Coulomb’s Law

Coulomb’s Law

Coulomb’s law is a quantitative statement about the force between two point charges. When the linear size of charged bodies are much smaller than the distance separating them, the size may be ignored and the charged bodies are treated as point charges. Coulomb measured the force between two point charges and found that it varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges. Thus, if two point charges q1, q2 are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by 

F =|q1×q2|/r²

How did Coulomb arrive at this law from his experiments? Coulomb used a torsion balance* for measuring the force between two charged metallic spheres. When the separation between two spheres is much larger than the radius of each sphere, the charged spheres may be regarded as point charges.

Basic Properties of Electric Charge

1.5 Basic Properties of Electric Charge

We have seen that there are two types of charges, namely positive and negative and their effects tend to cancel each other. Here, we shall now describe some other properties of the electric charge. 

If the sizes of charged bodies are very small as compared to the distances between them, we treat them as point charges. All the  charge content of the body is assumed to be concentrated at one point  in space.

1.5.1 Additivity of charges

We have not as yet given a quantitative definition of a charge; we shall follow it up in the next section. We shall tentatively assume that this can be done and proceed. If a system contains two point charges q1 and q2, the total charge of the system is obtained simply by adding algebraically q1 and q2 , i.e., charges add up like real numbers or they are scalars like the mass of a body. If a system contains n charges q1, q2, q3, …, qn, then the total charge of the system is q1 + q2 + q3 + … + qn . Charge has magnitude but no direction, similar to mass. However, there is one difference between mass and charge. Mass of a body is always positive whereas a charge can be either positive or negative. Proper signs have to be used while adding the charges in a system. For example, the  total charge of a system containing five charges +1, +2, –3, +4 and –5,  in some arbitrary unit, is (+1) + (+2) + (–3) + (+4) + (–5) = –1 in the  same unit.

1.5.2 Charge is conserved

We have already hinted to the fact that when bodies are charged by rubbing, there is transfer of electrons from one body to the other; no new charges are either created or destroyed. A picture of particles of electric charge enables us to understand the idea of conservation of charge. When we rub two bodies, what one body gains in charge the other body loses. Within an isolated system consisting of many charged bodies, due to interactions among the bodies, charges may get redistributed but it is found that the total charge of the isolated system is always conserved. Conservation of charge has been established experimentally.

It is not possible to create or destroy net charge carried by any isolated system although the charge carrying particles may be created or destroyed in a process. Sometimes nature creates charged particles: a neutron turns into a proton and an electron. The proton and electron thus created have equal and opposite charges and the total charge is zero before and after the creation. 

1.5.3 Quantisation of charge

Experimentally it is established that all free charges are integral multiples of a basic unit of charge denoted by e. Thus charge q on a body is always given by

q = ne

where n is any integer, positive or negative. This basic unit of charge is the charge that an electron or proton carries. By convention, the charge on an electron is taken to be negative; therefore charge on an electron is written as –e and that on a proton as +e. 

The fact that electric charge is always an integral multiple of e is termed as quantisation of charge. There are a large number of situations in physics where certain physical quantities are quantised. The quantisation of charge was first suggested by the experimental laws of electrolysis discovered by English experimentalist Faraday. It was experimentally demonstrated by Millikan in 1912. 

In the International System (SI) of Units, a unit of charge is called a coulomb and is denoted by the symbol C. A coulomb is defined in terms the unit of the electric current which you are going to learn in a subsequent chapter. In terms of this definition, one coulomb is the charge flowing through a wire in 1 s if the current is 1 A (ampere), (see Chapter 2 of Class XI, Physics Textbook , Part I). In this system, the value of the basic unit of charge is 

e = 1.602192 × 10–19 C

Thus, there are about 6 × 1018 electrons in a charge of –1C. In electrostatics, charges of this large magnitude are seldom encountered and hence we use smaller units 1 µC (micro coulomb) = 10–6 C or 1 mC (milli coulomb) = 10–3 C. 

If the protons and electrons are the only basic charges in the universe, all the observable charges have to be integral multiples of e. Thus, if a body contains n1 electrons and n2 protons, the total amount of charge on the body is n2 × e + n1 × (–e) = (n2 – n1) e. Since n1 and n2 are integers, their difference is also an integer. Thus the charge on any body is always an integral multiple of e and can be increased or decreased also in steps of e. 

The step size e is, however, very small because at the macroscopic level, we deal with charges of a few µC. At this scale the fact that charge of a body can increase or decrease in units of e is not visible. In this respect, the grainy nature of the charge is lost and it appears to be continuous.

This situation can be compared with the geometrical concepts of points and lines. A dotted line viewed from a distance appears continuous to us but is not continuous in reality. As many points very close to each other normally give an impression of a continuous line, many  small charges taken together appear as a continuous charge  distribution.

At the macroscopic level, one deals with charges that are enormous compared to the magnitude of charge e. Since e = 1.6 × 10–19 C, a charge of magnitude, say 1 µC, contains something like 1013 times the electronic charge. At this scale, the fact that charge can increase or decrease only in units of e is not very different from saying that charge can take continuous values. Thus, at the macroscopic level, the quantisation of charge has no practical consequence and can be ignored. However, at the microscopic level, where the charges involved are of the order of a few tens or hundreds of e, i.e., they can be counted, they appear in discrete lumps and quantisation of charge cannot be ignored. It is the magnitude of scale involved that is very important.

Mathematics MCQ Class 12

CBSE Class 12 2023 : Maths Important MCQs for CBSE Board Exam 2023

In this post, CBSE Board 12th Maths Exam 2023 Important MCQs Question Answers have been given.

The purpose of these practice questions and answers is to help the students prepare for their exams and perform well in them.

It is advised to read this resource before the actual paper as it can give students an idea about the type of questions they can expect in the exam.

1 The number of equivalence relations that can be defined in the set A= {1,2,3} which containing the elements (1,2) is

(a) 0

(b) 1

(c) 2

(d) 3

2 The number of one-to-one functions that can be defined from the set {1,2,3,4,5} to {a, b}

a) 5

b) 0

c) 2

d) 3

3 What is the simplified form of cos-1(4x3 − 3x)

(a) 3sin−1 x

(b) 3cos−1 x

(c) π − 3 sin−1 x

(d) π − 3 cos−1 x

4 The number of all possible matrices of order2x3 with entry 1 or 2

1)16

2) 64

3) 6

4) 24

5 If the order of matrix P is 2x3 and the order of matrix Q is 3x4 , find the order of PQ.

1) 2x4

2) 2x2

3) 4x2

4) 3x3

6 Let A is a non – singular matrix of order 3 x 3 then | A ( adj A )| is equal to

a)| A |

b)| A |2

c)| A |3

d)3| A |

7 The function f(x) =[x] is continuous at

a)4

b)-2

c)1

d)1.5

8 The bottom of a rectangular swimming tank is 25 m by 40 m water is pumped into the tank at the rate of 500 cubic meters per minute. Find the rate at which the level of water in the tank is rising?

A) 1⁄4 m/min

B) 2/3 m/min

C) 1/3 m/min

D) 1⁄2 m/min

9 Area of region bound by circle x2+y2=1

a)2 sq units

b) sq units

c)3 sq units

d)4 sq units

10 Integrating factor of the differential equation dy/dx + y tan x – sec x = 0 is:

(A) cosx 

(B) secx 

(C) ecosx

(D) esecx

11 If points A (60 î+ 3 ĵ), (40 î– 8 ĵ) and C ( aî- 52ĵ) are collinear, then ‘a’ is equal to

a) 40

b) -40

c) 20

d) -20

12 Write direction cosines of a line parallel to z-axis.

(a) 1,0,0

(b) 0,0,1

(c) 1,1,0

(d) -1,-1,-1

13 Find the foot of the perpendicular drawn from the point (2,-3,4) on the y-axis.

(a) (2,0,4)

(b) (0.3.0)

(c) (0,-3,0)

(d) (-2,0,-4)

14 If α,β, Υ are the angles that a line makes with the positive direction of x,y,z axis respectively then the direction cosines of the line are

(a) cosα,sinβ, cosΥ

(b) cosα,cosβ, cosΥ

(c) sinα,sinβ, sinΥ

(d) 1, 1, 1

15 The feasible region of the inequality x+y≤1 and x–y≤1 lies in......... quadrants.

(a)Only I and II 

(b)Only I and III

(c)Only II and III

(d)All four

16 Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6, the probability of getting a sum 3 is

a)1/18

b)5/18

c)1/5

d)2

17 The solution set of the inequality 3x + 5y < 4 is

a)an open half-plane not containing the origin.

b)an open half-plane containing the origin.

c)the whole XY-plane not containing the line 3x + 5y = 4.

d)a closed half plane containing the origin.

18 If A is a square matrix of order 3 and |A| = 5, then |adjA| =

(a) 5

(b) 25

(c) 125

(d) ⅕

19 The area of a triangle with vertices (2, −6), (5,4) and (k, 4) is 35 square units then , k is

A.12

B. −2

C. −12, −2

D. 12, −2

20 The vector having initial and terminal points as (2,5,0) and (-3,7,4) respectively is

A. 5î+ 2ĵ− 4k̂

B. −î+ 12ĵ+ 4k̂

C. −5î+ 2ĵ+ 4k̂

D.−5î+ 12ĵ+ 4k̂

21. Let A = { 1 , 2, 3 } and consider the relation R = {(1 , 1), (2 , 2), ( 3 , 3), (1 , 2), (2 , 3), (1,3)} Then , R is

(a) Reflexive but not symmetric

(b) Reflexive but not transitive

(c) Symmetric and transitive

(d)Neither symmetric nor transitive

22. If A is a skew – symmetric matrix , then A² is

(a) Symmetric

(b) Skew – symmetric

(c) A² = A

(d) A² ≠ A

23. The number of arbitrary constants in the general solution of a differential equation of fourth order are :

(a) 0

(b) 2

(c) 3

(d) 4

24. If y = log x/ (1+x) , then dy/ dx is equal to

(a) x/ (1+x)²

(b) 1/ (1+x) ²

(c) 2x/ (1+x)²

(d) 1/ x(1+x)

25. Corner points of the region for an LPP are (0, 2), (3, 0), (6, 0),(6, 8), and (0, 5) . Let F = 4x + 6y be the objective function. The minimum value of F occurs at

(a) (0,2) only

(b)(3,0) only

(c) The midpoint of the line segment joining the points (0,2) and ( 3, 0) only

(d) Any point on the line segment joining the points (0,2) and (3, 0) only.

26. f(x) = xˣ has a stationary point at

(a) x = e

(b) x = 1/ e

(c) x = 1

(d) x = √e

27. If A and B are events such that P ( A/ B) = P (B/ A) , then ………

(a) P(A) = P(B)

(b) P(A) ≠ P(B)

(c) P(A) + P(B) = 1

(d) None of these

28. If R = {(x , y ): x + 2y = 8 } is a relation N , then the range of R is :

(a){3, 2 , 1}

(b) { 3 , 2}

(c) {2 , 8 , 1}

(d) {3}

29. Let f ∶ R → R be defined as f(x) = 3x. Choose the correct answer.

(a) f is one – one onto

(b) f is many-one onto

(c) f is one – one but not onto

(d) f is neither one-one nor onto

30. The principal value of [tan-1 √3 − cot-1, (−√3)] is :

(a)π

(b) − π/2

(c) 0

(d) 2√3

31. If cos (sin-1 2/ 5 + cos-1 x) = 0 , then x is equal to

(a) 1/ 5

(b)2/ 5

(c) 0

(d) 1

32. The value of ∫ (cos 2x)/ (sinx +cosx)² dx is

(a) log | cos x + sin x | + C

(b) log | cos x − sin x | + C

(c) log | cos x + sin x |² + C

(d) log | cos x + sin x |-2 + C

33. The derivative of 3ˣ  w.r.t x is :

(a) log 3

(b) x. 3ˣ-1

(c) 0

(d) 3ˣ log 3

34. A is a square matrix of order 2, then adj(adj A) is :

(a) O

(b) I

(c) A-1

(d) A

35. The number of arbitrary constants in the general solution of a differential equation of fourth order is :

(a) 1

(b) 2

(c) 3

(d) 4

 

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1. Chemical Reactions and Equations

Chemical Equations 

When a magnesium ribbon is burnt in oxygen, it gets converted to magnesium oxide. This description of a chemical reaction in a sentence form is quite long. It can be written in a shorter form. The simplest way to do this is to write it in the form of a word-equation.

The word-equation for the above reaction would be –

Magnesium + Oxygen → 

(reactants)

Magnesium Oxide 

(product)

The substances that undergo chemical change in the reaction, magnesium and oxygen, are the reactants. The new substance is magnesium oxide, formed during the reaction, as a product.

A word-equation shows change of reactants to products through an arrow placed between them. The reactants are written on the left-hand side (LHS) with a plus sign (+) between them. Similarly, products are written on the right-hand side (RHS) with a plus sign (+) between them. The arrowhead points towards the products, and shows the direction of the reaction.

Writing a Chemical Equation

Chemical equations can be made more concise and useful if we use chemical formulae instead of words. A chemical equation represents a chemical reaction. If you recall formulae of magnesium, oxygen and magnesium oxide, the above word-equation can be written as –

Mg + O2 → MgO

Count and compare the number of atoms of each element on the LHS and RHS of the arrow. Is the number of atoms of each element the same on both the sides? If yes, then the equation is balanced. If not, then the equation is unbalanced because the mass is not the same on both sides of the equation. Such a chemical equation is a skeletal chemical equation for a reaction. Equation is a skeletal chemical equation for the burning of magnesium in air.

Balanced Chemical Equations

Recall the law of conservation of mass that you studied in Class IX; mass can neither be created nor destroyed in a chemical reaction. That is, the total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants.

In other words, the number of atoms of each element remains the same, before and after a chemical reaction. Hence, we need to balance a skeletal chemical equation. Is the chemical Eq balanced? Let us learn about balancing a chemical equation step by step.

The word-equation for Activity may be represented as –

Zinc + Sulphuric acid → Zinc sulphate + Hydrogen

The above word-equation may be represented by the following chemical equation –

Zn + H2SO4 → ZnSO4 + H2 

Let us examine the number of atoms of different elements on both sides of the arrow.

As the number of atoms of each element is the same on both sides of the arrow, Eq.is a balanced chemical equation.

Let us try to balance the following chemical equation –

Fe + H2O → Fe3O4 + H2 

Step I: To balance a chemical equation, first draw boxes around each formula. Do not change anything inside the boxes while balancing the equation.

Step II: List the number of atoms of different elements present in the unbalanced equation.

Step III: It is often convenient to start balancing with the compound that contains the maximum number of atoms. It may be a reactant or a product. In that compound, select the element which has the maximum number of atoms. Using these criteria, we select Fe3O4 and the element oxygen in it. There are four oxygen atoms on the RHS and only one on the LHS.

To balance the oxygen atoms –

To equalise the number of atoms, it must be remembered that we cannot alter the formulae of the compounds or elements involved in the reactions. For example, to balance oxygen atoms we can put coefficient ‘4’ as 4 H2O and not H2O4 or (H2O)4. Now the partly balanced equation becomes–

Step IV: Fe and H atoms are still not balanced. Pick any of these elements to proceed further. Let us balance hydrogen atoms in the partly balanced equation.

To equalise the number of H atoms, make the number of molecules of hydrogen as four on the RHS.

The equation would be –

Step V: Examine the above equation and pick up the third element which is not balanced. You find that only one element is left to be balanced, that is, iron.

To equalise Fe, we take three atoms of Fe on the LHS.

Step VI: Finally, to check the correctness of the balanced equation, we count atoms of each element on both sides of the equation.

The numbers of atoms of elements on both sides of Eq. (1.9) are equal. This equation is now balanced. This method of balancing chemical equations is called hit-and-trial method as we make trials to balance the equation by using the smallest whole number coefficient.

Step VII: Writing Symbols of Physical States Carefully examine the above balanced Eq. (1.9). Does this equation tell us anything about the physical state of each reactant and product? No information has been given in this equation about their physical states.

To make a chemical equation more informative, the physical states of the reactants and products are mentioned along with their chemical formulae. The gaseous, liquid, aqueous and solid states of reactants and products are represented by the notations (g), (l), (aq) and (s), respectively. The word aqueous (aq) is written if the reactant or product is present as a solution in water.

The balanced Eq. (1.9) becomes

3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g) (1.10)

Note that the symbol (g) is used with H2O to indicate that in this reaction water is used in the form of steam.

Usually physical states are not included in a chemical equation unless it is necessary to specify them.

Sometimes the reaction conditions, such as temperature, pressure, catalyst, etc., for the reaction are indicated above and/or below the arrow in the equation. For example 

Using these steps, can you balance Eq. (1.2) given in the text earlier?