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Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 2 Animal Kingdom

Plus One Animal Kingdom One Mark Questions and Answers

Question 1.
Fill in the blanks.

Answer:
i) B – Nereis
ii) B – Ctenoplanna
C – Ctenophora
iii) A – radula

Question 2.
Write any two members of the Phylum Aschelminthes which are found parasitic on Human beings.
Answer:

1. Ascaris (Roundworm)
2. Wuchereria (Filaria worm)

Question 3.
In some animal groups, the body is found divided into compartments with at least some organs/ organ repeated. This characteristic feature is named
(a) Segmentation
(b) Metamerism
(c) Metagenesis
(d) Metamorphosis
Answer:
(b) Metamerism

Question 4.
Given below are types of cells present in some animals. Each one is specialized to perform a single specific function except
(a) Choanocytes
(b) Interstitial cells
(c) Gastrodermal cells
(d) Nematocytes
Answer:
(b) Interstitial cells

Question 5.
Which one of the following sets of animals share a four chambered heart?
(a) Amphibian, Reptiles, Birds
(b) Crocodiles, Birds, Mammals
(c) Crocodiles, Lizards, Turtles
(d) Lizards, Mammals, Birds
Answer:
(b) Crocodiles, Birds, Mammals

Question 6.
Which of the following pairs of animals has non glandular skin.
(a) Snake and Frog
(b) Chameleon and Turtle
(c) Frog and Pigeon
(d) Crocodile and Tiger.
Answer:
(c) Frog and Pigeon

Question 7.
Birds and mammals share one of the following characteristics as a common feature.
(a) Pigmented skin
(b) Alimentary canal with some modification
(c) Viviparity
(d) Warm blooded nature
Answer:
(d) Warm blooded nature

Question 8.
Note the relationship between the first two words and find a suitable word for the fourth place,

1. Coelenterata: radial symmetry, platyhelminthes, _______
2. Lizard: Poikilothermous, crow, _________

Answer:

1. bilaterally symmetrical
2. Homoiothermous

Question 9.

1. Annelida: Parapodia :: __________ : Comb plates
2. _________: Water vascular system :: Coelenterata : cnidoblast

Answer:

1. Ctenophora
2. Echinodermata

Question 10.
Malpighian tubule is the excretory organ of which phylum?
(a) Phylum porifera
(b) Phylum arthropoda
(c) Phylum Coelenterata
(d) Phylum mollusca
Answer:
(b) Phylum Arthropoda

Question 11.
A chordate animal having flame cells as the excretory organ.
Answer:
Amphioxus

Question 12.
From the pictures given below, find out the poikilothermic animals.

Answer:
B and D are poikilothermic animals.

Question 13.
Segmentation in the body is first observed in which of the following:
(a) Platyhelminthes
(b) Aschelminthes
(c) Annelida
(d) Arthropoda
Answer:
(c) Annelida

Question 14.
Shark has to swim continuously, otherwise, it will sink down. Give reason.
Answer:
Due to absence of air bladder.

Plus One Animal Kingdom Two Mark Questions and Answers

Question 1.
Arrange the phylum in order.
Arthropoda → Platyhelminthes → Porifera → Ctenophora → Cnidaria → Mollusca → Annelida → Echinodermata → Aschelminthesip
Answer:

• Presence of milk producing mammary gland.
• Presence of hair on skin.
• External ear or pinnae is present.
• Different types of teeth are present in the jaw.

Question 2.
Complete the blanks.

Character	Phylum
a. Body is flat
b. Body has similar segments
c. Body has jointed appendages
d. Body is round

Answer:

Character	Phylum
a. Body is flat	Platyhelminthes
b. Body has similar segments	Annelida
c. Body has jointed appendages	Arthropoda
d. Body is round	Aschelminthes

Question 3.
Triploblastic animals are more complex than diploblastic animals. Do you agree with this statement? Justify.
Answer:
Triploblastic animals have more cell layers so they have the possibility of greater degree of cellular specialisation.

Question 4.
All vertebrates are chordates but all chordates are not vertebrates. Justify.
Answer:
Notochord is present in all vertebrates but vertebral column is present only in vertebrates and not in all chordates.

Question 5.
Copy and complete the table.

Chondrichthyes	Osteichthyes
a _____________	Seen in all water forms
b. Endoskeleton is cartilage	_____________
c _______________	Body covered by cycloid scales
d ______________	Mouth is terminal

Answer:

Chondrichthyes	Osteichthyes
a. Marine form	Seen in all water forms
b. Endoskeleton is cartilage	Endoskeleton is bony.
c. Body is covered by placoid scale	Body covered by cycloid scales
d. Mouth is ventral	Mouth is terminal

Question 6.
Arrange the phylum in order.
Arthropoda → Platyhelminthes → Porifera → Ctenophora → Cnidaria → Mollusca → Annelida → Echinodermata → Aschelminthes
Answer:
Porifera → Cnidaria → Ctenophora→ Platyhelminthes → Aschelminthes → Annelida → Arthropoda → Mollusca → Echinodermata

Question 7.
Categorise and classify the following organisms and arrange them in a table with separate columns and provide appropriate headings.
Exocoetus, Physalia, Ascaris, Apis, Locusta, Corvus, Pita, Hydra, Sepia, Ancylostoma.
Answer:

Question 8.

1. Identify the phylum which exhibit metagenesis or alternation of generation.
2. What is meant by alternation of generation?

Answer:

1. Cnidaria
2. Alternation of sexual and asexual forms of organism: ie., Polyp asexually produce medusa, medusa sexually product polyp.

Question 9.
Name the following.

1. Phylum in which flatworms are included
2. Excretory organs of Annelids.
3. Largest phylum.
4. An oviparous mammal.

Answer:

1. Platyhelminthes
2. Nephridia
3. Arthropoda
4. Platypus

Question 10.
Apis, Prawn, Locust, Spider
Following animals have different habit and habitat. But they have many common characters.

1. Mention the common characters.
2. Identify their phylum.

Answer:

1. Joint footed animals
   • Metameric segmentation
   • Chitinous exoskeleton
2. Arthropoda

Question 11.
A list of animals are given below. Arrange them according to increase in complexity of organization. Scorpion, Earthworm, Liver fluke, Pigeon, Seaanemon, Sycon, Elephant, Anabas.
Answer:
Sycon, Seaanemon, Liver fluke, Earthworm, Scor¬pion, Anabas, Pegeon, Elephant.

Question 12.
Nithin is Studying in Std. XI. He collected some specimens during the field trip conducted by the Science Club of his School. Help Nithin to Classify the Animal in respective Phylum.
Prawn, Slug worm, Butterfly, Pila, Grass Hopper, Crab
Answer:

Arthropoda	Mollusca
Prawn	Slugworm
Butterfly	Pila
Grass hopper
Crab

Question 13.
During a field trip Raju has collected some organisms with the following characters. Help him to identify the phyla of those organisms.

1. Metamerically segmented body.
2. Body covered with calcareous shell.
3. Dorso-ventrally flattened leaf like body.
4. Body divided into head, thorax and abdomen.

Answer:

1. Annelida
2. Mollusca
3. Platyhelminthes
4. Arthropoda

Question 14.
Categorise the following fishes into Osteichthyes and Chondrichthyes?

1. Exocoetus
2. Trygon

Answer:

1. Exocoetus – Oesteichthyes
2. Trygon – Chondrichthyes

Question 15.
Arrange the following terms in two columns correctly. Malpighian tubules, radula, metamerism, Bioluminescence, choanocytes, nematocytes, Phylum-coelenterate, phylum-Arthropoda, phylum- ctnophora, phylum-Mollusca, Phylum-Porifera.
Answer:

Malpighian	Phylum – Arthropoda
Radula	Phylum – Mollusca
Metamerism	Phylum – Annelida
Bioluminescence	Phylum – Ctenophora
Choanocytes	Phylum – Porifera
Nematocytes	Phylum – Coelenterata

Question 16.
Match the following

Bidders canal	Earthworm
Typhlosole	Catla
Air bladder	Shark
Placoid scale	frog

Answer:

Bidders canal	frog
Typhlosole	Earthworm
Air bladder	Catla
Placoid scale	Shark

Question 17.
Observe the given organisms

1. Place these animals in proper phylum.
2. Segmentation in the body is first observed in which of the above phylum?

Answer:

1.  i) Arthropoda,
   ii) Porifera,
   ii) Annelida
2. Annelida

Question 18.
Observe the table given below and fill the blank columns A, B, C and from the animals given in brackets. (Ascaris, Starfish, Fasciola, Earthworm)

Answer:

Question 19.
1. Identify the animal given below.

2. Write one well marked property of the above animal.
Answer:

1. Pleurobrachia
2. Bioluminescence is well marked property of pleurobrachia

Question 20.
Representatives of some vertebrate classes are introducing themselves. Write down the name of the class in which they belong.

1. My gills are covered by operculum. I have bony endoskeleton.
2. I give birth to young ones. My body is covered by hair.
3. My skin is glandular? Have trilocular heart.
4. I live only in marine water. My endoskeleton is made up of cartilage.

Answer:

1. Osteichthyes
2. Mammalia
3. Amphibia
4. Chondrichthyes

Question 21.
During classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him. Justify.
Answer:
NO. Sponges are asymmetrical and body is formed of loose aggregate of cells. Cells are not organised to from tissues and organs. Cnidarians are radially symmetrical and tissue grade of organisation. So cnidarians are more complex than sponges.

Question 22.
Due to the absence of air bladder, fishes belonging to the class Chondrichthyes have to swim constantly. How important is the presence of air bladder in these fishes?
Answer:
Due to the absence of air bladder in chondrichthyes, they have to swim constantly to avoid sinking. If air bladder is present which regulates buoyancy.

Question 23.
Match the column A, B &C in the table given below:

Answer:

Question 24.
Identify the phylum whose larvae are bilaterally symmetrical, but adults are radially symmetrical.

1. Annelida
2. Arthropoda
3. Mollusca
4. Echinodermata

Mention two salient features of the phylum.
Answer:
4. Echinodermata
Salient features Presence of Echinodermata:

• Water vascular system
• Spiny bodies, Endoskeleton of Calcareous ossicles

Question 25.
Write the name of phylum.

1. Diploblastic, tissue grade of organisation, radially symmetrical, polymorphic animals.
2. Soft bodied, Unsegmented, Bilaterally symmetrical animals with open type circulation.
3. Triploblastic, bilaterally symmetrical, true coelomic animals with metamerism, closed circulation.
4. Triploblastic, chitinous exoskeleton and open circulation.

Answer:

1. Cnidaria
2. Mollusca
3. Annelida
4. Arthropoda

Question 26.
Observe the table given below and fill the blank columns a, b, c, and d from the animals given in brackets.

(Hydra, Shark, Spongilla, Obelia)
Answer:
(a) Hydra/Obelia
(b) Shark
(c) Spongilla
(d) Hydra/Obelia

Question 27.
Copy and complete the table.

Answer:

Question 28.
You are provided with two fishes Catla (Bony fish) and Shark (Cartilagenous fish). Prepare a table showing difference in:
(a) Position of mouth
(b) Air bladder
(c) Scales
(d) Fertilization
Answer:

Calta (Bony fish)	Shark (Cartilaginous fish)
(a) Mouth is terminal	Mouth is ventral
(b) Air bladder present	Air bladder absent
(c) Cycloid scales	Placoid scales
(d) External Fertilization	Internal Fertilization

Question 29.
Prepare a list of some animals that are found parasitic on human beings.
Answer:
Tapeworm (Taenia), Ascaris (Roundworm), Wuchereria (Filaria worm), Ancylostoma(Hookworm)

Question 30.

1. Identify the 2 forms of Cnidarians.
2. Mention any 2 difference between them.

Answer:

1. A – polyp
   B – medusa
2. Polyp: Asexual, sessile, mouth upwards Medusa: Sexual, Free swimming, Mouth downwards

Plus One Animal Kingdom Three Mark Questions and Answers

Question 1.
Classify the given organisms and arrange them in the order of their phylum.
Limulus, Corvus, Spongilla, Ascaris, Physalia, Nereis, Catla, Sepia, Echinus, Taenia, Pleurobrachia, Tiger, Viper, Toad.
Answer:

Non chordata	Chordata
Limulus – Arthropoda	Corvus
Spongilla – Porifera	Catla
Ascaris – Aschelminthes	Tiger
Physalia – Cnidaria	Viper
Nereis-Annelida	Toad
Sepia – Mollusca
Echinus – Echinodermata
Taenia – Platyhelminthes
Pleurobrachia – Ctenophora

Question 2.
Match the following.

(a) Operculum	i. Ctenophora
(b) Parapodia	ii. Mollusca
(c) Scales	iii. Porifera
(d) Comb plates	iv. Reptilia
(e) Radula	v. Annelida
(f) Hairs	vi. Cyclostomata
(g) Choanocytes	vii. Mammalia
(h) Gill slits	viii. Osteichthyes

Answer:

(a) Operculum	viii. Osteichthyes
(b) Parapodia	v. Annelida
(c) Scales	iv. Reptilia
(d) Comb plates	i. Ctenophora
(e) Radula	ii. Mollusca
(f) Hairs	vii. Mammalia
(g) Choanocytes	iii. Porifera
(h) Gill slits	vi. Cyclostomata

Question 3.
Observe the diagram

1. Identify the phylum of this hypothetical organism.
2. List out the features that helps in identifying it.
3. Write about the fate of notochord in Urochordata, Cephalochordata and Chordata.

Answer:

1. Chordata
2. Notochord, Dorsal nerve cord, Pharyngeal gill slits, Post anal tail
3. In Urochordatarfiotochond is present only in larval tail. In Cephalochondata, notochord extends from head to tail region is persistent throughout their life. In vertebrata, the notochord is replaced by a cartilagenous or bony vertebral column in the adults.

Question 4.
Select the following items into their appropriate phylum.
Radula, Parapodia, Comb plate, Nephridia, Choanocytes, Flame cells.
Answer:

• Radula – Mollusca
• Parapodia – Annelida
• Comb plate – Ctenophora
• Nephridia – Annelida
• Choanocytes – Porifera
• Flame cells – Platyhelminthes

Question 5.

(a) Fill and complete the chart given below.

(b) Write any two fundamental characters of the phylum chordata.
(c) Classify Tetrapoda into classes:
Answer:
Male accessory ducts store and transport the sperms from testis to the outside through urethra. Male accessory glands secrete seminal Plasma, Which is rich in fructose, citrate, prostaglandins and certain enzymes. The secretion of cowper’s glands lubricate the penis.

Question 6.
Write down the functions of the following (any two) structures and assign their phyla.

1. Radula
2. Flame cells
3. Parapodia

Answer:

1. Radula: File like rasping organ for feeding.
   Phylum: Mollusca
2. Flame cells: Osmoregulation and excretion
   Phylum: Platyhelminthes
3. Parapodia: help in swimming
   Phylum: Annelida

Question 7.
1. Identify the organism A and B.

2. Which features of these organism enable you to identify them?
Answer:

1. A-Tapeworm
   B – Earthworm
2. Features of these organism:
   • Tapeworm: Scolex is present, suckers and hooks are present. No true segmentation.
   • Earthworm: True segmentation, Absence of Scolex, hook and suckers, Clitellum is present.

Question 8.

1. Which of the following animals exhibit metagenesis? (Ascaris Obelia Earthworm Crab)
2. To which phylum does it belong?
3. Write any two features of the phylum.

Answer:

1. Obelia
2. Cnidaria
3. Two features of the phylum:
   • Presence of Cnidoblasts
   • Cnidarians exhibit two basic forms called polyp and medusa.

Question 9.
Pick out the appropriate one from the term given within bracket and put against the corresponding phylum.

1. Porifera
2. Coelenterata
3. Platyhelminthes
4. Annelida
5. Arthropoda
6. Mollusca
7. Echinodermata
8. Chordata

(Hirudin, Flame cell, Choanocytes, Cnidoblast, Jointed legs, Radula, Notochord and Dermal Ossicles)
Answer:

1. Porifera – Choanocytes
2. Coelenterata – Cnidoblast
3. Platyhelminthes – Flame cells
4. Annelida – Hirudin
5. Arthropoda – Jointed legs
6. Mollusca – Radula
7. Echinodermata -Dermal ossicles
8. Chordata – Notochord

Question 10.

 

1. Identify the organism (A) and (B) and which class do they belong?
2. On which basis do you classify these animals?

(Hint: Write any 2 Identifying characters)
Answer:

1. Organism in (A) and (B)
   • A – Bony fish – Osteichthyes
   • B – Cartilagenous fish – Chondrichthyes

2.

Osteichthyes	Chondrichthyes
Mouth is terminal	Mouth is ventral
Operculum is present	Operculum absent

Question 11.
From a fish market, you got a fish, on a close watching you friend says it is a cartilaginous fish.

1. Which characters helped him to identify it as a cartilaginous fish, (any four characters.)
2. Name the class it belongs.

Answer:

1. Characters:
   • a – Gillslits are separate and without operculum
   • b – Placoidscale
   • c – Mouth is located ventrally
   • d – heterocercal caudal fin
2. Chondrichthyes

Question 12.
Presence or absence of a cavity between the body wall and the gut wall is very important in classification.

 

1. Identify the different types of body cavities.
2. Give examples to each

Answer:

1. Different types of Body Cavities:
   • a – Coelomate
   • b – Pseudocoelomate
   • c – Acoelomate
2. Examples:
   • Coelomate – Chordates
   • Pseudocoelomate – Aschelminthes
   • Acoelomate – Platyhelminthes

Question 13.
From the following general characters find out corresponding/Class with an Example

1. Exclusively marine, triploblastic, spines on the skin, radially symmetrical in the adult and bilaterally symmetrical in the larval stage.
2. Marine, they migrate towards freshwater for spawning, then their larvae return to ocean after metamorphosis.
3. Triploblastic, bilaterally symmetrical, coelomate and metamerically segmented animals.

Answer:

1. Echinodermata
   • eg: Starfish
2. Class – Cyclostomata
   • eg: Petromyzon
3. Annelida
   • eg: Earthworm

Question 14.
Arrange the following points in a two-column table and give suitable heading for each column.

1. Notochord present
2. Post anal tail absent
3. Pharynx Perforated by gill slits
4. Notochord absent
5. Posts anal tail present
6. Gill slits are absent

Answer:

Chordata	Non Chordata
Notochord present	Post anal tail present
Pharynx perforated by gill slits	Notochord absent
Post anal tail present	Gill slits absent

Question 15.
Observe the figure and answer the questions.

1. Identify the structure.
2. Name the phylum which possess this structure.
3. How this structure help the organism?

Answer:

1. Cnidoblastornematocyst
2. Cnidaria
3. Capture of Prey
   • Defense organ.

Question 16.
Note the relationship between first two words and suggest suitable words for the 4th place.

1. Planaria: Flartfe cells: Earthworm: _________
2. Jaw present: Gnathostomata:: Jaw absent: ___________

Answer:

1. Nephridia
2. Agnatha

Question 17.
Give reasons for the following.

1. Respiratory and circulatory system are absent in parasitic platyhelminthes and Aschelminthes.
2. Arthropods are the most successful invertebrate.
3. Body of endoparasites are covered with cuticle.

Answer:

1. Parasitic Plalyhelminthes and Aschelminthes lives in anaerobic condition. So respiratory and circulatory systems are absent in these parasitic forms.
2. Arthropods are most successful, because of the presence of unique chitinous cuticle.
3. In these parasites, the cuticle helps in escaping from the action of digestive enzymes.

Question 18.
Arthropodes are organisms with chitinous exoskeleton. Suppose exoskeleton is absent in arthropods. List the difficulties arthropodes has to face.
Answer:

1. Their body will be dried up due to evaporation
2. They couldn’t escape from predators.
3. They couldn’t live in all environments.

Question 19.
Suppose during your field visit for collection from a rocky seashore you have got some live specimens such as sea urchin, sea cucumber, sea anemone. Is it possible to keep them on an aquarium in your school. Give reasons for your answer.
Answer:
No. It is not possible.
Marine animals cannot live on freshwater because it leads to endosmosis and death occurs.

Question 20.

1. Why nematocysts are more concentrated on the oral end and tentacles of cnidarians?
2. What are the difficulties that coelenterate have to face if nematocysts were absent in body.

Answer:

1. Nematocysts are concerned with defence and offence. Tentacles are usually used for defence, offence and food collection. Hence nematocysts are more concentrated in the oral end.
2. The major difficulties cnidarians has to face in the absence of nematocysts are for food collection and escaping from enemies.

Question 21.

1. Which of the following show the body cavity of earthworm?
2. Identify the names of germlayers ‘a’ and ’b’.

 

Answer:

1. Figure C is the body cavity of earthworm. Because it is a true coelom
2. The names of germlayers
   • a-ectoderm
   • b-endQderm

Question 22.
Identify the characters listed below and put (✓) mark on appropriate places

Answer:

Question 23.

1. Identify the organisms.
2. Which feature of these organisms enable you to identify them?

Answer:

1. The organisms:
   • A-Tape worm
   • B – Ascaris
   • C – Earthworm
2. Features:
   • Tapeworm – Body is dorso-ventrally flattened and segmented.
   • Ascaris – Body is cylindrical and vermiform
   • Earthworm – Body is divided into similar segments and clitellum is present.

Question 24.
During a classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him. Jusftfy.
Answer:
1. Sponges: Cellular grade of organisation and cell aggregate body plan.

2. Cnidarians: Tissue level of organisation and blind sac body plan. So Cnidarians are more complex than sponges.

Question 25.
While comparing the digestive system of a roundworm and flatworm, a boy noted some differences. List out the differences.
Answer:
1. Flatworms: Digestive system incomplete, has only a single opening, ingestion and egestion occurs through the same opening.

2. Round worm: Complete digestive system, it has both mouth and anus. A muscular pharynx is also present in their digestive system.

Question 26.

1. Identify the invertebrate whose mouth is on ventral or lower side and anus is on dorsal or upper side.
2. Mention the phylum.
3. Comment on its General characters.

Answer:

1. Starfish
2. Echinodermata
3. Its General characters are:
   • Water vascular system
   • Tube feet
   • Coelomate
   • Spiny body

Question 27.
Match the following.

Answer:

• Arthropoda – Bombyx – Joint footed animals
• Annelida – Earthworm – Little rings
• Echinodermata – Antedon – Spiny bodies animals
• Mollusca – Pila – Soft bodies animals

Question 28.
Animals showing metameric segmentation are included under annelida. Body of tapeworm has numerous segments, but the animal is not included under annelida. How will you account for it?
Answer:
Tapeworm show false segmentation. In true segmentation number of segments is fixed and age of the segments are all same.

Question 29.
Complete the given flow chart showing the flow of water through canal system in sponges.

Answer:

Question 30.
Mention a single word for the following.

1. Sexes are not separate.
2. Body wall with three layers.
3. File like rasping organs in Mollusca.
4. Segmentation of animal body.
5. The property of a living organism to emit light.
6. Alternation of generation in Cnidarians.

Answer:

1. Hermaphrodite
2. Triploblastic
3. Radula
4. Metamerism
5. Bioluminescence
6. Metagenesis

Question 31.
The birds are well adapted for flying.

1. Write the general characters of Aves.
2. Give some of the flight adaptations seen in birds.

Answer:

1. General characters of Aves:
   • Presence of feathers
   • Presence of beak
   • Forelimbs are modified into wings.
   • Warm blooded
   • Respiration by lungs with air sacs.
   • Oviparous
2. Flight adaptations:
   • Forelimbs are modified into wings
   • Long bones are hollow with air cavities (Pneumatic)
   • Air sacs connected to lungs supplement respiration
   • Constant body temperature

Plus One Animal Kingdom NCERT Questions and Answers

Question 1.
What is the difference between direct and indirect development?
Answer:
1. Direct Development:
In direct development the young animal resembles an adult. There is no intermediate stage.

2. Indirect Development:
In indirect development there is intermediate stage, like larval stage. For example, frog before being developed into adult passthrough a tadpole stage. This is the case of indirect development.

Question 2.
What are the peculiar features that you find in parasitic platyhelminthes?
Answer:
In parasitic platyhelminthes hooks and suckers are present. Suckers help the parasite, in sucking the blood from the host.

Question 3.
What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
Answer:
Arthropods are the first phylum to have well developed systems to carry out different activities. There is distinct system for respiration, locomotion and reproduction. Their survival capacity is great because of elaborate system. This has helped them survive in diverse conditions. They can live in water, on land and in air.

This can be one of the reasons why arthropods are the largest group among the animal kingdom. Another reason is their early development compared to animals of higher phylum.

Question 4.
Water vascular system is the characteristic of which group of the following:
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
Answer:
(c) Echinodermata

Question 5.
“All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.
Answer:
All chordates have notochord present in some stage of life.
The difference between vertebrates and nonvertebrates is as follows:
In vertebrates the notochord is present in the embryonic stage. This is replaced by a vertebral column during the adult stages.

Question 6.
How important is the presence of air bladder in Pisces?
Answer:
Presence of air-bladder in Pisces helps in buoyancy. This means that members of pisces don’t have to keep on swimming to remain floating.

Question 7.
What are the modifications that are observed in birds that help them fly?
Answer:
Following modification in birds help them fly:

1. Pneumatic or hollow bones make for a light weight skeleton.
2. Fore limbs are modified into wings to assist in flight.
3. Excertion of urine and faeces is through single opening facilitating weight reduction.
4. Aerodynamic body helps in flying.

Question 8.
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Answer:
Usually number of eggs produced by oviparous mothers is greater than number of young ones produced by viviparous mothers. The main reason for this is the need of resources required for development of the embryo.
In oviparous the major part of development of the embryo takes place outside the uterus. This makes lesser burden on the mother.

On the other hand in viviparous animals the development takes place inside the uterus so lesser number of young ones can be successfully incubated. Moreover, once eggs are outside they are at risk of getting eaten by some predator because of their immobility, so need of more eggs is there to ensure continuity of progeny.

Plus One Animal Kingdom Multiple Choice Questions and Answers

Question 1.
Calcareous skeleton is found in
(a) echinoderms
(b) some sponges
(c) mollusca
(d) all the above
Answer:
(d) all the above

Question 2.
Which cannot be the character of cnidaria
(a) musculoepithelial cells
(b) gastrovascular cavity
(c) nerve cells and process
(d) organ grade organization
Answer:
(d) organ grade organization

Question 3.
A non-matching set in the following is
(a) sepia – cuttle fish
(b) octopus – devilfish
(c) limulus – king crab
(d) ancylostoma – pinworm
Answer:
(d) ancylostoma – pinworm

Question 4.
A character common to Echinoderms and chordates
(a) marine
(b) benthonic
(c) deuterostome
(d) none of the above
Answer:
(c) deuterostome

Question 5.
Largest animal in the world that feed on smallest plankton is the
(a) dolphin
(b) killer whale
(c) blue whale
(d) sea cow
Answer:
(c) blue whale

Question 6.
Which is common to amphibian, reptelia and fishes
(a) nucleated RBC
(b) dermal scales
(c) poikelothermic condition
(d) both a and d
Answer:
(d) both a and d

Question 7.
Which of the following has pseudocoelomate tube within a tube body plan
(a) hydra
(b) planaria
(c) ascaris
(d) pheretima
Answer:
(c) ascaris

Question 8.
Ink gland associated with alimentary canal is found in
(a) sepia
(b) earthworm
(c) starfish
(d) cockroach
Answer:
(a) sepia

Question 9.
Which is common to all tetrapods
(a) epidermal scales
(b) red coloured blood
(c) 12 pairs of cranial nerve
(d) ureotelism
Answer:
(b) red coloured blood

Question 10.
Which one of the following is not a characteristic of phylum annelida?
(a) closed circulatory system
(b) segmentation
(c) pseudocoelom
(d) ventral nerve cord
Answer:
(c) pseudocoelom

Question 11.
Respiratory pigment of mollusc is
(a) haemocyanin
(b) haemoglobin
(c) haemoerythrin
(d) both a and b
Answer:
(a) haemocyanin

Question 12.
Select the character that can be attributed to chondrithytes
(a) persistened notochord
(b) placoid scales
(c) poikelothermic body
(d) all the above
Answer:
(d) all the above

Question 13.
The central cavity of sponge is called
(a) spongocoel
(b) coelocentron
(c) canal system
(d) spongilla
Answer:
(c) canal system

Question 14.
Flame cell are excretory organ of
(a) hydra
(b) cockroach
(c) planaria
(d) frog
Answer:
(c) planaria

Question 15.
Pneumatic skeleton is a feature of
(a) amphibians
(b) reptiles
(c) fishes
(d) birds
Answer:
(d) birds

Question 16.
The number of gills present in osteichthyes is
(a) 2 pairs
(b) 6-15 pairs
(c) 5 pairs
(d) 4 pairs
Answer:
(d) 4 pairs

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

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

Plus One Structural Organisation in Animals One Mark Questions and Answers

Question 1.
Which one of the following types of cell is involved in making of the inner walls of large blood vessels?
(a) Cuboidal epithelium
(b) Columnar epithelium
(c) Squamous epithelium
(d) stratified epithelium
Answer:
(c) Squamous epithelium

Question 2.
To which one of the following categories does adipose tissue belong?
(a) Epithelial
(b) Connective
(c) Muscular
(d) Neural
Answer:
(b) Connective

Question 3.
Which one of the following is not a connective tissue?
(a) Bone
(b) Cartilage
(c) Blood
(d) Muscles
Answer:
(d) Muscles

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

Question 5.
Match the following with reference to Cockroach and choose the correct option.

A. Phallomere	i. Chain of developing ova
B. Gonopore	ii. Bundles of sperm
C. Spermatophore	iii. Opening of the ejaculatory dust
D. Ovarioles	iv. The external genitalia

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

Question 6.
Complete the following sentences:

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

Answer:

1. Ranatigrina
2. Open

Question 7.
Name the respiratory organs of insects.
Answer:
Trachea

Question 8.
Note the relationship between the first two words and suggests a suitable word.

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

Answer:

1. Myelin sheath
2. Fibres

Question 9.

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

Answer:

1. Osteocyte
2. Clitellum

Question 10.
‘Neuroglia make up more than one half the volume of neural tissue in our body’. Write the function of neuroglial cells.
Answer:
Neuroglia cells protect and support neurons in our neural system.

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

Question 12.
Name the cells present in areolar tissue.
Answer:
Fibroblasts, Macrophages, Mast cells

Question 13.
_____________ make up more than one half the volume of neural tissue in our body.
Answer:
Neuroglia

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

Question 15.
Heart is made of
1. Epithelial tissue and Muscular tissue.
2. Muscular tissue and neural tissue.
3. Musculartissue and connective tissue.
4. All the four types of tissues.
Answer:
4. All the four types of tissues.

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

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

Question 18.
Intestine of earthworm contain a special organ which help in absorption. Identify the organ.
Answer:
Typhlosole

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

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

Plus One Structural Organisation in Animals Two Mark Questions and Answers

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

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

Answer:

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

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

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

Answer:

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

Question 3.
Give one word for the following.

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

Answer:

1. Ligaments
2. Neuroglial cell

Question 4.
Match the following.

A	B
Cuboidal epithelium	Fallopian tube
Squamous epithelium	Ducts of glands
Columnar epithelium	alveoli
Ciliated epithelium	intestine

Answer:

A	B
Cuboidal epithelium	Ducts of glands
Squamous epithelium	alveoli
Columnar epithelium	intestine
Ciliated epithelium	Fallopian tube

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

Answer:

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

Question 6.
Observe the diagram

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

Answer:

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

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

Answer:

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

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

Answer:

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

Question 10.
Match the following.

Column A	Column B
Haemocoelom	Frog
Typhlosole	Frog
Cloaca	Earthworm
Pseudocoelom	Cockroach

Answer:

Column A	Column B
Haemocoelom	Cockroach
Earthworm	Earthworm
Cloaca	Frog
Pseudocoelom	Ascaris

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

Answer:
A. Chondocyte

B. Collagen fibers

Question 12.
Circulating fluid of cockroach is colourless.

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

Answer:

1. hemolymph
2. hemocytes

Question 13.
Match the following:

A	B
Intestinal mucosa	Skeletal muscle
Biceps muscle	Smooth muscle
Tendon	Brush bordered columnar epithelium
Intestinal muscle	Connective tissue

Answer:

A	B
Intestinal mucosa	Brush bordered columnar epithelium
Biceps muscle	Skeletal muscle
Tendon	Connective tissue
Intestinal muscle	Smooth muscle

Question 14.
Give one word for the following.

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

Answer:

1. Adipose tissue
2. Areolar tissue
3. Tendons

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

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

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

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

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

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

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

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

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

Answer:

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

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

Question 21.
Match the following

i	ii
Hypopharynx	Food grinding
Nephridium	Food absorption
Gizzard	Bone cell
Typhlosole	Phallomere
Osteocyte	Tongue
Genitalia	Earthworm
	Frog

Answer:

i	ii
Hypopharynx	Tongue
Nephridium	Earthworm
Gizzard	Food grinding
Typhlosole	Food absorption
Osteocyte	Bone cell
Genitalia	Phallomere

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

Answer:
Frog female reproductive system Part A is Oviduct

Question 23.
Observe the given diagram.

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

Answer:

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

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

Answer:

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

Question 26.
Observe the given diagrams.

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

Answer:

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

Question 27.
Write the significance of

1. Blood
2. Neuroglia

Answer:

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

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

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

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

Answer:

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

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

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

Answer:

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

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

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

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

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

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

Question 33.
Match the following.

Bidder’s canal	Cockroach
Labium	Earthworm
Typhlosole	Frog

Answer:

Bidder’s canal	Frog
Labium	Cockroach
Typhlosole	Earthworm

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

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

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

Answer:

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

Animals	Excretory organs
Earthworm	_______________
________________	Malpighian tubules
Frog	________________

Answer:

Animals	Excretory organs
Earthworm	Nephridia
Coackroach	Malpighian tubules
Frog	Kidney

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

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

Answer:

1. Earthworm
2. Frog
3. Cockroach

Question 39.
Identify the statements as True or False.

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

Answer:

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

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

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

Answer:

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

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

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

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

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

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

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

Answer:

1. Stern items
2. Malpighian tubules

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

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

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

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

Answer:

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

Question 48.

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

answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

Question 58.
Name the portal systems found in frog

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

Answer:

1. Hepatic portal system
2. Renal portal system

Question 59.

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

Answer:

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

Question 60.
Answer in one word.

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

Answer:

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

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

Plus One Structural Organisation in Animals Three Mark Questions and Answers

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

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

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

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

Question 3.
Answer in one word or one line.

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

Answer:

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

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

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

Answer:

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

Question 5.
Observe the following diagram carefully.

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

Answer:

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

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

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

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

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

Answer:

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

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

Answer:

Question 9.
Give reasons.

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

Answer:

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

Plus One Structural Organisation in Animals NCERT Questions and Answers

Question 1.
Answer in one word or one line:

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

Answer:

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

Question 2.
Answer the following:

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

Answer:

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

Question 3.
Distinguish between the following:

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

Answer:

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

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

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

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

1. Chondriocytes
2. Axons
3. Ciliated epithelium

Answer:

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

Question 6.
Distinguish between

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

Answer:

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

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

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

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

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

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

Question 7.
Mark the odd one in each series:

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

Answer:

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

Plus One Structural Organisation in Animals Multiple Choice Questions and Answers

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Students can Download Chapter 1 Sets Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 1 Sets

Plus One Maths Sets Three Mark Questions and Answers

Question 1.
There are 200 individuals with a skin disorder, 120 had been exposed to the chemical A, 50 to chemical B and 30 to both chemical A and B, Find the number of individuals exposed to

1. Chemical A but not chemical B. (1)
2. Chemical B but not.chemical C. (1)
3. Chemical A or chemical B. (1)

Answer:
1. Given; n(U) = 200; n(A) = 120;
n(B) = 50; n(A∩B) = 30
n (Chemical A but not chemical B)
= n(A ∩ B’) = n(A) – n(A ∩ B) = 120 – 30 = 90

2. n (Chemical B but not chemical A)
= n(A’ ∩ B) = n(B) – n(A ∩ B) = 50 – 30 = 20

3. n (Chemical A or chemical B)
= n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 120 + 50 – 30 = 140.

Question 2.
In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.
Answer:
Let A – Apple juice; O – Orange juice be the sets.
Given; n(U) = 400; n(A) = 100;
n(O) = 150; n(A ∩ O) = 75
n (neither apple juice nor orange juice)
= n(A’ ∩ O’) = n((A ∪ O)’)
= n(U) – n(A ∪ O)
= 400 – [n(A) + n(O) – n(A ∩ O)]
= 400 – [100 + 150 – 75] = 400 – 175 = 225.

Question 3.
In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speaks at least one of these two languages?
Answer:
Let F – French; S – Spanish be the sets.
Given; n(F) = 50; n(S) = 20;w(F ∩ S) = 10
n (speaks at least one of these two languages)
= n(F ∪ S) = n(F) + n(S) – n(F ∩ S)
= 50 + 20 – 10 = 60.

Question 4.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis, how many like tennis only and not cricket? How many like tennis?
Answer:
Let C – Cricket; T – Tennis be the sets.
Given;
n(C’ ∪ T) = 65; n(C) = 40; n(C ∩ T) = 10
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 65 = 40 + n(T) – 10 ⇒ n(T) = 35
n (tennis only and not cricket)
= n(T ∪ C’) = n(T) – n(T ∩ C) = 35 – 10 = 25.

Question 5.
Let A and B be two sets such that n( A) = 20, n(A ∪ B) = 42, n(A ∩ B) = 4. Find

1. n(B) (1)
2. n(B – A) (1)
3. n(A – B) (1)

Answer:

1. n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
   ⇒ 42 = 20 + n(B) – 4 ⇒ n(B) = 26
2. n(B – A) = n(B) – n(A ∩ B) = 26 – 4 = 22
3. n(A – B) = n(A) – n(A ∩ B) = 20 – 4 = 16.

Question 6.
A = {x: x is a natural number less than 8}

1. Write in roster form. (1)
2. Write a subset of A containing all even numbers in A. (1)
3. Which of the following could not be the number of elements of power set of a set [2, 8, 10, 16]? (1)

Answer:

1. A = {1, 2, 3, 4, 5, 6, 7}
2. {2, 4, 6} or {2, 4, 6, 7}
3. 10. (since other are powers of 2.)

Plus One Maths Sets Four Mark Questions and Answers

Question 1.
Observe the Venn diagram.

1. Write in roster form. (1)
2. Verify that (A – B) ∪ (A ∩ B) = A (2)
3. Find (A ∩ B)’ (1)

Answer:

1. A = {1, 3, 4, 8} ; B = {2, 3, 5}
2. A – B = {1, 4, 8}; A ∩ B = {3}
   ⇒ (A – B) ∪ (A ∩ B) = {1, 3, 4, 8}
   Hence; (A – B) ∪ (A ∩ B) = A
3. (A ∩ B)’ = {1, 2, 4, 5, 6, 7, 8, 9}

Plus One Maths Sets Practice Problems Questions and Answers

Question 1.
Write the following sets in roster form.

1. A = {x: x xis an integer and -3 < x < 7}
2. B = {x: x ∈ N; x ≤ 6}
3. C = {x : x is a vowel in English alphabet}
4. D = {x : x is a two-digit natural number such that the sum of its digits is 8}
5. E = {x: x ∈ Z; \(-\frac{1}{2}<x<\frac{9}{2}\)}

Answer:

1. A = {-2, -1, 0, 1, 2, 3, 4, 5, 6}
2. B = {1, 2, 3, 4, 5, 6}
3. C = {a, e, i, o, u}
4. D = {17, 71, 26, 62, 35, 53}
5. E = {1, 2, 3, 4}

Question 2.
Write the following sets in Set builder form.

1. A = {3, 6, 9, 12}
2. B = {2, 4, 8, 16, 32}
3. C = \(\left\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}\right\}\)
4. D = {5, 25, 125, 625}
5. E = {2, 4, 6……..} (1 score each)

Answer:

1. A = {x: x = 3n, n ∈ N, n< 4 }
2. B = {x: x = 2ⁿ; n ∈ N; n < 5 }
3. C = {x: x = \(\frac{n}{n+1}\); n ∈ N, n ≤ 6}
4. D = {x: x = 5ⁿ; n ∈ N, n ≤ 4 }
5. E = {x: x is an even number}

Question 3.
Write the following in interval form.

1. {x: x∈ R, -4 < x ≤ 6}
2. {x: x∈ R, 0 ≤ x < 7 }
3. {x: x∈ R, 3 ≤ x ≤ 4 } (1 score each)

Answer:

1. (-4, 6]
2. [0, 7)
3. [3, 4]

Question 4.
Write the following in set builder form.

1. [0, 10]
2. [-2, 7)
3. (3, 4)

Answer:

1. {x: x∈ R, 0 ≤ x ≤ 10 }
2. { x: x ∈ R, -2 ≤ x < 7 }
3. {x: x ∈ R, 3 < x < 4 }

Question 5.
Find Set A, B and Universal set U (1 score each)

Answer:
A = {e, f, d}; B = {a, b, c, d} and U = {a, b, c, d, e, f, g, h}

Question 6.
Write all subset of the following

1. {1, 2}
2. {a, b, c}
3. Φ (1 score each)

Answer:

1. Φ, {1}, {2}, {1, 2}
2. Φ, {a}, {b}, {c} ,{a, b}, {a, c}, {b, c}, {a, b, c}
3. Φ

Question 7.
Let A = {1, 2, {3, 4}, s, d, θ} , Which of the following statements are true/false and why?

1. 3 ∈ A
2. {1, {3, 4}} ∈ A
3. {1, 2, 3} ⊂ A
4. Φ ∈ A
5. 1 ⊂ A (1 score each)

Answer:

1. False
2. True
3. False
4. False
5. False

Question 8.
If A = {1, 2, 4, 6, 7, 8}; B = {2, 5, 7, 9, 10} and C = {4 , 5, 9, 10}. Find

1. A ∪ B
2. B ∪ C
3. A ∪ C
4. A ∩ B
5. B ∩ C
6. A∪ B ∪ C
7. A ∩ B ∩ C
8. (A ∩ B) ∪ (C ∩ A) (1 score each)

Answer:

1. A ∪ B = {1, 2, 4, 5, 6, 7, 8, 9, 10}
2. B ∪ C = {2, 4, 5, 7, 9, 10}
3. A ∪ C = {1, 2, 4, 5, 6, 7, 8, 9, 10}
4. A ∩ B = {2, 7}
5. B ∩ C = {5, 9, 10}
6. A ∪ B ∪ C = {1, 2, 4, 5, 6, 7, 8, 9, 10}
7. A ∩ B ∩ C = Φ
8. (A ∩ B) ∪ (C ∩ A) = {2, 7} u {4} = {2, 4, 7}

Question 9.
If A = {2, 4, 6, 7, 8, 12}; B = {2, 7, 9, 10} and C = {5, 9, 10, 12}. Find

1. A – B
2. B – C
3. A – C
4. B – A
5. C – A
6. (A ∪ B) – C
7. A – {B ∩ C)
8. (A ∩ B) – (C ∩ A) (1 score each)

Answer:

1. A – B = {4, 6, 8, 12}
2. B – C = {2, 7}
3. A – C = {2, 4, 6, 7, 8}
4. B – A = {9, 10}
5. C – A = {5, 9, 10}
6. (A ∪ B) – C = {2, 4, 6, 7, 8, 9, 10, 12} – {5, 9, 10, 12} = {2, 4, 6, 7, 8}.
7. A – (B ∩ C) = {2, 4, 6, 7, 8, 12} – {9, 10} = {2, 4, 6, 7, 8, 12}
8. (A ∩ B) – (C ∩ A) = {2, 7} – {12} = {2, 7}

Question 10.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}; A = {2, 7, 9, 10} B = {5, 9, 10, 12} and C = {1, 4, 5, 7, 11}. Find

1. A’
2. B’
3. A’ – C
4. (B – A)’
5. B’ ∩ C’
6. (A ∪ B)’
7. A’ ∩ B’ (1 score each)

Answer:

1. A’ = { 1, 3, 4, 5, 6, 8, 11, 12}
2. B’ = { 1, 2, 3, 4, 6, 7, 8, 11}
3. A’ – C = {1, 3, 4, 5, 6, 8, 11, 12} – {2, 3, 6, 8, 9, 10, 12} = {1, 4, 5, 11}
4. (B – A)’ = {5, 12}’ = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11}
5. B’ ∩ C’ = {1, 2, 3, 4, 6, 7, 8, 11} – {2, 3, 6, 8, 9, 10, 12} = {2, 3, 6, 8}
6. (A ∪ B)’ = {2, 5, 7, 9, 10, 12}’ = {1, 3, 4, 6, 8, 11}
7. A’ ∩ B’ = {1, 3, 4, 5, 6, 8, 11, 12} ∩ {1, 2, 3, 4, 6, 7, 8, 11} = {1, 3, 4, 6, 8, 11}

Question 11.
If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements and Y has 32 elements, how many elements does X ∩ Y have?
Answer:
Given; n(X ∪ Y) = 50; n(X) = 28; n(Y) = 32
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 50 = 28 + 32 – n(X ∩ Y)
⇒ n(X ∩ Y) = 60 – 50 = 10

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

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

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

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

Question 2.
Between steel and diamond, which is more elastic?
Answer:
Steel.

Question 3.
What is the value of Y for a perfectly elastic body?
Answer:
Infinity.

Question 4.
Young’s modulus of the wire depends on
(a) length of the wire
(b) diameter of the wire
(c) material of the wire
(d) mass hanging from the wire
Answer:
(c) material of the wire
Young’s modulus of wire depends only on the nature of the material of the wire.

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

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

Question 7.
The stress required to double the length of a wire of Young’s modulus Y is
(a) Y/2
(b) 2Y
(c) Y
(d) 4Y
Answer:
(c) Let L be the length of a wire.
If the length of a wire is doubled, the longitudinal strain will be

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

Question 8.
What is the bulk modulus of a perfectly rigid body?
Answer:
Infinity

Question 9.
What is shear modulus of liquid?
Answer:
Zero

Question 10.
What is rigidity modulus of liquid?
Answer:
Zero.

Question 11.
Identify the type of modulus in twisting of cylinder.
Answer:
Shear modulus

Question 12.
Why work is required to be done to stretch a wire?
Answer:
Work is required to be done against the inter molecular forces of attraction.

Question 13.
Name the material which is famous for a large elastic after effect.
Answer:
Glass

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

Question 1.
Find stress required to double the length of a wire of Young’s modulus Y
Answer:
Let L be the length of a wire.
If the length of a wire is doubled, the longitudinal strain will be

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

Question 2.
When wire is bent back and forth, it becomes hot. Why?
Answer:
When wire is bent back and forth, the deformations are beyond elastic limit. A part of work done is converted in to heat energy. Hence wire becomes hot.

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

Question 1.
The stress strain graph of two bodies A and B are given in the figure

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

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

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

3.

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

Question 2.
Hook’s law is said to be that fundamental law in elasticity.

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

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

2.

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

Question 1.
When an external force deforms a solid, internal restoring forces are developed in the body giving rise to stress and strain.

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

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

2.

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

Question 1.

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

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

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

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

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

Question 2.
The modulus of elasticity of rubber is higher than that of steel.

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

Answer:
1. No

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

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

Question 3.
We have common experience that a thin thread breakes if pulled at its ends. Also a thin metallic wire, when pulled at the ends elongates and then contracts, when released.

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

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

2. Stress strain curve

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

r = 0.005m
∴ diameter = 0.01m.

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

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

Answer:
1. The values:

• infinity
• Zero

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

3.

Plus One Physics Mechanical Properties of Solids NCERT Questions and Answers

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


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

Dividing (1) by (2), we get

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

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

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

= 7.5 × 10¹⁰Nm^-2.

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

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

3.92 × 10^-7m
≃ 4 × 10^-7m.

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

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

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

Students can Download Chapter 8 Arrays Questions and Answers, Plus One Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Science Chapter Wise Questions and Answers Chapter 8 Arrays

Plus One Arrays One Mark Questions and Answers

Question 1.
From the following which is not true for an array.
(a) It is easy to represent and manipulate array variable
(b) Array uses a compact memory structure
(c) Readability of program will be increased
(d) Array elements are dissimilar elements
Answer:
(d) Array elements are dissimilar elements

Question 2.
Consider the following declaration.
int mark(50) Is it valid? If no give the correct declaration
Answer:
It is not valid. The correct declaration is as follows, int mark[50]. Use square brackets instead of parenthesis

Question 3.
Consider the following declaration.
int mark[200]
The index of the last element is ________.
Answer:
199

Question 4.
Consider the following declaration int mark[200]. The index of the first element is ________.
Answer:
0

Question 5.
Consider the following int age[4] = {15,16,17,18};
From the following which type of initialisation is this.
(a) direct assignment
(b) along with variable declaration
(c) multiple assignment
(d) None of these
Answer:
(b) along with variable declaration

Question 6.
From the following which is used to read and display array elements.
(a) loops
(b) if
(c) switch
(d) if else ladder
Answer:
(a) loops

Question 7.
Write down the corresponding memory consumption in bytes

1. int age[10] = ——-
2. charname[10] = ——-
3. int age[10][10]= ——–

Answer:

1. 2 × 10 = 20 bytbs (2 bytes for one integer)
2. 1 × 10 = 10 (one byte for each character)
3. 2 × 10 ×10 = 200 (2 × (100 elements))

Question 8.
A two dimensional array is having ________ number of subscript
(a) one
(b) two
(c) three
(d) none of these
Answer:
(b) two

Question 9.
Consider the following
int age[10][10];
In this array how many elements will contain.
Answer:
10 × 10 = 100 elements

Question 10.
Consider the following int age[4] = {12, 13, 14};
cout<<age[3];
What will be the output?
(а) 14
(b) 12
(c) 13
(d) 0
Answer:
(d) 0

Question 11.
The elements of 2 dimensional array can be read using _________ loop.
Answer:
nested loop

Question 12.
________ is the process of reading/visiting elements of an array.
Answer:
traversal

Question 13.
Anjali wants to read the 10 marks that already stored in an array and find the total. This process is known as ________.
(a) insertion
(b) deletion
(c) traversal
(d) linear search
Answer:
(c) traversal

Question 14.
Mani wants to check a number from 100 numbers already stored in an array. This process is called ___________.
(a) insertion
(b) deletion
(c) searching
(d) None of these
Answer:
(c) searching

Question 15.
Divide and conquer method used in ____________ search.
Answer:
binary

Question 16.
“Each element of the array is composed with value to be searched from the beginning of the array”. This method is adopted by _____________ search.
Answer:
linear search

Question 17.
____________ search requires a sorted array as input.
Answer:
binary search

Question 18.
___________ searching is slower for larger array.
Answer:
linear searching

Question 19.
____________ is a simple sorting algorithm.
(a) bubble sort
(b) selection sort
(c) linear search
(d) binary search
Answer:
(a) bubble sort

Question 20.
Adjacent elements are checked and inter changed in _____________ sort.
Answer:
bubble sort

Question 21.
The array is divided into sorted part and unsorted part in ____________ sort.
selection.

Question 22.
The elements of an array of size ten are numbered from _____ to _______.
Answer:
0 to 9

Question 23.
Element mark[6] is which element of the array?
(a) The sixth
(b) the seventh
(c) the eighth
(d) impossible to tell
Answer:
(b) the seventh

Question 24.
When a multidimensional array is accessed, each array index is ________.
(a) Separated by column.
(b) Surrounded by brackets and separated by commas.
(c) Separated by commas and surrounded by brackets.
(d) Surrounded by brackets,
Answer:
(d) surrounded by brackets

Question 25.
You have used a 2D array with the Name Mat representing a matrix. Write the C++ expression to access the 3rd element in the 2nd row.
Answer:
mat[1] [2];

Question 26.
Write a C++ statement that defines a string variable called ‘name’ that can hold a string of upto 20 characters.
Answer:
char name[21j;

Question 27.
Given some array declaration. Pick the odd man out.
Float a[+40], int num[0-10], double [50], char name[50], amount[20] of float.
Answer:
char name[50]. It is a valid array decalaration the remaining are not valid.

Question 28.
__________ is a collection of elements with same datatype.
Answer:
Array

Question 29.
int num[10];
The above C++ statement declares an array named num that can store maximum ________ integer numbers. (SEP-2015 (IMP) (1)
(а) 9
(b) 10
(c) N
(d) none of these
Answer:
(b) 10

Question 30.
Declare a two dimensional array to store the elements of a matrix with order 3 × 5. (MARCH-2016) (1)
Answer:
int m[3] [5]; or float m[3] [5]

Question 31.
_________ search method is an example for ‘divide and conquer method’. (SEP-2016) (1)
Answer:
Binary

Question 32.
Read the following C++ statement:
int AR[10]
How many bytes will be allocated for this array? (1)
Answer:
To store an integer 4 bytes needed so AR[10] needs 10 × 4 = 40 bytes

Question 33.
Find the value of score [4] based on the following declaration statement. int score [5] = {988,87,92,79,85}; (MARCH-2017) (1)
Answer:

Hence (158)₁₀ = (9E)₁₆

Plus One Arrays Two Mark Questions and Answers

Question 1.
Whether the statement char text[] = “COMPUTER”; is True/False? Justify.
Answer:
It is a single dimensional array. If the user doesn’t specify the size the operating system allocates the number of characters + one (for null character for text) bytes of memory. So here OS allocates 9 bytes of memory.

Question 2.
Given a word ” Mathematics”. How will you locate this word in a standard dictionary. What are the steps used for this purpose
Answer:
In the standard dictionary the words are arranged in sorted order so binary search method is more suitable to find the word “MATHEMATICS”.

Question 3.
How can you find out the word “MATHEMATICS” from a text book. Which is the most convenient method? Justify your answer.
Answer:
In a text book the words are not arranged in sorted order so linear search method is convenient.

Question 4.
Consider the statement char str[] = “PROGRAM” What will be stored in last location of this array. Justify.
Answer:
The last location is the null character(\0) because each string must be appedend by a null character.

Question 5.
Explain different array operations in detail.
Answer:

1. Traversal: All the elements of an array is visited and processed is called traversal
2. Search: Check whether the given element is present or not
3. Sorting: Arranging elements in an order is called sorting

Question 6.
Read the following C++ statement: int MAT[5][4];

1. How many bytes will be allocated for this array?
2. Suppose MAT[4][4] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the’ sum of all the elements in the array. (MARCH – 2015) (1)

Answer:

Question 7.
Declare an array of size 5 and intitialize it with the numbers 8, 7, 2, 4 and 6. (MARCH-2016) (2)
Answer:
int n[5] = {8, 7, 2, 4, 6};

Question 8.
Predict the output of the following C++ program.
#include <iostream.h>
int main()
{
int array[] = {1, 2, 4, 6,7,5};
for (int n =1; n<=5; n++)
array [n] = array [n – 1 ];
for (n = 0; n <= 5; n++)
cout<< array [n];
return 0;
}
Answer:
1, 1, 1, 1, 1, 1

Question 9.
Suppose M[5][5] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the sum of the diagonal elements. (SCERT SAMPLE – II) (2)
Answer:
for (i = 0; i < 5, i++)
for (j = 0; j < 5; j++)
if (i == j)
S = S + M[i][j];

Question 10.

1. Write C++ program for sorting a list of numbers using Bubble Sort Method. (2)
2. Write the different passes of sorting the following numbers using Bubble Sort.
   32, 21, 9, 17, 5

Answer:
1. Example for Bubble sort is given below

2. Bubble sort (ascending order):
Consider the first 2 elements of the numbers and compare it. 32 is greater than 21 then interchange both of them. Then the numbers are
21, 32, 9, 17, 5
Next compare 32 and 9. Here 32 is greater so interchange both of them. Then the numbers are
21, 9, 32, 17, 5
Next compare 32 and 7. Then interchange and the numbers are
21, 9, 17, 32, 5
Next compare 32 and 5. Then interchange and the numbers are
21, 9, 17, 5, 32.
After the first phase the largest number is on the right side. Similarly do the remaining. At last we will get the result as follows.
5, 9, 17, 21, 32

Question 11.
Suppose M[5] [5] is a 2D array that contains the elements of a square matrix. Write C++ statements to find the sum of the diagonal elements. (MARCH-2017)
Answer:

Plus One Arrays Three Questions and Answers

Question 1.
Suppose you are given Total mark of 50 students in a class

1. How will you store these values using ordinary variable?
2. Is there any other efficient way to store these values? Give reason.

Answer:
We have to declare 50 variables individually to store total marks of 50 students. It is a laborious work. Hence we use array, it is an efficient way to declare 50 variables. With a single variable name we can store multiple elements.
eg: int mark[50]. Here we can store 50 marks. The first mark is in mark[0], second is in mark[1]…. etc the fiftieth mark is in mark[49],

Question 2.
Four friends are in a queue for watching a movie. How can you locate a particular person from that queue? Write all the steps sequentially.
Answer:
Locate a particular person from a queue can be carried out by using two ways.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Question 3.
Total mark of 50 students in a class are given in an array. A bonus of 10 marks is awarded to all of them. Write the program code for making such a modification.
Answer:

Question 4.
In School Assembly, try to arrange all the students in your class according to the increasing order of their heights separately for male/female students. Smallest one will be in front and tallest one at the last position. Write sequential steps for doing this, in two different methods.
Answer:
Arranging elements from smallest to largest (ascending) or largest to smallest(descending) is called sorting. There are two sorting methods
1. Bubble sort:
It is a simple sorting method. In this sorting considering two adjacent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest(in the case of descending sorting) will be the end of the array. This process continues.

2. Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. First smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Question 5.
Using the concept obtained from algorithms in question number (9), write complete program for the two types of sorting.
Answer:

Question 6.
Collect the heights of 12 students from your class in which 7 students are male and others are female students. Suppose these male and female students be seated in two separate benches and you are given a place which is used for sitting these 12 students in linear form. How will you combine and make them sit without mixing male/female students. Write a program for the same.
Answer:

Question 7.
Collect the heights of 12 students from your class in which 7 students are male and others are female students. Suppose you are given the heights according to a sorted order separately for male/female students. How will you combine these group according to the same order in linear form. Write a program for the same.
Answer:

Question 8.
Write the program code for counting the number of vowels from your school name.
Answer:

Question 9.
Write the program code for counting the number of words from the given string.” Directorate of Higher Secondary Examination”
Answer:

Question 10.
Given a word like “ECNALUBMA” Write the program code for arranging it in into a meaningful word.
Answer:

Question 11.
Explain the needs for arrays Array
Answer:
Array is collection of same type of elements. With the same name we can store more elements. The elements are distinguished by using its index or subscript. To store 50 marks of 50 students we have to declare 50 variables, it is a laborious work.

Hence the need for arrays arise. By using array this is very easy as follows int mark[50]. Here the index of first element is 0, then 1, 2, 3, etc upto 49.

Question 12.
Explain different types of arrays.
Answer:
There are two tyes of array
1. Single dimensional:
It contains only one index or subscript. The index starts from 0 and ends with size-1.
eg: int n[50];
char name[10];

2. Multi-dimensional:
It contains more than one index or subscript. The two dimensional array contains two indices, one for rows and another for columns. The row index starts from 0 and end at row size-1 and column index starts at 0 and ends at colunn size-1.
eg: int n[10][10] can store 10 × 10 = 100 elements. The index of the first element is n[0][0] and index of the 100th element is n[9][9],

Question 13.
Write a program to read the 5 marks of a students and display the marks and total
Answer:

Question 14.
Write a program to read 3 marks of 5 students and find the total and display it.
Answer:

Question 15.
Differentiate linear search and binary search
Answer:
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is re¬turned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Question 16.
Differentiate bubble sort and selection sort.
Answer:
1. Bubble sort:
It is a simple sorting method. In this sorting considering two adjascent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest (in the case of descending sorting) will be the end of the array. This process continues.

2. Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. First smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Question 17.
Write a program to read a string and a character and find the character by using linear search.
Answer:

Question 18.
Write a program to read marks of 10 students and read a mark and check whether it is in the array or not using binary search
Answer:

Question 19.
Write a program reverse a string without using a function.
Answer:

Question 20.
Write a program to read a string and find the no. of vowels .consonents and special characters.
Answer:

Question 21.
Given a word “COMPUTER”, write a C++ program to reverse the word without using any string functions.
Answer:
#include<iostream>
using namespace std;
int main(()
{
char a[9] = “COMPUTER”;
inti;
for (i=8; i>=0; i- -)
{
cout<<a[i];
}
}

Question 22.
Write a program to accept marks of 10 students and find out the largest and smallest mark from the list.

OR

C++ program to store the scores of 10 batsmen of a school cricket team and find the largest and smallest score.
Answer:

Question 23.
Write the names of two searching methods in arrays. Prepare a chart that shows the comparisons of two searching methods. (MARCH-2015)
Answer:
It is the process of finding the position of the given element.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.

Question 24.
Write a C++ program to illustrate array traversal. (SAY-2015 (IMP) (3)
Answer:
#include<iostream>
using namespace std;
int main()
{
int i, mark[50];
for(i=0;i<50; i++)
{
cout<<“Enter mark “<<i + 1<<“: cirt>>mark[i];
}
cout<<“\nThe mark of 50 students after adding bonus mark 10 is given below \n”; for(i=0;i<50; i++)
cout<<mark[i] + 10<<endl;
}

Question 25.
Define an array. Also write an algorithm for searching an element in the array using any one method that you are familiar with. (MARCH-2016) (3)
Answer:
An array is a collection of elements with same data type. The index of first element is 0 (zero) and the index of last element is size -1. There are 2 methods linear search and binary search.

Searching: It is the process of finding the position of the given element.
1. Linear search:
In this method each element of the array is compared with the element to be searched starting from the first element. If it finds the position of the element in the array is returned.

2. Binary search:
It uses a technique called divide and conquer method. It can be performed only on sorted arrays. First we check the element with the middle element. There are 3 possibilities. The first possibility is the searched element is the middle element then search can be finished.

The second possibility is the element is less than the middle value so the upper bound is the middle element. The third possibility is the element is greater than the middle value so the lower bound is the middle element. Repeat this process.8.3 Two dimensional (2D) arrays.

Some occasions we have to store 6 different marks of 50 students. For this we use 2D arrays. An array with two subscripts is used, eg. int mark[r][c]; Here r is the row and c is the column.

Question 26.

1. Finding sum of all elements in an array is an example of _________ operation.
2. If 24, 45, 98, 56, 76, 24, 15 are the elements of an array, illustrate the working of selection sort for arranging the elements in descending order. (SCERT SAMPLE -1) (3)

Answer:
1. Traversal

2. In selection sort the array is divided into two parts the sorted part and unsorted part. First smallest elemant in the unsorted part in searched and exchanged with the first elemant now there is 2 parts sorted part and unsorted part. This process continues.

Question 27.
Read the following C++ statement:

1. How many bytes will be allocated for this array?
2. Write algorithm to sort this array using bubble sort method. (SCERT SAMPLE – II) (3)

Answer:
1. To store an integer 4 bytes needed so AR[10] needs 10 × 4 = 40 bytes.

2. Sorting – Arranging elements of an array in an order(ascending or descending).
(a) Bubble sort:
It is a simple sorting method. In this sorting considering two adjascent elements if it is out of order, the elements are interchanged. After the first iteration the largest(in the case of ascending sorting) or smallest(in the case of descending sorting) will be the end of the array. This process continues.

(b) Selection sort:
In selection sort the array is divided into two parts, the sorted part and unsorted part. First smallest element in the unsorted part is searched and exchanged with the first element. Now there is 2 parts sorted part and unsorted part. This process continues.

Question 28.
Explain ‘Call by Value’ and ‘Call by Reference’ methods of function calling with the help of a suitable example. (SEP-2016) (3)
Answer:
Two types are call by value and call by reference.
1. Call by value:
In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
Example

2. Call by reference:
In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.
Example

Question 29.
What is an array? Write C++ program to declare and use a single dimensional array for storing the computer science marks of all students in your
class. (SEP-2016) (3)
Answer:
An array is a collection of elements with same data type or with the same name we can store many elements, the first or second or third etc can be distinguished by using the index(subscript). The first element’s index is 0, the second elements index is 1, and so on.

Question 30.
Write an algorithm for arranging elements of an array in ascending order using bubble sort. (MARCH-2017) (3)
Answer:

Plus One Arrays Five Mark Questions and Answers

Question 1.
Write a program to perform matrix addition, multiplication, subtraction.
Answer:

Students can Download Chapter 12 Organic Chemistry: Some Basic Principles and Techniques Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques One Mark Questions and Answers

Question 1.
Which of the following does not contain fused benzene ring?
a) Naphthalene
b) Anthracene
c) Diphenyl
d) Phenanthrene
Answer:
c) Diphenyl

Question 2.
Hemolytic fission of a covalent bond results in the formation of ___________ .
CH₂=CH – C ≡ CH₂OH
Answer:
Free radicals

Question 3.
The IUPAC name of is
a) 2-pentyl-4-en-1-ol
b) 4-penten-2-yn-1-ol
c) 1-pentene-3-yn-5-ol
d) 5-Hydroxy-1-pentene-3-yne
Answer:
b) 4-penten-2-yn-1-ol

Question 4.
Lassigne’s solution on treating with sodium nitro prusside solution gives a violet colour indication the presence of ___________ in the organic compound.
Answer:
sulphur

Question 5.
In Kjeldahl’s method, nitrogen present is estimated as ___________ .
Answer:
NH₃

Question 6.
In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
a) Na₄[Fe(CN)₆]
b) Fe₄[Fe(CN)₆]₃
c) Fe₂[Fe(CN)₆]
d) Fe₃[Fe(CN)₆]₄
Answer:
The Prussian blue colour i§ due to the formation Fe₄[Fe(CN)₆]₂. Thus, option (b) is correct,

Question 7.
Which of the following carbocation is most stable?

Answer:
The order of stability of carbocation is : 3º>2º>1º

Thus, option (b) is correct.

Question 8.
The best and latest technique for isolation, purification and separation of organic compounds is:
a) Crystallisation
b) Distillation
c) Sublimation
d) Chromatography
Answer:
Chromatography. Thus, option (d) is correct.

Question 9.
The following reaction is classified as:
CH₂CH₂l + KOH (aq) → CH₃CH₂OH + Kl
a) Electrophilic Substitution
b) Nucleophilic substitution
c) Elimination
d) Addition.
Answer:
This is an example of nucleophilic substitution reaction since the nucleophile l – is replaced by the nucleophile OH^– ion. Thus, option (b) is correct.

Question 10.
Which of the carbocation is more stable?

Answer:

Question 11.
Absolute alcohol cannot be obtained by fractional distillation because
Answer:
constant boiling azeotrope mixture is formed with water

Question 12.
Lassaigne’s test fails in
a) NH₂NH₂
b) H₂NCONH₂
c) C₆H₂NHNH₂
d) NH₂OH
Answer:
a) Hydrazine NH₂NH₂

Question 13.
Beilstein test is for the detection of __________ .
Answer:
Halogens

Question 14.
Glycerine can be purified by __________ .
Answer:
Distillation under reduced pressure.

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques Two Mark Questions and Answers

Question 1.
2-Butene exhibits geometrical isomerism.
1. Represent the cis-trans isomers of 2-Butene.
2. Explain ozonolysis with a suitable example.
Answer:
1.

2. When an alkene is allow to react with ozone, an ozonide is obtained. This on hydrolysis gives aldehyde/ ketone. The whole process is called ozonolysis. Eg:

Question 2.
Draw the structures of the molecules represented by the IUPAC names, Pent-3-en-1-ol and 2- Nitrocyclohexene.
Answer:
a) CH₃ – CH = CH – CH₂ – CH₂ – OH

Question 3.
Addition of HBr to propene yields 2-bromopropane, what happens if benzoyl peroxide is added to the above reaction.
Answer:
When propene is allowed to react with HBr in the presence of peroxide, 2 bromo propane is obtained as the minor product and this is called peroxide effect.

Question 4.
Write the IUPAC name of

Answer:
i) 2-Pentanone
ii) 3-Methyl-1-pentanal

Question 5.
Write any two necessary condition for a compound to be aromatic. Convert Acetylene to benzene.
Answer:
Cyclic, Planar and should contain (4n+2) n electrons.

Question 6.
What are hybridisation states of each carbon atom in the following compounds?
CH₂ = C = O, CH₃CH = CH₂,(CH₃)₂CO,
CH₂ = CHCN, C₆H₆
Answer:
The hybridisation of each carbon is written as superscript on the carbon atom in the molecule.

Question 7.
Which is expected to be more stable, O₂ NCH₂CH₂O or
Answer:

has -l- effect and hence it tends to disperse the -ve charge on the O-atom. In contrast, CH₃CH₂ exerts + l – effect. It, therefore, tends to intensity the -ve change and hence destabilizes it.

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques Three Mark Questions and Answers

Question 1.
a) Write the structure of the following compounds.
i) Hexane-2, 4-dione
ii) 3-Bromo-4 methyl hexane-2-ol
iii) 2-Bromo-4-methyl hept-5-enal
Answer:
a) i) Hexane 2, 4-dione

Question 2.
Isomers are compounds with same formula and different properties.
a) Write all the possible structural isomers of C₅H₁₂.
b) Give the IUPAC name of above isomers.
Answer:

b) 1) Pentane
2) 2-Methyl butane
3) 2,2-Dimethyl propane

Question 3.
a) Give the IUPAC names of:

b) How is nitrogen detected by Lassaignes test?
c) Name a suitable technique for separation of the components from a mixture of benzene (b.p.353 K) and aniline (b.p.-457 K)
Answer:
a) (i) 2-Chloro 3-methyl hexane
(ii) 2-Methylbutanal

b) About 2 mL of the extract is boiled with about 1 mL of freshly prepared ferrous sulphate solution. One or two drops of con.H₂SO₄ are added to the solution. Presence of nitrogen is indicated by the appearance of a blue colouration.

c) Distillation

Question 4.
i) Draw the structure of propanone. Write the hybridisation of each carbon in propanone.
ii) Arrange the following carbocations in the increasing order of their stability. Justify.
CH₃⁺, CH₃CH₂⁺, (CH₃)₂CH⁺
iii) What is the method used to separate a mixture of o-Nitrophenol from p-Nitrophenol? Which property is utilized for separation?
Answer:

I^st carbon of Propanone is sp³ hybridisation.
II^nd carbon of Propanone is sp² hybridisation.
III^rd carbon of Propanone is sp³ hybridisation.

iii) Chromatography; Difference in absorption rate of different substances..

Question 5.
1. Give the IUPAC name of the following compound.

2. How can you detect the presence of nitrogen in an organic compound?
3. Arrange the following in the increasing order of stability.
(CH₃)₂CH⁺, CH₃-CH₂⁺, (CH₃)₃C⁺
Answer:
1. 2-chloro, 2-methyl propane

2. Presence of nitrogen can be detected by Lassigne’s test. The sodium fusion extract is boiled with FeSO₄ and then acidified with con.H₂SO₄. The formation of Prussian blue colour conforms the presence of nitrogen.
3.

Question 6.
Predict the product in the following reactions and identify the rules:

Answer:
a) When propene is allowed to react with HBr in the presence of peroxide, 2-Bromo propane is obtained as the minor product. (Peroxide effect, Kharasch effect or Anti MarkownikofFs rule of addition).

Question 7.
a) Draw the structures of the following compounds.
i) 3-hexenoic acid
ii) 2-chloro-2-methyl butanol
iii) 4-nitro-1-pent-l-yne
Answer:

Question 8.
Write the IUPAC names of the following:

Answer:
i) 2,2,4-Trimethyl pentane
ii) 2-Methyl 1-butene
iii) Propyl benzene

Question 9.
Draw the structure of the following molecules.
i) 3, 4-Dimethylhept-3-ene
ii) Neo-pentane
iii) 3-Nitrocyclohexene
Answer:

Question 10.
What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Answer:
a) Structural isomers (actually position isomers as well as metamers)
b) geometrical isomers,
c) resonance contributors because they differ in the position of electrons but not atoms:

Question 11.
Expain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and phosphorus.
Answer:
The organic compound is fused with sodium metal to convert these elements (which are present in the covalent form) to ionic form. For example, Sulphur is changed to Na₂S, nitrogen to NaCN and phosphorus to Na₃PO₄ The presence of sulphide ions, cyanide ions and phosphate ions can thus be confirmed by using suitable reagents.

Plus One Chemistry Organic Chemistry: Some Basic Principles and Techniques Four Mark Questions and Answers

Question 1.
1) Write the structural formula of
a) 4-Ethyl-1-fluoro-2-nitrobenzene
b) 2,3,6-Trimethyl octane.
c) 1,2-Dibromo benzene.
2) Categorize the following as nucleophile and electrophile
a) HS^–
b) BF₃
c) NO₂⁺
d) C₂H₅O^–
e) (CH₃)₃N
f)NH₂^–
Answer:

Question 2.
1. What is chromatography? Name the different types of chromatography.
2. What is Lassaigne’s test?
Answer:
1. Chromatography is a valuable method for the separations, purification and identification of the constituents of a mixture. Chromatography is classified into 2 types.
a) Adsorption chromatography
b) Partition chromatography

2. Lassigne’s test is used to determine the presence of nitrogen, halogens and sulphur present in the organic compound.

Question 3.
1. Write IUPAC names of the products obtained by addition of HBr to Hex-1-ene.
i) in the absence of peroxide
ii) in the presence of peroxide
2. Complete the following reaction:

Answer:
1.

2. 2HCHO + H₂O₂

Question 4.
1. What is metamerism? Give example for metamers.
2. What are free radicals? How are they formed?
Answer:
1. It is the isomerism which arises due to different alkyl chains on either side of the functional group in the molecule.
e.g. Methoxypropane(CH₃OC₃H₇) and ethoxyethane (C₂H₅OC₂H₅) are metamers.

2. Free radicals are highly reactive species containing unpaired electrons. They are formed by homolytic cleavage of covalent bond. e.g. \(\dot { C } \)H₃

Question 5.
1. How will you prepare butane?
2. Explain Markownikoff’s rule for the addition reaction using a suitable example.
Answer:
1. When an alkyl halide (ethyl chloride) is allowed to react with metalic sodium in presence of dry ether,

2. Markownikoff’s rule of addition:
When a hydrogen hallide is added to an unsymmetrical alkane the halogen atom will goes to the double bond carbon containing lesser number of hydrogen atom, eg:

Question 6.
Identify the isomerism exhibited by the following compounds.

1. CH₃CH₂CH₂OH and (CH₃)₂CHOH
2. CH₃CH₂CHO and CH₃COCH₃
3. CH₃CH2₂CH₂CH₃andCH₃OCH₂CH₂CH₃
4. CH₃(CH₂)₃ CH₃ and (CH₃)₄C

Answer:

1. Position isomerism
2. Functional group isomerism
3. Metamerism
4. Chain isomerism

Question 7.
i) Predict the products of

ii) Classify the following compounds into aromatic and non-aromatic.

(iii) Which of the following compounds will show geometrical isomerism?
a) CH₃ CH=CHCH₃
b) (CH₃)₂₃ C=CH₂
Answer:

Question 8.
Write the IUPAC name

b) Write the structure of the following.
i) 2-Chloro-2-methyl butanol
ii) 4-Nitro-1-pentene
Answer:
a) i) 4-Methyl pentanal
ii) Cyclohexanol

Question 9.
a) Write IUPAC names of the products obtained by addition reactions of HBrto hex-1-ene:
i) In the absence of peroxide.
ii) In the presence of peroxide.
b) How will you convert:
i) Benzene to toluene.
ii) Benzene to nitrobenzene
Answer:

Question 10
a) Explain the term
(i) Inductive effect
(ii) Nucleophile
b) Write the structural formula of the following.
i) 2,5,6-Trimethyloctane
ii) 2,4-Dimethylpentane
c) Suggest the suitable technique for separation of organic compounds given in the data.
i) Aniline-water mixture
ii) Glycerol from spent-lye
Answer:
a) Inductive effect:
If more electronegative atom X linked to a carbon atom then the bonded electron pair will be shifted more towards X. So X acquires a small negative charge and carbon get a small charge. So this carbon atom attracts the a bonded electron pair towards its from nearest carbon atom. Therefore the change is transferred from one carbon to other.

Nucleophile :
Nucleus loving species negatively charge species like \(\overline { OH } \), \(\overline { X } \) etc. are called nucleophiles.
b) 2,5,6-Trimethyloctane

c) i) Fractional distillation.
ii) Distillation under reduced pressure.

Question 11.
Compounds having the same molecular formula exhibit different properties is called isomerism. Explain different types of isomerism with examples.
Answer:
Chain isomerism: Consider the molecular formula C₅H₁₂ the following 3 isomers are possible for this compound.
They are

Position isomerism:
This isomerism arises as a result of the difference in the position of double bond triple bond and functional group.
The following chain isomers are possible for the molecular formula C₄H₈.
CH₂ = CH – CH₂ – CH₃ 1-butene
CH₃ – CH = CH – CH₃ 2-butene

Metamerism:
This isomerism arises as a result of the difference in the alkyl group present on either side of the functional group.
Consider the molecular formula C₄H₁₀O. The following two metamers are possible for this molecular formula.
CH₃ – O – CH₂ -CH₂ -CH₃ Methyl propyl ether
CH₃ – CH₂ – O – CH₂ – CH₃ Diethyl ether

Functional group isomerism:
Consider the molecular formula C₂H₆O the following two functional isomers are possible for this compound.
1) CH₃-CH₂-OH Ethanol
2) CH₃ – O – CH₃ Di methyl ether

Question 12.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
Answer:
Sodium extract is boiled with nitric acid to decompose NaCN and Na₂S, if present,

If the sodium extract is not boiled with nitric acid then NaCN and Na₂S formed will react with AgN03 and hence will interfere with the test as shown below:

Students can Download Chapter 8 Redox Reactions Questions and Answers, Plus One Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions

Plus One Chemistry Redox Reactions One Mark Questions and Answers

Question 1.
In which of the following, oxidation number of chlorine is +5?
a) Cl^–
b) ClO^–
c) ClO₂^–
d) ClO₃^–
Answer:
d) ClO₃^–

Question 2.
An oxidising agent is a substance which can
a) Gain electrons
b) Lose an electronegative radical
c) Undergo decrease in the oxidation number of one of its atoms
d) Undergo any one of the above changes
Answer:
d) Undergo any one of the above changes

Question 3.
The arrangement of metals in the order of decreasing tendency to lose electrons is called _________ .
Answer:
Activity series

Question 4.
When KMnO₄, reacts with acidified FeSO₄
a) Only FeSO₄ is oxidised
b) Only KMnO₄ is oxidised
c) FeSO₄ is oxidised and KMnO₄ is reduced
d) KMnO₄ is oxidised and FeSO₄ is reduced
Answer:
c) FeSO₄ is oxidised and KMnO₄ is reduced

Question 5.
In the disproportionation reaction, which of the following statements is not true?
a) The same species is simultaneously oxidised as well as reduced
b) The reacting species must contain an element having at least three oxidation states
c) The element in the reacting species is present in the lowest oxidation state
d) The element in the reacting species is present in the intermediate oxidation state
Answer:
c) The element in the reacting species is present in the lowest oxidation state

Question 6.
Find the oxidation state of oxygen in OF₂.
Answer:
The oxidation number of fluorine in its compounds is always taken as -1.
In OF₂
X+ (-1 × 2) = 0
X = +2

Question 7.
The oxidation numbers of chlorine atoms in bleaching powder is _________ .
Answer:
-1

Question 8
SO₂ can act as
a) Oxidising agent only
b) Reducing agent only
c) Both oxidising and reducing agents
d) Acid and a reducing agent only
Answer:
c) Both oxidising and reducing agents

Question 9.
In the reaction
2KMnO₄ +16HCl → 5Cl₂ + MnCl₂ + 2KCl + 8H₂O the reduction product is _________ .
Answer:
MnCl₂

Question 10.
The strongest reducing agent is .
a) K
b) Ba
c) Li
d) Na
Answer:
c) Li

Question 11.
Oxidation state of oxygen in H₂O₂ is _________ .
Answer:
-1

Plus One Chemistry Redox Reactions Two Mark Questions and Answers

Question 1.
Balance the following equation using oxidation number method:
MnO₂ + Cl^– → Mn²⁺ + Cl₂
Answer:
Assigning oxidation numbers:

Equating the increase and decrease in oxidation number:
MnO₂ + 2Cl^– → Mn²⁺ + Cl₂
Balancing hydrogen and oxygen atoms:
MnO₂ + 2Cl^– + 4H⁺ → Mn²⁺ + Cl₂ + 2H₂O

Question 2.
Balance the following equation using half reaction method:
Cu + NO₃^– → Cu²⁺ + NO₂
Answer:
Separating into half reactions:
Oxidation half: Cu → Cu²⁺
Reduction half: NO₃^– → NO₂
Balancing oxygen and hydrogen atoms:
NO₃^– + 2H⁺ → NO₂ + H₂O
Balancing charge by adding electrons and making the number of electrons equal in the two half reactions:
Cu → Cu²⁺ + 2e^–
2NO₃^– + 4H⁺ + 2e^– → 2NO₂ + 2H₂O
Adding the two half reactions to achieve the overall reaction:
Cu + 2NO₃ + 4H⁺ → Cu²⁺ + 2NO₂ + 2H₂O

Question 3.
Complete the following ionic equations:

1. Al³⁺ + 3e^– → …………….
2. MnO₄^2- → + e^–
3. K → K⁺ + ……………
4. Fe²⁺ → Fe³⁺ +

Answer:

1. Al³⁺ + 3e → Al
2. MnO₄^2- → MnO₄^–+ e^–
3. K → K⁺ + e^–
4. Fe²⁺ → Fe³⁺ + e^–

Question 4.
Find the oxidation number of P in the following compounds:

1. Na₂PO₄
2. H₃P₂O₇
3. PH₃
4. H₃PO₄

Answer:

1. Na₂PO₄, Oxidation state of P = +6
2. H₄P₂O₇, Oxidation state of P = +5
3. PH₃, Oxidation state of P = -3
4. H₃PO₄, Oxidation state of P = +5

Question 5.
Choose the correct oxidation number of sulphur in the compounds in column Afrom column B.

Column A	Column B
Na2SO4	-2
H2sO3	+7
H2S	+6
H2S2O7	+4

Answer:

Column A	Column B
Na2SO4	+6
H2SO3	+4
H2S	-2
H2S2O8	+7

Question 6.
Explain oxidation number and valency.
Answer:
Valency of an atom is its combining capacity and is denoted by a number without sign. The valency of an element is always a whole number.

Oxidation number is a net charge which an atom has or appears to have when the other atoms from the molecule are removed as ions assuming that the shared pair of electrons is with more electronegative atom.

Question 7.
Some rules related to oxidation number are given below. Correct the mistakes.

• Oxidation number of alkali metals and alkaline earth metals is +2.
• Oxidation number of hydrogen is always +1.
• Algebraicsum of oxidation number of all the atoms in an ion is not equal to the charge on the ion.

Answer:

• Oxidation number of alkali metals is +1.
• Oxidation number of alkaline earth metals is +2.
• Oxidation number of H is +1 except in metallic hydrides.

Question 8.
Match the following:

Oxidation number of Cl in Cl2O7	Cu
Oxidant	Zn
Stannous Chloride, SnCl2	+7
Oxidation number of C in diamond	Get reduced easily
The metal which can’t displace H from dil.HCl	Zero
	Reducing agent for mercuric chloride

Answer:

Oxidation number of Cl in Cl2O7	+7
Oxidant	Get reduced I easily
Stannous Chloride, SnCl2	Reducing agent for mercuric chloride
Oxidation number of C in diamond	Zero
The metal which can’t displace H from dil.HCl	Cu

Question 9.
1. Calculate the oxidation number of oxygen in OF₂ and KO₂.
2. When Zn rod is dipped in blue CuSO₄ solution ‘ the blue colour of CuSO₄ fades due to displacement reaction. Write the reaction and identify the following:
i) The substance oxidised and the substance reduced.
ii) The oxidant and the reductant.
Answer:
1. OF₂: x + (-1 × 2) = 0
x – 2 = 0
x = +2
KO₂: (+1 × 1) + 2x = 0
1 + 2x = 0
2x = -1
x = –\(\frac{1}{2}\)

2. Zn(s) + CuSO₄ (aq) → ZnSO₄(aq) + Cu(s)
i) Substance oxidised – Zn
Substance reduced – Cu
ii) Oxidant-Cu
Reductant – Zn

Question 10.
a) Calculate the oxidation number of C in CH₄ and in CH₃Cl.
b) The sum of oxidation numbers of all atoms in a molecule is …………
Answer:
a) CH₄:
x + (1 × 4) = 0
x + 4 = 0
x = -4
Oxidation number of C in CH₄ is -4.

CH₃Cl:
x + (3 × 1) + -1 = 0
x + 3 – 1= 0
x + 2 = 0
x = -2
Oxidation number of C in CH₃Cl is -2.

b) Zero

Question 11.
1. Write the oxidation state of each element and identify the oxidising agent and reducing agent in the following reaction:
H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)
2. Fill in the blanks and classify the following reactions into oxidation and reduction:
i) Mn⁷⁺ + 5e^– → ……………
ii) Sn⁴⁺ + …………… → Sn²⁺
iii) Na → Na⁺ + ……………
iv) Fe³⁺ +…………… → Fe²⁺
Answer:
1.
Reducing agent – H₂S
Oxidising agent -Cl₂

2. i) Mn⁷⁺ + 5e^– → Mn²⁺
ii) Sn⁴⁺ + 2e^– → Sn²⁺
iii) Na → Na+ + e^–
iv) Fe³⁺ + e^– → Fe²⁺
Oxidation: Reaction (iii)
Reduction: Reactions (i), (ii) and (iv)

Question 12.
Dihydrogen undergoes redox reactions with many metals at high temperature.
a) Write the reaction between hydrogen with sodium.
b) Comment, whether the product formed, is covalent compound or ionic compound.
c) Which is the reducing agent in this reaction?
Answer:
1. 2Na + H₂ → 2NaH
2. Ionic compound is formed. When alkali metals react with hydrogen ionic hydrides are formed.
3. Na is the reducing agent.

Question 13.
1. Is it possible to keep copper sulphate solution in zinc pot? Why?
2. Assign oxidation numbers of the underlined elements.
i) NaH₂\(\underline { P } \)O₄
ii) NaH\(\underline { S } \)O₄
Answer:
1. No. Zn being more reactive will displace Cu from CuSO₄. Thus Cu will be deposited on the vessel.

2. i) NaH₂\(\underline { P } \)O₄
+1 +(+1 × 2) + x +(-2 × 4) = 0
1 + 2 + x – 8 = 0
x – 5 = 0
x = +5
ii) NaH\(\underline { S } \)O₄
+1 + 1 + x +(-2 × 4) = 0
+1 + 1 + x – 8 = 0
x – 6 = 0
x = +6

Question 14.
Identify the substance oxidised, reduced, oxidising agent and reducing agent in the reaction:
2Cu₂O + Cu₂S → 6Cu + SO₂
Answer:

In this reaction, Cu is reduced from +1 state to zero. oxidation state and S is oxidised from -2 state to +4 state. Cu₂O helps S in Cu₂S to increase its oxidation number. Therefore, Cu(l) is the oxidising agent. S of Cu₂S helps Cu both in Cu₂S itself and Cu₂O to decrease its oxidation number. Therefore, S of Cu₂S is the reducing agent.

Question 15.
Explain the following in terms of electron transfer concept:

1. Oxidation
2. Reduction
3. Oxidising agent
4. Reducing agent

Answer:

1. Oxidation: Loss of electron(s) by any species.
2. Reduction: Gain of electron(s) by any species.
3. Oxidising agent: Any species which accepts electrons).
4. Reducing agent: Any species which donates electron^).

Question 16.
Represent the following compounds using Stock notation:
Cu₂O, SnCl₄, MnO, Fe₂O₃, V₂O₅
Answer:

Question 17.
In a redox reaction, oxidation and reduction occur simultaneously.
a) Write the classical concept of oxidation and reduction.
b) Identify the species undergoing oxidation and reduction in the following reaction:
H₂S(S) + Cl₂(g) → 2HCl(g) + S(s)
Answer:
1. Oxidation:
addition of oxygen/electronegative element to a substance or removal of hydrogen/ electropositive element from a substance.

Reduction:
removal of oxygen/electronegative element from a substance or addition of hydrogen/ electropositive element to a substance.

2.Oxidised species:
H₂S. This is because a more electronegative element, Cl is added to H or a more electro positive element, H has been removed from S.

Reduced species:
Cl. This is due to addition of more electropositive element H to it.

Plus One Chemistry Redox Reactions Three Mark Questions and Answers

Question 1.
An equation is given below:
HNO₃+ l2 → HlO₃ + NO₂ + H₂O

• Find the oxidising agent and reducing agent.
• Balance the equation using half reaction method.

Answer:
Oxidising agent = HNO₃
Reducing agent = l₂
Skeletal equation:

Balancing the charge on the half reactions by adding electrons and equalising the number of electrons:

Question 2.
1. Define redox reactions.
2. Predict whether the following reaction is a redox reaction or not? Justify.
Cr₂O₇^2- + H₂O → 2CrO₄^2- + 2H⁺
Answer:
1. Redox reactions are those reactions are those reactions in which reduction and oxidation occur simultaneously. These reactions involve change in oxidation state of the interacting species.

2. No.
Because no element undergoes change in oxidation number.

Question 3.
a) Find out the oxidising agent and reducing agent in the following reaction:
Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)
b) Balance the following redox reaction in acid medium using oxidation number method.
Cr₂O₇^2- + Fe²⁺ → Cr³⁺ + Fe³⁺
Answer:
1. Oxidising agent – Ag
ReducingAgent – Cu

2. Assigning oxidation number:

Question 4.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₄, Cr₂O₇^2- and NO₃^–.
Answer:
H₂SO₄
(2 × +1) + x+ (4 × -2) = 0
+2 + x – 8 = 0
x – 6 = 0
x = +6

Cr₂O₇^2-
2x + 7 ×-2 = -2
2x = -2 + 14
2x = 12
∴ x = +6

NO₃^–
x + 3 × -2 = -1
x = -1 + 5
x = +4

Question 5.
1. Assign oxidation numbers
(i) P in NaH₂PO₄
(ii) Mn in KMnO₄
(iii) B in NaBH₄
(iv) S in H₂SO₄
2. Identify the oxidising and reducing agents in the following reaction:
CuO + H₂ → Cu + H₂O
Answer:
1. i) NaH₂PO₄
Na⁺¹H₂⁺¹PO₄^-2
1+2 + x- 8 = 0
3 + x – 8 = 0
x – 5 = 0
x =+ 5

ii) K⁺¹MnO₄^-2
1+ x – 8 = 0
x – 7 = 0
x = +7

iii) Na⁺¹BH₄⁺¹
1 + x + 1 × 4 = 0
x + 5 = 0
x = -5

iv) H₂⁺¹SO₄^-2
2 + x – 8 = 0
x – 6 = 0
x = +6

2.

Question 6.
A copper rod is dipped in silver nitrate solution.

1. What are the observations?
2. Write the displacement reaction.
3. Identify the species getting oxidised and reduced.

Answer:

1. The colour of the solution changes to blue. Silver is deposited on the copper rod.
2. Cu(s) +2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
3. Oxidised species – Cu Reduced species – Ag⁺

Question 7.
1. Identify the oxidising and reducing agent in the reaction:
CuS + O₂ → Cu + SO₂
2. Determine the oxidation number of the underlined element in the following:

Answer:

Question 8.
1. Identify the substance oxidised, substance reduced, oxidising agent and reducing agent in the reaction:
Cl₂ + 2l^– → 2Cl^– +l₂

2. Calculate the oxidation number of underlined elements in the following compounds:
i) K₂\(\underline { Cr } \)₂O₇
ii) H\(\underline { H } \)O₃
Answer:
1. Cl₂ is reduced, therefore Cl₂ is the oxidising agent. I’ is oxidised, therefore I” is the reducing agent.

2. i) K₂\(\underline { Cr } \)₂O₇
(+1 × 2) + 2x +(-2 × 7) = 0
+2 + 2x – 14 =0
2x – 12 =0
2x = 12
x = +6
ii) H\(\underline { H } \)O₃
(+1 × 1) + x + (-2 × 3) = 0
1+ x – 6 = 0
x – 5 = 0
x = +5

Question 9.
Determine oxidation number of the elements underlined in each of the following.

Answer:

Plus One Chemistry Redox Reactions Four Mark Questions and Answers

Question 1.
Permanganate ion (MnO₄^–) reacts with bromide ion (Br^–) in basic medium to give manganese dioxide
(MnO₂) and bromate ion (BrO₃^–).
a) Write the balanced ionic equation for this reaction.
b) Identify the oxidising agent and reducing agent in this reaction.
Answer:

Question 2.
A redox reaction involves oxidation and reduction.
a) What do you understand by electrode potential?
b) Define a redox couple.
c) Explain the set-up for Daniell cell with a diagram.
d) Write the electrode reactions and overall cell reaction which occur in the Daniel cell.
Answer:
a) The potential difference between metal and its own ion is called electrode potential.

b) A redox couple is defined as the combination of oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction.

c) Take copper sulphate solution in a beaker and put a copper strip. Take zinc sulphate solution in another beaker and put a zinc rod. The two redox couples are represented as Zn²⁺/Zn and Cu²⁺/Cu. Put the beaker containing copper sulphate solution and beaker containing zinc sulphate side by side. Connect two solution by a salt bridge. The Zn and Cu rods are connected by a metalic wire with a provision for ammeter and switch. Transfer of electrons now does not take place directly from Zn to Cu²⁺, but through metallic wire. The electricity from solution in one beaker to solution in the other beaker flows by the migration of ions through the salt bridge.

Question 3.
Redox reactions are those in which oxidation and reduction takes place. Explain the different types of redox reactions with suitable examples.
Answer:
Combination Reactions: The reactions in which two substances combine together to form a new compound are called combination reactions. These can be denoted as A+ B → C where either A or B or both A and B should be in the elemental form.

Decomposition reactions:
The reactions in which a compound breaks up into two or more substances at least one of which is in elemental form are called decompositions reactions.

Displacement reactions:
The reactions of the type X + YZ → XZ + Y in which an atom or ion in a compound is displaced by an ion (atom) of another element, such that X and Y are in elemental form are called displacement reactions. They are of two categories:
1. Metal displacement reactions: Reactions in which a more electropositive metal displaces a less electropositive metal from its compound.

2. Non-metal displacement reactions: These are reactions in which a non-metal is displaced by another metal or non-metal.

Disproportionation reactions: These are special type of redox reactions in which an element in one oxidation state is simultaneously oxidised and reduced. Here one of the reactants should contain an element that should exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction.
e.g. The decomposition of hydrogen peroxide.

Here the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O₂ and to -2 state in H₂O.

Plus One Chemistry Redox Reactions NCERT Questions and Answers

Question 1.
Fluorine reacts with ice and results in the change :

Justify that this reaction is a redox reaction.
Answer:
In the given reaction O.N. of F₂ changes from zero to -1 in HF and HOF whereas O.N. of oxygen change from -2 in H₂O to zero in HOF. Thus, F₂ is reduced, whereas oxygen is oxidised and, therefore, it is a redox reaction.

Question 2.
Write formulas for the following compounds:

1. Mercury (II) chloride
2. Nickel (II) sulphate
3. Tin (IV) oxide
4.  Thallium (I) sulphate
5. Iron (III) sulphate
6. Chromium (III) oxide

Answer:

1. Hg(II)Cl₂
2. Ni(II)SO₄
3. Sn(IV)O₂
4. Tl₂(I)SO₄
5. Fe₂(III)(SO₄)₃
6. Cr₂(III)O₃

Question 3.
The compound AgF₂ is unstable. However, if formed, the compound acts as a very strong oxiding agent. Why?
Answer:
In AgF₂, oxidation state of Ag is + 2 which is very unstable. Since Ag can exist in a stable state of + 1 it quickly accepts an electron to form the more stable + 1 oxidation state.
Ag²⁺ + e^– → Ag⁺

Question 4.
Consider the reactions:

Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
The average O.N. of S in S₂O₃^2- is + 2 while in S₄O₆^2- it is + 2.5. The O.N. of S in SO₄^2- is+6. Since Br₂ is a stronger oxidising agent that l₂, it oxidises S of S₂O₃^2- to a higher oxidation state of + 6 and hence forms SO₄^2- ion. l₂, however, being a weaker oxidising agent oxidises S of S₂O₃^2- ion to a lower oxidation of + 2.5 in S₄O₆^2- ion.

Question 5.
Why does the following reaction occur?
XeO₆^4-(aq) + 2F^–(aq) + 6H⁺(aq) → XeO₃(s) + F₂(g) + 3H₂O(I)
What conclusion about the compound Na₄XeO₆ (of which XeO₆^4- is a part) can be drawn from the reaction?
Answer:
The balanced equation along with O.N. of the elements above their symbols will be as:

In the equation the, O.N. of Xe decreases from + 8 in XeO₆^4- to + 6 in XeO₃ while that of F increases from – 1 in F^– to 0 in F₂. Therefore, XeO₆^4- is reduced while F^– is oxidised. This reaction occurs because Na₄XeO₆ (0r XeO₆^4-) is stronger oxidising agent than F₂.

Students can Download Chapter 12 Thermodynamics Questions and Answers, Plus One Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Chapter Wise Questions and Answers Chapter 12 Thermodynamics

Plus One Physics Thermodynamics One Mark Questions and Answers

Question 1.
If the pressure and the volume of certain quantity of ideal gas are halved, then its temperature
(a) is doubled
(b) becomes one-fourth
(c) remains constant
(d) is halved
Answer:
(b) becomes one-fourth
According to ideal gas law

Question 2.
The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T) is

where k_B is the Boltzmann constant and N is the number of molecules.
Answer:
c) PV = k_BNT
According to ideal gas equation,
PV = Nk_BT.

Question 3.
Which of the following laws of thermodynamics forms the basis for the definition of temperature?
(a) First law
(b) Zeroth law
(c) Second law
(d) Third’law
Answer:
(b) Zeroth law
The Zeroth law of thermodynamics gives the definition of temperature.

Question 4.
Which of the following is a true statement?
(a) The total entropy of thermally interacting systems is conserved
(b) Carnot engine has 100% efficiency
(c) Total entropy does not change in a reversible process.
(d) Total entropy in an irreversible process can either increase or decrease.
Answer:
(c) Total entropy does not change in a reversible process.

Question 5.
In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas
(a) the temperature will decrease,
(b) the volume will increase
(c) the pressure will remain constant
(d) the temperature will increase
Answer:
(a) From first law of thermodynamics
dQ = dU + dW
dQ = dU (if dW = 0)
Since, dQ < 0
dU < 0
or U_final < U_intial
Hence, temperature will decrease.

Question 6.
An electric fan is switched on in a closed room will the air of the room be cooled?
Answer:
No. Infact speed of air molecules will increase and this results in increase in temperature.

Question 7.
When an iron nail is hammered, if becomes hot. Why?
Answer:
The kinetic energy of hammer gets converted in to heat energy which increases temperature of iron nail.

Question 8.
Identify the thermodynamic process in which temperature of system may increase even when no heat is supplied to the system.
Answer:
Adiabatic process.

Question 9.
Which thermodynamic variable is defined by first law of thermodynamics?
Answer:
Internal energy

Question 10.
The door of an operating refrigerator kept open in a closed room. Will it make the room cool?
Answer:
No. The room will be slightly warmed.

Plus One Physics Thermodynamics Two Mark Questions and Answers

Question 1.
A Carnot engine working between 300K and 600K has a work output of 800J per cycle. Find the amount of energy consumed per cycle

• 800J
• 400J
• 1600J
• 1200J
• 3200J

Answer:
Efficiency of carrots engine

Efficiency also can be written as

∴ input power = 1600 J.

Question 2.
A carnote’s engine is made to work between 200°C and 0°C first and then between 0°C and minus 200°C. Compare the values of efficiencies in the two cases.
Answer:
Case -1:

= 1 – 0.577
= 0.42 = 42%
Case – 2:
η = 1 – \(\frac{73}{473}\)
T2 = -200 = 73k
= 1 – 0.26 = 0.73
= 73%
T₁ = 0 = 273k.

Question 3.
Match the following

Answer:

Question 4.
What is the value of specific heat capacity of gas in

1. Isothermal process
2. Adiabatic process

Answer:

1. Infinite
2. Zero

Question 5.
Is the function of refrigerator against the second law of thermodynamics? Explain.
Answer:
No. The refrigerator transfers heat from the inside space to outer atmosphere, at the expense of external work supplied by the compresser fed by electric supply.

Plus One Physics Thermodynamics Three Mark Questions and Answers

Question 1.
Heat is supplied to a system, but its internal energy does not increase

1. Which process is involved in this case? (1)
2. Obtain an expression for the work done in the above process (2)

Answer:
1. Isothermal process.

2. Work done by adiabatic process
Let an ideal gas undergoes adiabatic charge from (P₁, V₁, T₁) to (P₂, V₂, T₂). The equation for adiabatic charge is
PV^γ = constant = k
ie; P₁V₁^γ = P₂V₂^γ = k _____(a)
The work done by

from equation (a) P₁V₁^γ = P₂V₂^γ = k

Substituting ideal gas equation.

Question 2.
A heat engine is a device which effectively converts heat energy into mechanical energy.

1. State the law which describes this principle. (1)
2. Derive an expression for the efficiency of a Carnot engine. (2)

Answer:
1. Kelvin – Plank statement:
No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of heat into work.

Clausius statement:
No process is possible whose sole result is the transfer of heat from a colder object to hotter object.

2. Carnot’s cycle:
The Carnot cycle consists of two isothermal processes and two adiabatic processes.

Let the working substance in Carnot’s engine be the ideal gas.
Step 1: The gas absorbs heat Q₁ from hot reservoir at T₁ and undergoes isothermal expansion from (P₁, V₁, T₁) to (P₂, V₂, T₁).

Step 2: Gas undergoes adiabatic expansion from (P₂, V₂, T₁) to (P₃, V₃, T₂)

Step 3: The gas release heat Q₂ to cold reservoir at T₂, by isothermal compression from (P₃, V₃, T₂) to (P₄, V₄, T₂).

Step 4: To take gas into initial state, work is done on gas adiabatically [(P₄, V₄, T₂) to (P₁, V₁, T₁)]. Efficiency of Carnot’s engine:

Question 3.
A thermo flask contains coffee. It is violently shaken. Considering the coffee as a system:

1. does its temperature rise?
2. has heat been added to its?
3. has internal energy changed?

Explain your answers.
Answer:

1. Yes
2. Heat is added to coffee
3. Internal energy is changed
4. When we shake, the mechanical energy is added to the liquid contained in a flask.

Question 4.
It is predicted by the meteorologists that global warming will result in the flooding of oceans due to the melting of ice caps on the earth.

1. Name the thermodynamic process involved in the melting of ice.
2. Determine the heat energy required to melt 5kg of ice completely at 0°C. Latent heat of fusion of water is 3336 × 10³ J Kg^-1.
3. During the melting process, what change will occur to its internal energy?

Answer:

1. Isothermal process
2. ∆Q = mL = 5 × 33.36 × 10⁵ = 16.68 × 10⁶J
3. Internal energy increases

Question 5.
Water is heated in an open vessel.

1. This process is
   • isothermal
   • isobaric
   • isochoric
   • adiabatic
2. Which law of thermodynamics is suitable to explain the transfer of heat here?
3. Draw the heat-temperature graph of ice below 0°C heated up to steam above 100°C.

Answer:
1. Isobaric
2. First law of thermodynamics
3.

Question 6.

1. The molar heat capacity of oxygen at constant volume is 20J mol^-1K^-1. What do you mean by molar heat capacity at constant volume (C_V)?
2. The difference between C_P and C_V is always a constant. Give a mathematical proof.

Answer:
1. Molar specific heat at constant volume is the heat required to raise the temperature of I mol substance by IK at constant volume.

2. According to 1^st law of thermodynamics
∆Q = ∆U + PAV
If ∆Q heat is absorbed at constant volume (∆V = 0).

From ideal gas equation for one mole PV = RT. Differentiating w.r.t. temperature (at constant pressure

Equation (4) – Equation (1), we get
C_P – C_V = R.

Question 7.
A heat engine is a device which converts heat energy into work.

1. What is the working substance in an ideal heat engine.
2. A Carnot engine is working between in melting point and steam point. What is its efficiency?

Answer:
1. Ideal gas.

2. T₂ = 0°C = 273K
T₁ = 100°C = 373K
Efficiency η = I – \(\frac{T_{2}}{T_{1}}=I-\frac{273}{373}\)
η = 0.27
η = 27%.

Question 8.
A gas is taken in a cylinder. The walls of the cylinder is insulated from the surroundings

1. The gas in a cylinder is suddenly compressed. Which thermo dynamic process involves in this statement. Explain the thermo dynamic process.
2. If gas is suddenly compressed to 1/4^th of its, original volume. Calculate the rise in temperature. The initial temperature was 27°C and γ = 1.5.

Answer:
1. Adiabatic. A thermo dynamic process in which no heat enters or leaves the system.

2. T. = 27°C = 27 + 273 = 300k

= 300 × 4^{1.5 – 1}
= 300 × 4^1/2
= 600K
rise in temperature = 600K – 300K = 300K.

Question 9.
Raju brought a motor pump from a shop. The efficiency of the motor pump is printed on the label as 60%.

1. The efficiency of a water pump is 60%. What is . meant by this?
2. Can you design an engine of 100% efficiency? Justify your answer.

Answer:
1. Efficiency means that, 60% of the total energy received is converted into useful work.

2. The efficiency of heat engine η = 1 – \(\frac{T_{2}}{T_{1}}\). The efficiency will be 100% if T₂ = OK, both those conditions cannot be attained practically. So art engine can’t have 100% efficiency.

Question 10.
Categorise into reversible and irreversible process

1. Waterfall
2. rusting of iron
3. electrolysis
4. slow compression
5. diffusion of gas
6. melting of ice
7. dissolving NaCl in water
8. flow of heat from hot body to cold body
9. slow compression of spring
10. heat produced by friction.

Answer:
Reversible process → (3), (4), (6) and (9)
Irreversible process → (1), (2), (5), (7), (8) and (10)

Plus One Physics Thermodynamics NCERT Questions and Answers

Question 1.
A geyser heats water flowing at the rate of 3.0 litres per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 10⁴ Jg^-1?
Answer:
Volume of water heated = 3.0 litre per minute
mass of water heated, m = 3000 g per minute
increase in temperature, ∆T = 77°C – 27°C = 50°C
specific heat of water, c= 4.2Jg^{-1 °}C^-1
amount of heat used, Q = mc ∆T
or Q = 3000 g min^-1 × 4.2Jg^-1
°C^-1 × 50°C
= 63 × 10⁴J min^-1
rate of combustion of fuel = \(\frac{63 \times 10^{4} \mathrm{Jmin}^{-1}}{4.0 \times 10^{4} \mathrm{Jg}^{-1}}\)
= 15.75gmin^-1.

Question 2.
What amount of heat must be supplied to 2.0 x 10^-2 kg of nitrogen (at room temperature) to raise its temperature by 45°C at constant pressure? (Molecular mass of N₂ = 28, R = 8.3 J mol^-1 K^-1).
Answer:
m = 2 × 10³ kg^-1 = 20g,
∆T = 45°C, M = 28
R = 8.3 J mol^-1 K^-1, M = 28
Number of moles, n = \(\frac{m}{M}=\frac{20}{28}=\frac{5}{7}\)
Since nitrogen is diatomic,
∴ C_P = \(\frac{7}{2}\)
R = \(\frac{7}{2}\) × 8.3 J mol^-1K^-1
Now, ∆Q = nC_P ∆T
= \(\frac{5}{2}\) × \(\frac{7}{2}\) × 8.3 × 45J = 933.75J.

Question 3.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas, is compressed to half its original volume?
Answer:
P₂V₂^γ = P₁V₁^γ

Question 4.
A steam engine delivers 5.4 × 10⁸ J of work per minute and services 3.6 × 10⁹J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Answer:
Work done per minute, output = 5.4 × 10⁸J
Heat absorbed per minute, input = 3.6 × 10⁹J
Efficiency, η = \(\frac{5.4 \times 10^{8}}{3.6 \times 10^{9}}\) = 0.15
%η = 0.15 × 100 = 15
Heat energy wasted/ minute = Heat energy absorbed/minute – Useful work done/minute
= 3.6 × 10⁹ – 5.4 × 10⁸
= (3.6 – 0.54) × 10⁹ = 3.06 × 10⁹J.

Question 5.
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joule per second, at what rate is the internal energy increasing?
Answer:
Heat supplied, ∆Q = 100W = 10OJs^-1
Useful work done, ∆W = 75J s^-1
Using first law of thermodynamics
∆Q = ∆U+ ∆W
∆U = ∆Q – ∆W
= 100Js^-1 – 75Js^-1 = 25Js^-1.

Question 6.
A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.
Answer:
T₁ = 36°C = (36 + 273) K = 309 K
T₂ = 9°C = (9 + 273) K = 282 K
Coefficient of performance = \(\frac{T_{2}}{T_{1}-T_{2}}\)

Question 7.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0°C). Show that it is 22.4 litres.
Answer:
PV = µRT or V = \(\frac{\mu \mathrm{RT}}{\mathrm{P}}\)

= 22.4 × 10^-3 m³ = 22.4 litre.

Students can Download Chapter 1 Biological Classification Questions and Answers, Plus One Botany Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

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

Plus One Botany Biological Classification One Mark Questions and Answers

Question 1.
In Whit takers, five-kingdom classification eukaryotes are distributed among
(a) two kingdoms
(b) three kingdoms
(c) four kingdoms
(d) all the five kingdoms
Answer:
(c) four kingdoms

Question 2.
Cyanobacteria are classified under which of the following kingdoms?
(a) Monera
(b) Protista
(c) Plantae
(d) Algae
Answer:
(a) Monera

Question 3.
Main component of cell wall of fungi is
(a) cellulose
(b) chitin
(c) pectin
(d) silica
Answer:
(b) chitin

Question 4.
Dinoflagellates are mostly
(a) marine and saprophytic
(b) freshwater and saprophytic
(c) marine and photosynthetic
(d) terrestrial and
Answer:
(c) marine and photosynthetic

Question 5.
Which of the following kingdoms do viruses belong to
(a) monera
(b) Protista
(c) fungi
(d) none of these
Answer:
(d) none of these

Question 6.
Observe the relationship between the first pair and fill up the blanks.

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

Answer:
Ethylene.

Question 7.
Fill in the blanks.

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

Answer:

1. Ascomycetes
2. N₂ fixation

Question 8.
Who proposed Five kingdom classification?
Answer:
R .H. Whittaker

Question 9.
Find out the correct sequence of taxonomical category.

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

Answer:
2. species → genus → order → phylum

Question 10.
In the five-kingdom system of Whittaker, how many kingdoms are eukaryotes?
Answer:
Four kingdoms

Question 11.
Observe the relationship between the first pair and fill up the blanks.

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

Answer:

1. Archaebacteria
2. Ascomycetes

Question 12.
Find out the odd one.
a. Diatom, Gonyaulax, Yeast, Euglena, Plasmodium
Answer:
Yeast

Question 13.
Vinod observed blooms in a polluted water body, his friend Kumar said that it might be nitrogen-fixing Nostoc or Anabaena. Can you suggest which type of cell can fix atmospheric nitrogen in these organisms?
Answer:
Heterocyst

Question 14.
Observe the relationship of the terms in the first pair and fill in the blanks:

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

Answer:

1. Bacillus
2. Ascomycetes

Question 15.
Difference between Virus and Viroid.
(a) Absence of protein coat in viroid but present in virus
(b) Presence of low molecular weight RNA in virus but absent in viroid
(c) Both a and b
(d) None of the above
Answer:
(a) Absence of protein coat in viroid but present in virus

Question 16.
Viruses are non-cellular organisms but replicate themselves once they infect the host cell. To which of the following kingdom do viruses belong to?
(a) Monera
(b) Protista
(c) Fungi
(d) None of the above
Answer:
(d) None of the above

Question 17.
A virus is considered as a living organism and an obligate parasite when inside a host cell. But virus is not classified along with bacteria or fungi. What are the characters of virus that are similar to nonliving objects?
Answer:
Viruses are acellular and can be crystallized.

Plus One Botany Biological Classification Two Mark Questions and Answers

Question 1.
The seven taxonomic categories are given below. Arrange them in the correct sequence starting from the smallest taxon.
Class → species → kingdom → order → family → division → genus.
Answer:
Species → genus → family → order → class → division → kingdom.

Question 2.
“Two kingdom classification is inadequate one”. Comment on it.
Answer:

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

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

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

Question 4.
Name the following;

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

Answer:

1. Euglena
2. Slime mould

Question 5.
State two economic importance of

1. Heterotrophic bacteria
2. Archaebacteria

Answer:

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

Question 6.
What is the nature of cell walls of diatoms?
Answer:
Cell walls are made up of silica with two overlapping shells fit together like a soapbox.

Question 7.
Find out what do the terms algal blooms and red tides signify?
Answer:

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

Question 8.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify.
Answer:
1. Algal bloom’: When colour of water changes due to profuse growth of coloured phytoplanktons, it is called algal bloom.

2. ‘Red tides: Redness of the red sea is due to the luxuriant growth of Trichodesmium erythrium, a member of cyanobacteria (blue-green algae)’

Question 9.
How are viroids different from virus?
Answer:
Viroids are free RNA without protein coat. Viruses have protein coat which encloses either RNA or DNA.

Question 10.
Justify the physiological relationship between the algal and fungal component of lichen.
Answer:
The fungus holds water, provides protection and ideal housing to the alga. The alga supplies carbohydrate food for the fungus. If the alga is capable of fixing nitrogen, it supplies fixed nitrogen to fungus. This association is called symbiosis.

Question 11.
Bacteria reproduce by various methods. Mention the type of reproduction given in the diagram. What are the other methods of reproduction occur in bacteria?

Answer:
Binary fission
The other methods are sporulation and sexual reproduction.

Question 12.
Biological classification is essential. Comment.
Answer:
The animals and plants vary greatly in their form, structure and mode of life. To find out an organism of known characters from the vast number of organism is simply impossible. So classification is important to divide into groups and subgroups.

Question 13.
Match the following:

a. Produces a plant disease	p. Saccharomyces cere visae
b. is edible- light blight of potato.	q. Phytophthora infestans
c. is a source of antibiotic	r. Agaricus campestris
d. is used in the manufacture of ethanol	s. Penicillium notatum

Answer:

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

Question 14.
Plants are autotrophs. Can you think of some plants that are heterotrophs?
Answer:
Generally all plants are autotrophs but plants like loranthus and cuscuta absorbs water & nutrients from other plants so they are called as heterotrophs.

Question 15.
What are the characteristic features of Euglenoides?
Answer:
They have protein sheath is called pellicle instead of cell wall. They have two flagella – One long and other short. They are photosynthetic in the presence of light and behave as heterotrophs in the absence of sunlight.

Question 16.
Give 4 difference between Ascomycetes and Basidiomycetes:
Answer:

Ascomycetes	Basidiomycetes
1. Mycelium consists of branched multicellular septate hyphae.	1. Mycelium may be primary, secondary (or) tertiary
2. The fruiting bodies are ascocarps	2. Fruiting bodies are basidiocarps.
3. Sexual reproduction leads to the formation of ascus	3. Formation of basidia formation of ascus.

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

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

Answer:

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

Question 18.
What do the terms phycobiont and mycobiont signify?
Answer:
Algal component of lichen is called phycobiont. It prepares food for fungus. Fungal partner is called mycobiont. It provides shelter and absorbs mineral nutrients for algae.

Question 19.
Prepare a comparative account of different classes of kingdom fungi by considering following statements.
Answer:

1. Mode of nutrition
2. Mode of reproduction

Question 20.
The two-kingdom classification is introduced by Linnaeus. Why is the two kingdom classification inadequate?
Answer:
There was no place of viruses and bacteriophages which can neither be considered as prokaryotes not eukaryotes.

In this classification, eukaryotes were put together with prokaryotes and non-photosynthetic fungi along with photosynthetic plants.

Question 21.
How is the five-kingdom classification advantageous over the two kingdom classification?
Answer:
In this classification main criteria used by R H Whittaker include cell structure, thallus organisation, mode of nutrition, reproduction and phylogenetic relationships. These characters were not considered in two kingdom classification.

Question 22.
Are chemosynthetic bacteria-autotrophic or heterotrophic?
Answer:
Autotrophic, because they get energy from the oxidation of inorganic compounds. So the released energy is stored in the ATP molecules.

Question 23.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Answer:
Cyanobacteria and other photosynthetic bacteria have thylakoids suspended freely in the cytoplasm (i.e., they are not enclosed in membrane), and they have bacteriochlorophyll

Question 24.
With respect to fungal sexual cycle, choose the correct sequence of events.
Answer:

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

Question 25.
What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement?
Answer:
It is due to the presence of special nitrogen-fixing cell called heterocyst present between the filaments. So it helps to increase N2 content in the soil.

Question 26.
Methane is the main component of biogas and it is produced by bacteria.

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

Answer:

1. Methanogens
2. Archaebacteria

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

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

Answer:

1. Deuteromycetes
2. Halophiles

Question 28.
Name the kingdom in which euglena belongs. Give the special type nutrition.
Answer:
Kingdom Protista, Mixotrophic nutrition (ie, both autotrophic and heterotrophic).

Question. 29
Some bacteria are different from others and they have the ability to survive in extreme conditions. Name it.
Answer:
Archaebacteria (halophiles, thermoacidophiles and methanogens).

Question 30.
Mycoplasma are included in five kingdom classification but not viruses. Why?
Answer:
Because mycoplasmas are living cellular organisms but viruses are acellular particles.

Question 31.
In which division of protista chief producers in ocean belongs. Give the cell wall composition of such organisms.
Answer:
Chrysophytes, silicified cell wall.

Question 32.
Nitrobactor and nitrosomonas are free living nitrogen fixers and chemoautotrophs but their functions are different. Do you agree. Give reasons.
Answer:
Yes. Nitrobactor converts nitrite into nitrate while nitrosomonas converts ammonia into nitrites.

Question 33.
Name the classes fungi shows exogenous and endogenous spore production. In which fruiting bodies they are found.
Answer:

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

Question 34.
Rust and smut diseases are caused by the members of basidiomycetes. Name it.
Answer:
Smut disease- Ustilago, Rust disease-Puccinia.

Question 35.
What are the events takes place in slime mould during favourable and unfavourable season?
Answer:
During favourable condition the cells aggregate and form plasmodium while in unfavourable season plasmodium differentiates and produce fruiting bodies that bear spores at tip.

Question. 36
Suppose you accidentally find an old preserved permanent slide without a label. In your effort to identify it, you place the slide under microscope and observe the following features

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

What would you identify it as? Can you name the kingdom it belongs to?
Answer:
Dinoflagellates, Kingdom protista

Question. 37
What would you identify it as? Can you name the kingdom it belongs to?
Answer:
Dinoflagellates, Kingdom protista

Question. 38
Why lichens are called as dual organisms?
Answer:
Lichens are said to be dual organisms because they show a symbiotic association between a fungus and alga.

Question 39.
Name the asexual, reproductive structure of penicillium and yeast.
Can penicilium reproduce through sexual method? If the yes or no Give reason.
Answer:
Conidia – penicilium, buds – yeast
Yes, It is done by the production of ascospores in asci of Ascocarp.

Question 40.
Organise a discussion in your class on the topic virus. Are viruses living or non-living?
Answer:
They are filterable and may becrystalised. They are inert outside their specific host and able to reproduce inside the living host cell, so they are considered as living. They use the protein synthesising machinery of the host.
Eg. AIDS virus, mumps virus etc.

Question 41.
How are viroids different from viruses?
Answer:

Virus	Viroid
1. Their size is smaller than bacteria	1. Their size is smaller than viruses
2. Protein coat is Present	2. Protein coat is absent
3. Genetic material may be DNA or RNA	3. Genetic material is only RNA
4. They cause AIDS, smallpox etc.	4. They cause potato spindle tuber diseases

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

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

Answer:
1. Types of bacteria

• Methanogens
• Halophiles
• Thermo acidophiles

2. Ceil wall structure

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

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

Answer:

1. Dikaryotic mycelium
2. Ascomycetes, Basidiomycetes

Question 44.
Classify the pathogenic microorganisms and disease in different groups based on the following symptoms mosaic disease, citrus canker .potato spindle tuber disease, sleeping sickness, malaria.
Answer:
mosaic disease-virus, citrus canker-Bacteria, potato spindle tuber disease -viroids, sleeping sickness- Trypanosoma, malaria-Plasmodium vivax.

Plus One Botany Biological Classification Three Mark Questions and Answers

Question 1.
Describe briefly the four major groups of protozoa.
Answer:
Protozoans are heterotrophs act either as predators
or parasites. They are of four groups

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

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

Groups	Fungi
Phycomycetes	Trichoderma
Ascomycetes	Neurospora
Basidiomycetes	Albugo
Deuteromycetes	Mucor
	Agaricus
	Ustilago
	Alternaria
	Claviceps

2. Write the distinguishing characters of ascomycetes and basidiomycetes
3. The characteristic features of members of monera are given below.

Organisms lack cell wall, live without oxygen, smallest living cell and causes diseases. Identify the organism by analysing the above characters.
Answer:
1. specific classes.

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

2. In ascomycetes, Asexual mode of reproduction is prominent by conidiospores. In Basidiomycetes asexual spores are not found. Sexual spores are arranged in ascus with Ascospores in ascomycetes, whereas sexual spores are arranged in basidium in basidiomycetes.

3. Mycoplasma

Question 3.
Give a brief account of virus with respect to their structure and nature of genetic material. Also, name four common viral diseases?
Answer:
Viruses are organism having inert crystalline structure outside the living cell. They have genetic material RNA or DNA.which is either single-stranded/double-stranded. It is enclosed by protein capsid with subunits called capsomeres.

The viral genetic material takes control over the host cell mechanism during infection. Some common viral diseases are mumps, herpes, smallpox and influenza in animals and mosaic disease in plants.

Question. 4
In which groups are the following found- Sporangiophore, Conidia, zygospore and ascospore.
Answer:

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

Plus One Botany Biological Classification NCERT Questions and Answers

Question 1.
What is the nature of cell walls in diatoms?
Answer:
The cell walls in diatoms are embedded with silica, which makes them indestructible. They form two thin overlapping shells which fit together as in a soapbox. Thus diatoms have left behind large amounts of cell wall deposits in their habitat.

Question 2.
How are viroids different from viruses?
Answer:
Viroids are free RNAs without the protein coat, while virus have a protein coat encapsulating the RNA.

Question 3.
Describe briefly the four major groups of Protozoa.
Answer:
Four major groups of Protozoa are as given below:
1. Amoeboid Protozoa:
They are found in freshwater, seawater or moist soil. They have pseudopodia, like amoeba, hence the name ameoboid protozoa.

2. Flagellated Protozoans:
They have flagella helps in locomotion. Some are parasite. Eg. Trypanosoma causes sleeping sickness.

3. Ciliated Protozoa:
They have thousands of cilia present all over the body. The cilia helps in locomotion and steering of food into the gullet.

4. Sporozoans:
Many protozoans have an infectious spore-like stage in the life cycle. The spore-like stage helps them get transferred from one host to another host.

Question. 4
Plants are autotrophic. Can you think of some plants that are partially heterotrophic?
Answer:
Certain insectivorous plants, like bladderwort and venus fly trap, are partially heterotrophic.

Question. 5
What do the terms phycobiont and mycobiont signify?
Answer:
Lichens are good examples of symbiotic life of algae and fungi. Phycobiont is the name of the part composed of algae and Mycobiont is the name of the part composed of fungi. Fungi provide minerals and support to the alage, while algae provide nutrition to the fungi.

Question 6.
What are the characteristic features of Euglenoids?
Answer:
Features of Euglenoids.

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

Question 7.
Give a brief account of viruses with respect to their structure and nature of genetic material. Also name four common viral diseases.
Answer:
Virus Structure:
Outside a host cell, virus is a crystalline structure, composed of protein. Inside the crystal, there is genetic material, which can be either RNA or DNA. No virus has both RNA and DNA. Viruses, infecting plants, have single-stranded RNA. Viruses, infecting animals, have either single or double-stranded RNA or double-stranded DNA.

The protein coat is called capsid. Capsid is made of smaller subunits, called capsomeres, it protects nucleic acid. Diseases caused by Virus; AIDS, Mumps, Influenza, Herpes.

Question 8.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify.
Answer:
Dinoflagellates can be of different colours depending on the type of pigment present. The red dinoflagellate sometimes multiplies at a very rapid rate. This is called as algal bloom. This gives a red appearance to the part of affected sea. This is also known as ‘red tide’. Toxins released by them can kill other marine species.

Plus One Botany Biological Classification Multiple Choice Questions and Answers

Question 1.
The life form used as indicators of pollution
(A) Lichens
(B) Protozoa
(C) Algae
(D) Agaricus
Answer:
(A) Lichens

Question 2.
Kingdom monera comprises
(A) Amoeba, Bacteria,Trypanosoma
(B) Bacteria, Viruses,Virolds
(C) Archaebacteria, Eubacteria, Mycoplasma
(D) Mycoplasma, Viruses, Bacteria
Answer:
(C) Archaebacteria, Eubacteria, Mycoplasma

Question 3.
Who discovered two-kingdom classification
(A) Ivanowsky
(B) Stanley
(C) leuwernhoek
(D) Linnaeus
Answer:

Question 4.
Asexual reproduction takes place by Zoospores in
(A) Pythium
(B) Agaricus
(C) Rhizopus
(D) Ustilago
Answer:
(D) Ustilago

Question 5.
Identify the organism used as bioweapon
(A) Bacillus thuringiensis
(B) Bacillus anthracis
(C) Pseudomonas citri
(D) Rhizobium tumefacient
Answer:
(B) Bacillus anthracis

Question 6.
Reserve food in the form of glycogen and cell wall made up of chitin are characteristic of
(A) Protists
(B) bacteria
(C) Fungi
(D) protozoa
Answer:
(C) Fungi

Question 7.
The fruiting body of club fungi is
(A) Basidium
(B) Ascus
(C) Ascocarp
(D) Basidiocarp
Answer:
(D) Basidiocarp

Question 8.
RNA without protein coat are found in
(A) bacteria
(B) protozoa
(C) viruses
(D) viroides
Answer:
(D) viroides

Question 9.
The phycobiont and mycobiont are found in
(A) bacteria
(B) lichen
(C) viroides
(D) fungi
Answer:
(B) lichen

Question 10.
The organism which causing sleeping sickness belongs to
(A) Protists
(B) bacteria
(C) Fungi
(D) viruses
Answer:
(A) Protists

Question 11.
In which of the following groups are neurospora and Penicillium included?
(A) Phycomycetes
(B) Basidiomycetes
(C) Zygomycetes
(D) Ascomycetes
Answer:
(D) Ascomycetes

Question 12.
Occurrence of Dikaryon phase is characteristic feature of
(A) Bacteria
(B) Fungus
(C) Slime moulds
(D) Cyanobacteria
Answer:
(B) Fungus

Question13.
Methane producers are belongs to
(A) Archaebacteria
(B) Cyanobacteria
(C) Eubactenia
(D) Actinomycetes
Answer:
(A) Archaebacteria

Question 14.
Heterocyst are found in
(A) Nitrosomonas
(B) cyanobacteria
(C) fungi
(D) protozoa
Answer:
(B) cyanobacteria

Question 15.
Colletotrichum falcatum is a fungus causing the following disease
(A) Smut of wheat
(B) Wilt disease of cotton
(C) Red rot of sugar cane
(D) Late blight of potato
Answer:
(C) Red rot of sugar cane

Students can Download Chapter 4 Principle of Mathematical Induction Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers

Question 1.
For all n ≥ 1 , prove that
1² + 2² + 3² +……….+ n² > \(\frac{n^{3}}{3}\)
Answer:
Let p(n): 1² + 2² + 3² + n²
Put n = 1 ⇒ p(1) = 1 > \(\frac{1}{3}\) which is true.
Assuming that true for p(k)
p(k): 1² + 2² + 3² +……….+ k² > \(\frac{k^{3}}{3}\)
Let p(k + 1): 1² + 2² + 3² +…….+ k² + (k + 1)² > \(\frac{k^{3}}{3}\) + (k + 1)²


Hence by using the principle of mathematical induction true for all n ∈ N.

Question 2.
For all n ≥ 1 , prove that 1 + 2 + 3 +…….+ n < \(\frac{1}{8}\)(2n + 1)²
Answer:
Let p(n): 1 + 2 + 3 +…….+ n ,
Put n = 1 ⇒ p(1) = 1 < \(\frac{9}{8}\) which is true.
Assuming that true for p(k)
p(k): 1 + 2 + 3 +…….+ k < \(\frac{1}{8}\)(2k + 1)²
Let p(k +1): 1 + 2 + 3 +……..+ k + (k +1) < \(\frac{1}{8}\) (2k + 1)² + (k + 1)

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 3.
For all n ≥ 1, prove that p(n): 2³ⁿ – 1 is divisible by 7.
Answer:
p(1): 2³⁽¹⁾ – 1 = 8 – 1 = 7 divisible by 7, hence true. Assuming that for p(k)
p(k) : 2^3k – 1 is divisible by 7.
2^3k – 1 = 7M
P(k + 1): 2^{3(k + 1)} – 1 = 2^{3k + 3} – 1
= 2^3k2³ – 1 = 2^3k × 8 – 1
= 2^3k × 8 – 8 + 7 = 8(2^3k – 1) + 7
= 8(7M) + 7
Hence divisible by 7. Therefore by using the principle of mathematical induction true for all n ∈ N.

Question 4.
For all n ≥ 1, prove that p(n): n³ + (n + 1)³ + (n + 2)³ is divisible by 9.
Answer:
p(1): 1 + 2³ + 3³ = 1 + 8 + 27 = 36 divisible by 9, hence true. Assuming that true for p(k)
p(k): k³ + (k + 1)³ + (k + 2)³ is divisible by 9.
⇒ k³ + (k + 1)³ + (k + 2)³ = 9M
p(k +1 ): (k + 1)³ + (k + 2 )³ + (k + 3)³
= (k +1)³ + (k + 2)³ + k³ + 9k² + 27k + 27
= [(k + 1)³ + (k + 2)³ + k³] + 9[k² + 3k + 3]
= 9M + 9[k² + 3k + 3]
Hence p(k + 1)divisible by 9. Therefore by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers

Question 1.
For all n ≥ 1, prove that

Answer:

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 2.
For all n ≥ 1, prove that

Answer:
Let

Assuming that true for p(k)

Let p(k + 1):

Hence by using the principle of mathematical induction true for all n ∈ N.

Question 3.
For all n ≥ 1 , prove that 1.2.3 + 2.3.4 +………+ n(n + 1)(n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\).
Answer:
Let p(n): 1.2.3 + 2.3.4 +……..+ n(n + 1)(n + 2),
Put n = 1
p(1) = \(\frac{1(1+1)(1+2)(1+3)}{4}\) = 6 which is true.
Assuming that true for p(k)
p(k): 1.2.3 + 2.3.4 +……..+ k(k + 1)(k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\),
Let p(k + 1)


Hence by using the principle of mathematical induction true for all n ∈ N.

Question 4.
For all n ≥ 1, prove that

Answer:


Hence by using the principle of mathematical induction true for all n ∈ N.

Question 5.
For all n ≥ 1 , prove that p(n): n(n + 1 )(n + 5) is divisible by 3.
Answer:
p(1): 1(1 + 1)(1 + 5) = 12divisible by 3, hence true. Assuming that true for p(k)
p(k): k(k + 1)(k + 5) is divisible by 3.
k(k + 1)(k + 5) = 3M
p(k + 1): (k + 1)(k + 2)(k + 6)
= (k + 1)(k² + 8k + 12)
= (k + 1)(k² + 5k + 3k +12)
= (k + 1)[k(k + 5) + 3(k + 6)]
= [k(k + 1)(k + 5) + 3(k + 1)(k + 6)]
= [3M + 3(k + 1)(k + 6)]
= 3[M + (k + 1)(k + 6)]
Hence divisible by 3. Therefore by using the principle of mathematical induction true for all n ∈ N.

Question 6.
For all n ≥ 1 , prove that p(n): 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24.
Answer:
p(1): 2.7¹ + 3.5¹ – 5 = 14 + 15 – 5 = 24 divisible by 24, hence true. Assuming that true for p(k)
p(k): 2.7^k + 3.5^k – 5 is divisible by 24.
⇒ 2.7^k + 3.5^k – 5 = 24M
p(k + 1): 2.7^{k + 1} + 3.5^{k + 1} – 5
= 2.7^k.7 + 3.5^k.5 – 5
= 2.7^k.(6 + 1) + 3.5^k.(4 + 1) – 5
= 12.7^k + 2.7^k + 12.5^k + 3.5^k – 5
= 12(7^k + 5^k) + (2.7^k + 3.5^k) – 5
= 12(7^k + 5^k) + 24M
7^k And 5^k are odd numbers, therefore (7^k + 5^k) will be an even and will be divisible by 24, Hence p(k + 1)divisible by 24. Therefore by using the principle of mathematical induction true for all n ∈ N.