Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 4 Vasanavikrithi Text Book Questions and Answers, Summary, Notes.
Students can Download Chapter 5 Law of Motion Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Summary
Laws Of Motion Class 11 Notes Chapter 5 Introduction
In this chapter we are going to learn about the laws that governs the motion of bodies.
Inertia:
The inability of a body to change by itself it’s state of rest or uniform motion along a straight line is called inertia.
Examples of inertia:
1. When a fast moving bus is suddenly stopped, a standing passenger tends to fall in the forward direction.
Explanation
The passenger has the same velocity as that of the bus. When the bus stops suddenly the lower part of his body is brought to rest suddenly because of the friction between his feet and floor of the bus. But the upper part continues to move because of its inertia.
2. When a bus suddenly takes off, a standing passenger tends to fall in the backward direction. This is because the lower part of the body gets a speed when the bus picks up speed and upper part continues to be at rest because of its inertia.
3. Consider a person sitting inside a stationary train and tossing a coin. The coin falls into his own hand. If he repeats the experiment when the train is moving with uniform speed, then also the coin falls into his own hand.
4. Cleaning a carpet by beating is in accordance with law of inertia.
5. Rabbit chased by a dog runs in zigzag manner. This is to take advantage of the large inertia of the massive dog.
6. A person chased by an elephant runs in a zigzag manner or in a circle. This is to take the advantage of the large inertia of the massive elephant.
Newton’s Laws:
Newton built on Galileo’s ideas and laid the foundation of mechanics in terms of three laws.
Laws Of Motion Class 11 Notes Pdf Chapter 5 Newton’s First Law Of Motion
Everybody continues in its state of rest or of uniform motion along a straight line unless it is compelled by an external unbalanced force to change that state:
Note: Newton’s first law of motion brings the idea of inertia. Inertia of a body is measured by the mass of the body. Heavier the body, greater is the force required to change its state and hence greater is its inertia.
Class 11th Physics Chapter 5 Notes Newton’s Second Law Of Motion
Linear Momentum (\(\vec{p}\)):
Momentum of a body is defined as the product of its mass m and velocity \(\vec{v}\)
Explanation
Momentum of a body can be produced or destroyed by the application of force on it. Therefore, momentum of a body is measured by the force required to stop the body in unit time.
Force required to stop a moving body depends upon
1. Mass of the body:
When a ball and a big stone are allowed to fall from the same height, we find that a greater force is required to stop the big piece of stone than the ball. Thus larger the mass of a body, greater is its linear momentum.
2. Velocity of the body:
A bullet thrown with the hand can be stopped easily than the same bullet fired from the gun. Therefore, langerthe velocity of a body, greater is its linear momentum.
Note: Momentum is a vector quantity. Its unit is Kgms-1
Newton’s Second Law of motion:
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Mathematically this can be written as
Laws Of Motion Class 11 Notes Pdf Download Chapter 5 Question 1.
Derive F = ma from Newton’s Second law.
Answer:
Consider a body of mass ‘m’ moving with a momentum \(\vec{p}\). Let \(\vec{F}\) be the force acting on it for time internal Dt. Due to this force the momentum is changed from \(\vec{p}\) to p + Dp. Then according to Newtons second law, we can write
Where K is a constant proportionality. When we take the limit ∆t → 0, we can write
Unit of force:
Unit of force is newton. 1N = 1Kgms-2
Force in terms of the components:
We know force is a vector, Hence we can write as
Impulsive force:
The forces which act on bodies for short time are called impulsive forces.
Example:
Impulse:
An impulse force does not remain constant, but changes from zero to maximum. This impulsive force is not easy to measure, because it changes with time. In such a case, we measure the total effect of the force called impulse.
The impulse of a force is the product of the average force and the time for which it acts.
Relation between impulse and momentum:
We know from Newtons second law
F = \(\frac{\Delta p}{\Delta t}\)
R.H.S. is the impulse and L.H.S. is change of momentum ie; change of momentum = impulse.
Laws Of Motion Notes Class 11 Chapter 5 Question 2.
When we jump to hard soil there is greater discomfort than when we jump to loose soil. Why?
Answer:
F = \(\frac{\Delta p}{\Delta t}\). When we jump to hard soil, Dt is small and F is large. When we jump to loose soil it takes more time for the body to come to rest. Therefore, Dt is large and F will be small.
Plus One Physics Laws Of Motion Chapter 5 Question 3.
A cricketer draws his hand while catching a cricket ball. Why?
Answer:
When cricketer draws his hand, the Dt will increase. Hence F acting on the hand will decrease.
Newtons Third Law Of Motion
Statement:
To every action, there is always an equal and opposite reaction.
Explanation: When a book is placed on the table, the weight of the book acts on the table downwards. The table exerts an equal force on the book in the upward direction. If the force applied by the book on the table is action, the force applied by the table on the book is reaction.
Newton’s Laws Of Motion Class 11 Notes Chapter 5 Question 4.
If action and reaction are equal and opposite, why they do not cancel?
Answer:
Though action and reaction are equal and opposite, they do not cancel each other because action is on one body and reaction is on another body.
Consider a pair of bodies A and B. According to the
third law FAB = – FBA
(force on A by B) = – (force on B by A).
Plus One Physics Laws Of Motion Notes Chapter 5 Conservation Of Momentum
Second law and third law lead to conservation of linear momentum.
Statement:
When there is no external force on a body (or system), the total momentum remains constant.
Proof in the case of a single body:
According to Newtons second law, F = \(\frac{d p}{d t}\). if F = 0, we get p = constant. Which means that momentum of a body remains constant, if there is no external force acting on it.
Conservation of momentum in the case of firing a gun:
Consider a gun of mass M and bullet of mass ‘m’ at rest. On firing the gun exerts a force F on the bullet and bullet exerts an equal force -F in the opposite direction. Because of this action and reaction (due to firing), the gun acquires a momentum Pg and bullet acquires a momentum Pb.
Momentum before firing
The bullet and gun are at rest. Hence momentum before firing = M × 0 + m × 0
Momentum before firing = 0 ________(1)
Momentum after firing
According to Newtons second law, the change in
momentum of bullet. ∆Pb = Pb – 0 = F∆t ______(2)
Since initially both are rest,
Dp = final momentum – initial
momentum Similarly the change in momentum of gun
∆pg = pg – 0 = -F∆t _______(3)
∴ Total momentum after firing = pb + pg
= F∆t + – F∆t.
Total momentum after firing = 0 _______(4)
from eq (1) and eq (4), we get,
Total momentum before firing = Total momentum after firing.
Conservation of momentum in the case of two colliding bodies:
Consider two bodies A and B with initial momenta PA and PB. After collision, they acquire momenta P1A and P1g respectively.
According to Newton’s second law, the change in momentum of A due to the collision with B,
Similarly the change in momentum of B due to the collision with A, FBA∆t = P1B – PB
[Where Dt is time for which the two bodies are in contact].
According Newton’s third law, we can write
FAB = -FBA
Total momentum before collision = Total momentum after collision.
Note: Conservation linear momentum is always satisfied for elastic collision and inelastic collision.
Class 11 Physics Chapter 5 Notes Equilibrium Of A Particle
Equilibrium of a particle in mechanics refers to the situation, when the net external force on the particle is zero.
Common Forces In Mechanics
There are two types of forces commonly used in mechanics,
1. Contact forces:
A contact force on an object arises due to contact with some other object. Example : Friction, viscosity, air resistance etc.
2. Non contact forces:
A non contact force on an object arises due to non contact with some other object
Example: Gravitational force
Friction:
Friction is the force that develops at the surfaces of contact of two bodies and impedes (opposes) their relative motion.
There are different types of friction.
Laws of static Friction:
a. Laws of Kinetic friction:
b. Coefficient of static friction:
The force of static friction (fs)max is directly proportional to the normal reaction N
(fs)max α N
Where ms is called coefficient of static friction.
Definition of ms
Coefficient of static friction is the ratio of the force of the maximum static friction to the nprmal reaction.
c. Coefficient of Kinetic friction:
The force of kinetic friction is directly proportional to the normal reaction N.
i e (fk)max α N
Where µk is called coefficient of Kinetic friction.
Definition of µk
Coefficient of Kinetic friction is the ratio of the force of Kinetic friction to the normal reaction.
d. Angle of friction:
Angle of friction is the angle whose tangent gives the coefficient of friction.
Proof:
Consider a body placed on a surface. Let N be the normal reaction and flimit is the limiting friction. Let ‘θ’ be the angle between Resultant vector and normal reaction. From the triangle OBC,
∴ tanθ = µ
Angle of repose:
The angle of repose is the angle of the inclined plane at which a body placed of it just begins to slide.
Explanation
considers body placed on a inclined plane. Gradually increase the angle of inclination till the body placed on its surface just begins to slide down. If α is the inclination at which the body just begins to slide down, then α is called angle of repose.
The limiting friction F acts in upward direction along the inclined plane. When the body just begins to move, we can write
F = mg sin α ______(1)
from the figure normal reaction,
N = mg cos α ______(2)
dividing eq (1) by eq (2)
Note: Angle of repose is equal to angle of friction.
Rolling friction:
Why rolling friction is less than kinetic friction?
When a body rolls over a plane, there is just one point of contact between the body and plane. The relative motion between point and plane is zero. Hence in this ideal situation, kinetic friction becomes zero.
Advantages of friction
Disadvantages of friction
Steps to reduce friction
Circular Motion
When a body moves along circumstances of a circle, there is an acceleration towards it’s centre. This acceleration is called centripetal acceleration. The force providing this acceleration is called centripetal force.
Centripetal force f = \(\frac{\mathrm{mv}^{2}}{\mathrm{R}}\)
1. Motion of a car on a level road:
Consider a vehicle moving overa level curved road. The two forces acting on it are
The normal reaction can’t produce sufficient centripetal force required for circular motion. The centripetal force for circular motion is provided by friction. This friction opposes the motion of the car moving away from the circular road. Hence condition for circular motion can be written as Centripetal force ≤ force of friction
The maximum speed of circular motion of the car
vmax = \(\sqrt{\mu_{s} \mathrm{rg}}\)
Question 5.
Why surface of the road is kept inclined to the horizontal?
Answer:
Consider a vehicle moving along a level curved road. The vehicle will have a tendency to slip outward. This outward slip is prevented by frictional force. But friction causes unnecessary wear and tear. More over, for typical value of µ and R the maximum speed v = \(\sqrt{\mu_{s} \mathrm{rg}}\) rg will be very small.
These defects can be avoided if we raise the outer edge of the road slightly above the inner edge. This process is called banking of curve. The angle made by the surface of the road with the horizontal is called the angle of banking.
2. Motion of a car on a banked road:
Consider a vehicle along a curved road with angle of banking q. Then the normal reaction on the ground will be inclined at an angle q with the vertical.
The vertical component can be divided into N Cosq (vertical component) and N sinq (horizontal component). Suppose the vehicle has a tendency to slip outward. Then the frictional force will be developed along the plane of road as shown in the figure. The frictional force can be divided into two components. Fcosq (horizontal component) and F sinq (vertical component).
From the figure are get
N cos q = F sinq + mg
N cosq – F sinq = mg ______(1)
The component Nsinq and Fsinq provide centripetal force. Hence
Dividing both numerator and denominator of L.H.S by N cosq. We get
This is the maximum speed at which vehicle can move over a banked curved road.
Optimum speed:
Optimum speed is the speed at which a vehicle can move over a curved banked road without using unnecessary friction.
When a car is moved with optimum speed Vo, m can be taken as zero.
putting m = 0 in the above equation we get
Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 3 Muhyadheen Mala Text Book Questions and Answers, Summary, Notes.
Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 2 Anukampa Text Book Questions and Answers, Summary, Notes.
Students can Download Chapter 5 Complex Numbers and Quadratic Equations Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Plus One Maths Chapter Wise Questions And Answers Pdf Chapter 5 Question 1.
If z1 = 2 – i, z2 = 1 + i
Answer:
1. |z1 + z2 + 1| = |2 – i + 1 + i + 1| = 4
|z1 – z2 + i| = |2 – i – 1 – i + i| = |1 – i|
\(=\sqrt{1+1}=\sqrt{2}\)
2.
Hsslive Maths Textbook Answers Plus One Chapter 5 Question 2.
Find the square root of -15 – 8i.
Answer:
Let x + iy = \(\sqrt{-15-8 i}\)
Then (x + iy)2 = -15 – 8i
⇒ x2 – y2 + 2xyi = – 15 – 8i
Equating real and imaginary parts, we have
x2 – y2 = -15 ______(1)
2xy = – 8
We know the identity
(x2 + y2)2 = (x2 – y2)2 + (2xy)2
= 225 + 64
= 289
Thus, x2 + y2 = 17 _______(2)
From (1) and (2), x2 = 1 and y2 = 16 or x = ±1 and y = ±4
Since the product xy is negative, we have
x = 1, y = -4 or, x = -1, y = 4
Thus, the square roots of -15 – 8i are 1 – 4i and -1 + 4i.
Plus One Maths Questions And Answers Chapter 5 Question 1.
Consider the complex number \(\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}\)
Answer:
1.
2.
The complex number lies in the first quadrant;
⇒ θ = α = \(\frac{5 \pi}{12}\)
Plus One Maths Chapter Wise Questions And Answers Chapter 5 Question 2.
Answer:
1.
2. 27x2 – 10x + 1 = 0
Complex Numbers Class 11 Extra Questions Chapter 5 Question 3.
Answer:
1. Given; 4x + i(3x – y) = 3 – 6i
⇒ 4x = 3; 3x – y = -6
2. 21x2 – 28x + 10 = 0
Complex Numbers And Quadratic Equations Chapter 5 Question 4.
Consider the complex number z = \(\frac{1+i}{1-i}\)
1. Write z in a + ib form.
2.
In the figure radius of the circle is 1. Write the polar form of the complex number represent by the points P and Q. (2)
3. Find the square root of i. (2)
Answer:
1.
2. Polar form of the point P is \(1\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)\)
Polar form of the point Q is \(1\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\)
3. i = 0 + i ⇒ \(\sqrt{i}\) = x + iy ⇒ i = x2 + y2 + 2xyi x2 + y2 = 0; 2xy = 1
(x2 + y2)2 = (x2 – y2)2 + 4x2y2
(x2 + y2)2 = 0 + (1)2 = 1
x2 + y2 = 1; x2 + y2 = 0
Complex Numbers And Quadratic Equations Class 11 Pdf Question 1.
Answer:
1.
2.
The complex number lies in the third quadrant;
3. ix2 – x + 12i = 0
Complex Numbers And Quadratic Equations Class 11 Solutions Question 1.
Express each of the following in a + ib form. (1 score each)
Answer:
1. (2 – 4i) + (5 + 3i) = (2 + 5) + (-4 + 3)i = 7 – i
2. (1 – i) – (-1 + 6i) = 1 – i + 1 – 6i = 2 – 7i
3. 3(7 + 7i) + i(7 + 7i) = 21 + 21i + 7i – 7 = 14 + 28i
4.
Question 2.
Express each of the following in a + ib form. (1 score each)
Answer:
1.
2.
3. i99 = i96 + 3 = i96i3 = -i
4. i111 + i222 + i333 + i108 + i220 + 2 + i332 + 1
= i3 + i2 + i1 = -i – 1 + i = -1
5. (7 – i)(2 + 7i) = 7 × 2 – 2i + 7 × 7i – i × 7i
= 14 – 2i + 49i + 7 = 21 + 47i
6. (-1 – i)(4 + 2i) = -4 – 4i – 2i + 2 = – 2 – 6i
7. (5 – 3i)2 = 52 – 2 × 5 × 3i + (3i)2
= 25 – 30i – 9 = 16 – 30i
8.
Question 3.
Find the multiplicative inverse of the following; (1 score each)
Answer:
1. Multiplicative inverse = \(\frac{1}{3-4 i}\)
2. Multiplicative inverse = \(\frac{1}{2-3 i}\)
3. Multiplicative inverse = \(\frac{1}{\sqrt{5}+3 i}\)
Question 4.
Express each of the following in a + ib form. (2 score each)
Answer:
Question 5.
Convert the following into polar form. (2 score each)
Answer:
1. Given; 1 + i = r(cosθ + isinθ)
r = \(\sqrt{1+1}=\sqrt{2}\)
tanα = \(\left|\frac{1}{1}\right|\) = 1 ⇒ α = \(\frac{\pi}{4}\)
The complex number lies in the first quadrant;
2. -1 + i = r(cosθ + isinθ)
r = \(\sqrt{1+1}=\sqrt{2}\)
tanα = \(\left|\frac{1}{-1}\right|\) = 1 ⇒ α = \(\frac{\pi}{4}\)
The complex number lies in the second quadrant;
3.
The complex number lies in the fourth quadrant;
4.
The complex number lies in the fourth quadrant;
Students can Download Chapter 3 Recording of Transactions – I & II Notes, Plus One Accountancy Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Summary:
Plus One Accountancy Chapter 3 Meaning of source documents:
Various business documents such as invoice, bills, cash memos, vouchers, which form the basis and evidence of a business transaction recorded in the books of account are called source documents.
Hsslive Plus One Accountancy Notes Meaning of accounting equation:
A statement of equality between debits and credits signifying that the assets of a business are always equal to the total liabilities and capital.
Plus One Accountancy Notes Rules of debit and credit:
An account is divided into two sides. The left side of an account is known as debit and the credit. The rules of debit and credit depend on the nature of an account. Debit and Credit both represent either increase or decrease, depending on the nature of an account.
These rules are summarised as follows:
Accountancy Class 11 Chapter 3 Notes Books of original entry:
The transactions are first recorded in these books in a chronological order. Journal is one of the books of original entry. The process of recording entries in the journal is called journalising.
Format of Journal
Chapter 3 Accounts Class 11 Notes Ledger:
A book containing all accounts to which entries are transferred from the books of original entry. Posting is process of transferring entries from books of original entry to the ledger.
Accountancy Class 11 Chapter 3 Solutions Journalising Format of a Ledger
Plus One Accountancy Chapter 3 Notes Special Journals:
Special journals are also called day books or subsidiary books. Transactions that cannot be recorded in any special journal are recorded in journal is called the “Journal Proper.”
The special-purpose journals are:
(a) Cash Book
A book used to record all cash receipts and payments. Cash book may be single column cash book, doulbe column cash book and three column cash book.
Single Column Cash book
This is cash book containing only one column for cash and prepared as cash account in ledger.
Format of Single Column Cash Book
Double Column Cash book:
This is cash book containing one more column for bank along with the cash column, it serves the purpose of cash and bank account.
Format of Double Column Cash Book
(b) Petty Cash Book:
A book used to record small cash payments
(c) Purchase Book / Purchase Journal:
A special journal in which only credit purchases are recorded.
Accounts Class 11 Chapter 3 Notes Format of Purchase Day Book
(d) Purchase Return Book:
A book in which return of purchased goods on credit is recorded.
Accounts Chapter 3 Class 11 Notes Format of Purchase Return Book
(e) Sales Book / Sales Journal:
A special journal in which only credit sales are recorded.
Format of Sales Day Book
(f) Sales Return Book:
A special book in which return of goods sold on credit is recorded.
Format of Sale Return Book
Balancing the Accounts:
Accounts in the ledger are periodically balanced, generally at the end of the accounting period with the object of ascertaining the net position of each amount.
Balancing of an account means that the two sides are totaled and the difference between them is shown on the side which is shorter in order to make their totals equal. The words ‘balance carried down (c/d)’ are written against the amount of the difference between the two sides.
The amount of balance is brought down (b/d) in the next accounting period indicating that it is a continuing account until finally settled or closed. In case the debit side exceeds the credit side.
The difference is written on the side, if the credit side exceeds the debit side, the difference between the two appears on the debit side and is called debit and credit balance respectively. The accounts of expenses losses, gains and revenues are not balanced but are closed by transferring to trading and profit and loss account.
Students can Download Chapter 1 Indian Economy on the Eve of Independence Notes, Plus One Economics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Low level of economic development under the colonial rule:
The British rule started in India in 1757 and came to an end in 1947. The Indian economy underwent rapid changes under British rule. The economic policies pursued by the colonial government in India were concerned more with the protection and promotion of the economic interests of their home country than the development of the Indian economy. The twin objectives of British rule in India were
Plus One Economics Notes Chapter 1 Agricultural Sector:
Agricultural Sector was the backbone of the Indian economy.
During the British colonial rule India remained fundamentally an agrarian economy. Around eighty percent of India’s population lived in villages. Agriculture was stagnant and it was the main source of livelihood of the population. People depended directly or indirectly on agriculture and its productivity was very slow. The agricultural sector stagnated during British rule.
Major reasons for agricultural stagnation were:
Economics Plus One Notes Chapter 1 Industrial Sector:
India’s industrial sector could not make progress during British rule. Their aim was to collect raw materials from India and sell their final products in India.
By the second half of the nineteenth century, modem industry began to take root in India. Initially, cotton industries in Maharashtra and Gujarat (Bombay presidency) and the jute industry in Bengal were established. Then industries of fertilizers, rayon, rubber, cement, sugar, pepper, etc., were established in some regions of the country. The setting up of Tata Iron and Steel Company (TISCO) in 1907 was a landmark in the industrialization of India. Jemshedji Tata established TISCO in Jamshedpur in Bihar. During the British rule hardly any capital goods industries were established in the country.
Plus One Economics Chapter 1 Foreign Trade:
Though India exported value-added products before the British period, we started exporting primary products during their rule. Consequently, India became an exporter of primary products such as raw silk, cotton, wool, sugar, indigo, jute, etc. and an importer of finished consumer goods like cotton, silk and woollen clothes and capital goods like light machinery produced in the factories of Britain.
The most important characteristic of India’s foreign trade, throughout the colonial period was the generation of a large export surplus.
Plus One Economics Chapter 1 Notes Demographic Condition:
Various details about the population of British India were first collected through a census in 1881. Through Suffering from certain limitations, it revealed the unevenness in India’s population growth. Subsequently, every ten years such census operations were carried out. Before 1921, India was in the first stage of demographic transition. The second stage of transition began after 1921.
Economics Notes Plus One Chapter 1 Occupational Structure:
Occupational structure refers to the distribution of working persons across different industries and sectors. Broadly we divide occupations into three types. Agriculture, animal husbandry, forestry, fisheries, etc., are collectively known as ‘primary’ activities. Manufacturing industries, both small and large scale, are known as ‘secondary’ activities. Transport, communication, banking, financial services, etc., are ‘tertiary’ activities.
Hsslive Economics Plus One Chapter 1 Infrastructure:
Infrastructural facilities developed in India during the British period. Infrastructure means some kind of permanent installation, which are used over a long period of time for the supply of basic inputs like railway lines, roads, dams, canal systems, power stations, pipelines, hospitals, educational institutions like schools, colleges, etc. Basic infrastructure facilities such as railways, ports, water transport, and telegraph did develop during the British rule. The real intention behind such a development was to serve the various colonial interests of Britain.
Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Chapter 1 Peeli Kannukal Text Book Questions and Answers, Summary, Notes.
Students can Download Chapter 2 Structure of Atom Notes, Plus One Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Plus One Chemistry Chapter 2 Notes Pdf Introduction
The atomic theory of matter was first proposed by John Dalton. His theory, called Dalton’s atomic theory, regarded the atom as the ultimate particle of matter.
Sub-Atomic Particles
Discovery Of Electron
The experiments of Michael Faraday in discharge tubes showed that when a high potential is applied to a gas taken in the discharge tube at very low pres-sures, certain rays are emitted from the cathode. These rays were called cathode rays.
The results of these experiments are summarised below:
1. The cathode rays start from cathode and move towards the anode.
2. In the absence of electrical or magnetic field, these rays travel in straight lines.ln the presence of electrical or magnetic field, they behave as negatively charged particles, i.e.,they consist of negatively charged particles, called electrons.
3. The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.Thus, we can conclude that electrons are the basic constituent of all the atoms.
Charge To Mass Ratio Of Electron
In 1897, the British physicist J.J. Thomson measured the ratio of electrical charge (e) to the mass of electron (m<sub>e</sub>) by using cathode ray tube and applying electrical and magnetic field perpendicular to each other as well as to the path of electrons.
From the amount of deviation of the particles from their path in the presence of electrical or magnetic field, the value of e/m was found to be 1.75882 × 1011 coulomb per kg or approximately 1.75288 × 10<sup>8</sup> cou-lomb per gram. The ratio e/m was found to be same irrespective of the nature of the gas taken in the dis-charge tube and the material used as the cathode.
Structure Of Atom Class 11 Notes Charge Of The Electron
Millikan (1868-1953) devised a method known as Oil drop experiment (1906-14), to determine the charge on the electrons. He found the charge on the electron to be – 1.6 × 10-19C.
Mass of the electron (m)
Discovery Of Protons And Neutrons
Electrical discharge earned out in the modified cathode ray tube led to the discovery of canal rays. The characteristics of these positively charged particles are listed below:
The smallest and lightest positive ion was obtained from hydrogen and was called proton. Later, electrically neutral particles were discovered by Chadwick (1932) by bombarding a thin sheet of beryllium by α – particles when electrically neutral particles having a mass slightly greater than that of the protons was emitted. He named these particles as neutrons.
Atomic Models
Structure Of Atom Class 11 Notes Hsslive Thomson Model Of Atom
J.J. Thomson was the first to propose a model of the atom. According to him, the atom is a sphere in which positive charge is spread uniformly and the electrons are embedded in it so as to make the atom electrically neutral. This model is also known as “plumpudding model’. But this model was soon discarded as it could not explain many of the experimental observations.
Hsslive Structure Of Atom Notes Rutherford’s Nuclear Model of Atom
Rutherford and his students (Hans Geiger and Ernest Marsden) bombarded very thin gold foil with α – particles. The experiment is known as α -particle scattering experiment. On the basis of the observations, Rutherford drew the following conclusions regarding the structure of atom :
1. Most of the space in the atom is empty as most of the α -particles passed through the foil undeflected.
2. A few α – particles were deflected. Since the α – particles are positively charged, the deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as Thomson had presumed. The positive charge has to be concentrated in a very small volume that repelled and deflected the positively charged α – particles.
3. Calculations by Rutherford showed that the volume occupied by the nucleus is negligibly small as compared to the total volume of the atom.
On the basis of above observations and conclusions, Rutherford proposed the nuclear model of atom (after the discovery of protons). According to this model:
1.The positive charge and most of the mass of the atom was densely concentrated in extremely small region. This very small portion of the atom was called nucleus by Rutherford.
2. The electrons move around the nucleus with a very high speed in circular paths called orbits. Thus, Rutherford’s model of atom resembles the solar system in which the nucleus plays the role of sun and the electrons that of revolving planets.
3. Electrons and the nucleus are held together by electrostatic forces of attraction.
Chemistry Notes For Class 11 Chapter 2 Atomic Numberand Mass Number
’ Knowing the atomic number Z and mass number A of an element, we can calculate the number of protons, electrons and neutrons present in the atom of the element.
Atomic Number (Z) = Number of protons = Number of electrons
Mass Number (A) – Atomic number (Z) = Number of neutrons
Isotopes, Isobars And Isotones
Isotopes are atoms of the same element having the same atomic number but different mass numbers. They contain different number of neutrons. For ex-ample, there are three isotopes of hydrogen having mass numbers 1,2 and 3 respectively. All the three isotopes have atomic number 1. They are represented as \(_{ 1 }^{ 1 }{ H }\), \(_{ 1 }^{ 2 }{ H }\) and \(_{ 1 }^{ 3 }{ H }\) and named as hydrogen or protium, deuterium (D) and tritium (T) respectively. Isobars are atoms of different elements which have the same mass number. For example, \(_{ 6 }^{ 14 }{ C }\) and \(_{ 7 }^{ 14 }{ N }\) are isobars.
Isotones may be defined as atoms of different elements containing same number of neutrons. For example \(_{ 6 }^{ 13 }{ C }\) and \(_{ 7 }^{ 14 }{ N }\) are isotones.
Developments Leading To The Bohr’S Model Of Atom
Neils Bohr improved the model proposed by Rutherford. Two developments played a major role in the formulation of Bohr’s model of atom. These were:
Wave Nature Of Electromagnetic Ra-Diation
Light is the form of radiation and it was supposed to be made of particles known as corpuscules.
As we know, waves are characterised by wavelength (λ), frequency (υ) and velocity of propagation (c) and these are related by the equation
c = vλ or v = \(\frac { c }{ \lambda } \)
The wavelengths of various electro magnetic radia-tions increase in the order.
γ rays < X-rays< uv rays < visible < IR < Microwaves < Radio waves
Particle Nature Of Electro Magnetic Radiation: Planck’S Quantum Theory
Planck suggested that atoms and molecules could emit (or absorb) energy only in discrete quantities and not in a continuous manner, a belief popular at that time. Planck gave the name quantum to the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation. The energy (E) of a quantum of radiation is proportional to its frequency (υ) and is expressed by the equation E = hυ
Class 11 Chemistry Chapter 2 Notes Photoelectric Effect
When a metal was exposed to a beam of light, electrons were emitted. This phenomenon is called photoelectric effect. Obseravations of the photoelectric effect experiment are the following:
The kinetic energies of these electrons increase with the increase of frequency of the light used.
Using Plank’s quantum theory Einstein explained photoelectric effect. When a light particle, photon with sufficient energy strikes an electron instantaneously to the electron during the collision and the electron is ejected without any time lag. Greater the energy of photon greater will be the kinetic energy of ejected electron and greater will be the frequency of radiation.
If minimum energy to eject an electron is hv0 and the photon has an energy equal to hv. Then kinetic en-ergy of photoelectron is given by, hv=hv0 + 1/2 mev2 where me is the mass of electron and hv0 is called the work function.
Duel Behaviour Of Electromagnetic Ra-Diation
Light has dual behaviour that is it behaves either as a wave or as a particle. Due to this wave nature, it shows the phenomena of interference and diffraction.
Evidence For The Quantized Electronic Energy Levels : Atomic Spectra
It is observed that when a ray of white light is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength. Since ordinary white light consists of waves with ail the wave-lengths in the visible range, a ray of white light is spread out into a series of coloured bands called spectrum. In a continuous spectrum light of different colours merges together. For example violet merges into blue, blue into green and soon.
Chapter 2 Class 11 Chemistry Notes Emission and absorption spectra
The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum. Atoms, molecules or ions that have absorbed radiation are said to be “excited”.
A continuum of radiation is passed through a sample which absorbs radiation of certain wavelengths. The missing wavelength which corresponds to the radiation absorbed by the matter, leave dark spaces in the bright continuous spectrum. The study of emission or absorption spectra is referred to as spectroscopy Line spectra or atomic spectra is the spectra where emitted radiation is identified by the appearance of bright lines in the spectra.
Line spectrum of Hydrogen
The hydrogen spectrum consists of several series of lines named after their discoverers. Balmershowed in 1885 on the basis of experimental observations that if spectral lines are expressed in terms of wavenumber (\(\overline { v } \)), then the visible lines of the hydrogen spectrum obey the following formula :
\(\overline { v } \) = 109,677 \(\left[\frac{1}{2^{2}}-\frac{1}{n^{2}}\right] \mathrm{cm}^{-1}\)
where n = 3, 4, 5, ………….
The series of lines described by this formula are called the Balmer series.
The value 109,677cm-1 is called the Rydberg constant for hydrogen. The first 5 series of lines correspond to n1 = 1, 2, 3, 4, 5 are known as Lyman, Balmer, Paschen, Bracket and Pfund series respectively. Line specrum becomes more complex for heavier atoms.
Chapter 2 Chemistry Class 11 Notes Bhor’S Model For Hydrogen Atom
Bhors model for hydrogen atom says that
1. the energy of an electron does not change with time.
The diagram shows the Lyman, Balmer and Paschen series of transitions for hydrogen atom.
2. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ∆E, is given by :
\(v=\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}\)
E1 and E2 are the energies of the lower and higher allowed energy states respectively.
The angular momentum of an electron in a given stationary state can be expressed as in equation,
Chemistry Chapter 2 Class 11 Notes Bohr’s theory for hydrogen atom:
1. The stationary states for electron are numbered n = 1,2,3. These integral numbers are known as Principal quantum numbers.
2. The radii of the stationary states are expressed as:
rn = n² a0
where a0 = 52.9 pm
3. The most important property associated with the electron, is the energy of its stationary state. It is
given by the expression, \(E_{n}=-R_{H}\left(\frac{1}{n^{2}}\right)\)
where RH is called Rydberg constant and its value is 2.18 × 10-18 J. The energy of the lowest state, also called as the ground state, is
E1 = -2.18 × 10-18 \(\left(\frac{1}{1^{2}}\right)\) = -2.18 × 10-18 J. The energy of the stationary state for n = ∝, will be :
E2 = -2.18 × 10-18 J\(\left(\frac{1}{2^{2}}\right)\) = -0.545 × 10-18 J.
When the electron is free from the influence of nucleus(n = ∞), the energy is taken as zero. When the electron is attracted by the nucleus and is present in orbit n, the energy is emitted and its energy is lowered. That is the reason for the presence of negative sign and depicts its stability relative to the reference state of zero energy and n = ∞
4. Bohr’s theory can also be applied to the ions containing only one electron, similar to that present in hydrogen atom. For example, He<sup>+</sup> Li<sup>2+</sup>, Be<sup>3+</sup> and so on. The energies of the stationary states associated with these hydrogen-like species are given by the expression,
Structure Of Atom Class 11 Notes Pdf Explanation of Line Spectrum of Hydrogen
The frequency (v) associated with the absorption and emission of the photon can be evaluated by using equation,
Class 11 Chapter 2 Chemistry Notes Limitations of Bohr’s Model
Bohr’s model was too simple to account for the following points:
1. It fails to account for the finer details (doublet, that is two closely spaced lines) of the hydrogen atom spectrum. This model is also unable to explain the spectrum of atoms other than hydrogen Further, Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or an electric field (Stark effect).
2. It could not explain the ability of atoms to form molecules by chemical bonds.
Towards Quantum Mechanical Model Of The Atom
Two important developments which contributed significantly in the formulation of a more suitable and general model for atoms were:
Structure Of Atom Class 10 Notes Pdf Dual Behaviour of Matter
The French physicist, de Broglie proposed that matter, like radiation, should also exhibit dual behaviour i. e., both particle and wavelike properties. This means that just as the photon, electrons should. also have momentum as well as wavelength. de Broglie, from this analogy, gave the following relation between wavelength (λ) and momentum (p) of a material particle.
\(\lambda=\frac{h}{m v}=\frac{h}{p}\)
Heisenberg’s Uncertainty Principle
Werner Heisenberg a German physicist in 1927, stated uncertainty principle which is the consequence of dual behaviour of matter and radiation. It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as in equation,
∆x is the uncertainty in position and ∆p<sub>x</sub> (or ∆v<sub>x</sub>) is the uncertainty in momentum (or velocity) of the particle. If the position of the electron is known with high degree of accuracy (∆x is small), then the velocity of the electron will be uncertain ∆v<sub>x</sub> is large]. On the other hand, if the velocity of the electron is known precisely ( ∆v<sub>x</sub> is small), then the position of the electron will be uncertain (∆x will be large). Thus, if we carry out some physical measurements on the electron’s position or velocity, the outcome will always depict a fuzzy or blur picture.
Significance of Uncertainty Principle
Heisenberg Uncertainty Principle rules out existence of definite paths or trajectories of electrons and other similar particles. The trajectory of an object is determined by its location and velocity at various moments. If we know where a body is at a particular instant and if we also know its velocity and the forces acting on it at that instant, we can tell where the body would be sometime later. We, therefore, conclude that the position of an object and its velocity fix its trajectory. The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects.
Reasons for the Failure of the Bohr Model
In Bohr model, an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus. The wave character of the electron is not considered in Bohr model. Further, an orbit is a clearly defined path and this path can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Bohr.model of the hydrogen atom, therefore, not only ignores dual behaviour of matter but also contradicts Heisenberg uncertainty principle. There was no point in extending Bohr model to other atoms. In fact, an insight into the structure of the atom was needed which could account for wave-particle duality of matter and be consistent with Heisenberg uncertainty principle. This came with the advent of quantum mechanics.
Quantum Mechanical Model Of Atom
Quantum mechanics is a theoretical science that deals with the study of motions of microscopic objects such as electrons.
In quantum mechanical model of atom, the behaviour of an electron in an atom is described by an equation known as Schrodinger wave equation. Fora system, such as an atom or molecule whose energy does not change with time, the Schrodinger equation written as Hψ = Eψ where H is a mathematical operator, called Hamiltonian operator, E is the total energy and ψ is the amplitude of the electron wave called wave function.
Hydrogen Atom And The Schrodinger Equation
The wave function ψ as such has no physical significance. It only represents the amplitude of the electron wave. However ψ² may be considered as the probability density of the electron cloud. Thus, by determining ψ² at different distances from the nucleus, it is possible to trace out or identify a region of space around the nucleus where there is high probability of locating an electron with a specific energy.
According to the uncertainty principle, it is not possible to determine simultaneously the position and momentum of an electron in an atom precisely. So Bohr’s concept of well defined orbits for electron in an atom cannot hold good. Thus, in quantum mechanical mode, we speak of probability of finding an electron with a particular energy around the nucleus. There are certain regions around the nucleus where probability of finding the electron is high. Such regions are called orbitals. Thus an orbital may be defined as the region in space around the nucleus where there is maximum probability of finding an electron having a specific energy.
Orbitals and Quantum Numbers
Orbitals in an atom can be distinguished by their size, shape and orientation. An orbital of smaller size means there is more chance of finding the electron near the nucleus. Similarly, shape and orientation mean that there is more probability of finding the electron along certain directions than along others. Atomic orbitals are precisely distinguished by what are known as quantum numbers. Each orbital is designated by three quantum numbers labelled as n, l and m<sub>l</sub>
The principal quantum number n’ is a positive integer with value of n= 1, 2, 3 ……………
The principal quantum number determines the size and to large extent the energy of the orbital.
The principal quantum number also identifies the shell. With the increase in the value of ‘n’, the number of allowed orbital increases and are given by ‘n²’ Ait the orbitals of a given value of ‘n’ constitute a single shell of atom and are represented by the following letters
n= 1 2 3 4 ………………
Shell = K LM N ………………
Size of an orbital increases with increase of principal quantum number ‘n’. Since energy of the orbital will increase with increase of n.
Azimuthal quantum number, ‘F is also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital. For a given value of n, l can have n values ranging from 0 to (n – 1), that is, for a given value of n, the possible value of l are: l = 0, 1, 2, ……….. (n – 1)
Each shell consists of one or more subshells or sub-levels. The number of subshells in a principal shell is equal to the value of n. For example h the first shell (n = 1), there is only one sub-shell which corresponds to l = 0. There are two sub-shells (l= 0, 1) in the second shell (n = 2), three l= 0, 1, 2) and so on. Each sub-shell is assigned an azimuths! quantum number (l). Sub-shells corresponding to different values of l are represented by the following symbols.
l : 0 1 2 3 4 5 …………….
Notation for sub-shell : s p d f g h …………….
Magnetic orbital quantum number. ‘m<sub>l</sub>’ gives information about the spatial orientation of the or bital with respect to standard set of co-ordinate axis. For any sub-shell (defined by T value) 21+ 1 values of m,are possible and these values are given by:
m, = -l, -(l-1), (l-2)… 0, 1… (l-2), (l-1), l Thus for l = 0, the only permitted value of m,= 0, [2(0) + 1 = 1, one s orbital].
Electron spin ‘s’:
George Uhlenbeck and Samuel Goudsmit proposed the presence of the fourth quantum number known as the electron spin quantum number (m<sub>s</sub>). Spin angular momentum of the electron — a vector quantity, can have two orientations relative to the chosen axis. These two orientations are distinguished by the spin quantum numbers ms which can take the values of +½ or -½. These are called the two spin states of the electron and are. normally represented by two arrows, ↑ (spin up) and ↓ (spin down). Two electrons that have different m<sub>s</sub> values (one +½ and the other -½) are said to have opposite spins. An orbital cannot hold more than two electrons and these two electrons should have opposite spins.
Shapes of Atomic Orbitals
The orbital wave function or V for an electron in an atom has no physical meaning. It is simply a mathematical function of the coordinates of the electron.
According to the German physicist, Max Bom, the square of the wave function (i.e., ψ²) at a point gives the probability density of the electron at that point.
For 1 s orbital the probability density is maximum at the nucleus and it decreases sharply as we move away from it. The region where this probability I density function reduces to zero is called nodal surfaces or simply nodes. In general, it has been found that ns-orbital has (n – 1) nodes, that is, number of nodes increases with increase of principal quantum number n.
These probability density variation can be visualised . in terms of charge cloud diagrams.
Boundary surface diagrams of constant probability density for different orbitals give a fairly good representation of the shapes of the orbitals. In this representation, a boundary surface or contour surface is drawn in space for an orbital on which the value of probability density |ψ|² is constant. Boundary ‘ surface diagram for a s orbital is actually a sphere centred on the nucleus. In two dimensions, this sphere looks like a circle. It encloses a region in which probability of finding the electron is about 90%. The s-orbitals are spherically symmetric, that is, the probability of finding the electron at a given distance is equal in all the directions.
unlike s-orbitals, the boundary surface diagrams of p orbitals are not spherical. Instead, each p orbital consists of two sections called lobes that are on either side of the plane that passes through the nucleus. The probability density function is zero on the plane where the two lobes touch each other. The size, shape and energy of the three orbitals are identical. They differ, however, in the way the lobes are oriented. Since the lobes may be considered to lie along the x, y or z-axis, they are given the designations 2px, 2py, and 2pz. It should be understood, however, that there is no simple relation between the values of m, (-1, 0 and+1) and the x, y and z directions. For our purpose, it is sufficient to remember that, because there are three possible values of m, there are, therefore, three p orbitals whose axes are mutually perpendicular. Like s orbitals, p orbitals increase in size and energy with increase in the principal quantum number
The number of nodes are given by (n -2), that is number of radial node is 1 for 3p orbital, two for 4p orbital and so on.
For l = 2, the orbital is known as d-orbital and the minimum value of principal quantum number (n) has to be 3 as the value of l cannot be greater than n-1. There are five m; values (-2, -1, 0, +1 and +2) for l = 2 and thus there are five d orbitals. The five d-orbitals are designated as dxy, dyz, dxz, dx²-y² and dz². The shapes of the first fourd-orbitals are similarto each other, where as that of the fifth one, dz², is different from others, but all five 3d orbitals are equivalent in energy. The d orbitals for which n is greater than 3 (4d, 5d…) also have shapes similar to 3d orbital, but differ in energy and size.
Besides the radial nodes (i.e., probability density function is zero), the probability density functions for the np and nd orbitals are zero at the plane (s), passing through the nucleus (origin). For example, in case of pz orbital, xy-plane is a nodal plane, in case of dxy orbital, there are two nodal planes passing through the origin and bisecting the xy plane containing z-axis. These are called angular nodes and number of angular nodes are given by T, i.e., one angular node for p orbitals, two angular nodes for cf orbitals and so on. The total number of nodes are given by (n-1), i.e., sum of I angular nodes and (n-l-1) radial nodes.
Energies Of Orbitals
The order of energy of orbitals in single electron sys-tem are given below:
1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f The orbitals having same energy are called degenerate.
Filling Of Orbitals In Atom
Aufbau principle: According to this principle in the ground state of an atom, an electron will occupy the orbital of lowest energy and orbitals are occupied by electrons in the order of increasing energy.
Pauli’s exclusiohn principle : Pauli’s exclusion principle states that ‘no two electrons in an atom can have the same values for all the four quantum numbers’
Since the electrons in an orbital must have the same n, I and m quantum numbers, if follows that an orbital can contain a maximum of two electrons provided their spin quantum numbers are different. This is an important consequence of Pauli’s exclusion principle which says that an orbital can have maximum two electrons and these must have opposite spins.
Hund’s rule of maximum multiplicity :
This rule states that electron pairing in orbitals of same energy will not take place until each available orbital of a given subshell is singly occupied (with parallel spin).
The rule can be illustrated by taking the example of carbon atom. The atomic number of carbon is 6 and its electronic configuration is 1s²2s²2p². The two electrons of the 2p subshell can be distributed in the following three ways.
According to Hund’s rule, the configuration in which the two unpaired electron occupying 2px, and 2py orbitals with parallel spin is the correct configuration of carbon.
Exceptional configurations of chromium and copper
The electronic configuration of Cr (atomic number 24) is expected to be [Ar] 4s² 3d4, but the actual configuration is [Ar] 4s¹ 3d5. Similarly, the actual configuration of Cu (At. No. 29) is [Ar] 4s¹ 3d10 instead of the expected configuration [Ar] 4s² 3d9.
This is because of the fact that exactly half filled or completely filled orbitals (i.e., d5, d10, f7, f14) have lower energy and hence have extra stability.
Students can Download Chapter 1 Biological Classification Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations
Plus One Botany Chapter Wise Questions And Answers Pdf Question 1.
In Whit takers, five-kingdom classification eukaryotes are distributed among
(a) two kingdoms
(b) three kingdoms
(c) four kingdoms
(d) all the five kingdoms
Answer:
(c) four kingdoms
Plus One Botany Chapter Wise Questions And Answers Question 2.
Cyanobacteria are classified under which of the following kingdoms?
(a) Monera
(b) Protista
(c) Plantae
(d) Algae
Answer:
(a) Monera
Plus One Botany Questions And Answers Question 3.
Main component of cell wall of fungi is
(a) cellulose
(b) chitin
(c) pectin
(d) silica
Answer:
(b) chitin
Plus One Botany Chapter Wise Previous Questions And Answers Question 4.
Dinoflagellates are mostly
(a) marine and saprophytic
(b) freshwater and saprophytic
(c) marine and photosynthetic
(d) terrestrial and
Answer:
(c) marine and photosynthetic
Plus One Biology Chapter Wise Questions And Answers Pdf Question 5.
Which of the following kingdoms do viruses belong to
(a) monera
(b) Protista
(c) fungi
(d) none of these
Answer:
(d) none of these
Hsslive Plus One Botany Chapter Wise Questions And Answers Question 6.
Observe the relationship between the first pair and fill up the blanks.
Answer:
Ethylene.
Plus One Botany Chapter Wise Previous Year Questions And Answers Question 7.
Fill in the blanks.
Answer:
Plus One Botany Chapter Wise Questions And Answers Hsslive Question 8.
Who proposed Five kingdom classification?
Answer:
R .H. Whittaker
Plus One Botany Previous Questions Chapter Wise Pdf Question 9.
Find out the correct sequence of taxonomical category.
Answer:
2. species → genus → order → phylum
Biological Classification Important Questions Question 10.
In the five-kingdom system of Whittaker, how many kingdoms are eukaryotes?
Answer:
Four kingdoms
Plus One Botany Previous Question Papers Chapter Wise Question 11.
Observe the relationship between the first pair and fill up the blanks.
Answer:
Biology Classification Questions And Answers Question 12.
Find out the odd one.
a. Diatom, Gonyaulax, Yeast, Euglena, Plasmodium
Answer:
Yeast
Hsslive Botany Previous Questions And Answers Question 13.
Vinod observed blooms in a polluted water body, his friend Kumar said that it might be nitrogen-fixing Nostoc or Anabaena. Can you suggest which type of cell can fix atmospheric nitrogen in these organisms?
Answer:
Heterocyst
Botany Chapter Wise Questions And Answers Question 14.
Observe the relationship of the terms in the first pair and fill in the blanks:
Answer:
Biological Classification Previous Year Questions Question 15.
Difference between Virus and Viroid.
(a) Absence of protein coat in viroid but present in virus
(b) Presence of low molecular weight RNA in virus but absent in viroid
(c) Both a and b
(d) None of the above
Answer:
(a) Absence of protein coat in viroid but present in virus
Question 16.
Viruses are non-cellular organisms but replicate themselves once they infect the host cell. To which of the following kingdom do viruses belong to?
(a) Monera
(b) Protista
(c) Fungi
(d) None of the above
Answer:
(d) None of the above
Question 17.
A virus is considered as a living organism and an obligate parasite when inside a host cell. But virus is not classified along with bacteria or fungi. What are the characters of virus that are similar to nonliving objects?
Answer:
Viruses are acellular and can be crystallized.
Question 1.
The seven taxonomic categories are given below. Arrange them in the correct sequence starting from the smallest taxon.
Class → species → kingdom → order → family → division → genus.
Answer:
Species → genus → family → order → class → division → kingdom.
Question 2.
“Two kingdom classification is inadequate one”. Comment on it.
Answer:
Question 3.
Five-kingdom classification of organism was given by R.H.Whittaker. State the criteria followed by Whittaker for his classification.
Answer:
Question 4.
Name the following;
Answer:
Question 5.
State two economic importance of
Answer:
Question 6.
What is the nature of cell walls of diatoms?
Answer:
Cell walls are made up of silica with two overlapping shells fit together like a soapbox.
Question 7.
Find out what do the terms algal blooms and red tides signify?
Answer:
Question 8.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify.
Answer:
1. Algal bloom’: When colour of water changes due to profuse growth of coloured phytoplanktons, it is called algal bloom.
2. ‘Red tides: Redness of the red sea is due to the luxuriant growth of Trichodesmium erythrium, a member of cyanobacteria (blue-green algae)’
Question 9.
How are viroids different from virus?
Answer:
Viroids are free RNA without protein coat. Viruses have protein coat which encloses either RNA or DNA.
Question 10.
Justify the physiological relationship between the algal and fungal component of lichen.
Answer:
The fungus holds water, provides protection and ideal housing to the alga. The alga supplies carbohydrate food for the fungus. If the alga is capable of fixing nitrogen, it supplies fixed nitrogen to fungus. This association is called symbiosis.
Question 11.
Bacteria reproduce by various methods. Mention the type of reproduction given in the diagram. What are the other methods of reproduction occur in bacteria?
Answer:
Binary fission
The other methods are sporulation and sexual reproduction.
Question 12.
Biological classification is essential. Comment.
Answer:
The animals and plants vary greatly in their form, structure and mode of life. To find out an organism of known characters from the vast number of organism is simply impossible. So classification is important to divide into groups and subgroups.
Question 13.
Match the following:
| a. Produces a plant disease | p. Saccharomyces cere visae |
| b. is edible- light blight of potato. | q. Phytophthora infestans |
| c. is a source of antibiotic | r. Agaricus campestris |
| d. is used in the manufacture of ethanol | s. Penicillium notatum |
Answer:
Question 14.
Plants are autotrophs. Can you think of some plants that are heterotrophs?
Answer:
Generally all plants are autotrophs but plants like loranthus and cuscuta absorbs water & nutrients from other plants so they are called as heterotrophs.
Question 15.
What are the characteristic features of Euglenoides?
Answer:
They have protein sheath is called pellicle instead of cell wall. They have two flagella – One long and other short. They are photosynthetic in the presence of light and behave as heterotrophs in the absence of sunlight.
Question 16.
Give 4 difference between Ascomycetes and Basidiomycetes:
Answer:
| Ascomycetes | Basidiomycetes |
| 1. Mycelium consists of branched multicellular septate hyphae. | 1. Mycelium may be primary, secondary (or) tertiary |
| 2. The fruiting bodies are ascocarps | 2. Fruiting bodies are basidiocarps. |
| 3. Sexual reproduction leads to the formation of ascus | 3. Formation of basidia formation of ascus. |
Question 17.
Observe the cyanobacteria given below and answer the following.
Answer:
Question 18.
What do the terms phycobiont and mycobiont signify?
Answer:
Algal component of lichen is called phycobiont. It prepares food for fungus. Fungal partner is called mycobiont. It provides shelter and absorbs mineral nutrients for algae.
Question 19.
Prepare a comparative account of different classes of kingdom fungi by considering following statements.
Answer:
Question 20.
The two-kingdom classification is introduced by Linnaeus. Why is the two kingdom classification inadequate?
Answer:
There was no place of viruses and bacteriophages which can neither be considered as prokaryotes not eukaryotes.
In this classification, eukaryotes were put together with prokaryotes and non-photosynthetic fungi along with photosynthetic plants.
Question 21.
How is the five-kingdom classification advantageous over the two kingdom classification?
Answer:
In this classification main criteria used by R H Whittaker include cell structure, thallus organisation, mode of nutrition, reproduction and phylogenetic relationships. These characters were not considered in two kingdom classification.
Question 22.
Are chemosynthetic bacteria-autotrophic or heterotrophic?
Answer:
Autotrophic, because they get energy from the oxidation of inorganic compounds. So the released energy is stored in the ATP molecules.
Question 23.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Answer:
Cyanobacteria and other photosynthetic bacteria have thylakoids suspended freely in the cytoplasm (i.e., they are not enclosed in membrane), and they have bacteriochlorophyll
Question 24.
With respect to fungal sexual cycle, choose the correct sequence of events.
Answer:
Question 25.
What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement?
Answer:
It is due to the presence of special nitrogen-fixing cell called heterocyst present between the filaments. So it helps to increase N2 content in the soil.
Question 26.
Methane is the main component of biogas and it is produced by bacteria.
Answer:
Question 27.
Based on the relationship, fill in the blanks.
Answer:
Question 28.
Name the kingdom in which euglena belongs. Give the special type nutrition.
Answer:
Kingdom Protista, Mixotrophic nutrition (ie, both autotrophic and heterotrophic).
Question. 29
Some bacteria are different from others and they have the ability to survive in extreme conditions. Name it.
Answer:
Archaebacteria (halophiles, thermoacidophiles and methanogens).
Question 30.
Mycoplasma are included in five kingdom classification but not viruses. Why?
Answer:
Because mycoplasmas are living cellular organisms but viruses are acellular particles.
Question 31.
In which division of protista chief producers in ocean belongs. Give the cell wall composition of such organisms.
Answer:
Chrysophytes, silicified cell wall.
Question 32.
Nitrobactor and nitrosomonas are free living nitrogen fixers and chemoautotrophs but their functions are different. Do you agree. Give reasons.
Answer:
Yes. Nitrobactor converts nitrite into nitrate while nitrosomonas converts ammonia into nitrites.
Question 33.
Name the classes fungi shows exogenous and endogenous spore production. In which fruiting bodies they are found.
Answer:
Question 34.
Rust and smut diseases are caused by the members of basidiomycetes. Name it.
Answer:
Smut disease- Ustilago, Rust disease-Puccinia.
Question 35.
What are the events takes place in slime mould during favourable and unfavourable season?
Answer:
During favourable condition the cells aggregate and form plasmodium while in unfavourable season plasmodium differentiates and produce fruiting bodies that bear spores at tip.
Question. 36
Suppose you accidentally find an old preserved permanent slide without a label. In your effort to identify it, you place the slide under microscope and observe the following features
What would you identify it as? Can you name the kingdom it belongs to?
Answer:
Dinoflagellates, Kingdom protista
Question. 37
What would you identify it as? Can you name the kingdom it belongs to?
Answer:
Dinoflagellates, Kingdom protista
Question. 38
Why lichens are called as dual organisms?
Answer:
Lichens are said to be dual organisms because they show a symbiotic association between a fungus and alga.
Question 39.
Name the asexual, reproductive structure of penicillium and yeast.
Can penicilium reproduce through sexual method? If the yes or no Give reason.
Answer:
Conidia – penicilium, buds – yeast
Yes, It is done by the production of ascospores in asci of Ascocarp.
Question 40.
Organise a discussion in your class on the topic virus. Are viruses living or non-living?
Answer:
They are filterable and may becrystalised. They are inert outside their specific host and able to reproduce inside the living host cell, so they are considered as living. They use the protein synthesising machinery of the host.
Eg. AIDS virus, mumps virus etc.
Question 41.
How are viroids different from viruses?
Answer:
| Virus | Viroid |
| 1. Their size is smaller than bacteria | 1. Their size is smaller than viruses |
| 2. Protein coat is Present | 2. Protein coat is absent |
| 3. Genetic material may be DNA or RNA | 3. Genetic material is only RNA |
| 4. They cause AIDS, smallpox etc. | 4. They cause potato spindle tuber diseases |
Question 42.
Some bacteria are specialised and live in extreme habitat.
Answer:
1. Types of bacteria
2. Ceil wall structure
Question 43.
The two nuclei per cell can be seen in fungal cell but it later fuse in some members.
Answer:
Question 44.
Classify the pathogenic microorganisms and disease in different groups based on the following symptoms mosaic disease, citrus canker .potato spindle tuber disease, sleeping sickness, malaria.
Answer:
mosaic disease-virus, citrus canker-Bacteria, potato spindle tuber disease -viroids, sleeping sickness- Trypanosoma, malaria-Plasmodium vivax.
Question 1.
Describe briefly the four major groups of protozoa.
Answer:
Protozoans are heterotrophs act either as predators
or parasites. They are of four groups
Question 2.
Different types of fungi are given
1. Classify them into their specific classes.
| Groups | Fungi |
| Phycomycetes | Trichoderma |
| Ascomycetes | Neurospora |
| Basidiomycetes | Albugo |
| Deuteromycetes | Mucor |
| Agaricus | |
| Ustilago | |
| Alternaria | |
| Claviceps |
2. Write the distinguishing characters of ascomycetes and basidiomycetes
3. The characteristic features of members of monera are given below.
Organisms lack cell wall, live without oxygen, smallest living cell and causes diseases. Identify the organism by analysing the above characters.
Answer:
1. specific classes.
2. In ascomycetes, Asexual mode of reproduction is prominent by conidiospores. In Basidiomycetes asexual spores are not found. Sexual spores are arranged in ascus with Ascospores in ascomycetes, whereas sexual spores are arranged in basidium in basidiomycetes.
3. Mycoplasma
Question 3.
Give a brief account of virus with respect to their structure and nature of genetic material. Also, name four common viral diseases?
Answer:
Viruses are organism having inert crystalline structure outside the living cell. They have genetic material RNA or DNA.which is either single-stranded/double-stranded. It is enclosed by protein capsid with subunits called capsomeres.
The viral genetic material takes control over the host cell mechanism during infection. Some common viral diseases are mumps, herpes, smallpox and influenza in animals and mosaic disease in plants.
Question. 4
In which groups are the following found- Sporangiophore, Conidia, zygospore and ascospore.
Answer:
Question 1.
What is the nature of cell walls in diatoms?
Answer:
The cell walls in diatoms are embedded with silica, which makes them indestructible. They form two thin overlapping shells which fit together as in a soapbox. Thus diatoms have left behind large amounts of cell wall deposits in their habitat.
Question 2.
How are viroids different from viruses?
Answer:
Viroids are free RNAs without the protein coat, while virus have a protein coat encapsulating the RNA.
Question 3.
Describe briefly the four major groups of Protozoa.
Answer:
Four major groups of Protozoa are as given below:
1. Amoeboid Protozoa:
They are found in freshwater, seawater or moist soil. They have pseudopodia, like amoeba, hence the name ameoboid protozoa.
2. Flagellated Protozoans:
They have flagella helps in locomotion. Some are parasite. Eg. Trypanosoma causes sleeping sickness.
3. Ciliated Protozoa:
They have thousands of cilia present all over the body. The cilia helps in locomotion and steering of food into the gullet.
4. Sporozoans:
Many protozoans have an infectious spore-like stage in the life cycle. The spore-like stage helps them get transferred from one host to another host.
Question. 4
Plants are autotrophic. Can you think of some plants that are partially heterotrophic?
Answer:
Certain insectivorous plants, like bladderwort and venus fly trap, are partially heterotrophic.
Question. 5
What do the terms phycobiont and mycobiont signify?
Answer:
Lichens are good examples of symbiotic life of algae and fungi. Phycobiont is the name of the part composed of algae and Mycobiont is the name of the part composed of fungi. Fungi provide minerals and support to the alage, while algae provide nutrition to the fungi.
Question 6.
What are the characteristic features of Euglenoids?
Answer:
Features of Euglenoids.
Question 7.
Give a brief account of viruses with respect to their structure and nature of genetic material. Also name four common viral diseases.
Answer:
Virus Structure:
Outside a host cell, virus is a crystalline structure, composed of protein. Inside the crystal, there is genetic material, which can be either RNA or DNA. No virus has both RNA and DNA. Viruses, infecting plants, have single-stranded RNA. Viruses, infecting animals, have either single or double-stranded RNA or double-stranded DNA.
The protein coat is called capsid. Capsid is made of smaller subunits, called capsomeres, it protects nucleic acid. Diseases caused by Virus; AIDS, Mumps, Influenza, Herpes.
Question 8.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify.
Answer:
Dinoflagellates can be of different colours depending on the type of pigment present. The red dinoflagellate sometimes multiplies at a very rapid rate. This is called as algal bloom. This gives a red appearance to the part of affected sea. This is also known as ‘red tide’. Toxins released by them can kill other marine species.
Question 1.
The life form used as indicators of pollution
(A) Lichens
(B) Protozoa
(C) Algae
(D) Agaricus
Answer:
(A) Lichens
Question 2.
Kingdom monera comprises
(A) Amoeba, Bacteria,Trypanosoma
(B) Bacteria, Viruses,Virolds
(C) Archaebacteria, Eubacteria, Mycoplasma
(D) Mycoplasma, Viruses, Bacteria
Answer:
(C) Archaebacteria, Eubacteria, Mycoplasma
Question 3.
Who discovered two-kingdom classification
(A) Ivanowsky
(B) Stanley
(C) leuwernhoek
(D) Linnaeus
Answer:
Question 4.
Asexual reproduction takes place by Zoospores in
(A) Pythium
(B) Agaricus
(C) Rhizopus
(D) Ustilago
Answer:
(D) Ustilago
Question 5.
Identify the organism used as bioweapon
(A) Bacillus thuringiensis
(B) Bacillus anthracis
(C) Pseudomonas citri
(D) Rhizobium tumefacient
Answer:
(B) Bacillus anthracis
Question 6.
Reserve food in the form of glycogen and cell wall made up of chitin are characteristic of
(A) Protists
(B) bacteria
(C) Fungi
(D) protozoa
Answer:
(C) Fungi
Question 7.
The fruiting body of club fungi is
(A) Basidium
(B) Ascus
(C) Ascocarp
(D) Basidiocarp
Answer:
(D) Basidiocarp
Question 8.
RNA without protein coat are found in
(A) bacteria
(B) protozoa
(C) viruses
(D) viroides
Answer:
(D) viroides
Question 9.
The phycobiont and mycobiont are found in
(A) bacteria
(B) lichen
(C) viroides
(D) fungi
Answer:
(B) lichen
Question 10.
The organism which causing sleeping sickness belongs to
(A) Protists
(B) bacteria
(C) Fungi
(D) viruses
Answer:
(A) Protists
Question 11.
In which of the following groups are neurospora and Penicillium included?
(A) Phycomycetes
(B) Basidiomycetes
(C) Zygomycetes
(D) Ascomycetes
Answer:
(D) Ascomycetes
Question 12.
Occurrence of Dikaryon phase is characteristic feature of
(A) Bacteria
(B) Fungus
(C) Slime moulds
(D) Cyanobacteria
Answer:
(B) Fungus
Question13.
Methane producers are belongs to
(A) Archaebacteria
(B) Cyanobacteria
(C) Eubactenia
(D) Actinomycetes
Answer:
(A) Archaebacteria
Question 14.
Heterocyst are found in
(A) Nitrosomonas
(B) cyanobacteria
(C) fungi
(D) protozoa
Answer:
(B) cyanobacteria
Question 15.
Colletotrichum falcatum is a fungus causing the following disease
(A) Smut of wheat
(B) Wilt disease of cotton
(C) Red rot of sugar cane
(D) Late blight of potato
Answer:
(C) Red rot of sugar cane
Students can Download Chapter 2 Relations and Functions Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Plus One Maths Relations And Functions Previous Questions And Answers Question 1.
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by R = {(a, b): a, b ∈ A, b is exactly divisible by a}
Answer:
Plus One Maths Chapter Wise Questions And Answers Pdf Question 2.
Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}
Answer:
R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9),(5, 10)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}
Relations And Functions Class 11 Important Questions Pdf Question 3.
A function f is defined as f(x) = 2x – 5, Write down the values of f(0), f(7), f(-3).
Answer:
Given; f(x) = 2x – 5
f(0) = -5;
f(7) = 2(7) – 5 = 14 – 5 = 9
f(-3) = 2(-3) – 5 = -6 – 5 = -11
Hsslive Maths Textbook Answers Plus One Question 4.
Find the range of the following functions.
Answer:
Plus One Maths Chapter Wise Questions And Answers Question 1.
Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
Answer:
1. A × (B ∩ C) ={1, 2} × Φ = Φ
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
(A × B) ∩ (A × C) = Φ
Hence; A × (B ∩ C) = (A × B) ∩ (A × C)
2. A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × D = {1, 2, 3, 4} × {5, 6, 7, 8}
= {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3,6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Hence A × C is a subset of B × D.
Relations And Functions Class 11 Important Questions Question 2.
The arrow diagram given below shows a relation R from P to Q. Write the relation in roster form, set-builder form. Find its domain and range.
Answer:
R – {(9, -3), (9, 3), (4, -2), (4, 2), (25, -5), (25, 5)}
R = {{x, y) : y2 = x}
Domain of R = {9, 4, 25}
Range of R = {5, 3, 2, -2, -3, -5}
Question 3.
Find the domain of the following.
Answer:
1. Given; f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\)
The function is not defined at points where the denominator becomes zero.
x2 – 8x +12 = 0 ⇒ (x – 6)(x – 2) = 0 ⇒ x = 2, 6
Therefore domain of fis R – {2, 6}.
2. Given; f(x) = \(\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\)
The function is not defined at points where the denominator becomes zero.
x2 – 5x + 4 = 0 ⇒ (x – 4)(x -1) = 0 ⇒ x = 1, 4
Therefore domain of f is R – {1, 4}.
Plus One Maths Questions And Answers Question 4.
Let f(x) = \(=\sqrt{x}\) and g(x) = x be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and \(\left(\frac{f}{g}\right)(x)\).
Answer:
(f + g)(x) = f(x) + g(x) = \(=\sqrt{x}\) + x
(f – g)(x) = f(x) – g(x) = \(=\sqrt{x}\) – x
(fg)(x) = f(x) × g(x) = \(=\sqrt{x}\) × x = \(x^{\frac{3}{2}}\)
Plus One Maths Relations And Functions Question 5.
Let f(x) = x2 and g(x) = 2x + 1 be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and \(\left(\frac{f}{g}\right)(x)\).
Answer:
(f + g)(x) = f(x) + g(x) = x2 + 2x + 1
(f – g)(x) = f(x) – g(x) = x2 – 2x – 1
f(fg)(x) = f(x) × g(x)
= x2(2x +1) = 2x3 + x2
Relations And Functions Questions And Answers Pdf Question 6.
A = {1, 2}, B = {3, 4}
Answer:
1. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
2. Any Subset of A × B (say R={(1, 3),(2, 4)})
3.
Relations And Functions Class 11 Important Questions With Solutions Question 1.
Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find
Answer:
1. A × (B ∩ C) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}
2. (A × B) ∩ (A × C)
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} ∩ {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5) , (2, 6), (3, 4), (3, 5), (3, 6)}
= {(1, 4), (2, 4), (3, 4)}
3. A × (B ∪ C) = {1, 2, 3} × {3, 4, 5, 6}
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
4. (A × B) ∪ (A × C)
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} ∪ {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
Question 2.
Find the domain and range of the following
Answer:
i) Given; f(x) = -|x|
D(f) = R, R(f) = (-∞, 0]
ii) Given; f(x) = \(\sqrt{9-x^{2}}\)
x can take values where 9 – x2 > 0
⇒ x2 ≤ 9 ⇒ -3 ≤ x ≤ 3 ⇒ x ∈ [-3, 3]
Therefore domain of f is [-3, 3]
Put \(\sqrt{9-x^{2}}\) = y, where y ≥ 0
⇒ 9 – x2 = y2⇒ x2 = 9 – y2
⇒ x = \(\sqrt{9-x^{2}}\)
⇒ 9 – y2 ≥ 0 ⇒ y2 ≤ 9 ⇒ -3 ≤ y ≤ 3
Therefore range of fis [0, 3].
iii) Given; f(x) = |x – 1|
Domain of f is R
The range of |x| is [0, ∞) , then the range of
f(x) = |x -1| is [0, ∞)
iv) Given; f(x) = \(\sqrt{x-1}\)
x can take values where x – 1 ≥ 0
⇒ x ≥ 1 ⇒ x ∈ [1, ∞]
Therefore domain of fis [1, ∞]
The range of \(\sqrt{x}\) is [0, ∞), then the range of
f(x) = \(\sqrt{x-1}\) is [0, ∞).
Question 1.
If (x + 1, y – 2) = (3, 1), find the values of x and y.
Answer:
(x + 1, y – 2) = (3, 1) ⇒ x + 1 = 3, y – 2 = 1 ⇒ x = 2, y = 3.
Question 2.
If \(\left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)\), find the values of x and y.
Answer:
Question 3.
If G = {7, 8}; H = {2, 4, 5}, find G × H and H × G.
Answer:
Question 4.
if A = {-1, 1} find A × A × A
Answer:
A × A ={-1, 1} × {-1, 1}
= {(-1, -1), (-1, 1), (1, -1), (1, -1)}
A × A × A
= {(-1, -1), (-1, -1), (1,-1), (1, -1)} × {-1, 1}
= {(-1, -1, -1), (-1, 1, -1), (1, -1, -1), (-1, 1, -1), (-1, -1, 1), (-1, 1, 1), (1, -1, 1), (-1, 1, 1)}.
Question 5.
Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.
Answer:
2, 3, 5, 7 are the prime number less than 10.
R = {(2, 8),(3, 27),(5, 125),(7, 343)}
Question 6.
If f(x) = x2, find \(\frac{f(1.1)-f(1)}{(1.1-1)}\)?
Answer:
Question 7.
Let \(\left\{\left(x, \frac{x^{2}}{1+x^{2}}\right), x \in R\right\}\) be a real function from R to R. Determine the domain and range of f.
Answer:
Domain of f is R.
Let \(\frac{x^{2}}{1+x^{2}}\) = y ⇒ x2 = y(1 + x2)
⇒ x2 = y + yx2 ⇒ x2 – yx2 = y
⇒ x2(1 – y) = y
⇒ y ≥ 0, 1 – y > 0
⇒ y ≥ 0, y < 1 ⇒ 0 ≤ y ≤ 1
Therefore range of f is [0, 1).
Question 8.
Graph the following real functions. (each carries 2 scores)
Answer:
1. f(x) = |x – 2| = \(\left\{\begin{aligned}x-2, & x \geq 2 \\-x+2, & x<2 \end{aligned}\right.\)
2. f(x) = x2
3. f(x) = x3
4. f(x) = \(\frac{1}{x}\)
5. f(x) = (x – 1)2
6. f(x) = 3x2 – 1
7. f(x) = |x| – 2
Question 9.
Consider the relation, R = {(x, 2x – 1)/x ∈ A) where A = (2, -1, 3}
Answer:
1. x = 2 ⇒ 2x – 1 = 2(2) – 1 = 3
x = -1 ⇒ 2x – 1 = 2(-1) – 1 = -3
x = 3 ⇒ 2x – 1 = 2(3) – 1 = 5
R = {(2, 3), (-1, -3), (3, 5)}
2. Range of R = {3, -3, 5}
Question 10.
Let A = {1, 2, 3, 4, 6} and R be a relation on A defined by R = {(a, b): a, b ∈ A, b is exactly divisible by a}
Answer:
Question 11.
Consider the real function
Answer:
1. Given; f(x) = 1 ⇒ 1 = \(\frac{x^{2}+2 x+3}{x^{2}-8 x+12}\)
⇒ x2 – 8x + 12 = x2 + 2x + 3
⇒ 10x = 9 ⇒ x = \(\frac{9}{10}\)
2. Find the value for which denominator is zero.
⇒ x2 – 8x + 12 = 0
⇒ (x – 6)(x – 2) = 0 ⇒ x = 6, 2
Therefore domain of f is R – {2, 6).
Question 12.
If f(x) = x3 + 5x and g(x) = 2x +1, find (f + g)(2) and {fg)(1).
Answer:
(f + g)(2) = f(2) + g(2) = (2)3 + 5(2) + 2(2) + 1
= 8 + 10 + 4 + 1 = 23
(fg)(1) = f(1)g(1) = (1 + 5)(2 + 1) = 6 × 3 = 18.
Question 13.
Let A = {1, 2, 3, 4, 5} and R be a relation on A defined by R = {(a, b):b = a2}
Answer:
Question 14.
Draw the graph of the function
f(x) – |x| + 1, x ∈ R
Answer:
Question 15.
Draw the graph of the function.
f(x) = x3, x ∈ R
Answer:
Students can Download Chapter 4 Bank Reconciliation Statement Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Plus One Accountancy Bank Reconciliation Statement Question 1.
A bank reconciliation statement is prepared with the balance
(a) Passbook
(b) Cashbook
(c) Both pass book and cash book
(d) None of these
Answer:
(c) Both pass book and cash book
Plus One Accountancy Chapter 4 Questions And Answers Question 2.
Passbook is a copy of
(a) Copy of customers account
(b) Bank column of cash book
(c) Cash column of cash book
(d) Copy of receipts and payments
Answer:
(c) Copy of customers’ accounts.
Bank Reconciliation Statement Problems And Solutions Question 3.
………… is a statement showing the causes of the difference between the cash book and passbook balance.
Answer:
Bank Reconciliation statement.
Bank Reconciliation Questions And Solutions Question 4.
Bank Reconciliation Statement is prepared by ………….
Answer:
Businessman/Debtor/Account holder
Bank Reconciliation Statement Problems And Solutions Pdf Question 5.
Credit balance in the passbook means a/an ………… to the depositor.
Answer:
Asset
Bank Reconciliation Statement Questions And Answers Question 6.
Normally, the cash book shows a ………. balance, passbook shows …………. balance.
Answer:
Debit, Credit
Bank Reconciliation Statement Questions Question 7.
Favourable balance as per the cash book means …………… balance in the bank column of the cash book.
Answer:
Debit
Bank Reconciliation Questions For Class 11 Question 1.
What is a Bank Reconciliation Statement?
Answer:
A statement prepared to reconcile the bank balance as per cash book with the balance as per passbook or bank statement, by showing the items of difference between the two accounts. By the preparation of bank reconciliation statement, one of the balances (either the cash book balance or passbook balance) may be equalized with the other.
Bank Reconciliation Statement Questions And Answers Pdf Question 2.
State the need for the preparation of bank reconciliation statement.
Answer:
It is generally experienced that when a comparison is made between the bank balance as shown in the firms cash book, the two balances do not tally, to reconcile (tally) the two balances of cash book and passbook, bank reconciliation statement is prepared.
Bank Reconciliation Statement Solutions Question 1.
From the following particulars, prepare the Bank Reconciliation Statement of Asha & Co. as on 31.3.2012.
Answer:
Bank Reconciliation Statement as on 31.3.2012
Bank Reconciliation Statement Class 11 Question 2.
The cash book of Reji showed a debit balance of Rs. 4,900 on may 31,2009. On comparing the cash book with a passbook, the following were found.
Answer:
Bank Reconciliation Statement of Reji as on 31/05/09
Bank Reconciliation Statement Problems Question 3.
The Bank overdraft of Smith Ltd. on December 31, 2010, as per cash book is Rs. 18,000. From the following information, ascertain the adjusted cash balance and prepare bank reconciliation statement.
Rs.
Answer:
Adjusted Cashbook (Bank column)
Bank Reconciliation Statement Questions And Answers Class 11 Question 4.
Prepare Bank Reconciliation Statement of Mr. Syam on 31, March 2009 from the following details.
Answer:
Bank Reconciliation Statement of Mr. Syam as on 31/03/09
Questions On Bank Reconciliation Statement Question 5.
Balance as per passbook of Mr Kumar is 3,000.
Prepare a bank reconciliation statement.
Answer:
Bank reconciliation Statement of Mr Kumar as on …………
Bank Reconciliation Problems And Solutions Question 6.
The passbook of Mr Mohit current account showed a credit Balance of ₹20,000 on dated December 31, 2005. Prepare a Bank Reconciliation Statement, with the following information.
Answer:
Bank Reconciliation Statement of Mr Mohit as on December 31st, 2010
Question 7.
On 1st January 2010, Rakesh had an overdraft of ₹8,000 as showed by his cash book. Cheques amounting to ₹2,000 had been paid in by him but were not collected by the bank by January 1, 2010.
He issued cheques of ₹ 800 which were not presented to the bank for payment up to that day. There was a debit in his passbook of ₹60 for interest and ₹100 for buffet charges. Prepare bank reconciliation statement for comparing both the balance.
Answer:
Bank Reconciliation Statement of Mr Rakesh as on 1st January 2010.
Question 8.
Prepare bank reconciliation statement.
Answer:
Bank Reconciliation Statement as on 31st December, 2010.
Question9.
Prepare a bank reconciliation statement from the following particulars and show the balance as per cash book.
Answer:
Bank Reconciliation Statement as on 31st March, 2010
Question 1.
What are the various reasons for the difference between balance as per cash book and passbook?
Answer:
There are several reasons that contribute for the disagreement of the balance as shown by cash book and passbook. They are as under:
1. Cheques issued but not presented for payment:
When the trader issues a cheque, he credits its amount immediately in his cash book. The same will be entered in the passbook only on presenting the cheque and making payment by the bank. If the cheque is not presented for payment before the date of preparation of bank reconciliation statement, the balance as per Pass Book will be more than the balance as per Cash Book.
2. Cheque paid in for collection but not collected:
On deposition cheques into the bank for collection, the trader debits the same amount in the bank account. The bank credit the amount in the passbook only on getting the amount collected. Such uncleared cheques make the cash book balance to be more than the passbook balance.
3. Direct payment by a customer to the bank:
Customers of the trader occasionally make some payments directly into the trader’s bank account. The trader may come to know of an only later. But the banker gives immediate credit to the trader on receipt of the amount. If it remains unrecorded in the cash book, the balance as per pass book will be more than the balance as per cash book.
4. Interest on deposit credited by the banker:
At regular intervals, banks allow interests on the deposit balance of the trader and credit the amount in the passbook. The same usually remains unrecorded in the cash book. In such a case, the passbook balance will be more than the cash book balance.
5. Interest, dividend, rent, etc. collected by bank:
Bank collects interest, dividend, rent, etc. on behalf of the customer and credits the same to his account. The trader comes to know of it only on a later date. If such collection remains unrecorded in the cash book, the passbook balance will be more than the cash book balance.
6. Payment made on behalf of the customer:
The banker makes payment for rent, insurance, etc., for the customer as per standing instructions. The banker debit the trader’s account with such payments. The trader comes to know of it only later. Due to such payments that remain unrecorded in cash book, the balance as per pass book will be less than the balance as per cash book.
7. Bank charges as per Pass Book:
Bank charges and commission for collection of cheques, bills, etc., are debited in the passbook. The corresponding credits are often not given in the cash book. As these items are not entered in the cash book, its balance will be more than that of the passbook.
8. Bills Receivable discounted, but dishonored:
When a trader discounts bills of exchange, the banker credits the trader’s account with the amount due. The same amount is debited by the trader in cash book. If such a bill is later dishonored, the banker immediately debits it in the passbook. But the same remains unrecorded in the cash book. This cause the balance as per cash book to be more than the passbook balance.
9. Interest on overdraft debited in passbook:
Periodically the bank calculates interest due by the trader on his overdraft and debits the amount in the passbook. Corresponding credit is often not made by the trader in his cash book. It leads to difference in the balance as per cash book and passbook.
10. Credit instruments credited by bank but not recorded in cash book:
Bills of exchange, promissory notes and other credit instruments collected by bank are credited in the passbook. But if they remain unrecorded in the cash book it may lead to disagreement between the balance as per the two books.
11. There may also be instance of cheque recorded as paid in for collection but failed to be deposited into the bank, by which the cash book balance will be more than the balance as per passbook.
General Instructions to Candidates
Read the following lines and answer the questions given below:
Look to this day
For it is life, the very life of life,
In its brief course lie all the
varieties and realities of your existence.
The bliss of growth,
The glory of action,
The splendor of beauty;
For yesterday is but a dream
And tomorrow is only a vision,
But today well lived makes
Every yesterday a dream of happiness,
And every tomorrow a vision of hope,
Look well therefore to this day!
Such is the salutation of the dawn.
Complete the sentences, choosing the right option.
Question 1.
The‘theme’of the poem is about
a. the importance of life
b. the importance of concentrating on the reality of the present
c. in the importance of thinking about the past and the future
d. the importance of having a dream and a vision
Question 2.
The expression ‘brief course’ means
a. the course of study
b. a short report
c. lasting for only a short timed
. in a few words
Question 3.
What is the implied meaning of the lines given below? ,
‘Look well therefore to this day!
Such is the solution of the dawn.’
Write your answer in a paragraph.
Question 4.
Choose appropriate words from the box to complete the passage.
In developing countries, only 30% of the buildings are constructed (a) ………….. with the regulations laid down for ensuring safety and security, (b) ………….. the lack of a master plan and the inferior quality of materials used for construction also aggravate the casualties arising out of disasters, (c) ………….. both private and public buildings should be constructed according to the guidelines prescribed by law (d) ………….. constructions should strictly adhere to the master plan approved by the authorities.
Questions 5 and 6: Read the following excerpt from the story, ‘His First Flight’ and answer the questions.
He waited a moment in surprise, wondering why she did not come nearer and then maddened by hunger, he dived at the fish with a loud scream; he fell outwards and downwards into space. His mother had swooped upwards. As he passed beneath ‘ her, he heard the swish of her wings.
Question 5.
Who is ‘she’ referred to here?
Question 6.
What prompted the young seagull to fly finally?
Question 7.
Read the following excerpt from ‘Is society Dead?’,
there are four errors in the passage. Identify the errors and correct them. What we do get from this? The chance to slip away for a while from everydayness, to give your lives its own soundtracks, to still the monotony of the commute, to listen more closely and carefully on music that can lift you up and can keep you go.
Question 8.
Rewrite the following conversation between Gupta and the waitress (The Price of flowers) in reported speech.
Gupta: Do you know the girl who was sitting over there?
Waitress: No, sir, I do not know her, to speak of. I’ve noticed she has lunch here on Saturdays.
Gupta: Doesn’t she come on any other day?
Waitress: I never see her on other days?
Questions (9 -14): Answer any five of the following questions in not more than 80 words. Each question carries 4 scores.
Question 9.
Stephen Hawking has overcome his crippling disease to become the ‘supernova’ of world physics. On the occasion of the release of a book on Stephen Hawking, you are asked to deliver a speech describing him to the audience. Prepare the script of the speech.
Question 10.
India has faced a number of natural disasters like floods, landslides etc. We may not be able to avert many of these disasters, but We can definitely mitigate their impact. Prepare an essay on the various stages involved in the effective management of disasters.
Question 11.
If you could bring yourself to tell mother only once, after looking into the crystal, that Frank is all right, that he is alive will be too much of a lie? Will it be very, very wrong? As she spoke, tears streamed from her eyes. (From Price of Flowers) What trait of Maggie’s character is evident here? Sketch the character of Maggie in a short paragraph.
Question 12.
Your class is conducting a group discussion on ‘ The Relevance of Gandhian ideas in the 21st century. ‘Imagine that you are also participating in it. What would you speak? Prepare a script for your presentation. (Word Limit: 50 to 60 words)
Question 13.
Analyze the news headlines given below.
In the light of the news headlines, prepare an article about the relevance of Gandhian ideas on nonviolence, cleanliness and equal rights.
Question 14.
‘What kind of a country is India, Sir?’ asked Mrs. Clifford, as she resumed her cooking. What impression of India is created by Mr. Gupta in the mind of Mrs. Clifford
Questions(15 -19): Answer any four of the following questions in not less than 100 words. Each question carries 5 scores.
Question 15.
Imagine that you are a tourist guide working with the tourism department of Fiji. Make an introductory speech to a group of tourists from Europe to welcome them to Namuana
Question 16.
Here is an advertisement for the ‘Walkers Club’
In the advertisement, you see that the box for Don’ts has been left blank. Write four points to be included in that box.
(Need not to copy the whole advertisement. Write only the points)
Question 17.
Nowadays? our rivers are rapidly getting polluted. Many reasons are cited but seldom are measures adopted to prevent the pollution of rivers. Write a letter to the editor of a local newspaper highlighting the gravity of the issue.
Question 18.
A group of foreigners from France visit your locality. They intend to visit a few tourist destinations in our state. Prepare a short write-up about one or two famous tourist centers in your district.
[Hints : Name -location – major attractions – how to reach there]
Question 19.
Captain }ovis was interviewed by a news reporter before the take-off of Le-Horla. There are the few questions that the reporter puts forth. Read the questions and prepare the likely responses.
Reporter Captain Jovis Reporter Captain Jovis Reporter Captain Jovis Reporter
Reporter: Sir, this is your first experience in a hot air balloon. How do you feel?
Captain Jovis: Yes, I am really excited as well as nervous.
Reporter: Will you please introduce your supporting crew?
Captain Jovis: 1) ……………………………………………….
Reporter: What all are the arrangements made for the flight?
Captain Jovis: 2) ……………………………………………….
Reporter: Is guiding a balloon an easy task?
Captain Jovis: 3) ……………………………………………….
Questions (20 -23): Answer any three of the following each in not more than 250 words. Each question carries 8 scores.
Question 20.
Study the graph given below about the money spent on fast food by people of different age groups over the last 20 years.
Interpret the data and prepare an analytical report in the form of a short essay.
Question 21.
The following information is about the famous deaf German composer Ludwig Van Beethoven. Prepare a profile of him.
Birth: 16th December 1770 in Germany
Parents: John Van Beethoven and Maria Magdalena Keverich
Childhood: Fond of music – father teaches him music day and night – first public performance at the age of seven (March 26, 1778)- publishes first work in 1782 (age 12)
Turning point: Deafness begins in 1796 decides to commit suicide realizes his abilities – decides to explore, discover and to pass on.
Achievements: Becomes an outstanding instrumentalist, playing both piano and violin- publishes first three sonatas (October 14, 1783) becomes an excellent keyboard composer (1795) – sets new standards for conveying emotions and passion through music – creates melody and harmonies-composes ten symphonies (1800 -1824).
Death: 26 March 1827- funeral rites at the church of Holy Trinity -about 30000 people attends the funeral
Question 22.
Go through the points listed below.
Prepare a write-up on this ritual/ practice, highlighting the plot, characters, local culture, social/ historical aspects, the relevance of the ritual in the current scenario and its impact on society.
Question 23.
Read the poem given, below and write a note of appreciation.
Good Deeds by Sasikanth Nishanth Sarma
The World is a double-edged knife
And we have only one life
Enjoy it, live it well.
In good company do well.
Always do good deeds
Sow in yourself character’s seeds
Cultivate mind with manure of thought
For ages, the experiences brought.
Feed your life with good deeds
So that there is no place for weeds.
To harm, that oft leads
The decay of character’s seeds
The one who wants to succeed
He must always heed
And must seek
The good qualities
And not the quantities
A wise man can only plead
But you have to pay heed
What are you needs?
Then decide what to do
And what not to do
Answers
Answer 1.
The theme of the poem is about the importance of concentrating on the reality of the present.
Answer 2.
The expression ‘brief course’ means ‘last-ing only for a short time’.
Answer 3.
Focus on today’s happiness. The dawn addresses you for the sake of having the short span of today’. It is today or the present that you should count; not the past or the future. This day is reality for you. On this day is your bliss.
Answer 4.
a. in accordance with
b. similarly
c. therefore
d. moreover
Answer 5.
The young seagull’s mother.
Answer 6.
Hunger prompted the young seagull to fly.
Answer 7.
What do we get from this? The chance to slip away from the every day ness, to give your lives its own soundtracks, to still the monotony of the commute, to listen more closely and carefully to music that can lift you up and can keep you going.
Answer 8.
Gupta asked the waitress whether she knew the girl who had been sitting over there.
The waitress replied that to speak of, she did not know her and that she had noticed she had lunch there on Saturdays.
Gupta again asked whether she didn’t come on any other day.
The waitress then answered that she never saw her on other days.
Answer 9.
Honorable dignitaries on and off the dais, and my dear friends, Almost everyone wishes to be successful and all of us feel proud of the successful people. But no one ever tries to see how the so-called successful people have become successful. Success is not a matter of luck, nor is it easily attained. It needs a lot of effort and determination. It needs the power to dare challenges.
The life of Stephen Hawking, the world-renowned astrophysicist teachers us this. Let me take this opportunity to introduce him to you. Stephen Hawking has earned an international reputation as the most brilliant theoretical physicist since Einstein. He is an insightful, absorbing, and inspiring person of extraordinary courage who dared his challenges and the threat- ‘ ending disease with his strong willpower.
His life and works bear witness to his brilliant mind.
He was not a born prodigy. Many times he met with borderline results. At age of 21, he was diagnosed with Amyotrophic Lateral Sclerosis. It was devastating news for Hawking and his family. A few events, however, prevented him from becoming completely pessimistic.
In a sense, Hawking’s disease helped him become the noted scientist he is today. With the sudden realization that he might not even live long enough to earn his Ph.D., Hawking poured himself into his work and research and despite his devastating illness, he has done groundbreaking work in physics and cosmology, ’and his several books have helped to make science accessible to everyone.
Today, we are releasing his book A Brief History of Time.’ The book articulates the physicist’s personal search for science’s Holy Grail: a single unifying theory that can combine cosmology with quantum mechanics to explain how the universe began. Dear sir, my hearty congratulations and I wish that you may continue giving such valuable contributions in the future too.
I really wonder, how he could do all these and become the supernova of physics even against your handicaps. Of course, his life will be a source of inspiration to us all. I wish everyone has the courage to face life boldly like this, and then the world will be a different one.
Thank you.
Answer 10.
Effective management of Disasters
Disasters, both natural and manmade, have become a challenge for all humanity and people become vulnerable as the intensity and frequency increase. It affects both developing and developed countries equally.
There are both direct and indirect impacts for disasters like destruction and death, failure of lifeline support systems, severe stress experienced by health care and hospitals, disturbance in commercial and economic activities etc. which make the situation drastic. The poor sections are the most affected usually.
There are many causes for disasters. Defects in the construction of a building are the major reasons for high risk due to disasters. The scenario in the developing countries is worse as only 30% of buildings are constructed in accordance with rules. Lack of master- plan and inferior quality of building materials also aggravate the situation.
We cannot avert or avoid disasters. But we can adopt measures to mitigate the gravity. Destruction can be minimized if private and public buildings adhere to master -plan and constructed in accordance with rules and good quality materials are used. Existing buildings also should be technically assessed and people concerned are to be informed. Purpose of management in India is not prevention, but reduction of impact.
State government plays a major role and should ensure effective functioning of the state-level committee. The central government has got only a facilitating role i.e to coordinate crisis management committee and provide support like defense services, air dropping, rescuing, searching, transport of relief goods, conveyance etc. Rehabilitation of victims is an integral part of disaster management.
Disasters are non-routine events and therefore need non-routine responses for effective management. Proper coordination among different departments is needed to bring speedy relief. Rescue teams should have additional skills and should be equipped with latest technology.
Answer 11.
The full name of Maggie was Alice Marga-ret Clifford. She was thirteen or fourteen years old. Her clothes betrayed her poverty. Her hair hung in a heavy stream down her back. Her eyes were very large. They had a sad expression. Maggie worked as a typist in the civil service stores. She lived with her old widowed mother. Her only brother Frank was in the Indian army. She loved her brother very much. Maggie is ambitious and she is not happy with her job as a typist. She wants to do a job where her brain can be used. When she gets a better job she will rent a better house and take her mother there. This shows how much she cares for her mother. When her mother is in a critical condition she even asks Mr. Gupta to lie about her brother so that the mother feels better by hearing that her son is alive and doing well in India.
Answer 12.
I would say that some of the Gandhian ideas are quite relevant even in the twentieth century. The biggest idea of Gandhi was non-violence. Every day we hear of wars and killings. People are killing one another in the name of religion. Even India is not free from religious intolerance. Gandhi taught us that though the names of God is different- some people call him Ishwar and some call him Allah. Both are the names of the same Almighty. So there is no need to fight, as we all – Hindus, Muslims, Sikhs, Buddhists and Christians are the children of the same father. Gandhi taught people to practice truth. He also taught us the dignity of labor. His ideas of truth, non- violence, hard work, respect of human rights etc are still very much relevant in the 21st century.
Answer 13.
Gandhi provided the world with his timeless philosophy. It was not meant for the Independence of India only. Nonviolence is itself normative. It applies to any situation. Though Gandhi is dead, his philosophy remains alive. He is a living power, more powerful in death than life. Through Gandhi, we can now understand that the philosophy of non-violence can be applied in economics, politics, religion, society etc. Gandhi’s Satyagraha has been applied to bring about useful changes. Many Institutes have been founded to propagate Gandhi’s philosophy of Non-violence. On 2nd October 2014, the Indian Prime Minister, Narendra Modi, launched a nationwide cleanliness campaign on the occasion of Mahatma Gandhi birth anniversary. The concept of Swachh Bharat is to provide sanitation facilities to every family, including toilets, solid and liquid waste disposal systems, village cleanliness and safe and adequate drinking water supply. Mahatma said that women must realize their full status and play their part as equal as men.
Answer 14.
Mr. Gupta described to Mrs. Clifford that India was a beautiful country. It is not cold like England, but somewhat hot. Of course, there are tigers and snakes in India, but they live in the jungles. If they came to places where people live, they would be killed. There are fevers in some places in India. But they differ with places and seasons. This impression of India given by Mr. Gupta removed from her the fear she had about her sons living conditions in India.
Answer 15.
Ladies and gentlemen, you are welcome to this beautiful island. I am guided Fernandez. I wish you all a nice time here. Fiji is a multi-cultured island nation, having the cultural tradition of the people of Oceanic, European, South -Asian, and East Asian origins. lt consists of nearly 320 islands in the South West Pacific Ocean. It is 3152 km away from Sydney, Australia. Nearly a hundred of these islands are inhabited. You can enjoy the little ports and beautiful seacoasts. You can enjoy boat journeys and also tasty dishes of various kinds of fish. So ladies and gentlemen, let us enjoy ourselves.
Answer 16.
Don’t eat while walking.
Don’t think about serious matters while walking.
Don’t talk while you walk. ’
Avoid the use of mobile phone when you are walking fast.
Answer 17.
Rupali Tharun,
Pranavam,
5/15 High Street, Alappuzha.
30th May 2018
The Editor,
The City Chronicle,
Thiruvananthapuram.
Dear Sir,
Through the columns of your esteemed daily, I would like to draw the attention of the •public and the authorities to a very important concern of Kerala. The water in most of our rivers is highly polluted. The pollution of river water is caused by the dumping of untreated sewage and industrial wastes. In addition to sewage and industrial wastes, the pollution is also caused by other human activities like bathing, washing of clothes, animals etc. The industries also discharge chemical effluents into the river water. The toxicity of these chemical effluents kills the fish in many parts of the river. Pollutions also lead to scarcity of pure drinking water.
I shall be happy if you could project this problem through the valuable pages of your newspaper. Hope, the Government would take note of the matter and take some drastic steps to adopt measures to prevent water pollution.
Yours truly,
Sd /-
Rupali Tharun.
Answer 18.
Thusharagiri is a beautiful spot which tourists must visit. Its location is in the Western Ghats, below the Wayanadan hills. It is 55 kilometers away from Kozhikode city. There are roads leading to Thusharagiri from Kozhikode, Kap- pad beach and Wayand. The attractions of Thusharagiri are the three waterfalls, the Thanni Muthassi and the panoramic beauty of the Wayanad forests. Tourists can bath in the cool and pure water. For those who want to enjoy trekking, there is the facility, accompanied by guides. Good food also is available here.
Answer 19.
Answer 20.
The given graph provides us with a clear picture of the money spent on fast food by people of different age groups over the last 20 years. It is startling to know that our young people are spending a lot of money for buying these junk food nowadays.
In 1995, 3000 rupees per year had been spent by the youths whereas in 2010 they had spent around 7000 rupees per year for fast food. The people in the age group 25- 35 had spent only around 4000 rupees per year in 2000 for junk food. Their consumption through the years is not that much alarming. In 2010 they had spent around 5000 rupees per year. In the case of the age group between 35-45 in 1995 fast food consumption was the lowest, ie. only 1500 rupees per year.
From these data, we can conclude that our youngsters are addicted to fast food. On the contrary, our middle-aged people seem to be not excited in the thought of consuming fast food recently.
Answer 21.
Ludwig van Beethoven
The world-renowned German music composer and pianist Ludwig van Beethoven was born on 16th December 1770 in Germany. His parents were John Van Beethoven and Maria Magdalena Kever- ich. In childhood, Beethoven was fond of music. His father was his first music teacher. He taught his son music day and night. Little Beethoven’s first public performance was at the age of seven. Later he published his first composition in 1782 at the age 12. Unfortunately, deafness at-tacked this young genius in 1796. He decided to commit suicide. But soon he recognized his talents and determined to explore much more in the field of music.
Beethoven, well-known as a pianist and violinist, published his first three Sonatas in 1783. He conveyed new standards and emotions through music by creating melodies and harmonies. Altogether he composed ten symphonies during the period between 1800 and 1824. The great musician passed away on March 26, 1827. About 3000 people attended his funeral.
Answer 22.
Lantern Festival
Lantern festival signals the end of the new year festival period. People get together watching lanterns and fireworks, guessing lantern riddles and performing folk dance during this festival.
A common legend dealing with the origin of the lantern festival speaks of a secured bird that flew down to the earth from heaven. After it landed on earth it was hunted and killed by a hunter and some villagers. This angered the emperor in heaven. So he planned a firestorm to destroy the village. However, a Wiseman suggested that every family should hang red lanterns outside their homes, set up bonfires on the screen and light firecrackers for 3 days. Even today this festival is celebrated.
Answer 23.
Good Deeds – Sasikanth Nishanth Sarma. As the title of the poem suggests, the poem is about the good deeds that we do and the good results they produce.
The world is a mixture of happiness and sorrow. The poet reminds the readers that there is only one life and it should be enjoyed. At the same time, we should take care to keep company with good people. We should saw the seeds of virtues and cultivate the mind with healthy thoughts obtained from the experiences for ages. Keep on doing good deeds so that you do not have time to do bad things. If you do any harm to anyone, it will be a blemish on your character. One, who wants to succeed in life, must always pay attention to acquire good qualities. It is not the quantity but the quality that matters. A wise man can only give advice. But it is you who should know what your needs are. Then you decide what to do and what not to do.
It is a moral poem written in twenty-two lines. The poem is written in rhyming couplets. Every two lines rhyme. Knife/ life, well/dwell are examples. The poem is written in simple language.
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