Tuesday, August 27, 2024

Electromagnetic Induction _BSc

Electromagnetic Induction _BSC Electromagnetic InductionFaraday's Law"A change in the magnetic field within a closed loop of wire induces an electromotive force (emf) in the loop, causing an electric current to flow if the circuit is closed."
Let us consider a conductor moving with velocity V, over the another conductor. The motion of the conductor is perpendicular to the magnetic field B. The magnetic field in to the paper is denoted by ⊙ the magnetic field out of the paper is denoted by ⨂ (or towards the reader). When the conductor moves over the magnetic field B force is exerted on the electron present on the conductor. The direction of the force is given by Fleming's left hand rule. When the force F pushes the electrons from b to a, then the top part becomes positive and bottom part becomes negative. There exist an induced emf between b and a. Let that induced emf be denoted by Vba the potential and electric field, we can write E=-dv
dr
dv= -E.drv= -E.drWhere, E be the induced electric field over the distance dr moved by the force F. We know the force experienced by the elecctron (Charge) is given by,F= q ( v×B )F
q
=v × B
And induced electric field is E = F
q
So, E =v × B Then the potential isVba=-Edr=-(V×B)drThis potential gives rise to the electric current in the circuit. Therefore, one can easily change the limit.Vind =-(V×B)dr =-(dr
dt
×B)dr
=-d
dt
(r×dr)B
=-d
dt
dsB
where ds=r×dr gives surface area sweptout by the conductor. i.e. Vind =-d
dt
B.ds
or, Edr=-d
dt
Bds
....................(1)
But from the stokes theoremEdr=(V×E)ds ...............................................(2)From equation (1) and (2)(𝛻×E)ds=-d
dt
(Bds)
So, (𝛻×E)ds=(-𝜕B
𝜕t
)ds
(𝛻×E) =-𝜕B
𝜕t
.......................................(3)
Again, Vind=Eind. dr = - d
dt
B.ds =-d
dt
𝜙B
ie, Vind=-d𝜙B
dt
where, 𝜙B= B.dsThe motional induced voltage is equal to the negative rate of change of the magnetic flux, In this case, B is constant in time. The induced voltage develops in the conductor by making time variation of the magnetic field.Thus we can write, Vind= -∂𝜙
∂t
Which is called Faraday law of elecromagnetic induction. Lenz's law:
According to Lenz's law, the direction of induced current in a coil is such that it always opposes the cause which produces it. This law follows the law of conservation of energy i.e. Energy neither can be created nor can be destroyed but can change from one form to another. In lenz's law when the magnet is moved towards or away from coil, emf is induced in the coil at the expense of mechanical energy spent by the external agent. Hence in this way mechanical energy is converted to electrical energy. This show the Lenz's law is in accordance with the law of conservation of energy. Skin EffectIn electromagnetism, skin effect is the tendency of an alternating electric current (AC) to become distributed within a conductor such that the current density is largest near the surface of the conductor and decreases exponentially with greater depths in the conductor. It is caused by opposing eddy currents induced by the changing magnetic field resulting from the alternating current. The electric current flows mainly at the skin of the conductor, between the outer surface and a level called the skin depth.
Skin depth depends on the frequency of the alternating current; as frequency increases, current flow becomes more concentrated near the surface, resulting in less skin depth. Skin effect reduces the effective cross-section of the conductor and thus increases its effective resistance. Ballastic GalvanometerA ballistic galvanometer is a type of sensitive galvanometer.Unlike the current measuring galvanometer the moving part has large moment of inertia thus giving it a long oscillationperiod. It measure the charge discharge through it. It can be either moving coil or moving magnet type.The working principle of ballistic galvanometer is that it depends on the deflection of coil which is directly proportional to the charge passes through it. The galvanometer measures the majority of charge passes through it instead of current. The ballistic galvanometer consists of copper wire which is wound on nonconducting frame of galvanometer. For increasing the magnetic flux the iron core places within the coil. The lower portion of the coil connect the spring. This spring provides the restoring torque to the coil.
Consider a rectangular coil having N numbers of turn placed in a uniform magnetic field. Let 𝑙 be the length and 𝑏 be the breadth of the coil. Then the area of the coil is given by 𝐴 = 𝑙 × 𝑏 ............. (𝑖)When current passes through the coil the torque act on it. The magnitude of torque developed is 𝜏 = BINA .......................(𝑖𝑖)Let the current flow through the coil for the very short duration (𝑑𝑡) is 𝜏𝑑𝑡 = BINA𝑑𝑡 ......................(𝑖𝑖𝑖)If the current passing through the coil in time t second the equation (𝑖𝑖𝑖) becomes𝜏dt = BINA dt =BNAIdt = BANq Here, I = q
t
q=I.t
So, Idt = q is the total charge passes through the coil. The moment of inertia is given by 𝐼 and angular velocity by 𝜔 . By definition angular momentum is given by 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑀𝑜𝑚𝑒𝑛��𝑢𝑚 = 𝐼𝜔 ................. (𝑣)The angular momentum of the coil is equal the force acting on the coil. Thus from equation (𝑖𝑣) and (𝑣) we get 𝐼𝜔 = BAN𝑞 .............................(𝑣𝑖)The kinetic energy (𝐾) deflects the coil through the angle 𝜃 and this defection restored through the spring.restoring torque = 1
2
C𝜃2....................................(vii)
and, Kinetic energy = 1
2
2....................................(vii)
The restoring torque is equal to deflection. Thus we can write 1
2
C𝜃2=1
2
2..............................(viii)
C𝜃2=2................................(ix)The periodic oscillation is given byT = 2𝜋I
c
T2=4𝜋2I
c
T2c
4𝜋2
=I...............................(xi)
Multiplying Equation (𝑖𝑥) with (𝑥𝑖) we get thatT2c2𝜃2
4𝜋2
=I2ω2
Tc𝜃
2𝜋
=Iω......................(xii)
Putting the value of 𝐼𝜔 from equation 𝑖𝑣 in the above equation we get Tc𝜃
2𝜋
=BANq
q=Tc𝜃
2𝜋BAN
putting k= Tc
2𝜋BAN
q= k𝜃or, q 𝜃Therefore the amount of charge passing through the coil is directly proportional to the deflection of the coil. This is the principle of Ballistic Galvanometer. Search CoilExplain the Principle of Search coil. Describe how is it used to measure magnetic field. The search coil is a small flat coil of fine insulated wire with a large number of turns. When the coil is placed in a varying magnetic field an e.m.f is induced in it which is directly proportional to the flux density of the field. If this e.m.f. is measured the strength of the field may be found. The experimental arrangement for measuring field between the pole-pieces of a strong magnet. The search coil C is placed in the gap between the pole-pieces with its plane normal to the field. It is connected in series with a B.G., and secondary coil S of a solenoid. The primary coil P is connected through a reversing-key K to a battery, an ammeter A and a rheostat. Initially the primary circuit is kept off.
When the flux linked with search coil changes it produces emf in the circuit and galvanometer shows deflection θ. The flux linked with the search coil is NBA. The change in the flux d𝜙=NABInduced emf in the time dt=NAB
dt
Current I = ϵ
dt
= NAB
Rdt
The charge flowing in time dt= q = Idt=NAB
R
If charge flows through galvanometer then NAB
R
= k𝜃(1+𝜆
2
)
.........................(1)
where k and 𝜆 are ballistic constant and logarithm decrement of ballistic galvanometer respective. For determining k and 𝜆 the known current I is passed throught primary coil which is connected to rheostat. This produces the field inside the secondary coil. The flux linked with secondary coil 𝜙= 𝜇0npNsAsI where, np is number of turns per unit length in primary coil Ns is number of turns in secondary coil As is area of each turn of secondary coilWhen the direction of current in primary coil is suddenly reversed with the help of key, the total flux changed = 𝜇0npNsAsI-(-𝜇0npNsAsI) = 2 𝜇0npNsAsIAs above, the charge due to change in flux = 2 𝜇0npNsAsI
R
Let 𝜃1 be the deflection(throw) observed in the ballistic galvanometer when this charge passes through it then 2 𝜇0npNsAsI
R
=k𝜃1(1+𝜆
2
)
....................................(ii)
Dividing (10 by (2) we get, NAB
2 𝜇0npNsAsI
=𝜃
𝜃1
B = 2 𝜇0npNsAsI
NA
𝜃
𝜃1
Hence, B is determined. Flux MeterIt is used for measuring magnetic flux. It is a special type of ballistic galvanometer. In it the restoring torque is made negligibly small and the electromagnetic damping large. Construction. It consists of a coil D suspended by a single silk fiber. The coil is suspended between the pole-pieces N and S of permanent magnet. C is the usual iron core. The upper end of the fiber is attached to a flat spiral spring E. Spring E saves the fiber from mechanical shocks. Frame F, F carries a pointer which moves on a graduated scale. The current enters and leaves the coil by two fine silver spirals S1 and S2 which are connected to the terminals T1 and T2. A search coil of known cross-sectional area and number of turns is connected to the terminals T1 and T2.
N = No. of turns in fluxmeter coil. A = area of cross-section of fluxmeter coil. dθ/dt = ω = angular velocity of flux meter coil. B = magnetic field induction due to permanent magnet NS. Φ = flux from which search coil is withdrawn. It is to b e measured of fluxmeter.We know, Φ = NBAθ = kθ where k is a constant.Hence the total change in magnetic flux linked with the search coil is directly proportional to the deflection of the moving – coil and independent of time during which the change in flux takes place. Importance of Flux meter (TU)Flux meter is a modified moving coil ballistic galvanometer specially designed for the measeurement of magnetic flux or megnetic field strength.1. It is used to measure the magnetic flux and hence magnetic flux density.2. The modern flux meter has an AC mode which enables us to measure the fields over a wide range using simple sensing coil.3. It helps in analysing the material because of its feature of high resolution and low drift. Earth Inductor Earth inductor is an instrument used for the measurement of magnetic elements of earths. The purpose of an earth inductor is to generate an induced emf by virtue of the rotation of coil in earth’s magnetic field. Earth inductor is simply a circular coil consisting of a large number of turns of insulated Cu-wire wound over a wooden or ebonite frame. The coil is fitted in wooden frame. The coil can be rotated quickly through 180 about an axis passing through 180 about an axis passing through the centre of the coil and lying in its plane (i.e., about diameter AB). The frame itself can be rotated about an axis CD | to AB. Thus the axis of the rotation of the coil can be adjusted vertical or horizontal as desired. The ends of the coil are connected by means of flexible wired to two binding screws on the frame.
Consider a coil of N turns and cross-sectional area A placed in a uniform magnetic field of flux density B. At any instant t, the coil makes an angle 𝜃 to B.Then the deflection in the coil is given by2NBA
R
=k𝜃1(1+𝜆
2
)
Where K is the ballistic reduction factor and λ is the logarithmic decrement.The earth inductor can be used for the following purposes. Self and mutual inductances
When a change current (a.c) passing through a coil then the magnetic flux ɸ linked changes. The changes in the magnetic flux linked causes to induce emf in the coil itself. Therefore the phenomenon by which emf is induction in a coil by passing changing current through itself is called self induction.From faraday's second law of electromagnetic induction, the self inducted emf (𝜖) is proportional to the rate of change of flux 𝜙 linked with the coil i.e. 𝜖 𝛼 d𝜙
dt
………(1)
d𝜙
dt
is also propotional to the dI
dt
i.e. d𝜙
dt
𝛼 dI
dt
.. (2)
from equation (1) and (2) we have 𝜖 𝛼 dI
dt
or, 𝜖=-LdI
dt
where L is proportionality constant called coefficient of self induction.
Or, L=-e
dI
dt
if, -dI
dt
=1
L=𝜖Hence, Inductance (L) is also define as the emf induced for unit rate of decrease of current through itself.Self Inductance of Solenoid, toroid and two parallel conductors. 1. SolenoidA solenoid is a helical winding of an insulated wire on a cylinder, usually in eross-section. In this solenoid we are required to find the coefficient of self induction. For this let's consider a solenoid of length l having numbers of turns N as shown in figure. Let I be the current flowing through the solenoid, then magnetic field intensity inside it at any point is given by,B=𝜇0nI ...........................................(i) [a
where, n=N
l
is the
number of turns per
unit length
]
Magnetic flux (𝜙) is given by =N(BA) =NBA....................(ii) [where, A area of each turn] From equation (i) and (ii), we geta
𝜙=N(𝜇0nI)A
=N𝜇0NIA
l
[n=N
l
]
=𝜇0N2IA
l
By the definition of self inductance (L)=𝜙
I
a
L=𝜇0N2IA
I
=𝜇0N2A
l
L=𝜇0N2A
l
This is the required expression for the coefficient of selfinduction of solenoid.If the solenoid is wound over a core of constant permeability (𝜇)L=𝜇N2A
l
2. ToroidFinding the magnetic field inside a toroidal coil is a good example of Amper's, law. The current encloses by the dashed line is just the number of loops times current in each loop. Ampers law when gives the magnetic field at the centerline of toroid as B×2𝜋r=𝜇0NIor, B=𝜇0NI
2𝜋r
Now, From Faraday's law of electromagnetic induction.or, 𝜀=-Nd𝜙
dt
=-NdBA
dt
=-dLI
dt
It means, LI=NBAor, L=NBA
I
Substuting B then,or, L=NA
I
×𝜇0NI
2𝜋r
=𝜇0N2A
2𝜋r
Self inductance of toroid is 𝜇0N2A
2𝜋r
3. Two parallel conductorsLet us consider two long parallel wires PQ and RS separated by a distance d, the current flowing from P to Q in the wire PQ and R to S in the wire RS is I. Let us consider a point T in between two long parallel wires PQ and RS at a distance x from PQ then, the distance of the point T from anoher wire RS is (d-x).
The magnetic field at the point T due to the wire PQ is B1=𝜇0I
2𝜋x
and the magnetic field at the point T due to the wire RS isB2=𝜇0I
2n(d-x)
Hence, the total magnetic field at this point T is a
B=B1+B2
=𝜇0I
2𝜋x
+𝜇0I
2𝜋(d-x)
=𝜇0I
2𝜋
[1
x
+1
d-x
]
The total magnetic flux inside these wire (from one wire to anuther wire)𝜙=Bdx Since x is variable, B also should te variablea
Hence the total flux 𝜙=𝜇0I
2𝜋
[1
x
+1
d-x
]dx
𝜙=𝜇0I
2𝜋
[lnx-ln(d-x)]d0
=𝜇0I
𝜋
lnd
Hence, the induced emf is ϵ=-d𝜙
dt
=H0
𝜋
ln d(dt
dt
)=-Ldl
dt
Where L =𝜇aln d
𝜋
If the two long parallel wire on the two sides of the medium of permeability 𝜇 thenL=𝜇
𝜋
ln d
Mutual Induction
The induction of an emf in the coil due to the flow of current in the neighboring coil is called mutual induction. If S1 and S2 are the two coils place near to each other, the changing current I1 is set up in the coil S1,then the magnetic field around it also changes due to which flux linked with S2 is also changing if 𝜙2 is the flux linked with S2 then, then, the magnetic flux inside the solenoid S1is given by B1= 𝜇0n1I1 =𝜇0N1
l
I1
where, n1=N1
l
As, B=𝜙
A
the magnetic flux linked with each turns of solenoid S1is = B1Aand the total flux solenoid S2 is = N2B1Aie. 𝜙2=N2𝜇0N1
l
I1A
or, 𝜙2 = 𝜇0N1N2A
l
I1
or, 𝜙2= M12I1Where, M12=𝜇0N1N2
l
A
is the mutual inductance, when current fluctates in solenoid S1 and emf is induced in solenoid S2.
Simillarly, when current fluctated in solenoid S2 thenM21=𝜇0N1N2
l
A
so, M12=M21=Mi.e, Mutual inducttance of second coil with respect to first coil is equal to mutual inductance of first coil with respect to second coil. This is called Reciporcity Theorem.so,
M =𝜇0N1N2
l
A
Relation Between Self Induction and Mutual InductionLet us consider two toroid coil (1) and (2) with efffective length 'l' and number of turns 'N' and crosss section area 'A' in each coil. I1 and I2 be the current in coil (1) and (2).
The magnetic field due to coil (1) is B1=𝜇0NI1
l
And magnetic field due to coil (2) isB2=𝜇0NI2
l
The flux linked with the coil ( 1 ) due to current I1 in it is𝜙1=B1NA=𝜇0N2I1A
l
The flux linked with the coil (2) due to current I2 in it is𝜙2=B2NA=𝜇0N2I2A
l
The self inductance of coil (1) isL1=𝜙1
I1
=𝜇0N2I1A
I1l
=𝜇0N2A
l
Similarly, self inductance of second coil (2)L2=𝜙2
I2
=𝜇0N2A
l
Here L1L2=𝜇0N2A
l
×𝜇0N2A
l
or, (𝜇0N2A
l
)2= L1L2
𝜇0N2A
l
=L1L2
Flux linked with coil (2) due to current in coil (1)a
𝜙21=B1NA=𝜇0NI1
l
.NA
=𝜇0N2AI1
l
And the mutual inductance between the two coilM=𝜙21
I1
=𝜇0N2A
l
=L1L2
M=L1L2
this is the relation between inductance and self inductance.
Energy Stored in a InductorConsider an inductor of inductance L having initially zero current. It is assumed that an inductor has zero resistance so that there is no dissipation of energy with in inductor. Let I be the current at any instant of time so that dI/dt is the rate of change of current. Here current is increasing. The voltage between the terminals a and b of the inductor at this instant is Vab=Ldi
dt
and the rate P at which energy is being delivered to the inductor is given by
P=vabI=LdI
dt
I =LIdI
dt
The energy supplied to the inductor in small amount of time is small which is written asa
dU=Pdt ( P = dU
dt
)
=LIdI
dt
dt
dU= LI dI
To obtain total amount of energy stored in the inductor, we have to integrate it from 0 to I. Total energy stored in an inductor, a
i.e U=L.I.dI
or, U=L[I2
2
]I0=1
2
LI2
U=1
2
LI2
Transformer ransformers is a device used to convert high volt, low current AC into low volt high current AC or vice versa.
The transformer which converts high voltage AC into low voltage is called step down transformers. The transformers which converts low voltage AC into high voltage is called step up transformers. Transformer is based on the principle of mutual induction. When a magnetic flux is linked with a coil changes emf is induced in the nearby coil. It consists of two coild, primary and secondary of insulated copper wire on the laminated soft iron core. The soft iron core consists of thin strips of iron coated with vanish to insulated from each other and put together to form a single. The primary coil is connected with the source of AC and the secondary is connected with the load. When the current is passed through the primary coil, magnetic flux linked with the coil changes and emf is induced in the coil.If 𝜙 is the flux linked with the primary coil and Ep is itself induction then the primary voltage. Ep is given by:Ep=-NPd𝜙
dt
(i)
Where, Np is the number of turns in the primary coil. If there is no loss of flux then the rate of flux linked with the secondary coil will be same.Then the induced emf is secondary coil,Or, ES=NSd𝜙
dt
(ii)
Where, NS is the number of turns in secondary coil. Dividing equation (i) and (ii), we get,Or, EP
Es
=Np
Ns
= k
Where, Np
Ns
= k, is called transformer ratio
If NP/NS>1 then EP>ES and the transformer is step down.If NP/NS then ES>EP and the transformers in step up.If Lp and Ls is the coefficient of self - induction of primary and secondary coil respectively.For an ideal transformer, there is no any power loss.ie. Input power = Output power.Then,a
IpEp=ISES
Or, Ep
Es
=Is
Ip
.
Hence, the output voltage is increased, the output current will decrease and vice - versa. Therefore, for a long distance transmission line, high voltage is used. If the secondary voltage is high then current will be below and hence the energy loss (I2Rt) due to resistance of wire is low. Cause of Energy loss in Practical Transformer1. Copper LossesEnergy lost in windings of the transformer is known as copper loss. When current flows through copper wires, there is loss in power. This can be reduced by using thick wires for windings.2. Flux lossesIn the actual transformer, the coupling between primary and secondary coil is not perfect. So certain amount of electrical energy supplied to the primary coil is wasted.3. Iron losses a)Eddy current losses:When a changing magnetic flux links with the iron core of the transformer, eddy currents are set-up. These eddy currents in the iron core produce heat which leads to wastage of energy. This can be reduced by using laminated core.b. Hysteresis lossesWhen alternating current passes through the primary coil of the transformer, the iron core of the transformer is magnetized and demagnetized over a complete cycle. Some energy is lost during magnetizing and demagnetizing the iron core. The energy loss in a complete cycle is equal to the area of the hysteresis loop. This can be minimized by using the suitable material having narrow hysteresis loop for the core of a transformer.4. Losses due to vibration of coreA transformer produces humming noise due to magnetostriction effect. Some electrical energy is lost in the form of mechanical energy to produce vibration in the core.

Sunday, August 25, 2024

Activity 1 : Faraday's Law

 [to be done in practical sheets]

Activity 1: TO STUDY FARADAY LAW OF ELECTROMAGNETIC INDUCTION

REQUIREMENT: Phet simulation

THEORY
[Write in details about Faraday law of electromagnetic induction in detalis with figure]
.......................................
................................
.........................
.........................

PROCEDURE
[this is instruction for you, you should convert these instructions in to third person passive form]
1. Visit phet simulation website and open the Faraday's law simulation.
2. Observe the screen and find the different control icons to change the  number of loops in the coil, to change the polarity of magnet etc.
3. Pick up the magnet and insert in the coil and observe the deflection in voltmeter ( check the voltmeter and field lines box).
4. Next change the polarity of the magnet and observe the change in direction of deflection in voltmeter.
5. Change the speed of movement of magnet and observe the difference in amount of voltage induced.
6. Repeat the observation for amount of voltage by inserting the coil with same speed in  few turns and then coil of many turns.


OBSERVATION

Table 1: Observation of cause of electromagnetic induction

S.N

Action

Observation

Remarks

1

Magnet inserted in the coil

Field lines changes in the coil

Deflection occurs

2

Magnet kept at rest inside the coil

Field lines doesn't change

Deflection doesn't occurs

3.

Magnet taken out from coil

………………..

…………………….


Result 1. ………………………………………………………………………………….

Table 2: Observation for direction of current induced 

S.N

Action

Observation                  

Remarks

1

North pole of magnet inserted in coil

Deflection is in ........(right/left) side

Current in induced in one direction

2

South pole of magnet inserted in coil

Deflection is in ........ side

Deflection occurs in opposite direction




Result 2: Direction of induced current depends upon the direction of magnetic field.

Table 3: Observation for amount of current induced current with the speed of motion

S.N

Action

Observation

Remarks

1

Magnet moved slowly    

Deflection is ......... unit

.........(more/less) current is induced

2

Magnet moved quickly

Deflection is .......unit

........(more/less) current is induced




Result 3: Amount of current induced is directly proportional to ........................................


Table 4: Observation for amount of current induced current number of turns of coil

S.N

Action

Observation

Remarks

1

Magnet moved in coil of less turn 

Deflection is ......... unit

.........(more/less) current is induced

2

Magnet moved in coil of more turns

Deflection is .......unit

........(more/less) current is induced




Result 4: ........................................


CONCLUSION
The above experiment on simulation proves Faraday law of electomagnetic induction. Which can be concluded in following points:
1.......................
2.
3.
4



Wednesday, June 5, 2024

Meiosis

Meiosis Cell Division


Prophase-I occurs over a long duration and involves several complicated changes in meiotic cell division. It is important because genetic recombination and variation in sexually reproducing organism occurs due to the events of this phase.
Leptotene
(i) The chromatin network opens out and threads become clear.
(ii) The chromosomes are thin, slender and long.
(iii) Chromosome number is diploid.
Solution Image
Zygotene
(i) Corresponding chromosomes become intimately associated.
(ii) The process of pairing is known as synapse. It is so exact that pairing is not merely between corresponding chromosomes but between corresponding individual units.
(iii) The chromosomes become shorter and thicker.
Solution Image
Pachytene or Pachynema
(i) The synaptic chromosomes become very intimately associated.
(ii) The pair of chromosomes becomes short and thick.
(iii) Crossing over occurs at this stage. Chiasmata are clearly seen. 
Solution Image
Diplotene
(i) Homologous chromosomes start separating from one another.
(ii) Chiasmata tend to slip out of the chromosomes. This is known as terminalisation of chiasmata.
(iii) Chromosomes start separating out but the separation is not complete.
(iv) Nuclear membrane and nucleolus start disappearing. 
Solution Image
Diakinesis
(i) The bivalents condense further and get randomly distributed.
(ii) The separation of paired chromosomes is almost complete.
(iii) Terminalisation of chiasmata is almost complete.
(iv) Nuclear membrane and nucleolus disappear. 
Solution Image

Thursday, March 14, 2024

English Listining Nine

 Welcome, You will find here the study materials of class 9

English

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Chapter 17

Chapter 18

 Optional Science Book Janak Publication


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