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Kerala State Board New Syllabus Plus One Maths Notes Chapter 8 Binomial Theorem.

Kerala Plus One Maths Notes Chapter 8 Binomial Theorem

Binomial theorem gives the expansion of (a + b)n for a rational number ‘n’. In this Unit, we study the binomial theorem for positive integral indices only.

Expanding Binomial Calculator is a free online tool.

I. Binomial Theorem
The expansion of a binomial for any positive integral ‘n’ is given by the binomial theorem.

There are (n + 1) terms in the expansion of (a + b)n.

The sum of the indices of ‘a’ and ‘b’ in every term of the expansion is ‘n’.

The general term in the expansion is tr+1 = nCr an-r br

Middle term in the expansion:
1. If ‘n’ is even, \(\left(\frac{n}{2}+1\right)^{t h}\) term.

2. If ‘n’ is odd, \(\left(\frac{n+1}{2}\right)^{t h}\) and \(\left(\frac{n+1}{2}+1\right)^{b_{t}}\) term.

Students can Download Chapter 4 Presentation of Data Questions and Answers, Plus One Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus One Economics Chapter Wise Questions and Answers Chapter 4 Presentation of Data

Plus One Economics Presentation of Data One Mark Questions and Answers

Plus One Economics Chapter Wise Questions And Answers Pdf Question 1.
Which of the following comes under geometric diagram?
(a) Histogram
(b) Bar diagram
(c) Ogives
(d) Frequency polygon
Answer:
(b) Bar diagram

Plus One Economics Chapter Wise Questions And Answers Question 2.
Which of the following comes under frequency diagrams?
(a) Bar diagram
(b) Histogram
(c) Pie diagram
(d) All the above
Answer:
(b) Histogram

Plus One Economics Chapter Wise Questions And Answers Malayalam Question 3.
To draw time-series graph, time is presented on:
(a) X-axis
(b) Y-axis
(c) any of two
Answer:
(a) X-axis

Plus One Statistics Chapter Wise Questions And Answers Question 4.
Name the types of graphs.
Answer:

1. One dimensional graph
2. Two-dimensional graph
3. Three-dimensional graph
4. Pictograms

Plus One Economics Chapter Wise Questions And Answers Pdf Download Question 5.
State whether true or false.

1. The width of bars in a bar diagram need not be equal.
2. The width of rectangles in a histogram should essentially be equal.
3. Histograms can only be formed with continuous classification of data.
4. Histogram and column diagram are the same method of presentation of data.
5. Mode of a frequency distribution can be drawn graphically with the help of histogram,
6. The median of a frequency distribution cannot be drawn from the Ogive.

Answer:

1. true
2. false
3. true
4. true
5. true
6. true

Plus One Economics Presentation of Data Two Mark Questions and Answers

Hsslive Economics Plus One Chapter Wise Questions And Answers Question 1.
Which of the following is a cumulative frequency curve?
Answer:
(a) Bar diagram
(b) Histogram
(c) Ogive
(d) Pie diagram
Answer:
(c) Ogive

Plus One Economics Questions And Answers Question 2.
Distinguish between captions and stubs.
Answer:
Captions refers to the column headings and stubs refers to the row heading.

Histogram And Frequency Polygon Questions With Answers Question 3.
Match the following.

Answer:

Plus One Economics Presentation of Data Three Mark Questions and Answers

Presentation Of Data Class 11 Question 1.
What kind of diagrams are more effective in representing the following?

1. Monthly rainfall in a year
2. Composition of the population of Delhi by religion
3. Components of cost in a factory

Answer:

1. Simple bar diagram
2. Sub-divided or component bar diagram
3. Pie diagram

Pie Chart Class 11 Economics Question 2.
Name different types of diagrams.
Answer:
The different types of diagrams are:
1. Geometric diagram

• Bar diagrams
• Pie diagram

2. Frequency diagram

• Histogram
• Frequency polygon
• Frequency curve -Ogive

3. Arithmetic line graph

Question 3.
“Diagrams and graphs help us visualize the whole meaning of numerical complex data at a single glance”. Comment.
Answer:
One of the most convincing and appealing ways in which statistical results may be presented is through diagrams and graphs. The special feature of graphs and diagrams is that they do away with figures altogether. Diagrams and graph is a statistical method which can be used for simplifying the complexity of quantitative data and t make them easily intelligible.

It presents dry and uninteresting statistical facts in the shape of attractive and appealing pictures and charts. They are important methods of visual aids and are appealing t the eye and mind of the observer.

Question 4.
“There are generally three forms of diagrammatic presentation of data” explain.
Answer:
There are various methods to present data. But generally, three forms of presentation of data are there
which are noted below:

1. Geometric diagram
2. Frequency diagram
3. Arithmetic line graph

1. Geometric Diagram:
Bar diagram and pie diagram come in the category of geometric diagram for presentation of data. The bar diagrams are of three types-simple, multiple and component bar diagrams.

2. Frequency Diagram:
Data in the form of grouped frequency distributions are generally represented by frequency diagrams like histogram, frequency polygon, frequency curve, and ogive

3. Arithmetic Line Graph:
An arithmetic line graph is also called time-series graph and is a method of diagrammatic presentation of data. A line graph by joining these plotted points, thus, obtained is called arithmetic line graph or time-series graph.

Question 5.
Explain Ogive?
Answer:
Cumulative frequency of any class is equal to the sum of the frequencies of all the classes preceding that class and its own frequency e.g., frequencies are 10, 7, 12, 17 and 22. Cumulative frequencies are 10, 10 + 7 = 17, 17 + 12 = 29, 29 + 17 = 46 and 46 + 22 = 68.
Cumulative frequency of the last class = Total frequency.

For drawing an Ogive, cumulative frequency (i.e. number of values) is taken on the Y-axis and limits of class intervals on the X-axis.
Ogive is of two types:

1. less than
2. more than

In a “less than” type Ogive, we plot the upper limit of each class along the X-axis and in a “more than” type Ogive, we plot the lower limit of each class along the X-axis. Along the Y-axis, we plot the cumulative frequencies at the end of each class. Ogive can be drawn even if the class interval are unequal or open end. Ogives are performed over frequency curves for comparative study.

Question 6.
Illustrate how classes can be formed while presenting the data?
Answer:
Classes can be formed in two ways:

1. Exclusive type
2. Inclusive type

1. Exclusive Type:
When the class intervals are so fixed that the upper limit of one class is the lower limit of the new class, it is known as exclusive method of classification.

In this method, higher value of the variable in the class is not included in that class i.e.,

2. Inclusive Type:
In this method, the students getting say 39% marks will be included in class 30 – 39 itself i.e.,

Plus One Economics Presentation of Data Four Mark Questions and Answers

Question 1.
Choose the correct answer
a. Bar diagram is a

1. one-dimensional diagram
2. two-dimensional diagram
3. diagram with no dimension
4. none of the above

b. Data represented through a histogram can help in finding graphically the

1. mean
2. mode
3. median
4. all the above

c. Ogives can be helpful in locating graphically the

1. mode
2. mean
3. median
4. none of the above

d. Data represented through arithmetic line graph help in understanding

1. long term trend
2. cyclicity in data
3. seasonality in data
4. all the above

Answer:
a. 1. one-dimensional diagram
b. 3. mode
c. 3. median
d. 1. long term trend

Question 2.
Point out major parts of a statistical table.
Answer:

1. Table number
2. Title
3. Headnote
4. Stub
5. Box head or caption
6. Body or field
7. Footnote
8. Source note

Question 3.
Give the rules for constructing tables.
Answer:
The rules of constructing diagrams are:

• Every diagram should be titled.
• It should suit the size of the paper
• It should be neat and attractive
• It should be neatly indexed
• It should contain footnotes
• The details in diagram should be self-explanatory

Plus One Economics Presentation of Data Five Mark Questions and Answers

Question 1.
Explain the advantages of diagrammatic presentation.
Answer:
The advantages of diagrammatic presentation are given below.

1. Diagram give a clear picture of data
2. Comparison can be made easy
3. Diagrams can be used university at any place
4. It saves time and energy
5. The data can be remembered easily

Question 2.
Show how pie diagram is drawn for the following data?

Answer:

Question 3.
Give steps in the preparation of pie diagram.
Answer:
A pie diagram is also a component diagram, but unlike a component bar diagram, a circle whose area is proportionally divided among the components it represents. It is also called a pie chart. The circle is divided into as many parts as there are components by drawing straight lines from the centre to the circumference.

The following steps in the preparation of pie diagram are given below:

• Convert each component as percentage of the total.
• Multiply the percentage by 360/100 = 3.6 to convert into degree.
• Starting with the twelve o’clock position on the circle draw the largest component circle
• Draw other components in clockwise succession in descending order of magnitude except for each all components

Like all others and miscellaneous which are shown last:

• Use different columns or shades to distinguish between different components
• Explain briefly the different components either within the components in the figure or outside by arrow.

Plus One Economics Presentation of Data Eight Mark Questions and Answers

Question 1.
Write short notes on the following

1. pie diagrams
2. frequency curves
3. frequency polygon
4. ogive
5. arithmetic line graph

Answer:
1. Pie Diagram:
A pie diagram is also a component diagram, but unlike a component bar diagram, a circle whose area is proportionally divided among the components it represents. It is also called a pie chart. The circle is divided into as many parts as there are components by drawing straight lines from the centre to the circumference. Pie charts usually are not drawn with absolute values of a category.

The values of each category are first expressed as percentage of the total value of all the categories. A circle in a pie chart, irrespective of its value of radius, is thought of having 100 equal parts of 3.6° (3607100) each. To find out the angle, the component shall subtend at the centre of the circle, each percentage figure of every component is multiplied by 3.6°.

2. Frequency Polygon:
A frequency polygon is a plane bounded by straight lines, usually four or more lines. Frequency polygon is an alternative to histogram and is also derived from histogram itself. A frequency polygon can be fitted to a histogram for studying the shape of the curve. The simplest method of drawing a frequency polygon is to join the midpoints of the topside of the consecutive rectangles of the histogram.

3. Frequency Curve:
The frequency curve is obtained by drawing a smooth freehand curve passing through the points of the frequency polygon as closely as possible. It may not necessarily pass through all the points of the frequency polygon but it passes through them as closely as possible

4. Ogive:
Ogive is also called cumulative frequency curve. As there are two types of cumulative frequencies, for example, less than type and more than type, accordingly there are two ogives for any grouped frequency distribution data. Here in place of simple frequencies as in the case of frequency polygon, cumulative frequencies are plotted along y-axis against class limits of the frequency distribution.

For less than give the cumulative frequencies are plotted against the respective upper limits of the class intervals whereas for more than ogives the cumulative frequencies are plotted against the respective lower limits of the class interval. An interesting feature of the two ogives together is that their intersection point gives the median

5. Arithmetic Line Graph:
An arithmetic line graph is also called time-series graph and is a method of diagrammatic presentation of data. Init, time (hour, day/date, week, month, year, etc.) is plotted along x-axis and the value of the variable (time series data) along y-axis. A line graph by joining these plotted points, thus, obtained is called arithmetic line graph (time series graph). It helps in understanding the trend, periodicity, etc. in a long term time series data.

Question 2.
3 Forms of presentation of data

1. Textual
2. Tabular
3. Diagrams & graphs Prepare a flow chart.

Answer:

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.

Kerala Plus One Physics Notes Chapter 5 Law of Motion

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.

• Newtons first law
• Newtons second law
• Newtons third law

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
2. velocity of the body.

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:

• In hitting a ball with a bat
• In firing a gun

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 F_AB = – F_BA
(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 P_g and bullet acquires a momentum P_b.
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. ∆P_b = P_b – 0 = F∆t ______(2)
Since initially both are rest,
Dp = final momentum – initial
momentum Similarly the change in momentum of gun
∆p_g = p_g – 0 = -F∆t _______(3)
∴ Total momentum after firing = p_b + p_g
= 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 P_A and P_B. After collision, they acquire momenta P¹_A and P¹_g 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, F_BA∆t = P¹_B – P_B

[Where Dt is time for which the two bodies are in contact].
According Newton’s third law, we can write
F_AB = -F_BA

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
2. Non contact forces

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.

• Static friction: The opposing force that comes into play when one body tends to move over the surface of another (but the actual motion has yet not started)
• Limiting friction (f_s): The maximum value of static friction is called limiting friction.
• Kinetic friction (f_k)(or) dynamic friction: Kinetic friction or dynamic friction is the opposing force that comes into play when one body is actually moving’overthe surface of another body.
• Sliding friction: The opposing force that comes into play when one body is actually sliding over the surface of the other body is called sliding friction.
• Rolling friction: The opposing force that comes into play when one body is actually rolling over the surface of the other body is called rolling friction.

Laws of static Friction:

• The force of maximum static friction is directly proportional to the normal reaction
• The force of static friction is opposite to the direction in which the body tends to move.
• The force of static friction is parallel to the surfaces in contact.
• The force of maximum static friction is independent of the area of contact (as long as the normal reaction remains constant).
• The force of static friction depends only on the nature of surfaces in contact.

a. Laws of Kinetic friction:

1. The force of Kinetic friction is proportional to normal reaction.
2. The force of Kinetic friction is opposite to the dh rection in which the body moves.
3. The force of Kinetic friction is parallel to the surfaces in contact.
4. The force of Kinetic friction is independent of the area contact (as long as the normal reaction remains constant)
5. The force of Kinetic friction depends on the nature of surface.
6. Force of Kinetic friction is almost independent of the speed.
7. Force of Kinetic friction is less than force of static friction.

b. Coefficient of static friction:
The force of static friction (f_s)_max is directly proportional to the normal reaction N
(f_s)_max α N

Where m_s is called coefficient of static friction.
Definition of m_s
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 (f_k)_max α N

Where µ_k is called coefficient of Kinetic friction.
Definition of µ_k
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 f_limit 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

• Friction helps us to walk on the ground.
• Friction helps us to hold objects.
• Friction helps in striking matches.
• Friction helps in driving automobiles.
• Friction is helpful in stopping a vehicle etc.

Disadvantages of friction

• Friction produces wear and tear.
• Friction leads to wastage of energy in the form of heat.
• Friction reduces the efficiency of the engine etc.

Steps to reduce friction

• Polishing the surfaces in contact
• Use of lubricants
• Ball bearing placed between moving parts of machine.

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. For a stone rotated in a circle by a string, the centripetal force is provided by the tension in the string.
2. The centripetal force for motion of a planet around the sun is the gravitational force on the planet due to sun.
3. For a car on circular road, the centripetal force is provided by the friction between tire and road.

1. Motion of a car on a level road:

Consider a vehicle moving overa level curved road. The two forces acting on it are

• Weight (mg) vertically down
• The reaction (N)

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
v_max = \(\sqrt{\mu_{s} \mathrm{rg}}\)

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 V_o, m can be taken as zero.
putting m = 0 in the above equation we get

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.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 5 Complex Numbers and Quadratic Equations

Plus One Maths Complex Numbers and Quadratic Equations Three Mark Questions and Answers

Plus One Maths Chapter Wise Questions And Answers Pdf Chapter 5 Question 1.
If z₁ = 2 – i, z₂ = 1 + i

1. Find | z₁ + z₂ + 1| and |z₁ – z₂ + i| (2)
2. Hence find \(\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}\right|\) (1)

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 Complex Numbers and Quadratic Equations Four Mark Questions and Answers

Plus One Maths Questions And Answers Chapter 5 Question 1.
Consider the complex number \(\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}\)

1. Express in a + ib form. (2)
2. Convert into polar form. (2)

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.

1. Express the complex number \(\frac{2-i}{(1-i)(1+2 i)}\) in the form a + ib (2)
2. Solve the equation 27x² – 10x + 1 = 0 (2)

Answer:
1.

2. 27x² – 10x + 1 = 0

Complex Numbers Class 11 Extra Questions Chapter 5  Question 3.

1. For what value of x and y 4x + i(3x – y) = 3 – 6i (2)
2. Solve the equation 21x² – 28x + 10 = 0 (2)

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

Plus One Maths Complex Numbers and Quadratic Equations Six Mark Questions and Answers

Complex Numbers And Quadratic Equations Class 11 Pdf Question 1.

1. Express the complex number \(\frac{3-\sqrt{-16}}{1-\sqrt{-9}}\) in the form a + ib (2)
2. Represent the complex number \(\frac{5+i \sqrt{3}}{-4+2 \sqrt{3 i}}\) in the polar form. (2)
3. Solve the equation ix² – x + 12i = 0 (2)

Answer:
1.

2.

The complex number lies in the third quadrant;

3. ix² – x + 12i = 0

Plus One Maths Complex Numbers and Quadratic Equations Practice Problems Questions and Answers

Complex Numbers And Quadratic Equations Class 11 Solutions Question 1.
Express each of the following in a + ib form. (1 score each)

1. (2 – 4i) + (5 + 3i)
2. (1 – i) – (-1 + 6i)
3. 3(7 + 7i) + i(7 + 7i)
4. \(\left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+\frac{5}{2} i\right)\)

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)

1. (-5i)(\(\frac{1}{8}\)i)
2. (-i)(2i)(-\(\frac{1}{8}\)i)³
3. i⁹⁹
4. i¹¹¹ + i²²² + i³³³
5. (7 – i)(2 + 7i)
6. (-1 – i)(4 + 2i)
7. (5 – 3i)²
8. (\(\frac{1}{3}\) + 3i)³

Answer:
1.

2.

3. i⁹⁹ = i^{96 + 3} = i⁹⁶i³ = -i

4. i¹¹¹ + i²²² + i³³³ + i¹⁰⁸ + i^{220 + 2} + i^{332 + 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)

1. 3 – 4i
2. 2 – 3i
3. \(\sqrt{5}\) + 3i

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)

1. 1 + i
2. -1 + i
3. \(\sqrt{3}\) – i
4. \(\frac{5-\sqrt{3} i}{4+2 \sqrt{3} i}\)

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.

Kerala Plus One Accountancy Notes Chapter 3 Recording of Transactions – I & II

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:

• Cash Book
• Petty Cash Book
• Purchase Book
• Purchase Return Book
• Sales Book
• Sales Return Book
• Journal Proper

(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.

Kerala Plus One Economics Notes Chapter 1 Indian Economy on the Eve of Independence

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

1. To use India as a supplier of raw materials for British Industries.
2. To convert India into a market for the finished products produced in Britain.

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:

1. The exploitative land settlement system followed by British rulers
2. Use of low level of technology
3. Rural indebtedness
4. Low agricultural productivity
5. Use of limited chemical fertilizer
6. Inadequate irrigation facilities

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.

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.

Kerala Plus One Chemistry Notes Chapter 2 Structure of Atom

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_e) 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 × 10⁸ 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:

• unlike cathode rays, the e/m ratio of the particles depend upon the nature of gas present in the cathode ray tube.
• Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.
• The behaviour of these particles in the magnetic or electrical field is opposite to that observed for cathode rays.

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:

1. electromagnetic radiation possess both wave like and particle like properties(Dual character)
2. Experimental results regarding atomic spectra which can be explained only by assuming quantized electronic energy levels in atoms.

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:

• There is no time lag belween the striking of light beam and the ejection of electrons from the metal surface.
• The number of electrons ejected is proportional to the intensity or brightness of light.
• For each metal, there is a characteristic minimum frequency, u₀ (also known as threshold frequency) below which photoelectric effect is not observed. At a frequency u>u₀, the ejected electrons come out with certain kinetic energy.

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 m_ev² where m_e 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:
r_n = 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 R_H 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⁺ Li²⁺, Be³⁺ 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:

1. Dual behaviour of matter
2. Heisenberg uncertainty principle

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_x (or ∆v_x) 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_x is large]. On the other hand, if the velocity of the electron is known precisely ( ∆v_x 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_l

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_l’ 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_s). 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_s 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 2p_x, 2p_y, and 2p_z. 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 d_xy, d_yz, d_xz, d_x²-y² and d_z². The shapes of the first fourd-orbitals are similarto each other, where as that of the fifth one, d_z², 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 p_z orbital, xy-plane is a nodal plane, in case of d_xy 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

Kerala Plus One Botany Chapter Wise Questions and Answers Chapter 1 Biological Classification

Plus One Botany Biological Classification One Mark Questions and Answers

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.

1. Thermoacidophiles: Archaebacteria in hot spring
2. Ripening of fruits: …………….

Answer:
Ethylene.

Plus One Botany Chapter Wise Previous Year Questions And Answers Question 7.
Fill in the blanks.

1. Rhizopus: Phycomycetes
   Yeast: ………..
2. Holdfast: Anchorage
   Heterocyst: ……….

Answer:

1. Ascomycetes
2. N₂ fixation

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.

1. Order → Kingdom → species → phylum
2. species → genus → order → phylum

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.

1. Nostoc : Eubacteria:: methanogens: ………….
2. Yeast: ………………..:: Rhizopus: Phycomycetes :

Answer:

1. Archaebacteria
2. Ascomycetes

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:

1. Vibrio: Comma shaped
   ……….: Rod-shaped
2. Agaricus: Basidiomycetes
   Penicillium: ………….

Answer:

1. Bacillus
2. Ascomycetes

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.

Plus One Botany Biological Classification Two Mark Questions and Answers

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:

1. It does not include organisms showing both plant and animal character.
2. It does not take into the consideration of nature of nucleus.

Question 3.
Five-kingdom classification of organism was given by R.H.Whittaker. State the criteria followed by Whittaker for his classification.
Answer:

1. Nature of cell
2. Nature of nucleus
3. Mode of nutrition

Question 4.
Name the following;

1. A protist which can live both as an autotroph and as a heterotroph.
2. Name a protist group which consists of saprophytes.

Answer:

1. Euglena
2. Slime mould

Question 5.
State two economic importance of

1. Heterotrophic bacteria
2. Archaebacteria

Answer:

1. Major decomposers that help in the curdling of milk, production of antibiotic, fixing nitrogen and cause diseases like tetanus, typhoid, cholera etc.
2. Archaebacteria: production of biogas.

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:

• Algal bloom: Excessive growth of blue-green algae causes pollution of water bodies with characteristic odour.
• Red tide: Dinoflagellates like gonyaulax are red in colour which imparts red colour to seawater.

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:

Answer:

• a – Phytophthora infestans – light blight of potato.
• b- Agaricus campestris
• c – Penicillium notatum
• d – Saccharomyces cere visae

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:

Question 17.
Observe the cyanobacteria given below and answer the following.

1. Name the cyanobacteria, and the kingdom it belongs.
2. Label’s ‘P’ and mention its functions.

Answer:

1. Nostoc-kingdom-Monera
2. Heterocyst – To fix nitrogen from the atmosphere.

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:

1. Mode of nutrition
2. Mode of reproduction

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:

1. Karyogamy, Plasmogamy and Meiosis
2. Meiosis, Plasmogamy and Karyogamy
3. Plasmogamy, Karyogamy and Meiosis
4. Meiosis, Karyogamy and Plasmogamy

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.

1. Name the bacteria.
2. Identify the group in which it belongs.

Answer:

1. Methanogens
2. Archaebacteria

Question 27.
Based on the relationship, fill in the blanks.

1. Sac fungi: Ascomycetes
   Imperfect fungi: …………
2. Thermoacidophiles: Archaebacteria in hot springs
   …………………: Archaebacteria in Salty areas

Answer:

1. Deuteromycetes
2. Halophiles

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:

• Exogenous-Basidiomycetes. Its fruiting body is basidiocarp.
• Endogenous-Ascomycetes. Its fruiting body is ascocarp.

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

1. Unicellular
2. Well defined nucleus
3. Biflagellate-one flagellum lying longitudinally and the other transversely.

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:

Question 42.
Some bacteria are specialised and live in extreme habitat.

1. Name the types of bacteria are specified in the above statement.
2. Which is the part of bacteria modified to live in that condition?

Answer:
1. Types of bacteria

• Methanogens
• Halophiles
• Thermo acidophiles

2. Ceil wall structure

Question 43.
The two nuclei per cell can be seen in fungal cell but it later fuse in some members.

1. Name such type of fungal hyphae or mycelium.
2. Identify the classes of fungi.

Answer:

1. Dikaryotic mycelium
2. Ascomycetes, Basidiomycetes

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.

Plus One Botany Biological Classification Three Mark Questions and Answers

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

1. Amoeboid protozoans: They capture their prey by using pseudopodia. They live in freshwater. Some are parasites eg: entamoeba.
2. Flagellated protozoans: They are free-living or parasites. They cause diseases, eg: Trypanosoma-sleeping sickness.
3. ciliated protozoans: They possess cilia in their body surface for locomotion. They have gullet for food intake. Eg: Paramecium
4. Sporozoans: They are spore-producing organism that causes diseases eg: plasmodium causing malaria.

Question 2.
Different types of fungi are given
1. Classify them into their specific classes.

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.

• Phycomycetes – Mucor, Albugo
• Ascomycetes – Neurospora, Claviceps
• Basidiomycetes-Agaricus, Ustilago
• Deuteromycetes – Altemaria, Trichoderma.

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:

• Conidia are spores found in ascomycetes.
• These are haploid asexual spores produced in chains exogenously.
• Zygospores are the diploid resting spores found in mucor.
• Ascospores are haploid sexual spores found in sac-like structure (ascus).
• Sporangiophore is an aerial branch produced by hyphae in mucor that bear sporangia.

Plus One Botany Biological Classification NCERT Questions and Answers

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.

• No cell wall.
• Protein-rich layer, called pellicle, which makes flexible body.
• Two flagella of different lengths.
• Autotrophs in sunlight, heterotrophs in the absence of sunlight. Example: Euglena.

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.

Plus One Botany Biological Classification Multiple Choice Questions and Answers

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.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 2 Relations and Functions

Plus One Maths Relations and Functions Three Mark Questions and Answers

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}

1. Write R in roster form. (1)
2. Find the domain of R. (1)
3. Find the range of R. (1)

Answer:

1. R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 4), (2, 2), (4,4), (6,6), (3,3), (3,6)}
2. Domain of R = {1, 2, 3, 4, 6}
3. Range of R = {1, 2, 3, 4, 6}

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.

1. f(x) = 2 – 3x, x ∈ R, x>0 (1)
2. f(x) = x² + 2, x is a real number. (1)
3. f(x) = x, x is a real number. (1)

Answer:

1. Given; f(x) = 2 – 3x is a first degree polynomial function, therefore the range is R.
2. Given; f(x) = x² + 2, The range of x2 is [0, ∞) , then the range of f(x) = x² + 2 is [2, ∞)
3. Given; f(x) = x is the identity function, therefore the range is R.

Plus One Maths Relations and Functions Four Mark Questions and Answers

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

1. A × (B ∩ C) = (A × B) ∩ (A × C) (2)
2. A × C is a subset of B × D (2)

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.

1. f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\) (2)
2. f(x) = \(\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\) (2)

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}

1. Write A × B
2. Write relation from A to B in roster form. (1)
3. Represent all possible functions from A to B (Arrow diagram may be used) (2)

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.

Plus One Maths Relations and Functions Six Mark Questions and Answers

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

1. A × (B ∩ C) (1)
2. (A × B) ∩ (A × C) (2)
3. A × (B ∪ C) (1)
4. (A × B) ∪ (A × C) (2)

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, ∞).

Plus One Maths Relations and Functions Practice Problems Questions and Answers

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:

• G × H ={(7, 2), (7, 4), (7, 5), (8, 2), (8, 4), (8, 5)}
• H × G ={(2, 7), (2, 8), (4, 7), (4, 8), (5, 7), (5, 8)}

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)

1. f(x) = |x – 2|
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

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}

1. Write R in roster form. (1)
2. Write the range of R. (1)

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}

1. Write R in the roster form. (1)
2. Find the domain and range of R. (1)

Answer:

1. R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (4, 4), (5, 5), (6, 6)}
2. Domain = {1, 2, 3, 4, 6}; Range = {1, 2, 3, 4, 6}

Question 11.
Consider the real function

1. \(f(x)=\frac{x^{2}+2 x+3}{x^{2}-8 x+12}\)
2. Find the value of x if /(x) = 1
3. Find the domain of f.

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}

1. Write R in the roster form.
2. Find the range of R.

Answer:

1. R ={(1, 1), (2, 4)}
2. Range = {1, 4}

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.

Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 4 Bank Reconciliation Statement

Plus One Accountancy Bank Reconciliation Statement One Mark Questions and Answers

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

Plus One Accountancy Bank Reconciliation Statement Two Mark Questions and Answers

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.

Plus One Accountancy Bank Reconciliation Statement Five Mark Questions and Answers

Bank Reconciliation Statement Solutions Question 1.
From the following particulars, prepare the Bank Reconciliation Statement of Asha & Co. as on 31.3.2012.

1. Credit balance as per passbook is Rs. 10,000.
2. Bank collected a cheque of Rs. 500 on behalf of Asha & Co. but wrongly credited it to Asha’s account.
3. Bank recorded a cash book deposited of Rs. 1589 as Rs. 1598.
4. Withdrawal column of the passbook undercast by Rs.100.
5. The credit balance of Rs. 1500 as on the passbook was recorded in the debit balance.
6. The payment of a cheque of Rs. 350 was recorded twice in the passbook.
7. The passbook showed a credit balance for a cheque of Rs. 1000 deposited by Asha & Co.

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.

1. Cheques deposited into bank for collection, but not collected till date Rs. 720.
2. Cheques issued, but not paid by bank Rs. 650.
3. Direct payment by a customer to the bank not recorded in cash book Rs. 520.
4. Interest on deposit credited in passbook Rs. 310. Discounted bill dishonoured, entered on in the passbook Rs. 400.
5. Bank charges debited in passbook Rs. 75. Prepare Bank Reconciliation Statement of Reji as on May 31, 2009.

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.

1. Unpresented cheque 6000
2. Uncleared cheque 3400
3. Bank charges debited in the passbook only 1000
4. Bill collected and credited in the passbook only 1600
5. Cheque of Biju traders dishonored 1000
6. Cheque issued to Varma & Co. not yet entered in the cash book 600

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.

1. On 31/3/2009 his passbook show a credit balance of Rs. 1,150 Which was different from his cash book balance.
2. On comparison it was found that the cheque of Rs. 1,500 issued on 27^th March was paid by the bank oh 4^th April.
3. Cheques amounting to Rs. 1,700 were deposited . on 28^th March but in the passbook only Rs.
   700 was credited.
4. A customer made direct deposit in the bank on 31^st March amounting to Rs. 500, this was not recorded in the cash book.
5. A discounted bill receivable of Rs. 700 was returned dishonored to the bank on 29^th March. This entry was made in the cash book on 3^rd April.

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.

1. Cheque paid into bank but not yet cleared. Ram Kumar ₹1,000 Kishorekumar ₹500 1
2. Bank charges ₹300
3. Cheque issued but not presented Hameed₹2,000 Kapoor₹500
4. Interest entered in the passbook but not entered in the Cashbook ₹100

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.

1. A cheque of ₹400 drawn on his saving account has been shown on the current account.
2. He issued two cheques of ₹300 and ₹500 on December 25, but only the 1^st cheque, was presented for payment.
3. One cheque issued by Mr. Mohit of ₹ 500 on December 25, but it was not presented for payment whereas it was recorded twice in the cash book.

Answer:
Bank Reconciliation Statement of Mr Mohit as on December 31^st, 2010

Question 7.
On 1^st 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 1^st January 2010.

Question 8.
Prepare bank reconciliation statement.

1. Overdraft shown as per cash book on December 31, 2010, ₹10,000.
2. Bank charges for the above period also debited in the passbook, ₹100.
3. Interest on overdraft for six months ending December 31, 2010, ₹380 debited in the passbook.
4. Cheques issued but not encashed prior to December 31, 2010, amounted to₹2,150.
5. Interest on Investment collected by the bank and credited in the passbook, ₹600.
6. Cheques paid into bank but not cleared before December31,2010were₹1,100.

Answer:
Bank Reconciliation Statement as on 31^st December, 2010.

Question9.
Prepare a bank reconciliation statement from the following particulars and show the balance as per cash book.

1. Balance as per pass book on December 31, 2010, overdrawn₹20,000.
2. Interest on bank overdraft not entered in the cash book ₹2,000.
3. ₹200 insurance premium paid by bank has not been entered in the cash book.
4. Cheques drawn in the last week of December 2010, but not cleared till date for ₹3,000 and ₹3,500.
5. Cheques deposited into bank in November 2010, but yet to be credited on dated December 31, 2010, ₹6,000.
6. Wrongly debited by bank, ₹500.

Answer:
Bank Reconciliation Statement as on 31^st March, 2010

Plus One Accountancy Bank Reconciliation Statement Eight Mark Questions and Answers

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.

Kerala Plus One English Model Question Papers Paper 2

General Instructions to Candidates

• There is a ‘Cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘Cool off time’ to get familiar with questions and to plan your answers.
• Read the instructions carefully.
• Read the questions carefully before answering.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Give equations wherever necessary.
• Electronic devices except nonprogrammable calculators are not allowed in the Examination Hall.

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.

• Lantern festival signals the end of the new year festival period.
• People get together watching lanterns and fireworks, guessing lantern idles and performing a folk dance.
• Long ago, a sacred bird from heaven accidentally got lost and fell into the mortal world.
• An ignorant hunter thought the bird was a kind of fierce animal and killed it.
• The emperor planned a firestorm in retaliation.
• The villagers sought a way out.
• A wise man told them -hang red lanterns outside your homes – make bonfires and light firecrackers for 3 days.
• People, even today, light up lanterns and fireworks.

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.

1. Well, this is Lieutenant Mallet, this is M.Etierine Beer, and this is M.Paul Bessant and this is Guy de Maupassant.
2. The basket for the traveler to sit in was attached to the balloon and barometers were brought. The two trumpets, the eatables, the overcoats and raincoats and all the small articles that could go with the men in the flying basket were arranged.
3. Guiding the balloon is not an easy task. Winds and storms may change its course

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.

Plus One English Previous Year Question Papers and Answers

Kerala State Board New Syllabus Plus One Malayalam Textbook Answers Unit 4 Uravu Text Book Questions and Answers, Summary, Notes.

Kerala Plus One Malayalam Textbook Answers Unit 4 Uravu

Uravu Questions and Answers

Students can Download Chapter 3 Morphology of Flowering Plants Notes, Plus One Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Botany Notes Chapter 3 Morphology of Flowering Plants

Define Morphology:
It is study of external features of a plant i.e presence of roots, stems, leaves, flowers and fruits.

Morphology Of Flowering Plants Class 11 Notes The Root:
Root system:
In dicotyledonous plants, the direct elongation of the radicle leads to the formation of primary root which bears lateral roots that are secondary, tertiary, etc.

1. Tap root system:
The radical elongate and forms primary root or tap root that bears number of lateral roots. It is found in dicot root. eg: the mustard plant.

2. Fibrous root system:
In monocotyledonous plants, the primary root is short lived from there thin fibre roots originates. eg: wheat plant.

3. Adventitious roots:
In grass, Monstera and the banyan tree, roots arise from parts of the plant other than the radicle.

Anatomy Of Flowering Plants Class 11 Notes Hsslive Regions of the Root:
1. Region of Root cap:
It is the covering of root apex that protects the tender apex.

2. Region of Meristem:
This is the region just behind the the root cap that is capable of active cell division.

3. Region of elongation:
The cells proximal to this region undergo rapid elongation and enlargement for the growth

4. Region of maturation:
This is proximal to the region of elongation gradually differentiate and mature

5. Region root hairs:
From the region of maturation root hairs arise. These root hairs absorb water and minerals from the soil.

Plus One Botany Chapter Morphology Of Flowering Plants Modifications of Root:
Roots are modified for

1. Mechanical support
2. Storage of food
3. Respiration

1. Mechanical support:
(a) Prop roots:
In banyan tree, adventitious roots are modified and provide mechanical support

(b) Stilt roots:
In maize and sugarcane adventitious roots are supporting and coming out from the lower nodes of the stem. .

2. Storage of food:
In carrot, turnips tap roots are modified for food storage. In sweet potato adventitious roots are swollen and store food.

3. Respiration:
Rhizophora growing in swampy areas, many roots come out of the ground and grow vertically upwards. Such roots are called pneumatophores.
Function:
It help in the process of respiration.

Plus One Morphology Of Flowering Plants The Stem:
Salient features:

1. The stem arise from the plumule of the embryo of a germinating seed.
2. The stem bears nodes and internodes.
3. The region of the stem where leaves are born are called nodes while internodes are the portions between two nodes.

Function:
Support leaves, flowers and fruits. It also conducts water, minerals and do photosynthesis.

Plus One Botany Morphology Of Flowering Plants Modifications of Stem:
1. Storage of food:
Underground stems of potato, ginger, turmeric, zaminkand, Colocasia are modified to store food

2. Climbing:
Stem tendrils which develop from axillary buds, are slender and spirally coiled that help the plants to climb. eg: in gourds (cucumber, pumpkins, watermelon) and grapevines.

3. Protection:
Axillary buds of stem are modified into woody, straight and pointed thorns. eg: in Citrus, Bougainvillea Thev nrotect Dlants from browsina animals

4. Photosynthesis:
Some stems are lattened (Opuntia), or fleshy cylindrical (Euphorbia) structures. They contain chlorophyll and carry out photosynthesis.

5. Vegetative propagation:

• In grass and strawberry, etc. stem spread to new niches and when older parts die new plants are formed.
• In mint and jasmine lateral branch arises from the base of the main axis and after growing aerially and arch downwards to touch the ground.
• In Pistia and Eichhornia the lateral branch with short internodes and each node bearing a rosette of leaves and a tuft of roots
• In banana, pineapple and Chrysanthemum, the lateral branches originate from the basal underground portion of the main stem, grow horizontally beneath the soil and then come out obliquely upward giving rise to leafy shoots.

Plus One Botany Chapter 3 Notes The Leaf:
Salient features:

1. It is the flattened structure develops at the node and bears a bud in its axil.
2. The axillary bud later develops into a branch.
3. They are the most important vegetative organs for photosynthesis.
4. some plants leaf base bear two lateral stipules.
5. In monocotyledons, the leaf base expands into a sheath covering the stem.
6. In some leguminous plants the leaf base become swollen, which is called the pulvinus.
7. The lamina or the leaf blade is the green expanded part of the leaf with veins and veinlets.
8. The middle prominent vein, which is known as the midrib.

Function of veins:
Veins act as channels of transport for water, minerals and food materials.
A typical leaf consists of three main parts:

• Leaf base
• petiole
• lamina

Morphology Of Flowering Plants Class 11 Pdf Venation:
1. Reticulate Venation:
The arrangement of veins and the veinlets in the lamina of leaf. Veinlets repeatedly branched to form a network. eg: dicotyledonous plants.

2. Parallel Venation:
When the veins run parallel to each other within a lamina. eg: monocotyledons plants.

Morphology Of Flowering Plants Pdf Download Types of Leaves:

1. Siimple leaf: In this lamina is entire or the incisions do not touch the midrib.
2. Compound leaf: In this incisions of the lamina reach up to the midrib breaking it into a number of leaflets.

Can you see the bud in the axil of leaflet of compound leaf?
A bud is not present in the axil of leaflets of the compound leaf.

Two types of compound leaves:

1. Pinnately compound leaf: Number of leaflets are present on a common axis, the rachis, which represents the midrib of the leaf eg neem.
2. Palmately compound leaf: In this leaflets are attached at the tip of petiole, eg silk cotton.

Morphology Of Flowering Plants Notes Pdf Phyllotaxy:
It is the arrangement of leaves on the stem or branch Three types of phyllotaxy in plants alternate, opposite and whorled.

1. Alternate type: A single leaf arises at each node in alternate manner. eg: china rose, mustard and sunflower.
2. Opposite type: A pair of leaves arise at each node and lie opposite to each other. eg: Calotropis and guava.
3. Whorled type: More than two leaves arise at a node. eg: Alstonia.

Morphology Of Flowering Plants Class 11 Notes Pdf Download Modifications of Leaves:

1. Food storage: The fleshy leaves of onion and garlic store food.
2. Protection: The spines are developed in cacti act as organ of defence
3. Climbing: In peas Leaves are modified into tendrils for climbing
4. Photosynthesis: In Australian acacia, the leaves are small and short-lived. The petioles in these plants expand become green and synthesise food
5. Insect capture: In pitcher plant and venus-fly trap (insectivorous plants) leaves are modified for Capturing insects.

Morphology Of Flowering Plants Pdf The Inflorescence:
The arrangement of flowers on the floral axis is termed as inflorescence. Two major types of inflorescences are

1. Racemose
2. Cymose.

How will you differentiate recemose inflorescence from cymose?
In racemose type, the main axis continues to grow, the flowers are arranged in an acropetal succession. In cymose type, the main axis terminates in a flower. The flowers arranged in a basipetal order.

Morphology Of Flowering Plants Class 11 The Flower:
The flower is the reproductive unit in the angiosperms. It consists of different kinds of whorls (calyx, corolla, androecium and Gynoecium) arranged successively on the swollen end of the stalk or pedicel, called thalamus or receptacle.

What are the accessory and reproductive organs?
Calyx and corolla are accessory organs,while androecium and gynoecium are reproductive organs.

Did you see single accessory organ of a flower:
In lily plant, the calyx and corolla are not distinct, it is called as perianth. This is the single accessory organ of a flower. When a flower has both androecium and gynoecium, it is bisexual. A flower having either only stamens or only carpels is unisexual.
1. Actinomorphic Flower:
A flower can be divided into two equal radial halves in any radial plane passing through the centre, eg: mustard, datura, chilli.

2. Zyqomorphic:
A flower can be divided into two similar halves only in one particular vertical plane, eg: pea, gulmohur, bean, Cassia.

3. Asymmetric (irregular):
A flower cannot be divided into two similar halves by any vertical plane passing through the centre, as in canna. Afloweristrimerous, tetramerous or pentamerous when the floral appendages are in multiple of 3, 4 or 5, respectively.

Bracteate and ebracteate flower:
Flowers with reduced leaf found at the base of the pedicel, are called bracteate and those without bracts, ebracteate.

Classification of flower:
It is based on the position of calyx, corolla and androecium in respect of the ovary on thalamus
1. Hypogynous flower:
The position of gynoecium is highest when compared to other. The ovary in such flowers is said to be superior, eg: mustard, china rose and brinjal.

2. Perigynous flower:
The position of gynoecium is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level. The position of ovary is half inferior, eg: plum, rose, peach.

3. Epigynous flowers:
The margin of thalamus grows upward and other parts of flower arise above the ovary. The position of ovary is inferior. eg: guava and cucumber, and the ray florets of sunflower.

Morphology In Plants Class 11 Notes Parts of a Flower:
Each flower has four floral whorls -calyx, corolla, androecium and gynoecium.

Morphology Of Flowering Plants Class 11 Notes Pdf Calyx:
Act as protective whorl:
The calyx is the outermost whorl of the flower and segments are called sepals. The sepals are green, leaf like and protect the flower in the bud stage. The calyx may be gamosepalous (sepals united) or polysepalous (sepals free).

Plus One Botany Morphology Of Flowering Plants Notes Corolla:
Act as an attractive whorl:
Petals are brightly coloured that attract insects for pollination. Corolla may be free (polypetalous) or united (gamopetalous).

Aestivation:
The mode of arrangement of sepals or petals in floral bud is known as aestivation. The main types of aestivation are:

1. Valvate
2. Twisted
3. Imbricate
4. Vexillary.

1. In valvate sepals or petals in a whorl just touch one another at the margin, without overlapping, Eg -Calotropis
2. In twisted the one margin of the appendage overlaps the next one and so on. g. china rose, lady’s finger and cotton
3. In imbricate the margins of sepals or petals overlap one another but not in any particular direction. eg: Cassia and gulmohur

Can vou find out aestivation type in papillionaceous corolla?
There are five petals, the largest (standard) overlaps the two lateral petals (wings) which in turn overlap the two smallest anterior petals (keel). This is vexillary aestivation; Eg pea and bean flowers.

Androecium:
Structure of stamen:
Androecium is composed of stamens. Stamen is the male reproductive organ consists of a filament and an anther. Each anther is usually bilobed and each lobe has two chambers, the pollen-sacs. The pollen grains are produced in pollen-sacs. A sterile stamen is called staminode.

Differentiate between epibetalous and epiphvllous condition:
When stamens are attached to the petals, they are epipetalous as in brinjal, or epiphyllous when attached to the perianth as in the flowers of lily.

Free and fused nature of stamens:
The stamens in a flower remain free called as polyandrous. If the stamens are united into one bundle called as monoadelphous. eg: china rose, or two bundles called as diadelphous eg: pea, or into more than two bundles called as polyadelphous eg: citrus.

Variation in the length of filaments:
eg: Salvia and mustard.

Gynoecium:
Structure of carpel/pistil:
Gynoecium is the female reproductive part of the flower and is made up of one or more carpels. Acarpel consists of three parts namely stigma, style and ovary. Ovary is the enlarged basal part, on which lies the elongated tube, the style.

The style connects the ovary to the stigma. The stigma is the receptive surface for pollen grains. Each ovary bears one or more ovules attached to a flattened, cushion-like placenta.

Free and fused nature of carpel:
If carpels are free they are called apocarpous, eg lotus and rose. If carpels are fused they are called syncarpous. eg: mustard and tomato.

What happens to ovule and ovary after fertilization?
After fertilisation, the ovules develop Into seeds and the ovary matures into a fruit.

Placentation:
The arrangement of ovules within the ovary is known as placentation. Different types of placentation are marginal, axile, parietal, basal, central and free central.

The Fruit:
It is a ripened ovary developed after fertilisation.
Parthenocarpic fruit:
If a fruit is formed without fertilisation of the ovary, it is called a parthenocarpic fruit. The fruit consists of a wall called pericarp and seeds.

Meaning of Drupe:
Fruit that develops from monocarpellary superior ovaries and are one seeded. Eg:- mango and coconut.

Different layers of pericarp:
In mango the pericarp is well differentiated into an outer thin epicarp, a middle fleshy edible mesocarp and an inner stony hard endocarp. In coconut fruit is drupe, the mesocarp is fibrous.

The Seed:
Seed consists of a seed coat and an embryo. The embryo is made up of a radicle, an embryonal axis, one cotyledons as in wheat, maize or two cotyledons as in gram and pea.

Structure of a Dicotyledonous Seed:
The seed coat has two layers, the outer testa and the inner tegmen. The hilum is a scar on the seed coat .Above the hilum is a small pore called the micropyle. It consists of an embryonal axis and two cotyledons. The cotyledons are fleshy and contains reserve food materials. At the two ends of the embryonal axis are present the radicle and the plumule.

What is non endospermic seed?
In plants such as bean, gram and pea, the endosperm is not present in mature seeds and such seeds are called non endospermous.

Structure of Monocotyledonous Seed:
Monocotyledonous seeds are endospermic but it is non-endospermic in orchids. The seed coat is membranous and fused with the fruit wall. The outer covering of endosperm separates the embryo by a proteinous layer called aleurone layer The embryo is situated in one end of the endosperm.

It consists of Vy one large and shield shaped cotyledon known as scutellum and a short axis with a plumule and a radicle. The plumule is enclosed in sheaths called coleoptile and radicle are enclosed in sheaths called as coleorhiza.

Some Technical Description Of A Typical Flowering Plant:
In the floral formula, Br stands for bracteate K stands for calyx, C for corolla, P for perianth, A for androecium and G for Gynoecium, for superior ovary and for inferior ovary, for male, for female, for bisexual plants, for for actinomorphicand forforzygomorphic nature of flower

Description Of Some Important Families:
Fabaceae (Papilonoideae ):
It is a subfamily of family Leguminosae.

Vegetative Characters:

• Stem: Erect or climber
• Leaves: alternate, pinnately compound or simple; leaf base, pulvinate; stipulate; venation reticulate.

Economic importance:
Sources of pulses (gram, arhar.sem, moong, soyabean; Edible oil (soyabean, groundnut); dye (indigofera); Fibres (sunhemp); Fodder (Sesbania, Trifolium), Ornamentals (lupin, sweet pea) Medicine (muliathi).

Solanaceae (‘potato family’):
Vegetative Characters:
1. Stem:
herbaceous rarely woody, aerial; erect, cylindrical, branched, solid or hollow, hairy or glabrous, underground stem in potato (Solatium tuberosum).

2. Leaves:
alternate, simple, rarely pinnately compound, exstipulate; venation reticulate

Floral Characters:

• Inflorescence: Solitary, axillary or cymose as in Solanum
• Flower: bisexual, actinomorphic
• Calyx: sepals five, united, persistent, valvate aestivation
• Corolla: petals five, united; valvate aestivation
• Androecium: stamens five, epipetalous
• Gynoecium: bicarpellary, syncarpous; ovary superior, bilocular, placenta swollen with many ovules
• Fruits : berry or capsule
• Seeds: many, endospermous.

Economic Importance:
source of food (tomato, brinjal, potato)spice (chilli); Medicine (belladonna, ashwagandha) Fumigatory (tobacco); ornamentals (petunia).

Lilaceae (‘Lily family’):
Vegetative characters:
Perennial herbs with underground bulbs/corms/rhizomes. Leaves mostly basal, alternate, linear, exstipulate with parallel venation.

Floral characters:

• Inflorescence: solitary/cymose; often umbellate clusters.
• Flower: bisexual; actinomorphic
• Perianth: tepal six (3 + 3), often united into tube valvate aestivation.
• Androcium: stamen six, 3 + 3
• Gynoecium: tricarpellary, syncarpous, ovar ovules; axile placentation.
• Fruit: capsule, rarely berry
• Seed: endospermous

Economic Importance:
Ornamentals (tulip, Gloriosa), Source of medicine (Aloe), Vegetables (Asparagus), and colchicine (Colchicum autumnale).

Students can Download Chapter 3 Structural Organisation in Animals Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 3 Structural Organisation in Animals

Plus One Structural Organisation in Animals One Mark Questions and Answers

Plus One Zoology Chapter Wise Previous Questions And Answers Question 1.
Which one of the following types of cell is involved in making of the inner walls of large blood vessels?
(a) Cuboidal epithelium
(b) Columnar epithelium
(c) Squamous epithelium
(d) stratified epithelium
Answer:
(c) Squamous epithelium

Plus One Zoology Previous Question Papers Chapter Wise Question 2.
To which one of the following categories does adipose tissue belong?
(a) Epithelial
(b) Connective
(c) Muscular
(d) Neural
Answer:
(b) Connective

Plus One Zoology Previous Question Papers Chapter Wise Hsslive Question 3.
Which one of the following is not a connective tissue?
(a) Bone
(b) Cartilage
(c) Blood
(d) Muscles
Answer:
(d) Muscles

Plus One Zoology Structural Organisation Of Animals Question 4.
Which one of the following statements is true for cockroach?
(a) The number of ovarioles in each ovary are ten.
(b) The larval stage is called caterpillar
(c) Anal styles are absent in females
(d) They are ureotelic
Answer:
(c) Anal styles are absent in females

Plus One Zoology Chapter Wise Questions And Answers Pdf Question 5.
Match the following with reference to Cockroach and choose the correct option.

(a) A – iii, B – iv, C – ii, D – i
(b) A – iv, B – iii, C – ii, D – i
(c) A – iv, B – ii, C – iii, D – i
(d) A – ii, B – iv, C – iii, D – i
Answer:
(b) A – iv, B – iii, C – ii, D – i

Plus One Zoology Chapter Wise Questions And Answers Question 6.
Complete the following sentences:

1. The most common species of frog found in India is
2. Blood Vascular system of cockroach is

Answer:

1. Ranatigrina
2. Open

Zoology Chapter Wise Questions And Answers Question 7.
Name the respiratory organs of insects.
Answer:
Trachea

Plus One Zoology Chapter Wise Questions And Answers Pdf Download Question 8.
Note the relationship between the first two words and suggests a suitable word.

1.  Dendrons: Dendrites
   Axon: _________
2. Mast cells: Histamine
   Fibroblast: ________

Answer:

1. Myelin sheath
2. Fibres

Hsslive Plus One Zoology Chapter Wise Questions And Answers Question 9.

1. Which among the following is the bone cell _____________.
   (Leucocyte, Chondrocyte, Osteocyte, Thrombocyte)
2. In earthworm, segments 14-16 are covered by a prominent dark band of glandular tissue called __________. (Setae, Clitellum, Typlosol, None of these)

Answer:

1. Osteocyte
2. Clitellum

Plus One Zoology Chapter Wise Questions And Answers Pdf Hsslive Question 10.
‘Neuroglia make up more than one half the volume of neural tissue in our body’. Write the function of neuroglial cells.
Answer:
Neuroglia cells protect and support neurons in our neural system.

Plus One Zoology Animal Kingdom Questions Question 11.
From the following segments, find out the segments in which, Clitellum is situated in a mature earthworm.
(a) 12-16
(b) 14 -16
(c) 10-14
(d) 16-18
Answer:
(b) 14 -16

Zoology Previous Year Question Paper Plus One Question 12.
Name the cells present in areolar tissue.
Answer:
Fibroblasts, Macrophages, Mast cells

Structural Organisation In Animals And Plants Question 13.
_____________ make up more than one half the volume of neural tissue in our body.
Answer:
Neuroglia

Hsslive Zoology Plus One Previous Questions Question 14.
Given below are different types of epithelial tissues. Find out the tissue that are present in the tubular parts of nephrons in kidneys.
Answer:
Cuboidal

Plus One Zoology Previous Question Papers And Answers Question 15.
Heart is made of
1. Epithelial tissue and Muscular tissue.
2. Muscular tissue and neural tissue.
3. Musculartissue and connective tissue.
4. All the four types of tissues.
Answer:
4. All the four types of tissues.

Structural Organisation In Plants Question 16.
Write the cellular components of blood.
Answer:
Red blood cells (RBC), White blood cells (WBC) and platelets.

Plus One Zoology Questions And Answers Question 17.
In earthworm, setae are absent in the segments.
(a) First and last
(b) Clitellum only
(c) First, last and clitellum
(d) Peristomium and Clitellum
Answer:
(c) First, last and Clitellum.

Function Of Hepatic Caeca In Cockroach Question 18.
Intestine of earthworm contain a special organ which help in absorption. Identify the organ.
Answer:
Typhlosole

Question 19.
Name the structures which help in locomotion in Earthworm.
Answer:
Setae

Question 20.
In cockroach the fertilized eggs are encased in capsule called __________
Answer:
Oothecae

Plus One Structural Organisation in Animals Two Mark Questions and Answers

Question 1.
Regarding the blood vascular system of frog, some statements are given below. Check whether these are True or False and make corrections if necessary.

1. Open type.
2. Sinus venosus join the right atrium.
3. RBC is non-nucleated.
4. Circulation is achieved by the pumping action of the heart.

Answer:

1. False – Closed type
2. True
3. False – RBC is nucleated
4. True

Question 2.
You are given two cockroaches and the teacher asked to display the mouthparts of male cockroach.

1. How will you identify the male cockroach
2. Name the mouthparts of cockroach.

Answer:

1. In males styles are present and the abdomen is narrow and V shaped. In females abdomen is broad and anal styles are absent.
2. Labrum, mandibles, Hypopharynx, maxilla and Labium.

Question 3.
Give one word for the following.

1. Tissue attach one bone to another.
2. The structures protect and support neurons in neural system.

Answer:

1. Ligaments
2. Neuroglial cell

Question 4.
Match the following.

Answer:

Question 5.
Copy the diagram and label the numbered parts.

Answer:

1. Pharynx
2. Gizzard
3. Intestinal caecum
4. Typhlosole

Question 6.
Observe the diagram

1. Name the above connective tissue. Where they found in our body?
2. Name any two principle cell found in it.

Answer:

1. Areolar connective tissue, present beneath the skin.
2. Fibroblast and macrophage

Question 7.
Observe the pool of connective tissues carefully. Classify them under three headings.

Answer:

Question 8.
Anatomical structures of 3 animals are given below. Identify the 3 animals and arrange the structures into 3 separate groups.  (Typhlosole, Malpighian tubules, Blood glands, Cloaca, Phallic gland, Kidney, Collateral Gland)
Answer:

Question 9.
Copy the diagram and label the numbered parts.

Answer:

1. Testis
2. Vasa efferentia
3. Fat bodies
4. Urino genital duct

Question 10.
Match the following.

Answer:

Question 11.
Name the labeled parts A, B in the diagram given below

Answer:
A. Chondocyte

B. Collagen fibers

Question 12.
Circulating fluid of cockroach is colourless.

1. Name the circulating fluid.
2. Name the cell concerned in it.

Answer:

1. hemolymph
2. hemocytes

Question 13.
Match the following:

Answer:

Question 14.
Give one word for the following.

1. Tissue specialised to store fats.
2. Tissue contains fibroblasts, macrophages and mast cells.
3. Tissue attach skeletal muscles to bones.

Answer:

1. Adipose tissue
2. Areolar tissue
3. Tendons

Question 15.
You are given permanent slides of a bone and cartilage. How will you identify them using a microscope.
Answer:

1. Bone: Presence of Haversian system.
   Presence of Haversian canal in the centre.
2. Cartilage: Matrix contain spaces called lacunae.
   Lacunae contain chondrocyte.

Question 16.
Frogs maintain ecological balance. As a biological student do you agree with this statement? Why?
Answer:
Frogs maintain ecological balance because they serve as an important link of food chain and food web in the ecosystem.

Question 17.
Ratheesh mounted the mouthparts of cockroach in the following manner. If there is any error, correct it.

Answer:
A) Labium
B) Labrum
C) Mandibles
D) Hypopharynx
E) Maxilla.

Question 18.
Observe the diagram given below and label the parts
A, B, C, D.

Answer:
A – Liver
B – Stomach
C – Intestine
D – Rectum

Question 19.
While viewing the fallopian tube through CT scan we can find large number of cilia.

1. Identify the tissue present in the fallopian tube.
2. Mention the function of cilia.

Answer:

1. Ciliated epithelium
2. Ovum is moved by the help of cilia.

Question 20.
‘An earthworm respires through skin and excretes through nephridia’ Rewrite the above sentence beginning with ‘A cockroach ………………..’ making appropriate
changes.
Answer:
A cockroach respires through trachea and excretes through malpighian tubules.

Question 21.
Match the following

Answer:

Question 22.
Identify the diagram and label the part marked A.

Answer:
Frog female reproductive system Part A is Oviduct

Question 23.
Observe the given diagram.

1. Identify X and Y.
2. Write any two characters of X and Y.

Answer:

1. X – Striated muscle
   Y – Cardiac muscle
2. Striated muscle – Skeletal muscle, voluntary muscle, striations present Cardiac muscle – Muscles of heart, branched muscle fibres, intercalated disc.

Question 24.
Observe the table given below and fill in the blanks from the list given below:
(Pancreas, Liver, Salivary gland, Tear gland, Adrenal gland, Thyroid gland)

Answer:

Question 25.
There are some physiological similarities between typhlosole and villi. Substantiate your answer.
Answer:
Both typhlosole and villi are part of intestine of earthworm and human beings respectively. Typhlosole and villi increase the effective area of absorption in the intestine.

Question 26.
Observe the given diagrams.

1. Identify A and B.
2. Write any one function of each.

Answer:

1. A – Non-striated muscle (smooth muscle)
   B – Ciliated epithelium
2. A – Found in the wall of internal organs and concerned with the movement of these organs. B – Ciliated epithelium is to move particles or mucus in a specific direction over the epithelium.

Question 27.
Write the significance of

1. Blood
2. Neuroglia

Answer:

1. Blood – The main circulating fluid that help in the transport of various substances.
2. Neuroglia – Protect and support neurons. Neuroglia make up more than one half the volume of neural tissue in our body.

Question 28.
Classify the given terms into three columns and give appropriate headings.
Striated, heart muscle, intercalated disc, biceps, visceral organs, no striation.
Answer:

Question 29.
Where do you find the following structures in human body.

1. Chondrocytes
2. Axon
3. Osteocytes
4. Plasma

Answer:

1. Cartilage
2. Neuron
3. Bone
4. Blood

Question 30.
Write the suitable world for the 4^th place.

1. Muscle to bone – Tendon; Bone to Bone – _____________
2. Bone – osteocytes; Cartilage – __________
3. Cardiac Muscle – Involuntary; Striated Muscle – __________
4. Contractile cell – Muscle; Excitable cell – ______________

Answer:

1. Ligament
2. Chondrocytes
3. Voluntary
4. Neuron

Question 31.
Give an account on Nephridia of earthworm.
Answer:
Excretory organs in earthworm is nephridia. There are three types:

1. Septal nephridia: Present on both the sides of intersegmental septa of segments 5 to the last that open into intestine.

2. Integumentary nephridia: attached to lining of the body wall of segment 3 to the last that open on the body surface.

3. Pharyngeal nephridia: present as three paired tufts in the 4^th, 5^th, and 6^th segments.

Question 32.
Flow chart showing flow of food through the alimentary canal of an earthworm given. Correct the flow chart if there is any mistake.
Mouth→Pharynx → Stomach → intestine → Gizzard → Oesophagus.
Answer:
Mouth → Pharynx → Oesophagus → Gizzard → Stomach → Intestine

Question 33.
Match the following.

Answer:

Question 34.
You are given an earthworm for dissection, how will you identify its anterior, posterior, dorsal and ventral sides.
Answer:
The dorsal surface of the body is marked by a dark median mid dorsal line along the longitudinal axis of the body. The ventral surface is distinguished by the presence of genital openings. Anterior end consists of the mouth and the prostomium. The posterior end is slightly rounded.

Question 35.
Blood cell of earthworm lack hemoglobin. But blood of earthworm is red in colour. How do you account for it?
Answer:
Blood cell of earth worm lack hemoglobin. But blood of earthworm is red in colour because hemoglobin is dissolved in blood plasma.

Question 36.
Match the related items from B and C with Column A.

Answer:

Question 37.
Copy and complete the animals on the table and the excretory organs.

Answer:

Question 38.
Identify the animals from the hints given below.

1. Invertebrate, Burrowing in moist soil, Nephridia, Genital pore.
2. Chordata, Amphibian, Kidney, Cloaca
3. Invertebrate, Insecta, Malpighian tubules, gonapophysis.

Answer:

1. Earthworm
2. Frog
3. Cockroach

Question 39.
Identify the statements as True or False.

1. First segment of earthworm is Prostomium.
2. Peristomium contain mouth.
3. Clitellum is on segments 14-16.
4. Earthworm has only a single female genital aperture.

Answer:

1. False (First segment is peristomium)
2. True
3. True
4. Tare

Question 40.
Earthworm feeds on humus, which contain large amount of humic acid.

1. How does the earthworm neutralises the humic acid?
2. Name the region of alimentary canal where humic acid is neutralised.

Answer:

1. Calciferous gland, present in the stomach, neutralise the humic and present in humus.
2. In stomach.

Question 41.
Write a note on typhlosole.
Answer:
The special structure present in the intestine of earthworm between 26-35 segment is called typhlosole. It is an internal median fold of dorsal wall of intestine. Typhlosole increases the effective area of absorption in the intestine.

Question 42.
Blood glands are important in earthworm. Give reason.
Answer:
Blood glands are present on the 4^th, 5^th and 6^th segments of earthworm. They produce blood cells and hemoglobin which is dissolved in blood plasma.

Question 43.
Earthworm is a hermaphrodite animal. Where are its sex organs situated?
Answer:
Two pairs of testes present in the 10^th and 11^th segments. One pair of ovaries is attached at the inter-segmental septum of the 12^th and 13^th segment.

Question 44.
Earthworms are known as ‘Friends of farmers’. Give reason.
Answer:
Earthworms are known as ‘friends of farmers’ because they make burrows in the soil and make it porous which helps in respiration and penetration of the developing plant roots. The process of increasing fertility of soil by the earthworm is called vermicomposting. They are also used as bait in the game fishing.

Question 45.
Write the missing word in 4^th place.

1. Dorsally – Tergites; Ventrally – ……………
2. Earthworm – Nephridia; Cockroach – …………….

Answer:

1. Stern items
2. Malpighian tubules

Question 46.
Cockroach show sexual dimorphism. Write any two morphological differences between male and female cockroaches.
Answer:

1. In females, the 7^th sternum is boat-shaped and there is presence of a brood pouch.
2. In males, there is presence of anal style and genital pore in the 9^th segment.

Question 47
While dissecting a cockroach to display its digestive system, a student observed certain thick white tubules. Teacher explained it as part of respiratory system.

1. Name the tubules.
2. Draw a flow chart showing the flow of air through this system.

Answer:

1. Tracheal tubules
2. Spiracle → Trachea → Tracheoles → Tissue

Question 48.

1. Identify the diagram given below.
2. Label the parts a to e.

answer:

1. Digestive system of cockroach
2. parts from a to e
   • a – Oesophagus
   • b – Crop
   • c – Hepatic caeca
   • d – Mesentrorr
   • e – Malpighian tubules

Question 49.
Blood of cockroach is colourless. Give reason.
Answer:
Blood of cockroach is composed of colourless plasma and hemocytes. No respiratory pigments present in the blood of cockroach. So blood is colourless.

Question 50.
Cockroach is uricotelic animal. Give reason.
Answer:
Cockroach excrete nitrogenous waste in the form of uric acid. So cockroach is known as uricotelic animal.

Question 51.
Name the four structure helps in excretion in cockroaches.
Answer:

1. Malpighian tubule
2. fat body
3. nephrocytes
4. urecose gland.

Question 52.
If the head of cockroach is cut off, it will still alive for as long as one week. Give reason.
Answer:
The head of cockroach holds a bit of a nervous system while the rest is situated along the ventral part of its body. So, if the head of cockroach is cut off it will still live for long as one week.

Question 53.
Name the sensory organs of cockroaches.
Answer:
Antennae, Compound eyes, anal cerci, maxillary palps, Labial palps.

Question 54.
Compound eyes are more effective than vertebrate eye. Give reason.
Answer:
Each compound eyes in cockroach consists of about 200 ommatidia. With the help of several ommatidia, a cockroach can receive several images of an object. This kind of vision is known as mosaic vision. Compound eye is more effective than vertebrate eye because it can detect the movement of other animals more easily and efficiently.

Question 55.
Frog show sexual dimorphism. Write the morphological difference present only in male frog.
Answer:

1. Presence of vocal sacs
2. Presence of copulatory pad on the 1^st digit of the forelimb.

Question 56.
Prepare a flow chart showing the flow of food in the alimentary canal of frog. Compare it with alimentary canal of cockroach.
Answer:
1. In frog:
Mouth → oesophagus → stomach → Intestine → Rectus → Cloacal aperture

2. In coackroach:
Mouth → Oesophagus → Crop → Gizzard → Mesentron → ileum → Colon → Rectum → Anus

Question 57.
Cloaca is a common opening of 3 tubes or tracts.

1. Name the tracts which opens to cloaca.
2. Name the animal which possess cloaca.

Answer:

1. Tracts which opens to cloaca
   • Alimentary canal
   • Urinary canal
   • Reproductive tract
2. Frog

Question 58.
Name the portal systems found in frog

1. Between liver and intestine.
2. Between kidney and lower parts of the body.

Answer:

1. Hepatic portal system
2. Renal portal system

Question 59.

1. Frog is ureotelic animal. How?
2. Name the different parts of the excretory system- off rog.

Answer:

1. Frog excrete nitrogenous waste in the form of urea. So they are called ureotelic animal,
2. A pair of kidneys, ureters, cloaca and urinary bladder.

Question 60.
Answer in one word.

1. Covering of eyes in frog.
2. Number of cranial nerves in frog.
3. Membranous covering of Heart of frog.
4. Brain of frog is enclosed in a bony structure named ____________
5. Larval stage of frog.

Answer:

1. Nictitating membrane
2. Ten pairs
3. Pericardium
4. Cranium
5. Tadpole

Question 61.
Frogs are beneficial for mankind. Justify.
Answer:
Frogs are beneficial for mankind because they eat insects and protect the crop. Frogs maintain ecological balance because these serve as an important link of food chain and food web in the ecosystem. In some countries the muscular legs of frog are used as food by man.

Plus One Structural Organisation in Animals Three Mark Questions and Answers

Question 1.
Compare the circulatory system and blood of frog with that of an earthworm.
Answer:
In Frog blood circulatory system is closed type. Frog have a lymphatic system also. Heart is three chambered – two atria and one ventricle. Blood is composed of plasma and cells. The blood cells are RBC, WBC and platelets. RBC’s are nucleated and contain red coloured pigment namely hemoglobin. In earthworm blood circulatory system is closed type.

Blood glands are present on the 4^th, 5^th and 6^th segments. They produce blood cells and hemoglobin. Hemoglobin is dissolved in the plasma. Blood cells are phagocytic in nature.

Question 2.
The mounting of mouthparts of cockroach are given below. If there is any error, correct it.

Answer:
a. Labrum
b. Mandible
c. Hypopharynx
d. Maxilla
e. Labium

Question 3.
Answer in one word or one line.

1. Give the common name of Periplaneta Americana.
2. How many spermathecae are found in earthworm?
3. What is the position of a ovaries in cockroach?
4. How many segments are present in the abdomen of cockroach.
5. Where do you find Malpighian tubules?
6. Development through nymphal stage.

Answer:

1. Cockroach
2. 4 pairs of spermathecae are located 6^th – 9^th segments.
3. 2^nd – 6^th abdominal segments
4. 10 segments
5. At the junction of midgut and hindgut.
6. Paurometabolous

Question 4.
“All glands in our body do not have ducts of their own.” Do you agree with this statement.

1. Justify your answer using example.
2. Name three types of cell junctions in the epithelium.

Answer:

1. No. There are two types of glands in our body. Exocrine gland and Endocrine gland.
   • Exocrine gland – ducted gland
     • eg: Salivary gland
   • Endocrine gland – ductless gland
     • eg: Thyroid gland
2. Three types of junctions:
   • Tight junction
   • Adhering junction
   • Gap junction

Question 5.
Observe the following diagram carefully.

1. Identify and compare two diagrams.
2. Mention its location and function.

Answer:

1. A – squamous epithelium
   B – Columnar epithelium
2. Its locations and functions are
   • Both squamous and columnar epithelium are simple epithelium. Squamous epithelium is made up of a single thin layer of flattened cells with irregular boundaries. The columnar epithelium is composed of a single layer of tall and slender cells.
   • Squamous epithelium is found in the walls of blood vessels and air sacs of lungs and are involved in the functions like forming a diffusion boundary.
     Columnar epithelium is found in the lining of stomach and intestine and help in secretion and absorption.

Question 6.
Compare the excretory System in Cockroach and Frog.
Answer:
1. Cockroach:
Excretion is performed by malpighian tubules. They absorb nitrogenous waste products and convert them into uric acid which is excreted out through the hind gut. Therefore, cockroach is called Uricotelic. In addition, the fat body nephrocytes and urecose gland also help in excretion.

2. Frog:
Excretory system consists of a pair of kidneys. Ureter, Urinary bladder and cloaca. Excretory wastes are carried by blood into the kidney where it is separated and excreted. The frog excrete urea and thus a ureotelic animal.

Question 7.
Mention the mode of respiration in a frog while it is

1. In water
2. On land
3. When under aestivation or hibernation.

Answer:

1. Gills
2. Lungs
3. Skin (cutaneous respiration)

Question 8.
Arrange the columns A, B, and C in the table below and match them properly.

Answer:

Question 9.
Give reasons.

1. Simple epithelium functions as lining for the body cavities, ducts, tubes etc. where as compound epithelium is protective in function.
2. Cockroach does not possess respiratory pigment.
3. Frog in ureotelic where as tadpole is ammonotelic.

Answer:

1. Simple epithelium is composed of a single layer of cells. The compound epithelium consists of two or more cell layers and has protective function.
2. Cockroach respiratory organ is trachea. These tracheal tubules carry oxygen from the air to all the parts of their body. So they does not require any respirations pigment.
3. Ammonotelism demands an excess of water in the body. Tadpole is aquatic and there is availability of large amount of water. So tadpole is ammonotelic. Whereas Ureotelism requires only a moderate quantity of water. Frog is semiaquatic. So frog is ureotelic.

Plus One Structural Organisation in Animals NCERT Questions and Answers

Question 1.
Answer in one word or one line:

1. Give the common name of Periplanata americana.
2. How many spermathecae are found in earthworm?
3. What is the position of ovaries in cockroach?
4. How many segments are present in the abdomen of cockroach?
5. Where do you find Malpighian tubules?

Answer:

1. Cockroach
2. 4 pair of spermathecae are found in earthworm
3. 2 large ovaries are found lying laterally in the 2^nd to the abdominal segment.
4. The abdomen of cockroach consists of 10 segments.
5. Malpighian tubules are the main excretory organs of the cockroach.

Question 2.
Answer the following:

1. What is the function of nephridia?
2. How many types of nephridia are found in earthworm based on their location?

Answer:

1. Nephridia is the excretory organ of the earthworm or pheretima.
2. There are three types of nephridia in the earthworm:
   • septal nephridia, present on both the sides of intersegmental septa of segment 15 to the last that open into intestine,
   • integumentary nephridia, attached to lining of the body wall of segment 3 to the last that open on the body surface and pharyngeal nephridia, present as three paired tufts in the 4^th, 5^th and the segments.

Question 3.
Distinguish between the following:

1. Prostomium and peristomium
2. Septal nephridium and pharyngeal nephridium

Answer:

1. Prostomium is the frontmost part of the earth-worm. This is not called a true segment as it doesn’t contain typical organs of an annelida. Peristomium is from where the true segment of the earthworm body starts.
2. Septal nephridia, present on both the sides of intersegmental septa of segment 15 to the last that open into intestine.

Pharyngeal nephridia, present as three paired tufts in the 4^th, 5^th and 6^th segments Both are same structurally and functionally.

Question 4.
What are the cellular components of blood?
Answer:
Red Blood Cells and White Blood Cells are the cellular components of the blood.

Question 5.
What are the following and where do you find them in animal body?

1. Chondriocytes
2. Axons
3. Ciliated epithelium

Answer:

1. Chondriocytes are cells of cartilage.
2. Axons it is the tail like structure of a neuron
3. Ciliated epithelium is found in the inner lining of bronchioles, Cilia help trap and clear dust and foreign particles.

Question 6.
Distinguish between

1. Simple epithelium and compound epithelium
2. Cardiac muscle and striated muscle
3. Dense regular and dense irregular connective tissues
4. Adipose and blood tissue
5. Simple gland and compound gland

Answer:

1. Simple epithelium is composed of one layer of cells, while compound epithelium is composed of more than one layer of cells.

2. Cardiac muscles are present in the cells of heat only. They have contractile property which helps in the pumping action of the heart. Striated muscles are present near articulatory joints. Their role is to facilitate movement of organs like hands and feet.

3. Dense Connective Tissue. Fibres and fibroblasts are compactly packed in the dense connective tissues. Orientation of fibres show a regular or irregular pattern and are called dense regular and dense irregular tissues. In the dense regular connective tissues, collagen fibres are present in rows between may parallel bundles of fibres.

Tendons, which attach skeletal muscles to bones and ligments which attach one bone to another are examples of this tissue. Dense irregular connective tissue has fibroblasts and many fibres (mostly collagen) that are oriented differently. This tissue is present in the skin.

4. Adipose tissue is a type of loose connective tissue located mainly beneath the skin. The cells of this tissue are specialised to store fats. The excess of nutrients which are not used immediately are converted into fats and are stored in this tissue. Blood is a fluid connective tissue. The main function of blood is to transport gases, nutrients and easte products in the body.

5. Simple gland is composed of single cell, while compound gland is composed of multiple cells.

Question 7.
Mark the odd one in each series:

1. Areolar tissue, blood, neuron, tendon
2. RBC, WBC, platelets, cartilage
3. Exocrine, endocrine, salivary gland, ligament
4. Maxilla, mandible, labrum, antennae
5. Protonema, mesothorax, metathorax, coxa

Answer:

1. Neuron is not a connective tissue
2. Cartilage is not part of blood
3. Ligament is not part of gland
4. Antennae is not a masticating part of cockroach
5. Protonema is thread like chain of cells, all others are about morphological structure of cockroach.

Plus One Structural Organisation in Animals Multiple Choice Questions and Answers

Question 1.
Carotene pigment is found in the cells of
(a) dermis
(b) epidermis
(c) adipose cell
(d) Both (b) and (c)
Answer:
(d) Both (b) and (c)

Question 2.
The lining of intestine and kidneys in human is
(a) keratinised
(b) brush bordered
(c) ciliated
(d) None of these
Answer:
(b) brush bordered

Question 3.
The type of cell junction, which facilitates cell to cell communication is
(a) tight junction
(b) adhering junction
(c) gap junction
(d) desmosomes
(e) brush borders
Answer:
(c) gap junction

Question 4.
Keratinised dead layer of skin is made up of
(a) stratified squamous
(b) simple cuboidal
(c) simple columnar
(d) stratified columnar
Answer:
(a) stratified squamous

Question 5.
Endothelium of blood vessels is made up of
(a) simple cuboidal epithelium
(b) simple squamous epithelium
(c) simple columnar epithelium
(d) simple non-ciliated columnar
Answer:
(b) simple squamous epithelium

Question 6.
In which one of the following preparations are you likely to come across cell junctions most frequently?
(a) Ciliated epithelium
(b) Thrombocyte
(c) Tendon
(d) Hyaline cartilage
Answer:
(a) Ciliated epithelium

Question 7.
Bowman’s glands are located in the
(a) proximal end of uriniferous tubules
(b) anterior pituitary
(c) female reproductive system of cockroach
(d) olfactory epithelium of our nose
Answer:
(d) olfactory epithelium of our nose

Question 8.
Compound squamous epithelium is found in
(a) stomach
(b) intestine
(c) trachea
(d) pharynx
Answer:
(d) pharynx

Question 9.
Find out the wrongly matched pair.
(a) Squamous epithelium – Skin of frog
(b) Columnar epithelium – Peritoneum of body cavity
(c) Ciliated epithelium – Bronchioles
(d) Stratified squamous – Oesophagus epithelium
(e) Glandular epithelium Salivary gland
Answer:
(b) Columnar epithelium – Peritoneum of body cavity

Question 10.
The type of tissue lining the nasal passage, bronchioles and fallopian tubes is
(a) columnar ciliated epithelium
(b) cuboidal epithelium
(c) neurosensory epithelium
(d) germinal epithelium
(e) stratified columnar epithelium
Answer:
(a) columnar ciliated epithelium

Question 11.
The nasal chamber of rabbit has three thin twisted bony plates called conchae. They are lined by
(a) striated cuboidal epithelium
(b) simple cuboidal epithelium
(c) simple squamous epithelium
(d) simple ciliated columnar epithelium
Answer:
(d) simple ciliated columnar epithelium

Question 12.
The cavities of alveoli of lungs are lined by
(a) cuboidal epithelium
(b) columnar epithelium
(c) stratified cuboidal epithelium
(d) squamous epithelium
Answer:
(d) squamous epithelium

Question 13.
The type of epithelial cells, which line the inner surface of fallopian tubes, bronchioles and small bronchi, are known as
(a) squamous epithelium
(b) columnar epithelium
(c) ciliated epithelium
(d) cubical epithelium
Answer:
(c) ciliated epithelium

Question 14.
Agranulocytes are
(a) lymphocytes and monocytes
(b) eosinophils and basophils
(c) lymphocytes and eosinophils
(d) basophils and monocytes
Answer:
(a) lymphocytes and monocytes

Question 15.
________ acts as a shock absorber to cushion when tibia and femur came together.
(a) Ligament
(b) Cartilage
(c) Tendon
(d) Disc
Answer:
(b) Cartilage

Question 16.
Erythropoiesis starts in
(a) kidney
(b) liver
(c) spleen
(d) red bone marrow
Answer:
(b) liver

Question 17.
Largest single mass of lymphatic tissue in the body is
(a) lung
(b) spleen
(c) liver
(d) kidney
Answer:
(b) spleen

Question 18.
Hypochronic microcytic anaemia and leucopenia are caused by the deficiency of respectively.
(a) pyridoxine and riboflavin
(b) pyridoxine and folacin
(c) biotin and folacin
(d) biotin and cyanocobalamin
Answer:
(b) pyridoxine and folacin

Students can Download Chapter 9 Mechanical Properties of Solids Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 9 Mechanical Properties of Solids

Plus One Physics Mechanical Properties of Solids One Mark Questions and Answers

Plus One Physics Mechanical Properties Of Solids Chapter 9 Question 1.
If longitudinal strain fora wire is 0.03 and its poisson’s ratio is 0.5, then its lateral strain is
(a) 0.003
(b) 0.0075
(c) 0.015
(d) 0.4
Answer:
(c) 0.015
Poisson’s ratio σ = \(\frac { Lateral\quad strain }{ Longitudina\quad strain } \)
Lateral strain = Longitudinal strain × σ
= 0.03 × 0.5 = 0.015.

Mechanical Properties Of Solids Question Bank Chapter 9 Question 2.
Between steel and diamond, which is more elastic?
Answer:
Steel.

Mechanical Properties Of Solids Important Questions Chapter 9 Question 3.
What is the value of Y for a perfectly elastic body?
Answer:
Infinity.

Mechanical Properties Of Solids Neet Questions And Answers Pdf Chapter 9 Question 4.
Young’s modulus of the wire depends on
(a) length of the wire
(b) diameter of the wire
(c) material of the wire
(d) mass hanging from the wire
Answer:
(c) material of the wire
Young’s modulus of wire depends only on the nature of the material of the wire.

Mechanical Properties Of Solids Class 11 Important Questions Pdf Question 5.
Two solid spheres of the same material have the same radius but one is hollow while the other is solid. Both spheres are heated to same temperature. Then
(a) the solid sphere expands more
(b) the hollow sphere expands more
(c) expansion is same for both
(d) nothing can be solid about their relative expansion if their masses are not given
Answer:
(c) expansion is same for both.

Plus One Physics Chapter Wise Questions And Answers Chapter 9 Question 6.
For most materials the Young’s modulus is n times the rigidity modulus, where n is
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3
For most materials, the modulus of rigidity, η is one third of the Young’s modulus, Y.
η = \(\frac{1}{3}\)Y or Y = 3η
∴ n = 3.

Hsslive Plus One Physics Chapter Wise Questions And Answers Chapter 9 Question 7.
The stress required to double the length of a wire of Young’s modulus Y is
(a) Y/2
(b) 2Y
(c) Y
(d) 4Y
Answer:
(c) Let L be the length of a wire.
If the length of a wire is doubled, the longitudinal strain will be

Young’s modulus Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ Y = Stress (∵ Strain = 1).

Plus One Physics Chapter Wise Questions And Answers Pdf Hsslive Chapter 9 Question 8.
What is the bulk modulus of a perfectly rigid body?
Answer:
Infinity

Mechanical Properties Of Solids Neet Questions Chapter 9 Question 9.
What is shear modulus of liquid?
Answer:
Zero

Mechanical Properties Of Solids Pdf Chapter 9 Question 10.
What is rigidity modulus of liquid?
Answer:
Zero.

Mechanical Properties Of Solids Class 11 Questions And Answers Pdf Question 11.
Identify the type of modulus in twisting of cylinder.
Answer:
Shear modulus

Plus One Physics Important Questions And Answers Pdf Chapter 9 Question 12.
Why work is required to be done to stretch a wire?
Answer:
Work is required to be done against the inter molecular forces of attraction.

Mechanical Properties Of Solids Questions And Answers Chapter 9 Question 13.
Name the material which is famous for a large elastic after effect.
Answer:
Glass

Plus One Physics Mechanical Properties of Solids Two Mark Questions and Answers

Mechanical Properties Of Solids Neet Questions Pdf Chapter 9 Question 1.
Find stress required to double the length of a wire of Young’s modulus Y
Answer:
Let L be the length of a wire.
If the length of a wire is doubled, the longitudinal strain will be

Young’s modulus Y = \(\frac{\text { Stress }}{\text { Strain }}\)
∴ Y = Stress (∵ Strain = 1).

Hss Live Plus One Physics Chapter Wise Questions And Answers Chapter 9 Question 2.
When wire is bent back and forth, it becomes hot. Why?
Answer:
When wire is bent back and forth, the deformations are beyond elastic limit. A part of work done is converted in to heat energy. Hence wire becomes hot.

Plus One Physics Mechanical Properties of Solids Three Mark Questions and Answers

Class 11 Physics Chapter Mechanical Properties Of Solids Chapter 9 Question 1.
The stress strain graph of two bodies A and B are given in the figure

1. Which of the material has greater youngs modulus?
2. Which of the two material is preferable to be used as a rope in a crane?
3. Redraw the graph and mark the regions in the graphs where the elastic force is strictly conservative

Answer:
1. Youngs modulus, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
The slope of above graph gives youngs modulus. The body A has larger slope. Hence youngs modulus of ‘A ‘is greater than ‘B’.

2. The fracture point is greater for A. Hence A is stronger. Due to high strength, the material A is used as rope in a crane.

3.

Up to the point C, the elastic force is strictly conservative.

Mechanical Properties Of Solids Solutions Chapter 9 Question 2.
Hook’s law is said to be that fundamental law in elasticity.

1. State Hook’s law of Elasticity.
2. Name the different types of modulus of elasticity with their equations.

Answer:
1. For small deformations, the stress and strain are proportional to each other, ie: stress ∝ strain stress = k × strain. Where k is the proportionality constant and is known as modulus of elasticity.

2.

Plus One Physics Mechanical Properties of Solids Four Mark Questions and Answers

Mechanical Properties Of Solids Class 11 Solutions Chapter 9 Question 1.
When an external force deforms a solid, internal restoring forces are developed in the body giving rise to stress and strain.

1. Define stress and strain.
2. Draw the stress-strain diagram and mark the positions of elastic limit and the regions of elastic and plastic behaviors.

Answer:
1. Stress: It is the restoring force developed per unit area of cross section.
Strain: The ratio of change in dimension to original dimension is called strain.

2.

Plus One Physics Mechanical Properties of Solids Five Mark Questions and Answers

Plus One Physics Chapter Wise Questions And Answers Pdf Chapter 9 Question 1.

1. Distinguish between perfectly plastic and perfectly elastic materials.
2. What is the quantity obtained from the slope of a stress-strain graph and what is the area under the curve?
3. Two wires have their lengths in the ratio 1:3 and radii in the ratio 2:1. What will be the ratio of elongations for the same linear stress?

Answer:
1. When deforming forces are applied on a body its volume or shape will change. On removing the deforming forces it regains its original size and shape. It is called elastic body.

On removing the deforming forces. The body which cannot retain the original shape and size is called plastic body.

2. Modulus of elasticity. Area under stress strain graph gives workdone.

3. we know l₁: l₂, 1 : 3
r₂ : r₁, 2 : 1

Mechanical Properties Of Solids Questions Pdf Chapter 9 Question 2.
The modulus of elasticity of rubber is higher than that of steel.

1. Do you agree with this statement? Why?
2. Give your explanation.
3. Prove that the elastic potential energy density of a stretched wire is half the product of stress and strain.

Answer:
1. No

2. Y = \(\frac{F l}{A \Delta l}\)
∆l is higher for rubber than steel. Hence youngs modulus is smaller for rubber than steel.

3. Workdone in stretching a wire.
= \(\frac{1}{2}\) × stretching force × extension
= \(\frac{1}{2}\) Fe
Let ‘a’ be the area of cross section and ‘L’ the length of the wire. The volume ‘aL’
Workdone for a volume ‘aL’ = \(\frac{1}{2}\)
Workdone per unit volume

Important Questions From Mechanical Properties Of Solids Class 11 Question 3.
We have common experience that a thin thread breakes if pulled at its ends. Also a thin metallic wire, when pulled at the ends elongates and then contracts, when released.

1. What do you mean by elasticity?
2. Draw a the stress-strain graph of a metal and mark elastic limit, plastic region, and fracture point.
3. A lift is tied with thick iron wires and its mass is 1000kg. What should be the minimum diameter of the wire if the maximum acceleration of the lift is 1.2 m/s² and maximum safe stress of the wire is 1.4 × 10⁸ N/m² (g = 9.8m/s²)

Answer:
1. Elasticity of a body is the property of the body by virtue of which the body gains its original shape and size when the deforming force is removed.

2. Stress strain curve

3. When the lift goes upward with as acceleration a, then tension in the wires,
T = m (g + a)
= 1000(9.8 + 1.2)
= 11000N
Let r be the minimum radius of the wire
Then maximum stress = \(\frac{T}{\pi r^{2}}\)
But the maximum stress the wire can withstand = 1.4 × 10⁸

r = 0.005m
∴ diameter = 0.01m.

Question 4.
The property of material bodies to regain its original size on the remaoval of deforming force is called elasticity.

1. What are the values of
   • Young’s modulus for a perfectly rigid body
   • Rigidity modulus of a liquid
2. Why do we prefer steel to copper in the manufacture of spring?
3. How much should the pressure on a littre of water be changed to compress it by 0.10%?
   (B = 2.2 × 10⁹N/m²)

Answer:
1. The values:

• infinity
• Zero

2. Young’s modulus of steel is more than that of copper. Hence steel is more elastic than copper.

3.

Plus One Physics Mechanical Properties of Solids NCERT Questions and Answers

Question 1.
A steel wire of length 4.7m and cross sectional area 3 × 10^-5m² stretches by the same amount as a copper wire of length 3.5m and cross sectional area of 4 × 10^-5m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Answer:
For steel
l₁ = 4.7m, a₁ = 3 × 10^-5m²
If F newton is the stretching force and ∆l metre the extension in each case, then


For copper
l₂ = 3.5m, a₂ = 4 × 10^-5m²

Dividing (1) by (2), we get

Question 2.
Following figure shows the strain-stress curve for a given material. What are

1. Young’s modulus and
2. approximate yield strength for this material?

Answer:
1. It is clear from the graph that for a stress of 150 × 10⁶Nm^-2, the strain is 0.002
∴ Young’s modulus of the material

= 7.5 × 10¹⁰Nm^-2.

2. It is clear from the graph that A corresponds to the yield point. So, the approximate yield strength of them material is 300 × 10⁶Nm^-2 or 3 × 10⁸Nm^-2.

Question 3.
The edge of an aluminium cube is 10cm long. One face of the cube is firmly fixed to a vertical wal. A mass of 100kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25GPa. What is the vertical deflection of this face?
Answer:

3.92 × 10^-7m
≃ 4 × 10^-7m.

Question 4.
A steel cable with a radius 1.5cm supports a chair lift. If the maximum stress is not to exceed 10⁸Nm^-2, what is the maximum load the cable can support?
Answer:
Maximum load = Maximum Stress × Cross sectional area
= 10⁸Nm^-2 × \(\frac{22}{7}\) × (1.5 × 10^-2)^-2N
= 7.07 × 10⁴N.

Question 5.
Compute the fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10atm.
Answer:
Given : Bulk modulus of elasticity of glass
= 37 × 10⁹Nm² and 1 atm
= 1.013 × 10⁵Pa
p = 10 atm = 10 × 1.013 × 10⁵Pa
Fractional change in volume = Volumetric strain

Question 6.
Determine the volume of contraction of a solid copper tube 10cm on an edge, when subjected to a hydraulic pressure of 7 × 10⁶Pa. Bulk modulus of copper =140 × 10⁹Pa.
Answer:
V = L³ = (10cm)³ = (0.1m)³
p = 7 × 10⁶Pa, K = 140 × 10⁹Pa