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.