Category: CBSE

CBSE Class 9 Maths Lab Manual – Quadratic Polynomials

Objective
Learning geometrical representation of the factorization of the following quadratic polynomials:

1. x² + 5x + 6
2. x² – 5x + 6

Prerequisite Knowledge

1. Knowledge of quadratic equations.
2. Splitting of the middle term of a quadratic polynomial as ax² + bx+ c = a(x+p)(x – q)
   where -p + q = \(-\frac { b }{ a }\), -pq = \(\frac { c }{ a }\)
3. Area of a rectangle = l x b
4. Area of a square = (side)²

Materials Required
Glazed papers (blue, green, orange, yellow and pink), white sheet of paper, geometry box, ruler, pair of scissors and gluestick.

Procedure

1. Every x² represents the area of pink square of side x-units.
   Therefore, to represent 2x², use two pink squares of side x units each. Take x as 3 units.
   
2. Every x represents area of a green rectangular strip of dimensions (1 x x).
   [For 5x, use 5 green strips each of dimensions (1 x x)]
   
3. Every (-x) is represented by a blue rectangular strip of dimensions (1 x x).
   For (-5x), use 5 blue strips each of dimensions (1 x x).
   
4. All positive integers are represented by yellow unit squares and all negative integers are shown by orange squares.
   

Case I
Let us consider the expression x² + 5x + 6 which is of the form (ax² + bx + 2).

1. The polynomial x² + 5x + 6
   ⇒ x² + 3x + 2x + 6 can be factorized as (x+ 3)(x+2).
2. To present x², draw a pink square of xunits [fig. (i) ].
   
3. To represent 3x, draw three rectangular strips of green colour of dimension (1 x x) [fig. (ii)].
   
4. To represent 2x, draw two green rectangular strips of dimensions (1 x x) [fig. (iii)].
   
5. To represent 6, draw 6 yellow unit squares [fig. (iv)].
   
6. Cut all the strips from the glazed paper.
7. Now, paste all the strips together on the white sheet of paper as shown in fig.(v).
   

Case II
Consider the expression x² – 5x + 6 and factorize it x² – 3x – 2x + 6 = (x – 3)(x – 2).

1. Cut a pink square of dimension xunits (say 8 units).
2. To represent 6, cut six yellow squares of dimension 1 unit.
3. To represent -5x {(-3x) + (-2x)}, cut five blue strips of dimension (1 x x).
4. Paste the pink square strips and all the yellow squares on a white sheet paper as shown in fig. (vi).
   
5. Now, paste all the five blue strips over the pink polygon as shown in fig. (vii).
   

Observation and Calculation
Case I
x² + 5x + 6
area of 5 green strips = 5x=2x+3x
area of pink square = x²
area of 6 yellow unit squares = 6
total area of rectangle obtained = x² + 3x + 2x + 6 = x² + 5x + 6 = (x+3)(x+2)

Case II
x² – 5x + 6
area of 5 blue rectangular strips = 5x (negative)
area of a pink square = x²
area of 6 yellow unit squares = 6
total area of pink rectangle obtained after pasting all strips
= (x – 2)(x – 3)
= x² – 2x – 3x + 6
= x² – 5x + 6x
∴ x² – 5x + 6 = (x – 3)(x – 2)

Result
We verified the factors of two quadratic polynomials geometrically by paper cutting and pasting.

Learning Outcome
Above method gives us the geometrical interpretation of the factorization of quadratic expressions of the form ax² + bx + c or ax² – bx + c.

Remarks

1. Pasting of blue strips over pink area means reducing pink area.
2. The pink portion so obtained represents the factors of the given quadratic expression.
3. Students may take different colour combinations.

Activity Time
By using paper cutting and pasting method, represent the factors of following quadratic expressions:

1. x² – x – 6
2. 2x² + 5x + 2

Viva Voce

Question 1.
How many linear factors can be in a quadratic polynomial ?
Answer:
2 linear factors.

Question 2.
Find two numbers whose sum is 1 and product is -12.
Answer:
-3 and 4.

Question 3.
Factorize: x² + 7x + 12.
Answer:
(x + 3)(x + 4).

Question 4.
Find two numbers whose sum is 0 and product is -6.
Answer:
√6 and -√6

Question 5.
What is the degree of a quadratic polynomial ?
Answer:
The degree of a quadratic polynomial is 2.

Question 6.
Give one example of a binomial.
Answer:
x + 5y.

Question 7.
Is 2 + x² + x a polynomial ?
Answer:
Yes.

Question 8.
Whatis the degree of ax² + bx + c?
Answer:
2.

Question 9.
Write the product of (2x – 1)(x + 1).
Answer:
2x² + x – 1

Question 10.
Is y² + \(\frac { 2 }{ y }\) + 5 a polynomial?
Answer:
No, as power of y in \(\frac { 2 }{ y }\) is -1

Multiple Choice Questions

Question 1.
Factorize the quadratic polynomial x²+ 6x + 8 :
(i) (x + 4)(x + 2)
(ii) (x – 4)(x – 2)
(iit) (x + 5)(x + 3)
(iv) none of these

Question 2.
Write the factors of x² – 6x + 8 :
(i) (x – 4)(x – 2)
(ii) (x – 4)(x + 2)
(iii) (x + 4)(x – 2)
(iv) none of these

Question 3.
Factorize x² – 5x + 6
(i) (x – 3)(x + 2)
(ii) (x – 3)(x – 2)
(iii) (x + 3)(x + 2)
(iv) none of these

Question 4.
Write the quadratic polynomial for (x— 1) (x— 2)
(i) x² + 3x + 2
(ii) x² + 3x – 2
(iii) x² – 3x + 2
(iv) none of these

Question 5.
What is the degree of 3x² + 2x + 1 ?
(i) 1
(ii) 3
(iii) 2
(iv) none of these

Question 6.
What will be the degree of a biquadratic polynomial ?
(i) 2
(ii) 3
(iii) 4
(iv) none of these

Question 7.
How many zeroes , are possible of a quadratic polynomial ?
(i) 2
(ii) 3
(iii) 1
(iv) none of these

Question 8.
Is x – 2 a factor of x² – 6x + 5 ?
(i) yes
(ii) no
(iii) can’t say
(iv) none of these

Question 9.
Is (x – 4)(x – 5) = 7x a quadratic polynomial ?
(i) yes
(ii) no
(iii) linear
(iv) none of these

Question 10.
Is (2x² – 4) = 2x² + 5x a quadratic polynomial ?
(i) yes
(ii) no
(iii) can’t say
(iv) none of these

Answers

1. (i)
2. (i)
3. (ii)
4. (iii)
5. (iii)
6. (iii)
7. (i)
8. (ii)
9. (i)
10. (ii)

Math Lab ManualMath Labs with ActivityMath LabsScience LabsScience Practical Skills

CBSE Class 10 Science Practical Skills – Stomata

BASIC BUILDING CONCEPTS
Stomata are tiny pores present on both upper and lower surfaces of the leaves but cannot be observed by an unaided eye. A compound microscope is required to study the minutest details of a stoma(singular).
We should know how to use a compound microscope in order to observe minute things while conducting experiments in the Biology labs.

Compound Microscope
A compound microscope is an instrument which is used to observe the enlarged and magnified image of the small objects.
A compound microscope consists of two systems of lenses called the objective lens and ocular lens (eyepiece) to obtain higher magnification.

Parts of a Compound Microscope

I. Mechanical Parts

1. Base
   It is a horse shoe-shaped structure that provides sta-bility and support to the microscope.
2. Pillar
   It is a vertical pro-jection fixed with hinge to the foot (base).
3. Arm
   It is curved and sup¬ports body tube and is used for handling the microscope.
4. Inclination joint
   It is the point where arm is attached to pillar and this permits tilting of microscope to adjust to eye level.
5. Stage
   Stage is rectangular flat metallic plate fixed to the lower end of arm. It has a hole in centre for the light to reach the object. It is used for keeping the object to be observed.
6. Clips
   Two clips on stage help in holding the slide.
7. Diaphragm
   It is present below the stage and regulates the entry of light on to the slide. Diaphragm may be disc diaphragm and iris diaphragm.
8. Body tube
   It is hollow tube which is attached to upper part of the arm. It can move up and down with the help of screws.
9. Nosepiece
   It is the circular metallic piece attached to lower end of body tube. It is fitted with three objective lens having different magnifications.
10. Coarse adjustment screw
    This can move body tube up and down for focussing
11. Fine adjustment screw
    This is small-size screw needed for fine adjustments.

II. Optical Parts

1. Mirror
   It is used for reflecting light into the microscope and is located at lower end of the arm, below the stage. It is concave on one side and plane on the other.
2. Eyepiece
   It is lens fitted at the top of body tube. It can be of 10 X or 15 X magnification.
3. Objective lenses
   These are attached to nosepiece. Usually, objectives with magnification lOx (low power) and 45 X (high power) are available to the students.

How to use Microscope

1. Take a compound microscope and clean it with the help of cotton cloth.
2. Move the objective lens (low power) and set it right above the central hole in the stage.
3. Adjust the mirror, while looking in the microscope to set the light. Adequate light should be available to illuminate the microscope.
4. Take a clean prepared slide and clip it on to the stage.
5. Move the coarse adjustment screw of microscope to focus object on the slide.
6. Now, once slide is focussed and observed under low power (10 X), then move the nosepiece and adjust the high power (45 X) without moving the coarse adjustment.
7. Focussing at high power (45 X) can be made sharp by use of fine adjustment screw.

Mounting a slide

1. Take a clean glass slide and wipe it with cotton cloth to remove any dust ).
2. To locate the centre, place the slide on a paper and draw its outline. Remove the slide and draw the diagonals to mark the centre in the outline of slide. Now, place the slide on this paper over the outline drawn and hence centre point can be observed.
3. In a watch glass, take some stain and place the material in it for staining. After a minute, wash the material with water so that excess stain get removed.
4. Now, place a drop of glycerine on the centre of slide and transfer the material on to drop with the help of brush.
5. Place the coverslip on the slide in such a position that its one side touches the liquid on the slide.
   Hold the other side edge with the help of needle, obliquely placed. Now, gently lower down the needle till the coverslip touches the slide and covers the material. In this way air bubbles will be eliminated.
   

AIM
To prepare a temporary mount of a leaf peel to show stomata.

MATERIALS REQUIRED
Leaf of Tradescantia or periwinkle or balsam, slide, coverslip, forceps, needles, cotton cloth, brush, blotting paper, watch glass, blade, dropper, glycerine, safranin and compound microscope.

THEORY
Stomata (singular: stoma) are tiny pores present on the surface of the leaves. Though they are found on both upper and lower epidermis of leaf but they are more in number on the lower epidermis. Each stoma is bounded by two kidney-shaped guard cells. These guard cells possess a nucleus and a number of chloroplasts. The walls of guard cells are differentially thickened and elastic, i.e. they are thickened on inner side and thinner and more elastic on outer side.
The opening and closing of stomata is operated by the change in the turgidity of guard cells. The stomata help in exchange of oxygen, carbon diodide and water vapour between atmosphere and plant.

PROCEDURE

1. Take a freshly plucked leaf, and remove the peel from its lower surface by tearing it.
2. Put the leaf peel in a watch glass containing water so that the peel does not dry.
3. To the watch glass containing leaf peel, add 1-2 drops of safranin to stain the peel.
4. With the help of a brush, place the leaf peel in the centre of a clean slide.
5. Put a drop of glycerine on the slide over the peel.
6. Now, with the help of a needle, gently place a coverslip over the peel.
7. Remove the excess glycerine with a blotting paper.
8. Observe the slide, first, under the lower magnification (i.e., 10 X) of a compound microscope and then, under the higher magnification (i.e., 45 X).

OBSERVATIONS

1. Epidermis or leaf peel consists of a number of cells which are irregular in outline and are arranged in single layer with no intercellular spaces.
2. Tiny pores called stomata are seen in the epidermal cells.
3. Each stoma consists of two kidney-shaped guard cells.
4. Each guard cell has a nucleus and many chloroplasts.
   

RESULT
Minute apertures called stomata are seen in the temporary mount of leaf peel. Each stoma is enclosed by two kidney-shaped guard cells. These guard cells differ from other epidermal cells in having chloroplast.

PRECAUTIONS

1. Peel should be taken from freshly plucked leaf.
2. Peel should not be allowed to dry.
3. Leaf peel should not be over stained.
4. The slide should not be dirty. –
5. Use a brush to transfer the leaf peel from watch glass to slide.
6. Peel should be placed in centre of slide.
7. Curling of peel should be avoided while placing it on slide.
8. The epidermal peel should be small in size.
9. Place the coverslip gently to avoid entry of air bubbles.
10. Excess stain and glycerine should be removed with blotting paper.

INTERACTIVE SESSION

Question 1.
Why is it that we most commonly use safranin as a stain for staining sections of plants?
Answer:
Safranin (red coloured stain) stains the lignin and suberin and other plant materials easily.

Question 2.
Why should the leaf peel be mounted in glycerine?
Answer:
The leaf peel should be mounted in glycerine so that the cells of peel do not dry up.

Question 3.
Why is it preferred to take an epidermal peel from lower surface of leaf?
Answer:
In monocots, the distribution of stomata on lower and upper surface of leaf is almost similar. But in case of dicots, the number of stomata are more on the lower surface than upper surface. Hence, it is preferred to take epidermal peel from lower surface of leaf.

Question 4.
What are guard cells?
Answer:
Guard cells are the cells surrounding the stomatal pore. Each guard cell consists of a nucleus and many chloroplasts.

Question 5.
What is the shape of guard cells in

1. Dicots
2. Monocots?

Answer:
The shape of guard cells in

1. Dicots — Kidney-shaped
2. Monocots — Dumb-bell-shaped.

Question 6.
Name the instrument used for knowing the relative size of stomata.
Answer:
Porometer.

Question 7.
Where can you find chloroplasts in the stomata?
Answer:
Chloroplasts are present in the guard cells.

Question 8.
Give an example of a plant where stomata is absent.
Answer:
Hydrilla.

Question 9.
How are guard cells different from the rest of the epidermal cells?
Answer:
Guard cells are nucleated and have peculiar shape (either kidney-shaped as in dicots or dumb¬bell-shaped as in monocots) and possess many chloroplasts. Guard cell walls show differential thickenings. Epidermal cells on the other hand lack chloroplast and are irregular in shape.

Question 10.
What is the most important function of stomata?
Answer:
Functions of stomata include gaseous exchange and transpiration.

Question 11.
What is transpiration?
Answer:
The loss of water in the form of vapour from leaves of plant is called transpiration.

Question 12.
State the type of modification in xerophytic plants with respect to structure of stomata.
Answer:
They possess sunken stomata.

Question 13.
Name the plant hormone that helps in opening of stomata.
Answer:
Cytokinin.

Question 14.
Name the plant hormone which causes stomatal closure.
Answer:
Abscisic acid (ABA).

Question 15.
Name the inorganic ions which help in opening and closing of stomata.
Answer:
K+ ions (potassium ions).

Question 16.
Why do guard cells bend, when they swell up?
Answer:
The guard cells bend because of the differential thickenings of its inner and outer walls. The inner wall of guard cell is thicker and less stretchable as compared to the outer wall.

Question 17.
State the most important functions of leaf.
Answer:
Important functions of leaf include

• Photosynthesis
• Respiration
• Transpiration.

Question 18.
State the most important significance of transpiration.
Answer:
Transpiration helps in upward movement of water (against gravity) in plants.

NCERT LAB MANUAL QUESTIONS

Question 1.
What is the function of guard cells in stomata?
Answer:
Guard cells control the opening and closing of stomata.

Question 2.
Why is the number of stomata greater on the lower surface of a leaf?
Answer:
If more stomata are present on upper surface of leaf, then there would be more transpiration. Hence, to avoid more transpiration the number of stomata is greater on the lower surface of leaf.

Question 3.
Why are stomata absent in roots?
Answer:
Stomata are present on aerial parts of plant for transpiration to occur. Roots are present in soil which would block them.

Question 4.
What is the shape of guard cells in stoma of grass leaf?
Answer:
Dumb-bell shaped.

Question 5.
Do guard cells have rigid or elastic walls? Justify your answer.
Answer:
When the guard cells become turgid, their thin walls get extended and thick walls become concave. This results in opening of stomata. On the other hand, when guard cells lose water and become flaccid, the thin and thick walls revert back to original position resulting in the closure of stomatal pore. This shows that guard cells have elastic walls.

PRACTICAL BASED QUESTIONS

Multiple Choice Questions/VSA (1 Mark)

Question 1.
While preparing a temporary mount of stomata, four students used different stains as given below: [CBSE Sample Paper 2009]

The correct stain was used by the student
(a) A
(b) B
(c) C
(d) D

Question 2.
To prepare a good temporary mount of the petunia leaf peel showing many stomata, the student has to get the peel from the [Foreign 2009]
(a) tip of the leaf
(b) upper surface of the leaf
(c) lower surface of the leaf
(d) point of attachment of the leaf to its petiole

Question 3.
Why do we place the leaf peels in water ?
(а) Guard cells become turgid
(б) Peels remain green
(c) Peels don’t dry up
(d) To wash the leaf peels.

Question 4.
Precautions to be taken by students while preparing a temporary mount of a leaf peel to show its stomata is/are
(a) the slide should not be dirty
(b) peel should be taken from freshly plucked leaf
(c) leaf peel should not be over stained
(d) all of these

Question 5.
While preparing a temporary stained mount of a leaf epidermal peel, the extra glycerine is removed by [CBSE 2012, 2011, Delhi 2007]
(a) washing with water
(b) washing with calcium chloride solution
(c) soaking with blotting paper
(d) absorbing with cotton wool

Question 6.
In order to complete the diagram of stomatal
apparatus given below, nuclei should be drawn in the parts marked. [Delhi 2007C]

(a) A and B
(b) A and C
(c) B and C
(d) A, B and C

Question 7.
In the following diagram of the stomatal apparatus, which parts are incorrectly labelled? [AI 2007C]

(a) I and IV
(b) II and III
(c) IV and V
(d) I and V

Question 8.
A well stained leaf peel preparation when focussed under high power of the microscope would show [AI 2007]
(a) epidermal cells, guard cells each with one nucleus and many chloroplasts and stomata.
(b) epidermal cells, stomata, guard cells with many nuclei and one chloroplast each
(c) stomata and guard cells without nuclei or chloroplasts
(d) stomata but no guard cells or epidermal cells

Question 9.
A well stained leaf peel mount when observed under the high power of a microscope shows nuclei in [CBSE 2012, AI 2008]
(a) only epidermal cells
(b) only guard cells
(c) guard cells and epidermal cells
(d) guard cells, epidermal cells and stoma

Question 10.
In the slide of an epidermal peel, the parts which appear pink coloured after staining with safranin are [Foreign 2008]
(a) stomata only
(b) nuclei only
(c) cell membrane and cytoplasm
(d) all parts in the peel

Question 11.
In the diagram of stomata shown below, the labelling by four students was tabulated by the teacher in the table given below. Whose labelling was correct?

(a) A
(b) B
(c) C
(d) D

Question 12.
Students observed the epidermal peel of a leaf under the high power of a microscope. The following are the sketches made by them.

The correct sketch is
(a) A
(b) B
(c) C
(d) D

Question 13.
Epidermal cells are different from guard cells with respect to
(a) thickenings of wall
(b) shape of cell
(c) cytoplasmic contents
(d) all of these

Question 14.
The steps involved in making a slide of epidermal peel of leaf are given as follows:
I. Pull out a thin peel from the lower surface of the leaf.
II. Place a drop of glycerine on the slide.
III. Stain the peel in safranin.
IV. Place the stained peel on the glycerine.
V. Remove the extra stain by washing with water.
VI. Place the cover slip over the peel.
Which one is the correct sequence of steps to be followed?
(a) I, II, III, IV, V, VI
(b) I, III, V, II, IV, VI
(c) I, III, IV, II, V, VI
(d) I, II, IV, III, V, VI

ANSWER KEY

1. (c)
2. (c)
3. (c)
4. (d)
5. (c)
6. (c)
7. (b)
8. (a)
9. (c)
10. (d)
11. (c)
12. (c)
13. (d)
14. (b)

Short Answer Questions (2 Marks)

Question 1.
Tradescantia leaf is usually taken for preparing temporary mount of leaf peel to observe stomata. Why? Name any two other leaves that can be used for this experiment.
Answer:
Tradescantia provides good leaf peel. Therefore, it is commonly used for this experiment. Lily leaf and Bryophyllum leaf can also be used for preparing temporary mount of leaf peel to observe stomata.

Question 2.
How can you take out peel from lily leaf? Why is it necessary to keep the peel in water?
Answer:
Take a fresh lily leaf and simply fold it in the centre. Remove the peel from the lower surface. Leaf peel should be put in a watch glass containing water so that the peel does not dry. If peel gets dried up, shape of the cell boil becomes distorted.

Question 3.
Name the cells that form stoma. What is their shape?
Answer:
Two guard cells constitute a stoma. In dicotyledons they are kidney-shaped but in monocotyledons they are dumbbell shaped.

Question 4.
List the various steps of observing a slide under the microscope.
Answer:

• Take a clean prepared slide and clip it on
• Look in the microscope and adjust the mirror to illuminate the slide.
• Move the coarse adjustment screw of microscope to focus the slide (object) under low power.
• Move the nosepiece and adjust the high power objective lens.
• Focus the slide (object) at high power by using fine adjustment screw. the stage.

Question 5.
When a student observes a temporary mount of leaf peel under a microscope, he observes two different types of cells in leaf peel. Name these two different types of cells. On what basis can a student differentiate between these two cells.
Answer:
Two different types of cells are epidermal cells and guard cells.
Epidermal cells lack chloroplast and are irregular in shape while guard cells have chloroplast. The guard cells can be kidney-shaped or dumb-bell shaped.

Science Practical SkillsScience LabsMath LabsMath Labs with Activity

CBSE Class 9 Science Practical Skills – Animal Kingdom

EXPERIMENT

AIM
To observe and draw the given specimens – earthworm, cockroach, bony fish and bird. For each specimen record.

1. One specific feature of its phylum.
2. One adaptive feature with reference to its habitat.

MATERIALS REQUIRED
Preserved specimens of earthworm, cockroach, bony fish and bird, pencil, practical notebook.

THEORY
Different organisms live in different habitats. Each animal has certain features which enable it to live in its habitat. The special features which enable plants and animals to be successful in a particular environment are called adaptations. Adaptations are due to certain modifications that are observed in organism at morphological, anatomical and physiological levels. By observing simple morphological features different adaptations found in organisms can be studied.

PROCEDURE

1. Observe carefully the preserved specimen.
2. Note the characteristics of organisms.
3. Draw well labelled diagrams in practical notebook.

OBSERVATIONS
1. EARTHWORM (Pheretima)
It belongs to phylum Annelida.

FEATURES OF PHYLUM

1. The body of the animal is divided into segments (internally and externally) called metameres.
2. Chitinous setae which act as locomotory organ are present.

Adaptive Features
Earthworm lives inside a burrow in the soil. It has following adaptations:

1. Its body is cylindrical and elongated.
2. The anterior end is pointed and bears mouth. Over the mouth is a small sensory lobe called prostomium. It helps in sensing the soil while digging.
3. It feeds on soil or humus and excretes fine soil and undigested material as ring-shaped castings through anus. This helps in making burrows.
4. Skin is always kept moist with the help of mucus glands.
   

2. COCKROACH (Periplaneta)
It belongs to phylum Arthropoda.

FEATURES OF PHYLUM

1. The body is divided into segments with a distinct head.
2. It possesses jointed appendages with three pairs of legs.

ADAPTIVE FEATURES
Cockroaches live in warm, dark and wet places. They are found in sewers, basements, in drains etc. They are adapted to nocturnal habit. Following are adaptations found in cockroaches.

1. The body is covered by chitinous cuticle which is impervious to water.
2. On the lateral side of body spiracles are present which help in respiration.
3. Head is triangular in shape and bears two compound eyes and two antennae. Antennae are sensory in function.
   

3. Bony fish
It belongs to phylum Chordata, sub phylum Vertebrata and class Pisces.

Features of Phylum chordata

1. They have notochord.
2. They have post anal tail.

Features of Class pisces

1. They are aquatic animals.
2. They have four pairs of gills which are covered by operculum.
3. They have terminal mouth.

Adaptive Features
Bony fish adaptive to live in water (aquatic adaptations).

1. They have streamlined body which helps in swimming.
2. Their body is covered by scales.
3. They have dorsal, pelvic, pectoral and tail fins for locomotion. Dorsal and pelvic fins help in balancing and movement in water, pectoral fins act as brakes and tail fins help in changing direction.
4. Their eye is covered by nictitating membrane to protect them from water.
5. They usually have air bladder which helps in buoyancy.
   

4. BIRD
It belongs to phylum Chordata, sub phylum Vertebrata and class Aves.

Features of Phylum chordata

1. They have notochord.
2. They have post anal tail.

Features of Class aves

1. Their forelimbs are modified into wings.
2. They have beak.
3. They are oviparous.

Adaptive Features
Birds are adaptive to aerial mode of life (to live in air). Following adaptations are found in them.

1. They have streamlined body to provide minimum resistance to air current and help in flying.
2. They have feathers on the body which provide insulation.
3. Birds have air cavities in their bones (pneumatic bones) which reduce the weight of the body and help in flying.
4. Their eyes are covered with nictitating membrane.
   

 

PRECAUTIONS
 Always first observe the specimen carefully and then draw a well labelled diagram in your notebook.

INTERACTIVE  SESSION

Examiner :
What do you mean by adaptations?
Examinee:
Adaptations are modifications in organism which help it to adjust in its habitat.

Examiner :
State any two aquatic adaptations.
Examinee:

1. Streamlined body,
2. Fins for locomotion.

Examiner :
Name the sensory organ in fish.
Examinee:
Lateral line.

Examiner :
What is the function of tail fins of fish?
Examinee:
Tail fin helps fish in changing direction.

Examiner :
Why is it that fish can see clearly in water but not humans? ‘
Examinee:
Eyes of fish are covered with nictitating membrane which does not allow water to enter eyes but it is not so in case of humans.

Examiner :
What is vermicomposting?
Examinee:
Earthworms add manure to soil in form of its castings, this is vermicomposting.

Examiner :
Explain the feature of skin in earthworm which makes it suitable for respiration.
Examinee:
In earthworm, the skin is rich in mucous glands which keeps the skin always moist.

Examiner :
Give the technical term for respiration through skin.
Examinee:
Cutaneous respiration.

Examiner :
Why is it said that amphibians have dual mode of life?
Examinee:
Amphibians can live both on land and in water and have organs to support this mode of life.

Examiner :
Differentiate between the type of body segments present in earthworm and cockroach.
Examinee:
In earthworm, body is segmented both internally and externally into metameres whereas cockroach is segmented only externally.

Examiner :
What are food habits of cockroach?
Examinee:
Cockroach is an omnivore, with cutting and biting mouth parts.

Examiner :
Why is earthworm known as farmer’s friend?
Examinee:

1. It ploughes the soil during burrowing by turning it upside down.
2. Castings of earthworm enrich the soil.

Examiner :
The exoskeleton of cockroach is made up of what?
Examinee:
Chitin.

Examiner:
Name a vestigial organ of humans which is of great importance for birds and fish.
Examinee:
Nictitating membrane on eye helps the eyes to remain open and aids in clear vision in air and water for birds and fish, respectively.

Examiner:
Apart from helping the birds to fly, what is other important feature of feathers on the body of birds?
Examinee:
Feathers also provide insulation.

Examiner:
What is the difference between the endoskeleton of bony fish and cartilaginous fish?
Examinee:
In case of bony fish, endoskeleton is made up of bones whereas in cartilaginous fish, it is made up of cartilage.

Examiner:
Give any two examples of bony fish.
Examinee:
Labeo, catla.

Examiner:
What is meant by “Birds are Oviparous”?
Examinee:
Oviparous means egg laying. Birds lay eggs.

Examiner:
What are upper and lower jaw in birds modified into?
Examinee:
In birds, upper and lower jaw is modified into horny beak, which lack teeth.

Examiner:
Fore limbs of birds are modified for flying, what are hind limbs modified for?
Examinee:
Hind limbs are modified for walking, perching or swimming.

NCERT LAB MANUAL QUESTIONS

Question 1:
Name th phyla to which earthworm,cockroach,bony fish and bird belong
Answer:

Question 2:
In which body segments of an earthworm is the clitellum found?
Answer:
The clitellum is found in 14-16 segment of an earthworm.

Question 3:
How does a cockroach adapt itself to a wide range of habitats?
Answer:
It undergoes hibernation to overcome unfavourable conditions.

Question 4:
Mention two adaptive characters of a bony fish besides the possession of a streamlined body and air bladder.
Answer:

1. Presence of scales on body.
2. Presence of nictitating membrane.

Question 5:
Feathers are an adaptive feature of birds. How are they helpful to them?
Answer:
Feathers provide insulation and help in flying.

PRACTICAL BASED QUESTIONS
Multiple Choice Questions/VSA

Question 1:
Earthworm is
(a) bisexual with self-fertilisation
(b) bisexual with cross-fertilisation
(c) unisexual with cross-fertilisation
(d) none of the above

Question 2:
Wings of a bird are modified form of
(a) hindlimbs
(b) forelimbs
(c) both (a) and (6)
(d) none of these

Question 3:
Abdomen of cockroach contains
(a) 10 segments
(6) 8 segments
(c) 20 segments
(d) 5 segments.

Question 4:
Characteristics of phylum Arthropoda are
(a) jointed legs
(b) chitinous plates
(c) compound eyes
(d) all of these

Question 5:
A feature absent in annelids is
(a) clitellum
(b) pseudocoelom
(c) metameric segmentation
(d) nephridia

Question 6:
Museum specimens are preserved in
(a) acetone
(b) water
(c) alcohol
(d) formalin

Question 7:
The first segment of earthworm is called
(a) peristomium
(b) prostomium
(c) clitellum
(d) none of these

Question 8:
Setae in earthworm is made up of
(a) lignin
(b) chitin
(c) cellulose
(d) pectin

Question 9:
The distinguishing characteristic that places animals into the phylum chordata is
(a) head with a pointed mouth
(ib) post anal tail
(c) presence of locomotory organs
(d) bilaterally symmetrical body

Question 10:
The structure present in male cockroach, but absent in female cockroach is [NCT 2008]
(a) Brood pouch
(b) Compound eye
(c) Anal style
(d) Anal cerci

Question 11:
Phylum possessing jointed appendages or legs is
(a) Arthropoda
(b) Porifera
(c) Mammalia
(d) Annelida

Question 12:
Which one of following belong to Class Pisces (True-fish)?
(a) Star fish
(b) Bony fish
(c) Silver fish
(d) Jelly fish

Question 13:
In birds, the upper and lower jaws are modified into
(a) beak with 2 teeth
(b) beak with 24 teeth
(c) beak with no teeth
(d) beak with 32 teeth

Question 14:

Two pictures of animals A and B are shown. Which characteristic features of their bodies are specific features of their respective phyla?
(a) Antennae of A and segments of B
(b) Three pairs of legs of A and scales of B
(c) Jointed appendages of A and metamerically segmented body of B
(d) Wings of A and slender body of B

Question 15:
A student observed the posterior part of a male cockroach in the laboratory. The following sketch

(a) anal cerci
(b) anal styles
(c) brood pouch
(d) antennae

 

ANSWER KEY
Multiple Choice Questions/VSA

1. (b)
2. (b)
3. (a)
4. (d)
5. (b)
6. (d)
7. (a)
8. (b)
9. (b)
10. (c)
11. (a)
12. (b)
13. (c)
14. (c)
15. (b)

Science Practical SkillsScience LabsMath Lab ManualMath Labs with ActivityMath Labs

CBSE Notes for Class 8 Computer in Action – Introduction to Microsoft Access 2013

Supermarkets use computers for storing lists of items, their prices, quantities available and quantities sold. Have you ever wondered how they store, manage and quickly retrieve data about so many products? Your elders can book railway tickets online. Where is the data about trains, their timings and seat allocation stored? How is it that when a person books a ticket, the reserved and the available seats are automatically shown?
If a person is asked to memorise such records, it will be an almost impossible task. To deal with such problems, data is stored and managed in a computer using databases. Thesedays, large amounts of data are required to be stored at all the places such as hotels, hospitals, supermarkets, railway stations, airports and offices. In this chapter, we will learn more about database and its related terminologies.

DATABASE AND DBMS
A database is an organised collection of related data so that it can be easily accessed, managed and updated. A Database Management System (DBMS) is a software program that enables us to create, modify and extract data from a database. DBMS can be based on different data models such as network, hierarchical and relational.
A DBMS based on the relational data model stores data in the form of tables and is called Relational Database Management System (RDBMS). In this chapter, we will learn about Microsoft Access, which is an example  of RDBMS. Some other examples of RDBMS Oracle and Microsoft SQL Server.

Advantages of a Database
Some of the advantages of storing data in a database as compared to files are listed here.

• Reduced Data Redundancy: The duplication of data is referred to as data redundancy. In contrast to non-database systems, which maintain multiple copies of the same data at different locations, DBMS stores data at a central place.
  The user is not required to maintain multiple copies of the same data. Thus, DBMS prevents data duplication.
• Reduced Update on Errors and Increased Consistency: When the data is stored at multiple locations, there are chances that modifications are not carried out at all the places. Multiple-mismatching copies of the same data are known as data inconsistency. DBMS ensures data consistency by storing data at one place and ensuring that there is no duplication of data.
• Improved Data Access to Users: A DBMS stores data at a centralised location and facilitates sharing of data among multiple users according to their requirements. For example, users from all over the country access the database for booking railway tickets.
• Improved Data Security: One of the most important advantages of DBMS is data security. In DBMS, user IDs can be created with various levels of security.
  Users have limited rights and permissions. Only authorised users can access the data. Some may have the privilege of changing the data while others can only view the data.
• Maintaining Standards: A DBMS ensures that the stored data follows the organisation’s own standards or national/ international standards. This ensures greater data integrity. This also aids in sharing data between different systems.

MICROSOFT ACCESS 2013
Microsoft Access is the. most popular RDBMS that comes as a part of the Microsoft Office suite. Access provides a graphical user interface for managing data. The databases created in Access 2013 are saved with the extension .accdb. Databases in Access 2013 are composed of four main objects—tables, queries, forms and reports. These objects allow us to enter, store, analyse and compile the data.
Let us learn more about these objects.

Objects in Access
The main objects that can be created in Access are as follows.

• Tables: They are used to store data in the form of rows (records) and columns (fields). A table is also referred to as a relation. We will learn more about tables in this chapter.
• Queries: Query let us find and work on the data resulting from one or multiple tables based on specified conditions.
• Forms: Forms provide a user interface that lets the users enter and change in the tables.
• Reports: If forms are for input, then reports are for output. Reports are used to display the data stored in database tables in a professional format for printing purposes. Figure 2.1 illustrates the relationship between the objects in Access.

Starting Access
To start Access, type Access 2013 in the Search box and click on Access 2013 option. The Microsoft Access start up window opens as shown in Figure 2.3.

CREATING A DATABASE
Follow the given steps to create a blank database in Access.

Step 1: Click on the Blank desktop database option.
Step 2: Type a name for the database in the File Name textbox.
Step 3: Click on the Browse button next to the File Name textbox and choose a location for the database.
Step 4: Click on the Create button. A blank database gets created. Access creates the database with an empty table named Table1. We will learn about working with tables after familiarising ourselves with the various components of the Access window shown in Figure 2.6.

COMPONENTS OF THE ACCESS WINDOW
Let us get familiar with the various components of the Access window (Fig. 2.7).

• Title Bar: The Title Bar appears on the top of the window and displays the name of the document on which you are currently working.
• Quick Access Toolbar  : The Quick Access Toolbar is present on the top left corner of the Access window. It has buttons for commands that are used frequently. By default, the following buttons are present on it.
  1. Save button: to save your work
  2. Undo button: to undo the previous action/rollback the action done
  3. Redo button: to redo the action that was undone.

  We can customise the Toolbar and add more buttons to it by clicking on the arrow next to the Redo button.

• Ribbon: The Ribbon contains multiple tabs, each with several groups of commands. Some tabs appear only when you work with certain objects such as Forms. These tabs are called contextual tabs.
• Navigation pane: The Navigation pane displays the names of all the objects in the database. The objects are grouped on the basis of their type, for example, Tables, Queries and so on. We can double-click on an object name in the Navigation Pane to open it. We can right-click on an object in the Navigation pane to perform various operations such as renaming and deleting an object.

• Navigation ButtonsThe Navigation buttons display the current record number in an object (in this example, Tables ► Employee) on which you are working. It has buttons to move back and forth between the various records.
• Object Tabs: The objects that you have opened in a database appear in a tabbed form. When you click on an object tab, the contents of that object are displayed in the Work Area. Click on the cross button (X) on the right end of the bar to close the object displayed in the current tab.

• Status Bar: The Status bar appears at the bottom of the window. The view buttonsare present on the right side of the Status bar. Usually two common buttons that are present are Datasheet view and Design view.

TABLES IN ACCESS
The tables are the building blocks of a database and are used for storing data. A table is made up of rows and columns. A database can have one or more tables.
Consider the following table that stores data about the employees in an organisation.Components of a Table
The important components of a table are:
1. Field or Attribute: The columns in a table are known as fields or A field is a named unit of information. A field stores one type of information about all the objects or items. Every field has a data type that determines the type of values that can be stored under it. For example, the above table has four fields or attributes. The field Ename stores text values and the field Esalary stores numeric values. Let us learn more about field data types.
Field Data types
The various data types available in Access are given in Table 2.2.
After setting the data type for a field, we can set its properties. Let us learn more about field properties. 
Field Properties
After we have decided on the data type of the table, we can set field properties. The Field properties let us have more control on data that can be entered in a particular field. Some of the field properties are listed in Table 2.3.

2. Record: The rows in a table are known as records A record stores complete information about an object or an item. For example, Table Employee (Table 2.1) has four records. Each record contains the complete information about an employee.
3. Primary key:A primary key is a field or combination of fields that uniquely identifies the records in a table. A primary key field cannot have repetitive values and cannot be left blank. In Table 2.1, Eno field can be made the primary key as every employee has a unique employee number.

Views of a Table
We can work on a table in two views—Design view and Datasheet view. Let us learn more about them.
Design view
In the Design view of the table, you can enter the field names, their data types and description. You can also set the field properties. The Design view window is divided into two panes—

1. Field Grid Pane: Field Grid
   Pane is used for entering field names and their data types. You can also give an optional description about each field in this pane.
2. Field Properties Pane: Field Properties Pane is used to set properties for the fields in the table.

Datasheet view
You use the Datasheet view to enter data in the table. When you double-click the table name in the database window, the table opens in the Datasheet view.

Switching between Views
You can quickly switch from one view to another by following either of the two options:

• Click on the View option in the Views group on the Home tab to change to the desired view.
  (OR) 
• Click on the Design viewor Datasheet viewbuttons at the bottom-right corner of the Status bar.

CREATING A TABLE
There are various ways to create tables in Access. In this chapter, you will be learning to create tables in the Design view. This view lets you design the structure of a table by specifying the field names, their data types and properties.
The steps to create a table in the Design view are:
Step 1: Click on the Create tab.
Step 2: Click on the Table Design option in the Tables group. A new table is inserted in the database and it opens in the Design View. A new tab named Design appears in the Ribbon.

Step 3: Enter the name for the field in the Field Name column.
Step 4: Enter the name for the field in the Field Name column.
Step 5: Use the Field Properties pane to set the properties for the fields.
Repeat steps 3 to 5 for all the fields in the table.

Step 6: Place the cursor on the field to be set as the Primary key and click onthe Primary Key option in the Tools group on the Design tab. You will notice that a key symbolappears next to the field.
Step 7: Click on the Save button on the Quick Access toolbar or Click on the File tab and choose the Save option. Save the table with an appropriate name.The table object is now added to the Navigation Pane.

ADDING RECORDS OR DATA IN A TABLE
After creating the table in the Design view, we can add records to it in the Datasheet view. The steps to enter records in a table are as follows:
Step 1: Open the table in the Datasheet view. (Double-click on the table in the Navigation Pane to open it in the Datasheet view)
Step 2: Click on the New option in Records group on the Home tab. A new record is inserted in the table.
Step 3: Type the values for various fields in a record. When you start entering data, a new record gets inserted automatically at the end of a table.

DELETING RECORDS IN A TABLE
Step 1: Open the table in the Datasheet view and follow these steps to delete a record. Select the record that needs to be deleted.
Step 2: Click on the Delete optionin the Records group on the Home tab.

SORTING RECORD IN A TABLE
Open the table in the Datasheet view and follow these steps to sort records.
Step 1: Select the field on the basis of which you want to sort the records.
Step 2: Click on the Ascending or Descending option in the Sort & filter group on the Home tab. The records get sorted.

MODIFYING THE TABLE DESIGN
We can make changes to the design of a table after it has been created. Let us learn how to insert and remove a field from a table.
Inserting a New Field
Open the table in Design view and follow the given steps to insert a field in the table.
Step 1: Place the cursor on the field before which you want to insert a new field.
Step 2: Click on the Insert Rows option in the Tools group on the Design tab. A new row is inserted. Enter the field name, data type and properties for the new field.

Deleting a Field
Open the table in Design view and follow the given steps to delete a field in the table.
Step 1: Click the field that has to be removed.
Step 2: Click on the Delete Rows option in the Tools group on the Design tab.
Step 3: Before permanently deleting a field, Access displays a warning box confirming whether you want to delete the field permanently. Click on the Yes button.
The field is deleted from the table.

In this article, we are providing Fog Extra Questions and Answers PDF Class 10 English First Flight CBSE, Extra Questions for Class 10 English was designed by subject expert teachers.

Fog Extra Questions and Answers Class 10 English First Flight

Extract Based Question [3 Marks each]

Read the following extract carefully and answer the questions that follow.
Question 1.
The fog comes on little cat feet. It sits looking over harbour and city on silent haunches and then moves on.
(a) What does the poet mean by little cat feet?
(b) What do haunches mean?
(c) How is the fog looking over the harbour and the city?
(d) “It sits looking….” what is the poetic device used here? [CBSE 2012]
Answer:
(a) Litde cat feet here represents the silent and careful steps of a cat. The way fog comes, resembles the steps of a cat.
(b) Haunches mean hips.
(c) The fog looks over the harbour and the city by sitting on its haunches like a cat.
(d) Metaphor is used here.

Short Answer Type Questions [2 Marks each]

Question 1.
How does the poet compare fog to a living being?
Answer:
The poet compares the fog to a cat. The silent steps of a cat and the way it sits on its haunches is very similar to the way fog comes and surrounds the city and looks over it.

Question 2.
What image does the poet give to the fog? What are the similarities between that image and fog?
Answer:
The poet looks at fog as a living creature and compares it to a cat. The fog moves like a cat on little cat feet and sits on haunches like a cat.

Long Answer (Value Based) Type Questions [8 Marks each]

Question 1.
Difficulties come but they are not to stay forever. They come and go. Comment referring to the poem ‘Fog’.
Answer:
Difficulties, when faced by people, tend to leave them hopeless and shattered. It takes a lot of courage to overcome any problem and to solve it. If we take a clue from the poem and compare difficulties to fog, we find that just like fog, difficulties also come and go.
One need not be hopeless and lose courage when problems come, one should rather think of it as fog, meaning that it has not come to stay but will always leave, like fog.

Question 2.
Nature has many wonders that people take for granted and never pay attention to. How is the poem ‘Fog’ different from this perspective?
Answer:
The poet has taken utmost pleasure in nature and natural phenomenon like fog. Fog is so special to the poet that he cared enough not only to write about it, but also thinks of its resemblance with other things in the world. In his close attention to fog, he found fog resembles a cat, in the way it moves and sits on its haunches. The fact that such a resemblance was found by the poet shows how connected he is to various things in nature. This poem serves as a motivation for people, who take nature for granted to find such interesting comparisons and similarities around them.

For More Resources

• NCERT Solutions for Class 10 English
• CBSE Class 10 English First Flight Chapter Wise Question Bank
• CBSE Class 10 English Footprints Without Feet Chapter Wise Question Bank
• CBSE Class 10 English First Flight Extra Questions
• CBSE Class 10 English Footprints Without Feet Extra Questions
• CBSE Revision Notes for Class 10 English First Flight
• CBSE Revision Notes for Class 10 English Footprints Without Feet

CBSE previous Year Solved  Papers  Class 12 Biology Outside Delhi 2013

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET-I

SECTION-A

Question.1. Name an organism where cell division in itself is mode of reproduction.
Answer : The division of cell in itself is a mode of reproduction found in amoeba and paramoecium.

Question.2. When does a human body elicit a anamnestic response ?
Answer : When our body attacked by pathogens for the second time the memory cells which were formed during the first attack produces a highly intensified secondary or anamnestic response.

Question.3. Name any two disease the ‘Himgiri’ variety of wheat is resistant to.
Answer : The ‘Himgiri’ variety of wheat is resistant to leaf and stripe rust and hill bunt disease.

Question.4. State the role oi transposons in silencing of mRNA in eukaryotic cells.
Answer : Role of transposons : Silencing of a gene is done in order to prevent translation of mRNA, where transposons act as a complementary RNA that is used to stop translation.

Question.5. Why are green algae not likely to be found in the deepest  strata of the ocean ?
Answer : Green algae not likely to be found in the deepest strata of the ocean because deep inside the sea presence of sufficient light for photosynthesis and brackish water are not available so green algae are not present at this level, instead algae inhabits littoral zone of water.

Question.6. State what does‘standing crop’of a trophic level represent.
Answer : ‘Standing crop’ of a trophic level represent certain j mass of a living material at a particular time.

Question.7. Why is the use of unleaded petrol recommended for motor vehicles equipped with catalytic converters ?
Answer : The use of unleaded petrol- recommended for 1 motor vehicles equipped with catalytic converters because lead in petrol inactivates the catalysts which convert harmful pollutants (CO, unburnt hydrocarbons, nitric oxide) to lesser ,harmful pollutants (C02, HzO, N2).

Question.8. Name the type of biodiversity represented by the following :
(i) 1000 varieties of mangoes in India
(ii) Variations in terms of potency and concentration of reserpine in Rauwolfia vomitoria growing in different
regions of Himalayas.
Answer : (i) Genetic Biodiversity : (ii) Genetic Biodiversity

SECTION – B

Question.9. In angiosperm, zygote is diploid while primary endosperm cell is triploid. Explain.
Answer : (i) Fertilisation of haploid egg cell by one haploid male gamete to form diploid zygote also called syngamy.

(iii) Fertilisation of two (diploid) polar nuclei by the other haploid male gamete to form triploid primary endosperm nucleus also called triple fusion.

Question.10. A cross between red flower bearing plant and a white flower bearing plant of Antirrhinum produced all plants having pink flowers. Work out a cross to explain how this is possible.
Answer:

Phenotypic ratio: 1:2:1
Genotypic ratio: 1:2:1
R(Red) factor is not completely dominant over r(white) factor is incomplete dominance.

Question.11. List the two main propositions of Oparin and Haldane.
Answer : The two main propositions of Oparin and Haldane are : (i) The first form of life could have come from preexisting non-living organic molecules (e.g., RNA, protein etc.) i.e., first abiogenesis and biogenesis later.
(ii) The first form of life was preceded by chemical evolution i. e., formation of organic molecules from inorganic molecules like CH4, NH3 etc.

Question.12. Write the events that take place when a vaccine for any disease is introduced into the human body.
Answer : Vaccines are a non-virulent form of pathogens, when administered into the body, the body start making antibodies against the antigens present in the vaccine. , The vaccine also generate memory – B and T – cells, that recognize the pathogen quickly on. subsequent second exposure and wipe out the invaders with a massive production of antibodies.
OR
Why is a person with cuts and bruises following an accident administered tetanus antitoxin ? Give reasons. Answer : Tetanus antitoxins neutralize and provide passive immunity to the bacterial toxin. The antitoxin contain antibody against pathogen it attach and inactivate pathogen.

Question.13. Name the bacterium responsible for the large holes seen in ‘Swiss Cheese’. What are these holes due to ?
Answer : Propionibacterium sharmanii is responsible for the large holes seen in Swiss Cheese. The large holes are due to production of a large amount of C02 by a bacterium.

Question.14. Name the source of the DNA polymerase used in PCR technique. Mention why it is used.
Answer : Thermus aquaticus because it is heat stable DNA polymerase. Polymerase chain reaction (PCR) is a method in which the desired gene is synthesised in vitro in following steps:
(a) Denaturation : The double-stranded DNA is denatured ’ by applying high temperature of 95°C for 15 seconds. Each separated single stranded now acts as template for DNA synthesis.
(b) Annealing : Two sets of primers are added which anneal to the 3′ end of each separated strand. Primers act as initiators of replication.
(c) Extension : DNA polymerase extends the primers by adding nucleotides complementary to the template provided. in the reaction. A thermostable DNA polymerase (Taq polymerase) is used in the reaction which can tolerate the high temperature of the reaction. All these steps are repeated many times to obtain several copies of desired DNA.

Question.15. Write any four ways used to introduce a desired DNA segment into bacterial cell in recombinant technology experiment.
Answer : The four ways used to introduce a desired DNA segment into bacterial cell in recombinant. technology experiment as following:
(i) Chemical Method : Poration by divalent cation such as calcium.
(ii) Micro injection : Recombinant DNA is directly injected into the nucleus of an animal cell.
(iii) Biolistic or gene gun: Plant cells are bombarded with hi gh velocity micro-particles of gold or tungsten coated with DNA.
(iv) Disarmed pathogen vectors : when allowed to infect the cell, transfer the recombinant DNA into the host.

Question.16. Why is proinsulin so called ? How is insulin different from it?
Answer : Proinsulin is a protein molecule and like a pro¬enzyme. It contain an extra strech of C peptide so it need to be processed to become fully mature and functional hormone like insulin, it is a mature hormone and is produced by the beta cells. Proinsulin is different from insulin because it serves as a precursor hormone to insulin.

Question.17. Where would you expect more species biodiversity in tropics or in polar region ? Give reasons in support of your answer.
Answer : High species or biodiversity lies in tropical areas because tropics are :
(i) Undistributed habitats since millions of year in comparison to temperate and polar region which face frequent glaciation. It favours speciation, as speciation is product of time.
(ii) Less seasonal variation than polar areas.
(iii) High availability of solar radiatipns than polar area, which harbours more plant species.

Question.18. “It is possible that a species may occupy more than one trophic level in the same ecosystem at the same time.” Explain with the help of one example.
Answer : Yes, as the trophic level of a species represents the functional role of organism in energy flow which is determined by the food intake. The availability of food is depends on what the organism want to eat, so have more than one trophic level at a time.
Ex. Sparrow is a Primary consumer, when eating seeds where as Secondary consumer, when eating insects.

SECTION-C

Question.19. Explain the steps in the formation of an ovum from an oogonium in humans.
Answer : The steps in the formation of an ovum from an oogonium in humans involves in oogenesis process. It can be divided into three stages :
(a) Multiplication phase (b) Growth phase
(c) Maturation phase
(a) Multiplication phase : In this stage primordial germ cells or ovum mother cells are repeatedly divided by mitosis to form large number of diploid oogonia. This process completes in embryo stage of female in most higher animals.
(b) Growth phase : In this process oogonia grow in size and form primary oocytes. The growth phase is the longest phase oogenesis (except humans). During growth phase size of egg increases many times.
(c) Maturation phase : Oogenesis takes place in the ovaries. In contrast to males the initial steps in egg production occur prior to birth. By the time the foetus is 25 weeks old, all the oogonia that she will ever produce, are already formed by mitosis. Hundreds of these diploid cells develop into primary oocytes, begin the first steps of the first meiotic division, proceed up to diakinesis, and then stop any further development. The oocytes grows much larger and complete the meiosis I, forming a large secondary oocyte and a small polar body that receives very little amount of cytoplasm but one full set of chromosomes.

OR
Suggest and explain any three Assisted Reproductive Technologies (ART) to an infertile couple.
Answer : Three assisted reproductive technologies (ART) to an infertile couple are :
(i) In vitro fertilisation (IVF-fertilisation outside the body in almost similar conditions as that in the body) followed by embryo transfer (ET) is one of such methods. In this method, popularly known as,, test tube baby programme, ova from the wife/donor (female) and sperms from the husband/donor (male) are collected and are induced to form zygote under simulated conditions-in the laboratory. The zygote or early embryos (with upto 8. blastomeres) could then be transferred into the fallopian tube (ZIFT- zygote intra fallopian transfer) and embryos with more than 8 blastomeres, into the uterus (IUT —intra uterine transfer), to complete its further development.
(ii) Embryos formed by in-vivo fertilization (fusion of gametes within the female) also could be used for such transfer to assist those females who cannot conceive. Transfer of an ovum collected from a donor into the fallopian tube (GIFT – gamete infra fallopian transfer) of another female who cannot produce one, but can provide suitable environment for fertilisation and further development is another method attempted.
(iii) Intra cytoplasmic sperm injection (ICSI) is another ‘ specialised procedure to form an embryo in the laboratory in which a sperm is directly injected into the ovum.
Infertility cases either due to inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates, could be corrected by Artificial Insemination (AI) technique. In this technique, the semen collected either from the husband or a healthy donor is artificially introduced either into the vagina or into the uterus (IUI-Intrauterine Insemination) of the female.

Question.20. Why a human females are rarely haemophilic ? Explain. How do haemophilic patients suffer ?
Answer : Haemophilia is sex linked recessive disease; it is transmitted from unaffected female carrier to a male child with haemophilia. Y has no allele for this. If male inherits Xh from the mother, he will be haemophilic (with the genotype XhY). If female inherits XhXh, one from the carrier tnother and one from her haemophilic father, then she can be haemophilic. Simple cut will results to increased bleeding time in haemophilie patients.

Question.21. In a maternity clinic, for some reasons the authorities are not able to hand over the two new-borns to their respective real parents. Name and describe the technique that you would suggest to sort out the matter.
Answer: DNA Fingerprinting or DNA test is the technique that suggested to describe the parental identification of these two new born babies in a maternity clinic.
The procedure of finger printing is as follows :
(i) Isolation of DNA,
(ii) Digestion of DNA by restriction endonucleases,
(iii) Separation of DNA fragments by electrophoresis,
(iv) Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon,
(v) Hybridisation using labelled VNTR probe, and
(vi) Detection of hybridised DNA fragments by autoradio-graphy. Half of the band of child will resemble to father and half to mother.

Question.22. Explain the increase in the number of melanic (dark winged) moths in the urban areas of post – industrialisation period in England.
Answer: In England, before industrial revolution the environ-ment was unpolluted. The white-winged moths were more and lichens on the barks of trees were pale. The white-winged moths could easily camouflage, while the dark winged were spotted out by the birds for food. Hence, they could not * survive. After industrial revolution the lichens became dark (due to soot-deposit). This favoured the dark-winged moths while the white-winged were picked by birds. The population of the former which was naturally selected increased.

Question.23. Describe how biogas is generated from activated sludge. List the components of biogas.
Answer : Biogas .can be produced by anaerobic digestion or fermentation of biodegradable materials. Bio wastes are collected and slurry of dung is mixed, a floating cover is placed over the slurry. The slurry having gas outlet is placed which keeps on rising as the gas is produced in the tank due to microbial activity of methanogens like Methanobacterium. Anaerobic fermentation of waste biomass can be visualized in three stages:
(i) The facultative anaerobic microbes degrade the complex polymers to simple monomers by enzymatic action. The Polymer like cellulose, hemicellulose, proteins and lipids get degraded into monomers but lignins and inorganic salts are left as residue because they do not degrade.
(ii) In second stage, monomers are converted in to organic acids by microbial action under partially aerobic conditions which are finally converted to acetic acid.
(iii) In third stage acetic acid is oxidized in to methane by the activity of anaerobic methanogenic bacteria. These bacteria are commonly found in the anaerobic sludge during sewage treatment. In this whole process digestion of cellulose takes place at very slow rate so that it is the “rate limiting factor in biogas production.”

Question.24. Name the pest that destroys the cotton balls. Explain the role of Bacillus thuringiensis in protecting the cotton crop against the pest to increase the yield.
Answer : Cotton ballworm is die pest that destroy the cotton balls. Bt toxin protein is produced by a soil bacterium called Bacillus thuringiensis in inactive prototoxin and crystalline form. The prototoxin form does not kill the bacteria. It becomes active and toxic when it is consumed by insects such as lepidopterans (armyworm), coleopterans (beetles) and dipterans (flies/ mosquitoes) due to presence of alkaline pH in the gut. The activated toxin (delta endotoxins) binds to the epithelial cells in the midgut of an insect and creates pores that cause lyses and swelling, eventually killing the insect.

Question.25. (a) Write the importance of measuring the size of a population in a habitat or an ecosystem.
(b) Explain with the help of an example how the percentage cover is a more meaningful measure of population size than more numbers.
Answer : (a) The importance of measuring the size of a population cover is more meaningful measure of population size than numbers. The size of the population tell us alot about its status in the habitat.
(b) Percentage cover is more meaningful measure of popula-tion size than more numbers because the’ relative abundance of a species is not only determined by number of individual w but by both i. e., the relative abundance in number and relative abundance in biomass.
Ex. In unit area the number of a grass species individuals or relative abundance in number is high but not in relative abundance of biomass. If the same area has one or two ficus benghalensis (Bargad) tree as it is very low in relative abundance in number while high in relative abundance of biomass.

Question.26. Differentiate between two different types of pyramids of biomass with the help of example of each.
Answer : Difference between two types of-pyramids of biomass:
Pyramid of Biomass is graphic representation of amount of biomass per unit area in the trophic levels with producers at the base & top carnivore at the apex. Biomass is maximum in producers. Only 10% of biomass is passed to next level. This is in accordance to the 10% law by Lindeman, (1942). Thus,.the biomass at higher trophic levels become smaller and smaller.
The pyramid of biomass is upright in grassland ecosystem.

For an aquatic system, the pyramid of biomass may be inverted or spindle-shaped. This is because the diatoms & other phytoplankton have a small standing crop, but they have a high annual productivity & high turn over rate. The reason or small standing crop is their short life span.

Question.27. (a) Describe the endosperm development in coconut.
(b) Why is tender coconut considered a healthy source of nutrition ?
(c) How are pea seeds different from castor seeds with respect to endosperm ?
Answer : Endosperm is ,a nutritive triploid tissue formed from mitotic divisions in primary endosperm nucleus (PEN). The cells of this tissue are filled with reserve food material & are used for the nutrition of the developing embryo.
In cococut, the type of endosperm formed is cellular endosperm, For this, the PEN divides many times and each division is followed by wall formation.

SECTION – D

Question.28 (a) Draw a L.S. of a pistil showing pollen tube entering the embryo-sac in an angiosperm and level any six parts other than stigma, style and ovary.
(b) Write the changes a fertilized ovule undergoes within the ovary in an angiosperm plant.
OR
(a) Draw a diagrammatic sectional view of a human seminiferous tubules and label sertoli cells, primary spermatocyte, spermatogonium and spermatozoa in it.
Answer: (a)

(b) Fertilised ovule forms seed which contains embryonic plant, reserve food and protective coat. A seed contains/ consists of two parts—seed contains/consists
(i) Seed Coat: It is the outer covering of the seed, functioning as a protective coat. It is formed from the integuments of the ovule, which hardens after fertilisation. The outer seed coat is called Testa & is formed from outer integument. The inner seed coat is called Tegumen is formed from inner integument. The micropyle remains as a small pore in the seed coat. It helps in the entry of oxygen & H20 into the seed during germination. Above the micropyle, hilum is visible as a depression, being remanant of the attachment point.

(ii) Embryo : The embryo is made up of embryonal axis Cotyledon/s and endosperm. Embryonal axis is the central axis, also called ‘Tigellum. One end of Tigellum bears Radicle (future root) & the other end bears plumule (future shoot). Cotyledon is present at the node of embryo axis. It is a fleshy structure, used for storage of food. Endosperm is the nutritive layer. Present inside the seed coat providing nutrition to growing embryo.

OR
(a) Spermatogenesis is under the control of endocrine hormones.

(i) Hypothalamus produces Gonadotropin Releasing Hormone (GnRH).
(ii) GnRH acts on anterior pituitary to produce gonadotropins,
ICSH (Interstitial Cell stimulating Hormone) & FSH. * ICSH acts on interstitial or leydig cells which produce testosterone.
(iii) FSH stimulates sertoli cells to develop ABP (Androgen Binding Protein) which help in concentrating testosteron in seminiferous tubules.
(iv) Excess testosterone inhibits LH/ICSH production by anterior pituitary & subsequendy GriRH production by Hypothalamus.
(v) Sertoli cells also produce a glycoprotein called Inhibition which suppresses FSH synthesis by anterior pituitary and GnRH by Hypothalamus.
This is called Negative Feedback control for release of testosterone.

Question.29. (a) Write the conclusion drawn by Griffith at the end of his experiment with streptococcus pneumonia.
(b) How did O. Avery, C. MacLeod and M. McCarty prove the DNA was the genetic material ?
Answer : (a) Griffith transformation experiment : Griffith performed his experiment in 1928 on Streptococcus pneumoneae bacteria which cause pneumonia in mice.
He used two strains of bacteria.
(i) Rough strain : Non capsulate non-virulent rough colonies on culture media.
(ii) Smooth strain : Capsulated virulent form smooth colonies on media.
Experiment: (i) Mice + Smooth strain bacteria —> Dead mice.
(ii) Mice + Rough strain bacteria —> Living mice.
(iii) Mice + Heat killed bacteria + Rough bacteria —> Dead Bacteria.
On the basis of third experiment he proposed that rough bacteria absorb some heat stable material from dead smooth bacteria and transformed into smooth bacteria which killed the mice.
(b) O Avery, C. Macleod and M. McCarty prove the DNA was the genetic material by purifying biochemicals (proteins, DNA, RNA etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells. They discovered that DNA alone from S bacteria caused R bacteria to become transformed. They also discovered that protein – digesting enzymes (proteases) and RNA – digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. They concluded that DNA is the hereditary material.
OR
(a) Explain the mechanism of sex-determination in humans.
(b) Differentiate between male heterogamety and female heterogamety with the help of an example of each.
Answer : (a) Sex determination in humans :
(i) The males have 22 pairs autosomes and A pair of XY-
chromosome.
(ii) The females have 22 pairs autosomes and a pair of XX- chrornosomes
(iii) In male, 50% of sperms carry X- chromosome and other 50% carry Y – chromosomes.
(iv) In females, all ova contains X-chromosomes.
(v) The sex of an individual is determined by the type of sperm fertilizing the ovum.
(vi) If the ovum is fertilized by Y-chromosome, the zygote (XY) develops info a male and if the ovum in fertilized by X-chromosome, Zygote (XX) develops into a female.

(b) There are two types of sex determining mechanisms, i.e., XO type and XY type. But in both cases males produce two different types of gametes,
(i) Either with or without X-chromosome or,
(ii) Some gametes with X-chromosome and some with Y-chromosome.
Such types of sex determination mechanism is designated to be an example of male heterogamety. In some other organisms, e.g., birds a different mechanism of sex determination is observed. In this case the total number of chromosome is same in both males and females. But two . different types of gametes in terms of the sex chromosomes, are produced by females, i.e., female heterogamety. The two different sex chromosomes of a female bird has been designated to be the Z and W chromosomes. In these organisms the females have one Z and one W chromosome, ‘ whereas males have a pair of Z-chromosomes besides the autosomes.

Question.30. A person in your colony has recently been diagnosed with AIDS. People/Residents in the colony want him to leave the colony for the fear spread of AIDS.
(a) Write your views on the situation, living reasons.
(b) List the possible preventive measures that you would suggest to the residents of your locality in a meeting organized by you so that they understand the situation.
(c) Write the symptoms and the causative agent of AIDS.
Answer: (a) AIDS is infectious but not contagious it does not spread by shaking hand and use of common utensils so there is no need of fear to live with AIDS patient.
(b) Making blood (from blood banks) safe from HIV, ensuring the use of only disposable needles and syringes in public and private hospitals and clinics, free distribution of condoms, controlling drug abuse, advocating safe sex and promoting regular check-ups for HIV in susceptible populations, are some such steps taken up.
(c) Symptoms and the causative agent of AIDS :
AIDS is caused by the Human Immuno Deficiency Virus (HIV), a member of a group of viruses called retrovirus, which have an envelope enclosing the RNA genome. These target the T lymphocytes, due to which the person starts suffering from infections that could have been otherwise overcome such as those due to bacteria especially Mycobacterium, viruses, fungi and even parasites like Toxoplasma. The patient becomes so immuno-deficient that he/she is unable to protect himself/herself against these infections.

SET-II

SECTION-A

Question.3. Write the basis on which an organism occupies a space in its community/natural surroundings.
Answer: It is based on the feeding relationships of that organism with other organisms and source of their nutrition or food.

SECTION-B

Question.8. Name an algae that reproduces asexually through zoospores. Why are these reproductive units so called.
Answer : Algae that reproduces asexually .through zoospores is “Chlamydomonas”. They are microscopic and motile, due to flagella so known as zoospores.

SECTION-C 

Question.12. “Stability of a community depends on its species richness”. Write how did David Tilman show this experimentally.
Answer : (i) David Tilman s long term ecosystem experiment – shows that plots with more species show less year-to-year variation in total biomass.
(ii) He also showed in his experiment that, increased diversity contributed to higher productivity.

Question.14. Name the haploid cells present in an unfertilized mature embryosac of a flowering plant. Vfite the total number of cells in it.
Answer : In an unfertilized mature embryo sac of a flowering
k plant 6 haploid cells are present, three antipodal cells, two synergids and one egg cell.

Question.15. In a typical monohybrid cross the F2-population ratio is written as 3 : 1 for phenotype but expressed as 1 : 2 : 1 for genotype. Explain with the help of an example.
Answer:

Genotype 1 : 2 : 1- One is homozygous dominant and 2 are heterozygous but dominant and one is homozygous recessive .

Question.16.Mention the contribution of S. L. Miller’s experiment on Origin of Life.
Answer : Miller carried out the experiment as a proof of oparin-Haldane hypothesis. The experiment was as follows :
(i) Miller sealed a minture of water vapour (H20), NH3, CH4 8H2 in a spark chamber, which was provided with electrodes to provide electrical discharges.
(ii) The electrical sparks were of 75,000 volts and the ratio of CH4 : NH3 : H2 as 2 : 1 : 2 and water vapour at 800°C.
(iii) The spark chamber was connected to another falsk with arrangement for boiling H20 for evaporation.
(iv) The other end of spark chamber was connected to a condenser for condensation and collection of equeous solution.
(v) A trap was connected with flask for boiling HzO.
(vi) The apparatus was a controlled one, without any energy source.
(vii) After 18 days, he analysed the products after cooling (hem and proposed that HGN was formed from methane and NH3 and reacted with other compounds of gas to born amino acids like Alanine, Glycine and As partic Acid.
He gare the conclusion that similar synthesis could have occured in the prinitire atmsopheric condition.

Question.20. (a) Explain “birth rate” in a population by taking a suitable example.
(b) Write the other two chlracteristics which only a population shows but an individual cannot.
Answer : (a) It is the average number of new individuals added per unit population per year due to births, hatchings and germinations. If in a pool there were 20 lotus plants last year and plants reproduced and gave rise to 8 new plants, the total population is now 28. Birth rate will be equal to 8/20=0.4 offspring per lotus per year (b) (i) Death rate : per capita death
(ii) Sex Ratio : An individual is either a male or female, but a population has a sex ratio, e.g. 60% of the population of females and 40% of males.

SECTION-D

Question.25. A burglar in a huff forgot to wipe off his blood – stains from the place of crime where he was involved in a theft. Name the technique which can help in identifying the burglar from the blood stains. Describe the technique.
Answer: The technique which help in identifying the burglar from the blood stain is DNA fingerprinting. The procedure of DNA fingerprinting is as follows :
(i) Isolation of DNA : Isolation of DNA from the blood stain .
(ii) Cutting, sizing, and sorting : Special enzymes called restriction enzymes are used to cut the DNA at specific places. The DNA pieces are sorted according to size by a sieving technique called electrophoresis. This technique is the biotechnology equivalent of screening sand through progressively finer mesh screens to determine particle sizes. The DNA fragments contain VNTRs (Variable Nun her of Tandem Repeats)
(iii) Transfer of DNA to nylon : The distribution of DNA pieces is transferred to a nylon sheet by placing the sheet on the gel and soaking them overnight.
(iv) Probing : Adding radioactive or coloured probes to the nylon sheet produces a pattern called the DNA fingerprint. Each probe typically sticks in only one or two specific places on the nylon sheet.
(v) Radiograph : The -final DNA fingerprint is built by using several probes (5-10 or more) simultaneously. These places are marked as dark bands when X-ray film is developed. This process is called auto radiography.

Question.28. (a) Write the specific features of the genetic code AUG.
(b) Genetic codes can be universal and degenerative. Write about them, giving one example of each.
(c) Explain aminoacylation of tRNA.
Answer: (a) Features of the genetic code AUG are as follows:
(i) AUG has dual function.
(ii) AUG codes for Methionine and acts a initiator codon.
(b) (i) Universal: From bacteria to human UUU would code for Phenylalanine. Some exceptions are there in mitochondrial codons and in some protozoans.
(ii) Degenerate : Some amino acids are coded by more than one codon. E.g. Phenylalanine is coded by UUU and UUC.
(c) Aminoacylation of tRNA : (i) In the presence of enzyme Amino-acyl-tRNA synthetase from DHU loop of tRNA, specific amino acid binds with ATP.

(ii) The AA-AMP-E complex formed in first step reacts with specific tRNA. Thus, amino acid is transferred to tRNA. The enzyme & AMP are then released.

OR
(a) Differentiate between dominance and co-dominance.
(b) Explain co-dominance taking an example of human blood groups in population.
Answer:


(b) In humans, the ABO blood groups are controlled by a gene called gene ‘I’. It has three alleles IA, IBand i Hence, referred to as multiple allelism. A person possesses any two of the three alleles. IA and IB dominate over i. But with each other, IAand IB are co-dominant.

SET-III

SECTION-A

Question.1. What is detritus food chain made up of ? How do they meet their energy and nutritional requirements ?
Answer : Detritus food chain made up of decomposers which are heterotrophic organisms, mainly fungi and bacteria. They get energy and nutrients by decomposing dead organic matter or detritus.

Question.3. Name the phenomenon and one bird where the female gamete directly develops into a new organism.
Answer : Phenomenon : Parthenogenesis Bird : Turkey

SECTION-B

Question.10. What is meant by “alien species” invasion ? Name one plant one animal alien species that are a threat to our Indian species.
Answer : An alien species whose introduction does or is likely to pose threat to the survival of many native species and cause their extinction.
Plants ; Parthenium, Lantana, Eicchomia Animal: African catfish, Clarias gariepinus

SECTION-C

Question.14. Work out a cross to find the genotype of a tall pea plant. Name the type of cross.
Answer : (i) Test cross is used to find out the genotype of any trait. In this cross F hybrid of pure tall plant and a pure dwarf plant is crossed with a dwarf plant, e.g., Tt x tt Plants of F, Tt (Tall) x tt (Recessive)

The progeny consists of tall and dwarf plants in the ratio of 1:1
(ii) If the dominant plant are homozygous i.e. TT, then the progeny will have all tall plants :

Question.15. Write the Oparin and Haldanes hypothesis about the origin of life on earth. How does meteorite analysis favour this hypothesis.
Answer : Oparin-Haldane Hypothesis : Alexander I Oparin (1894-1980) a Russian biochemist &J.B.S. Haldane (1892-,1964), a British Scientist, put forward the concept that the first living organism evolved from non-living material. According to Oparin & Haldane (1929). Spontaneous generation of early molecules might have taken place through a series of chemical reactions from the earth’s primordial soup in a reducing atmosphere. The compounds of soup could be expected to react with one another producing a variety of chemical substances like amino acids, sugars, N2 bases. These precursor molecules then combined resulting in the appearance of proteins, polysacharides & nucleic acids. Energy required for these reactions was provided by UV radiations, cosmic rays, electric discharges etc.

Question.19 (i) Explain DNA polymorphism as the basis of genetic mapping of human genome.
(ii) State the role of VNTR in DNA fingerprinting.
Answer : (i) Polymorphism (variation at genetic level) are tlie result of mutation. Variation in allele sequence is DNA
polymorphism i.e., if more than one variant (allele) at a locus occurs in human population with a frequency greater than 0.01. In other term we can say that if an inheritable mutation is observed in a population at high frequency, it is referred as DNA polymorphism. There is a variety of different types of polymorphism ranging from single nucleotide change to very large scale changes. For evolution and specification, such polymorphism are important. DNA polymorphism is also used in genetic and physical maps on the human genome. In genetic mapping, information on polymorphism of restriction endonuclease recognition sites are used.
(ii) Role of VNTR in DNA Finger printing : Radiolabelled VNTR (Variable Number of Tandem Repeat) is used as a probe in DNA finger printing. A tandem repeat is a short sequence of DNA that is repeated in a head-to-tail fashion at a specific chromosomal locus. Tandem repeats are interspersed throughout the human genome. Some sequences are found at only one site – a single locus – in the human genome. For many tandem repeats, the number of repeated units vary between individuals. Such loci are termed VNTRs.

SECTION-D

Question.30. How does the process of natural selection affect Hardy- Weinberg equilibrium ? Explain. List other four factors that disturb the equilibrium.
Answer : Population or Mendelian population is a group of individuals present in ageographical area which share acommon pool. Gene frequency is the percentage of an allele in relation to the total alleles of a gene in an interbreeding population. All the genes & their alleles together constitute gene pool. Normally the alleles tend to maintain an equilibrium with reference to one another over the generations. It is referred as genetic equilibrium. Such a population is referred as non¬evolving populaiton. G.H. Harady & withelm weinberg proposed a principle independently in 1908, about the genetic structure of a non-evolving population. It is known as Hardy-weinberg equilibrium. The allelic frequencies in non-evolving population are stable and remain constant from generation to generation.
The Hardy-Weinberg Equilibrium can change by process of natural selection. Natural or survival of the fittest is a major factor that adds variation in the population, change the gene frequies of the gene pool leading to evolution & formation of a new distinctive gene pool. It operates through differential or non-random reproduction. If same type of selection continues for a few generation; the gene popl will undergo change the alleles having the advanage of surviving. Thus it leads to change of gene pool.
OR
(a) Explain Mendel’s Law of independent assortment by taking a suitable example.
(b) How did Morgan show the deviation in inheritance pattern in Drosophila with respect to this law.
Answer : (a) Dihybrid cross is based on Law of Independent Assortment. This law states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.
The image shows a true-breeding plant with the dominant traits of green pod colour (GG) and yellow seed colour (YY) being cross-pollinated with a true-breeding plant with yellow pod colour (gg) and green seeds (yy). The resulting offspring are all heterozygous for green pod colour and yellow seeds (GgYy). If the offspring are allowed to self pollinate, a 9 : 3 : 3 : 1 ratio will be seen in the next generation. About 9 plants will have green pods and yellow seeds, 3 will have green pods and green seeds, 3 will have yellow pods and yellow seeds and 1 will have a yellow pod and green seeds.

(b) Morgan and his group observed that when the two genes in a dihybrid cross were located on the same chromosome, the proportion of parental gene combinations in the progeny were much higher than the non-parental or new combinations -(recombinants) of genes.

CBSE Class 12 Case Studies In Business Studies – Financial Markets

FINANCIAL MARKET
Financial Market: Definition
A financial market is a market for the creation (new issue of securities) and exchange (sale of existing securities) of financial assets.

Financial Markets: Purpose
Financial market serves as an intermediary between the surplus sector (households which have savings) and deficit sector (business firms which needs funds).

Functions of Financial Market

• It performs the allocative function
  by mobilisation of savings and channelising them into the most productive avenues.
• It helps to determine the price for the financial asset in a particular financial market through the market forces of demand and supply.
• It provides liquidity to the financial assets by providing ready markets wherein the securities can be easily converted into cash or vice versa.
• It provides a common platform for exchange of securities thereby reducing the cost of transactions by saving time, effort and money spent by the buyers and sellers in locating each other.

Financial Intermediation: Definition
Financial intermediation refers to the process through which allocation of funds is done by the savers in the household sectors through two main mechanisms; banks and financial markets.
Types / Segments of Financial Market

• Capital Market
• Money Market

CAPITAL MARKET
Capital Market: Definition
Capital Market refers to facilities and institutional arrangements through which long term funds, both debt and equity, are raised and invested.
Capital market is a segment of financial market.

Features of Capital Market

• It is a market for long term funds.
• The main participants in capital market are banks, financial institutions, corporate bodies, foreign investors and retail investors.
• Since the cost of securities may be low, investment can be made in the capital market with less capital.
• The securities in capital market enjoy good liquidity.
• The instruments in capital market carry high risk as the expected return on them is high.

Main Instruments in Capital Market

• Equity Shares
• Preference Shares
• Debentures
• Bonds

Types / Segments of Capital Market

• Primary Market
• Secondary Market

Constituents of the Capital Market

• Development Banks
• Commercial Banks
• Stock Exchanges

PRIMARY MARKET
Primary Market: Definition
Primary market is also known as new issue market as the securities are issued for the first time by the companies through this market.
Primary Market is a segment of capital market.

Features of Primary Market

• It is the new issue market.
• Only buying of securities takes place.
• Prices of the securities are determined by the company.
• It involves dealings between the company and investors.
• There is no fixed location of primary market.

Instruments of Primary Market

• Shares
• Debentures
• Bonds

Methods of Floatation in Primary Market

• Offer through Prospectus (The company approaches the members of the general public directly by issuing a prospectus)
• Offer For Sale (The company approaches members of the general public indirectly through intermediaries like issuing houses, stock brokers etc.)
• Private Placement (The company can raise finance by allotting securities to selective individuals and institutions only)
• e-IPOs (The investors may subscribe to the securities of a company online)
• Rights Issue (It is a pre-emptive right given only to the existing shareholders to subscribe to the securities of the company as per its terms and conditions)

SECONDARY MARKET
Secondary Market: Definition
It is a market for old or existing securities
It is a segment of capital market.

Features of Secondary Market / Stock Exchanges

• It is the market for old/existing securities.
• Both buying and selling of securities takes place.
• Prices of the securities are determined by the forces of demand and supply.
• It involves dealings between two investors.
• Stock exchanges exist at fixed location.

MONEY MARKET
The money market is a market for short term debt instruments whose period of maturity is upto one year.
It is a segment of financial market.

Features of Money Market

• It is a market for short term funds.
• The main participants are institutional investors.
• Since the cost of securities may be high, investment in the money market may require huge capital outlay.
• The money market enjoys high liquidity as The Discount Finance House of India works as a compulsory market maker for it.
• The instruments in money market carry low risk as the expected return is low on

INSTRUMENTS IN MONEY MARKET

Call Money	Call money is a short term money market instrument through which one bank may borrow money from another bank to maintain its cash reserve ratio as per the guidelines of RBI. It’s maturity period may range from a single day to a fortnight. The rate at which the interest is paid on call money is called call rate.
Commercial Paper	Commercial Papers are the short term money market instrument issued by large and credit worthy companies. The instrument is in the form of an unsecured promissory note and is freely transferable by endorsement. Its maturity period may range from a fortnight to a year. It is sold at discount and redeemed at par. It is used for bridge financing.
Certificate of Deposit	Certificate of deposit is a short term money market instrument issued by commercial banks and development financial institutions. It is an unsecured and negotiable instrument in bearer form. It may be issued to individuals, corporations and companies when banks need cash to meet credit needs
Treasury Bills	Treasury bills are short term money market instrument which are issued by Reserve bank of India on behalf of the Government of India. They are issued in the form of promissory notes and are very safe instruments. They are sold at discount and redeemed at par. They are also known as zero-coupon-discount bonds. The minimum value of their purchase is ? 25,000 and in multiples thereof.
Commercial Bill	Commercial bill is a short term money market instrument. A business firm may draw a bill of exchange in favour of another in lieu of credit purchases. On acceptance by the drawee (buyer) it becomes a trade bill. When the trade bill is accepted by a commercial-bank for discounting it is called commercial bill. A bill of exchange is a freely negotiable instrument. It is usually drawn for a period of 3 months.

 STOCK EXCHANGE
Stock Exchange: Definition
According To Securities Contracts (Regulation) Act 1956, “Stock Exchange means any body of individuals, whether incorporated or not, constituted for the purpose of assisting, regulating or controlling the business of buying and selling or dealings in securities.”

Functions of Stock Exchange

• Ensures liquidity and marketability of existing securities by a providing a ready and continuous market for the sale and purchase of securities.
• Helps in determining the prices of the securities through the forces of demand and supply.
• It promotes the habit of saving and investment among the general public.
• It provides a legal framework for fair and safe dealings.
• It helps the companies in raising finance thus facilitating capital formation and economic growth.
• It provides scope for healthy speculation in a controlled and restricted way.

Dematerialisation: Definition
Dematerialisation refers to the process of holding securities in electronic form.

Depository: Definition
Depository is the organisation with which an investor has to open a D-Mat account to hold securities in electronic form.
In India there are two depositories:

• National Secuities Depository Limited (NSDL)
• Central Depository Services Limited (CDSL)

The depository participant serves as a link between the investor and the depository i.e. either NSDL or CDSL.

Screen Based Trading: Definition
Screen-based trading refers to the process of buying or selling securities on-line.

Advantages of Screen-based Trading

• As the investors get an access to the stock market during real time, there is complete transparency and in the dealings.
• It provides a common platform for exchange of securities thereby increasing the efficient transactions by saving time, effort and money.
• This virtual market has a very wide reach hence it increases its liquidity.

STEPS IN THE TRADING AND SETTLEMENT PROCEDURE

1. The investor approaches a registered broker or sub-broker for trading.
   The investor has to sign a broker-client agreement and a client registration form.
   • The investor has to furnish certain details and information about himself including PAN number which is mandatory, date of birth, bank account details, income details etc.
   • A broker acts as an intermediary between the buyers and sellers.
   • After the completion of the above formalities, the broker opens a trading account in the name of the investor.
2. The investor has to open a demat account with a depository participant and a bank account for trading transactions in cash.
3. The investor then places an order to buy or sell shares with the broker:
   • By giving clear instructions about the desired quantity and price.
   • The broker will then make the investor aware about the feasibility of the order.
   • The broker will issue an order confirmation slip to the investor.
4. The broker will then execute the order through screen based trading by considering the best available deal.
5. The broker will issue a trade confirmation slip to the investor.
6. Within 24 hours after the deal is executed the broker issues a contract note.
   • A contract note contains details about the deal i.e. the number of securities bought/sold, price, date and time of transaction etc.
     The contract note includes a unique order code generated by the stock exchange for that transaction.
   • A contract note is a legal which may be used to settle the claims between the investor and the broker.
7. Since the settlement cycle is T +2 therefore, within two days of receiving the Contract Note, the investor has to pay cash or deliver shares sold as the case may be. The broker can then forward it to the exchange. This is called pay-in-day.
8. On the T+2 day, cash will be paid or shares will be delivered as the case may be by the exchange to the other broker. This is called pay-out-day. Then, in case of sale of shares, the broker has to make the payment to the investor within 24 hours.
9. However, in the case of purchase of securities, the amount will be transferred electronically to the investor’s demat account.

SECURITIES AND EXCHANGE BOARD OF INDIA (SEBI)
The Securities and Exchange Board of India (SEBI) is the regulator for the securities market in India.

• It was established in the year 1988 by the Government of India. It was to function under the overall administrative control of Ministry of Finance of the Government of India.
• It was given statutory powers on 30th January 1992 through an ordinance.
• The ordinance was later on replaced by an Act of Parliament known as the SEBI Act, 1992.

The Organisation Structure of SEBI

• The various activities undertaken by SEBI are now divided into five operational departments.
• Each department is headed by an executive director.
• The head office of SEBI is located at Mumbai.
• Besides, regional offices have been set up in Kolkata, Chennai and Delhi to attend to consumers complaints and maintain liaison with the issuers, intermediaries and stock exchanges in the concerned regions.
• SEBI has also formed Primary Market Advisory Committee and Secondary Market Advisory Committee to assist in the process of SEBI’S policies formation.
• These two committees consist of the market players, the investor’s association recognised by SEBI and the eminent persons in the capital market.

Objectives of SEBI

• To prevent trading malpractice in thesecurities markets.
• To protect the rights and interest of investors, and to guide and educate them.
• To regulate and develop a code of conduct and fair practices by intermediaries like brokers, merchant bankers, etc. with a view to making them competitive and professional.
• To regulate stock exchanges and the securities market to promote their orderly functioning.

Purpose and Role of SEBI
SEBI has to be responsive to the needs of three groups, which constitue the market namely:

• The issuers of securities so as to provide them a platform for raising capital in an easy, effective and efficient manner.
• The investors so as to protect their interests in securities by keeping them abreast about the developments through true and appropriate information.
• The market intermediaries in order to provide them a framework so as to enable them to perform their functions effectively and efficiently.

Objectives of Advisory Committees Formed by SEBI

• To advise SEBI on matters relating to regulation of intermediaries for ensuring investor protection in the primary market.
• To advise SEBI on issues related to development of primary market in India.
• To advise SEBi on matters required to be taken by for changes in legal framework to introduce simplification and transparency in the primary market.
• To advise SEBI on disclosure requirements for companies.
• To advise the board in matters relating to the development and regulation of secondary market in the country.

FUNCTIONS OF SEBI
Protective Functions of SEBI

• SEBI prohibits fraudulent and unfair trade practices in the securities market
• Promotion of fair practices and code of conduct in securities market
• Undertaking steps for investor protection
• Controlling insider trading and imposing penalties for such malpractices.

The Preamble of the Securities and Exchange Board of India describes the basic functions of the Securities and Exchange Board of India as “…to protect the interests of investors in securities and to promote the development of, and to regulate the securities market and for matters connected therewith or incidental thereto”

Developmental Functions of SEBI

• Ensuring training of intermediaries of securities market
• Conducting research and publishing information useful to all market participants
• Facilitating flexibility in the working of capital markets.

Regulatory Functions of SEBI

• Registration and regulation of of brokers, sub-brokers and other players in the financial market.
• Registration of collective investment schemes and Mutual Funds.
• Conducting enquiries and audits of stock exchanges & intermediaries.
• Regulation portfolio exchanges, underwriters, merchant bankers and the dealings in the stock exchanges.
• Regulation of take over bids by the companies

MIND MAP

LATEST CBSE QUESTIONS

Question 1.
Meca Ltd. a reputed automobile manufacturer needs Rupees ten crores as additional capital to expand its business. Atul Jalan, the CEO of the company wanted to raise funds through equity. On the other hand the Finance Manager, Nimi Sahdev said that the public issue may be expensive on account of various mandatory and non-mandatory expenses. Therefore, it was decided to allot the securities to institutional investors.
Name the method through which the company decided to raise additional capital. (CBSE, Delhi 2017)
Answer:
Private placement is method through which the company decided to raise additional capital.

Question 2.
These days, the development of a country is also judged by its system of transferring finance from the sector where it is in surplus to the sector where it is needed most. To give strength to the economy, SEBI is undertaking measures to develop the capital market. In addition to this there is another market in which unsecured and short-term debt instruments are actively traded everyday. These markets together help the savers and investors in directing the available funds into their most productive investment opportunity.

1. Name the function being performed by the market in the above case.
2. Also, explain briefly three other functions performed by this market. (CBSE, Delhi 2017)

Answer:

1. Mobilisation of funds is the function being performed by the financial market in the above case. It performs the allocative function by mobilisation of savings and channelising them into the most productive avenues.
2. The other three functions being performed by the financial market are outlined below:
   • It helps to determine the price for the financial asset in a particular financial market through the market forces of demand and supply.
   • It provides liquidity to the financial assets by providing ready markets wherein the securities can be easily converted into cash or vice versa.
   • It provides a common platform for exchange of securities thereby reducing the cost of transactions by saving time, effort and money spent by the buyers and sellers in locating each other.

Question 3.
These days, the development of a country is also judged by its system of transferring finance from the sector where it is in surplus to the sector where it is needed the most. To give strength to the economy, SEBI is undertaking measures to develop the capital market. In addition to this, there is another market in which unsecured and short-term debt instruments are actively traded every day. These markets together help the savers and investors in directing the available funds into their most productive investment opportunity.

1. Name the function being performed by the market in the above case.
2. Name the market segment other than the capital market segment in which unsecured and short-term debt instrument are traded. Also, give any three points of difference between the two. (CBSE, OD 2017)

Answer:

1. Mobilisation of funds is the function being performed by the financial market in the above case. It performs the allocative function by mobilisation of savings and channelising them into the most productive avenues.
2. Money market is the other segment of market.
   The difference between capital market and money market is given below:

   S.No	Basis	Capital Market	Money Market
   1.	Participants	The main participants in capital market are banks, financial institutions , corporate bodies, foreign investors and retail investors.	The main participants are institutional investors.
   2.	Investment outlay	Since, the cost of securities may be low, investment can be made in the capital market can be with less capital.	Since, the cost of securities may be high, investment in the money market requires huge capital outlay.
   3.	Liquidity	The securities in capital market enjoy good liquidity.	The securities in money market enjoy high liquidity as The Discount Finance House of India works as a compulsory market maker.

Question 4.
ABC Ltd. issued prospectus for the subscription of its shares for Rs. 500 crores in 2008. The issue was oversubscribed by 20 times. The company issued shares to all the applicants on pro-rata basis. Later SEBI inspected the prospectus and found some misleading statement about the management of the company in it. SEBI imposed a penalty of Rs. 5 crores and banned its three executive directors for dealing in securities market for three years. Identify the function and its type performed by SEBI in the above case. (CBSE, Sample Paper, 2017)
Answer:
Protective function has been performed by SEBI in the above case. And the type of Protective function is Prohibition of fraudulent and unfair trade practices.

Question 5.
“Money market is essentially a market for short-term funds’. In the light of this statement state any three features of money market.
(CBSE, Sample Paper, 2017)
Answer:
The three features of money market are described below:

• It is a market for short term funds whose maturity period is upto one year.
• The main participants are institutional investors.
• Since the cost of securities may be high, investment in the money market requires huge capital outlay.

Question 6.
“Unicon Securities Pvt. Ltd” was established to deal in securities. It was registered as a stock broker with National Stock Exchange (NSE) and Bombay Stock Exchange (BSE) to trade in securities listed at these exchanges. It is also a depository participant with CDSL and NSDL. In the first three years, it developed its business successfully. After that the composition of Board of Directors changed. Some customers complained to the customer care centre of the company that shares purchased by them and for which the payment has been duly made, were not transferred to their D’mat Accounts by “Unicon securities Pvt. Ltd” . The executive of customer care centre promised the aggrieved customers that their shares will be transferred to their respective D’mat Accounts very soon. But the company delayed the matter and didn’t transfer the shares of the customers to their D’mat Accounts. This eroded investors confidence and multiplied, their grievances.

1. Identify the step of trading procedure in a stock exchange which has not been followed by “Unicon Securities Pvt. Ltd” .
2. Name the Apex statutory body of capital market to whom customer can complain to redress their grievances.
3. Write two values not followed by Unicon Security Pvt. Ltd.(CBSE, Sample Paper, 2017)

Answer:

1. The step of trading procedure in a stock exchange which has not been followed by “Unicon Securities Pvt. Ltd” is settlement i.e. the delivery of shares through the D’mat Account of the broker to D’mat account of the investors.
2. The Securities and Exchange Board of India (SEBI) is the Apex statutory body of capital market to whom customer can complain to redress their grievances.
3. The two values not followed by Unicon Security Pvt. Ltd are’:
   • Truthfulness
   • Fair practices

Question 7.
Reshu’s father has gifted her the shares of a large cement company with which he had been working. The securities were in physical form. She already has a bank account and does not possess any other forms of securities.
She wished to sell the shares and approached a registered broker for the purpose. Mention one mandatory detail which she will have to provide with the broker. (CBSE, Sample Paper, 2016)
Answer:
Reshu will have to give her Permanent Account Number (PAN) to the broker as it is mandatory as per law.

Question 8.
Squib Ltd. is a large creditworthy company operating in the Kashmir Valley. It is an export- oriented unit, dealing in exclusive embroidered shawls. The floods in the Valley have created many problems for the company. Many craftsmen and workers have been dislocated and raw material has been destroyed. The firm is therefore, unable to get an uninterrupted supply of raw materials and the duration of the production cycle has also increased. To add to the problems of the organisation, the suppliers of raw materials who were earlier selling on credit are asking the company for advance payment or cash payment on delivery. The company is facing a liquidity crisis. The CEO of the company feels that taking a bank loan is the only option with the company to meet its short-term shortage of cash.
As a finance manager of the company, name and explain the alternative to bank borrowings that the company can use to resolve the crisis. (CBSE, Sample Paper 2016)
Answer:
Commercial Papers may be used by Squib Ltd. as it is a popular short term instrument which is issued by large and credit worthy companies. The instrument is an unsecured promissory note and is freely transferable by endorsement. Its maturity period may range from a fortnight to a year. It is sold at discount and redeemed at par.

Question 9.
Mr. Sanjay Nehra was the Chairman of Taran Bank. The bank was earning good profits. Shareholders were happy as the bank was paying regular dividends. The market price of their shares was also steadily rising. The bank was about to announce taking over the ‘Vena Bank.’ Mr. Sanjay Nehra knew that the share price of Taran Bank would rise on this announcement. Being a part of the bank, he was not allowed to buy shares of the bank. He called one of his rich friends Sudhir and asked him to invest Rs.5 crores in shares of his bank promising him the capital gains.
As expected, the share prices went up by 40% and the market price of Sudhir’s shares was now ? 7 crores. He earned a profit of Rs. 2 crores. He gave Rs. 1 crore to Mr. Sanjay Nehra and kept Rs. 1 crore with himself. On regular inspection and by conducting enquiries of the brokers involved, the Securities and Exchange Board of India (SEBI) was able to detect this irregularity. The SEBI imposed a heavy penalty on Mr. Sanjay Nehra.
By quoting the lines from the above paragraph, identify and state any two functions that were performed by SEBI in the above case
Answer:
The two functions performed by SEBI in the given case are stated below:

• Regulatory function is being performed by SEBI: “On regular inspection and by conducting inquires of the brokers involved.”
• Protective function is performed by SEBI: “The SEBI imposed a heavy penalty on Mr. Sanjay Nehra.”

Question 10.
Mr. Vikas Mehra was the Chairman of IBM Bank. The bank was earning good profits. Shareholders were happy as the bank was paying regular dividends. The market price of their shares was also steadily rising. The bank was about to announce the taking over of ‘UK Bank’. Mr. Vikas Mehra knew that the share price of IBM Bank, would rise on this announcement. Being a part of the bank, he was not allowed to buy shares of the bank. He
called one of his rich friends Mukand and asked him to invest Rs. 4 crores in the shares of his bank promising him the capital gains.
As expected, after the announcement, the share prices went up by 50% and the market price of Mukand’s shares was now Rs. 6 crores. Mukand earned a profit of Rs. 2 crores. He gave Rs. 1 crore to Vikas Mehra and kept Rs. 1 crore with him. On regular inspection and by conducting enquiries of the brokers involved, the Securities and Exchange Board of India (SEBI) was able to detect this irregularity. SEBI imposed a heavy penalty on Vikas Mehra.
Quoting lines from the above paragraph, identify and state any two functions performed by the SEBI in the above case. (CBSE, OD 2016)
Answer:
The two functions performed by SEBI in the given case are stated below:

• Regulatory function is being performed by SEBI: “on regular inspection and conducting inquires of the brokers involved.”
• Protective function is performed by SEBI: “The SEBI imposed heavy penalty on Mr.Vikas Mehra.”

Question 11.
Supriya’s grandmother who, was unwell, called her and gave her a gift packet. Supriya opened the packet and saw many crumpled share certificates inside. Her grandmother told her that they had been left behind by her late grandfather.
As no trading is now done in physical form, Supriya wants to know the process by adopting which she is in a position to deal with these certificates.

1. Identify and state the process.
2. Also, give two reasons to Supriya why dealing with shares in physical form has been stopped. (CBSE, Sample Paper 2015)

Answer:

1. Dematerialisation refers to the process of holding securities in electronic form.
2. The two problems of holding shares in physical form are as follows:
   • When the shares certificates are held in physical form, there is danger of loss or theft.
   • There is risk of forgery, as the buyer may be delivered fake certificates .

Question 12.
Mission Coach Ltd. is a large creditworthy company that manufactures coaches for the Indian Railways. It now wants to export these coaches to other countries and decides to invest in new hi-tech machines. Since the investment is large, it requires long-term finance. It decides to raise funds by issuing equity shares. The issue of equity shares involves huge floatation cost. To meet the expenses of floatation cost, the company decides to tap the money market.

1. Name and explain the money-market instrument the company can use for the above purpose.
2. What is the duration for which the company can get funds through this instrument?
3. State any other purpose for which this instrument can be used. (CBSE, OD 2015)

Answer:

1. Commercial Papers can be used for Bridge financing by Mission Coach Ltd. as it is issued by large and credit worthy companies. The instrument is in the form of an unsecured promissory note and is freely transferable by endorsement. It is sold at discount and redeemed at par.
2. Its maturity period may range from a fortnight to a year.
3. It is also used to meet the short term seasonal and working capital requirements of a business enterprise.

Question 13.
Ganesh Steel Ltd. is a large and creditworthy company that manufactures steel for the Indian market. It now wants to cater to the Asian market and decides to invest in new hi-tech machines. Since the investment is large, it requires long-term finance. It decides to raise funds by issuing equity shares. The issue of equity shares involves huge floatation cost. To meet the expenses of floatation cost, the company decides to tap the money market.

1. Name and explain the money-market instrument the company can use for the above purpose.
2. What is the duration for which the company can get funds through this instrument?
3. State any other purpose for which this instrument can be used. (CBSE, Delhi 2015)

Answer:

1. Commercial Papers can be used for Bridge financing by Ganesh Steel Ltd. as they are issued by large and credit worthy companies. The instrument is in the form of an unsecured promissory note and is freely transferable by endorsement. It is sold at discount and redeemed at par.
2. Its maturity period may range from a fortnight to a year.
3. It is also used to meet the short term seasonal and working capital requirements of a business enterprise.

ADDITIONAL QUESTIONS

Question 1.
Incorporated in 1990, Raju Dairy Ltd., is one of the leading manufacturers and marketers of dairy-based branded foods in India. In the initial years, its operations were restricted only to collection and distribution of milk. But, over the years it has gained a reasonable market share by offering a diverse range of dairy based products including fresh milk, flavoured yogurt, ice creams, butter milk, cheese, ghee, milk powders etc. In order to raise capital to finance its expansion plans, Raju Dairy Ltd. has decided to approach capital market through a mix of Offer for sale of Rs. 4 crore shares and a public issue of Rs. 2 crore shares.
In context of the above case:

1. Name and explain the segment of capital market being approached by the company.
2. Identify the two methods of floatation used by the company to raise the required capital. Give one difference between them.

Answer:

1. Primary market is the segment of capital market being approached by the company. It is also known as the new issue market as the securities are issued for the first time by the companies through this market.
2. The two methods of floatation used by the company to raise the required capital are – Issue through prospectus and Offer for sale.
   In case of issue through prospectus, the company approaches the members of the general public directly by issuing a prospectus whereas in case of Offer for sale, the company approaches members of the general public indirectly through intermediaries like issuing houses, stock brokers etc.

Question 2.
The SEBI has imposed a penalty of Rs. 7,269.5 crore on Pearls Agrotech Corporation Limited (PACL) and its four directors — Tarlochan Singh, Sukhdev Singh, Gurmeet Singh and Subrata Bhattacharya who had mobilised funds from the general public through illegal collective investment schemes in the name of purchase and development of agriculture land. While imposing the penalty, the biggest in its history, Securities and Exchange Board of India (SEBI) said the company deserved “maximum penalty” for duping the common man. Its Prevention of Fraudulent and Unfair Trade Practices Regulations provides for “severe to severe penalties” for dealing with such violations. As per SEBI norms, it can impose a penalty of Rs. 25 crore or three times of the profit made by indulging in fraudulent and unfair trade practices and in the present case, the regulator has imposed a fine equivalent to three times of the illicit gains.
In the context of the above case:

1. State the objectives of setting up SEBI.
2. Identify the type of function performed by SEBI by quoting lines from the paragraph.

Answer:

1. The objective of setting up SEBI are outlined below:
   • To prevent trading malpractice in the securities markets.
   • To protect the rights and interest of investors, and to guide and educate them.
   • To regulate and develop a code of conduct and fair practices by intermediaries like brokers, merchant bankers, etc. with a view to making them competitive and professional.
   • To regulate stock exchanges and the securities market to promote their orderly functioning.
2. Protective function is performed by SEBI: “The SEBI has imposed a penalty of Rs. 7,269.5 crore on Pearls Agrotech Corporation Limited.”

Question 3.
Harsh works as a manager in a software company. He opened a Demat account with a broking house in order to trade in securities with the money he received as his first performance bonus. Since then he has been very active in stock trading under the guidance of a stock broker. However, when he was hospitalised for a few days this year, his wife received several calls from the his stock broker for permission to transact on Harsh’s behalf. Though she told him to wait till her husband had recovered, the stock broker went ahead and executed the transactions. When Harsh got home from hospital, he discovered that the unauthorised transactions had led to a loss for him.
Harsh complained to the broking house, but they claimed he had authorised the transactions. Keeping in view, the guidelines issued by the National Stock Exchange that he had read in the national newspaper Harsh demanded proof and threatened to file a complain. Since, the broking house had no evidence that the deals had been authorised they made good the loss that Harsh had incurred due to the transactions.
In the context of the above case:

1. What is a Demat account?
2. Who is acting as the depository participant for Harsh?
3. Name the document that is legally enforceable and helps to settle the claims between the investor and the broker.

Answer:

1. A Demat account is an account used for holding securities in electronic form.
2. The Broking house is acting as the depository participant for Harsh.
3. A contract note is a legally enforceable document that helps to settle the claims between the investor and the broker.

Question 4.
Make Good Technologies Ltd. is one of the top suppliers of security software products and solutions in India with a market share of over 20% in the retail segment. Its customers includes people all sections of the society i.e. both households and corporates. Its unique threat detection system works to detect security threats including virus attacks in real time to protect users’ IT assets across varied platforms and devices. The company has an established track record of growth and financial performance. At present the company operates only through its website. The company now intends to launch a range of computer accessories and plans to market it by opening its own retail outlets. So, the board of directors of the company have decided to only raise capital for the first time through an issue of shares, but at the same time they do not wish to get into the hassles of launching a public issue of shares.
In context of the above case:

1. Name and explain the way through which the company can raise finance by allotting securities to selective individuals and institutions only.
2. Can the company also raise capital through a right issue? Why or why not? Give a reason to justify your answer.

Answer:

1. The company can raise finance by allotting securities to selective individuals and institutions only through Private Placement. It is a relatively economical way of raising money as it helps to save time, cost and money involved in the process of issuing securities.
2. No, the company cannot raise capital through a right issue as it is issuing securities for the first time. A rights issue is a pre-emptive right given only to the existing shareholders to subscribe to the securities of the company as per its terms and conditions.

Question 5.
After doing a course in online trading, Arsh started an online portal for stock trading under the name ‘Investment Guru’. He met his school friend Ajay after a long time in a bank where Ajay had come to open a D-Mat account. Arsh urged Ajay to invest in the forthcoming IPO of a blue chip companies whereas Ajay was inclined to buy existing securities of the other companies to build his investment portfolio.
In context of the above case:

1. Identify the two different types of capital market being referred to by quoting lines from the para.
2. State any four differences between the two different types of capital markets as identified in part (1).

Answer:

1. The two different types of capital market being referred to are-
   • Primary Market: “Arsh urged Ajay to invest in the forthcoming IPO of a blue chip companies.”
   • Secondary Market: “Ajay was inclined to buy existing securities of the other companies to build his investment portfolio.”
2. Differences between Primary Market and Secondary Market:
   

   S. No	BASIS	PRIMARY MARKET	SECONDARY MARKET
   1.	Meaning	It is the new issue market.	It is the market for old securities.
   2.	Aspects	Only buying of securities takes place.	Both buying and selling of securities takes place.
   3.	Price	Prices of the securities are determined by the company.	Prices of the securities are determined by the forces of demand and supply.
   4.	Parties involved	It involves dealings between the company and investors	It involves dealings between the two investors.

Question 6.
Ketan won a cash prize of Rs. 20,000 in the National level Robotics Competition. On the advise of his father, he visits a nearby bank to open a Fixed deposit account in his name with the prize money. His sister Suhasini accompanied him to the bank. On reaching the bank, he notices big banners which are placed within the premises containing information about the various arrangements through which corporates may raise their capital through the bank. Being a finance graduate, Suhasini explains to Ketan that banks play the role of the financial intermediary by helping in the process of channelizing the savings of the households into the most profitable business ventures.
In context of the above case: .

1. Name another financial intermediary that helps in the process of channelizing the savings of the households into the most productive use.
2. Also, outline any two functions of another financial intermediary as identified in part (1).

Answer:

1. Financial markets is the other financial intermediary that helps in the process of channelizing the savings of the households into the most productive use.
2. The two functions of financial market are as follows:
   • It helps to determine the price for the financial asset in a particular financial market through the market forces of demand and supply.
   • It provides liquidity to the financial assets by providing ready markets wherein the securities can be converted into cash or vice versa easily.

Question 7.
Ragu works as a waiter in a five star hotel in Mumbai. While serving the customer he overhears him at the table saying that the he has made profits higher than expected by investing in securities market. So, Ragu also decides to make a nominal investment from his savings in the stock market in pursuit of higher gains.
In context of the above case:
As a financial consultant, apprise him of the steps involved in the working of a Demat system.
Answer:
Ragu will have to initiate the following steps for trading through a Demat system:

• He will have to first open a Demat account with a depository participant (DP) who may either be a bank, broker or financial services company by furnishing certain details and information about himself including PAN number, date of birth, bank account details, income details etc.
• If he plans to buy shares through a public offer he will have to give details of his Demat account, bank account etc. On allotment the shares will be directly credited to his account.
• If he decides to buy shares otherwise, he will have to instruct his broker with the details about the name of the company, number of shares, price etc. The transaction will be executed through the depository participant and he will have to make payment for them within T + 2 days.
• On contrary, whenever he decides to sell shares, he will have to instruct his broker with the details about the name of the company, number of shares, price etc. The transaction will be executed through the depository participant and his account will be debited accordingly. He will receive the payment in T + 2 days.

Question 8.
The Bombay Stock Exchange (BSE) is Asia’s first stock exchange and the world’s 11th largest stock exchange. It became the first stock exchange to be recognized by the Indian Government under the Securities Contracts Regulation Act. Its automated, screen-based trading platform called BSE On-Line Trading (BOLT) had a capacity of 8 million orders per day. The BSE has also introduced a centralized exchange-based internet trading system, BSEWEBx.co.in to enable investors anywhere in the world to trade on the BSE platform.
In context of the above case:

1. Name the organisation that regulates the working of stock exchanges in India.
2. State any three functions performed by stock exchanges.
3. Give any two advantages of screen-based trading.

Answer:

1. Securities And Exchange Board of India (SEBI) regulates the working of stock exchanges in India.
2. Three functions performed by stock exchanges are as follows:
   • Ensures liquidity and marketability of existing securities by a providing a ready and continuous market for the sale and purchase of securities.
   • Helps in determining the prices of the securities through the forces of demand and supply.
   • It provides a legal framework for fair and safe dealings.
3. Two advantages of screen-based trading are as follows:
   • As the investors get access to the stock market during real time, there is complete transparency in the dealings.
   • It provides a common platform for exchange of securities thereby increasing the efficient transactions by saving time, effort and money.

Question 9.
Ragunath Ahuja is one of the Promoter-Director of Vishwas Ltd. The company is engaged in the real estate sector, which has recently witnessed a steady fall in its revenue and the value of its assets due to a downward trend persisting in the market in specific and the economy in general. The periodical financial results of the company were to be declared in a fortnight time. Ragunath Ahuja, being an insider, had access to unpublished price sensitive information related to it. Consequently, he sells a major portion of his holdings in an anticipation of a fall in the market price of the shares of the Company subsequent to the announcement of periodical financial results of the company. Moreover, he doesn’t inform The Securities and Exchange Commission (SEC) about the dealings.
On conducting a probe, Securities and Exchange Board Of India (SEBI) finds Ragunath Ahuja guilty of insider trading. As per law, company directors, officials or any individual with a stake of 10% or more in the company are considered to be insiders and they are required to report their insider transactions within two business days of the date the transaction occurred.
In context of the above case:

1. State the purpose of setting up SEBI.
2. Identify the type of function performed by SEBI by quoting lines from the paragraph.

Answer:

1. SEBI has to be responsive to the needs of three groups, which constitute the market namely:
   • The issuers of securities so as to provide them a platform for raising capital in an easy, effective and efficient manner.
   • The investors so as to protect their interests in securities by keeping them abreast about the developments through true and appropriate information.
   • The market intermediaries in order to provide them a framework so as to enable them perform their functions effectively and efficiently.
2. Regulatory function is being performed by SEBI: “On conducting a probe Securities and Exchange Board of India (SEBI) finds Ragunath Ahuja guilty of insider trading.”

Question 10.
Sumita is a professor in a reputed business institute. While explaining the procedure of stock exchange trading, she shared with her students that many years back she had bought 200 shares of a leading automobiles company. As per the settlement procedure she paid for the shares and received the share certificates in physical form. However, when she had sent those certificates to the company to get them endorsed in her name, she was informed by the company that those certificates were duplicate.
Therefore, in order to protect the investors from many such malpractices, now only screen-based trading is done and dematerialisation is compulsory.
In context of the above case:

1. What is screen based trading?
2. Give the meaning of ‘dematerialisation’. State any two of its advantages.

Answer:

1. Screen-based trading refers to the process of buying or selling securities online.
2. Dematerialisation refers to the process of holding securities in electronic form. The two advantages of Dematerialisation as follows:
   • The securities in the demat account can be offered as security to raise loans.
   • Since the shares certificates are not held in physical form, there is no danger of loss, theft or forgery.

Question 11.
Madhav’s is one of the India’s most trusted brands in Indian sweets and snacks segment. The company has manufacturing plants in Kota, Kanpur, New Delhi, and Mumbai. Madhav’s has its own retail chain stores and a range of restaurants in these cities. Now, the company plans to extend its business in 12 more cities in India. In order to raise the funds, its directors have decided to float a public issue through prospectus. Besides, it intends to raise money to meet the floatation costs in terms of brokerage, underwriting commission, advertising etc.
In context of the above case:

1. What is the other name used for the funds required to meet floatation costs?
2. Describe briefly the short term instrument popularly used by the companies to raise for the funds required to meet floatation costs. Who can issue them?
3. Distinguish between the two types of financial markets that the company intends to approach to meet its financial needs.

Answer:

1. Bridge financing is the other name used for the funds required to meet floatation costs.
2. Commercial Papers issued by large and credit worthy companies. The instrument is in the form of an unsecured promissory note and is freely transferable by endorsement. It is sold at discount and redeemed at par. Its maturity period may range from a fortnight to a year. It is also used to meet the short term seasonal and working capital requirements of a business enterprise. For example it is used for the purpose of bridge financing.
3. Capital Market and Money Market.
   Differences between Primary Market and Secondary Market:
   Differences between Capital Market and Money Market:

   S. No	Basis	Capital Market	Money Market
   1.	Duration	It is a market for long term funds.	It is a market for short term funds whose maturity period is upto one year.
   2.	Participants	The main participants in capital market are banks, financial institutions, corporate bodies, foreign investors and retail investors.	The main participants are institutional investors.
   3.	Investment outlay	Since, the cost of securities may be low, investment can be made in the capital market can be with less capital.	Since the cost of securities may be high, investment in the money market requires huge capital outlay.
   4.	Liquidity	The securities in capital market enjoy good liquidity.	The securities in money market enjoy high liquidity as The Discount Finance House of India works as a compulsory market maker.
   5.	Risk and return	The instruments in capital market carry high risk as the expected return is high on them.	The instruments in money market carry low risk as the expected return is low on them.

Question 12.
During navratras ,Varun finalises a deal to buy a new house. So, he visits a nearby branch of ‘Subh Bank’ to withdraw Rs. 10 lakhs from his account in order to pay the token money to the seller. In the bank he observes that a large number of customers are present to make cash with drawls, probably because it is an auspicious time to make purchases. After sometime, he overhearsone of the bank staff members telling his colleague that, “Today ‘Subh Bank’ is likely to fall short of cash and to make up for the deficit and maintain its cash reserve ratio it will have to approach another bank.”
In context of the above case:

1. Identify the instrument that ‘Subh Bank’ will use to meet its short term requirements of funds.
2. State any three feature of the instrument as identified in part (1).

Answer:

1. Call money is the instrument used by ‘Subh Bank’ to meet its short term requirements of funds.
2. Three features of call money are as follows:
   • Call money is an instrument through which one bank may borrow money from another bank to maintain the cash reserve ratio as per the guidelines of RBI.
   • Its maturity period may be from a single day to a fortnight.
   • The rate at which the interest is paid on call money is called call rate.

Question 13.
The stock market regulator, Securities and Exchange Board of India (SEBI), has initiated a certification programme for all market intermediaries. Under this programme, people associated with stock markets in any way, will have to obtain a qualifying certificate from the regulator.The National Institute of Securities Market (NISM), a trust formed by SEBI, is tasked with the certification programme.
In the context of the above case:

1. Identify the type of function performed by SEBI.
2. Outline any two reasons for setting up SEBI.

Answer:

1. Developmental function is being carried out by SEBI by starting a certification programme for all market intermediaries.
2. SEBI was set up as a regulatory body by the government for the following reasons:
   • To curb malpractices in the financial market.
   • To enhance the confidence of the investors by ensuring fair, efficient and transparent dealings.

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CBSE Class 10 Science Practical Skills – Properties of Acids and Bases

Basic Building Concepts

• Acids are sour in taste while bases are bitter in taste.
• When an acid reacts with a metal, hydrogen gas is evolved and a corresponding salt is formed.
• When a base reacts with a metal, along with the evolution of hydrogen gas a salt is formed which has a negative ion composed of metal and oxygen.
• When an acid reacts with a metal carbonate, it gives the corresponding salt, carbon dioxide gas and water.
• Bases do not react with metal carbonate.

Aim
(a) To study the properties of acids (HCl) by their reaction with

1. Litmus solution(Blue/Red)
2. Zinc metal(Zn)
3. Solid sodium carbonate(Na₂CO_s).

Materials Required
Test tubes, test tube stand, test tube holder, cork, droppers, beaker, litmus (red and blue), Zn metal granules, solid sodium carbonate, dil. HCl.

You can also download NCERT Science Class 10 to help you to revise complete syllabus and score more marks in your examinations.

Theory

1. HCl is an acid. Acids turn blue litmus red and do not affect red litmus.
2. On reacting with zinc metal, it forms a salt, zinc chloride (ZnCl₂) and hydrogen gas (H₂) is liberated.
   Zn(s) + 2HCl(aq) → ZnCl₂(aq) +H₂↑
   Hydrogen gas burns in air with a pop sound.
   2H₂(g) + O₂(g) → 2H₂0(l)
3. HCl reacts with sodium carbonate (aqueous/solid) to liberate carbon dioxide(CO₂) which turns lime water milky. When excess of CO₂ is passed through the solution, the milkiness disappears.
   

Procedure And Observation Table

Result

1. HCI turns blue litmus red.
2. It releases H₂ gas on reaction with Zn metal.
3. It liberates C0₂ gas on reaction with Na₂C0₃.

Precautions

1. HCI is corrosive, therefore, handle it with care.
2. Use small quantities of chemicals.
3. A fine jet should be taken for burning hydrogen.
4. Use clean zinc metal, otherwise the reaction will occur very slowly.
5. Wash your hands properly with soap after the experiment.

Aim

(b) To study the properties of bases(NaOH) by their reaction with

1. Litmus solution(Blue/Red)
2. Zinc metal(Zn)
3. Solid sodium carbonate(Na₂CO₃)

Materials Required
Dil. NaOH, test tubes, test tube holder, test tube stand, red and blue litmus solutions, zinc metal, sodium carbonate and droppers.
Theory

1. NaOH is a base. Bases turn red litmus blue and do not affect blue litmus.
2. On reacting with zinc metal, NaOH forms sodium zincate and hydrogen gas is liberated. The hydrogen gas burns with a pop sound.
   
3. NaOH does not react with solid or aqueous Na₂CO₃.

Procedure And Observation Table

Result

1. NaOH turns red litmus blue.
2. On reaction with Zn metal, it releases H₂
3. It does not react with Na₂CO_g.

Precautions

1. Handle NaOH with care.
2. Use small quantities of chemicals.
3. A fine jet should be taken for burning hydrogen.
4. Zn granules should be clean.
5. Wash your hands properly after completing experiment.

INTERACTIVE SESSION

Question 1:
What is the aim of your experiment?
Answer:
The aim of my experiment is to study the properties of acids and bases.

Question 2:
What are acids?
Answer:
Acids are the substances which furnish hydrogen ions in a solution

Question 3:
What are bases?
Answer:
Bases are the substances which furnish hydroxide ions in a solution.

Question 4:
How does an acid affect blue and red litmus?
Answer:
An acid turns blue litmus red and has no effect on red litmus.

Question 5:
How does a base affect blue and red litmus?
Answer:
It turns red litmus blue and no change is observed on blue litmus.

Question 6:
Why do metals displace hydrogen from dilute acids?
Answer:
Metals displace hydrogen from dilute acids because they are more reactive than hydrogen. Metals lose electrons which the hydrogen ions from acids accept and hydrogen gas is liberated.

Question 7:
Do all metals displace hydrogen gas from dilute acids?
Answer:
No, only the metals which are above hydrogen in the reactivity series are able to displace hydrogen from dilute acids.

Question 8:
What would be the affect of milk of magnesia on litmus?
Answer:
It would turn red litmus blue as it is an antacid and basic in nature.

Question 9:
What would happen when dil. HCl and dil. NaOH (same concentrations) are mixed in equal amounts? Write the equation also.
Answer:
Neutralisation reaction occurs. As a result of this salt and water will be formed.
NaOH + HCl → NaCl + H₂0

Question 10:
Write the reaction occurring between HCl and Na₂CO₃.
Answer:
2HCl + Na₂CO₃ -> 2NaCl + H₂O + CO₂

Question 11:
How do you test CO₂ gas?
Answer:
When CO₂ is passed through lime water, it turns lime water milky.

Question 12:
Why does hydrogen gas burn with an explosion?
Answer:
Hydrogen gas has the highest calorific value with low ignition temperature and, therefore, burns with an explosion.

Question 13:
Write the reaction which occurs between Zn metal and dil. H₂SO₄.
Answer:
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

Question 14:
What is the difference between a strong and a weak acid?
Answer:
Strong acids are those which completely dissociate into ions.
For example,                HCl(aq) → H⁺(aq) + Cl^– (aq),
whereas weak acids are those which partially dissociate into ions.
For example,                CH₃COOH(aq)\(\rightleftharpoons\) CH₃COO^–(aq) + H⁺(aq)

Question 15:
Which out of hydrochloric acid and acetic acid would have a lower pH?
Answer:
HCl would have a lower pH as it has a higher concentration of hydrogen ions.

Question 16:
Which out of dil. H₂SO₄ and lime water would turn red litmus blue?
Answer:
Lime water being basic in nature will turn red litmus blue.

NCERT LAB MANUAL QUESTIONS

Question 1:
What will be the colour of a blue litmus paper on bringing it in contact with a drop of dil. hydrochloric acid?
Answer:
The colour of blue litmus turns red on bringing it in contact with a drop of dil. hydrochloric acid.

Question 2:
What will happen to a lighted candle if it is brought near the mouth of a gas jar containing hydrogen gas?
Answer:
The candle extinguishes with a pop sound and the gas burns with a pale blue flame.

Question 3:
Which gas is produced when zinc metal reacts with hydrochloric acid?
Answer:
Hydrogen gas is produced when zinc metal reacts with hydrochloric acid.
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Question 4:
Which gas is liberated when sodium carbonate reacts with hydrochloric acid?
Answer:
When sodium carbonate reacts with hydrochloric acid, CO₂ gas is liberated.
Na₂C0₃(s) + 2HCl(aq) → 2NaCl(aq) + H₂0(l) + CO₂(g)

Question 5:
What will be the colour of a blue litmus paper on bringing it in contact with a drop of dil. NaOH?
Answer:
Blue litmus paper does not undergo any change in colour on bringing it in contact with a drop of dil. NaOH as dil. NaOH is a base and changes colour of red litmus to blue and has no effect on blue litmus.

PRACTICAL BASED QUESTIONS

Multiple Choice Questions/VSA

1. 10 ml of HCl and 10 ml of NaOH solutions are contained in two separate beakers labelled I and II respectively. On adding zinc granules to both, it is observed that at room temperature [CBSE 2012, AI 2009]
(a) gas is evolved vigorously in both.
(b) gas is evolved vigorously in beaker I but not so in beaker II.
(c) gas is evolved in beaker II but not in beaker I.
(d) no gas is evolved in either of the two beakers.

2. A student took two test tubes containing 2 ml of dilute hydrochloric acid and added zinc granules to test tube (A) and solid sodium carbonate to test tube (B) as shown below: [CBSE 2012, Foreign 2009]

The correct observation would be
(a) rapid reaction in both the test tubes
(b) slow reaction in (A) and rapid reaction in (B)
(c) rapid reaction in (A) but a slow reaction in (B)
(d) no reaction in any of the test tubes

3. A blue litmus paper was first dipped in dil. HCl and then in dil. NaOH solution. It was observed that the colour of the litmus paper [CBSE 2011]
(a) changed to’red
(b) changed first to red and then to blue
(c) changed blue to colourless
(d) remained blue in both the solutions

4. When a student added zinc granules to dilute HCl, a colourless and odourless gas was evolved, which was tested with a burning matchstick, it was observed that
(a) the matchstick continued to burn brilliantly
(b) the matchstick burnt slowly with a blue flame
(c) the matchstick extinguished and the gas burnt with pop sound
(d) the matchstick burnt with an orange flame

5. The milkiness in lime water is obtained due to the formation of
(a) Ca(HSO₃)₂
(b) Ca(HCO₃)₂
(c) CaO
(d) CaCO₃

6. When excess carbon dioxide is passed through lime water, its milkiness disappears due to the formation of
(a) Ca(HCO₃)₂
(b) CaCO₃
(c) NaHCO₃
(d) Na₂CO₃

7. Zn metal shows amphoteric behaviour. It can be shown by
(а) its reaction with dil. NaOH only
(b) its reaction with dil. HCl only
(c) its reaction with both dil. HCl and dil. NaOH
(d) it does not react with either dil. HCl or dil. NaOH

8. Which one of the following will be required to identify the gas evolved when dilute hydrochloric acid reacts with zinc metal? [Delhi 2007]
(а) Blue litmus solution
(b) Red litmus solution
(c) A burning splinter
(d) Lime water

9. Four students performed the reactions of a dil. hydrochloric acid and a solution of sodium hydroxide with zinc metal and solid sodium carbonate separately. They reported the possible reaction by ( ✓) and no reaction by (). [CBSE 2011]

In which of the following sets all observations are correct?
(a) A
(b) B
(c) C
(d) D

10. A teacher gave two test tubes, one containing water and the other containing sodium hydroxide solution, to the students and asked them to identify the test tube containing sodium hydroxide solution. Which one of the following can be used for correctly identifying the test tube containing sodium hydroxide solution? [CBSE 2011, AI 2007]
(a) Blue litmus
(b) Red litmus
(c) Sodium carbonate solution
(d) Dilute hydrochloric acid

11. A student added dil. HC1 to a test tube conta­ining zinc granules and made the following observations:
I. The zinc surface becomes dull and black.
II. A gas evolves which bums with a pop sound.
III. The solution is colourless.
The correct observations are
(a) I and II
(b) I and III
(c) II and III
(d) I, II and III

12. Which one of the following set-ups is the most appropriate for the evolution of hydrogen gas and its identification? [Delhi 2008]

(a) I
(b) II
(c) III
(d) IV

13. A solution in test tube ‘A’ turns red litmus blue, evolves hydrogen on reaction with Zn and does not react with sodium carbonate, whereas solution in test tube ‘B’ turns blue litmus red, liberates hydrogen on reaction with Zn and evolves CO₂ with Na₂CO₃. Identify ‘A’ and ‘B’.
(a) ‘A’ is an acid, ‘B’ is a base
(b) ‘A’ is a base, ‘B’ is an acid
(c) Both ‘A’ and ‘B’ are bases
(d) Both ‘A’ and ‘B’ are acids

14. Given below are certain chemical properties of substances.
I. It turns blue litmus red II.
II. It turns red litmus blue
III. It reacts with zinc and a gas evolves
IV. It reacts with solid sodium carbonate to give brisk effervescence
Which out of these properties are shown by dilute hydrochloric acid?
I and II only
I and III only
I, III and IV only
II, III and IV only

Short Answer Questions
1. What are acids? Give any two characteristics of acids.

2. Give any two points of difference between an acid and a base.

3. What are bases? How does a base like sodium hydroxide react with zinc?

4. What is litmus solution? What is its colour in (i) neutral (ii) acidic and (iii) basic solutions?

5. What will happen:
(a) if a lighted candle is brought near the mouth of gas jar containing hydrogen gas?
(b) if carbon dioxide gas is passed through lime water?

ANSWER KEY

Multiple Choice Questions/VSA
1. (b)
2. (a)
3. (b)
4. (c)
5. (d)
6. (a)
7. (c)
8. (c)
9. (d)
10. (b)
11. (d)
12. (b)
13. (b)
14. (c)

Short Answer Questions
1. Acids are substances which contain hydrogen and which when dissolved in water give hydrogen ions in the solution.
Characteristics of acids:

1. They are sour in taste.
2. They turn blue litmus red but they do not affect red litmus.

2.

3. A base is a substance which when dissolved in water gives hydroxide (OH^–) ions in the solution. Sodium hydroxide reacts with zinc to liberate hydrogen gas along with the formation of sodium zincate.

4. Litmus solution is a purple coloured dye which is extracted from lichen. It is the most commonly used indicator in the laboratory.
Neutral litmus solution has purple colour. In acidic solution, it turns red whereas in basic solution, it turns blue.

5. (a) The candle extinguishes with a pop sound and the gas burns with a pale blue flame.
(b) CO₂ turns lime water milky.

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CBSE Class 9 Maths Lab Manual – Algebraic Identity (a + b)² = a² + 2ab + b²

Objective
To verify the identity (a + b)² = a² + 2ab + b² by paper cutting and pasting.

Prerequisite Knowledge

1. Area of a square = (side)².
2. Area of a rectangle = l x b.

Materials Required
A sheet of white paper, three sheets of glazed paper (different colours), a pair of scissors, gluestick and a geometry box.

Procedure
Take distinct values of a and b, say a = 4 units, b = 2 units

1. Cut a square of side a (say 4 units) on a glazed paper (blue).
2. Cut a square of side b (say 2 units) on glazed paper (pink).
3. Now, cut two rectangles of length a (4 units) and breadth b (2 units) from third glazed paper (red).
   
4. Draw a square PQRS of (a+ b) = (4 + 2), 6 units on white paper sheet as shown in fig. (i).
5. Paste the squares I and II and two rectangles III and IV on the same white squared paper. Arrange all the pieces on the white square sheet in such a way that they form a square ABCD fig. (ii)
   
   

Observation

1. Area of the square PQRS on the white sheet paper.
   (a+b)² = (4+2)² = 6 x 6 = 36 sq. units ……….(i)
2. Area of two coloured squares I and II
   area of Ist square = a² = 4² = 16 sq.units
   area of IInd square = b² = 2² = 4 sq.units
3. Area of two coloured rectangles III and IV = 2(a x b) = 2(4 x 2) = 16 sq. units
   Now, total area of four quadrilaterals (calculated)
   = a² + b² + 2(ab)
   = 16+4+16
   = 36 sq. units ……….(ii)
   Area of square ABCD = Total area of four quadrilaterals = 36 sq. units
   Equating (i) and (ii)
   Area of square PQRS = Area of square ABCD
   i.e., (a+b)² = a² + b² + 2ab

Result
Algebraic identity (a+b)² = a² + 2ab + b² is verified.

Learning Outcome
The identity (a+b)² = a² + 2ab + b² is verified by cutting and pasting of paper. This identity can be verified geometrically by taking other values of a and b.

Activity Time
Divide a square plot into four parts (quadrilaterals) such that two parts have same area and other two are squares.

Viva Voce

Question 1.
What is the degree of polynomial x² + 4x + 2?
Answer:
2

Question 2.
Write the simplification of (2x + 1)².
Answer:
4x²+4x+1.

Question 3.
What is the coefficient of x² in (3x + 1)² ?
Answer:
9.

Question 4.
Write (x² + \(\frac { 1 }{ { x }^{ 2 } }\) + 2) in square form.
Answer:
(x + \(\frac { 1 }{ x }\))²

Question 5.
Factorize x² + 10x +25.
Answer:
(x+5)²

Question 6.
Write the product of (7x + 3) (7x + 3).
Answer:
49x² + 42x + 9.

Question 7.
What do you mean by zeroes of the polynomial ?
Answer:
The values of x for which the given polynomial vanishes.

Question 8.
Is \(-\frac { 1 }{ 3 }\) the zero of the polynomial 3x + 1?
Answer:
Yes

Question 9.
If f(y) = y² – y + 1 find f(1).
Answer:
f(1) = (1)² – (1) + 1 = 1.

Question 10.
(a + b)² is binomial or trinomial ?
Answer:
Trinomial.

Multiple Choice Questions

Question 1.
How will you find the value of 101² by using identity (a + b)²:
(i)(100+1)²
(ii) 100² + 1²
(iii) (100-1)²
(iv) none of these

Question 2.
Which algebraic identity can be used to find the value of 912?
(i) (a+ b)²
(ii) a² – b²
(iii) a³ + b³
(iv) none of these

Question 3.
Simplify: (2x +y)² (By using identity only).
(i) 2x² + y² + 4x
(ii) 4x² +y² + 2xy
(iii) 4x² +y² + 4xy
(iv) none of these

Question 4.
Find the factors of x² + 9y² + 6xy:
(i) (x+3y)(x+3 y)
(ii) (x-3 y)(x+3 y)
(iii) (x-3y)(x-3y)
(iv) none of these

Question 5.
Write the factors of 25x² + 20x +4:
(i) (5x + 2) (5x + 2)
(ii) (5x- 4) (4 + 5x)
(iii) (4 + 5x)(4 – 5x)
(iv) none of these

Question 6.
What will be area of a square of side (x + 5) ?
(i) x² + 25
(ii) (x + 5)²
(iii) x + 52
(iv) none of these

Question 7.
Find the factors of 3 + 2√3 x + x² ?
(i) (x + √3)²
(ii) x² + 3
(iii) x + (√3)²
(iv) none of these

Question 8.
The expression 4x² + 12x + 9 represents an area of square, write the dimensions of a square.
(i) (2x + 3) by (2x + 3)
(it) (2x – 3) by (2x + 3)
(iii) 2x by 3
(iv) none of these

Question 9.
If (a + b)² = 25, a² = 4, 2ab = 12, then what will be the value of a and b?
(i) a = -2, b = 3
(ii) a = 2, b = -3
(iii) a = 2, b = 3
(iv) none of these

Question 10.
Write the factors of 169 + 26y + y²
(i) (y – 13)²
(ii) (13 + y)²
(iii) (13 – y)²
(iv) none of these

Answers

1. (i)
2. (i)
3. (iii)
4. (i)
5. (i)
6. (ii)
7. (i)
8. (i)
9. (iii)
10. (ii)

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