Electric Charges and Fields

Class XII

 Electric Charges 

Gold Leaf Electroscope

Conductors And Insulators

Electric Charge

Electric charge is a scalar quantity. The SI unit of charge is coulomb. It is generally denoted by Q.

There are two types of charges. These charges are named by American Scientist Benjamin Franklin.

1. Positive Charge

2. Negative Charge

When glass rod is rubbed with silk. Something is acquired by both glass rod and silk. This is called Electric Charge.

The glass rod acquires Positive Charge.

When two glass rods are rubbed with silk and brought close to each other. They repel each other.

From this it is clear that

Like or Same charges repel to each other whereas opposite or unlike charges attract to each other.

When a plastic rod is rubbed with woolen cloth. The plastic rod acquires Negative Charge.

Hence, Two Positive or Negative Charges repel each other whereas Positive and Negative charges attract to each other.

The charges acquired after rubbing are lost when the charged the bodies are brought in contact.

If a body has charge then it is said to be charged. Gold Leaf Electroscope is an apparatus which is used to detect the charge on a body.

Gold Leaf Electroscope

The apparatus has a metal rod, two thin gold leaves and charged body. All of these are fitted in transparent glass window box. The figure is given below.

When charged body touches the metal knob at the top of the rod. The charge flows on the leaves and they diverge. The degree of divergence indicates the amount of charge.

Conductors and Insulators

Those substances which allow electricity to pass through them easily are said to Conductors. Conductors have electric charges (electrons). These electrons comparatively free to move inside the material.

There are following examples of conductors.

Metals, human and animal bodies and earth.

Those substances which offer high resistance to passage of electricity through them are said to Insulators.

There are following examples of insulators.

Non-metals, glass,porcelain, plastic, nylon, wood offer high resistance to passage of electricity through them.

Mathematics

Circle

Class X

Learning Points

1. Circle
2. Tangent to a Circle
3. Number of Tangents from a Point on a Circle

Circle

Circle is a collection of points. These points are located at a certain distance from a fixed point. The fixed point is known as centre. The certain distance is known as radius. A circle has centre, radius, diameter, circumference, Tangent, secant, chord. These are basic terminology of a circle.

Any lines which passing through one and only one point of circle is said to be tangent. Any lines which join two point of a circle is said to be secant. Any lines which passing through a circle without touching it is said to be non - intersectiong line.

Tangent to a Circle

Any lines which passing through one and only one point of circle is said to be tangent. Tangent is a Latin word derived from tangere means touching.

There is one and only one tangent at certain point of a circle. 

Tangent is a special types of secant in which two ends point coincides. Therefor all tangents are secant but all secant are not a tangent.

Theorem 10.1

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given:-

There is a circle. O is the centre. P is a point on the circle. The tangent XY passing through P.

To Prove that :-

XY is perpendicular on OP.

Construction:-

Point Q is on XY outside of the circle. Join OQ.

Proof:

OQ is greater than OP.

If Q lies inside the circle.

XY will be secant.

OP is the shortest distance from centre O.

We know that the shortest distance from a point on line is perpendicular.

Therefore,

XY is perpendicular on OP. Proved.

1. By theorem above, we can also conclude that at any point on a circle there can be one and only one tangent.

2. The line containing the radius through the point of contact is also sometimes called the ‘normal’ to the circle at the point.

In next part I shall post "Number of Tangents from a Point on a Circle". Please do wait for this.

Electro Chemistry

Class XII

Electro - Chemistry

"The study of chemical reactions in which ion solution, electrolysis and electric cells are included is said to be Electro - Chemistry."

Or

"The study of production of electricity from energy released during spontaneous chemical reaction and the use of electrical energy to bring about non-spontaneous chemical transformations."

Batteries and fuel cells convert chemical energy into electrical energy. This energy is used in various types instruments and devices. This energy produces less pollution. This energy is eco friendly.

Electro Chemical Cell

The cell which converts chemical energy into electrical energy is said to be Electro Chemilal Cell.

The device in which the concentration of Zinc ion and Copper ion is unity and electric potential is 1.1 the device is said to galvanic or voltaic cell.

The galvanic cell is an electrochemical cell that converts the chemical energy of spontaneous redox reaction into electrical energy in this device the Gibbs energy of spontaneous redox reaction is converted into electrical work which may be used for running a motor or other electrical gadgets like heater, fan geyser, etc.

The potential difference between the two electrodes and as soon as the switch is in the on position the electrodes the electrons follows from negative to positive electrode.. The direction of current flow is opposite to that of electron flow.

The potential difference between the two electrodes of a galvanic cell is called the cell potential. The unit of cell potential is volts.

The cell potential is the difference between the electrode potentials of the cathode and anode.
This is said to be the cell Electromotive Force. It is generally denoted by emf. The unit of emf is volts.

प्रमेय 6.6 कक्षा X

"दो समरूप त्रिभुजों के क्षेत्रफलों का अनुपात इनकी संगत भुजाओं के अनुपात के वर्ग के बराबर होता है"।

दियाः-

दो त्रिभुज क्रमशः ABC एवं PQR समरूप हैं।

सिद्द करना है किः-

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/QR)²

रचनाः-

AM एवं PN क्रमशः BC एवं QR पर लम्ब डाला।

उपपत्तिः-

ar(ABC) = BCxAM/2

ar(PQR) =  QRxPN/2
ar(ABC)/ar(PQR) = BCxAM/QRxPN ............................ (i)

अब, त्रिभूज ABM एवं PQN में विचार करने पर
<B = <Q {त्रिभूज ABC समरूप PQR}
<M = <N {प्रत्येक समकोण}
<BAN = <QPN {शेष कोण}
AAA समरूपता से त्रिभूज ABM एवं PQN समरूप है।
AB/PQ = AM/PN ............................. (ii)

अब (i) तथा (ii) से
ar(ABC)/ar(PQR) = BCxAB/QRxPQ .............................. (iii)

त्रिभुज ABC एवं PQR समरूप है।
इसलिए
AB/PQ = BC/QR = AC/PR .............................. (iv)

अब (iii) तथा (iv) से
ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/QR)² सिद्द हो गया।

Theorem 6.6 Class X

Theorem 6.6

Statement

"The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides"

Given:-

Two similar triangles ABC and PQR are given.

To prove that:-

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)²

Construction:-

Draw AM and PN perpendicular on BC and QR respectively.

Proof:-

ar(ABC) = BCxAM/2

ar(PQR) = QRxPN/2

Therefore,

ar(ABC)/ar(PQR) = BCxAM/QRxPN .............................. (i)
Now, consider in triangles ABM and PQN are similar under AAA similarity criterion.
<B = <Q { Triangles ABC and PQR are similar}
<M = <N {Each right angle}
<BAM = <QPN {Remain angle}

Therefore,
AB/PQ = AM/PN ............................. (ii)
Since, Triangles ABC and PQR are similar.

Now, from (i) and (ii);
ar(ABC)/ar(PQR) = BCxAB/QRxPQ ................................... (iii)

AB/PQ = BC/QR = AC/PR {Triangles ABC and PQR are similar}................ (iv)

Now, from (iii) and (iv)

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² Proved

Osmotic Pressure and Calculation of Molecular Weight of Solute

Class XII
Chapter : Solutions

Osmotic Pressure and Calculation of Molecular Weight of Solute

"The process of flow of the solvent is called osmosis".

The flow of the solvent from solvent side to solution side across a semipermeable membrane can be stopped if some extra pressure is applied on the solution. 

"Pressure that just stop the flow of solvent is called osmotic pressure of the solution."

The osmotic pressure of a solution is the excess pressure that must be applied to a solution to prevent osmosis, to stop the passage of the solvent molecules through a semipermeable membrane into the solution.

" Osmotic pressure is proportional to the molarity, C of the solution at a given temperature T."

П = CRT

П  represent osmotic pressure
R represent gas constant

Therefore,
П  = (n₂/V)RT
Here V is the volume of a solution in litres containing n₂ moles the solute.
Let the weight of the solute of w₂ grams of solute, of molar mass M₂ is present in the solution.

Therefore,

n₂ = w₂/M₂
ПV =  w₂RT/M₂

M₂ = w₂RT/ПV 

Resolution

Class XI

Chapter 4 

Motion in a Plane

Resolution

          Let a and b be two non-zero vectors in a plane with different direction and let A be another vector in the same plane. A can be expressed as a sum of two vectors - one obtained by multiplying a by a real number and the other obtained by multiplying b by another real number. To see this, let O and P be the tail and head of the vector A. Then, through O, draw a straight line parallel to a, and through P, a straight line parallel to b. Let them intersect at Q. 

Therefore,

A = OP = OQ + QP

But since OQ is parallel to a, and QP is parallel to a, and QP is parallel to b , we can write:

OQ = ℷa and  𝜇b

Where  ℷ and  𝜇 are real numbers

Therefore,

A =  ℷa + 𝜇b

A has been resolved into two component vectors ℷa and  𝜇b along a and b respectively.

                   

           Resolve a vector A in terms of component vectors that lie along unit vectors i and j.

Consider a vector A that lies in x-y plane as shown in Figure. Draw a lines from the head of A perpendicular to the coordinate axes and get vectors A₁ and A₂ such that 

A₁ + A₂ = A
Since A₁ is parallel to i and A₂ is parallel to j 
A₁ = Aхⅰ
A₂ = Ayj
Therefore
Aхⅰ + Ayj = A

Aх and Ay are called the x and y components of  the vector A and the angle ፀ it makes with the x axis.
Aх = ACosፀ 
Ay = A Sinፀ 

If A and  ፀ are given                                                                 

Aх² + Ay² = A²Cos² ፀ   + A²Sin² ፀ  = A²

Tan ፀ  = Ay/Aх

Same procedure can be used to resolve vector A into three components along x, y and z axes in three dimension. If α, β and γ are the angles between x, y and z axes respectively.

Aх = ACosα
Ay = ACosβ 
Az = ACosγ

Therefore,
A = Aхi + Ayj + Azk

The magnitude of vector A
A = ✓Aх² + Ay² + Az²
r = xi + yj + zk 
Here x, y, and z are the components of r along x, y and z axes respectively.

Relative Velocity

Class XI Chapter 3

Relative Velocity

"The consideration of velocity of an object with respect to another object is said to be Relative Velocity."

Consider about two objects A and B moving in straight line with average velocities Va and Vb respectively.

Let the positions of the objects are at time t = 0 are Xa(0) and Xb(0) respectively.

After time T positions of the objects will be

Xa(t) = Xa(0) + Va.T

Xb(t) = Xb(0) + Vb.T

We know that 

Distance = Velocity x Time

Displacement from object A to object B 

= Xb(t) - Xa(t) = Xb(0) + Vb.T - Xa(t) - Va.T

or, Displacement from A to B = [Xa(0) - Xb(0)] + [Vb - Va].T

 The displacement A to B changes steadily by the amount Va - Vb.

This (Vb - Va) is also known as velocity of B relative to velocity of A.

Therefore, Relative velocity of B to A

Vba = Vb - Va.

Position - time graph of two objects with equal velocities.

When velocities are equal

Va = Va

Displacement from A to B = [Xa(0) - Xb(0)] 

The relative velocity will be zero and stay at constant distance.

Position - Time graph of two objects with unequal velocities.

When Va greater than Vb

The relative velocity of B to A will be Negative.

One graph is steeper than the other and they meet each other at a common point.

When Va and Vb have opposite sign.

 The magnitude of Vba will be greater than Va or Vb.

These three cases arises in Relative velocity.

Colligative Properties and Calculation of Molecular Mass

Chemistry Class XIIChapter Solutions

Colligative Properties and Calculation of Molecular Mass

"That property of a solution depend upon numbers of the present particles in the solution  is known as Colligative Properties."

As per Raoult's Law
"The relative lowering the vapour pressure is equal to the mole fraction of solute in the solution".

Let the vapour pressure of pure solvent and solvent in the solution are P⁰ and Ps respectively.

Lowering of vapour pressure =  P⁰ - Ps
Relative lowering of vapour pressure =  (P⁰ - Ps)/P⁰

Mole fraction of solute in the solution = n/(n + N)

Here,
n is the number of moles of the solute in the solution.
N is the number of moles of pure solvent.

According to Raulot's Law;

 (P⁰ - Ps)/P⁰ =  n/(n + N)
When n is far far less than N.

 (P⁰ - Ps)/P⁰ = n/N ...................... (i)
We know that
n = w/m = Weight of solute/ Molecular weight of solute
N = W/M = Weight of solvent/ Molecular weight of solvent

Now, from (i)

 (P⁰ - Ps)/P⁰ = wM/mW
Therefore, molecular weight of the solute

m = P⁰wM/W(P⁰ - Ps)

The above expression calculate the molecular mass of the solute.

पाइथागोरस प्रमेय

पाइथागोरस प्रमेय

"समकोण  त्रिभुज में कर्ण का वर्ग अन्य दो भुजाओं के वर्गों के योगफल के बराबर होता है"।

दिया है ः-

ABC एक समकोण त्रिभुज है  जिसमें AC कर्ण है . AB एवं BC अन्य दो भुजाएं है।

सिद्ध करना है कि :-
(AC)² = (AB)² + (BC)²

रचना :-
AC पर BD लम्ब डाला

प्रमाण:-
समकोण त्रिभुज ABC एवं ADB में विचार करने पर
<BAC = <BAD            (उभयनिष्ठ)
<ABC = <ADB           (प्रत्येक समकोण)
<ACB = <ABD           (शेष कोण)
त्रिभुज ABC एवं ADB समरूप AAA समरूपता से
AC/AB = AB/AD
ACxAD = (AB)²  .....................  (i)

समकोण त्रिभुज ABC एवं BDC में विचार करने पर
<ACB = <BCD            (उभयनिष्ठ)
<ABC = <CDB           (प्रत्येक समकोण)
<CAB = <DBC           (शेष कोण)
त्रिभुज ABC एवं BDC समरूप AAA समरूपता से

AC/BC = BC/DC
ACxDC = (BC)²   .................... (ii)

अब (i) तथा (ii) को जोडने पर
ACxAD + ACxDC = (AB)² + (BC)²
AC(AD + DC) = (AB)² + (BC)²
चित्र से स्पष्ट है कि
AD + DC = AC
अतः  ACxAC = (AB)² + (BC)²

(AC)² = (AB)² + (BC)²
सिद्ध हो गया।

Solution

Solution 

              A homogeneous mixture of two or more substances is said to be solution. These substances are classified as Solvent and Solute. The solution in which there are two substances is said to be binary solution. One solvent and other is solute.
All binary solutions are solution but all solutions are not binary.Solution are classified as the following solution which are as follow

1. Solid
2. Liquid
3. Gasseous

         The solution in which solid is as solvent and solid, liquid and gas anyone of these is as a solute is said to be solid solution.There are following types of solid solution

• Solid to Solid

• Solid to Liquid

• Solid to Gas

       The solution in which liquid is as solvent and solid, liquid and gas anyone of these is as a solute is said to liquid solution.There are following types of liquid solution

• Liquid to Solid

• Liquid to Liquid

• Liquid to Gas

     The solution in which gas is as solvent and solid, liquid and gas anyone of these is as a solute is said to gaseous solution.There are following types of gaseous solution

• Gas to Solid

• Gas to Liquid

• Gas to Gas

  Aqueous or Aqueous Solution

        The solution in which water is used as solvent is said to be aqueous solution

Non aqueous Solution

        The solution in which water is not used as solvent is said to be Non aqueous solution.

The amount of solute dissolved in a solvent.

 As per, there are following types of solution

• Unsaturated

• Saturated

• Supersaturated

Unsaturated

               The solution in which more solute can be dissolved without rising temperature is said to be unsaturated solution. 

Saturated 

              The solution in which more solute cannot be dissolved further at a given temperature is said to be saturated solution. 

Supersaturated

            The solution in which having more solute than that would be necessary to saturate it at given temperature is said to be Supersaturated.

Continue ..........................................

Matter Surrounding Us

Matter

Anythings which have both mass and volume is said to be matter or occupy space and have mass.

Air, Gas, Water, Stone, Clouds, Stars, Plants and Animals, Food stuffs etcetera are the examples of matter.

Early Indian philoshgers classified matter in the form of five basic elements the Panch Tatva 

which are

1. Air
2. Earth
3. Water
4. Fire 
5. Sky

According to them everything, living or non-living, was made of these five basic elements.

As per Modern Age, classification of matter based on Physical Properties and Chemical Nature of matter.

As per Physical Properties,

 Matters are made up of particles. Particles are very small. They are small beyond of our imagination.

Estimating how small are the particles of matter. With every dilution, though the colour becomes light, it is still visible.

There are following characteristics of particles of matter.

1. Particles of matter have space between them.

For example, Lemonade (nimbu paani) Particles of lemon evenly distributed in water. Particles of one type of matter get into the spaces between particles of the other. This show that there is enough space between particles of matter.

Due to which a diver is able to cut through water in a swimming pool.

2. Particles of matter are continuously moving.

Particles of matters possess kinetic energy. As the temperature rises, particles move faster. The kinetic energy of the particles increases with the increase of temperature.

Due to which smell of hot sizzling food reaches us several metres away, but to get the smell from cold food we have to go close. 

3. Particles of matter attract each other.

Particles of matter have force acting between them. The force keep the particles together. The strength of this force of attraction is different for different matter.

States of Matter

There are three states of a matter which are as follows

1. Solid 
2. Liquid 
3. Gas

Continue ....................................................................................

Newton's Third Law of Motion

Statement

To every action there is always an equal and opposite reaction

Proof
Consider about two moving bodies in an isolated system. Masses of the bodies are M1 and M2. Velocities are V1 and V2.

Due to interaction, their velocities change. Consequently there will be change in their momenta in time ∆t. Suppose, change in momentum of the first body = ∆p1 and change in momentum of the second body = ∆p2.

From the law of conservation of linear momentum, we know that

∆p1 + ∆p2 = 0
∆p2 = -∆p1
Now, dividing both sides by ∆t
∆p2/∆t = - ∆p1/∆t
Take limit both sides when ∆t tends to zero.
dp2/dt = - dp1/dt
F2 = - F1
Force acting on M2 = - Force acting on M1
Action = - Reaction
To every action there is always an equal and opposite reaction. Proved.
This represents Newton's third law of motion.

Thales Theorem

Statement


"If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio "
Given
To Prove that
Construction
Join B to E and C to D. Draw perpendicular EN on AB and DM on AC.
Proof


ABC is a triangle. DE parallel to BC.

AD/DB = AE/EC

Consider in triangle ADE and BDE.
ar(ADE) = 1/2 ADxEN
ar(BDE) = 1/2DBxEN
Therefore,
ar(ADE)/ar(BDE) = AD/DB ......... (i)
Again, consider in triangle ADE and CDE.
ar(ADE) = 1/2 AExDM
ar(CDE) = 1/2 ECxDM
Therefore,
ar(ADE)/ar(CDE) = AE/EC ............ (ii)
Triangles BDE and CDE have same base DE and DE parallel to BC.
Therefore,
ar(BDE) = ar(CDE)
Now from (i) and (ii)
AD/DB = AE/EC Proved.

Phythagoras Theorem

Pythagoras Theorem 

Statement

" In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides"

Given

ABC is a right triangle. AC, AB and BCare hypotenuse and the other two sides respectively.

To Prove that

(AC)² = (AB)² + (BC)²

Construction

Draw BD perpendicular on AC.

Proof

Consider in right triangle ABC and ADB
<BAC = <BAD   (Common)
<ABC = < ADB  (Each right angle)
<ACB = <ABD   ( Remain angle)

Hence,
Triangle ABC and ADB are similar under AAA similarity.

Therefore,
AC/AB = AB/AD
ACxAD = (AB)²  ........  (i)

Now, consider in right triangle ABC and BDC
<BAC = <CBD   (Remain angle)
<ABC = < CDB  (Each right angle)

<ACB = <BCD   ( Common angle)

Hence,
Triangle ABC and BDC are similar under AAA similarity.

Therefore,
AC/BC = BC/DC
ACxDC =  (BC)² ......... (ii)

Now, add (i) and (ii)
ACxAD + ACxDC = (AB)² + (BC)²
AC(AD + DC) = (AB)² + (BC)²

From figure it is clear that
AD + DC = AC
Therefore,
ACxAC = (AB)² + (BC)²
(AC)² = (AB)² + (BC)² Proved.

Conservation of Linear Momentum

Conservation of Linear Momentum

          Momentum

    The product of mass and velocity of a moving body is said to be momentum. 

It is generally denoted by P. 

The SI unit of momentum is kilogram metre per second or kgms⁻¹. 

The dimension of momentum is [MLT⁻¹]

          Law of conservation of Linear Momentum

"The vector sum of the linear momenta of all the particles in an isolated system remains constant in the absence of any external force"

          Proof:

    Consider about an isolated system having n particles. Let the masses of the particles be M1, M2, M3 ..... Mn and their velocities V1, V2, V3 ....... Vn respectively. 

The vector sum of the linear momenta of all the particles in the system is given by,

P  = M1V1 + M2V2 + M3V3 + ....................... + MnVn .......................... (i)

Let the total mass of the system is M and velocity of the centre of mass of the system is Vcm

Form of Equation (i) will be

P = MVcm 

Now derivative both sides with respect to t

dP/dt = M(Vcm/dt)

dP/dt = Macm (acm is the acceleration of centre of mass of system)

dP/dt = Fext (Fext is external force applied on the system)

When Fext is Zero

dP/dt = 0

We know that derivative of a constant term is zero.

Therefore,

P = Constant

Hence,

P =  M1V1 + M2V2 + M3V3 + ....................... + MnVn  = Constant.

Circular Motion and Physical Equation

Consider about a circular path of radius r. A body moves from A to B with uniform speed v in time t. The body subtends an angle 𝜃 at the centre of the circular path as angular displacement and an arch s as linear displacement.

Figure as regard this is drawn below.

Mathematical calculation:-

Change in velocity along X axis = vcos𝜃 - v
Change in velocity along Y axis = vsin𝜃 - 0
When 𝜃 is very small
cos𝜃 = 1 and sin𝜃 = 𝜃
Therefore,
Required change in velocity along X axis = 0
Required change in velocity along Y axis = v𝜃
The rate of change of velocity per unit time along Y axis is said to be centripetal acceleration.
It is generally denoted by a.

The rate of change of angular displacement per unit time is said to angular velocity w.
The SI unit of angular velocity is radian per second.
Angular velocity = w = 𝜃/t
Centripetal Acceleration = a = v𝜃/t
Therefore,
a = vw ------------------ (i)

In one complete rotation of the body in the circular path. Time taken to move the body to one complete rotation is T.

w = 2π/T ----------------- (ii)
v = 2πr/T ----------------- (iii)

Therefore,
From (ii) and (iii)
w/v = 1/r
w = vr

Now putting the value of w in (i)
a = v²/r
 This is the expression of centripetal acceleration.
If the mass of the body is m.
Centripetal Force = F = mv²/r

Required Physical Equation in circular motion are

1. Linear Displacement = l = 𝜃r
2. Linear Velocity = v = wr
3. Centripetal Acceleration = a = v²/r
4. Centripetal Force =  F = mv²/r

Velocity - Time Graph And Physical Equation

Consider about a body which moves from A to B in time t with uniform velocity. The velocity of the body at A and B is u and v respectively.

The slope of the velocity time graph = Acceleration = a
= Slope of AB
= BD/AD
From figure it is clear that
OA = u
BC = v
OC = t
OA = CD
BD = BC - CD
BC = v - u
Therefore,
a = (v - u)/t
v = u + at
t = (v - u)/a
u = v - at

The area described by body to move from A to B
= Distance covered by body = S
= Area of Trapezium
= (OA + BC)xOC/2
= (u + v)x t/2                ------------------- (i)
Now, putting the value of v.
(u + u + at)xt/2
S  = ut + 1/2 at^2

Now, from (i)
S = (u + v)(v - u)/2a
2aS = v^2 - u^2
Therefore,
v^2  = u^2 + 2aS

The required physical equation of velocity time graph are

a = (v -  u)/t
S = ut + 1/2at^2
v^2 = u^2 + 2aS