Resonance Study Material Class 10

1  Pre-requisite : Before going through this chapter, you

should be thorough with the basic concepts of the chapter explained in X NCERT.

CONSERVATION OF ELECTRIC CHARGE

Whenever two bodies are charged by rubbing, one gets positively charged and the other gets negatively charged. The net charge on the two bodies, however, remains zero–the same as that before rubbing. In other

words, charge is conserved. It can neither be created nor be destroyed. The only thing that happens on rubbing is that charged particles (electrons) get transferred from one body to the other.

In some phenomena, charged particles are created. But even then the conservation of charge holds. For example, a free neutron converts itself into an electron and the proton taken together is also zero. So, there is no change in the conversion of a neutron to an electron and a proton.

COULOMB’S LAW

Charles Augustine de Coulomb studied the interaction forces of charged particles in detail in 1784. He used a torsion balance. On the basis of his experiments he established Coulomb’s law. According to this law the

magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of the two charges and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.

In mathematical terms, the force that each of the two point charges q₁ and q₂ at a distance r apart exerts on

the other can be expressed as–

F = 1₂2 r

q q k

This force is repulsive for like charges and attractive for unlike charges.

Where k is a constant of proportionality. k =

0 4

1

 , here 0 is absolute permittivity of free space.

The force is directed along the line joining the centres of the two charged particles.

For any other medium except air, free space or vacuum coulomb’s law reduces to

F = 1 2₂

q q

1

4



r

 = Permittivity of the medium

and = ₀_r

r = relative Permittivity or dielectric constant of the

me-dium.

Coulomb’s law is based on physical observation and

it is not logically derived from any other concept.

ILLUSTRATIONS

1. Find out the electrostatics force between two point

charges placed in air (each of +1 C) if they are separated by 1m . Sol. F_e = 1₂2 r q kq = ₂ 9 1 1 1 10 9   = 9×109 N

Note : From the above result we can say that 1 C

charge is too large to realize. In nature, charge is usually of the order of C

2. A particle of mass m carrying charge q₁ is revolving around a fixed charge –q2 in a circular path of radius

r. Calculate the period of revolution and its speed also. Sol. 0 4 1  2 2 1 r q q = mr2 = ₂ 2 T mr 4 ' T2 ₌ 2 1 2 2 0

q

q

)

mr

4

(

r

)

4

(





and also we can say that

2 0 2 1 r 4 q q  =

r

mv

PROPERTIES OF ELECTRIC FIELD INTENSITY

(i) It is a vector quantity. Its direction is the same

as the force experienced by positive charge.

(ii) Electric field due to positive charge is always

away from it while due to negative charge always towards it.

(iii) Its S.. unit is Newton/Coulomb.

(iv) Electric force on a charge q placed in a region

of electric field at a point where the electric field intensity is E is given by FqE.

Electric force on point charge is in the same direction of electric field on positive charge and in opposite direction on a negative charge.

ELECTRICITY

(v) It obeys the superposition principle, that is, the

field intensity at a point due to a system of charges is vector sum of the field intensities due to individual point charges. 3 2 1

E

E

E

E















(vi) It is produced by source charges. The electric

field will be a fixed value at a point unless we change the distribution of source charges.

3. Five point charges, each of value q are placed on five vertices of a regular hexagon of side L. What is the magnitude of the force on a point charge of value –q

coulomb placed at the centre of the hexagon?

q q C q q D E O -q q B A F

L

Sol. If there had been a sixth charge +q at the remaining

vertex of hexagon force due to all the six charges on –

q at O would be zero (as the forces due to individual charges will balance each other), i.e.,

0 FR 

Now if f is the force due to sixth charge and F due to remaining five charges.

F + f = 0 i.e. _F = –f  or, |F| = |f| = 0 4 1  L2 q q = ₂ 2 0 L q 4 1  Net F = F_CO = ₂ 2

L

q

4

1





4. Calculate the electric field intensity which would be just sufficient to balance the weight of a particle of charge –10 c and mass 10 mg.

Sol. As force on a charge q in an electric field E is

F_q = qE

So according to given problem

q E W A Fe | W | | F |_q   i.e., |q|E = mg i.e., E = | q | mg = 10 N/C., in downward direction. ELECTROSTATIC EQUILIBRIUM

The position where the resultant force on a charged particle becomes zero is called equilibrium position.

( a ) St a ble Equilibrium :

A charge is initially in equilibrium position and is displaced by a small distance. If the charge tries to return back to the same equilibrium position then this equilibrium is called position of stable equilibrium.

( b) Uns t a b le Equ ilibri um :

If charge is displaced by a small distance from its equilibrium position and the charge has no tendency to return to the same equilibrium position. Instead it goes away from the equilibrium position.

( c) Ne ut ral Equilibrium :

If charge is displaced by a small distance and it is still in equilibrium condition then it is called neutral equilibrium.

5. Two equal positive point charges 'Q' are fixed at points B(a, 0) and A(–a, 0). Another test charge q0 is also

placed at O(0, 0). Show that the equilibrium at 'O' is (i) stable for displacement along X-axis.

(ii) unstable for displacement along Y-axis.

Sol. (i) Initially F_AO + F_BO = 0 |F_AO|  = |F_BO| = 2 0

a

KQq

When charge is slightly shifted towards + x axis by a small distance x, then.

| F

3

Therefore the particle will move towards origin (its original position) hence the equilibrium is stable. (ii) When charge is shifted along y axis

After resolving components net force will be along y axis so the particle will not return to its original position so it is unstable equilibrium. Finally the charge will move to infinity.

ELECTRIC LINES OF FORCE (ELOF)

The line of force in an electric field is an imaginary line, the tangent to which at any point on it represents the direction of electric field at the given point.

(a) Properties :

(i) Line of force originates out from a positive

charge and terminates on a negative charge. If there is only one positive charge then lines start from positive charge and terminate at . If there

is only one negative charge then lines start from

 and terminates at negative charge.

(ii) The electric intensity at a point is the number

of lines of force streaming through per unit area normal to the direction of the intensity at that point. The intensity will be more where the density of lines is more.

(iii) Number of lines originating (terminating) from

(on) is directly proportional to the magnitude of the charge.

(iv) ELOF of resultant electric field can never intersect

with each other.

(v) Electric lines of force produced by static

charges do not form close loop.

( vi ) E l e c t r i c l i n e s o f f o r c e e n d o r s t a r t

perpendicularly on the surface of a conductor.

(vii) Electric lines of force never enter into

conductors.

6. If number of electric lines of force from charge q are 10 then find out number of electric lines of force from 2q charge.

Sol. No. of ELOF  charge 10  q

 20  2q

So number of ELOF will be 20.

7. A charge + Q is fixed at a distance of d in front of an infinite metal plate. Draw the lines of force indicating the directions clearly.

Sol. There will be induced charge on two surfaces of

conducting plate, so ELOF will start from +Q charge and terminate at conductor and then will again start from other surface of conductor.

ELECTRIC FLUX

Consider some surface in an electric field _E. Let us select a small area element _dS on this surface. The electric flux of the field over the

area element is given by dE =

 

ds . E

Direction of _dS is normal to the surface. It is along

n

ˆ

or dE = EdS cos 

or dE = (E cos ) dS

dS

E

or dE = En dS

where E_n is the component of electric field in the direction of _dS.

If the electric field is uniform over that area then

E = E S  



(a) Phys ical Meaning :

The electric flux through a surface inside an electric field represents the total number of electric lines of force crossing the surface in a direction normal the surface. It is a property of electric field

(b) Unit :

(i) The SI unit of electric flux is Nm2 _C–1 _{(gauss) or J} m2 _C–1_.

(ii) Electric flux is a scalar quantity. (It can be positive, negative or zero)

8. The electric field in a region is given by,

j E 5 4 i E 5 3 E  ₀ ₀ with E0 = 2.0 × 10 3 _{N/C. Find the}

flux of this field through a rectangular surface of area 0.2m2 _{parallel to the Y} –Z plane. Sol. E = E S    =        E j 5 4 i E 5 3 0 0   .

_

_

ELECTRIC POTENTIAL ENERGY

Consider a charge Q placed at a point P as shown in figure. If another charge q of the same sign is now brought from a very far away distance (infinity) to point O near P, then charge q will experience a force of repulsion due to charge Q. If charge q is still pushed towards P, work is done. This work done is the potential energy of the system of these two charges.

Q

P

q

O

r

Thus, the electric potential energy of a system of charges is defined as the amount of work done in bringing the various charges from infinite separation to their present positions to form the required system. It is denoted by U. For the system of two charges separated by distance r as shown in figure, the electric potential energy is given by :

U = r kQq

Electric potential energy is the from of energy, therefore it is measured in joule (J).

SUPER CONDUCTOR AND ITS APPLICATIONS

Prof. K. Onnes in 1911 discovered that certain metals and alloys at very low temperature lose their resistance considerably. This phenomenon is known as super-conductivity. As the temperature decreases, the resistance of the material also decreases, but when the temperature reaches a certain critical value (called critical temperature or transition temperature), the

resistance of the material completely

disappears i.e. it becomes zero. Then the material behaves as if it is a super-conductor and there will be flow of electrons without any resistance whatsoever. The critical temperature is different for different material. It has been found that mercury at critical temperature

4.2 K, lead at 7.25 K and niobium at critical temperature 9.2 K become super-conductor.

Applications of super conductors :

(i) Super conductors are used for making very strong

electromagnets.

(ii) Super conductivity is playing an important role in

material science research and high energy particle physics.

(iii) Super conductivity is used to produce very high

speed computers.

(iv) Super conductors are used for the transmission of

5 CELL

It converts chemical energy into electrical energy. Electrochemical cells are of three types :

(a) Primary cell (b) Secondary cell

(c) Fuel cell

(a) Primary Cell :

It is an electrochemical cell, which cannot be recharged, but the chemicals have to be replaced after a long use. The reactions taking place in the cell are irreversible.

Eg. : Daniel cell, Lechlanche cell, Dry cell etc. ( b) Sec onda ry C e ll :

Electrical energy can be converted into chemical energy and chemical energy can be converted into electrical energy in these cells, i.e. secondary cells can be recharged after use. Chemical reaction taking place in these cells are reversible.

Example : Lead accumulator, Edison cell (alkali cell)

and iron nickel cell.

(c) Electro Motive Force of a Cell (E.M.F.) :

It is the maximum potential difference between the two electrodes of the cell when no current is drawn from the cell or cell is in the open circuit.

(d) Potential Difference of a Cell :

It is the difference of potential between two terminals of the cell when current is drawn from it or the cell is in closed circuit.

(e) Int erna l Re sista nce of a Cell :

It is the resistance offered to the flow of current inside the cell i.e. internal resistance is the resistance offered to the flow of current by electrolyte. Internal resistance decreases with the increase of the area of plates and also with the decrease of the distance between plates.

Determination of internal resistance of a cell :

Connect a voltmeter to a cell through key K₁. Also connect a resistor R to cell through K₂. First put in key K₁. The reading shown by voltmeter gives us the e.m.f. of the cell since negligible current flows through cell due to high resistance of the voltmeter. Insert key K₂

also so that current flows through resistor R. If r is the internal resistance of the cell and V is the reading shown by voltmeter, then

I = r R E   E = I (R + r)  E = IR + Ir

Here, IR = V the potential difference

So, E = V + r r = I V – E ...(i)  V = IR or I = _R V

So for equation (i)

r = V V)R – (E ...(ii) GROUPING OF CELLS

(a) Cells in Series :

B E ,r1 1 E ,r2 2E ,r3 3 E rn n

A B

E ,req eq Equivalent EMF

E

Equivalent internal resistance r_eq = r₁ + r₂ + r₃ + r₄ + ... r_n

If n cells each of emf E, are arranged in series and if r is internal resistance of each cell, then total emf = nE

R E,r E,r E,r E,r

I Upto n

A B

So current in the circuit, I =

nr R

nE



There may by two cases :

(i) If nr << R, then I =

R nE

= n × current due to one cell. So, series combination is advantageous.

(ii) If nr >> R, then I =

r E

= current due to one cell.

So, Series combination is not advantageous. Note : If polarity of m cells is reversed, then equivalent

e.m.f. = (n–2m) E while the equivalent resistance is

still nr + R, so current in R will be

i = R nr E ) m 2 n (  

(b) Ce lls in Para llel :

If m cells each of emf E and internal resistance r be connected in parallel and if this combination is connected to an external resistance then the emf of the circuit is E.

Internal resistance of the circuit =

m r . and I = r mR mE m r R E   

There may by two cases :

(i) If mR << r, then I =

r mE

= m × current due to one cell. So, Parallel combination is advantageous.

(ii) If mR >> r, then I =

R E

= current due to one cell.

So, parallel combination is not advantageous.

If emf and internal resistances of each cell are different, then, E_eq = n 2 1 n n 2 2 1 1 r 1 r 1 r 1 r E r E r E / ... / / / ... / /    

for two cells E =

2 1 1 2 2 1 r r r E r E  

(Use emf with polarity)

E

E

E

E

(c) Cells in Multiple Arc : n = number of rows

m = number of cells in each row.

mn = N (total number of identical cells) :

The combination of cells is equivalent to single cell of

emf = mE and internal resistance =

n mr Current I = n mr R mE 

For maximum current, nR = mr

or R =

n mr

= internal resistance of the equivalent battery. I_max = R 2 mE r 2 nE  .

using mn = N in above equation we get number of

rows n =

R Nr

9. 9 cells, each having the same emf and 3 ohm internal

resistance, are used to draw maximum current through an external resistance of 3 ohm. find the combination of cells.

Sol. For the condition of maximum current number of rows

n = R Nr so n = 3 3 9 = 3

so combination will be like 3 rows and 3 cells in each row.

BATTERY

Battery is an arrangement that creates a constant potential difference between its terminals. It is a combination of a number of cells in series.

The impact of battery :

With the discovery of voltaic cell, it was soon realised that if one constructs a number of cells and joins the negative terminal of one with the positive terminal of the other and so on, then the emf (which is the potential difference between the electrodes in an open circuit) of the combination of cells will be the sum of the emf’s

of the individual cells. This observation led to a burst of scientific activity in 1802. Humphrey Davy, an English chemist, made a battery of 60 pairs of zinc and copper plates. The large emf thus produced, was used to get high current, which could melt iron and platinum wires. By 1807, he had a battery of almost 300 plates with which he was able to decompose chemical salts. This led to the discovery of new elements.

7

By 1808, Davy had assembled 2,000 pairs of plates. W ith this battery, he created electric arcs and succeeded in extracting the elements like barium, calcium and magnesium from their compounds. Thus, electricity took a front seat in exploring the nature of matter.

ELECTRICAL RESISTANCE

The property of a substance by virtue of which it opposes the flow of electric current through it is termed as electrical resistance. Electrical resistance depends on the size, geometry, temperature and internal structure of the conductor.

We known that, v_d = _ m eE =  m eV I = Anevd = Ane  m eV I =  m V Ane2    ₂ Ane m I V  R =   ₂ Ane m I V  R =  2 Ane m R = A    =  RA =  2 ne m

 is called resistivity (it is also called specific

resis tance), and =

 2 ne m =  1 ,  is cal led

conductivity. Therefore current in conductors is proportional to potential difference applied across

its ends. This is Ohm's Law. Units: 1 1

m 

   also called siemens m–1_.

10. If a copper wire is stretched to make its radius

decrease by 0.15%, Find the percentage increase in resistance (approximately).

Sol. Due to stretching resistance changes are in the ratio 4 2 1 1 2 r r R R          or

R

r



EFFECT OF STRETCHING OF A WIRE ON RESISTANCE

In stretching, the density of wire usually does not change. Therefore

Volume before stretching = Volume after stretching

2 2 1 1A  A   and 2 1 1 2 1 2 A A R R    

If information of lengths before and after stretching

is given, then use

1 2 2 1 A A    2 1 2 1 2 R R           

If information of radius r₁ and r₂ is given then use

2 1 1 2 A A    2 2 1 1 2 A A R R          4 2 1 r r          CONDUCTIVITY :

(a) Reciprocal of resistivity of a conductor is called

its conductivity. It is generally represented by _.

(b)







1

(c) Unit :

ohm

.

metre

EFFECT OF TEMPERATURE ON RESISTANCE

AND RESISTIVITY

The resistance of a conductor depends upon the temperature. As the temperature increases, the random motion of free electrons also increases. If the number density of charge carrier electrons remains constant as in the case of a conductor, then the increase of random motion increases the resistivity. The variation of resistance with temperature is given by the following relation





0

t

R

1

t

t

where R_t and R₀ are the resistance at t0_{C and 0}0_C

respectively and _ and



constant







This constant  is called as temperature coefficient of resistance of the substance.

If R₀ = 1 ohm, t = 10_{C, then}



_

_

Thus, the temperature coefficient of resistance is equal to the increase in resistance of a conductor having a resistance of one ohm on raising its temperature by 10_{C. The temperature coefficient of}

resistance may be positive or negative.

From calculations it is found that for most of the metals the value of _ is nearly / C

273 1 0

. Hence substituting  in the above equation

        273 t 1 R Rt 0 273 T R 273 t 273 R₀  ₀       

where T is the absolute temperature of the conductor.

 Rt T

Thus, the resistance of a pure metallic wire is directly proportional to its absolute temperature.

The graph drawn between the resistance R_t and temperature t is found to be a straight line

Rt

tºC

R0

The resistivity or specific resistance varies with temperature. This variation is due to change in resistance of a conductor with temperature. The dependence of the resistivity with temperature is represented by the following equation.







With the rise of temperature the specific resistance or resistivity of pure metals increases and that of semi-conductors and insulators decrease.

The resistivity of alloys increases with the rise of temperature but less than that of metals.

On applying pressure on pure metals, its resistivity decreases but on applying tension, the resistivity increases.

The resistance of alloys such as eureka, manganin etc., increases in smaller amount with the rise in temperature. Their temperature coefficient of resistance is negligible. On account of their high resistivity and negligible temperature coefficient of resistance these alloys are used to make wires for resistance boxes, potentiometer, meter bridge etc., The resistance of semiconductors, insulators, electrolytes etc., decreases with the rise in temperature. Their temperature coefficients of resistance are negative.

On increasing the temperature of semi conductors a large number of electrons get free after breaking their bonds. These electrons reach the conduction band from valence band. Thus conductivity increases or resistivity decreases with the increase of free electron density.

11. A wire has a resistance of 2 ohm at 273 K and a

resistance of 2.5 ohm at 373 K. What is

the temperature coefficient of resistance of the material? Sol.









Wheatstone bridge is an arrangement of four resistors in the shape of a quadrilateral which can be used to measure unknown resistance in terms of the remaining three resistances.

The arrangement of Wheatstone bridge is shown in figure below. Out of four resistors, two resistances R₁,

R₂ and R₃, R₄ are connected in series and are joined in parallel across two points a and c. A battery of emf E is connected across junctions a and c and a galvanometer (G) between junction b and d. The keys K₁ and K₂ are used for the flow of current in the various branches of bridge.

Principle of Wheatstone Bridge :

When key K₁ is closed, current i from the battery is divided at junction a in two parts. A part i₁ goes through

R₁ and the rest i₂ goes through R₃. When key K₂ is closed, galvanometer shows a deflection.

9

The direction of deflection depends on the value of potential difference between b and d. When the value of potential at b and d is same, then no current will flow through galvanometer. This condition is known as the condition of balanced bridge or null deflection condition. This situation can be obtained by choosing suitable values of the resistances. Thus, in null deflection state, we have :

V_a– Vb = Va– Vd

or i₁ R₁ = i₂ R₃ ...(i)

Similarly :

V_b– Vc = Vd– Vc

or i₁ R₂ = i₂ R₄ ...(ii)

On dividing equation (i) by (ii), we get

2 1 1 1 R i R i = 4 2 3 2 R i R i or 4 3 2 1 R R R R  ...(iii)

Equation (iii) states the condition of balanced bridge. Thus, in null deflection condition the ratio of resistances of adjacent arms of the bridge are same. The resistor of unknown resistance is usually connected in one of the arm of the bridge. The resistance of one of the remaining three arms is adjusted such that the galvanometer shows zero deflection. If resistance of unknown resistor is R₄. Then

R₄ = (R₃) _       1 2 R R

For better accuracy of the bridge one should choose resistances R₁, R₂, R₃ and R₄ of same order.

GALVANOMETER

Galvanometer is a simple device, used to detect the current, to find direction of current and also to compare the currents.

With the help of galvanometer we make two important devises known as Ammeter and voltmeter as discussed below.

( a ) Amme te r :

Ammeter is an electrical instrument which measures the strength of current in ‘ampere’ in a circuit. Ammeter

is a pivoted coil galvanometer which is always connected in series in circuit so that total current (to be measured) may pass through it. For an ammeter of good quality, the resistance of its coil should be very low so that it may measure the strength of current accurately (without affecting the current passing through the circuit). The resistance of an ideal ammeter is zero (practically it should be minimum). So, to minimize the effective resistance of an ammeter, a low value resistance (shunt) as per requirement is connected in parallel to the galvanometer to convert it to ammeter of desired range.

In electric circuit, the positive terminal of an ammeter is connected to positive plate and negative terminal is connected to negative plate of battery.

Desired value of shunt depends on the range (measurable maximum current) of ammeter converted from galvanometer.

If pivoted galvanometer of resistance G is to measure current i (as an ammeter) then from figure.

G i ig ig i is S i_g G = (i – ig) S or S =

₍

_i

_i

₎

G

i



Where i_g is an amount of current required for full deflection in galvanometer. By using a low value of resistance S (shunt) in parallel to the galvanometer (resistance G), the effective resistance of converted ammeter R_A =

S) (G

GS

 becomes very low..

 NOTE :

Shunt : If anyhow, the flowing current through

galvanometer becomes more than its capacity, the coil has possibility of burning due to heat produced by flowing current. Secondly, its pointer may break up due to impact with ‘stop pin’ as its proportional deflection

as per amount of flowing current.

In order to minimize these possibilities a low resistance wire (or strip) is connected in parallel with galvanometer, which is known as shunt.

(b) Volt me ter :

It is an electrical instrument which measures the potential difference in ‘volt’ between two points of

electric circuit. It’s construction is similar as that of

ammeter. The only difference between ammeter and voltmeter is that ammeter has its negligible (approximately zero) resistance so that it may measure current of circuit passing through it more accurately giving the deflection accordingly, while the voltmeter passes negligible current through itself so that potential difference developed due to maximum current passing

through circuit may be measured. Therefore, an appropriate value of high resistance is required to be connected in series of galvanometer to convert it into a voltmeter of desired range.

Voltmeter is connected in parallel to the electric circuit. If a galvanometer of resistance G is to be converted into a voltmeter of range V, then required value of high resistance R_H will be V = i_g (R_H + G) or R_H =

_















V

I

Connecting this value of high resistance in the series of galvanometer, it will be converted to a voltmeter of range V. After connecting high resistance R_H in series

of galvanometer of resistance G, the effective resistance of voltmeter becomes R_V = (R_H + G) very high (high in

comparison to G).

Ideal voltmeter has infinite resistance of its own. When ideal voltmeter is connected parallel to a part of an electric circuit, it passes zero amount of current through itself from the circuit so that measurement of potential difference across the points of connection may be perfectly accurate.

KIRCHHOFF'S LAWS

(a) Kirchhoff’s Current Law (Junction law) :

This law is based on law of conservation of charge. It states that "The algebraic sum of the currents meeting at a point of the circuit is zero" or total current entering a junction is equal total current leaving the junction.

in = out.

It is also known as KCL (Kirchhoff's current law).

( b ) K i rch h o f f ’s Vo l t a ge L aw ( L o o p l a w ) : “The algebraic sum of all the potential differences

along a closed loop is zero. IR + EMF =0”. The

closed loop can be traversed in any direction. While traversing a loop if potential increases, put a positive sign in expression and if potential decreases put a negative sign.

V1 V2 + V3 V4 = 0. Boxes may contain resistor

or battery or any other element (linear or nonlinear). It is also known as KVL

12. Figure shows, current in a part of electrical circuit, what

will be the value of current (i) ?

2 A 1 A 1.3 A i 3 A 2 A P Q R S Sol. 2 A 1 A 1.3 A i 3 A 2 A P Q R S i 1 i 2 i 3

From KCL, current at junction P, i₁ = 2 + 3 = 5 A From KCL, current at junction Q, i₂ = i₁ + 1 = 5 + 1 = 6 A From KCL, current at junction R, i₃ = i₂– 2 = 6 – 2 = 4 A

From KCL, current at junction S, i = i₃– 1.3 = 4 – 1.3

= 2.7 A

13. In the circuit shown, calculate the value of R in ohm

that will result in no current through the 30 V battery.

Sol. Applying KVL in loop CEFDC

50 – iR – 20 i = 0 i = R 20 50 

20

R

50V

10

i

i

_E

F

D

i

B

A

C

Potential drop across R = Potential drop across AB iR = 30  R 20 50  .R = 30  R = 30 

11

EXERCISE

Charge and coulomb’s Law :

1. Conductivity of superconductor is :

(A) infinite (B) very large

(C) very small (D) zero

2. Two charges of +1 C & + 5 C are placed 4 cm apart,

the ratio of the force exerted by both charges on each other will be

-(A) 1 : 1 (B) 1 : 5

(C) 5 : 1 (D) 25 : 1

3. A body has 80 microcoulomb of charge. Number of additional electrons on it will be :

(A) 8 x 10–5 _{(B) 80 x 10}15

(C) 5 x 1014 _{(D) 1.28 x 10}–17

4. Which of the following relation is wrong ? (A) Q = It (B) 1 ampere = Second 1 Coulomb 1 (C) V = Wq (D) V = q W

5. Two particles having charges q₁ and q₂ when kept at a certain distance, exert force F on each other. If distance is reduced to half, force between them becomes : (A) 2 F (B) 2F (C) 4F (D) 4 F 6. 25 4

Coulomb of charge contains... electrons :

(A) 1015 _{(B) 10}18

(C) 1020 _{(D) none of these}

7. 5 charges each of magnitude 10–5 _{C and mass 1 kg}

are placed (fixed) symmetrically about a movable central charges of magnitude 5 × 10–5_{C and mass 0.5}

kg as shown. The charges at P₁ is removed. The

acceleration of the central charge is : (KVPY/2009)

P

P

P

P

P

O

(A) 9 m s–2 _{upwards (B) 9 m s}–2 _downwards

(C) 4.5 m s–2 _{upwards (D) 4.5 m s}–2 _downwards

8. 12 positive charges of magnitude q are placed on a circle of radius R in a manner that they are equally spaced. A charge +Q is placed at the centre. If one of the charges q is removed, then the force on Q is :

(KVPY/2010) (A) zero (B) 2 0R 4 qQ

 away from the position of the removed

charge. (C) 2 0R 4 qQ 11

 away from the position of the removed

charge.

(D) 2

0R

4 qQ

 towards the position of the removed

charge.

9. In a neon discharge tube 2.8 × 1018 Ne+ ions move to

the right per second while 1.2 ×1018 electrons move to

the left per second. Therefore , the current in the

discharge tube is : (IJSO/Stage-I/2011)

(A) 0.64 A towards right (B) 0.256 A towards right (C) 0.64 A towards left (D) 0.256 A towards left

Electric filed and Potential :

10. If Q = 2 coloumb and force on it is F = 100 newton,

then the value of field intensity will be:

(A) 100 N/C (B) 50 N/C

(C) 200 N/C (D) 10 N/C

11. In the electric field of charge Q, another charge is

carried from A to B. A to C, A to D and A to E, then work done will be -B Q + A C D E centre

(A) minimum along path AB. (B) minimum along path AD. (C) minimum along path AE. (D) zero along all the paths.

12. A negatively charged particle initially at rest is placed in

an electric field that varies from point to point. There are no other fields. Then : (KVPY/2008) (A) the particle moves along the electric line of force passing through it.

(B) the particle moves opposite to the electric line of force passing through it.

(C) the direction of acceleration of the particle is tangential to the electric line of force at every instant. (D) the direction of acceleration of the particle is normal to the electric line of force at every instant.

13. Two charges +q and –q are placed at a distance b

apart as shown in the figure below. (KVPY/2009)

b/2 b +q –q B A P C

The electric field at a point P on the perpendicular bisector as shown as :

(A) along vector

A







14. Two charges +Q and _2Q are located at points A and B

on a horizontal line as shown below :

The electric field is zero at a point which is located at a

finite distance : (KVPY/2011)

(A) On the perpendicular bisector of AB (B) left of A on the line

(C) between A and B on the line (D) right of B on the line

Resistance :

15. There are three resistance 5, 6and 8connected in

parallel to a battery of 15 V and of negligible resistance. The potential drop across 6resistance is :

(A) 10 V (B) 15 V

(C) 20 V (D) 8 V

16. In the given circuit, the equivalent resistance between

points A and B will be.

(A)

3 8

R (B) 4R

(C) 6R (D) 10R

17. Resistance of a conductor of length 75 cm is 3.25 .

What will be the length of a similar conductor whose resistance is 13.25 ?

(A) 305.76 cm (B) 503.76 cm

(C) 200 cm (D) 610 cm

18. A piece of wire of resistance 4 is bent through 1800

at its mid point and the two halves are twisted together, then resistance is :

(A) 1  (B) 2 

(C) 5  (D) 8 

19. In how many parts (equal) a wire of 100  be cut so that a resistance of 1  is obtained by connecting

them in parallel ?

(A) 10 (B) 5

(C) 100 (D) 50

20. If a wire of resistance 1 is stretched to double its

length, then the resistance will become : (A) 2 1  (B) 2  (C) 4 1  (D) 4 

21. Two copper wires, one of length 1 m and the other of

length 9 m, are found to have the same resistance. Their diameters are in the ratio :

(A) 3 : 1 (B) 1 : 9

(C) 9 : 1 (D) 1 : 3

22. Reading of ammeter in ampere for the following circuit

is :

(A) 0.8 (B) 1

13 23. Two resistors are joined in series, then their equivalent

resistance is 90 . When the same resistors are

joined in parallel, the equivalent resistance is 20 .

The resistances of the two resistors will be :

(A) 70 , 20  (B) 80 , 10 

(C) 60 , 30  (D) 50 , 40 

24. In the ladder network shown, current through the

resistor 3 is 0.25 A. The input voltage ‘V’ is equal

to (A) 10 V (B) 20 V (C) 5 V (D) 2 15 V

25. The reading of voltmeter is

(A) 50V (B) 60 V (C) 40V (D) 80 V 26. 2 5 25 10 1.4A 1.4A A (A) 0.4 (B) 1 (C) 0.6 (D) 1.2

27. Three identical bulbs are connected in parallel with a

battery. The current drawn from the battery is 6 A. If one of the bulbs gets fused, what will be the total current drawn from the battery ?

(A) 6A (B) 2A

(C) 4A (D)

28. A uniform wire of resistance R is uniformly compressed

along its length, until its radius becomes n times the original radius. Now, the resistance of the wire becomes :

(A) R/n (B) R/n4

(C) R/n2 _{(D) n R}

29. The resistance of a wire of cross-section ‘a’ and length ‘ ’ is R ohm. The resistance of another wire of the

same material and of the same length but cross-sec-tion ‘4a’ will be

(A) 4R (B) R

4

(C) R

16 (D) 16 R

30. In the following circuit the value of total resistance

be-tween X and Y in ohm is :

r to r r r r r r r r X Y  (A) (1 +

₃

₃

31. Wires A and B are made from the same material. Wire

A has length 12m and weight 50 g, while wire B is 18 m long and weighs 40 g. Then the ratio (R_A / R_B) of their

resistances will be : (IAO/Jr./Stage-I/2008)

(A) 16 / 45 (B) 4 / 5

(C) 8 / 15 (D) 4 / 9

32. In case of the circuit arrangement shown below, the

equivalent resistance between A and B is :

(IAO/Jr./Stage-I/2009)

A

B

W (D) None of the above

33. The net resistance between points P and Q in the circuit shown in fig. is

(A) R/2 (B) 2R/5

34. A wire of resistance 10.0 ohm is stretched so as to

increase its length by 20%. Its resistance then would

be : (IAO/Sr/Stage-I/2008)

(A) 10.0 ohm (B) 12.0 ohm

(C) 14.4 ohm (D) 10.2 ohm

35. In the circuit shown below, all the resistances are equal,

each equal to R. The equivalent resistance between

points A and C is : (IAO/Sr./Stage-I/2009)

R

R

R

R

A

D

C

B

(C) R /2 (D) none of the above

36. A battery or 10 V and negligible internal resistance is

connected across the diagonally opposite corner of a cubical network consisting of 12 resistors each of resistance 1 . The total current 1 in the circuit external

to the network is : (KVPY/2007)

A

10V

(A) 0.83 A (B) 12 A

(C) 1 A (D) 4 A

37. Figure (a) below shows a Wheatstone bridge in which

P, Q, R, S are fixed resistances, G is a galvanometer and B is a battery. For this particular case the galvanometer shows zero deflection. Now, only the positions of B and G are interchanged,. as shown in figure (b). The new deflection of the galvanometer.

(KVPY/2010)

(A) is to the left. (B) is to the right. (C) is zero.

(D) depends on the values of P, Q, R, S

38. In the circuit arrangement shown, if the point A and B are

joined by a wire the current in this wire will be :

(IJSO/Stage-I/2011)

A

B 24Volt

(A) 1A. (B) 2A.

(C) 4A. (D) zero.

39. In the following circuit, each resistor has a resistance

of 15  and the battery has an e.m.f. of 12 V with

negligible internal resistance. (IJSO/Stage-II/2011)

When a resistor of resistance R is connected between D & F, no current flows through the galvanometer (not shown in the figure) connected between C & F. Calculate the value of R.

(A) 10  (B) 15 

(C) 5  (D) 30 

40. The circuit given below is for the operation of an industrial fan. The resistance of the fan is 3 ohms. The regulator provided with the fan is a fixed resistor and a variable resistor in parallel. (IJSO/Stage-II/2011)

Under what value of the variable resistance given below, Power transferred to the fans will be maximum? The power source of the fan is a dc source with internal resistnace of 6 ohm.

(A) 3  (B) 0

15 41.When all the resistances in the circuit are 1each,

then the equivalent resistance across points A & B will

be : (IJSO/Stage-II/2011)

(A) 5/6  (B) 1/2 

(C) 2/3  (D) 1/3 

42. A cylindrical copper rod has length L and resistance R.

If it is melted and formed into another rod of length 2L.

the resistance will be : (KVPY/2011)

(A) R (B) 2R

(C) 4R (D) 8R

43. There are four resistors of 12 ohm each. Which of the following values is/are possible by their combinations

(series and / or parallel) ? (IJSO/Stage-I/2008)

(A) 9 ohm (B) 16 ohm

(C) 12 ohm (D) 30 ohm

44. In case of the circuit shown below, which of the following

statements is/are true ? (IJSO/Stage-I/2009)

• + – • R1 R2 R3 A B 2 1 3 4

(A) R₁, R₂ and R₃ are in series. (B) R₂ and R₃ are in series. (C) R₂ and R₃ are in parallel.

(D) The equivalent resistance of the circuit is R₁+ 3 2 3 2 R R R R 

45. A current i reaching at a point in a circuit gets branched and flows through two resistors R₁ and R₂. Then, the current through R₁ varies as : (IAO/Jr./Stage-I/2007)

(A) R₁ (B) R₂

(C) (R₁+ R₂) (D) 1l (R₁ + R₂)

46. In the circuit shown below, (IAO/Sr./Stage-I/2007)

10V

Y

X

(A) current flowing in the circuit is 200 mA (B) power supplied by the battery is 2 watt (C) current from X to Y is zero

(D) potential difference across 10 is equal to zero 47. We are given n resistors, each of resistance R. The

ratio of the maximum to minimum resistance that can

be obtained by combining them is : (KVPY/2008)

(A) nn _{(B) n}

(C) n2 _{(D) log}n

Cell :

48. A cell of emf E is connected across a resistance R.

The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell is given as : (A) R(E – V) (B) R V E (C) E R ) V E (  (D) R V ) V E ( 

49. 24 cells, each having the same e.m.f. and 2 ohm

internal resistance, are used to draw maximum current through an external resistance of 3 ohm. The cells should be connected

(A) in series (B) in parallel

(C) in 4 rows, each row having 6 cells (D) in 6 rows, each row having 4 cells

50. The cells are joined in parallel to get the maximum

current when

(A) external resistance is very large as compared to the total internal resistance

(B) internal resistance is very large as compared to the external resistance

(C) internal resistance and external resistance are equal

(D) emf of each cell is very large

51. In secondary cells :

(A) Chemical changes can be reversed by heating electrodes

(B) Chemical changes can be reversed by passing electric current

(C) Current is produced by photo chemical reactions (D) None of these

52. Three types of electric cells which provide current are :

(A) Button cell, solar cell & secondary cell

(B) Solar cell, electrolytic cell, electro chemical cell (C) (A) and (B) both are correct

(D) Neither (A) nor (B) is correct

53. In which of the following cells, the potential difference

between the terminals of a cell exceeds its emf.

(A) a (B) b

(C) c (D) d

54. A cell, an ammeter and a voltmeter are all connected in

series. The ammeter reads a current I and the

voltmeter a potential difference V. If a torch bulb is connected across the voltmeter, then.

(IJSO/Stage-I/2009)

(A) both I and V will increase (B) both I and V will decrease (C) I will increase but V will decrease (D) I will decrease but V will increase

55. In the process of electrostatic induction.

(IJSO/Stage-II/2011)

(A) a conductor is rubbed with an insulator.

(B) a charge is produced by friction.

(C) negative and positive charges are separated.

(D) electrons are ‘sprayed’ on the object.

56. Consider the circuit below. The bulb will light up if : (KVPY/2009)

S

S

S

~

(A) S₁ S₂ and S₃ are all closed.

(B) S₁ is closed but S₂ and S₃ are open. (C) S₁ and S₃ are closed but S₂ is open. (D) none of these

Electric Energy and Power :

57. An electric iron of heating element of resistance 88 

is used at 220 volt for 2 hours. The electric energy spent, in unit, will be :

(A) 0.8 (B) 1.1

(C) 2.2 (D) 8.8

58. Two identical heater wires are first connected in series

and then in parallel with a source of electricity. The ratio of heat produced in the two cases is :

(A) 2 : 1 (B) 1 : 2

(C) 4 : 1 (D) 1 : 4

59. You are given three bulbs 25 W, 40 W and 60 W . Which

of them has the lowest resistance?

(A) 25 watt bulb (B) 40 watt bulb

(C) 60 watt bulb (D) insufficient data

60. If R₁ and R₂ are the filament resistances of a 200 W bulb and a 100 W bulb respectively designed to operate on the same voltage, then :

(A) R₁ = 2 R₂ (B) R₂ = 2 R₁

(C) R₂ = 4 R₁ (D) R₁ = 4 R₂

61. If two bulbs, whose resistance are in the ratio of 1 : 2,

are connected in series. The power dissipated in them has the ratio of :

(A) 1 : 1 (B) 1 : 2

(C) 2 : 1 (D) 1 : 4

62. When a voltage of 20 volt is applied between the two

ends of a coil, 800 cal/s heat is produced. The value of resistance of the coil is :(1 calorie = 4.2 joule) :

(A) 1.2  (B) 1.4 

(C) 0.12  (D) 0.14 

63. You are given two fuse wires A and B with current rating

2.5 A and 6 A respectively. Which of the two wires would you select for use with a 1100 W, 220 V room heater ?

(A) A (B) B

(C) A and B (D) none of these

64. An electric current of 2.0 A passes through a wire of

resistance 25 . How much heat (in joule) will be

developed in 1 minute ?

(A) 6 (B) 6000

(C) 50 (D) 10

65 . Two bulbs, one of 200W and the other of 100W, are

connected inseries with a 100 V battery which has no

internal resistance. Then, (KVPY/2009)

200W 100W

100V

(A) the current passing through the 200W bulb is more than that through the 100W bulb.

(B) the power dissipation in the 200W bulb is more than that In the 100 W bulb.

(C) the voltage drop across the 200W bulb is more than that across the 100W bulb.

(D) the power dissipation In the 100W bulb is more than that in the 200W bulb.

17 66. An electric heater consists of a nichrome coil and runs

under 220 V, consuming 1 kW power. Part of its coil burned out and it was reconnected after cutting off the burnt portion. The power it will consume now is :

(KVPY/2010)

(A) more than 1 kW.

(B) less that 1 kW, but not zero. (C) 1 kW.

(D) 0 kW.

67. In the following circuit, the 1 resistor dissipates

power P. If the resistor is replaced by 9. the power

dissipated in it is : (KVPY/2011)

(A) P (B) 3P

(C) 9P (D) P/3

68. A neon lamp is connected to a voltage a.c. source. The

voltage is gradually increased from zero volt. It is observed that the neon flashes at 50 V. The a.c. source is now replaced by a variable dc source and the experiment is repeated. The neon bulb will flash at :

(IAO/Sr./Stage-I/2008)

(A) 50V (B) 70V

(C) 100V (D) 35V

69. A certain network consists of two ideal and indentical voltage sources in series and a large number of ideal resistor. The power consumed in one of the resistor is 4W when either of the two sources is active and other is replaced by a short circuit. The power consumed by same resistor when both sources are simultaneously

active would be : (IJSO/Stage-II/2011)

(A) 0 or 16W (B) 4W or 8W

(C) 0 or 8W (D) 8W or 16W

Circuit and Other :

70. A galvanometer can be converted into a voltmeter by

connecting

(A) A high resistance in series with the galvanometer (B) A high resistance in parallel with the galvanometer (C) a low resistance in series with the galvanometer (D) a low resistance in parallel with the galvanometer

71. The circuit shown has 3 identical light bulbs A, B, C

and 2 identical batteries E₁, E₂ . When the switch is open, A and B glow with equal brightness. When the

switch is closed: (KVPY/2007)

A B C S E1 E2

(A) A and B will maintain their brightness and C will be dimmer than A and B.

(B) A and B will become dimmer and C will be brighter than A and B.

(C) A and B will maintain their brightness and C will not glow.

(D) A, B and C will be equally bright.

72. A student connects two lamps in the circuit shown.

The emf of the two batteries is different.

(IJSO/Stage-II/2011)

Which of the following statements are correct?

i. When keys 1, 2, 3 and 4 are closed, bulbs

A and B will both glow

ii. When key 2 and 4 are closed bulb A will glow

iii.When 1 and 4 are closed, bulb A will glow

iv.When 2, 3 and 4 are closed, both A and B will

glow

(A) only ii (B) only iv

(C) i, ii and iv (D) ii and iii

73. Figure below shows a portion of an electric circuit with

the currents in ampere and their directions. The magnitude and direction of the current in the portion

PQ is : (KVPY/2011)

(A) 0A (B) 3A from P to Q

MOLE CONCEPT

ATOMS

All the matter is made up of atoms. An atom is the smallest particle of an element that can take part in a chemical reaction. Atoms of most of the elements are very reactive and do not exist in the free state (as single atom).They exist in combination with the atoms of the same element or another element.

Atoms are very, very small in size. The size of an atom is indicated by its radius which is called "atomic radius" (radius of an atom). Atomic radius is measured in "nanometres"(nm).

1 metre = 109 _{nanometre or 1nm = 10}-9 _m.

Atoms are so small that we cannot see them under the most powerful optical microscope.

 Note :

Hydrogen atom is the smallest atom of all , having an atomic radius 0.037nm.

( a ) Sy m bols of Ele me n t s :

A symbol is a short hand notation of an element which can be represented by a sketch or letter etc.

Dalton was the first to use symbols to represent elements in a short way but Dalton's symbols for element were difficult to draw and inconvenient to use, so Dalton's symbols are only of historical importance. They are not used at all.

It was J.J. Berzelius who proposed the modern system of representing en element.

The symbol of an element is the "first letter" or the "first letter and another letter" of the English name or the Latin name of the element.

e.g. The symbol of Hydrogen is H. The symbol of Oxygen is O.

There are some elements whose names begin with the same letter. For example, the names of elements Carbon, Calcium, Chlorine and Copper all begin with the letter C. In such cases, one of the elements is given a "one letter "symbol but all other elements are given a "first letter and another letter" symbol of the English or Latin name of the element. This is to be noted that "another letter" may or may not be the "second letter" of the name. Thus,

The symbol of Carbon is C. The symbol of Calcium is Ca. The symbol of Chlorine is Cl.

The symbol of Copper is Cu (from its Latin name Cuprum)

It should be noted that in a "two letter" symbol, the first letter is the "capital letter" but the second letter is the _small letter_



English Name of

the Element Symbol

Hydrogen H Helium He Lithium Li Boron B Carbon C Nitrogen N Oxygen O Fluorine F Neon Ne Magnesium Mg Aluminium Al Silicon Si Phosphorous P Sulphur S Chlorine Cl Argon Ar Calcium Ca

Symbol Derived from English Names

Symbols Derived from Latin Names

English Name of

the Element Symbol

Latin Name of the Element Sodium Na Natrium Potassium K Kalium ( b ) S i g n i f i c a n c e o f T h e S y m b o l o f a n El e me nt :

(i) Symbol represents name of the element. (ii) Symbol represents one atom of the element. (iii) Symbol also represents one mole of the element.

That is, symbol also represent 6.023 × 1023 atoms of

the element.

(iv) Symbol represent a definite mass of the element

i.e. atomic mass of the element.

Example

:

(i) Symbol H represents hydrogen element.

(ii) Symbol H also represents one atom of hydrogen

element.

(iii) Symbol H also represents one mole of hydrogen

atom.

(iv) Symbol H also represents one gram hydrogen

atom.

IONS

An ion is a positively or negatively charged atom or group of atoms.

Every atom contains equal number of electrons (negatively charged) and protons (positively charged). Both charges balance each other, hence atom is electrically neutral.

(a) Cation :

If an atom has less electrons than a neutral atom, then it gets positively charged and a positively charged ion is known as cation.

e.g. Sodium ion (Na+_{), Magnesium ion (Mg}2+_{) etc.}

A cation bears that much units of positive charge as there are the number of electrons lost by the neutral atom to form that cation.

e.g. An aluminium atom loses 3 electrons to form

aluminium ion, so aluminium ion bears 3 units of positive charge and it is represented as Al3+_.

(b) Anion :

If an atom has more number of electrons than that of neutral atom, then it gets negatively charged and a negatively charged ion is known as anion.

e.g. Chloride ion (Cl¯), oxide ion (O2-) etc.

An anion bears that much units of negative charge as there are the number of electrons gained by the neutral atom to form that anion.

e.g. A nitrogen atom gains 3 electrons to form nitride ion, so nitride ion bears 3 units of negative charge and it is represented as N3-_.

 Note :

Size of a cation is always smaller and anion is always greater than that of the corresponding neutral atom.

(c) Monoatomic ions and polyatomic ions :

(i) Monoatomic ions : Those ions which are formed

from single atoms are called monoatomic ions or simple ions.

e.g. Na+_{, Mg}2+ _etc.

(ii) Polyatomic ions : Those ions which are formed

from group of atoms joined together are called polyatomic ions or compound ions.

e.g. Ammonium ion (NH₄+_{) , hydroxide ion (OH}–_{) etc.}

which are formed by the joining of two types of atoms, nitrogen and hydrogen in the first case and oxygen and hydrogen in the second.

(d) Valency of ions :

The valency of an ion is same as the charge present on the ion.

If an ion has 1 unit of positive charge, its valency is 1 and it is known as a monovalent cation. If an ion has 2 units of negative charge, its valency is 2 and it is known as a divalent anion.

LIST OF COMMON ELECTROVALENT POSITIVE RADICALS

LIST OF COMMON ELECTROVALENT NEGATIVE RADICALS

Monovalent Electronegative Bivalent

Electronegative Trivalent Electronegative Tetravalent Electronegative 1. Fluoride F– 1. Sulphate SO4 2- _{1. Nitride N} 3-1. Carbide C 4-2. Chloride Cl– 2. Sulphite SO3 2-2. Phosphide P 3-3. Bromide Br– _{3. Sulphide S} 2-3. Phosphite PO3 3-4. Iodide I 4. Thiosulphate S2O3 2-4. Phosphate PO4 3-5. Hydride H– 5. Zincate ZnO2 2-6. Hydroxide OH– _{6. Oxide O} 2-7. Nitrite NO2 – _{7. Peroxide O} 2 2-8.Nitrate NO3 – _{8. Dichromate Cr} 2O7

2-9. Bicarbonate or Hydrogen carbonate HCO3

– _{9. Carbonate CO}

3

2-10. Bisulphite or Hydrogen sulphite HSO3

– _{10. Silicate SiO}

3

2-11. Bisulphide or Hydrogen sulphide HS– 12. Bisulphate or Hydrogen sulphate HSO4

– 13. Acetate CH COO3

–

 Note :

Cation contains less no. of electrons and anion contains more no. of electrons than the no. of protons present in them.

LAWS OF CHEMICAL COMBINATION

The laws of chemical combination are the experimental laws which led to the idea of atoms being the smallest unit of matter. The laws of chemical combination played a significant role in the development of Dalton’s atomic theory of matter.

There are two important laws of chemical combination. These are:

(i) Law of conservation of mass (ii) Law of constant proportions

(a) Law of Conservation of Mass or Matter :

This law was given by Lavoisier in 1774 . According to the law of conservation of mass, matter can neither be created nor be destroyed in a chemical reaction.

Or

The law of conservation of mass means that in a chemical reaction, the total mass of products is equal to the total mass of the reactants. There is no change in mass during a chemical reaction.

Suppose we carry out a chemical reaction between A and B and if the products formed are C and D then, A + B  C + D

Suppose 'a' g of A and 'b' g of B react to produce 'c' g of C and 'd' g of D. Then, according to the law of conservation of mass, we have,

a + b = c + d

Example :

W hen Calcium Carbonate (CaCO₃) is heated, a chemical reaction takes place to form Calcium Oxide (CaO) and Carbon dioxide (CO₂). It has been found by experiments that if 100 grams of calcium carbonate is decomposed completely, then 56 grams of Calcium Oxide and 44 grams of Carbon dioxide are formed.

Since the total mass of products (100g ) is equal to the total mass of the reactants (100g), there is no change in the mass during this chemical reaction. The mass remains same or conserved.

( b ) L a w of C o n s t a n t Pr o p o r t i o n s / L a w o f D e finit e Proport ions :

Proust, in 1779, analysed the chemical composition (types of elements present and percentage of elements present ) of a large number of compounds and came to the conclusion that the proportion of each element in a compound is constant (or fixed). According to the law of constant proportions: A chemical compound always consists of the same elements combined together in the same proportion by mass.

 Note :

The chemical composition of a pure substance is not dependent on the source from which it is obtained.

Example :

Water is a compound of hydrogen and oxygen. It can be obtained from various sources (like river, sea, well etc.) or even synthesized in the laboratory. From whatever source we may get it, 9 parts by weight of water is always found to contain 1 part by weight of hydrogen and 8 parts by weight of oxygen. Thus, in water, this proportion of hydrogen and oxygen always remains constant.

 Note :

The converse of Law of definite proportions that when same elements combine in the same proportion, the same compound will be formed, is not always true.

(c ) La w of Multiple Proportions :

According to it, when one element combines with the other element to form two or more different compounds, the mass of one element, which combines with a constant mass of the other, bears a simple ratio to one another.

Simple ratio means the ratio between small natural numbers, such as 1 : 1, 1 : 2, 1 : 3

e.g.

Carbon and oxygen when combine, can form two oxides that are CO (carbon monoxide), CO₂ (carbon dioxide).

In CO,12 g carbon combined with 16 g of oxygen. In CO₂,12 g carbon combined with 32 g of oxygen. Thus, we can see the mass of oxygen which combine with a constant mass of carbon (12 g) bear simple ratio of 16 : 32 or 1 : 2

 Note :

The law of multiple proportion was given by Dalton in 1808.

Sample Problem :

1. Carbon is found to form two oxides, which contain 42.8% and 27.27% of carbon respectively. Show that these figures illustrate the law of multiple proportions.

Sol. % of carbon in first oxide = 42.8

% of oxygen in first oxide = 100 - 42.8 = 57.2 % of carbon in second oxide = 27.27

 % of oxygen in second oxide = 100 - 27.27 = 72.73

For the first oxide

-Mass of oxygen in grams that combines with 42.8 g of carbon = 57.2

 Mass of oxygen that combines with 1 g of carbon = 1.34

42.8 57.2

 g

For the second oxide

-Mass of oxygen in grams that combines with 27.27 g of carbon = 72.73

 Mass of oxygen that combines with 1 g of carbon =

2.68 27.27 72.73

 g

Ratio between the masses of oxygen that combine with a fixed mass (1 g) of carbon in the two oxides = 1.34 : 2.68 or 1 : 2 which is a simple ratio. Hence, this illustrates the law of multiple proportion.