CBSE Class 11 – Page 29

NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification

January 5, 2022June 18, 2021 by Raju

NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification.

Question 1.
Discuss how classification systems have undergone several changes over a period of time?
Solution:
The classification was born instinctively out of a need to all organisms for our own use since the dawn of civilization. Aristotle was the earliest to attempt a more scientific basis for classification. He used simple morphological characteristics to classify plants into trees, herbs and shrubs. He also divided animals into Animals with red blood and those who do not have red blood. Linnaeus proposed a two-kingdom system of classification with Plantae and Animalia including plants and animals respectively.

The above system did not distinguish eukaryotes and prokaryotes, unicellular and multicellular, photosynthetic and non-photosynthetic organisms. A need was felt for including besides gross morphology, other characteristics like cell structure, nature of cell wall, mode of nutrition, habitat, method of reproduction, evolutionary relationships etc., Recently R.H. Whittaker proposed a five-kingdom classification to answer above the Five kingdoms are

Monera
Protista
Fungi
Plantae
Animalia.

Question 2.
State two economically important uses of
(a) heterotrophic bacteria
(b) archaebacteria
Solution:
(a) Heterotropic bacteria : These bacteria are natural scavengers. The souring of milk into lactic acid and alcohol to vinegar is brought about by some saprophytic bacteria, e.g., Lactic acid bacteria and acetic acid bacteria respectively.
A number of antibiotic are extracted from actinomycetes especially from the genus Streptomyces e.g. Streptomycin, Chloramphenicol, Oilorotetracycline, Erythromycin, Terramycin etc.
(b) Archaebacteria live as symbionts in the rumen of herbivorous animals.
Methanogens are present in the guts of several ruminant animals such as cows and buffaloes and they are responsible for the production of methane (biogas) from the dung of these animals.

Question 3.
What is the nature of the cell wall in diatoms?
Solution:
In diatoms, ail walls form two thin overlapping shells which fit together as in a soapbox. The walls are embedded with silica and thus the walls are indestructible. Thus diatoms have left behind a large amount of cell wall deposits in their habitat.

Question 4.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify?
Solution:
Algal bloom : When colour of water changes due to profuse growth of coloured phytoplankton, it is called algal bloom.
Red tides : Redness of the red sea is due to luxurient growth of Trichodesrium erythrium, a member of cynobacteria (blue green alage).

Question 5.
How are viroids different from viruses?
Solution:
Viroids are simpler than viruses, consisting of
a single RNA molecule that is not covered by protein capsid. The genetic material of viruses are surrounded by protein coat.

Question 6.
Describe briefly the four major groups of protozoa.
Solution:
The four major group of protozoa are flagellated protozoan, amoeboid protozoan, sporozoan, ciliated protozoan. The main characters of these group are as follows:
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 1

Question 7.
Plants are autotrophic. Can you think of some plants that are partly hetrotrophic?
Solution:
Bladderwort and venus fly trap are examples of insectivorous plants and Cuscuta is a parasite. These are plants which are partially heterotrophic.

Question 8.
What do the terms phycobiont and mycobiont signify?
Solution:
Lichens shows symbiotic association between algae and fungi. The fungal component of lichen is called mycobiont and the algal component is called phycobiont.

Question 9.
Give a comparative account of the classes of kingdom fungi under the following:
(a) Mode of nutrition
(b) Mode of reproduction
Solution:
Kingdom fungi has four classes, these are Phycomycetes, ascomycetes, basidiomycetes and Deuteromycetes. The comparison between these classes are as follows :
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 2

Question 10.
What are the characteristic features of Euglenoids?
Solution:
Euglenoids show the following characteristic features:

They store carbohydrates in the form of paramylon.
Since euglenoids are green and holophytic like other plants.
Few are non-green and saprophytic, some are holotropic.
They bear a red-pigmented eyespot and a gullet near the base of flagellum.
All the euglenoids have one or two flagella which help in swimming.
Absence of cell-wall but contain flexible pellicle made up of protein.
Freshwater, free-living found in ponds and ditches.

Question 11.
Give a brief account of viruses with respect to their structure and the nature of genetic material. Also, name four common viral diseases.
Solution:
Viruses have the following characteristics:
(i) All plant viruses have single-stranded RNA and all animal viruses have either single or double-stranded RNA or double-stranded DNA.
(ii) Protein vims also contain genetic material RNA or DNA. A vims is a nucleoprotein and the genetic material is infectious, These are obligate parasites, self replicating, non-cellular organisms.
(iii) Vimses are smaller than bacteria and their genetic material is surrounded by protein I coat called capsid. Capsid is made up of small subunits called capsomeres.
Four common viral diseases are :
(a) Cough and cold
(b) Mumps
(c) Influenza
(d) Smallpox

Question 12.
Organise a discussion is your class on the topic are viruses living or non-living?
Solution:
Vimses are link between living and non-living. They possess some living characters and some non-living characters. Crystallization is a non-living character but it can reproduce inside living body.
Actually vimses are metabolically inert when outside the host-cell. They reproduce using the metabolic machinery of the host cell.

VERY SHORT ANSWER QUESTIONS

Question 1.
Who wrote the books ‘Species Plantarum’ and ‘Systema Naturae?
Solution:
Carolus Linnaeus.

Question 2.
Name the two kingdoms of the living world proposed by Linnaeus.
Solution:
Plantae and Animalia

Question 3.
What are protists?
Solution:
Protists are unicellular, eukaryotic organisms.

Question 4.
Which organism was earlier placed in the plant as well as animal kingdoms and why?
Solution:
Euglena because it has locomotory organelle, flexible pellicle, contractile vacuole and reproduce by binary fission like animals and chloroplasts and pyrenoids like plants.

Question 5.
Name the 5 kingdoms of organisms in the order of their supposed evolution.
Solution:
Monera, Protista, Fungi, Animalia and Plantae.

Question 6.
Mention 2 traits in which fungi resemble animalia.
Solution:
Heterotrophy and glycogen as reserve food.

Question 7.
Define
(a) Plasmogamy
(b) Karyogamy
Solution:
(a) Plasmogamy – Fusion of protoplasms between two motile or non-motile gametes.
(b) Karyogamy – Fusion of two nuclei.

Question 8.
What is a retrovirus? Give an example
Solution:
Retrovirus is organisms that have RNA s as genetic material. For example HTV

Question 9.
Give two salient features of slime moulds.
Solution:
The two salient features of slime moulds are:

These do not have a cell wall
These have pseudopodia for movement

Question 10.
What is called the jokers of microbiology and why?
Solution:
Jokers of microbiology are mycoplasma as they have no cell wall and no definite shape.

Question 11.
Give the names of two diseases caused by Protozoans
Solution:
Two diseases caused by protozoans are
(1) Amoebiasis
(2) Malaria

SHORT ANSWER QUESTIONS

Question 1.
Cyanobacteria play a major role in our ecology. Discuss.
Solution:
Cyanobacteria, also known as ‘blue-green algae’ help in carbon fixation in a major way on the ocean surface.

They are helpful in nitrogen fixation in paddy fields leading to a better harvest. About 80% of photosynthesis on ocean surface is done by cyanobacteria. So, it can be said that they play a major role in our ecology.

Question 2.
What is the role of methanogens?
Solution:
Methanogens are type of bacteria which live in the gut of ruminating animals.

They assist those animals in digestion and the byproduct of that digestive process is methane.

More number of livestock population results in increased methane level in the environment leading to global warming. So, indirectly methanogens can be responsible for global warming.

Question 3.
What are lichens? What are the roles of lichen in water pollution ?
Solution:
Lichens are symbiotic associations i.e. mutually useful associations, between algae and fungi.

The algal component is known as phycobiont and fungal component as mycobiont, which are autotrophic and heterotrophic, respectively.

Algae prepare food for fungi and fungi provide shelter and absorb mineral nutrients and water for its partner. Lichens are very good pollution indicators as they do not grow in polluted areas.

Question 4.
On what factors is the 5 kingdom classification of Whittaker based?
Solution:
The five kingdom classification is based upon the following factors :
(i) Complexity of cell structure – Prokaryotes or Eukaryotes
(ii) Complexity of organisms body – Unicellular or Multicellular
(iii) Mode of obtaining nutrition – Autotrophs or Heterotrophs
(iv) Phylogenetic relationships

Question 5.
Give the technical terms used for the following:
(a) Remains of an organism of a former geological age.
(b) Science of classification of organisms.
(c) Evolutionary history of a group of organisms.
(d) Organisms which synthesize their own food, using chemical energy.
Solution:
(a) Fossils
(b) Taxonomy
(c) Evolution
(d) Autotrophs

Question 6.
What are the kinds (shapewise) bacteria found in nature. Name the pathogen with the disease caused
Solution:

Question 7.
Why is
(i) Basidiomycetes called club fungi?
(ii) Ascomycetes called sac fungi?
Solution:
(i) After sexual reproduction basidium is formed which form the shape a club and this chin these fungi are called Club Fungi.
(ii) In sexual reproduction ascospores are formed in a sac like asci and thus this fungi is called sac fungi.

LONG ANSWER QUESTIONS

Question 1.
Give an account of early work in taxonomy.
Solution:

Since the dawn of civilisation, there have been many attempts to classify living organisms.
It was done instinctively not using criteria that were scientific but borne out of a need to use organisms for our own use – for food, shelter and clothing.
Aristotle was the earliest to attempt a more scientific basis for classification. He used simple morphological characters to classify plants into trees, shrubs and herbs.
He also divided animals into two groups, those which had red blood and those that did not.
In Linnaeus’ time a Two Kingdom system of classification with Plantae and Animalia kingdoms was developed that included all plants and animals respectively.
Classification of organisms into plants and animals was easily done and was easy to understand, inspite, a large number of organisms did not fall into either category.
R.H. Whittaker (1969) proposed a Five Kingdom Classification.
The kingdoms defined by him were named Monera, Protista, Fungi, Plantae and Animalia.
The main criteria for classification used by him include cell structure, thallus organisation, mode of nutrition, reproduction and phylogenetic relationships.

Question 2.
Differentiate briefly characteristics of kingdom Plantae and Animalia.
Solution:

Kingdom Plantae includes all eukaryotic chlorophyll-containing organisms commonly called plants.
A few members are partially heterotrophic such as the insectivorous plants or parasites.
Bladderwort and Venus fly trap are examples of insectivorous plants and Cuscuta is a parasite.
The plant cells have an eukaryotic structure with prominent chloroplasts and cell wall mainly made of cellulose.
Plantae includes algae, bryophytes, pteridophytes, gymnosperms and angiosperms.
The animal kingdom is characterised by v heterotrophic eukaryotic organisms that are multicellular and their cells lack cell walls.
They directly or indirectly depend on plants for food. They digest their food in an internal cavity and store food reserves as glycogen or fat.
Their mode of nutrition is holozoic – by ingestion of food. They follow a definite growth pattern and grow into adults that have a definite shape and size.
Higher forms show elaborate sensory and neuromotor mechanism. Most of them are capable of locomotion.

Question 3.
Give the economic importance of diatoms. Diatoms are used
Solution:
(1) as a cleaning agent in tooth pastes and metal polishes.
(2) Adding to make sound proof rooms.
(3) In Alteration of sugar, alcohol and antibiotics
(4) as put in paints to ad the paint visibility at night
(5) as an insulating material in Refrigerators, fumances etc.

Question 4.
What are the distinguishing characters of kingdom fungi?
Solution:
The distinguishing characters of kingdom fungiare as follows :
(i) Fungi are non-vascular, non-seeded, non-flowering, eukaryotic achlorophyllous (nongreen), heterophic (heterophytic) spore bearing, thalloid, multicellular decomposers and mineralisers of organic wastes and help in recycling of matter in the biosphere.
(ii) In true fungi the plant body is thallus. It may be non-mycelial or mycelial.
a. Non mycelial: The non-mycelial forms are unicellular; however they may form a pseudomycelium by budding,
b. Mycelial: In mycelial form plant body is made up of thread like structures called hyphae. Hyphae are usually branched tube like structure bounded by a cell-wall of chitin. The hyphae may be septate (higher fungi) or aseptate (lower fungi).
Septate hyphae may be of 3 kinds, uninucleate (monokaryotic hyphae), with binucleate cells (dikaryotic hyphae) ormultinucleate. Some fungi are aseptate and known as coenocytic fungi, with hundreds of nuclei in continuous cytoplasmic mass.
(iii) The cell shows eukaryotic organization but lack chloroplast and Golgi bodies. The genetic material is DNA and mitosis is intracellular (karyochorisis).
(iv) Fungi lack chlorophyll, hence, they do not prepare food by photosynthesis. Thus they can grow everywhere, where organic material is available.
(v) Fungi are heterotrophs that acquire their nutrient by absorption. They store their food in the form of glycogen.
(vi) The primitive fungi have oogamous type of sexual reproduction where as most advanced ones do not have sexual reproduction.

Question 5.
Compare the main features of Monera with Protista.
Solution:

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 2 Biological Classification, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 2 Biological Classification, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 1 The Living World

January 5, 2022June 18, 2021 by Raju

NCERT Solutions for Class 11 Biology Chapter 1 The Living World

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 1 The Living World.

Question 1.
Why are living organisms classified?
Solution:
Classification of living organisms grouped them in special categories, which is based on observable characters. It makes their study easy and convenient. For example, Mammals are those who possess mammary glands, the hair on the body, external pinnae, etc.

Question 2.
Why are classification systems changing every now and then?
Solution:
The classification system changes when more information becomes available about the organisms. Additional information are updated from time to time about different organisms at this stage there is a need arises to make changes in the classification system.

Question 3.
What different criteria would you choose to classify people that you meet often?
Solution:
Classification means the arrangement of organisms into groups on the basis of their affinities or relationships. The branch of biology that deals with the study of principles and procedures of biological classification are called taxonomy. Some fundamental elements of taxonomy are discussed below.

Nomenclature: It is the science of providing distinct and proper names to organisms. It is the determination of the correct name as per established universal practices and rules.

Classification: It deals with the mode of arranging organisms or group^ of organisms into categories according to a systematic plan or ah order. The categories used in the classification of animals are Class, Order, Family, Genus, and Species. Each category is a unit and is also called a taxon (PI. Taxa).

Identification: It is the determination of the correct name and place of an organism in a system of classification. It determines that the particular organism is similar to some other organism of known identity. This implies assigning an organism to a particular taxonomic group. Suppose there are three plants say x, y, z. AH represent different species. Another plant w resembles y. The recognition of the plant was identical to the already known plant y is its identification.
One of the important features of systematics is the naming of living organisms. The organisms have been given two types of names i.e

common or vernacular names
Scientific names.

Question 4.
What do we learn from the identification of individuals and populations?
Solution:
Identification of individuals and populations determines their exact place or position in the set plan of classification.

Question 5.
Given below is the scientific name of mango. Identify the correctly written name.
(a) Mangifera Indica
(b) Mangifera indica
Solution:
(b) Mangifera indica

Question 6.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Solution:
“Taxon is a unit of classification or a rank or a level of hierarchy in system of classification. The following chart gives taxonomical categories showing a hierarchical arrangement in ascending order.

Kingdom
↑
phylum or Division
Class
↑
Order
↑
Family
↑
Genus
↑
Species

Question 7.
Can you identify the correct sequence of taxonomical categories?
(a) Species → Order → Phylum → Kingdom
(b) Genus → Species → Order → Kingdom
(c) Species → Genus → Order → Phylum
Solution:
(c) Species → Genus → Order → Phylum

Question 8.
Try to collect all the currently accepted meanings for the word ‘species’. Discuss with your teacher the meaning of species in the case of higher plants and animals on one hand and bacteria on the other hand.
Solution:

Species is one of the basic units of biological classification. A species is often defined as a
group of organisms capable of interbreeding aid in producing fertile offspring.
Sometimes more precise or differing measures such as similarity of DNA, morphology o,^ecological niche are used to define the basis of species.
In case of animals, the name of species is defined by the specific name or the specific epithet. For example, gray wolves belong to the species Canis lupus, golden Jackals to Cam’s aureus etc.
Both of them belong to same genus Canis, but species name varies. But species name of plant is only called species epithet.
The ‘specific name’ in botany is always the combination of genus name and species epithet such as saccharum in Acer saccharum (Sugar maple).
But bacteria are grouped under four categories based on their shape – spherical, rod-shaped, comma and spiral shaped and species of bacteria is according to their shapes. Thus the meaning of species in higher organism and bacteria are different.

Question 9.
Define and understand the following terms:
(i) Phylum (ii) Class (iii) Family (iv) Order (v) Genus
Solution:
(i) Phylum: A phylum is a group of related classes having some common features, e.g., protozoa.
(ii) Class: A class is a group of related orders, for e.g., order Rodentia, Lagomorpha and Carnivora all having hair and milk glands are placed in class Mammalia.
(iii) Family: A family is a group of related genera. The genus Felis of cats and the genus Panthera of lion, tiger and leopard are placed in the family Felidal.
(iv) Order: An order is a group of related families. The family Felidae of cats and the »family Coridal of dogs are assigned to the order Carnivora. Cats and dogs have large canine teeth and are flesh-eaters.
(v) Genus: A genus is a group of species alike in the broad features of their organization but different in detail. As per the rules of binomial nomenclatures, a species can not be named without assigning it to a genus.
NCERT Solutions for Class 11 Biology Chapter 1 The Living World 1

Question 10.
How is key helpful in the identification and classification of organisms?
Solution:
Keys are contrasting pairs of characters (couplet), it represents the choice made between two opposite options. This results in acceptance of only one and rejection of the other. Each statement in the key is called a lead. Separate taxonomic keys are required for each taxonomic category such as family, genus, and species for identification purposes.

Question 11.
Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal.
Solution:
NCERT Solutions for Class 11 Biology Chapter 1 The Living World 2

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the basic Unit of classification.
Solution:
Species.

Question 2.
Who introduced the hierarchy in taxonomy?
Solution:
Linnaeus

Question 3.
Who is the father of taxonomy?
Solution:
Carolus Linnaeus.

Question 4.
What is meant by cytotaxonomy?
Solution:
Classification based on chromosome number.

Question 5.
Who devised the binomial nomenclature?
Solution:
Carolus Linnaeus

Question 6.
What is a type specimen?
Solution:
Establishment of the name of the new species on the basis of the original specimen is called type specimen

Question 7.
In which language binomial nomenclature is written?
Solution:
Latin

Question 8.
What term is used to describe organisms without a well-developed nucleus?
Solution:
Prokaryote

Question 9.
Is inter-specific breeding possible?
Solution:
Yes, both.

Question 10.
What are DNA viruses / RNA viruses?
Solution:
Viruses that possess DNA as the genetic material are called DNA viruses.

Question 11.
What is speciation?
Solution:
Formation of a new species from an existing one by the appearance of mutation.

Question 12.
What are correlated characters?
Solution:
The common features the species have to qualify for inclusion in a genus are called correlated characters

Question 13.
Why classification of plants and animals is necessary?
Solution:
Classification divides millions of plant and animal species into convenient groups that make their study easier

Question 14.
What is cohort or order?
Solution:
The cohort is a unit of classification higher than the 6. family

Question 15.
Give an example of symbiotic bacteria.
Solution:
Rhizobium leguminosarum

Question 16.
Give botanical and zoological names of the following:
(1) Pea
(2) Wheat
(3) Man
(4) Potato
Solution:
(1) Pea → Pisumsatinum
(2) Wheat → Triticumaextivum
(3) Man → Homo sapiens
(4) Potato → Solanum tuberosum

SHORT ANSWER QUESTIONS

Question 1.
Write a note on bacteriophages. (Dharwar. 2004, Belgaum. 04,2005)
Solution:
The viruses that infect bacteria are called bacteriophages. They were discovered by Twort. They are Tadpole shaped. They have DNA as their genetic material. They are distinguished into T – odd phages as well as T – even phages.

Question 2.
What is a taxonomic aid?
Solution:
A taxonomic aid is storage of record of either live or dead specimens of flora or fauna, which helps scientists in taking reference to study classification

Question 3.
Give the classification of man.
Solution:
Common Name – Human
Scientific Name – Homo sapiens
Genera – Homo
Families – Hominidae
Orders – Primata
Classes – Mammalia
Phyla/Division – Chordate

Question 4.
What is a museum? How many kinds of museums are found?
Solution:
Museum in an institution where artistic and educational materials are exhibited to the public. The material available for observation and study is called a collection.
Kinds of Museums:

Art Museum
History Museum
Applied Science Museum
Natural Science Museum

Question 5.
Give a reason for the following.
Bacteria are the Natural Scavengers ‘ (D.Kannada 2006)
Solution:
because they bring about the decomposition of organic debris and clean the earth’s surface.

Question 6.
What is the role of characteristics of living beings in classification?
Solution:
A group of common features of living beings are placed under a common category of classification and when uncommon under a different category. It means more systematic a process for further study, research, protection and recording.

Question 7.
What is the significance of a HERBARIUM?
Solution:
HERBARIUM:- A book, case, or room containing an orderly collection of dried plants is called Herbarium. It develops interest in Nature for the activists in it. It can be used to gain knowledge and be updated about plants and their scientific names and even compare various samples. It is a small scale it can be proactive to do. One can make projects too from it for schools, colleges and research institutions.

Question 8.
Explain the role of blue-green algae in soil fertility.
Solution:
Blue-green algae like Nostoc, Anabaena fix atmospheric nitrogen. Heterocyst contains nitrogens enzyme that helps in nitrogen fixation. Nitrogen-fixing blue-green algae are inoculated in the rice field to increase soil fertility.

LONG ANSWER QUESTIONS

Question 1.
Write a short note on Binomial Nomencia? ture and guidelines for Binomial nomenclature.
Solution:
Binomial Nomenclature was introduced by Carolus Linnaeus. In this method every organism is given a scientific name, which has two parts, the first is the name of the genus (generic name) and the second is the name of the species (specific epithet) e.g.: Homo sapiens In the above examples, Homo is a generic name, while sapiens is the name of the species belonging to Homo.

Guidelines:

scientific names are generally in Latin or derived from Latin irrespective of their origin
The scientific names are written in italics or underlined (when handwritten)
The first word denotes the name of the genus and the second word denotes the specific epithet
The generic name starts with a capital letter, while the specific name starts with a small letter (If a specific name starts with a capital letter it denotes the name of a person or place)
The name of the author is written in an abbreviated form after the specific name. e.g.: Homo sapiens Linn.

Question 2.
What is the difference between living and nonliving?
Solution:
Question 3.
Explain the binomial system of nomenclature.
Solution:
Binomial nomenclature system was developed by Linnaeus. Binomial nomenclature is the system of providing organisms with appropriate and distinct names consisting of two words, first generic and second specific. The first or 4.

generic word is also called genus. It is like a noun and its first letter is written in capital form.The second word or specific epithet represents the species.
It is like an adjective. Its first letter is written in small form except occasionally when it denotes a person or place. The two word name is appended with the name of the taxonomist who discovered the organism and provided with its scientific name, e.g., Ficus bengalensis L., Mangifera indica Linn, The name of taxonomist can be written in full or in abbreviated form.
There are several technical names which have three words, e.g., Homo sapien sapiens, Acacia nilotica indica, Gerilla gorilla. Here the first word is generic, the second specific while the third word represents variety (mostly in botanical literature) or subspecies (mostly in zoological literature).
If the same scientific name is to be written time and again, the name of the genus can be abbreviated, e.g., F. bengalensis.

Question 4.
What is the role of zoological parks in wildlife conservation?
Solution:

In the early stages, the zoological parks were considered as places of relaxation and enjoyment for public, however, there has been a change in the objective of the purposefulness of these parks.
The establishment of zoological parks help in providing knowledge about different native and exotic wild mammals, birds, reptiles, fish and flora to the public in general and school children in particular.
Since the key to wildlife conservation lies in the education of the masses and the involvement of voluntary organisations, zoological parks are very useful in spreading knowledge on the wildlife wealth of the country.
These are also important centres for organising seminars, training and researches on the management of wildlife species and for study of their social behaviour, breeding and ecological species.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 1 The Living World, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 1 The Living World, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 15 Waves

January 5, 2022June 18, 2021 by Veer

NCERT Solutions for Class 11 Physics Chapter 15 Waves

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 15 Waves

Question 1.
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 cm. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end ?
Answer:
Tension in the string, T = 200 N ; Mass of string, M = 2.50 kg ;
Length, l= 20.0 cm = 20 x 10^-2 m
NCERT Solutions for Class 11 Physics Chapter 15 Waves 1

Question 2.
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top? Given that the speed of sound in air is 340 m s^-1. (g = 9.8 m s^-2)
Answer:
Here, h=300 m, g = 9.8 m s^-2 and velocity of sound,υ= 340 m s^-1Let t₁ be the time taken by the stone to reach the surface of the pond.
NCERT Solutions for Class 11 Physics Chapter 15 Waves 2
.’. Total time after which sound of splash is heard = t₁+ t₂= 7.82 + 0.88 = 8.7 s.

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 m s^-1.
Answer:
Mass, M= 2.10 kg, Length, l = 12.0 m.
NCERT Solutions for Class 11 Physics Chapter 15 Waves 3

Question 4.
Use the formula
NCERT Solutions for Class 11 Physics Chapter 15 Waves 4
to explain why the speed of sound in air
(a) is independent of pressure.
(b) increases with humidity.
(c) increases with temperature
Answer:

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f(x,t) where x and t must appear in the combination x – υt or x + υt, i.e. y = F(x ± υt.) Is the converse true ? Examine if the following functions for y can possibly represent a travelling wave:
(a) (x – υt)²
(b) log [(x +υt )/x₀]
(c)exp [-(x + υt)/x₀]
(d) 1/(x + υt)
Answer:
The converse of the above statement is not true. For a function to represent a travelling wave, the function must be finite and well defined at any point and at any time. Only the function (C) satisfies this condition and is a wave function.

Question 6.
A bat emits the ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound?
Speed of sound in air is 340 m s^-1 and in water 1486 m s-1.
Answer:
Here v = 1000 kHz = 1000 x 10³ Hz = 10⁶ Hz
Velocity of sound in air, υ_a = 340 ms^-1; the velocity of sound in water, υ_w = 1486 ms^-1(a) For reflected sound, medium remains the same
NCERT Solutions for Class 11 Physics Chapter 15 Waves 7

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1 ? The operating frequency of the scanner is 4.2 MHz.
Answer:
Here,υ = 1.7kms^_1 = 1.7 x 1000 ms^_1 = 1700 ms^_1Frequency, v= 4.2 MHz = 4.2 X 10⁶ Hz
NCERT Solutions for Class 11 Physics Chapter 15 Waves 8
Question 8.
A transverse harmonic wave on a string is described by
y (x, t) = 3.0 sin (361 + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave?
(b) If it is travelling, what are the speed and direction of its propagation?
(c) What are its amplitude and frequency ?|
(d) What is the initial phase at the origin?
(e) What is the least distance between two successive crests in the wave?
Answer:
(a) It is traveling which is propagating from right to left. Comparing the given equation with y (x, t) = r sin (ωt+kr+φ)
NCERT Solutions for Class 11 Physics Chapter 15 Waves 9

Question 9.
For the wave described in Exercise 8, plot the displacement (y) versus (f) graphs for x = 0, 2, and 4 cm. What is the shape of these graphs? In which aspects do the oscillatory motion in travelling waves differ from one point to another: amplitude, frequency, or phase?
Answer:
The transverse harmonic wave is
NCERT Solutions for Class 11 Physics Chapter 15 Waves 10
The graph obtained is sinusoidal. Similar graphs are obtained for x = 2 cm and x = 4 cm. The oscillatory motion in the travelling wave only differs in respect of phase. The amplitude and frequency of oscillatory motion remain the same in all the cases.

Question 10.
For the traveling harmonic wave
y (x, t) = 2.0 cos 2π(10t – 0.0080 x + 0.35)
where x and y are in cm and t in s. Calculate the phase difference between the oscillatory motion of two points separated by a distance of
(a)4 m
(b) 0.5 m
(c) λ/2
(d) 3λ/4.
Answer:
Comparing the equation given in the question, with the standard equation of travelling wave:
NCERT Solutions for Class 11 Physics Chapter 15 Waves 11

Question 11.
The transverse displacement of a string (clamped at its both ends) is given by
NCERT Solutions for Class 11 Physics Chapter 15 Waves 12
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 x 10^-2 kg.
Answer the following:
(a) Does the function represent a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves traveling in opposite directions. What are the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.
Answer:
(a) The function does not represent a travelling wave. It represents a stationary wave.
(b) Using the relation
NCERT Solutions for Class 11 Physics Chapter 15 Waves 13

Question 12.
(1) For the wave on a string described in Exercise 11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers, (2) What is the amplitude of a point 0.375 m away from one end?
(a) All the points on the wave have
(b) same frequently everywhere except at the nodes.
(c) same phase everywhere in a loop except at the nodes.
(d) different amplitude.
(2) Here, x = 0.375 m, 1 = 0
NCERT Solutions for Class 11 Physics Chapter 15 Waves 15

Question 13.
Given below are some functions of x and y to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent
(1) a travelling wave,
(2) a stationary wave or
(3) none at all:
(a) y = 2 cos (3x) sin (10t)
(b)
(c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t)
(d) y = cos x sin t- cos 2x sin 2t
Answer:
(a) It represents a stationary wave.
(b) It does not represent either a travelling wave or a stationary wave.
(c) It is a representation of the travelling wave.
(d) It is a superposition of two stationary waves.

Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x 10^-2 kg and its linear density is
4.0 x 10^-2 kg m^_1. What is
(a) the speed of a transverse wave on the string, and
(b) the tension in the string?
Answer:
Here, v = 45 Hz ; mass of wire, M = 3.5 x 10^-2 kg
NCERT Solutions for Class 11 Physics Chapter 15 Waves 17

Question 15.
A meter-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when this tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in the air at the temperature of the experiment. The edge effects may be neglected.
Answer:
Let v be the fundamental frequency of the closed-end organ pipe for the length
l_x = 25.5 cm
NCERT Solutions for Class 11 Physics Chapter 15 Waves 18

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
Answer:
l = 100 cm = 1 m f = 2.53 kHz
Since the rod is champed at the middle, a node is formed and since the rod is oscillating at a fundamental frequency of 2.53 kHz antinodes are formed at both ends as shown in the figure.
∴ l = λ/4 × 2 = λ/2 ⇒ λ = 2l
λ = 2 × 1 = 2 cm
We know that
v = fλ = 2.53 × 10³ × 2
= 5.06 × 10 ³ ms^-1

Question 17.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 ms^-1).
Answer:

This is the first harmonic. The second harmonic is 3v i.e., 1275 Hz, the third harmonic is 5v i.e., 2125 Hz etc. Therefore, only the first harmonic of frequency 425 is rasonantly excited by a 430 Hz source.If both the ends of the pipe are open, the fundamental frequency is given by

The second, third, fourth… etc. harmonics have frequencies 2v’, 3v’, 4v’……. i.e., 1700 Hz, 2550 Hz, 3400 Hz etc.No harmonic can be excited by the 430 Hz source in this case.

Question 18.
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Answer:
Let v₁and v₂ be the frequencies of strings A and B respectively
Then, v₁= 324 Hz, v₂ = ?
Number of beats, b = 6
v₂ = v₁± b = 324 ± 6
i.e., v₂ = 330 Hz or 318Hz.
Since the frequency is directly proportional to square root of tension, on decreasing the tension in the string A, its frequency v₁; will be reduced i.e., number of beats will increase if v₂ = 330 Hz. This is not so because a number of beats becomes 3. Therefore, it is concluded that the frequency v₂ = 318 Hz because on reducing the tension in string A, its frequency may be reduced to 32l Hz, thereby giving 3 beats with v₂ = 318 Hz.

Question 19.
Explain why (or how) :
(a) in a sound wave, a displacement node is a pressure antinode and vice versa.
(b) bats can ascertain distances, directions, nature and sizes of the obstacles without any “eyes”.
(c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes.
(d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) The shape of a pulse gets distorted during propagation in a dispersive medium.
Answer:
(a) In a sound wave, a decrease in displacement i..e. displacement node causes an increase in the pressure there i.e, a pressure antinode is formed. Also, an increase in displacement is due to the decrease in pressure.

(b) Bats emit ultrasonic waves of high frequency from their mouths. These waves after being reflected back from the obstacles on their path are observed by the bats. These waves give them an idea of distance, direction, nature and size of the obstacles.

(c) The quality of a violin note is different from the quality of sitar. Therefore, they emit different harmonics which can be observed by human ear to differentiate between the two notes.

(d) This is due to the fact that gases have only the bulk modulus of elasticity whereas solids have both, the shear modulus as well as the bulk modulus of elasticity.

(e) A pulse of sound consists of a combination of waves of different wavelength. In a dispersive medium, these waves travel with different velocities giving rise to the distortion in the wave.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (1) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 ms^-1, (b) recedes from the platform with a speed of 10 ms^-1 ? (2) What is the speed of sound in each case ? The speed of sound in still air can be taken as 340 ms^-1.
Answer:
Here, frequency of whistle, v = 400 Hz,
Speed of sound, v = 340 ms^-1; Speed of train, v_s = 10 m s^-1,
(1) (a) When the train approaches the platform i.e., the observer at rest,
NCERT Solutions for Class 11 Physics Chapter 15 Waves 21

Question 21.
A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station at a speed of 10 ms^-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 ms^-1? The speed of sound in still air can be taken as 340 ms^-1.
Answer:
Frequency of source of the sound, v = 400 Hz.
Speed of sound, υ= 340 ms^-1; Speed of wind, υ_m = 10 ms^-1.
Effective speed of sound = Speed of sound + Speed of air
= υ + υ_m = 340 + 10 = 350 ms^-1.
The source, as well as the listener, are both at rest. Therefore, there is no relative motion.
NCERT Solutions for Class 11 Physics Chapter 15 Waves 22
Since the source of sound is at rest, the wavelength A does not change
.’. wavelength = 0.875 m.
Also, speed of sound = 340 + 0 = 340 ms^-1The situations are altogether different, because both the observer and the source are in motion with respect to the medium in this case.

Question 22.
A traveling harmonic wave on a string is described by
y (x,t) = 7.5 sin (0.0050 x + 12t + π/4)
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2s, 5s, and 11s.
Answer:
The traveling harmonic wave is
NCERT Solutions for Class 11 Physics Chapter 15 Waves 23

(b) All the point, located at distance nλ, (where n is an integer) from the point x = 1 have the same velocity and displacement.

Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium, (a) Does the pulse have a definite (1) frequency, (2) wavelength, (3) speed of propagation ? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz? ¹Answer:
1. The pulse does not have a definite frequency and definite wavelength but has a definite speed of propagation.
2. The frequency of the note is not 1/20 Hz or 0.05 Hz. It is only the frequency of repetition of the whistle.

Question 24.
One end of a long string of linear mass density 8.0 x 10^-3 kg m^-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y- direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
Answer:
Here, mass per unit length, m = linear mass density = 8 x 10^-3 kg m³.
Tension in the string, T = 90 kg = 90 x 9.8 = 882 N
Frequency, v = 256 Hz ; Amplitude, r = 5.0 cm = 5 x 10^-2 m. Hz
The wave produced in the string has velocity υ given by
NCERT Solutions for Class 11 Physics Chapter 15 Waves 25

Question 25.
A SONAR system fixed in a submarine operates at a frequency of 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 ms^-1.
Answer:
Here, frequency of SONAR(source )=40.0 kHz = 40 x10³

Question 26.
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of the S wave is about 4.0 kms^-1, and that of P wave is 8.0 kms^-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in a straight line, how far away does the earthquake occur?
Answer:
Let υ₁ and υ₂ be the velocities of two waves respectively and t₁and t₂ is the time taken by them to travel to the position of a seismograph.
NCERT Solutions for Class 11 Physics Chapter 15 Waves 27

Question 27.
A bat is fitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall ?
Answer:
Here, frequency of sound emitted by bat, v = 40 kHz
Velocity of bat υ_s= 0.03 u, where 0 is the velocity of sound.
The bat is moving towards the flat wall. This is the case of source in motion and the observer at rest. Therefore, the frequency of sound reflected at the wall is
NCERT Solutions for Class 11 Physics Chapter 15 Waves 28
The frequency v’ is reflected by the wall and is again received by the bat moving towards the wall. This is the case of an observer moving towards the source with velocity
υ₀= 0.03υ.
.’. frequency observed by bat,

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NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

January 5, 2022June 18, 2021 by Veer

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displayed from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Answer:
(b) and (c).
(Swimmer does not repeat trip after regular interval of time. Similarly, arrow released from a bow is not periodic).

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) the rotation of earth about its axis.
(b) motion of an oscillating mercury column in a U-tube.
(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Answer:
(b) and (c) represent SHM in particular, (a) represents the period motion, (d) represents a superposition of a number of simple harmonic motions and is a periodic motion not necessarily S.H.M.

Question 3.
Figure depicts’four x-t plots for linear motion of a particle. Which of the plots represents periodic motion ? What is the period of motion (in case of periodic motion) ?
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 1
Answer:
Fig (b) and (d) represent periodic motions and the time period of each of these is 2 second, (a) and (c) are non periodic motions.

Question 4.
Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion ? Give period for each case of periodic motion : (ω is any positive constant).
(a) sin ωt – cos ωt
(b) sin² ωt
(c) 3 cos (π/4 — 2ωt)
(d) cos ωt + cos 3ωt + cos 5ωt
(e) exp (- ω²t² )
(f) 1 +ωt +ω²t²
Answer:

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 3

Question 5.
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A, (b) at the end B, (c) at the mid-point of AB going towards A, (d) at 2 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from A going towards A.
Answer:
A and B form the extreme positions and the center of line AB (let it be O) forms the mean position.

(a) At the end A, e., extreme position, velocity is zero, acceleration and force are directed towards O and are positive.

(b) At the end B, e., second extreme position, velocity is zero whereas the acceleration and force are directed towards the point O and are negative.

(c) At the mid point 0, while going towards A, velocity is negative and maximum. The acceleration and force are both zero.

(d) At a distance 2 cm away from B going towards A, velocity is negative. Since acceleration and force are also directed towards O, both of them are negative.

(e) At a distance 3 cm away from A going towards B, velocity is positive. Since the acceleration and force are directed towards 0 from A, both of these are positive.

(f) At a distance of 4 cm away from A and going towards A, velocity is directed along BA, therefore, it is positive.Since acceleration and force are directed towards OB, both of them are positive.

Question 6.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion ?
(a) a = 0.7x (b) a = – 200 x² (c) a = -10x (d) a = 100³Answer:
Only the relation (c) represents the simple harmonic motion because for S.H.M., a a – x.

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (cos ωt + Φ)
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is co cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is n s^_1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (at + a), what are the amplitude and initial phase of the particle with the above initial conditions ?
Answer:

Question 8.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance,,when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ?
Answer:
Length of scale = 20 cm = 0.20 m
Total scale reading = 50 kg
∴ F = mg = 50 x 9.8 = 490 N
x= 20 cm = 0.20 m
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 6

Question 9.
A spring with a spring constant 1200 N m^-1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Determine
(1) the frequency of oscillations,
(2) maximum acceleration of the mass, and
(3) the maximum speed of the mass.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 7
Answer:
Here k = 1200 N m^_1, m = 3 kg, A = 2.0 cm = 2 x 10^-2 m

Question 10.
In Exercise 9, let us take the position of mass when the spring is unstreched as x= 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase ?
Answer:
Here, the maximum displacement = Amplitude A = 2 cm
(a) If the time t = 0 at x = 0, the displacement can be expressed as the sine function of t.
x(t) = A sin ωt
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 9

Question 11.
Following figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e clockwise or anti-clockwise) are indicated on each figure.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 10
Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P in each case.
Answer:

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case (or is in cm and t is in s) :
(a) x = – 2 sin (3t + π/3)
(b) x = cos (π/6 – t)
(c) x = 3 sin (2nt + tc/4)
(d) x = 2 cos πt
Answer:

Question 13.
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 13
(a) What is the maximum extension of the spring in the two cases ?
(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released free, what is the period of oscillation in each case ?
Answer:
(a) Let y be the maximum extension produced in the spring in fig. (a).
Then F = Ay (in magnitude) ∴y = \(\frac { F }{ K } \)
In fig. (b), the force on one mass acts as the force of reaction due to the force on the other mass. Therefore, each mass behaves as if it is fixed with respect to the other.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 14

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rev/min,
what its maximum speed ?
Answer:

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 ms^-2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 ms^-2)
Answer:

Question 16.
Answer the following questions :
(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle :

A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum ?

(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation ,a more involved analysis shows that T is grater than
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 19
Think of a qualitative argument to appreciate this results.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity
Answer:
In case of a spring, k does not depend upon m. However, in case of a simple pendulum, k is directly proportional to
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 20
(c) The wrist watch uses an electronic system t>r spring system to give correct time, which does not change with acceleration due to gravity. Therefore, watch gives the correct time.

(d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero e., the pendulum will not vibrate at all.

Question 17.
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ?
Answer:
In this case, the bob of the pendulum is under the action of two accelerations.
(1) Acceleration due to gravity ‘g’ acting vertically downwards.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 21
Question 18.
A cylindrical piece of cork of base area A and height h floats in a liquid of density ρ. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 22
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Let l be the height of the cork which is inside the liquid. Then, at equilibrium,
weight of the cork = upthrust of the portion of cork inside the liquid .
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 23
Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Answer:
The suction pump creates the pressure difference, thus mercury rises in one limb of the U-tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be expressed as:
Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.
Let P = density of the mercury.
L = Total length of the mercury column in both the limbs.
A = internal cross-sectional area of U-tube. m = mass of mercury in U-tube = LAP.
Assume, the mercury be depressed in left limb to F by a small distance y, then it rises by the same amount in the right limb to position Q’.
.’. Difference in levels in the two limbs = P’ Q’ = 2y.
:. Volume of mercury contained in the column of length 2y = A X 2y
.•. m – A x 2y x ρ.
If W = weight of liquid contained in the column of length 2y.
Then W = mg = A x 2y x ρ x g
This weight produces the restoring force (F) which tends to bring back the mercury to its equilibrium position.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 24
Question 20.
An air chamber of volume V has a neck area of cross-section a into which a ball of mass m can move up and down without any friction (Fig.). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig.]
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 25
Answer:
When the ball is lying at rest in the long neck, the pressure of air below the ball inside the chamber is equal to the atmospheric pressure outside. If the ball is slightly depressed below through a distance y, then change in volume of air inside the chamber,
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 26

Question 21.
You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
Answer:
(a) Here, mass M = 3000 kg, displacement x = 15 cm = 015 m, g = 10 m/s². There are four spring systems. If k is the spring constant of each spring, then total spring constant of all the four springs in parallel is
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 27

Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Answer:
Let the SHM be represented as
y = r sin ωt, where r – amplitude…..(1)
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 28

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its center. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = – αθ, where J is the restoring couple and 0 the angle of twist).
Answer:
Here, mass of disc, m = 10 kg,
radius r =15 cm = 0.15 m Tortional spring constant a = ?
Time period of tortional oscillations, T = 1.5 s
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 30

Question 24.
A body describes a S.H.M. with an amplitude of 5 cm and period of 0.2 s. Find the acceleration and velocity of the body, when displacement is (a) 5 cm, (b) 3 cm, (c) 0 cm.
Answer:
A = 5 cm, T = 0.2 s.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 31

Question 25.
A mass attached to a spring is free to oscillate with angular velocity to in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the center with velocity υ₀ at time
t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters oo, x₀ and υ₀ .
Answer:
According to the law of conservation of energy
Total energy at x₀ = Total energy at extreme position.
NCERT Solutions for Class 11 Physics Chapter 14 Oscillations 32

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NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

January 5, 2022June 18, 2021 by Veer

NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

Question 1.
A geyser heats water flowing at the rate of 3.0 liters 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 x 10⁴ J/g ?
Answer:

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:
Nitrogen is a diatomic gas. Therefore its molar specific heat at constant pressure is
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics 2

Question 3.
Explain why
(a) Two bodies at different temperatures T₁ and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car type increases during driving.
(d) The climate of a harbor town is more temperate than that of a town in a desert at the same latitude.
Answer:
(a) Two bodies would settle at the mean temperature (T₁+ T₂)/2 only if they have equal thermal capacities.
(b) A coolant of high specific heat can withdraw more heat out of the chemical or nuclear plant than a coolant of ordinary specific heat.
(c) During driving, the tyres of the car are heated up due to friction between tyres and road. Since heating of tyres causes an increase in temperature, the pressure in the tyres increases (P a T at constant volume).
(d) The relative humidity in a harbour town is more than the relative humidity in a desert town. Therefore, the climatic conditions of a harbour town do not reach extremes of hot or cold making it a temperate zone.

Question 4.
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:
Hydrogen is a diatomic gas. Therefore, for hydrogen, ratio of two specific heats is
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics 3

Question 5.
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken form state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? [Take 1 cal = 4.19 J]
Answer:
When the state of the gas changes adiabatically from A to B, amount of work done is used to increase the internal energy of the gas.
Increase in internal energy, ΔU = 22.3 J.
In the second case, when state A changes to state B, heat absorbed by the system
ΔQ = 9.35 cal = 9.35 x 4.19J = 39 18 J
Now, using the first law of thermodynamics,
ΔQ =ΔU + ΔW
=> ΔW =ΔQ-ΔU = 39 18-22.3 = 16.88 J

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :
(a) What is the final pressure of the gas in A and B?
(b) What is the change in the internal energy of the gas?
(c) What is the change in the temperature of the gas?
Do the intermediate states of the system (before settling to the final equilibrium state) lie of its P-V-T surface ?
Answer:
(a) Since the final temperature and initial temperature remain the same (isothermal process).
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics 4
(b) Since the temperature of the system remains unchanged, the change in internal energy is zero.

(c) The system being thermally insulated, there is no change in temperature (because of free expansion)

(d) The expansion is a free expansion. Therefore, the intermediate states are non-equilibrium states and the gas equation is not satisfied in these states. As a result, the gas can not return to an equilibrium state which lie on the P-V-T surface.

Question 7.
A steam engine delivers 5.4 x 10⁸ J of work per minute and services 3.6 x 10⁹ J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Answer:
Output i.e., useful work done per minute = 5.4 x 10⁸J
Input i.e., heat absorbed per minute = 3.6 X 10⁹ J
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics 5
Question 8.
An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 joules per second, at what rate is the internal energy increasing?
Answer:
Rate of heat supplied to the system is 100 W = 100 Js^-1
Rate of work done by the system is 75Js^-1.
From the first law of thermodynamics, we have
∆ Q = ∆ U + ∆ W ……..(1)
If the above parameters are observed for a time‘∆ t’,
\(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\Delta \mathrm{U}}{\Delta \mathrm{t}}+\frac{\Delta \mathrm{W}}{\Delta \mathrm{t}}\) …….(2)
where \(\frac{\Delta Q}{\Delta t}\) is the rate of heat supplied to the system for the given time, \(\frac{\Delta U}{\Delta t}\) is the rate of increase in internal energy of the system for the given time and \(\frac{\Delta W}{\Delta t}\) is the
rate of work done by the system in the given time.
∴ From equation (2), we have
100 = \(\frac{\Delta U}{\Delta t}\) + 75
∴ \(\frac{\Delta U}{\Delta t}\) = 25 Js^-1
Therefore the internal energy increases at rate of 25 Joules per second.

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig.
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics 6
Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the work done by the gas from D to E to F.
Answer:
As is clear from above Fig.,
Change in pressure, dP = EF = 5.0 – 2.0 = 3.0 N m^-2Change in volume, dW = DF = 600 – 300 = 300 m³Work done by the gas from D to E to F = area of ADEF
NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics 7

Question 10.
A refrigerator is to remove heat from the eatables kept inside at 10°C. Calculate the coefficient of performance, if room temperature is 36°C.
Answer:
Here, T₁ = 36°C = 36 + 273 = 309 K
T₂ = 10°C = 10 + 273 = 283 K

We hope the NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics, drop a comment below and we will get back to you at the earliest