AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 5 Locomotion and Reproduction in Protozoa Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 5 Locomotion and Reproduction in Protozoa

Very Short Answer Type Questions

Question 1.
Draw a labeled diagram of the T.S of the flagellum.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa VSAQ Q1

Question 2.
List any two differences between a flagellum and a cilium.
Answer:

FlagellumCilium
1. Flagellum helps in locomotion only.1. Cilium helps in locomotion and feeding and acts as sensory structures.
2. Flagellum produces undular movement.2. Cilium produces pendular movement.
3. Flagellum is about 150µ in length.3. Cilium is small in size 5-10µ in length.

Question 3.
What are dynein arms? What is their significance?
Answer:
‘A’ tube of each peripheral doublet bears paired arms along its length called dynein arms made up of protein dynein.
The dynein arms of the ‘A’ tubule face the tubule ‘B’ of the adjacent doublet.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 4.
What is a kinety?
Answer:
In the ciliate protozoans, a longitudinal row of kinetosomes together with kinetodesmata constitute a unit called kinety.

Question 5.
Distinguish between synchronous and metachronous movements.
Answer:
Synchronous movement: Cilia in a transverse row beat simultaneously in one direction. It is called synchronous movement.
Metachronous movement: The sequential movement of cilia, in a longitudinal row, one after the other in one direction is called metachronous movement.

Question 6.
Why do we refer to the offspring, formed by the asexual method of reproduction, as a clone?
Answer:
As a result of the asexual method, the offsprings are, not only identical to one another but also exact copies of their parent. The term ‘clone’ is used to describe such morphologically and genetically similar individuals.

Question 7.
Distinguish between proter and opisthe.
Answer:
During transverse binary fission in Paramecium two daughters, individuals are formed. The anterior one is called proter and the posterior is called opisthe.

Question 8.
How is sexual reproduction advantageous in evolution?
Answer:
Sexual reproduction results the advantageous in evolution in genetic recombination occurs in sexual reproduction.

Question 9.
Distinguish between lobopodium and filopodium. Give an example to each of them.
Answer:
Lobopodium: The blunt and finger-like tubular pseudopodia containing both ectoplasm and endoplasm is called lobopodium.
Ex: Amoeba proteus
Filopodium: The slender filamentous pseudopodia with pointed tips, composed of only ectoplasm are called Filopodium.
Ex: Euglypha

Question 10.
Define conjugation with reference to ciliates. Give two examples.
Answer:
Conjugation is a temporary union between two senile ciliates that belong to two different ‘mating types’ for the exchange of nuclear material and its reorganization. – Wichterman.
Ex: Paramecium and Vorticella.

Short Answer Type Questions

Question 1.
Name the system that controls the fastest swimming movement of protozoans and write its components.
Answer:
Ciliary locomotion is the fastest swimming movement of protozoans, Hence, ciliates are the fastest protozoans.
Cilia are small hair-like structures found in ciliate protozoans like Paramecium.

Infraciliary system: It is located just below the pellicle in the ectoplasm of a ciliate. It includes kinetosomes, kinetodesmal fibrils, and kinetodesmata. The kinetosomes are present at the bases of cilia in transverse and longitudinal rows. The kinetodesmal fibrils are connected to the kinetosomes and run along the right side of each row of kinetosomes as a ‘cord of fibres’ called kinetodesmata. A longitudinal row of kinetosomes together with kinetodesmata constitute a unit called ‘kinety’.

All the kineties together form an infraciliary system, which is connected to a ‘motorium’, located near the cytopharynx. The infraciliary system and motorium form the ‘neuromotor system’ that controls and coordinates the movement of cilia.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 2.
Write the mechanism of bending the flagellum and explain effectively recovery strokes.
Answer:
The bending movement of a flagellum is brought about by the sliding of microtubules past each other due to the functioning of ‘dynein arms’ utilizing ATP. A flagellum pushes the fluid medium at right angles to the surface of its attachment, by its bending movement.
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q2
Bending movement of flagella and cilia: Dynein arms show a complex cycle of movements using energy provided by ATP (dynein arms are the sites of ATPase activity in the cilia and flagella). The dynein arms of each doublet attach to an adjacent doublet and pull the neighbouring doublet. So the doublets slide past each other in opposite directions. The arms release and reattach a little further on the adjacent doublet and again ‘puli’. As the doublets of a flagellum or cilium are physically held in place by the radial spokes, the doublets cannot slide past much. Instead, they curve and cause the bending of flagellum or cilium. Such bending movements of flagella and cilia play an important role in flagellar and ciliary locomotion.

The flagellar movement of many organisms is a sidewise-lash which consists of two strokes namely the effective or propulsive stroke and the recovery stroke.

  • Effective stroke: Flagellum becomes rigid and starts bending to one side beating against the water. This beating against water is at right angles to the body axis and the organism moves forwards.
  • Recovery stroke: Flagellum becomes comparatively soft so as to offer the least resistance to water and moves back to its original position. It is called ‘recovery stroke’.

Question 3.
What are lateral appendages? Based on their presence and absence, write the various types of flagella giving atleast one example for each type.
Answer:
Lateral appendages: Some flagella bear one or two or many rows of short, lateral hair-like fibrils called lateral appendages. They are of two types namely ‘mastigonemes’ and ‘flimmers’.
Types of Flagella: Based on the presence or absence and/or the number of rows of lateral appendages, five types of flagella are recognized.
(a) Stichonematic: This flagellum bears one row of lateral appendages on the axoneme.
E.g. Euglena and Astasia.
(b) Pantonematic: This flagellum has two or more rows of lateral appendages on the axoneme.
E.g. Peranema and Monas.
(c) Acronematic: This type of flagellum does not bear lateral appendages and the terminal part of the axoneme is naked without the outer sheath at its tip.
E.g. Chlamydomonas and Polytoma
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q3
(d) Pantacronematic: This type of flagellum is provided with two or more rows of lateral appendages and the axoneme ends in a terminal naked filament.
E.g. Urceolus.
(e) Anematic or simple: In this type of flagellum, lateral appendages and terminal filament are absent. Hence, it is called anematic (a-no; nematic-threads)
E.g. Chilomonas and Gryptomonas.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 4.
Describe the process of transverse binary fission in paramecium.
Answer:
Transverse binary fission is performed by Paramecium. Binary fission is the most common method of sexual reproduction in protozoans. During favourable conditions, Paramecium stops feeding after attaining its maximum growth.

At first, the micronucleus divides by mitosis and the macronucleus divides into two daughter nuclei by amitosis. The oral groove disappears. After karyokinesis, a transverse constriction appears in the middle of the body, which deepens and divides the parent cell into two daughter individuals, the anterior proper and the posterior opisthe. The proper receives the anterior contractile vacuole, cytopharynx, and cytosome from its parent individual. It develops a posterior contractile vacuole and a new oral groove.
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q4
The opisthe receives the posterior contractile vacuole of its parent. It develops a new anterior contractile vacuole, cytopharynx, cytostome and a new oral groove. Binary fission is completed in almost two hours, in favourable conditions and Paramecium can produce four generations of daughter individuals by binary fission in a day. The transverse binary fission is also called homothetogenic fission because the plane of fission is at right angles to the longitudinal axis of the body. As it occurs at right angles to the kineties, it is also called perkinetal fission.

Question 5.
Describe the process of longitudinal binary fission in Euglena.
Answer:
Binary fission is the most common method of asexual reproduction in protozoans. Longitudinal binary fission is performed by Euglena. In this type of binary fission, the body divides into two halves longitudinally, hence called longitudinal binary fission.
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q5
During the process of binary fission, the nucleus, basal granules, chromatophores, and cytoplasm undergo division. The nucleus divides by mitosis into two daughter nuclei. Then the kinetosomes and the chromatophores also divide. At first, a longitudinal groove develops in the middle of the anterior end. This groove extends gradually towards the posterior end until the two daughter individuals are separated. One daughter Euglena retains the parental flagella. The other daughter individual develops new flagella from the newly formed basal granules. The stigma, paraflagellar body, and contractile vacuole of the parent disappear. They develop afresh in both the daughter Euglenae. The longitudinal binary fission is known as symmetrogenic division because the two daughters Euglenae resemble each other like mirror images.

Question 6.
Write a short note on multiple fission.
Answer:
Multiple fission: It is the division Of the parent body into many smaller individuals (Multi-many; Fission- splitting). Normally multiple fission occurs during unfavourable conditions. During multiple fission, the nucleus first undergoes repeated mitotic divisions without cytokinesis. This causes the formation of many daughter nuclei. Then the cytoplasm also divides into as many bits as there are nuclei. Each cytoplasmic bit encircles one daughter nucleus. This results in the formation of many smaller individuals from a single-parent organism. There are different types of multiple fissions in protozoans such as Schizogony, malegametogony, sporogony in plasmodium, sporulation in Amoeba, etc.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 7.
Give an account of pseudopodia.
Answer:
Locomotion in protozoans is performed by cellular extensions such as pseudopodia found in rhizopods organisms. The pseudopodia are temporary extensions of cytoplasm that develop in the direction of the movement. These temporary structures are useful to move on the substratum as our legs do, hence the name ‘pseudopodia’. There are four kinds of pseudopodia in protozoans.

  1. Lobopodia: blunt + finger like Pseudopodia. Ex: Amoeba, Entamoeba.
  2. Filopodia: fiber like pseudopodia contain ectoplasm. Ex: Euglypha.
  3. Reticulopodia: net like pseudopodia. Ex: Elphidium.
  4. Axopodia or heliopodia: Sun ray-like pseudopodia. Ex: Actinophiys.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q7

Question 8.
Give an account of the ultrastructure of an axoneme.
Answer:
Ultrastructure of flagellum: The axial filament or axoneme shows 9+2 organisation under the electron microscope. Two central singlets are enclosed by a fibrous inner sheath. Nine peripheral doublets form a cylinder between the inner sheath and the outer sheath. The “A” microtubule of each doublet is connected to the inner sheath by radial spokes. It also has pairs of arms all along the length and is directed towards the neighbouring doublet.

These arms are made of a protein called dynein. These arms create the sliding force. The peripheral doublets are surrounded by an outer membranous sheath called a protoplasmic sheath, which is the extension of the plasma membrane. Some flagella bear lateral appendages called flimmers or mastigonemes along the length of the axoneme above the level of the pellicle. Each flagellum arises from a basal granule that lies below the cell surface in the ectoplasm.
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q8

Question 9.
Draw a neat labelled diagram of Euglena.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q9

Question 10.
Draw a neat diagram of paramecium and label its important structures/components.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q10

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 4 Animal Diversity-II: Phylum Chordata Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 4 Animal Diversity-II: Phylum Chordata

Very Short Answer Type Questions

Question 1.
List out the characters shared by chordates and echinoderms.
Answer:
Chordates and echinoderms are enterococci, deuterostomes, and bilaterally symmetrical.

Question 2.
Write four salient features of cyclostomes.
Answer:

  1. Cyclostomes are jawless aquatic forms.
  2. The body is scaleless, long, slender, and eel-like in shape.
  3. Endoskeleton is cartilaginous.
  4. Vertebrae are represented by imperfect neural arches in some.
  5. The mouth is circular and suctorial, Hence there are called Cyclostomes. Ex: Petromyzon.

Question 3.
What is the importance of endostyle in lancelets and ascidians?
Answer:
Endostyle is useful for accumulating and moving food particles to the oesophagus in lancelets and ascidians.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 4.
Name the type of caudal fin and scales that are present in a Shark and Catla respectively.
Answer:
The caudalfin in shark is heterocercal and scales are placoid. The caudal fin is catla is homocercal and scales are cycloid.

Question 5.
What is the importance of air bladder in fish?
Answer:
Fishes have an ‘air bladder’ acting as a ‘hydrostatic organ’ helping the fish float easily at the desired level without much expenditure of energy.

Question 6.
How do you justify the statement ‘heart in fishes is a branchial heart’?
Answer:
The heart of fish is two-chambered and is described as a branchial heart as it supplies blood only to the gills.

Question 7.
What are claspers? Which group of fishes possesses them?
Answer:
Claspers are formed from the posterior portion of pelvic fins in male cartilaginous fish. They serve as intermittent organs used to channel semen into the female’s cloaca during mating.
Ex: Chondrichthyes fishes possess Claspers.

Question 8.
How does the heart of an amphibian differ from that of a reptile?
Answer:
The heart of an amphibian is three-chambered. The heart of a reptile is incompletely four-chambered.

Question 9.
Name the structures that appeared for the first time in amphibians, in the course of evolution.
Answer:
The two pairs of pentadactyl limbs appeared for the first time in amphibians in the course of evolution.

Question 10.
How do you distinguish a male frog from a female frog?
Answer:
The male frog can be distinguished by the presence of sound amplifying vocal sacs and a copulatory pad on the first digit of each forelimb.

Question 11.
What is a ‘force pump’ in a frog? Why is named so?
Answer:
In frogs, during pulmonary respiration, the buccopharyngeal cavity acts like a ‘force pump’. Due to the elevation of the buccopharyngeal cavity the air forces the glottis to open and enter the lungs.

Question 12.
What are corporabigemina? Mention their chief function.
Answer:
Midbrain is represented by a pair of optic lobes called corpora bigemina. The optic lobes are associated with the sense of sight.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 13.
Distinguish between mesorchium and mesovarium.
Answer:
The testes are attached to the kidneys and dorsal body wall by a double fold of the peritoneum called mesorchium.
The ovaries are attached to the kidneys and dorsal body wall by a double fold of the peritoneum called mesovarium.

Question 14.
Distinguish between milt and spawn.
Answer:
During amplexus, the mass of eggs and the mass of sperm released by the female and male are called spawn and milt.

Question 15.
What are the ‘Golden ages’ of the first jawed vertebrates and the first amniotes?
Answer:
The Devonian period is considered the ‘golden age of first jawed vertebrates (Fishes).
The Mesozoic era is considered the golden age of amniotes (Reptiles).

Question 16.
Name two poisonous and non-poisonous snakes found in south India.
Answer:
Poisonous Snakes:

  1. Naja naja (Cobra)
  2. Bungarus (Krait)
  3. Vipera russelli (Chain viper)

Non-poisonous Snakes:

  1. Ptyas (rat snake)
  2. Tropidonotus (Pond or grass snake).

Question 17.
In which features does the skin of a reptile differ from that of a frog?
Answer:

  • The skin of reptiles is rough and dry, covered by horny epidermal scales, and shields.
  • The skin of a frog is thin, scaleless, and moist.

Question 18.
Describe a cat and a lizard on the basis of their chief nitrogenous wastes excreted.
Answer:
Based on the nitrogenous wastes excreated by lizards are Uricotelic and cats are ‘Ureotelic’ animals.

Question 19.
Name the four extraembryonic membranes.
Answer:
The extraembryonic membranes namely amnion, allantois, chorion, and yolk sac.

Question 20.
What are Jacobson’s organs? What is their function?
Answer:
Jacobson’s organs are the specialized olfactory structures, that are highly developed in lizards and snakes.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 21.
What are pneumatic bones? How do they help birds?
Answer:
The main bones in birds are extensions of air sacs without bone marrow are called pneumatic bones. These are helpful in flying birds.

Question 22.
What is a ‘wishbone? What are the skeletal components that form it?
Answer:
In birds, both the clavicles are fused with the interclavicular to form a ‘V-shaped bone, called fiircula or ‘wish hone’ or ‘Merrythought bone’.

Question 23.
What is continuous oxygenation of the blood? How is it made possible in birds?
Answer:
The lungs of birds are compact, spongy, undistensible lungs associated with air sacs. Air sacs facilitate continuous air supply is called ‘continuous oxygenation of the blood.

Question 24.
Distinguish between the crop and the gizzard in birds.
Answer:

  • The Oesophagus of birds is often dilated into a crop for the storage of food.
  • The stomach is usually divided into glandular proventriculus and muscular gizzard a grinding mill.

Question 25.
Distinguish between altricial and precocial hatchlings.
Answer:

  • Altricial: Young ones of flying birds’ hatchlings are altricial.
  • Precocial: Young ones of flightless bird hatchlings are precocial.

Question 26.
In which group of animals do we find three ear ossicles on each side and what are their names from the innermost to the outermost?
Answer:
The middle ear possesses three ear ossicles in the Mammalia group of animals. They are malleus, incus and stapes.

Question 27.
How does a mature RBC of a mammal differ from that of other vertebrates?
Answer:

  • In mammals, mature RBC is enucleated and biconcave.
  • In other vertebrates RBC is nucleate.

Question 28.
Name the characteristic type of vertebrae found in reptiles, birds, and mammals.
Answer:

  • Reptiles’ vertebrae are procoelons.
  • Birds’ vertebrae are heterologous.
  • Mammalian vertebrae are amphiplatyan.

Question 29.
Name the three meninges. In which group of animals do you find all of them?
Answer:
Mammals have three meninges. They are the outer dura mater, middle arachnoid mater, and inner diameter.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 30.
Name the vertebrate groups in which ‘renal portal system1 is absent.
Answer:
The renal portal system is absent in aves (birds) in vertebrate animals.

Short Answer Type Questions

Question 1.
Give three major differences between chordates and non-chordates and draw a sketch of a chordate’s body showing those features.
Answer:
The major differences between chordates and non-chordates.

ChordatesNon-chordates
1. Notochord is present.1. Notochord is absent.
2. Central nervous system is dorsal hollow, single and non-ganglionated.2. Central nervous system is ventral, solid, double, and ganglionated.
3. Pharynx is perforated by gill slits.3. Gill slits are absent.
4. Heart is ventral.4. Heart is dorsal (if present).
5. A post-anal tail is present.5. Post-anal tail is absent.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q1

Question 2.
Name the four ‘hallmarks’ of chordates and explain the principal function of each of them.
Answer:
The hallmarks of chordates: All the chordates exhibit four fundamental characteristics. They are Notochord, Dorsal tubular nerve cord, Pharyngeal slits or clefts, and Post-anal tail.

Notochord: It is a flexible rod-like structure situated along the mid-dorsal line between the gut and the nerve cord. It is derived from the embryonic chorda mesoderm. It is firm but flexible. It is present throughout life in the lancelets and cyclostomes. It is present in the tail of the tadpole larva of an ascidian, It is present in embryonic stages, but is replaced partly or wholly by the vertebral column in the adults of higher chordates. Remnants of notochord occur as nucleipulposi in the intervertebral discs of mammals.

Dorsal tubular nerve cord: A single, hollow tubular, and fluid-filled nerve cord is situated above the notochord and below the dorsal body wall. It is derived from the ectoderm of the embryo. In the higher chordates, it gets enlarged to form a distinct brain at the anterior end the rest of it becomes the spinal cord.

Pharyngeal slits of clefts: These are slit openings present on the Pharyngeal wall and meant for the exit of the water from the pharyngeal cavity. They are present throughout life in the protochordate, fishes, and some amphibians. These are present in larval stages in amphibians. They develop by in-pushing of ectoderm and corresponding out pursing of the endoderm. In land vertebrates, the gills become vestigial and nonfunctional and are restricted to embryonic stages only.

Post-anal tail: Chordates have a tail extending posteriorly to the anus. It is lost in many species during late embryonic development. It contains skeletal elements and muscles, coelom and visceral organs are absent in it.

Question 3.
Describe the features of a tunicate that reveals its chordate identity.
Answer:

  1. The body of these animals is covered by cellulose-like covering tunicin, hence called tunicates.
  2. These possess Notochord in the tail region during the larval stage, hence called Urochordata.
  3. These are sedentary or pelagic marine forms.
  4. The notochord is present only in larval tails and degenerated in adults.
  5. Open type of blood vascular system with blood pigment vanadium.
  6. Indirect development with tadpole larva.
  7. Adults (mostly) show degenerate characters.
  8. The nervous system is represented in the adult by a single dorsal ganglion.
  9. They are bisexual or hermaphrodites.
  10. Ex: Ascidia, Salpa, Doliolum, Pyrosoma and Oikopleura.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q3

Question 4.
Compare and contrast sea squirts and lancelets.
Answer:
Sea squirts: These are included in class – Ascadiaceae of subphylum – Urochordata. These are sessile. These are solitary or colonial. The body is enclosed in a permanent test and un-segmented. All these are marine and occur from the surface water to greater depths. Coelom in the absent, pharynx is large and is perforated by numerous gill slits. Branchial aperture in anterior and atrial aperture is dorsal. The digestive tract is ‘complete’. The circulations system is of an open type, the heart is the tubular and the ventral heart. These are bisexual. Development generally includes a free-swimming tadpole larva. Notochordcontinued to the tail hence the name Urochordata.
Ex: Ascidia
AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q4

Lancelets: Cephalochordates are also called Lancelets. These are marine animals and are small fish without paired fins. These are typical chordates because they possess the notochord, tubular nerve cord, and pharyngeal slits throughout this life. The coelom is enterocoelic respiration mostly across the external body surface. The circulatory system is of a closed type, the heart, blood corpuscles, and respiratory pigments are absent. Excretion by protonephridia fertilization is external and development is indirect.
Ex: Branchiostoma (amphioxus or Lancelet)
AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q4.1

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 5.
List out eight characteristics that help distinguish a fish from the other vertebrates.
Answer:
General characters:

  1. Fishes are completely aquatic poikilothermic (cold-blooded) animals.
  2. The body of a fish is usually streamlined and differentiated into the head, trunk, and tail.
  3. The exoskeleton consists of mesodermal scales or bony plates. A few are scaleless.
  4. The endoskeleton may be cartilaginous or bony. Skull is monocondylic. Vertebrae are amphicoelous. Centrum is concave at both anterior and posterior faces.
  5. Locomotion is assisted by unpaired (median and caudal) fins along with paired (pectoral and pelvic) fins.
  6. The mouth is ventral or terminal. Teeth are usually acrodont, homodont, and polyphyodont.
  7. The exchange of respiratory gases is performed by the gills. The heart is ‘two-chambered’.
  8. Kidneys are mesonephric. Fishes are mostly ammonotelic and some are ureotelic. (cartilaginous fishes).
  9. Cranial nerves are 10 pairs, Meninx Primitiva is the only ‘meninx’ enveloping the central nervous system.
  10. The internal ear consists of three semicircular canals. Lateral-line sensory system (to detect movement and vibration in the surrounding water) is well-developed.
  11. Eyes are without eyelids and each eyeball is protected by a nictitating membrane.
  12. Sexes are separate. Fertilization is internal or external. Development may be direct or indirect.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q5

Question 6.
Compare and contrast cartilaginous and bony fishes.
Answer:

Cartilaginous fishesBony fishes
1. These are marine farms.1. These live in all kinds of aquatic habits.
2. Endoskeleton made by cartilaginous.2. Endoskeleton made of bone.
3. Body covered by placoid scales.3. Body covered by cosmid, ganoid, cycloid or ctenoid scales.
4. Caudal fin is heterocercal.4. Caudal fin is homocercal.
5. Operculum absent.5. Operculum present.
6. Air bladder absent.6. Air baldder present.
7. Gills are lamelliform and are five to seven on each side.7. Gills are filamentous and are four on each side.
8. These are ureotelic.
Ex: Scoliodon, Pristic, Torpedo.
8. These are mostly ammonotelic.
Ex: Catla, Labeo, Exocetus, Hippocampus.

Question 7.
Describe the structure of the heart of the frog.
Answer:
The blood vascular system consists of the heart, blood vessels, and blood. The heart is a muscular organ situated in the upper part of the body cavity. It has two separate atria and a single undivided ventricle. It is covered by a double-layered membrane called the pericardium. A triangular chamber called sinus venosus joins the right atrium on the dorsal side. It receives blood through three vena cavae (caval veins). The ventricle opens into the conus arteriosus on the ventral side. The conus arteriosus bifurcates into two branches and each of them divides into three aortic arches namely carotid, systemic and pulmocutaneous. Blood from the heart is distributed to all parts of the body by the branches of the aortic arches. Three major veins collect blood from the different parts of the body and carry it to the sinus venosus.
AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q7

Question 8.
Write eight salient features of the class – Amphibia.
Answer:
General characters of Amphibia:

  1. They are the first tetrapods and lead a dual mode of life, i.e. on land and in freshwater.
  2. The body is divided into distinct ‘head’ and ‘trunk’. The tail may or may not be present.
  3. Skin is soft, scale-less (except for the members of Apoda), moist and glandular.
  4. The body bears two pairs of equal or unequal pentadactyle limbs (caecilians are limbless).
  5. Skull is dicondylic as in mammals. Vertebrae are mostly precocious (centrum is concave at its anterior face only) in the anurans, amphicoelous in the caecilians, and usually opisthocoelous (centrum is concave at its posterior face) in the urodeles. Sternum appeared for the first time in the amphibians.
  6. The mouth is large; teeth are acrodont, homodont, and polyphyodont.
  7. Respiratory gaseous exchange is mostly cutaneous; pulmonary and buccopharyngeal respiration also occurs. Branchial respiration is performed by larvae and some adult urodeles.
  8. The heart is three-chambered with sinus venosus and conus arteriosus. Three pairs of aortic arches and well-developed portal systems are present; erythrocytes are nucleated.
  9. Kidneys are mesonephric; ureotelic.
  10. Meninges are the inner pia mater and outer dura mater; cranial nerves are 10 pairs.
  11. The middle ear consists of a single ear ossicle, the columella Auris which is the modified ‘hyomandibula’ of the fishes.
  12. Tympanum, lacrimal and harderian glands appeared for the first time in the amphibians.
  13. Sexes are separate and fertilization is mostly external. Development is mostly indirect.
  14. e.g. Bufo (toad), Rana (frog), Hyla (tree frog), Salamandra (salamander), Ichthyophis (limbless amphibian), Rhacophorus (flying frog).

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q8

Question 9.
Describe the male reproductive system of a frog with the help of a labelled diagram.
Answer:
Male Reproductive System of frog: The male reproductive system consists of a pair of yellowish and ovoid testes, which are attached to the kidneys and dorsal body wall by a double fold of peritoneum called mesorchium. Each testis is composed of innumerable seminiferous tubules which are connected to form 10 to 12 narrow tubules, the vasa efferentia. They enter the kidneys and open into the Bidders canal which is connected to the ureter through transverse canals of the kidney. The urinogenital ducts of both sides open into the cloaca.
AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q9

Question 10.
Write short notes on organs of special senses in frogs.
Answer:
Special senses: Frog has sense organs such as the organs of touch, taste, smell, sight, and hearing. The well-organized structures among them are eyes, and internal ears, and the rest are ‘cellular aggregations’ around nerve endings. The receptors of touch occur in the skin. Organs of taste are called taste buds that lie on small papillae of the tongue. The organs of smell are a pair of nasal chambers.

The organs of sight are a pair of eyes located in the orbits of the skull. Eyes are protected by eyelids. The upper eyelid is immovable. The lower eyelid is folded into a transparent nictitating membrane, which can be drawn across the surface of the eye. The retina of the eye contains both rods and cones. Cones provide ‘colour vision’ and rods are helpful in ‘dim light vision’.

The ear is useful for hearing and balance. It consists of a middle ear closed externally by a large tympanic membrane (ear drum) and a columella that transmits vibrations to the inner ear. The inner ear consists of a utriculus with three semicircular canals and a small sacculus.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 11.
List out the salient features of Exo and endoskeleton in reptiles.
Answer:
The exoskeleton of reptiles occurs in the form of horny epidermal scales, shields, and claws.
Endoskeleton:

  1. Skull is monocondylic and many have temporal fossae.
  2. Each half of the lower jaws is formed by six bones.
  3. Vertebrae are mostly procoelous.
  4. The first two cervical vertebrae are specialized into the atlas and axis.
  5. The vertebral column is distinguished into cervical, thoracic, lumbar, sacral, and caudal regions.
  6. Most living reptiles possess two sacral vertebrae.
  7. Interclavicular is associated with the pectoral girdle.
  8. Ribs are single-headed except in crocodilians.

Question 12.
List out the extant orders of the Class – Reptilia. Give two examples for each Order.
Answer:

  1. Chelonia – Chelone (marine green turtle), Testudo (terrestrial form), Trionyx (freshwater form)
  2. Rhynchocephalia – Sphenodon (a ‘living fossil’, endemic to New Zealand)
  3. Crocodilia – Crocodylus pulustris (Indian crocodile or mugger), Alligator (alligator), Gavialis gangeticus (Indian gavial or gharial)
  4. Squamata
    • Lizards – Hemidactylus (wall lizard), Chameleon, Draco (flying lizard)
    • Snakes
      • Poisonous Snakes: Naja naja (cobra), Ophiophagus hannah (King cobra), Bungarus (krait), Daboia/Vipera russelli (chain viper)
      • Non-Poisonous Snakes: Ptyas (rat snake), Tropidonotus (grass snake or pond snake)

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q12

Question 13.
What are the modifications that are observed in birds that help them in flight?
Answer:
So many modifications are observed in birds that help them in flight.

  1. Exo and endo skeletons and body structure features might have contributed to their successful aerial mode of life.
  2. The exoskeleton consists of epidermal feathers. Feathers are unique to birds. They are useful for flight, particularly the Quill feathers help in flight.
  3. The body is boat-shaped and streamlined.
  4. Four limbs are modified into wings.
  5. Many bones are neumatic with extensions of air sacs.
  6. All modern flying birds are provided with powerful breast muscles (flight muscles) chiefly the pectoralis major and pectoralis minor.
  7. Lungs are associated with air and seas.

Question 14.
What are the features peculiar to ratite birds? Give two examples of ratite birds.
Answer:
Ratite birds:

  1. These are modern flightless running birds.
  2. They are ‘discontinuous’ in their distribution like the lungfishes and marsupials.
  3. They are characterized by the presence of reduced wings.
  4. Feathers are without an interlocking mechanism.
  5. Rectrices are absent or irregularly arranged.
  6. Prren gland is absent.
  7. Pygostyle is rudimentary or absent.
  8. The sternum is like without a keel.
  9. Clavicles are absent, and syrinx is absent.
  10. The male animal has a penis.
  11. Young ones are precocial.
  12. Ex: Struthio camelus – (African ostrich); Dromaeus (Emu) Kiwi.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q14

Question 15.
Mention the most important features of the nervous system and sense organs in mammals.
Answer:

  • The nervous system and sense organs are well developed in mammals.
  • Mammals have relatively large brains when compared to that other animals in relation to body size.
  • The four optic lobes constitute corpora quadrigemina.
  • The two halves of the cerebrum are connected by the corpus callosum.
  • The central nervous system is enveloped by three meninges.
  • Eyes have movable eyelids with eyelashes.
  • The external ear has a large pinna middle ear and possesses three ear ossicles.
  • They are malleus, incus, and stapes, Cochlea of the internal ear is spirally coiled and bears the organ of Corti which is the receptor of sound.
  • Skin is one of the sense organs.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 16.
Write short notes on the following features of the eutherians.

  1. Dentition
  2. Endoskeleton

Answer:

  1. Dentition: The Dental formula of eutherians is i 3/3; c 1/1j pm 4/4, m 3/3; dentition is the codon, heterodont, diphyodont.
  2. Endoskeleton: Skull is dycondylic. Most mammals have seven cervical vertebrae; vertebrae are of the amphiplatyan type, sacral vertebrae are two or five, and ribs are double-headed.

Question 17.
Give an example for each of the following.

  1. A viviparous fish
  2. A fish possessing electric organs
  3. A fish possessing poison sting
  4. An organ that regulates buoyancy in the body of a fish
  5. An oviparous animal with milk-producing glands.

Answer:

  1. Scoliodon fish is viviparous fish.
  2. Torpedo fish is possessing electric organs.
  3. Dasyatis/Trygon fish possess poison sting.
  4. The air bladder regulates buoyancy in the body of a fish.
  5. Ornithorhynchus anatinus (Duck-billed platypus) is an oviparous animal with milk-producing glands.

Question 18.
Mention two similarities between
(a) Aves and mammals
(b) A frog and a crocodile
(c) A lizard and a snake
Answer:
(a) Aves and mammals:

  • Aves and mammals are Triploblaste and bilaterally symmetrical.
  • The heart is four-chambered in both.

(b) A frog and a crocodile:

  • Erythrocytes are nucleated in both.
  • Frogs and crocodiles are uriotelic animals.

(c) A lizard and a snake:

  • Lizards and snakes are reptilian animals.
  • The heart is incompletely four-chambered.
  • Jacobson’s organs, the highly developed specialized olfactory structures are present.

Question 19.
Name the following animals.

  1. A limbless amphibian
  2. The largest of all living animals
  3. An animal possessing dry and cornified skin
  4. ‘National animal’ of India.

Answer:

  1. Ichthyophis is a limbless amphibian.
  2. Balaenoptera musculus (Blue whale) is the largest of all living animals.
  3. Crocodylus is an animal possessing dry and cornified skin.
  4. Panther Tigris (tiger) is the National animal of India.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 20.
Write the generic names of the following.

  1. An oviparous mammal
  2. Flying fox
  3. Blue whale
  4. Kangaroo

Answer:

  1. An oviparous mammal’s generic name is Ornithorhynchus (Duckbilled platypus).
  2. Flying fox’s generic name is Pteropus.
  3. The blue whale’s generic name is Balaenoptera musculus
  4. Kangaroo generic name is Macropus

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 3 Animal Diversity-I: Invertebrate Phyla Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 3 Animal Diversity-I: Invertebrate Phyla

Very Short Answer Type Questions

Question 1.
What physical feature pertaining to the organism and its medium do you notice in a sponge body form in which sponges can be/were identified as animals and not plants? What do you call the region in the sponge body in which you noticed that feature?
Answer:
Sponges are primitive multicellular and sessile animals and have a cellular level of organisation. The body wall is composed of two layers separated by matrix mosohyl, and are heaving canal system for transport of water through Ostia, having a cavity in the body called a spongocoel hence the sponge are animals and are not plants.

Question 2.
What are the different structures that make up the internal skeleton of a sponge? What are the chemicals involved in the formation of these structures?
Answer:
The internal skeleton of a sponge is made up of different types of spicules.
Calcareous spicules made up of CaCO3.
Ex: Sycon
Siliceous spicules – are made up of Silicon dioxide – glass.
Ex: Euplectella
Spongin fibres.
Ex: Spongilla

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 3.
What are the functions of the canal system of sponges?
Answer:
The functions of the canal system of a sponge are gathering of food, respiratory exchange of gases, and removal of wastes.

Question 4.
What are the two chief morphological ‘body forms’ of cnidarians? What are their chief functions?
Answer:
The body form of Cnidarians is polyp and medusa. Polyp produces medusae by asexual reproduction. Medusae produce polyps by sexual reproduction.

Question 5.
What is metagenesis? Animals belonging to which phylum exhibit metagenesis?
Answer:
Cnidarians show two basic body forms called polyp and medusa. Cnidarians which exist in both forms exhibit alternation of generations called metagenesis.

Question 6.
What is the cnidarian group with quantitatively/relatively large mesoglea? What is the significance of such a well-developed mesoglea pertaining to the aquatic life of that group?
Answer:
The Scyphozoa of cnidarian animals have large mesoglea, it is the significance of these animals.

Question 7.
What is the chief difference between the hydrozoans and the rest, of the cnidarians regarding the germinal layer (s) in which its ‘defencive structures or cells of defence occur?
Answer:
The defencive structures Cnidocytes or Cnidoblasts occur only in the ectoderm, in the hydrozoans in the rest of the Cnidarians the cnidocytes occur in both ectoderm and endoderm.

Question 8.
What are the excretory cells of flatworms called? What is the other important function of these specialized cells?
Answer:
The excretory cells of flatworms are flame cells. Another important function of these specialized cells is osmoregulation.

Question 9.
Distinguish between amphids and phasmids.
Answer:
Amphids: These are the cuticular depressions present on the lips surrounding the mouth in the nematodes such as Aphasmidia animals and serve as Chemoreceptors.
Phasmids: These are the well-developed sensory organs and they occur in some nematodes such as phasmidia animals.

Question 10.
What is the essential difference between a ‘flat worm’ and a ’round worm’ with reference to the perivisceral area of the ‘bodies’.
Answer:
With the reference to the perivisceral area of the body, the flatworms have dorso-ventrally flattened bodies. The body is not segmented, but some of the animals exhibit pseudometamerism. In the Nematoda the body is circular in cross-section, hence the name roundworms, body is not segmented.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 11.
How do you account for the origin of the perivisceral space in the body of a nematode and an annelid?
Answer:
The perivisceral space in the body of a nematode is circular in cross-section. Hence the name ‘wound worms’ body unsegmented. In an annelid, the body is segmented by septa into segments or metameres (annulus – little rings) some of them (Nereis) possess lateral appendages parapodia.

Question 12.
What is metamerism? What is the essential difference between the mode of formation of individual morphological body units of a tapeworm and those of an earthworm?
Answer:
The body is divided into segments like units called metameres. Like the divination in known as metamerism.
Ex: Earthworm.
In tapeworm body segments are pseudometameres.
In earthworms, body segments are true segments or metameres.

Question 13.
How do you distinguish a ‘hirudineaun’ from the rest of the annelids, based on the morphological features pertaining to metamerism? How does the coelom of a leech differ from the coelom of an earthworm with reference to its contents?
Answer:
In hirudinean like Leach, the body is with a definite number of segments. The segments are externally sub-divided into annute, internal segmentation’ is absent.
In Leech coelom is filled with a characteristic tissue called botryoidal tissue. In earthworms, the coelom is filled with coelomic fluid.

Question 14.
What do you call the locomotor structures of Nereis? Why is Nereis called a polychaete?
Answer:
Locomotor structures of Nereies are parapodia. The parapodia bear many setae that help in locomotion hence the name Polychaeta.

Question 15.
What is botryoidal tissue?
Answer:
The coelom of Leech is filled with a characteristic tissue called botryoidal tissue, it is resembling a bunch of grapes. They range from excretion to storage of iron, calcium, and revascularization in areas of injury.

Question 16.
What is the difference between the epidermis of a Nematoda and that of an annelid? How does a nematode differ from an annelid with reference to the musculature of the body wall?
Answer:
The epidermis of Nematoda is syncytial and the epidermis of annelid animals is informed by one cell thick ectodermal epithelial cells.

Question 17.
What do you call the first and second pairs of cephalic appendages of a scorpion?
Answer:
The first and second pairs of cephalic appendages of a Scorpion are Chelicerae and Pedipalpi.

Question 18.
What is the uniqueness of the first two pairs of cephalic appendages of a crustacean compared to those of the other extant arthropods?
Answer:
In crustaceans, cephalic appendages are two pairs of antennae (antennules and antennae). It is the unique feature of Crustaceans compared to those of the other extent arthropods.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 19.
What is the sub-phylum to which ‘ticks’ and ‘mites’ belong? How do you distinguish them from insects with reference to their walking legs?
Answer:
Ticks and mites belongs to the sub-phylum Chelicerata and class Arachnida. These have four pairs of walking legs.

Question 20.
What are the respiratory structures of Limulus and Palamnaeus respectively?
Answer:
The respiratory structures of Limulus are book gills, and in palamnaeus are book-lungs.

Question 21.
What ara antennae? What is the arthropod group without antennae?
Answer:
Antennae are the sensory organs, of the animals of sub-phylum-Mandibulata of arthropod bear antennae.

Question 22.
What do you call the perivisceral cavity of an arthropod? Where from is it derived during development?
Answer:
The perivisceral cavity of an Arthropoda is a haemocoel, it is not true coelom, but derived from mostly the embryonic blastocoel.

Question 23.
Which arthropod, you have studied, is called a living fossil? Name its respiratory organs.
Answer:
The arthropod animal Limulus is called a living fossil, it is respiratory organs are book-gills.

Question 24.
How do you identify a Chiton from its external appearance? How many pairs of gills help in the respiration of Chiton?
Answer:
Chiton is bilaterally symmetrical and dorsoventrally flattened. Shell is dorsal and consists of eight transverse plates. Poat is ventral elongated and flat. Gills are 6 to 88 pairs helps in respiration.

Question 25.
What is the function of the radula? Give the name of the group of mollusks that do not possess a radula.
Answer:
The buccal cavity contains a file-like rasping organ called radula for feeding, except for the bivalves and tusk of Molluscs.

Question 26.
What is the other name for the gill of a mollusc? What is the function of osphradium?
Answer:
The other name for the gill of a mollusc is Ctenidia. The main function of Osphradium is to test the purity of water.

Question 27.
What is Aristotle’s lantern 7 Give one example of an animal possessing it?
Answer:
In the mouth of the sea Urchin a complex five Jawed masticatory apparatus called Aristotle’s Lantern.
Ex: Echinus.

Question 28.
What is the essential difference between the Juveniles and adults of echinoderms, symmetry-wise?
Answer:
The adult echinoderms are radially symmetrical (pentamerous radial symmetry), but Juveniles (Larvae) are bilaterally Symmetrical.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 29.
What are blood glands in pheretima?
Answer:
Blood glands are present in the 4th, 5th, and 6th segments of pheretima. They produce blood cells and haemoglobin which is dissolved in the plasma.

Question 30.
What are spermathecae on the body of pheretima?
Answer:
In Pheretima there are four pairs of spermothecae are located in the segments 6th to 9th as one pair in each segment. This receives and stores spermatozoa during copulation.

Short Answer Type Questions

Question 1.
Write short notes on the salient features of the anthozoans.
Answer:

  1. Anthozoans are commonly referred to as sea anemones.
  2. Anthozoa includes sea anemones, corals, and sea pens.
  3. All are marine forms. These are solitary or colonial.
  4. They are sedentary and only have polypoid information.
  5. Coeienteron is divided into several compartments by vertical septa called mesenteries.
  6. Mesoglea contains connective tissue.
  7. Cnidocytes occur both in the ectoderm and endoderm and are cellular and contain amoebocytes.
  8. Germ cells are derived from the endoderm. Ex: Adamsia (sea anemone), Gorgonia (sea fan), Pennatula (sea pen).

Question 2.
What is the class to which the flukes belong? Write short notes on the chief characters of the group.
Answer:
Flukes belong to the class Trematoda of Phylum-Platyhelminthes.

  1. Trematoda organs are commonly called flukes.
  2. These are parasitic on other animals.
  3. The body is covered by a thick cuticle and bears two suckers, an oral and a ventral.
  4. The mouth is anterior and the intestine is bifurcated.
  5. These are bisexual (monoecious).
  6. Life history is complex with many hosts and different types of stages – miracidium, sporocyst, redia, cercaria, etc. Ex: Fasciola (Liver fluke), Schistosoma (blood fluke).

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 3.
What are the salient features exhibited by Polychaetes?
Answer:

  1. These are commonly known as bristle worms.
  2. All are marine. Many are burrowing, others are free swimming or crawling or tubicolous.
  3. Head is distinct with sensory structures like eyes, antennae, palps, and cirri.
  4. Clitellum is absent.
  5. Each segment has a pair of lateral appendages called parapodia in which bundles of setae are arranged.
  6. Animals are unisexual. Most segments bear glands. Gonoducts are absent.
  7. Gametes are shed into the coelom.
  8. Fertilization is external.
  9. Development includes a trochophore larva. Ex: Nereis

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla SAQ Q3

Question 4.
How do the hirudineans differ from the polychaetes and oligochaetes?
Answer:

  1. Definite number of body segments are present in hirudinean but many segments are present in polychaetes and earthworms.
  2. Locomotion in leeches is by suckers but body setae in oligochaetes and parapodia in polychaetes. Parapodia also help in respiration.
  3. Temporary clitellum during the breeding season is present in leeches but clitellum is absent in polychaetes and permanent clitellum is present in oligochaetes.
  4. Hirudineans are bisexuals, oligochaetes are bisexual and polychaetes are unisexual animals.
  5. Coelom is reduced on leeches, but coelom is spacious in oligochaetes and polychaetes.
  6. Development is direct in leeches and earthworms but indirect in polychaetes.
  7. Nutrient tissue called botryoidal tissue fills the coelom in hirudinean.
  8. Anterior and posterior suckers are present in hirudineans. Such suckers are absent in polychaetes and oligochaetes. Ex: Pheretima, Tubifex

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 5.
What are the chief characteristics of crustaceans?
Answer:

  1. This includes prawns, crabs, lobsters, crayfishes, etc.
  2. Mostly marine, a few are fresh water and some are adapted to terrestrial life.
  3. In most species, the head and thorax fuse to form a cephalothorax.
  4. Cephalic appendages are five pairs – first antennae (antennules) second antennae, mandibles, first maxillae and second maxillae.
  5. Thoracic and abdominal appendages are typically biramous
  6. Respiration is by gills.
  7. Excretory organs are green glands or antennal glands.
  8. Sense organs include statocysts, compound eyes, and antennae.
  9. Gonopores are paired.
  10. Development is direct or indirect involving several larval stages. The basic larva is nauplius. Ex: Palaemon (Prawn); Cancer (Crab).

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla SAQ Q5

Question 6.
Mention the general characters of Arachnida.
Answer:

  1. This includes scorpions, spiders, ticks, and mites.
  2. Primarily they are all terrestrial.
  3. Prosoma bears one pair of pre-oral chelicerae, one pair of post-oral pedipalps, and four pairs of walking legs.
  4. In spiders each chelicera bears a fang into which the poison gland opens.
  5. Abdominal appendages are modified into book lungs, spinnerets, pectines, etc.
  6. Telsun is usually absent. It is present as a sting in scorpions.
  7. Respiration is by book lungs or tracheae.
  8. Excretory organs are coaxial glands and malpighian tubules.
  9. Scorpions are viviparous.
  10. Development is direct. Ex: Palamnaeus (scorpion); Aranea (spider).

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla SAQ Q6

Question 7.
Compare briefly a centipede and a millipede.
Answer:

CentipedeMillipede
1. Centipedes are commonly called a hundred leggers.1. Millipedes are commonly called a thousand leggers.
2. Body is divided into the head and trunk.2. Body is divided into the head, thorax, and abdomen.
3. Centipedes are carnivorous in nature.3. Millipedes are herbivorous in nature.
4. In each segment one pair of appendages are present.4. In each segment 2 pairs of appendages are present.
5. Single genital aperture occurs at the posterior end of the trunk.5. Single genital aperture opens in the anterior part of the trunk.
6. Head bears one pair of antennae, one pair of mandibles, and two pairs of maxillae. Ex: Scolopendra.6. Head bears paired antennae, mandibles, and maxillae. The maxillae are fused to form a plate-like under lip, the gnathochilarium-a masticatory structure. Ex: Spirostreptus.

Question 8.
Cephalopods show several unique or advanced features when compared to the other molluscs. Discuss briefly.
Answer:

  1. The class Cephalopoda includes cuttlefishes, squids, octopuses, nautilus, etc.
  2. The Head is discrete and bears very conspicuous eyes.
  3. Shell is either present (e.g.: Sepia) or absent (e.g.: Octopus). When present it may be multi-charactered and external (e.g.: Nautilus) or internal (e.g.: Loligo).
  4. The foot is modified into eight to ten arms (tentacles) present around the mouth and siphons.
  5. Some Cephalopods (e.g: Sepia) possess an ink gland as a defensive adaptation.
  6. Ctenidia are two or four in number – dibranchiate. e.g.: Sepia and tetrabranchiates. (e.g.: nautilus)
  7. The brain is complex and is protected by a cartilaginous cranium.
  8. Eyes are superficially similar to those of vertebrates.
  9. Development is direct. Ex: Architeuthis (giant squid).

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 9.
Which class of Mollusca represents the primitive molluscs? What are their chief features?
Answer:
The primitive molluscs are represents the class Aplacophora of Phylum-Mollusca.
These are primitive forms with ‘worm-like’ bodies. These are marine forms without mantle, shell, foot, and nephridia. The Head is poorly developed. A rasping organ radula is present in the buccal cavity. Cuticle contains calcareous spicules. Eyes, statocysts, and tentacles are absent. The heart consists of a single auricle and a ventricle. A pair of gonads are present in some, there is a mid-ventral groove that is homologous to the foot of the other molluscs.
Ex: Neonmia, Chaetoderma.

Question 10.
What are the salient features of the echinoids?
Answer:

  1. It includes sea urchins, heart urchins, sand dollars, etc. The body is ovoid or discoidal and covered by movable spines.
  2. Arms are absent, tube feet are arranged in five bands, and bear suckers.
  3. Ossicles of the body unite to form a rigid test or corona or case.
  4. Pedicellaria is “three jawed”.
  5. Anus and madreporite are aboral in position.
  6. Ambulacral grooves are closed.
  7. A complex five-jawed masticatory apparatus called Aristotle’s lantern is present just inside the mouth. It is absent in heart urchins.
  8. Life history includes a larval form called echinopluteus.
  9. Specialized gills called peristomial gills as present in sea urchins. Eg: Salmacis (Sea urchin), Echino Cardium (Heart urchin), Clypeastoer (Cake Urchin).

Question 11.
Mention the salient features of Holothuroidea.
Answer:
Holothuroidea: This class includes sea cucumbers. Body elongated in the oro-aboral axis. Arms, spines, and pedicellariae are absent skin are soft and leathery (Coriaceous). The dermis contains microscopic, isolated ossicles. The madreporite is internal, suspended in the perivisceral coelom. Tube feet are provided with suckers. The mouth is surrounded by retractile feeding tentacles, which are modified tube feet, chief gas exchange organs are a pair of respiratory trees that arise from the wall of the cloaca and form branched tubes in the perivisceral coelom. The development includes auricularia and doliolaria larvae.
Ex: Cueumaria, synaptic, Thyone.

Question 12.
What is the function of nephridia?
Answer:

  • The nephridia of pheretima are ectodermal in origin and are metanephridia.
  • Several types of nephridia occur in pheretima but are fundamentally similar in structure.
  • Which opens outside through the nephridiopore – The nephridia tolled open nephridia. Ex: Septal nephridia.
  • Those who do not have nephridiopore are called closed-type nephridia. Ex: Pharyngeal nephridia.
  • Those open at the outer surface are called exonephridia.
  • The nephridia play an important role in osmoregulation.
  • Earthworms mostly excrete urea as the excretory product and are described as ureotelic animals.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 13.
How many types of nephridia occur in pheretima and how do you distinguish them?
Answer:
In Pheretima three types of nephridia are present.

  1. Septal nephridia: The septal nephridia are present on the intersegmental septum from 15 & 16 segments onwards to last and are opened into the alimentary canal.
  2. Integumentary nephridia: The integumentary nephridia attached to the inner body wall from the 3rd segment to the last. They open to the exterior on the body surface by nephridiopores.
  3. Pharyngeal nephridia: The pharyngeal nephridia present three paired tufts in the segments 4tfl, 5th, and 6th. They open into the buccal cavity and pharynx.

Question 14.
Give an account of the hearts in the circulatory system of pheretima.
Answer:
Hearts in Pheretima: The dorsal blood vessel and the ventral blood vessel are connected by a pair of pulsatile hearts, in each of the seventh, ninth, twelfth, and thirteenth segments. Of these four pairs, the anterior two pairs connect only the dorsal blood vessel to the ventral blood vessel. Hence they are ’ called lateral hearts. The posterior two pairs connect both the dorsal blood vessel and the supra-oesophageal blood vessel with the ventral blood vessel. Hence, they are called lateral oesophageal hearts. These two types of hearts also differ in the number and arrangement of their valves. Four pairs of valves are present in each lateral heart, while three pairs of valves are present in each lateral oesophageal heart. Hearts allow the blood to flow into the ventral blood vessel only.
AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla SAQ Q14

Long Answer Type Questions

Question 1.
Draw a labelled diagram of the reproductive organs of Pheretima.
Answer:
Reproductive organs of pheretima: Pheretima is a hermaphrodite (bisexual). There are two pairs of testes. One pair each present in the 10th and 11th segments. Their vasa deferentia run up to the 18th segment where they join the prostatic ducts. Two pairs of seminal vesicles present in the 11th and 12th segments are sacs in which spermatogonia mature into spermatozoa. The common prostatic and spermatic ducts open to the exterior by a pair of male genital pores on the ventrolateral sides of the 18th segment. Two pairs of accessory glands’ are present one pair each in the 17th and 19th segments. Four pairs of spermathecae are located in the segments 6th to 9th (one pair in each segment). They receive and store spermatozoa (spermatophores) during copulation.
AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla LAQ Q1
One pair of ovaries is attached to the posterior face of the inter-segmental septum of the 12th and 13th segments. Oviducal funnels are present beneath the ovaries and they continue into oviducts (14th segment). They join together and open to the exterior on the ventral side of the 14th segment by a single median female genital pore.

Question 2.
Describe the digestive system and process of digestion in pheretima.
Answer:
The digestive system in pheretima: The alimentary canal is a straight tube and runs from the first to the last segment of the body. The mouth opens into the buccal cavity (1-3 segments) which leads into the muscular pharynx (4th segment). A small narrow tube, oesophagus (5-7 segments), continues into a muscular gizzard (8th segment). It helps in grinding the small particles of food in the decaying leaves (grinding mill). The stomach extends from segments 9 to 14. The food of earthworms is decaying leaves and other organic matter mixed with the soil. Calciferous glands, present in the stomach, neutralise the humic acid present in the humus of the soil. The intestine starts from the 15th segment and continues till the last segment.

A pair of short and conical intestinal caeca project from the intestine in the 26th segment. An internal median fold of the dorsal wall of the intestine called typhiosole, helping in increasing the area of absorption, is poorly developed in Pheretima (between the 26th and the rectum which occupies the last 23 to 28 segments). The alimentary canal opens to the exterior by a small rounded aperture called the anus. The ingested soil rich in organic matter passes through the digestive tract where digestive enzymes break down complex food into smaller absorbable units. These simpler molecules are absorbed through intestinal membranes and are utilized for various metabolic activities.
AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla LAQ Q2
Process of digestion: Digestion in earthworms is extracellular. Earthworm obtains their nourishment from the organic debris (detritus) present in the soil. So it is called a detritivore. The pharynx is ejected due to the inside out of the buccal chamber. The pharynx, with the help of its radial-dilator muscles, works as a suction pump in feeding.

The organic food along with the swallowed soil particles is sucked into the pharynx, where it mixes with the salivary secretion. The mucin in the saliva lubricates the gut wall for the easy passage of food and also helps in the formation of the bolus. The proteolytic enzyme in the saliva partly digests the proteins. Then the food reaches the gizzard. Its circular muscle and the thick cuticle grind the food into fine particles. In this state, the food is easily acted upon by the digestive enzymes in the stomach and intestine.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

The intestinal juice of an earthworm is comparable to the pancreatic juice of higher animals. All these enzymes like proteases, amylases, and lipases act upon the finely ground food and digest the organic matter in it. Proteases digest proteins into amino acids, amylases digest carbohydrates into glucose and lipases digest lipids into fatty acids and glycerol.

The digested food is absorbed by the intestinal epithelium in the typhlosolar region. The extensive capillary network of blood vessels of the intestine plays a vital role in absorption. The typholosole helps in increasing the area of absorption. The undigested food then passes to the rectum, where water is absorbed from the undigested food. Then the undigested matter is egested out through the anus in the form of worm castings.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 2 Structural Organisation in Animals Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 2 Structural Organisation in Animals

Very Short Answer Type Questions

Question 1.
The body of sponges does not possess a tissue level of organisation, though it is made up of thousands of cells. Comment on it.
Answer:
Even though the sponge’s body is made up of thousands of cells, they exhibit a cellular grade of organization, due to the absence of sensory and nerve cells spongs do not possess a tissue level of organisation.

Question 2.
What is the tissue level of organisation among animals? Which metazoans exhibit this organisation?
Answer:
The tissue level of organisation is the lowest level of organisation among the eumetazoans, exhibited by diploblastic animals like cnidarians.

Question 3.
Animals exhibiting which level of organisation lead the relatively more efficient way of life when compared to those of the other levels of organisation? Why?
Answer:
Animals exhibit an organ-system level of organisation of the animals and are exhibited by the triploblastic animals.
Organ level of organisation is a further advancement over the tissue level in the evolution of levels of organisation.

Question 4.
What is monaxial heteropolar symmetry? Name the group of animals in which it is the principal symmetry.
Answer:
Monaxial heteropolar symmetry: When any plane passing through the central axis of the body divides an organism into two identical parts is called Monaxial heteropolar symmetry or radial symmetry. It is the principal symmetry of diploblastic animals such as Cnidarians and ctenophores.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 5.
Radial symmetry is an advantage to sessile or slow-moving organisms. Justify this statement.
Answer:
Animals showing radial symmetry live in water and they can respond equally to stimuli that arrive from all directions. Thus, radial symmetry is an advantage to sessile or slow-moving animals.

Question 6.
What is cephalization? How is it useful to its possessors?
Answer:
Cephalization: Concentration of nerve (Brain) and sensory cells at the anterior end of the body is called Cephalization. As a result of cephalization, these animals can sense the new environment and are more efficient than the other animals in seeking food, locating mates, and avoiding or escaping from predators.

Question 7.
Mention the animals that exhibited a ‘tube-within-a-tube’ organisation for the first time? Name their body cavity.
Answer:
Cnidarians and some flat warms are the first animals to exhibit a ‘tube-within-a-tube’ organisation. The body cavity is pseudocolor.

Question 8.
Why is the true coelom considered a secondary body cavity?
Answer:
During the embryonic development of the coelomates, the blastocoel is replaced by a true coelom derived from the mesoderm. So, the true coelom is also called ‘the secondary body cavity.

Question 9.
What are retroperitoneal organs?
Answer:
Certain organs such as the kidneys of the vertebrates are covered by the parietal peritoneum only on their ventral side. Such a peritoneum is called the ‘retroperitoneum’ and the organs lined by it are called ‘retroperitoneal organs.

Question 10.
If the mesentoblast cell is removed in the early embryonic development of protostomes what would be the fate of such animals?
Answer:
In the protostomes, the mesentoblast cell of the early embryo divides to form mesodermal blacks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal black leads to the formation of Schizocoelom. If the mesentoblast cell is removed in the early embryonic development of protostomes it will cause no ceolome in their animals.

Question 11.
What is enterocoelom? Name the enterocoelomate phyla in the animal kingdom.
Answer:
Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called ‘enterocoelomates’.
Echinoderms, hemichordates, and chordates are enterocoelomates.

Question 12.
Stratified epithelial cells have a limited role in secretion. Justify their role in our skin.
Answer:
The main function of stratified epithelial cells is to provide protection against chemicals and mechanical stress. Hence the stratified epithelial cells have a limited role in secretion.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 13.
Distinguish between exocrine and endocrine glands with examples.
Answer:
Exocrine glands are provided with ducts. Secrete mucus, saliva, earwax, oil, milk, digestive enzymes, and other cell products.
Endocrine glands are ductless and their products are hormones that are not sent out via ducts but are carried to the target organs by blood.
Ex: Pituitary gland.

Question 14.
Distinguish between holocrine and apocrine glands.
Answer:
Apocrine glands in which the apical part of the gland cell in pinched off along with the secretory product.
Ex: Mammary glands
Holocrine glands, in which the entire cell disintegrates to discharge the contents.
Ex: Sebaceous glands

Question 15.
Mention any two substances secreted by mast cells and their functions.
Answer:
Mast cells secrete heparin – an anticoagulant, histamine, bradykinin – vasodilators, and serotonin – vasoconstrictor. Vasodilators cause inflammation in response to injury and infection.

Question 16.
Distinguish between a tendon and a ligament.
Answer:
Tendons are the collagen fibrous tissue of dense regular connective tissue which attaches the skeletal muscles to bones.
Ligaments are also the collagen fibers tissue of dense regular connective tissue which attach bones to other bones.

Question 17.
Distinguish between brown fat and white fat.
Answer:
White fat: It is the predominant type in adults, the adipocyte has a single large lipid droplet. White fat is metabolically not active.
Brown fat: It is found in fetuses and infants. Adipocyte of Brown fat has several small ‘lipid droplets’ and are metabolically active and generates heat to maintain body temperature required by infants.

Question 18.
What is the strongest cartilage? In which regions of the human body, do you find it?
Answer:
The fibrous cartilage is the strongest of all types of cartilage. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

Question 19.
Distinguish between osteoblasts and osteoclasts.
Answer:
Osteoblasts are immature bone cells that secrete the organic components of the matrix and also play an important role in the mineralization of bone and become Osteocytes. Osteoclasts are phagocytic cells involved in the resorption of bone.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 20.
Define Osteon.
Answer:
In compact bone structure, a Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or Osteon.

Question 21.
What are Volkmann’s canals? What is their role?
Answer:
In compact bone structure, the Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals.

Question 22.
What is a Sesamoid bone? Give an example.
Answer:
Sesamoid bones are formed by ossification in tendons.
Eg: Patella (Knee cap) and Pisiform bone of the wrist of a mammal.

Question 23.
What is lymph? How does it differ from plasma?
Answer:
Lymph is a colourless fluid. It lacks RBC, platelets, and large plasma proteins, but has more number of leucocytes. It is chiefly composed of plasma and lymphocytes.

Question 24.
What is the hematocrit value?
Answer:
The percentage of the total volume occupied by RBCs in the blood is called the hematocrit value.

Question 25.
What are intercalated discs? What is their significance?
Answer:
The dark lines across cardiac muscle are called intercalated discs (IDS). These discs are highly characteristic of the cardie muscle.

Question 26.
“Cardiac muscle is highly resistant to fatigue”. Justify.
Answer:
The cardiac muscle is highly resistant to fatigue because it has numerous acrosomes, many molecules of myoglobin (Oxygen storing pigment), and a copious supply of blood which facilitate continuous aerobic respiration. The muscles are immune to fatigue and work tirelessly from the embryonic state until death.

Question 27.
Distinguish between ‘nucleus’ and ‘ganglion’ with respect to the nervous system.
Answer:
A group of cell bodies in the Central Nervous System is called a ‘nucleus’, and in the Peripheral Nervous System, it is called a ‘ganglion’.

Question 28.
Distinguish between tracts and nerves with respect to the nervous system.
Answer:
Groups of axons (nerve fibers) in the central nervous system (CNS) are called tracts’ and in the peripheral nervous system (PNS) they are called ‘nerves’.

Question 29.
Name the glial cells that form the myelin sheath around the axons of the central nervous system and peripheral nervous system respectively.
Answer:
In the central nervous system, the glial cells are called ‘Oligodendrocytes’, in the peripheral nervous system, the glial cells are called Schwann cells.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 30.
Distinguish between white matter and Greymatter of ‘CNS’.
Answer:
Myelinated nerve fibers occur in the white matter of the CNS, and in most peripheral nerves, and Non-myelinated axons are commonly found in the grey matter of the CNS and autonomous nervous system.

Question 31.
What are microglia and what is their origin add a note on their function.
Answer:
Microglial cells are the Neurogila (supporting cells) of cells of CNS which are phagocytic cells, of mesodermal origin.

Question 32.
What are pseudounipolar neurons? Where do you find them?
Answer:
In a unipolar neuron, the soma or cyton is found in the dorsal root ganglion they are called pseudounipolar neurons. These are found in spinal nerves.

Short Answer Type Questions

Question 1.
Describe the four different levels of organization in metazoans.
Answer:
The levels of organisation in metazoans are as follows:
1. Cellular level of organisation: It is the lowest level of organisaiton among the metazoans and ‘ is exhibited by the sponges. Different types of cells are functionally isolated due to the absence of sensory and nerve cells. There is no uniformity of labor among the cells and they don’t form tissues.

2. Tissue level of organisation: This is the lowest level of organisation among the eumetazoans, exhibited by diploblastic animals like the cnidarians. In these animals, the cells which perform the same function are arranged into tissues due to the presence of nerve cells and sensory cells.

3. Organ level of organisation: An aggregation of different kinds of tissues that are specialized for a particular function is called an organ. Organ level of organisation is a further advancement over the tissue level in the evolution of levels of organisation. It is the first time appeared in the members of platyhelminths.

4. Organ-system level: It is the highest level of organisation among the animals and is exhibited by triploblastic animals such as the flatworms, nematodes, annelids, arthropods, molluscs, echinodermates, and chordates. High specilized sensory and nerve cells bring about a higher level of coordination and integration among the various organ systems to lead an efficient way of life.

Question 2.
In which group of bilaterians do you find solid bauplan? Why is it called so?
Answer:
The bilaterian’s body plan is the solid bauplan when only one plane that passes through the identical axis divides an organism into two identical parts, it is called bilateral symmetry. It is the principal type of symmetry in triploblastic animals.

Bilaterally symmetrical animals are more efficient than the other animals in seeking food locating mates and in avoiding or escaping from predators, because of cephalization. As a result of cephalization, bilaterally symmetrical animals can sense the new environment into which they enter and respond more efficiently and quickly.

Question 3.
Mention the advantages of coelom over pseudocolor.
Answer:
Advantages of coelom over pseudocoelom:
1. Visceral organs of eucoelomates are muscular (because of their association with mesoderm) and so they can contract and relax freely independent of the muscular movements of the body wall in the coelomic space, e.g.: peristaltic movements of the alimentary canal.
2. Gametes are released into the coelom in some invertebrates (which do not have glnoducts) and in the female vertebrates.
3. Coelomic fluid receives excretory products and stores them temporarily before their elimination.
4. In the eucoelomates, the mesoderm comes into contact with the endoderm of the alimentary canal, and it causes ‘regional specialization of the gut, such as the development of gizzard, stomach, etc., This is referred to as ‘primary induction’. In the case of the pseudocoelomates, due to the absence of such a contract between the gut and the mesoderm, the wall of the gut does not show complex and highly specialized organs.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 4.
Describe the formation of schizocoelom and enterocoelom.
Answer:
Schizocoelom: During embryonic development, a specialized cell called 4d blastomere is formed. The cells formed from these cells divide and redivide and develop blocks of mesoderm in blastocoel. The blocks fuse and form the mesodermal band which later on splits to form the Schizocoelom. This type of coelom is present in Annelida, Arthropoda, and Mollusca.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q4

Enterocoelomates: Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates, and chordates are enterocoelomates. In these animals, mesodermal ouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q4.1

Question 5.
Describe briefly the three types of intercellular junctions of epithelial tissues.
Answer:
The specialized ‘junctions’ provide both structural and functional links between the individual cells of an epithelium. They show different types of junctions so as to serve the specific needs of that tissue. They are
(A) Tight junctions: These junctions between epithelial cells prevent ‘leakages’ of body fluids. For example, they prevent leakages of water into the surrounding cells in our sweat glands. The plasma membranes of adjacent cells are tightly pressed against each other and are bound together by specific proteins.

(B) Desmosomes: Muscle cells are provided with “desmosomes” (anchoring junctions) which act as ‘rivets’ binding the cells together into strong sheets. Intermediate filaments made of protein ‘keratin’ anchor desmosomes in the cytoplasm.

(C) Gap junctions: They provide continuous cytoplasmic channels between adjacent cells. Various types of ions, sugar molecules amino acids, etc. can pass from a cell to an adjacent cell through ‘gap junctions. They occur in many types of tissues including the ‘cardiac muscles’ where they allow rapid conduction of impulses or depolarization.

Question 6.
Give an account of glandular epithelium.
Answer:
Glandular epithelium: Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types.
Unicellular glands: Consisting of isolated glandular cells such as goblet cells of the gut.
Multicellular glands: Consisting of clusters of cells such as salivary glands. On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands.
Exocrine glands: These glands provided with ducts secrete mucus, saliva, earwax, oil, milk, digestive enzymes, etc.
Endocrine glands: These glands are ductless and their products are hormones, which are carried to the target organs by blood.
Based on the mode of secretion, Exocrine glands are further divided into three types.
1. Merocrine glands: Which release the secretory granules without the loss of other cellular material.
Ex: Pancreas
2. Apocrineglands: The apical part of the cell is pinched off along with the secretory product.
Ex: Mammary glands
3. Holocrine glands: The entire cell disintegrates to discharge the contents.
Ex: Sebaceous gland

Question 7.
Give a brief account of the cells of areolar tissue.
Answer:
Areolar tissue: It is one of the most widely distributed connective tissue in the body. It forms the packing tissue in almost all organs. Areolar tissue forms the subcutaneous layer of the skin. It has cells and fibres. Cells of the areolar tissue are fibroblasts, mast cells, macrophages, adipocytes, and plasma cells.
1. Fibroblasts are the most common cells which secrete fibres. The inactive cells are called fibrocytes.
2. Mast cells secrete heparin (an anticoagulant), histamine, bradykinin (vasodilators), and serotonin vasoconstrictor). Vasodilators cause inflammation in response to injury and infection.
3. Macrophages are amoeboid cells, Phagocytic in function, and act as internal scavengers. They are derived from the monocytes of blood. Tissue-fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
4. Plasma cells are derived from the B-lymphocytes and produce antibodies.
5. Adipocytes are specialized cells for the storage of fats.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q7

Question 8.
Describe the three types of cartilage.
Answer:
Cartilage is a solid, but semi-rigid (flexible) connective tissue. Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.

Cartilage is a solid, but semi-rigid (flexible) connective tissue It resists compression. Matrix is firm but somewhat pliable. It has collagen fibres, elastic fibres (only in the elastic cartilage), and matrix-secreting cells called chondroblasts. These cells are enclosed in fluid-filled spaces called lacunae. Chocolates are the inactive cells of cartilage. Cartilage is surrounded by a fibrous connective tissue sheath called perichondrium. Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.

1. Hyaline cartilage: It is bluish-white, translucent, and glass-like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of cartilage. It is found in the walls of the nose, larynx, trachea, and bronchi.

2. Elastic cartilage: It is yellowish due to elastic fibres. Matrix has an abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes, and epiglottis.

3. Fibrous cartilage: Matrix has bundles of collagen fibres. The perichondrium is absent. It is the strongest of all types of cartilage. It occurs in the Intervertebral discs and pubic symphysis of the pelvis.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q8

Question 9.
Explain the Haversian system.
Answer:
The compact bone consists of several structural units called Osteons or Haversian systems arranged around and parallel to the bone marrow cavities.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q9
The Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein, and a lymphatic vessel. The Haversian canal is surrounded by concentric lamellae. Small fluid-filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae. Canaliculi connect the lacunae with one another and with the Haversian canal.

Each lacuna encloses one osteocyte (an inactive form of osteoblast). The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the perio¬steum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 10.
Write short notes on lymph.
Answer:
Lymph: Lymph is a colourless fluid. It lacks RBC, platelets, and large plasma proteins, but has more number of leucocytes. It is chiefly composed of plasma and lymphocytes. When compared to the tissue fluid, it contains very small amounts of nutrients and oxygen but has abundant CO2 and other metabolites.

The most important site of the formation of lymph is the interstitial space. As blood passes through the blood capillaries, some portion of blood that includes water, solutes, and proteins of low molecular weight passer: through the walls of capillaries, into the interstitial spaces due to hydrostatic pressure at the arteriolar ends. This fluid forms the interstitial fluid (tissue fluid). Most of the interstitial fluid is returned directly to the capillaries due to osmotic pressure at the venular ends.

A little amount of this tissue fluid passes through a system of lymphatic capillaries (lymph capillaries of the intestinal villi are called ‘lacteals’), vessels, and ducts and finally, reaches the blood through the subclavian veins. The extracellular ’tissue fluid’ that passes into the lymph capillaries and lymph vessels is called ‘lymph’. The lymphatic system represents an ‘accessory route’ by which interstitial fluid flows from tissue spaces into blood.

Question 11.
Describe the structure of a skeletal muscle.
Answer:
Skeletal (striped and voluntary) muscle:
It is usually attached to skeletal structures by ‘tendons’. In a typical muscle such as the ‘biceps’ muscle, skeletal muscle fibre is surrounded by a thin connective tissue sheath, the endomysium. A bundle of muscle fibres is called a fascicle. It is surrounded by a connective tissue sheath called perimysium. A group of fascicles forms a ‘muscle’ which is surrounded by an epimysium (outermost connective tissue sheath). These connective tissue layers may extend beyond the muscle to form a chord-like tendon or sheet-like aponeurosis.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q11
A skeletal muscle fibre is a long, cylindrical, and unbranched cell. It is a multinucleated cell with many oval nuclei characteristically in the ‘peripheral’ cytoplasm (a syncytium formed by the fusion of cells). Sarcoplasm has many myofibrils which show alternate dark and light bands. So it is called striped or striated muscle.

Question 12.
Describe the structure of a cardiac muscle.
Answer:
Cardiac (striped and involuntary) muscle: The cardiac muscle is striated like the skeletal muscle (shows sarcomeres). Cardiac muscle is found in the ‘myocardium’ of the heart of vertebrates. The cardiac muscle cells or the myocardial cells’ are short, cylindrical, mononucleate, or binucleate cells whose ends branch and form junctions with other cardiac muscle cells. Each myocardial cell is joined to adjacent myocardial cells by ‘electrical synapses’ or ‘gap junctions. They permit ‘electrical impulses to be conducted along the long axis of the cardiac muscle fibre. The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q12

Question 13.
Give an account of the supporting cells of nervous tissue.
Answer:
Nervous tissue is of two types.
1. Nervous
2. Supporting cell or neuroglia

Neuroglia (supporting cells): These are the supporting and non-conducting cells that provide a microenvironment suitable for neuronal activity. Unlike neurons, they continue to divide throughout life. Neuroglial cells of the CNS include oligodendrocytes (that form myelin sheath as mentioned above); astrocytes (star-shaped cells) that form an interconnected network and bind neurons and capillaries (helping in providing the blood-brain barrier); ependymal cells, which are ciliated cells that line the cavities of the brain and spinal cord to bring movements in the cerebrospinal fluid; microglial cells, which are phagocytic cells, of mesodermal origin. Neuroglial cells of the peripheral nervous system include the satellite cells and Schwann cells. Satellite cells surround the cell bodies in ganglia, and Schwann cells form neurolemma around axons.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 14.
Describe the structure of a multipolar neuron.
Answer:
Multipolar neurons have one axon and two or more dendrites. Most neurons in our body are multipolar neurons.
A neuron usually consists of a cell body with one too many dendrites and a single axon.

Neurons: Neurons are the ‘functional units of nervous tissue. These are electrically excitable cells that receive, initiate, and conduct/transmit impulses. When a neuron is stimulated, an electric disturbance (action potential) is generated which swiftly travels along its plasma membrane. A neuron usually consists of a “cell body” with one to many dendrites and a single axon.

Cell body: It is also called perikaryon, cyton, or soma. It contains abundant granular cytoplasm and a large spherical nucleus. The cytoplasm has Nissl bodies (they represent RER, the sites of protein synthesis), neurofibrils, and lipofuscin granules.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q14
Dendrites: Several short, branched processes that arise from the cyton are called dendrites. They also contain Nissl bodies and neurofibrils. They conduct nerve impulses towards the cell body (afferent processes).

Axon: An axon is a single, long, cylindrical process that originates from a region of the cyton called the axon hillock. The Plasmalemma of an axon is called axolemma, and the cytoplasm is called axoplasm, which contains neurofibrils. However, Nissl bodies are absent. An axon may give rise to collateral branches. Distally it branches into many fine filaments called telodendrites, (axon terminals), which end in bulb-like structures called synaptic knobs or terminal boutons. Synaptic knobs possess ‘synaptic vesicles’ containing chemicals called neurotransmitters.

Question 15.
Write short notes on (a) Platelets and (a) Synapse.
Answer:
(a) Blood platelets (Thrombocytes): These are colourless non- nucleated, round or oval biconvex discs. The number of platelets per cubic mm of blood is about 2,50,000-4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average lifespan of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q15
(b) Synapse: It is the place in between the two neurons or inter-neuronal or neuromuscular junctions. Nerve cells do not have direct contact with each other. There is a microscopic gap of about 200-300 A° present between the nerve cells called a synapse. The nerve cell present before synapse is called the presynaptic neuron and the behind is called the post-synaptic neuron. A neurotransmitter substance called acetylcholine is secreted in the synapse by presynaptic neurons’ telodendrites. Acetylcholine helps in the conduction of nerve impulses from one neuron to another neuron.

Long Answer Type Questions

Question 1.
What is coelom? Explain the different types of coelom with suitable examples and neat labelled diagrams.
Answer:
The body cavity, which is lined by mesoderm, is called coelom more elaborately, coelom is a fluid-filled space between the body wall and visceral organs and lined by mesodermal epithelium peritoneum. Based on the body cavity triploblastic animals can be classified into acoelomentes, pseudo-coelo mates, and coelomates.

Acoelomate bilaterians: The bilaterian animals in which the body cavity is absent are called acoelomates, e.g. Platy-helminthes (lowest bilaterians). In these animals, the mesenchyme derived from the thrid germinal layer, called mesoderm, occupies the entire blasto coel, between the ectoderm and the endoderm, so that the adults have neither the primary cavity (blasto-coelom) nor the secondary cavity (coelom). As there is no body cavity, the acoelomates exhibit a solid body plan.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q1

Pseudocoelomate bilaterians: In some animals, the body cavity is not lined by mesodermal epithelia. Such animals are called Pseudocoelomates. They include the members of phylum Aschelminthes (Nematoda, Rotifera, and some minor phyla). During embryonic development mesoderm (mesenchyme) occupies only a part of the blastocoel adjoining the ectoderm. The unoccupied portion of the blastocoel persists as pseudocoelom.

Eucoelomate bilaterians: Coelom or ‘true coelom’ is a fluid-filled cavity, that lies between the body wall and the visceral organs and is lined by mesodermal epithelium, the peritoneum. The portion of the peritoneum that underlines the body wall is the parietal peritoneum or somatic peritoneum. The portion of the peritoneum that covers the visceral organs is the splanchnic peritoneum or visceral peritoneum. In coelomates, the visceral organs are suspended in the coelom by the peritoneum.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q1.1
During the embryonic development of the eucoelomates, the blastocoel is replaced by the true coelom derived from the mesoderm. So, the true coelom is also called the ‘secondary body cavity. Based on the mode of formation of coelom, the eucoelomates are classified into two types:

I. Schizocoelomates: Animals in which the body cavity is formed by the ‘splitting of mesoderm’ are called schizocoelomates. Annelids, arthropods, and mollusks are schizocoelomates in the animal kingdom. All the schizocoelomates are protostomians and they show ‘holoblastic’, ‘spiral’, and ‘determinate’ cleavage. The 4d blastomere or mesentoblast cell of the early embryo divides to form mesodermal blocks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal block leads to the formation of Schizocoelom.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q1.2

II. Enterocoelomates: Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates, and chordates are the enterocoelomates. In these animals, mesodermal pouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q1.3

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 2.
What is symmetry? Describe the different types of symmetry in the animal kingdom with suitable examples.
Answer:
Symmetry: The concept of symmetry is fundamental in understanding the organisation of an animal. Symmetry in animals is the balanced distribution of paired body parts. The body plan of a vast majority of metazoans exhibits some kind of symmetry. However, most of the sponges and snails show asymmetry (lack of symmetry). The symmetry of an animal and its mode of life are correlated.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2
Asymmetry: The animals, which cannot be cut into two equal parts (antimeres) in any plane passing through the centre of the body is called asymmetrical, e.g.: most sponges and adult gastropods. In asymmetrical animals, the body lacks a definite form.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2.1
Symmetry: The regular arrangement of body parts in a geometrical design relative to the axis of the body is called symmetry. In a symmetrical animal, paired body parts are arranged on either side of the plane passing through the principal axis, such that they are equidistant from the plane. The unpaired body parts are located mostly on the plane, passing through the principal axis.
Basically, the symmetry in animals is of two kinds:
(i) Radial symmetry
(ii) Bilateral symmetry

(i) Radial Symmetry or Monaxial heteropolar Symmetry: When any plane passing through the central axis (oro-aboral axis/principal axis) of the body divides an organism into two identical parts, it is called radial symmetry. The animals with radial symmetry are either sessile or planktonic or sluggish forms. It is the principal symmetry of the diploblastic animals such as the cnidarians and ctenophores symmetrical (as it is five-angled, it is also called pentamerous radial symmetry). Radially symmetrical animals have many planes of symmetry, whereas pentamerous radially symmetrical animals have five planes of symmetry.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2.2

(ii) Bilateral symmetry: When only one plane (median sagittal plane) that passes through the central axis (anterior-posterior axis) divides an organism into two identical parts, it is called bilateral symmetry. It is the ‘principal type of symmetry’ in triploblastic animals. Among the triploblastic animals, some gastropods become secondarily asymmetrical though they have primarily bilaterally symmetrical larvae.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2.3
Bilaterally symmetrical animals are more efficient than other animals in seeking food, locating mates, and in avoiding or escaping from predators, because of cephalization (concentration of nerve and sensory cells at the anterior end). As a result of cephalization, bilaterally symmetrical animals can sense the new environment into which they enter and respond more efficiently and quickly.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2.4

Question 3.
Classify and describe the epithelial tissues on the basis of structural modification of cells with examples.
Answer:
Epithelium (epi-upon; thelia – growing) forms the outer covering of the body and the living of internal organs and cavities. There are two types of epithelial tissues namely ‘simple epithelia1 and ‘compound epithelia’ based on the number of layers or strata.

Simple epithelium is composed of a single layer of cells and forms the lining of body cavities, ducts, and vessels. It helps in the diffusion, absorption, filtration, and secretion of substances. On the basis of the shape of the cells, it is further divided into three types:
(i) Simple squamous epithelium (Pavement epithelium): It is composed of a single layer of flat and tile-like cells, each with a centrally located ‘ovoid nucleus’. It is found in the endothelium of blood vessels, mesothelium of body cavities (pleura, peritoneum, and pericardium), wall of Bowman’s capsule of the nephron, lining of alveoli of lungs, etc.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3

(ii) Simple cuboidal epithelium: It is composed of a single layer of cube-like cells with centrally located spherical nuclei. It is found in germinal epi – thelium, proximal, and distal convoluted tubules of the nephron. Cuboidal epithelium of proximal convoluted tubule of nephron has ‘microvilli’.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.1

(iii) Simple columnar epithelium: It is composed of a single layer of tall and slender cells with oval nuclei located near the base. It has mucus-secreting ‘goblet cells’ in some places. It is of two types:
(a) Ciliated columnar epithelium: Columnar epithelial cells have cilia on their free surface. It is mainly present in the inner surface of hollow organs like fallopian tubes, ventricles of the brain, the central canal of the spinal cord, bronchioles, etc.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.2

(b) Non-ciliated columnar epithelium: Columnar cells are without cilia. It is found in the lining of the stomach and intestine. Microvilli are present in the columnar epithelium of the intestine to increase the surface area of absorption.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.3

(II) Compound epithelium (stratified epithelium): It is made up of more than one layer of cells. Its main function is to provide protection against chemical and mechanical stress. It covers the dry surface of the skin as stratified, keratinized, squamous epithelium. It covers the moist surface of the buccal cavity, pharynx, esophagus, and vagina as stratified non-keratinized squamous epithelium. It forms the inner lining of the larger ducts of salivary glands, sweat glands, and pancreatic ducts as stratified cuboidal epithelium. It forms the wall of the urinary bladder as transitional epithelium.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.4

(III) Glandular epithelium: Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types – unicellular glands consisting of isolated glandular cells such as goblet cells of the gut, and multicellular glands, consisting of clusters of cells such as salivary glands. On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands. Exocrine glands are provided with ducts; that secrete mucus, saliva, earwax (cerumen), oil, milk, digestive enzymes, and other cell products. In contrast, endocrine glands are ductless and their products are ‘hormones’, which are not sent out via ducts, but are carried to the target organs by blood.

Based on the mode of secretion, exocrine glands are further divided into
(i) merocrine glands (e.g.: pancreas) which release the secretory granules without the loss of other cellular material.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.5
(ii) Apocrine glands (e.g.: mammary glands) in which the apical part of the cell is pinched off along with the secretory product and
(iii) Holocrine glands (e.g.: sebaceous glands), in which the entire cell disintegrates to discharge the contents.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 4.
Describe the various types of connective tissue properly with suitable examples.
Answer:
Connective tissue proper is of two types as
(A) Loose connective tissue proper
(B) Dense connective tissue proper

(A) Loose connective tissue: Cells and fibres are loosely arranged in a semi-fluid ground substance there are three types of loose connective tissues-areolar tissue adipose tissue and reticular tissue.
(i) Areolar tissue: It is one of the most widely distributed connective tissues in the body. It forms the packing tissue in almost all organs. Areolar tissue forms the subcutaneous layer of the skin. It has cells and fibres. Cells of the areolar tissue are the fibroblasts, mast cells, macrophages, adipocytes, and plasma cells.
1. Fibroblasts are the most common cells with secreted fibres. The inactive cells are called
fibrocytes.
2. Mast cells secrete heparin (an anticoagulant), histamine, bradykinin (vasodilators), and serotonin (vasoconstrictor), vasodilators cause inflammation in response to injury and infection.
3. Macrophages are amoeboid cells, phagocytic in function, and act as internal scavengers. They are derived from the monocytes of blood. ‘Tissue fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
4. Plasma cells are derived from the B-lymphocytes and produce antibodies.
5. Adipocytes are specialized cells for the storage of fats.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q4

(ii) Adipose tissue: It is specialized for fat storage. It consists of a large number of adipocytes and few fibres. The adipose tissue which is found beneath the skin provides. Adipose tissue is of two types: white adipose tissue, and brown adipose tissue. Excess nutrients which are not used immediately are converted into fats and stored in this tissue.
White adipose tissue (WAT): It is the predominant type in adults, and the adipocyte has a single large lipid droplet (monocular). White fat is metabolically not active.
Brown adipose tissue (BAT): It is found in fetuses and infants. Adipocyte of BAT has several small ‘lipid droplets’ (multilocular) and numerous mitochondria. Brown fat is metabolically active and generates ‘heat’ to maintain the body temperature required by infants.

(iii) Reticular tissue: It has specialized fibroblasts called reticular cells. They secrete ‘reticular fibers’ that form an interconnecting network. It forms the ‘supporting framework’ of lymphoid organs such as bone marrow, spleen, and lymph nodes and forms the reticular lamina of the ‘basement membrane’.

(B) Dense connective tissue: This tissue consists of more fibres, but fewer cells. It has very little ground substance. Based on the arrangement of fibres, the dense connective tissue is of three types.
(i) Dense regular connective tissue: In this tissue, collagen fibres are arranged parallel to one another in bundles. Tendons that attach the skeletal muscles to bones and ligaments which attach bones to other bones are examples of this type of connective tissue.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q4.1

(ii) Dense irregular connective tissue: In this type of connective tissue, bundles of collagen fibres are irregularly arranged. Periosteum, endosteum, pericardium, heart valves, joint capsule, and deeper region of the dermis of skin contain/are made up of this type of connective tissue.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q4.2

(iii) Elastic connective tissue: It is mainly made of yellow elastic fibres, capable of considerable extension and recoil. This tissue can recoil to its original shape, when the forces of stretch are released. It occurs in the wall of arteries, vocal cords, trachea, bronchi, and ‘elastic ligaments’ present between vertebrae.

In addition to the above-mentioned connective tissues, mucous connective tissue occurs as foetal or embryonic connective tissue. It is present in the umbilical cord as Wharton’s jelly.

Question 5.
What is skeletal tissue? Describe the various types of skeletal tissue.
Answer:
Skeletal tissue (supporting tissue): It forms the endoskeleton of the vertebrates. It supports the body, protects various organs, provides a surface for the attachment of muscles, and helps in locomotion. It is of two types:
(A) Cartilage (Gristle): Cartilage is a solid, but semi-rigid (flexible) connective tissue. It resists compression. Matrix is firm but somewhat pliable. It has collagen fibres, elastic fibres (only in the elastic cartilage), and matrix-secreting cells called chondroblasts. These cells are enclosed in fluid-filled spaces called lacunae. Chondrocytes are the inactive cells of cartilage. Cartilage is surrounded by a fibrous connective tissue sheath called perichondrium.

Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.
1. Hyaline cartilage: It is bluish-white, translucent, and glass-like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of cartilage. It is found in the walls of the nose, Larynx, trachea, and bronchi.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q5

2. Elastic cartilage: It is yellowish due to elastic fibres. Matrix has an abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes, and epiglottis.

3. Fibrous cartilage: Matrix has bundles of collagen fibres. The perichondrium is absent. It is the strongest of all types of cartilage. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

(B) Bone (osseous) tissue: Bone is highly calcified (mineralized), solid, hard and rigid connective tissue. It is the major component of the endoskeleton of most adult vertebrates.

Bone has an outer fibrous connective tissue sheath called periosteum, the inner connective tissue sheath that lines the marrow cavity called the endosteum, a non-living extracellular matrix, living cells, and bone marrow. Bone cells include osteoblasts, osteocytes, and osteoclasts. Osteoblasts (immature bone cells) secrete the organic components (collagen fibres) of the matrix and also play an important role in the ‘mineralization of bone’ and become osteocytes (mature bone cells). Osteocytes are enclosed in fluid-filled lacunae. Osteoclasts are phagocytic cells involved in the resorption of bone.

Types of bones based on the method of formation:
(i) Cartilage bones (replacing bones or endochondral bones) are formed by ossification within the cartilage e.g. bones of limbs, girdles, and vertebrae.
(ii) Investing bones (membrane bones or dermal bones) are formed by the ossification in the embryonic mesenchyme e.g. most of the bones of the cranium.
(iii) Sesamoid bones are formed by ossification in tendons e.g.: patella (knee cap) and pisiform bone of the wrist of a mammal
(iv) are formed by ossification in the soft tissues, e.g.: Oscordis (inside the heart of ruminants), and Ospenis (inside the glans-penis of many mammals such as rodents, bats, and carnivores).

Types of bones based on the structure:
1. Spongy bone (Cancellous bone or trabecular bone): It occurs in the epiphyses and metaphyses
of long bones. It looks spongy and contains columns of bone called ‘trabeculae’ with irregular interspaces filled with red bone marrow.
2. Compact bone: The diaphysis of a long bone is made up of ‘compact bone. It has a dense continuous lamellar matrix between the periosteum and endosteum.

Structure of a compact bone: Diaphysis (shaft) is a part of a long bone that lies in between expanded ends. The diaphysis is covered by a dense connective fibrous tissue called the periosteum. The diaphysis of a long bone has a hollow cavity called marrow cavity which is lined or surrounded by the endosteum. In between periosteum and endosteum, the matrix of the bone is laid down in the form of ‘lamellae’. Outer circumferential lamellae are located immediately beneath the periosteum; inner circumferential lamellae are located around the endosteum.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q5.1
Between the outer and inner circumferential lamellae, there are many Haversian systems (osteons – units of bone). The spaces between the Haversian systems are filled with interstitial lamellae. Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein and a lymphatic vessel. Haversian canal is surrounded by concentric lamellae. Small fluid filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae.

Canaliculi connect the lacunae with one another and with Haversian canal. Each lacuna encloses one osteocyte (inactive form of osteoblast). The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmanns canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 6.
Give an account of the “formed elements” of Blood.
Answer:
Blood is a red-coloured, opaque, and slightly alkaline fluid. It is composed of blood plasma and formed elements or blood cells – the RBC, WBC, and platelets.
Formed elements of Blood cells: The blood corpuscles (RBC and WBC) constitute 45% of the total blood by volume.

(i) Red blood corpuscles (Erythrocytes): Erythrocytes of mammals are circular (elliptical in camels and Liamas), biconcave and enucleate. The biconcave shape provides a large surface area-to-volume ratio, thus providing more area for the exchange of gases. These are 7.8 pm in diameter. The number of RBCs per cubic millimeter of blood is about 5 million in a man and 4.5 million in a woman.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6

(ii) White blood corpuscles (Leucocytes): These are nucleate, colourless, complete cells. They are spherical or irregular in shape and are capable of exhibiting amoeboid movement into the extravascular areas by diapedesis. WBC are two main types: 1) Granulocytes and 2) Agranulocytes.

Granulocytes: They possess cytoplasmic granules that may take three different types of stains, neutral or acidic, or basic. The nucleus of the granulocytes is divided into lobes and assumes different shapes, hence, these are also called polymorphonuclear leucocytes. Based on the staining properties these are of three tvoes.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6.1

Basophils: They constitute about 0.4% of the total leucocytes. The nucleus is divided into irregular lobes. Cytoplasmic granules are ‘fewer’ and ‘irregular’ in shape. They take basic stains. They produce heparin, histamine, etc. They supplement the function of mast cells when needed.

Eosinophils (acidophils): They constitute about 2.3% of the total leucocytes. The nucleus is distinctly bilobed. The cytoplasm has large granules which stain with acidic dyes such as ‘eosin’.

Neutrophils: They constitute about 62% of the total leucocytes. The nucleus is many lobed (2-5). Specific cytoplasmic granules are small and abundant. They stain with ‘neutral dyes’.

Agranulocytes: Cytoplasmic granules are absent in agranulocytes. divided into lobes. These are of two types:
(a) Lymphocytes: They constitute about 30% of the total leucocytes. They are small, spherical cells with large spherical nuclei and scanty peripheral cytoplasm. There are functionally two types of lymphocytes – ‘B’ lymphocytes, which produce ‘antibodies’ and T lymphocytes which play a key role in the immunological reactions of the body. Some lymphocytes live only a few days while others survive for many years.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6.2

(b) Monocytes: They constitute about 5.3% of the leucocytes. The nucleus is kidney-shaped (reniform). These are the largest, motile phagocytes. They engulf bacteria and cellular debris. They differentiate into macrophages when they enter the connective tissues.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6.3

(iii) Blood platelets (Thrombocytes): These are colourless non-nucleated, round, or oval biconvex discs. The number of platelets per cubic mm of blood is about 2,50,000-4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average lifespan of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6.4

Question 7.
Compare and contrast the three types of muscular tissues.
Answer:
Muscular tissue is mesodermal in origin. Muscles show three essential properties such as excitability, conductivity, and contractility. The study of muscular tissues is known as mycology. Muscular tissue has elongated cells called ‘muscle fibers’ (myocytes) which are surrounded by a connective tissue sheath. The extracellular matrix is absent. The plasma membrane of a muscle fiber is called sarcolemma.

The cytoplasm of a muscle fibre is called Sarco plasm, the endoplasmic reticulum, the sarcoplasmic reticulum, and the mitochondria, the Sarcosomes. The cytoplasm of a muscle fibre has several myofibrils. Each myofibril has thick (myosin) and thin (actin) myofilaments. The regular arrangement of myosin and actin filaments is responsible for the alternate dark and light bands of a ‘striated muscle’. Sarcoplasm also contains ATP, phosphocreatine, glycogen, and myoglobin. Muscles are of three types – skeletal, smooth, and cardiac.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q7

Skeletal (striped and voluntary) muscle: It is usually attached to skeletal structures by ‘tendons’. A typical muscle such as the ‘biceps’ muscle/skeletal muscle fibre is surrounded by a thin connective tissue sheath, the endomysium. A bundle of muscle fibres is called a fascicle.

It is surrounded by a connective tissue sheath called perimysium. A group of fascicles forms a ‘muscle’ that is surrounded by an epimysium (outermost connective tissue sheath). These connective tissue layers may extend beyond the muscle to form a chord-like tendon or sheet¬like aponeurosis.

A skeletal muscle fibre is a long, cylindrical, and unbranched cell. It is a multinucleated cell with many oval nuclei characteristically in the “peripheral” cytoplasm (a syncytium formed by the fusion of cells). Sarcoplasm has many myofibrils which show alternate dark and light bands. So it is called striped or striated muscle.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q7.1

Smooth (unstriped and involuntary muscle): It is located in the walls of the visceral organs such as blood vessels, trachea, bronchi, stomach, intestine, excretory and genital ducts, and so this is also called ‘visceral muscle’. As cross striations are absent, it is called ‘smooth muscle. It is also found in the iris and ciliary body of the eye and in the skin as ‘arrector pili muscles that are attached to hair follicles.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q7.2
Usually, smooth muscles are arranged in ‘layers’/’sheets’. A smooth muscle fibre is a spindle-shaped (fusiform), uninucleate cell. Myofibrils do not show alternate dark and light bands due to the irregular arrangement of actin and myosin fibres. They do not work under conscious control, and so they are called involuntary muscles. Smooth muscles exhibit ‘slow’ and ‘prolonged’ contractions. They may remain contracted for long periods without fatigue (show sustained involuntary contractions called ‘spasms’). The contraction of smooth muscles is under the control of the autonomous nervous system.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Cardiac (striped and involuntary) muscle: The cardiac muscle is striated like the skeletal muscle (shows sarcomeres). Cardiac muscle is found in the ‘myocardium’ of the heart of vertebrates. The cardiac muscle cells or the ‘myocardial cells’ are short, cylindrical, mononucleate, or binucleate cells whose ends branch and form junctions with other cardiac muscle cells. Each myocardial cell is joined to adjacent myocardial cells by ‘electrical synapses’ or gap junctions. They permit ‘electrical impulses’ to be conducted along the long axis of the cardiac muscle fibre. The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q7.3

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 1 Diversity of Living World Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 1 Diversity of Living World

Very Short Answer Type Questions

Question 1.
Define the term metabolism. Give any one example.
Answer:
The sum total of all the chemical reactions occurring in the bodies of organisms constitutes metabolism.
Ex: Photosynthesis is one of the metabolic processes in living organisms.

Question 2.
How do you differentiate between growth in a living organism and a non-living object?
Answer:
Growth is one of the fundamental characteristics of living beings growth in living beings is growth from the inside, whereas growth in non-living things is by the accumulation of material on the surface.

Question 3.
What is biogenesis?
Answer:
Life comes only from life is called biogenesis. Living organisms produce young ones of their kind using molecules of heredity.

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 4.
Define the term histology. What is it otherwise called?
Answer:
Histology is the study of the microscopic structure of different tissues. It is also called Micro anatomy.

Question 5.
Distinguish between embryology and ethology.
Answer:
Embryology: It is the study of events that lead to fertilization, cleavages, early growth, and differentiation of a zygote into an embryo.
Ethology: The study of animal behaviour based on systematic observation, with special attention to physiological, ecological, and evolutionary aspects is called ethology.

Question 6.
In a given area, remains of animals that lived in the remote past are excavated for study. Which branch of science is it called?
Answer:
The branch of science Palaeontology deals with that. In a given area, remains of animals that lived in the remote past (fossilized remains) are excavated for study.

Question 7.
Zoos are tools for ‘classification’ Explain.
Answer:
Zoos are places where wild animals are taken out of their natural habitat and are placed in protected environments under human care. This enables us to learn about the animal’s external features, habits, behaviour, etc. These observations enable us to systematize the organism and position it in the animal world.

Question 8.
Where and how do we preserve skeletons of animals dry specimens etc?
Answer:
The Skeletons and dry specimens are preserved in Museums and are usually stuffed and preserved.

Question 9.
What is trinominal nomenclature? Give an example.
Answer:
The trinominal nomenclature is the extension of the binominal system of nomanclature. It permits the designation of subspecies with a three-worded name called ‘trinomen’.
Ex: Homo Sapiens Sapiens, Corvus splendns spelendns.

Question 10.
What is meant by tautonymy? Give two examples.
Answer:
The practice of naming animals or organisms, in which the generic name and species name are the same, is called Tautonymy.
Ex: Axis axis – spotted dear
Naja naja – The Indian Cobra

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 11.
Differentiate between Protostomia and Deuterostomia.
Answer:
Protostomia (Gr. mouth first) are the organisms in which blastopore develops into the mouth.
Deuterostomia (Gr. second mouth) are the organisms in which blastopore develops into the anus, the mouth is formed later.

Question 12.
‘Echinoderms are enterocoelomates’. Comment.
Answer:
The animals of phyla Echinodermata have a true coelom, which is an ‘enterocoel’. It is formed from the archenteron.

Question 13.
What does ICZN stand for?
Answer:
ICZN stands for ‘International Code of Zoological Nomenclature which specifies the mandatory rules to be followed for the naming of animals by the International congress (ICZ) in 1898.

Question 14.
Give the names of any four protostomian phyla.
Answer:
The phylum Platyhelminthes, Nematoda, Annelida, Arthropoda, and Mollusca are the protostomian phyla.

Question 15.
Nematoda is a protostomian but not a coelomate justify the statement.
Answer:
Animals of group Nematoda are protostomian but they have no true coelom/secondary body cavity as it is not lined by mesodermal epithelial layers. Pseudocoel is a remnant of the embryonic blastocoel. Hence they are protostonian. Pseudocoelomata, but not coelomates.

Question 16.
What is ecological diversity? Mention the different types of ecological diversities.
Answer:
Diversity at a higher level of organization, i.e., at the ecosystem level is called ‘Ecological diversity.
The other ecological diversities are Alpha, Beta, and Gama diversities.

Question 17.
Define species richness.
Answer:
The more the number of species in an area (unit area) the more species richness.

Question 18.
Mention any two products of medicinal importance obtained from Nature.
Answer:
Anticancer drugs Vinblastin from the plant Vinco rosa and Digitalin from the plant for gloves are obtained from nature.

Question 19.
Invasion of an Alien species leads to the extinction of native species. Justify this with two examples.
Answer:
When alien species are introduced into a habitat, they turn invasive and establish themselves at the cost of the native species.
Ex: Nail perch introduced into lake Victoria, in east Africa led to the extinction of 200 species of Cichlid fish in the lake. The illegal introduction of exotic African catfish for aquaculture purposes in posing a threat to the native catfish.

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 20.
List out any four sacred groves in India.
Answer:
The following are the Sacred Groves in India.

  1. Khasi and Jaintia Hills – Meghalaya
  2. Aravalli Hills – Rajasthan and Gujarat
  3. Western Ghat region – Karnataka and Maharashtra
  4. Sarguja, Bastar – Chhattisgarh
  5. Chanda – Madhya Pradesh

Question 21.
Write the full form of IUCN. In which book threatened species are enlisted.
Answer:
IUCN – International Union for the Conservation of Nature and Natural Resources.
All the threatened species are enlisted in the Red Data Book Published by IUCN.

Short Answer Type Questions

Question 1.
Explain the phylogenetic system of biological classification.
Answer:
Phylogenetic classification is an evolutionary classification based on how a common ancestry was shared. Cladistic classification summarizes the ‘genetic distance’ between all species in this ‘Phylogenetic tree’. In Cladistic classification characters such as analogous characters (characters shared by a pair of organisms due to convergent evolution e.g. wings in sparrows and patagia (wing-like structures) in flying squirrels) and homologous characters (characters shared by a pair of organisms, inherited from a common ancestor e.g. wing of sparrows and finches) are taken into consideration. Ernst Haeckel introduced the method of representing Phylogeny by ‘tree’ branching diagrams.

Question 2.
Explain the hierarchy of classification.
Answer:
Human beings are not only interested in knowing more about different kinds of organisms and their diversities, but also the relationships among them. This branch of study is referred to as systematics. Systematics is the branch of science that deals with the vast diversity of life. It also reveals the trends and evolutionary relationships of different groups of organisms. These relationships establish the phylogeny of organisms. A key part of systematics is taxonomy. The taxonomic hierarchy includes seven obligate categories namely kingdom, phylum, class, order, family, genus, and species, and other intermediate categories such as subkingdom, grade, division, subdivision, subphylum, superclass, subclass, superorder, suborder, superfamily, subfamily, subspecies, etc.

Question 3.
What is meant by classification? Explain the need for classification.
Answer:
Classification is defined as the process by which anything is grouped into convenient categories based on some easily observable characteristics. It is impossible to study all living organisms. So, it is necessary to devise some means to make this possible. This process is called classification. The scientific term used for these categories is ‘TAXA’. Taxa can indicate categories at different levels, e.g. Animalia, Chordata, Mammalia, etc. represent taxa at different levels.

Hence based on characteristics, all living organisms can be classified into different taxa: This process of classification is called taxonomy. External and internal structures, along with the structure of cells, developmental processes, and ecological information of organisms are essential and they form the basis of modern taxonomic studies, Hence characterization, identification, nomenclature, and classification are the processes that are basic to taxonomy. To understand the interrelationships among the diversified animal groups, a systematic classification is necessary.

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 4.
Define species. Explain the various aspects of ‘species’.
Answer:
Species: Species is the ‘basic unit’ of classification. Species is a Latin word meaning ‘kind’ or ‘appearance’. John Ray in his book ‘Historia Generalis Plantarum’ used the term ‘species’ and described it on the basis of common descent (origin from common ancestors) as a group of morphologically similar organisms. Linnaeus considered species, in his book ‘Systema Naturae’, as the basic unit of classification. Buffon, in his book ‘Natural History, proposed the idea of the evolution of species which is the foundation for the biological concept of evolution. This biological concept of species (dynamic nature of species) became more popular with the publication of the book “The Origin of Species” by Charles Darwin.

Buffon’s biological concept of species explains that species is an interbreeding group of similar individuals sharing the common ‘gene pool’ and producing fertile offspring. Species is considered as a group of individuals which are:

  1. Reproductively isolated from the individuals of other species – a breeding unit.
  2. Sharing the same ecological niche – an ecological unit.
  3. Showing similarity in the karyotype – a genetic unit.
  4. Having similar structure and functional characteristics – an evolutionary unit.

Question 5.
What is genetic diversity and what are the different types of genetic diversity?
Answer:
Genetic diversity is the diversity of genes within a species. A single species may show high diversity at the genetic levels over its distributional range. For e.g. Rauwolfia vomitoria, a medical plant growing in the Himalayas ranges shows great genetic variation, which might be in terms of potency and concentration of the active chemical (reserpine extracted from it is used in treating high blood pressure) that the plant produces. India has more than 50,000 different strains of rice and 1,000 varieties of mangoes. Genetic diversity increases with environmental variability and is advantageous for its survival.

Question 6.
What are the reasons for greater biodiversity in the tropics?
Answer:
Reasons for greater biodiversity in the tropics:
Reason 1: Tropical latitudes have remained relatively undisturbed for millions of years and thus had a long ‘evolutionary time’. The as long duration available in this region for speciation led to species diversification. (Note: The temperate regions were subjected to frequent glaciations in the past).

Reason 2: Tropical climates are relatively more constant and predictable than that temperate regions. A constant environment promotes niche specialization (how an organism responds, and behaves with the environment and with other organisms of its biotic community) and this leads to greater species diversity.

Reason 3: Solar energy, resources like water, etc., are available in abundance in this region. They contribute to higher productivity in terms of food production, leading to greater diversity.

Question 7.
What is the ‘evil quartet’?
Answer:
The following are the ‘four major causes (The Evil Quartet) for accelerated rates of species extinction in the world.
Habitat loss and Fragmentation: These are the most important reasons for the loss of biodiversity.

  • Deforestation leads to species extinction in forests.
    e.g: tropical rainforests once covered 14% of the earth’s land surface now not more than 4%.
  • Conversion of forest land to agricultural land.
    e.g: the amazon rainforest, called the lungs of our planet, harbouring innumerable species is cut and cleared to cultivate soybeans or convert to grasslands for raising beef cattle.
  • Pollution enhances the degradation of habitats and threatens the survival of many species as pollutants change the quality of the environment.
  • Fragmentation of habitat leads to population decline.
    e.g: mammals and birds requiring large territories and certain animals with migratory habits are badly affected.

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 8.
Explain in brief ‘Biodiversity Hot Spots’.
Answer:
Biodiversity hot spots: A Biodiversity hot spot is a Biogeographic Region that is both a significant reservoir of biodiversity and is threatened with destruction.
The concept of biodiversity originated from Norman Myers. There ate about 34 biodiversity hot spots in the world. As these regions are threatened by destruction habitat loss is accelerated.
e.g.: (I) the Western Ghats and Srilanka
(II) Indo Burma
(III) Himalayas in India.

Ecologically unique and biodiversity-rich regions are legally protected as in

  • Biosphere Reserves – 14
  • National Parks – 90
  • Sanctuaries – 448

Biosphere Reserves: An area that is set aside, minimally disturbed for the conservation of the resources of the biosphere is the ‘Biosphere reserve. The latest biosphere reserve (17th biosphere reserve in India) is Seshachalam hills.

National Parks: A National Park is a natural habitat strictly reserved for the protection of natural life. National Parks, across the country, offer a fascinating diversity of terrain, flora, and fauna. Some important National Parks in India are – Jim Corbett National Park (the first National Park in India located in Uttarakhand), Kaziranga National Park (Assam), Kasu Brahmananda Reddy National Park, MahavirHarinaVanasthali National Park (AP). Keoladeo Ghana National Park (Rajasthan), etc.

Sanctuaries: Specific endangered faunal species are well protected in wildlife sanctuaries which permits eco-tourism (as long as animal life is undisturbed). Some, important Sanctuaries in India (AP) include-Koringa Sanctuary, Eturnagaram Sanctuary, and Papikondalu Sanctuary.

Question 9.
Explain the ‘Rivet Popper’ hypothesis.
Answer:
What if we lose a few species? Will it affect man’s life? Paul Ehrlich experiments Rivet popper, hypothesis, taking an aeroplane as an ecosystem, explains how the removal of one by one ‘rivets’ (species of an ecosystem) of various parts can slowly damage the plane (ecosystem)-shows how important a ‘species’ is in the overall functioning of an ecosystem. Removing a rivet from a seat or some other relatively minor important parts may not damage the plane, but the removal of a rivet from a part supporting the wing can result in a crash. Likewise, the removal of a ‘critical species’ may affect the entire community and thus the entire ecosystem.

Question 10.
Write short notes on In-situ conservation.
Answer:
In-situ conservation (On-site conservation): In-situ conservation is the process of protecting an animal species in its natural habitat. The benefit is that it maintains recovering populations in the surrounding where they have developed their distinctive properties. Conservationists identified certain regions by the name ‘Biodiversity hot spots’ for maximum protection as they are characterized by very high levels of species richness & high degree of endemism. By definition ‘A biodiversity hot spot’ is a ‘Biogeographic Region’ with a significant reservoir of biodiversity that is under threat of extinction from humans. They are Earth’s biologically ‘richest’ and ‘most threatened’ Terrestrial Ecoregions.

AP Inter 2nd Year Physics Important Questions Chapter 16 Communication Systems

Students get through AP Inter 2nd Year Physics Important Questions 16th Lesson Communication Systems which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 16th Lesson Communication Systems

Very Short Answer Questions

Question 1.
What are the basic blocks of a communication system?
Answer:
Basic blocks in a communication system are

  1. Transmitter
  2. Receiver
  3. Channel

Question 2.
What is ‘World Wide Web” (WWW) ?
Answer:
Tern Berners -Lee invented the World Wide Web.
It is an encyclopedia of knowledge accessible to every one round the clock through out the year.

AP Inter 2nd Year Physics Important Questions Chapter 16 Communication Systems

Question 3.
Mention the frequency range of speech signals.
Answer:
Speech signals frequency range is 300 Hz to 3100 Hz.

Question 4.
What is sky wave propagation ?
Answer:
In the frequency range from a MHz upto about 30 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation.

Question 5.
Mention the various parts of the ionosphere ?
Answer:
Parts of ionosphere are

  1. D Part of stratosphere (65-70 km day only),
  2. E Part of stratosphere (100 km day only),
  3. F1 Part of mesosphere (170 km – 190 km),
  4. F2 Part of thermosphere [300 km at night 250 – 400 km during day time].

AP Inter 2nd Year Physics Important Questions Chapter 16 Communication Systems

Question 6.
Define modulation. Why is it necessary ? [A.P. 17; A.P., T.S. Mar. 16, T.S. Mar. 15, Mar. 14]
Answer:
Modulation : The process of combining low frequency audio signal with high frequency carrier wave is called modulation.
Necessary: Low frequency signals cannot transmit directly. To reduce size of the antenna and to avoid mixing up of signal from different transmitters modulation is necessary.

Question 7.
Mention the basic methods of modulation. [T.S. Mar. 15, 17, A.P. Mar. 16]
Answer:
The basic methods of modulation are :

  1. Amplitude modulation (AM)
  2. Frequency modulation (FM)
  3. Phase modulation (PM)

Question 8.
Which type of communication is employed in Mobile Phones ? [A.P. Mar. 15]
Answer:
Space wave mode of propagation is employed in mobile phones.

Textual Examples

Question 1.
A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna Is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth 6.4 × 106 m.
Solution:
dm = \(\sqrt{2 \mathrm{Rh}_{\mathrm{T}}}+\sqrt{2 \mathrm{Rh}_{\mathrm{R}}}\)
dm = \(\sqrt{2 \times 64 \times 10^5 \times 32}+\sqrt{2 \times 64 \times 10^5 \times 50 \mathrm{~m}}\)
=64 × 102 × \(\sqrt{10}\) +8 × 103 × \(\sqrt{10}\)m = 144 × 102 × \(\sqrt{10}\)m
= 45.5 km.

AP Inter 2nd Year Physics Important Questions Chapter 16 Communication Systems

Question 2.
A message signal of frequency 10 kHz and peak voltage of 10 volts Is used to modulate a carrier of frequency 1 MHz and peak voltage of 20 volts. Determine
(a) modulation index
(b) the side bands produced.
Solution:
a) Modulation index = \(\frac{10}{2}\) = 0.5
b) The side bands are at (1000 + 10 kHz) = 1010 kHz and (1000 – 10 kHz) = 990 kHz.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Students get through AP Inter 2nd Year Physics Important Questions 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits

Very Short Answer Questions

Question 1.
What is an n-type semiconductor? What are the majority and minority charge carriers in it?
Answer:

  • If a pentavalent impurity is added to a pure tetravalent semiconductor, it is called an n-type semiconductor.
  • In an n-type semiconductor majority, of charge carriers are electrons and the minority charge carriers are holes.

Question 2.
What are intrinsic and extrinsic semiconductors? [A.P. Mar. 15]
Answer:

  • Pure form of semiconductors are called intrinsic semiconductors.
  • When impure atoms are added to increase their conductivity, they are called extrinsic semiconductors.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
What is a p-type semiconductor ? What are the majority and minority charge carriers in it ? [A.P. & T.S. Mar. 17]
Answer:
If a trivalent impurity is added to a tetravalent semiconductor, it is called p-type semi-conductor.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 1
In p-type semiconductor majority charge carriers are holes and minority charge carriers are electrons.

Question 4.
What is a p-n junction diode ? Define depletion’ layer.
Answer:
When an intrinsic semiconductor crystal is grown with one side doped with trivalent element and on the other side doped with pentavalent element, a junction is formed in the crystal. It is called p-n junction diode.

A thin narrow region is formed on either side of the p-n junction, which is free from charge carriers is called depletion layer.

Question 5.
How is a battery connected to a junction diode in i) forward and ii) reverse bias ?
Answer:
i) In p-n junction diode, if p-side is connected to positive terminal of a cell and n-side to negative terminal, it is called forward bias.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 2
ii) In a p-n junction diode, p-side is connected to negative terminal of a cell and n-side to positive terminal, it is called reverse bias.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 3

Question 6.
What is the maximum percentage of rectification in half wave and full wave rectifiers ?
Answer:

  1. The percentage of rectification in half-wave rectifier is 40.6%.
  2. The percentage of rectification in full-wave rectifiers is 81.2%.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 7.
What is Zener voltage (Vz) and how will a Zener diode be connected in circuits generally ?
Answer:

  1. When Zener diode is in reverse biased, at a particular voltage current increases suddenly is called Zener (or) break down voltage.
  2. Zener diode always connected in reverse bias.

Question 8.
Write the expressions for the efficiency of a full wave rectifier and a half wave rectifier.
Answer:

  1. Efficiency of half wave rectifier (η) = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
  2. Efficiency of full wave rectifier (η) = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)

Question 9.
What happens to the width of the depletion layer in a p-n junction diode when it is
i) forward-biased and ii) reverse biased ?
Answer:
When a p-n junction diode is forward bias, thickness of depletion layer decreases and in reverse bias, thickness of depletion layer increases.

Question 10.
Draw the circuit symbols for p-n-p and n-p-n transistors. [Mar. 14]
Answer:
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 4

Question 11.
Define amplifier and amplification factor.
Answer:

  1. Rising the strength of a weak signal is known as amplification and the device is called amplifier.
  2. Amplification factor is the ratio between output voltage to the input voltage.
    A = \(\frac{\mathrm{V}_0}{\mathrm{~V}_{\mathrm{i}}}\)

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 12.
In which bias can a Zener diode be used as voltage regulator ?
Answer:
In reverse bias Zener diode can be used as voltage regulator.

Question 13.
Which gates are called universal gates ? [T.S. Mar. 15]
Answer:
NAND gate and NOR gates are called universal gates.

Question 14.
Write the truth table of NAND gate. How does it differ from AND gate ?
Answer:
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 5
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 6

Short Answer Questions

Question 1.
What is a rectifier ? Explain the working of half wave and full Wave rectifiers with diagrams. [A.P. Mar. 17]
Answer:
Rectifier: It is a circuit which converts ac into d.c. A p-n junction diode is used as a rectifier.
Half-wave rectifier:
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 7

  1. A half wave rectifier can be constructed with a single diode. The ac input signal is fed to the primary coil of a transformer. The output signal is taken across the load resistance RL.
  2. During positive half cycle, the diode is forward biased and current flows through the diode.
  3. During negative half cycle, diode is reverse biased and no current flows through the load resistance.
  4. This means current flows through the diode only during positive half cycles and negative half cycles are blocked. Hence in the output we get only positive half cycles.
  5. Rectifier efficiency is defined as the ratio of output dc power to the input ac power.
    η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}=\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
    Where rf = Forward resistance of a diode; RL = Load resistance
    The maximum efficiency of half wave rectifier is 40.6%.

Full wave rectifier : The process of converting an alternating current into a direct current is called rectification.
The device used for this purpose is called rectifier.

  1. A full wave rectifier can be constructed with the help of two diodes D1 and D2.
  2. The secondary transformer is centre tapped at C and its ends are connected to the p regions of two diodes D1 & D2.
  3. The output voltage is measured across the load resistance RL.
  4. During positive half cycles of ac, the diode D1 is forward biased and current flows through the load resistance RL. At this time D2 will be reverse biased and will be in switch off position.
  5. During negative half cycles of ac, the diode D2 is forward biased and the current flows through RL. At this time D1 will be reverse biased and will be in switch off position.
  6. Hence positive output is obtained for all the input ac signals.
  7. The efficiency of a rectifier is defined as the ratio between the output dc power to the input ac power.
    η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}=\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
    The maximum efficiency of a full wave rectifier is 81.2%.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 2.
What is a junction diode ? Explain the formation of depletion region at the junction. Explain the variation of depletion region in forward and reverse-biased condition.
Answer:
p-n junction diode: When a p-type semiconductor is suitablyjoined to n-type semiconductor, a p-n junction diode is formed. ’
The circuit symbol of p-n junction diode is shown in figure.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 8
Formation of depletion layer at the junction: When p-n junction is formed, the free electrons on n-side diffuse over to p-side, they combine with holes and become neutral. Similarly holes on p-side diffuse over to n-side and combihe with electrons become neutral.

This results in a narrow region formed on either side of the junction. This region is called depletion layer. Depletion layer is free from charge carriers.

The n-type material near the junction becomes positively charged due to immobile donor ions and p-type material becomes negatively charged due to immobile acceptor ions. This creates an electric field near the junction directed from n-region to p-region and cause a potential barrier.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 9
The potential barrier stops further diffusion of holes and electrons across the* junction. The value of the potential barrier depends upon the nature of the crystal, its temperature and the amount of doping.

Forward bias :
“When a positive terminal of a battery is connected to p-side and negative terminal is connected to n-side; then p-n junction diode is said to be forward bias”.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 10
The holes in the p-region are repelled by the positive polarity and move towards the junction. Similarly electrons in the n-region are repelled by the negative polarity and move towards the junction.

As a result, the width of the depletion layer decreases. The charge carriers cross the junction apd electric current flows in the circuit.
Hence in forward bias resistance of diode is low. This position is called switch on position.
Reverse bias:
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 11
“When the negative terminal of the battery is connected to p-side and positive terminal of the battery is connected to n-side of the p-n junction, then the diode is said to be reverse bias”.

The holes in the p-region are attracted towards negative polarity and move away from the junction. Similarly the electrons in the n-region are attracted towards positive polarity and move away from the junction.

So, width of the depletion layer and potential barrier increases. Hence resistance of p-n junction diode increases. Thus the reverse biased diode is called switch off position.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
What is a Zener diode ? Explain how it is used as a voltage regulator.
Answer:
Zener diode : Zener diode is a heavily doped germanium (or) silicon p-n junction diode. It works on reverse bias break down region.
The circuit symbol of zener diode is shown in figure.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 12
Zener diode can be used as a voltage regulator. In- general zener diode is connected in reverse bias in the circuits.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 13

  1. The zener diode is connected to a battery, through a resistance R. The battery reverse biases the zener diode.
  2. The load resistance RL is connected across the terminals of the zener diode.
  3. The value of R is selected in such away that In the absence of load RL maximum safe current flows in the diode.
  4. Now consider that load is connected across the diode. The load draws a current.
  5. The current through the diode falls by the same amount but the voltage drops across the load remains constant.
  6. The series resistance ‘R’ absorbs the output voltage fluctuations, so as to maintain constant voltage across the load.
  7. The voltage across the zener diode remains constant even if the load RL varies. Thus, zener diode works as voltage regulator.
  8. If I is the input current, IZ and IL are zener and load currents.
    I = IZ + IL; Vin = IR + VZ
    But V0ut = VZ
    ∴ Vout = Vin – IR

Question 4.
Explain the working of LED and what are its advantages over conventional incandescent low power lamps.
Answer:
Light emitting diode (LED) : It is a photoelectronic device which converts electrical energy into light energy.

It is a heavily doped p-n junction diode which under forward bias emits spontaneous radiation. The diode is covered with a transparent cover so that the emitted light may not come out. Working : When p-n junction diode is forward biased, the movement of majority charge carriers takes place across the junction. The electrons move from n-side to p-side through the junction and holes move from p-side to n-side.

As a result of it, concentration of majority carriers increases rapidly at the junction.

When there is no bias across the junction, therefore there are excess minority carriers on either side of the junction, which recombine with majority carriers near the junction.

On recombination of electrons and hole, the energy is given out in the form of heat and light.
Advantages of LED’s over incandescent lamp :

  1. LED is cheap and easy to handle.
  2. LED has less power and low operational voltage.
  3. LED has fast action and requires no warm up time.
  4. LED can be used in burglar alarm system.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 5.
Define NAND and NOR gates. Give their truth tables. [T.S. Mar. 17]
Answer:
NAND gate : NAND gate is a combination of AND gate and NOT gate.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 14
NAND gate can be obtained by connecting a NOT gate in the output of an AND gate. NAND gates are called universal gates.

  1. If both inputs are low, output is high.
    A = 0, B = 0, X = 1
  2. If any input is low, output is high.
    A = 0, B = 1, X = 1
    A = 1, B = 0, X =1
  3. If both inputs are high, output is low.
    A = 1, B = 1, X = 0
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 15

NOR gate : NOR gate is a combination of OR gate and NOT gate when the output of OR gate is connected to NOT gate. It has two (or) more inputs and one output.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 16

  1. If both inputs are low, output is high.
    A = 0, B = 0, X = 1
  2. If any input is high, the output is low.
    A = 0, B = 1, X = 0
    A = 1, B = 0, X = 0
  3. If both inputs are high, the output is low.
    A = 1, B = 1, X = 0
    NOR gate is also a universal gate.
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 17

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 6.
Explain the working of a solar cell and draw its I-V characteristics.
Answer:
Solar cell is a p-n junction device which converts solar energy into electric energy.
It consists of a silicon (or) gallium – arsenic p-n junction diode packed in a can with glass window on top. The upper layer is of p- type semiconductor. It is very thin so that the incident light photons may easily reach the p-n junction.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 18
Working : When light (E = hv) falls at the junction, electron – hole pairs are generated near the junction. The electrons and holes produced move in opposite directions due to junction field. They will be collected at the two sides of the junction giving rise to a photo voltage between top and bottom metal electrodes. Top metal acts as positive electrode and bottom metal acts as a negative electrode. When an external load is connected across metal electrodes a photo current flows.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 19
I-V characteristics : I – V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes. Because it does not draw current.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 20
Uses : They are used in calculators, wrist watches, artificial satellites etc.

Question 7.
Explain the operation of a NOT gate and give its truth table. [IPE 15, T.S.]
Answer:
NOT gate: NOT gate is the basic gate. It has one input and one output. The NOT gate is also called an inverter. The circuit symbol of NOT gate is shown in figure.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 21

  1. If input is low, output is high.
    A = 0, X = \(\overline{0}\) = 1
  2. If input is high, output is low.
    A = 1, X = \(\overline{1}\) = 0
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 22

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 8.
Draw an OR gate using two diodes and explain its operation. Write the truth table and logic symbol of OR gate.
Answer:
OR Gate : It has two input terminals and one output terminal. If both inputs are low the output is low. If one of the inputs is high, or both the inputs are high then the output of the gate is high. The truth tables of OR gate.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 23
In truth table logic function is written as A or B ‘OR’ logic function is represented by the symbol ‘plus’.
Q = A + B
Logic gate ‘OR’ is shown given below.

Implementation of OR gate using diodes :
Let D1 and D2 be two diodes.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 24
A potential of 5V represents the logical value 1.
A potential of OV represents the logical value 0.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 25
When A = 0, B = 0 both the diodes are reverse biased and there is no current through the resistance. So, the potential at Q is zero i.e., Q = 0. When A = 0 or B = 0 and the other equal to a potential of 5 V the diode whose anode is at a potential of 5 V is forward – biased and that diode behaves like a closed switch. The output potential then becomes 5 V i.e., Q = 1. When both A and B are 1, both the diodes are forward-biased and the potential at Q is same as that at A and B which is 5 V i.e., Q = 1. The output is same as that of the OR gate.

Question 9.
Sketch a basic AND circuit with two diodes and explain its operation. Explain how doping increases the conductivity in semiconductors ?
Answer:
AND gate : It has two input terminals and one output terminal.

  • If both the inputs are low (or) one of the inputs is low.
    • The output is low in an AND gate.
  • If both the inputs .are high
    • The output of the gate is high.
  • Note : If A and B are the inputs of the gate and the output is ‘Q’ then ‘Q’ is a logical function of A and B.
    AND gate Truth Tables
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 26
    The logical function AND is represented by the symbol dot so that the output, Q = A.B and the circuit symbol used for the logic gate AND is shown in Fig.
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 27
    The logical function AND is similar to the multiplication.

Implementation of AND gate using diodes : Let D1 and D2 represents two diodes. A potential of 5 V represents the logical value 1 and a potential of 0 V represents the logical value zero (0). When A = 0, B = ,0 both the diodes D1 and D2 are forward-biased and they behave like closed switches. Hence, the output Q is same as that A or B (equal to zero.) When A = 0 or B = 0, D1 or D2 is forward – biased and Q is zero. When A = 1 and B = T both the diodes are reverse – biased and they behave like open switches. There is no current through the resistance R making the potential at Q equal to 5V i.e., Q = 1. The output is same as that of an AND gate.

Doping increases the conductivity in Semiconductors: If a pentavalent impurity (Arsenic) is added to a pure tetravalent semiconductor it is called n-type semiconductor. Arsenic has 5 valence electrons, but only 4 electrons are needed to form covalent bonds with the adjacent Germanium atoms. The fifth electron is very loosely bound and become a free electron. Therefore excess electrons are available for conduction and conductivity of semiconductor increases.

Similarly when a trivalent impurity Indium is added to pure Germanium it is called p-type semi-conductor. In this excess holes in addition to those formed due to thermal energy are available for conduction in the valence band and the conductivity of semiconductor increases.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 10.
Explain how transistor can be used as a switch ?
Answer:
To understand the operation of transistor as a switch.

  1. As long as Vi is low and unable to forward bias the transistor, V0 is high (at Vcc).
  2. If Vi is high enough to drive the transistor into saturation then V0 is low, very near to zero.
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 28
  3. When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on.
  4. We can say that a low input to the transistor gives a high output and high input gives a low output.
  5. When the transistor is used in the cutoff (or) saturation state it acts as a switch.
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 29

Problems

Question 1.
In a half wave rectifier, a p-n junction diode with internal resistance 20 ohm is used. If the load resistance of 2 ohm is used in the circuit, then find the efficiency of this half wave rectifier.
Solution:
Internal resistance (rf) = 20Ω
RL = 2kΩ = 2000 Ω
η = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.406 \times 2000}{20+2000} \times 100=\frac{812 \times 100}{2020}\)
η = 40.2%.

Question 2.
A full wave p-n junction diode rectifier uses a load resistance of 1300 ohm. The internal resistance of each diode is 9 ohm. Find the efficiency of this full wave rectifier.
Solution:
Given that RL = 1300 Ω
rf = 9Ω
η = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.812 \times 1300}{9+1300} \times 100 ; \eta=\frac{8120 \times 13}{1309}\)
η = 80.64%.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
Calculate the current amplification factor β(beta) when change in collector current is 1mA and change in base current is 20μA.
Solution:
Change in collector current (∆IC) = 1mA = 10-3 A
Change in base current (∆IB) = 20 μA = 20 × 10-6 A
β = \(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{10^{-3}}{20 \times 10^{-6}}\); β = \(\frac{1000}{20}\)
β = 50

Question 4.
For a transistor amplifier, the collector load resistance RL = 2k ohm and the input resistance Ri = 1 k ohm. If the current gain is 50, calculate voltage gain of the amplifier.
Solution:
RL = 2kΩ = 2 × 103
Ri = 1kΩ = 1 × 103
β = 50.
Voltage gain (AV) = β × \(\frac{\mathrm{R}_{\mathrm{L}}}{\mathrm{R}_{\mathrm{i}}}=\frac{50 \times 2 \times 10^3}{1 \times 10^3}\)

AV = 100.

Textual Examples

Question 1.
C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors ?
Solution:
The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out and electron from these atoms(i.e., ionisation energy Eg) will be least for Ge, following by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.

Question 2.
Suppose a pure Si crystal has 5 × 1028 atoms m-3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n. = 1.5 × 1016 m-3.
Solution:
Note that thermally generated electrons (ni ~ 1016 m-3) are negligibly small as compared to those produced by doping.
Therefore, ni ≈ ND.
Since nenh = ni2, The number of holes, nh = (1.5 × 1016)2 / 5 × 1028 × 16-6
nh = (2.25 × 1032)/(5 × 1022) ~ 4.5 × 109m-3

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction ?
Solution:
No ! Any slab, howsoever flat, will have roughness much large than the inter-atomic crystal spacing(~2 to 3 A°) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.

Question 4.
The V-I characteristics of a silicon diode are shown in the Fig. Calculate the resistance of the diode at (a) ID = 15 mA and (b) VD = -10 V.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 30
Solution:
Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.
a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V
rfb = ∆V/∆I = 0.1V/10mA = 10Ω
b) From the curve at V = -10 V, I = -1 µA.
Therefore rrb = 10V/1µA = 1.0 × 107

Question 5.
In a Zener regulated power supply a Zener diode’with Vz = 6.0 V is used for regulation. The load current is to be 4.0 mA and the unregulated input is 10,0 V. What should be the value of series resistor Rs?
Solution:
The value of Rs should be such that the current through the Zener diode is much larger than the load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e., Iz = 20 mA. The total current through RS is , therefore, 24 mA. The voltage drop across RS is 10.0 – 6.0 = 4.0 V This gives RS = 4.0V/(24 × 10-3) A = 167Ω. The nearest value of carbon resistor is 150 Ω. So, a series resistor of 150 Ω is appropriate. Note that slight variation in the value of the resistor does not matter, what is important is that the current IZ should be sufficiently larger than IL.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 6.
The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). What is the reason then to operate the photodiodes in reverse bias ?
Solution:
Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n> > p). On illumination, let the excess eletrons and holes generated be ∆n and ∆p, respectively,
n’ = n + ∆n
p’ = p + ∆p
Here n’ and p’ are the electron and hole concentrations at any particular illumination and n and p are carriers concentration when there is no illumination. Remember ∆n = ∆p and n > > p. Hence, the fractional change in the majority carriers (i.e., ∆n/n) would be much less than that in the minority carriers (i.e,, ∆p/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes-are preferably used in the reverse bias condition for measuring light intensity.

Question 7.
Why are Si and GaAs are preferred materials for solar cells ?
Solution:
The solar radiation spectrum received by us is shown in figure.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 31
The maxima is near 1.5 eV. For photo-excitation, hv > Eg. Hence, semiconductor with band gap ~1.5 eV or lower is likely to give better solar conversion efficiency. Silicon has Eg ~ 1.1 eV while for GaAS it is ~ 1.53 eV. In fact, GaAs is better On spite of its higher band gap) than Si because of its relatively higher absorption coefficient. If we choose materials like CdS or CdSe(Eg ~ 2.4 eV), we can use only the high energy component of the solar energy for photo-conversion and a significant part of energy will be of no use.

The question arises: why we do not use material like pbS(Eg ~ 0.4 eV) which satisfy the condition hv > Eg for v maxima corresponding to the solar radiation spectra ? if we do so, most of the solar radiation will be absorbed on the top-layer of solar cell mid will not reach in or near the depletion region. For effective electron-hole separation, due to the junction field, we want the photo-generation to occur in the junction region only.

Question 8.
From the output charactristics shown in fig, calculate the values of βac and βdc of the transistor when VCE is 10 V and IC = 4.0 mA.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 32
Solution:
βac = \(\left(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right)_{\mathrm{V}_{\mathrm{CE}}}\) ; βdc = \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}\)
For determining βac and βdc at the stated values of VCE and IC one can proceed as follows. Consider any two characteristics for two values of IB which lie above and below the given value of IC. Here IC = 4.0 mA. (Choose characteristics for IB = 30 and 20μA.) At V CE = 10V
we read the two values of Ic from the graph. Then
∆IB = (30 – 20)μA = 10μA. ∆IC
= (4.5 – 3.0) mA = 1.5 mA
Therefore, βac =1.5 mA/ 10μA = 150
For determining βdc either estimate the value of IB corresponding to IC = 4.0 mA at VCE = 10V or calculate the two values of βdc for the two characteristics chosen and find their mean.
Therefore, for IC = 4.5 mA and IB = 30 μA
βdc = 4.5 mA/30 μA = 150 and for IC = 3.0 mA/ and IB = 20 μA
βdc = 3.0 mA / 20 μA= 150
Hence, βdc = (150 + 150)/ 2 = 150

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 9.
In Fig. the VBB supply can be varied from OV to 5.0 V. The Si transistor has βdc = 250 and RB = 100 kΩ, RC = 1 KΩ, VCC = 5.0V. Assume that when the transistor is saturated, VCE = 0V and VBE = 0.8V. Calculate
(a) the minimum base current, for which the transistor will reach saturation. Hence,
(b) determine V1 for when the transistor is ‘switched on’,
(c) find the ranges of V1 for which the transistor is ‘switched of and ‘switched on’.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 33
Solution:
Given at stauration
VCE = OV, VBE = 0.8V
VCE = VCC – ICRC
IC = VCC / RC = 5.0V/1/0 kΩ = 5.0mA
Therefore, IB = IC
= 5.0 mA/250 = 20μA
The input voltage at which the transistor will go into saturation is given by
VIH = VBB = IBRB + VBE
= 20μA × 100 kΩ + 0.8V = 2.8V
The value of input voltage below which the transistor remains cutoff is given by
VIL = 0.6V, VIH = 2.8V
Between 0.0V and 0.6V the transistor will be in the ‘switched off-state. Between 2.8V and 5.0V, it will be in ‘switched on’ state.
Note that the transistor is in active state when IB varies from 0.0mA to 20mA. In this range, IC = βIB is valid. In the saturation range IC ≤ βIB.

Question 10.
For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0 kΩ is 2.0 V. Suppose the current amplification factor of the transistor is 100, what should be the value of RB in series with VBB supply of 2.0 V if the dc base current has to be 10 times the signal current. Also calculate the dc drop across the collector resistance. (Refer to Fig)
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 34
Solution:
The output ac voltage is 2.0 V. So, the ac collector current iC= 2.0/2000 = 1.0 mA. The signal current through the base is, therefore given by iB = iC/β = 1.0 mA/100 = 0.010 mA. The dc base current has to be 10 × 0.010 = 0.10 mA
From VBB = VBE+ IB RB RB = (VBB – VBE)/IB. Assuming VBE = 0.6V
RB = (2.0 – 0.6)/0.10 = 14kΩ
The dc collector current IC = 100 × 0.10 = 10 mA.

Question 11.
Justify the output waveform (Y) of the OR gate for the following inputs A and B given in fig.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 35
Solution:
Note the following :

  • At t ≤ t1; A = 0, B = 0; Hence Y = 0
  • For t1 to t2; A = 1, B = 0; Hence Y = 1
  • For t2 to t3; A = 1, B = 1; Hence Y = 1
  • For t3 to t4; A = 0, B = 1; Hence Y = 1
  • For t4 to t5; A = 0, B = 0; Hence Y = 0
  • For t5 to t6; A = 1, B = 0; Hence Y = 1
  • For t > t6; A = 0,B = 1; Hence Y = 1

Therefore the waveform Y will be as shown in the Fig.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 36

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 12.
Take A and B input waveforms similar to that in Example 11. Sketch the output waveform obtained from AND gate.
Solution:

  • For t ≤ t1; A = 0, B = 0; Hence Y = 0
  • For t1 to t2; A = 1, B = 0; Hence Y = 0
  • For t2 to t3; A = 1, B = 1; Hence Y = 1
  • For t3 to t4; A = 0, B = 1; Hence Y = 0
  • For t4 to t5; A = 0, B = 0; Hence Y = 0
  • For t5 to t6; A = 1, B = 0; Hence Y = 0
  • For t > t6; A = 0,B = 1; Hence Y = 0

Based on the above, the output waveform for AND gate can be drawn as given below.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 37

Question 13.
Sketch the output Y from a NAND gate having inputs A and B given below :
Solution:

  • For t ≤ t1; A = 1, B = 1; Hence Y = 0
  • For t1 to t2; A = 0, B = 0; Hence Y = 1
  • For t2 to t3; A = 0, B = 1; Hence Y = 1
  • For t3 to t4; A = 1, B = 0; Hence Y = 1
  • For t4 to t5; A = 1, B = 1; Hence Y = 0
  • For t5 to t6; A = 0, B = 0; Hence Y = 1
  • For t > t6; A = 0,B = 1; Hence Y = 1
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 38

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Students get through AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei

Very Short Answer Questions

Question 1.
The half life of 58Co is 72 days. Calculate Its average life. [Board model Paper]
Answer:
Thalf = 0.693 × TMean ⇒ TMean = \(\frac{T_{\text {half }}}{0.693}=\frac{72}{0.693}\) = 103.8 days.

Question 2.
Why do all electrons emitted during p-decay not have the same energy?
Answer:
When a neutron is converted into a proton, an electron and neutron are emitted along with it.
\({ }_1^1 \mathrm{n} \longrightarrow{ }_1^1 \mathrm{H}+{ }_{-1}^0 \mathrm{e}+\mathrm{v}\)
In β – decay proton remains in the nucleus, but electron and neutron are emitted with constant energy. The energy of neutron is not constant. So, ail electrons do not have same energy.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 3.
Neutrons are the best projectiles to produce nuclear reactions. Why ?
Answer:
Neutrons are uncharged particles. So they do not get deflected by the electric and magnetic fields. Hence Neutrons are considered as best projectiles in nuclear reaction.

Question 4.
Neutrons cannot produce ionization. Why ?
Answer:
Because neutrons are uncharged particles and cannot produce ionization.

Question 5.
What are delayed neutrons ?
Answer:
Neutrons are emitted in the fission products after, sometime are called delayed neutrons.

Question 6.
What are thermal neutrons ? What is their importance ?
Answer:
Neutrons having kinetic energies approximately 0.025 eV are called as slow neutrons or thermal neutrons. 235U undergoes fission only when bombarded with thermal neutrons.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 7.
What is the value of neutron multiplication factor in a controlled reaction and in an uncontrolled chain reaction ?
Answer:
In controlled chain reaction K = 1
In uncontrolled chain reaction K > 1

Question 8.
What is the role of controlling rods in a nuclear reactor ?
Answer:
In nuclear reactor, controlling rods are used to absorb the neutrons. Cadmium, boron materials are used in the form of rods in reactor. These control the fission rate.

Question 9.
Why are nuclear fusion reactions called thermo nuclear reactions ?
Answer:
Nuclear fusion occurs at very high temperatures. So it is called as thermo nuclear reaction.

Question 10.
Define Becquerel and Curie.
Answer:
Becquerel: 1 disintegration or decay per second is called Becquerel. It is SI unit of activity.
1 disintegration or decay
i.e., Becquerel = \(\frac{1 \text { disintegration or decay }}{\text { second }}\)
Curie : 3.7 × 1010 decays per second is called Curie.
1 Curie : 1 Ci = \(\frac{3.7 \times 10^{10} \text { decays }}{\text { second }}\) = 3.7 × 1010Bq.

Question 11.
What is a chain reaction ?
Answer:
Chain reaction : The neutrons produced in the fission of a nucleus can cause fission in other neighbouring nuclei producing more and more neutrons to continue the fission until the whole fissionable material is disintegrated. This is called chain reaction.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 12.
What is the function of moderator in a nuclear reactor ?
Answer:
They are used to slow down the fast moving neutrons produced during the fission process.
e.g.: Heavy water, Beryllium.

Question 13.
What is the energy released in the fusion of four protons to form a helium nucleus ?
Answer:
26.7 MeV energy is released.

Short Answer Questions

Question 1.
Write a short note on the discovery of neutron.
Answer:

  1. Bothe and Becker found that when beryllium is bombarded with a – particles of energy 5 MeY which emitted a highly penetrating radiation.
  2. The equation for above process can be written as
    \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{13} \mathrm{C}\) + γ – (radiation energy)
  3. The radiations are not effected by electric and magnetic fields.
  4. In 1932, James Chadwick, had subjected nitrogen and argon to the beryllium radiation. He interpreted the experimental results by assuming that the radiation is of a new kind of particles which has no charge and its mass is equal to proton. These neutral particles were named as neutrons. Thus the neutron was discovered.
  5. The experimental results can be represented by the following equation.
    \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 \mathrm{n}+\mathrm{Q}\)
    AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 1

Question 2.
What are nuclear forces ? Write their properties.
Answer:
The forces which hold the nucleons together in nucleus are called nuclear forces.
Properties of Nuclear forces:

  1. Nuclear forces are attractive forces between proton and proton (P – N, proton and neutron (P – N) and neutron and neutron (N – N).
  2. Nuclear forces are independent of charge. It was found that force between proton and proton is same as force between neutron and neutron.
  3. These forces are short range forces i.e., these forces will act upto a small distance only. Generally the range of nuclear forces is upto few Fermi (10-15 m).
  4. These forces are non central forces, i.e., they do not act along the line joining the two nucleons.
  5. These forces are exchange forces. The force between two nucleons is due to exchange of n-mesons.
  6. These forces are spin dependent. These forces are strong when the spin of two nucleons are in same direction and they are weak when they are in opposite direction.
  7. Nuclear forces are saturated forces i.e., the force between nucleons will extend upto the immediate neighbouring nucleons only.
  8. These are the strongest forces in nature. They are nearly 1038 times stronger than gravitational forces and nearly 100 times stronger than Coulombic forces.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 3.
Define half life period and decay constant for a radioactive substance. Deduce the relation between them.
Answer:
Half life period (T) : Time taken for the number of radio active nuclei to disintegrate to half of its original number of nuclei is called Half life period.

Decay constant (λ) : The ratio of the rate of radioactive decay to the number of nuclei present at that instant.
It is a proportional constant and is denoted by ‘λ’.
∴ λ = \(\frac{-\left(\frac{d N}{d t}\right)}{N}\)

Relation between half the period and decay constant:

  1. The radioactive decay law N = N0 e-λt states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time. Here A is called decay constant.
  2. If N0 is the number of nuclei at t = 0 and N is the radioactive nuclei at any instant of time’t’.
  3. Substituting N = \(\frac{\mathrm{N}_0}{2}\) at t = T in N = N0 e-λt
  4. \(\frac{\mathrm{N}_0}{2}\) = N0 e-λT
    eλT = 2
    λT= ln 2
    T = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log _{10}^2}{\lambda}\)
    ∴ T = \(\frac{0.693}{\lambda}\)

Question 4.
What is nuclear fission ? Give an example to illustrate it.
Answer:
Nuclear fission : The process of dividing a heavy nucleus into two or more smaller and stable nuclei due to nuclear reaction is called nuclear fission.
Ex: The fission reaction is \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_0^1 \mathrm{n}+\mathrm{Q}\)
Where Q is the energy released.
Q = (Total mass of reactants – Total mass of product) C2
= [(Mass of \({ }_{92}^{235} \mathrm{U}\) + Mass of \({ }_0^1 \mathrm{n}\)) – (Mass of \({ }_{56}^{141} \mathrm{Ba}\) + Mass of \({ }_{36}^{92} \mathrm{Kr}\) + Mass of three neutrons)] C2
= (235.043933 – 140.9177 – 91.895400 – 2 × 1.008665) amu × C2.
= 0.2135 × 931.5 MeV = 198.9 MeV = 200 MeV

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 5.
What is nuclear fusion ? Write the conditions for nuclear fusion to occur.
Answer:
Nuclear fusion : The process of combining lighter nuclei to produce a larger nucleus is known as nuclear fusion.
E.g : Hydrogen nuclei (1H1) are fused together to form heavy Helium (2He4) along with 25.71 MeV energy released.
Conditions for nuclear fusion :

  1. Nuclear fusion occurs at very high temperatures such as 107 kelvin and very high pressures. These are obtained under the explosion of an atom bomb.
  2. Higher density is also desirable so that collisions between light nuclei occur quite frequently.

Question 6.
Distinguish between nuclear fission and nuclear fusion.
Answer:
Nuclear fission

  1. In this process heavy nucleus is divided into two fragments along with few neutrons.
  2. These reactions will takes place even at room temperature.
  3. To start fission atleast one thermal neutron from outside is compulsory.
  4. Energy released per unit mass of participants is less.
  5. In this process neutrons are liberated.
  6. This reaction can be controlled.
    Ex: Nuclear reactor.
  7. Atom bomb works on principle of fission reaction.
  8. The energy released in fission can be used for peaceful purpose.
    Ex: Nuclear reactor and Atomic power stations.

Nuclear fusion

  1. In this process lighter nuclei will join together to produce heavy nucleus.
  2. These reactions will takes place at very high temperature such as 107 Kelwin.
  3. No necessary of external neutrons.
  4. Energy released per unit mass of participants is high. Nearly seven times more than fission reaction.
  5. In this process positrons are liberated.
  6. There is no control on fusion reaction.
  7. Hydrogen bomb works on the principle of fusion reaction.
  8. The energy released in fusion cannot be used for peaceful purpose.

Long Answer Questions

Question 1.
Define mass defect and binding energy. How does binding energy per nucleon vary with mass number ? What is its significance ?
Answer:
1. Mass defect (∆M) : The difference in mass of a nucleus and its constituents is called the mass defect. The nuclear mass M is always less than the total mass, Em, of its constituents.
Mass defect, (∆M) = [Zmp + (A – Z)mn – M]

2. Binding energy: The energy required to break the nucleus into its constituent nucleons is called the binding energy.
Binding Energy, (Eb) = ∆MC2 = [Zmp + (A – Z)mn – M] 931.5 MeV
Nuclear binding energy is an indication of the stability of the nucleus.
Nuclear binding energy per nucleon Ebn = \(\frac{E_{\mathrm{b}}}{\mathrm{A}}\)

3. The following graph represents how the binding energy per nucleon varies with the mass number A.
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 2
4. From the graph that the binding energy is highest in the range 28 < A < 138. The binding energy of these nuclei is very close to 8.7 MeV

5. With the increase in the mass number the binding energy per nucleon decreases and consequently for the heavy nuclei like Uranium it is 7.6 MeV

6. In the region of smaller mass numbers, the binding energy per nucleon curve shows the characteristic minima and maxima.

7. Minima are associated with nuclei containing an odd number of protons and neutrons such as \({ }_3^6 \mathrm{Li},{ }_5^{10} \mathrm{~B},{ }_7^{14} \mathrm{~N}\) and the maxima are associated with nuclei having an even number of protons and neutrons such as \({ }_2^4 \mathrm{He},{ }_6^{12} \mathrm{C},{ }_8^{16} \mathrm{O}\).

Significance:
8. The curve explains the relationship between binding energy per nucleon and stability of the nuclei.

9. Uranium has a relatively low binding energy per nucleon as 7.6 MeV Hence to attain greater stability Uranium breaks up into intermediate mass nuclei resulting in a phenomenon called fission.

10. On the other hand light nuclei such as hydrogen combine to form heavy nucleus to form helium for greater stability, resulting in a phenomenon called fusion.

11. Iron is the most stable having binding energy per nucleon 8.7 MeV, and it neither undergoes fission per fusion.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 2.
What is radioactivity ? State the law of radioactive decay. Show that radioactive decay is exponential in nature. [T.S. Mar. 16]
Answer:
1. Radioactivity : The nuclei of certain elements disintegrate spontaneously by emitting alpha (α), beta (β) and gamma (δ) rays. This phenomenon is called Radioactivity or Natural radioactivity.

2. Law of radioactivity decay : “The rate of radioactive decay \(\left(\frac{d N}{d t}\right)\) (or) the number of
nuclei decaying per unit time at any instant, is directly proportional to the number of nuclei (N) present at that instant is called law of radioactivity decay”.

3. Radioactive decay is exponential in nature : Consider a radioactive substance. Let the number of nuclei present in the sample at t = 0, be N0 and let N be the radioactive nuclei remain at an instant t.
\(\frac{\mathrm{dN}}{\mathrm{dt}}\) ∝ N ⇒ \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = – λN
dN = – λ Ndt …………………….. (1)
The proportionality constant λ is called decay constant or disintegration constant. The negative sign indicates the decrease in the number of nuclei.

4. From eq. (1) \(\frac{\mathrm{dN}}{\mathrm{N}}\) = – λ dt ……………… (2)

5. Integrating on both sides
\(\int \frac{\mathrm{dN}}{\mathrm{N}}=-\lambda \int \mathrm{dt}\)
ln N = – λt + C …………….. (3)
Where C = Integration constant.

6. At t = O; N = N0. Substituting in eq. (3), we get ln N0 = C
∴ ln N = -λt + ln N0
ln N – ln N0 = – λt
ln (\(\frac{\mathrm{N}}{\mathrm{N}_0}\)) = – λt
∴ N = N0 e-λt
The above equation represents radioactive decay law.

7. It states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time.

Sample Problems

Question 1.
The half life of radium is 1600 years. How much time does lg of radium take to reduce to 0.125g. [IPE 2016 (TS)]
Answer:
Half – life of radium = 1600 years;
\(\frac{N}{N_0}=\frac{1}{2^n} \Rightarrow \frac{0.125}{1}=\frac{1}{2^n} \Rightarrow \frac{125}{1000}=\frac{1}{2^n} \Rightarrow \frac{1}{8}=\frac{1}{2^n} \Rightarrow \frac{1}{2^3}=\frac{1}{2^n} \Rightarrow n=3\)
∴ Time taken = Half life × no. of Half lives = 1600 × 3 = 4800 years

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 2.
Plutonium decays with a half-life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ?
Answer:
Half life of plutonium, T = 24000 years; Stored time of plutonium, t = 72000 years
no. of half lives, n = \(\frac{t}{T}=\frac{72000}{24000}\) = 3; Fraction of plutonium remains = \(\frac{N}{N_0}=\frac{1}{2^n}=\frac{1}{2^3}=\frac{1}{8}\)

Question 3.
Explain the principle and working of a nuclear reactor with the help of a labelled diagram. [A.P. & T.S. Mar. 17, A.P. Mar. 16 15, Mar. 14]
Answer:
Principle : A nuclear reactor works on the principle of achieving controlled chain reaction in natural Uranium 238U enriched with 235U, consequently generating large amounts of heat.
A nuclear reactor consists of (1) Fuel (2) Moderator (3) Control rods (4) Radiation shielding (5) Coolant.
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 3
1. Fuel and clad : In reactor the nuclear fuel is fabricated in the form of thin and long cylindrical rods. These group of rods treated as a fuel assembly. These rods are surrounded by coolant, which is used to transfer of heat produced in them. A part of the nuclear reactor which is used to store the nuclear fuel is called the core of the reactor. Natural uranium, enriched uranium, plutonium and uranium – 233 are used as nuclear fuels.

2. Moderator : The average energy of neutrons released in fission process is 2 MeV. They are used to slow down the velocity of neutrons. Heavy water or graphite are used as moderating materials in reactor.

3. Control Rods : These are used to control the fission rate in reactor by absorbing the neutrons. Cadmium and boron are used as controlling the neutrons, in the form of rods.

4. Shielding : During fission reaction beta and gamma rays are emitted in addition to neutrons. Suitable shielding such as steel, lead, concrete etc., are provided around the reactor to absorb and reduce the intensity of radiations to such low levels that do not harm the operating personnel.

5. Coolant : The heat generated in fuel elements is removed by using a suitable coolant to flow around them. The coolants used are water at high pressures, molten sodium etc.

Working : Uranium fuel rods are placed in the aluminium cylinders. The graphite moderator is placed in between the fuel cylinders. To control the number of neutrons, a number of control rods of cadnium or beryllium or boron are placed in the holes of graphite block. When a few 235U nuclei undergo fission fast neutrons are liberated. These neutrons pass through the surrounding graphite moderator and loose their energy to become thermal neutrons. These thermal neutrons are captured by 235U. The heat generated here is used for heating suitable coolants which is turn heat water and produce steam. This steam is made to rotate steam turbine and there by drive a generator of production for electric power.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 4.
Explain the source of stellar energy. Explain the carbon – nitrogen cycle, proton – proton cycle occuring in stars.
Answer:
Scientists proposed two types of cyclic processes for the sources of energy in the sun and stars. The first is known as carbon-nitrogen cycle and the second is proton-proton cylce.

1. Carbon-Nitrogen Cycle : According to Bethe carbon-nitrogen cycle is mainly responsible for the production of solar energy. This cycle consists of a chain of nuclear reactions in which hydrogen is converted into Helium, with the help of Carbon and Nitrogen as catalysts. The nuclear reactions are as given below.
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 7

2. Proton – Proton Cycle: A star is formed by the condensation of a large amount of matter at a point in space. Its temperature rises to 2,00,000°C as the matter contracts under the influence of gravitational attraction. At this temperature the thermal energy of the protons is sufficient to form a deuteron and a positron. The deuteron then combines with another proton to form lighter nuclei of helium \({ }_2^3 \mathrm{He}\). Two such helium nuclei combine to form a helium nucleus latex]{ }_2^4 \mathrm{He}[/latex] and two protons releasing a total amount of energy 25.71 MeV The nuclear fusion reactions are given below.
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 4

Problems

Question 1.
Compare the radii of the nuclei of mass numbers 27 and 64.
Solution:
A1 = 27; Asub>2 = 64
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right]^{1 / 3}\) [∵ R = R0A1/3]
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{27}{64}\right]^{\frac{1}{3}}=\frac{3}{4}\)
∴ R1 : R2 = 3 : 4

Question 2.
Find the energy required to split \({ }_8^{16} \mathrm{O}\) nucleus into four α-particles. The mass of an a-particle is 4.002603u and that of oxygen is 15.994915u.
Solution:
The energy required to split O = [Total mass of the products – Total mass of the reactants] c2
Mass of four \({ }_2^4 \mathrm{He}\) – Mass of \({ }_8^{16} \mathrm{O}\)] × c2
= [(4 X 4.002603) – 15.994915] u × c2
= [16.010412 – 15.994915] u × c2
= (0.015497) 931.5 MeV = 14.43 MeV

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 3.
A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Solution:
Fraction of substance decays
= \(\frac{\text { Quantity remains }}{\text { Initial quantity }}\)
= \(\frac{1}{2^n}=\frac{1}{32}=\frac{1}{2^5}\)
∴ n = 5
Duration of time = 25 days
We know (n) = \(\frac{\text { Duration of time }}{\text { Half life time }}\)
∴ Half life time = \(\frac{\text { Duration of time }}{\mathrm{n}}\)
\(\frac{25}{5}\) = 5 days

Question 4.
One gram of radium is reduced by 2 milli- gram in 5 years by a-decay. Calculate the half¬life of radium.
Solution:
Initial (original) mass (N0) = 1 gram
Reduced mass – 2 mg = 0.002 grams
Final mass (N)= 1 – 0.002 = 0.998 grams
t = 5 years
e-λt = \(\frac{\mathrm{N}}{\mathrm{N}_0}\) ⇒ eλt = \(\frac{\mathrm{N}_0}{\mathrm{~N}}\) ⇒ λt = loge[latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]
λt = 2.303 log [latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]
λt = 2.303 log [latex]\frac{1}{0.998}[/latex]
= 2.303 log (1.002)
= 2.303 × 0.000868
= 0.001999
λ = \(\frac{0.001999}{5}\) = 0.0003998
T = \(\frac{0.693}{\lambda}=\frac{0.693}{0.0003998}\) = 1733.3 years

Question 5.
If one microgram of \({ }_92^{235} \mathrm{U}\) is completely destroyed in an atomhomb, how much energy will be released ?
Solution:
m = 1 μg = 1 × 10-6 g = 1 × 10-6 × 10-3 kg
= 10-9 kg
c = 3 × 108 m/s
E = mc2 = 1 × 10-9 × 9 × 106 = 9 × 107 J

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 6.
200 MeV energy is released when one nucleus of 235U undergoes fission. Find the number of fissions per second required for producing a power of 1 megawatt.
Solution:
E = 200 MeV
P = 1 × 106 W
P = \(\frac{\mathrm{nE}}{\mathrm{t}} \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{10^6}{200 \times 10^6 \times 1.6 \times 10^{-19}}\)
= \(\frac{1}{32}\) × 1018
∴ P = 0.03125 × 1018
= 3.125 × 106

Textual Examples

Question 1.
Given the mass of iron nucleus as 55.85u and A = 56, find the nuclear density ?
Solution:
mFe = 55.85, u = 9.27 × 10-26 kg
Nuclear density = \(\frac{\text { mass }}{\text { volume }}\)
= \(\frac{9.27 \times 10^{-26}}{(4 \pi / 3)\left(1.2 \times 10^{-15}\right)^3} \times \frac{1}{56}\)
= 2.29 × 1017 kg m-3
The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

Question 2.
Calculate the energy equivalent of 1 g of substance.
Solution:
Energy, E = 10-3 × (3 × 108)2 J
E = 10-3 × 9 × 1016 = 9 × 1013 J
Thus, if one gram of matter is converted to energy, there is a release of an enormous amount of energy.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 3.
Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of \({ }_8^{16} \mathrm{O}\) in MeV/c2.
Solution:
1 u = 1.6605 × 10-27 kg
To convert it into energy units, we multiply it by c2 and find, that energy. equivalent
= 1.6605 × 10-27 × 2.9979 × 108 kg m2/s2
= 1.4924 × 10-10 J
= \(\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}}\) eV
= 0.9315 × 109 eV = 931.5 MeV
or, 1 u = 931.5 MeV/c2.
For \({ }_8^{16} \mathrm{O}\), ∆M = 0.13691 u = 0.13691 × 931.5 MeV/c2 = 127.5 MeV/c2
The energy needed to separate \({ }_8^{16} \mathrm{O}\) into hs constituents is thus 127.5 MeV/c2.

Question 4.
The half-life of \({ }_{92}^{238} \mathrm{U}\) undergoing a – decay is 4.5 × 109 years. What is the activity of 1 g sample of \({ }_{92}^{238} \mathrm{U}\) ?
Solution:
T1/2 = 4.5 × 109 y
= 4.5 × 109 y × 3.16 × 107 s/y.
= 1.42 × 1017 s
One k mol of any isotope contains Avogadro’s number of atoms, and so 1 g of \({ }_{92}^{238} \mathrm{U}\) contains
\(\frac{1}{238 \times 10^{-3}}\) k mol × 6.025 × 1026 atoms/kmol
= 25.3 × 1020 atoms.
The decay rate R is
R = λN
= \(\frac{0.693}{\mathrm{~T}_{1 / 2}}\) N = \(\frac{0.693 \times 25.3 \times 10^{20}}{1.42 \times 10^{17}}\) S-1
= 1.23 × 104 S-1
= 1.23 × 104 Bq

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 5.
Tritium has a half-life of 12.5 y undergoing beta decay. What fraction of sample of pure tritium will remain undecayed after 25 y.
Solution:
By definition of half-life, half of the initial i sample will remain undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei would have decayed. Hence, one fourth of the sample of the initial pure s tritium will remain undecayed.

Question 6.
We are given the following atomic masses:
\({ }_{92}^{238} \mathrm{U}\) = 238.05079 u
\({ }_{2}^{4} \mathrm{He}\) = 4.00260 u
\({ }_{90}^{234} \mathrm{Th}\) = 234.04363 u
\({ }_{1}^{1} \mathrm{H}\) = 1.00783 u
\({ }_{91}^{237} \mathrm{Pa}\) = 237.05121 u
Here the symbol Pa is for the element protactinium (Z = 91).
a) Calculate the energy released during the alpha decay of \({ }_{92}^{238} \mathrm{U}\).
b) Show that \({ }_{92}^{238} \mathrm{U}\) cannot spontaneously emit a proton.
Solution:
a) The alpha decay of \({ }_{92}^{238} \mathrm{U}\) is given by
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 6
The energy released in this process is given by
Q = (MU – MTh – MHe) c2
Substituting the atomic masses as given in the data, we find
Q = (238.05079 – 234.04363 – 4.00260)u × c2
= (0.00456 u) c2
= (0.00456 u) (931.5 MeV/u)
= 4.25 MeV

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

b) If \({ }_{92}^{238} \mathrm{U}\) spontaneously emits a proton, the decay process would be
\({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{91}^{237} \mathrm{~Pa}+{ }_1^1 \mathrm{H}\)
The Q for this process to happen is
= (MU – MPa – MH)c2
(238.05079 – 237.05121 – 1.00783) u × c2
=(- 0.00825 u) c2
= – (0.00825 u) (931.5 MeV/u)
= -7.68 MeV
Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of 7.68 MeV to a \({ }_{92}^{238} \mathrm{U}\) nucleus to make it emit a proton.

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Students get through AP Inter 2nd Year Physics Important Questions 13th Lesson Atoms which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 13th Lesson Atoms

Very Short Answer Questions

Question 1.
What is the physical meaning of negative energy of an electron’ ?
Answer:
The ‘negative energy of an electron’ indicates that the electron is bound to the nucleus due to force of attraction.

Question 2.
Sharp lines are present in the spectrum of a gas. What does this indicate ?
Answer:
Sharp lines in the spectrum of gas, indicates bright lines against dark background.

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 3.
Name a physical quantity whose dimensions are the same as those of angular momentum.
Answer:
Planck’s constant.

Question 4.
What is the difference between α – particle and helium atom ?
Answer:
Alpha particle

  1. It is a + 2e charged Helium nucleus.
  2. It contains 2 protons and 2 neutrons.

Helium atom

  1. It has no charge.
  2. It contains 2 protons, 2 electrons and 2 neutrons.

Question 5.
Among alpha, beta and gamma radiations, which get affected by the electric field ?
Answer:
Alpha and Beta radiations are get affected by the electric field.

Question 6.
What do you understand by the phrase ground state atom ?
Answer:
If the electron is present in the ground state, it is called ground state atom.

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 7.
Why does the mass of the nucleus not have any significance in scattering in Rutherford’s experiment ?
Answer:
The size of the atom is 10-10 m and size of the nucleus is 10-15 m. Hence atom has large empty space. So the mass of nucleus has no significance in Rutherford’s scattering experiment.

Question 8.
The Lyman series of hydrogen spectrum lies in the ultraviolet region. Why ? [A.P. Mar. 15]
Answer:
The calculated values of wavelengths lie in the ultraviolet region of the spectrum will agree with the values of wavelengths observed experimentally by Lyman.

Question 9.
Write down a table giving longest and shortest wavelengths of different spectral series.
Answer:
Wavelength limits of some spectral series of hydrogen.
AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms 1

Question 10.
Give two drawbacks of Rutherford’s atomic model.
Answer:
Drawbacks of Rutherford’s atomic model:

  1. As the revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. But matter .is stable, we cannot expect the atom collapse.
  2. The atoms should emit continuous spectrum, but what we observe is only a line spectrum.

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 11.
If the kinetic energy of revolving electron in an orbit is K, what is its potential energy and total energy ?
Answer:
For an electron revolving round the nucleus, total energy is always negative and it is numerically equal to kinetic energy.
∴ Total energy = -Kinetic energy = -K
Potential energy is always negative and PE = 2 × TE = -2K

Short Answer Questions

Question 1.
What is impact parameter and angle of scattering ? How are they related to each other ?
Answer:

  1. Impact parameter (b) : Impact parameter is defined as the perpendicular distance of the initial velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
  2. Scattering angle (θ) : The scattering angle (θ) is the angle between the asymtotic direction of approach of the α – particle and the asymtotic direction in which it receeds.
  3. The relation between b and θ is b = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Ze}^2}{\mathrm{E}}\) cot \(\frac{\theta}{2}\) where E = K.E of α – particle = \(\frac{1}{2}\) mυ2.

Question 2.
Explain the distance of closest approach and impact parameter.
Answer:
Distance of closest approach :

  1. Suppose an α-particle with initial kinetic energy (K.E) is directed towards the centre of the nucleus of an atom.
  2. On account of Coulomb’s repulsive force between nucleus and alpha particle, kinetic energy of alpha particle goes on decreasing and in turn, electric potential energy of the particle goes on increasing.
  3. At certain distance d’ from the nucleus, K. E of α-particle reduces to zero. The particle stops and it cannot go closer to the nucleus. It is repelled by the nucleus and therefore it retraces its path, turning through 180°.
  4. Therefore, the distance d is known as the distance of closest of approach.
    AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms 2
    The closest distance of approach,
    d = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{Z e^2}{\left(\frac{1}{2} m v^2\right)}\)
  5. Impact parameter (b) : Impact parameter is defined as the ⊥r distance of the initial velocity vector of the α – particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
    AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms 3

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 3.
Describe Rutherford atom model. What are the draw backs of this model ?
Answer:
Rutherford atom model: The essential features of Rutherford’s nuclear model of the atom or planetary model of the atom are as follows :

  1. Every atom consists of tiny central core, called the atomic nucleus, in which the entire positive charge and almost entire mass of the atom are concentrated.
  2. The size of nucleus is of the order of 10-15m, which is very small as compared to the size of the atom which is of the order of 10-10 m.
  3. The atomic nucleus is surrounded by certain number of electrons. As atom on the whole is electrically neutral, the total negative charge of electrons surrounding the nucleus is equal to total positive charge on the nucleus.
  4. These electrons revolve around the nucleus in various circular orbits as do the planets around the sun. The centripetal force required by electron for revolution is provided by the electrostatic force of attraction between the electrons and the nucleus.

Drawbacks : According to classical E.M. theory,

  1. the revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. As matter is stable, we cannot expect the atoms to collapse.
  2. since the frequency of radiation emitted is the same as the frequency of revolution, the atom should radiate a continuous spectrum, but what we observe is only a line spectrum.

Question 4.
What are the limitations of Bohr’s theory of hydrogen atom ? [A.P. Mar. 17; Mar. 14]
Answer:
Limitations of Bohr’s theory of Hydrogen atom :

  1. This theory is applicable only to simplest atom like hydrogen, with z = 1. The theory fails in case of atoms of other elements for which z > 1.
  2. The theory does not explain why orbits of electrons are taken as circular, while elliptical orbits are also possible.
  3. Bohr’s theory does not say anything about the relative intensities of spectral lines.
  4. Bohr’s theory does not take into account the wave properties of electrons.

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 5.
Write a short note on Debroglie’s explanation of Bohr’s second postulate of quantization. [T.S. Mar. 17]
Answer:
Debroglie’s explanation of Bohr’s second postulate of quantization :

  1. The second postulate of Bohr atom model says that angular momentum of electron orbiting around the nucleus is quantized i.e., mυr = \(\frac{\mathrm{nh}}{2 \pi}\) where n = 1, 2, 3,….
  2. According to Debroglie, the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave.
  3. When a string fixed at two ends is plucked, a large number of wavelengths are excited and standing wave is formed.
  4. It means that in a string, standing waves form when total distance travelled by a wave down the string and back is an integral number of wavelengths.
  5. According to Debroglie, a stationary orbit is that which contains an integral number of Debrogile waves associated with the revolving electron.
  6. For an electron revolving in nth circular orbit of radius rn, total distance covered = circumference of the orbit = 2πrn
    ∴ For permissible orbit, 2πrn = nλ
  7. According to Debrogile, λ = \(\frac{h}{m v_n}\)
    Where υn is speed of electron revolving in nth orbit
    ∴ 2πrn = \(\frac{\mathrm{nh}}{\mathrm{m} v_{\mathrm{n}}}\)
    nrn = \(\frac{\mathrm{nh}}{2 \pi}=\mathrm{n}\left(\frac{\mathrm{h}}{2 \pi}\right)\)
    i.e., angular momentum of electron revolving in nth orbit must be an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\), which is the quantum condition proposed by Bohr in second postulate.

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 6.
Explain the different types of spectral series in hydrogen atom. [T.S. Mar. 16, A.P. Mar. 15]
Answer:
The atomic hydrogen emits a line spectrum consisting of five series.

  1. Lyman series : v = Rc \(\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\) where n = 2, 3, 4
  2. Balmer series : v = Rc \(\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) where n = 3, 4, 5.
  3. Paschen series : v = Rc \(\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\) where n = 4, 5, 6
  4. Brackett series: v = Rc \(\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\) where n = 5, 6, 7
  5. Pfund series: v = Rc \(\left(\frac{1}{5^2}-\frac{1}{n^2}\right)\) where n = 6, 7, 8

Long Answer Questions

Question 1.
State the basic postulates of Bohr’s theory of atomic spectra. Hence obtain an expression for the radius of orbit and the energy of orbital electron in a hydrogen atom. [A.P. Mar. 16]
Answer:
a) Basic postulates of Bohr’s theory are

  1. The electron revolves round a nucleus in an atom in various orbits known as stationary orbits. The electrons cannot emit radiation when moving in their own stationary levels.
  2. The electron can revolve round the nucleus only in allowed orbits whose angular momentum is the integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)
    i.e., mυnrn = \(\frac{\mathrm{nh}}{2 \pi}\) …………… (1)
    where n = 1, 2, 3 …………..
    If an electron jumps from higher energy (E2) orbit to the lower energy (E1) orbit, the difference of energy is radiated in the form of radiation.
    i.e., E = hv = E2 – E1 ⇒ v = \(\frac{E_2-E_1}{h}\) …………… (2)

b) Energy of emitted radiation: In hydrogen atom, a single electron of charge – e, revolves around the nucleus of charge e in a circular orbit of radius rn.
1) K.E. of electron : For the electron to be in circular orbit, centripetal force = The electrostatic force of attraction between the electron and nucleus.
AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms 4
3) Radius of the oribit: Substituting the value of (6) in (2),
\(\frac{\mathrm{m}}{\mathrm{r}_{\mathrm{n}}}\left(\frac{\mathrm{n}^2 \mathrm{~h}^2}{4 \pi^2 \mathrm{r}_{\mathrm{n}}^2 \mathrm{~m}^2}\right)=\frac{\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}^2}\)
rn = \(\frac{n^2 h^2}{4 \pi^2 m K e^2}\) ……………. (1)
∴ rn = 0.53 n2

4) Total energy (En) : Revolving electron possess K.E. as well as P.E.
AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms 5

Problems

Question 1.
The radius of the first electron orbit of a hydrogen atom is 5.3 x HTum. What is the radius of the second orbit ?
Solution:
∴ rn ∝ n2
\(\) ⇒ r2 = 4r1
∴ r2 = 4 × 5.3 × 10-11 = 2.12 × 10-10m.

Question 2.
The total energy of an electron in the first excited state of the hydrogen atom is -3.4eV. What is the potential energy of the electron in this state?
Solution:
In 1st orbit, E = -3.4 eV
Total energy E = \(\frac{\mathrm{KZe}^2}{2 \mathrm{r}}-\frac{\mathrm{KZe}^2}{\mathrm{r}}\)
\(\frac{\mathrm{KZe}^2}{\mathrm{r}}\) = U(say)
E = \(\frac{\mathrm{U}}{2}\) – u = \(\frac{\mathrm{-U}}{2}\)
U = -2E
∴ U = -2 × -3.4 = 6.8 eV.

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 3.
The total energy of an electron in the first excited state of hydrogen atom is -3.4eV. What is the kinetic energy of the electron in this state ?
Solution:
In Hydrogen like atom, we know that
K = – Total energy E
Here E = – 3.4eV
∴ K = -(-3.4) = 3.4 eV . . ,

Question 4.
Prove that the ionisation energy of hydrogen atom is 13.6 eV.
Solution:
n = 1 corresponds to ground state.
E = \(\frac{-13.6}{\mathrm{n}^2}\) eV
E = \(\frac{-13.6}{\mathrm{1}^2}\) eV
E = -13.6 eV
The minimum energy required to free the electron from the ground state of hydrogen atom = 13.6 eV.
∴ Ionisation energy of hydrogen atom = 13.6 eV

Question 5.
Calculate the ionization energy for a lithium atom.
Solution:
For 3Li7 atom, Z = 3, n = 2 [∵Li = 1s2 2s1]
En = \(\frac{13.6 \mathrm{Z}^2}{\mathrm{n}^2}\) eV
= \(\frac{13.6 \times(3)^2}{4}\) = 30.6 eV
∴ Ionization energy of Lithium = 30.6eV

Question 6.
The wavelength of the first member of Lyman series is 1216 A. Calculate the wavelength of second member of Balmer series.
Solution:
AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms 6

Textual Examples

Question 1.
In the Rutherford’s nuclear model of the atom, the nucleus (radius about 10-15m) is analogous to the sun about which the electron move in orbit (radius ≈ 10-10m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is ? The radius of earth’s orbit is about 1.5 × 1011m. The radius of sun is taken as 7 × 108 m.
Solution:
The ratio of the radius of electron’s orbit to the radius of nucleus is (10-10 m)/(10-15 m) = 105, that is, the radius of the electron’s orbit is 105 times larger than the radius of nucleus. If the radius of the earth’s orbit around the sun were 105 times larger than the radius of the sun, the radius of the earth’s orbit would be 105 × 7 × 108 m = 7 × 1013 m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun.

It implies that an atom contains a much greater fraction of empty space than our solar system does.

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 2.
In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV α-particle before it comes momentarily to rest and reverses its direction?
Solution:
The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an α-particle and a gold nucleus is conserved. The system’s initial mechanical energy is Ei; before the particle and nucleus interact, and it is equal to its mechanical energy Ef when the α-particle momentarily stops. The initial energy Et is just the kinetic energy K of the incoming α-particle. The final energy Ef is just the electric potential energy U of the system. The potential energy U can be calculated from Equation.
F = \(\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}^2}\)
Let d be the centre-to-centre distance between the α-particle and the gold nucleus when the α-particle is at its stopping point. Then we can write the conservation of energy
Ei = Ef as
K = \(\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{d}}=\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{~d}}\)
Thus the distance of closest approach d is given by
d = \(\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{~K}}\)
The maximum kinetic energy found in α-particles of natural origin is 7.7 MeV or 1.2 × 10-12 J. Since 1/4πε0 = 9.0 × 109N m2/C2. Therefore with e = 1.6 × 10-19C, we have,
d = \(\frac{(2)\left(9.0 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^2 \mathrm{Z}}{1.2 \times 10^{-12} \mathrm{~J}}\)
= 3.84 × 10-16 Zm
The atomic number of foil material gold is Z = 79, so that
d (Au) = 3.0 × 10-14m = 30 fm. (1 fm (i.e. fermi) = 10-15m.)
The radius of gold nucleus is, therefore, less than 3.0 × 10-14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the a-particle. Thus, the α-particle reverses its motion without ever actually touching the gold nucleus.

Question 3.
It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.
Solution:
Total energy of the electron in hydrogen atom is – 13.6 eV = -13.6 × 1.6 × 10-19J
= -2.2 × 10-18 J.
Thus from equation, E = – \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) we have
– \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) = 2.2 10-18 J
This gives the orbital radius
r = – \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^2}{(2)\left(-2.2 \times 10^{-18} \mathrm{~J}\right)}\)
= 5.3 × 10-11 m.
The velocity of the revolving electron can be computed from Equation r = –\(\frac{e^2}{4 \pi \varepsilon_0 r m v^2}\)
with
m = 9.1 × 10-31kg,
υ = \(\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{mr}}}\) = 2.2 × 106 m/s

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 4.
According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.
Solution:
From Examjple 3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10-11 m is 2.2 × 10-6 m/s. Thus, the frequency of the electron moving around the proton is
v = \(\frac{v}{2 \pi \mathrm{r}}=\frac{2.2 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}\)
≈ 6.6 × 1015 Hz.
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 1015Hz.

Question 5.
A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite.
Solution:
From equation, we have
nrn = nh/2π
Here m = 10 kg and rn = 8 × 106 m. We have the time period T of the circling satellite as 2h. That is T = 7200 s.
Thus the velocity υn = 2π rn/T
The quantum number of the orbit of satellite
n = (2π rn)2 × m(T × h)
Substituting the values,
n = (2π × 8 × 106m)2 × 10/(7200 s × 6.64 × 10-34 J s)
= 5.3 × 1045
Note that the quantum number for the satellite motion is extremely large. In fact for such large quantum numbers the results of quantisation conditions tend to those of classical physics.

AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms

Question 6.
Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.
Solution:
The Rydberg formula is
hc/λif = \(\left(\frac{1}{n_{\mathrm{f}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}\right)\)
The wavelengths of the first four lines in the Lyman series correspond to transitions from ni = 2, 3, 4, 5 to nf = 1. We know that
\(\frac{\mathrm{me}^4}{8 \varepsilon_0^2 \mathrm{~h}^2}\) = 13.6 eV = 21.76 × 10-19 J
Therefore,
AP Inter 2nd Year Physics Important Questions Chapter 13 Atoms 7
Substituting, ni = 2, 3, 4, 5 we get λ21 = 1218 Å, λ31 = 1028 Å, λ41 = 974.3 Å, and λ51 = 951.4 Å.

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter

Students get through AP Inter 2nd Year Physics Important Questions 12th Lesson Dual Nature of Radiation and Matter which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 12th Lesson Dual Nature of Radiation and Matter

Very Short Answer Questions

Question 1.
What are “cathode rays”? [A.P. Mar. 17]
Answer:
Cathode rays are streams of fast-moving electrons or negatively charged particles.

Question 2.
What important fact did Millikan’s experiment establish?
Answer:
Millikan’s experiment established that electric charge is quantized. That means the charge on anybody (oil drop) is always an integral multiple of the charge of an electron, i.e., Q = ne.

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter

Question 3.
What is “work function” ? [T.S. Mar. 17, 15]
Answer:
The minimum energy required to liberate an electron from photometal surface is called work function, Φ0.

Question 4.
What is “photoelectric effect” ?
Answer:
When light of sufficierit energy is incident on the photometal surface, electrons are emitted. This phenomenon is called photoelectric effect.

Question 5.
Give examples of “photosensitive substances”. Why are they called so ?
Answer:
Examples of photosensitive substances are Li, Na, K, Rb and Cs etc.
The work function of alkali metals is very low. Even the ordinary visible light, alkali metals can produce photoelectric emission. Hence they are called photosensitive substances.

Question 6.
Write down Einstein’s photoelectric equation. [A.P. Mar. 15]
Answer:
Einstein’s photoelectric equation is given by Kmax = \(\frac{1}{2}\) mv2max = hυ – Φ0

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter

Question 7.
Write down de-Broglie’s relation and explain the terms therein. [T.S. & A.P. Mar. 16]
Answer:
The de-Broglie wave length (λ) associated with a moving particle is related to its momentum (p) is λ = \(\frac{h}{p}=\frac{h}{m v}\), where h is planck’s constant.

Question 8.
State Heisenberg’s Uncertainity Principle. [Mar. 14]
Answer:
Uncertainity principle states that “it is impossible to measure both position (∆x) and momentum of an electron (∆p) [or any other particle] at the same time exactly”, i.e., ∆x . ∆p ≈ h where ∆x is uncertainty in the specification of position and ∆p is uncertainty in the specification of momentum.

Question 9.
The photoelectric cut off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ? [Mar. 11]
Answer:
Cut off voltage, V0 = 1.5 V; Maximum kinetic energy, (KE)max = eV0 = e × 1.5 = 1.5eV

Question 10.
An electron, an α-particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength ? [T.S. Mar. 15]
Answer:
For a particle,
de Broglie wavelength, λ = h/p
Kinetic energy, K = p2/2m
Then, λ = h/\(\sqrt{2 \mathrm{mK}}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses.A proton AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 1 is 1836 times massive than an electron and an a-particle \(\left({ }_2^4 \mathrm{He}\right)\) four times that of a proton.
Hence, α-particle has the shortest de Broglie wavelength.

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter

Qeustion 11.
What is-the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts ? [A.P. Mar. 15]
Solution:
Accelerating potential V = 100 V The de Broglie wavelngth λ is
λ = h/p = \(\frac{1.227}{\sqrt{\mathrm{V}}}\) nm
λ = \(\frac{1.227}{\sqrt{100}}\) nm = 0.123 nm
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

Question 12.
If the wave length of electro magnetic radiation is doubled, what happens to energy of photon ? [IPE 2015 (TS)]
Answer:
E = \(\frac{\mathrm{hc}}{\lambda}\) ⇒ E ∝ \(\frac{1}{\lambda}\), since wave length of photon is doubled, its energy becomes halved.

Question 13.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10-15 V s. Calculate the value of Planck’s constant.
Solution:
Given, slope of graph tan θ = 4.12 × 10-15 V – s;
AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 2
For slope of graph, tan θ = \(\frac{\mathrm{V}}{\mathrm{v}}\)
We know that hv = eV
\(\frac{\mathrm{V}}{\mathrm{v}}=\frac{\mathrm{h}}{\mathrm{e}} \Rightarrow \frac{\mathrm{h}}{\mathrm{e}}\) = 4.12 × 10-15; h = 4.12 × 10-15 × 1.6 × 10-19 = 6.592 × 10-34 J – s.

Question 14.
The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Solution:
Given, P = 1.388 × 103 W/m2; λ = 550 nm = 550 × 10-9 m ,
h = 6.63 × 10-34 J-s; c = 3 × 108 m/s
Energy of each photon E = \(\frac{\text { hc }}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}}\) = 3.616 × 10-19 J
No. of photons incident on the earth’s surface, N = \(\frac{\mathrm{P}}{\mathrm{E}}=\frac{1.388 \times 10^3}{3.66 \times 10^{-19}}\)
∴ N = 3.838 × 1021 photons/m2 – s.

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter

Question 15.
Show that the wavelength of electromagnetic radiation is equal to the de-Brogile wavelength of its quantum (photon). [Mar. 14]
Answer:
Wave length of electromagnetic wave of frequency v and velocity C is given by,
λ = \(\frac{\mathrm{C}}{\mathrm{v}} \Rightarrow \lambda=\frac{\mathrm{C}}{\mathrm{v}} \times \frac{\mathrm{h}}{\mathrm{h}}=\frac{\mathrm{h}}{\left(\frac{\mathrm{hv}}{\mathrm{C}}\right)}=\frac{\mathrm{h}}{\mathrm{p}}\)    (∵ \(\frac{\mathrm{hv}}{\mathrm{C}}\) = p)
Hence, we can say wavelength of electromagnetic radiation is equal to the de-Brogile wavelength.

Sample Problem :

Question 1.
Calculate the (a) momentum and (b) dE-Brogile wavelength of the electrons accelerated through a potential difference of 56 V. [Mar. 14]
Answer:
a) Mass of the electron, m = 9 × 10-31 kg;
Potential difference, V = 56V
Momentum of electron, mv = \(\sqrt{2 \mathrm{eVm}}=\sqrt{2 \times\left(1.6 \times 10^{-19}\right) \times 56 \times 9 \times 10^{-31}}\) = 4.02 × 10-24kg ms-1

b) de-Broglie wavelength, λ = \(\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.62 \times 10^{-34}}{4.04 \times 10^{-24}}\) = 1.64 × 10-10m

Short Answer Questions

Question 1.
Describe an experiment to study the effect of frequency of incident radiation on ‘stopping potential’.
Answer:
Experimental study of the effect of frequency of incident radiation on stopping potential:

  1. The experimental set up is shown in fig.
    AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 3
  2. Monochromatic light of sufficient energy (E = hv) from source ‘s’ is incident on photosensitive plate ‘C’ (emitter), electrons are emitted by it.
  3. The electrons are collected by the plate A (collector), by the electric field created by the battery.
  4. The polarity of the plates C and A can be reversed by a commutator.
  5. For a particular frequency of incident radiation, the minimum negative (retarding) potential V0 given to the plate A for which the photo current stops or becomes zero is called stopping potential.
  6. The experiment is repeated with different frequencies, and their different stopping potential are measured with voltmeter.
  7. From graph, we note that
    AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 4

    • The values of stopping potentials are different for different frequencies.
    • The value of stopping potential is more negative for radiation of highef incident frequency.
    • The value of saturation current depends on the intensity of incident radiation but it is independent of the frequency of the incident radiation.

Question 2.
What is the deBroglie wavelength of a ball of mass 0.12 Kg moving with a speed of 20 ms-1? What can we infer from this result ?
Answer:
Given, m = 0.12 kg; υ = 20 m/s; h = 6.63 × 10-34 J-s;
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.63 \times 10^{-34}}{0.12 \times 20}=\frac{6.63 \times 10^{-34}}{2.4}\)
∴ λ = 2.762 × 10-34 m = 2762 × 10-21 Å.
The wave length of ball is very very small. Hence, its motion can be observed.

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter

Question 3.
What is the effect of (i) intensity of light (ii) potential on photoelectric current ?
Answer:
(i) Effect of intensity of light on photoelectric current:
1) When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ0) is increased then the number of photoelectrons emitted increases i.e., the value of photoelectric current (i) increases, ie., i ∝ I.
AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 5
ii) The effect of potential on photoelectric current:

  1. On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
    AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 6
  2. On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential.
  3. Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.

Sample Problem :

Question 1.
The work function of caesium metal is 2.14 eV When light of frequency 6 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) stopping potential and (c) maximum speed of the emitted photoelectrons ? [A.P. Mar. 16]
Solution:
Given, Φ0 = 2.14 eV; v = 6 × 1014 Hz
a) KEmax = hv – Φ0 = \(\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}\) – 2.14
∴ KEmax = 0.35 eV

b) KEmax = eV0 ⇒ 0.35 eV = eV0
∴ V0 = 0.35 V

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 7

Long Answer Questions

Question 1.
How did Einstein’s photoelectric equation explain the effect of intensity and potential on photoelectric current ? How did this equation account for the effect of frequency of ‘ incident light on stopping potential ?
Answer:

  1. Einstein postulated that a beam of light consists of small energy packets called photons or quanta.
  2. The energy of photon is E = hv. Where ‘h1 is Planck’s constant; v is frequency of incident light (or radiation).
  3. If the absorbed energy of photon is greater than the work function (Φ0 = hυ0), the electron is emitted with maximum kinetic energy i.e., kmax = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = eV0 = hv – Φ0. This equation is known as Einstein’s photoelectric equation.
  4. Effect of intensity of light on photoelectric current:
    When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ0) is increased then the number of photoelectrons emitted decreases i.e., the value of photoelectric current (i) increases, ie., i ∝ I.
    AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 5
  5. The effect of potential on photoelectric current:
    • On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
      AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 6
    • On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential (v0).
    • Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.
  6. The effect of frequency of incident radiation on stopping potential:
    On increasing the frequency of incident light, the value of stopping potential goes on increasing gradually as shown in fig. That means kmax increases eV0 also increases.
    AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 8
  7. From the graph, we note that
    • For a given photosensitive metal, the cut off potential (v0) varies linearly with the frequency of the incident radiation.
    • For a given photosensitive metal, there is a certain minimum cut off frequency υ0 (called threshold frequency) for which the stopping potential is zero.
      AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 9
  8. From the graph we note that
    • The value of cut-off potential is different for radiation of different frequency.
    • The value of stopping potential is more negative for radiation of higher incident frequency.
  9.  From above experiments, it is found that, if the incident radiation is of higher frequency than that of threshold frequency, the photoelectric emission is. possible.

Textual Examples

Question 1.
Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10-3 W. (a) What is the energy of a photon in the light beam ? (b) How many photons per second, on an average, are emitted by the source ?
Solution:
a) Each photon has an energy
E = hv = (6.63 × 10-34 J s) (6.0 × 1014 Hz)
= 3.98 × 10-19 J

b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = NE.
Then N = \(\frac{\mathrm{P}}{\mathrm{E}}=\frac{2.0 \times 10^{-3} \mathrm{~W}}{3.98 \times 10^{-19} \mathrm{~J}}\)
= 5.0 × 1015 photons per second.

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter

Question 2.
The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. [A.P. Mar. 16]
Solution:
a) For the cut-off or threshold frequency, the energy hv0 of the incident radiation must be equal to work function Φ0, so that
v0 = \(\frac{\phi_0}{\mathrm{~h}}=\frac{2.14 \mathrm{eV}}{6.63 \times 10^{-34} \mathrm{Js}}\)
= \(\frac{2.14 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}\) = 5.16 × 1014 Hz
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.

b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy e V0 by the retarding potential V0. Einstein’s Photoelectric equation is
eV0 = hv – Φ0 = \(\frac{\mathrm{hc}}{\lambda}\) – Φ0
or λ = hc/(eV0 + Φ0)
= \(\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{(0.60 \mathrm{eV}+2.14 \mathrm{eV})}\)
= \(\frac{19.89 \times 10^{-26} \mathrm{~J} \mathrm{~m}}{(2.74 \mathrm{eV})}\)
λ = \(\frac{19.89 \times 10^{-26} \mathrm{~J} \mathrm{~m}}{2.74 \times 1.6 \times 10^{-19} \mathrm{~J}}\) = 454 nm

Question 3.
The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellow-green colour and about 760 nm for red colour.
(a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum ? (Take h = 6.63 × 10-34 J s and 1 eV = 1.6 × 10-19 J.)
(b) From which of the photosensitive materials with work functions listed in table and using the results of (i), (ii) and (iii) of (a) can you build a photoelectric device that operates with visible light ?
AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter 10
Solution:
a) Energy of the incident photon,
E = hv = hc/λ
E = (6.63 × 10-34 J s) (3 × 108 m/s)/λ.
= \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{\lambda}\)
i) For violet light,
λ1 = 390 nm (lower wavelength end)
Incident photon energy,
E1 = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{390 \times 10^{-9} \mathrm{~m}}\)
5.10 × 10-19 J = \(\frac{5.10 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\)
= 3.19 eV

ii) For yellow-green light,
λ2 = 550 nm (average wavelength)
Incident photon energy,
E2 = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{550 \times 10^{-9} \mathrm{~m}}\)
= 3.62 × 10-19 J = 2.26 eV.

iii) For red light,
λ3 = 760 nm (higher wavelength end)
Incident photon energy,
E3 = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{760 \times 10^{-9} \mathrm{~m}}\)
= 2.62 × 10-19 J = 1.64 eV

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter

b) For a photoelectric device to operate, we require incident light energy E to be equal to or greater than the work function Φ0 of the material. Thus, the photoelectric device will operate with violet light (with E = 3.19 eV) photosensitive material Na (with Φ0 = 2.75 eV), K (with Φ0 = 2.30 eV) and Cs (with Φ0 = 2.14 eV). It will also operate with yellow-green light (with E = 2.26 eV) for Cs (with Φ0 = 2.14 eV) only. However, it will not operate with red light (with E = 1.64 eV) for any of these photosensitive materials.

Question 4.
What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4 × 106 m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s ?
Solution:
a) For the electron :
Mass m= 9.11 × 10-31 kg, speed υ = 5.4 × 106 m/s. Then, momentum
P = mυ = 9.11 × 10-31 (kg) × 5.4 × 106 (m/s)
P = 4.92 × 10-24 kg m/s
de Broglie wavelength, λ = h/p
= \(\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{4.92 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}\)
λ = 0.135 nm

b) For the ball:
Mass m’ = 0.150 kg, speed υ’ = 30.0 m/s.
Then momentum
p’ = m’υ’ = 0.150 (kg) × 30.0 (m/s)
p’ = 4.50 kg m/s
de Broglie wavelength λ’ = h/p’.
= \(\frac{6.63 \times 10^{\pm 34} \mathrm{Js}}{4.50 \times \mathrm{kg} \mathrm{m} / \mathrm{s}}\)
λ’ = 1.47 × 10-34 m
The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10-19 times the size of the proton, quite beyond experimental measurement.

Question 5.
An electron, an a-particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength ? [T.S. Mar. 15]
Solution:
For a particle,
de Broglie wavelength, λ = h/p
Kinetic energy, K = p2/2m
Then, λ = h /\(\sqrt{2 \mathrm{mK}}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses. A proton \(\left({ }_1^1 \mathrm{He}\right)\) is 1836 times massive than an electron and an a-particle \(\left({ }_2^4 \mathrm{He}\right)\) four times that of a proton
Hence, α-particle has the shortest de Broglie wavelength.

AP Inter 2nd Year Physics Important Questions Chapter 12 Dual Nature of Radiation and Matter

Question 6.
A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10-4. Calculate the particle’s mass and identify the particle.
Solution:
de Broglie wavelength of a moving particle, having mass m and velocity υ :
λ = \(\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}\)
Mass, m = h/A.
For an electron, mass me = h/λe υe
Now, we have υ/υe = 3 and
λ/λe = 1.813 × 10-4
Then, mass of the particle,
m = me \(\left(\frac{\lambda_{\mathrm{e}}}{\lambda}\right)\left(\frac{v_{\mathrm{e}}}{v}\right)\)
m = (9.11 × 10-31 kg) × (1/3) × (1/1.813 × 10-4)
m = 1.675 × 10-27 kg.
Thus, the particle, with this mass could be a proton or a neutron.

Question 7.
What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts? [A.P. Mar. 15]
Solution:
Accelerating potential V = 100 V The de Broglie wavelength λ is
λ = h/p = \(\frac{1.227}{\sqrt{\mathrm{V}}}\) nm
λ = \(\frac{1.227}{\sqrt{100}}\) nm = 0.123 nm
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves

Students get through AP Inter 2nd Year Physics Important Questions 11th Lesson Electromagnetic Waves which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 11th Lesson Electromagnetic Waves

Very Short Answer Questions

Question 1.
Give anyone use of infrared rays. [T.S. Mar. 17; A.P. Mar. 16]
Answer:

  1. Infrared radiation plays an important role in maintaining the Earth warm.
  2. Infrared lamps are used in physical therapy.
  3. Infrared detectors are used on Earth Satellites.
  4. These are used in taking photographs during conditions of fog, smoke, etc.

Question 2.
How are infrared rays produced? How they can be detected?
Answer:
Infrared rays can be produced by vibrations of atoms and molecules. These waves can be detected by Thermopile, Bolometer, IR photographic film.

AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves

Question 3.
How are radio waves produced ? How can they detected ?
Answer:
Radio waves can be produced by rapid acceleration and deceleration of electrons in aerials (conductors). These can be detected by receivers of aerials.

Question 4.
If the wave length of E.M radiation is doubled, what happens to the energy of photon ? [IPE 2016 (TS)]
Answer:
If the wave length of electromagnetic radiation is doubled, then energy will be halved because energy is inversely proportional to.wavelength of electromagnetic waves.
E = hυ = hc/λ ⇒ E ∝ 1/λ (∵ hc is a constant)

Question 5.
What is the principle of production of electromagnetic waves ?
Answer:
If the charge is accelerated both the magnetic field and electric field will change with Space and time, then electromagnetic waves are produced.

Question 6.
What is the ratio of speed of infrared rays and ultraviolet rays in vaccum ?
Answer:
The ratio of speed of infrared rays and ultraviolet rays in vacuum is 1 : 1.
All electromagnetic waves travel with same speed 3 × 108 m /s in vaccum.

AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves

Question 7.
What is the relation between the amplitudes of the electric and magnetic fields in free space for an electromagnetic wave ?
Answer:
E0 = CB0
Where E0 = Amplitude of electric field.
B0 = Amplitude of magnetic field.
C = velocity of light.

Question 8.
What are the applications of microwaves ? [A.P. Mar. 17; T.S. Mar. 15]
Answer:

  1. Microwaves are used in Radars.
  2. Microwaves are used for cooking purposes.
  3. A radar using microwave can help in detecting the speed of automobile while in motion.

Question 9.
Microwaves are used in Radars, why ? [Mar. 14]
Answer:
As microwaves are of smaller wavelengths, hence they can be transmitted as a beam signal in a particular direction. Microwaves do not bend around the comers of any obstacle coming in their path.

Question 10.
Give two uses of infrared rays.
Answer:

  1. Infrared rays are used for producing dehydrated fruits.
  2. They are used in the secret writings on the ancient walls.
  3. They are used in green houses to keep the plants warm.

AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves

Question 11.
How are microwaves produced ? How can they detected ? [A.P. Mar. 16; IPE 15]
Answer:
Microwaves can be produced using Klystron valve or Magnetrons.
Microwaves can be detected using point contact diodes.

Question 12.
The charging current for a capacitor is 0.6 A. What is the displacement current across its plates ?
Answer:
i = charging current for a capacitor = 0.6 A
i = id = ε0 = \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
∴ i = id = 0.6 A
∴ Displacement current (id) = 0.6 A.

Question 13.
What physical quantity is the same for X-rays of wavelength 10-10m, red light of wavelength 6800 Å and radiowaves of wavelength 500in ?
Answer:
The speed in vaccum is same for all the given wavelengths, which is 3 × 108 m/s.

Question 14.
A radio can tune into any station in the 7.5 MHz to 12MHz band. What is the corresponding wavelength band ?
Answer:
λ1 = \(\frac{3 \times 10^8}{7.5 \times 10^6}\) = 40 m
λ2 = \(\frac{3 \times 10^8}{12 \times 10^6}\) = 25 m
Thus wavelength band is 40m to 25m.

AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves

Question 15.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vaccum is B0 = 510 nT. What is the amplitude of the electric field part of the wave ?
Answer:
Here, B0 = 510nT = 510 × 10-9T
E0 = CB0 = 3 × 108 × 510 × 10-9 = 153 NC-1.

Question 16.
Define displacement current
Answer:
Displacement current (Id) is equal to ε0 times to the rate of change of electric flux. Displacement current is not the current produced due to charge carried. But it is due to varying electric flux. It is the current in the sense that it produces a magnetic field.
Id = ε0 \(\frac{\mathrm{d} \phi_{\varepsilon}}{\mathrm{dt}}\)

Short Answer Questions

Question 1.
State six characteristics of electromagnetic waves.
Answer:
Characteristics of electromagnetic waves :

  1. Electromagnetic waves are produced by accelerated charges.
  2. Electromagnetic waves are transverse in nature.
  3. Electromagnetic waves donot require material medium for their propagation.
  4. Electromagnetic waves obey principle of superposition of waves.
  5. Velocity of E.M waves in vaccum depends on permittivity and permeability of free space.
  6. Electromagnetic waves carry energy and momentum.
  7. Electromagnetic waves exert pressure when they strike a surface.

AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves

Question 2.
What is Greenhouse effect and its contribution towards the surface temperature of earth ?
Answer:
Green house effect: Temperature of the earth increases due to the radiation emitted by
the earth is trapped by atmospheric gases like CO2, CH4, N2, chlorofluoro carbons etc., is called green house effect.

  1. Radiation from the sun enters the atmosphere and heat the objects on the earth. These heated objects emit infrared rays.
  2. These rays are reflected back to Earth’s surface and trapped in the Earth’s atmosphere. Due to this temperature of the earth increases.
  3. The layers of carbon dioxide (CO2) and low lying clouds prevent infrared rays to escape Earth’s atmosphere.
  4. Since day-by-day the amount of carbondioxide in the atmosphere increases, more • infrared rays are entrapped in the atmosphere.
  5. Hence the temperature of the Earth’s surface increases day by day.

Problems

Question 1.
A plane electromagnetic wave travels in vaccum along z-direction. What can you say about the directions of its electric and magnetic field vectors ? If the frequency of the wave is 30 MHz. What is its wavelength ?
Answer:
Electric and magnetic fields \(\overline{\mathrm{E}}\) and \(\overline{\mathrm{B}}\) of an electromagnetic wave must be perpendicular to the propagation of electromagnetic wave. Hence they lie in X – Y plane mutually perpendicular to each other.
Frequency of wave, v = 30MHz = 30 × 106Hz.; Velocity of light, C = 3 × 108m/s
Wavelength of the wave, λ = \(\frac{C}{V}=\frac{3 \times 10^8}{30 \times 10^6}\) = 10m

AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves

Question 2.
A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator ?
Answer:
According to Maxwell, a charged particle oscillating with a frequency produces electro-magnetic waves of same frequency. Hence frequency of EM waves produced is, 109Hz.

Textual Examples

Question 1.
A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, It is connected for charging in series with a resistor R = 1 M Q across a 2V battery (fig)- Calculate the magnetic field at a point P halfway between the centre and the periphery of the plates, after t = 10-3 s. (The charge on the capacitor at time τ is q (t) = CV [1 – exp (-t/τ)], where the time constant τ is equal to CR.).
Solution:
The time constant of the CR circuit is τ = CR = 10-3s. Then, we have
q(t) = CV [1 – exp (-t/τ) ]
= 2 × 10-9 [1 – exp (-t /10-3)
The electric field in between the plates at time t is
E = \(\frac{q(t)}{\varepsilon_0 A}=\frac{q}{\pi \varepsilon_0}\); A = π (1)2 m2 = area of the plates.
AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves 1
Consider now a circular loop of radius (1/2)m parallel to the plates passing through R The magnetic field B at all points on the loop is along the loop arid of the same value.
The flux ΦE through this loop is
The flux ΦE = E × area of the loop
= E × π × (\(\frac{1}{2}\))2 = \(\frac{\pi \mathrm{E}}{4}=\frac{\mathrm{q}}{4 \varepsilon_0}\)
The displacement current
id = e0 \(\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\) = \(\frac{1}{4} \frac{\mathrm{dq}}{\mathrm{dt}}\) = 0.5 × 10-6 exp (-1)
at t = 10-3s. Now, applying Ampere-Maxwell law to the loop, we get
B × 2π × (\(\frac{1}{2}\)) = m0.(ic + id) = m0(0 + id) = 0.5 × 10-6 m0 exp(-1)
or, B = 0.74 × 10-13 T.

Question 2.
A plane electromagnetic wave of frequency 25 MHz travels in free space along the x – direction. At a particular point in space and time, E = \(6.3 \hat{\mathbf{j}}\) V/m. What is B at the point ?
Solution:
Using Eq. B0 = [E0/c] the magnitude of B is
B = \(\frac{\mathrm{E}}{\mathrm{c}}\)
= \(\frac{6.3 \mathrm{~V} / \mathrm{m}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 2.1 × 10-8 T
To find the direction, we note that E is along y-direction and the wave propagates along x- axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction. Since, \((+\hat{\mathrm{j}})\) × \((+\hat{\mathrm{k}})\) = i, B is along the z-direction. Thus. B = 2.1 × 10-8 \(\hat{\mathrm{k}}\) T

AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves

Question 3.
The magnetic field in a plane electromagnetic wave is given by
By = 2 × 10-7 sin (0.5 × 103 × + 1.5 × 1011 t)T.
a) What is the wavelength and frequency of the wave ?
b) Write an expression for the electric field.
Solution:
a) Comparing the given equation with
By = B0 sin \(\left[2 \pi\left(\frac{\mathrm{x}}{\lambda}+\frac{\mathrm{t}}{\mathrm{T}}\right)\right]\)
We get, λ = \(\frac{2 \pi}{0.5 \times 10^3}\) m = 1.26 cm,
and \(\frac{1}{\mathrm{~T}}\) = v = (1.5 × 1011)/2π = 23.9 GHz

b) E0 = B0C = 2 × 10-7 T × 3 × 108 m/s = 6 × 101 V/m
The electric field component is perpendicular to the direction of propagation and the di-rection of magnetic field. Therefore, the electric field component along the z-axis is obtained as Ez = 60 sin (0.5 × 103x + 1.5 × 1011t) V/m.

Question 4.
Light with an energy flux of 18 W/cm2 falls on a non reflecting surface at normal incidence. If the surface has an area of 20 cm2 find the average force exerted on the surface during a 30 minute time span.
Solution:
The total energy falling on the surface is
U = (18 W/cm2) × (20 cm2) × (30 × 60)
= 6.48 × 105 J
Therefore, the total momentum delivered (for complete absorption) is
P = \(\frac{\mathrm{U}}{\mathrm{C}}=\frac{6.48 \times 10^5 \mathrm{I}}{3 \times 10^8 \mathrm{m} / \mathrm{s}}\) = 2.16 × 10-3 kg m/s
The average force exerted on the surface is
F = \(\frac{\mathrm{p}}{\mathrm{t}}=\frac{2.16 \times 10^{-3}}{0.18 \times 10^4}\) = 1.2 × 10-6 N

AP Inter 2nd Year Physics Important Questions Chapter 11 Electromagnetic Waves

Question 5.
Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3m. Assume that the efficiency of the bulb is 2.5% and it is a point source.
Solution:
The bulb, as a point source, radiates light in all directions uniformly. At a distance of 3m, the surface area of the surrounding sphere is A = 4 πr2 = 4π (3)2 = 113m2
The intensity at this distance is
I = \(\frac{\text { Power }}{\text { Area }}=\frac{100 \mathrm{~W} \times 2.5 \%}{113 \mathrm{~m}^2}\) = 0.022 W.m2
Half of this intensity is provided by the electric field and half by the magnetic field.
\(\frac{1}{2} I=\frac{1}{2}\left(\varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2 \mathrm{C}\right)\)
= \(\frac{1}{2}\) (0.022 W/m2)
Erms = \(\sqrt{\frac{0.022}{\left(8.85 \times 10^{-12}\right)\left(3 \times 10^8\right)}}\) V/m = 2.9 V/m
The value of E found above is the root mean square value of the electric field. Since the electric field in a light beam is sinusoidal, the peak electric field, E0 is
E0 = \(\sqrt{2}\)Erms = \(\sqrt{2}\) × 2.9 V/m
= 4.07 V/m
Thus, you see that the electric field strength of the light that you use for reading is fairly large. Compare it with electric field strength of TV or FM waves, which is of the order of a few microvolts per metre.
Now, let us calculate the strength of the magnetic field. It is
Brms = \(\frac{E_{\mathrm{rms}}}{\mathrm{C}}=\frac{2.9 \mathrm{Vm}^{-1}}{3 \times 10^8 \mathrm{~ms}^{-1}}\) = 9.6 × 10-9 T.
Again, since tbs field in the light beam is sinusoidal, the peak magnetic field is B0 = \(\sqrt{2}\) Brms = 1.4 × 10-8 T. Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak.

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Students get through AP Inter 2nd Year Physics Important Questions 10th Lesson Alternating Current which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 10th Lesson Alternating Current

Very Short Answer Questions

Question 1.
A transformer converts 200 V ac into 2000 V ac. Calculate the number of turns in the ‘ secondary if the primary has 10 turns. [T.S. Mar. 16]
Answer:
\(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)
Vp = 200V, Vs = 2000V, Np = 10
Ns = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}\) × Np = \(\frac{2000}{200}\) × 10
Ns = 100.

Question 2.
What type of transformer is used in a 6V bed lamp ? [A.P. Mar. 17]
Answer:
Step down transformer is used in 6V bed lamp.

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Question 3.
What is the phenomenon involved in the working of a transformer ? [Mar. 16(A.P.) Mar. 14]
Answer:
Transformer works on the principle of mutual induction.

Question 4.
What is transformer ratio ?
Answer:
The ratio of secondary e.m.f to the primary e.m.f. (or) number of turns in secondary to the number of turns in the primary is called the transformer ratio.
Transformer ratio = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)

Question 5.
Write the expression for the reactance of i) an inductor and (ii) a capacitor.
Answer:

  1. Inductive reactance (XL) = ωL
  2. Capacitive reactance (XC) = \(\frac{1}{\omega C}\)

Question 6.
What is the phase difference between A.C emf and current in the following: Pure resistor, pure inductor and pure capacitor. [T.S. Mar. 15]
Answer:

  1. In pure resistor A.C. e.m.f and current are in phase with each other.
  2. In pure inductor, current lags behind the e.m.f. by an angle of \(\frac{\pi}{2}\) (or) 90°.
  3. In pure capacitor, current leads the e.m.f by an angle \(\frac{\pi}{2}\) (or) 90°.

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Question 7.
Define power factor. On which factors does power factor depend ?
Answer:
The ratio of true power and apparent power (virtual power) in an a.c circuit is called as power factor of the circuit.
Power factor (cosΦ) = \(\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{rms}}}\) [∵ Prms = Vrms Irms]
Power factor depends on r.m.s voltage, r.m.s current and average power (P).

Question 8.
What is meant by wattless component of current ?
Answer:
Average power (P) = Vrms(Irms sinΦ) cos\(\frac{\pi}{2}\)
The average power consumed in the circuit due to (Irms sinΦ) component of current is zero. This component of current is known as wattless current. (Irms sinΦ) is the wattless component of current.

Question 9.
When does a LCR series circuit have minimum impedance ?
Answer:
In LCR series circuit, Impendence (Z) = \(\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}\)
At a particular frequency, ωL = \(\frac{1}{\omega C}\)
The impedance is minimum (Z = R)
This frequency is called resonant frequency.

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Question 10.
What is the phase difference between voltage and current when the power factor in LCR series circuit is unity ?
Answer:
In LCR series circuit power factor (cosΦ) = 1
Phase difference between voltage and current is zero. (Φ = 0)

Short Answer Questions

Question 1.
State the principle on which a transformer works. Describe the working of a transformer with necessary theory.
Answer:
Transformer is a device to convert a low alternating current of high voltage into high alternating current of low voltage and vice versa.
Principle : It works on the principle of mutual induction between two coils.
Working : When an alternating emf is applied across the primary coil, the input voltage changes with time. Hence the magnetic flux through the primary also changes with time.
AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current 1
This changing magnetic flux will be linked with secondary through the core. An emf is induced in the secondary.

Theory: Let N1 and N2 be the number of turns in the primary and secondary. Let VP and VS be the emf s across the primary and secondary.
\(\frac{V_S}{V_p}=\frac{\text { Output emf }}{\text{Input emf}}=\frac{-N_2 \frac{d \phi}{d t}}{-N_1 \frac{d \phi}{d t}}=\frac{N_2}{N_1}\)
∴ \(\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{N}_2}{\mathrm{~N}_1}\) = Transformer ratio
Efficiency of transformer :
It is the ratio of output power to the input power.
η = \(\frac{\text { Outputpower }}{\text { Input power }}\) × 100

Problems

Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
Solution:
(a) We are given P = 100 W and V = 220V The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω

(b) The peak voltage of the source is υm = \(\sqrt{2}\)V = 311 V

(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Solution:
The inductive reactance,
XL = 2πvL = 2 × 3.14 × 50 × 25 × 10-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A

Question 3.
The instantaneous current and instantaneous voltage across a series circuit containing resistance and inductance are given by i = \(\sqrt{2}\) sin (100t – π/4)A and υ = 40 sin (100t) V. Calculate the resistance ?
Solution:
i = \(\sqrt{2}\) sin (100t – π/4)A (∵i = i0sin(ωt – Φ))
υ = 40 sin(100t)V (∵ V = V0sin(ωt ))
i0 = \(\sqrt{2}\) , V0 = 40, ω = 100, Φ = π/4
R = \(\frac{\mathrm{V}_0}{\mathrm{i}_0}\) cosΦ = \(\frac{40}{\sqrt{2}}\) cos\(\frac{\pi}{4}\), R = \(\frac{40}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\), R = 20 Ω

Question 4.
In an AC circuit, a condenser, a resistor and a pure inductor are connected in series across an alternator (AC generator). If the voltages across them are 20 V, 35 V and 20 V respectively, find the voltage supplied by the alternator.
Solution:
VC = 20V, VR = 35V, VL = 20V
V = \(\sqrt{V_R^2+\left(V_L^2-V_C^2\right)}\) ; V = \(\sqrt{(35)^2+\left(20^2-20^2\right)}\) ; V = \(\sqrt{35^2}\); V = 35 Volt.

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Question 5.
What is step up transformer ? How it differs from step down transformer ?
Solution:
The ratio of number of turns in the secondary coil to the number of turns in the primary coil is called transformer ratio.
T = \(\frac{N_S}{N_p}=\frac{\text {No. of turns in the secondary }}{\text {No. of turns in the primary }}\)
If NS > NP, then the transformer is called step up transformer.
If NS < NP, then the transformer is called step down transformer.

Textual Examples

Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
Solution:
(a) We are given P = 100 W and V = 220V.
The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω

(b) The peak voltage of the source is υm = \(\sqrt{2}\)V = 311 V

(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.

Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Solution:
The inductive reactance,
XL = 2πvL = 2 × 3.14 × 50 × 25 × 10-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Question 3.
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced ?
Solution:
When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.

Question 4.
A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current ?
Solution:
The capacitive reactance is
Xc = \(\frac{1}{2 \pi v C}=\frac{1}{2 \pi(50 \mathrm{~Hz})\left(15.0 \times 10^{-6} \mathrm{~F}\right)}\)
= 212 Ω
The rms current is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}=\frac{220 \mathrm{~V}}{212 \Omega}\)
= 1.04 A
The peak current is
im = \(\sqrt{2}\)I = (1.41)(1.04A) = 1.47A
This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Question 5.
A light bulb and an open coil inductor are connected to an ac source through a key as shown in the figure.
AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current 2
The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons.
Solution:
As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Question 6.
A resistor of 200Ω and a capacitor of 15.0 μF are connected in series to a 220V, 50 Hz ac source,
(a) Calculate the current in the circuit;
(b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve the paradox.
Solution:
Given
R = 200Ω. C = 15.0 μF = 15.0 × 10-6F
V = 220V, v = 50Hz
(a) In order to calculate the current, we need the impedance of the circuit. It is
Z = \(\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{C}}^2}=\sqrt{\mathrm{R}^2+(2 \pi v C)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+\left(2 \times 3.14 \times 50 \times 10^{-6} \mathrm{~F}\right)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+(212 \Omega)^2}\) = 291.5Ω
Therefore, the current in the circuit is
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{220 \mathrm{~V}}{291.5 \Omega}\) = 0.755A

(b) Since the current is the same throughout the circuit, we have
VR = IR = (0.755 A) (200Ω) = 151V
VC = IXC = (0.755A) (212.3Ω) = 160.3V
The algebraic sum of the two voltages, VR and VC is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox ? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem:
VR+C = \(\sqrt{\mathrm{V}_{\mathrm{R}}^2+\mathrm{V}_{\mathrm{C}}^2}\) = 220 V
Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

Question 7.
a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.
Solution:
a) We know that P = IV cosΦ where cosΦ is the power factor. To supply a given power at a given voltage, if cosΦ is small, we have to increase current accordingly. But this will lead to large power loss (IR) in transmission.

b) Suppose in a circuit, current I lags the voltage by an angle Φ.
Then power factor cosΦ = R/Z
We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram in the figure how this can be achieved. Let us resolve I into two components, IP
AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current 3
along the applied voltage V and Iq perpendicular to the applied voltage. Iq is called the wattless component since corresponding to this component of current, there is no power loss. IP is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.

It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current Iq by an equal leading wattless current I’q. This can be done by connecting a capacitor of appropriate value in parallel so that Iq and I’q cancel each other and P is effectively IP V.

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Question 8.
A sinusoidal voltage of peak value 283 V , and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω. L = 25.48 mH. and C = 796μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
Solution:
a) To find the impedance of the circuit, we first calculate XL and XC.
XL = 2πvL
= 2 × 3.14 × 50 × 25.48 × 10-3Ω = 8Ω
XC = \(\frac{1}{2 \pi v \mathrm{C}}\)
= \(\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}\) = 4Ω
Therefore,
z = \(\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}=\sqrt{3^2+(8-4)^2}\)
= 5Ω

b) Phase difference, Φ = tan-1\(\frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\)
= tan-1\(\left(\frac{4-8}{3}\right)\) = -53.1°

c) The power dissipated in the circuit is
P = I2R
AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current 4
Therefore, P = (40A)2 × 3Ω = 4800W
= 4.8 kW

d) Power factor = cos Φ = cos 53.1° = 0.6.

Question 9.
Suppose the frequency of the source in the previous example can be varied,
(a) What is the frequency of the source at which resonance occurs ?
(b) Calculate the impedance, the current, and the power dissipated at the resonant condition.
Solution:
(a) The frequency at which the resonance occurs is
ω0 = \(\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{25.48 \times 10^{-3} \times 796 \times 10^{-6}}}\)
= 222.1 rad/s
vr = \(\frac{\omega_0}{2 \pi}=\frac{221.1}{2 \times 3.14}\) Hz = 35.4Hz

b) The impedance Z at resonant condition is equal to the resistance
Z = R = 3Ω
The rms current at resonance is ,
as V = \(\frac{v_{\mathrm{m}}}{\sqrt{2}}\)
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{\mathrm{V}}{\mathrm{R}}=\left(\frac{283}{\sqrt{2}}\right) \frac{1}{3}\) = 66.7 A
The power dissipated at resonance is
P = I2 × R = (66.7)2 × 3 = 13.35 kW
You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 8.

AP Inter 2nd Year Physics Important Questions Chapter 10 Alternating Current

Question 10.
At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work ?
Solution:
The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.

Question 11.
Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time.
Solution:
Let q0 be the initial charge tin a capacitor. Let the charged capacitor be connected to an inductor of inductance L. this LC circuit will sustain an oscillation with frequency
\(\omega\left(2 \pi v=\frac{1}{\sqrt{\mathrm{LC}}}\right)\)
At an instant t, charge q on the capacitor and the current i are given by :
q(t) = q0 cos ωt
i(t) = -q0 co sin ωt
Energy stored in the capacitor at time ‘t’ is
UE = \(\frac{1}{2}\) C V2 = \(\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) cos2 (ωt)
Energy stored in the inductor at time ‘t’ is
UM = \(\frac{1}{2}\) L i2
= \(\frac{1}{2}\) L q02 ω2sin2 (ωt)
= \(\frac{Q_0^2}{2 C} \sin ^2(\omega t)\) [∵  ω = 1/ \(\sqrt{L C}\)]
Sum of energies
UE + UM = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) [cos2 ωt + sin2ωt) = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\)
This sum is constant in time as q0 and C, both are time-independent.