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Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 12 Biotechnology Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 12 Biotechnology

Multiple choice questions

Question 1.
Recombinant DNA technique was established by ………………..
(a) Herbert Boyer
(b) Stanley Cohen
(c) Peter Lobban
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 2.
Modern biotechnology includes ………………..
(a) r-DNA technology and PCR
(b) microarrays, cell culture and fusion
(c) production of curds, cheese, wine
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 3.
How many of the following statements are correct with reference to electrophoresis.
1. DNA is positively charged and it moves towards the negative electrode.
2. DNA is negatively charged and it moves towards the anode.
3. Longer DNA fragments move faster during electrophoresis.
4. Shorter DNA fragments move slowly during electrophoresis.
(a) One
(b) TWo
(c) Three
(d) Four
Answer:
(a) One

Question 4.
PCR technique was developed by ………………..
(a) Stanley Cohen
(b) Herbert Boyer
(c) W. Arber
(d) K. Mullis
Answer:
(d) K. Mullis

Question 5.
……………….. technique is used for in vitro gene cloning or gene multiplication.
(a) Electrophoresis
(b) SDS-PAGE
(c) Spectroscopy
(d) PCR
Answer:
(d) PCR

Question 6.
During denaturation step of polymerase chain reaction, the reaction mixture is heated at ………………..
(a) 40-60°C
(b) 70-75°C
(c) 90-98°C
(d) 36°C
Answer:
(c) 90-98 °C

Question 7.
Annealing temperature of primer is ………………..
(a) 40-60°C
(b) 72°C
(c) 90-98°C
(d) 36°C
Answer:
(a) 40-60 °C

Question 8.
During PCR new strand of DNA is synthesized by ………………..
(a) thermostable Taq DNA polymerase
(b) ligase
(c) phosphorylase
(d) RE
Answer:
(a) thermostable Taq DNA polymerase

Question 9.
Restriction enzymes were discovered by ………………..
(a) K. Mullis
(b) S. Cohen
(c) H. Boyer
(d) Smith, Nathan and Arber
Answer:
(d) Smith, Nathan and Arber

Question 10.
The molecular scissors of DNA are ………………..
(a) ligases
(b) polymerases
(c) endonucleases
(d) transcriptases
Answer:
(c) endonucleases

Question 11.
……………….. generates DNA fragments with sticky ends.
(a) Eco RI
(b) Bam HI
(c) Hind II
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 12.
……………….. is Type III endonulcease.
(a) Hapal
(b) Eco R I
(c) Bgll
(d) Eco K
Answer:
(a) Hapal

Question 13.
Select the incorrect statement.
(a) Type II restriction endonucleases have separate activities for cleaving and methylation.
(b) Type I restriction endonucleases function simultaneously as endonuclease and methylase.
(c) Type II restriction endonucleases cut DNA at specific non-pallindromic sequences.
(d) Thousands of Type II restriction endonucleases have been discovered.
Answer:
(c) Type II restriction endonucleases cut DNA at specific non-palindromic sequences

Question 14.
Ti plasmids are present in ………………..
(a) Bacillus thuringiensis
(b) Agrobacterium tumejaciens
(c) Haemophilus influenza
(d) Escherichia coli
Answer:
(b) Agrobacterium tumefaciens

Question 15.
Plasmid DNA containing foreign DNA is called ………………..
(a) chimeric DNA
(b) passenger DNA
(c) recombinant DNA
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

Question 16.
Bacterial host cell takes up naked r-DNA by process of ………………..
(a) transduction
(b) transfection
(c) transformation
(d) all of these
Answer:
(c) transformation

Question 17.
Virosomes are ………………..
(a) liposome
(b) inactivated HIV
(c) liposome + inactivated HIV
(d) none of these
Answer:
(c) liposome + inactivated HIV

Question 18.
Recombinant protein used in the treatment of emphysema is ………………..
(a) a 1-Antitrypsin
(b) relaxin
(c) Interleukin-1 receptor
(d) urokinase
Answer:
(a) a 1-Antitrypsin

Question 19.
Tissue plasminogen activator and urokinase are used in the treatment of ………………..
(a) blood clots
(b) atherosclerosis
(c) emphysema
(d) asthama
Answer:
(a) blood clots

Question 20.
Bacillus is involved in the production of ……………….. vaccine that melts in mouth.
(a) polio
(b) flu
(c) chicken pox
(d) none of these
Answer:
(b) flu

Question 21.
First transgenic plant produced is …………………..
(a) Bt cotton
(b) tlavr savr tomato
(c) wheat
(d) tobacco
Answer:
(d) tobacco

Question 22.
Insect resistant GMO plants contain ………………..
(a) cowpea trypsin inhibitor gene
(b) cry gene
(c) chalone isomerase gene
(d) (a) or (b)
Answer:
(d) (a) or (b)

Question 23.
‘Cry’ genes are present in ………………..
(a) Agrobacterium tumifaciens
(b) Bacillus thuringiensis
(c) Rhizobium species
(d) Escherichia coli
Answer:
(b) Bacillus thuringiensis

Question 24.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because ………………..
(a) bacteria are resistant to toxin
(b) toxin is immature
(c) toxin is inactive
(d) bacteria encloses toxin in a special case
Answer:
(c) toxin is inactive

Question 25.
The gene coding for a-amylase inhibitor isolated from ……………….. is transferred to tobacco.
(a) mung bean
(b) adzuki bean
(c) E.coli
(d) daffodils
Answer:
(b) adzuki bean

Question 26.
The enzyme affecting the shelf life of Jlavr savr tomato is ………………..
(a) galactosidase
(b) transacetylase
(c) permease
(d) polygalactouranase
Answer:
(d) polygalactouranase

Question 27.
Ferritin, an iron storage protein, isolated from ……………….. and Phaseolus is transferred to ………………… to increase its iron content.
(a) soybean, rice
(b) rice, wheat
(c) maize, soybean
(d) rice, soybean
Answer:
(a) soybean, rice

Question 28.
To improve oil content and oil quality, ……………….. genes are transferred to soybean, oil palm, rapeseed and sunflower.
(a) Arabidopsis
(b) cry
(c) tobacco
(d) canola
Answer:
(a) Arabidopsis

Question 29.
Artemecin is an ……………….. drug developed from transgenic plants.
(a) anticancer
(b) antimalarial
(c) antibacterial
(d) antifungal
Answer:
(b) antimalarial

Question 30.
The transgenic cow born in Scotland, could produce a human protein in her milk for human therapeutics.
(a) Dolly
(b) Molly
(c) Tracy
(d) none of these
Answer:
(c) Tracy

Question 31.
Indian patent does not allow ………………..
(a) process patent
(b) product patent
(c) biopatent
(d) both (b) and (c)
Answer:
(b) product patent

Question 32.
Biopatent are awarded for ………………..
(a) strains of microbes and cell lines
(b) DNA sequences
(c) GMO
(d) All of these
Answer:
(d) All of these

Match the columns

Question 1.

Column A	Column B
(1) Sir Edward Sharpey Schafer	(a) Chemically synthesized DNA sequence of insulin
(2) Hakura	(b) Discovery of insulin
(3) Gilbert and Villokomaroff	(c) PCR
(4) K. Mullis	(d) Insulin production using r-DNA technology

Answer:

Column A	Column B
(1) Sir Edward Sharpey Schafer	(b) Discovery of insulin
(2) Hakura	(a) Chemically synthesized DNA sequence of insulin
(3) Gilbert and Villokomaroff	(d) Insulin production using r-DNA technology
(4) K. Mullis	(c) PCR

Question 2.

Restriction enzyme	Recognition sequence
(1) Alu I	(a) 5′ G-A-A-T-T-C 3′ 3′ C-T-T-A-A-G 5’
(2) Bam HI	(b) 5′ G-T-C-G-A-C3′ 3′ C-A-G-C-T-G 5′
(3) Eco RI	(c) 5′ A-G-C-T 3′ 3′ T-C-G-A5′
(4) Hind II	(d) 5′ G-G-A-T-T-C 3′ 3′ C-C-T-A-A-G 5′

Answer:

Restriction enzyme	Recognition sequence
(1) Alu I	(c) 5′ A-G-C-T 3′ 3′ T-C-G-A5′
(2) Bam HI	(d) 5′ G-G-A-T-T-C 3′ 3′ C-C-T-A-A-G 5′
(3) Eco RI	(a) 5′ G-A-A-T-T-C 3′ 3′ C-T-T-A-A-G 5’
(4) Hind II	(b) 5′ G-T-C-G-A-C3′ 3′ C-A-G-C-T-G 5′

Classify the following to form Column B as per the category given in Column A

(i) Provide tissues for human transplants
(ii) Cancer research
(iii) E. coli hygromycin resistant gene
(iv) Supply of factor IX.

Column A: (Transgenic animal)	Column B : (Application)
(a) Transgenic mice	——————
(b) Transgenic cattle	——————
(c) Pig clones	——————
(d) Transgenic fish	——————

Answer:

Column A: (Transgenic animal)	Column B : (Application)
(a) Transgenic mice	(ii) Cancer research
(b) Transgenic cattle	(iv) Supply of factor IX
(c) Pig clones	(i) Provide tissues for human transplants
(d) Transgenic fish	(iii) E. coli hygromycin resistant gene

Very short answer questions

Question 1.
Who used the term biotechnology for the first time and for what purpose?
Answer:
Karl Ereky in 1919 first used the term biotechnology to describe a process for large scale production of pigs.

Question 2.
Which is the oldest form of biotechnology ?
Answer:
Making curds or bread is the oldest form of biotechnology. Preparation of wine and other alcoholic beverages using microbial fermentation is also old biotechnology.

Question 3.
Modern biotechnology is based on which two core techniques?
Answer:
Modern biotechnology is based on two core techniques – Genetic engineering and chemical engineering.

Question 4.
What is the use of Chemical engineering?
Answer:
Chemical engineering technology is used to maintain sterile environment for manufacturing products like vaccines, antibodies, enzymes, organic acids, vitamins, therapeutics, etc.

Question 5.
What are the different techniques used to characterize macromolecules on the basis of their molecular weight?
Answer:
The techniques used to characterize macromolecules on the basis of molecular weight are gel permeation, osmotic pressure, ion exchange chromatography, spectroscopy, mass spectrometry, electrophoresis, etc.

Question 6.
What are the different types of electrophoresis ?
Answer:
Different types of electrophoresis are agarose gel electrophoresis, PAGE (Polyacrylamide Gel Electrophoresis), SDS – PAGE (Sodium dodecyl Sulphate-PAGE).

Question 7.
What is the use of polymerase chain reaction?
Answer:
Polymerase chain reaction is used for in vitro gene cloning or gene multiplication to produce a billion copies of the desired segment of DNA and RNA, with high accuracy and specificity, in few hours.

Question 8.
What are the requirements of polymerase chain reaction?
Answer:
Polymerase chain reaction requires thermal cycler, DNA containing the desired segment to be amplified, deoxyribonuclueoside triphosphates (dNTPs), excess of two primer molecules, heat stable DNA polymerase and appropriate quantities of Mg⁺⁺ions.

Question 9.
Enlist various biological tools required for transformation of recombinant DNA?
Answer:
Biological tools used for transformation of recombinant DNA are enzymes, cloning vectors (vehicle DNA) and competent host (cloning organisms).

Question 10.
Enlist various enzymes used in recombinant DNA technology?
Answer:
Various enzymes used in recombinant DNA technology are lysozymes, nucleases (exonucleases, endonucleases, restriction endonucleases), DNA ligases, DNA polymerases, alkaline phosphatases, reverse transcriptases, etc.

Question 11.
How do bacteria protect their own DNA from restriction enzymes?
Answer:
The bacteria protect their own DNA from restriction enzymes by methylating the bases at susceptible sites. This chemical modification blocks the action of the enzyme.

Question 12.
What is restriction?
Answer:
Restriction is the process by which the DNA strand is cut into restriction fragments with the help of restriction endonuclease enzymes or REs.

Question 13.
What are molecular scissors? Why are they so called?
Answer:
The enzyme restriction endonucleases are called molecular scissors because they can cut the DNA molecule at a specific point.

Question 14.
What is palindrome in DNA?
Answer:
Palindrome is a DNA sequence which when read on opposite strands of DNA (3′ to 5′ or 5′ to 3′) it reads same.

Question 15.
What are sticky ends?
Answer:
Sticky ends are short extensions of cleaved DNA molecule which can form hydrogen bonded base pairs with other complementary sticky ends.

Question 16.
Give examples of restriction endo¬nucleases that cut DNA at non- pallindromic sequences.
Answer:
Hpal, MboII

Question 17.
Give example of Type I restriction endonucleases.
Answer:
Eco KI

Question 18.
Give example of Type II restriction endonucleases.
Answer:
Eco RI, Bg II

Question 19.
Give the role of plasmids in bacterial cells.
Answer:
Plamids in bacterial cells carry genes for antibiotic resistance (R plasmids) and F plasmids are involved in conjugation.

Question 20.
What are plasmids?
Answer:
The small, extra chromosomal double stranded, circular forms of DNA which are capable of autonomous replication are called plasmids.

Question 21.
What is the source of gene to be cloned in vector?
Answer:
Gene to be cloned is obtained from gene library or by amplification.

Question 22.
What is chimeric DNA?
Answer:
Chimeric DNA is the combination of vector DNA and foreign DNA.

Question 23.
What is meant by transformed cells?
Answer:
The competent host cells which have taken up r-DNA Eire called transformed cells.

Question 24.
What are the different techniques by which foreign DNA can be transferred to host cell without using a vector?
Answer:
Foreign DNA can be transferred to host cell without using a vector by techniques like electroporation, microinjection, lipofection, shot gun, ultrasonification, biolistic method, etc.

Question 25.
Which marker genes are present in plasmid PBR 322?
Answer:
Markers genes in PBR322 plasmid are ampicillin resistant gene and tatracyclin resistant gene.

Question 26.
What is c-DNA?
Answer:
DNA produced by reverse transcription of m-RNA is known as c-DNA.

Question 27.
What are the objectives of CCMB?
Answer:
The objectives of CCMB are to conduct high quality basic research and training in frontier areas of modern biology and promote centralized national facilities for new and modern techniques in the interdisciplinary areas of biology.

Question 28.
Give examples of health problems which develop due to interaction between genetic and environmental factors?
Answer:
Health problems like high cholesterol and high blood pressure develop due to interaction between genetic and environmental factors.

Question 29.
Give examples of diseases which are caused due to single gene defects.
Answer:
Human genetic diseases like sickle-cell anaemia, thalassemia, Tay-sach’s disease, cystic fibrosis, Huntington’s chorea, haemophilia, alkaptonuria, albinism, etc. . are caused by single gene defects.

Question 30.
What is gene therapy?
Answer:
Gene therapy is the treatment of genetic disorders by replacing, altering or supplementing a gene that is absent or abnormal and whose absence or abnormality is responsible for the disease.

Question 31.
What is the use of virosomes?
Answer:
Virosomes are used in gene delivery.

Question 32.
Enlist the diseases where clinical trials of somatic cell gene therapy have been employed?
Answer:
The clinical trials of somatic cell gene therapy have been employed for the treatment of disorders like cancer, rheumatoid arthritis, SCID, Gaucher’s disease, familial hypercholesterolemia, haemophilia, phenylketonuria, cystic fibrosis, sickle-cell anaemia, Duchenne muscular dystrophy, emphysema, thalassemia, etc.

Question 33.
Which gene codes for Bt toxin?
Answer:
‘Cry’ gene codes for Bt toxin.

Question 34.
The a-amylase inhibitor gene transferred to tobacco from azuki bean acts against which pests?
Answer:
The a-amylase inhibitor gene transferred to tobacco from azuki bean acts against Zabrotes subjasciatus and Callosobruchus chinensis.

Question 35.
Which gene has been introduced in j sugarbeet for synthesis of fructants?
Answer:
1-sucroese sucrose fructosyl transferase gene has been introduced in sugarbeet for synthesis of fructants.

Question 36.
What percent of world population is affected by iron deficiency?
Answer:
30% of the population is affected by iron deficiency.

Question 37.
What causes softening of tomatoes during ripening?
Answer:
The enzyme polygalacturonase in tomato, breaks down pectin in the middle lamella of cell wall. This results in softening of fruits J during ripening.

Question 38.
What is superglue?
Answer:
Superglue is a biochemical glue for body repairs during surgery.

Question 39.
How is superglue produced?
Answer:
Superglue is produced by tobacco plants which contain genes encoding for adhesive proteins which allow marine mussels to stick to rocks.

Question 40.
Which oncogenes are being analyzed in transgenic mice to find out their role in development of breast cancer?
Answer:
myc and ras oncogenes are being analyzed in transgenic mice to find out their role in development of breast cancer.

Question 41.
How much milk is provided by Holstein cow on an average?
Answer:
Holstein cow provides about 6000 litres of milk per year.

Question 42.
Which gene is introduced in sheep to increase meat production?
Answer:
Human growth hormone gene is introduced in sheep for promoting growth and meat production.

Question 43.
Enlist desirable traits present in transgenic chicken?
Answer:
Desirable traits in transgenic chicken are low levels of fat and cholesterol, high protein containing eggs, in vivo resistance to viral and coccidial diseases, better feed efficiency and better meat quality.

Question 44.
Clones of which animals can provide tissues and organs for human transplants?
Answer:
The pig clone can provide animal organs and tissues for human transplants (xenotransplantation).

Question 45.
Which genes have been introduced in transgenic fish?
Answer:
Transgenic fish are transfected with E.coli hygromycin resistance gene, growth hormone and chicken crystalline protein.

Question 46.
Why does the Indian Government has set up the Genetic Engineering Approval Committee (GEAC)?
Answer:
The Indian Government has set up the Genetic Engineering Approval Committee (GEAC) to make decisions regarding the validity of research involving GMOs and addresses the safety of GMOs introduced for public use.

Question 47.
What is the duration of a biopatent?
Answer:
Duration of biopatents is five years from the date of the grant or seven years from the date of filing the patent application, whichever is less.

Question 48.
What was the first biopatent awarded for?
Answer:
First biopatent was awarded for genetically engineered bacterium ‘Pseudomonas’ used for clearing oils spills.

Question 49.
What is Texmati?
Answer:
Texmati is a trade name of “Basmati rice line and grains” for which Texas based American company Rice Tec Inc was awarded a patent by the US Patent and Trademark Office (USPTO) in 1997.

Give definitions of the following

Question 1.
Biotechnology
Answer:
Biotechnology is defined as ‘the development and utilization of biological forms, products or processes for obtaining maximum benefits to man and other forms of life’.
OR
According to OECD (Organization for Economic Cooperation and Development, 1981), biotechnology is defined as the application of scientific and engineering principles to the processing of materials by biological agents to provide goods and service to the human welfare’.

Question 2.
Genetic engineering/r-DNA technology
Answer:
Genetic engineering is defined as the manipulation of genetic material towards a desired end and in a directed and predetermined way, using in vitro process.
OR
According to John E. Smith (1996), genetic engineering as ‘the formation of new combination of heritable material by the insertion of nucleic acid molecule produced by whatever means outside the cells, into any virus, bacterial plasmid or other vector system so as to allow their incorporation into a host organism in which they do not occur naturally but in which they are capable of continued propagation’.

Question 3.
Nucleases
Answer:
Enzymes that cut the phosphodiester bonds of polynucleotide chains are called as nucleases.

Question 4.
Vector
Answer:
Vectors are DNA molecules that carry a foreign DNA segment and replicate inside the host cell.

Question 5.
Plasmid
Answer:
Plasmids are small, extra chromosomal, double stranded circular forms of DNA that replicate autonomously.

Question 6.
Transformation
Answer:
Insertion of a vector into the target bacterial cell is called transformation. (Learn this as well)

Question 7.
Transfection
Answer:
Insertion of a vector into the eukaryotic cells is called transfection. (Learn this as well)

Question 8.
Transduction
Answer:
Inserting a viral vector in cloning procedures is called transduction. (Learn this as well)

Question 9.
Gene library
Answer:
Gene libarary is a collection of different DNA sequences from an organism where each sequence has been cloned into a vector for ease of purification, storage and analysis.

Question 10.
Genomic library
Answer:
Genomic library is a collection of clones that represent the complete genome of an organism.

Question 11.
c-DNA library
Answer:
c-DNA library is a collection of clones containing c-DNAs inserted into suitable vector like a phage or plasmid.

Question 12.
Passenger DNA
Answer:
Passanger DNA is the foreign DNA which is inserted into a cloning vector.

Question 13.
Gene therapy
Answer:
Gene therapy is the treatment of genetic disorders by replacing, altering or supplementing a gene that is absent or abnormal and whose absence or abnormality is responsible for the disease.

Question 14.
Genetically modified organisms
Answer:
Genetically modified organisms are those whose genetic material has been artificially manipulated in a laboratory through genetic engineering to create combinations of plant, animal, bacterial, and viral genes that do not occur in nature or through traditional crossbreeding methods.

Name the following

Question 1.
Restriction endonucleases which produce fragments with sticky ends.
Answer:
Bam HI and Eco RI.

Question 2.
Restriction endonucleases which produce fragments with blunt ends.
Answer:
Alu I, Hind III.

Question 3.
Most commonly used vectors.
Answer:
Plasmid vectors (pBR 322, pUC, Ti plasmid) and bacteriophages (lambda phage, M13 phage.

Question 4.
Bacteriophages used as vectors.
Answer:
Ml3, lambda virus

Question 5.
Constructed plasmids.
Answer:
pBR 322, pBR 320, pACYC 177

Question 6.
Most commonly used plasmid in r-DNA technology.
Answer:
pBR 322

Question 7.
Plasmid vector for plants.
Answer:
Ti plasmid

Question 8.
Soil bacterium that causes a plant disease called crown gall.
Answer:
Agrobacterlum tumejaciens

Question 9.
Bacteria as competent host.
Answer:
Bacillus Haemophilus, Helicobacter pyroli and E. coli.

Question 10.
Most commonly used host in transformation experiments.
Answer:
E. coli

Question 11.
Cloning organisms used in plant biotechnology.
Answer:
Agrobacterium tumejaciens.

Question 12.
Human protein produced by r-DNA technology to treat anaemia.
Answer:
Erythropoeitin

Question 13.
Human protein produced by r-DNA technology to treat asthma.
Answer:
Interleukin 1 receptor

Question 14.
Human protein produced by r-DNA technology to treat antherosclerosis.
Answer:
Platelet derived growth factor

Question 15.
Human protein produced by r-DNA technology to treat parturition.
Answer:
Relaxin

Question 16.
Human protein produced by r-DNA technology to treat cancer.
Answer:
Interferons, tumour necrosis factor, inter-leukins, macrophage activating factor.

Question 17.
Recombinant protein used in the treatment of haemophiliaA.
Answer:
Factor VIII

Question 18.
Recombinant protein used in the treatment of haemophiliaB.
Answer:
Factor IX

Question 19.
Transgenic food crop used to reduce vitamin A deficiency diseases.
Answer:
Golden rice

Question 20.
Human proteins expressed in cow milk.
Answer:
Human lactoferrin, human alpha lactalbumin, human serum albumin, human bile salt.

Question 21.
Bacterial genes concerned with biosynthesis of cystein.
Answer:
cys E, cys M

Question 22.
Transgenic fish.
Answer:
Atlantic salmon, catfish, goldfish, Tilapia, zebra-fish, common carp, rainbow trout.

Give significance or functions of the following

Question 1.
Transgenic plants.
Answer:

1. Transgenic plants Eire genetically engineered to carry various desirable traits.
2. They are resistant to bacterial and viral diseases (e.g. tomato, potato, etc), insect pests (Bt cotton), herbicides (e.g. maize and wheat) and abiotic stresses.
3. Transgenic plants can be used as bioreactors or factories (molecular farming) for production of novel drugs like interferons, humanized antibodies against infective agents like HIV, amino acids and immunotherapeutic drugs.
4. Some transgenic plants have improved nutritional qualities, e.g. golden rice and golden mustard are biofortified with vitamin A.
5. Transgenic plants like Flavr savr tomatoes have been engineered to have more shelf life.
6. Transgenic plants also produce edible vaccines, e.g. Potato, tomato, etc.

Question 2.
Transgenic animals.
Answer:

1. Transgenic animals are genetically modified animals that are used in a wide range of fields.
2. In the field of medical research, transgenic animals like mice are used to identify the functions of specific factors through over- or under-expression of a modified gene (the inserted transgene). They are designed to study how genes contribute to the development of disease. These animals are used to investigate development of diseases like cancer, cystic fibrosis, rheumatoid arthritis, etc.
3. In toxicology field, they are used for detection of toxicants. They are used as responsive test animals.
   Transgenic animals are used to evaluate a specific genetic change in molecular biology studies.
4. In pharmaceutical industry, transgenic animals are used for targeted production of pharmaceutical proteins, drug production and product efficacy testing.
5. They are also used in study of mammalian developmental genetics.

Transgenic farm animals like cattle, sheep, v poultry, pigs, etc. exhibit many desirable traits.

1. Improved quantity and quality of meat
2. Improved quality and quantity of milk
3. More egg production
4. Better quality and quantity of wool
5. Disease resistance
6. Production of low-cost pharmaceuticals and biologicals

Distinguish between the following

Question 1.
Old or classical biotechnology and Modem biotechnology.
Answer:

Old or classical biotechnology	Modem biotechnology
1. It is mainly based on fermentation technology.	1. It is based on genetic engineering and chemical engineering.
2. It does not involve alteration or modification of genetic material of organisms to develop specific Product.	2. It involves the alteration or modification of genetic material of organisms to develop specific product.
3. It does not involve the use of r-DNA technology, polymerase chain reaction (PCR), microarrays, cell culture and fusion, and bioprocessing to develop products.	3. It involves the use of r-DNA technology, polymerase chain reaction (PCR), microarrays, cell culture and fusion, and bioprocessing to develop specific products.
4. There is no ownership of old biotechnology.	4. It involves ownership of technology.
5. Examples : Preparation of curd, ghee, soma, vinegar, yogurt, cheese making, wine making, etc.	5. Examples : Transgenic organisms.

Question 2.
Genomic library and c-DNA library.
Answer:

Genomic library	c-DNA library
1. It is a collection of clones that represent the complete genome of an organism.	1. It is a collection of clones containing c-DNAs inserted into suitable vectors like phages or plasmids.
2. DNA fragments to be cloned are obtained by cutting genomic DNA by restriction enzymes.	2. c-DNAs are produced by the process of reverse transcription, using complete m-RNA complement obtained from a tissue or an organism.

Question 3.
Germ line gene therapy and somatic cell gene therapy.
Answer:

Germ line gene therapy	Somatic cell gene therapy
1. Germ line gene therapy involves modification of genome of germ cells like sperms, eggs, early embryos.	1. Somatic cell gene therapy involves modification of genome of somatic cells like bone marrow cells, hepatic cells, fibroblasts, endothelium, pulmonary epithelial cells, central nervous system and smooth muscle cells of wall of blood vessels.
2. It allows transmission of the modified genetic information to the next generation.	2. It does not allow transmission of the modified genetic information to the next generation.
3. Its application in human beings is not encouraged because of technical and ethical reasons.	3. It is a feasible option and clinical trials are carried out for treatment of various diseases.

Give reason

Question 1.
The genetic engineering is alternatively called recombinant DNA technology or gene cloning.
Answer:

1. Genetic engineering involves the manipulation of genetic material in a directed, predetermined, in vitro way to develop a desired end product.
2. It manipulates the genes for improvement of living organisms,
3. It involves repairing of the defective genes, replacing of defective genes by healthy genes or normal genes and artificially synthesizing of a totally new gene.
4. Genetic engineering also involves transfer of a new gene, transfer of genes to a new location or into a new organism, gene cloning, combining of genes from two organisms.
5. This results in alteration of the genotype and desired products can be developed.
6. Therefore, the genetic engineering is alternatively called recombinant DNA technology or gene cloning.

Question 2.
Bacteria have restriction enzymes.
Answer:

1. Restriction endonucleases or restriction enzymes in bacteria help them to recognize and destroy various viral DNAs that might enter the cell.
2. They cut the phosphodiester back bone at highly specific sites on both strands of DNA.
3. Thus, these enzymes restrict the potential growth of the virus and protect bacteria.
4. Hence, bacteria produce restriction enzymes.

Question 3.
All the fragments of a genome are cloned for storing them in genomic library.
Answer:

1. Genomic DNA is fragmented at the time of preparing genomic library.
2. It is not known which fragment has the desired gene.
3. Therefore all the fragments have to be cloned to store the copies of each separately.
4. Screening for the desired gene is later done through complementation or using DNA probes.
5. Therefore, all the fragments of a genome are cloned for storing them in genomic library.

Question 4.
The establishment of genomic library is more meaningful in prokaryotes than in eukaryotes.
Answer:

1. The prokaryotic genome does not contain repetitive DNA.
2. Eukaryotic DNA genome contains introns, regulatory genes and repetitive DNA.
3. Hence, the establishment of genomic library is more meaningful in prokaryotes than in eukaryotes.

Question 5.
Flavr savr tomato has longer shelf life.
Answer:

1. Flavr savr is genetically modified type of tomato.
2. It is developed by inserting antisense gene which retards ripening.
3. Due to the presence of this gene a cell wall degrading enzyme called polygalactouronase is produced in lesser amounts.
4. Owing to the above reason, Flavr savr tomato has longer shelf life.

Question 6.
Pigs are regarded as the most suitable animals to be bred for heart transplant.
Answer:

1. A pig’s heart is about the same size as a human heart.
2. Pig heart valves are used in human heart surgery for over a decade.
3. Hence, pigs are regarded as the most suitable animals to be bred for heart transplant.

Question 7.
Patent jointly issued by Delta and Pineland Company and U. S. department of agriculture under the title ‘control of plant gene expression’ was not granted by Indian government.
Answer:

1. Patent jointly issued by Delta and Pineland Company and U. S. department of agriculture under the title ‘control of plant gene expression’ is based on a gene that produces a protein toxic to plant and thus prevents seed germination.
2. Because of such patents, financially powerful corporations would acquire monopoly over biotechnological process.
3. Thus it would pose a threat to global food security.
4. Therefore this patent was considered morally unacceptable and fundamentally unequitable and it was not granted by the Indian government.

Write short notes

Question 1.
Electrophoresis.
Answer:

1. Electrophoresis is a technique that involves migration and separation of charged particles under the influence of electric field.
2. It is used for the separation of charged molecules like DNA, RNA and proteins, by application of an electric field.
3. Movement of charged particles is determined by particle size, shape and charge.
4. DNA is negatively charged and hence it moves towards the positive anode.
5. It results in size separation of DNA fragments as small fragments of DNA molecules move faster.
6. Different types of electrophoresis are agarose gel electrophoresis, PAGE and SDA PAGE.

Question 2.
Recognition sequences or restriction sites.
Answer:

1. Restriction endonucleases have the ability to recognize specific sequences in DNA and cleave it.
2. They are 4 to 8 nucleotides long and characterized by a particular type of internal symmetry.
3. The specific site at which restriction endonuclease cuts the DNA is called recognition site or restriction site.
4. Each restriction endonuclease recognizes its specific recognition sequence.
5. Restriction cutting may result in DNA fragments with blunt ends or cohesive or sticky ends or staggered ends (having short, single stranded projections).

For example, recognition sequence of by the enzyme EcoRI is
3′ —– -CTTAA-G —–5′
5′ —– -G A A T T C —–3′
It is as palindrome, i.e. When read on opposite strand of DNA (3′ to 5′ or 5′ to 3′) it reads same.
When the enzyme EcoRI recognizes this sequence, it breaks each strand at the same site in the sequence i.e. between the A and G residues.

Question 3.
Plasmids as cloning vectors.
Answer:

1. Plasmids are small, extra-chromosomal, double stranded circular forms of DNA that replicate autonomously.
2. They are seen in bacterial cells, yeast and animal cell.
3. Plasmids are considered as replicons as they are capable of autonomous replication in suitable host.
4. The most commonly used vectors in r-DNA technology are plasmids as they replicate in E. coli.

Plasmid as a cloning vector should have a replication origin, a marker gene for antibiotic resistance, control elements like promoter, operator, ribosome binding site, etc. and a region where foreign DNA can be inserted. Naturally plasmids do not have all these features. Hence, they are constructed by inserting gene for antibiotic resistance. pBR 322, pBR320, paCYC177 are the constructed plasmids.

Ti plasmid (for tumor-inducing) of Agrobacterium tumefaciens is an important vector for carrying new DNA in many plants. It contains a transposon, called T DNA, which inserts copies of itself into the chromosomes of infected plant cells. The transposon, with the new DNA, can be inserted into the host cell’s chromosomes. A plant cell containing this DNA, can then be grown in culture or induced to form a new, transgenic plant.

Question 4.
Bt cotton.
Answer:

1. Bt cotton is well known example of insect resistant transgenic plant which is engineered with a gene from B. thuringiensis.
2. ‘cry’ gene present in B. thuringiensis produces a protein that forms crystalline inclusions in bacterial spores.
3. When insect ingests it, because of high pH and the proteinase enzymes present in insect’s midgut, the crystalline inclusions are hydrolyzed to release the core toxic fragments.
4. This toxin causes midgut paralysis and disruption of midgut cells of insect.
5. Bt toxin acts against many species of Lepidoptera, Diptera and Coleoptera insects.

Question 5.
Golden rice.
Answer:

1. Golden rice is a transgenic plant developed by Swiss researchers.
2. It contain genes from the soil bacterium Erwinia and either maize or daffodil plants.
3. These plants are biofortified to have high content of vitamin A.
4. The golden colour is due to vitamin A.
5. Consumption of golden rice and golden mustard can reduce occurrence of vitamin A deficiency diseases (VAD).

Question 6.
Insulin.
Answer:

1. Insulin is a peptide hormone produced by β -cells of islets of Langerhans of pancreas.
2. Insulin is essential for the control of blood sugar levels.
3. Disease Diabetes mellitus is caused due to inability to make insulin.
4. Insulin was discovered by Sir Edward Sharpey Schafer (1916) while studying Islets of Langerhans.
5. Hakura et al (1977) chemically synthesized DNA sequence of insulin for two chains A and B and separately inserted into two PBR322 plasmid vector.

Gilbert and Villokomaroff, 1978 produced insulin production using r-DNA technology.

1. The recombinant plasmids, containing insulin gene inserted by the side of β-galactosidase, were transferred into E. coli host.
2. The host produced penicillinase pnd pre-pro insulin.
3. Insulin is later separated by trypsin treatment.

Question 7.
Transgenic cattle.
Answer:

1. Transgenic cattle are used for food production and for the production of human therapeutic proteins.
2. Transgenic cattle engineered with additional copies of bovine beta or kappa casein, show 8 to 20% increase in beta casein and a two-fold increase in kappa casein.
3. Various human proteins like Human lactoferrin, human alpha lactalbumin, human serum albumin and human bile salt stimulated lipase are expressed in the milk of transgenic cattle.
4. Transgenic cows produce factor IX (plasma thromboplastin component), used in the treatment of haemophilia.
5. Tracy, the transgenic cow born in Scotland, could produce a human protein in her milk for human therapeutics.
6. Human antibody products are made using transgenic cows that are immunized with a vaccine containing the disease agent. Antibodies are currently used for treatment of infectious diseases, cancer, transplanted organ rejection, autoimmune diseases and for use as antitoxins.

Question 8.
Transgenic fish.
Answer:

1. The commercially important fish like Atlantic salmon, catfish, goldfish, Tilapia, zebra-fish, common carp, rainbow trout, etc. are transfected with growth hormone, chicken crystalline protein and E.coli hygromycin resistance gene.
2. Transgenic fish showed increased cold tolerance and improved growth.

Short Answer Questions

Question 1.
What are the two phases of the development of biotechnology in terms of its growth?
Answer:
Two phases of the development of biotechnology in terms of its growth are as follows:
Traditional biotechnology (old biotechnology) : It is based on fermentation technology that uses microorganisms in the preparation of curd, ghee, soma, vinegar, yogurt, cheese making, wine making, etc.

Modern biotechnology (new biotechnology) :

1. During 1970 ‘recombinant DNA technology was developed and then established by Stanley Cohen and Herbert Boyer in 1973.
2. This technique alters or modifies genetic material to develop of new products.
3. The combination of biology and production technology based on genetic engineering evolved into modern biotechnology.
4. Modern biotechnology is based on two core techniques-genetic engineering and chemical engineering.

Question 2.
What are the different techniques and devices used in r-DNA technology?
Answer:
(1) Several techniques are used in r-DNA technology to isolate and characterize the macromolecules like DNA, RNA, proteins.
(2) The techniques used on the basis of molecular weight are gel permeation, osmotic pressure, ion exchange chromatography, spectroscopy, mass spectrometry, electrophoresis, etc.

(3) Electrophoresis:

• It is used for the separation of charged molecules like DNA, RNA and proteins, by application of an electric field.
• Different types of electrophoresis : Agarose gel electrophoresis, PAGE, SDA PAGE.

(4) Polymerase chain reaction (PCR) : It is used for in vitro gene cloning or gene multiplication to produce a billion copies of the desired segment of DNA or RNA, with high accuracy and specificity, in few hours.

Question 3.
What are the basic requirements of PCR technique?
Answer:
The basic requirements of PCR technique are as follows:

1. DNA containing the desired segment to be amplified.
2. Excess of forward and reverse primers which are synthetic oligonucleotides of 17 to 30 nucleotide.
3. They are complementary to the sequences present in DNA.
   dNTPs which are of four types such as dATB dGTB dTTP and dCTR
4. A thermostable DNA polymerase (e.g. Taq DNA polymerase enzyme) that can withstand a high temperature of 90-98°C.
5. Appropriate quantities of Mg⁺⁺ ions.
6. Thermal cycler, a device required to carry out PCR reactions.

Question 4.
What are the two types of nucleases? What is their function?
Answer:

1. The two types of nucleases are exonucleases and endonucleases.
2. Exonucleases remove nucleotides from the ends of the DNA.
3. Endonucleases are those enzymes that have ability to make cuts at specific positions within the DNA molecule.
4. Of the endonucleases, restriction endonucleases serve as the molecular scissors in genetic engineering experiments.
5. They are used for cutting DNA molecules at specific predetermined sites. This is needed for gene cloning or recombinant DNA technology.

Question 5.
Explain different types of restriction enzymes?
Answer:
Different types of restrictions enzymes are as follows:

1. Type I – They function . simultaneously as endonuclease and methylase e.g. EcoK.
2. Type II – They exhibit separate cleaving and methylation activities. They are more stable and are used in r-DNA technology e.g. EcoRI, Bgll. They cut DNA at specific sites within the pallindrome. Thousands of type II restriction enzymes have been discovered.
3. Type III – They cut DNA at specific non- palindromic sequences e.g. Hpal, MboII.

Question 6.
With the help of a suitable example, illustrate palindrome.
Answer:

1. Palindrome is a sequence which when read on opposite strands of DNA (3′ to 5’ or 5’ to 3’), reads same.
2. When the enzyme EcoRI recognizes this sequence, it breaks each between the A and G residues.
3. In palindrome, the base sequence of second half in DNA represents the mirror image of the base sequence of the first half.
4. Palindromes are actually groups of letters which form the same word when read in both forward and backward directions.

For example, recognition sequence of by the enzyme EcoRI is a palindrome.
3′ —— – C T T A A G—–5′
5′ —— – G A A T T C—–3′

Same restriction enzyme must be used to cut vector and donor DNA, because it will produce fragments with the same complementary sticky ends, making it bond formation possible between them.

Question 7.
Explain how Ti plasmid of Agrobacterium tumefaciens acts as a vector for transferring genes to plants?
Answer:

1. Agrobacterium tumefaciens is a soil bacterium that causes crown gall disease in plants.
2. This disease involves the formation of tumor in the plant.
3. Ti plasmid in A. tumefaciens contains a transposon called T DNA.
4. T DNA inserts copies of itself into the chromosomes of infected plant cells.
5. The transposons, with the foreign DNA, can be inserted into the host cell’s chromosomes.
6. A plant cell containing this DNA, can then be grown in culture or induced to form a new, transgenic plant.

construction of Genomic library:

1. When genomic library is constructed, the entire genome or DNA is isolated from a particular organism.
2. This DNA is fragmented using suitable restriction endonucleases.
3. These separated fragments are later inserted into cloning vectors.
4. This develops recombinant vectors.
5. Such recombinant vectors are transferred into suitable organisms such as bacteria or yeast. Each host cell then contains one fragment.
6. These transformed organisms are cultured and their clones are thus produced. These clones are stored in the genomic library.

Question 8.
Explain with example how transformed host cells are selected from non- transformed?
Answer:

1. Transformed recombinant cells are selected using marker genes.
2. For example, markers genes in pBR 322 plasmid are ampicillin resistant gene and tetracyclin resistant gene.
3. When pstl restriction enzyme is used, ampicillin resistant gene gets knocked off from the plasmid and recombinant cells become sensitive to ampicillin.

Question 9.
Give the types of human proteins and hormones produced by recombinant DNA techniques.
Answer:

1. Blood proteins produced by recombinant DNA technique are Erythropoeitin, Tissue plasminogen activator, urokinase, Factor VIII, Factor IX, etc.
2. Hormones : Insulin, Epidermal growth factor.

Question 10.
What are edible vaccines? How are they produced?
Answer:

1. Edible vaccine is an edible plant part engineered to produce an immunogenic protein, which when consumed gets recognized by immune system.
2. Immunogenic protein of certain pathogens are active when administered orally.
3. The gene encoding for immunogenic protein is isolated and inserted in a suitable vector.
4. Recombinant vector is then transferred to plant genome.
5. Expression of this gene in specific parts of the plant results in the synthesis of immunogenic proteins.
6. When animals or mainly humans consume these plant parts, they get vaccinated against certain pathogen.

Question 11.
Give an example of ‘melt in mouth’ vaccine and state the advantages of such vaccines.
Answer:

1. Example of ‘melt in the mouth’ vaccine that can be administered by placing it under tongue, is the production of flu vaccine by Bacillus which melts in the mouth and get delivered into the blood stream.
2. Advantages of edible oral vaccines are that they can be easily administered, can be easily stored and they are of low cost.

Question 12.
What is meant by recalcitrant seeds? How such plants can be conserved?
Answer:

1. Recalcitrant seeds are those whose survival is educed upon drying (reduction in moisture below a certain level) and freezing and thus are difficult to store.
2. It involves subcellular damage of seeds which results in loss of viability, when dried.
3. Plants which produce recalcitrant sees could be conserved using tissue culture technique.

Question 13.
What are the different ways in which gene therapy is used?
Answer:
Gene therapy is being used as follows:

1. Replacement of missing or defective genes.
2. Delivery of genes that speed the destruction of cancer cells.
3. Supply of genes that cause cancer cells to revert back to normal cells.
4. Delivery of bacterial or viral genes as a form of vaccination.
5. Delivery of DNA to antigen expression and generation of immune response.
6. Supply of gene for impairing viral replication.
7. Provide genes that promote or impede the growth of new tissue.
8. Deliver genes that stimulate the healing of damaged tissue.

Question 14.
What are the different types in which genes could be delivered during gene therapy?
Answer:
Genes can be delivered by three ways:

1. Ex vivo delivery : In this type of gene delivery, viral or non-viral vectors are used to introduce the desired gene in the cells isolated from patient, e.g. Parkinson’s disease, a neurological disorder.
2. In vivo delivery : In this method, therapeutic genes are directly delivered at the target sites of the cells of diseased tissue. Intravenous infusion genes are injected directly into tumor in the treatment of cancer.
3. Use of virosomes (Liposome + inactivated HIV), bionic chips.

Question 15.
What are transgenic plants? Explain with any two examples.
Answer:
The genetically engineered crop plants carrying desirable traits are called transgenic plants.
Examples of transgenic plants:

1. Bt Cotton : Bt cotton is a transgenic plant. Bt toxin gene has been cloned and introduced in many plants to provide resistance to insects without the need of insecticides.
2. Golden rice : It is a genetically engineered rice with higher beta carotene (provitamin A) content.
3. Flavr savr tomato : It is developed by inhibiting synthesis of polygalactournonase by inserting antisense gene. This type of tomato has a longer shelf life.

Question 16.
Give any two examples of insect resistant transgenic crops.
Answer:
Examples of insect resistant transgenic crops are as follows:
(1) BT crops:

• Insect resistant transgenic plants contain either a gene from B. thuringiensis or the cowpea trypsin inhibitor gene.
• ‘cry’ gene present in B. thuringiensis produces a protein that forms crystalline inclusions in bacterial spores. When insect ingests it, because of high pH and the proteinase enzymes present in insect’s midgut, they are hydrolyzed to release the core toxic fragments.
• This toxin causes midgut paralysis and disruption of midgut cells of insect.
• Bt toxin activity has been against many species of insects within the orders of Lepidoptera, Diptera and Coleoptera.

(2) Transgenic tobacco:

• The gene of a-amylase inhibitor (aAl-Pv), isolated from adzuki bean (Phaseolus vulgaris) is transferred to tobacco.
• This gene works against pests like Zabrotes subfasciatus and Callosobruchus chinensis.

Question 17.
Give any two examples of biofortified transgenic crops.
Answer:
Examples of biofortified transgenic crops are as follows:

1. Golden rice and Golden mustard These are transgenics rich in vitamin A.
2. Arabidopsis genes are transferred to soybean, oil palm, rapeseed and sunflower for improvement in oil content and oil quality.
3. Ferritin, an iron storage protein, isolated from soybean and Phaseolus is transferred to rice to increase its iron content.
4. Plants deficient in amino acids like methionine, lysine and tryptophan have been engineered to improve protein content.

Question 18.
What factors are responsible for losses during storage and transport of crops? Explain, with example, how genetic engineering can reduce these losses?
Answer:

1. Diseases and pests, bruising on soft fruits and vegetables, heat and cold storage, over-ripeness, loss of flavours and odours, etc. lead to great deal of losses during storage and transport of crops.
2. Most of these changes are caused due to endogenous enzyme activities which could be slowed down using genetic engineering.

For example, shelf life of Flavr savr tomatoes has been increased using genetic engineering techniques.
(a) The enzyme polygalacturonase breaks down pectin in the cell wall, leading to softening of fruit during ripening of tomatoes.
(b) In genetically modified Flavr savr tomatoes, polygalactouronase enzyme is inhibited by antisense genes. These tomatoes can remain on the vine until mature and be transported in a firm solid state.

Question 19.
How transgenic animals are produced.?
Answer:

1. Transgenic animals are produced using recombinant DNA technology.
2. Foreign DNA is introduced in transgenic animals using r-DNA technology.
3. It is then transmitted through the germ line so that every cell if the animal contains the same modified genetic material.
4. This involves cloning of desired gene and introduction of cloned gene into fertilized eggs, successful implantation of modified eggs into receptive female and obtaining progeny carrying cloned genes.

Question 20.
Write any two scientific and commercial values of transgenic animals in favour of human beings.
Answer:
(1) Scientific value of transgenic animals:

• Transgenic mice are used medical research to identify the functions of specific factors through over – or under-expression of a modified gene (the inserted transgene). They are designed to study how genes contribute to the development of disease. These animals are used to investigate development of diseases like cancer, cystic fibrosis, rheumatoid arthritis, etc.
• In toxicology field, they are used for detection of toxicants. They are used as responsive test animals.

(2) Commercial value of transgenic animals:

• Transgenic cattle are used for production of human therapeutic proteins such as human lactoferin, human serum albumin, etc.
• Better quality and quantity of wool by transgenic sheep.

Question 21.
How sheep are genetically altered to produce wool of better quality?
Answer:

1. Bacterial genes, cys E and cys M, are identified, cloned and introduced in sheep.
2. These genes are involved in biosynthesis of cysteine.
3. Cysteine is involved in formation of keratin protein found in wool.
4. Thus, transgenic sheep produce more quantity and better quality of wool.

Question 22.
What is meant by ethics?
Answer:

1. Ethics is a discipline concerned with moral values or principles.
2. It deals with certain sets of standards which regulate behaviour of community.
3. It is concerned with socially accepted norms of moral duty, conduct and judgment.
4. Ethical concepts differ according to culture and traditions.
5. They also change with time and get influenced by progress in science and technology.

Question 23.
What are the adverse effects of Biotechnology on the Environment?
Answer:
The adverse effects of biotechnology on the environment are as follows:

1. Unintended hybrid strains of weeds and other plants can develop resistance to herbicides through cross-pollination. E.g. Crops of Round Up-ready soybeans which are used in agriculture, possibly confer Round Up resistance to neighbouring plants.
2. Bt corn has adversely affected non target species – Monarch butterfly. It may also prove harmful to neutral or even beneficial species.

Question 24.
Discuss various health concerns regarding the use of GMO crops.
Answer:
Various concerned related to health regarding the use of GMO crops are as follows:

1. GMO crops may develop some allergies, e.g. A gene from the Brazil nut was transferred to soybean in order to increase methionine content. But this transgenic soybean has caused allergies in some people which are known to suffer from nut allergies (“Biotech Soybeans”).
2. GMO technology is a recent development and its the long-term effects on health cannot be anticipated at this point.
3. Potential effects of transgenic proteins, which were never been ingested earlier, on the human body are yet unknown.
4. The use of GMOs may lead to the development of antibiotic and vaccine- resistant strains of diseases.

Question 25.
What is a patent?
Answer:

1. Patent is a special right granted to the inventor by the government.
2. A patent consists of three parts – grant (an agreement with the inventor), specification (subject matter of invention) and claims (scope of invention to be protected).
3. Patent is a personal property of inventor and it can be sold like any other property.

Question 26.
What is meant by traditional knowledge? What is its importance?
Answer:

1. Traditional knowledge is a deep understanding of ecological processes and the ability to obtain useful products from the local habitat in a sustainable way.
2. Most traditional knowledge is handed down through generations. This helps in the development of modern, commercial applications. This saves the makers time, money and effort.

Traditional knowledge includes:

1. Knowledge about food, crop varieties and agricultural/farming practice.
2. Sustainable management of natural resources and conservation of biological diversity.
3. Biologically important medicines.

Chart or Table based Questions

Question 1.

RE	Source	End products
Alu I	————-	Blunt ends
————	Bacillus amyloliquefaciens H	Sticky ends
Eco R I	————–	Sticky ends
Hind II	H. influenza Rd	————–

Answer:

RE	Source	End products
Alu I	Arthobacter luteus	Blunt ends
Bam H I	Bacillus amyloliquefaciens H	Sticky ends
Eco R I	E. coli Ry 13	Sticky ends
Hind II	H. influenza Rd	Blunt ends

Question 2.

Substance	Potential benefit	Crop	Transgene
Provitamin A	Anti-oxidant	————–	Phytoene synthase, Lycopene cyclase
Fructans	————–	Sugarbeet	—————-
Vitamin E	—————-	Canola	γ -tocopherol methyl transferase
Flavonoids	Anti-oxident	Tomato	————–
————–	Iron fortification	Rice	Ferritin, metallothioein, phytase

Answer:

Substance	Potential benefit	Crop	Transgene
Provitamin A	Anti-oxidant	Rice	Phytoene synthase, Lycopene cyclase
Fructans	Low calories	Sugarbeet	I – sucrose : sucrose fructosyl transferase
Vitamin E	Anti-oxidant	Canola	γ -tocopherol methyl transferase
Flavonoids	Anti-oxident	Tomato	Chalone isomerase
Iron	Iron fortification	Rice	Ferritin, metallothioein, phytase

Question 3.

Organization	Expand
OECD	——————
GEAC	——————
USDA	——————
USPTO	——————
CSIR	—————–

Answer:

Organization	Expand
OECD	Organization for Economic Cooperation and Development
GEAC	Genetic Engineering Approval Committee
USDA	US Department of Agriculture
USPTO	US Patent and Trademark Office
CSIR	Council of Scientific and Industrial Research

Diagram Based Questions

Question 1.
(a) Name the reaction shown in the given diagram.
(b) What are the three steps in this reaction?
Answer:

(a) Polymerase chain reaction
(b) Three steps of polymerase chain reaction are denaturation of DNA, annealing of primer and extension of primer.

Question 2.
(a) Which enzyme has recognition sequence shown in the diagram given below?
(b) What is meant by palindrome?
(c) Enzyme in the given diagram cuts DNA to produce ———- ends.

Answer:
(a) Enzyme is EcoRI.
(b) Palindrome is a DNA sequence which when read in opposite direction (3’ to 5’ or 5’ to 3’) it reads same.
(c) Enzyme in the given diagram cuts DNA to produce sticky ends.

Question 3.
Draw a labelled diagram of a plasmid showing ori, ampr and a region into which foreign DNA can be inserted.
Answer:

Question 4.
Draw a diagram showing steps in r-DNA technology.
Answer:

Long Answer Questions

Question 1.
Explain with examples how transgenic plants can be used as factories or bioreactors?
Answer:

1. Transgenic plants are potential factories or bioreactors for biochemicals and biopharmaceuticals like starch, sugar, lipids, proteins, hormones, antibodies, vaccines or enzymes.
2. Various fine chemicals, perfumes, adhesive compounds industrial lubricants, etc. can be isolated from plants.
3. Plants can be the source of biodegradable plastic and ‘renewable’ energy to replace fossil fuels.
4. Transgenic plants useful for production of novel drugs like interferons, edible vaccines, antibodies, amino acids, immunotherapeutic drugs, etc. Thus, they are like bioreactors for molecular farming.

Examples:

1. The gene for Human growth hormone has been inserted into the chloroplast DNA of tobacco plants.
2. Humanized antibodies against HIV, Respiratory syncytial virus (RSV), Herpes simplex virus (HSV), the cause of “cold sores” are developed using transgenic plants.
3. Superglue’ is a biochemical glue for body repairs during surgery. It is produced by tobacco plants which contain genes encoding for adhesive proteins which allow marine mussels to stick to rocks.
4. Protein antigens to be used in vaccines : e.g. Patient-specific antilymphoma vaccines. B-cell lymphomas are clones of malignant B cells expressing a unique antibody molecule on their surface.
5. Transgenic plants cam be used as factories for producing oil having nutritional value like cod-liver oil. These plants are engineered with a oil encoding gene from marine algae.
6. Transgenic plants produce the antimalarial drug, Artemisinin.
7. Genetically engineered opium poppy can be used to produce powerful painkillers.
8. Transgenic plants like potatoes, tomatoes, bananas, soybeans, alfalfa and cereals can be used as edible vaccine.

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 11 Enhancement of Food Production Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 11 Enhancement of Food Production

Multiple choice questions

Question 1.
Means of in situ germplasm conservation are ………………….
(a) forests
(b) Natural Reserves
(c) botanical gardens, seed banks
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 2.
Means of ex situ germplasm conservation are ………………….
(a) forests and seed banks
(b) natural Reserves
(c) botanical gardens, seed banks
(d) botanical garden and forests
Answer:
(c) botanical gardens, seed banks

Question 3.
Germplasm includes ………………….
(a) only improved varieties of crop
(b) all cultivated varieties and wild relatives of a particular crop
(c) all hybridized varieties only
(d) only mutant varieties of a crop
Answer:
(b) all cultivated varieties and wild relatives of a particular crop

Question 4.
Taichung Native-1 is a variety of rice from ………………….
(a) China
(b) Korea
(c) Malaysia
(d) Taiwan
Answer:
(d) Taiwan

Question 5.
Which of the following is not a fungal disease ?
(a) Late blight of potato
(b) Brown rust of wheat
(c) Red rot of sugar cane
(d) Black rot of crucifers
Answer:
(d) Black rot of crucifer

Question 6.
…………………. variety of wheat is resistant to Hill bunt disease.
(a) Himgiri
(b) Pusa swarnim
(c) Kalyan sona
(d) Pusa A-4
Answer:
(a) Himgiri

Question 7.
Regina-II variety of …………………. is resistant to bacterial rot.
(a) wheat
(b) cabbage
(c) cauliflower
(d) Brassica
Answer:
(b) cabbage

Question 8.
The nectar-less cotton having smooth leaves has resistance against ………………….
(a) bollworms
(b) jassids
(c) aphids
(d) stem borers
Answer:
(a) bollworms

Question 9.
………………… gave concept of in vitro cell culture.
(a) Haberlandt
(b) Frank
(c) Yabuta and Sumiki
(d) None of these
Answer:
(a) Haberlandt

Question 10.
Tissue culture requirements are ………………….
(a) pH of nutrient medium 5 to 5.8
(b) Sterilized glassware, nutrient medium, explants, inoculation chamber
(c) Temperature 18 °C to 20 °C
(d) All of these
Answer:
(d) All of these

Question 11.
Hybrid maize with double the quantity of amino acids …………………. have been developed.
(a) lysine and tryptophan
(b) alanine and aspartic acid
(c) glutamic and proline
(d) histidine and cystine
Answer:
(a) lysine and tryptophan

Question 12.
Inbreeding increases ………………….
(a) homozygosity
(b) heterozygosity
(c) heterosis
(d) hemizygosity
Answer:
(a) homozygosity

Question 13.
Which hormone is used for MOET method?
(a) GH
(b) LH
(c) FSH
(d) ICSH
Answer:
(c) FSH

Question 14.
Find the odd one out.
(a) Sahiwal
(b) Sindhi
(c) Gir
(d) Holstein
Answer:
(d) Holstein

Question 15.
Pullorum is a …………………. disease.
(a) viral
(b) bacterial
(c) fungal
(d) parasitic
Answer:
(b) bacterial

Question 16.
Propolis is ………………….
(a) bee glu
(b) royal jelly
(c) bee venom
(d) none of these
Answer:
(a) bee glu

Question 17.
Select the incorrect pair.
(a) Apis dorsata – Rock bee
(b) Apis indica -Indian bee
(c) Apis florae – Little bee
(d) Apis mellifera – Wild bee
Answer:
(d) Apis mellifera – Wild bee

Question 18.
fishery takes place in Sundarban area of ………………….
(a) Estuarine, West Bengal
(b) Marine, Odisha
(c) Fresh water, West Bengal
(d) None of these
Answer:
(a) Estuarine, West Bengal

Question 19.
Which stage in the life cycle of silk moth secretes silk?
(a) Caterpillar
(b) Egg
(c) Pupa
(d) Adult
Answer:
(a) Caterpillar

Question 20.
Lac insect is a native of ………………….
(a) China
(b) India
(c) Africa
(d) Europe
Answer:
(b) India

Question 21.
Alcoholic fermentation is brought about by ………………….
(a) Lactobacillus
(b) Saccharomyces
(c) Trichoderma
(d) Streptomyces
Answer:
(b) Saccharomyces

Question 22.
…………………. and …………………. are alcoholic beverages produced without distillation.
(a) Wine, rum
(b) Wine, beer
(c) Whisky, brandy
(d) Brandy, beer
Answer:
(b) Wine, beer

Question 23.
Organic acid can be produced directly from glucose or formed as end products from ………………….
(a) pyruvate
(b) ethanol
(c) gluconic acid
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 24.
The microbial source of vinegar is ………………….
(a) Aspergillus niger
(b) Rhizopus arrhizus
(c) Acetobacter aceti
(d) Streptomyces venezuelae
Answer:
(c) Acetobacter aceti

Question 25.
Statins are produced by ………………….
(a) Monascus purpureus
(b) Streptococcus
(c) Aspergillus niger
(d) None of these
Answer:
(a) Monascus purpureus

Question 26.
…………………. is used as a ‘clot buster’.
(a) Pectinase
(b) Statin
(c) Invertase
(d) Streptokinase
Answer:
(d) Streptokinase

Question 27.
Aspergillus niger is used to manufacture ………………….
(a) pectinase, gluconic acid and vitamin C
(b) pectinase, gluconic acid and vitamin B₁₂
(c) invertase, acetic acid and vitamin C
(d) pectinase, citric acid and invertase
Answer:
(a) pectinase, gluconic acid and vitamin C

Question 28.
The first Gibberellin was isolated by ………………….
(a) Frank
(b) Skoog
(c) Yabuta and Sumiki
(d) None of these
Answer:
(c) Yabuta and Sumiki

Question 29.
Once the BOD of waste water is reduced, it is passed into a ………………….
(a) settling tank
(b) primary sedimentation tank
(c) anaerobic sludge digesters
(d) aeration tank
Answer:
(a) settling tank

Question 30.
During biogas production species used to bring about hydrolysis or solubilization is ………………….
(a) Pseudomonas
(b) Rhizopus
(c) Methanococcus
(d) Methanobacillus
Answer:
(a) Pseudomonas

Question 31.
…………………. bacteria are used as herbicides.
(a) Pseudomonas spp., Xanthomonas spp., Agrobacterium spp.
(b) Bacillus thuringiensis, B. papilliae, B. lentimorbus
(c) Pseudomonas spp., Bacillus thuringiensis, B. papilliae
(d) Xanthomonas spp., Agrobacterium spp., Bacillus thuringiensis
Answer:
(a) Pseudomonas spp., Xanthomonas spp., Agrobacterium spp.

Question 32.
The weed Senecio jacobeac is controlled by ………………….
(a) Cactoblastis cactorum
(b) Xanthomonas spp
(c) Bacillus thuringiensis
(d) tyrea moth
Answer:
(d) tyrea moth

Question 33.
Nosema locustae is …………………. pathogen.
(a) bacteria
(b) fungal
(c) protozoan
(d) viral
Answer:
(c) protozoan

Question 34.
Which of the following bacterial pathogen is not used as herbicide ?
(a) Pseudomonas
(b) Xanthomonas
(c) Agrobacterium
(d) Azotobacter
Answer:
(d) Azotobacter

Question 35.
Identity free living bacterial biofertlizer ………………….
(a) Rhizobium
(b) Azotobacter
(c) Nostoc
(d) Bacillus thuringiensis
Answer:
(b) Azotobacter

Question 36.
The ectomycorrhizae form …………………. on the root surface.
(a) root tuber
(b) mantle
(c) root hair
(d) arbuscles
Answer:
(b) mantle

Match the columns

Question 1.

Column I (Biofortified crop)	Column II (Nutrient Enrichment)
(1) Maize	(a) Five times more iron
(2) Rice	(b) Twice the amount of lysine and tryptophan
(3) Wheat Atlas-66	(c) Enriched in vitamin A and minerals
(4) Carrots, spinach	(d) High protein content

Answer:

Column I (Biofortified crop)	Column II (Nutrient Enrichment)
(1) Maize	(b) Twice the amount of lysine and tryptophan
(2) Rice	(a) Five times more iron
(3) Wheat Atlas-66	(d) High protein content
(4) Carrots, spinach	(c) Enriched in vitamin A and minerals

Question 2.

Column I (Organic Acids]	Column II (Microbial source)
(1) Citric acid	(a) Rhizopus arrhizus
(2) Fumaric acid	(b) Acetobacter aceti
(3) Gluconic acid	(c) Aspergillus niger
(4) Acetic acid	(d) Aspergillus niger

Answer:

Column I (Organic Acids]	Column II (Microbial source)
(1) Citric acid	(c) Aspergillus niger
(2) Fumaric acid	(a) Rhizopus arrhizus
(3) Gluconic acid	(d) Aspergillus niger
(4) Acetic acid	(b) Acetobacter aceti

Classify the following to form Column B as per the category given in Column A.

Question 1.
i. Alternaria crassa
ii. Agrobacterium spp.
iii. Cactoblastis cactorum
iv. Beavueria bassiana

Column A (Biocontrol agents)	Column B (Host)
Microbial pesticide	————
Mycoherbicide	————
Insect as herbicide	————–
Bacterial herbicide	—————

Answer:

Column A (Biocontrol agents)	Column B (Host)
Microbial pesticide	Beavueria bassiana
Mycoherbicide	Alternaria crassa
Insect as herbicide	Cactoblastis cactorum
Bacterial herbicide	Agrobacterium spp.

Question 2.
i. Hairy leaves in wheat
ii. Nectar-less cotton having smooth leaves
iii. Hairy leaves in cotton
iv. Solid stem in wheat

Resistance to insects	Morphological characters
Jassids	————
Cereal leaf beetle	————
Stem borers	————–
Bollworms	—————

Answer:

Resistance to insects	Morphological characters
Jassids	Hairy leaves in cotton
Cereal leaf beetle	Hairy leaves in wheat
Stem borers	Solid stem in wheat
Bollworms	Nectar-less cotton having smooth leaves

Very Short Answer Questions

Question 1.
What are the different methods of plant breeding?
Answer:
Different methods of plant breeding are introduction, selection, hybridization, mutation breeding, polyploidy breeding, tissue culture, r-DNA technology and SCP (Single cell protein).

Question 2.
Which variety of sugar cane having high sugar content and better yield is cultivated in South India?
Answer:
Saccharum ojficinarum variety of sugar cane has high sugar content and better yield. It is cultivated in South India.

Question 3.
What are the desirable characteristics in hybrid varieties millets developed in India?
Answer:
Hybrid varieties of millets developed in India are high yielding and resistant to water stress.

Question 4.
Give examples of natural physical mutagens.
Answer:
Natural physical mutagens are high temperature, high concentration of CO₂, X-rays, UV rays.

Question 5.
Give examples of chemical mutagens.
Answer:
Chemical mutagens are nitrous acid, EMS (Ethyl – Methyl – Sulphonate), mustard gas, colchicine, etc.

Question 6.
How are seedlings or seeds mutated?
Answer:
Seedlings or seeds are mutated by irradiating them by CO-60 or exposing them to UV bulbs, X-ray machines, etc.

Question 7.
What are the effects of mutagens?
Answer:
Effects of mutagens are gene mutations and chromosomal aberrations.

Question 8.
Which biochemical characters are responsible for resistance against stem borers in maize?
Answer:
High aspartic acid, low nitrogen and sugar content are responsible for resistance against maize stem borers.

Question 9.
What does the plant tissue culture medium consists of?
Answer:
The plant tissue culture medium consists of water, all essential minerals, sources for carbohydrates, proteins and fats, growth hormones like auxins and cytokinins, vitamins. Agar is added to solidify nutrient medium for callus culture.

Question 10.
What are the types of tissue culture based on nature of explants?
Answer:
Cell culture, organ culture, pollen or anther culture, meristem culture and embryo culture are the types of tissue culture based on nature of explants.

Question 11.
What are the types of tissue culture based on the type of in vitro growth?
Answer:
Callus culture and suspension culture are the type of tissue culture based on the type of in vitro growth.

Question 12.
What is the necessity of subculturing?
Answer:
Both the callus and suspension cultures die in due course of time when nutrients get consumed. Therefore, a part of callus or suspension of cells is transferred to the flask containing new nutrient medium. This is subculturing, which is necessary to ensure continuous nutrient supply essential for continuous growth.

Question 13.
Enlist substances that are used as substrate for the production of SCP.
Answer:
Agricultural waste like corn cobs, sugar cane molasses, wood shavings, sawdust, paraffin, N-alkanes, human and animal wastes are the substrates used for the production of SCR.

Question 14.
How biofortified crops are produced?
Answer:
Biofortified crops are produced by conventional selective breeding practices and r-DNA technology.

Question 15.
What per cent of world livestock population is present in India and China and what is the productivity?
Answer:
India and China have 70% of world livestock populations, but the productivity is only 25% of the world farm produce.

Question 16.
What is silage made up of?
Answer:
Silage is fermented fodder made up of legumes, grasses, maize and jowar.

Question 17.
What is the supplementary food to silage?
Answer:
Silage is supplemented with oilcakes, minerals, vitamins and salts.

Question 18.
What is layer?
Answer:
The hen which is reared to obtain eggs is called a layer.

Question 19.
What is broiler?
Answer:
The hen which is reared to obtain meat is called a broiler.

Question 20.
What are the allied professions to poultry?
Answer:
The allied professions to poultry are processing of eggs and meat, marketing of poultry products, compounding and sale of poultry feed, poultry equipment, pharmaceuticals, feed additives, etc.

Question 21.
Which areas are suitable for bee keeping?
Answer:
The areas having sufficient wild shrubs, cultivated crops of sunflower, mustard, safflower, chilly, cabbage, cucumber, legumes, etc. and fruit orchards of apple, mangoes, citrus, etc. are suitable for bee keeping.

Question 22.
What is indicated by yellow spots?
Answer:
Yellow spots indicate shrinking of female lac insect.

Question 23.
What is indicated by orange spots on the eggs of lac insect?
Answer:
Orange spots indicate that eggs are about to hatch.

Question 24.
What is the main function of a fermenter?
Answer:
The main function of a fermenter is to provide a controlled environment for growth of specific microorganisms or a defined mixture of microorganisms, to obtain the desired product.

Question 25.
What is known as Brewer’s yeast?
Answer:
Saccharomyces cerevisiae var. ellipsoids is commonly known as Brewer’s yeast.

Question 26.
Which fermenter is used for large scale preparation of alcohol?
Answer:
Tubular tower fermenter is used for large scale preparation of alcohol.

Question 27.
What is the use of gluconic acid?
Answer:
Gluconic acid is used in medicine for solubility of Ca⁺⁺

Question 28.
What is the use of citric acid?
Answer:
Citric acid is used in confectionary.

Question 29.
What is the use of fumaric acid?
Answer:
Fumaric acid is used in resins as wetting agents.

Question 30.
Give examples of diseases which are treated using antibiotics.
Answer:
Diseases like plague, whooping cough, diphtheria and leprosy are treated using antibiotics.

Question 31.
What is the mechanism of actions of statins?
Answer:
Statins produced by yeast Monascus purpureus are blood cholesterol lowering agents. They are competitive inhibitors of the enzyme that catalyzes synthesis of cholesterol.

Question 32.
Enlist the various enzymes produced using microorganisms.
Answer:
Amylase, cellulase, protease, lipase, pectinase, streptokinase, invertase enzymes are produced using microorganisms.

Question 33.
Gibberellin was first isolated from which plant?
Answer:
Gibberellin was first isolated from rice plant infected by fungus Gibberella Jujikuroi.

Question 34.
How many different types of Gibberellins have been isolated?
Answer:
About 15 different types of Gibberellins have been isolated.

Question 35.
Give the chemical composition of biogas.
Answer:
The biogas consists of methane (50-60%), CO₂ (30 to 40%), H₂S (0-3%) and other gases like CO, N₂, H₂ in traces.

Question 36.
Which substrates are used for biogas production?
Answer:
Substrates like cattle dung (most commonly used substrate, a rich source of cellulose from plants), plant wastes, animal wastes, domestic wastes, agriculture waste, municipal wastes, forestry wastes, etc. are used for biogas production.

Question 37.
Which are the most commonly used models of biogas plants ?
Answer:
Models of biogas plants developed by KVTC (Khadi and Village Industries Commission) and IARI (Indian Agricultural Research Institute) are the most commonly used in India.

Question 38.
Which bacteria transform acetic acid into biogas?
Answer:
The acetic acid is transformed into biogas by methanogenic bacteria like Methanococcus, Methanobacterium and Methanobacillus.

Question 39.
What are the four groups of biocontrol agents?
Answer:
The four groups of biocontrol agents are bacteria, fungi, viruses and protozoans.

Question 40.
What is mycoherbicide ?
Answer:
The pathogenic fungus which kills or inhibits the growth of a weed is called mycoherbicide.

Question 41.
What are the three types of bacterial biofertilizers on the basis of function?
Answer:
On the basis of function, bacterial biofertilizers are of three types – nitrogen fixing, phosphate solubilizing and compost making biofertilizers.

Question 42.
What are the eight different types of mycorrhiza as per recent classification?
Answer:
Nowadays, mycorrhiza are classified into 8 different types – ectomycorrhizae, endomycorrhizae, ectendomycorrhizae, orchidaceous mycorrhizae, ericoid mycorrhizae, arbutoid mycorrhizae, monotrapoid mycorrhizae and ophioglossoid mycorrhizae.

Question 43.
What is heterocyst?
Answer:
Cynobacteria possess specialized colourless cells called heterocysts which are the sites of nitrogen fixation.

Question 44.
Give the role of heterocyst.
OR
Give the importance of heterocyst in cyanobacteria.
Answer:
The heterocysts are the sites of nitrogen fixation in cyanobacteria.

Question 45.
Who discovered mycorrhizae?
Answer:
Mycorrhizae were discovered by Frank (1885).

Give definitions of the following

Question 1.
Food
Answer:
Food is defined as solid or liquid substance, which is swallowed, digested and assimilated in the body, keeping us well.

Question 2.
Plant breeding
Answer:
Plant breeding is the improvement or purposeful manipulation in the heredity of crops and the production of new superior varieties of existing crop plants.

Question 3.
Germplasm collection
Answer:
Germplasm collection is the entire collection having all the diverse alleles for all genes in a given crop.

Question 4.
Mutation
Answer:
Mutation is defined as the sudden heritable change in the genotype, which is caused naturally.

Question 5.
Tissue culture
Answer:
Tissue culture is growing isolated cells, tissues, organs ‘in vitro’ on a solid or liquid nutrient medium, under aseptic and controlled conditions of light, humidity and temperature, for achieving various objectives.

Question 6.
Explant
Answer:
The part of plant used in tissue culture is known as explant.

Question 7.
Totipotency
Answer:
An inherent ability of living plant cell to grow, divide, redivide and give rise to a whole plant is known as totipotency.

Question 8.
Callus
Answer:
Callus is defined as a mass of undifferentiated cells, formed by division and redivision of the cells of explant.

Question 9.
Single cell protein
Answer:
Single cell protein is defined as a crude or a refined edible protein, extracted from pure microbial cultures or from dead or dried cell biomass.

Question 10.
Biofortiflcation
Answer:
Biofortification is a method of developing crops having higher quantity and quality of vitamins, minerals and fats, to overcome problem of malnutrition.

Question 11.
Animal husbandry
Answer:
Animal husbandry is an agricultural practice of breeding and raising livestock.

Question 12.
Inbreeding
Answer:
Breeding of closely related individuals for 4 to 6 generations is known as inbreeding.

Question 13.
Outbreeding
Answer:
Breeding of unrelated animals either of the same breed but having no common ancestors for 4 to 6 generations (outcrossing) or of different breeds (crossbreeding) or even of different species (interspecific hybridization), is known as outbreeding.

Question 14.
Outcrossing
Answer:
Breeding of animals of the same breed but having no common ancestors for 4 to 6 generations is known as outcrossing.

Question 15.
Crossbreeding
Answer:
Breeding of superior male of one breed with superior female of another breed is known as crossbreeding.

Question 16.
Interspecific hybridization
Answer:
Breeding of animals of two different but related species is known as interspecific hybridization.

Question 17.
Apiculture
Answer:
Apiculture is an artificial rearing of honey bees to obtain bee products like honey, wax, pollens, bee venom, propolis (bee glue) and royal jelly and to use honey bees as pollinating agents for crop plants.

Question 18.
Antibiotics
Answer:
Antibiotics are organic substances produced in small amounts by certain microbes to kill or inhibit the growth of other microbes.

Question 19.
Biotechnology
Answer:
Biotechnology is defined as the applications of scientific and engineering principles for the processing of materials by biological agents to provide goods and service to humans or for human welfare.

Question 20.
Biocontrol or biological control:
Answer:
Biocontrol is the natural method of eliminating and controlling insects, pests and other disease-causing agents by using their natural, biological enemies.

Question 21.
Biocontrol agents
Answer:
Biocontrol agents are the organisms like insects, bacteria, fungi, viruses and protozoans which are employed for biocontrol.

Question 22.
Fertilizers
Answer:
Fertilizers Eire the nutrients necessary for plant growth and which increase the productivity of cultivated plants.

Question 23.
Biofertilizers
Answer:
Biofertilizers are commercial preparation of ready-to-use live bacterial, cyanobacterial (mostly N₂ fixing) or fungal formulations which enhance the nutrient quality of soil.

Name the Following

Question 1.
Hybrid wheat varieties in India.
Answer:
Sonalika and Kalyan Sona

Question 2.
Semi-dwarf rice varieties in India.
Answer:
Jaya, Padma and Ratna

Question 3.
Sugar cane varieties developed at Coimbatore, Tamil Nadu.
Answer:
CO-419, 421, 453

Question 4.
Hybrid varieties of millets developed in India.
Answer:
Ganga-3 (maize), CO-12 (Jowar), Niphad (Bajra)

Question 5.
Fungal disease of plants.
Answer:
Brown rust of wheat, Red rot of sugar cane, Late blight of potato

Question 6.
Bacterial disease of plants.
Answer:
Black rot of crucifers

Question 7.
Viral disease of plants.
Answer:
Tobacco mosaic disease

Question 8.
Mutant variety of rice.
Answer:
Jagannath

Question 9.
Mutant variety of wheat.
Answer:
NP 836 (rust resistant)

Question 10.
Mutant variety of cotton.
Answer:
Indore-2 (resistant to bollworm)

Question 11.
Mutant variety of cabbage.
Answer:
Regina-II

Question 12.
The most preferred tissue culture medium.
Answer:
MS (Murashige and Skoog) medium.

Question 13.
High yielding varieties of banana used in Maharashtra.
Answer:
Shrimati, Basarai, G-9

Question 14.
The fungi used for the production of SCP.
Answer:
Aspergillus niger, Trichoderma viride, Saccharomyces cerevisiae, Candida utilis.

Question 15.
Algae used for the production of SCR
Answer:
Spirulina spp, Chlorella pyrenoidosa

Question 16.
Bacteria used for the production of SCR
Answer:
Methylophilus methylotrophus, Bacillus megasterium

Question 17.
A new breed of sheep developed from crossing of Bikaneri ewe and Marino rams in Punjab.
Answer:
Hisardale

Question 18.
Indian breeds of cows.
Answer:
Sahiwal, Sindhi, Gir

Question 19.
Exotic breeds of cows.
Answer:
Jersey, Brown Swiss, Holstein

Question 20.
Breeds of bufaaloes in India.
Answer:
Jaffarabadi, Mehsana, Murrah, Nagpuri, Nili, Surati.

Question 21.
Common breeds of cattle and poultry in the farms, found in your area.
Answer:
Khillari, Gir (breeds of cows), Aseel, Kadaknath (poultry)

Question 22.
American poultry breeds.
Answer:
Plymouth Rock, New Hampshire, Rhode Island Red

Question 23.
Asiatic poultry breeds.
Answer:
Brahma, Cochin and Langshan

Question 24.
Mediterranean poultry breeds.
Answer:
Leghorn, Minorca

Question 25.
English poultry breed.
Answer:
Australorp

Question 26.
Indian poultry breeds.
Answer:
Chittagong, Aseel, Brahma and Kadaknath.

Question 27.
Best layer.
Answer:
Leghorn

Question 28.
Best broilers.
Answer:
Plymouth rock, Rhode Island Red, Aseel, Brahma and Kadaknath

Question 29.
Viral diseases of poultry.
Answer:
Ranikhet, Bronchitis, Avian influenza (bird flu), Bird flu

Question 30.
Bacterial diseases of poultry.
Answer:
Pullorum, Cholera, Typhoid, TB, CRD (chronic respiratory disease), Enteritis.

Question 31.
Fungal diseases of poultry.
Answer:
Aspergillosis, Favus and Thrush.

Question 32.
Parasitic diseases of poultry.
Answer:
Lice infection, roundworm, caecal worm infections, etc.

Question 33.
Protozoan diseases of poultry.
Answer:
Coccidiosis

Question 34.
Domesticated species of bees.
Answer:
Apis mellifera, Apis indica

Question 35.
Lac insect.
Answer:
Trachardia lacca

Question 36.
Silk moth.
Answer:
Bombyx mori

Question 37.
The common fresh water fish.
Answer:
Labeo rohita (rohu), Catla (catla), Cirrihanus mrigala (mrigala) and other carps.

Question 38.
The common marine fish.
Answer:
Harpadon (Bombay duck), Sardinella (sardine), Rastrelliger (mackerel) and Stromateus (pomphret).

Question 39.
Estuaries found in Maharashtra and where these estuaries are located.
Answer:
Thane creek, Manori creek, Rajapuri creek, Kalbadevi Estuary in Ratnagiri, Damanganga estuary and Narmada estuary

Question 40.
Different fish found at an estuary.
Answer:
Clupeids, mullets, catfishes, perches, Mugil cephalus gar fishes, halfbeaks, eels, flatfishes, sharks, rays, oysters and migratory fishes include Hilsa ilisha, Polynemus spp., Pampana, Tachysurus spp, Pangasius spp., etc.

Question 41.
Best Silk.
Answer:
Mulberry silk

Question 42.
Silk of inferior quality.
Answer:
Tussar silk, Eri silk

Question 43.
Distilled alcoholic beverages.
Answer:
Whisky, brandy, rum

Question 44.
Traditional drink made by fermenting the sugar sap extracted from palm plants and coconut palm.
Answer:
Toddy

Question 45.
The wine of Goa, made by fermenting fleshy pedicels of cashew fruits.
Answer:
Fenny

Question 46.
The microbes used in fermentation of dhokla.
Answer:
Lactobacilli

Question 47.
Microbes used in the production of vitamin B₂.
Answer:
Neurospora gossypii, Eremothecium ashbyi

Question 48.
Microbes used in the production of vitamin B₁₂.
Answer:
Pseudomonas denitrtficans

Question 49.
Microbes used in the production of vitamin C.
Answer:
Aspergillus niger

Question 50.
Symbiotic N₂ fixing microorganisms.
Answer:
Rhizobium, Anabaena, Frankia.

Question 51.
Free-living or Non-Symbiotic N₂ fixing microorganisms.
Answer:
Azotobacter, Nostoc, Clostridium, Beijerinkia, Klebsiella, etc.

Question 52.
Phosphate solubilizing biofertilizers.
Answer:
Pseudomonas striata, Bacillus polymyxa, Agrobacterium, Microccocus, Aspergillus spp., etc.

Question 53.
Nitrogen fixing cyanobacteria.
Answer:
Anabaena, Nostoc, Plectonema, Oscillatoria.

Question 54.
Cyanobacteria associated with lichens.
Answer:
Anabaena, Nostoc and Tolypothrix

Question 55.
Cyanobacteria associated with plants like Azolla and Cycas.
Answer:
Anabaena

Question 56.
The aquatic fern commonly used in paddy field as a biofertilizer.
Answer:
Azolla.
Give functions or significance

Question 1.
Hybridization
Answer:

1. Hybrdization is an effective means of combining together the desirable characters of two or more varieties.
2. New genetic combinations of already existing characters and new genetic variations can be created by hybridization.
3. Hybridization exploits and utilizes hybrid- vigour.

Question 2.
Food
Answer:

1. Food is organic, energy rich, non-poisonous, edible and nourishing substance.
2. It provides nutrients for growth and development of body.
3. It provides energy for metabolic reactions.
4. It keeps us alive, strong and healthy.

Question 3.
Antibiotics
Answer:

1. Antibiotics are secondary metabolites of therapeutic importance, produced in small amounts by certain microbes like bacteria, fungi and a few algae.
2. They inhibit growth of other microbial pathogens like fungi and bacteria.
3. Thus, they are antifungal and antibacterial in nature.
4. Antibiotics are used in treatment of deadly diseases like plague, whooping cough, diphtheria, leprosy, etc.

Distinguish between the following

Question 1.
Callus culture and Suspension culture.
Answer:

Callus Culture	Suspension culture
1. Solid nutrient medium is used in callus culture.	1. Liquid nutrient medium is used in suspension culture.
2. No shaker or agitator is needed.	2. Shaker or agitator is needed.
3. Cells of explants divide and redivide to form callus.	3. Callus is not formed.
4. Callus a mass of undifferentiated cells.	4. Suspension culture consists of single isolated cells or small groups of cells.
5. It shows slow growth.	5. It shows faster growth.

Question 2.
Outcrossing and Crossbreeding.
Answer:

Outcrossing	Crossbreeding
1. Breeding of animals of the same breed but having no common ancestors for 4 to 6 generations is known as outcrossing.	1. Breeding of superior male of one breed with superior female of another breed is known as crossbreeding.
2. Progeny is known as outeross.	2. Progeny is known as hybrid.
3. New breeds are not developed by outcrossing.	3. New breeds or hybrids are formed by crossbreeding.
4. An outcross helps to overcome inbreeding depression.	4. Hybrids are subjected to inbreeding and new stable breeds are developed by crossbreeding.

Question 3.
Inorganic fertilizers and Organic fertilizers / biofertilizers.
Answer:

Inorganic fertilizers	Organic fertilizers / biofertilizers
1. They are non-renewable nutritional resources.	1. They are renewable nutritional resources.
2. Inorganic fertilizers are synthetic and are in the form of chemicals.	2. They are biological in origin.
3. They are mixtures of mineral salts of NPK in definite proportions.	3. Organic fertilizers are farmyard manure, green manure and compost. Whereas, bio fertilizers are live bacterial, cyanobacterial (mostly N<sub>2</sub> fixing) or fungal formulations which enhance the nutrient quality of soil.
4. Excessive use of inorganic fertilizers results in pollution of soil, groundwater and air.	4. They do not cause pollution.
5. They are not part of sustainable agriculture.	5. They are part of organic farming and sustainable agriculture.

Question 4.
Ectomycorrhizae and Endomycorrhizae.
Answer:

Ectomycorrhizae	Endomycorrhizae
1. Mycelium of ectomycorrhizal fungi form a sheath called mantle on the surface of roots.	1. Most of the endomycorrhizal hyphae penetrate in the root cortex.
2. Few hyphae form hartig-net in the intercellular spaces of root cortex.	2. Fungal hyphae do not produce hartig-net.
3. Arbuscles and vesicles are not formed.	3. Arbuscles and vesicles are formed.

Give reasons

Question 1.
Why are honey bees called as best pollinators?
Answer:

1. About 80% of insect pollination is carried by honey bees.
2. They pollinate various crops like sunflower, mustard, safflower, chilly, cabbage, cucumber, legumes, fruits like apple, mango, citrus, etc.
3. They increase the productivity of crops.
4. Hence, honey bees are important pollinators.

Question 2.
Regular visit of veterinary doctor to dairy farm is mandatory.
Answer:
Daily visit of veterinary doctor to dairy farm is mandatory to diagnose health problems, diseases and for their rectification.

Question 3.
Apis mellifera and Apis indica are known as domesticated species.
Answer:

1. Apis mellifera and Apis indica are suitable for apiculture.
2. Hence, they are known as domesticated species.

Question 4.
Cages of silkworm larvae must be well managed and protected.
Answer:

1. Silkworm larvae may get infected by protozoans, viruses and fungi.
2. Ants, crows, other birds and predators may attack these larvae.
3. Hence, cages of silkworm larvae must be well managed and protected.

Question 5.
Buttermilk is used in the dough of dhokla.
Answer:

1. Buttermilk contains the lactobacilli.
2. These lactobacilli bring about the fermentation of gram flour.
3. CO₂ produced during fermentation increases the volume of the dough.
4. It escapes during the steaming of dough, making dhokla porous and spongy.
5. Hence, buttermilk is used in the dough of dhokla.

Question 6.
We include mushrooms in our diet.
Answer:

1. Mushrooms are directly used as food.
2. They produce large, fleshy fruiting bodies which are edible.
3. They are low calorie, sugar free, fat free, but rich in proteins, vitamins, minerals and amino acids.
4. Hence, we include mushrooms in our diet.

Question 7.
Vitamins are to be consumed through food or tablets or capsules.
Answer:

1. Vitamins B and K are produced in the body.
2. But some vitamins like C, D and E are not produced in the body.
3. If our body does not get these vitamins in the required quantity, then the deficiency of these vitamins may result in various diseases.
4. Therefore, vitamins are to be consumed through food or tablets or capsules.

Question 8.
Enzymes are essential for survival of living organisms.
Answer:

1. Enzymes are proteins which act as biocatalysts.
2. They speed up the chemical reactions without undergoing any change themselves.
3. They catalyze reactions more quickly and efficiently at body temperature.
4. Thus they play key role in metabolic reactions.
5. Hence, enzymes are essential for survival of living organisms.

Question 9.
Biogas plants are more often built in rural areas.
Answer:

1. Biogas is a non-conventional and renewable source of energy obtained by microbial fermentation.
2. Cattle dung (the main substrate), domestic wastes, agricultural waste, agro industrial wastes, forestry wastes, etc. are utilized as substrates for production of biogas.
3. Biogas is eco-friendly and does not cause pollution, can be used as domestic fuel.
4. As the raw material for its production is readily available, the biogas plants are more often built in rural areas.

Question 10.
Why are healthy root nodules pink in colour?
Answer:

1. Rhizobium has symbiotic relationship with roots of leguminous plants.
2. It infects root cortex and form root nodules.
3. Root nodules are the site of nitrogen fixation.
4. Enzyme nitrogenase which catalyzes nitrogen fixation, gets inhibited by oxygen.
5. But root nodule contain a pigment called leghaemoglobin which acts as oxygen scavanger and protects nitrogenase from getting inhibited.
6. Leghaemoglobin is pink in colour.
7. Hence, healthy root nodules are pink in colour.

Give Short Notes

Question 1.
Callus culture.
Answer:

1. In callus culture, nutrient medium is solidified using agar-agar is used.
2. Shaker or agitator is not required.
3. Sterilized explant is placed on solid nutritive medium.
4. The cells of explants absorb nutrients and start multiplying.
5. This results in the formation of callus.
6. Callus is a mass of undifferentiated cells, formed by division of the cells of explants.
7. Growth hormones, auxins and cytokinins are provided to callus in specific proportion to induce formation of organs.
8. If auxins are in more quantity, roots are formed (rhizogenesis) and if the cytokinins are in more quantity, shoot formation takes place (caulogenesis).
9. Thus new plantlets are formed.
10. Callus culture required subculturing to ensure its continuous growth.

Question 2.
Suspension culture.
Answer:

1. In suspension culture, small groups of cells or a single cell are transferred to liquid nutritive medium as explants.
2. The liquid medium is constantly agitated by using shakers (agitators).
3. The agitation serves the purpose of aeration, mixing of medium and prevents the aggregation of cells.
4. Generally the suspension culture shows a high proportion of single isolated cells and small clumps of cells.
5. Suspension culture grows much faster than callus culture.
6. Suspension culture is used for cell biomass production which can be utilized for biochemical isolation, regeneration of new plants, etc.

Question 3.
Applications of micropropagation.
Answer:

1. Micropropagation involves in rapid multiplication of genetically similar plants (clones).
2. A large number of plantlets are obtained within a short period and in a small space.
3. Plants are obtained throughout the year, under controlled conditions, independent of seasons.
4. As micropropagation results in the formation of clones, desirable characters (genotype and sex) of superior variety can be maintained for many generations.
5. The rare plant and endangered species are multiplied and conserved using this technique.
6. With the help of somatic hybrids (cybrids), new variety can be obtained in short time span.
7. Micropropagation is involved in commercial production of ornamental plants like v orchids, Chrysanthemum, Eucalyptus, etc. and fruit plants like banana, grapes, Citrus, etc.

Question 4.
Applications of tissue culture.
Answer:
Applications of tissue culture are as follows:

1. Production of healthy plants from diseased plants using apical meristems as explants.
2. Production of stress resistant plants.
3. Production of haploid plantlets by pollen culture.
4. Production of secondary metabolites such as alkaloids, enzymes, hormones, etc.
5. Multiplication of rare and endangered plants.
6. Production of somaclonal variants.
7. Use of micropropagation techniques to produce large number of genetically identical plants.
8. Protoplast culture
9. Tissue culture has applications in forestry, agriculture, horticulture, genetic engineering and physiology.

Question 5.
Single cell protein (SCP).
Answer:

1. Single-cell protein is a crude or a refined edible protein, extracted from pure microbial cultures or from dead or dried cell biomass.
2. Microorganisms like algae, fungi, yeast and bacteria with high protein content in their biomass, are grown using waste and inexpensive substrates.
3. Substrates used for growing microbial biomass are wood shavings, sawdust, corn cobs, paraffin, N-alkanes, sugar cane molasses, even human and animal wastes.
4. SCP is rich in proteins, vitamins, vitamin B complex, minerals and fats.
5. It can be used as fodder for achieving fattening of calves, pigs, in breeding fish and even in poultry and cattle farmimg.
6. Fungi like Aspergillus niger, Trichoderma viride, Saccharomyces cerevisiae, Candida utilis, algae like Spirulina spp, Chlorella pyrenoidosa, bacteria like Methylophilus, methylotrophus and Bacillus megasterium are used for the production of SCR

Question 6.
Advantages of Single Cell Protein.
Answer:

1. Microbes that are used as SCP have very high protein contents in their biomass – 43% to 85% (W/W basis).
2. SCP is a good source of vitamins, vitamin B complex, fats, amino acids, minerals, crude fibres, etc.
3. As microorganisms multiply fast, large quantity of biomass can be produced in a short duration.
4. The microbes can be grown using waste materials and inexpensive substrates. This decreases pollution.
5. The microbes can be genetically modified to vary the amino acid composition.
6. SCPs can be used as fodder for achieving fattening of calves, pigs, in breeding fish, poultry and cattle farming.

Question 7.
Inbreeding.
Answer:

1. Inbreeding is the mating of two closely related individuals of the same breed for 4 to 6 generations.
2. During inbreeding superior males and superior females of the same breed are identified. The superior males and superior females from this progeny are identified and used for further mating.
3. Due to inbreeding, homozygosity is increased and harmful recessive genes are exposed. Inbreeding is done when a pure line of an animal is expected.
4. Inbreeding helps in accumulation of superior genes and elimination of harmful or less desirable genes.
5. Continued inbreeding usually reduces fertility and productivity. This is called inbreeding depression.

Question 8.
Fish farming.
Answer:
Fish farming is the practice of culturing the edible and commercially important fish species in the ponds, lakes or reservoirs. Fish farming helps in boosting the productivity and the economy of the nation.

For maintaining a fish farm, following aspects are taken into consideration:

• selection of the site
• excavation of the pond
• managing hatchery
• nursery
• looking after rearing ponds and
• stocking ponds besides managing the water source, supplying fertilizer and supplementary feed, etc.

Question 9.
Microbes in industrial vitamin production.
Answer:
(1) Vitamins are nitrogenous organic compounds, required in minute quantities for normal growth and development of the body.

(2) The microbes are involved in the industrial production of vitamins like thiamine (vitamin B₁, riboflavin (vitamin B₂), pyridoxine, folic acid, pantothenic acid, biotin, vitamin B₁₂, ascorbic acid (Vitamin C), beta-carotene (provitamin A) and ergosterol (provitamin D).

(3) Examples of some vitamins produced by fermetaion using different microbial sources are-

• Vitamin B₂ – Neurospora gossypii, Eremothecium ashbyi
• Vitamin B₁₂ – Pseudomonas denitrijicans
• Vitamin C – Aspergillus niger

Question 10.
Industrial uses of enzymes.
Answer:

1. Enzymes are used to improve the quality of the fabrics in the textile industry.
2. In the pulp and paper industry, they are involved in biomechanical pulping and bleaching.
3. In food industry, enzymes are used in the fermentation processes to produce bread and alcoholic beverages such as wine and beer.
4. They are used in the extraction of carotenoids and olive oil.
5. Enzymes are also used in cosmetics, animal feed and agricultural industries.
6. Lipases are used to remove oil stains and to increase the brightness in detergent industry. They have superior cleaning properties.
7. Streptokinase is used as a ‘clot buster’ for clearing blood clots in the blood vessels in heart patients.

Question 11.
Microorganisms in sewage.
Answer:

1. Sewage contains pathogenic bacteria, viruses, fungi, protozoa which cause dysentery, cholera, typhoid, polio and infectious hepatitis, etc. It also contains nematodes and algae.
2. Their number and type of microorganisms in sewage depends on the composition and source of sewage.
3. Millions of bacteria per ml may be present in raw untreated sewage.
4. Bacteria in sewage include coliforms, fecal Streptococci, anaerobic spore forming bacilli and other bacteria found in the intestinal tract of humans.
5. The sewage also contains soil bacteria.
6. During sewage decomposition, initially aerobic and facultative anaerobic organisms predominate which are followed by strict anaerobic especially methogenic bacteria that produce methane (CH₄) and CO₂.

Question 12.
Rhizobia as a biofertilizer
Answer:

1. Rhizobia [Singular – Rhizobium) are rod¬shaped, motile, aerobic, gram negative, non- spore forming, nitrogen-fixing bacteria.
2. They contain Nod genes and Nif genes.
3. They live in symbiotic association with leguminous roots.
4. They form nodule on the roots of leguminous plants and multiply inside the nodule.
5. Nodules are the site of nitrogen fixation. They are pink in colour and contain enzyme nitrogenase and oxygen scavenger leghaemoglobin.
6. Rhizobia fix atmospheric nitrogen into organic forms, which can be used by plants as nutrients. The host plant in return provides food and water to the bacterium (Rhizobia).
7. Rhizobia are host specific e.g. leguminosarum is specific to pea R. phaseoli is specific to beans.
8. In a laboratory, pure cultures of specific Rhizobium species are raised which are used for the preparation of biofertilizer.

Question 13.
Azotobacter as a biofertilizer
Answer:

1. Azotobacter is a free living, nitrogen fixing, aerobic, non-photosynthetic, non-nodule forming bacterium.
2. It is associated with roots of grasses and certain plants.
3. It is used as a biofertilizer for non-leguminous plants like rice, cotton, vegetables, etc.

Question 14.
Azospirillumas as a biofertilizer
Answer:

1. Azospirillum acts as a biofertilizer for non- leguminous plants like cereals, millets, cotton, oilseed, etc.
2. It is a free living, aerobic nitrogen fixing bacterium associated with roots of corn, wheat and jo war.
3. It fixes the nitrogen (20-40 kg N/ha).

Question 15.
Anabaena as a biofertilizer
Answer:

1. Anabaena is a cyanobacterial biofertilizer.
2. It is multicellular, filamentous nitrogen fixing organism that exits as plankton.
3. It can fix nitrogen both in free living conditions as well as by forming symbiotic associations.
4. Anabaena forms symbiotic relationship with Azolla, Anthoceros and Cycas.
5. It is found in dorsal leaf lobe in Azolla, thallus of Anthoceros and in coralloid roots of Cycas.
6. It has specialized colourless cells known as heterocysts.
7. Heterocysts are the sites for nitrogen fixation.

Question 16.
Benefits of Biofertilizers.
Answer:

1. Biofertilizers increase soil fertility.
2. They are low cost and can be used by marginal farmers.
3. They do not cause pollution.
4. BGA secret growth promoting substances, organic acids, proteins and vitamins.
5. Azotobacter supplies nitrogen and antibiotics in the soil.
6. Use of biofertilizers improves physico-chemical properties of soil-like texture, structure, pH, water holding capacity of soil by providing nutrients and organic matter.
7. The use of chemical fertilizers gets reduced and the pollution also becomes less.

Short Answer Questions

Question 1.
What are the objectives of plant breeding?
Answer:
Objectives of plant breeding are as follows:

1. To increase crop yield,
2. To improve quality of produce.
3. To increase tolerance to environmental stresses.
4. To develop varieties of plants resistant to pathogens and insect pest.
5. To alter the life span.

Question 2.
What are the different types of hybridization in plants?
Answer:
The different types of hybridization in plants are as follows:

1. Intravarietal : It is the hybridization between plants of same variety.
2. Intervarietal : It is the hybridization between two varieties of the same species.
3. Interspecific : It is the hybridization between two species of the same genus.
4. Intergeneric : It is the hybridization between two genera of the same family.
5. Wide/distant crosses : These are the crosses between distantly related parental plants.

Question 3.
How aseptic conditions are maintained in tissue culture?
Answer:

1. Glassware is sterilized by using detergents and hot air oven.
2. Nutrient medium is autoclaved under constant pressure of 15 lb/sq inch, continuously for 20 minutes to sterilize it.
3. Explant is treated with 20% ethyl alcohol and 0.1% HgCl₂.
4. Sterilization of inoculation chamber (Laminar air flow) is done using UV ray tube for 1 hour before actual inoculation of explant on the sterilized nutrient medium.

Question 4.
What are the objectives of biofortification ?
Answer:
Objectives of biofortification are as follows:

1. Improvement in protein content and quality.
2. Improvement in oil content and quality.
3. Improvement in vitamin content.
4. Improvement in micronutrient content and quality.
5. To overcome the problem of malnutrition.

Question 5.
What are the objectives of animal breeding?
Answer:
Objectives of animal breeding are as follows:

1. To increase the yield of animals.
2. To improve the production of milk, quality of milk product, quality of meat or maximum yield of eggs per year, etc.
3. To develop breeds with desirable characters.

Question 6.
What is artificial insemination? What are its advantages?
Answer:

1. Artificial insemination is the technique used for controlled breeding experiments.
2. Superior males of a particular commercial breed are selected.
3. Semen from such superior males is collected and injected into the genital tract of female.
4. This insemination is either done immediately or semen is frozen and used later on.
5. In frozen semen, sperms can remain alive for long duration. They are also convenient for transport.
6. Artificial insemination is preferred as it is easy and helpful to overcome several problems of normal mating.

Question 7.
What is Multiple Ovulation Embryo Transfer (MOET) technology? Where is it used?
OR
Explain the technique of multiple ovulation embryo transfer (MOET) in animal breeding.
Answer:

1. Multiple Ovulation Embryo Transfer (MOET) technology is used to increase chances of successful production of hybrids.
2. In this technique, cow is administered with Follicle Stimulating Hormone (FSH) which induces follicular maturation and then the super ovulation is brought about.
3. By such technique in each cycle, 6 to 8 eggs mature simultaneously.
4. The cow is then either mated with a superior bull or she is artificially inseminated.
5. The blastocysts at 8 to 32 cell stage are recovered non-surgically and transferred to surrogate mothers.
6. The genetic mother who gave the egg is then again subjected to another round of super ovulation.
7. This technology is used for cattle, sheep, rabbits, buffaloes, etc.
8. The MOET is used to produce high milk yielding breeds of female and high quality meat yielding bulls with lean meat containing less lipids. It helps in increasing favourable herd size in a short period.

Question 8.
As a dairy owner what measures will you adopt to improve the quality of milk?
Answer:

1. In order to improve the quality of milk, following measures should be taken at every stage of dairy farming:
2. Good breeds having high yielding potential should be selected.
3. The breeds selected should be suitable for the local climatic conditions.
4. The breeds selected should have proper resistance to diseases.
5. Cattles should be well looked after with proper care.
6. The feed should be of suitable quality and quantity. Feed includes silage made from legumes, grasses, maize and jowar. Silage should be supplemented with oilcakes, minerals, vitamins and salts.
7. Utmost care should be taken about cleanliness and hygiene of the cattle as well as the handlers who handle the cattle.
8. This is especially important during milking, storage and transport of milk and its products. Mechanized processes should be adopted as far as possible as they reduce chance of direct contact of produce with the handlers.
9. The shed must be clean and well maintained. Similarly the dairy should be spacious with adequate facilities for feeding, watering and light.
10. Help of veterinary doctor should be sought from time to time for the identification of health problems, diseases and rectification.
11. Transportation of milk, processing, marketing and distribution play a vital role in dairy industry. If all the above care is taken then the quality of milk will surely improve.

Question 9.
Explain in short the poultry management.
Answer:
For the management of poultry, following aspects are to be taken care of:

1. Selection of proper and disease free breed, suitable and safe farm conditions.
2. Proper feeding practice and the quality of feed and water.
3. Hygiene and health care of the birds.
4. Management of layers is done by selecting high yielding chicken. Their farms are kept clean, dry and well ventilated. They are given proper feed at proper times. Other aspects such as debeaking, etc. are also taken care of.
5. In the farm, importance is given to infrastructure such as proper and adequate lighting, placing waterer at places, looking after sanitation, culling and vaccination.
6. Management of broiler similarly includes selection of breed, housing, temperature, ventilation, lighting, observing the floor space and broiler feed.

Question 10.
What is apiculture? What is its importance?
Answer:

1. Apiculture is artificial rearing of honey bees to obtain various bee products.
2. Various products such as honey, wax, pollen, bee venom, propolis (bee glue) royal jelly, etc. are obtained from this cottage industry. Honey is an important item as ayurvedic medicine and in food due to its nutritional value.
3. Bees also help in the cross pollination of various crops. Hence in pastures, wild shrubs, fruit orchards and cultivated crops, honeybees play an important role as pollinators.
4. Bee keeping in crop fields increases the productivity of both honey and the crop.
5. Apiculture itself is a means for employment for rural youth. It is an age old cottage industry which can be done along with agriculture.

Question 11.
What are the requirements of bee keeping?
Answer:

1. Bee keeping requires bee hive boxes which consist of comb foundation sheets.
2. In addition, the bee veil, smoker, bee brush, gloves, gumshoes, uncapping knives, swarm net, queen excluder, overall hive tool, etc. are required.
3. Bee keeping requires familiarity with the habits of bees, selection of suitable location, catching and hiving of swarms, management of hives during different seasons, handling and collection of honey, bee wax and other products.
4. Successful bee keeping also requires periodic inspection for cleanliness of hive boxes, activity of bees and queen, condition of brood, provision of water.

Question 12.
What are the main divisions of fishery?
Answer:

1. Fishery can be capture fishery and culture fishery.
2. Three main divisions of capture fishery are : Inland fishery, estuarine fishery and marine fishery.
3. Inland fishery : It is culturing and capturing of from fresh water bodies. It is carried out on about 40 to 50 lakh acres of fresh water bodies such as rivers, ponds, lakes and dams.
4. Marine fishery : It includes capturing fish from sea water. India has a coastline of about 7500 km.
   Estuarine fishery : It includes capture of fish from estuaries.
5. Culture fishery is either of polyculture or of monoculture type. In polyculture, different species are cultured simultaneously at the same time in the same pond. In monoculture, only a single species is cultured.

Question 13.
Can “fishery” be a sustainable job option?
Answer:

1. Fishery can never be a job. It is a livelihood or can be an occupation.
2. Fish is a renewable resource. If managed properly, it can be a sustainable resource.
3. The sustainability is dependent upon the availability of fishes and other edible organisms. But due to climate change, pollution and overexploitation of fish resources, these are depleting rapidly, it is estimated by the scientists that by 2050, no fish will be left in the seas.
4. However, if aquaculture is done to raise fishes, partly it can be sustaining. But there . are many environmental regulations that can hamper the business of aquaculture.

Question 14.
Describe sericulture in brief.
Answer:

1. Sericulture is the practice of rearing silkworms for the production of silk.
2. The silkworm (Bombyx mort) is reared for obtaining best quality of silk called mulberry silk. Tussar silk and Eri silk are other varieties of silk which are inferior to the mulberry silk.
3. Larvae of silkworm are fed on the mulberry leaves. Quality and quantity of silk depends on the quality of mulberry leaves.
4. These larvae are reared, developed and well looked after by the skilful labour keeping a constant watch.
5. Silkworm larvae may be infected by protozoans, viruses and fungi. Ants, crows, birds, and other predators are ready to attack these insects, hence the cages of these larvae must be managed to prevent predators attack.
6. Silk is obtained from the cocoon of the silkworm.
7. Sericulture is an age old practice and can be started with low investment and small space. It requires scientific knowledge and skill. Disabled, older and handicapped people also can practise it.

Question 15.
What are the different stages found in the life cycle of silkworm?
Answer:

1. Stages of development in the life cycle of silkworm are egg, larva, pupa and adult.
2. The larva is the silkworm caterpillar.
3. The adult (imago) stage is the silkworm moth.

Question 16.
Describe the process of cocoon formation.
Answer:

1. The eggs of silkworm hatch into larvae.
2. The larvae develop into a caterpillar.
3. Caterpillar feeds on fresh mulberry leaves.
4. After its growth and moulting, the silkworm secretes a silk fibre to form cocoon.
5. The silk is a continuous filament comprising fibroin protein, secreted from salivary glands of silkworm and a gum called sericin, which cements the filaments.
6. The silk solidifies when it contacts the air.
7. The silkworm spins approximately one mile of filament and completely encloses itself in a cocoon in about two or three days.

Question 17.
Which process is involved in silk production from cocoon?
Answer:

1. The silk is a continuous filament comprising fibroin protein, secreted from salivary glands of silkworm and a gum ailed sericin.
2. To remove the sericin, which cements,the filaments, cocoons are placed in hot water.
3. It frees the silk filaments and readies them for reeling. This is known as the degumming process.
4. The sillworm pupa gets killed in hot water.
5. Single filaments are combined to form thread.
6. The threads are plied to form yarn.
7. After drying, the raw silk is packed according to quality.

Question 18.
Give the economic importance of lac.
OR
State the economic importance of ‘lac culture’.
Answer:
Lac is used for the following purposes:

1. For making bangles.
2. For making different types of toys.
3. It is used in wood works.
4. Polish is made from lac.
5. Inks can be prepared from lac.
6. Lac is largely used for silvering mirrors.

Question 19.
Tribal people from India have a great contribution in production of lac. But it needs certain trees. Name at least two such trees which give food yield of lac. How is lac purified?
Answer:

1. Like her, peepal, palas, kusum, babul, etc. are used for feeding lac insects during the practice of lac culture.
2. Natural lac is contaminated and hence it is washed and purified. This helps to obtain shellac in pure form.

Question 20.
Give an account of alcoholic beverages.
Answer:

1. Alcoholic beverages are the products of alcoholic fermentation of particular substrates.
2. Tubular tower fermenters are used to produce alcoholic beverages on a large scale.
3. Beer is produced from barley by fermentation. For the production of beer, strains of Saccharomyces cerevisiae are used.
4. Wine is prepared from grapes.
5. Whisky is prepared by fermenting mixed grains of wheat, barley and corn followed by the distillation of the products of fermentation.
6. Liquors like beer, wine are produced without distillation.
7. Whisky, rum and brandy are distilled alcoholic beverages.
8. Toddy is prepared by fermenting the sugar sap extracted from palms and coconut palms.
9. Fenny is fermented by fleshy pedicels of cashew fruits.

Question 21.
Industrial production of which substances involves fermentation by microbes? What is the nature of these substances and which factors determine the synthesis of specific products?
Answer:

1. During fermentation of substrates, various useful products like alcoholic beverages, organic acids, vitamins, growth hormones, enzymes, antibiotics and other molecules of medical significance are produced.
2. They are secondary metabolites produced during idio phase and are not required for their growth.
3. A specific secondary metabolite is produced depending on the type of microorganism and the type of substrate.

Question 22.
Can antibiotics kill viruses?
Answer:

1. Antibiotics work by targeting cell wall or other metabolic pathways in bacteria.
2. But viruses do not have cell walls and they do not carry out any metabolic reaction when outside the host.
3. Hence, antibiotics cannot kill viruses.

Question 23.
What are gibberellins? Give the applications of gibberellins.
Answer:
Gibberellins are growth hormones produced by higher plants and fungi.

Applications of gibberellins are as follows:

1. Gibberellins induce parthenocarpy in fruits like pear and apple.
2. Gibberellins promote growth by stem elongation.
3. They break the dormancy of seeds.
4. They induce flowering in long day plants in short day conditions.
5. They are used to enlarge the size of grapes.

Question 24.
What is sewage? Describe its composition.
Answer:

1. Sewage is the waste matter carried off on drainage.
2. Composition of sewage varies depending upon its industrial source, e.g. textile, chemicals, pharmaceuticals, dairy, canning, brewing, meat packing, tannery, oil refineries and meat industries, etc.
3. Sewage consists of human excreta, animal dung, household waste, slaughter house waste, dissolved organic matter, algae, nematodes, pathogenic bacteria, viruses and protozoa, discharged waste water from hospitals, industries (contains toxic dissolved organic and inorganic chemicals), tannery and pharmaceutical waste, etc.
4. Bacteria in sewage include coliforms, fecal Streptococci, anaerobic spore forming bacilli and other bacteria found in human intestinal tract.
5. Sewage consists of about water (99.5% to 99.9%) and inorganic and organic matter (0.1 to 0.5%) in suspended and soluble form.

Question 25.
State whether BOD will be high or low
(a) in water after primary treatment
(b) in water after secondary treatment.
Answer:
(a) After primary treatment, the primary effluent present in the primary sedimentation tank, still contains large amount of dissolved organic matter. Hence, the BOD is high.

(b) Secondary treatment is a biological treatment. Primary effluent is passed into aeration tank, where aerobic bacteria and fungi form flocks and consume the major part of organic matter in the effluents. Hence, the BOD is reduced.

Question 26.
Enlist the advantages of biogas.
Answer:
Advantages of biogas are as follows:

1. Biogas is a cheap, safe, non-conventional and renewable source of energy.
2. It can be easily generated, stored and transported.
3. Biogas burns with a blue flame without producing smoke.
4. Biogas is of great help in improving ‘ the sanitation of the surrounding.
5. Biogas is an eco-friendly gas. It does not cause pollution and imbalance of the environment.
6. Leftover sludge can be used as fertilizer.
7. It is used as domestic and industrial fuel. Biogas can be used for domestic lighting, street lighting, cooking, small scale industries, etc.

Question 27.
Explain how Bacillus thuringiensis acts as a bio-control agent.
Answer:

1. Bacillus thuringiensis (Bt) is an effective biocontrol agent against butterfly, caterpillars.
2. Dried spores of Bacillus thuringiensis are mixed with water and sprayed onto vulnerable plants such as Brassicas and fruit trees.
3. When insect larvae eat the leaves, they get killed by the toxin (cry protein) released in their gut by bacteria.

Question 28.
Explain how Trichoderma acts as a bio-control agent.
Answer:

1. Trichoderma is an effective biocontrol agent against soil borne fungal plant pathogens.
2. It is a free-living fungus found in the root ecosystem (rhizosphere).
3. It produces substances like viridin, gliotoxin, gliovirin, etc. that inhibit the soil borne pathogens which infect root and rhizomes to cause rot disease.

Question 29.
What are bioherbicides? Give any two examples.
Answer:

1. Bacteria, fungi and insects which kill the dicot herbs which acts as weeds in the fields of monocot cereal crops, are known as bioherbicides.
2. Pseudomonas spp. and Xanthomonas spp. kill several weeds.
3. Fungus Alternaria crassa controls water hyacinth.

Question 30.
What is composting? Which microorganisms are found in active compost?
Answer:

1. Composting is a natural process in which organic matter is converted into a dark rich compost or humus.
2. During composting, microorganisms break down organic matter into compost.
3. Microorganisms found in active compost are bacteria, fungi, actinobacteria, protozoa and rotifers.

Question 31.
What are cyanobacteria? Give any two examples cyanobacteria as biofertilizers.
Answer:

1. Cyanobacteria are aerobic, photosynthetic, N₂ fixing microorganisms.
2. They are aquatic or terrestrial.
3. They may be free-living or symbiotic.
4. They may be heterocystous or non- heterocystous. Heterocyst is the site of nitrogen fixation.
5. e.g. Free living cyanobacteria are Anabaena, Nostoc, Tolypothrix, Plectonema, Oscillatoria.
6. Anabaena and Nostoc have symbiotic relationship with lichens.
7. Anabaena is also symbiotically associated with Azolla and Cycas.

Chart based or table based questions

Question 1.
Complete the table given below.

Crop	Variety	Resistant to disease
Wheat	————–	Leaf and stripe rust, Hill bunt
————-	Pusa swarnim	White rust
Cauliflower	————–	Black rot and Curl blight black rot
Chilli	Pusa sadabahar	—————

Answer:

Crop	Variety	Resistant to disease
Wheat	Himgiri	Leaf and stripe rust, Hill bunt
Brassica	Pusa swarnim	White rust
Cauliflower	Pusashubhra	Black rot and Curl blight black rot
Chilli	Pusa sadabahar	Chilli mosaic virus, Tobacco mosaic virus, leaf curl

Question 2.
Complete the table given below.

Crop	Variety	Insect pest
Brassica	Pusa Gaurav	————-
————-	Pusa sem 2 m Pusa sem 3	Jassids, aphids, fruit borer
Okra	————–	Shoot and fruit borer

Answer:

Crop	Variety	Insect pest
Brassica	Pusa Gaurav	Aphids
Flat bean	Pusa sem 2 m Pusa sem 3	Jassids, aphids, fruit borer
Okra	Pusa Sawani, Pusa A-4	Shoot and fruit borer

Diagram based questions

Question 1.
a. Identify A and B in the following diagram.
b. What is organogenesis?
c. What is meant by hardening?

a. A : Explant, B : Callus

b. When auxins are provided to callus in more quantity compared to cytokinins, it gives rise to roots (rhizogenesis) and when cytokinins are provided in more quantity, there is development of shoot (caulogenesis). This induction of organ formation in callus by providing growth hormones in proper proportion is known as organogenesis.

c. Plantlets produced in tissue culture laboratory are transferred to polythene bags containing sterilized soil and are kept on low light and high humidity conditions for suitable period of time. This is known as hardening.

Question 2.
a. Identify honey bees A, B and C in the given diagram.
b. They belong to which species?

a. A : Worker honey bee, B : Queen honey bee, C : Drone of honey bee
b. They belong to species Apis mellifera.

Question 3.
Identify A, B, C and D in the given diagram.

Answer:
A : Honey super
B : Queen excluder
C : Hive bodies
D : Entrance reducer

Question 4.
Identify fish A, B, C and D in the given diagram.

Answer:
A : Rohu fish,
B : Mrigal fish,
C : Grass carp and
D : Silver carp

Question 5.
a. Identify A, B and C in the given diagram.
b. The given diagram represent life cycle of _____

a. A : Mature caterpillar, B : Cocoon and C : Adult female
b. Silk moth (Bombyx mori)

Question 6.
a. Identify stages A and B in the given diagram.
b. Larvae are also called …………..
c. The diagram represent life cycle of ……………

Answer:
a. A : Hatching, B : Pupa
b. Crawlers
c. Lac insect

Question 7.
Draw a labelled diagram of Tubular tower fermenter.
Answer:

Question 8.
Draw a labelled diagram of biogas plant.
Answer:

Question 9.
a. Identify A in the given diagram.
b. Name the bacteria which form ‘A’ in roots of leguminous plants.

Answer:
a. A : Root nodules
b. Rhizobia form root nodules in leguminous plants.

Question 10.
Draw a labelled diagram of T.S. of root nodule.
Answer:

Question 11.
Identify and describe the plant in the given diagram.

Answer:
The plant in the given diagram is Azolla.
Azolla is an aquatic fern that has symbiotic association with nitrogen fixing cyanobacterium Anabaena.
It propagates vegetatively and spreads in rice fields very rapidly.
It has a floating rhizome with small overlapping bilobed leaves and roots.
Azolla provides habitat to Anabaena.
The leaf shows dorsal and ventral lobe.
7. Anabaena filaments are present in the aerenchyma of dorsal lobe.

Question 12.
a. Identify A, B and C in the given diagram.
b. Name the plant which has symbiotic relationship with ‘A’.

Answer:
a. A : Anabaena, B : Photosynthetic zone. C : Dorsal lobe.
b. Azolla m

Long Answer Questions

Question 1.
Explain concept of outbreeding and its types.
Answer:
(1) Outbreeding involves breeding of two unrelated animals.
(2) It is of three types, viz. outcrossing, cross-breeding and interspecific hybridization.

(3) Outcrossing:

• Outcrossing involves mating of animals of same breed, which do not have a common ancestors on either side of mating partners up to 4 to 6 generations.
• The progeny obtained from such mating is called an outcross.
• Outcrossing is done to overcome inbreeding depression.

(4) Crossbreeding:

• In crossbreeding superior males of one breed are mated with superior females of another breed.
• New animal breeds of desirable characters are developed by this method.
• Example : Hisardale breed of sheep is developed in Punjab by crossing Bikaneri ewes and Marino rams.

(5) Interspecific hybridization:

• It involves breeding of animals of two different but related species.
• It is used to produce animals with desirable characters from both the parents.
• e.g. Mule is a breed obtained from horse and donkey.
• It may not be always successful.

Question 2.
Which dairy products are prepared using microorganisms? How?
Answer:

1. Dairy products prepared using microorganisms are curds, yogurt, butter milk and cheese.
2. The starter or inoculum used in preparation of dairy products contains millions of lactic acid bacteria (LAB).
3. Curd is prepared by inoculating milk with Lactobacillus acidophilus. It ferments lactose sugar of milk into lactic acid. Lactic acid causes coagulation and partial digestion of milk protein casein. Thus, milk is changed into curd. It also checks growth of disease causing microbes.
4. Yogurt is produced by curdling milk with the help of Streptococcus thermophilus and Lactobacillus bulgaricus.
5. Buttermilk is the acidulated liquid left after churning of butter from curd, is called buttermilk.
6. During the preparation of cheese, the milk is coagulated with LAB. The curd formed is filtered and whey is separated. The solid mass is then ripened with growth of mould that develops flavour in it. Characteristic texture, flavour and taste of cheese are developed by different specific microbes.

The ‘Roquefort cheese is ripened by blue green mold Penicillium roquejortii. Camembert cheese is ripened by blue-green mold P. camembertii. The large holes in Swiss cheese are developed due to production of a large amount of CO₂ by a bacterium known as Propionibacterium shermanii.

Question 3.
What are biofertilizers? Explain what are the different types of biofertilizers with suitable examples.
Answer:

1. Biofertilizers are commercial ready to use live bacterial, cyanobacterial or fungal formulations.
2. When they are applied to plants, in soil or in composting pits, soil fertility increases. Biofertilizers are cost effective and eco-friendly.

Types of biofertilizers as follows:
1. Bacterial biofertilizers:

• Nitrogen fixing bacterial biofertilizers : They convert atmospheric nitrogen into compounds of nitrogen like ammonia, nitrites and nitrates. The nitrogen fixing bacteria Rhizobium forms symbiotic association with roots of leguminous plants. Free living nitrogen fixing bacteria are Azotobacter and Azospirillum.
• Phosphate solubilizing biofertilizers : They are the bacteria which solubilize the insoluble inorganic phosphate compound, e. g. Pseudomonas striata, Bacillus polymyxa, Agrobacterium, Microccocus, Aspergillus spp., etc.
• Bacteria are also involved in composting.

2. Cyanobacterial biofertilizers:

• They are nitrogen fixing biofertilizers.
• Heterocysts are the site of nitrogen fixation.
• Free living cyanobacteria are Anabaena, Nostoc, Tolypothrix, Plectonema, Oscillatoria.
• Anabaena and Nostoc have symbiotic relationship with lichens
• Anabaena is also symbiotically associated with Azolla and Cycas.

3. Fungal biofertilizers:

• Fungal biofertilizers are mycorrhizae which form symbiotic association with roots of higher plants. There may be ectomycorrhiza and endomycorrhiza (VAM).
• Ectomycorrhizae increase the absorptive surface area of rots and increase uptake of water and nutrients.
• Plants with endomycorrhizae grow well even in less irrigated lands.

4. Microbes involved in composting:
During composting, microorganisms break down organic matter into compost. E.g. bacteria, fungi, actinobacteria, protozoa and rotifers.

Question 4.
Describe in details different types of mycorrhizae.
Answer:
Mycorrhizae are fungi that form symbiotic association with the roots of higher plants in humid forests.
There are two types as follows:
(1) Ectomycorrhizae:

• Mycelium of these fungi form mantle on the surface of the roots.
• Due to this absorptive surface area of roots increases and uptake of water and nutrients (N, P Ca and K) improves.
• The plant vigour, growth and yield increase.
• Some hyphae may penetrate into the root and form hartig-net in ‘the intercellular spaces of root cortex.

(2) Endomycorrhizae:

• Fugal hypahe of endomycorrhizae penetrate the root cortex and form branched arbuscules intracellularly. They also form vesicles mostly in the intercellular spaces.
• Hence, they are called Vesiculo Arbuscular Mycorrhizae (VAM). Nowadays they are described as AM fungi.
• The plants associated with VAM grow luxuriantly in less irrigated lands.
• VAM increase the productivity of field.

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 7 Thermal Properties of Matter Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 7 Thermal Properties of Matter

Question 1.
State various units of heat and relate them to SI unit of heat.
Answer:

1. CGS unit of heat: erg and it is related to SI unit as, 1 J = 10⁷ erg
2. Thermodynamic unit of heat: calorie (cal) and it is related as 1 cal = 4.184 J

Question 2.
What is thermal equilibrium?
Answer:

1. When two bodies at different temperatures come into the contact with each other, they exchange heat.
2. After some time, temperature of two bodies become equal and heat transfer between them stops.
3. The two bodies are then said to be in thermal equilibrium with each other.

Question 3.
State true of false. Correct the statement and rewrite if false.
i. Heat transfer takes place between the body and the surrounding medium until the body and the surrounding medium are at the same temperature.
ii. Whenever two bodies are in contact, there is a transfer of heat.
Answer:

1. True.
2. False.

Whenever two bodies at different temperature are in contact, there is a transfer of heat.

Question 4.
Give reason: Temperature is said to be a measure of average kinetic energy of the atoms/molecules of the body.
Answer:

1. Matter consists of particles which are in continuous vibrational motion and thereby possess kinetic energy.
2. When external energy is provided to these particles, internal energy of particles increases.
3. This increase in internal energy is in the form of increased kinetic energy of atoms/molecules and raises temperature of body (except at melting or boiling point of the body).
4. Greater the kinetic energy, faster the atoms/ molecules move and temperature of body becomes higher.
5. Thus, temperature of body is directly proportional to its kinetic energy.
   Hence, temperature is said to be a measure of average kinetic energy of the atoms and molecules of the body.

Question 5.
Why do solid particles possess potential energy?
Answer:
The solid particles possess potential energy due to the interatomic forces that hold the particles together at some mean fixed positions.

Question 6.
What happens when heat is supplied to a solid at its melting point?
Answer:

1. When heat is supplied to a solid at its melting point, average kinetic energy of constituent particles does not change.
2. As a result, temperature of body remains constant.
3. Supplied energy is used to weaken the bonds between constituent particles.
4. While order of magnitude of average distance between the molecules remains almost same as solid, substance melts, i.e., changes into liquid state, at melting point.

Question 7.
Why do solids have definite shape and volume?
Answer:
The solid particles possess potential energy due to the interatomic forces that hold the particles together at some mean fixed positions.
Hence, solids have definite shape and volume.

Question 8.
Why is density of liquid at melting point nearly same as density of solids at melting point?
Answer:

1. During change of state from solid to liquid, mass of substance does not change.
2. Also, mean distance between particles during change of state does not alter at melting point.
3. Density depends upon mass and volume, in turn, on mean distance between particles.
   Hence, density of liquid at melting point is nearly same as density of solids at melting point.

Question 9.
State true or false. If false correct the statement and rewrite.
Due to weakened interatomic bonds liquid do not possess definite volume but have definite shape.
Answer:
False.
Due to weakened bonds liquids do not possess definite shape but have definite volume.

Question 10.
What happens when heat is supplied to liquid its freezing point (melting point)?
Answer:

1. On heating, the atoms/molecules in liquid gain kinetic energy and temperature of the liquid increases.
2. If liquid is continued to heat further, at the boiling point, the constituents overcome the interatomic/molecular forces.
3. The mean distance between the constituents increases so that the particles are farther apart.
4. At boiling point, the liquid gets converted into gaseous state.

Question 11.
Why, according to kinetic theory of gases, gases have neither definite volume nor shape?
Answer:
As per kinetic theory of gases, for an ideal gas, there are no forces between the molecules of a gas. Hence, gases neither have a definite volume nor shape.

Question 12.
Match the pairs.

Answer:
i – b,
ii – c,
iii – a

Question 13.
Distinguish between an adiabatic wall and diathermic wall.
Answer:

	Adiabatic wall	Diathermic wall
i.	An ideal wall or partition separating two systems such that no heat exchange can take place between the systems is called adiabatic wall.	A wall that allows exchange of heat energy between two systems is said to be diathermic wall.
ii.	It is a perfect thermal insulator.	It is not a perfect thermal insulator.
iii.	It does not exist in reality.	Partition like thin sheet of copper acts as diathermic wall.
iv.	It is generally represented as thick cross-shaded (slanting lines region).	It is represented as a thin dark region.

Question 14.
State and explain zeroth law of thermodynamics.
Answer:
Statement: If two bodies A and B are in thermal equilibrium and also A and C are in thermal equilibrium then B and C are also in thermal equilibrium.
Explanation:

1. Consider two sections of a container separated by an adiabatic wall containing two different gases as system A and system B.
2. Systems A and B are independently brought in thermal equilibrium with a system C.
3. When the adiabatic wall separating systems A and B is removed, there will be no transfer of heat from system A to system B or vice versa.
4. This indicates that systems A and B are also in thermal equilibrium.
5. This means, if systems A and B are separately in thermal equilibrium with a system C, then A and B are also mutually in thermal equilibrium.

Question 15.
What is thermometry? What is thermometer?
Answer:
Thermometry is the science of temperature and its measurement. The device used to measure temperature is a thermometer.

Question 16.
State the principle used to measure the temperature of a system using a thermometer.
Answer:
When two or more systems/bodies are in thermal equilibrium, their temperatures are same. This principle is used to measure the temperature of a system by using a thermometer.

Question 17.
Explain how thermal equilibrium is attained between thermometer and the patient holding thermometer, in mouth.
Answer:

1. Thermometer indicating lower temperature is held in mouth by patient.
2. As body of patient is at higher temperature, heat energy is transferred from patient to thermometer.
3. When temperature of thermometer becomes same as temperature of patient, heat exchange stops and thermal equilibrium is attained between thermometer and body of patient.

Question 18.
Define the following terms.
i) Ice point
ii) Steam point
Answer:

1. Ice point: The temperature at which pure water freezes at one standard atmospheric pressure is called as ice point or freezing point.
2. Steam point: The temperature at which pure water boils into steam or steam changes to liquid water at one standard atmospheric pressure is called as steam point or boiling point.

Question 19.
Explain Celsius and fahrenheit scale of temperature. Give relation between the two scales with the help of the graph.
Answer:

1. Celsius scale:
   • The ice point (melting point of pure ice) is marked as 0 °C (lower point) and steam point (boiling point of water) is marked as 100 °C (higher point).
   • Both are taken at one atmospheric pressure.
   • The interval between these points is divided into loo equal pans. Each of these parts is called as one degree celsius and it is written as 1 oc.
2. Fahrenheit scale:
   • The ice point (melting point of pure ice) is marked as 32 °F and steam point (boiling point of water) is marked as 212 °F.
   • The interval between these two reference points is divided into 180 equal parts. Each part is called as degree fahrenheit and is written as 1 °F.
3. Relation between fahrenheit temperature and celsius temperature:
   \(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}\) = \(\frac{T_{C}-0}{100}\)
   Where, T_F = temperature in fahrenheit scales.
   T_C = temperature in celsius scale.
   The graph of T_F versus T_C is as shown
   

Question 20.
What is thermometer? Explain with examples, the thermometric property used in a thermometer.
Answer:

1. An instrument designed to measure temperature is called as thermometer.
2. Any property of a substance which changes sufficiently with temperature can be used as a basis of constructing a thermometer and is known as the thermometric property.
3. There are different types of thermometers.
   • In a constant volume gas thermometer, the pressure of a fixed volume of gas (measured by the difference in height) is used as the thermometric property.
   • The liquid-in-glass thermometer depends on the change in volume of the liquid with temperature. Small change in temperature changes the volume of liquid considerably. Two such liquids are mercury and alcohol. Mercury thermometers are used for measurement of temperature range -39 °C to 357 °C while alcohol thermometers are used only to measure temperatures near ice point (melting point of pure ice).
   • The resistance thermometer uses the change of electrical resistance of a metal wire with temperature.
   • Normally in research laboratories, a thermocouple is used to measure the temperature. A thermocouple is a junction of two different metals or alloys eg.: copper and iron joined together.
   • When two such junctions at the two ends of two dissimilar metal rods are kept at two different temperatures, an electromotive force is generated that can be calibrated to measure the temperature.
4. Thermometers are calibrated so that a numerical value may be assigned to a given temperature. The standard fixed points are melting point of ice and boiling point of water.

Question 21.
State characteristics of thermometer.
Answer:

1. Thermometer must be sensitive, i.e., a noticeable change in the thermometric property should be observed for a very small change in temperature.
2. It has to be accurate.
3. It should be easily reproducible.
4. It is important that the system attains thermal equilibrium with the thermometer quickly.

Question 22.
Explain relation between unknown temperature T and thermodynamic property P_T at temperature T.
Answer:
If the values of a thermometric property are P₁ and P₂ at the ice point (0 °C) and steam point (100 °C) respectively and the value of this property is P_T at unknown temperature T, then T is given by the following equation.
T = \(\frac{100\left(P_{T}-P_{1}\right)}{P_{2}-P_{1}}\)

Question 23.
State true or false. If false correct the statement and rewrite.
Ideally, there should be no difference in temperatures recorded on two different thermometers.
Answer:
True.

Question 24.
List an advantage and a disadvantage of constant volume gas thermometer.
Answer:
Advantage: There is no difference in temperatures recorded on two different constant volume gas thermometers. Hence, it is very accurate.
Disadvantage: Constant volume gas thermometer is bulky instrument. Hence, it is not easily portable.

Question 25.
Give short note on liquid-in-glass thermometer.
Answer:

1. Liquid-in-glass thermometer depends on the change in volume of the liquid with temperature.
2. When the bulb is heated, the liquid in a glass bulb expands upward in a capillary tube.
3. The liquid is such that it is easily seen and expands (or contracts) rapidly and by a large amount over a wide range of temperature.
4. Most commonly used liquids are mercury and alcohol as they remain in liquid state over a wide range. Mercury freezes at -39 °C and boils at 357 °C; alcohol freezes at -115 °C and boils at 78 °C.

Question 26.
What are thermochromic liquids? Give two examples.
Answer:

1. Thermochromic liquids are ones which change colour with temperature.
2. These liquids are very sensitive to temperature, especially in range of room temperature.
3. Hence, only specific liquids display distinct colour variations at normal temperature.
   Examples: Titanium dioxide and zinc oxide are white at room temperature but when heated change to yellow.

Question 27.
Write a note on resistance thermometer.
Answer:

1. Resistance thermometer uses the change of electrical resistance of a metal wire with temperature.
2. It measures temperature accurately in the range -2000 °C to 1200 °C is best for steady temperatures.
3. It is bulky and hence not easily portable.

Solved Examples

Question 28.
If the temperature in the room is 29 °C, what is its temperature in degree fahrenheit?
Solution:
Given: T_C = 29 °C
To find: Temperature in degree fahrenheit (T_F)

Answer:
Temperature in the room in degree fahrenheit is 84.2 °F.

Question 29.
Average room temperature on a normal day is 27 °C. What is the room temperature in °F?
Solution:
T_C = 27 °C
Room temperature in °F
Calculation: From formula,

Answer:
Room temperature in °F is 80.6 °F.

Question 30.
Normal human body temperature in fahrenheit is 98.4 °F. What is the body temperature in °C?
Solution:
T_F = 98.4 °F
Formula: Body temperature in °C (T_C)

Body temperature in °C is 36.89 °C.

Question 31.
The length of a mercury column in a mercury-in-glass thermometer is 25 mm at the ice point and 180 mm at the steam point. What is the temperature when the length is 60 mm?
Solution:
Here the thermometric property P is the length of the mercury column.
Using equation,
T = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{1}\right)}{\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right)}\)
For P₁ = 25 mm,
P₂ = 180 mm,
P_T = 60 mm
T = \(\frac{100(60-25)}{(180-25)}\)
= 22.58 °C
The temperature corresponding to the length of 60 mm is 22.58 °C.

Question 32.
A resistance thermometer has resistance 95.2 Ω at the ice point and 138.6 Ω at the steam point. What resistance would be obtained if the actual temperature is 27 °C?
Solution:
Here the thermometric property P is the resistance. If R is the resistance at 27 °C,
Using equation,
T = \(\frac{100\left(\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{1}\right)}{\left(\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{1}\right)}\)
For T = 27 °C, P₁ = 95.2 Ω, P₂ = 138.6 Ω .
∴ 27 = \(\frac{100(\mathrm{R}-95.2)}{(138.6-95.2)}\)
∴ R = \(\frac{27 \times(138.6-95.2)}{100}\) + 95.2
= 11.72 + 95.2
= 106.92Ω
The resistance obtained would be 106.92 Ω.

Question 33.
Explain the need for thermodynamic (absolute) scale.
Answer:

1. The two fixed point scale, Celsius scale and Fahrenheit scale had a practical shortcoming for calibrating the scale.
2. It was difficult to precisely control the pressure and identify the fixed points, especially for the boiling point as the boiling temperature is very sensitive to changes in pressure.
3. Hence, a one fixed point scale was adopted to define a temperature scale.
4. This scale is called the absolute scale or thermodynamic scale.

Question 34.
What is triple point of water? State its physical significance.
Answer:

1. The triple point of water is that point where water in a solid, liquid and gas state co-exists in equilibrium and this occurs only at a unique temperature and a pressure.
2. To know the triple point, one has to see that three phases coexist in equilibrium and no one phase in dominating. This occurs for each substance at a single unique combination of temperature and pressure.
3. Thus, if three phases of water solid ice, liquid water and water vapour coexist, the pressure and temperature are automatically fixed.
4. Internationally, triple point of water has been assigned as 273.16 K at pressure equal to 6.11 × 10² Pa or 6.11 × 10^-3 atmosphere, as the standard fixed point for calibration of thermometers.
5. The physical significance of triple point of water is that, it represents unique condition and it is used to define the absolute temperature.

Question 35.
Write a short note on absolute scale of temperature.
Answer:

1. The absolute scale of temperature, is so termed since ills based on the properties of an ideal gas and does not depend on the property of any particular substance.
2. The zero of this scale is ideally the lowest temperature possible although it has not been achieved in practice.
3. It is termed as Kelvin scale after Lord Kelvin with its zero at -273.15 °C and temperature intervals same as that on the Celsius scale. It is written as K (without °).

Question 36.
Draw a neat and well labelled diagram to show comparison of kelvin, Celsius and fahrenheit temperature scales.
Answer:

Question 37.
Answer the following:
i) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
ii) The absolute temperature (Kelvin scale) T is related to temperature t_c on the Celsius scale by t_c = T – 273.15. Why do we have 273.15 in this relation and not 273.16?
Answer:
i) The triple point of water has been assigned a fixed value of 273.15 K. This number represents a unique value associated with a unique condition of temperature and pressure in which all the three phases of water co-exist. On the other hand, melting point of ice and boiling point of water do not have a unique set of values as they are subject to changes in pressure and volume. For this reason, triple point of water is a standard fixed point in modem thermometry.

ii) On Celsius scale, the melting point of ice at normal pressure has a value 0 °C. The value corresponding to this value on the absolute scale is 273.15 K. The value 273.16 K denotes the triple point of water which has a value,
273.16 – 273.15 = 0.01 °C on Celsius scale as per the given relation.

Question 38.
State Charles’ law and give its formula.
Answer:
Charles’ law:
At constant pressure, volume of a fixed mass of gas is directly proportional to its absolute temperature.
Mathematically,
V ∝ T … ( at constant pressure)
∴ V = kT
where k is constant of proportionality
∴ \(\frac{\mathrm{V}}{\mathrm{T}}\) = k = constant
For two gases, \(\frac{V_{1}}{T_{1}}\) = \(\frac{V_{2}}{T_{2}}\) = constant.

Question 39.
State Pressure law and give its formula.
Answer:
Pressure (Gay Lussac’s) law:
At constant volume, pressure of a fixed mass of gas is directly proportional to its absolute temperature.
Mathematically,
P ∝ T .. .(at constant volume)
∴ P = kT
Where, k is constant of proportionality P
∴ \(\frac{\mathrm{P}}{\mathrm{T}}\) = k = constant
For two gases, \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\) = constant

Question 40.
State Boyle’s law and give its formula.
Answer:
Boyle’s law:
At constant temperature, the volume of a fixed mass of a gas is inversely proportional to its pressure.
Mathematically,
V ∝ \(\frac{1}{\mathrm{P}}\).. ..(at constant temperature)
V = k × \(\frac{1}{\mathrm{P}}\)
where, k is constant of proportionality.
PV = k = constant For two gases,
P₁V₁ = P₂V₂ = constant

Question 41.
Derive ideal gas equation PV = nRT.
Answer:

1. The relation between three variables of a gas i.e., pressure, volume and absolute temperature is called as ideal gas equation.
   From Boyle’s law,
   V ∝ \(\frac{\mathrm{T}}{\mathrm{P}}\), at constant temperature ….(1)
   From Charles’ law,
   V ∝ T, at constant pressure … .(2)
2. Combining equations (1) and (2) we get,
   ∴ V ∝ \(\frac{\mathrm{T}}{\mathrm{P}}\)
   ∴ \(\frac{\mathrm{PV}}{\mathrm{T}}\)= constant
3. For one mole of a gas,
   \(\frac{\mathrm{PV}}{\mathrm{T}}\) = R or PV = RT … (3)
   where R is the constant of proportionality.
4. Equation (3) is called ideal gas equation. The value of constant R is same for all gases. Therefore, R is called as universal gas constant. R = 8.31 JK^-1mol^-1.
5. For ‘n’ moles of gas, i.e. if the gas contains ‘n’ moles, equation (3) can be written as,
   PV = nRT

Solved Examples

Question 42.
Express T = 24.57 K in Celsius and fahrenheit.
Solution:
Given: T_K = 24.57 K
To find: Temperature in Celsius (T_C),
Temperature in fahrenheit (T_F)

Calculation:
From formula,
∴ T_C = T_K – 273.15
= 24.57 – 273.15
= -248.58 °C

Temperature in celsius (T_C) is – 248.58 °C
Temperature in fahrenheit (T_F) is -415.44 °F

Question 43.
Calculate the temperature which has the same value on fahrenheit scale and kelvin scale.
Solution:
Given: T_K = T_F = x
To find: Temperature at which Kelvin and Fahrenheit scales coincide (x)
Formula: \(\frac{\mathrm{T}_{\mathrm{F}}-32}{180}\) = \(\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\)
Calculation: From formula.
∴ \(\frac{x-32}{180}\) = \(\frac{x-273.15}{100}\)
∴ 5(x – 32) = 9(x – 273.15)
∴ 5x – 160 = 9x – 2458.35
∴ 4x = 2298.35
∴ x = \(\frac{2298.35}{4}\) = 574.6
∴ x = 574.6 °F
The temperature at which kelvin and fahrenheit scales coincide is 574.6 °F.

Question 44.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K, respectively. Express these temperatures on
the celsius and fahrenheit scales. (NCERT)
Solution:
Given: For neon, T_K = 24.57 K
For carbon dioxide, T_K = 216.55 K
To find:
i) Triple point of neon on celsius (T_CN) and fahrenheit scale (T_FN)
ii) Triple point of carbon dioxide on Celsius (T_CC) and fahrenheit scale (T_FC)

Formulae:
i) T_K – 273.15 = T_C
ii) \(\frac{\mathrm{T}_{\mathrm{K}}-273.15}{100}\) = \(\frac{\mathrm{T}_{\mathrm{c}}-32}{180}\)
Calculation: From formula (i),
T_C = T_K – 273.15
For neon, T_CN = 24.57 – 273.15
∴ T_CN = -248.58 °c
For carbon dioxide.
T_CC = 216.55 – 273.15
∴ T_CC = -56.60 °c
From formula (ii),
T_F = \(\frac{9}{5}\)(T_k – 273.15) + 32
For neon, T_k = 24.57 K
∴ T_FN = \(\frac{9}{5}\)[24.57 – 273.15] + 32
∴ T_FN = -415.44 °F
For CO₂, T_K = 216.55 K
∴ T_FC = \(\frac{9}{5}\)(216.55 – 273.15) + 32
∴ T_FC = -69.88 °F
i) Triple point of neon on celsius scale is -248.58 °c and on a fahrenheit scale is -415.44 °F.
ii) Triple point of carbon-dioxide on celsius scale is -56.60 °C and on fahrenheit scale is -69.88 °F.

Question 45.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B?
Solution:
Triple point of water is, T = 273.16 K
Since the absolute scales measure the triple point as 200 A and 350 B.
∴ 200A = 350B = 273.16 K
∴ 1A = \(\frac{273.16}{200}\)K and 1B = \(\frac{273.16}{350}\)K
If T_A and T_B are the temperatures on the two scales, then
\(\frac{273.16}{200}\)T_A = \(\frac{273.16}{350}\)T_B
∴ T_A = \(\frac{200}{350}\) T_B = \(\frac{4}{7}\)T_B
\(\frac{\mathbf{T}_{\mathbf{A}}}{\mathbf{T}_{\mathbf{B}}}\) = \(\frac{4}{7}\)
The relation between T_A and T_B is T_A : T_B = 4 : 7

Question 46.
The pressure reading in a thermometer at steam point is 1.367 × 10³ Pa. What is pressure reading at triple point knowing the linear relationship between temperature and pressure?
Solution:
Given: P = 1.367 × 10³ Pa at steam point (T) i.e., at 273.15 + 100 = 373.15 K.
Linear relationship between temperature and pressure means that.
P ∝ T ⇒ P₁T₁ = P₂T₂
To find: Pressure reading (P_triple)
Formula: P_triple = 273.16 × \(\left(\frac{\mathrm{P}}{\mathrm{T}}\right)\)
where P_triple and P are the pressures at temperature of triple point (273.16 K) and T (375.15 K) respectively.
Calculation: From formula,
∴ P_triple = 273.16 × \(\left(\frac{1.367 \times 10^{3}}{373.15}\right)\)
= 1000 × 10³ Pa
Pressure reading is 1.000 × 10³ Pa.

Question 47.
When the pressure of 0.75 litre of a gas at 27 °C is doubled, its temperature rises to 111°C. Calculate the final volume of a gas.
Solution:
Given: V₁ = 0.75 litre = 750 cm³,
T₁ = 27 + 273.15 = 300.15 K,
T₂ = 111 + 273.15 = 384.15 K,
P₂ = 2P₁
To find: Final volume (V₂)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula,

∴ V₂ = 480 cm³
The final volume of the gas is 480 cm³.

Question 48.
A certain mass of a gas at 20 °C is heated until both its pressure and volume are doubled. Calculate the final temperature.
Solution:
T₁ = 20 + 273.15 = 293.15 K,
P₂ = 2P₁, and V₂ = 2V₁
To find: Final temperature (T₂)
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: From formula,
\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}\) = \(\frac{\left(2 \mathrm{P}_{1}\right)\left(2 \mathrm{~V}_{1}\right)}{\mathrm{T}_{2}}\)
∴ \(\frac{1}{\mathrm{~T}_{1}}\) = \(\frac{4}{\mathrm{~T}_{2}}\)
∴ T₂ = 4 × 293.15 = 1172.6 K
= 1172.6 – 273.15 = 899.45 °C
∴ T₂ = 1172.6 K or 899.45 °C
The final temperature of the gas is 1172.6 K or 899.45 °C.

Question 49.
Explain how substances expand on the basis of vibrational motion of atoms.
Answer:

1. The atoms in a solid vibrate about their mean positions.
2. When heated, they vibrate faster and force each other to move a little farther apart. This results into expansion.
3. The molecules in a liquid or gas move with certain speed.
4. When heated, they move faster and force each other to move a little farther apart. This results
   in expansion of liquids and gases on heating.
5. The expansion is more in liquids than in solids; gases expand even more.

Question 50.
What is thermal expansion? List types of thermal expansion.
Answer:

1. A change in the temperature of a body causes change in its dimensions.
2. The increase in the dimensions of a body due to an increase in its temperature is called thermal expansion.
3. There are three types of thermal expansion:
   • Linear expansion
   • Areal expansion
   • Volume expansion

Question 51.
Define linear expansion of solids.
Answer:
The expansion in length of a solid due to thermal energy is called linear expansion.

Question 52.
Define coefficient of linear expansion of solid. State its unit and dimensions.
Answer:

1. The coefficient of linear expansion of a solid is defined as increase in the length per unit original length at 0 °C per degree rise in temperature.
   It is denoted by α and given by,
   α = \(\frac{l_{2}-l_{1}}{l_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
   where l₁ = initial length at temperature T₁ °C
   l₂ = final length at temperature T₂ °C
2. Unit: °C^-1 or K^-1
3. Dimensions: [L⁰M⁰T⁰K^-1]

Question 53.
Derive an expression for the coefficient of linear expansion in solid.
Answer:
i) If the substance is in the form of a long rod of length l, then for small change ∆T, in temperature, the fractional change ∆l/l, in length is directly proportional to ∆T.

[Note: Linear expansion ∆l is exaggerated for explanation.]
\(\frac{\Delta l}{l}\) ∝ ∆T
∴ \(\frac{\Delta l}{l}\) = α∆T … (1)
where, α is called the coefficient of linear expansion of solid.

ii) Rearranging terms in equation (1),
α = \(\frac{\Delta l}{l \Delta \mathrm{T}}\)
= \(\frac{l_{\mathrm{T}}-l_{0}}{l_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)
where,
l₀ = length of rod at 0 °C,
l_T = length of rod when heated to T °C,
T₀ = 0 °C (initial temperature)
T = final temperature,
∆l = l_T – T₀ = change in length,
∆T = T – T₀ = rise in temperature.

iii) If l₀ = 1 m and T – T₀ = 1 °C, then α = l_T – l₀ (numerically).

iv) As magnitude of α varies negligibly with temperature, it is assumed to be constant for a particular material.
Hence, it is not essential to take initial temperature as 0 °C.
This modifies equation (2) into,
α = \(\frac{l_{2}-l_{1}}{l_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
where l₁ = initial length at temperature T₁ °C
l₂ = final length at temperature T₂ °C

Question 54.
State true or false. If false correct the statement and rewrite.
i) Coefficient of linear expansion is same for all substances.
ii) Metals have high values for the coefficient of linear expansion, than non-metals.
Answer:

1. False.
   Coefficient of linear expansion is different for different substances.
2. True.

Question 55.
Define areal expansion of solids.
Answer:
The increase in the surface area of a solid, on heating is called areal expansion or superficial expansion of solids.

Question 56.
Define coefficient of superficial (areal) expansion of a solid, state its unit and dimensions.
Answer:

1. Coefficient of superficial (areal) expansion of a solid is defined as the increase in area per unit original area at 0 °C per degree rise in temperature.
   It is denoted by β and given by,
   β = \(\frac{A_{2}-A_{1}}{A_{1}\left(T_{2}-T_{1}\right)}\)
   where, A₁ = Area of solid sheet at T₁ °C,
   A₂ = Area of solid sheet at T₂ °C
2. Unit: 0°C^-1 or K^-1
3. Dimensions: [L⁰M⁰T⁰K^-1

Question 57.
Derive expression for coefficient of areal expansion.
Answer:
i) If a substance is in the form of a plate of area A, then for small change ∆T in temperature, the fractional change in area, ∆A/A in figure given below, is directly proportional to ∆T.

[Note: Areal expansion ∆A is exaggerated for explanation]
\(\frac{\Delta \mathrm{A}}{\mathrm{A}}\) ∝ ∆T
∴ \(\frac{\Delta \mathrm{A}}{\mathrm{A}}\) = β∆T … (1)
where β is called the coefficient of areal expansion of solid.

ii) Rearranging terms in equation (1),
β = \(\frac{\Delta \mathrm{A}}{\mathrm{A} \Delta \mathrm{T}}\)
= \(\frac{A_{T}-A_{0}}{A_{0}\left(T-T_{0}\right)}\)
where,
A₀ = volume at 0 °C,
A_T = volume when heated to T °C,
T₀ = 0 °C (initial temperature),
T = final temperature,
∆A = A_T – A₀ = change in area,
∆T = T – T₀ = rise in temperature.
iii) If A₀ = 1 m², T – T_o = 1 °C, then β = A_T – A₀ (numerically).
iv. As, β does not vary significantly with temperature. Hence, if A₁ is the area of a metal plate at T₁ °C and A₂ is the area at higher temperature at T₂ °C, then
β = \(\frac{A_{2}-A_{1}}{A_{1}\left(T_{2}-T_{1}\right)}\)

Question 58.
Define volume expansion of solids.
Answer:
The increase in volume due to heating is called volume expansion or cubical expansion.

Question 59.
Define coefficient of cubical (volume) expansion of a solid. State its unit and dimensions.
Answer:
i) Coefficient of cubical (volume) expansion:
Coefficient of cubical (volume) expansion of a solid is defined as increase in volume per unit original volume at O °C per degree rise in temperature.
It is denoted by γ and is given by,
γ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
where,
V₁ = Volume of solid at T₁ °C,
V₂ = Volume of solid at T₂ °C
ii) Dimensions: [L⁰M⁰T⁰K^-1]

[Note: Units and dimension of areal expansion in solid (β) and cubical expansion in solid (γ) are same as that of linear expansion in solid (α).

Question 60.
Derive an expression for coefficient of cubical expansion.
Answer:
i) If the substance is in the form of a cube of volume V, then for small change ∆T in temperature, the fractional change, ∆/V in volume is directly proportional to ∆T.

[Note: Volume expansion ∆V is exaggerated for explanation.)
ii) γ = \(\frac{\Delta \mathrm{V}}{\mathrm{V} \Delta \mathrm{T}}\) = \(\frac{\mathrm{V}_{\mathrm{T}}-\mathrm{V}_{0}}{\mathrm{~V}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)
where, V₀ = volume at 0 °C,
V_T = volume when heated to T °C,
T₀ = 0 °C (initial temperature),
T = final temperature.
∆V = V_T – V₀ = change in volume,
γ = V_T – T₀ = rise in temperature.

iii) If V₀ = 1 m³, T – T₀ = 1 °C, then γ = V_T – V₀ (numerically).

iv) If V₁ is the volume of a body at T₁ °C and V₂ is the volume at higher temperature T₂ °C, then
γ₁ = \(\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
γ₁, is the coefficient of volume expansion at temperature T₁ °C.

Question 61.
How does coefficient of volume expansion depend upon temperature?
Answer:

1. As compared to coefficient of linear expansion (α) and coefficient of areal expansion (β), γ changes more with temperature.
2. It is constant only at high temperatures.

Question 62.
Do coefficient of areal expansion and coefficient of volume expansion depend upon nature of material?
Answer:
Yes, coefficient of areal expansion and coefficient of volume expansion depend upon nature of material.

Question 63.
Explain expansion in fluids.
Answer:

1. Since fluids possess definite volume and take the shape of the container, they exhibit only change in volume significantly.
2. Equations valid for cubical or volume expansion of fluids are:
   γ = \(\frac{\Delta V}{V \Delta T}=\frac{V_{T}-V_{0}}{V_{0}\left(T-T_{0}\right)}\)
   where, V₀ = volume at 0 °C,
   V_T = volume when heated to T °C,
   T₀ = 0 °C (initial temperature),
   T = final temperature,
   ∆V = V_T – V₀ = change in volume,
   ∆T = T – T₀ = rise in temperature.
   and γ₁ = \(\frac{V_{2}-V_{1}}{V_{1}\left(T_{2}-T_{1}\right)}\)
   where, V₁ is volume of body at T₁ °C, V₂ is volume of body at higher temperature T₂ °C and γ₁ is coefficient volume expansion at T₁ °C.
3. As fluids are kept in containers, while dealing with the volume expansion of fluids, expansion of the container also needs to be considered.
4. If expansion of fluid results in a volume greater than the volume of the container, the fluid overflows if the container is open.
5. If the container is closed, volume expansion of fluid causes additional pressure on the walls of the container.

Question 64.
Use the data given in the table below:

Materials	γ(K-1)
Invar	2 × 10 -6
Steel	(3.3 – 3.9) × 10-5
Aluminium	6.9 × 10“-5
Mercury	18.2 × 10-5
Water	20.7 × 10-5
Paraffin	58.8 × 10-5
Gasoline	95.0 × 10-5
Alcohol (ethyl)	110 × 10-5

What conclusions can be drawn using the data?
Answer:

1. Coefficient of volume expansion (γ) is a characteristic of the substance.
2. It (γ) has higher order of magnitude for liquids than that of solids.

Question 65.
Explain how behaviour of water is different than solids and liquids when heat is supplied.
Answer:

1. Normally solids and liquids expand on heating. Hence their volume increases on heating.
2. Since the mass is constant, it results in a decrease in the density on heating.
3. Water expand on cooling from 4 °C to 0 °C.
4. Hence its density decreases on cooling in this temperature range.

Question 66.
Derive relation between coefficient of linear expansion (α) and coefficient of areal expansion (β).
Answer:
Consider a square plate of side l₀ at 0 °C and h at T °C.

1. l_T = l_o (1 + αT) .
   If area of plate at 0 °C is A_o, A_o = \(l_{0}^{2}\)
   If area of plate at T °C is A_T,
   A_T = \(l_{\mathrm{T}}^{2}\) = \(l_{0}^{2}\)(1 + αT)²
   or A_T = A₀(1 + αT)² …. (1)
   Also,
   A_T = A₀(1 + βT) … (2)
   
2. Using Equations (1) and (2), A₀(1 + αT)² = A₀(1 + βT)
   ∴ 1 + 2αT + α²T² = 1 + βT
3. Since the values of a are very small, the term α²T² is very small and may be neglected, ∴ β = 2α
4. The result is general because any solid can be regarded as a collection of small squares.

Question 67.
Derive relation between coefficient of linear expansion (α) and coefficient of cubical expansion (γ).
Answer:

1. Consider a cube of side l_o at 0 °C and l_T at T°C.
   ∴ l_T = l_o(l + αT)
   If volume of the cube at 0 °C is V₀, V₀ = \(l_{0}^{3}\)
   If volume of the cube at T °C is
   V_T, V_T = \(l_{\mathrm{T}}^{3}\) = \(l_{0}^{3}\)(1 + αT)³
   V_T = V₀ (1 + αT)³ ….(1)
   Also,
   V_T = V₀(1 + γT) ….(2)
   
2. Using Equations (1) and (2),
   V_o(1 + αT)³ = V₀(1 + γT)
   ∴ 1 + 3αT + 3α²T² + α³T³ = 1 + γT
3. Since the values of a are very small, the terms with higher powers of a may be neglected.
   ∴ γ = 3α
4. The result is general because any solid can be regarded as a collection of small cubes.

Question 68.
State the relation between α, β and γ and write their meaning.
Answer:
Relation between α, β and γ is given by,
α = \(\frac{\beta}{2}\) = \(\frac{\gamma}{3}\)
where, α = coefficient of linear expansion.
β = coefficient of superficial expansion.
γ = coefficient of cubical expansion.

Solved Examples

Question 69.
The length of a rail on a railway line is 25 m at 10 °C. During summer, maximum temperature attained in the region is 50 °C. Find the minimum gap between the rails, (a = 1.2 × 10^-5/°C)
Solution:
Given: L₁ = 25 m, T₁ = 10 °C, T₂ = 50 °C,
α = 1.2 × 10^-5/°C
To find: Minimum gap between rails (L₂ – L₁)
Formula: L₂ – L₁ = L₁α(T₂ – T₁)
Calculation: From formula,
L₂ – L₁ = 25 × 1.2 × 10^-5 × (50 – 10)
= 25 × 1.2 × 10^-5 × 40
= 1.2 × 10^-2 m
∴ L₂ – L₁ = 1.2 cm
The minimum gap between the rails is 1.2 cm.

Question 70.
The length of a metal rod at 27 °C is 4 cm. The length increases to 4.02 cm when the metal rod is heated upto 387 °C. Determine the coefficient of linear expansion of the metal rod.
Solution:
Given: T₁ = 27 °C, T₂ = 387 °C
L₁ = 4 cm = 4 × 10^-2 m
L₁ = 4.02 cm = 4.02 × 10^-2 m
To find: Coefficient of linear expansion

Coefficient of linear expansion is 1.389 × 10^-5/°C

Question 71.
The length of a metal rod is 150 cm at 25 °C. Find its length when it is heated to 150 °C. (α_steel = 2.2 × 10⁵ /°C)
Solution:
Given. L₁ = 150 cm, T₁ = 25 °C, T₂ = 150 °C,
α_steel = 2.2 × 10⁵ /°C
To find: Length of rod (L₂)
Formula: α = \(\frac{L_{2}-L_{1}}{L_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
L₂ – L₁ = L₁ α(T₂ – T₁)
∴ L₂ = L₁[1 + α(T₂ – T₁)]
= 150[1 + 2.2 × 10⁵ × (150 – 25)]
= 150(1 + 2.2 × 10⁵ × 125)
= 150(1 + 0.00275)
= 150× 1.00275 = 150.4125
∴ L₂ = 150.4cm
Length of the rod at 150 °C) is 150.4 cm.

Question 72.
Length of a metal rod at temperature 27 °C is 4.256 m. Find the temperature at which the length of the same rod increases to 4.268 m. (α for iron = 1.2 × 10⁵ K^-1)
Solution:
Given: T₁ = 27°C, L₁ = 4.256 m, L₂ = 4.268 m,
α = 1.2 × 10⁵ K^-1
To find: Temperature (T₂)

= antilog [log 1114.912 – log 4.256]
= antilog [3.0468 – 0.6290]
= antilog [2.4 1781
= 2.617 × 10²
= 261.7°C
Required temperature is 261.7 °C.

Question 73.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10^-5 °C^-1.
Solution:
Given: d₁ = 4.24 cm, ∆T = 227 – 27 = 200 °C,
α = 1.70 × 10^-5 °C^-1
To find: Change in diameter (∆d)
Formula: α = \(\frac{d_{2}-d_{1}}{d_{1} \Delta T}=\frac{\Delta d}{d_{1} \Delta T}\)
Calculation: From formula.
∆d = α x d₁ x αT
= 1.70 × 10 × 4.24 × 200
∴ ∆d = 1.44 × 10^-2 cm
The change in diameter of the hole is 1.44 × 10^-2 cm.

Question 74.
A thin aluminium plate has an area 286 cm2 at 20 °C. Find its area when it is heated to 180 °C.
(β for aluminium = 4.9 × 10^-5 °C)
Solution:
Given: T₁ = 20°C, T₂ = 180°C, A₁ = 286 cm²
β = 4.9 × 10^-5 °C
To find: Final area (A₂)
Formula: β = \(\frac{A_{2}-A_{1}}{A_{1}\left(T_{2}-T_{1}\right)}\)
Calculation: From formula,
A₂ = A₁ [1 + β(T₂ – T₁)]
= 286[1 + 4.9 × 10^-5 (180 – 20)]
= 286[1 + 4.9 × 10^-5 × 160]
= 286 [1 + 784.0 × 10]
= 286 [1 + 0.00784]
= 286 [1.00784]
∴ A₂ = 288.24 cm²
Its area when its heated is 288.24 cm².

Question 75.
The surface area of the metal plate is 2.4 × 10^-2 m² at 20°C. When the plate is heated to 185 °C, its area increases by 0.8 cm². Find the coefficient of areal expansion of metal.
Solution:
Given: T₁ = 20 °C, A₁ = 2.4 × 10^-2 m²,
T₂ = 185°C,
∆A = 0.8cm² = 0.8 × 10^-4 m²
To find: Coefficient of areal expansion (β)

The coefficient of areal expansion of the metal is 2.02 × 10^-5/ °C.

Question 76.
A liquid occupies a volume of 2 × 10^-4 m³ at 0 °C. Calculate the increase in its volume If It is heated to 80 °C. [The coefficient of cubical expansion of the liquid is 4 × 10^-4 K^-1]
Solution:
Given: V₀ = 2 × 10^-4 m³, T₀ = 0°C,T = 80°C,
γ_r = 4 × 10^-4 K^-1,
T – T₀ = 80 – 0 = 80°C
To find: Increase in volume (∆V)
Formula. ∆V = V₀γ_r(T – T₀)
Calculation: From formula.
∆V = (2 × 10^-4) (4 × 10^-4) × 80
∴ ∆V = 6.4 × 10^-6 m³
Increase m volume of the liquid is 6.4 × 10^-6 m³.

Question 77.
A liquid at O °C is poured in a glass beaker of volume 600 cm3 to fill it completely. The beaker is then heated to 90 °C. How much liquid will overflow?
(γ_liquid = 1.75 × 10^-4/ °C, γ_glass = 2.75 × 10^-5/ °C)
Solution:
Given: V₁ = 600 cm³, T₁ = 0 °C, T₂ = 90 °C
γ_liquid = 1.75 × 10^-4/°C,
γ_glass = 2.75 × 10^-4/°C
To find: Volume of liquid that overflows
Formula: γ = \(\frac{\mathrm{V}_{2}-\mathrm{V}_{1}}{\mathrm{~V}_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}\)
Calculation: From formula,
Increase is volume = V₂ – V₁
= γ V₁(T₂ – T₁)
Increase in volume of beaker
= γ_glass × V₁ (T₂ – T₁)
= 2.75 × 10^-5 × 600 × (90 – 0)
= 2.75 × 10^-5 × 600 × 90
= 148500 × 10^-5 cm³
∴ Increase in volume of beaker = 1.485 cm³
Increase in volume of liquid
= γ_liquid × V₁ (T₂ – T₁)
= 1.75 × 10^-4 × 600 × (90 – 0)
= 1.75 × 10^-4 × 600 × 90
= 94500 × 10^-4 cm³
∴ Increase in volume of liquid = 9.45 cm³
∴ Volume of liquid which overflows
= (9.45 – 1.485) cm³
= 7.965 cm³
Volume of liquid that overflows is 7.965 cm³.

Question 78.
The surface area of an iron plate is 80 cm² at 20 °C. Find its surface area at 120 °C. (α_iron = 1.25 × 10^-5 / °C)
Solution:
Given: A₁ = 80 cm², T₁ = 20 °C, T₂ = 120 °C,
α_iron = 1.25 × 10^-5 / °C
To find: Surface area (A₂)
Formula: A₂ = A₁ [1 + β (T₂ – T₁)]
Calculation: β_iron = 2 × α_iron = 2.5 × 10^-5 / °C
From formula,
A₂ = 80[1 + 2.5 × 10^-5 (120 – 20)]
∴ A₂ = 80.2 cm²
Surface area of the iron plate at 120 °C is 80.2 cm².

Question 79.
A sheet of brass is 50 cm long and 8 cm broad at 0 °C. If the surface area at 100 °C is 401.57 cm² find the coefficient of linear expansion of brass.
Solution:
Given: l = 50 cm, b = 8cm.
∴ A₁ = l × b = 50 x 8 = 400 cm²,
T₁ = 0°C, T₂ = 100°C,
A₂ = 401.57 cm²
To find: coefficient of linear expansion (α)

Coefficient of linear expansion of brass is 1.962 × 10^-5/°C.

Question 80.
On heating a glass block of 10.000 cm³ from 25 °C to 40 °C, its volume increases by 4 cm3. Calculate coefficient of linear elipansion of glass.
Solution:
Given: V = 10,000 cm³, ∆V = 4 cm³,
∆T = 40 – 25 = 15°C,
To find: Coefficient of linear expansion (α)

Coefficient of linear expansion of the glass block is 8.89 × 10^-6 /°C.

Question 81.
Explain in detail what is specific heat or specific heat capacity of a substance.
Answer:

1. Specific heat capacity is defined as the amount of heat per unit mass absorbed or given out by the substance to change its temperature by one uni! (one degree) Le., 1 °C or 1 K.
2. The amount of heat (∆Q) required to change the temperature of a substance is directly proportional to:
   • the mass of the substance (m).
   • change in temperature of the substance (∆T).
     ∴ ∆Q ∝ m and ∆Q ∝∆T
     ∴ ∆Q ∝ m∆T
     ∴ ∆Q = sm∆T …………….. (1)
     where ‘s’ is specific heat or specific heat capacity of a substance.
     From equation (1).
     s = \(\frac{\Delta Q}{m \Delta T}\)
     If m = 1 kg and ∆T = 1 °C,then s = ∆Q.
3. Unit: S.I. unit of specific heat is J kg^-1 °C^-1 or J kg^-1 K^-1 and C.G.S. unit is erg g^-1 K^-1 or erg g^-1 °C^-1.
4. Example: The specific heat of water is 4.2 J kg^-1 °C^-1
   It means that 4.2 J of energy must be added to 1 kg of water to rise its temperature by 1 °C.
5. The specific heat capacity is a property of the substance.
6. Specific heat capacity weakly depends on temperature of object. Except for very low temperatures. the specific heat capacity is almost constant for all practical purposes.

Question 82.
State the heat equation.
Answer:
Heat received or given out (Q)
= mass (m) temperature change (∆T) × specific heat capacity (s).
or Q = m × ∆T × s

Question 83.
Write a note on: Molar specific heat.
Answer:

1. If the amount of substance is specified in terms of moles (μ) instead of mass (m) in kg. then the specific heat is called molar specific heat (C).
2. It is given by, C = \(\frac{1}{\mu} \frac{\Delta Q}{\Delta \mathrm{T}}\)
3. The SI unit of molar specific heat capacity is J/mol °C or J/mol K.
4. Like specific heat, molar specific heat also depends on the nature of the substance and its temperature.

Question 84.
Give reason: Water is used as a coolant in automobile radiators.
Answer:

1. Water has the highest specific heat capacity compared to other substances.
2. As a result, water requires higher amount of energy to get heated.
3. This allows water to absorb heat readily while increasing its temperature minimally.
   Hence, water is used as a coolant in automobile radiators.

Question 85.
Give reason: Water is preferred as heater in hot water hag than other liquids.
Answer:

1. Water has the highest specific heat capacity compared to other substances.
2. This means certain mass of water heated to certain temperature contains more heat than the same mass of any other liquid heated to same temperature.
3. As a result, water takes longer time to cool than any other liquid heated to same temperature.
   Hence, water is preferred as heater in hot water bag than other liquids.

Question 86.
Explain why specific heat capacity of a gas at constant pressure is greater than that at constant volume.
Answer:

1. When the gas is heated at constant volume. there is no work done against external pressure.
2. Hence, all the supplied heat is used in raising the temperature of the gas.
3. But when the gas is heated at constant pressure, volume of the gas changes.
4. Due to this, part of the supplied heat is used by the gas to expand against external pressure and remaining part of heat supplied is used to raise the temperature.
5. Because of this, for the same rise in temperature, the heat to be supplied at constant pressure is greater than that for heating at constant volume.
   Hence, specific heat capacity of a gas at constant pressure is greater than specific heat capacity at constant volume.

Question 87.
Define principal and molar specific heat of a gas at constant volunie and constant pressure.
Answer:

1. Principal specific heat:
   • Principal specific heat of a gas at constant volume (s_v):
     Principal specific heat of a gas at constant volume is defined as the quantity of heat absorbed or released for rise or fall temperature of unit mass of a gas through 1 K (or 1 °C), when its volume is kept constant.
   • Principal specific heat of a gas at constant pressure (s_p): Principal specfic heat of a gas at constant pressure is defined as the quantity of heat absorbed or released for rise or fall the temperature of unit mass of a gas through 1 K (or 1 °C), when its pressure is kept constant.
2. Molar specific heat:
   • Molar specific heat of a gas at constant volume (C_V): Molar specific heat of a gas at Constant volume is defined as the quantity of absorbed or released for rise or fall the temperature of one mole of the gas through 1 K (or 1 °C), when its volume is kept constant.
   • Molar specific heat of a gas at constant pressure (C_P): Molar specific heat of a gas at constant pressure is defined as the quantity of heat absorbed or released for rise or fall the temperature of one mole of the gas through 1 K (or 1 °C), when its pressure is kept constant.

Question 88.
State the relation between principal specific heat capacity and molar specific heat capacity for a gas.
Answer:
Molar specific heat capacity = Molecular weight × principal specific heat capacity.
i.e. C_V = μ × S_P and C_V = μ × S_V
where, μ is the molecular weight of the gas.
[Note: Symbols (S_V) and (S_P) are used as per standard convention.]

Question 89.
What is heat capacity?
Answer:

1. Heat capacity or thermal capacity of a body is the quantity of heat needed to raise or lower the temperature of the whole body by 1 o°C (or 1 K).
2. Heat capacity can be written as Heat received or given out
   = mass × 1 × specific heat capacity
   Heat capacity = Q = m × s
   Heat capacity (thermal capacity) is measured in J/°C.

Solved Examples

Question 90.
If the temperature of 4 kg mass of a material of specific heat capacity 300 J/ kg °C rises from 20 °C to 30 °C. Find the heat received.
Solution:
m = 4 kg, s = 300 J/kg °C
∆T = 30 – 20 = 10 °C
To find: Heat received (Q)
Formula: Q = ms∆T
Calculation: From formula,
Q = 4 × 300 × 10
∴ Q = 12000 J
Heat received is 12000 J.

Question 91.
How much heat is required to raise temperature of 750 g of copper pot from 20 to 50 °C?
(The specific heat of copper is 0.094 kcal/kg °C)
Solution:
Given: s = 0.094 kcal/kg°C,
m = 750 g = 0.750kg,
T₁ = 20°C and T₂ = 50°C
Rise in temperature,
∆T = T₂ – T₁ = 50 – 20 = 30°C
To find: Heat required (Q)
Formula: Q = m × s × ∆T
Calculation: From formula,
Q = 0.750 × 0.094 × 30
∴ Q = 2.115 kcal
Heat required to raise the temperature of copper pot is 2.115 kcal.

Question 92.
Calculate the difference in the temperatures between the water at the top and bottom of a water fall 200 m high. Specific heat of water is 4200 J kg^-1 °C^-1.
Solution:
Given: s = 4200 J kg^-1 °C^-1, h = 200 m
To find: Difference in temperatures (∆T)
Formulae:
i) Q = ms∆T
ii) P.E. = mgh
Calculation: From formulae (j) and (ii)
When water falls from top to bottom. assuming no loss in energy, potential energy is converted into heat energy.
∴ Q = P.E.
∴ ms∆T = mgh
∴ s∆T = gh
∴ ∆T = \(\frac{\mathrm{gh}}{\mathrm{s}}=\frac{9.8 \times 200}{4200}\)
∴ ∆T = 0.467 °C
The difference in temperatures between the water at top and bottom is 0.467 °C.

Question 93.
Find thermal capacity for a copper block of mass 0.2 kg, if specific heat capacity of copper is 290 J/kg °C.
Solution:
Given: m = 0.2 kg, s = 290 J/kg °C
To find: Thermal capacity
Formula: Thermal capacity = m × s
Calculation: From formula,
Thermal capacity = 0.2 × 290
= 58 J/ °C
Thermal capacity is 58 J/ °C.

Question 94.
What is calorimetry?
Answer:
Calorimetry is an experimental technique for quantitative measurement of heat exchange.

Question 95.
Explain construction of calorimeter with the help of a labelled diagram.
Answer:

1. A device in which heat measurement can be made is called calorimeter.
2. It consists of a cylindrical vessel and stirrer as well as lid of the same material like copper or aluminium.
3. The vessel is kept inside a wooden jacket which contains heat insulating materials like glass. wool etc. to prohibit any transfer of heat into or out of the calorimeter.
   

Question 96.
State the principle behind working of calorimeter.
Answer:
Calorimeter being isolated system works on the principle of conservation of energy, where heat gained equals heat lost.

Question 97.
Explain the technique “method of mixtures’.
Answer:

1. Method of mixtures is a technique used to determine specific heat capacity of a material using calorimeter.
2. In this technique a sample ‘A’ of the substance is heated to a high temperature which is accurately measured.
3. The sample ‘A’ is then placed quickly in the calorimeter containing water.
4. The contents are stirred constantly until the mixture attains a final common temperature.
5. The heat lost by the sample ‘A’ will be gained by the water and the calorimeter.
6. The specific heat of the sample ‘A’ of the substance can be calculated as follows:
   • Let,
     m₁ = mass of the sample ‘A’
     m2₂ = mass of the calorimeter and the stirrer
     m₃ = mass of the water in calorimeter
     s₁ = specific heat capacity of the substance of sample ‘A’
     s₂ = specific heat capacity of the material of calorimeter (and stirrer)
     s₃ = specific heat capacity of water
     T₁ = initial temperature of the sample ’A’
     T₂ = initial temperature of the calorimeter stirrer and water
     T = final temperature of the combined system
   • Using heat equation,
     Heat lost by the sample ‘A’ = m₁s₁ (T₁ – T)
     Heat gained by the calorimeter and the stirrer = m₂s₂ (T – T₂)
     Heat gained by the water = m₃s₃ (T – T₂)
   • c. Assuming no loss of heat to the surroundings, the heat lost by the sample goes into the calorimeter, stirrer and water,
     ∴ m₁s₁(T₁ – T) = m₂s₂(T – T₂) + m₃s₃(T – T₂) ………….. (1)
   • Knowing the specific heat capacity of water and copper material of the calorimeter and the stirrer, specific heat capacity (si) of material of sample ‘A’ can be calculated.
     e. Rearranging terms of equation (1),
     s₁ = \(\frac{\left(\mathrm{m}_{2} \mathrm{~s}_{2}+\mathrm{m}_{3} \mathrm{~s}_{3}\right)\left(\mathrm{T}-\mathrm{T}_{2}\right)}{\mathrm{m}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}\right)}\)
7. One can find specific heat capacity of water or any liquid using the following expression, if the specific heat capacity of the material of calorimeter and sample is known
   s₃ = \(\frac{\mathrm{m}_{1} \mathrm{~s}_{1}\left(\mathrm{~T}_{1}-\mathrm{T}\right)}{\mathrm{m}_{3}\left(\mathrm{~T}-\mathrm{T}_{2}\right)}-\frac{\mathrm{m}_{2} \mathrm{~s}_{2}}{\mathrm{~m}_{3}}\)

Question 98.
In method of mixtures, why is it essential that density of solid sample be greater than the liquid in calorimeter?
Answer:

1. In method of mixtures, solid sample gives away heat to liquid, and heat exchange between, solid, liquid and calorimeter is considered as isolated.
2. If density of solid sample is lesser than liquid in calorimeter, sample will float on liquid.
3. This will cause partial heat loss to air inside the calorimeter and standard heat equations of calorimeter will no longer be applicable.
   Hence, in method of mixtures it is essential that density of solid sample be greater than liquid in calorimeter.

Solved Examples

Question 99.
A sphere of aluminium of 0.06 kg is placed for sufficient time in a vessel containing boiling water so that the sphere is at 100 °C. It is then immediately transferred to 0.12 kg copper calorimeter containing 0.30 kg of water at 25 °C. The temperature of water rises and attains a steady state at 28 °C. Calculate the specific heat capacity of aluminium.
(Specific heat capacity of water, s_w = 4.18 × 10³ J kg^-1 K^-1, specific heat capacity of copper, s_Cu = 0.387 × 10³ J kg^-1 K^-1)
Solution:
Given: Mass of aluminium sphere = m₁ = 0.06 kg
Mass of copper calorimeter = m₂ = 0.12 kg
Mass of water in calorimeter = m₃ = 0.30 kg
Specific heat capacity of copper
= S_Cu = s₂ = 0.387 × 10³ J/kg K = 387 J/kg K
Specific heat capacity of water
= S_w = s₃ = 4.18 × 10³ J/kg K = 4180 J/kg K
Initial temperature of aluminium sphere
= T₁ = 100 °C
Initial temperature of calorimeter and water
= T₂ = 25 °C
Final temperature of the mixture = T = 28 °C
To find: Specific heat capacity of aluminium (s_al)

= 903.08 J/kg K
Specific heat capacity of aluminium is 903.08 J/kg K.

Question 100.
A copper sphere of 100 g mass is heated to raise its temperature to 100 °C and is released in water of mass 195 g and temperature 20 °C in a copper calorimeter. If the mass of calorimeter is 50 g, what will be the maximum temperature of water? (Given: specific heat of copper = 0.1 cal/g °C and specific heat of calorimeter = 0.1 cal/g °C)
Solution:
Let copper sphere, water and calorimeter attain final temperature T °C.
We have,

Heat lost by copper sphere
Q = m_sphere × S_sphere × ∆T
= 100 × 0.1 × (100 – T)
Heat gained by water in calorimeter
Q₁ = m_water × S_water × ∆T
= 195 × 1 × (T – 20)
Heat gained by calorimeter
Q₂ = m_calorimeter × m_calorimeter × ∆T
= 50 × 0.1 × (T – 20)
According to principle of heat exchange,
Q = Q₁ + Q₂
∴ 10 × (100 – T) = 195 × (T – 20) + 5 × (T – 20)
∴ 1000 – 10T = 200(T – 20)
∴ 210 T = 5000
∴ T ≈ 23.8 °C
Maximum temperature of water will be 23.8 °C.

Question 101.
What is a change of state? When does it occur?
Answer:

1. Matter normally exists in three states: solid. liquid and gas. A transition from one of these states to another is called a change of state.
2. This change can occur when exchange of heat takes place between the substance and its surroundings.

Question 102.
Explain the following temperature vs time graph obtained during process of boiling water.

Variation of temperature with time
Answer:

1. The given temperature v/s time graph demonstrates the behaviour of water when heated continuously and uniformly.
2. Line segment AB indicates temperature of ice remaining constant at 0 °C for certain period of time.
   • This means, amount of heat (latent heat of fusion) supplied to ice is entirely used for changing its state from solid to liquid.
   • Thus, line segment AB denotes conversion of ice at 0 °C into water at 0 °C.
3. Line segment BC indicates continuous rise in temperature of water from 0 °C to 100 °C.
   • At point C, boiling point of water is
     reached and heat energy (latent heat of vaporisation) supplied further is used to convert water into steam.
   • During this transformation, temperature remains unchanged as represented by line segment CD.
   • Thus, line segment CD denotes conversion of water at 100 °C into steam at 100 °C.
4. Beyond point D, thermometer again shows rise in temperature.

Question 103.
Write a note on latent heat of substance.
Answer:

1. Latent heat of a substance is the quantity of heat required to change the state of unit mass of the substance without changing its temperature.
2. Mathematically, if mass m of a substance undergoes a change from one state to the other then the quantity of heat absorbed or released is given by, Q = mL
   where. L is known as latent heat.
3. It is characteristic of the substance.
4. Its SI unit is J/kg.
5. The value of L depends on the pressure and is usually quoted at one standard atmospheric pressure.

Question 104.
Explain the following terms.
i) Latent heat of fusion
ii) Latent heat of vaporisation
Answer:

1. Latent heat of fusion:
   • The quantity of heat required to convert unit mass of a substance from its solid state to the liquid state, at its melting point, without any change in its temperature is called its latent heat of fusion.
   • The S.I. unit of latent heat of fusion is J/kg and its C.G.S. unit is cal/’g.
2. Latent heat of vaporisation:
   • The quantity of heat required to convert unit mass of a substance from its liquid state to vapour state, at its boiling point, without any change in its temperature is called its latent heat of vaporisation.
   • The S.I. unit of latent heat of vaporization is J/kg and its C.G.S unit is cal/g.

Question 105.
Explain why latent heat of vaporisation is much larger than latent heat of fusion.
Answer:

1. The energy required to completely separate the molecules or atoms in liquids is greater than the energy needed to break the rigidity (rigid bonds between the molecules or atoms) in solids.
2. Also, when the liquid is converted into vapour, it expands. Work has to be done against the surrounding atmosphere to allow this expansion.
   Hence, latent heat of vaporisation is larger than latent heat of fusion.

Question 106.
A plot of temperature versus heat energy for a given quantity of water is shown below. What can be inferred studying it?

Temperature versus heat for water at one standard atmospheric pressure (not to scale)
Answer:
Inferences:

1. When heat is added (or removed) during a change of state, the temperature remains constant.
2. Also the slopes of the phase lines are not all the same, which indicates that specific heats of the various states are not equal.
3. For water, the latent heat of fusion and vaporisation are L_f = 3.33 × 10⁵ J kg^-1 and L_v = 22.6 × 10⁵ J kg^-1 respectively, i.e., 3.33 × 10⁵ J of heat is needed to melt 1 kg of ice at 0 °C and 22.6 × 10⁵ J of heat is needed to convert 1 kg of water to steam at 100 °C.
4. This means, steam at 100 °C carries 22.6 × 10⁵ J kg^-1 more heat than water at 100 °C.

Question 107.
Compare change of state from solid to liquid and from liquid to vapour.
Answer:

	Solid to liquid	Liquid to vapour
i.	The change of state from solid to liquid is called melting and from liquid to solid is called solidification.	The change of state from liquid to vapour is called vaporisation while that from vapour to liquid is called condensation.
ii.	Both the solid and liquid states of the substance co-exist in thermal equilibrium during the change of states from solid to liquid or vice versa.	Both the liquid and vapour states of the substance coexists in thermal equilibrium during the change of state from liquid to vapour.
iii.	The temperature at which the solid and the liquid states of the substance are in thermal equilibrium with each other is called the melting point of solid or freezing point of liquid. The freezing point describes the liquid to solid transition while melting point describes solid to liquid transition.	The temperature at which the liquid and the vapour states of the substance coexist is called the boiling point of liquid. This is also the temperature at which water vapour condenses to form liquid.
iv.	It is the characteristic of the substance and also depends on pressure.	It is characteristic of substance and depends on pressure.

Question 108.
What is normal melting point?
Answer:
The melting point of a substance at one standard atmospheric pressure is called its normal melting point.
Example: Normal melting point of water is 0°C.

Question 109.
State true or false. If false correct the statement and rewrite.
Normal freezing point ice is 32 °C.
Answer:
False.
Normal freezing point ice is 0 °C or 32 °F.

Question 110.
What is normal boiling point?
Answer:
The boiling point of a substance at one standard atmospheric pressure is called its normal boiling point.
Example: Normal boiling point of water is 99.97 °C,

Question 111.
Distinguish between boiling and evaporation of liquid.
Answer:

	Boiling of liquid	Evaporation of liquid
i.	Boiling of liquid takes place at boiling point which is fixed for a given pressure and unique for a given liquid.	Evaporation of liquid can take place at any temperature.
ii.	It occurs throughout the liquid.	It occurs only at surface of liquid.
iii.	The process does not depend on area of liquid surface exposed.	The process depends upon area of liquid surface exposed. Higher the exposed surface area, higher the rate of evaporation.
iv.	Source of energy is needed.	Energy is taken from surrounding.
V.	Boiling does not reduce temperature of liquid.	When evaporation takes place, temperature of liquid decreases.
vi.	During boiling process, bubbles are formed in liquid.	During evaporation, no bubbles are formed in liquid.

Question 112.
Explain evaporation in terms of kinetic energy of liquid molecules.
Answer:

1. Molecules in a liquid are moving about randomly.
2. The average kinetic energy of the molecules decides the temperature of the liquid.
3. However, all molecules do not move with the same speed.
4. Some with higher kinetic energy may escape from the surface region by overcoming the interatomic forces.
5. This process can take place at any temperature and is termed as evaporation.

Question 113.
Explain the dependence of evaporation on temperature of liquid.
Answer:

1. If the temperature of the liquid is higher, more is the average kinetic energy.
2. This implies that the number of fast moving molecules is more.
3. Hence the rate of losing such molecules to atmosphere will be higher.
4. Thus, higher is the temperature of the liquid, greater is the rate of evaporation.

Question 114.
Why does evaporation gives a cooling effect to the remaining liquid?
Answer:

1. In the process of evaporation, faster moving molecules escape from surface of liquid overcoming the interatomic forces.
2. Since faster molecules are lost, the average kinetic energy of the liquid is reduced.
3. As a result, the temperature of the liquid is lowered.
   Hence, evaporation gives a cooling effect to the remaining liquid.

Question 115.
Explain two applications of evaporation in details.
Answer:

1. Drying of clothes:
   • Clothes dry faster when hanged exposing more surface area than when kept folded.
   • Due to more surface area, water in clothes gets evaporated faster, drying clothes quickly.
2. Using a spirit swab on skin before injecting gives cooling effect:
   • Before giving an injection to a patient, normally a spirit swab is used to disinfect the region.
   • A cooling effect is experienced by skin of patient due to evaporation of the spirit as explained before.

Question 116.
Write a note on sublimation.
Answer:

1. All substances do not pass through the three states: solid-liquid-gas.
2. There are certain substances which normally pass from the solid to the vapour state directly and vice versa.
3. The change from solid state to vapour state without passing through the liquid state is called sublimation and the substance is said to sublime.
   Examples: Dry ice (solid CO₂) and iodine.
4. During the sublimation process, both the solid and vapour states of a substance coexist in thermal equilibrium.
5. Most substances sublime at very low pressures.

Question 117.
What is a phase? Give an example.
Answer:
A phase is a homogeneous composition of a material.
Example: Graphite and diamond are two phases of carbon.

Question 118.
What is a phase diagram?
Answer:
A pressure-temperature (P-T) diagram particularly convenient for comparing different phases of a substance is called as a phase diagram.

Question 119.
Study phase diagrams given below and answer the following questions.

i)Explain vaporisation curve (l – v).
ii) Explain fusion curve (l – s).
iii) Explain sublimation curve (s – v).
iv) Explain triple point.
Answer:

1. • The curve labelled l – v represents those points where the liquid and vapour phases are in equilibrium.
   • It is a graph of boiling point versus pressure.
   • The l – v curve of water correctly shows that at a pressure of 1 atmosphere, the boiling point of water is 100 °C and the boiling point gets lowered for a decreased pressure.
   • The l – v curve for CO₂ yields that CO₂ cannot exist as a liquid under normal atmospheric pressure conditions.
   • The curve l – s represents the points where the solid and liquid phases coexist in equilibrium.
   • It is a graph of the freezing point versus pressure.
   • At one standard atmosphere pressure, the freezing point of water is 0 °C which can be depicted using l – s curve of water.
   • At a pressure of one standard atmosphere water is in the liquid phase if the temperature is between 0 °C and 100 °C but is in the solid or vapour phase if the temperature is below 0 °C or above 100 °C.
   • Also, l – s curve for water slopes upward to the left i.e., fusion curve of water has a slightly negative slope.
   • This is true only of substances that expand upon freezing.
   • However, for most materials like CO2, the l – s curve slopes upwards to the right i.e., fusion curve has a positive slope. The melting point of C02 is -56 °C at higher pressure of 5.11 atm.
   • The curve labelled s – v is the sublimation point versus pressure curve.
   • Water sublimates at pressure less than 0.0060 atmosphere, while carbon dioxide, which in the solid state is called dry ice, sublimates even at atmospheric pressure at temperature as low as -78 °C.
   • The temperature and pressure at which the fusion curve, the vaporisation curve and the sublimation curve meet and all the three phases of a substance coexist is called the triple point of the substance.
   • The triple point of water is that point where water in solid, liquid and gaseous states coexist in equilibrium and this occurs only at a unique temperature and pressure.
   • The triple point of water is 273.16 K and 6.11 × 10^-3 Pa and that of CO₂ is -56.6 °C and 5.1 × 10 ^-5 Pa.

Question 120.
What is critical temperature of a gas?
Answer:
In order to liquefy a gas, it must be cooled to a certain temperature. This temperature is called critical temperature.

Question 121.
Compare gas and vapour.
Answer:

	Gas	Vapour
i.	A substance that is in gaseous phase above its critical temperature is called a gas.	A substance that is in gaseous phase below its critical temperature is call vapour.
ii.	Gas cannot be liquified only by pressure alone.	Vapour can be liquified simply by increasing pressure.
iii.	Gas exerts pressure.	Vapour exerts pressure

Solved Examples

Question 122.
Calculate the amount of heat energy to be supplied to convert 2 kg of ice at 0 °C completely into water at 0 °C if latent heat of fusion for ice is 80 cal/g.
Solution:
Given:
Mass (m) = 2 kg = 2 × 10³ g. latent heat of fusion for ice (L) = 80 cal/g.
To find: Heat energy (Q)
Formula: Q = mL
From formula,
Q = 2 x× 10³ × 80 = 160000 cal
= 160 kcal
Heat energy to be supplied is 160 kcal.

Question 123.
When 0.1 kg of ice at 0 °C is mixed with 0.32 kg of water at 35 °C in a container. The resulting temperature of the mixture is 7.8 °C. calculate the heat of fusion of ice (s_water = 4186 J kg^-1 K^-1).
Solution:
Given:
m_ice = 0.1 kg, m_water = 0.32 kg,
T_ice = 0 °C, T_water = 35 °C, T_F = 7.8 °C
s_water = 4186 J kg^-1 K^-1
To find: Heat of fusion (L_f)

Formula: i) Heat lost by water
Q₁ = m_water × s_water × (T_water – T_F)
ii) Heat required to melt ice
Q₂ = m_ice L_F
iii) Heat required to raise temperature of molten ice (water now) to find temperature
Q₃ = m_ices_water (T – T_ice)
Calculation: From formula (j),
Q₁ = 0.32 × 4186 × (35 – 7.8)
= 36434.944 J
From formula (ii),
Q₂ = 0.1 × L_f
From formula (iii),
Q₃ = 0.1 × 4186 × (7.8 – 0) = 3265.08J
According to principle of heat conservation,
heat lost = heat gained
Q₁ = Q₂ + Q₃
∴ 36434.944 = 0.1 L_f + 3265.08
∴ L_f = \(\frac{36434.944-3265.08}{0.1}\)
= 331698.64 J kg^-1
Rounding off to correct significant figure,
L_f = 3.31699 × 10⁵ J kg^-1
Heat of fusion of ice 3.3 1699 × 10 ⁵ J kg^-1.

Question 124.
If 80 g steam of temperature 97 °C is released on an ¡ce slab of temperature 0 °C, how much ice will melt? How much energy will be transferred to the ice when the steam will be transformed to water?
(Given: Latent heat of melting the ice = L_melt =80 cal/g ; Latent heat of vaporisation of water L_vap = 540 cal/g)
Solution:
Mass of steam (m_s) = 80 g,
Change in temperature (∆T)
= 97 – 0 = 97 °C
We know that: Latent heat of melting of ice = L_melt = 80 cal/g
Latent heat of vaporisation of water = L_vap = 540 cal/g
Specific heat of water c_w = 1 cal /g °C
To find: i) Energy transferred (Q)
ii) Mass of ice that melts (m_i)

Formula: i) Heat released during conversion of steam into water at 97 °C (Q₁)= m_s × L_vap
ii) Heat released during decrease of temperature of water from 97°C to 0°C (Q₂) = m_s × c_w × ∆T
iii) Heat gained by ice (Q)= m_i × L_melt

From formula (i),
Q₁ = 80 × 540 cal
From formula (ii),
Q₂ = 80 × 1 × (97 – 0) = 80 × 97 cal
According to principle of heat conservation,
Total heat gained by ice
Q = Q₁ + Q₂
= 80 × 540 + 80 × 97
= 80 × (540 + 97)
= 80 × 637
= 50960 cal
This energy would cause m; mass of ice to melt,
From formula (iii),
∴ m_i × L_melt = 50960
∴ m_i = \(\frac{50960}{80}=\frac{80 \times 637}{80}\) = 637 g
Energy transferred to ice is 50960 cal and it will melt 637 g of ice.

Question 125.
Name three modes of heat transfer.
Answer:
Three modes of heat transfer are conduction, convection and radiation.

Question 126.
Define conduction. State conditions for conduction of heat.
Answer:
Conduction is the process by which heat flows from the hot end to the cold end of a solid body without any net bodily movement of the particles of the body.
Conditions for conduction:

1. The two points should be at different temperatures.
2. There should be a medium between the two points.

Question 127.
Explain process of conduction in solid.
Answer:

1. Heat passes through solids by conduction only.
2. When one end of a rod is heated, the molecules near the hot end receive the thermal energy and start oscillating with larger amplitudes.
3. In doing so, they collide with the neighbouring molecules and transfer a part of their energy to these molecules.
4. The molecules which receive the energy vibrate with increased amplitudes and collide with the neighbouring molecules. Thus, energy of thermal motion is transferred by molecular collisions down the rod.
5. As the distance of a molecule from the hot end increases, its amplitude of oscillation decreases and hence there is continuous decrease in temperature.
6. This transfer of heat continues till two ends of the object are at the same temperature.
7. In metals, mainly free electrons conduct the heat energy.

Question 128.
Define good conductors and bad conductors.
Answer:
Good conductors:
The substances which conduct heat easily are called good conductors of heat.
All metals are good conductor.
eg: Steel, silver, Aluminium etc.

Insulators:
The substances which do nor conduct heal easily are called insulators or bad condiciors of heal.
eg.: Glass. wood, air, paper. etc.
[Note: In general, good conductors of heat are also good conductors of electricity, while bad conductors of hear are had conductors of electricity.]

Question 129.
Explain why metals are good conductors of heat and electricity.
Answer:

1. Metals like iron, copper. aluminium etc, contain free electrons in their atoms.
2. These free electrons assist the atoms in transfer of thermal energy as well as electrical energy.
3. Therefore, metal are good conductors of heat and electricity.

Question 130.
What is thermal conductivity?
Answer:

• Thermal conductivity of a solid is a measure of the ability of the solid ¡o conduct heat through it.
• Thus good conductors of heat have higher thermal conductivity than bad conductors.

Question 131.
Give reason: Hot water when poured in glass beaker, it cracks.
Answer:

1. When hot water is poured in a glass beaker the inner surface of the glass expends on heating.
2. Since glass is a bad conductor of heat, the heat from inside does not reach the outside surface so quickly.
3. Hence the outer surface does not expand thereby causing a crack in the glass.

Question 132.
Explain mechanism of thermal conduction and temperature gradient.
Answer:

1. When one end of a metal rod is heated, the heat flows by conduction from hot end to the cold end.
   
2. As a result, the temperature of every section of the rod starts increasing.
3. Under this condition, the rod is said to be in a variable temperature state.
4. After some time, the temperature at each section of the rod becomes steady i.e., does not change.
5. Temperature of each cross-section of the rod now becomes constant though not the same. This is called steady state condition.
6. Under steady state condition, the temperature at points within the rod decreases uniformly with distance from the hot end to the cold end.
7. The fall of temperature with distance between the ends of the rod in the direction of flow of heat, is called temperature gradient.
   ∴ Temperature gradient = \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{x}}\)
   Where, T₁ = temperature of hot end
   T₂ = temperature of cold end
   x = length of the rod

Question 133.
State SI unit and dimensions of temperature gradient.
Answer:
S.I unit: = °C /m or K/m
Dimensions: [L^-1M⁰T⁰K¹]

Question 134.
State S.I. unit and dimensions of coefficient of thermal conductivity.
Answer:
SI unit of coefficient of thermal conductivity is J s^-1 m^-1 °C^-1 or J s^-1 m^-1 K^-1 and its dimensions are [L¹M¹T³K^-1].

Question 135.
Explain how SI unit of coefficient of thermal conductivity be obtained as W/m °C or W/m K.
Answer:

1. Consider equation, \(\frac{Q}{t}=\frac{k A\left(T_{1}-T_{2}\right)}{x}\)
2. The quantity Q/t, denoted by P_cond, is the time rate of heat flow (i.e. heat flow per second) from the hotter face to the colder face, at right angles to the faces.
3. Its SI unit is watt (W).
4. SI unit of k can therefore be written as W m^-1°C^-1 or W m^-1 K^-1.
   [Note: Above equation, using calculus can be written as, \(\frac{d Q}{d t}\) = – kA \(\frac{d T}{d x}\), where \(\frac{d T}{d x}\) is the temperature gradient. The negative sign indicates that heat flow is in the direction of decreasing temperature. If A = 1 m² and \(\frac{d T}{d x}\) = 1, then \(\frac{d Q}{d t}\) = k.]

Question 136.
Define coefficient of thermal conductivity in terms of temperature gradient.
Answer:
Coefficient of thermal conductivity of a material is defined as the rate of flow of heat per unit area per unit temperature gradient when the heat flow is at right angles to the faces of a thin parallel-sided slab of material.

Question 137.
Define conduction rate.
Answer:
Conduction rate (P_cond) is the amount of energy transferred per unit time through a slab of area A and thickness x, the two sides of the slab being at temperatures T₁, and T₂ (T₁ > T₂),
and is given P_cond = \(\frac{Q}{t}\) = kA \(\frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{x}}\)

Question 138.
Explain the analogy between electrical resistance and thermal resistance.
Answer:

1. Electrical resistance is ratio of \(\frac{V}{I}\) where, V is electrical potential difference between two ends of conductor and I is current or rate flow of charge.
2. Consider expression for conduction rate,
   P_cond = kA \(\frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{x}}\)
   ⇒ \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{P}_{\mathrm{cond}}}=\frac{\mathrm{x}}{\mathrm{kA}}\) ……………… (1)
3. Comparing equation (1) with \(\frac{V}{I}\), (T₁ – T₂) is temperature difference between two ends and P_cond is rate of flow of heat.
4. Ratio \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{P}_{\text {cond }}}\) is called as thermal resistance (R_T) of material.
5. Using (1), thermal resistance R_T = \(\frac{\mathrm{x}}{\mathrm{kA}}\)
6. Thermal resistance depends on the material and dimensions (length / breadth) of object.

Question 139.
What is thermal resistivity? What does it depend upon?
Answer:
i. Thermal resistivity (ρ_T) is the reciprocal of thermal conductivity (k).
ii. It is characteristic of a material.

Question 140.
Complete the table.

Answer:

Question 141.
Define convection.
Answer:
The process by which heat is transmitted through a substance from one point to another due to actual bodily movement of the heated particles of the substance is called convection.

Question 142.
Describe the mechanism of heat transfer by convection in liquids and gases.
Answer:

1. Consider liquid being heated in a vessel from below.
2. The liquid at the bottom of the vessel is heated
   first and consequently its density decreases i.e., liquid molecules at the bottom are separated farther apart.
3. These hot molecules have high kinetic energy and rise upward to cold region while the molecules from cold region come down to take their place.
4. Thus, each molecule at the bottom gets heated and rises then cool and descends.
5. This action sets up the flow of liquid molecules called convection currents.
6. The convection currents transfer heat to the entire mass of liquid via actual physical movement of the liquid molecules.
7. Similar process takes place in case of a gas.

Question 143.
Give two applications of convection.
Answer:

1. Heating and cooling of rooms:
   • The mechanism of heating a room by a heater is entirely based on convection.
   • The air molecules in immediate contact with the heater are heated up.
   • These air molecules acquire sufficient energy and rise upward.
   • The cool air at the top being denser moves down to take their place. This cool air in turn gets heated and moves upward.
   • In this way, convection currents are set up in the room which transfer heat to different parts of the room.
   • The same principle but in opposite direction is used to cool a room by an air-conditioner.
2. Cooling of transformers:
   • Due to current flowing in the windings of the transformer, enormous heat is produced.
   • Therefore, transformer is always kept in a tank containing oil.
   • The oil in contact with transformer body heats up, creating convection currents.
   • The warm oil comes in contact with the cooler tank, gives heat to it and descends to the bottom. It again warms up to rise upward.
   • This process is repeated again and again. The heat of the transformer is thus carried away by convection to the cooler tank.
   • The cooler tank, in turn loses its heat by convection to the surrounding air.

Question 144.
Distinguish between free convection and forced convection.
Answer:

	Free convection	Forced convection
i.	When a hot body is in contact with air under ordinary conditions, like air around a firewood, the air removes heat from the body by aprocess called free or natural convection.	The convection process can be accelerated by employing a fan to create a rapid circulation of fresh air. This is called forced convection.
ii.	Land and sea breezes are formed as a result of free convection currents in air.	Heat convector, air conditioner, heat radiators in IC engine etc. operate using forced convection.

Question 145.
Define radiation.
Answer:
The transfer of heat energy from one place to another via emission of electromagnetic (EM) energy (in a straight line with the speed of light) without heating the intervening medium is called radiation.

Question 146.
Compare conduction, convection and radiation.
Answer:

Solved Examples

Question 147.
The temperature difference between two sides of an iron plate. 2 cm thick, is 10 °C. Heat is transmitted through the plate at the rate of 600 kcal per minute per square metre at steady state. Find the thermal conductivity of iron.
Solution:
Given: \(\frac{\mathrm{Q}}{\mathrm{At}}\) = 600 kcal/min m² = \(\frac{600}{60}\) kcal/s m²
= 10 kcal/s m²
x = 2cm = 2 × 10^-2 m
T₁ – T₂ = 10°C
To Find.- Thermal conductivity (k)
Formula: Q = \(\frac{\mathrm{kA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right) \mathrm{t}}{\mathrm{x}}\)
Calculation: From formula.
∴ k = \(\frac{\mathrm{Q}}{\mathrm{At} \mathrm{t}} \frac{\mathrm{x}}{\mathrm{T}_{1}-\mathrm{T}_{2}}=\frac{10 \times 2 \times 10^{-2}}{10}\)
= 0.02 kcal / m s
Thermal conductivity is 0.02 kcal / m s °C.

Question 148.
Calculate the rate of loss of heat through a glass window of area 1000 cm² and thickness of 4 mm. when temperature inside is 27 °C and outside is – 5 °C. Coefficient of thermal conductivity of glass is 0.022 cal /s cm °C.
Solution:
Given: A = 1000 cm² 1000 × 10 m²
k = 0.22 cal / s cm °C
= 0.22 × 10² cal/ m °C
x = 4mm = 0.4 × 10^-2 m
T₁ = 27°C, T₂ = -5°C

= 1.76 × 10³ cal/s = 1.76kcal / s
Rate of loss of heat is 1.76 kcal / s.

Question 149.
Heat is conducted through a copper plate at the rate of 460 cal/s-cm². Calculate the temperature gradient when the steady state is reached. (k_copper = 92 cal/m-s °C)
Solution:

Temperature gradient of the copper plate is 5°C/m.

Question 150.
Two parallel slabs of metals A and B of thickness 5 cm and 3 cm respectively are joined together. The outer face of the metal A is maintained at 100 °C and that of metal B is maintained at 40 °C. If the thermal conductivities of metal A and B are 0.045 kcal/m-s K and 0.015 kcal/m-s K respectively, find the temperature of the interface of two plates.
Solution:
For metal A:
T₁ = 100 °C, T₂ = θ
dx₁= 5 cm = 5 × 10^-2 m,
k₁ = 0.045 kcal/m-s K

∴ 9(100 – T) = 5(T – 40)
∴ 900 – 9T = 5T – 200
∴ 14T = 1100
∴ T = 78.57°C
Temperature of the interlace of two plates is 78.57 °C.

Question 151.
What is the rate of energy loss in watt per square metre through a glass window 5 mm thick if outside temperature in -20 °C and inside temperature is 25 °C? (k_glass = 1 W/m K)
Solution:
Given:
k_glass = 1W/m K, T₁ = 25 °C, T₂ = -20 °C
T₁ – T₂ = 25- (-20)°C = 45 °C
X = 5 mm = 5 × 10^-3 m
As one degree celsius equates to one kelvin, temperature difference of 450 C equals 45 K.
To find: Rate of energy loss per square metre \(\left(\frac{\mathrm{P}_{\text {cond }}}{\mathrm{A}}\right)\)
Formula: P_cond = \(\frac{Q}{t}=k A \frac{T_{1}-T_{2}}{x}\)
Calcula lion: From formula,
∴ The energy loss per square metre,
\(\frac{\mathrm{P}_{\text {oond }}}{\mathrm{A}}=\mathrm{k} \frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{x}}=\frac{1 \times 45}{5 \times 10^{-3}}\)
= 9 × 10³ W/m²
Rate of energy loss per square metre is 9 × 10³ W/m².

Question 152.
A metal sphere cools at the rate of 1.6 °C/min when its temperature is 70 °C. At what rate will it cool when its temperature is 40 °C? The temperature of surroundings is 30 °C.
Solution:
Given: T₁ = 70°C, T₂ = 40 °C, T₃ = 30 °C

= 0.4 (40 – 30)
= 0.4 °C/mm
Rate of cooling is 0.4 °C/mm.

Question 153.
A body cools at the rate of 0.5 °C/s when it is at 50 °C above the surrounding temperature. What is its rate of cooling when ¡t is at 30 °C above the surrounding temperature?
Solution:

Divide equation (2) by (1),
\(\frac{\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{2}}{0.5}=\frac{\mathrm{C}(30)}{\mathrm{C}(50)}\)
∴ \(\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)_{2}\) = 0.5 × \(\frac{30}{50}\) = 0.3 °C/s
The rate of cooling of the body at 30 °C above the surrounding temperature is 0.3 °C/s.

Apply Your Knowledge

Question 154.
A metre scale made up of aluminium (α_a = 24 × 10^-6 /°C) measures length of a steel rod (α_s = 12 × 10^-6 /°C) at room temperature as 50.00 cm. Now the temperature of the room is increased by 100 °C. What can be said about the measured length of the rod at new temperature?
Answer:
At new temperature T °C,
change in length of rod is,
∆L_s = L₀(α_s∆T) = 50.00(12 × 10^-6 × 100)
= 0.06 cm.
Hence, the actual length of rod at T °C,
L_s = 50.06 cm
Due to change in temperature, along with rod, the scale will also increase in length.
For aluminium, at T °C,
∆L_a = L₀(α_a∆t) = 100.00 × 24 × 10^-6 × 100
= 0.24 cm
∴ Length of the scale will be,
L_a = 100.24 cm
As, the expansion in scale is more than that in rod, the reading recorded by the scale at t°C will be less than 50 cm.

Question 155.
A mercury thermometer calibrated to measure temperature in Fahrenheit scale is kept in liquid phosphorus at 150 °F. The liquid phosphorus is then heated continuously until it reaches its boiling point measured by the thermometer to be 500 °F. Find the percentage fractional change in the density of mercury during the whole process. (γ_Hg = 10^-4/°F)
(Assume that no heat is lost to the surrounding during the process.)
Answer:
Given:
T₂ = 500°F, T₁ = 150°F
γ_Hg = 10^-4 /°F
As the liquid phosphorus is heated, the mercury in the thermometer also gets heated.
Due to thermal expansion,
V₂ = V₁(1 + γ∆t)
= V₁ [1 + 10^-4 × (500 – 150)]
= 1.035 V₁
Now, initial density of mercury is,
ρ₁ = \(\frac{\mathrm{m}}{\mathrm{V}_{1}}\)
After heating,
ρ₂ = \(\frac{\mathrm{m}}{\mathrm{V}_{2}}=\frac{\mathrm{m}}{1.035 \mathrm{~V}_{1}}=\frac{\rho_{1}}{1.035}\)
⇒ ρ₂ < ρ₁
∴ change in density of mercury is,
\(\frac{\rho_{1}-\rho_{2}}{\rho_{1}}=\frac{\rho_{1}(1-0.9662)}{\rho_{1}}\) = 0.0338
∴ Percentage fractional change = 3.38%

Question 155.
When ‘m’ g of ice is added to ‘M’g of water at 20 °C, state the conditions for m and M for which
i) temperature of the mixture remains 0° C.
ii) temperature of the mixture exceeds 0 °C. (Specific heat of water = s_w = 4.2 × 10³ J kg^-1 °C^-1, latent heat of fusion = L = 3.36 × 10⁵ J kg^-1 )
Answer:
The heat lost by water in going from 20 °C to 0°C,
Q₁ = Ms_w ∆T = \(\frac{\mathrm{M}}{1000}\) × 4.2 × 10³ × (20) = 84M J
Now, heat required to convert m g of ice into water at 0 °C,
Q₂ = mL = \(\frac{\mathrm{M}}{1000}\) × 3.36 × 10⁵ = 336m J

1. For temperature of mixture to be 0 °C,
   Q₂ > Q₁
   ⇒ 336m > 84M
   ⇒ m > \(\frac{\mathrm{M}}{4}\)
2. For temperature of mixture to exceed 0 °C,
   Q₂ < Q₁ ⇒ m < \(\frac{\mathrm{M}}{4}\)

Quick Review

Multiple Choice Questions

Question 1.
Heat is transferred between two (or more) systems or a system and its surrounding by virtue of
(A) temperature difference.
(B) material difference.
(C) amount of heat difference.
(D) mass difference.
Answer:
(A) temperature difference.

Question 2.
On celsius scale, the two fixed points are marked as
(A) 0°C and 232°C
(B) 32°C and 100°C
(C) 0°C and 100°C
(D) 100°C and 180°C
Answer:
(C) 0°C and 100°C

Question 3.
If the temperature in a room is 30 °C, temperature in degree fahrenheit is
(A) 22 °F
(B) 62 °F
(C) 86 °F
(D) 96 °F
Answer:
(C) 86 °F

Question 4.
If the temperature on Fahrenheit scale is 140 °F, then the same temperature on kelvin scale will be
(A) 60.15 K
(B) 213.15 K
(C) 333.15 K
(D) 413.15 K
Answer:
(C) 333.15 K

Question 5.
In the gas equation, PV = RT, V stands for volume of
(A) any amount of gas .
(B) one gram mole of gas.
(C) one gram of a gas.
(D) one litre of a gas.
Answer:
(B) one gram mole of gas.

Question 6.
1 litre of an ideal gas at 27 °C is heated at constant pressure so as to attain temperature 297 °C. The final volume is approximately
(A) 1.2 litre
(B) 1.9 litre
(C) 19 litre
(D) 2.4 litre
Answer:
(B) 1.9 litre

Question 7.
How much should the pressure be increased in order to decrease the volume of a gas by 10% at a constant temperature?
(A) 7%
(B) 8%
(C) 10%
(D) 11.11%
Answer:
(D) 11.11%

Question 8.
Two rods of same material are equal in length, but one has cross-sectional area double the other. If they are heated through the same temperature then
(A) thick rod expands more.
(B) thin rod expands more.
(C) both rods will expand equally.
(D) none of these.
Answer:
(C) both rods will expand equally.

Question 9.
Two iron bars of same length with unequal radii are heated for the same rise in temperature. The linear expansion will be
(A) more in thin bar.
(B) more in thick bar.
(C) same for both.
(D) less in thick bar.
Answer:
(A) more in thin bar.

Question 10.
Which of the following has minimum coefficient of linear expansion?
(A) Gold
(B) Copper
(C) Platinum
(D) Invar steel
Answer:
(D) Invar steel

Question 11.
The coefficient of cubical expansion of a solid is the increase in volume per unit original volume at 0 °C per degree rise in _________.
(A) pressure
(B) volume
(C) temperature
(D) area
Answer:
(C) temperature

Question 12.
A disc has an area of 0.32 m2 at 20 °C, what will be its area at 100 °C? (α = 2 × 10^-6 / °C)
(A) 0.12 m²
(B) 0.32 m²
(C) 0.51 m²
(D) 0.71 m²
Answer:
(B) 0.32 m²

Question 13.
The coefficient of linear expansion of iron is 1.1 × 10^-5 per K. An iron is 10 m long at 27 °C. Length of the rod will be decreased by 1.1 mm when the temperature of the rod changes to
(A) 0°C
(B) 10 °C
(C) 17 °C
(D) 20 °C
Answer:
(C) 17 °C

Question 14.
The length of an aluminium rod is 120 cm at 20 °C. What is its length at 80 °C, if coefficient of linear expansion of aluminium is 2.5 × 10^-5/°C?
(A) 130.18 cm
(B) 120.18 cm
(C) 110.18 cm
(D) 100.18 cm
Answer:
(B) 120.18 cm

Question 15.
A metal rod having a coefficient of linear expansion of 2 × 10^-5 /°C has a length of 100 cm at 20 °C. The temperature at which it is shortened by 1 mm is
(A) -40 °C
(B) -30 °C
(C) -20 °C
(D) -10 °C
Answer:
(B) -30 °C

Question 16.
Iron sheet 50 cm × 20 cm is heated through 100 °C. If a = 12 × 10^-6 / °C, the change in area is
(A) 2.4 cm²
(B) 3.4 cm²
(C) 4.2 cm²
(D) 5.3 cm²
Answer:
(A) 2.4 cm²

Question 17.
A liquid with coefficient of volume expansion γ is filled in a container of a material having the coefficient of linear expansion α. If the liquid over flows on heating then
(A) γ = 3α
(B) γ < 3α (C) γ > 3α
(D) γ = 3α²
Answer:
(B) γ < 3α

Question 18.
The volume of a metal block changes by 0.18% when it is heated through 20 °C. Its coefficient at cubical expansion will be
(A) 9 × 10^-5 / °C
(B) 3 × 10^-5 / °C
(C) 18 × 10^-5 / °C
(D) 36 × 10^-5 / °C
Answer:
(A) 9 × 10^-5 / °C

Question 19.
The volume of liquid is 830 m³ at 30 °C and 850 m³ at 90 °C. The coefficient of volume expansion of liquid is
(A) 2 × 10^-4 per °C
(B) 8 × 10^-4 per °C
(C) 4 × 10^-4 per °C
(D) 2.5 × 10^-4 per °C
Answer:
(C) 4 × 10^-4 per °C

Question 20.
The superficial expansivity is 1 / x times the cubic expansivity. The value of x is
(A) 2/3
(B) 3/2
(C) 2
(D) 3
Answer:
(B) 3/2

Question 21.
The unit of molar specific heat is
(A) JK^-1 mole^-1
(B) JK mole^-1
(C) J^-1 K^-1mole^-1
(D) JK^-1 mole
Answer:
(A) JK^-1 mole^-1

Question 22.
The S.I. unit of latent heat is
(A) J^-1 kg
(B) J kg^-1
(C) J k^-1 °C
(D) J^-1 kg °C
Answer:
(B) J kg^-1

Question 23.
The slowest mode of transfer of heat is
(A) conduction
(B) convection
(C) radiation
(D) specific heat
Answer:
(A) conduction

Question 24.
The quantity of heat which crosses unit area of a metal plate during conduction depends on
(A) the density of the metal.
(B) the temperature gradient perpendicular to the area.
(C) the temperature to which the metal is heated.
(D) the area of the metal plate.
Answer:
(B) the temperature gradient perpendicular to the area.

Question 25.
Which of the following is not the unit of thermal conductivity?
(A) J/m s °C
(B) K cal/m s K
(C) Watt/m °C
(D) J/m² s °C
Answer:
(D) J/m² s °C

Question 26.
The most desirable combination for the material of a cooking pot is
(A) high specific heat and high conductivity.
(B) low specific heat and high conductivity.
(C) high specific heat and low conductivity.
(D) low specific heat and low conductivity.
Answer:
(B) low specific heat and high conductivity.

Question 27.
While measuring thermal conductivity of a liquid, we keep the upper part hot and lower part cold, so that
(A) radiation may start.
(B) radiation may stop.
(C) convection may start.
(D) convection may be stopped.
Answer:
(D) convection may be stopped.

Question 28.
Convention currents in air in day time is from
(A) land to sea
(B) sea to land
(C) sea to sky
(D) land to land
Answer:
(B) sea to land

Question 29.
One end of a metal rod one metre long is kept in ice and the other end is at 100 °C. What is temperature gradient throughout the rod?
(A) 10 °C/m
(B) 100 °C/m
(C) 50 °C/m
(D) 1 °C/m
Answer:
(B) 100 °C/m

Question 30.
__________ amount of heat is required to raise the temperature of 100 g of kerosene from 10 °C to 30 °C (Given: specific heat of kerosene is 0.51 kcal/kg °C)
(A) 0.102 kcal
(B) 1.02 kcal
(C) 10.2 kcal
(D) 102 kcal
Answer:
(B) 1.02 kcal

Question 31.
One end of copper rod is in contact with water at 100 °C and the other end in contact with ice at 0 °C. The length of the rod is 100 cm. At a point which is at a distance of 35 cm from the cold end, temperature is (assuming steady state heat flow)
(A) 35 °C
(B) 65 °C
(C) 56 °C
(D) 53 °C
Answer:
(A) 35 °C

Question 32.
In Newton’s law of cooling, the rate of fall of temperature
(A) is constant
(B) increases
(C) decreases
(D) doubles
Answer:
(C) decreases

Question 33.
The metal sphere cools at 1 °C/min, when its temperature is 50 °C. If the temperature of environment is 30 °C, its rate of cooling at 35 °C is
(A) 0.25 °C/min
(B) 0.5 °C/min
(C) 0.75 °C/min
(D) 0.4 °C/min
Answer:
(A) 0.25 °C/min

Competitive Corner

Question 1.
A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminum rod is:
(α_Cu = 1.7 × 10^-5 K^-1 and α_Al = 2.2 × 10^-5 K^-1)
(A) 88 cm
(B) 68 cm
(C) 6.8 cm
(D) 1 13.9 cm
Answer:
(B) 68 cm
Hint:
L_Cu α_Cu ∆T = L_Al α_Al ∆T
∴ 88 × (1.7 × 10^-5) = L_Al(2.2 × 10^-5)
∴ L_Al = \(\frac{88 \times 1.7}{2.2}\)
∴ L_Al = 68 cm

Question 2.
The unit of thermal conductivity is:
(A) W m K^-1
(B) W m^-1 K^-1
(C) JmK ^-1
(D) Jm^-1K^-1
Answer:
(B) W m^-1 K^-1
Hint:
Q = kA \(\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\) t
Q = quantity of heat conducted.
A = area of cross section
t = time for which heat is passed
\(\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)\) = temperature gradient
∴ K = \(\frac{Q}{A t\left(\frac{\Delta T}{\Delta x}\right)}\)
∴ Unit of k = W m^-1 K^-1

Question 3.
A deep rectangular pond of surface area A, containing water (density = p, specific heat capacity = s), is located in a region where the outside air temperature is at a steady value of -26°C. The thickness of the frozen ice layer in this pond, at a certain instant is x. Taking the thermal conductivity of ice as K, and its specific latent heat of fusion as L, the rate of increase of the thickness of ice layer, at this instant, would be given by
(A) 26K/ρx(L + 4s)
(B) 26K/ρx(L – 4s)
(C) 26K/(ρx² L)
(D) 26K/(ρxL)
Answer:
(D) 26K/(ρxL)

Question 4.
An object kept in a large room having air temperature of 25°C takes 12 minutes to cool from 80°C to 70°C. The time taken to cool for the same object from 70°C to 60°C would be nearly.
(A) 15 min
(B) 10mm
(C) 12mm
(D) 20mm
Answer:
(A) 15 min
Hint:
By Newton’s law of cooling,

Question 5.
A thermally insulated vessel contains 150 g of water at 0 °C. Then the air from the vessel ispumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0 °C itself. The mass of evaporated water will be closest to :
(Latent heat of vaporization of water = 2.10 × 10⁶ J kg^-1 and Latent heat of Fusion of water = 3.36 × 10⁵ J kg^-1)
(A) 150g
(B) 20g
(C) 130g
(D) 35g
Answer:
(B) 20g
Hint:
m= 150 g = 0.15 kg
The heat required to evaporate ‘m’ grams of water,
∆Q_required = mL_v ………. ( 1)
(015 – m) is the amount of mass that converts into ice
∴ ∆Q_released = (0.15 – m) L_f …………. (2)
Now, amount of heat required = amount of heat released
∴ From (1) and (2),
mL_v = (0.15 – m)L_f
∴ m(L_f + L_v) = 0.15 L_f
∴ m = \(\frac{0.15 \mathrm{~L}_{\mathrm{f}}}{\mathrm{L}_{\mathrm{f}}+\mathrm{L}_{\mathrm{v}}}\)
∴ m = \(\frac{0.15 \times 3.36 \times 10^{5}}{2.10 \times 10^{6}+3.36 \times 10^{5}}\)
∴ m=0.0206kg ≈ 20g

Question 6.
A copper ball of mass 100 g is at a temperature T. It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, thetemperature of the system is found to be 75 °C. T is given by: (Given: room temperature = 30°C, specific heat of copper 0.1 cal/g °C)
(A) 1250°C
(B) 825°C
(C) 800°C
(D) 885°C
Answer:
(D) 885°C
Hint:
Heat lost by copper ball = Heat gained by calorimeter and water
∴m_bs_c∆T₁ = m_cc_c∆T₂ + m_ws_w∆T₂
∴ (100)(0.1)(T – 75) = (100)(0.1)(75 – 30) + (170)(1)(75 – 30)
10(T – 75) = 450 + 7650 = 8100
T – 75 = 810
T = 885 °C

Question 7.
Coefficient of linear expansion of brass and steel rods are α₁ and α₂. Lengths of brass and steel rods are l₁ and l₂ respectively. If (l₂ – l₁) is maintained same at all temperatures, which one of the following relations holds good?
(A) α₁²l₂ = α₂²l₁
(B) α₁l₁ = α₂l₂
(C) α₁l₂ = α₂l₁
(D) α₁l₂² = α₂l₁²
Answer:
(B) α₁l₁ = α₂l₂
Hint:
∆L = L(1 + α∆t)
i.e. ∆l₂ = l₂(1 + α₂∆t)
and ∆l₁ =1₁(1 + α₁∆t)
It is given,
l₂– l₁ = ∆tl₂ – ∆ll₁
∴ l₂ – l₁ = l₂ (1 + α₂∆t) – l₁ (1+ α₁∆t)
= l₂ + l₂α₂∆t – l₁ – l₁α₁∆t
∴ – l₂ + l₂ + l₂α₂∆t = -l₁ + l₁ + l₁α₁∆t
∴ l₂α₂∆t = l₁α₁∆t
i.e., l₂α₂ = l₁α₁

Question 8.
A pendulum clock loses 12 s a day if the temperature is 40 °C and gains 4 s a day if the temperature is 20 °C. The temperature at which the clock will show correct time, and the coefficient of linear expansion (a) of the metal of the pendulum shaft are respectively:
(A) 60 πC, a = 1.85 × 10^-4/πC
(B) 30 πC, a = 1.85 × 10^-3/πC
(C) 55 πC, a = 1.85 × 10^-2/πC
(D) 25 πC, a = 1.85 × 10^-5/πC
Answer:
(D) 25 πC, a = 1.85 × 10^-5/πC
Hint:
Period of pendulum, T = 2π\(\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\)
∴ T ∝ \(\sqrt{\mathrm{L}}\)
But, L = L₀ (1 + α∆t)
∴ T ∝ \(\sqrt{\mathrm{L}_{0}(1+\alpha \Delta \mathrm{t})}\)
As L₀ is constant,
⇒ T ∝ (1 + α ∆t)
Calculating fractional change in time period of pendulum, \(\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2}(\alpha \Delta \mathrm{t})\)
For the given pendulum,
T = 24 × 60 × 60 = 864000 s
When t1 = 40 °C, ∆T = 12 s,

Question 9.
A piece of ice falls from a height h so that it melts completely. Only one — quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is [Latent heat of ice is 3.4 × 10⁵ J/kg and g = 10 N/kg]
(A) 136 km
(B) 68km
(C) 34km
(D) 544km
Answer:
(A) 136 km
Hint:
When the piece of ice falls from the height h, it possesses potential energy, mgh.
This P.E. is converted to heat energy.
∴ Q = mgh
But only \(\frac{1^{\text {th }}}{4}\) of it is absorbed by ice which is used to change the state.
∴ \(\frac{\mathrm{mgh}}{4}\) = mL
∴ \(\frac{10 \times \mathrm{h}}{4}\) = 3.4 × 10⁵
∴ h = 13.6 × 10⁴ m = 136km

Question 10.
A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be
(A) T
(B) \(\frac{7}{4}\) T
(C) \(\frac{3}{2}\) T
(D) \(\frac{4}{3}\) T
Answer:
(C) \(\frac{3}{2}\) T
Hint:
By Newton’s law of cooling,

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 6 Mechanical Properties of Solids Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 6 Mechanical Properties of Solids

Question 1.
Explain the equilibrium state of solids at a given temperature.
Answer:

1. Solids are made up of atoms or a group of atoms placed in a definite geometric arrangement.
2. This arrangement is decided by nature so that the resultant force acting on each constituent due to others is zero. This is the equilibrium state of a solid at room temperature.
3. This equilibrium arrangement does not change with time but it can only change when an external stimulus, like compressive force is applied to a solid from all sides.
4. The constituents vibrate about their equilibrium positions even at very low temperature but cannot leave their fixed positions.
5. As a result, the solids possess a definite shape and size.

Question 2.
Explain the effects of applied force on a rigid body.
Answer:

1. When an external force is applied to a solid, the constituents are slightly displaced and restoring forces are developed in it.
2. These restoring forces try to bring the constituents back to their equilibrium positions so that the solid can regain its shape.
3. When the deforming forces are removed, the inter-atomic forces tend to restore the original positions of the molecules and thus the body regains its original shape and size.

Question 3.
Explain the concept of deforming force with the help of examples.
Answer:

1. When a force (within specific limit) is applied to a solid (which is not free to move), the size or shape or both change due to changes in the relative positions of molecules. Such a force is called deforming force.
2. The larger the deforming force on a body, the larger is its deformation.
3. Deformation could be in the form of change in length of a wire, change in volume of an object or change in shape of a body.
4. Examples:
   • When a deforming force such as stretching, is applied to a rubber band, it gets deformed (elongated) but when the force is removed, it regains its original length.
   • When a similar force is applied to a dough or a clay, it also gets deformed but it does not regain its original shape and size after removal of the deforming force.

Question 4.
Define plasticity.
Answer:
If a body does not regain its original shape and size and retains its altered shape or size upon removal of the deforming force, it is called a plastic body and the property is called plasticity.

Question 5.
When is a body said to be perfectly elastic? Give an example for perfectly elastic body and perfectly plastic body.
Answer:

1. If a body regains its original shape and size completely and instantaneously upon removal of the deforming force, then it is said to be perfectly elastic.

• • There is no solid which is perfectly elastic or perfectly plastic,
  • The best example of a near perfectly elastic body is quartz fibre and that of a plastic body is putty.

Question 6.
State SI unit and dimensions of strain.
Answer:
Strain is the ratio of two similar quantities. Hence strain is a dimensionless physical quantity and it has no units.

Question 7.
State and explain longitudinal stress (Tensile stress and Compressive stress).
Answer:
i. Stress produced by a deforming force acting along the length of a body or a rod is called longitudinal stress.

ii. Tensile stress: Consider a force F is applied along a length of a wire, or perpendicular to its cross-section A of a wire (or along its length).

[Note: Elongation of wire ∆l is exaggerated for explanation.]
This produces an elongation in the wire and the length of the wire increases, then

iii. Compressive stress: When a rod is pushed from two ends with equal and opposite forces.

[Note: Compression of wire ∆l is exaggerated for explanation.]
This restoring force per unit area is called compressive stress.
Compressive stress = \(\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{A}}\)

Question 8.
State and explain longitudinal strain (Tensile strain or Linear strain).
Answer:
The strain produced by a tensile deforming force is called longitudinal strain (tensile strain or linear strain,).

where, L = Original length,
∆l = Change in length.

Question 9.
State and explain volume stress / hydraulic stress.
Answer:
When a deforming force acting on a body produces change in its volume, the stress is called volume stress.
Volume stress/hydraulic stress:

[Note: Change in size is exaggerated for explanation.]

1. Let \(\vec{F}\) be a force acting perpendicular to the entire surface of the body.
2. It acts normally and uniformly all over the surface area A of the body.
3. Such a stress which produces change in size but no change in shape is called volume stress.
   Volume stress = \(\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{A}}\)
4. Volume stress produces change in size without change in shape of body, hence it is also called as hydraulic or hydrostatic volume stress.

Question 10.
Explain volume strain.
Answer:

1. A deforming force acting perpendicular to the entire surface of a body produces a volume strain.
2. 
   where ∆V = change in volume,
   V = original volume.

Question 11.
Define and explain shearing stress.
Answer:
The restoring force per unit area developed due to the applied tangential force is called shearing stress or tangential stress.
Shearing stress:

1. When a deforming force acting on a body produces change in the shape of a body shearing stress is produced.
2. Consider ABCD as a front face of a cube, a tangential force is applied to the cube so that the bottom of the cube is fixed and only the top surface is slightly displaced.
   

Question 12.
Explain shearing strain.
Answer:

1. The relative displacement of the bottom face and the top face of the cube is called shearing strain.
2. Shearing strain = \(\frac{\Delta l}{\mathrm{~L}}\) = tan θ
   where, ∆l = displaced length,
   L = Original length.
3. When ∆l is very small,
   tan θ ≈ θ and shearing strain = θ

Question 13.
State and explain Hooke’s law.
Answer:
Statement: Within elastic limit, stress is directly proportional to strain.
Explanation:

1. According to Hooke’s law,
   Stress-Strain
   
   This constant of proportionality is called modulus of elasticity.
2. Modulus of elasticity of a material is the slope of stress-strain curve in elastic deformation region and depends on the nature of the material.
3. The graph of strain (on X-axis) and stress (on Y-axis) within elastic limit is shown in the figure.
   

Question 14.
Define elastic limit.
Answer:
The maximum value of stress upto which stress is directly proportional to strain is called the elastic limit.

Question 15.
Define modulus of elasticity.
Answer:
The modulus of elasticity of a material is the ratio of stress to the corresponding strain.

Question 16.
State different types of modulus of elasticity.
Answer:

1. Young’s modulus (Y): It is the modulus of elasticity related to change in length of an object like a metal wire, rod, beam, etc., due to the applied deforming force.
2. Bulk modulus (K): It is the modulus of elasticity related to change in volume of an object due to applied deforming force.
3. Shear modulus or Modulus of rigidity (η): The modulus of elasticity related to change in shape of an object is called modulus of rigidity.

Question 17.
Explain the usefulness of Young’s modulus.
Answer:

1. Young’s modulus indicates the resistance of an elastic solid to elongation or compression.
2. Young’s modulus of a material is useful for characterization of an object subjected to compression or tension.

Question 18.
Within elastic limit, prove that Young’s modulus of material of wire is the stress required to double the length of wire.
Answer:

(i) Let, L = Initial length of wire
2 L = Final length of wire
∴ Increase in length = ∆l = 2L – L = L

(ii) Longitudinal strain of wire = \(\frac{\Delta l}{\mathrm{~L}}=\frac{\mathrm{L}}{\mathrm{L}}=1\)

(iii)

∴ Y = Longitudinal stress

(iv) Hence, Young’s modulus of material of wire is the stress required to double the length of wire.

Question 19.
What is bulk modulus? Derive an expression for bulk modulus.
Answer:
Definition:
Bulk modulus is defined as the ratio of volume stress to volume strain.
It is denoted by ‘K’.
Unit: N/m² or Pa in SI system
Dimensions: [L^-1M¹T^-2]
Expression for bulk modulus:

(i) If a sphere made from rubber is completely immersed in a liquid, it will be uniformly compressed from all sides.
Let, F = Compressive force,
dP = Change in pressure,
dV = Change in volume,
V = Original volume.

(ii) Volume stress = \(\frac{|\overrightarrow{\mathrm{F}}|}{\mathrm{A}}\) = dP

(iii) The negative sign indicates that there is a decrease in volume.
The magnitude of the volume strain is \(\frac{\mathrm{dV}}{\mathrm{V}}\)

Question 20.
Explain the usefulness of bulk modulus.
Answer:

1. Bulk modulus indicates the resistance of gases, liquids or solids to change their volume.
2. Materials with small bulk modulus and large compressibility are easier to compress.

Question 21.
State few applications of bulk modulus.
Answer:

1. When a balloon is filled with air at high pressure, its walls experience a force from within. It tries to expand the balloon and change its size without changing shape. When the volume stress exceeds the limit of bulk elasticity, the balloon explodes.
2. A gas cylinder explodes when the pressure inside it exceeds the limit of bulk elasticity of its material.
3. A submarine when submerged under water is under volume stress. The depth it can reach within water depends upon its limit of bulk elasticity.

Question 22.
Define compressibility. State its unit and dimensions.
Answer:

1. The reciprocal of bulk modulus of elasticity is called compressibility of the material.
   
2. Compressibility is the fractional decrease in volume, (-dV/V) per unit increase in pressure.
   Compressibility = \(\frac{-\mathrm{dV}}{\mathrm{V} \mathrm{dP}}\)
3. Unit: m²/N or Pa^-1 in SI system.
4. Dimensions: [L¹M^-1T²]

Question 23.
What is modulus of rigidity? Derive an expression for it.
Answer:
Definition: It is defined (is the ratio of shear stress to shear strain within elastic limit.
It is denoted by ‘η’.
Unit: N/m² or Pa in SI system.
Dimension: [L^-1M¹T^-2]

Expression for modulus of rigidity:

Consider a solid cube as shown in the figure.

Let, F = Tangential force
A = Cross sectional area
∆l = Relative displaced length
l = Original length
θ = Shear strain

[Note: Displacement of upper surface is exaggerated for explanation.]
The forces applied on the block is subjected to a shear stress,
Shearing stress = F/A
The comer angle which changes by a small amount θ (expressed in radian) is given by,
Shearing strain = \(\frac{\Delta l}{l}\) ≈ θ
From definition,

Question 24.
What does modulus of rigidity indicate?
Answer:
Modulus of rigidity indicates resistance offered by solid to change in its shape.

Question 25.
Explain the change of diameter of a wire when it is stretched and compressed.
Answer:
i. When a wire is fixed at one end and a force is applied at its free end so that the wire gets stretched, length of the wire increases and at the same time, its diameter decreases, i.e., the wire becomes longer and thinner as shown in figure.

[Note: Linear expansion ∆l is exaggerated far explanation.]

ii. If equal and opposite forces are applied to an object along its length inwards, the object gets compressed. There is a decrease in dimensions along its length and at the same time there is an increase in its dimensions perpendicular to its length. When length of the wire decreases, its diameter increases.

[Note: Compression of wire ∆l is exaggerated for explanation.]

Question 26.
Derive expression for Poisson’s ratio.
Answer:
i. Let,
L = original length
l = increase/decrease in length
D = original diameter
d = change in diameter

ii. The ratio of change in dimensions to original dimensions in the direction of the applied force is called linear strain.
Linear strain = \(\frac{l}{\mathrm{~L}}\) …. (1)

iii. The ratio of change in dimensions to original dimensions in a direction perpendicular to the applied force is called lateral strain.
Lateral strain = \(\frac{\mathrm{d}}{\mathrm{D}}\) ….(2)

iv. Poisson’s ration,

Question 27.
A wire of length 20 m and area of cross section 1.25 × 10^-4 m² is subjected to a load of 2.5 kg. (1 kg wt = 9.8 N). The elongation produced in wire is 1 × 10^-4 m. Calculate Young’s modulus of the material.
Solution:
L = 20 m, A = 1.25 × 10^-4m²,
F = mg = 2.5 × 9.8 N, l = 10^-4 m
Formula Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
To find: Young’s modulus (Y)
Calculation: From formula,
Y = \(\frac{2.5 \times 9.8 \times 20}{1.25 \times 10^{-4} \times 10^{-4}}\) = 3.92 × 10¹⁰Nm^-2
Answer: The Young’s modulus of the material is 3.92 × 10¹⁰ Nm^-2.

Question 28.
A wire of diameter 0.5 mm and length 2 m is stretched by applying a force of 2 kg wt. Calculate the increase in length of the wire, (g = 9.8 m/s², Y = 9 × 10¹⁰ N/m²)
Solution:
Given:
L = 2 m, F = 2 kg wt, d = 0.5 mm = 5 × 10^-4 m,
∴ r = \(\frac{\mathrm{d}}{2}\) = 2.5 × 10^-4 m.
Y = 9 × 10¹⁰ N/m², g = 9.8 m/s²
To find: Increase in length (l)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}=\frac{\mathrm{MgL}}{\pi \mathrm{r}^{2} l}\)
Calculation:
From formula,

Answer:
The increase in length of the wire is 2.218 × 10^-3 m.

Question 29.
A brass wire of length 4.5 m with cross-section area of 3 × 10^-5 m² and a copper wire of length 5.0 m with cross section area 4 × 10^-5 m² are stretched by the same load. The same elongation is produced in both the wires. Find the ratio of Young’s modulus of brass and copper.
Solution:
L_B = 4.5 m, A_B = 3 × 10^-5 m²
L_C = 5 m, A_C = 4 × 10^-5 m² l_B = _C, F_B = F_C
To find: Ratio of Young’s modulus \(\left(\frac{Y_{B}}{Y_{C}}\right)\)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
Calculation: For brass,

= 1.2
Answer:
The ratio of Young’s modulus of brass and copper is 1.2 : 1.

Question 30.
The length of wire increases by 9 mm when weight of 2.5 kg is hung from the free end of wire. If all conditions are kept the same and the radius of wire is made thrice the original radius, find the increase in length.
Solution:
Given; l₁ = 9mm = 9 × 10^-3m,
M = 2.5 kg, r₂ = 3r₁,
Y₁ = Y₂ = Y (material is same)
To find: Increase in length (l₂)
Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
Calculation: From formula,

Answer:
The longitudinal strain produced in 1st wire is

Question 31.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in the figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.(Y_s = 2.0 × 10¹¹ Nm^-2, Y_B = 0.91 × 10¹¹ Nm^-2) (NCERT)

Solution:
Given: D = 0.25 cm = 0.25 × 10^-2 m,
L_S = 1.5 m, L_B = 1 m,
Y_S = 2.0 × 10¹¹ Nm^-2,
Y_B = 0.91 × 10¹¹ Nm^-2
To find: Elongations of brass wire (l_B)
Elongations of steel wire (l_S)

Formula: Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)

Calculation: Since, A = \(\frac{\pi \mathrm{D}^{2}}{4}\)

Answer:
The elongation of the steel wire is 1.5 × 10^-4 m and that of brass wires is 1.32 × 10^-4 m.

Question 32.
One end of steel wire is fixed to a ceiling and load of 2.5 kg is attached to the free end of the wire. Another identical wire is attached to the bottom of load and another load of 2.0 kg, is attached to the lower end of this wire as shown in the figure. Compute the longitudinal strain produced in both the wires, if the cross-sectional area of wires is 10^-4m², (Y_steel = 20 × 10¹⁰N/m²)

Solution:
Given: M₁ = 2.5 kg, M₂ = 2kg, A = 10^-4m², Y_steel = 20 × 10¹⁰ N/m², L₁ = L₂ = L
To find: Longitudinal strain of 1^st wire (Strain₁).
Longitudinal strain of 2^nd wire (Strain₂)

Answer:
The longitudinal strain produced in 1^st wire is 1.225 × 10^-6 and in 2^nd wire is 2.205 × 10^-6

Question 33.
A steel wire having cross-sectional area 2 mm² is stretched by 10 N. Find the lateral strain produced in the wire. (Given: Y for steel = 2 × 10¹¹ N/m², Poisson’s ratio σ = 0.29)
Solution:
Given: A = 2 mm² = 2 × 10^-6 m²,
F = 10 N, Y_stee; = 2 × 10¹¹ N/m², σ = 0.29

To find: Lateral strain
Formulae:

Calculation: From formula (i),
longitudinal stress = \(\frac{10}{2 \times 10^{-6}}\)
= 5 × 10⁶ N/m²
From formula (ii),

From formula (iii),
lateral strain = σ × longitudinal strain
= 0.29 × 2.5 × 10^-5
∴ lateral strain = 7.25 × 10^-6
Answer:
Lateral strain produced in the wire is 7.25 × 10^-6.

Question 34.
A brass wire of radius 1 mm is loaded by a mass of 31.42 kg. What would be the decrease in its radius? (Y = 9 × 10¹⁰ N/m², Poisson’s ratio σ = 0.36)
Solution:
Given: R = 1 mm = 1 × 10^-3 m, M = 31.42 kg, Y = 9 × 10¹⁰ N/m², σ = 0.36
To find: Decrease in radius (r)

Formulae:

i. Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)
ii. σ = \(\frac{\mathrm{Lr}}{l \mathrm{R}}\)

Calculation: From formula (i),

Answer:
The decrease in radius of the brass wire would be 3.92 × 10^-7 m

Question 35.
A metal cube of side 1 m is subjected to a force. The force acts normally on the whole surface of cube and its volume changes by 1.5 × 10^-5 m³. The bulk modulus of metal is 6.6 × 10¹⁰ N/m². Calculate the change in pressure.
Solution:
Given: L = 1 m, dV = 1.5 × 10^-5 m³,
K = 6.6 × 10¹⁰ N/m².
To find: Change in pressure (dP)
Formula: K = V\(\frac{\mathrm{dP}}{\mathrm{dV}}\)
Calculation: From formula,

Answer:
The change in pressure is 9.9 × 10⁵ N/m².

Question 36.
Determine the volume contraction of solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ P.a. (Bulk modulus of copper = 140 × 10⁹ pa)
Solution:
Given: L = 10 cm =0.1 m, ∆P = 7 × 10⁶ Pa, K = 140 × 10⁹ pa
To find: Volume contraction (dV)
Formula: K = V × \(\frac{\mathrm{dP}}{\mathrm{dV}}\)
Calculation: From formula,
dV = \(\mathrm{L}^{3} \times \frac{\mathrm{dP}}{\mathrm{K}}\) = (0.1)³ × \(\frac{7 \times 10^{6}}{140 \times 10^{9}}\)
∴ dV = 5 × 10^-8 m³
Answer:
The volume contraction of solid copper cube is 5 × 10^-8 m³

Question 37.
Calculate the modulus of rigidity of a metal, if a metal cube of side 40 cm is subjected to a shearing force of 2000 N. The upper surface is displaced through 0.5 cm with respect to the bottom. Calculate the modulus of rigidity of the metal.
Solution:
Given: L = 40 cm = 0.4 m,
F = 2000 N = 2 × 10³N
l = 0.5 cm = 0.005 m, A = L² = 0.16 m²
To find: Modulus of rigidity (η)

Formulae:

i. θ = \(\frac{l}{\mathrm{~L}}\)
ii. η = \(\frac{\mathrm{F}}{\mathrm{A} \theta}\)

Calculation:

From formula (i),
θ = \(\frac{0.005}{0.4}\) = 0.0125
From formula (ii),
η = \(\frac{2 \times 10^{3}}{0.16 \times 0.0125}\) = 1 × 10⁶ N/m²
Answer:
The modulus of rigidity of the metal cube is 1 × 10⁶ N/m².

Question 38.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44500 N forces producing only elastic deformation. Calculate the resulting strain. (Rigidity modulus of copper = 42 × 10⁹ N m^-2)
Solution:
Given: A = 15.2 × 19.14 × 10^-6 m²,
F = 44500 N, η = 42 × 10⁹ N m^-2
To find: Strain (θ)
Formula: η = \(\frac{\mathrm{F}}{\mathrm{A} \theta}\)

Calculation:
From formula,

Answer:
The strain produced in the piece of copper is 3.64 × 10^-3

Question 39.
A copper metal cube has each side of length 1 m. The bottom edge of cube is fixed and tangential force 4.2 × 10⁸ N is applied to top surface. Calculate the lateral displacement of the top surface, if modulus of rigidity of copper is 14 × 10¹⁰ N/m².
Solution:
Given: L = h = 1 m, F = 4.2 × 10⁸ N, η = 1.4 × 10¹¹ N/m²
To find: Lateral displacement (x)
Formula: η = \(\frac{\mathrm{F}}{\mathrm{A} / \theta}\) = \(\frac{\mathrm{Fh}}{\mathrm{Ax}}\)
Calculation: From formula,
x = \(\frac{\mathrm{Fh}}{\mathrm{A\eta}}\)
∴ x = \(\frac{4.2 \times 10^{8} \times 1}{(1 \times 1) \times 1.4 \times 10^{11}}\) = 3 × 10^-3 m
∴ x = 3 mm
Answer:
The lateral displacement of top is 3 mm.

Question 40.
The area of the upper face of a rectangular block is 0.5 m × 0.5 m and the lower face is fixed. The height of the block is 1 cm. A shearing force applied at the top face produces a displacement of 0.0 15 mm. Find the strain and shearing force. (Modulus of rigidity: η = 4.5 × 10¹⁰ N/m²)
Solution:
Given: A = 0.5 m × 0.5 m = 0.25 m²,
h = 1 cm = 10^-2m,
x = 0.015 mm = 15 × 10^-6m
η = 4.5 × 10¹⁰ N/m²
To find: Strain (θ). Shearing force (F)

Formulae:

i. θ = \(\frac{\mathrm{x}}{\mathrm{h}}\)
ii. F = ηAθ

Calculations:

Using formula (i),
θ = \(\frac{15 \times 10^{-6}}{10^{-2}}\) = 1.5 × 10^-3
Using formula (ii),
F = 4.5 × 10¹⁰ × 0.25 × 1.5 × 10^-3
= 1.688 × 10⁷ N
Answer:
Shearing force is 1.688 × 10⁷ N and strain is 1.5 × 10^-3

Question 41.
Explain the behaviour of metal wire under increasing load.
Answer:
Consider a metal wire suspended vertically from a rigid support and stretched by applying load to its lower end. The load is gradually increased in small steps until the wire breaks. The elongation produced in the wire is measured during each step. Stress and strain are noted for each load and a graph is drawn by taking tensile strain along X-axis and tensile stress along Y-axis. It is a stress-strain curve as shown in the figure.

1. Proportional limit: The initial part of the graph is a straight line OA. This is the region in which Hooke’s law is obeyed and stress is directly proportional to stain. The straight line portion ends at A. The stress at this point is called proportional limit.
2. Yield point: If the load is further increased till point B is reached, stress and strain are no longer proportional and Hooke’s law is not valid. If the load is gradually removed starting at any point between O and B. The curve is retracted until the wire regains its original length. The change is reversible. The material of the wire shows elastic behaviour in the region OB. Point B is called the yield point. It is also known as the elastic limit.
3. Permanent Set: When the stress is increased beyond point B, the strain continues to increase. If the load is removed at any point beyond B, for example (at C), the material does not regain its original length. It follows the line CE. Length of the wire when there is no stress is greater than the original length. The deformation is irreversible and the material has acquired a permanent set.
4. Fracture point: Further increase in load causes a large increase in strain for relatively small increase in stress, until a point D is reached at which fracture takes place. The material shows plastic flow or plastic deformation from point B to point D. The material does not regain its original state when the stress is removed. The deformation is called plastic deformation.

Question 42.
Stress-strain curve for two materials A and B are shown in the figure. The graphs are drawn to the same scale.

1. Which material has greater Young’s modulus?
2. Which of the two is the stronger material? (NCERT)

Answer:

1. For a given strain, stress for A is more than that of B. Hence, Young’s modulus (= stress/strain) is greater for A than that of B.
2. Material A is stronger than B because A can bear greater stress before the breaking of the wire.

Question 43.
Figure shows the stress-strain curve for a given material. What are

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

Answer:

1. The graph, implies stress of 150 × 10⁶ N m^-2 corresponds to a strain of 0.002. Therefore, Young’s modulus is,
   
2. The yield strength for the material is less than 300 × 10⁶ N m^-2, i.e.. 3 × 10⁸ N m^-2 and greater than 2.5 × 10⁸ N m^-2.

Question 44.
Explain the following terms:

1. Ductile
2. Malleable

Answer:

1. Metals such as copper, aluminium, wrought iron, etc. have large plastic range of extension. They lengthen considerably and undergo plastic deformation till they break. They are called ductile.
2. Metals such as gold, silver which can be hammered into thin sheets are called malleable.

Question 45.
Define strain energy.
Answer:
The elastic potential energy gained by a wire during elongation b a stretching force is called as strain energy.

Question 46.
A steel wire is acted upon by a load of 10 N. Calculate the extension produced in the wire, if the strain energy stored in the wire is 1.1 × 10^-3 J.
Solution:
Given: F = 10 N, Strain energy =1.1 × 10^-3 J,
To find: Extension (l)
Formula: W = \(\frac{1}{2}\) × F × l
Calculation:
From formula,
l = \(\frac{2 \mathrm{~W}}{\mathrm{~F}}\) = \(\frac{2 \times 1.1 \times 10^{-3}}{10}\) = 2.2 × 10^-4 m.
Answer:
The extension produced in the wire is 2.2 × 10^-4 m.

Question 47.
Calculate the strain energy per unit volume in a brass wire of length 3 m and area of cross-section 0.6 mm² when it is stretched by 3 mm and a force of 6 kg-wt is applied to its free end.
Solution:
Given: L = 3 m, F = 6 kg wt = 6 × 9.8N = 58.8N, A = 0.6 mm² = 0.6 × 10^-6 m², l = 3 mm = 3 × 10^-3 m
To find: Strain energy per unit volume (u)

Formulae:

i. Stress = \(\frac{F}{A}\)
ii. Strain = \(\frac{l}{L}\)
iii. u = \(\frac{1}{2}\) × Stress × Strain

Calculation:

From formula (i),
Stress = \(\frac{F}{A}\) = \(\frac{58.8}{0.6 \times 10^{-6}}\) = 98 × 10⁶ N/m²
From formula (ii),
Strain = \(\frac{l}{L}\) = \(\frac{3 \times 10^{-3}}{3}\) = 10^-3
From formula (iii),
u = \(\frac{1}{2}\) × Stress × Strain
= \(\frac{1}{2}\) × 98 × 10⁶ × 10^-3
u = 49 × 10³ J/m³
Answer:
Strain energy per unit volume in the brass wire is 49 × 10³ J/m³

Question 48.
A steel wire of diameter 1 × 10^-3 m is stretched by a force of 20 N. Calculate the strain energy per unit volume. (Y_steel = 2 × 10¹¹ N/m²)
Solution:
Given: d = 1 × 10^-3 m, r = 5 × 10^-4 m, y_steel = 2 × 10¹¹ N/m²
To find: Strain energy per unit volume
Formula: Strain energy per unit volume
= \(\frac{1}{2}\) × \(\frac{\text { (stress) }^{2}}{\mathrm{Y}}\)
Calculation:
From formula.
Strain energy per unit volume

Answer:
The strain energy per unit volume of the steel wire is 1621 J.

Question 49.
A uniform steel wire of length 3 m and area of cross section 2 mm² is extended through 3 mm. Calculate the energy stored in the wire, if the elastic limit is not exceeded. (Y_steel = 20 × 10¹⁰ N/m²)
Solution:
Given: L = 3 m, A = 2 mm² = 2 × 10^-6 m²,
l = 3 mm = 3 × 10^-3 m,
Y_steel = 20 × 10¹⁰ N/m²
To find: Energy stored (U)
Formula: W = \(\frac{1}{2}\) × F × l
Calculation: Since, Y = \(\frac{\mathrm{FL}}{\mathrm{Al}}\)

Answer:
The energy stored in the steel wire is 0.6 J.

Question 50.
Explain the following terms:
i. Hardness
ii. Strength
iii. Toughness
Answer:

i. Hardness:

1. Hardness is the property of a material which enables it to resist plastic deformation.
2. Hard materials have little ductility and they are brittle to some extent.
3. The term hardness also refers to stiffness or resistance to bending, scratching, abrasion or cutting.
4. It is the property of a material which gives it the ability to resist permanent deformation when a load is applied to it.
5. The greater the hardness, greater is the resistance to deformation.
6. Hardness of material is different from its strength and toughness.
7. The most well known example for hard material is diamond, while metal with very low hardness is aluminium.

ii. Strength:

1. If a force is applied to a body, it produces deformation in it.
2. Higher is the force required for deformation, the stronger is the material, i.e., the material has more strength.
3. Steel has high strength whereas plasticine clay is not strong because it gets easily deformed even by a small force.

iii. Toughness:

1. Toughness is the ability of a material to resist fracturing when a force is applied to it.
2. Plasticine clay is relatively tough as it can be stretched and deformed due to applied force without breaking.

Question 51.
Explain the concept of frictional force.
Answer:

1. Whenever the surface of one body slides over another, each body exerts a certain amount of force on other body.
2. These forces are tangential to the surfaces. The force on each body is opposite to the direction of motion between two bodies.
3. It prevents or opposes the relative motion between two bodies.
4. It is common experience that an object placed on any surface does not move easily when a small force is applied to it.
5. This is because of certain force of opposition acting between the surface of the object and the surface on which it is placed.
6. To initiate any motion between pair of surfaces, we need a certain minimum force. Also, after the motion begins, it is constantly opposed by some natural force.
7. This mechanical force between two solid surfaces in contact with each other is called as frictional force.

Question 52.
Explain few examples of friction.
Answer:

1. A rolling ball comes to rest after covering a finite distance on playground because of friction.
2. Our foot ware is provided with designs at the bottom of its sole so as to produce force of opposition to avoid slipping. It is difficult to walk without such opposing force. When we try to walk fast on polished flooring at home with soap water spread on it. There is a possibility of slipping due to lack of force of friction.
3. Relative motion between solids and fluids (i.e. liquids and gases) is also opposed naturally by friction, eg.: a boat on the surface of water experiences opposition to its motion.

Question 53.
Explain the origin of friction.
Answer:

1. If smooth surfaces are observed under powerful microscope, many irregularities and projections are observed.
2. Friction arises due to interlocking of these irregularities between two surfaces in contact.
3. The surfaces can be made extremely smooth by polishing to avoid irregularities but then case also, friction does not decrease but may increase.
4. Hence the interlocking of irregularities is not the real cause of friction.
5. According to modem theory, cause of friction is the force of attraction between molecules of two surfaces in actual contact in addition to the. force due to the interlocking between the two surfaces.
6. When one body is in contact with another body, the real microscopic area in contact is very small due to irregularities in contact.
   
7. Due to small area, pressures at points of contact is very high. Hence there is strong force of attraction between the surfaces in contact.
8. When the surfaces in contact become more and more smooth, the actual area of contact goes on increasing.
9. Due to this, the force of attraction between the molecules increases and hence the friction also increases.

Question 54.
State the following terms:

1. Cohesive force
2. Adhesive force

Answer:

1. When two Surfaces are of the same material, the force of attraction between them is called cohesive force.
2. When two surfaces are of different materials, the force of attraction between them is called adhesive force.

Question 55.
Explain the concept of static friction.
Answer:

1. Consider a wooden block placed on a horizontal surface as shown in the figure and small horizontal force F is applied to it.
   
2. The block does not move with this force as it cannot overcome the frictional force between the block and horizontal surface.
3. In this case the force of static friction is equal to F and balances it.
4. The frictional force which balances applied force when the body is static is called force of static friction. In other words, static friction prevents sliding motion.
5. If we keep increasing F, a stage will come
   • For F < F_max, the force of static friction is equal to F.
   • when for F = F_max, the object will start moving.
   • For F ≥ F_max, the kinetic friction comes into play.
6. Static friction opposes impending motion i.e. the motion that would take place in absence of frictional force under the applied force.

Question 56.
Define limiting force of friction.
Answer:
Just before the body starts sliding over another body, the value of frictional force is maximum, it is called as limiting force of friction.

Question 57.
Explain the concept of kinetic friction.
Answer:

1. Once the sliding of block on the surface starts the force of friction decreases.
2. The force required to keep the body sliding steadily is thus less than the force required to just start its sliding.
3. The force of friction that comes into play when a body is in steady state of motion over another surface is called kinetic fòrce of friction.
4. Friction between two surfaces in contact when one body is actually sliding over the other body, is called kinetic friction or dynamic friction.

Question 58.
Why ball bearings are used in machines?
Answer:

1. For same pair of surfaces, the force of static friction is greater than the force of kinetic friction while the force of kinetic friction is greater than force of rolling friction.
2. As ball bearing undergo rolling friction and rolling friction is the minimum, ball bearings are used to reduce friction in parts of machines to increase its efficiency.

Question 59.
The coefficient of static friction between a block of mass 0.25 kg and a horizontal surface is 0.4. Find the horizontal force applied to it.
Solution:
Given: μ_s = 0.4, m = 0.25 kg, g = 9.8 m/s²
To find: Force (F_L)
Formula: F_L = μ_sN = μ_s(mg)
Calculation: From formula,
F_L = 0.4 × 0.25 × 9.8 = 0.98 N
Answer:
The horizontal force applied to it is 0.98 N.

Question 60.
Calculate the force required to move a block of mass 20 kg resting on a horizontal surface, if μ_s = 0.3 and g = 9.8 m/s².
Solution:
Given: m = 20 kg, μ_s = 0.3, g = 9.8 m/s²
To find. Force required (F)
Formula: F_S = μ_sN = μ_smg
Calculation: From formula,
F_S = 3 × 20 × 9.8 = 58.8N
Answer:
The force required to move the block is 58.8 N.

Question 61.
A block of mass 40 kg resting on a horizontal surface just begins to slide when a horizontal force of 120 N is applied to it. Once the motion starts, It can be maintained by a force of 80 N. Determine the coefficients of static friction and kinetic friction (g = 9.8 m/s²)
Solution:
Given: F_L= 120N, F_k = 80N, m = 40 kg, g = 9.8 m/s²

To find:

i. Coefficient of static friction (μ_s)
ii. Coefficient of kinetic friction (μ_k)

Formulae:

i. μ_s = \(\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{N}}\)
ii. μ_k = \(\frac{\mathrm{F}_{\mathrm{k}}}{\mathrm{N}}\)

Calculation:

From formula (i) we get.
N = mg = 40 × 9.8 = 392 N
∴ μ_s = \(\frac{F_{L}}{N}\) = \(\frac{120}{392}\) = 0.306
From formula (ii) we get,
∴ μ_k = \(\frac{F_{k}}{N}\) = \(\frac{80}{392}\) = 0.204
Answer:

1. The coefficient of static friction is 0.306.
2. The coefficient of kinetic friction is 0.204.

Question 62.
A 20 kg metal block is placed on a horizontal surface. The block just begins to slide when horizontal force of 100 N is applied to it. Calculate the coefficient of static friction. If coefficient of kinetic friction is 0.4, then find minimum force to maintain its uniform motion.
Solution:
Given: m = 20 kg, F_L = 100 N, μ_k = 0.4

To find:

i. Coefficient of static friction (μ_s)
ii. Minimum force required (F_k)

Formulae:

i. μ_s = \(\frac{F_{L}}{N}\) = \(\frac{\mathrm{F}_{\mathrm{L}}}{\mathrm{mg}}\)
ii. μ_s = \(\frac{F_{k}}{N}\)

Calculation:

From formula (i),
μ_s = \(\frac{100}{20 \times 9.8}\) = 0.5102
From formula (ii),
F_k = μ_kN = μ_k × mg = 0.4 × 20 × 9.8
∴ F_k = 78.4 N
Answer:

1. The coefficient of static friction is 0.5102.
2. The minimum force required is 78.4 N.

Question 63.
The Mariana trench is located in the Pacific Ocean and at one place it is nearly 11 km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches the bottom? (K = 1.6 × 10¹¹ N/m²)
Answer:
Given: V = 0.32 m³, K = 1.6 × 10¹¹ N/m²,
P_W= 1.1 × 10⁸ Pa, P_atm = 1.01 × 10⁵ Pa
∴ dP = P_W – P_atm = 1.1 × 10⁸ – 1.01 × 10⁵ ≈ 1.1 × 10⁸ Pa
As, bulk modulus is given as,
K = V × \(\frac{d P}{d V}\)
∴ dV = \(\frac{\mathrm{V} \times \mathrm{dP}}{\mathrm{K}}\) = \(\frac{\left(1.1 \times 10^{8}\right) \times 0.32}{1.6 \times 10^{11}}\)
∴ dV = 2.2 × 10^-4 m³
The change in the volume of the ball when it reaches the bottom will be 2.2 × 10^-4 m³.

Question 64.
Two spheres A and B having same volume are dropped from same height in ocean. Sphere A is made up of brass and sphere B is made up of steel. Will there be same change in volume of spheres at a certain depth inside water? What will be the ratio of change in volumes of the two spheres at this depth?
Answer:

1. Brass and copper have different elastic moduli. Hence, there won’t be same change in the volume of spheres at a certain depth inside water.
2. As two spheres are dropped from same height and are at same depth in water pressure exerted on them remains same.
   ∴ dP_A = dP_B = dP
   V_A = V_B = V
3. If dV_A and dV_B be the change in volume of two spheres A and B then,
   dV_A = \(\frac{\mathrm{V} \mathrm{dP}}{\mathrm{K}_{\mathrm{A}}}\) and dV_B = \(\frac{\mathrm{V} \mathrm{dP}}{\mathrm{K}_{\mathrm{B}}}\)
4. Ratio of change in volume,
   
   where, K_A and K_B are bulk modulus of material of spheres A and B respectively.

Question 65.
What is the basis of deciding the thickness of metallic ropes used in crane to lift the heavy weight?
Answer:

1. The thickness of the metallic ropes used in cranes is decided on the basis of the elastic limit of the material of the rope and the factor of safety.
2. To lift a load upto 10⁴ kg, the rope is made for a factor of safety of 10.
3. It should not break even when a load of (original load × factor of safety = 10⁴ × 10) 10⁵ kgf i.e., 10⁵ × 9.8 N is applied on it.
4. If ‘r’ is the radius of the rope, then maximum stress = \(\frac{10^{5}}{\pi r^{2}}\).
5. The maximum stress (ultimate stress) should not exceed the breaking stress (5 to 20 × 10⁸ N/m²) as well as elastic limit of steel Y_s(30 × 10⁷ N/m²).
   
6. In order to have flexibility, the rope is made up of large number of thin wires twisted together.

Question 66.
Multiple choice questions

Which one of the following substances possesses the highest elasticity?
(A) Rubber
(B) Glass
(C) Steel
(D) Copper
Answer:
(C) Steel

Question 1.
S.I unit of stress is
(A) Newton/ metre
(B) Newton/ metre²
(C) Newton²/metre
(D) Newton/metre³
Answer:
(B) Newton/ metre²

Question 2.
A wire is stretched to double of its length. The strain is
(A) 2
(B) 1
(C) zero
(D) 0.5
Answer:
(B) 1

Question 3.
A and B are two steel wires and the radius of A is twice that of B. If they are stretched by the same load, then the stress on B is ____.
(A) four times that of A
(B) two times that of A
(C) three times that of A
(D) same as that of A
Answer:
(A) four times that of A

Question 4.
Two wires of the same material have radii r_A and r_B respectively. The radius of wire A is twice the radius of wire B. If they are stretched by same load then stress on wire B is ___
(A) equal to that of A
(B) half that of A.
(C) two times that of A.
(D) four times that of A.
Answer:
(D) four times that of A.

Question 5.
The length of a wire increases by 1% by a load of 2 kg wt. The linear strain produced in the wire will be
(A) 0.02
(B) 0.001
(C) 0.01
(D) 0.002
Answer:
(C) 0.01

Question 6.
A wire of length ‘L’, radius ‘r’ when stretched with a force ‘F’ changes in length by l’. What will be the change in length of a wire of same material having length ‘2L’ radius ‘2r and stretched by a force ‘2F’?
(A) l/2
(B) l
(C) 2l
(D) 4l
Answer:
(B) l

Question 7.
A force of 100 N produces a change of 0.1% in a length of wire of area of cross section 1 mm². Young’s modulus of the wire is ____
(A) 10⁵ N/m²
(B) 10⁹ N/m²
(C) 10¹¹ N/m²
(D) 10¹² N/m²
Answer:
(C) 10¹¹ N/m²

Question 8.
A copper wire (Y = 1 × 10¹¹ N/m²) of length 6 m and a steel wire (Y = 2 × 10¹¹ N/m²) of length 4 m each of cross-section 10^-5 m² are fastened end to end and stretched by a tension of 100 N. The elongation produced in the copper wire is
(A) 0.2 mm
(B) 0.4 mm
(C) 0.6 mm
(D) 0.8 mm
Answer:
(C) 0.6 mm

Question 9.
When a force is applied to the free end of a metal wire, metal wire undergoes
(A) longitudinal and lateral extension.
(B) longitudinal extension and lateral contraction.
(C) longitudinal contraction and lateral extension.
(D) longitudinal and lateral contraction.
Answer:
(B) longitudinal extension and lateral contraction.

Question 10.
A force of 1 N doubles the length of a cord having cross-sectional area 1 mm². The Young’s modulus of the material of cord is
(A) 1 N m^-2
(B) 0.5 × 10⁶ N m^-2
(C) 10⁶ N m^-2
(D) 2 × 10⁶ N m^-2
Answer:
(C) 10⁶ N m^-2

Question 11.
In equilibrium the tensile stress to which a wire of radius r is subjected by attaching a mass ‘m’ is ____.
(A) \(\frac{\mathrm{mg}}{\pi \mathrm{r}}\)
(B) \(\frac{\mathrm{mg}}{2 \pi \mathrm{r}}\)
(C) \(\frac{\mathrm{mg}}{\pi \mathrm{r}^{2}}\)
(D) \(\frac{\mathrm{mg}}{2 \pi \mathrm{r}^{2}}\)
Answer:
(C) \(\frac{\mathrm{mg}}{\pi \mathrm{r}^{2}}\)

Question 12.
When load is applied to a wire, the extension is 3 mm, the extension in the wire of same length but half the radius by the same load is
(A) 0.75 mm
(B) 6 mm
(C) 1.5 mm
(D) 12.0 mm
Answer:
(D) 12.0 mm

Question 13.
The S.I. unit of compressibility is ____.
(A) \(\frac{\mathrm{m}^{2}}{\mathrm{~N}}\)
(B) Nm²
(C) \(\frac{\mathrm{N}}{\mathrm{m}^{2}}\)
(D) \(\frac{\mathrm{kg}}{\mathrm{m}^{3}}\)
Answer:
(A) \(\frac{\mathrm{m}^{2}}{\mathrm{~N}}\)

Question 14.
When the pressure applied to one litre of a liquid is increased by 2 × 10⁶ N/m². Its volume decreases by 1 cm³. The bulk modulus of the liquid is
(A) 2 × 10⁹ N/m²
(B) 2 × 10³ dyne/cm²
(C) 2 × 10³ N/m²
(D) 0.2 × 10⁹ N/m²
Answer:
(A) 2 × 10⁹ N/m²

Question 15.
A cube of aluminium of each side 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be
(A) 0.02
(B) 0.1
(C) 0.005
(D) 0.002
Answer:
(D) 0.002

Question 16.
A wire has Poisson’s ratio of 0.5. It is stretched by an external force to produce a longitudinal strain of 2 × 10^-3. If the original diameter was 2 mm, the final diameter after stretching is
(A) 2.002 mm
(B) 1.998 mm
(C) 0.98 mm
(D) 1.999 mm
Answer:
(B) 1.998 mm

Question 17.
The compressibility of a substance is the reciprocal of ___.
(A) Young’s modulus
(B) Bulk modulus
(C) Modulus of rigidity
(D) Poisson’s ratio
Answer:
(B) Bulk modulus

Question 18.
For which of the following is the modulus of rigidity highest?
(A) Aluminium
(B) Quartz
(C) Rubber
(D) Water
Answer:
(B) Quartz

Question 19.
An elongation of 0.2% in a wire of cross-section 10^-4 m² causes a tension of 1000 N. Then its Young’s modulus is
(A) 6 × 10⁸ N/m²
(B) 5 × 10⁹ N/m²
(C) 10⁸ N/m²
(D) 10⁷ N/m²
Answer:
(B) 5 × 10⁹ N/m²

Question 20.
Within the elastic limit, the slope of graph between stress against strain gives ____
(A) compressibility
(B) Poisson’s ratio
(C) modulus of elasticity
(D) extension
Answer:
(C) modulus of elasticity

Question 21.
Solids which break or rupture before the elastic limits are called
(A) brittle
(B) ductile
(C) malleable
(D) elastic
Answer:
(A) brittle

Question 22.
Substances which break just after their elastic limit is reached are ___.
(A) ductile
(B) brittle
(C) malleable
(D) plastic
Answer:
(B) brittle

Question 23.
Which of the following substances is ductile?
(A) glass
(B) high carbon steel
(C) Steel
(D) copper
Answer:
(D) copper

Question 24.
The Young’s modulus of a material is 10¹¹ Nm^-2 and its Poisson’s ratio is 0.2. The modulus of rigidity of the material is
(A) 0.42 × 10¹¹ N/m²
(B) 0.42 × 10¹⁴ N/m²
(C) 0.42 × 10¹⁶ N/m²
(D) 0.42 × 10¹⁸ N/m²
Answer:
(A) 0.42 × 10¹¹ N/m²

Question 25.
Two pieces of wire A and B of the same material have their lengths in the ratio 1:2, and their diameters are in the ratio 2:1. If they are stretched by the same force, their elongations will be in the ratio
(A) 2 : 1
(B) 1 : 4
(C) 1 : 8
(D) 8 : 1
Answer:
(C) 1 : 8

Question 26.
Young’s modulus of material of wire is ‘Y’ and strain energy per unit volume is ‘E’, then the strain is

Answer:
(C) \(\sqrt{\frac{2 \mathrm{E}}{\mathrm{Y}}}\)

Question 27.
Strain energy per unit volume is given by

Answer:
(A) \(\frac{1}{2} \times \frac{(\text { stress })^{2}}{\mathrm{Y}}\)

Question 28.
The strain energy per unit volume of the wire under increasing load is
(A) \(\frac{1}{2}\) × (stress)² × strain
(B) \(\frac{1}{2}\) × stress × (strain)²
(C) 0.5 × stress × strain
(D) 0.5 × (strain)² × \(\frac{1}{Y}\)
Answer:
(C) 0.5 × stress × strain

Question 29.
The energy stored per unit volume of a strained wire is
(A) \(\frac{1}{2}\) × (load) × (extension)
(B) \(\frac{1}{2}\) × \(\frac{Y}{(\text { strain })^{2}}\)
(C) \(\frac{1}{2}\) × Y × (strain)²
(D) Stress × strain
Answer:
(C) \(\frac{1}{2}\) × Y × (strain)²

Question 30.
If work done in stretching a wire by 1 mm is 2 J, the work necessary for stretching another wire of same material, but with double the radius and half the length by 1 mm in joule is
(A) 1/4
(B) 4
(C) 8
(D) 16
Answer:
(D) 16

Question 31.
When the load on a wire is slowly increased from 3 to 5 kg wt, the elongation increases from 0.61 to 1.02 mm. The work done during the extension of wire is
(A) 0.16 J
(B) 0.016 J
(C) 1.6 J
(D) 16 J
Answer:
(B) 0.016 J

Question 32.
A body lies on a table. Its weight is balanced by the ___.
(A) frictional force
(B) normal force
(C) force causing motion on the body
(D) surface of the table
Answer:
(B) normal force

Question 33.
The friction that exists between the surface of two bodies in contact when one body is sliding over the other, is called ___.
(A) rolling friction
(B) friction
(C) kinetic friction
(D) static friction.
Answer:
(C) kinetic friction

Question 34.
Limiting force of static friction does NOT depend on
(A) actual area of contact.
(B) geometrical area of contact.
(C) interlocking between surfaces in contact.
(D) intermolecular forces of attraction between molecules of the two surfaces.
Answer:
(C) interlocking between surfaces in contact.

Question 35.
In case of coefficient of static friction (µ_s), kinetic friction (µ_k) and rolling friction (µ_r), which of the following relation is true
(A) µ_s > µ_k > µ_r
(B) µ_s > µ_r > µ_k
(C) µ_r > µ_s > µ_k
(D) µ_r > µ_k > µ_s
Answer:
(A) µ_s > µ_k > µ_r

Question 36.
A body of weight 50 N is placed on a smooth surface. If the force required to move the body on the surface is 30 N, the coefficient of friction is
(A) 0.6
(B) 0.4
(C) 0.3
(D) 0.8
Answer:
(A) 0.6

Question 37.
A wooden block of 50 kg is at rest on the table. A force of 70 N is required to just slide the block. The coefficient of static friction is
(A) \(\frac{6}{7}\)
(B) \(\frac{5}{7}\)
(C) \(\frac{7}{35}\)
(D) \(\frac{1}{7}\)
Answer:
(D) \(\frac{1}{7}\)

Question 38.
When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + l). The elastic potential energy stored in the extended wire is:
(A) \(\frac{1}{2}\)Mgl
(B) \(\frac{1}{2}\)MgL
(C) Mgl
(D) MgL
Answer:
(A) \(\frac{1}{2}\)Mgl
Hint: elastic potential energy: \(\frac{1}{2}\) × F × L = \(\frac{1}{2}\)Mgl

Question 67.
A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1 π ms^-2, What will be the tensile stress that would be developed
(A) 6.2 × 10⁶ N m^-2
(B) 5.2 × 10⁶ N m^-2
(C) 3.1 × 10⁶ N m^-2
(D) 4.8 × 10⁶ N m^-2
Answer:
(C) 3.1 × 10⁶ N m^-2

Question 68.
A boy’s catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross-section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms^-1. Neglect the change in the area of cross-section of the cord while stretched. The Young’s modulus of rubber is closest to:
(A) 10⁶ N m^-2
(B) 10⁴ N m^-2
(C) 10⁸ Nm^-2
(D) 10³ N m^-2
Answer:
(A) 10⁶ N m^-2
Hint: d = 6 mm = 0.006 m, l = 42 cm = 0.42 m, ∆l = 20 cm = 0.2 m, m = 0.02 kg, v = 20 m/s
Energy stored in the catapult,

Question 69.
The stress-strain curves are drawn for two different materials X and Y. It is observed that the ultimate strength point and the fracture point are close to each other for material X but are far apart for material Y. We can say that materials X and Y are likely to be (respectively),
(A) Plastic and ductile
(B) Ductile and brittle
(C) Brittle and ductile
(D) Brittle and plastic
Answer:
(C) Brittle and ductile
Hint: Ductile materials have a fracture strength lower than the ultimate Tensile strength (i.e., the points are far apart.) whereas in brittle materials, the fracture strength is equivalent to ultimate tensile strength (i.e., the points are close.)
∴ Material X is brittle and Y is ductile in nature.

Question 70.
The compressibility of water is ‘o’ per unit atmospheric pressure. The decrease in its volume ‘V’ due to atmospheric pressure ‘P’ will be
(A) \(\frac{\sigma \mathrm{V}}{\mathrm{P}}\)
(B) \(\frac{\sigma P}{\mathrm{~V}}\)
(C) σPV
(D) \(\frac{\sigma}{\mathrm{PV}}\)
Answer:
(C) σPV

Question 71.
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by ∆l on applying a force F, how much force is needed to stretch the second wire by the same amount?
(A) 9 F
(B) 6 F
(C) 4 F
(D) F
Answer:
(A) 9 F
Hint: As material is same, Young’s modulus of two wires is same.
Also, volume of both wires is same.
V₁ = V₂

Question 72.
A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, \(\left(\frac{\mathrm{dr}}{\mathrm{r}}\right)\),is:
where negative sign indicates volume decreases with increase in pressure.
(A) \(\frac{\mathrm{mg}}{\mathbf{3} \mathrm{Ka}}\)
(B) \(\frac{\mathrm{mg}}{\mathrm{Ka}}\)
(C) \(\frac{\mathrm{Ka}}{\mathrm{mg}}\)
(D) \(\frac{\mathrm{Ka}}{3 \mathrm{mg}}\)
Answer:
(A) \(\frac{\mathrm{mg}}{\mathbf{3} \mathrm{Ka}}\)
Hint:
Bulk modulus is given as,
K = \(\left(\frac{-\mathrm{dP}}{\mathrm{dV} / \mathrm{V}}\right)\)
where negative sign indicates volume decreases with increase in pressure.
∴ Fractional decrease in volume will be,
\(\frac{\mathrm{dV}}{\mathrm{V}}\) = \(\frac{\mathrm{dP}}{\mathrm{K}}\)
If area of cross-section of cylinder is a, then,

Question 73.
Two metal wires ‘P’ and ‘Q’ of same length and material are stretched by same load. Their masses are in the ratio m₁ : m₂. The ratio of elongations of wire ‘P’ to that of’Q’ is

Answer:
(C) m₂ : m₁
Hint:

Question 74.
The increase in energy of a metal bar of length ‘L’ and cross-sectional area ‘A’ when compressed with a load ‘M’ along its length is (Y = Young’s modulus of the material of metal bar)
(A) \(\frac{\mathrm{FL}}{2 \mathrm{AY}}\)
(B) \(\frac{\mathbf{F}^{2} \mathbf{L}}{\mathbf{2 A Y}}\)
(C) \(\frac{\mathrm{FL}}{\mathrm{AY}}\)
(D) \(\frac{F^{2} L^{2}}{2 A Y}\)
Answer:
(B) \(\frac{\mathbf{F}^{2} \mathbf{L}}{\mathbf{2 A Y}}\)
Hint:
U = \(\frac{1}{2}\) × F × l
= \(\frac{1}{2}\) × F × \(\frac{\mathrm{FL}}{\mathrm{AY}}\) = \(\frac{\mathrm{F}^{2} \mathrm{~L}}{2 \mathrm{AY}}\)