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NCERT Exemplar Class 12 Biology Chapter 1 Reproduction in Organisms are part of NCERT Exemplar Class 12 Biology. Here we have given NCERT Exemplar Class 12 Biology Chapter 1 Reproduction in Organisms. https://www.cbselabs.com/ncert-exemplar-problems-class-12-biology-reproduction-organisms/

NCERT Exemplar Class 12 Biology Chapter 1 Reproduction in Organisms

Multiple Choice Questions
Single Correct Answer Type

1. A few statements describing certain features of reproduction are given below
i. Gametic fusion takes place
ii. Transfer of genetic material takes place
iii. Reduction division takes place
iv. Progeny have some resemblance with parents
Select the options that are true for both asexual and sexual reproduction from the options given below:
(a) i and ii (b) ii and iii
(c) ii and iv (d) i and iii
Answer. (c) Transfer of genetic material and progeny have some resemblance with parents are the phenomenon common in’both asexual and sexual reproduction while gametic fusion and reduction division takes place in sexual reproduction only.

2. The term ‘ clone ’ cannot be applied to offspring formed by sexual reproduction because
(a) Offspring do not possess exact copies of parental DNA
(b) DNA of only one parent is copied and passed on to the offspring
(c) Offspring are formed at different times
(d) DNA of parent and offspring are completely different
Answer. (a)
• In asexual reproduction, a single individual (parent) is capable of producing offspring which are not only identical to one another but are also exact copies of their parent. The term clone is used to describe such morphologically and genetically similar individuals.
• In sexual reproduction because of the fusion of male and female gametes (either by same individual or by different individual of the opposite sex), sexual reproduction results in offspring that are not identical to the parents or amongst themselves.

3. Amoeba and Yeast reproduce asexually by fission and budding respectively, because they are
(a) Microscopic organisms
(b) Heterotrophic organisms
(c) Unicellular organisms
(d) Uninucleate organisms
Answer. (c) Many single-celled organisms reproduce by binary fission (e.g., Amoeba, Paramecium), where a cell divides into two halves and each rapidly grows into an adult.
In yeast, the division is unequal and small buds are produced that remain attached initially to the parent cell which eventually gets separated and mature into new yeast organism (cells). Budding is also found in Hydra.

4. A few statements with regard to sexual reproduction are given below
i. Sexual reproduction does not always require two individuals
ii. Sexual reproduction generally involves gametic fusion
iii. Meiosis never occurs during sexual reproduction
iv. External fertilisation is a rule during sexual reproduction
Choose the correct statements from the options below:
(a) i and iv , (b) i and ii
(c) ii and iii (d) i and iv
Answer. (b)
• Sexual reproduction requires male and female gametes (either by same individual or by different individual of the opposite sex).
• Sexual reproduction generally involves gametic fusion.
• Meiosis occurs during sexual reproduction in dipoloid organisms.
• External fertilisation is not a rule during sexual reproduction, internal fertilization also takes place

5. A multicellular, filamentous alga exhibits a type of sexual life cycle in which the meiotic division occurs after the formation of zygote. The adult filament of this alga has
(a) Haploid vegetative cells and diploid gametangia
(b) Diploid vegetative cells and diploid gametangia
(c) Diploid vegetative cells and haploid gametangia
(d) Haploid vegetative cells and haploid gametangia
Answer. (d) Adult filament of a multicellular, filamentous alga have haplontic life cycle in which the meiotic division occurs after the formation of zygote. So, the filament of this alga have haploid vegetative cells and haploid gametangia.

6. The male gametes of rice plant have 12 chromosomes in their nucleus. The chromosome number in the female gamete, zygote and the cells of the seedling will be, respectively,
(a) 12,24,12 . (b) 24,12,12
(c) 12,24,24 (d) 24,12,24
Answer. (c) Gametophytic structure (n) of rice plant contain 12 chromosomes and sporophytic structure (2n) of rice contain 24 chromosomes.
Female gamete (n) =12,
Zygote (2n) = 24,
The cells of the seedling (2n) = 24.

7. Given below are a few statements related to external fertilization. Choose the correct statements.
i. The male and female gametes are formed and released simultaneously.
ii. Only a few gametes are released into the medium.
iii. Water is the medium in a majority of organisms exhibiting external fertilization.
iv. Offspring formed as a result of external fertilization have better chance of survival than those formed inside an organism.
(a) iii and iv (b) i and iii
(c) ii and iv (d) i and iv .
Answer. (b) In most aquatic organisms, such as a majority of algae and fishes as well as amphibians, syngamy occurs in the external medium (water), i.e., outside the body of the organism. This type of gametic fusion is called external fertilisation. Organisms exhibiting external fertilisation show great synchrony between the sexes and release a number of gametes into the surrounding medium (water) in order to enhance the chances of syngamy. This happens in the bony fishes and frogs where a large number of offspring are produced. A major disadvantage is that the offspring are extremely vulnerable to predators threatening their survival up to adulthood.

8. The statements given below describe certain features that are observed in the pistil of flowers.
i. Pistil may have many carpels
ii. Each carpel may have more than one ovule
iii. Each carpel has only one ovule
iv. Pistil have only one carpel
Choose the statements that are true from the options below:
(a) i and ii (b) i and iii
(c) ii and iv (d) iii and iv
Answer. (a)
• Pistil may have many carpels (multicapillary pistil like Papaver)
• Each carpel may have more than one ovule (like Watermelon,.Papaya etc.)

9. Which of the following situations correctly describe the similarity between an angiosperm egg and a human egg?
i. Eggs of both are formed only once in a lifetime
ii. Both the angiosperm egg and human egg are stationary
iii. Both the angiosperm egg and human egg are motile transported
iv. Syngamy in both results in the formation of zygote
Choose the correct answer from the options given below:
(a) ii and iv (b) iv only
(c) iii and iv (d) i and iv
Answer. (b) Syngamy in both results in the formation of zygote is similarity between an angiosperm egg and a human egg.

10. Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because
(a) Nodes are shorter than intemodes
(b) Nodes have meristematic cells
(c) Nodes are located near the soil
(d) Nodes have non-photosynthetic cells
Answer. (b) Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because nodes have meristematic cells. Examples of vegetative propagules: (i) Leaf buds of bryophyllum, (ii) Eyes of potato, (iii) Bulbifof Agave, (iv) Offset of water hyacinth, (v) Rhizome of ginger.

11. Which of the following statements, support the view that elaborate sexual reproductive process appeared much later in the organic evolution?
i. Lower groups of organisms have simpler body design
ii. Asexual reproduction is common in lower groups
iii. Asexual reproduction is common in higher groups of organisms
iv. The high incidence of sexual reproduction in angiosperms and vertebrates
Choose the correct answer from the options given below:
(a) i, ii and iii (b) i, iii and iv
(c) i, ii and iv (d) ii, iii and iv
Answer. (c) Elaborate sexual reproductive process appeared much later in the organic evolution because of
• Lower groups of organisms have simpler body design.
• Asexual reproduction is common in lower groups of organisms.
• High incidence of sexual reproduction in angiosperms and vertebrates.

12. Offspring formed by sexual reproduction exhibit more variation than those formed by asexual reproduction because
(a) Sexual reproduction is a lengthy process
(b) Gametes of parents have qualitatively different genetic composition
(c) Genetic material comes from parents of two different species
(d) Greater amount of DNA is involved in sexual reproduction
Ans. (b)
• Offspring formed by sexual reproduction exhibit more variation than those formed by asexual reproduction because gametes of parents have qualitatively different genetic composition.
• In asexual reproduction due to involvement of only one parent, so there is no chance of variation.

13. Choose the correct statement from amongst the following:
(a) Dioecious (hermaphrodite) organisms are seen only in animals.
(b) Dioecious organisms are seen only in plants.
(c) Dioecious organisms are seen in both plants and animals.
(d) Dioecious organisms are seen only in vertebrates.
Answer. (c) Dioecious organisms are seen in both plants (like papaya) and animals (like cockroach).

14. There is no natural death in single celled organisms like Amoeba and bacteria because
(a) They cannot reproduce sexually
(b) They reproduce by binary fission
(c) Parental body is distributed among the offspring
(d) They are microscopic
Answer. (c) There is no natural death in single celled organisms like Amoeba and bacteria because the parental body is distributed among the offspring.

15. There are various types of reproduction. The type of reproduction adopted by an organism depends on
(a) The habitat and morphology of the organism
(b) Morphology of the organism
(c) Morphology and physiology of the organism
(d) The organism’s habitat, physiology and genetic make up
Answer. (d) The organism’s habitat, its internal physiology and several other factors (genetic make up) are collectively responsible for how it reproduces. When offspring is produced by a single parent with or without the involvement of gamete formation, the reproduction is asexual.

16. Identify the incorrect statement.
(a) In asexual reproduction, the offspring produced are morphologically and genetically identical to the parent.
(b) Zoospores are sexual reproductive structures.
(c) In asexual reproduction, a single parent produces offspring with or without the formation of gametes.
(d) Conidia are asexual structures in Penicillium.
Answer. (b) Zoospores are asexual reproductive structures.

17.Which of the following is a post-fertilisation event in flowering plants?
(a) Transfer of pollen grains
(b) Embryo development
(c) Formation of flower
(d) Formation of pollen grains
Answer.

18. The number of chromosomes in the shoot tip cells of a maize plant is 20. The number of chromosomes in the micro spore mother cells of the same plant shall be
(a) 20 (b) 10 (c) 40 (d) 15
Answer. (a) Shoot tip cells of a maize plant is a sporophytic structure (2n) and microspore mother cells of maize plant is also a sporophytic structure (2n). So, microspore mother cells (MMC) contain 20 chromosomes.

Very Short Answer Type Questions
1. Mention two inherent characteristics of Amoeba and yeast that enable them to reproduce asexually.
Answer. a. They are unicellular organisms.
b. They have a very simple body structure.

2. Why do we refer to’offspring formed by asexual method of reproduction as clones?
Answer. Offspring formed by asexual reproduction are called clones because they are morphologically and genetically similar to the parent.

3. Although potato tuber is an underground part, it is considered as a stem. Give two reasons.
Answer. a. The tuber has nodes and intemodes (as stem),
b. Leafy shoots appear from the nodes.

4. Between an annual and a perennial plant, which one has a shorter juvenile phase? Give one reason.
Answer. An annual has a shorter juvenile phase. Since its entire life cycle has to be completed in one growing season, its juvenile phase is shorter.

5. Rearrange the following events of sexual reproduction in the sequence in which they occur in a flowering plant: embryogenesis, fertilisation, gametogenesis, pollination.
Answer. Gametogenesis, Pollination, Fertilisation, Embryogenesis

6. The probability of fruit set in a self-pollinated bisexual flower of a plant is far greater than a dioecious plant. Explain.
Answer. There is assured fruit set in self pollinated bisexual flower even in the absence of pollinators. In dioecious plants, there is male and female flowers present on different plants, so external pollinating agent is required for pollination.

7. Is the presence of large number of chromosomes in an organism a hindrance to sexual reproduction? Justify your answer by giving suitable reasons.
Answer. Presence of large number of chromosomes in an organism is not a hindrance to sexual reproduction. Butterfly has 380 chromosomes but it can reproduce sexually.

8. Is there a relationship between the size of an organism and its life span? Give two examples in support of your answer.
Answer. Life spans of organisms are not necessarily correlated with their sizes. The sizes of crows and parrots are not very different yet their life spans show a wide difference. Live span of crow is 15 year and of parrot is 140 years. A mango tree has a much shorter life span as compared to a peepal tree.

9. In the figure given below, the plant bears two different types of flowers marked ‘A’ and ‘B\ Identify the types of flowers and state the type of pollination that will occur in them.

Answer. ‘A’ is chasmogamous flower while ‘B’ is cleistogamous flower. A bisexual flower which normally open is called chasmogamous flower. Cleistogamous flowers do not open at all.
Cleistogamous flowers are invariably autogamous as there is no chance of cross-pollen landing on the stigma.
In a normal flower which opens and exposes the anthers and stigma complete autogamy is rather rare. Chasmogamous flower may show autogamy, geitonogamy or xenogamy.

10. Give reasons as to why cell division cannot be a type of reproduction in multicellular organisms.
Answer. Cell division cannot be a type of reproduction in multicellular organisms because cell division only increases the number of cells in an organism which leads to the growth of body.

11. In the figure given below, mark the ovule and pericarp.

Answer.

12. Why do gametes produced in large numbers in organisms exhibit external fertilisation?
Answer. Organisms exhibiting external fertilisation release a number of gametes into the surrounding medium (water) in order to enhance the chances of syngamy because there are few’ chances of fusion between male and female gametes.

13. Which of the followings are monoecious and dioecious organisms?
a. Earthworm ——————–
b. Chara ——————–
c. Marchantia ——————-
d. Cockroach ——————–
Answer. a. Earthworm—Monoecious
b. Chara—Monoecious
c. Marchantia—Dioecious
d. Cockroach—Dioecious

14. Match the organisms given in Column ‘A’ with the vegetative propagules given in column ‘B’.

Answer. Bryophyllum—leaf buds Agave—bulbils Potato—eyes
Water hyacinth—offset

15. What do the following parts of a flower develop into after fertilisation?
a. Ovary
b. Ovules
Answer. a. Ovary—Fruit
b. Ovules—Seeds

Short Answer Type Questions
1. In haploid organisms that undergo sexual reproduction, name the stage in the life cycle when meiosis occurs. Give reasons for your answer.
Answer. Meiosis takes place during its post-zygotic stage. Since the organism is haploid, meiosis cannot occur during gametogenesis.

2. The number of taxa exhibiting asexual reproduction is drastically reduced in higher plants (angiosperms) and higher animals (vertebrates) as compared with lower groups of plants and animals. Analyse the possible reasons for this situation.
Answer. Both angiosperms and vertebrates have a more complex structural organisation. They have evolved very efficient mechanism of sexual reproduction. Since asexual reproduction does not create new genetic pools in the offspring and consequently hampers their adaptability to external conditions, these groups have resorted to reproduction by the sexual method.

3. Honeybees produce their young ones only by sexual reproduction. Inspite of this, in a colony of bees we find both haploid and diploid individuals. Name the haploid and diploid individuals in the colony and analyse the reasons behind their formation.
Solution.
• The colony of honey bees has three types of members: (i) Diploid queen are fertile females, (ii) Worker bees are sterile females and (iii) Drones are haploid males.
• An offspring formed from the union of a sperm and an egg develops as a female (queen or worker), and an unfertilized egg develops as a male (drone) by means of parthenogenesis. This means that the males have half the number of chromosomes than that of a female.

4. With which type of reproduction do we associate the reduction division? Analyse the reasons for it.
Answer. Reduction division (meiosis) is associated with sexual reproduction. The reasons for this are:
a. Since sexual reproduction involves the fusion of two types of gametes (male and female), they must have haploid number of chromosomes.
b. The cell (meiocyte) which gives rise to gametes often has diploid number of chromosomes and it is only by reducing the number by half that we can get haploid gametes.
c. Reduction division also ensures maintenance of constancy of chromosome number from generation to generation.

5. Is it possible to consider vegetative propagation observed in certain plants like Bryophyllum, water hyacinth, ginger etc., as a type of asexual reproduction? Give two/three reasons.
Answer. Vegetative propagation is considered as a type of asexual reproduction because
(i) This is uniparental.
(ii) Clone formation takes place.
(iii) There is no fertilisation.

6. ‘Fertilisation is not an obligatory event for fruit production in certains plants’. Explain the statement.
Answer. Yes, it is observed in parthenocarpic fruits. The ‘seedless fruits’ that are available in the market such as pomegranate, grapes etc. are in fact good examples. Flowers of these plants are sprayed by a growth hormone that induces fruit development even though fertilisation has not occurred. The ovules of such fruits, however, fail to develop into seeds.

7. In a developing embryo, analyse the consequences if cell divisions are not followed by cell differentiation.
Answer. During embryogenesis, zygote undergoes cell-division (mitosis) and cell differentiation. While cell divisions increase the number of cells in the developing embryo; Cell differentiation helps groups of cells to undergo certain modifications to form specialised tissues and organs to form an organism.
If cell divisions are not followed by cell differentiation then there will be no formation of tissues or organs, so a new organisms cannot be formed.

8. List the changes observed in an angiosperm flower subsequent to pollination and fertilisation.
Answer. Post-fertilisation modifications

9. Suggest a possible explanation why the seeds in a pea pod are arranged in a row, whereas those in tomato are scattered in the juicy pulp.
Answer. In a fruit, seed arrangement depends on type of placentation. Pea and tomato shows different placentation. Pea shows marginal placentation while tomato shows axile placentation.

10. Draw the sketches of a zoospore and a conidium. Mention two dissimilarities between them and alt least one feature common to both structures.
Answer.

11. Justify the statement ‘Vegetative reproduction is also a type of asexual reproduction’.
Answer. Vegetative propagation is also a type of asexual reproduction because
(i) This is uniparental.
(ii) Clone formation takes place.
(iii) There is no fertilisation.
(iv) There is no gamete formation.

Long Answer Type Question
1. Enumerate the differences between asexual and sexual reproduction. Describe the types of asexual reproduction exhibited by unicellular organisms.
Answer.

The types of asexual reproduction exhibited by unicellular organisms:
• Many single-celled organisms reproduce by binary fission (e.g., Amoeba, Paramecium), where a cell divides into two halves and each rapidly grows into an adult.
• In yeast, the division is unequal and small buds are produced that remain attached initially to the parent cell which eventually gets separated and mature into new yeast organism (cells).

2. Do all the gametes formed from a parent organism have the same genetic composition (identical DNA copies of the parental genome)? Analyse the situation with the background of gametogenesis and provide or give suitable explanation.
Answer. The gametes of a parent do not have the same genetic composition because they do not have identical copies of DNA. In the pachytene and diplotene stages of meiosis-I, the phenomenon of crossing over and chiasma formation take place between homologous chromosomes. This shifts segments of DNA from one chromatid to another (homologous chromosomes) in a random manner resulting in several new combinations of DNA sequences. As a result, when meiotic division is completed, gametes possess DNA with varying degree of variations.

3. Although sexual reproduction is a long drawn, energy-intensive complex form of reproduction, many groups of organisms in Kingdom Animalia and Plantae prefer this mode of reproduction. Give at least three reasons for this.
Answer. a. Sexual reproduction brings about variation in the offspring.
b. Since gamete formation is preceded by meiosis, genetic recombination occurring during crossing over (meiosis-I), leads to a great deal of variation in the DNA of gametes.
c. The organism has better chances survival in a changing environment.

4. Differentiate between (a) oestrus arid menstrual cycles; (b) ovipary and vivipary. Cite an example for each type.
Answer. Differences between oestrus and menstrual cycles

5. Rose plants produce large, attractive bisexual flowers but they seldom produce Suits. On the other hand a tomato plant produces plenty of fruits though they have small flowers. Analyse the reasons Tor failure of fruit formation in rose.
Answer. Failure of fruit formation in rose may be due to several reasons. Some of the likely reasons are
a. Rose plants may not produce viable pollen.
b. Rose plants may not have functional egg.
c. Rose plants may have abortive ovules.
d. Being hybrids, the meiotic process may be abnormal resulting in non-viable gametes.
e. There may be self-incompatibility.
f. There may be internal barriers for pollen tube growth and/or fertilisation.

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NCERT Exemplar Class 8 Maths Chapter 5 Understanding Quadrilaterals and Practical Geometry are part of NCERT Exemplar Class 8 Maths. Here we have given NCERT Exemplar Class 8 Maths Chapter 5 Understanding Quadrilaterals and Practical Geometry.

NCERT Exemplar Class 8 Maths Chapter 5 Understanding Quadrilaterals and Practical Geometry

Multiple Choice Questions
Question. 1 If three angles of a quadrilateral are each equal to 75°, then, the fourth angle is(a) 150° (b) 135°
(c) 45° (d) 75°
Solution.

Question. 2 For which of the following, diagonals bisect each other?
(a) Square (b) Kite
(c) Trapezium (d) Quadrilateral
Solution. (a) We know that, the diagonals of a square bisect each other but the diagonals of kite, trapezium and quadrilateral do not bisect each other.

Question. 3 In which of the following figures, all angles are equal?
(a) Rectangle (b) Kite
(c) Trapezium (d) Rhombus
Solution. (a) In a rectangle, all angles are equal, i.e. all equal to 90°.

Question. 4 For which of the following figures, diagonals are perpendicular to each other?
(a) Parallelogram (b) Kite
(c) Trapezium (d) Rectangle
Solution. (b) The diagonals of a kite are perpendicular to each other.

Question. 5 For which of the following figures, diagonals are equal?
(a) Trapezium (b) Rhombus
(c) Parallelogram (d) Rectangle
Solution. (d) By the property of a rectangle, we know that its diagonals are equal.

Question. 6 Which of the following figures satisfy the following properties?
All sides are congruent
All angles are right angles.
Opposite sides are parallel.

Solution. (c) We know that all the properties mentioned above are related to square and we can observe that figure R resembles a square.

Question. 7 Which of the following figures satisfy the following property?Has two pairs of congruent adjacent sides.

Solution. (c) We know that, a kite has two pairs of congruent adjacent sides and we can observe that figure R resembles a kite.

Question. 8 Which of the following figures satisfy the following property?
Only one pair of sides are parallel.

Solution. (a) We know that, in a trapezium, only one pair of sides are parallel and we can observe that figure P resembles a trapezium.

Question. 9 Which of the following figures do not satisfy any of the following properties?
All sides are equal.
All angles are right angles.
Opposite sides are parallel.

Solution. (a) On observing the above figures, we conclude that the figure P does not satisfy any of the given properties.

Question. 10 Which of the following properties describe a trapezium?
(a) A pair of opposite sides is parallel
(b) The diagonals bisect each other
(c) The diagonals are perpendicular to each other
(d) The diagonals are equal
Solution. (a) We know that, in a trapezium, a pair of opposite sides are parallel.

Question. 11 Which of the following is a propefay of a parallelogram?
(a) Opposite sides are parallel
(b) The diagonals bisect each other at right angles
(c) The diagonals are perpendicular to each other
(d) All angles are equal
Solution. (a) We,know that, in a parallelogram, opposite sides are parallel.

Question. 12 What is the maximum number of obtuse angles that a quadrilateral can have?
(a) 1 (b) 2
(c) 3 (d) 4
Solution. (c) We know that, the sum of all the angles of a quadrilateral is 360°.
Also, an obtuse angle is more than 90° and less than 180°.
Thus, all the angles of a quadrilateral cannot be obtuse.
Hence, almost 3 angles can be obtuse.

Question. 13 How many non-overlapping triangles can we make in a-n-gon (polygon having n sides), by joining the vertices?
(a)n-1 (b)n-2
(c) n – 3 (d) n – 4
Solution. (b) The number of non-overlapping triangles in a n-gon = n – 2, i.e. 2 less than the number of sides.

Question. 14 What is the sum of all the angles of a pentagon?
(a) 180° (b) 360° (c) 540° (d) 720°
Solution. (c) We know that, the sum of angles of a polygon is (n – 2) x 180°, where n is the number of sides of the polygon.
In pentagon, n = 5
Sum of the angles = (n – 2) x 180° = (5 – 2) x 180°
= 3 x 180°= 540°

Question. 15 What is the sum of all angles of a hexagon?
(a) 180° (b) 360° (c) 540° (d) 720°
Solution. (d) Sum of all angles of a n-gon is (n – 2) x 180°.
In hexagon, n = 6, therefore the required sum = (6 – 2) x 180° = 4 x 180° = 720°

Question. 16 If two adjacent angles of a parallelogram are (5x – 5) and (10x + 35), then the ratio of these angles is
(a) 1 : 3 (b) 2 : 3 (c) 1 : 4 (d) 1 : 2
Solution.

Question. 17 A quadrilateral whose all sides are equal, opposite angles are equal and
the diagonals bisect each other at-right angles is a .
(a) rhombus (b) parallelogram (c) square (d) rectangle
Solution. (a) We know that, in rhombus, all sides are equal, opposite angles are equal and diagonals bisect each other at right angles.

Question. 18 A quadrilateral whose opposite sides and all the angles are equal is a
(a) rectangle (b) parallelogram (c) square (d) rhombus
Solution. (a) We know that, in a rectangle, opposite sides and all the angles are equal.

Question. 19 A quadrilateral whose all sides, diagonals and angles are equal is a
(a) square (b) trapezium (c) rectangle (d) rhombus
Solution. (a) These are the properties of a square, i.e. in a square, all sides, diagonals and angles are equal.

Question. 20 How many diagonals does a hexagon have?
(a) 9 (b) 8 (c) 2 (d) 6
Solution.

Question. 21 If the adjacent sides of a parallelogram are equal, then parallelogram is a
(a) rectangle (b) trapezium (c) rhombus (d) square
Solution. (c)We know that, in a parallelogram, opposite sides are equal.
But according to the question, adjacent sides are also equal.
Thus, the parallelogram in which all the sides are equal is known as rhombus.

Question. 22 If the diagonals of a quadrilateral are equal and bisect each other, then the quadrilateral is a
(a) rhombus (b) rectangle (c) square (d) parallelogram
Solution. (b) Since, diagonals are equal and bisect each other, therefore it will be a rectangle.

Question. 23 The sum of all exterior angles of a triangle is
(a) 180° (b) 360° (c) 540° (d) 720°
Solution. (b) We know that the sum of exterior angles, taken in order of any polygon is 360° and triangle is also a polygon.
Hence, the sum of all exterior angles of a triangle is 360°.

Question. 24 Which of the following is an equiangular and equilateral polygon?
(a) Square (b) Rectangle (c) Rhombus (d) Right triangle
Solution. (a) In a square, all the sides and all the angles are equal.
Hence, square is an equiangular and equilateral polygon.

Question. 25 Which one has all the properties of a kite and a parallelogram?
(a) Trapezium (b) Rhombus (c) Rectangle (d) Parallelogram
Solution. (b) In a kite
Two pairs of equal sides.
Diagonals bisect at 90°.
One pair of opposite angles are equal.
In a parallelogram Opposite sides are equal.
Opposite angles are equal.
Diagonals bisect each other.
So, from the given options, all these properties are satisfied by rhombus.

Question. 26 The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. The smallest angle is
(a) 72° (b) 144° (c) 36° (d) 18°
Solution.

Question. 27 In the trapezium ABCD, the measure of \(\angle D\) is
(a) 55° (b) 115° (c)135° (d) 125°

Solution.

Question. 28 A quadrilateral has three acute angles. If each measures 80°, then the measure of the fourth angle is
(a) 150° (b) 120° (c) 105° (d) 140°
Solution.

Question. 29 The number of sides of a regular polygon where each exterior angle has a measure of 45° is
(a) 8 (b) 10 (c) 4 (d) 6
Solution.

Question. 30 In a parallelogram PQRS, if \(\angle P\) = 60°, then other three angles are
(a) 45°, 135°, 120° (b) 60°, 120°, 120°
(c) 60°, 135°, 135° (d) 45°, 135°, 135°
Solution.

Question. 31 If two adjacent angles of a parallelogram are in the ratio 2 : 3, then the measure of angles are
(a) 72°, 108° (b) 36°, 54° (c) 80°, 120° (d) 96°, 144°
Solution. (a) Let the angles be 2x and 3x.
Then, 2x + 3x = 180° [ adjacent angles of a parallelogram are supplementary]
=> 5x = 180°
=> x = 36°
Hence, the measures of angles are 2x = 2 x 36°= 72° and 3x = 3×36°= 108°

Question. 32 IfPQRS is a parallelogram then \(\angle P\) – \(\angle R\) is equal to
(a) 60° (b) 90° (c) 80° (d) 0°
Solution. (d) Since, in a parallelogram, opposite angles are equal. Therefore, \(\angle P\) – \(\angle R\) = 0, as \(\angle P\) and \(\angle R\) are opposite angles.

Question. 33 The sum of adjacent angles of a parallelogram is
(a) 180° (b) 120° (c) 360° (d) 90°
Solution. (a) By property of the parallelogram, we know that, the sum of adjacent angles of a parallelogram is 180°.

Question. 34 The angle between the two altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is 30°. The measure of the obtuse angle is
(a) 100° (b) 150° (c) 105° (d) 120°
Solution.

Question. 35 In the given figure, ABCD and BDCE are parallelograms with common base DC. If \(BC\bot BD\), then \(\angle BEC\) is equal to
(a) 60° (b) 30° (c) 150° (d) 120°

Solution.

Question. 36 Length of one of the diagonals of a rectangle whose sides are 10 cm and 24 cm is
(a) 25 cm (b) 20 cm (c) 26 cm (d) 3.5 cm
Solution.

Question. 37 If the adjacent angles of a parallelogram are equal, then the parallelogram is a (a) rectangle (b) trapezium (c) rhombus (d) None of these
Solution. (a) We know that, the adjacent angles of a parallelogram are supplementary, i.e. their sum equals 180° and given that both the angles are same. Therefore, each angle will be of measure 90°. .
Hence, the parallelogram is a rectangle.

Question. 38 Which of the following can be four interior angles of a quadrilateral?
(a) 140°, 40°, 20°, 160° (b) 270°, 150°, 30°, 20°
(c) 40°, 70°, 90°, 60°    (d) 110°, 40°, 30°, 180°
Solution. (a) We know that, the sum of interior angles of a quadrilateral is 360°.
Thus, the angles in option (a) can be four interior angles of a quadrilateral as their sum is 360°.

Question. 39 The sum of angles of a concave quadrilateral is
(a) more than 360° (b) less than 360°
(c) equal to 360° (d) twice of 360°
Solution. (c) We know that, the sum of interior angles of any polygon (convex or concave) having n sides is(n -2) x 180°.
.-.The sum of angles of a concave quadrilateral is (4 – 2) x 180°, i.e. 360°

Question. 40 Which of the following can never be the measure of exterior angle of a regular polygon? (a) 22° (b) 36° (c)45° (d) 30°
Solution. (a) Since, we know that, the sum of measures of exterior angles of a polygon is 360°, i.e. measure of each exterior angle =360°/n ,where n is the number of sides/angles.
Thus, measure of each exterior angle will always divide 360° completely.
Hence, 22° can never be the measure of exterior angle of a regular polygon.

Question. 41 In the figure, BEST is a rhombus, then the value of y – x is
(a) 40° (b) 50° (c) 20° (d) 10°

Solution.

Question. 42 The closed curve which is also a polygon, is

Solution. (a) Figure (a) is polygon as no two line segments intersect each other.

Question. 43 Which of the following is not true for an exterior angle of a regular polygon with n sides?

Solution. (d) We know that, (a) and (b) are the formulae to find the measure of each exterior angle, when number of sides and measure of an interior angle respectively are given and (c) is the formula to find number of sides of polygon when exterior angle is given.
Hence, the formula given in option (d) is not true for an exterior angle of a regular polygon with n sides.

Question. 44 PQRS is a square. PR and SQ intersect at 0. Then, \(\angle POQ\) is a (a) right angle (b) straight angle (c) reflex angle (d) complete angle
Solution.

Question. 45 Two adjacent angles of a parallelogram are in the ratio 1 : 5. Then, all the angles of the parallelogram are
(a) 30°, 150°, 30°, 150° (b) 85°, 95°, 85°, 95° .
(c) 45°, 135°, 45°, 135° (d) 30°, 180°, 30°, 180°
Solution. (a) Let the adjacent angles of a parallelogram be x and 5x, respectively.
Then, x + 5x = 180° [ adjacent angles of a parallelogram are supplementary] => 6x = 180°
=> x = 30°
The adjacent angles are 30° and 150°.
Hence, the angles are 30°, 150°, 30°, 150°

Question. 46 A parallelogram PQRS is constructed with sides QR = 6 cm, PQ = 4 cm and \(\angle PQR\) = 90°. Then, PQRS is a
(a) square (b) rectangle (c) rhombus  (d) trapezium
Solution. (b) We know that, if in a parallelogram one angle is of 90°, then all angles will be of 90° and a parallelogram with all angles equal to 90° is called a rectangle.

Question. 47 The angles P, Q, R and 5 of a quadrilateral are in the ratio  1:3 :7:9. Then, PQRS is a
(a) parallelogram  (b) trapezium with PQ \\ RS
(c) trapezium with  QR \\PS (d) kite
Solution.

Question. 48 PQRS is a trapezium in which PQ || SR and ZP = 130°, \(\angle Q\) = 110°. Then, \(\angle R\) is equal to.
(a) 70° (b) 50° (c)65° (d) 55°
Solution.

Question. 49 The number of sides of a regular polygon whose each interior angle is of 135° is (a) 6  (b) 7 (c) 8 (d) 9
Solution.

Question. 50 If a diagonal of a quadrilateral bisects both the angles, then it is a
(a) kite (b) parallelogram  (c) rhombus (d) rectangle
Solution. (c) If a diagonal of a quadrilateral bisects both the angles, then it is a rhombus.

Question. 51 To construct a unique parallelogram, the minimum number of measurements required is (a) 2 (b) 3 (c) 4 (d) 5
Solution. (b) We know that, in a parallelogram, opposite sides are equal and parallel. Also,
opposite angles are equal.
So, to construct a parallelogram uniquely, we require the measure of any two non-parallel sides and the measure of an angle.
Hence, the minimum number of measurements required to draw a unique parallelogram is 3.

Question. 52 To construct a unique rectangle, the minimum number of measurements required is (a) 4 (b) 3 (0 2 (d) 1
Solution. (c) Since, in a rectangle, opposite sides are equal and parallel, so we need the measurement of only two adjacent sides, i.e. length and breadth. Also, each angle measures 90°.
Hence, we require only two measurements to construct a unique rectangle.

Fill in the Blanks
In questions 53 to 91, fill in the blanks to make the statements true.
Question. 53 In quadrilateral HOPE, the pairs of opposite sides are————–.
Solution.

Question. 54 In quadrilateral ROPE, the pairs of adjacent angles are—————-.
Solution .

Question. 55 In quadrilateral WXYZ, the pairs of opposite angles are————–.
Solution.

Question . 56 The diagonals of the quadrilateral DEFG are———–and————–.
Solution.

Question. 57 The sum of all———— of a quadrilateral is 360°.
Solution. angles
We know that, the sum of all angles of a quadrilateral is 360°.

Question. 58 The measure of each exterior angle of a regular pentagon is————— .
Solution.

Question. 59 Sum of the angles of a hexagon is———————-.
Solution.

Question. 60 The measure of each exterior angle of a regular polygon of 18 sides is———.
Solution.

Question. 61 The number of sides of a regular polygon, where each exterior angle has a measure of 36°, is—————-.
Solution.

Question. 62

Solution. concave polygon
As one interior angle is of greater than 180°.

Question. 63 A quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure is—————–.
Solution. kite
By the property of a kite, we know that, it has two opposite angles of equal measure.

Question. 64 The measure of each angle of a regular pentagon is————–.
Solution.

Question. 65 The name of three-sided regular polygon is—————-.
Solution. equilateral triangle, as polygon is regular, i.e. length of each side is same.

Question. 66 The number of diagonals in a hexagon is—————-.
Solution.

Question. 67 A polygon is a simple closed curve made up of only————.
Solution. line segments ,
Since a simple closed curve made up of only line segments is called a polygon.

Question. 68 A regular polygon is a polygon whose all sides are equal and all———are equal.
Solution. angles
In a regular polygon, all sides are equal and all angles are equal.

Question. 69 The sum of interior angles of a polygon of n sides is———- right angles.
Solution.

Question. 70 The sum of all exterior angles of a polygon is————.
Solution. 360°
As the sum of all exterior angles of a polygon is 360°.

Question. 71 ————-is a regular quadrilateral.
Solution. Square
Since in square, all the sides are of equal length and all angles are equal.

Question. 72 A quadrilateral in which a pair of opposite sides is parallel is————-.
Solution. trapezium
We know that, in a trapezium, one pair of sides is parallel.

Question. 73 If all sides of a quadrilateral are equal, it is a————–.
Solution. rhombus or square
As in both the quadrilaterals all sides are of equal length.

Question. 74 In a rhombus, diagonals intersect at———– angles.
Solution. right
The diagonals of a rhombus intersect at right angles.

Question. 75 ———measurements can determine a quadrilateral uniquely.
Solution. 5
To construct a unique quadrilateral, we require 5 measurements, i.e. four sides and one angle or three sides and two included angles or two adjacent sides and three angles are given.

Question. 76 A quadrilateral can be constructed uniquely, if its three sides and———–angles are given.
Solution. two included
We cap determine a quadrilateral uniquely, if three sides and two included angles are given.

Question. 77 A rhombus is a parallelogram in which————sides are equal.
Solution. all
As length of each side is same in a rhombus.

Question. 78 The measure of——– angle of concave quadrilateral is more than 180°.
Solution. one
Concave polygon is a polygon in which at least one interior angle is more than 180°.

Question. 79 A diagonal of a quadrilateral is a line segment that joins two——– vertices of the quadrilateral.
Solution. opposite
Since the line segment connecting two opposite vertices is called diagonal.

Question. 80 The number of sides in a regular polygon having measure of an exterior angle as 72° is————— .
Solution. 5
We know that,the sum of exterior angles of any polygon is 360°.

Question. 81 If the diagonals of a quadrilateral bisect each other, it is a————.
Solution. parallelogram
Since in a parallelogram, the diagonals bisect each other.

Question. 82 The adjacent sides of a parallelogram are 5 cm and 9 cm. Its perimeter is—–.
Solution. 28 cm
Perimeter of a parallelogram = 2 (Sum of lengths of adjacent sides)
=2(5+ 9) = 2 x 14=28cm

Question. 83 A nonagon has————sides.
Solution. 9
Nonagon is a polygon which has 9 sides.

Question. 84 Diagonals of a rectangle are————.
Solution. equal
We know that, in a rectangle, both the diagonals are of equal length.

Question. 85 A polygon having 10 sides is known as————.
Solution. decagon
A polygon with 10 sides is called decagon.

Question. 86 A rectangle whose adjacent sides are equal becomes a ————.
Solution. square
If in a rectangle, adjacent sides are equal, then it is called a square.

Question. 87 If one diagonal of a rectangle is 6 cm long, length of the other diagonal is—–.
Solution. 6 cm
Since both the diagonals of a rectangle are equal. Therefore, length of other diagonal is also 6 cm.

Question. 88 Adjacent angles of a parallelogram are————.
Solution. supplementary
By property of a parallelogram, we know that, the adjacent angles of a parallelogram are supplementary.

Question. 89 If only one diagonal of a quadrilateral bisects the other, then the quadrilateral is known as————.
Solution. kite
This is a property of kite, i.e. only one diagonal bisects the other.

Question. 90 In trapezium ABCD with AB || CD, if \( \angle A\)= 100°, then \( \angle D\) =————.
Solution.

Question. 91 The polygon in which sum of all exterior angles is equal to the sum of interior angles is called————.
Solution. quadrilateral
We know that, the sum of exterior angles of a polygon is 360° and in a quadrilateral, sum of interior angles is also 360°. Therefore, a quadrilateral is a polygon in which the sum of both interior and exterior angles are equal.

True/False
In questions 92 to 131, state whether the statements are True or False.
Question. 92 All angles of a trapezium are equal.
Solution. False
As all angles of a trapezium are not equal.

Question. 93 All squares are rectangles.
Solution. True
Since squares possess all the properties of rectangles. Therefore, we can say that, all squares are rectangles but vice-versa is not true.

Question. 94 All kites are squares.
Solution. False
As kites do not satisfy all the properties of a square.
e.g. In square, all the angles are of 90° but in kite, it is not the case.

Question. 95 All rectangles are parallelograms.
Solution. True
Since rectangles satisfy all ”the”properties” of parallelograms. Therefore, we can say that, all rectangles are parallelograms but vice-versa is not true.

Question. 96 All rhombuses are square.
Solution. False
As in a rhombus, each angle is not a right angle, so rhombuses are not squares.

Question. 97 Sum of all the angles of a quadrilateral is 180°.
Solution. False
Since sum of all the angles of a quadrilateral is 360°.

Question. 98 A quadrilateral has two diagonals.
Solution. True
A quadrilateral has two diagonals.

Question. 99 Triangle is a polygon whose sum of exterior angles is double the sum of interior angles.
Solution. True
As the sum of interior angles of a triangle is 180° and the sum of exterior angles is 360°, i.e. double the sum of interior angles.

Question. 100

Solution. False
Because it is not a simple closed curve as it intersects with itself more than once.

Question. 101 A kite is not a convex quadrilateral.
Solution. False
A kite is a convex quadrilateral as the line segment joining any two opposite vertices inside it, lies completely inside it.

Question. 102 The sum of interior angles and the sum of exterior angles taken in an order are equal in case of quadrilaterals only.
Solution. True
Since the sum of interior angles as well as of exterior angles of a quadrilateral are 360°.

Question. 103 If the sum of interior angles is double the sum of exterior angles taken in an order of a polygon, then it is a hexagon.
Solution. True
Since the sum of exterior angles of a hexagon is 360° and the sum of interior angles of a hexagon is 720°, i.e. double the sum of exterior angles.

Question. 104 A polygon is regular, if all of its sides are equal.
Solution. False
By definition of a regular polygon, we know that, a polygon is regular, if all sides and all angles are equal.

Question. 105 Rectangle is a regular quadrilateral.
Solution. False
As its all sides are not equal.

Question. 106 If diagonals of a quadrilateral are equal, it must be a rectangle.
Solution. True
If diagonals are equal, then it is definitely a rectangle. –

Question. 107 If opposite angles of a quadrilateral are equal, it must be a parallelogram.
Solution. True
If opposite angles are equal, it has to be a parallelogram.

Question. 108 The interior angles of a triangle are in the ratio 1:2:3, then the ratio of its exterior angles is 3 : 2 : 1.
Solution.

Question. 109

Solution. False
As it has 6 sides, therefore it is a concave hexagon.

Question. 110 Diagonals of a rhombus are equal and perpendicular to each other.
Solution. False
As diagonals of a rhombus are perpendicular to each other but not equal.

Question. 111 Diagonals of a rectangle are equal.
Solution. True
The diagonals of a rectangle are equal.

Question. 112 Diagonals of rectangle bisect each other at right angles.
Solution. False
Diagonals of a rectangle does not bisect each other.

Question. 113 Every kite is a parallelogram.
Solution. False
Kite is not a parallelogram as its opposite sides are not equal and parallel.

Question. 114 Every trapezium is a parallelogram.
Solution. False
Since in a trapezium, only one pair of sides is parallel.

Question. 115 Every parallelogram is a rectangle.
Solution. False
As in a parallelogram, all angles are not right angles, while in a rectangle, all angles are equal and are right angles.

Question. 116 Every trapezium is a rectangle.
Solution. False
Since in a rectangle, opposite sides are equal and parallel but in a trapezium, it is not so.

Question. 117 Every rectangle is a trapezium.
Solution. True
As a rectangle satisfies all the properties of a trapezium. So, we can say that, every rectangle is a trapezium but vice-versa is not true.

Question. 118 Every square is a rhombus.
Solution. True
As a square possesses all the properties of a rhombus. So, we can say that, every square is a rhombus but vice-versa is not true.

Question. 119 Every square is a parallelogram.
Solution. True
Every square is also a parallelogram as it has all the properties of a parallelogram but vice-versa is not true.

Question. 120 Every square is a trapezium.
Solution. True
As a square has all the properties of a trapezium. So, we can say that, every square is a trapezium but vice-versa is not true.

Question. 121 Every rhombus is a trapezium.
Solution. True
Since a rhombus satisfies all the properties of a trapezium. So, we can say that, every rhombus is a trapezium but vice-versa is not true.

Question. 122 A quadrilateral can be drawn if only measures of four sides are given.
Solution. False
As we require at least five measurements to determine a quadrilateral uniquely.

Question. 123 A quadrilateral can have all four angles as obtuse.
Solution. False
If all angles will be obtuse, then their sum will exceed 360°. This is not possible in case of a quadrilateral.

Question. 124 A quadrilateral can be drawn, if all four sides and one diagonal is known.
Solution. True
A quadrilateral can be constructed uniquely, if four sides and one diagonal is known.

Question. 125 A quadrilateral can be drawn, when all the four angles and one side is given.
Solution. False
We cannot draw a unique-quadrilateral, if four angles and one side is known.

Question. 126 A quadrilateral can be drawn, if all four sides and one angle is known.
Solution. True
A quadrilateral can be drawn, if all four sides and one angle is known.

Question. 127 A quadrilateral can be drawn, if three sides and two diagonals are given.
Solution. True
A quadrilateral can be drawn, if three sides and two diagonals are given.

Question. 128 If diagonals of a quadrilateral bisect each other, it must be a parallelogram.
Solution. True
It is the property of a parallelogram.

Question. 129 A quadrilateral can be constructed uniquely, if three angles and any two included sides are given.
Solution. True
We can construct a unique quadrilateral with given three angles given and two included sides.

Question. 130 A parallelogram can be constructed uniquely, if both diagonals and the angle between them is given.
Solution. True
We can draw a unique parallelogram, if both diagonals and the angle between them is given.

Question. 131 A rhombus can be constructed uniquely, if both diagonals are given.
Solution. True
A rhombus can be constructed uniquely, if both diagonals are given.

Question. 132 The diagonals of a rhombus are 8 cm and 15 cm. Find its side.
Solution.

Question. 133 Two adjacent angles of a parallelogram are in the ratio 1 : 3. Find its angles.
Solution. Let the adjacent angles of a parallelogram be x and 8c.
Then, we have x + (3 x) = 180° [adjacent angles of parallelogram are supplementary]
=> 4 x = 180°
=> x = 45°
Thus, the angles are 45°, 135°.
Hence, the angles are 45°, 135, 45°, 135°. [ opposite angles in a parallelogram are equal]

Question. 134 Of the four quadrilaterals – square, rectangle, rhombus and trapezium-one is somewhat different from the others because of its design. Find it and give justification.
Solution. In square, rectangle and rhombus, opposite sides are parallel and equal. Also, opposite angles are equal, i.e. they all are parallelograms.
But in trapezium, there is only one pair of parallel sides, i.e. it is not a parallelogram. Therefore, trapezium has different design.

Question. 135 In a rectangle ABCD, AB = 25 cm and BC = 15 cm. In what ratio, does the bisector of \(\angle C\) divide AB?
Solution.

Question. 136 PQRS is a rectangle. The perpendicular ST from S on PR divides \(\angle S\) in the ratio 2 : 3. Find \(\angle TPQ\).
Solution.

Question. 137 A photo frame is in the shape of a quadrilateral, with one diagonal longer than the other. Is it a rectangle? Why or why not?
Solution. No, it cannot be a rectangle, as in rectangle, both the diagonals are of equal lengths.

Question. 138 The adjacent angles of a parallelogram are (2x – 4)° and (3x – 1)°. Find the measures of all angles of the parallelogram.
Solution. Since, the adjacent angles of a parallelogram are supplementary.
(2 x – 4)° + (3x – 1)° = 180°

Question. 139 The point of intersection of diagonals of a quadrilateral divides one diagonal in the ratio 1: 2. Can it be a parallelogram? Why or why not?
Solution. No, it can never be a parallelogram, as the diagonals of a parallelogram intersect each other in the ratio 1 : 1.

Question. 140 The ratio between exterior angle and interior angle of a regular polygon is 1 : 5. Find the number of sides of the polygon.
Solution.

Question. 141 Two sticks each of length 5 cm are crossing each other such that they bisect each other. What shape is formed by joining their end points? Give reason.
Solution. Sticks can be taken as the diagonals of a quadrilateral.
Now, since they are bisecting each other, therefore the shape formed by joining their end points will be a parallelogram.
Hence, it may be a rectangle or a square depending on the angle between the sticks.

Question. 142 Two sticks each of length 7 cm are crossing each other such that they bisect each other at right angles. What shape is formed by joining their end points? Give reason.
Solution. Sticks can be treated as the diagonals of a quadrilateral.
Now, since the diagonals (sticks) are bisecting each other at right angles, therefore the shape formed by joining their end points will be a rhombus.

Question. 143 A playground in the town is in the form of a kite. The perimeter is 106 m. If one of its sides is 23 m, what are the lengths of other three sides?
Solution. Let the length of other non-consecutive side be x cm.
Then, we have, perimeter of playground = 23 + 23+ x + x
=> 106 = 2 (23+ x)
=>46 + 2x = 106 2x = 106 – 46
=>2x = 60
=>x = 30 m
Hence, the lengths of other three sides are 23m, 30m and 30m. As a kite has two pairs of equal consecutive sides.

Question. 144 In rectangle READ , find \(\angle EAR\), \(\angle RAD\) and \(\angle ROD\).

Solution.

Question. 145 In rectangle PAIR, find \(\angle ARI\), ZRMI and \(\angle PMA\).

Solution.

Question. 146 In parallelogram ABCD, find \(\angle B\), \(\angle C\) and \(\angle D\).

Solution.

Question. 147 In parallelogram PQRS, 0 is the mid-point of SQ. Find \(\angle S\), \(\angle R\), PQ, QR and diagonal PR.

Solution.

Question. 148 In rhombus BEAM, find \(\angle AME\) and \(\angle AEM\).

Solution.

Question. 149 In parallelogram FIST, find \(\angle SFT\), \(\angle OST\) and \(\angle STO\).

Solution.

Question. 150 In the given parallelogram YOUR, \(\angle RUO\)= 120° and 0Y is extended to points, such that \(\angle SRY\) = 50°. Find \(\angle YSR\).

Solution.

Question.151 In kite WEAR, \(\angle WEA\) = 70° and \(\angle ARW\) = 80°. Find the remaining two angles.

Solution.

Question.152

Solution.

Question.153 In parallelogram LOST, SNLOL and \( SM\bot LT\). Find \(\angle STM\), \(\angle SON\) and \(\angle NSM\).

Solution.

Question. 154 In trapezium HARE, EP and RP are bisectors of \(\angle E\) and \(\angle R\), respectively. Find \(\angle HAR\) and \(\angle EHA\).

Solution.

Question. 155 In parallelogram MODE, the bisectors of \(\angle M\) and \(\angle O\) meet at Q. Find the measure of \(\angle MQO\).
Solution.

Question. 156 A playground is in the form of a rectangle ATEF. Two players are standing at the points F and B, where EF =EB. Find the values of x and y.

Solution.

Question. 157 In the following figure of a ship, ABDH and CEFG are two parallelograms. Find the value of x.

Solution.

Question. 158 A rangoli has been drawn on the floor of a house. ABCD and PQRS both are in the shape of a rhombus. Find the radius of semi-circle drawn on each side of rhombus ABCD.

Solution.

Question. 159 ABCDE is a regular pentagon. The bisector of angle A meets the sides CD at M. Find \(\angle AMC\)

Solution.

Question. 160 Quadrilateral EFGH is a rectangle in which J is the point of intersection of the diagonals. Find the value of x, if JF = 8x + 4 and EG = 24 x – 8.
Solution.

Question. 161 Find the values of x and y in the following parallelogram.

Solution.

Question. 162 Find the values of x and y in the following kite.

Solution.

Question. 164 Two angles of a quadrilateral are each of measure 75° and the other two angles are equal. What is the measure of these two angles? Name the possible figures so formed.
Solution.

Question. 165 In a quadrilateral PQRS, \(\angle P\) = 50°, \(\angle Q\) = 50°, \(\angle R\) = 60°. Find \(\angle S\). Is this quadrilateral convex or concave?
Solution.

Question. 166 Both the pairs of opposite angles of a quadrilateral are equal and supplementary. Find the measure of each angle.
Solution.

Question. 167 Find the measure of each angle of a regular octagon.
Solution.

Question. 168 Find the measure of an exterior angle of a regular pentagon and an exterior angle of a regular decagon. What is the ratio between these two angles?
Solution.

Question. 169 In the figure, find the value of x.

Solution.

Question. 170 Three angles of a quadrilateral are equal. Fourth angle is of measure 120°. What is the measure of equal angles?
Solution.

Question. 171 In a quadrilateral HOPE, PS and ES are bisectors of \(\angle P\) and \(\angle E\) respectively. Give reason.
Solution. Data insufficient.

Question. 172 ABCD is a parallelogram. Find the values of x, y and z.

Solution.

Question. 173 Diagonals of a quadrilateral are perpendicular to each other. Is such a quadrilateral always a rhombus? Give a figure to justify your answer.
Solution.

Question. 174 ABCD is a trapezium such that AB || CD, \(\angle A\): \(\angle D\) = 2:1, \(\angle B\) : \(\angle C\) = 7:5. Find the angles of the trapezium.
Solution.

Question. 175 A line / is parallel to Line m and a-transversal p intersects them at X, Y respectively. Bisectors of interior angles at X and Y intersect at P and Q. Is PXQY a rectangle? Give reason.
Solution.

Question. 176 ABCD is a parallelogram. The bisector of angle A intersects CD at X and bisector of angle C intersects AB at Y. Is AXCY a parallelogram? Give reason.
Solution.

Question. 177 A diagonal of a parallelogram bisects an angle. Will it also bisect the other angle? Give reason.
Solution. Consider a parallelogram ABCD.
Given, \(\angle 1\) = \(\angle 2\)

Question. 178 The angle between the two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 45°. Find the angles of the parallelogram.
Solution. Let ABCD be a parallelogram, where BE and BF are the perpendiculars through the vertex B to the sides DC and AD, respectively.

Question. 179 ABCD is a rhombus such that the perpendicular bisector of AB passes through D. Find the angles of the rhombus.[Hint Join BD. Then, AABD is equilateral.]
Solution. Let ABCD be a rhombus in which DE is perpendicular bisector of AB.

Question. 180 ABCD is a parallelogram. Point P and Q are taken on the sides AB and AD, respectively and 4he parallelogram PRQA is formed. If \(\angle C\)= 45°, find \(\angle R\).
Solution.

Question. 181 In parallelogram ABCD, the angle bisector of \(\angle A\) bisects BC. Will angle bisector of B also bisect AD? Give reason.
Solution.

Question. 182 A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find \(\angle BCF\)?
Solution.

Question. 183 Find maximum number of acute angles which a convex quadrilateral, a pentagon and a hexagon can have. Observe the pattern and generalise the result for any polygon.
Solution. If an angle is acute, then the corresponding exterior angle is greater than 90°. Now, suppose a convex polygon has four or more acute angles. Since, the polygon is convex, all the exterior angles are positive, so the sum of the exterior angle is at least the sum of the interior angles. Now, supplementary of the four acute angles, which is greater than 4 x 90° = 360°
However, this is impossible. Since, the sum of exterior angle of a polygon must equal to 360° and cannot be greater than it. It follows that the maximum number of acute angle in convex polygon is 3.

Question. 184 In the following figure, FD || BC || AE and AC || ED. Find the value of x.

Solution.

Question. 185 In the following figure, AB || DC and AD = BC. Find the value of x.

Solution.

Question. 186 Construct a trapezium ABCD in which AB || DC, \(\angle A\) = 105°, AD = 3 cm, AB = 4 cm and CD = 8 cm.
Solution.

Question. 187 Construct a parallelogram ABCD in which AB =4 cm, BC = 5cm and \(\angle B\) = 60°.
Solution.

Question. 188 Construct a rhombus whose side is 5 cm and one angle is of 60°
Solution.

Question. 189 Construct a rectangle whose one side is 3 cm and a diagonal is equal to 5 cm.
Solution.

Question. 190 Construct a square of side 4 cm.
Solution.

Question. 191 Construct a rhombus CLUE in which CL = 7.5 cm and LE = 6 cm.
Solution.

Question. 192 Construct a quadrilateral BEAR in which BE = 6 cm, EA = 7 cm, RB = RE = 5 cm and BA = 9 cm. Measure its fourth side.
Solution.

Question. 193 Construct a parallelogram POUR in which PO = 5.5 cm, OU = 7.2 cm and \(\angle O\) = 70°.

Question. 194 Draw a circle of radius 3 cm and draw its diameter and label it as AC. Construct its perpendicular bisector and let it intersect the circle at B and D. What type of quadrilateral is ABCD? Justify your answer.
Solution.

Question. 195 Construct a parallelogram HOME with HO = 6 cm, HE = 4 cm and OE = 3 cm.
Solution.

Question. 196 Is it possible to construct a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 5.4 cm, DA = 5.9 cm and diagonal AC = 8 cm? If not, why?
Solution. No,
Given measures are AS = 3 cm, SC = 4 cm,CD = 5.4 cm,
DA = 59cmand AC = 8cm
Here, we observe that AS + SC = 3 + 4 = 7 cm and AC = 8 cm
i.e. the sum of two sides of a triangle is less than the third side, which is absurd.
Hence, we cannot construct such a quadrilateral.

Question. 197 Is it possible to construct a quadrilateral ROAM in which RO = 4 cm, OA = 5 cm, \(\angle O\) = 120°,\(\angle R\) = 105° and \(\angle A\) = 135°? If not, why?
Solution.

Question. 198 Construct a square in which each diagonal is 5 cm long.
Solution.

Question. 199 Construct a quadrilateral NEWS in which NE = 7 cm, EW = 6 cm, \(\angle N\) = 60°, \(\angle E\)= 110° and \(\angle S\) = 85°
Solution.

Question. 200 Construct a parallelogram when one of its side is 4 cm and its two diagonals are 5.6 cm and 7 cm. Measure the other side.
Solution.

Question. 201 Find the measure of each angle of a regular polygon of 20 sides?
Solution.

Question. 202 Construct a trapezium RISK in which RI || KS, RI = 7 cm, IS = 5 cm, RK = 6.5 cm and \(\angle I\) = 60°.
Solution.

Question. 203 Construct a trapezium ABCD, where AB|| CD, AD = BC = 3.2 cm, AB = 6.4 cm and CD = 9.6 cm. Measure \(\angle B\) and \(\angle A\)

[Hint Difference of two parallel sides gives an equilateral triangle.]
Solution.

NCERT Exemplar Solutions

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NCERT Exemplar Class 12 Physics Chapter 9 Ray Optics and Optical Instruments are part of NCERT Exemplar Class 12 Physics. Here we have given NCERT Exemplar Class 12 Physics Chapter 9 Ray Optics and Optical Instruments. https://www.cbselabs.com/ncert-exemplar-problems-class-12-physics-ray-optics-optical-instruments/

NCERT Exemplar Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. A ray of light incident at an angle d on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is mad? of a material of refractive index 1.5, the angle of incidence is
(a) 7.5° (b) 5° (c) 15° (d) 2.5°

Question 2. A‘ short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is
(a) blue (b) green (c) violet (d) red
Solution: (d) As velocity of wave is given by the relation v = f λ . When light ray goes from one medium to other medium, the frequency of light remains unchanged. Hence v ∝ λ or greater the wavelength, greater the speed.
The light of red colour is of highest wavelength and therefore of highest speed. Therefore, after travelling through the slab, the red colour emerges first.

Question 3. An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image
(a) moves away from the lens with an uniform speed 5 m/s
(b) moves away from the lens with an uniform acceleration
(c) moves away from the lens with a non-uniform acceleration
(d) moves towards the lens with a non-uniform acceleration
Solution: (c)
In our problem the object approaches a convergent lens from the left of the lens with a uniform speed of 5 m/s, hence the image will move away from the lens with a non-uniform acceleration, the image moves slower in the beginning and faster later on will move from F to 2F and when the object moves from 2F to F,the image will move from 2F to infinity. At 2F, the speed of the object and image will be equal.

Question 4. A passenger in an aeroplane shall
(a) never see a rainbow
(b) may see a primary and a secondary rainbow as concentric circles
(c) may see a primary and a secondary rainbow as concentric arcs
(d) shall never see a secondary rainbow
Solution: (b) As aeroplane is at higher altitude, the passenger in an aeroplane may see a primary and a secondary rainbow like concentric circles.

Question 5. You are given four sources of light each one providing a light of a single colour—red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?
(a) The beam of red light would undergo total internal reflection.
(b) The beam of red light would bend towards the normal while it gets refracted through the second medium.
(c) The beam of blue light would undergo total internal reflection.
(d) The beam of green light would bend away from the normal as it gets refracted through the second medium.
Solution: (c)

Question 6. The radius of curvature of the curved surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will
(a) act as a convex lens only for the objects that lie on its curved side
(b) act as a concave lens for the objects that lie on its curved side
(c) act as a convex lens irrespective of the side on which the object lies
(d) act as a concave lens irrespective of side on which the object lies

Question 7. The phenomena involved in the reflection of radio waves by ionosphere is similar to
(a) reflection of light by a plane mirror
(b) total internal reflection of light in air during a mirage
(c) dispersion of light by water molecules during the formation of a rainbow
(d) scattering of light by the particles of air
Solution: (b) Radio waves are reflected by a layer of atmosphere called the Ionosphere, so they can reach distant parts of the Earth. The reflection of radio waves by ionosphere is due to total internal reflection. It is the same as total internal reflection of light in air during a mirage, i.e., angle of incidence is greater than critical angle.
Important point: The ionized part of the Earth’s atmosphere is known as the ionosphere. Ultraviolet light from the sun collides with atoms in this region knocking electrons loose. The creates ions, or atoms with missing electrons. This is what gives the Ionosphere its name- and it is the free electrons that cause the reflection and absorption of ratio waves.

Question 8. The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (figure). Which of the four rays correctly shows the direction of reflected ray?

(a) 1     (b) 2
(c) 3     (d) 4
Solution: (b) The ray PQ of light passes through focus F and incident on the concave mirror, after reflection, should become parallel to the principal axis and shown by ray 2 in the figure.
Important points:
We can locate the image of any extended object graphically by drawing any two of the following four special rays:
1. A ray initially parallel to the principal axis is reflected through the focus of the mirror (1).
2. A ray passing through the center of curvature is reflected back along itself (3).
3. A ray initially passing through the focus is reflected parallel to the principal axis (2).
4. A ray incident at the pole is reflected symmetrically.

Question 9. The optical density of turpentine is higher than that of water while its mass density is lower. Figure shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in figure, the path shown is correct?

(a) 1 (b) 2 (c) 3 (d) 4
Solution: (b)
Here, light ray goes from (optically) rarer medium air to optically denser medium turpentine, then it bends towards the normal, i.e., θ1 > θ2 whereas when it goes from to optically denser medium turpentine to rarer medium water, then it bends away the normal.

Question 10. A far is moving with a constant speed of 60 km h^-1 on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of 5 km h^-1.
In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct? 
(a) The speed of the car in the rear is 65 km h^-1
(b) In the side mirror, the car in the-rear would appear to approach with a speed of 5 km h-1 to the driver of the leading car
(c) In the rear view mirror, the speed of the approaching car would appear to decrease as the distance between the cars decreases
(d) In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases
Solution: (d)

Question 11. There are certain material developed in laboratories which have a negative refractive index figure. A ray incident from air (Medium 1) into such a medium (Medium 2) shall follow a path given by

Solution: (a) The materials with negative refractive index responds to Snell’s law just opposite way. If incident ray from air (Medium 1) incident on those material, the ray refract or bend same side of the normal as in option (a).

One or More Than One Correct Answer Type

Question 12. Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because
(a) the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge
(b) the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air
(c) some of the points of the object far away from the edge may not be visible because of total internal reflection
(d) water in a trough acts as a lens and magnifies the object
Solution: (a, b, c)
Key concept: The light from the pencil is refracted when it passes from the water into air, bending away from the normal as it moves from high to low refractive index.

When light from the submerged object before reaching to the observer gets, refracted from water surface, the rays bend away from normal and the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air. Also the apparent depth of the .points close to the edge are nearer the surface of the water compared to the points away from the edge.
As we move towards right, the angle of incident increases and becomes equal to critical angle. Hence some of the points of the object far away from the edge may not be visible because of total internal reflection.

Question 13. A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB of figure. When observed from the face AD, the pin shall

(a) appear to be near A
(b) appear to be near D
(c) appear to be at the centre of AD
(d) not be seen at all
Solution: (a, d) As long as angle of incidence on AD of the ray emanating from pin is less than the critical angle, the pin shall appear to be near A.

When angle of incidence on AD of the ray emanating from pins is greater than the critical angle, the light suffers from total internal reflection and cannot be seen through AD.

Question 14. Between the primary and secondary rainbow, these is a dark band known as Alexandar’s dark band. This is because
(a) light scattered into this region interfere destructively
(b) there is no light scattered into this region
(c) light is absorbed in this region
(d)angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately 42° and 50°
Solution: (a, d) The Alexandar’s dark band lies between the primary and secondary rainbows, formed due to light scattered into this region interfere destructively. The primary rainbows subtends an angle nearly 41° to 42° at observer’s eye, whereas secondary rainbows subtends an angle nearly 51° to 54° at observer’s eye w.r.t. incident light ray.
Hence, the scattered rays with respect to the incident light of the sun lies between approximately 42° and 50°.

Question 15. A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in
(a) a larger angle to be subtended by the object at the eye and hence, viewed in greater detail
(b) the formation of a virtual erect image
(c) increase in the field of view
(d) infinite magnification at the near point
Solution: (a, b)
Key concept: A magnifying glass is a single convex lens of lesser focal length.

When a magnifying glass is used, the object to be viewed can be brought closer to the eye than the normal near point. This results in a larger angle to be subtended by the object at the eye and hence, viewed in greater detail. Moreover, the formation of a virtual erect and enlarged image takes place.

Question 16. An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length 2 cm.
(a) The length of the telescope tube is 20.02 m
(b) The magnification is 1000
(c) The image formed is inverted
(d) An objective of a larger aperture will increase the brightness and reduce chromatic aberration of the image

Very Short Answer Type Questions

Question 17. Will the focal length of a lens for red light be more, same or less than that for blue light?
Solution:

Question 18. The near vision of an average person is 25 cm. To view an object with an angular magnification of 10, what should be the power of the microscope?
Solution:

Question 19. An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?
Solution:

Question 20. Three immiscible liquids of densities d₁> d₂> d₃ and refractive indices μ₁> μ₂> μ₃ are put in a beaker. The height of each liquid column is h/3. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.
Solution:

Question 21. For a glass prism (μ= √3 ), the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.
Solution:

Short Answer Type Questions

Question 22. A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f what will be the length of the image? You may take L << |v – f |.
Solution:

Question 23. A circular disc of radius R is placed co-axially and horizontally inside an opaque hemispherical bowl of radius a (figure). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index μ. and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

Solution:

Question 24. thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at (0, 0) and an object is placed at (-50 cm, 0). Find the coordinates of the image.
Solution:

Question 25. In many experimental set-ups, the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.
Solution:
Key Concept: This is also one of the methods for finding focal length of the length in laboratory and knows as “Displacement method”.

Question 26. A jar of height h is filled with a transparent  liquid of refraction index μ (figure).At the center of the jar on the bottom surface is a dot.Find the minimum diameter of a disc,such that when placed on the top surface symmetrically about the center,the dot is invisible.

Solution:

Long Answer Type Questions

Question 27. A myopic adult has a far point at 0.1 m. His power of accommodation is 4 D.
(i) What power lenses are required to see distant objects?
(ii) What is his near point without glasses?
(iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)
Solution: Key concepts:

Question 28. Show that for a material with refractive index μ ≥ √2, light incident at angle shall be guided along, a length perpendicular to the incident face.    
Solution: Let the ray incident on face AB at angle i, after refraction, it travels along PQ and then interact with face AC which is perpendicular to the incident face.

Question 29. The mixture of a pure liquid and a solution in a long vertical column (i.e., horizontal dimensions << vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d << h, the height of the column.
Solution:

Question 30.

Solution:

Question 31. An infinitely long cylinder of radius R is made of an unusual exotic material with refractive index -1 (figure). The cylinder is placed between two planes whose normals are along the y-direction. The centre of the cylinder O lies along they-axis.

A narrow laser beam is directed along the y-direction from the lower plate.
The laser source is at a horizontal
distance x from the diameter in the y-direction. Find the range of x such that light emitted from the lower plane does not reach the upper plane.
Solution:

Question 32. (i) Consider a thin lens placed between a source (S) and an observer (O)
(Figure). Let the thickness of the lens vary as w(b) =w₀– b²/α , where b is the vertical distance from the pole, w₀ is a constant. Using Fermat’s principle, i.e., the time of transit fora ray between the source and observer is an extremum find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.

Solution:

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NCERT Exemplar Class 12 Physics Chapter 8 Electromagnetic Waves are part of NCERT Exemplar Class 12 Physics. Here we have given NCERT Exemplar Class 12 Physics Chapter 8 Electromagnetic Waves. https://www.cbselabs.com/ncert-exemplar-problems-class-12-physics-electromagnetic-waves/

NCERT Exemplar Class 12 Physics Chapter 8 Electromagnetic Waves

Multiple Choice Questions (MCQs)
Single Correct Answer Type

Question 1. One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in
(a) visible region (b) infrared region
(c) ultraviolet region (d) microwave region
Solution:

Question 2.

Question 3. Light with an energy flux of 20 W/cm² falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm², the total momentum delivered (for complete absorption) during 30 min is
(a) 36 x 10^-5 kg-m/s (b) 36 x 10^-4 kg-m/s
(c) 108 x 10⁴ kg-m/s (d) 1.08 x 10⁷ kg-m/s
Solution:

Question 4. The electric field intensity produced by the radiations coming from a 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is

Solution: 

Question 5.

Solution: (d)
Key concept: A changing electric field produces a changing magnetic field and vice versa which gives rise to a transverse wave known as electromagnetic wave. The time varying electric and magnetic field are mutually perpendicular to each other and also perpendicular to the direction of propagation of this wave. The electric vector is responsible for the optical effects of an EM wave and is called the light vector.

Question 6. The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is

Question 7. An EM wave radiates outwards from a dipole antenna, with E₀ as the amplitude of its electric field vector. The electric field E₀ which transports significant energy from the source falls off as

Solution: (c) A diode antenna radiates the electromagnetic waves outwards. The amplitude of electric field vector (E₀) which transports significant energy from the source falls intensity inversely as the distance (r) from the antenna,

One or More Than One Correct Answer Type
Question 8.


Solution: (a, d) We are given that the electric field vector of an electromagnetic wave travels in a vacuum along z-direction as,

Question 9.

Solution: (a, b, c)

Question 10. A plane electromagnetic wave propagating along x-direction can have the following pairs of E and B.
(a) E_x, B_y (b) E_y, B_z
(c) B_x,E_y (d) E_z,B_y
Solution: (b, d)

Here in the question electromagnetic wave is propagating along x-direction. So, electro and magnetic field vectors should have either y-direction or 2-direction.

Question 11. A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. The electromagnetic waves produced
(a) will have frequency of 10⁹ Hz
(b) will have frequency of 2 x 10⁹ Hz
(c) will have wavelength of 0.3 m
(d) fall in the region of radio waves
Solution: (a, c, d)
Here we are given the frequency by which the charged particles oscillates about its mean equilibrium position, it is equal to 10⁹ Hz. The frequency of electromagnetic waves produced by a charged particle is equal to the frequency by which it oscillates about its mean equilibrium position.
So, frequency of electromagnetic waves produced by the charged particle is v= 10⁹ Hz.

Question 12. The source of electromagnetic waves can be a charge
(a) moving with a constant velocity
(b) moving in a circular orbit
(c) at rest
(d) falling in an electric field
Solution: (b, d)
Key concept:

• An electromagnetic wave can be produced by accelerated or oscillating charge.
• An oscillating charge is accelerating continuously, it will radiate electromagnetic waves continuously.
• Electromagnetic waves are also produced when fast moving electrons are suddenly stopped by a metal target of high atomic number.

Here, in option (b) charge is moving in a circular orbit.
In circular motion, the direction of the motion of charge is changing continuously, thus it is an accelerated motion and this option is correct.
In option (d), the charge is falling in electric field. If a charged particle is moving in electric field it experiences a force or we can say it accelerates. We know an accelerating charge particle radiates electromagnetic waves. Hence option (d) is also correct.
Also, we know that a charge starts accelerating when it falls in an electric field.
Important points:

• In an atom an electron is circulating around the nucleus in a stable orbit, although accelerating does not emit electromagnetic waves; it does so only when it jumps from a higher energy orbit to a lower energy orbit.
• A simple LC oscillator and energy source can produce waves of desired frequency

Question 13.  An EM wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure p on it. Which of the following are true?
(a) Radiation pressure is I/c if the wave is totally absorbed
(b) Radiation pressure is —I/c if the wave is totally reflected
(c) Radiation pressure is 2I/c if the wave is totally reflected
(d) Radiation pressure is in the range I/c < p < 2I/c for real surfaces

Very Short Answer Type Questions
Question 14. Why is the orientation of the portable radio with respect to broadcasting station important?
Solution: The electromagnetic waves are plane polarised, so the receiving antenna should be parallel to the vibration of the electric or magnetic field of the wave. So the receiving antenna should be parallel to electric/magnetic part of the wave. That is why the orientation of the portble radio with respect to broadcasting station is important.

Question 15. Why does microwave oven heats up a food item containing water molecules most efficiently?
Solution: The microwave oven heats up the food items containing water molecules most efficiently because the frequency of microwaves matches the resonant frequency of water molecules.

Question 16. The charge on a parallel plate capacitor varies as q=q₀  cos 2πvt. The plates are very large and close together (area = A, separation = d). Neglecting the edge effects, find the displacement current through the capacitor.
Solution:

Question 17. A variable frequency AC source is connected to a capacitor. How will the displacement current change with decrease in frequency?
Solution:

It means the displacement current decreases as the conduction current is equal to the displacement current.

Question 18. The magnetic field of a beam emerging from a fitter facing a flood light is given by B₀= 12 x 10^-8 sin (1.20 x 10⁷z- 3.60 x 10¹⁵ t) T What is the average intensity of the beam?
Solution:

Question 19.

Solution:

Question 20. Professor CV Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of EM waves was he exhibiting? Give one more example of this property.
Solution: The properties of an electromagnetic wave is same as other waves. Like other wave an electromagnetic wave also carries energy and momentum. Since, it carries momentum, an electromagnetic wave also exerts pressure called radiation pressure. This property of electromagnetic waves helped professor C V Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it.

Short Answer Type Questions
Question 21.

Solution:

Question 22. Electromagnetic waves with wavelength
(i) λ₁, is used in satellite communication.
(ii) λ₂, is used to kill germs in water purifier.
(iii) λ₃, is used to detect leakage of oil in underground pipelines.
(iv) λ₄, is used to improve visibility in runways during fog and mist conditions.
(a) Identify and name the part of electromagnetic spectrum to which these radiations belong.
(b) Arrange these wavelengths in ascending order of their magnitude.
(c) Write one more application of each.
Solution: (a) (i) In satellite communications, microwave is widely used. Hence λ₁, is the wavelength of microwave.
(ii) In water purifier, ultraviolet rays are used to kill germs. So, λ₂ is the wavelength of UV rays.
(iii) X-rays are used to detect leakage of oil in underground pipelines. So, λ₃ is the wavelength of X-rays.
(iv) Infrared rays are used to improve visibility on runways during fog and mist conditions. So, it is the wavelength of infrared waves.
(b) Wavelength of X-rays < wavelength of UV < wavelength of infrared < wavelength of microwave.
=> λ₃<λ₂<λ₄<λ₁

Question 23.

Solution:

Question 24. You are given a 2 μF parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?
Solution:

Question 25. Show that the radiation pressure exerted by an EM wave of intensity I on a surface kept in vacuum is I/C.
Solution: Let us consider a surface exposed to electromagnetic radiation. The radiation is falling normally on the surface. Further, intensity of radiation is I and area of surface exposed to radiation is A.

Question 26. What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of room, its intensity essentially remains constant.
What geometrical characteristic of LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?
Solution: We know intensity of light from a point source I α 1/r², r is the distance from point source.
As the distance is doubled, so the intensity becomes one-fourth the initial value. But in case of laser it does not spread, so its intensity remain same.
Some geometrical characteristics of LASER beam which are responsible for the constant intensity is
(i) Unidirection (ii) Monochromatic
(iii) Coherent light (iv) Highly collimated
These characteristics are missing in the case of normal light from the bulb.

Question 27. Even though an electric field E exerts a force qE on a charged particle yet electric field of an EM wave does not contribute to the radiation pressure (but transfers energy). Explain.
Solution: The electric field of an electromagnetic wave is an oscillation field. It exerts electric force on a charged particle, but this electric force averaged over an integral number of cycles is zero, since its direction changes every half cycle. Hence, electric field is not responsible for radiation pressure though it transfer the energy. In fact, radiation pressure appears as a result of the action of the magnetic field of the wave on the electric currents induced by the electric field of the same wave.

Long Answer Type Questions
Question 28.


Solution:

Question 29.

Solution:

Question 30.

(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current l^d.
(iii) Compare the conduction current I₀ with the displacement current I₀^d.
Solution:

Question 31.

Solution:

Question 32.

Solution: (i) Total energy carried by electromagnetic wave is due to electric field vector and magnetic field vector. In electromagnetic wave, E and B vary from point to point and from moment to moment.

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NCERT Exemplar Class 12 Physics Chapter 7 Alternating Current are part of NCERT Exemplar Class 12 Physics. Here we have given NCERT Exemplar Class 12 Physics Chapter 7 Alternating Current. https://www.cbselabs.com/ncert-exemplar-problems-class-12-physics-alternating-current/

NCERT Exemplar Class 12 Physics Chapter 7 Alternating Current

Multiple Choice Questions (MCQs)
Single Correct Answer Type

Question 1. If the rms current in a 50 Hz AC circuit is 5 A, the value of the current 1/300 s after its value becomes zero is

Solution: (b)
Key concept: Equation for i and V: Alternating current or voltage varying as sine function can be written as

Question 2. An alternating current generator has an internal resistance R_gand an internal reactance X_g It is used to supply power to a passive load consisting of a resistance R_g and a reactance X_L. For maximum power to be delivered from the generator to the load, the value of X_L is equal to
(a) zero (b) X_g
(c) -X_g (d) R_g
Solution: (c) For maximum power to be delivered from the generator (or internal reactance X_g) to the load (of reactance, X_L),
=> X_L + X_g = 0 (the total reactance must vanish)
=>X_L=-X_g

Question 3.

Solution:

Question 4. To reduce the resonant frequency in an L-C-R series circuit with a generator,
(a) the generator frequency should be reduced
(b) another capacitor should be added in parallel to the first
(c) the iron core of the inductor should be removed
(d) dielectric in the capacitor should be removed
Solution: (b)

Question 5. Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?
(a) R = 20 Ω, L = 1.5 H, C = 35μF
(b) R = 25 Ω, L = 2.5 H, C = 45 μF
(c) R=15Ω, L = 3.5H, C = 30 μF
(d) R = 25 Ω, L = 1.5 H, C = 45 μF
Solution: (c)


where R is the resistance, L is the inductance and C is the capacitance of the circuit.
For high Q factor R should be low, L should be high and C should be low. These conditions are best satisfied by the values given in option (c).
Important point: Be careful while writing formula for quality factor, this formula we used in this case is only for series L-C-R circuit.

Question 6. An inductor of reactance 1 Ω and a resistor of 2 Ω are connected in series to the terminals of a 6 V (rms) AC source. The power dissipated in the circuit is
(a) 8 W (b) 12 W
(c) 14.4 W (d) 18 W
Solution: (c) According to the problem, XL = 1 Ω , R = 2 Ω ,

Question 7. The output of a step-down transformer is measured to be 24 V when connected to a 12 W light bulb. The value of the peak current is

Solution: (a)

One or More Than One Correct Answer Type
Question 8. As the frequency of an AC circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?
(a) Inductor and capacitor (b) Resistor and inductor
(c) Resistor and capacitor (d) Resistor, inductor and capacitor
Solution: (a, d) Compare the given circuit by predicting the variation in their reactances with frequency. So, that then we can decide the elements.
Reactance of an inductor of inductance L is X_L = 2πvL, where v is the frequency of the AC circuit.

Question 9. In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?
(a) Only resistor (b) Resistor and an inductor
(c) Resistor and a capacitor (d) Only a capacitor
Solution: (c, d) This is the similar problem as we discussed above. In this problem, the current increases on increasing the frequency of supply. Hence, the reactance of the circuit must be decreased as increase in frequency. So, one element that must be connected is capacitor. We can also connect a resistor in series.
For a capacitive circuit,
X_C = 1/ωC =1/2πfC
When frequency increases, X_C decreases. Hence current in the circuit increases.

Question 10. Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?
(a) For a given power level, there is a lower current
(b) Lower current implies less power loss
(c) Transmission lines can be made thinner
(d) It is easy to reduce the voltage at the receiving end using step-down transformers
Solution: (a, b, d)

Question 11. For an L-C-R circuit, the power transferred from the driving source to the driven oscillator is P = I² Z cos Ф.
(a) Here, the power factor cos Ф > 0, P > 0
(b) The driving force can give no energy to the oscillator (P = 0) in some cases
(c) The driving force cannot syphon out (P < 0) the energy out of oscillator
(d) The driving force can take away energy out of the oscillator
Solution:

Question 12. When an AC voltage of 220 V is applied to the capacitor C
(a) the maximum voltage between plates is 220 V
(b) the current is in phase with the applied voltage
(c) the charge on the plates is in phase with the applied voltage
(d) the power delivered to the capacitor is zero
Solution: (c, d) If the alternating voltage is applied to the capacitor, the plate connected to the positive terminal of the source will be at higher potential and the plate connected to the negative terminal of source will be at lower potential. So the plates capacitor is charged.

Question 13.

Solution:

Very Short Answer Type Questions
Question 14. If an L-C circuit is considered analogous to a harmonically oscillating spring- block system, which energy of the L-C circuit would be analogous to potential energy and which one analogous to kinetic energy?
Solution: When a charged capacitor C having an initial charge q0 is discharged through an inductance L, the charge and current in the circuit start oscillating simple harmonically. If the resistance of the circuit is zero, no energy is dissipated as heat. We also assume an idealized situation in which energy is not radiated away from the circuit. The total energy associated with the circuit is constant.
The oscillation of the LC circuit are an electromagnetic analog to the mechanical oscillation of a block-spring system.
The total energy of the system remains conserved.

Question 15. Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance.

Solution:
Key concept: The element with infinite resistance will be considered as open circuit and the element with zero resistance will be considered as short circuited.

Question 16. Study the circuits (a) and (b) shown in figure and answer the following questions.

(a) Under which conditions would the rms currents in the two circuits be the same?
(b) Can the rms current in circuit (b) be larger than that in (a)?

Question 17. Can the instantaneous power output of an AC source ever be negative? Can the average power output be negative?

Question 18. In a series LCR circuit, the plot of I_max versus co is shown in figure. Find the bandwidth and mark in the figure.

Question 19. The alternating current in a circuit is described by the graph shown in figure. Show rms current in this graph.

Question 20. How does the sign of the phase angle Ф, by which the supply voltage leads the current in an L-C-R series circuit, change as the supply frequency is gradually increased from very low to very high values?

Short Answer Type Questions
Question 21. A device ‘X is connected to an AC source. The variation of voltage, current and power in one complete cycle is shown in figure.
(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device X.

Solution: (a) Power is the product of voltage and current (Power = P = VI).
So, the curve of power will be having maximum amplitude, equals to the product of amplitudes of voltage (V) and current (I) curve. Frequencies , of B and C are-equal, therefore they represent V and I curves. So, the curve A represents power.
(b) The full cycle of the graph (as shown by shaded area in the diagram) consists of one positive and one negative symmetrical area.

Hence, average power consumption over a cycle is zero.
(c) Here phase difference between V and I is π /2 therefore, the device ‘X’ may be an inductor (L) or capacitor (C) or the series combination of L and C.

Question 22. Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?
Solution: For a Direct Current (DC),
1 ampere = 1 coulomb/sec
Direction of AC changes with the frequency of source with the source frequency and the attractive force would average to zero. Thus, the AC ampere must be defined in terms of some property that is independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of AC.
So, r.m.s. value of AC is equal to that value of DC, which when passed through a resistance for a given time will produce the same amount of heat as produced by the alternating current when passed through the same resistance for same time.

Question 23. A coil of 0.01 H inductance and 1 ω resistance is connected to 200 V, 50 Hz AC supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current.

Question 24. A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used.

Question 25. Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.
Solution: Capacitor plates get charged and discharged when an AC voltage is applied across the plates. So the current through capacitor is as a result of charging charge. Because the frequency of the capacitive circuit increases, the polarities of the charged plates change more rapidly with time, giving rise to a’larger current. The capacitance reactance (X_C) due to a capacitor C varies
as the inverse of the frequency (f) (as X_C=1/2π fC) and hence approaches zero as v approaches infinity. The current is zero in a DC capacitive circuit, which corresponds to zero proportional and infinite reactance. Also, Since X_C is inversely proportional to frequency, capacitors tend to pass high-frequency current and to block low-frequency currents and DC (just the opposite of inductors).

Question 26. Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.
Solution: The inductive reactance is given by X_L = 2πfL, X_L is proportional to the frequency and current is inversely proportional to the reactance. An inductor opposes the flow of current through it by developing a back emf according to Lenz’s law. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice-versa.
Since, the induced emf is proportional to the rate of change of current.

Long Answer Type Questions
Question 27.

Question 28.

Question 29. Consider the L-C-R circuit shown in figure. Find the net current i and the phase of i. Show that i = V/Z. Find the-impedance Z for this circuit.

Solution: Key concept: In the circuit given above consists of a capacitor (C) and an inductor (L) connected in series and the combination is connected in parallel with a resistance R. Due to this combination there is an oscillation of electromagnetic energy.

Question 30.

Solution:

Question 31.

Solution:

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