take
the same time to reach the corresponding points on the
refracted wave front.
Then,
Physics Class 12th
20. Draw a circuit diagram of a C.E. transistor
æ sini sinr ö
AC sinr
t=
+ AF ç
÷
è V1
V2
V2 ø
Þ
ne
Y=A.B
(1)
D I c ´ Rout
D I b ´ Rin
D I c Rout
=
´
DIb
Rin
(ii) AC voltage gain, AV =
(1)
ll
ne
21. (i) Write three characteristic properties of
nuclear force.
(ii) Draw a plot of potential energy of a pair of
nucleons as a function of their separation.
Write two important conclusions that can be
drawn from the graph.
Sol. (i) Characteristics properties of nuclear force
(a) Nuclear forces act between a pair of neutrons,
a pair of protons and also between a
neutron-proton pair, with the same strength.
This, shows that nuclear forces are
independent of charge.
(b) The nuclear forces are dependent on spin or
angular momentum of nuclei.
(c) Nuclear forces are non-central forces. This
shows that the distribution of nucleons in a
nucleus is not spherically symmetric.
(1)
Potential energy (Mev)
(ii)
100
0
–100
r01
2
r (fm)
3
(Potential energy versus
distance)
(1)
From the plot, it is concluded that
(i) The potential energy is minimum at a
distance r0 (» 0.8 f m) which means that the
force is attractive for distances larger than
0.8 Fm and repulsive for the distance less than
0.8 Fm between the nucleons.
(ii) Nuclear forces are negligible, when the
distances between the nucleons is more than
10 Fm.
(1)
22. (i) Describe briefly three experimentally
observed features in the phenomenon of
photoelectric effect.
(ii) Discuss briefly how wave theory of light
cannot explain these features.
or
(i) Write the important properties of photons
which are used to establish Einstein’s
photoelectric equation.
599
(ii) Use this equation to explain the concept of
(a) threshold frequency and (b) stopping
potential.
Sol. (i) Three experimentally observed features in the
phenomenon of photoelectric effect.
(a) Intensity When intensity of incident light
increases as one photon ejects one electron,
the increase in intensity will increase the
number of ejected electrons. Frequency has
no effect on photoelectron.
(b) Frequency when the frequency of incident
photon increases, the kinetic energy of the
emitted electrons increases. Intensity has no
effect on kinetic energy of photoelectron.
(c) No time lag When energy incident photon is
greater than the work function, the
photoelectron is immediately ejected. Thus,
there is no time lag between the incidence of
light and emission of photoelectron. (½ ´ 3)
(ii) These features cannot be explained on the wave
number of light because wave nature of radiation
cannot explain the following.
(a) The instantaneous ejection of the
photoelectrons.
(b) The existence of threshold frequency for a
metal surface.
(c) The fact that kinetic energy of the emitted
electrons is independent of the intensity of
light and depends upon its frequency. (½ ´ 3)
or
(i) Important properties of photon’s which are used to
establish Einstein’s photoelectric equations.
(a) In interaction of radiation with matter,
radiation behaves as, if it is made up of
particles called photons.
hc ö
æ
(b) Each photon has energy E ç = hv = ÷ and
è
lø
æ hv hc ö
momentum p ç =
= ÷ , where c is the
è c
lø
speed of light h is planck’s constant, v and l
are frequency and wavelength of radiation
respectively.
(c) All photons of light of a particular frequency v
or wavelength l have the same energy
hc ö
æ
æ hv h ö
E ç = hv = ÷ and momentum p ç =
- ÷
è
è c lø
ø
l
whatever the intensity of radiation may be.
(½ ´ 3)
(ii) Since, Einstein’s photoelectric equation is given by,
1 2
K × E max = mv max
= hv - hf0
2
ll
600
(a) For a given material, there exist a certain
minimum frequency of the incident radiation
below, which no emission of photoelectron
takes place. This frequency is called threshold
frequency.
Above threshold frequency, the maximum
kinetic energy of the emitted photoelectron
or equivalent stopping potential is
independent of the intensity of the incident
light but depends only upon the frequency of
the incident light.
(1½)
(b) If the collecting plate in the photoelectric
apparatus is made at high negative potential,
then most of the high energetic electrons get
repelled back along the same path and the
photoelectric current in the circuit becomes
zero. So, for a particular frequency of incident
radiation, the minimum negative potential for
which the electric current becomes zero, is
called cutoff or stopping potential.
(1½)
Section D
23. One morning an old man walked bare-foot to
replace the fuse wire in kit kat fitted with the
power supply mains for his house. Suddenly he
screamed and collapsed on the floor. His wife
cried loudly for help. His neighbour’s son Anil
heard the cries and rushed to the place with shoes
on. He took a wooden baton and used it to switch
off the main supply.
Answer the following questions
(i) What is the voltage and frequency of mains
supply in India?
(ii) These days most of the electrical devices we
use require AC voltage. Why?
(iii) Can a transformer be used to step up DC
voltage?
(iv) Write two qualities displayed by Anil by his
action.
Sol. (i) In India, the voltage and frequency mains supply is
(1)
220 V and 60 Hz respectively.
(ii) Since, the power loss at 220 V supply is less that at
110 V. So due to this reason, most of the electrical
(1)
devices require AC voltage.
(iii) Since, the transformer requires an alternating
magnetic flux to operate correctly, transformers
cannot therefore be used to transform or supply
DC voltages or currents. Also, the magnetic field
must be changing to induce a voltage in the
secondary winding.
ne
Physics Class 12th
Even if in case, the transformers primary winding is
connected to a DC supply, the inductive reactance
of the winding. Would be zero as DC has no
frequency. So, the effective impedance of the
winding will therefore be very low and equal only to
the resistance of the copper used. Thus, the winding
will draw a very high current from the DC supply
causing it to overheat and eventually burnout,
V
(1)
because as we know I = .
R
(iv) Anil is very helpful and has a great presence of mind
as in such condition, he used the wooden baton to
(1)
switch off the main supply.
Section E
24. (i) Define electric flux. Write its SI unit.
“Gauss’s law in electrostatics is true for any
closed surface, no matter what its shape or
size is.” Justify this statement with the help of
a suitable example.
(ii) Use Gauss’s law to prove that the electric field
inside a uniformly charged spherical shell is
zero.
or
(i) Derive the expression for the energy stored in
parallel plate capacitor. Hence obtain the
expression for the energy density of the
electric field.
(ii) A fully charged parallel plate capacitor is
connected across an uncharged identical
capacitor. Show that the energy stored in the
combination is less than stored initially in the
single capacitor.
Sol. (i) Electric flux over an area in an electric field
represents the total number of electric field lines
crossing the area. The SI unit of electric flux is
Nm 2C -1 .
According to Gauss’s law in electrostatics, “The
surface integral of electrostatic field E produced by
any sources over any closed surface S enclosing a
volume V in vacuum, i.e. total electric flux over the
closed surface S in vacuum, is 1/ e 0 times the total
charge (q) contained inside S, i.e.
q
f E = ò E .ds =
e0
S
‘‘Gauss’s law in electrostatics is true for an closed
surface, no matter what its shape or size is.’’
So, In order to justify the above statement, suppose
in isolated positive charge q is situated at the centre
O of a sphere of radius r.

ll ne - Arihant Book

Transcription

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