Class 12

MP Board Class 12th Biology Important Questions Chapter 6 Molecular Basis of Inheritance

Molecular Basis of Inheritance Important Questions

Molecular Basis of Inheritance Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
The process of transfer of enetic information from DNA to RNA is :
(a) Transversion
(b) Transcription
(c) Translation
(d) Translocation.
Answer:
(b) Transcription

Question 2.
Transcription involves :
(a) Synthesis of RNA over DNA
(b) Joining of amino acids over polypeptides
(c) Synthesis of RNA over ribosomes
(d) Synthesis of DNA.
Answer:
(a) Synthesis of RNA over DNA

Question 3.
In operon model, RNA polymerase binds to :
(a) Structural gene
(b) Promotor gene
(c) Operator gene
(d) Regulator gene.
Answer:
(b) Promotor gene

Question 4.
The process of translation is :
(a) Ribosome synthesis
(b) Protein synthesis
(c) DNA synthesis
(d) RNA synthesis.
Answer:
(b) Protein synthesis

Question 5.
Operon model of gene expression is prokaryotes was proposed by :
(a) Meselson and stahl
(b) Wilkins and Franklin
(c) Beadle and Tatum
(d) Jacob and Monod.
Answer:
(d) Jacob and Monod.

Question 6.
Reverse transcription was discovered by :
(a) Beadle and Tatum
(b) Temin and Baltimore
(c) Watson and Crick
(d) Khorana.
Answer:
(b) Temin and Baltimore

Question 7.
Nitrogen base present in a codon are :
(a) 4
(b) 3
(c) 2
(d) 1.
Answer:
(a) 4

Question 8.
A non – sense/termination codon is :
(a) UUU
(b) GCG
(c) UAA
(d) CCC.
Answer:
(a) UUU

Question 9.
A unit of lac – operon which in the absence of lactose, suppresses the activity of operator gene is :
(a) Structural gene
(b) Regulator gene
(c) Repressor gene
(d) Promoter gene.
Answer:
(d) Promoter gene.

Question 10.
Sequence of nitrogenous bases on t RNA is known as :
(a) Anti codon.
(b) Terminating coon
(c) Repressor codon
(d) Initiate codon.
Answer:
(a) Anti codon.

Question 11.
Which is found in nucleosome :
(a) Histone molecule
(b) Luxary genes
(c) Nucleoplasmine
(d) House keeping genes
Answer:
(a) Histone molecule

Question 12.
Functional gene is :
(a) Gene battery
(b) Luxary genes
(c) Mylud gene
(d) House keeping genes.
Answer:
(d) House keeping genes.

Question 13.
Which RNA has very short life span :
(a) m RNA
(b) t RNA
(c) r RNA
(d) Sn RNA.
Answer:
(a) m RNA

Question 14.
Who terminate the transcription :
(a) Co – protein
(b) Sigma factor
(c) Raw Protein
(d) Omega factor.
Answer:
(c) Raw Protein

Question 15.
A Person which has a trisomy on 21 st chromosome is called :
(a) Klinefelter syndrome
(b) Down’s syndrome
(c) Turner syndrome
(d) None of these.
Answer:
(b) Down’s syndrome

Question 2.
Fill in the blanks:

1. ……………… is found in RNA, in the place of DNA’s thymine.
2. Formation of ……………… from DN A is called transcription.
3. In ……………… ribosomes are of 70s type.
4. ……………… is an exogenous gene that has been introduced into the genome of other organism.
5. The joining of DNA strands together by ………………
6. ……………… is the vector of genes.
7. ……………… enzyme is necessary for transcription.
8. ……………… discovered polytene chromosome.
9. Intron is known as ………………
10. ……………… nitrogenous protein which bind the DNA molecule with chromosome.
11. Highly condensed chromatin which is not available for transcription is called ………………
12. ……………… vims infects the bacteria.
13. ……………… is the outer covering of viruses.
14. ……………… type of RNA is found in ribosome.
15. ……………… is made by joining of peptide bond.

Answer:

1. Uracil
2. mRNA
3. Prokaryotes
4. Transgenic
5. Ligase
6. Plasmid
7. RNA Polymerase
8. Balbiani
9. Junk DNA
10. Histone
11. Hetero chromatin
12. Bacteriophage
13. Capsid
14. Ribosomal RNA
15. Polypeptide.

Question 3.
Match the followings :
I.

Answer:

1. (c)
2. (a)
3. (d)
4. (b)

II.

Answer:

(b)
(a)
(d)
(c)

III.

Answer:

1. (c)
2. (d)
3. (a)
4. (b)

Question 4.
Write the answer in one word/sentences:

1. Who discovered nucleic acid for the first time?
2. How many codons are coded to 20 types of amino acid?
3. Which codon is known as the starting of codon?
4. Name the RNA which function as enzyme.
5. Who proposed the operon model of gene regulation?
6. Name the organism which contain single stranded DNA.
7. Name the process in which to make R&IA from DNA.
8. Which bond is produce when sugar and phosphoric acid combine in DNA?
9. Name the enzyme which help in transcription.
10. How many nucleotides are found in a gene?
11. Name any one termination codon.
12. Genes which are not expressed their characters.
13. Who developed the DNA finger printing technique?
14. Name the gene which is control the activity of other gene?
15. Write the name of amino acid which starts the protein synthesis.

Answer:

1. Friedrich Meischer
2. 64
3. AUG
4. Ribozyme
5. Jacob and Monod
6. ϕ x 174
7. Transcription
8. Phosphodiester
9. RNApolymerase
10. 1000
11. UAA
12. Silent gene
13. Alec Jaffreys
14. Regulator gene
15. Methionine.

Molecular Basis of Inheritance Very Short Answer Type Questions

Question 1.
Structure formed by regulation + structural + operator + promoter gene.
Answer:
Operon.

Question 2.
What are the animals that have a foreign gene deliberately inserted into their genome?
Answer:
Transgenic animals.

Question 3.
What are the group of cells or organisms which have same hereditary characters?
Answer:
Clone.

Question 4.
By which the instructions of our DNA are converted into a functional product?
Answer:
Gene expression.

Question 5.
Write the name of sugar found in RNA.
Answer:
Ribose sugar.

Question 6.
Which codon is AUG?
Answer:
Anticodon.

Question 7.
Name the enzyme which takes part in transcription.
Answer:
RNA Polymerase.

Question 8.
Who tell that DNA is a heredity material?
Answer:
Alfred Hershey and Martha Chase.

Question 9.
Which bond is made in DNA when join the sugar and phosphoric acid?
Answer:
Phosphodiester bond.

Question 10.
Name the segment in which any nucleotide sequence within a gene that is removed by RNA splicing during maturation of the final RNA products.
Answer:
Intron.

Question 11.
Who gave the operon model?
Answer:
Jacob and Monod.

Question 12.
What do you mean by commaless genetic code?
Answer:
Between two codon has no internal punctuation.

Question 13.
Write the full name of Sn RNP.
Answer:
Small nuclear Ribonucleo Proteins.

Molecular Basis of Inheritance Short Answer Type Questions

Question 1.
Group the following as nitrogenous bases and nucleosides :
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Adenine, Guanosine, Thymine, Uracil and Cytosine are nitrogenous bases. (Adenine and Guanosine → Purine, Thymine, Uracil and Cytosine → Pyrimidine) Cytidine is a nucleoside.

Question 2.
If a double stranded DNA has 20 % of cytosine, calculate the percent of adenine in the DNA.
Answer:
According to Chargaff’s rule, the DNA molecule should has an equal ratio; Cytosine = 20 % therefore, Guanine = 20%
A + T = 100 – (G – C)
A + T = 100 – 40 since, both Adenine and Thymine are in equal amounts.
Thymine = Adenine = \(\frac { 60 }{ 2 }\) = 30%
So, quantity of Adenine is 30% in DNA helix.

Question 3.
What are oncogenes?
Answer:
Genes which are responsible for production of cancer in host by uncontrolled mitotic cell division are called as oncogenes.

Question 4.
What are Okazaki fragments and leadings strands?
Answer:
Okazaki fragments:
On second parental DNA template new complementary DNA strands are formed in smaller fragments starting from RNA primer. These short fragments are called Okazaki fragments.

Leading strands:
Second strand is formed on 5’ → 3’ strand of parental DNA in a continuous stretch in reverse direction 3’ → 5’ and is called as leading strand.

Question 5.
DNA nucleotides are formed by which molecule?
Answer:
Components of DNA Nucleotides:

Question 6.
What is peptide bond?
Answer:
The bond formed between the carboxylic group (- COOH) of one amino acid and amino group (- NH₂) of another amino acid is called as peptide bond. A molecule of water is released during the formation of peptide bond.

Question 7.
Write five characters of gene hypothesis.
Answer:
Sutton, Bridges, Muller and Morgan suggest these theory. The characters of gene of this theory are as follows:

1. Genes are situated on the chromosome.
2. They make the physiological character of organisms.
3. These are called functional unit of specific characters.
4. Genes have the capacity of self-transcription.
5. They perform mutation.
6. Characters go to one generation to other by parents.

Question 8.
What is gene expression? Explain by different methods of gene expression in animals.
Answer:
The mechanism at molecular level by which a gene is able to express itself in the phenotype of an organism is called gene expression.
Different methods of gene expression in animals are:

1. Transduction:
It is the process in which bacteriophages pick up pieces of DNA from one bacterial cell and transfer the same to another on infection.

2. Transformation:
It is the process by which DNA isolated from one type of cell when introduced into another, is able to bestow some of the properties of the former to the later.

Question 9.
What is proof reading and repair of DNA?
Answer:
Variety of environmental factors such as radiations, chemicals etc. may cause damage in DNA of a cell. The bacterial DNA polymerase III can do proof reading, in the sense that it can go back and remove the wrong base before it proceeds to add new bases in the 5′ → 3′ direction. It is called proof reading. Obviously, the survival of the cell depends on its availability of damages:

1. Monoadduct:
Which involve alterations in a single nitrogenous base.

2. Diadducts :
They are the alterations involving more than one nitrogenous base. Number of nucleases have been found to be involved in repair replication such as Exonucleases (defined as phosphodiesterases which require a terminus for hydrolysis and cut’off terminal nucleotides), Endonucleases (which are also phosphodiesterases which do not require a terminus for hydrolysis and break internal bonds). The endonucleases which act on the damaged DNA and cause repair or correction of this molecule are referred to as correctional nucleases. The following steps are said to be involved in the repair replication i.e., Incision, Excision, Reinsertion and joining of newly formed strands.

Question 10.
Write any four differences between DNA and RNA.
Answer:
Differences between DNA and RNA:
DNA:

• It contains deoxyribose sugar.
• It has adenine, thymine, cytosine and guanine as nitrogenous bases.
• It consists of two polynucleotide chains coiled into a double helix.
• It is main constituent of chromosome which is found in nucleus.

RNA:

• It contains ribose sugar.
• It has adenine, uracil, guanine and cytosine as nitrogenous bases.
• It consists of single polynucleotide chains which may get folded on it self to form double helix.
• It is main constituent of ribosome and generally found in cytoplasm.

Question 11.
Write the names of enzymes used in DNA replication.
Answer:
The names of enzymes used in DNA replication are as follows:

1. DNA helicase : For unwinding of two strands.
2. DNA gyrase : For relieving tension.
3. Primase : For formation of primer.
4. DNA polymerase : For DNA synthesis.
5. RNA primer : For initiation of the synthesis of DNA segments.
6. DNA ligase : For joining of DNA Okazaki segments.

Question 12.
What is transcription? Name the enzyme catalysing it.
Answer:
Transcription:
Formation of wRNA from DNA in the presence of enzyme is called transcription. It is the first stage of protein synthesis which is catalysed by RNA polymerase enzyme. The process of transcription involves in the following steps:

1. Exposing of the bases of DNA:
The two strands of DNA are separated due to presence of an unwinding protein and thus, their bases are exposed. The exposed chain of DNA functions as template for the synthesis of oiRNA in the presence of RNA polymerase enzyme.

2. Base pairing:
The ribonucleotides are jointed in a definite fashion on the exposed strand of DNA. G is bonded with ‘C’ ‘C’ bonded with ‘G’, ‘T’ bonded with ‘A’ and ‘A’ bonded with ‘T’ respectively.

3. Synthesis of RNA chain:
The new ribonucleotide bonded on DNA template are jointed with the help of RNA polymerase and thus, forming a new chain of RNA. Then this mRNA is separated from DNA and reaches the cytoplasm where it combines with ribosomes and thus, initiating the synthesis of protein.

Question 13.
What is translation? Explain it.
Answer:
Translation:
The translation step of protein synthesis involves translation of the language of nucleic acids (available in the form of mRNA) into language of protein. The sequence of bases in wRNA, decides the sequence of amino acids in proteins. Each amino acid is programmed by a triplet code. It consists of a sequence of three bases in the DNA and the complementary bases in /wRNA. The synthesis of protein occurs in three steps, initiation, elongation and termination. After the final step i.e., termination, the proteins are transported out of the cell or translocated within the cell. Thus, the transformation of nucleotides chain of RNA into polypeptide chain of protein is called as translation.

It is completed in following steps:

1. Activation of amino acids.
2. Binding of activated amino acids with?RNA.
3. Binding of mRNA with smaller unit of ribosome.
4. Initiation of polypeptide chain.
5. Elongation of polypeptide chain.
6. Termination of polypeptide chain.

Question 14.
Describe the evidence given by Griffith in support of DNA as genetic material. Explain it along with suitable diagram.
Answer:
Griffith had done transformation experiments in mice to prove that DNA is the genetic material. He took virulent strain of Diplococcus pneumonae (S – III) which causes pneumonia in mice and injected it into mice which resulted in the production of pneumonia in mice. He also injected a non – virulent strain of that bacteria in the body of mice and found that all the mice were unaffected. In third experiment he injected heat killed (S – III) strain and non-virulent strain R – II strain together in the body of mice and found that all the mice suffered from pneumonia and became dead.

After analysis it was found that these mice contained both the strains of Diplococcus pneumonae. Thus, this experiment proved that any substance of S – III strain is transferred into R – II strain due to which R – II strain become virulent. Later, McLeod, Avery and McCarty observed that DNA molecules are transferred from S – III to R – II strain and make virulent; Thus, it is proved that DNA is the genetic material.

Question 15.
Explain the semi-conservative method of replication of DNA.
Or
Explain the method of DNA duplication.
Answer:
Synthesis of new DNA strands:
The DNA polymerase plays an important role in adding the building blocks to the primer in a sequence as influenced by the template. Replication of DNA is not continuous. It takes place by semi – conservative method. The parent DNA unwinds sequentially in local areas. A nick in one strand of the helix provides two free ends in one strand and a swivel in the other to absorb the twist that occur in the unwinding process.

When the double stranded DNA gets unwound up to a point it will represent a Y – shaped replication fork. This unwinding exposes the internal bases for replication.The enzyme DNA polymerase (discovered by Kornberg in 1957) now starts adding the nucleotides complementary to the DNA templates in the direction 5′ → 3′. Since, the two strands run in anti – parallel manner, the synthesis of new strands will be in opposite direction. As the synthesis of new strand progresses the point of divergence of the fork will be seen moving further due to unwinding of the strands of parental DNA.

Second strand is formed on 5′ → 3′ strand of parental DNA in a continuous stretch in reverse direction 3′ 5′ and is called the leading strand. However, on the second parental DNA template, the new complementary DNA strands are formed in smaller fragments starting from the RNA primer.

These short fragments are called Okazaki fragments after the name of scientist who discovered them. Since these fragments are joined later to form the complete strand it is called the lagging strand. The enzyme which joins the Okazaki fragments with the help of the RNA primer to form the lagging strand is called polynucleotide ligase or the joining enzyme.

Question 16.
Give the functions of nucleotides.
Answer:
Functions of Nucleotides:

1. It works as a activated precursors of DNA and RNA.
2. They are perform the storage and conduction of energy to the form of ATR
3. It required for activation of intermediates in many biosynthetic pathway.
4. It works as Carrier of methyl group in the form of SAM.
5. It components of co – enzyme : NAD, FAD and Co A.
6. Some functions are as a vitamin.
7. They are control and coordinates different activities in our body.

Question 17.
Write four features of genetic code.
Answer:
According to Nirenberg, Khorana and Holley, genetic code is that sequence of nitrogenous bases of DNA in which genetic informations for the synthesis of protein are coded.

Characteristic features of genetic code:

1. The code is triplet:
The codon is a specific sequence of three nitrogenous bases of mRNA.

2. The code is commaless:
The sequence of bases read in blocks of three at a time form a particular position. There is no gap between two subsequent codons.

3. Code is degenerating:
Presence of more than one codon for one amino acid is called as degeneracy of codons, example Serine having three codons UCU, UCA, AGU.

4. Codes are universal:
Codons are similar in all organisms, example serine is coded by UCU codon in all the living beings.

5. Codes are non – ambiguous:
The position of genetic code in cellular medium is nonambiguous because a codon always codes only one amino acid. Sometimes a codor codes more than one amino acid, example in E. coli. UUV codon generally code phenylalanine, after treatment of their ribosome with streptomycin. It can also code isoleucine, leucine and serine.

6. Initiation and termination codon:
Codons responsible for the initiation of polypeptide chain are called as initiation codon, example AUG. Likewise codons responsible for the termination of polypeptide chain are called as chain termination codon, example UAA, U AG, UGA.

Question 18.
Define Codon and Anticodon.
Answer:
Codon:
A specific sequence of three consecutive nucleotides that is a part of the genetic code and that specifies a paticular amino acid in a protein or starts or stops protein synthesis example AUG codon which is situated on the wRNA, code methionine amino acid.

Anticodon:
A sequence of three adjacent nucleotides located on one end of transfer RNA. It bounds to the complementary coding triplet of nucleotides in wRNA during translation phase of protein synthesis. For example, the anticodon for Glycine is ccc that binds to the codon (which is GGE) ofwRNA.

Question 19.
Explain DNA duplication in short.
Answer:
Watson and Crick after giving the double helix model of DNA, also postulated the mechanism of DNA duplication, also known as replication. According to them, during duplication, the weak hydrogen bonds between the nitrogenous base of the nucleotides get separated, so that two polynucleotide chains of DNA also separate and uncoil. The chains thus, separated are complementary to one another. These strands act as template and because of the specificity of base pairing each nucleotide of separated chain attracts its complementary nucleotide from the cell cytoplasm.

Once the nucleotides are attached by their hydrogen bonds their sugar radicals write through their phosphate components completing the formation of a new polynucleotide chain. This results in the formation of two double helixes of DNA where each molecule has one old strand contributed by parent DNA and one synthesized new. This method of DNA duplication is known as semi-conservative method.

Question 20.
Describe the functions of nucleic acids.
Or
Explain the utility of nucleic acids.
Answer:
Utility of Nucleic acids:

1. Nucleic acids are the hereditary materials of organisms which involve in the transfer of hereditary characters from one generation to the next.
2. DNA controls the synthesis of enzymes which control the various activities of the body.
3. Nucleic acids also control protein synthesis.
4. Nucleic acids form maximum portion of chromatin network.
5. It causes mutation in living beings.
6. They form enzymes.

Question 21.
Explain the structure of RNA.
Answer:
RNA molecules are single stranded nucleic acids composed of nucleotides. Four types of bases are present in RNA. These nitrogenous bases joint in different manner and form the ribonucleoside. Ribonucleoside joins together and make a polyribonucleotide chain.
All four types of nucleoside and nucleotide are as follows:

Question 22.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acid synthesized from it (DNA or RNA) list the types of nucleic acid polymerases.
Answer:
These are two different types of nucleic acid polymerases:

1. DNA – dependent DNA polymerases
2. DNA – dependent RNA polymerases

The DNA dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA, whereas DNA dependent RNA polymerases use a DNA template strand for synthesizing RNA.

Question 23.
List two essential roles of ribosome during translation.
Answer:
Two essential roles of ribosome during translation are:

1. One of the RNA acts as a peptidyl transferase ribozyme for formation of peptide bonds.
2. Ribosome provides sites for attachment of TMRNA and charged fRNA for polypetide synthesis.

Molecular Basis of Inheritance Long Answer Type Questions

Question 1.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Hershey and Chase experiment:

1. They grew some bacteriophages on a medium that contained radioactive phosphorus and some in another medium that contained radioactive sulphur.

2. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein as phosphorus is present only in DNA.

3. Viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

4. It was found that bacteria which were infected with bacteriophages that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.

5. Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicated that proteins did not enter the bacteria from the viruses.

6. This was a clear cut proof that DNA is the genetic material that is passed from virus to bacteria.

Question 2.
Differentiate between the followings:

1. Repetitive DNA and Satellite DNA.
2. OTRNA and IRNA.
3. Template strand and Coding strand.

Answer:
1. Differences between Repetitive DNA and Satellite DNA :

Repetitive DNA :

• DNA in which certain base sequences are repeated many times are called repetitive DNA.
• Repetitive DNA sequences are transcribed.

Satellite DNA:

• DNA in which large protein of the gene is randomly repeated is called satellite DNA.
• Satellite DNA sequences are not transcribed.

2. Differences between mRNA and tRNA:

mRNA:

• It is linear.
• It carries coded information.
• mRNA undergoes additional processing, i.e., capping and tailing splicing
• Nitrogen bases are unmodified.

tRNA:

• It is clover – leaf shaped.
• It carries information for association with an amino acid and a anticodon for its in corporation in a polypeptide.
• It does not require any processing.
• Nitrogen bases may be modified.

3. Differences between Template strand and Coding strand:

• Template strand:
• It is the strand of DNA which takes part in transcription.
• The polarity is 3 ’ → 5′
• Nucleotide sequence is complementary to the one present in mRNA.

Coding strand:

• It is the stand that does not take part in transcription.
• The polarity is 5’ → 3’.
• The nucleotide sequence is same as the one present in mRNA except for presence of Thymine instead of Uracil.

Question 3.
Explain (in one or two lines) the function of the followings:

1. Promoter
2. tRNA
3. Exons.

Answer:
1. Promoter is an essential component of the transcription unit. It is located at the beginning of 5’-end. It provides a site for the attachment of transcription factors and RNA polymerase.

2. tRNA is a small sized RNA molecule that takes part in transcription. It physically picks up activated amino acids from the cytoplasm and carries (transfers) them to ribosomes, where they join together through peptide bonds and leave the /RNA to fetch more amino acids.

3. Exons are the coding sequences of DNA that are transcribed and translated.

Question 4.
Why is the human genome project called a mega project?
Answer:
Human genome project is called a mega project because:

1. Its aim was to determine the nucleotide sequence of complete human genome which was a task of enormous magnitude.
2. A total of 3 x 10⁹ base pairs were to be sequenced and the cost was about 9 billion US dollars.
3. It requires bioinformatics data base techniques and other contemporary devices for the analysis, storage and retrieval of information.
4. Many countries worked jointly to complete this timed project.

Question 5.
What is DNA Fingerprinting? Mention its application.
Answer:
DNA Fingerprinting:
Every human individual is characterised by unique print at the fingertips. The study of fingers, palm and sole print is called ‘dermatoglyphics’. Like prints of the fingertips, each individual has unique DNA fingerprint. Unlike the prints of finger, the DNA fingerprints can not be altered by surgery. The later is exactly similar in all the cells and tissues of an individual. It can not be changed by medical treatment.

The distinction of individuals on the basis of DNA fingerprint is due to sequence of nucleotides in whole genomic DNA. The technique to identify a person on the basis of his/her DNA specificity is called DNA fingerprinting. This was invented by Sir Alec Jeffreys in 1984 at Leicester University, U.K. In India, Dr. V. K. Kashyap and Dr. Lalji Singh started this technique at CCMB, Hyderabad.

DNA fingerprinting involves following steps:

1. The DNA of the organism to be tested is isolated, it is called host DNA
2. Host DNA is cleaved with the help of specific restriction enzymes into several fragments.
3. Double stranded DNA fragments are denatured to produce single stranded DNA by alkali treatment.
4. DNA segments are separated by electrophoresis.

Question 6.
Briefly describe the following:

1. Transcription
2. Polymorphism
3. Translation
4. Bioinformatics.

Answer:
1. Transcription:
It is the formation of RNA over the template of DNA. It forms single-stranded RNA which has a coded information similar to the sense or coding strand of DNA with the exception that thymine is replaced by uracil. One strand of DNA is used as template strand for the synthesis of a complementary strand of RNA called mRNA.

2. Polymorphism:
Genetic polymorphism means occurrence of genetic material in more than one form. It is of three major types, i.e., Allelic, SNP and RFLP.

(a) Allelic polymorphism:
Allelic polymorphism occurs due to multiple alleles of a gene. Allele possess different mutations which alter the structure and function of a protein formed by them as a result, change in phenotype may occur.

(b) SNP or Single Nucleotide Polymorphism:
Over 1 – 4 million single base DNA differences have been observed in human beings. According to SNP, every human being is unique. SNP is very useful for locating alleles, identifying disease associated sequence and tracing human history.

3. Translation:
The translation step of protein synthesis involves translation of the language of nucleic acids (available in the form of mRNA) into language of protein. The sequence of bases in mRNA, decides the sequence of amino acids in proteins. Each amino acid is programmed by a triplet code. It consists of a sequence of three bases in the DNA and the complementary bases in mRNA. The synthesis of protein occurs in three steps, initiation, elongation and termination. After the final step i.e., termination, the proteins are transported out of the cell or translocated within the cell. Thus, the transformation of nucleotides chain of RNA into polypeptide chain of protein is called as translation.

It is completed in following steps:

• Activation of amino acids.
• Binding of activated amino acids with tRNA.
• Binding of mRNA with smaller unit of ribosome.
• Initiation of polypeptide chain.
• Elongation of polypeptide chain.
• Termination of polypeptide chain.

4. Bioinformatics:
The science which deals with handling storing of huge information of genomics as databases, analysing, modelling and providing various aspects of biological information, especially the molecules connected with genomics and proteomics is called bioinformatics.

Question 7.
Explain the Watson and Crick model of DNA.
Answer:
The structure of DNA was proposed by Watson and Crick. It is twisted ladder like structure. It has got two coiled polynucleotides which are joined together by nitrogen bases with hydrogen bond in the centre. The longitudinal strands of DNA are made of sugars and phosphates of nucleotides. The horizontally placed nitrogen bases are of two types, purine and pyrimidine. Purines are adenine and guanine whereas pyrimidines are cytosine and thymine.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Biology Important Questions Chapter 7 Evolution

Evolution Important Questions

Evolution Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
Atmosphere of earth just before the origin of life consisted of:
(a) Water vapour, CH₄, NH₃and O₂
(b) CO₂, NH₃and CH₄
(c) CH₄, NH₃, H₂ and water vapour
(d) CH₄ O₃, O₂ and water vapour.
Answer:
(a) Water vapour, CH₄, NH₃and O₂

Question 2.
Primitive atmosphere is made up of mixture of these gases :
(a) CH₄
(b) NH₃
(C) Water vapour
(d) None of these.
Answer:
(d) None of these.

Question 3.
What is the meaning of Abiogenesis :
(a) Life is originated from non – living things
(b) Life is originated from living things
(c) Origin of virus and microorganisms
(d) None of these.
Answer:
(a) Life is originated from non – living things

Question 4.
Primitive atmosphere of earth did not posses :
(a) CH₄
(b) NH₃
(C) H₂
(d) O₂
Answer:
(a) CH₄

Question 5.
Which of the following was formed in Miller’s experiment:
(a) Microspheres
(b) Nucleic acid
(c) Amino acid
(d) UV – radiation.
Answer:
(d) UV – radiation.

Question 6.
What is the unit of evolution :
(a) Community
(b) Genus
(c) Order
(d) Species.
Answer:
(a) Community

Question 7.
Whith of the following has not life :
(a) Azoic era
(b) Mesozoic era
(c) Paleozoic era
(d) Cenozoic era.
Answer:
(a) Azoic era

Question 8.
Which of the following are not an analogous organ :
(a) Wings of birds and butterfly
(b) Eye of octopus and mammals
(c) Thom of Baugainvilla and tendril of Cucurbita
(d) Tuberous roots of sweet potato and tuber of potato.
Answer:
(d) Tuberous roots of sweet potato and tuber of potato.

Question 9.
The earliest era is :
(a) Cenozoic
(b) Mesozoic
(c) Paleozoic
(d) Precambrian.
Answer:
(c) Paleozoic

Question 10.
Which of the following is analogous organ :
(a) Which has structural symmetry
(b) Which has functional and structural symmetry
(c) Similar in functional structure
(d) Mostly inactive.
Answer:
(c) Similar in functional structure

Question 11.
Who gave the exact proof of evolution :
(a) Fossils
(b) Vestigeal organ
(c) Embryo
(d) Morphology.
Answer:
(a) Fossils

Question 12.
Which era is called of mammals and angiosperm’s era :
(a) Mesozoic
(b) Cenozoic
(c) Paleozoic
(d) Archaeozoic.
Answer:
(b) Cenozoic

Question 13.
Whose dental structure is similar to human :
(a) Homo erectus erectus
(b) Homo erectus pekinensis
(c) Homo habillis
(d) Homo sapiens neanderthalensis.
Answer:
(c) Homo habillis

Question 14.
Which of these cells are immortal:
(a) Germ cells
(b) Liver cells
(c) Kidney cells
(d) Neurons.
Answer:
(a) Germ cells

Question 2.
Fill in the blanks:

1. Earth is a member of ……………….
2. The great explosion that resulted into the formation of universe is ……………….
3. ‘Philosophic Zoologique’ is written by ……………….
4. ………………. is a link between reptiles and birds.
5. ………………. is the ancient ancestor of modem horses.
6. At present marsupiales (kangaroos) are found only in ……………….
7. ………………. is the first photosynthetic organism that originate on the earth.
8. ………………. is the process by which changes in the genetic composition of organism.
9. ………………. constructed geological time scale.
10. ………………. is the first prehistoric human.
11. ………………. was among the first fossils to be recognized as belonging to Homo sapiens.
12. ………………. is found since the 19th century in the Siwalik Hills of the Indian sub – continent.
13. Discoveries related to the genetic code and its function in protein synthesis is given by ……………….
14. ………………. was the father of the Paleontology.
15. ………………. is the name of Darwin’s ship.

Answer:

1. Solar system
2. Big bang
3. Lamarck
4. Archaeopteryx
5. Eohippus
6. Australia
7. Cyanobacteria
8. Organic evolution
9. Geovani Abduna
10. Homo habilis
11. Cro – magnon
12. Ramapithecus
13. Dr. Hargovind khorana
14. Leonardo da Vinci
15. Beagle.

Question 3.
Match the followings :
I.

Answer:

1. (d)
2. (e)
3. (b)
4. (c)
5. (a)

II.

Answer:

1. (d)
2. (a)
3. (e)
4. (c)
5. (b)

Question 4.
Write the answer in one word/sentences:

1. The matter which is a link between acellular and cellular system.
2. The island at which Darwin studies the organisms for his theory of evolution.
3. When the earth is originate?
4. Which is known as the link between living and non – living?
5. Where are originate the first animal on the earth?
6. Name the scientist who disapproved Lamarkism.
7. The whole process of the development at changes from embryo to adult organism.
8. Other name of natural selection.
9. Who proposed the Mutaion Theory of Evolution?
10. The organs which are different in their origin but have similar function.
11. Who proposed the theory of inheritance of acquired characters?
12. Who proposed the theory of natural selection ?
13. The remains of ancient organisms.

Answer:

1. Coacervates
2. Galapagos
3. 4.6 million years ago
4. Virus
5. In water
6. Weismann
7. Ontogeny
8. Darwinism
9. Hugo de Vries
10. Homologous organ
11. Lamarck
12. Darwin
13. Fossils.

Evolution Very Short Answer Type Questions

Question 1.
Who proposed the theory of natural selection?
Answer:
Darwin.

Question 2.
Who proposed the Recapitulation theory or Biogenetic law?
Answer:
Haeckel.

Question 3.
Astronomical distance measured in?
Answer:
Astronomical distance measured in Light years.

Question 4.
Name the theory by which earth is said to originate.
Answer;
The big – bang theory.

Question 5.
What is fossil?
Answer:
Fossils are the remains or impressions of ancient organisms preserved in sedimentary rocks or other media.

Question 6.
What is mutation?
Answer:
New species originate due to changes of hereditary characters are called mutation.

Question 7.
Name the scientist who tell the spontaneous theory is wrong.
Answer:
Louis Pasteurs.

Question 8.
Which era is called golden period of Dinosaurs?
Answer:
Mesozoic period is billed golden period of Dinosaurs.

Question 9.
In which ship Darwin studied the nature?
Answer:
Beagle.

Question 10.
Name any two vertibrates body organ which are homologus organs of human forelimb.
Answer:

1. Flipper of whale
2. Wing of birds.

Question 11.
What is the scientific name of modern man?
Answer:
Homo sapiens.

Question 12.
Who is the early man of the modern human?
Answer:
Cro – Magnon peoples are early human of modem human.

Question 13.
Who are the early human and sub – human.
Answer:
Ramapithecus is early human and Australopithecus is early human.

Question 14.
Which human form first started to walk on two legs?
Answer:
Australopithecus form first started to walk on two legs.

Question 15.
Which type of human was ‘Cro – Magnon’ on the basis of food intake?
Answer:
Cro – Magnon was carnivorous.

Question 16.
On the basis of evolution which human had brain size of 1400cc?
Answer:
Neanderthal.

Question 17.
Give the name of ape-like ancestors of humans.
Answer:
Ape – like ancestors of human’s are Dryopithecus.

Question 18.
Differenciate between Dryopithecus and Ramapithecus.
Answer:
Dryopithecus were apelike but Ramapithecus were mostly human like.

Evolution Short Answer Type Questions

Question 1.
Explain antibiotic resistance observed in bacteria in light of Darwinism selection theory.
Answer:
Darwinism theory of natural selection states that environment selects organisms with favourable variations and these organisms thus, survive and reproduce. It is observed when bacterial populations are exposed to certain antibiotic, the sensitive bacteria could not tolerate and hence, died due to the adverse environment. Whereas some bacteria that devel¬oped mutation became resistant to the particular antibiotic and survived.

As a result such resistant bacteria survive and multiply quickly as compared to other sensitive bacteria. So, the whole population is regained by multiplication of resistant variety and antibiotic resistant gene becomes widespread in the bacterial population.

Question 2.
Find out from newspapers and popular science articles any new fossil dis – coveries or controversies about evolution.
Answer:
Fossils of dinosaurs have revealed the evolution of reptiles in Jurassic period. As a result of this evolution of other animals such as birds and mammals has also been discovered. However, two unusual fossils recently unearthed in China have ignited a controversy over the evolution of birds confuciusomis is one such genus of primitive birds that were crow sized and lived during the Cretaceous period in China.

Question 3.
Attempt giving a clear definition of the term species.
Answer:
A species generally includes similar organism. Members of this group can show interbreeding. Similar group of genes are found in the members of same species and this group has capacity to produce new species. Every species has some cause of isolation which intruped the interbreeding with nearest reactional species which is refer as reproduptively isolated.

Question 4.
What is virus? Why is it treated as a link between living and non – iiving ?
Answer:
Viruses are simplest organism,of the earth, which consist of nucleic acid (DNA or RNA) surrounded by protein cover. It shows characteristics of living as well as non – living organisms.

(A) Living characters of virus:

1. Virus shows structural differentiations.
2. They contain hereditary material.
3. They exhibit mutation.
4. They spread plant and animal diseases.
5. Growth and development present.
6. They exhibit adaptaion.
7. They possess sensitivity.

(B) Non – living characters of virus:

1. Lack protoplasm and cell organelles.
2. Can be crystallized.
3. No metabolic activities seen.
4. Cannot reproduce outside living cells.
5. They lack enzymes.
6. Due to above reason, viruses are considered as link between living and non-living organisms, thus, it is the first life
7. originated in the earth.

Question 5.
What is oxygen revolution? Explain.
Answer:
Oxygen revolution:
Evolution of O₂ in photosynthesis during primitive environmental conditions is very important because, it is required in the evolution of organism and conversion of reducing environment into oxidizing environment hence, it is called oxygen revolution. Oxygen evolution should cause the following changes in the environment:

1. Oxygen evolution should cause the conversion of reducing envronment into oxidizing environment.
2. Ozone layer is formed 15 miles above from the earth which absorbs the ultraviolet light of the sunlight and thus, prevents
3. the entry of uv light in the atmosphere.
4. O₂ present in the environment dissociates methane (CH₄) into CO₂ and O₂. This CO₂ is used in photosynthesis.
5. NH₃ of the primitive environment is dissociated into H₂O and nitrogen.
   CH₄ + 2O₂ → CO₂ + 2H₂O
   4NH₃ + 3O₂ → 2N₂ + 6H₂O.

Question 6.
Explain the origin of the earth.
Answer:
Origin of the earth:

1. Earth was formed 4 – 5 billion years back.
2. Initially, the surface was covered with water vapour, methane, CO₂ and NH₃.
3. The UV rays of the sun broke water into hydrogen and oxygen.
4. Hydrogen escaped and oxygen combined with NH₃ and CH₄ to form water, CO₂ and other gases, also forming the ozone layer.
5. Cooling of water vapour led to rain which filled the depressions on earth’s surface, forming water bodies.

Question 7.
Mention the names of discoveries and principles given by the following scientists:

1. Louis Pasteur
2. A. I. Oparin
3. Urey and Miller
4. Francesco Redi
5. Faux.

Answer:

1. Louis Pasteur – He proved that air contains spores of microorganism and Biogenesis theory was supported by him.
2. A. I. Oparin – He presented the biochemical explanation of origin of life in his book “The Origin of Life on Earth”.
3. Urey and Miller – They supported the evidence of Oparin – Haldane theory of Origin of life.
4. Francesco Redi – He by conducting experiments proved that abiogenesis cannot exist but bioginesis theory can exist i.e., Life arises from pre – existing life.
5. Faux – He has been experimentally supporting the organic substances as described by Oparin.

Question 8.
What are homologous organs?
Or
What is homology?
Answer:
Organs which are similar in structure and origin but different in appearance and functions are called homologous organs and the phenomenon is called homology.

Examples:
Forelimbs of bat, wings of bat, hands of man, forelimbs of horse. These are the examples of homologous organs because, they are made up of similar bones, humerus, radius – ulna, carpals, metacarpals and fingers.

Question 9.
What do you mean by analogous?
Answer:
Analogous organs:
Organs which are different in origin and structure but performing similar functions are known as analogous organs and the phenomenon is called as analogy. Analogous organs do not indicate phylogeny.

Examples:
Wings of butterflies are made up of chitin, wings of birds made by produc¬tion of feathers on forelimbs and skin present between the fingers of bat are the examples of analogous orgAnswer:

Question 10.
What is the difference between homologous and analogous organs ? Give two examples of each of them.
Answer:
Differences between Homologous and Analogous organs :

Homologous organ:

• Organs which are similar in structure and origin but different in function are known as homologous organs.
• These structures indicate that organisms bearing them are evolutionary related.
• These organs are differ in outlook.
• They exhibit similarity in internal structure.
  Example: Forelimbs of frog, hand of human, wings of bat and forelimb of horse.

Analogous organ:

• Organs which are differing in structure and origin but similar in function are known as analogous organs.
• These structures indicate that organisms bearing them are evolutionary different.
• These organs are similar in outlook.
• They do not exhibit similarity in internal structure.
  Example: Wings of butterflies, bats and birds.

Question 11.
What do you mean by vestigial organs?
Or
Write the two names of vestigial organs of man.
Or
What are vestigial organs? Explain. Write four vestigial organs of the human body.
Answer:
Vestigial organs:
Organs that are reduced and have become functionless in an organism but were functional in their ancestors are called as vestigial organs.
Examples:

1. Vermiform appendix.
2. Coccygial vertebrae.
3. Nictitating membrane in the eyes of human.
4. Muscles of external ear (Pinna).

Question 12.
Write down the demerits of Darwinism.
Answer:
Some of the demerits of Darwinism are:

1. Darwinism stresses upon small fluctuating variations which has no role in evolution.
2. Does not satisfactorily explain effect of use and disused and presence of vestigial organs.
3. It did not differentiate somatic and germinal variations.
4. It explains survival of the fittest but not arrival of the fittest.

Question 13.
Differentiate between lamarckism and darwinism.
Answer:
Differences between Lamarckism and Darwinism:

Lamarckism:

• It is based on the theory of inheritance of acquired characters.
• According to this theory the length of neck of Giraffe increased only to facilitate it in eating leaves of higher trees.
• Increase in length of neck of Giraffe resulted in evolution of Giraffe with long necks.
• The theory is based on the use or misuse of the organs.

Darwinism:

• It is based on the theory of natural selection.
• Ancestors of Giraffe were both with long and short necks.
• Giraffe with long neck only could survive in struggle for existence.
• This theory is based on the inheritance of the characters in next generation.

Question 14.
Explain Lamarckism in short.
Or
Explain Lamarckism of organic evolution in brief.
Answer:
Lamarckism:
In 1809, Lamarck has proposed a theory to explain organic evolution, which is known, as the “Theory of inheritance of acquired characters.” According to Lamarck, organisms acquire certain characters during their lifetime due to changes in environment and these acquired characters are heritable. According to this theory, new species are originated as follows:

1. New requirements and wills are produced in the organisms due to the effect of changed environment.
2. New requirements and wills of organisms resulting in the production of new habits.
3. Changes in the habit bringing about modifications of the organ.
4. New habits resulting in the use or disuse of the organs.
5. Use of organ resulting in the development of acquired characters.
6. These acquired characters are heritable.
7. Inheritance of acquired characters resulting in the development of new species.

Question 15.
Write two incidences which influenced Darwin to set out theory of evolution.
Answer:
Two incidences which influenced Darwin are:

1. During the voyage of Beagle Darwin noticed distribution of a kind of bird called as finches and other organisms by looking at his observation of Galapagos island. He found variation in the beaks of finches adapted according to the environment.

2. Pegion keepers kills weak varieties of pegion generation after generation to obtain better variety of pegion.

3. Above two incidences influenced Darwin to set out theory of evolution (Natural selection).

Evolution Long Answer Type Questions

Question 1.
Draw a well – labelled diagram of Miller and Urey’s experiment.
Answer:

Experimental evidence of Chemical evolution or Miller’s experiment:

1. Experiment was performed by S.L. Miller and H.C. Urey in 1953.

2. Experimental set – up:
In a closed flask containing CH₄, H₂, NH₃and water vapour at 800°C, electric discharge was created. The conditions were similar to those in primitive atmosphere.

3. Observations:
After a week, they observed presence of amino acids and complex molecules like sugars, nitrogen bases, pigments and fats in the flask.

4. Conclusions:

• It provides experimental evidence for the theory of chemical origin.
• It showed that the first non-cellular form of life was created about 3 billion years ago.
• It showed that non-cellular biomolecules exist in the form of DNA, RNA, polysaccharides and protein.

Question 2.
Write an essay on modern concepts of origin of life.
Or
Explain the role of non – living in origin of life.
Answer:
Modern concept of origin of life:
The modem concept of origin of life was postulated by a Russian biochemist A.I. Oparin in 1936. According to this theory, after the formation of earth various chemicals played important role in the formation of atmosphere. Life originated and first organism came into existence from certain molecules when atmospheric conditions became suitable. According to Oparin, life originated in the following steps:

1. Formation of earth and its atmosphere:
Earth is believed to be originated some 4,500 million years ago by the condensation and cooling of the clouds of cosmic dust and gases called ylem. The heavier elements collected at the core and lighter elements around the core. Outermost layer contains H, C, O and N. Oxygen was found only in combination of other elements. These four elements reacted with each other forming H₂, H₂O, CH₄, NH₃, CO₂ and HCN.

2. Formation of small organic molecules:
The mixture of methane, ammonia, water and hydrogen comes in contact of solar energy. Cosmic rays and electric discharge could produce some simple organic compounds. These simple organic compounds formed in such a way and accumulated in primitive atmosphere and oceans were responsible for synthesis of complex micro-molecules as follows :

1. CH₄, H₂O → Sugars
→ Fatty acids
→ Glycerine

2. CH₄, H₂O, NH₃ → Amino acids

3. CH₄, HCN, H₂O, NH₃ → Nitrogenous bases (Purines and Pyrimidines)

3. Formation of polymers:
It is clearly understood from the above description that a large number of micro molecules such as hydrocarbons, amino acids, fatty acids, purine and pyrimidines and simple sugars accumulated in the oceans. When atmospheric water condensed on further cooling, the inorganic precursors collided, reacted and aggregated to form new molecules of increasing size and complexity. Thus, by polymerization macromolecules were formed. The chemical reactions for the formation of macro – molecules can be summarized as follows:

• Sugar + Sugar → Polysaccharides
• Fatty acid + Glycerine → Lipids
• Amino acid + Amino acid → Protein
• Nitrogenous base (Adenine) + Sugar + Phosphate → Adenosine phosphate
• Nitrogenous base + Sugar → Nucleoside
• Nucleoside + Phosphate → Nucleotide
• Nucleotide + Nucleotide → Nucleic acid.

4. Formation of molecular aggregates and primitive cells:
Over a vast of time, these molecules became associated with one temporary complex. Ultimately, it leads to the formation of a coacervate. A coacervate is a solution of high molecular weight of chemicals, i.e., proteins and carbohydrates, which become bounded by lipid membrane, which is selectively permeable.

The coacervate grows by absorbing molecules from their environment. The substances which got accumulated in the coacervates underwent reactions and resulted in the molecular reorganization of some proteins into enzymes. A coacervate having nucleoprotein surrounded by various nutritive organic substances and covered by surface membrane is considered to be the precell, which got later transformed into first living cell. The coacervate can reproduce by budding.

5. Evolution of complex biochemical reactions:
Primitive organism utilize chemical substances present in the environment as food hence, they are :

(a) Heterotrophic, chemosynthetic organisms appeared due to mutation and natural selection in heterotrophs.

(b) Blue – green algae evolved from chemosynthetic organisms by mutation and natural selection.

(c) The liberation of free oxygen into the atmosphere produced by the blue – green algae due to the process of photosynthesis. It finally changed the reducing atmosphere into an oxidizing one and therefore, all possibilities of further chemical evolution were finished.

Free living eukaryotes originated in the ocean from blue – green algae.

6. The origin of well – developed organisms:
From the simple eukaryotes which were like unicellular organisms of today various forms of life evolved during passage of time.

Question 3.
Write the process of formation of organic molecules in sea water on earth with the help of Miller and Urey’s experiment.
Answer:
The work of A.I. Oparin (1938 – 1965), H.Urey and Stanley Miller (1959) provided evidences in the favour of biochemical origin of life. They had prepared the atmosphere like that of primitive earth and as described by Oparin, they made the synthesis of organic compounds by the following methods:

1. Four elements H, C, O (not free O₂) and N react with each other to form H₂O, CH₄, NH₃, CO₂ and HCN on primitive earth.

2. From these four elements following organic molecules were formed in sea water of earth:
(i)

(ii) CH₄, H₂O, NH₃ → Amino acids.

(iii) CH₄, HCN, H₂O, NH₃ → Nitrogenous bases.

3. Macro – molecules of organic compounds were synthesized by these above prepared organic compounds.

• Sugar + Sugar → Polysaccharides (Carbohydrate).
• Fatty acids + Glycerol → Fats.
• Amino acid + Amino acid → Protein.
• Nitrogenous base + Sugar → Nucleoside.
• Nucleoside + Phosphoric acid → Nucleotide.
• Nucleotide + Nucleotide → Nucleic acid.
• Nitrogenous base (Adenine) + Sugar + Phosphate → Adenosine phosphate.

4. The above organic compounds and salts together constituted the first living being.

Question 4.
Name connecting link of reptiles and birds. Also write their characters.
Answer:
Archaeopteryx is the connecting link of birds and reptiles. Archaeopteryx was a bird. It is regarded as the connecting link between reptiles and birds, which suggests the path of evolution of the latter from the former. It is found as fossils. They are found during Jurassic period 140 million years. Archaeopteryx exhibits both reptiles and birds like characters.

1. Reptiles like characters:

• Bones were similar to that of reptiles in which air sacs were absent.
• Tail bearing vertebra.
• It had teeth in jaws, scales were present on the body.
• Metacarpals were free.
• Pelvic girdle recombines with the pelvic girdle of reptiles.

2. Birds like characters:

• Presence of feathers on body.
• Forelimbs were modified in wings.
• Skull large and monocondylar.
• Jaws were modified into beak.
• Hallux was backwarded and pointed.

Question 5.
Give a detailed account of theory of natural selection.
Or
Describe the Darwin’s opinion about the origin of new species of organisms.
Answer:
Charles Darwin (1809 – 1882) explained the theory of evolution in his book “Origin of species by natural selection.” Darwin undertook a long voyage for five years in the capacity of a naturalist on a British warship ‘Beagle’. He travelled to islands of Galapagos and collected evidences to explain evolution.

To explain origin of species he gave theory of natural selection as a mechanism for evolution.

The main points of Darwinism are given below:

1. Over production of offsprings:
Every living being has an inherent tendency to produce more offspring than that can survive.

2. Struggle for existence:
Though the offsprings are produced in large number yet their production remain almost constant. This is because of struggle for existence. There is struggle for food, space, breeding, etc. Moreover death of individuals due to diseases and predators, population are kept under control.

3. Survival of the fittest:
Darwin believed that any individual is successful in struggle for existence if it survives long enough to produces offsprings. Individuals who are fit in a particular environment can only survive.

4. Variation:
Due to constant struggle, the organisms change themselves in accordance with the new needs.

5. Natural selection:
In the struggle, for existence organisms having variations favourable to the environment would have more chance to survive and reproduce their own kind. Those which do not possess favourable variations would die or fail to reproduce.

6. Origin of species:
Any changes in environmental conditions cause natural selection to act upon the population and select the well adapted individuals. It results in changes of characters of the populations. By the inheritance of these changes in successive generations, new species are formed.

Question 6.
Organic Evolution is a continuous process, explain it. And giving any three evidences.
Answer:
Evolution is a complex phenomenon accounting for the present day diversity among organisms. But it has clearly maintained the basic unity among them since it occurred over a period of millions of years, no one would have seen/recorded evolution and hence scientists have provided various evidences to prove evolution. Some of the evidences of organic evolution are described below:

I. Evidences from Embryology:
Important activities that occur various animals are:

1. For the survival, all animals get energy and various substances from environment.
2. In all organisms energy is produced from ATP.
3. In all organisms, the duplication of DNA is similar.
4. In all organisms, protein synthesis is same and it is produced from ribosomes.
5. In all organisms respiration and steps of respiration is same.
6. All organisms multiplicate and reproduce, due to which they have basic similarities.
7. All organisms conduct hereditary characters on similar principles.

II. Evidence from Anatomy:
Anatomy of living organisms will be explained with different examples:

1. Homologous organs : Organs which are similar in structure and origin but different in function and appearance is known as homologous organs.

2. Analogous organs : Organs which are different in origin and structure but performing similar functions are known as analogous organs.

3. Vestigial organs : Organs that are reduced and have become functionless in an organism, but were functional in their ancestors are called vestigial organ.

III. Evidences from vestigial organs:
Organs of the body which are non – functional but they are functional in some other organisms are called vestigial organs: Morphological evidence of evolution is provided by the presence of vestigial organs of body which are often undesired, degenerated and non-functional. These might have been large and functional in some other animals or in ancestors of those which now possess it in rudimentary forms, example vermiform appendix in man, muscles of external ear (pinna) in man, nictitating membrane or plica, semilunaris in human eye, wisdom teeth (third pair of molars) tail bone (coccyx) in man, wings of ostrich, hindlimbs in snakes, etc.

Question 7.
What do you mean by organic evolution? How do fossils exhibit evidence to prove organic evolution?
Answer:
Descent with modification in organism is known as organic evolution.

Evidences of organic evolution from fossils record:
Fossils are treated as significant evidence of organic evolution. Fossils are the remains or impressions of ancient organisms preserved in the layers of rock and soil. Fossils only do not prove the theory of organic evolution, yet it evidently prove that gradually complexity increased in body organization. The complexity in the body of organization can be noticed as we study the upper layers. Thus, it can be concluded from above observations :

1. The crust of the earth and the organisms living on it underwent change in the course of time.
2. The organisms with simple structural organization originated earlier than the complex ones.
3. Some of the organisms lived on the earth for short time and became extinct. This was a result of drastic changes in the climate on the earth.

Hence, forth fossils produce bonafide record of such plants and animals which had shown their existence once upon a time and now are extinct or not present exactly in the same form, thus, producing strong evidence in favour of organic evolution.

Question 8.
Try to trace the various components of human evolution (Hint: Brain size and function, skeletal structure, dietary preference, etc.)
Answer:
The various components of human evolution are as follows:

1. Brain capacity.
2. Posture.
3. Food/Dietary preference and other important features.

Name, brain capacity, posture and food features:

1. Dryopithecus Africans:
Knuckle walker, walked similar to Gorillas and Chimpanzees (was more apelike), soft fruit and leaves; canines large, arm and legs are of equal size.

2. Ramapithecus:
Semi – erect (more manlike) ate seeds and nuts, canines were small while molars were large.

3. Australopithecus africanus:
Brain capacity 450 cm³, full erect prosture, height 1.05m, herbivorous (ate fruits), hunted with stone weapons, lived as trees, canines and incisors were small.

4. Homo habilus:
Brain capacity 735 cm³, fully erect posture, height, 1.5m, carnivorous, canines were small. They were first tool makers.

5. Homo erectus:
Brain capacity 800 – 1100 cm³, fully erect posture, height 1.5 – 1.8m, omnivorous. They used stone and bone tools for hunting games.

6. Homo neanderthalensis :
Brain capacity 1300 – 1600 cm³, fully erect posture, height 1.5 – 1.6m, omnivorous cave dwellers, used hiles to protect their bodies and buried their dead.

7. Homo sapiens fossils:
Brain capacity 1650 cm3, fully erect posture with height 1.8m, omnivorous. They had strong jaw with teeth close together. They were cave dwellers, made painting and carvings in the caves. They developed a culture and were called first modern man.

8. Homo sapiens sapiens:
Brain capacity 1200 – 1600 cm³, fully erect posture, height 1.5 – 1.8m, omnivorous. They are the living modem men with high intelligence. They developed art, culture, language, speech, etc. They cultivated crops and domesticated animals.

Question 9.
Find out through internet and popular science articles whether animals other than man has self – consciousness.
Answer:
There are many animals other than humans, which have self – consciousness. An example of an animal being self-conscious is dolphins. They are highly intelligent. They have a sense of self and, they also recognize others among themselves and others. They communicate with each other by whistles, tail – slapping, and other body movements. Not dolphins, there are certain other animals such as Crow, Parrot, Chimpanzee, Gorilla, Orangutan, etc., which exhibit self – consciousness.

Question 10.
List 10 modern day animals and using the internet resources link it to a corresponding ancient fossil. Name both.
Answer:
Modern and Ancient corresponding animals:

Question 11.
Practise drawing various animals and plants.
Answer:

Question 12.
Using various resources such as your school Library or the internet and discussions with your teacher, trace the evolutionary stages of any one animal, say horse.
Answer:
The evolution of horse started with Eohippus during Eocene period. It involved the following evolutionary stages:

1. Gradual increase in body size.
2. Elongation of head and neck region.
3. Increase in the length of limbs and feet.
4. Gradual reduction of lateral digits.
5. Enlargement of third functional toe.
6. Strengthening of the back.
7. Development of brain and sensory organs.
8. Increase in the complexity of teeth for feeding on grass.

1. Eohippus:
It had a short head and neck. It had four functional toes and a splint of 1 and 5 on each hind limb and a splint of 1 and 3 in each forelimb. The molars were short crowned that were adapted for grinding the plant diet.

2. Mesohippus:
It was slightly taller than Eohippus. It had three toes in each foot.

3. Merychippus:
It had the size of approximately 100 cm. Although it still had three toes in each foot, but it could run on one toe. The side toe did not touch the ground. The molars were adapted for chewing the grass.

4. Pliohippus:
It resembled the modern horse and was around 108 cm tall. It had a single functional toe with splint of 2nd and 4th in each limb.

Equus:
Pliohippus gave rise to Equus or the modern horse with one toe in each foot. They have incisors for cutting grass and molars for grinding food.

Question 13.
How far inter relationship among? The living organisms is helpful to understand the process of evolution?
Answer:
Organisms which are externally different shows some similarities. This is known as interrelationship. It proves that they might have evolved from the same ancestor. Inter relationship among the living organisms can be understood by following example:

1. All living organisms obtain energy and various materials from the environment.
2. All living organisms multiply and reproduce.
3. All living organisms show transmission of genetic information according to same principle.
4. All living organisms synthesize protein in the ribosome and steps of protein synthesis are same in all living organisms.
5. All organisms respire and steps of respiration in them are same.
6. In all organism method of DNA replication are same.
7. In all living organisms flow of information occurs by the help of nucleic acids present in the nucleus.
8. In all living organisms energy is obtained from ATP.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Hindi Swati Solutions गद्य Chapter 11 गीता और स्वधर्म (निबंध, आचार्य विनोबा भावे)

गीता और स्वधर्म अभ्यास

गीता और स्वधर्म अति लघु उत्तरीय प्रश्न

प्रश्न 1.
अर्जुन के स्वभाव में कौन-सी वृत्ति विद्यमान थी?
उत्तर:
कृत-निश्चय होकर और कर्तव्य भाव से समर-भूमि में खड़े अर्जुन के स्वभाव में क्षात्रवृत्ति विद्यमान थी।

प्रश्न 2.
किस सजा से अपराधी के सुधरने की आशा नष्ट हो जाती है?
उत्तर:
फाँसी की सजा अमानुषी होने के कारण मनुष्य को शोभा नहीं देती क्योंकि इससे अपराधी के सुधरने की आशा नष्ट हो जाती है।

प्रश्न 3.
लेखक किस सागर में डुबकी लगाकर बैठ जाता है?
उत्तर:
लेखक विनोबा भावे जब अकेले होते हैं तब गीता के अमृत सागर में गहरी डुबकी लगाकर बैठ जाते हैं।

प्रश्न 4.
किसने श्रीकृष्ण से अपना सारथ्य स्वीकार कराया? (2016)
उत्तर:
दुर्योधन के द्वारा संधि प्रस्ताव ठुकराने पर अर्जुन ने अनेक देशों के राजाओं को एकत्र करके श्रीकृष्ण से अपना सारथ्य स्वीकार कराया।

प्रश्न 5.
युद्ध में आप्त स्वजन सम्बन्धियों की कितनी पीढ़ियाँ एकत्र हुई थी?
उत्तर:
युद्ध भूमि के मध्य खड़े होकर अर्जुन ने देखा कि दादा, बाप, बेटे, पोते, आप्त-स्वजन सम्बन्धियों की चार पीढ़ियाँ एकत्र थीं।

प्रश्न 6.
अर्जुन के मन में कौन-सा भाव उत्पन्न हो गया था?
उत्तर:
युद्ध के लिए तत्पर स्वजन समूह को देखकर सदैव विजयी रहने वाले अर्जुन के मन में अहिंसा का भाव उत्पन्न हो गया था।

गीता और स्वधर्म लघु उत्तरीय प्रश्न

प्रश्न 1.
गीता के गगन में लेखक किन उपकरणों से उड़ान भरता है?
उत्तर:
जिस प्रकार एक पक्षी अपने दो पंखों की सहायता से नीले खुले गगन में स्वछन्द होकर यथाशक्ति दूर तक उड़ान भरता है। ठीक उसी प्रकार लेखक तर्क को त्यागकर श्रद्धा और प्रयोग,इन दो उपकरणों से गीता के गगन में यथाशक्ति उड़ान भरता रहता है।

प्रश्न 2.
अर्जुन ने अकेले ही किन-किन के दाँत खट्टे कर दिए थे? (2017)
उत्तर:
जिस समय पांडव अज्ञातवास में थे उस समय कौरवों की सेना ने उत्तर कुमार की गायों का हरण करने का प्रयास किया तब अर्जुन ने उत्तर कुमार का सारथी बनकर अकेले ही धीर भीष्म पितामह, आचार्य द्रोण और सूर्यपुत्र कर्ण के दाँत खट्टे कर दिए थे।

प्रश्न 3.
युद्ध टालने के लिए कौन-कौन सी दोनों बातें बेकार हो चुकी थीं?
उत्तर:
अर्जुन के स्वभाव में क्षात्रवृत्ति थी। वह कृत निश्चय होकर और कर्त्तव्य भाव से समर-भूमि में खड़ा था। लेकिन युद्ध के विनाशकारी परिणामों का उसे आभास था। अत: युद्ध को टालने के लिए अर्जुन ने पहली, कौरवों के पास कम से कम माँग का प्रस्ताव भेजा-दूसरी, कृष्ण को मध्यस्थ बना कर भेजा। लेकिन विनाशकारी बुद्धि वाले दुर्योधन ने एक भी बात को स्वीकार नहीं किया।

प्रश्न 4.
लेखक के अनुसार गीता के जन्म का उद्देश्य क्या था?
उत्तर:
युद्धभूमि में अर्जुन धर्म-सम्मूढ़ हो गया था, इस कारण उसके मन में स्वधर्म के विषय में मोह पैदा हो गया था। इस मोह को अर्जुन स्वयं स्वीकार करता है। अतः स्पष्ट है कि गीता का जन्म स्वधर्म में बाधक जो मोह है, उसके निवारणार्थ हुआ है। गीता के जन्म का प्रधान उद्देश्य यही था। गीता सुनने के बाद अर्जुन कृष्ण से कहता है-भगवन्, मेरा मोह नष्ट हो गया है, मुझे स्वधर्म का भान हो गया है।

प्रश्न 5.
लेखक के अनुसार गीता का मुख्य काम क्या-क्या है?
उत्तर:
यदि गीता के उपक्रम और उपसंहार को मिलाकर देखें तो स्वधर्म में बाधक मोह-ममत्व तथा स्वजनासक्ति दूर करना गीता का मुख्य काम है,मोह का निवारण करना ही गीता का काम है। गीता ही नहीं,सम्पूर्ण महाभारत का काम लोक हृदय के मोहावरण को दूर करना है। यही काम गीता ने किया जिसके परिणामस्वरूप अर्जुन का मोह नष्ट हो गया और उसे क्षात्रधर्म का ज्ञान हो गया।

गीता और स्वधर्म दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
विनोबा जी का गीता के साथ किस प्रकार का सम्बन्ध था?
उत्तर:
विनोबा जी का गीता के साथ जो सम्बन्ध था उसे बताना कठिन है क्योंकि वह अलौकिक था जिसे तर्क द्वारा नहीं बताया जा सकता है। विनोबा जी कहते हैं कि उनके शरीर का निर्माण तो माँ के दूध से हुआ किन्तु उनके हृदय और बुद्धि का पोषण गीता के दूध अर्थात् गीता के संदेशों से हुआ। इस सम्बन्ध को हम हार्दिक सम्बन्ध कह सकते हैं। हार्दिक सम्बन्धों में तर्क बहस,दलील या चर्चा का कोई महत्व नहीं होता। तर्क के स्थान पर श्रद्धा और प्रयोग इन दो बातों के आधार पर गीता को समझने का प्रयास किया जाता है।

गीता उनका प्राणतत्व है, आत्मा है। जब वह गीता के मोह निवारण पर चर्चा करते हैं तो मानो, वह गीता रूपी सागर पर तैर रहे होते हैं अर्थात् गीता के श्लोकों का शब्दार्थ कर रहे हैं परन्तु जब वह एकाकी होकर मोह निवारण का माप श्रद्धा व प्रयोग के आधार पर हृदय में मनन करते हैं तो उन्हें ऐसा लगता है, मानो वे अमृत-सागर में डुबकी (गोता) लगा रहे हैं जिससे उनकी आत्मा को परम सुख की प्राप्ति होती है। इसी आत्मा के परम सुख की प्राप्ति के रूप में ही विनोबा जी का सम्बन्ध गीता से है। यह सम्बन्ध अलौकिक होने के साथ-साथ हृदय के तारों से मधुर ईश्वरीय संगीत को सुनाने वाला है।

प्रश्न 2.
अर्जुन को महावीर क्यों कहा गया?
उत्तर:
विनोबा जी ने अर्जुन को महावीर कहा है, उनका यह कथन बिल्कुल सत्य है, क्योंकि अर्जुन में अठारह अक्षौहिणी सेना से अधिक शक्ति थी। वह सैकड़ों लड़ाइयों में अपना जौहर दिखा चुका था। उत्तर-गो ग्रहण के समय उसने अकेले ही भीष्म, द्रोण और कर्ण के दाँत खट्टे कर दिए थे। वह सदा विजय प्राप्त करने वाला और सब नरों में एक ही सच्चा नर था। वीरवृत्ति उसके रोम-रोम में भरी थी। उसके विचार कृत-निश्चय और कर्त्तव्य भाव से पूर्ण थे। क्षात्रवृत्ति उसके स्वभाव में थी। वीरवृत्ति का उत्साह उसकी प्रेरणा थी। युद्ध क्षेत्र में उसने असंख्य वीरों का संहार किया था।

उसका गांडीव हाथ से कभी नहीं गिरा था। उसके शरीर में कभी कम्पन नहीं हुआ था न आँखें गीली हुई थीं। वह सदैव युद्ध-प्रवृत्त ही था। युद्ध उसकी दृष्टि से उसका स्वभाव प्राप्त और अपरिहार्य रूप से निश्चित कर्त्तव्य था। पलभर के लिए उसके मन में परिवार के प्रति आसक्ति तथा मोह जाग्रत हुआ था। परन्तु कृष्ण ने उसके इस मोह को गीता सुनाकर नष्ट कर दिया था। गीता कर्मयोग तथा युद्धयोग दोनों का प्रतिपादन करती है जिसे अर्जुन ने पूर्ण रूप से अपनाया था। इन्हीं सब कारणों से अर्जुन को महावीर कहा गया।

प्रश्न 3.
युद्ध क्षेत्र में स्वजनों को देखकर अर्जुन के मन में कौन-कौन से भाव उत्पन्न होते है? (2014)
उत्तर:
पाण्डवों ने महाभारत के युद्ध को टालने के लिए दुर्योधन के पास कम-से-कम माँग का प्रस्ताव और श्रीकृष्ण जैसे मध्यस्थ को भेजा परन्तु दुर्योधन नहीं माना तो अर्जुन कृष्ण को अपना सारथी बनाकर रणांगन में पहुँचा और अपने रथ को दोनों सेनाओं के बीच खड़ा करने को कहा। सेनाओं के बीच खड़े होकर उसने सारे स्वजन समूह को देखा तो उसके हृदय में उथल-पुथल मच गई। उसे बहुत बुरा लगा। अर्जुन के मन में भाव उठे कि दादा, बाप, बेटे, पोते-आप्त,स्वजन-सम्बन्धियों की चार पीढ़ियाँ मरने-मारने के अन्तिम निश्चय से वहाँ एकत्र हुई हैं। इस प्रत्यक्ष दर्शन का उसके हृदय पर हृदय को विचलित कर देने वाला प्रभाव पड़ा।

आज तक उसने अनेक युद्धों में असंख्य वीरों का संहार किया है, परन्तु उस समय उसे बुरा नहीं लगा था। उसका गांडीव हाथ से नहीं गिरा था। शरीर में कंपन नहीं हुआ था, उसकी आँखें गीली नहीं हुई थीं। तो इसी समय ऐसा क्यों हुआ? उसके मन में स्वजनासक्ति के भाव उत्पन्न हुए। कर्त्तव्यविमुख होने के भावों के साथ तत्व ज्ञान (दर्शन) के भाव जाग्रत हुए। वह युद्ध को व्यर्थ समझकर पाप समझने लगा। युद्ध से कुल क्षय होगा, धर्म का लोप होगा। स्वैराचार मचेगा, व्यभिचार फैलेगा, अकाल आ पड़ेगा,समाज पर तरह-तरह के संकट आयेंगे। इस प्रकार के अनेक भाव अर्जुन के मन में उत्पन्न होते हैं।

प्रश्न 4.
न्यायाधीश के उदाहरण से लेखक क्या सीख देना चाहता है?
उत्तर:
एक न्यायाधीश ने सैकड़ों अपराधियों को फाँसी की सजा दी थी। परन्तु जब अपने पुत्र को खून के जुर्म में मृत्युदण्ड देने का समय आया तो पुत्र-स्नेह के कारण वह हिचकने लगा तथा बुद्धिवाद बघारने लगा-फाँसी की सजा बड़ी अमानुषी है, लज्जा की बात है, बड़ा कलंक है। उसकी यह बात आंतरिक नहीं थी। वह आसक्ति-जनित थी। इस प्रकार महावीर अर्जुन भी स्वजन आसक्ति में आकर अहिंसावादी तथा बुद्धिवादी बन गया और युद्ध में दोष बताने लगा। अतः लेखक बताना चाहता था कि यह अर्जुन का तत्व ज्ञान नहीं कोरा प्रज्ञावाद था। मनुष्य परिजनों के स्नेह में पड़कर कर्त्तव्य को भूल जाता है, वह मोह में पड़ जाता है। इसी मोह को दूर करना गीता का धर्म है। मनुष्य को मोह में न पड़कर अपना कर्तव्य पालन करना चाहिए। मोह मनुष्य को नीचे गिरा देता है। अर्जुन की गति न्यायाधीश जैसी हो गई थी।

प्रश्न 5.
श्रीकृष्ण ने अर्जुन की जिज्ञासाओं का कैसे समाधान किया?
उत्तर:
श्रीकृष्ण द्वारा मध्यस्थता करने तथा कम-से-कम मांग के प्रस्ताव को दुर्योधन ने स्वीकार नहीं किया तो युद्ध अनिवार्य हो गया। दोनों पक्षों की सेनाएँ युद्धभूमि में खड़ी थीं। तब अर्जुन के मन में अचानक जिज्ञासा जागी कि वह दोनों सेनाओं के चेहरे देखे। अर्जुन ने कृष्ण को अपनी जिज्ञासा बताई, कृष्ण ने अर्जुन का रथ दोनों सेनाओं के मध्य खड़ा कर दिया। अर्जुन ने अपनी चार पीढ़ियों को देखा और मोहग्रस्त हो युद्ध से विरक्त हो गया। इसी प्रकार स्वधर्म के विषय में उनके मन में मोह पैदा हो गया था। श्रीकृष्ण ने अर्जुन की जिज्ञासा व मोह को दूर किया। इसके लिए श्री कृष्ण ने अर्जुन को गीता सुनाई जो गीता महाभारत के मध्यभाग में एक ऊँचे दीपक के समान है तथा समस्त जिज्ञासाओं को हल करने में पूर्णतः सक्षम है।

प्रश्न 6.
गीता का निष्कर्ष क्या है?
उत्तर:
गीता की योजना महाभारत में की गई थी जो महाभारत के मध्य भाग में एक ऊँचे दीपक की तरह पूर्ण महाभारत को प्रकाशित करती है। गीता का निष्कर्ष केवल कर्म-योग ही नहीं, बल्कि युद्ध-योग का भी प्रतिपादन करता है। युद्ध-भूमि के मध्य खड़ा अर्जुन स्वजन आसक्ति के कारण क्षात्रवृत्ति से विमुख होकर युद्ध के दुष्परिणाम बघारने लगा क्योंकि सारे जन समूह को देखकर उसके हृदय में उथल-पुथल मच गई थी। कृष्ण अर्जुन के आसक्तिजनित मोह को जानते थे, इसलिए उन्होंने अर्जुन के मोह-पाश का उपाय शुरू किया। स्वधर्म में बाधक जो मोह है उसका निवारण करना ही गीता का उद्देश्य है। मोह, ममत्व, आसक्ति को दूर करके स्वधर्म का भान कराना ही गीता का निष्कर्ष है। गीता ही नहीं सम्पूर्ण महाभारत में लोक हृदय के मोहावरण को दूर करने के लिए यह इतिहास-प्रदीप जलाया गया है। मुख्य निष्कर्ष मोह निरसन ही है।

गीता और स्वधर्म भाषा-अध्ययन

प्रश्न 1.
दिए गए मुहावरों का वाक्यों में प्रयोग कीजिए

1. दाँत खट्टे कर देना
2. उथल-पुथल मचाना
3. आँखें गीली होना
4. जौहर दिखाना
5. उड़ान भरना
6.  डुबकी लगाना
7. जुनून उतर जाना।

उत्तर:

1. दाँत खट्टे करना – भारतीय सैनिकों ने 1971 के युद्ध में पाकिस्तानी सैनिकों के दाँत खट्टे कर दिए थे।
2. उथल-पुथल मचना – कलिंग युद्ध के मैदान में अनेक सैनिकों की कराहट को सुनकर सम्राट अशोक के हृदय में उथल-पुथल मच गई।
3. आँखें गीली होना – रेल दुर्घटना के वीभत्स दृश्य को देखकर उपस्थित जनसमूह की आँखें गीली हो गईं।
4. जौहर दिखाना – झाँसी वाली रानी ने अकेले ही ब्रिटिश सैनिकों के बीच जौहर दिखाया।
5. उड़ान भरना – जीवन में सफलता पाने के लिए कल्पना की उड़ान भरने के साथ परिश्रम करना भी आवश्यक है।
6. डुबकी लगाना – श्री विनोबा भावे ने गीता के अमृत सागर में डुबकी लगाकर जीवन। का लक्ष्य प्राप्त कर लिया।
7. जुनून उतर जाना – स्वजनों के मोह का जूनून उतरने पर ही अर्जुन महाभारत के युद्ध को जीत सका।

प्रश्न 2.
निम्नलिखित वाक्यों का भाव-विस्तार कीजिए-
(अ) जहाँ हार्दिक सम्बन्ध होता है, वहाँ तर्क की गुंजाइश नहीं होती। (2017)
उत्तर:
भाव विस्तार :
‘गीता और स्वधर्म’ शीर्षक में ‘श्री विनोबा भावे’ने बताया है कि गीता के साथ उनका हार्दिक सम्बन्ध है। इस कारण इस सम्बन्ध को साबित करने के लिए किसी भी प्रकार की दलील नहीं दी जा सकती और न ही बहस की जा सकती है। हार्दिक सम्बन्ध भावना से जुड़े होते हैं जिन्हें अनुभव किया जाता है, परन्तु दलीलों द्वारा नहीं बताया जा सकता। साथ ही यह सम्बन्ध सूक्ष्म होता है। इस सम्बन्ध में बड़ी शक्ति होती है,जो पाषाण हृदय को भी कोमल बना देती है। इस सम्बन्ध के द्वारा आत्मिक प्रसन्नता व शान्ति मिलती है।

(ब) इस आसक्ति जनित मोह ने उसकी कर्त्तव्य निष्ठा को ग्रस लिया और तब उसे तत्त्व ज्ञान याद आया।
उत्तर:
भाव विस्तार :
‘श्री विनोबा भावे’ जी ने अपने निबन्ध ‘गीता और स्वधर्म’ में अर्जुन के मोह व कृष्ण की गीता के विषय में बताया है। अर्जुन युद्धभूमि के मध्य रथ पर सवार था,तब उसने स्वजन समूह को देखा तो उसके हृदय में उथल-पुथल मच गई। उसके हृदय में परिजनों के प्रति मोह उत्पन्न हुआ, इस कारण वह अपने क्षात्रधर्म अर्थात् युद्ध कर्म को भूल कर प्रज्ञावादी बन गया। युद्ध को पाप व मानव जाति का कलंक कहने लगा। जो अर्जुन युद्ध प्रवृत्त था, युद्ध करना उसका स्वभाव था, अपरिहार्य रूप से उसका कर्त्तव्य था, उस कर्त्तव्य को अर्जुन भूल गया। युद्ध के दुष्परिणामों को बताने लगा यद्यपि जो सत्य थे परन्तु अर्जुन के मुख से सुशोभित नहीं हो रहे थे क्योंकि अर्जुन का कर्तव्य युद्ध में निहित था। उसका तत्व ज्ञान आत्मिक न होकर बौद्धिक था। कहीं न कहीं उस तत्त्व ज्ञान की जड़ में मोह था जिसने अर्जुन को कर्तव्य से गिरा दिया था।

गीता और स्वधर्म पाठ का सारांश

भूदान आन्दोलन के प्रणेता तथा किसानों के मसीहा ‘आचार्य विनोबा भावे’ द्वारा लिखित प्रस्तुत निबन्ध ‘गीता और स्वधर्म’ में आचार्य जी ने गीता के माध्यम से मनुष्य को अपनी बुद्धि के आधार पर न्यायपूर्ण कार्य करने का संदेश दिया है।

लेखक का गीता के साथ तर्कपूर्ण नहीं, आत्मिक रिश्ता है। उनके हृदय और बुद्धि का पोषण गीता के दूध से हुआ है। गीता पर उनकी अपार श्रद्धा है। गीता उनके लिए अमृत का सागर है। यह गीता श्रीकृष्ण ने युद्ध स्थल में युद्ध से विमुख हुए अर्जुन को कर्तव्य का ज्ञान कराने के लिए सुनाई थी। इसीलिए गीता महाभारत में दीपक के समान प्रकाश देती है।

माना जाता है कि कृष्ण ने अर्जुन को गीता दो कारणों से सुनाई होगी। एक, अर्जुन की नपुंसकता को दूर करने तथा दूसरी अर्जुन की अहिंसा-वृत्ति को दूर करने के लिए। लेकिन ये दोनों बातें ही ठीक नहीं हैं.क्योंकि अर्जन ‘महावीर’ थे,उन्होंने उत्तर-गो ग्रहण के समय अकेले ही भीष्म, द्रोण और कर्ण को पराजित किया था। वीर-वृत्ति उनके रोम-रोम में बसी थी। क्षात्रवृत्ति उनके स्वभाव में थी। युद्ध को टालने के लिए कम से कम माँग का प्रस्ताव और कृष्ण ने मध्यस्थता की थी परन्तु दुर्योधन सहमत नहीं था। इस कारण कृष्ण को अपना सारथी बनाकर वीर अर्जुन कुरुक्षेत्र में पहुँचे। वहाँ दोनों पक्षों की सेनाओं के मध्य अपने रथ में बैठे अपनी चार पीढ़ियों के मरण का उत्साह देखकर वह स्वजनाशक्ति में आ गए थे। वीरता का शौर्य दिखाने वाले उस महावीर अर्जुन का गांडीव कभी झुका नहीं था,न शरीर में कम्पन हुआ था और न कभी आँखें गीली हुई थीं।

वहीं अर्जुन जिसमें कभी अहिंसा-वृत्ति ने जन्म नहीं लिया था, उसके आसक्ति जनित मोह ने कर्तव्य को त्याग कर तत्व ज्ञान की शरण ली। कहने लगे कि युद्ध मानव जाति के विनाश का कारण होगा। उस समय अर्जुन की दशा उस न्यायाधीश जैसी थी जो अनेकों को मृत्यु दण्ड दे चुका था परन्तु जब अपने बेटे को खून के जुर्म में मृत्यु दण्ड देने का समय आया तो कहने लगा कि मृत्यु दण्ड अमानुषी सजा है, बड़ा कलंक है। यह कथन आसक्ति जनित है। अर्जुन के इस तत्व ज्ञान को कृष्ण जानते थे इसलिए अर्जुन के मोहपाश को नष्ट करने का उपाय किया और गीता में अर्जुन के इसी मोहपाश पर गदा प्रहार किया है।

अतः गीता का जन्म,स्वधर्म में बाधक जो मोह है,उसके निवारणार्थ हुआ है। गीता सुनाने के बाद कृष्ण ने अर्जुन से पूछा कि अर्जुन तेरा मोह गया? तब अर्जुन ने उत्तर दिया-हाँ भगवन! मेरा मोह नष्ट हो गया, मुझे स्वधर्म का भान हो गया। गीता ही नहीं सम्पूर्ण महाभारत का यही उद्देश्य है। व्यास जी ने महाभारत के आरम्भ में कहा है कि लोक हृदय के मोहावरण को दूर करने के लिए मैं यह इतिहास प्रदीप जला रहा हूँ।

गीता और स्वधर्म कठिन शब्दार्थ

तर्क = दलील। क्लैष्य = नपुंसकता। प्रवृत्त = संलग्न। परावृत्त = दूर भागना। जौहर = वीरता। ख्याति = यश। समर-भूमि = युद्ध-भूमि। क्षात्रवृत्ति = क्षत्रिय धर्म। रणांगण = युद्ध भूमि। स्वजनासक्ति = पारिवारिक जनों के प्रति मोह। प्रतिपादन = सिद्ध करना। क्षय = नष्ट। स्वैराचार = स्वेच्छाचार। नौबत = अवसर। अमानुषी = जंगलीपन। जुनून = धुन। आसक्ति = मोह। प्रज्ञावेद = बुद्धिवाद। अपरिहार्य = जो छोड़ा न जा सके। प्रहार = चोट। निरसन = निराकरण। सारथ्य = सारथी का कार्य। कर्त्तव्यच्युति = अपने कर्त्तव्य से विमुख। अवान्तर = अन्य।

गीता और स्वधर्म संदर्भ-प्रसंग सहित व्याख्या

(1) गीता का और मेरा सम्बन्ध तर्क से परे है। मेरा शरीर माँ के दूध पर जितना पला है, उससे कहीं अधिक मेरे हृदय और बुद्धि का पोषण गीता के दूध पर हुआ है। जहाँ हार्दिक सम्बन्ध होता है, वहाँ तर्क की गुंजाइश नहीं रहती। तर्क को काटकर श्रद्धा और प्रयोग, इन दो पंखों से ही मैं गीता-गगन में यथाशक्ति उड़ान भरता रहता हूँ।

संदर्भ :
प्रस्तुत गद्य अवतरण हमारी पाठ्य-पुस्तक के निबन्ध ‘गीता और स्वधर्म’ नामक पाठ से अवतरित है। इसके लेखक ‘आचार्य विनोबा भावे’ हैं।

प्रसंग :
इन पंक्तियों में गीता तथा विनोबा भावे के सम्बन्ध को तर्कहीन बताकर गीता के प्रति उनकी अपार श्रद्धा को व्यक्त किया गया है।

व्याख्या :
विनोबा भावे गीता पर प्रवचन देते हुए कहते हैं कि उनका और गीता का सम्बन्ध अलौकिक है जिसे शब्दों के द्वारा नहीं बताया जा सकता, न उस सम्बन्ध के विषय में कोई दलील या बहस की जा सकती है। गीता के साथ अपने सम्बन्ध को स्पष्ट करते हुए वे कहते हैं कि माँ के दूध व वात्सल्य से उनके शरीर का निर्माण हुआ परन्तु उनके हृदय व बुद्धि का विकास गीता के ज्ञान से हुआ। गीता के रहस्य ने ही उनके हृदय को पवित्र तथा बुद्धि को परिष्कृत किया है। जब किसी से हार्दिक सम्बन्ध हो जाते हैं तो वहाँ बहस या दलील व्यर्थ हो जाती है।

हृदय दलीलों को नहीं मानता, वह तो सत्य व कोमलता को स्वीकार करता है। जिस प्रकार एक पक्षी अपने दोनों पंखों की सहायता से स्वछन्द नीले आकाश में अपनी पूर्ण शक्ति के साथ उड़ान भरता हुआ प्रसन्न होता है ठीक उसी प्रकार विनोबा जी भी गीता में श्रद्धा रखते हुए तथा उसे अपने जीवन में साक्षात प्रयोग करते हुए गीता के अर्थ को या उसके रहस्य को समझते हैं और सुख का अनुभव करते हैं। श्रद्धा और प्रयोग उनके गहन अध्ययन के दो आधार रूपी पंख हैं।

विशेष :

1. विनोबा जी की गीता के प्रति श्रद्धा अभिव्यक्त की गई है।
2. संस्कृतनिष्ठ शुद्ध साहित्यिक खड़ी बोली का प्रयोग है।
3. गीता-गगन में रूपक अलंकार का प्रयोग है।
4. उदाहरण व गवेषणात्मक शैली का प्रयोग है।

(2) मैं प्राय: गीता के ही वातावरण में रहता हूँ। गीता मेरा प्राणतत्व है। जब मैं गीता के सम्बन्ध में किसी से बात करता हूँ, तब गीता-सागर पर तैरता हूँ और जब अकेला रहता हूँ तब उस अमृत-सागर में गहरी डुबकी लगाकर बैठ जाता हूँ।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत अंश में विनोबा जी गीता के प्रति अपनी श्रद्धा का वर्णन करते हैं। गीता के प्रति उनकी श्रद्धा तर्करहित हार्दिक है। उनके जीवन का हर पल गीता के मूल से जुड़ा है।

व्याख्या :
विनोबा भावे, आसक्ति रहित कर्म करने का संदेश देने वाली गीता से, आजीवन जुड़े रहे। गीता ने उन्हें जीवन प्रदान किया। गीता उनके प्राण-आत्मा थी। गीता की चर्चा करते समय वह उसके बाह्य रूप के सम्पर्क में रहते थे लेकिन एकान्त की अवस्था में वह उसके गूढ़ रहस्य को समझने में लगे रहते थे। परन्तु प्रतिपल वे गीता के सम्पर्क में रहते थे। एकाकी अवस्था में उन्होंने गीता रूपी सागर के निष्काम कर्म रूपी अमृत को पीया। गीता-सागर में डुबकी लगाने अर्थात् उसका गहन अध्ययन करने पर उन्हें सुख मिलता था। जैसे अमृत पीकर मनुष्य अमर हो जाते हैं, उसी प्रकार गीता के रहस्य को समझने पर वे अमर हो गये या कहें कि वे सुख-दुःख से परे हो गए। इस प्रकार गीता व विनोबा जी दो अलग रूप होते हुए एक हो गए थे। गीता ने उन्हें आनन्दपूर्ण जीवन प्रदान किया।

विशेष :

1. गीता के द्वारा ही मनुष्य मोक्ष पा सकता है।
2. प्राणतत्व जैसे गूढ़ शब्दों के कारण भाव कठिन हो गए हैं।
3. अलंकारिक व मुहावरेदार भाषा।
4. संस्कृत शब्दावली का प्रयोग।
5. साहित्यिक खड़ी बोली।

(3) अर्जुन, तो लड़ाई से परावृत्त हो रहा था, सो भय के कारण नहीं। सैकड़ों लड़ाइयों में अपना जौहर दिखाने वाला वह महावीर था। उत्तर-गो ग्रहण के समय उसने अकेले ही भीष्म, द्रोण और कर्ण के दाँत खट्टे कर दिये थे। सदा विजय प्राप्त करने वाला और सब नरों में एक ही सच्चा नर, ऐसी उसकी ख्याति थी। वीरवृत्ति उसके रोम-रोम में भरी थी।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत पंक्तियों में बताया गया है कि जब अर्जुन कुरुक्षेत्र की युद्ध-भूमि में खड़ी अट्ठारह अक्षोहिणी सेना के मध्य अपनी चार पीढ़ियों को खड़ा देखते हैं तो उनके मन में स्वजनासक्ति आ जाती है और वह युद्ध न करने की बात कृष्ण से कहते हैं। तब कृष्ण गीता के माध्यम से क्षात्रधर्म का सन्देश देकर उनमें उत्साह का संचार करते हैं।

व्याख्या :
श्री विनोबा जी बताते हैं कि अर्जुन ने युद्ध करना अस्वीकार किया। इसका कारण विशाल सेना की शक्ति को देखकर वह भयभीत नहीं था। अर्जुन महावीर था, उसने अनेक युद्धों में अपनी वीरता दिखाई थी तथा वह अनेक लड़ाइयाँ जीत चुका था। जिस समय पाण्डव अज्ञातवास में थे तब कौरव सेना, उत्तर कुमार की गायों का अपहरण करने आई थी। जिनमें भीष्म, द्रोण और सूर्यपुत्र कर्ण जैसे अजेय वीर थे, उस समय अर्जुन ने उत्तर कुमार का सारथी बनकर कौरवों की सेना को पराजित किया था।

यह अर्जुन की वीरता और निडरता का उदाहरण है। अत: अर्जुन शत्रु की सेना की शक्ति से नहीं डरा था। अर्जुन की प्रत्येक लड़ाई में विजय हुई थी, वह नर-श्रेष्ठ था। उसकी वीरता का यश दसों दिशाओं में फैला था। अर्जुन के सम्पूर्ण शरीर व रोम-रोम में वीरता की भावना भरी हुई थी। लेखक ने अर्जुन द्वारा युद्ध न किए जाने का कारण स्वजनासक्ति बताया है। उस आसक्ति को कृष्ण ने गीता सुना कर समाप्त किया था।

विशेष :

1. शुद्ध खड़ी बोली का प्रयोग।
2. ‘दाँत खट्टे करना’ तथा ‘रोम-रोम में भरा’ जैसे मुहावरों का प्रयोग।
3. उद्धरण शैली का प्रयोग।
4. ‘परावृत्त’ जैसे संस्कृत शब्दों का प्रयोग हुआ है।
5. अर्जुन की वीरता का परिचय दिया गया है।

(4) क्या अशोक की तरह उसके मन में अहिंसा-वृत्ति का उदय हुआ था? नहीं, यह तो केवल स्वजनासक्ति थी। इस समय भी यदि गुरु बंधु और आप्त सामने न होते, तो उसने शत्रुओं के मुण्ड गेंद की तरह उड़ा दिये होते। परन्तु इस आसक्ति जनित मोह ने उसकी कर्त्तव्यनिष्ठा को ग्रस लिया और तब उसे तत्वज्ञान याद आया।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
विनोबा जी ने बताया है कि युद्धभूमि में अपने परिजनों को देखकर अर्जुन मोह में पड़ गया, इस कारण वह अपने क्षात्रधर्म को भूल गया।

व्याख्या :
कुरुक्षेत्र के मैदान में अपने स्वजन समूह को देखकर अर्जुन के हृदय में हलचल मच गई। वह अचानक अहिंसावादी बन गया । अर्जुन की अहिंसा अशोक जैसी अहिंसा नहीं थी, उसकी अहिंसा का कारण परिवार के प्रति मोह था। अशोक युद्ध के वीभत्स परिणामों के कारण अहिंसावादी बना था। अशोक और अर्जुन की अहिंसा में अन्तर था। अर्जुन के सामने यदि उसके गुरु,भाईबन्धु, मित्र आदि न होकर शत्रु पक्ष की सेना होती तो वह विपक्ष को नष्ट कर देता। इस समय परिजनों के प्रति आसक्ति ने उसे मोह में डुबो दिया जिसका परिणाम था कि वह अपने क्षात्रधर्म अर्थात् कर्त्तव्य को भूल गया। वह अहिंसा को श्रेष्ठ मानने लगा तथा अपना तत्त्व-ज्ञान अर्थात् दर्शन बघारने लगा कि युद्ध पाप है,समाज का कलंक है, धर्म का नाश है, व्यभिचार की जड़ है, आदि। इसी मोह को नष्ट करने व क्षात्रधर्म याद कराने के लिए कृष्ण ने अर्जुन को युद्धभूमि में गीता सुनाई थी।

विशेष :

1. अर्जुन मोहवश अपने कर्तव्य-युद्ध को भूल गया।
2. शुद्ध साहित्यिक खड़ी बोली का प्रयोग है।
3. संस्कृत शब्दों की अधिकता है।
4. अशोक और अर्जुन की अहिंसा में अन्तर बताया है।
5. शैली में संक्षिप्तता तथा क्लिष्टता है।

(5) परन्तु सारी गीता में इस मुद्दे का कहीं भी जवाब नहीं दिया है, फिर भी अर्जुन का समाधान हुआ है। यह सब कहने का अर्थ इतना ही है कि अर्जुन में अहिंसा-वृत्ति नहीं थी, वह युद्ध प्रवृत्त ही था। युद्ध उसकी दृष्टि से उसका स्वभावप्राप्त और अपरिहार्य रूप से निश्चित कर्त्तव्य था। उसे वह मोह के वश होकर टालना चाहता था। और गीता का मुख्यत: इस मोह पर ही गदा प्रहार है।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
निबन्ध की अन्तिम पंक्तियों में बताया गया है कि युद्ध न करना अर्जुन का दर्शन नहीं बुद्धिवाद था। अतः कृष्ण ने अर्जुन के बुद्धिवाद पर ध्यान न देकर अर्जुन के मोह को तोड़ा।

व्याख्या :
कृष्ण अर्जुन की परिवार में निहित आसक्ति को जानते थे, वह यह भी जानते थे कि अर्जुन अहिंसावादी नहीं है, उसका स्वभाव युद्ध है, लेकिन इन तथ्यों को गीता में नहीं बताया गया है। कृष्ण ने तो केवल अर्जुन के मोह को नष्ट किया है, उसके कर्त्तव्य की याद दिलाई है। कृष्ण ने ऐसा इसलिए किया क्योंकि वे जानते थे कि अर्जुन क्षत्रिय है, युद्ध करना उसका स्वभाव है और एक क्षत्रिय अहिंसा को कभी नहीं अपनाता। युद्ध एक क्षत्रिय का कभी न त्यागने वाला कर्त्तव्य होता है। अर्जुन सच्चे अर्थों में एक वीर योद्धा था। युद्ध के परिणामों को जानते हुए भी वह युद्ध को नहीं त्याग सकता था। लेकिन परिवार के मोह ने उसे युद्ध से विमुख कर दिया था। इसी मोह को श्रीकृष्ण ने गीता के सन्देश के द्वारा नष्ट कर दिया। अन्त में, अर्जुन ने कहा कि उसका भ्रम अर्थात् मोह नष्ट हो गया है और वह अपने धर्म के लिए तैयार है।

विशेष :

1. कृष्ण के द्वारा अर्जुन का मोह भंग हुआ है।
2. संस्कृत के शब्दों के प्रयोग के साथ भावों की सरलता है।
3. लघु वाक्यों ने विचारों को स्पष्ट किया है।
4. शुद्ध साहित्यिक खड़ी बोली।
5. ‘मुद्दे’ व ‘जवाब’ जैसे उर्दू के शब्दों का प्रयोग।

MP Board Class 12th Hindi Solutions

MP Board Class 12th Biology Important Questions Chapter 15 Biodiversity and Conservation

Biodiversity and Conservation Important Questions

Biodiversity and Conservation Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Tera biodiversity is used for the first time by :
(a) V. G. Rosseu
(b) Linnaeus
(c) Odum
(d) Theophrastus.
Answer:
(a) V. G. Rosseu

Question 2.
When does wild life conservation Act come in the force in India :
(a) 1883
(b) 1972
(c) 1973
(d) 1972
Answer:
(b) 1972

Question 3.
When does IBWL is started :
(a) 1952
(b) 1981
(c) 1973
(d) 1972
Answer:
(a) 1952

Question 4.
Where is NBPR is situated :
(a) Delhi
(b) Kolkata
(c) Lucknow
(d) Mumbai
Answer:
(a) Delhi

Question 5.
The number of biosphere reserves in India is :
(a) 73
(b) 7
(c) 416
(d) 23
Answer:
(b) 7

Question 6.
Which National Park is associated with white tiger:
(a) Kanger
(b) Satpura
(c) Bandhavgarh
(d) Kanha
Answer:
(c) Bandhavgarh

Question 7.
First National Park of M. P. is :
(a) Shivpuri
(b) Bandhavgarh
(c) Kanha
(d) Kanger
Answer:
(c) Kanha

Question 8.
Where is “Plant Fossil” National Park is situated:
(a) Shivpuri
(b) Mandala
(c) Bastar
(d) Bhopal
Answer:
(b) Mandala

Question 9.
The National Park of M.P. which is marked as biosphere reserve is :
(a) Kanha
(b) Shivpuri
(c) Bandhavgarh
(d) Satpura.
Answer:
(b) Shivpuri

Question 10.
First Tiger Project of M.P. is :
(a) Bandhavgarh
(b) Kanha
(c) Sidhi
(d) Shivpuri.
Answer:
(b) Kanha

Question 2.
Fill in the blanks :

1. FRI is situated is …………………….
2. Kanjiranga sanctuary is associated with the conservation of …………………….
3. Red data book is related with the ……………………. conservation.
4.  ……………………. is the fust biosphere reserve of India.
5. List of the endangered species are available in ……………………. data book.
6.  ……………………. is the important componant of air for life.
7. Mineral is the example of ……………………. sources.
8. ……………………. percent part of the earth has water.
9. Biodiversity is the source of different types of …………………….
10. Species for which there are no living representative is called …………………….
11. Farming which is develop in fresh water is called …………………….
12. Botanical garden is an ……………………. method of plant conservation.
13. ……………………. conservation is the on site conservation.
14. ……………………. percent part of the mangrove plant are found in India.
15. Gir national park in India is famous for …………………….

Answer:

1. Dehradun
2. Rhinoceros
3. Endangered species
4. Nilgiri
5. Red
6. Oxygen
7. Non – renewable
8. 71
9. Germplasm
10. Extinct
11. Aquatic culture
12. Ex – situ
13. In – situ
14. 5
15. Asiatic lion.

Question 3.
Match the followings:
I.

Answer:

1. (b)
2. (c)
3. (d)
4. (a)

II.

Answer:

1. (d)
2. (a)
3. (b)
4. (e)
5. (c)

Question 4.
Write the answer in one word/sentences:

1. Where does Kanha National Park situated?
2. From which person Chipko movement is associated?
3. When does Wild life Conservation Act is emenced in force?
4. How many percent contribution of India is in world’s biodiversity?
5. How many species are found in India?
6. Which are known as the lungs of the earth?
7. How many national park in India?
8. How many wild 1 ife sanctuary in India?
9. What is the basic component of biodiversity?
10. Write the full name of IUCN.
11. Which are most helpful in nature balancing?
12. How many species of bamboo found in India?
13. Name any one book associated with the conservation of wild life.
14. Upper surface of the land is called.

Answer:

1. M.P
2. Sunder Lai Bahuguna
3. 1972
4. 8.1%
5. 1200 species
6. Amazon – rain forest
7. 90
8. 448
9. Gene
10. International Union for conservation of nature and natural resourses
11. Wild animal
12. 100 (about)
13. Red data book
14. Soil.

Biodiversity and Conservation Very Short Answer Type Questions

Question 1.
Name the three important components of biodiversity.
Answer:
Three important components of biodiversity are:

1. Genetic diversity.
2. Species diversity.
3. Ecosystem diversity.

Question 2.
What is forest?
Answer:
A large area of land covered with trees, shrubs and grasses.

Question 3.
As which day 21st March is observed?
Answer:
21st March is observed as “World Forest Day”.

Question 4.
What are extinct species?
Answer:
Species which are not found at present in the earth but survived in the past in this world are called as extinct species.
Example : Dinosaur.

Question 5.
What are mentioned in the Red Data Book of IUCN?
Answer:
The IUCN Red Data Book contains list of threatened species. It was founded in 1964. Correction was done in 2004.

Biodiversity and Conservation Short Answer Type Questions

Question 1.
How do ecologists estimate the total number of species present in the world?
Answer:
The diversity of living organisms present on the earth is very vast.According to an estimate by researchers, it is about seven millions. The total number of species present in the world is calculated by ecologists by statistical comparison between a species richness of a well studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the earth.

Question 2.
The species diversity of plants (22 %) is much less than that of animals (72 %). What may be the reason that animals achieved greater diversifications?
Answer:
Animals have achieved greater diversification than plants due to following reasons:

1. They are mobile and thus, can move away from their predators or unfavourable, environments. On the other hand plants are fixed and have fewer adaptations to obtain optimum amount of raw materials and sunlight therefore, they show lesser diversity.

2. Animals have well developed nervous system to receive stimuli against external factors and thus, can respond to them. On the other hand, plants do not exhibit any such mechanism, thus they show lesser diversity than animals.

Question 3.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:
Yes, there are various kinds of parasites and disease causing microbes that we deliberately want to eradicate from the earth. Since, these microorganisms are harmful to human beings, scientists are working hard to fight against them. Scientists have been able to eliminate smallpox virus from the world through the use of vaccinations. This shows that humans deliberately want to make these species extinct. Several other eradication programmes such as polio and hepatitis B vaccinations are aimed to eliminate these disease causing microbes.

Question 4.
Explain conservation of biodiversity.
Answer:
Conservation of biodiversity:
India is one of the richest (among the 12 mega centres of the world) countries in biological diversity. This rich biodiversity is due to a variety of climatic conditions prevailing on different ecological habits ranging from tropical, subtropical, temperate, alpine to desert. These varied conditions harbour a plethora of organisms, which forms an important natures wealth, responsible for socio – economic development of life in our country. But the biodiversity of organisms are under serious threat and there is an urgent need for biodiversity conservation on war foot level.

Question 5.
What is social forestry?
Answer:
The planning of social forestry started in India since 1976, which is related with the conservation of forests. This project ;s useful for local people in various ways, such as it fulfil their requirements, provides work to unemployed, use of wasteland and help to maintain O₂ and CO₂ balance in the atmosphere, etc. Thus, project is started by Indian government, the chief objectives of this project are as follows:

1. Plantation of useful plants in the forest.
2. Development of forests on personal lands by the cooperation of government.
3. To prevent the harmful effects of pollution by development of artificial forest.
4. Preservation of endangered wild animals.

Question 6.
Write importance of forests.
Answer:
Importance of forests:

1. Forests play a vital role in the life and economy of all tribes living in forests, by providing food, medicines and other products of commercial value.

2. Forests are large biotic communities. It provides shelter and sustenance to a larger number of diverse species of plants, animals and microorganisms.

3. Forests prevent soil erosion by wind and water. The trees provide shade which prevents the soil from drying during summer. Trees reduce the velocity of raindrops or wind striking the ground so that dislodging of the slil partiles is reduces. The root system of plants firmly binds the soil.

Question 7.
Write any five features of Indian forests.
Answer:
Indian forests are characterized by:

1. Indian forests are mainly tropical forests.
2. Himalayan forests are characterized by the presence of coniferous trees.
3. In few parts of our country having temperate forests.
4. Our forests contain a large number of useful varieties of plants and animals.
5. A great variations are present in Indian forests.

Question 8.
Enumerate any five reasons for the destruction of wildlife (animals).
Answer:
The main reason of destruction of wildlife (animals) are:

1. Entertainment, personal profit, earning money by wrong methods are the human illattitudes, unkindness toward wild animals has brought the animals at endanger level.
2. Huge reduction in the natural habitat of wild animal so it has reduced their living area due to urbanization, industrialization and deforestation.
3. Exhorbitant extraction and consumption is harmful for wild animals like skin of animals, Teeth of elephant etc.
4. Various types of pollution has force the reduction of wild animals.
5. Very loose and unpunishable wildlife act which has increase poaching, hunting of wild animals.

Question 9.
How is biodiversity important for functioning of ecosystem?
Answer:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods. If an ecosystem is rich in biodiversity, then the ecological balance would not get affected.

Various tropic levels are connected through food chains. If any one organism or all organisms of any one trophic level is illed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food. If all deer’s are dead, soon the tigers will also die.

Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 10.
What are sacred groves? What is their role in conservation?
Answer:
Sacred groves are forest patches for worship in several parts of India. All the trees and wildlife in them are again treated and given total protection. They are found in khasiandjointia hills in Meghalaya. Western Ghat regions of Karnataka and Maharashtra, etc. Tribals do not allow anyone to cut even a single branch of tree in these sacred groves thus, sacred groves have been free form all types of exploitations.

Question 11.
Among the ecosystem services are control of floods and soil erosion, how is this achieved by the biotic components of the ecosystem?
Answer:
Control of soil erosion:
Plant roots hold the soil particles tightly and do not allow the top soil to be drifted away be winds or moving water. Plants increase the porosity and fertility of the soil.

Control of floods:
It is carried out by retaining water and preventing run off rain water. Litter and humus of plants function as sponges thus, retaining the water which percolates down and get stored as underground water. Hence, the flood is controlled.

Question 12.
Give three hypothesis for explaining why tropics show greatest levels of species richness.
Answer:
There are three different hypothesis proposed by scientists for explaining species richness in the tropics:

1. Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity.
2. Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness.
3. Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

Biodiversity and Conservation Long Answer Type Questions

Question 1.
What is the significance of the slope of regression in a species – area relationship?
Answer:
The slope regression (z) has a great significance in order to find a species – area relationship. It has been found that in smaller areas (where the species-area relationship is analyzed), the value of slopes of regression is similar regardless of the taxonomic group or the region. However, when a similar analysis done in larger areas, then the slope of regression is much steeper.

Question 2.
What are the major causes of species loss in a geographical regions?
Answer:
The following are the major causes for the loss of biodiversity around the world:

1. Habitat loss and fragmentation:
Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanization. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species.

2. Over – exploitation:
Due to over – hunting and over – exploitation of various plants and animals by humans, many species have become endangered of extinct (such as; the tiger and the passenger pigeon).

3. Alien species invasions:
Accidental or intentional introduction of non-native species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in Lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake.

4. Co – extinction:
In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

Question 3.
What do you understand by Threatened species? Explain its types.
Answer:
Species which have been greatly reduced in their number or whose natural habitats have been, disturbed due to which these are near extinction and may become extinct if the causative factors continue are called threatened species. It is estimated that about 25,000 plant species and 1,000 vertebrate species and subspecies and many invertebrate species are threatened with extinction. It is believed that at least 10% of the living species are in danger.

The organisms which are near extinction are of following types:

1. Endangered (E) species:
The species which are facing danger of extinction and whose survival is unlikely if the causal factors continue to operate. These are the species whose number have been reduced to a critical level or whose habitats have been so drastically reduced that they are deemed Lto be in immediate danger of extinction. For example, Indian rhinoceros, Asiatic lion and the great Indian bustard, snow leopard etc.

2. Vulnerable (V) species:
These are the species having sufficient number of individuals in their natural habitats. However, in the near future, they might represent the category of endangered species if unfavourable factors in the environment continue to operate, e.g., Musk deer, black buck, golden langur, etc.

3. Rare (R) species:
These are species with small population in the world. At present these are not endangered and vulnerable but are at risk. These species are usually localize within geographical areas or habitats or are thinly scattered over a more extensive range, e.g., Indian elephant, Asiatic wild ass, gharial, wild yak etc.

4. Threatened (T) species:
The species which do not fall under the endangered or vulnerable categories but indications are available that such species may come under any of these two categories if appropriate measures are not taken to protect them.

Question 4.
What are National Parks? Explain any five National Parks found in India.
Answer:
National Park:
Natinal park is an area which is strictly reserved for the betterment of the wildlife and where activities like forestry, grazing or cultivation are not permitted.

Five national parks of India are:

1. Shivpuri national park – It is located at Shivpuri near Gwalior (Madhya Pradesh). Wildlife present in this park tiger, cheetal, sambhar.
2. Guindy deer national park – It is located near Chennai (Madras) in Tamil Nadu. Wildlife found here ore Alvino deer, black buck, cheetal and famous snake park.
3. Betla national park – It is located in Palamu at Bihar. Wildlife found in this national park are elephant, tiger.
4. Dachigham national park – It is located at Srinagar in Jammu and Kashmir. Wildlife leopard, black beer, brown bear, musk deer, hangul, scrow, etc.
5. Bandhavgarh national park – It is located at Shahdol in Madhya Pradesh. Wildlife such as white tiger, panther, cheetal, bison, nilgai, barking deer, wild boar, etc. are found here.

Question 5.
Write in brief, the reasons necessary for conservation of wild species.
Answer:
Necessity for wildlife conservation : The conservation of wildlife is required for the following reasons :

1. To maintain balance in nature:
The wildlife helps us in maintaining the balance of nature. Once this equilibrium is disturbed it leads to many problems. The destruction of carnivores or insectivores often leads to an increase in the herbivores which in turn affects the forest vegetation or crop.

2. Economic value:
The wildlife can be used commercially to earn money example Animal products like hides, ivory, fur etc. are of tremendous economic value. The collection and supply of dead or living specimens of wildlife for museums and zoos fetches good amount of money. Wildlife can increase our earning of foreign exchange if tourism is promoted properly.

3. Scientific value:
The preservation of wildlife helps many naturalists and behaviour biologists to study morphology, anatomy, physiology, ecology and behaviour biology of the wild animals under their natural surroundings.

4. Recreational value:
The wildlife of any country provides best means of sports and recreation. Bird – watching is a hobby of many people all over the world. A visit to the parks and sanctuaries is an enjoyable proposition for children as well as for adults.

5. Cultural value:
The wildlife of India is our cultural asset and has deep rooted impact on Indian art, sculpture, literature and religion. Indus valley civilization shows the use of animals symbols in their seals.

6. Preservation of human race:
The destruction of wildlife in an area may eventually lead to the end of human civilization.

Question 6.
Describe the National and International efforts prescribed for the conservation of forests.
Answer:
The forest conservation is started in India on national level since British government. In 1856, Lord Dalhousi had formulated a policy for the conservation of forest in Burma. In 1894, Indian government also prepared a forest policy on national level. The main points of this policy are:

1. Forest management
2. Proper use of forest land
3. Polity for, protected forests
4. Improved forest production

Indian government established national parks, sanctuaries and zoological parks, The F.A.O. of United Nations is also functioning on forest conservation on international level. This organization also provides financial help for this purpose. In 1952, Indian government also prepared India’s New National Forest Policy under the direction of F.A.O. Forest policy has been planned for :

1. Prevention of deforestation of hill plants.
2. Reforestation of grazing land.
3. Development of grazing land.
4. Plantation of economically useful forest trees.
5. Increase in the profit of government from forests.

Question 7.
Write a short note on wild animals in India.
Answer:
India as a country has a diverse range of wildlife. India is home for many species of wild animals. More than 25% land are dense forest in India and around 400 national parks. Some of the most important and popular wild animals in India are as follows:

(A)
Animals : In Indian forest, below mentioned animals are found:

1. Deer – Its many species are found in India example Musdeer, Sambhar deer, chital, etc.

2. Antelope – These are same as deer example Nilgai, Barasingha (Swamp deer), four homed antelope (Chousingha) etc.

3. Elephant – Elephants are large mammals of the family elephantidae. It is found in Kerela and North India.

4. Rhinoceros – It is found in Himalaya region and in the forest of Bengal and Assam. Humans are the biggest threat to the Indian rhinoceros as the have been hunted to the brink of extinction for their horns.

5. Wild Ass – Wild asses are not found in any part of the world. Now in India, it is found in the little Rann of Kutch in the Gujarat state of India.

6. Carnivorous animals – Some Indian wild carnivores are:

• Indian lion (Asiatic lion) : Now it is confined to forests in the fall.
• Cheetah : It is on the verge of extinction.
• Lion : Lion is a national animal, at present its population in our country are more than 3,000.
• Leopard : It is similar to cheetah but smaller than cheetah.

(B)
Birds:
Peacock, wild fowl, many types of duck, stork, pigeons, partridge, quail, vulture, kite, piquant, owl, indian paradise flycatcher (dudhraj) are found in forests of our country.

(C)
Reptiles:
Crocodiles, alligators, tortoise, lizards, snakes and other reptilians are found in Indian forest. Many vertebrates and invertebrates are also found in Indian forest.

Question 8.
What are the main rules of Indian forest act?
Answer:
The Indian Forest Act, 1927 was largely based on previous Indian forest acts implemented under the British. Things which are included in this act are as follows:

1. Forest arrangement – Due to this Act, give the protection and arrangement to forest and Midlife.
2. Appropriate use of forest land – Uses of extra land for forest animals and grow some plant these are useful in wild animals.
3. Act for protection of forest – This act, stop passion of deforestation and must be conservation of forest and wild animals.
4. Increasing of forest product – Due to this act try to increasing of forest product and discovered new information which are better for wildlife.

Question 9.
Write an essay on measures of forest conservation.
Answer:
Forest conservation:
Forest conservation and management are essential to maintain the forests in their natural state and also to prevent the depletion of wildlife and forest wealth. For the success of conservation it is necessary to know the cause of depletion and destruction of forests. Forests are generally destroyed by fire, improper cutting of trees and by animals.

Essentiality of forest conservation:
Forest is a complex system which is responsible for the ecological balance in nature. Deforestation causing natural imbalance and affects the biotic components of the environment resulting floods, drought, epidemics, environmental pollution. Many ecologically important species of plants and animals are lost due to which economically important substance like wood, medicines, resin, lac and various food materials will not be available for us.

Measures of forest conservation : The following measures or efforts are prescribed for reforestation:

1. Establishment of conserved forests and their conservation in proper way.
2. Reforestation on deforested land. Old and damaged plants would be replaced by new plants.
3. Proper management of forests.
4. By promoting public awareness about forests.
5. Replacement of burnt off areas of the forests.
6. Plantation of trees that increase forest productivity.
7. Forestation of plants on hills and wastelands and prevention of grazing by cattle.
8. Prevent forests from fire, diseases and insects.
9. Providing basic protection for all forests by law.
10. Regulating human activity in the forest such as grazing by cattle and collection of firewood and fodder etc.
11. Provide special attention for the conservation of endangered plant and animal species under the inspection of specialists.
12. Removal of undesirable trees and vegetation for the better growth of desirable species.
13. Forestation of industrially useful plants.
14. The government will arrange the management of useful forests.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Biology Important Questions Chapter 9 Strategies for Enhancement in Food Production

Strategies for Enhancement in Food Production Important Questions

 Strategies for Enhancement in Food Production Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Triticum aestivum wheat is:
(a) Haploid
(b) Diploid
(c) Tetraploid
(d) Hexaploid.
Answer:
(c) Tetraploid

Question 2.
Man – made cereal is :
(a) Potato
(b) Triticale
(c) Triticum
(d) Sugarcane.
Answer:
(b) Triticale

Question 3.
In cattles, anthrax is disease is caused by :
(a) Bacteria
(b) Fungi
(c) Virus
(d) Ticks.
Answer:
(a) Bacteria

Question 4.
The causal organism of haemorrhagic septicaemia as :
(a) Brucella avartus
(b) Bacillus sp.
(c) Pasturella boviseptica
(d) Clostridium.
Answer:
(c) Pasturella boviseptica

Question 5.
Foot and Mouth disease of cattles is caused by :
(a) Fungi
(b) Bacteria
(c) Virus
(d) Mycoplasma.
Answer:
(c) Virus

Question 6.
The disease caused after rainy season is :
(a) Black fever
(b) Haemorrhagic septicaemia.
(c) Ponkni
(d) Anthrax.
Answer:
(c) Ponkni

Question 7.
The vaccination for galghontu is given to animal in :
(a) January – February
(b) March – April
(c) May – June
(d) October – Nov ember.
Answer:
(c) May – June

Question 8.
The cause of plague disease of animals is :
(a) Fungi
(b) Bacteria
(c) Virus
(d) Mycoplasma.
Answer:
(c) Virus

Question 9.
Example of cereal plant is :
(a) Wheat
(b) Rice
(c) Maize
(d) All of these.
Answer:
(d) All of these.

Question 10.
Botanical name of common wheat is :
(a) Triticum aestivum
(b) T. vulgare
(c) T. durum
(d) T. monococcum
Answer:
(b) T. vulgare

Question 11.
Wheat and rice belong to family
(a) Graminae
(b) Papillionaceae
(c) Euphorbiaceae
(d) Compositae
Answer:
(a) Graminae

Question 12.
Padma and Jaya are the improved varieties of:
(a) Wheat
(b) Rice
(c) Gram
(d) Groundnut
Answer:
(b) Rice

Question 13.
Botanical name of Ratanjot is :
(a) Pongamia pinnata
(b) Ricinus communis
(c) Jatropha curcas
(d) None of these
Answer:
(c) Jatropha curcas

Question 14.
Fertilizer elements are:
(a) Nitrogen
(b) Phosphorus
(c) Potassium
(d) All of these
Answer:
(d) All of these

Question 15.
Fertilizer that supplies least % of nitrogen is :
(a) Urea %
(b) (NH₄) SO₄
(c) (NH₄) NO₃
(d) Organic nitrogen fertilizer.
Answer:
(d) Organic nitrogen fertilizer.

Question 16.
Fertilizer that supplies both nitrogen and phosphorus to soil:
(a) Ammonium sulphate
(b) Ammonium nitrate
(c) Urea
(d) Super phosphate
Answer:
(d) Super phosphate

Question 17.
Nitrogen fixing organism is :
(a) Rhizobium
(b) Azolla
(c) Azotobacter and Azospirillum
(d) All of these
Answer:
(d) All of these

Question 18.
VAM is :
(a) Bacteria
(b) Yeast
(c) Fungi
(d) Virus.
Answer:
(c) Fungi

Question 2.
Fill in the blanks :

1. The process of separating animals with desired characters for breeding purpose is called ………………………..
2. Breeding between two genetically different animals is called ………………………..
3. Anthrax fever of cattles is caused by bacterium ………………………..
4. Ranikhet disease of fowls was first observed in ……………………….. district.
5. Fowl pox disease is caused by ………………………..
6. No. of chromosomes in Triticum vulgare is ………………………..
7. Groundnut belongs to family ………………………..
8. Botanical name of Karanj is ………………………..
9.  ……………………….. is the first genetically engineered food.
10. Rhizobium, Azolla and Anabaena are the examples of ………………………..
11. Insecticides azadirachtin is obtained from leaves of ……………………….. plants.
12. Insects harming plants are called ………………………..
13.  ……………………….. bacterium is found in the roots of leguminous plants. (MP 2015)
14. Nostoc and Anabaena are called ………………………..
15. Bacteria responsible for N₂ fixation is ……………………….. (MP 2009 Set A)

Answer:

1. Selection
2. Outbreeding
3. Bacillus anthraxis
4. Ranikhet (Kumau)
5. Virus
6. 42
7. Papillionaceae
8. Pongamiapinmta
9. Flavr – Savr Tomato
10. Biofertilizers
11. Neem
12. Pest
13. Rhizobium
14. Cyanobacteria
15. Rhizobium.

Question 3.
Match the followings :
I.

Answer:

1. (c)
2. (d)
3. (e)
4. (a)
5. (b).

II.

Answer:

1. (d)
2. (e)
3. (a)
4. (c)
5. (b).

III.

Answer:

1. (d)
2. (c)
3. (a)
4. (b)

Question 4.
Write the answer in one word/sentences:

1. Name the type of vegetative propagation in the plants having bulb.
2. Name the product of hybridization of two different species.
3. Name the first man – made crop.
4. Give one example of millet.
5. Write botanical name of coconut. (MP 2013,17)
6. What is called breeding between two closely related animals?
7. In which animal Ranikhet disease occur ? (MP 2015)
8. Name the causal organism of anthrax disease of cattles.
9. Write the name of galghontu disease of cattles.
10. Which hormone is used for production of milk in cows? (MP2016)
11. Give the botanical name of any one cereal plants.
12. Name the soil best suited for rice cropping.
13. Give the botanical name of any oil seed crop.
14. Give the botanical name of Ratanjot.
15. Name the technique by which genetically modified plants are produced.

Answer:

1. Scaling
2. Hybrid
3. Triticale
4. Rye
5. Coccus nucifera
6. Inbreeding
7. Fowls
8. Bacillus anthraxis
9. Pasturella boviseptica
10. Stilbrestol hormone,
11. Triticum aestivum (Wheat)
12. Clay loam
13. Arachis hypogea (Groundnut)
14. Jatropha curcas
15. Transgenesis.

Strategies for Enhancement in Food Production Short Answer Type Questions

Question 1.
What is meant by the term ‘breed’? What are the objective of animal breeding?
Answer:
Breed referes to the group of animals having same ancestral characters, general appearance, size, etc.

Objectives of animal breeding:

1. To increase the quantity of yield.
2. To improve the desirable qualities of the produce.

Question 2.
Name the methods employed in animal breeding. According to you which of the methods is best? Why?
Answer:
The methods employed in animal breeding are :

1. Natural methods: These can be carried out by inbreeding and outbreeding methods.
2. Artificial methods : These are carried out by artificial insemination and Multiple Ovulation Embryo Transfer (MOET).

The artificial method of animal plant breeding is best as it ensures good quality of progeny. It is also economic and time saving process to obtain the desirable progeny.

Question 3.
What is apiculture? How is it important in our lives?
Answer:
Apiculture is the practice of bee – keeping for the production of various products such as honey-bee’s wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines. Other commercial products obtained from honeybees include bee’s wax and bee pollen. Bee’s wax is used for making cosmetics, polishes and is even used inseveral medicinal preparations. Therefore, to meet the increasing demand of honey, people have started practicing bee – keeping on a large scale. It has become an income generating activity for farmers since, it requires a low investment and is labour intensive.

Question 4.
Discuss the role of fishery in enhancement of food production.
Answer:
Fishery is an industry which deals with catching, processing and marketing of fishes and other aquatic animals that have a high economic value. Some commercially important aquatic animals are Prawns, Crabs, Oysters, Lobsters and Octopus. Fisheries play an important role in the Indian economy. This is because a large part of the Indian population is dependent on fishes as a source of food, which is both cheap and high in animal protein. A fishery is an employment generating industry especially for people staying in the coastal areas.

Question 5.
Briefly describe various steps involved in plant breeding.
Answer:
The major steps in breeding a new genetic variety of a crop are as follows:

1. Collection of variability.
2. Evaluation and selection of parents.
3. Cross – hybridization among the selected parents.
4. Selection and testing of superior recombinants.
5. Testing, release and commercialization of new cultivars.

Question 6.
Explain what is meant by biofortification.
Answer:
Biofortification is a process of breeding crops with higher levels of vitamins, minerals, proteins and fat content. This method is employed to improve public health. Breeding of crops with improved nutritional quality is undertaken to improve the content of proteins, oil, vitamins, minerals and micro – nutrients in crops. It is also undertaken to upgrade the quality of oil and proteins. An example of this is a wheat variety known as Atlas 66, which has high protein content in comparison to the existing wheat. In addition, there are several other improved varieties of crop plants such as rice, carrots, spinach etc. which have more nutritious value and more nutrients than the existing varieties.

Question 7.
Which part of the plant is best suited for making virus free plant and why?
Answer:
The terminal bud having apical meristem are the best suited parts of plant for making virus free plant because they are not infected by virus.

Question 8.
What are the major advantages of producing plants by nticropropagation?
Answer:
Major advantages of producing plants by micropropagation are:

1. Large number of plants can be grown in short – time
2. Disease – free plants can be obtained.
3. Plants that have lost the capacity to produce seeds can be grown
4. The plants where sexual reproduction is absent, may be hybridised by tissue culture.
5. Plants produced are genetically similar to the parent and have all its characteristics.

Question 9.
Find out what are the various components of the medium used for propagation of an explant in vitro are?
Answer:
The major components of the medium for in-vitro propagation are :

1. Water
2. Agar – agar
3. Sucrose
4. Inorganic salts
5. Vitamins
6. Amino acids
7. Growth hormones like Auxin, Cytokinins.

Question 10.
Name any five hybrid varieties of crop plants which have been developed in India.
Answer:

1. Cauliflower varieties – Pusa shubhra and Pusa snowball K – 1
2. Brassica varieties – Pusa swamim (Karan rai)
3. Wheat varieties – Himgiri
4. Rice varieties – Jaya and Ratna
5. Chilli varieties – Pusa Sadabahar.

Question 11.
What is scaling? What are its importance?
Answer:
Scaling:
It is a method of vegetative propagation which is applicable for growing plants having bulbs, (example Onion, Garlic). In this method, all the scales of bulb are separated and planted in the field (Soil) having all the necessary conditions for their growth. Here scales develop to produced small bulblets. About 3 to 5 bulblets developing from one scale. This method is applicable for the plants belonging to the family Liliaceae. example Garlic, Onion, Lilium, etc.

Question 12.
What is tissue culture? What are its objectives?
Answer:
Tissue culture:
“Tissue culture is an experimental process through which a mass of cells (callus) is produced from an explant tissue and the callus produced in this way can be used directly either to regenerate plantlets or to extract to some primary and secondary metabolites.”

When appropriate culture conditions were provided, cell masses could then proceed along various developmental pathway, to regenerate shoot and root organs and eventually whole plants.

Aims of Plant Tissue Culture:

1. To develop new plants from the plant organ other than seeds.
2. To produce hybrid varieties of plants.
3. To produce disease free plants from diseased plants.
4. To reduce the period of reproductive cycle.
5. Development of haploid plants.
6. To develop stress resistance plants.

Question 13.
What is callus culture? Give its technique.
Answer:
Callus culture:
Callus is the unorganised and undifferentiated mass of tissue. The plant body of higher plants is made of multicellular, well organised differentiated structures like root, stem, leaves etc. Sometimes the cells of these differentiated structures proliferate to form large mass of unorganized and undifferentiated cells which is called as callus and this process is called callus culture. In this methods, the isolated plant cells tissues or organs are cultured in nutrient medium in glass culture tubes or in petridishes under aseptic conditions. The cultured part is called as explant. The growth responses depend upon the source of the explant, composition of the nutrient medium and the suitable growth conditions.

Technique of callus culture : Callus culture involves the following steps :

1. First it is necessary to sterilize the plant organ from which an explant is taken. Sterilizing agents commonly used are mercuric chloride solution (0.1 w/v), sodium hypochloride (1.6 available chlorine) and a solution of bromine in water (1% w/v). Explants can be taken from seedlings (cotyledon, hypocotyl or root) or mature organs or from wood stem parts.

2. Wash the explant with distilled water, cut small pieces of it and transfer it on suitable culture medium. Agar agar is used for making culture medium. The culture media was supplemented with amino acids, vitamins, kinetin or other growth factor. An auxin and usually a cytokinin promote high growth rate of callus.

Question 14.
Why aeration is essential in the process of tissue culture?
Answer:
For any living being, respiration is a must. Through respiration only it can perform various activities. Oxygen is necessary for respiration.Tissue culture also needs oxygen, then only it can grow. Aeration provides oxygen to the plant parts, thus aeration is essential in the process of tissue culture.

Question 15.
What is inbreeding ? Give its advantages.
Answer:
The mating of more closely related individuals within the same breed for 4 – 6 generation is called inbreeding. It may lead to inbreeding depression i.e., the loss of fertility, vigour and productivity of the hybrid.

Question 16.
List the components which are present in honey. Name the three species of honeybee and write the chemical compositions of honey.
Answer:
There are three species of honeybee:

1. Apis indica
2. Trigona species
3. Malinopa species.

Uses of honey:

1. In the form of medicine.
2. In the form of nutritive food material.

Chemical composition :

Fructose – 41%
Glucose – 35%
Sucrose – 1.9%
Dextrine – 1.5%
Protein – 0.18%
Mineral salts – 3.3%
Water – 17.25%

Some amount of vitamins B, B6, Coline, Vitamin C and D.

Question 17.
What precautions should be kept in the process of incubation?
Answer:
Incubatory period is 21 days in fowl, precautions of incubation of hens are as follows:

1. The eggs chosen should be of good quality.
2. Egg size should be medium.
3. Colour of egg should be white.
4. Eggs should be washed in water.
5. Eggs should not be shaken.
6. In summers, eggs should not be kept more than 30 days.
7. At night, the fowl should be fed before incubation. Incubation of egg should be done by Indian fowl.

Question 18.
Give three significance of poultry farming.
Answer:

1. Poultry provide humans with companionship, food and fiber in the form of eggs, meat and feathers.
2. Poultry rearing and poultry farming is a good source of income.
3. It provides additional income and job opportunities.

Question 19.
Give the characteristics of meat providing chickens and give two examples of these.
Answer:
Characteristics of meat providing chickens are as follows:

1. Size is big.
2. They intake large quantity of nutrition.
3. Theii feathers are loose.
4. Their growth rate is low.
   Example: Australorps, Sussex.

Question 20.
Write the examples of three freshwater and three marine water fishes and give its significance.
Answer:
Examples of three freshwater fishes :

1. Rohu
2. Catla,
3. Seenghala.

Significance of Freshwater fishes :

1. Rohu (Labio rohita) : Its brain is rich in phospho protein which improve eye sight.
2. Catla {Catla catla) : Its brain is rich in phospho protein which improve eye sight.
3. Seenghala {Mystus seenghala) : It is rich in iron and copper which is good for circulatory system.

Examples of three marine water fishes:

1. Hilsa
2. Pomfret
3. Bombayduck.

Significance of Marine water fishes :

1. Hilsa (Ilishaa species): It is rich in omega – 3 fatty acid.
2. Pomfret {Sfromatus niger): High in vitamin D
3. Bombay duck {Harpodon): It is the source of calcium.

Question 21.
Fish meat is better than other animals, why?
Answer:

1. Fish meat is better because it is rich in protein.
2. Iodine is found in it which protects from goiter disease.
3. It has low fat so, it protects our heart.
4. Fat soluble vitamin A and D are found in much quantity.
5. It is digestible. So, it is better for children.

Question 22.
Name the developed breed of cow and buffalo. Explain why buffalo milk is better than cows milk.
Answer:
Developed varieties of Cow:

1. Holstein friesian
2. Jersey
3. Ayer Shayer and
4. Brown swiss.

Developed varieties of Buffalo:

1. Murra
2. Surti
3. Bhadavari and
4. Nagpuri.

Buffalo’s milk is better than Cows milk because :

1. Buffalo gives triple qunatity of milk than cow.
2. Buffalo milk is rich in fat.
3. It has much resistance power.

Question 23.
What is the significance of sheep and goat? Give the name of three Indian species of each.
Answer:
Zoological name of sheep is Ovis aries. It gives wool, meat and leather. Its Indian species are:

1. Lohi: It gives carpet quality of wool and meat production.
2. Bakharwal.
3. Patanwadi.

Zoological name of goat is Capra aegagrus. It gives us both meat and milk. The three, species are as follows :

1. Cashmere goat.
2. Sirohi.
3. Jamunapari.

Question 24.
Name the five species of chicks.
Answer:

1. Rhode Island Red
2. New Hampshired
3. Light Sussex
4. Australorps
5. White Leghorn.

Question 25.
How foot and mouth disease of cattles spreaded? Write symptoms and control measures of this disease.
Answer:
Foot and mouth disease is a highly contagious viral disease caused by Foot and Mouth Disease Virus (FMDV). It affects clovenhoofed animals including cattle, buffalo, camels, sheep, goats, deer and pigs. Since, it is highly infectious, can be spread by infected animals through contact with contaminated animals.

Symptoms of disease:

1. High fever lasting two to five days.
2. Animals become dull, eat less due to painful lesions in the mouth.
3. Weight loss.
4. Ulcer in mouth, foot and teats and causes wound thus, parts become painful and it fails to walk properly.

Question 26.
Give characteristics of egg laying chicken.
Answer:
Characteristics of egg laying chicken are as follows:

1. Their skin is soft.
2. A gap of 3 – 4 fingers is there between pubic and keel bones.
3. Their body is heavy in weight.
4. Their vent is soft and moist.
5. Their romb is well developed, red in colour and soft.

Strategies for Enhancement in Food Production Long Answer Type Questions

Question 1.
Write the name of causal organism, symptoms of disease and control measures of ranikhet disease of fowls.
Answer:
1. Ranikhet Disease : This disease was first discovered from the hills of Ranikhet (Kumau), hence called Ranikhet disease.

2. Causal organism: Virus.

3. Symptoms:

• Difficulty in breathing.
• Diarrhoea.
• Head, neck and legs are paralyzed.
• Loss of appetite.
• Increase in the temperature of the body.
• Secretion of mucilagenous substance from the mouth and nostrils.
• The colour of the body become voilet.
• Laying of egg is inhibited and eggs become ruptured.

4. Control and Treatment:

1. Isolation.
2. Died individuals should be immediatly burned.
3. Water should be disinfected.
4. Vaccination by ranikhet vaccine :

• F strain vaccine given to 1 to 3 days old chicks.
• Freeze dried chick embryo vaccine is given to 6 to 8 months old chicks.

Question 2.
Write the method and advantages of artificial breeding in cattles.
Or
What is artificial insemination? Write its importance.
Answer:
Artificial Hybridization:
It is a method in which sperms of male are injected into the reproductive tract of female ones. Where they fertilize egg to produce individuals with new characters.

Artificial hybridisation involves the following steps:

1. Selection of Parents:
It is the first step of natural hybridization in which male and female individuals with desired characters are selected. Usually healthy animals with desired characters are selected.

2. Collection of Semen:
It is the second step of artificial hybridization. In this step male individuals are stimulated mechanically or electrically so, that they release semen. This semen is collected in vials.

3. Preservation of Semen:
Collected semen is diluted with appropriate diluting liquid and is stored in refrigerators.

4. Introduction of Semen into vagina (= Insemination):
In this step, the desired semen is injected into the vagina of female individuals when they are heated. This technique was used for the first time by Spallanzani in 1780 in dogs. In India, this technique was used for the first time by Animal Research Institute, Eta Nagar (U.P.). More than 10 – 70% species of cattles are developed by this technique.

Precaution taken during insemination:

1. Insemination should be done in appropriate period.
2. Only high quality semen should be used.
3. Correct technique of insemination should be used.
4. The animals should be healthy during insemination.

Advantages of artificial insemination:

1. Few semen of healthy males are used for inseminating many female individuals.
2. The transportation of semen in ampules is easier.
3. The problem of availability of suitable males is solved by the development of this technique.
4. Cattles with desired characters should be used.

Question 3.
How can we provide high living standards for poultry farming?
Answer:
The steps for providing high living standards are as follows:

1. Protection
2. Provide proper places
3. Necessary facility
4. Cheap and comfortable
5. Clean
6. Facility of water and light
7. Nearness from market
8. Protection from parasites and other insects.

While the construction of poultry farms see that they are made of low prices and keep in mind the following things:

1. There is no moisture left in the farm.
2. Farm should be in a place where there is proper sunlight because microorganism can not grow in this situation.
3. There should be proper aeration facility because there are no sweat glands present in fowls as they release moisture by
4. the process of breathing.
5. There should be cleanliness in farm.
6. Fowl is a feable bird who has lots of enemies this is the reason they should be defended.

Question 4.
What is aquaculture? Describe various steps of aquaculture.
Answer:
The culture of useful aquatic organisms specially animals is called as aquaculture.

Steps involved in aquaculture (Fishery):
Four types of ponds are required for the culture of fished at commercial level.

1. Hatchery tank : For fish culture first of all fish seeds are sown in hatchery tank.
2. Nursing tank : After hatching, small fishes are transferred to the nursing tank.
3. Rearing tank : When fishes attain a size of 2 – 3 inches than they are transferred into the rearing tank.
4. Storage or Stocking tank : After attaining Suitable size, these fishes are transferred into the storage tank where these fishes are stored and supplied to the market.

Question 5.
What is single cell culture? Explain paper raft technique and its application.
Answer:
Single Cell Culture :
The culturing of a single cell on suitable sterilized culture media under controlled and aseptic conditions is called single cell culture. Establishment of a single cell culture provides an excellent opportunity to investigate the properties and potentialities of plant cells. Several workers have successfully isolated single cell division and even raised complete plants from single cell cultures.

Paper raft technique of single cell culture:
1. In this technique, the single cells are isolated with the help of micropipette and transferred on to the upper surface of filter paper resting on nurse callus. These active nurse callus and the culture medium piovides growth factors to the single cells.

2. The single cells then divides and as a result of proliferation in them, the colonies develop.

3. Colonies formed in nutrient medium are transferred to the subcultured fresh nutrient media. The callus developing from such single cells are called as single cell clones.

Question 6.
There is a scope for better prospects through aquaculture in the rural areas of Chhattisgarh. Discuss.
Answer:
In the village area of Chhattisgarh ponds are developed. In all villages and towns of Chhattisgarh many ponds can be seen. Canals, dams and water reservoirs are constructed for development purpose. These all sources can be used for progress of the state by scientific way.

Production of useful aquatic organisms is known as aquaculture. Various algae and aquatic animals such as fishes, prawn, crab, oyster etc. are cultured in pond Pisciculture, Prawn culture and Pearl culture are various main culture of aquaculture. Pisciculture is rearing and management of fishes. Prawn culture is production of prawn. Pearl culture is rearing and management of oyster for pearl production. All these productions has commercial importance.

Question 7.
How can crop varieties be made disease resistant to overcome food crisis in India? Explain, Name some disease resistance variety in India. Give the method of breeding for disease resistant variety.
Answer:
Crop varieties can be made disease – resistant by conventional breeding methods or by mutation breeding. The germplasm is screened for resistance source or mutations are introduced followed by hybridization of selected parents. The resulting hybrids are evaluted and tested finally, disease resistant varieties are released.

Disease resistant variety of:

1. Wheat to leaf and stripe rust – Himgiri.
2. Brassica to white rust – Pusa swamim.

Methods to breeding for disease resistance:

1. Hybridization.
2. Selection.

These steps are involved it:

1. Selections of breed from resistant source.
2. Hybridization of selected breed.
3. Selection of Hybrid.
4. Evaluation.
5. Testing of new variety and its production.

By hybridization and selection some fungi, bacteria and viruses can be disease resistant. These varieties given below :

Table : Disease resistant variety of crops.

Question 8.
How is plant breeding helpful to provide resistance to pests?
Answer:
Plant breeding for developing resistance to insect pests:

1. The host crop plants may be resistant to insect pests due to the morphological, biochemical or physiological characteristics.

2. Some characteristics that lead to pest resistance are :

Hairy leaves in plant example resistance to Jassids in cotton and cereal leaf beetle in wheat.
Solid stem in wheat exhibits non – preference by stem sawfly.
In cotton, smooth leaf and absence of nectar repel bollworms.
In maize, high aspartic acid, low nitrogen and sugar content protects them from stem borers.

3. The steps of breeding method is same as for the other agronomic traits.

4. Some varieties developed by hybridization and selection are as follows:

Question 9.
Explain in brief the role of animal husbandry in human welfare.
Answer:
1. Animal husbandry evolves new techniques and technologies for the management of livestock like buffaloes, cows, pigs, horses, cattle, sheep, camels, goats, etc., that are useful to humans.

2. These methods can also be applied rearing animals like bees, silkworm, prawns, crabs, fishes birds, pigs, cattle, sheep and camels for their products like milk, eggs, meat, wool, silk, honey, etc.

Question 10.
If your family owned a diary farm, what measures would you undertake to improve the quality and quantity of milk production?
Answer:
Dairy farm management deals with processes which aim at improving the quality and quantity of milk production. Milk production is primarily dependent on choosing improved cattle breeds, provision of proper feed for cattle, maintaining proper shelter facilities and regular cleaning of cattle. Choosing improved cattle breeds is an important factor of cattle management. Hybrid cattle breeds are produced for improved productivity.

Therefore, it is essential that hybrid cattle breeds should have a combination of various desirable genes such, as high milk production and high resistance to diseases, Cattle should also be given healthy and nutritious food consisting of roughage, fibre concentrates and high levels of proteins and other nutrients. Cattle’s should be housed in proper cattle-houses and should be kept in well ventilated roofs to prevent them from harsh weather conditions such as heat, cold and rain. Regular baths and proper brushing should be ensured to control diseases. Also, time – to – time check – ups by a veterinary doctor for symptoms of various diseases should be undertaken.

MP Board Class 12th Biology Important Questions

MP Board Class 12th Physics Important Questions Chapter 6 Electromagnetic Induction

Electromagnetic Induction Important Questions

Electromagnetic Induction Objective Type Questions

Question 1.
Choose the correct answer of the following:

Question 1.
The SI unit of magnetic flux is :
(a) Weber
(b) Gauss
(c) Oersted
(d) Tesla
Answer:
(a) Weber

Question 2.
The flux linked with a coil can be charged by :
(a) Keeping the coil in a time varying magnetic field
(b) Rotating the coil in a magnetic field
(c) By charging area of coil in the uniform magnetic field
(d) All the above.
Answer:
(d) All the above.

Question 3.
The direction of induced current can be determined by :
(a) Lenz’s law
(b) Flemming’s left hand rule
(c) Right hand palm rule
(d) Ampere’s swimming rule.
Answer:
(a) Lenz’s law

Question 4.
In the phenomenon of elctro magnetic induction :
(a) Electrical energy is transformed into mechanical energy
(b) Mechanical energy is transformed into electrical energy
(c) Electrical energy is transformed into thermal energy
(d) Mechanical energy is transformed into thermal energy.
Answer:
(b) Mechanical energy is transformed into electrical energy

Question 5.
The self inductance of a coil does not depend upon :
(a) Radius of coil
(b) Number of turns in the coil
(c) Lenght of coil
(d) Current through the coil.
Answer:
(d) Current through the coil.

Question 6.
If a core of soft iron is placed between the coil then their mutual inductance :
(a) Decreases
(b) Increases
(c) Remain unchanged
(d) Nothing can be said.
Answer:
(b) Increases

Question 7.
The eddy current are used in :
(a) Electrolysis
(b) Making a galvanometer dead beat
(c) Electroplating
(d) Increasing the sensitivity of galvanometer.
Answer:
(b) Making a galvanometer dead beat

Question 2.
Fill in the blanks :

1. Lenz’s law is used to find out direction of ……………
2. The SI unit of magnetic flux is ……………
3. The core of transformer is made laminated to reduce the loss of energy due to …………… currents.
4. To make the moving coil galvanometer …………… its coil is wound on the aluminium frame.
5. Magnetic flux is a …………… quantity.

Answer:

1. Induced current
2. Weber
3. Eddy
4. Dead beat
5. Scalar.

Question 3.
Match the Column :
I.

Answer:

1. (d)
2. (c)
3. (e)
4. (a)
5. (b)

II.

Answer:

1. (e)
2. (c)
3. (d)
4. (b)
5. (a)

Question 4.
Write the answer in one word/sentence :

1. What is the unit of magnetic flux?
2. What is unit of coefficient of self induction?
3. A straight conductor of length ‘l’ moving with a velocity of ‘V’ in a uniform magnetic field ‘B’ Perpendicular to it. Write the
4. formula for induced e.m.f. in it.
5. Write relation between Henry and Micro Henry.
6. Write the principle of alternating current generator.
7. Induced electric field is conservative field or non conservative field.
8. In which form the energy is stored in a coil?
9. By which law we determine the direction of induced e.m.f.?

Answer:

1. Weber
2. Henry
3. e = Blv
4. 1H = 10⁶pH
5. Electro magnetic induction
6. It is Non conservative field
7. Magnetic energy
8. Lenz’s law.

Electromagnetic Induction Very Short Answer Type Questions

Question 1.
When induced current flow in a closed circuit?
Answer:
When there is change in magnetic flux in a closed circuit, then induced current flow over it.

Question 2.
If a iron core is kept at the centre of a coil, then what will be the effect on coefficient of self induction?
Answer:
Coefficient of self induction will increase.

Question 3.
Lenz’s law is accordance to which law?
Answer:
Law of conservation of energy.

Question 4.
If any metal piece is kept in varying magnetic field, will eddy current generate in it?
Answer:
Yes, eddy current generate in the metallic piece.

Question 5.
Why coil with in the resistance box is double coiled?
Answer:
To avoid self – induction phenomenon.

Question 6.
When the magnitude of magnetic flux become maximum and minimum?
Answer:
When magnetic field is perpendicular to the plane, the magnetic flux is maximum and when it is parallel, its value is minimum.

Electromagnetic Induction Short Answer Type Questions

Question 1.
What is magnetic flux? Write its SI units.
Answer:
The number of magnetic lines of force passing normally through any area in a magnetic field, is called magnetic flux linked with that area. It is denoted by ϕ(phi).

Where, \(\vec { B }\) is the magnetic field, \(\vec { dS }\) is the normal to the surface and θ is the angle between \(\vec { B }\) and normal to surface. SI unit of fluxs weber.

Question 2.
Write Lenz’s law of electromagnetic induction.
Answer:
According to this law, the nature of induced e.m.f. is such that it opposes the cause which produces it. With the help of this law, the direction of induced e.m.f. can be determined.
∴ e = \(\frac {dϕ}{dt}\)
– ve sign is according to Lenz’s law.

Question 3.
A bird sitting on a wire carrying current at high voltage, flies away. Why?
Answer:
As soon as the current flows through the wire, an induced current is produced in the body of the bird, the direction of the current in the wings are opposite. Hence, they are repelled and the bird flies away.

Question 4.
In the given figure, which plate of the capacitor AB is positive and which plate is negative?
Answer:

According to Lenz’s law induced current flows from B to A, therefore plate A is positive and plate B is negative.

Question 5.
A metallic chain touching the ground is attached with the truck carrying explosive matter. Why?
Answer:
During the motion of the truck, the axle cut the magnetic flux of the earth, which causes an induced e.m.f. set up across its axle. This induced charges are earthed by the chain. Which prevent the explosive from the fire.

Question 6.
In metre bridge experiment, the battery’ key is pressed first then galvanometer. Why?
Answer:
If the galvanometer key is pressed first and then battery key, an induced current will be produced in the galvanometer circuit and galvanometer will give deflection even at the position of null point. Thus, the battery key is pressed first.

Question 7.
The resistances in the resistance box are double coiled. Why?
Answer:
In the double coil, the current flows in opposite directions, thus the magnetic fields produced cancel each other and hence induced current is not produced.

Question 8.
When any electric circuit is suddenly cut off, then sparks take place. Why?
Answer:
When the switch is off, the magnetic flux linked with the circuit becomes zero suddenly. Therefore, a strong induced current flows through the circuit in the direction of main current, which causes a high potential difference across the terminal and due to electric discharge, sparks are seen.

Question 9.
What is mutual induction?
Answer:
When the current changes in a coil, then the flux linked with another coil placed near the first changes, hence an induced current is produced in another coil. This phenomenon is known as mutual induction.

Question 10.
A coil in magnetic field is moved

1. Rapidly and
2. Slowly.

Are the e.m.f. and work done in the two cases equal or not?
Answer:
No, the e.m.f. induced in the first case will be larger due to larger rate of change of flux. Obviously, in the second case, the rate of change of flux is slow and hence induced e.m.f will be smaller.

Question 11.
Two coils A and B are kept perpendicular to each other as shown in figure. If flow of current in any one coil is change will induced current will generate in the second.
Answer:
When current is flows in use one coil, the induced flux produce will be parallel to second coil and there will be no change in magnetic flux. Hence no induced current will be produced in second coil.

Question 12.
A ring is hanging in a wall of room. When the north pole of the magnet is brought closer to the ring. Then what will be the direction of induced current if seen from the direction of magnet?
Answer:
The face of the ring which is toward the north pole of the magnet, in.it the direction of induced current will be anti clockwise. Hence this face become magnetic north pole and resist the magnetic pole to come near it.

Question 13.
Three coils are arranged as shown in the figure :

In which coil mutual induction will be maximum?
Answer:
In figure (a), the mutual induction will be maximum before the magnetic lines of force passing in figure (a) is more in comparison to other figure.

Question 14.
What is electromagnetic induction? State Faraday’s laws of electromag neticinduction.
Or
Write down the laws of electromagnetic induction. Find out the expression for the induced electromotive force.
Answer:
When a magnet and a conducting coil are in relative motion, an e.m.f. is induced in the coil. This e.m.f. is called induced e.m.f. If the coil is closed, a current flows in it, then this current is called induced current and the phenomenon is called electromagnetic induction.

Faraday’s laws of electromagnetic induction:
1. Whenever there is a change in magnetic flux linked with an electrical circuit, an e.m.f. is induced in the circuit, therefore an induced current flows in it. The induced current losts so long as there is change in flux.

2. The induced e.m.f. is directly proportional to the rate of change in magnetic flux.

Formula for induced e.m.f.:
Let Φ₁ weber be the flux linked with a closed electrical circuit at any instant of time t₁. After time ∆t i.e., at time t₂ seconds, let the flux be Φ2 weber.

This is the required formula.
Hence, the induced e.m.f. is directly proportional to the rate of change of flux.

Question 15.
Write Lenz’s law and explain how it can be used to determine the direction of induced current?
Answer:
Lenz’s law:
According to this law, the direc¬tion of induced current is such that it opposes the very cause producing it. With the help of this law, the direction of induced current can be determined as follows :

1. When N – pole of a bar magnet is brought closer to a coil, a current is induced to develop N – pole on the adjacent face of the coil to oppose the motion of N – pole towards the coil. This means the induced current must be anticlockwise as seen from the magnet, [see Fig.(a)].

2. When N – pole of the bar magnet is taken away from the coil, an induced current flows in the coil such that the face of the coil near the magnet becomes south pole which attracts the north pole and thus opposes the cause producing it (i.e., opposes the motion of the magnet away from the coil). So, the direction of the induced current in the coil must be clockwise as seen from the magnet, [see Fig. (b)].

Question 16.
Explain that Lenz’s law is in accordance with the law of conservation of eaefgy.
Answer:
When a magnetic north pole is brought near a plane of a coil, the magnetic pole induced in that plane is also north pole causing a repulsive force on the north pole of the magnet. Therefore, work will have to be done against this repulsive force to bring the magnet’s north pole closer to the coil. The work done (mechanical energy) is transformed into electri¬cal energy causing a current induced in the coil.

But, on moving the north pole of the magnet away from the coil, a south pole is developed on the face of the coil and an attractive force is developed on the face of the magnet. So, work has to be done in moving the magnet away from the coil. This results in flow of induced current in reverse direction. Thus, Lenz’s law is in accordance with the law of conservation of energy.

Question 17.
Explain self – induction and demonstrate the phenomenon by an experiment.
Answer:
Self – induction:
During the motion of the truck, the axle cut the magnetic flux of the earth, which causes an induced e.m.f. set up across its axle. This induced charges are earthed by the chain. Which prevent the explosive from the fire.

Experiment:
The circuit of the experiment is shown in the figure. It consists of an insulated copper coil L, wound on soft iron core. Cell E, rheostat Rh and a tapping key K are connected in series. A bulb B is connected in parallel with the coil. Now, the key is closed, the bulb glows slowly, then becomes bright and when it is opened, the bulb flashes bright and then goes off.

As the key is pressed, the flux linked with coil changes and an induced current flows in the opposite direction of main current, because of this the current grows slowly in the circuit.But when the key is opened, the flux suddenly decreases to zero, therefore a strong induced current flows in the same direction of circuit current makes the bulb very bright for a moment.

Question 18.
What is self – induction? What is meant by self-induction of coil? Explain. WritgTts unit.
Answer:
Self – inductance:
When the current flowing through a circuit, an induced e.m.f. is set up in the same circuit, which opposes the main current. This phenomenon is called self – induction.
When the current increases in the circuit, induced current flows in the opposite direc¬tion and when the current decreases in the circuit, the induced current flows in the same direction of circuit current.
Let ϕ be the flux linked with a circuit when a current (l) flows through it. Then,
ϕ ∝I
or ϕ = LI …(1)
Where, L is a constant, called self – inductance of the coil.
Now, if I = 1, then by eqn. (i), we get
ϕ = L
Thus, the self-inductance of a coil is numerically equal to the magnetic flux linked with the coil, produced by unit current.
By Faraday’s law, e = –\(\frac {dϕ}{dt}\)
Putting the value of ϕ from eqn. (i), e = –\(\frac {d(LI)}{dt}\)
or e = -L\(\frac {dI}{dt}\)
If \(\frac {dI}{dt}\) = 1
Then, e = – L
Therefore, the self – inductance of a coil is numerically equal to the opposing e.m.f. produced in the coil due to a unit rate of change of current in the circuit.
Unit : Unit of self-inductance is henry (H).

Question 19.
Explain mutual inductance. Give its unit. On what factors does it depend?
or
Write down the definition, unit and dimensional formula of mutual inductance.
Answer:
Mutual inductance is defined in two ways :
1. Let the flux linked with secondary coil be ϕ due to current l in primary coil.
∴ ϕ ∝ I
or ϕ = MI
Where, M is constant called mutual inductance.
If l = 1, then ϕ = M.
Hence, mutual inductance is defined by numerical value of flux linked with secondary coil, with unit current flows through primary coil.

2. When current changes in primary coil, then flux changes in secondary coil. Thus, by second law of Faraday
Induced e.m.fi, e = –\(\frac {dϕ}{dt}\)
or e = –\(\frac {d(MI)}{dt}\)
or e = ML\(\frac {dI}{dt}\)
If \(\frac {dI}{dt}\) = 1
Then, e = – M
Thus, the mutual inductance of two coils is numerically equal to the induced e.m.f. set up in secondary coil at that moment, when the rate of change of current in primary coil is unity.

Unit : Its unit is Henry in SI system.
Dimensional formula :

It depends upon:

1. Number of turns of both the coils : As the number of turns increases, mutual inductance also increases.
2. Area of cross – section of both the coils : As the area increases, the mutual inductance increases.
3. Material of core : In presence of soft – iron core, the inductance is greater than that of air.

Question 20.
Give differences between self – induction and mutual induction.
Answer:
Difference between self – induction and mutual induction :
Self – induction:

• This phenomenon happens in the same coil when the current flowing through the coil changes.
• One coil is used in this phenomenon.
• Induced current affects the main current.

Mutual induction:

• This phenomenon happens in the another coil placed nearby a coil in which the current is changing.
• Two coils are used in this phenomenon.
• The induced current produced in the second coil, therefore main current is not affected.

Question 21.
What are eddy currents? Show an experiment to demonstrate eddy currents.
Or
What are eddy currents? What are their disadvantages ? Write any two uses of eddy currents.
Answer:
Eddy currents:
When a metallic plate is moved in a magnetic field or placed in a changing magnetic field, then flux linked with the conductor changes, hence an induced current is produced in the plate. This induced current is called eddy current.

Demonstration:
A rectangular copper plate is free to oscillate about a horizontal axis, passing through O. It is placed between the pole pieces of an electromagnet NS. Oscillate the plate, when no current is flowing in the circuit, the plate will oscillate for a long time and due to air resistance finally its amplitude of oscillation decreases.

Now, oscillate and pass the current. It is observed that the oscillations of the plate will be damped soon. Just as eddies are produced on water surface, the change of flux through the metal plate develops induced current known as eddy currents.

Uses : Eddy currents are used in :

1. Making a galvanometer dead beat.
2. Induction furnace.
3. Electric break.
4. Induction motor, etc.

Disadvantages of eddy currents:
Production of eddy currents causes loss of electrical energy in the form of heat. In order to prevent this, soft-iron core is laminated in a transformer. Lamination increases resistance and decreases eddy currents. So, there is very little dissipation of electrical energy in the form of heat.

Question 22.
The self – inductance of two coils P and S are L₁ and L₂ respectively. If the coupling between them is ideal, then prove that the mutual inductance between these coils is M = \(\sqrt { { L }_{ 1 }{ L }_{ 2 } }\)
Answer:
Let coil P is having N₁ total number of turns, length of coil is l and I is the current flowing.

Question 23.
Obtain an expression for the self – inductance of two coils when they are joined in parallel.
Answer:
Consider two inductors of self – inductance L₁ and L₂and joined in parallel.
If I₁ and L₂, be the current flow through L₁and L₂ respectively, then
I = I₁ + I₂

Where 1 is the total current in the circuit differentiating both sides, we get
\(\frac {dI}{dt}\) = \(\frac { { dI }_{ 1 } }{ dt }\) + \(\frac { { dI }_{ 2 } }{ dt }\) … (2)
In parallel combination, the induced e.m.f, across each coil remains the same.

Putting the value of \(\frac {dI}{dt}\) From eqns. (1) and (3), we get

Equation (4) is the required expression

Electromagnetic Induction Long Answer Type Questions

Question 1.
Derive an expression for mutual inductance ¡n between two circular plane coils. Write the factors affecting mutual inductance between circular plane colts.
Answer:
Consider two circular plane coils P and S kept coaxially close to each other. Thenumber of turns in the primary coil P is n₁ and in the secondary coil S is n₂. The radius of primary coil P is r₁ and that of the secondary coil is r₂. Due to flow of current l in the primary coil, the magnetic field produced at its centre is
B = \(\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2\pi { n }_{ 1 }I }{ { r }_{ 1 } }\) = \(\frac { { \mu }_{ 0 }{ n }_{ 1 }I }{ 2{ r }_{ 1 } }\)
The magnetic flux linked with secondary coil S due to this magnetic field.
ϕ = nBA
=n₁ \(\frac { { \mu }_{ 0 }{ n }_{ 1 }I }{ 2{ r }_{ 1 } }\) πr²₂ … (1)
But, ϕ = MI … (2)
From eqns. (1) and (2),
\(\frac { { \mu }_{ 0 }{ n }_{ 1 }{ n }_{ 2 } }{ 2{ r }_{ 1 } }\).πr²₂
If some other medium is placed inside the coils in place of air or vacuum of perme – ability µ, then
M = µ\(\frac { { n }_{ 1 }{ n }_{ 2 } }{ 2{ r }_{ 1 } }\).πr²₂ … (4)

The mutual inductance between circular plane coils depends on following factors :

1. The mutual inductance increases on increasing the number of trims in primary coil and secondary coil.
2. On increasing the radius of primary coil r¹, the mutual inductance decreases.
3. On increasing the area of secondary coil, mutual inductance increases.
4. If a medium other than air is kept between the coil Mincreases.

Question 2.
Derive an expression for self – inductance of a long solenoid. On what factors does it depend?
Answer:
Expression for self – inductance of a long solenoid:
Let the length and radius of a solenoid be l and r respectively. Also n be the number of turns per unit length.

∴ Area of cross – section, A = πr²
and total number of turns = nl = N.
If current l flows through the solenoid, then the intensity of magnetic field inside the solenoid will be given by
B = µ₀nI, [∵B = \(\frac { { \mu }_{ 0 } }{ 4\pi }\)2πnl(cos0⁰ – cos180⁰)]
(for unit length if l > >r)
Where, µ₀ is the permeability of the medium.
∴ Magnetic flux linked with the total length of solenoid.

If the permeability of soft – iron core inside the solenoid is µ_r, then
L = \(\frac { { \mu }_{ 0 }{ \mu }_{ 1 }{ N }^{ 2 }A }{ l }\) … (3)
Equation (3) is the expression for self – inductance of a long solenoid. From this equation, it is clear that the self – inductance of a solenoid depends upon the following :

1. Area of cross – section of solenoid or radius of the solenoid : The self – inductance of a solenoid increases with radius and hence with the area of cross – section.

2. No. of turns : Self – inductance of a solenoid increases with number of turns.

3. Length of the solenoid : Self – inductance of a solenoid decreases on increasing the length of the solenoid.

4. Relative permeability of the core : The self – inductance of a solenoid increases on placing a core of higher permeability. This is the reason, that self-inductance of a solenoid with soft – iron core is greater than that of air core solenoid.

Equation (3) can be written in terms of radius as follows :
L = \(\frac { { \mu }{ N }^{ 2 }\pi { { r }^{ 2 } } }{ l }\)

Question 3.
Obtain an expression for mutual inductance of two long solenoids. What are the factors on which mutual inductance depends?
Answer:
Let S₁, and S₂ be the two long coaxial solenoids such that S₁, is completely surrounded by S₂. The lengths of both the solenoids are l. n₁ and n₂ are the number of turns per unit length respectively.

Case :
1. Let S₁ be taken as primary and S₂ as secondary solenoid. If I₁ is the current flowing through S₁ then the intensity of field inside S₁ will be
B₁ = μ₀n₁I₁
∴The magnetic field B₁ will superpose on solenoid S₂ and hence, magnetic flux linked with S₂ per turn = B₁A. Where, A is the area of cross – section of S₁.
Hence, net flux associated with S₂ will be
Φ₂₁ = n₂lB₁A
Where, n₂l is the total number of turns in solenoid S₂ and Φ₂₁ is magnetic flux on S₂ due to S₁
Substituting the value of B, from eqn. (1), we get
Φ₂₁ = μ₀n₁I₁n₂A = μ₀n₁n₂lAI₁ … (2)
But, Φ₂₁ = M₂₁I₁ … (3)
From eqns. (2) and (3),
M₂₁ = μ₀n₁n₂Al
This is the equation of mutual inductance of two solenoids. … (4)

2. Let S₂ be taken as primary and S₁ as secondary. If I₂ is the current through S₂, then the intensity of field inside S₂ will be
B₂ = μ₀n₂I₂ … (5)
∴ The magnetic field of S₂ i.e., B₂ will superpose on solenoid S, and hence, magnetic flux linked with each turn of S₁ = B₂A.
Hence, the total flux associated with S₁
Φ₁₂ =n₁lBn₂A … (6)
Where, n₁l is the total number of turns of solenoid S₁, and Φ₁₂ is magnetic flux on S₁, due to S₂.
Putting the value of B₂ from eqn. (5), we get
Φ₁₂ = μ₀n₂I₂An₁l = μ₀n₁n₂lAI₂ … (7)
But, Φ₁₂ = μ₁₂I₂ … (8)
From eqns. (7) and (8), we have
M₁₂ = μ₀n₁n₂Al … (9)
From eqns. (4) and (9), we have
M₂₁ = M₁₂
This is called reciprocity theorem.
It is clear that the mutual inductance of two solenoids remains the same whether cur¬rent flows in one or the other.
Thus, mutual inductance of two long solenoids is given by.
M = μ₀n₁n₂Al .
M = μ₀μ_r\(\frac { { N }_{ 1 } }{ l } \frac { { N }_{ 2 } }{ l }\)Al
μ₀μ_r\(\frac { { N }_{ 1 } }{ l } \frac { { N }_{ 2 } }{ l }\)Al

Mutual inductance depends upon :
1. Number of turns of both the coils : As the number of turns increases, the mutual inductance also increase.

2. Area of cross – section of both the coils : As the area increases, the mutual inductance increases.

3. Material of core : If the permeabilities of the cores in the solenoid are large, then mutual inductance increases. In presence of soft – iron core, the inductance is greater than that of air.

4. Length of the solenoid : For larger solenoids, mutual inductance is less.

Electromagnetic Induction Numerical Questions

Question 1.
The magnetic flux linked with a closed loop is given by the equation ϕ_B = 6t² + 7t +1, where ϕ_B is in milliweber and t is in second. Find the electromotive force induced at time t = 2 second.
Solution:
Formula: e = – \(\frac {dϕ}{dt}\)
Given : ϕ = 6t² + 7t +1
∴ = – latex]\frac {d}{dt}[/latex](6t² + 7t +1) = – 12t – 7
At t = 2 sec,
e = – 12 x 2 – 7
= – 31 milli volt.

Question 2.
An aircraft with a wing span of 40 m flies with a speed of 1080 km hr^-1 towards East where vertical component of Earth’s magnetic field is 1.75 x 10^-5 tesla. Find the e.m.f. developed between the tip of the wings. (NCERT)
Solution:
l = 40m, B_v, B_v = 1.75 x 10^-5T V = 1080\(\frac {Km}{hr}\)
= 1080 x \(\frac {5}{18}\) = 300 m/sec
e = B_vlv = 1.75 x 10^-5 x 40 x300
= 0.21 volt

Question 3.
The flux linked with a coil changes from 1 Wb to 0.1 Wb in 0.1 sec. Calculate the induced e.m.f.
Solution:
Formula : e = – \(\frac {dI}{dt}\)
Given : dl = (0.1 – 1) and dt = 0.1.
e = – \(\frac {(0.1 – 1) }{0.1}\)
or e = \(\frac {(0.9) }{0.1}\)

Question 4.
A current is flowing through a coil having 800 turns. The flux linked with it is 1.5 x 10-s Wb. Find the self-inductance of the coil.
Sol. Formula : = LI
Flux linked with one turn = 1.5 x 10^-5 Wb,
∴ Flux linked with 800 turns = 800 x 1.5 x 10^-5Wb
Substituting the values, we get
800 x 1.5 x 10^-5 = L x 1.5
∴ L = 800×10^-5
= 8.0 x 10^-3H
= 8.0 mH.

Question 5.
An air – core solenoid with length 30 cm area of cross – section 25 cm² and number of turns 500 carries a current of 2.5A. Current is switched off for 1 m sec. Find the average back emf induced across the ends of the open switch in the circuit. (NCERT)
Solution:
l = 30 cm = 0.3 m, A = 25 cm² = 25 x 10^-4m²
N = 500, dt = 10^-3sec., dl = 0 – 2.5 = – 2.5A
Back emf

MP Board Class 12th Physics Important Questions

MP Board Class 12th Maths Important Questions Chapter 6 Application of Derivatives

Application of Derivatives Important Questions

Application of Derivatives Objective Type Questions:

Question 1.
Choose the correct answer:

Question 1.
The rate of change of the area of a circle with respect to its radius r when r = 5 is:
(a) 10 π
(b) 12 π
(c) 8 π
(d) 11 π.
Answer:
(b) 12 π

Question 2.
Tangent line of a curve y² = 4x at y = x + 1 is:
(a) (1,2)
(b) (2, 1)
(c) (1, -2)
(d) (- 1, 2).
Answer:
(a) (1,2)

Question 3.
Approximate change in the volume of a cube of side x metre caused by increasing the side by 2%:
(a) 0.03 x³
(b) 0.02 x³
(c) 0.06 x³
(d) 0.09 x³
Answer:
(c) 0.06 x³

Question 4.
Point on the curve x² = 2y which lie minimum distance from the point (0,5):
(a) (2\(\sqrt{2}\),4)
(b) (2\(\sqrt{2}\),0)
(c) (0, 0)
(d) (2, 2).
Answer:
(a) (2\(\sqrt{2}\),4)

Question 5.
Minimum value of the function f(x) = x⁴ – x² – 2x + 6:
(a) 6
(b) 4
(c) 8
(d) None of these.
Answer:
(b) 4

Question 2.
Fill in the blanks:

1. The function f(x) = cosx, for 0 ≤ x ≤ π is …………………………..
2. The radius of circular plate is increasing at the rate of 0.2 cm /sec. when r = 10, then the rate of change of the area of the plate is ………………………………
3. The function y = x(5 – x) is maximum at x is equal to ……………………………
4. The minimum value of 2x + 3y is ………………………….. when xy = 6.
5. The maximum value of sin x + cos x is ……………………………
6. If line y = mx+ 1 is a tangent line to the curve y² = 4x. Then value of m will be ……………………………..
7. Slope of the tangent to the curve y = x² at the point (1, 1) is ………………………….
8. Using differential the approximate value of \(\sqrt{0.6}\) is …………………………….

Answer:

1. Decreasing
2. 4πcm²/sec
3. \(\frac{5}{2}\)
4. 12
5. \(\sqrt{2}\)
6. 1,
7. 2,
8. 0.8.

Question 3.
Write True/False:

1. For all real values of x, the function f(x) = e^x – e^-x is increasing function.
2. If the length of equal sides of an isosceles traingle be x then its maximum area will be \(\frac{1}{2}\) x²
3. Function f(x) = 3x² – 4x is increasing in the interval (- ∞, \(\frac{2}{3}\) )
4. Function f(x) = x – cot x is always decreasing.
5. Equation of the normal of the curve y = e^x at point (0, 1) is x + y = l.

Answer:

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

Application of Derivatives Short Answer Type Questions

Question 1.
The radius of a circle increases at the rate of 2cm/sec. At what rate the area increases when radius is 10cm?
Solution:
Given:
\(\frac { dr }{ dt } \) = 2cm/sec
Let the area of circle be A
Then,

∵ Area of circle A = πr²

= 40π sq.cm/second

Question 2.
The radius of air bubble is increasing at the rate of 1/2 cm per second. At what rate the volume of the air bubble is increasing when the radius is 1 cm?
Solution:
Let the radius of air bubble be r.
∴ Volume V = \(\frac{4}{3}\) πr³

Question 3.
The radius of a balloon is increasing at the rate of 10 cm/sec? At what rate surface area of the balloon is increasing when radius is 15cm?
Solution:
Let r be the radius of balloon at any time t and its surface be x then
A = 4πr²
Differentiating eqn.(1) w.r.t. t,

Hence, when radius of ballon is 15cm, then its area is increasing at rate of 1200π cm/sec.

Question 4.
Find those intervals in which the function f(x) = 2x³ – 15x² + 36x + 1 is increasing or decreasing?
Solution:
f(x) = 2x³ – 15x² + 36x + 1
⇒ f'(x) = 6x² – 30x + 36
= 6(x² – 5x + 6)
= 6(x – 2) (x – 3)
For increasing function of f(x),
f'(x) > 0
or (x – 2) (x – 3) > 0
⇒ x – 2 > 0 and x – 3 > 0
⇒ x > 2 and x < 3 ⇒ x > 3
Clearly the function is increasing in interval (3, ∞)
Again, (x – 2) (x – 3) > 0
or x – 2 < 0 and x – 3 < 0
⇒ x < 2 and x < 3
⇒ x < 2
Clearly the function is increasing in interval (-∞, 2)
Hence, the function is increasing in the interval (-∞, 2) ∪ (3, ∞)
Again for decreasing function of f(x),
f'(x) < 0
⇒ (x – 2) (x – 3) < 0
or x – 2 < 0 and x – 3 > 0
⇒ x < 2 and x > 3, which is impossible
or x – 2 > 0 and x – 3 < 0 ⇒ x > 2 and x < 3
⇒ 2 < x < 3
Hence, f(x) is decreasing function in the interval (2, 3).

Question 5.
(A) If x + y = 8, then find maximum value of xy?
Solution:
Let P = xy
⇒ x + y = 8
⇒ y = 8 – x
∴P = x(8 – x) = 8x – x²
⇒ \(\frac { dP }{ dx } \) = 8 – 2x
⇒ \(\frac { d^{ 2 }P }{ dx^{ 2 } } \) = -2
For maximum or minimum value
8 – 2x = 0
⇒ x = 4
Now, x = 4 then \(\frac { d^{ 2 }P }{ dx^{ 2 } } \) = -2, which is negative
∴ at x = 4, then y = 4
Maximum value of P, when x = 4, y = 4
= 4 × 4 = 16.

(B) If x + y = 10 then find maximum value of xy?
Solution:
Solve like Q.No.5(A).

Question 6.
The radius of a circle increases at the rate of 3cm/sec. At what rate the area increases when the radius is 10cm?
Solution:
Let r be the radius and A be the area of circle then,
A = πr²
\(\frac { dA }{ dt } \) = rate of change of area = ?
⇒ \(\frac { dA }{ dt } \) = 2πr \(\frac { dr }{ dt } \)
⇒ \(\frac { dA }{ dt } \) = 2π.(10).3
= 60π cm²/second.

Question 7.
The volume of a cube is increasing at the rate of 9 cm³/sec? If the edge of cube is 10cm then at what rate the surface area of cube is increasing? (NCERT)
Solution:
Let the edge of cube = a dm
Volufne = V= a³
Surface area of cube = s = 6a²

Surface area of cube

When a = 10cm, then \(\frac { ds }{ dt } \) = \(\frac { 36 }{ 10 } \) = 3.6 cm²/sec.

Question 8.
(A) A man 180 cm high walks away from a lamp post at a rate of 1.2metre per second. If the height of lamp post is 4.5 metre. Find the rate at which the length of his shadow increases?
Solution:
Let AB be the lamp post of the height 4.5 m, PQ be the position of man at time t, CQ = x and BQ = y
drrf

(B) A man of height 2 metre walking away from a lamp post at a rate of 5km/ sec. If the height of the lamp post 6m. Find the rate at which the length of his shadow is increasing? (NCERT)
Solution:
Solve like Q. No.8(A).
[Ans. \(\frac{5}{8}\) km/sec.]

Question 9.
A ladder 5 metre long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall, at the rate of 2 metre per second. How just is its height on the wall decreasing when the foot of the ladder is 4 metre away from the wall? (NCERT)
Solution:
Let at any time t, the bottom of the ladder be at a distance x metre from the wall and the height of the wall bey metre.
QA = xm, OB = ym, AB = 5m (given)
\(\frac { dx }{ dt } \) = 2m/sec.
In ∆OAB,

Question 10.
Find the intervals in which f(x) = 5x³ + 7x – 13 is increasing or decreasing?
Solution:
Given:
f(x) = 5x² + 7x – 13 (given)
∴ f'(x) = 10x + 7
If f(x) is increasing,
f'(x) > 0
⇒ 10x + 7 > 0
⇒ x > \(\frac{-7}{10}\)
Hence, f(x) is increasing in interval ( \(\frac{-7}{10}\) , ∞)
If f(x) os decreasing, then,
f'(x) < 0
⇒ 10x + 7 < 0
⇒ x < \(\frac{-7}{10}\)
∴ f(x) is decreasing in (-∞, \(\frac{-7}{10}\) ).

Question 11.
Find the intervals in which the function f(x) = 2x³ – 24x + 5 is increasing or decreasing?
Solution:
f(x) = 2x³ – 24x + 5
Differentiating w.r.t x,
f'(x) = 6x² – 24
⇒ f'(x) = 6(x² – 4)
⇒ f'(x) = 6(x + 2) (x-2) …………………. (2)
(A) If f(x) in is increasing, then
f'(x) > 0
6(x + 2)(x-2) > 0
∴ f(x), x ∈ (-∞,-2) ∪(2, ∞) is increasing.
(B) If f(x) is decreasing,
f'(x) < 0 ⇒ 6(x + 2) (x – 2) < 0
f(x), x ∈ (-2,2) decreasing.

Question 12.
Show that the function f(x) = x – cos x is always increasing?
Solution:
Given function f(x) = x – cos x
∴ f'(x) = 1 – (- sin x)
⇒ f'(x) = 1 + sin x
We know that -1 ≤ x ≤ 1
-1 + 1 ≤ 1 + sin x ≤ 1 + 1
0 ≤ 1 + sin x ≤ 2
Hence f'(x) = 1 + sin x is always positive for all values of x.
∴ f(x) = x – cos x is always increasing.

Question 13.
Find the least value of a such that the function given by f(x) = x² + ax + 1 is strictly increasing on (1, 2)? (NCERT)
Solution:
Given:
f(x) = x² + ax + 1
∴ f'(x) = 2x + a
When, x ∈ (1,2)
∴ 1 < x < 2
⇒ 2 < 2x < 4
⇒ 2 + a < 2x + a < A + a
⇒ 2 + a < f'(x) < 4 + a Since f(x) is increasing function ∴ f'(x) > 0
⇒ 2 + a > 0 ⇒ a > -2
Hence, the least value of a is -2.

Question 14.
Let I be any interval disjoint from [-1, 1] prove that the function f given by f(x) = x + \(\frac{1}{x}\) is strictly increasing on I? (NCERT)
Solution:
f(x) = x + \(\frac{1}{x}\)

x ∈ 1, x ∉ (-1, 1)
x < – 1 0r x > 1
⇒ x² – 1 > 0
⇒ \(\frac { x^{ 2 }-1 }{ x^{ 2 } } \) > 0
⇒ f'(x) > 0, x ∈ 1
∴ f(x) is increasing on I. Proved.

Question 15.
Find the interval in which the given function is increasing or decreasing:
f(x) = x⁴ – \(\frac { x^{ 3 } }{ 3 } \)?
Solution:
f(x) = x⁴ – \(\frac { x^{ 3 } }{ 3 } \)


Hence, the interval 0 and \(\frac{1}{4}\) on X – axis is divided in three intervals
(- ∞, 0), (0, \(\frac{1}{4}\) ), ( \(\frac{1}{4}\), ∞)
In interval (- ∞, 0)
f'(x) = x² (4x – 1),
⇒ f'(x) < 0 [x² = +ve] [4x – 1 > 0]
⇒ f'(x) < 0
∴ In interval (- ∞, 0) the function f(x) decreasing
In interval (0, \(\frac{1}{4}\) ):
f'(x) = x² (4x – 1),
⇒ f'(x) > 0 [∵ x² = positive] [4x – 1 > 0]
∴ In interval (0, \(\frac{1}{4}\) ) f(x) is decreasing
In interval ( \(\frac{1}{4}\), ∞):
f'(x) = x² (4x – 1), [∵ x² = positive] [4x – 1 > 0]
⇒ f'(x) > 0
In interval ( \(\frac{1}{4}\), ∞) in function f(x) increasing.

Question 16.
The perimeter of a rectangle is 100 cm. Find the length of sides of the rectangle for maximum area?
Solution:
Let length of rectangle be x and breadth be y.
∴Perimeter of rectangle = 2(x + y)
⇒ 2x + 2y = 100
⇒ x + y = 50
Let area of rectangle,
A = xy = x(50 – x) = 50x – x², [from eqn.(1)]
\(\frac { dA }{ dx } \) = 50 – 2x
and \(\frac { d^{ 2 }A }{ dx^{ 2 } } \) = -2
For maximum or minimum of A,
\(\frac { dA }{ dx } \) = 0
⇒ 50 – 2x = 0 or or x = 25
For any value of x, \(\frac { d^{ 2 }A }{ dx^{ 2 } } \) is -ve
For x = 25, area of rectangle is maximum
Put in eqn.(1),
y = 50 – x = 50 – 25 = 25.

Question 17.
Area of a rectangle is 25 sq.cm. Find its length and breadth when its perimeter is minimum?
Solution:
Let the length and breadth of rectangle be x and y units and A be the area.
xy = 25 ……………………….. (1)
Perimeter of rectangle,
P = 2(x + y)

For maximum or minimum \(\frac { dP }{ dx } \) = 0

Which is positive for x = 5
∴ For Minimum parimeter x = cm.
y = \(\frac{25}{x}\) = \(\frac{25}{5}\) = 5 cm.

Question 18.
Find the maximum value of sin x + cos x = \(\sqrt{2}\)? (NCERT)
Solution:
Given:
f(x) = sin x + cos x …………………. (1)
∴f'(x) = cos x – sin x …………………… (2)
and f”(x) = – sinx – cos x …………………….. (3)
For maxima or minima
f'(x) = 0
∴ cos x – sin x = 0
⇒ sin x = cos x
⇒ tan x = 1
∴ x = \(\frac { \pi }{ 4 } \), \(\frac { 3\pi }{ 4 } \), \(\frac { 5\pi }{ 4 } \)
Put x = \(\frac { \pi }{ 4 } \) in eqn (3)

Hence, at x = \(\frac { \pi }{ 4 } \) the given function is maximum. In the same way for x = \(\frac { 3\pi }{ 4 } \), \(\frac { 5\pi }{ 4 } \) ………………… the function is maximum.
Put x = \(\frac { \pi }{ 4 } \) in eqn.(1),
Maximum value f ( \(\frac { \pi }{ 4 } \) ) = sin \(\frac { \pi }{ 4 } \) + cos \(\frac { \pi }{ 4 } \)
= \(\frac { 1 }{ \sqrt { 2 } } \) + \(\frac { 1 }{ \sqrt { 2 } } \) = \(\frac { 2 }{ \sqrt { 2 } } \) = \(\sqrt{2}\). Proved.

Question 19.
Two positive numbers are such that x +y = 60 and xy³ is maximum. Prove (NCERT)
Solution:
Given:
x + y = 60 …………………….. (1)
Let p = xy³
⇒ P = (60 – y). y³
⇒ P = 60y³ – y⁴
⇒ \(\frac{dp}{dy}\) = 180 y² – 4y³ …………………….. (2)
Put \(\frac{dp}{dy}\) = 0
180 y² – 4y² = 0
⇒ 4y² (45 – y) = 0
∴ y = 45
Differentiating eqn. (2) w.r.t y,
\(\frac { d^{ 2 }p }{ dy^{ 2 } } \) = 360y – 12y²
= 12 y(30 – y)
Put y = 45,
\(\frac { d^{ 2 }p }{ dy^{ 2 } } \) = 12 × 45 (30 – 45) = – ve
Hence, at y = 45, p is maximum
From eqn.(1),
x + 45 = 60
⇒ x = 15
∴ The two numbers are 15 and 45.

Question 20.
Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x coordinate is 2? (NCERT)
Solution:
The equation of given curve,
y = x³ – x + 1

Slope of tangent at x = 2,

= 3 × 4 – 1 = 11.

Question 21.
Find the slope of the tangent to the curve y = \(\frac { x-1 }{ x-2 } \), x ≠ 2 at x = 10? (NCERT)
Solution:
The equation of given curve:


At x = 10 slope of tangent is:

Question 22.
Find the point at which the tangent to curve y = x³ – 3x² – 9x + 7 is parllel to X – axis? (NCERT)
Solution:
Equation of given curve y = x³ – 3x² – 9x + 7
\(\frac { dy }{ dx } \) = \(\frac { d }{ dx } \) (x³ – 3x² – 9x + 7)
⇒ \(\frac { dy }{ dx } \) = 3x² – 6x – 9
= 3[x² – 2x – 3]
= 3[x² – 3x + x – 3]
= 3[x(x – 3) + 1(x – 3)]
⇒ \(\frac { dy }{ dx } \) = 3(x – 3) (x + 1)
Tangent is parllel to X – axis.
∴\(\frac { dy }{ dx } \) = 0
3 (x – 3) (x + 1) = 0
⇒ x = 3,-1
When x = 3, then y = (3)³ – 3(3)² – 9 × 3 + 7
y = 27 – 27 – 27 + 7
y = – 20
When x = 3, then y = (-1)³ – 3(-1)² – 9(-1) + 7
y = – 1 – 3 + 9 + 7
y = 12
Hence, the point at which the tangent is parllel to X – axis are (3, -20) and (-1, 12).

Question 23.
Find the equation of the tangent to the parabolas y² = 4ax at the point (at², 2at)? (NCERT)
Solution:
Given:
y² = 4ax ……………………. (1)

Equation of tangent is:
y – y₁ = \(\frac { dy }{ dx } \) (x – x₁)
Where x₁ = at², y₁ = 2at, \(\frac { dy }{ dx } \) = \(\frac { 1 }{ t } \)
y – 2at = \(\frac { 1 }{ t } \) (x – at²)
⇒ yt – 2at² = x – at²
⇒ x – ty + at² = 0.

Question 24.
Find the equation of the tangent to the curve x^2/3 + y^2/3 = 2 at point (1, 1)? (NCERT)
Solution:
Equation of curve:
x^2/3 + y^2/3 = 2


Equation of tangent at point (x₁, y₁) is:

y – 1 = – (x – 1)
⇒ y – 1 = – x + 1
⇒ x + y – 2 = 0

Question 25.
Find the equation of tangent to the curve 2y + x² = 3 at point (1, 1)? (NCERT)
Solution:
Equation of given curve:
2y + x² = 3

Equation of tangent at point (x₁, y₁) is:

⇒ y – 1 = x – 1
⇒ x – y = 0.

Question 26.
(A) Find the equation of tangent to the cure x = cos t, y = sin at t = \(\frac { \pi }{ 4 } \)? (NCERT)
Solution:
Equation of 1^st curve:
x = cos t

x = cos t = cos \(\frac { \pi }{ 4 } \) = \(\frac { 1 }{ \sqrt { 2 } } \)
y = sin t = sin \(\frac { \pi }{ 4 } \) = \(\frac { 1 }{ \sqrt { 2 } } \)
Equation of tangent at (x₁, y₁) is:

(B) Find the equation of tangent and normal to the curve 16x² + 9y² = 145 at point (x₁, y₁), where x₁ = 2 and y₁ > 0? (CBSE 2018)
Solution:
Given:
16x² + 9y² = 145 ……………………. (1)

Put x = 2 in eqn.(1), we get
16.(2)² +9y² =145
⇒ 64 + 9y² =145
⇒ 9y² = 145 – 64 = 81
⇒ y² =9, [y ≠ -3 ∵ y₁ > 0]
⇒ y = 3
At point (2, 3)
\(\frac { dy }{ dx } \) = – \(\frac { 16 }{ 9 } \). \(\frac { 2 }{ 3 } \) = – \(\frac { 32 }{ 27 } \)
Equation of tangent of curve (1) at point (2, 3),

and equation of normal of curve (1) at point (2, 3) is,

Question 27.
Use differential to approximate (25)^1/3? (NCERT)
Solution:
Let y = x^1/3
Where, x = 27 and ∆x = -2


∆y is appromimately equal to dy.

Approximately value of (25)^1/3
= 3 + ∆y
= 3 – 0.074 = 2.926.

Question 28.
Use differential to approximate \(\sqrt { 36.6 } \)? (NCERT)
Solution:
Let y = \(\sqrt{x}\), where x = 36 and ∆x = 0.6

∆y is appromimately equal to b,

Approximate value of \(\sqrt { 36.6 } \)
= ∆y + 6
= 0.05 + 6 = 6.05

Question 29.
Use differential to approximate (15)^1/4? (NCERT)
Solution:
Let y = x^1/4
Where, x = 16 and ∆x = -1

∆y is appromimately equal to dy,

Approximately value of (15)^1/4
∆y + 2 = 2 – 0.031 = 1.969.

Question 30.
Use differential to approximate (26)^1/3? (NCERT)
Solution:
Let y = x^1/3
Where, x = 27 and ∆x = -1


∆y is appromimately equal to dy

Approximate value of (26)^1/3
= ∆y + 3
= 3 – 0.037 = 2.963

Question 31.
If the radius of the sphere is measured as 9cm with an error 0.03 cm then find the approximate error in calculating its surface area? (NCERT)
Solution:
Let the radius of circle = r
Given:
r = 9 cm, ∆r = 0.03 cm
Area of sphere

= (8πr) × 0.03
= 8π × 9 × 0.03 = 2.16 π cm²
The approximate error in caluculating the surface area is = 2.16 π cm²

Question 32.
If the radius of a sphere is measured as 7m with an error of 0.02m then find the approximate error in calculating its volume?
Solution:
Let the radius of sphere = r
and ∆r be the error in measuring the radius.
r = 7m, ∆r = 0.02 m (given)
Volume of sphere V = \(\frac{4}{3}\) πr³

The approximate error in calculating the volume is 3.92π m³

Question 33.
Find the approximate change in the volume V of a cube of sides x metre caused by increasing the side by 1%? (NCERT)
Solution:
Let x be the sides of cube.
Volume of cube V = x³, ∆x = 1% of x = \(\frac{x}{100}\)

Change in volume ∆V = ( \(\frac { dV }{ dx } \) ) ∆x

Approximate change in volume = 0.03 x³m³.

Question 34.
Find the approximate change in voume V of a cube of side x metre caused by increasing the side by 2%
Solution:
Let x be the side of cube,
∆x = 2% of x = \(\frac { x\times 2 }{ 100 } \) = 0.02 x
Volume of cube V = x³
\(\frac { dv }{ dx } \) = \(\frac { d }{ dx } \) x³ = 3x²
dV = ( \(\frac { dv }{ dx } \) ) ∆x = 3x² × 0.02 x = 0.06x³ m³
Thus the approximate change in voume = 0.06 x³m³.

MP Board Class 12 Maths Important Questions

MP Board Class 12th Economics Important Questions Unit 9 Government Budget and Economy

Government Budget and Economy Important Questions

Government Budget and Economy Objective Type Questions

Question 1.
Choose the correct answers:

Question 1.
The duration of Government budget is :
(a) 5 years
(b) 2 years
(c) 1 year
(d) 10 years.
Answer:
(c) 1 year

Question 2.
Budget is presented in the Parliament by :
(a) Prime Minister
(b) Home Minister
(c) Finance Minister
(d) Defence Minister.
Answer:
(c) Finance Minister

Question 3.
Budget speech in Lok Sabha is given by :
(a) President
(b) Prime Minister
(c) Finance Minister
(d) Home Minister.
Answer:
(c) Finance Minister

Question 4.
Professional tax is imposed by :
(a) Central Government
(b) State Government
(c) Municipal Corporation
(d) Gram Panchayat.
Answers:
(b) State Government

Question 5.
From the following which is included in the direct tax :
(a) Income Tax
(b) Gift Tax
(c) Both (a) and (b)
(d) Excise Tax.
Answer:
(c) Both (a) and (b)

Question 6.
Who issues 1 rupee note in India :
(a) Reserve Bank of India
(b) Finace Ministry of India
(c) State Bank of India
(d) None of these.
Answer:
(b) Finace Ministry of India

Question 2.
Fill in the blanks:

1. …………………… is a document containing income and expenditure of the government.
2. Income tax is …………………… tax.
3. …………………… tax is levied on the value of the goods.
4. Service tax is levied by the ……………………
5. budget is considered good for the country.
6. Finance bill contains …………………… proposals.
7. Government budget is presented on the last day of ……………………

Answer:

1. Budget
2. Direct
3. Advalorem
4. Central
5. Deficit
6. Tax
7. February.

Question 3.
State true or false :

1. Deficit budget is not considered as a good budget.
2. Electricity tax is levied by the State Government.
3. Budget speech is given by the Finance Minister.
4. Central excise duty is direct tax.
5. Interest payment is a planned item.
6. During deflation surplus budget is made.
7. Rail budget is generally not included in the annual budget.

Answer:

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

Question 4.
Match the following

Answer:

1. (b)
2. (d)
3. (a)
4. (e)
5. (c)

Question 5.
Answer the following in one word/sentence :

1. Write meaning of surplus budget.
2. Expenditure on education is considered as?
3. Who passes the budget presented by the Finance Minister every year?
4. Which tax was levied by the government on July 2017?
5. For how many years government makes budget?
6. What is the full form of G.S.T.?
7. Land Revenue is levied by whom?
8. What is the name gives to budget?

Answer:

1. More income and less expenditure
2. Developmental
3. Parliament
4. G. S. T
5. 1 year
6. Goods and Service Tax
7. By State Government
8. Master Finacial Scheme of Government.

Government Budget and Economy Very Short Answer Type Questions

Question 1.
What do you mean by government budget? How many types are there?
Answer:
A budget is the statement of financial plan of the government for a financial year (1st April to 31st March). It indicates the revenue expenditure estimates for the next financial year of the government.

Budget is of two types:

1. Capital budget
2. Revenue budget.

Question 2.
What do you mean by primary deficit?
Answer:
Primary deficit is the difference between fiscal deficit and interest payments. It indicates how much government borrowing is going to meet expenses other than interest payments. It is often used as the basic measure of fiscal responsibility.
Primary Deficit = Fiscal Deficit – Interest payments

Question 3.
What is tax?
Answer:
A tax is a compulsory contribution which is given by the people to the government in order to meet the expenditure on the welfare of the citizens.

Question 4.
What is deficit budget?
Answer:
When the expenditure of the government is more than the income of the government, it is called deficit budget.

Question 5.
What do you mean by supplementary budget?
Answer:
Supplementary budget is prepared for the temporary period. It is prepared during the period of emergency like flood, war, earthquake etc.

Question 6.
What do you mean by zero primary deficit?
Answer:
When government has to take loan only to fulfill the liability of interest it is called zero primary deficit.

Question 7.
What do you mean by vote on account?
Answer:
Special arrangement is made under which special power is vested with lok sabha to sanction some amount as advance till the financial budget is passed for the next year.

Question 8.
What is tax evasion?
Answer:
When people do not pay tax by hiding income it is called tax evasion.

Question 9.
What do you mean by surplus budget?
Answer:
When the income of the government is more than expenditure of the government in budget, it is called surplus budget.

Question 10.
What is a balanced budget?
Answer:
Budget in which income and expenditure of the government is equal is called balanced budget.

Question 11.
Write the tax multiplier.
Answer:
Tax multiplier = \(\frac {-c}{1 – c}\)

Question 12.
What do you mean by debt trap?
Answer:
Generally developing countries take loans from foreign countries to fulfill their projects and plans. Developing countries have to pay debt along with the interest on it. To pay this amount again government has to take loan. Thus, principal amount goes on increasing. These countries take loan from one country and pay loan of other country. Thus these countries go in the clutches of debt. This is called debt trap.

Question 13.
Write the main revenue sources.
Answer:
Public revenue:
By public revenue we mean all those income of the government which are essential for government expenditure. Public revenue can be divided into two parts :

1. Revenue receipts
2. Capital receipts.

1. Revenue receipts (Items of income) are of two types :

(A) Tax revenue : It includes all direct and indirect taxes which are imposed by the central government. For example, Income Tax, Corporation Tax, Production tax.

(B) Non – tax revenue :

• Interest receipts
• Dividends and Profits
• Foreign grants
• Fiscal services, Economic services, Subsidiary assistance.

Question 14.
Give the relationship between the Revenue deficit and the Fiscal deficit.
Answer:
Revenue deficit refers to the excess of Government’s revenue expenditure over revenue receipts whereas fiscal deficit is the difference between the government’s total expenditure and its total receipts excluding borrowings.

Question 15.
Explain why public goods must be provided by the government.
Answer:
Public goods must be provided by the government as they cannot be provided through market mechanism i.e., by transactions between individual consumer and producers.

Question 16.
Explain the relation between government deficit and government debt.
Answer:
Relation between government deficit and government debt: The government deficit and government debt are closely related. The government deficit in a flow concept, but it adds to the stocked debt. If the government continues to borrow year after year there is an accumulation of debt. It implies that government has to pay more and more by way of interest. These interest payment contribute to the debt. Thus, deficit is the cause and effect of debt.

Question 17.
Explain the concept of Deficit budget.
Answer:
Deficit budget: It is the budget in which government receipts are more than government expenditure.
Formula = Expected public income < Expected public expenditure.

Government Budget and Economy Short Answer Type Questions

Question 1.
What is a government budget? Write its objectives. (Delhi, Foreign 2013)
Answer:
“A budget is the statement of the financial plan of the government for a financial year” (1st April to 31st March). It indicates the revenue expenditure estimates for the next financial year of the government. The main objectives of budget are as follows :

1. Reallocation of financial resources.
2. Removal of inequality of income and wealth.
3. Stabilization of price level.
4. Management of public enterprises.
5. Expansion of employment opportunities.

Question 2.
How inequalities in income can be removed through budget?
Or
What is the role of budget in removing income inequalities? (Delhi, All India 2011, Foreign 2012)
Answer:
Fiscal policy implies the income and expenditure policy or the budgetary policy of the government. It is a branch of public finance which deals with the types of financial statements made by any government about its probable revenue and expenditure during a given year.

Through their fiscal policies government can play a significant role in reducing inequality of income and wealth as well as inequality of opportunity. Both tax and spending policies can alter the distribution of income over both short term and medium term. Government need to reform the policies by imposing more taxes oh riches sections and to reduce the burden of tax on poor with the motive of economic welfare of the poor people.

Question 3.
Explain in brief how economic stability can be obtained by government budget? (All India 2011)
Answer:
Government budget is used to prevent business fluctuations of inflation or deflation to achieve the objective of economic stability. The government aims to control the different phases of business fluctuations through the budgetary policy. Policies of surplus budget during inflation and deficit budget during deflation helps to maintain stability of prices in the economy.

Question 4.
What is the role of government in Reallocation of Resources? Explain. (Delhi 2012)
Answer:
Through the budgetary policy, Government aims to reallocate resources in’ accordance with the economic and social priorities of the government. Government can influence allocations of resources through:

1. Tax concessions or subsidies:
To encourage investment, government can give tax concessions subsidies etc. to the producers, example Government discourages the production of harmful consumption goods through heavy taxes and encourages the use of khadi products ‘by providing subsidies.

There is disequilibrium in the balance of payments when imports are more and exports are less. With demonitization of Indian rupee imports have become costlier, efforts can be made to make balance of payments favourable. But this cannot be done again and again because by this demand of our currency will decrease.

Question 5.
Explain four types of public expenditure.
Answer:
Following are the types of public expenditure :

1. Developmental expenditure:
Under it we include those expenditure which are done for the economic development and social welfare of people. It includes education, medicine, industry agriculture, transportation roads, canals, water welfare, electricity etc.

2. Non – development expenditure:
This expenditure includes that expenditure which is done on administration, on security and legal procedure of the government. It includes salaries of police department, military department, interest on loans, pension etc.

3. Plan expenditure:
It includes expenditure to be incurred during the year on programmes under the five year plan by planning commission. In this expenditure investors and consumers both are included. For example, economic activities like agriculture, industry transportation, communication etc. It also includes social welfare activities like education, family welfare, information and communication, drinking water, cleanliness, health etc.

4. Non – plan expenditure:
It includes all those expenditure other than plan expenditure. It includes debt given to State Government and others, expenditure on maintenance of property, expenditure on purchase of shares.

Question 6.
Differentiate between Direct Tax and Indirect Tax
Answer:

Question 7.
Differentiate between Revenue Expenditure and Capital Expenditure.
Answer:
Differences between Revenue Expenditure and Capital Expenditure :

Revenue Expenditure:

• Revenue expenditure is a current expenditure incured on civil administrations, defence forces, public health and education.
• It is of recurring type of expenditure. It is incurred regularly.
• It is called non – developmental expenditure.

Capital Expenditure:

• It refers to expenditure which leads to creation of assets or reduces liabilities.
• It is a non – recurring type of expenditure.
• It is called developmental expenditure.

Question 8.
Differentiate between Developmental and Non – Developmental Expenditure.
Answer:
Differences between Developmental and Non – Developmental Expenditure:

Developmental Expenditure:

• It is incurred on economic and social development of the country.
• Expenditure on agriculture, industries, transport, etc. are included in it.

Non – Developmental Expenditure:

• Its nature is non – developmental type.
• Expenditure on administrative services like police defence, grants to government, etc. are included in it.

Question 9.
What do you mean by budget? Write its characteristics.
Answer:
Origin of the word in India:
The word ‘budget’ has been derived from the French word ‘Bougette’, which means small bag. It symbolizes a bag containing the financial proposals. In England, the chanceller used to bring economic proposals and statements in a bag; In 1773, finance minister of Britain, Robert Walpole opened his leather bag for the budget proposal to take place in Parliament. From that it becomes popular in India.

Characteristics of budget : Following are the characteristics of budget:

1. Annual plan of income and expenditure : It is a description of annual income and expenditure of the government for the next financial year. It is a detailed description.
2. Fixed period : Budget is prepared before a fixed period. It is presented before the Parliament by finance minister on the last day of the month of February. It is related to the definite fixed period.
3. Financial discussion: In budget only financial discussion takes place.Other matters are not discussed here.
4. In advance : Budget is prepared in advance for the period during which it is to operate. It is an annual financial action plan for the next financial year.
5. Balance budget : Balanced budget is the symbol of economic stability of the country.

Question 10.
Write the objectives of budget.
Answer:
The objectives of budget are as follows :
1. Reallocation of financial resources:
The government reallocates the financial resources to achieve the desired goals. It is the primary responsibility of the government to build a sound socio – economic infrastructure regarding health housing, education and raising the standard of living of people. Government imposed more taxes on those items which are harmful for people on the other hand it may reduce the taxes which are useful from social point of view.

2. Removal of inequality of income and wealth:
Every nation tries to bring equality between the various sections of the people and to bridge the gap between poor and rich. Through subsidies, taxes government can remove their inequalities. The main objective of Budget is to fill the gap of rich and poor.

3. Stabilisation of price level:
It is the responsibility of the government to stabilise the price level in the country. Unnecessary fluctuation in the prices of goods and services adversely affect the economy. Through the budget the government can effectively control the price level and bring about stability in it.

4. Management of public enterprises:
The government looks after certain departments such as railways, post and telegraph, electricity, defence etc. The budget of the government makes special provisions for such public enterprises in order to safeguard the interest of public.

5. Expansion of employment opportunities:
Government throughout budget create the employment opportunities to people. Various employments oriented and productive programs can be implemented for this purpose.

Question 11.
Differentiate between Progressive Tax and Regressive Tax.
Answer:
Differences between Progressive Tax and Regressive Tax :

Progressive Tax:

• In progressive tax the rate of the tax increses as the taxable income increases.
• The burden of it is more on rich people.
• They are justified because they reduce inequalities of income.

Regressive Tax:

• In regressive tax the rate of the tax decreases as the taxable incomes increases.
• The burden of it is on poor people.
• These are not justified because they increase inequalities.

Question 12.
Write difference betw een Revenue Receipts and Capital Receipts. (Foreign 2013)
Answer:
Differences between Revenue Receipts and Capital Receipts :

Question 13.
Write a note on Surplus budget, Balance budget and Deficit budget.
Answer:
A budget is the statement of the financial plan of the government for a financial year. It indicates the revenue expenditure estimates for the next financial year. The different types of budget are :

1. Surplus Budget – When in budget the income of the government is more than expenditure of the government, it is called Surplus Budget.
2. Balance Budget – Balance budget is said to be balance when government revenue and expenditure are balanced.
3. Deficit Budget – If the expenditure of the government is more than the income of the government, it is called the Deficit Budget.

Question 14.
Does public debt impose a burden? Explain.
Answer:
Public debt does not impose a burden all the times. Only in the following situation, it imposes a burden. When government sectors to public debt, the government transfers the burden to reduce consumption on future generation, because the government may decide to pay off deut in future by raising taxes. Taxes reduce the savings and capital formation and growth. Thus, the debt acts as a burden on future generation.

Question 15.
(a) The fiscal deficit given the borrowing requirement of the government. Explain.
Answer:
Fiscal deficit gives the borrowing requirement of the government’s total expenditure and its total receipts excluding borrowing. The fiscal deficit has to be financed through borrowing. Thus, it indicates the total borrowing requirements of the government from all sources.

(b) Are fiscal deficits inflationary? (NCERT)
Answer:
It is not correct to that all deficits are necessarily inflationary. If the fiscal deficit results in higher demand and greater output, the fiscal deficit will not be inflationary. However, if the firms are unable to produce the higher quantities that are being demanded to due fiscal deficits, prices will rise resulting the inflationary.

Question 16.
We suppose that: C= 70 + 0.70YD, I = 90, G = 100, T = 0.10 Y
(a) Find the equilibrium income, (b) What are tax revenues at equilibrium income? Does the government have a balanced budget ?
Solution:
(a) Y = \(\frac {1}{1 – 0.70}\)(70 + 90 + 100)
Y = \(\frac {1}{0.30}\)(260)
Y = \(\frac {260}{0.30}\) = 866.66

(b) Tax revenue on balance budget (T) = 0.10 Y T = 0.10 (866.66) T = 86.66.

Question 17.
Suppose that for a particular economy investment is equal to 200, government purchases are 150, net taxes i.e., lump sum taxes minus transfer in 100 and consumption is given by C = 100 + 0.75. (a) What is the level of equilibrium multiplier? (b) Calculate the value of government expenditure multiplier and the tax multiplier, (c) If government expenditure increases by 200, And the change in equilibrium income.
Answer:
(a) The government directly affects the level of equilibrium income in two specific ways : 1. Purchase of goods and services and 2. Taxes and transfers. Taxes lower the consumption expenditure and disposable income. Hence, we use the following formula for calculating the level of equilibrium income:
Formula: Y = \(\frac {1}{1-c}\)(C – CT + CTR+ 1 + G)

(b) Govt expenditure multiplier:
\(\frac { ∆Y}{∆G}\) = \(\frac {1}{1-c}\)
\(\frac { ∆Y}{∆G}\) = \(\frac {1}{ 1-0.75}\) \(\frac {1}{0.25}\)
Tax multiplier \(\frac { ∆Y}{∆G}\) = \(\frac {-C}{1-c}\)
\(\frac { ∆Y}{∆G}\) = \(\frac {-0.75}{ 1-0.75}\) = \(\frac {-0.75}{ 0.25}\) = – 3

(c) If income of Govt, increases by 200 :

Question 18.
In the equation given in question 9, calculate the effect on output of a 10 increase in transfers and 10 increase in lumpsum taxes. Compare the effects of two.
Solution:
Transfer multiplier, = \(\frac {C}{1-C}\)
= \(\frac {0.80}{1-0.80}\)
= \(\frac {0.80}{0.20}\)
= 4
Increase in transfer = 10%
Hence, increase in output = 10%
Tax multiplier = 4
Decrease in output = 10 x 4 = 40%
or
Y = \(\frac {1}{0.20}\)(20 – 8 + 88 + 30 + 50)
Y = 5 x 188 = 940
Increase lump sum, its effect,
or Y = \(\frac {1}{1-0.80}\)(20 – 8 + 80 + 30 + 50)
Y = 5 x 172 = 860.
Hence, we find that effected change in transfer and change in taxes are equal. It is due to fact that their size of changes as well as multiplier are equal.

Question 19.
What do you understand by Goods and Service Tax (G.S.T.)? How is G.S.T. better than old system of taxation? Explain its types.
Answer:
(a) The Goods and Services Tax (G.S.T.) is a value Added Tax (VAT) levied on most goods and services sold for domestic consumption. The G.S.T. is paid by consumers, but it is remitted to the government by the businesses selling the goods and services:

G.S.T. in comparison with old taxation system :

1. In place of old taxation system, G.S.T. will become one tax economy.
2. The tax structure will be simplified with G.S.T.
3. G.S.T. will save both time and money.
4. The growth rate of economy will show a rapid increase with G.S.T.
5. But for the time being, the G.S.T. would be expected to increase the inflation rate in comparison to the old tax system.

For Goods and Services Tax, G.S.T., U.T.G.S.T. Act and S.G.S.T. tax are passed.

Government Budget and Economy Long Answer Type Questions

Question 1.
Explain the types of budget.
Answer:
The types of budget are as follows :
1. Central budget:
Central budget is prepared by Central Government. It is the numerical statement of income and expenditure done by central government. In India Central budget is presented in two parts:

• General budget
• Railway budget.

2. State budget:
State budget is prepared by State Government. State Government takes the help from Central Government.

3. Revenue budget and Capital budget:
In revenue budget we include the expenditure and income related to revenue of the government. On the other hand in capital budget the capital expenditure and capital income is included.

4. Supplementary budget:
Supplementary budget is prepared for the temporary period. It is prepared during the period of emergency like flood, earthquake and war etc. There is no fixed time for it.

5. Balanced budget and Imbalanced budget:
Balance budget is that budget where income and expenditure of the government is equal. Imbalanced budget is that budget where expenditure is more or less than income of the government. If the expenditure is more than income it is called deficit budget and if the income is more than the expenditure it is called surplus budget.

Question 2.
Explain the procedure of the budget.
Answer:
Under the budgetary procedure in India we can study the following heads:

1. Preparation of budget : The preparation of budget involves the following steps :

• Sending estimate forms by finance minister to all ministers and their department’s heads to get the estimates of revenue and expenditure required for the next financial year.
• Preparation of estimates by departmental heads : It includes revenue and expenditure of previous year budgetary estimate for next coming year.
• Preparation of consolidated estimate by the various ministers and sending them to finance minister.
  Scruting report estimates by A.G. of India and sending the same to finance ministry.

2. Presentation of the budget : In India the budget is presented in two parts :

• General budget and
• Rail budget.

Rail budget is always present before the general budget. General budget presented on the last day of the month of February at 5 p.m. generally 28th in February.

3. Discussion on the budget:
Budget is put before the Parliament for discussion. In the processes of passing the budget, the discussions on various items continues for 3 – 4 days. Presently different committees are formed. For the discussion and finance minister gives his final reply on the budget.

4. Voting on budget:
After the general discussion the budget is put for voting. The members of parliament give their speeches for and against the budget before voting. There after voting is held for passing the budget.

5. Appropriation bill:
The finance minister presents appropriation bill. According to the constitution no amount from the Reserve fund of India can be withdrawn without passing the appropriation bill in the Parliament.

6. Financial bill:
All the financial proposals for the coming year are included in a bill which is known as financial bill. This bill is generally presented immediately after the presentation of budget in the Lok Sabha. If this bill is not passed by the Parliament f ur government L not supposed to spend any amount.

7. Vote on account:
Special arrangement is made under which special power is vested with Lok shabha to snetion some amount as advance till the final budget is passed for the next year.

8. Implementation of budget:
After the budget is passed, it is implemented by the government for which the following steps are taken :

• Collection of revenue
• Preparation of accounts in
• Audit by A.G. (Auditor General).

Audit by comptroller and Auditor General of India presents their report before the Parliament which comprise the receipt of revenue, expenditure, loans taken.

Question 3.
Write the tax revenue sources of Public Revenue:
Answer:
Following are the sources of Public Revenue :
Tax:
Taxes are neither a fees nor a penalty because it is neither taken by the government to give any additional benefits nor taken as a fine. Taxes are compulsory contribution which are given by the people to the government in order to meet the expenditure on the Welfare of the citizens.

Types of taxes : Taxes are of following types :

• Progressive tax – In progressive tax the rate of my increases as the income increases
• Proportional tax – Proportional taxes are those taxes whose impact and incidence falls equally on both the sections of the society.
• Regressive tax – The rate of taxation decreases with the increase in income. The burden of text falls more heavily on poor than on rich.
• Direct taxes – These are those taxes in which the impact of the tax and incidence of tax is on the same person.

Question 4.
What do you understand by Non – tax sources of public revenue?
Answer:
Following are the non – tax sources of public revenue:

1. Fees, licence and permit:
Government gets non – tax revenue from fees licence fee and payment made for permits.

Fees – Registration fee for land, death and birth registration fees, passport fees are included in it.

Licence and permit – Fee is that fee which is paid by people to government after giving permission by the government to the people to do something. For example: Driving licence, import licence etc.

2. Fines:
Fines are collected by the govt, for breaking rules of the country. The main aim of this is not to collect money but to teach people lesson about it.

3. Income from public enterprise: Government may raise funds from many sources like Railways, Postal department, Steel Plants fertilizer Corporations, Indian Oil department etc. Government sells products of it and gets profit from it.

4. Gifts and grants: The government may raise the funds from others. During calamities like flood, earthquake etc. citizens and some NGO’s give help to the government . For example : W.H.O., UNESCO etc. they give assistant also during this period.

5. Notes issues:
Sometimes government prints extra notes and increase the treasury. Reserve Bank of India has got the right of printing notes and minting coins.

6. Stamp, registration and land revenue: The receipts of stamps, registration and land revenue is the another source of non – tax revenue. The stamp receipts of Supreme court is also the income of the central government.

7. Interest, receipts : Interest earned by  the government from the loans given to States, Union territories. Railways, Post and Telegraph are included in it.

8. Dividends: This includes the share from profits of various undertakings in which the central government has done investments example SAIL, ONGC etc.

9. Administrative receipts:
The central government provides a number of services to people like health, medical, education etc. The government earns certain income from these services. For example; court fees, registration fees.

Question 5.
Consider an economy described by the following functions. C = 20 + 0.80 Y, I = 30, G = 50, TR = 100.
(a) Find the equilibrium level of income and the autonomous expenditure multiplier in the model,
(b) If the government expenditure increases by 30, what is the impact on equilibrium income?
(c) If a lump sum tax of 30 added to pay for increase in government purchases, how will equilibrium income change?
Answer:
C = 20 + 0.80y
I = 30
G = 50
TR = 100
(a)
Y = \(\overline { C } \) + CY – (T – TR)
Y = 20 + 0.80 (Y + TR) + I + G
Y = 20 + 0.80 (Y + 100) + 30 + 50
Y = 0.8Y + 180
Y = \(\frac {180×100}{200}\)
= 900
The equilibrium level of income is 900. Autonomous expenditure multiplier.
= \(\frac {1}{1 – C}\)
= \(\frac {1}{1-0.08}\)
= \(\frac {1}{0.20}\)
= 5
Increase in equilibrium income = AG x Expenditure multiplier = 30 x 5 = 150
Tax multiplier = \(\frac {-C}{1 – C}\) = \(\frac {-0.08}{1-0.08}\) = \(\frac {-80}{0-20}\) = -4
Decrease in equilibrium income = ∆T x Tax Multiplier
= 30 x 4 = 120.
or
Y = \(\frac {1}{0.20}\)(180)
Y = 5 x 180 = 900
autonomus multiplier \(\frac { ∆Y}{∆G}\)= \(\frac {1}{1 – C}\)
or
\(\frac { ∆Y}{∆G}\) = \(\frac {1}{1-0.80}\) = \(\frac {1}{0.20}\) = 5

(b) Expenditure of govt, increase by 30 :
(∆Y) = \(\frac {1}{1 – C}\) ∆G
or
∆Y = \(\frac {1}{1-0.80}\) x 30
= \(\frac {1}{0.20}\) x 30 = 5 x30 = 150
New income = 900 + 150 = 1,050
So, it is clear that by 30 it become 150 to 1,050.

(c) Lump sum 30 is added then :
Change in balance income (∆Y) = \(\frac {-C}{1 – C}\) ∆T
or
∆Y = \(\frac {-0.08}{1-0.08}\) x 30
= \(\frac {-0.08}{0.20}\) x 30 = -4 x 30
= 1 – 120
= 900 – 120 = 780

MP Board Class 12th Economics Important Questions

MP Board Class 12th Maths Important Questions Chapter 9 Differential Equations

Differential Equations Important Questions

Differential Equations Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Degree of the differential equation
\(\frac { d^{ 2 }y }{ d^{ 2 }x } \) + x² \(\frac{dy}{dx}\) = e^x is:
(a) 1
(b) 2
(c) 3
(d) does not exist
Answer:
(c) 3

Question 2.
Solution of differential equation (1 + x) y dx + (1 – y) xdy = 0 is:
(a) log xy + x + y = c
(b) logxy + x – y = c
(c) log xy – x – y = c
(d) log xy – x + y = c.
Answer:
(b) logxy + x – y = c

Question 3.
The differential equation of all circles which passes through the origin whose center lie on the A – axis is:
(a) x² = y²+ xy \(\frac{dy}{dx}\)
(b) x² = y² + 3xy \(\frac{dy}{dx}\)
(c) y² = x² + 2xy \(\frac{dy}{dx}\)
(d) y² = x² – 2xy \(\frac{dy}{dx}\)
Answer:
(c) y² = x² + 2xy \(\frac{dy}{dx}\)

Question 4.
Solution of the differential equation \(\frac{dy}{dx}\) + y = e^x, y(o) = 0 is:
(a) y = e^-x(x – 1)
(b) y = xe^x
(c) y = xe^-x + 1
(d) y = xe^-x
Answer:
(d) y = xe^-x

Question 5.
The straight line which satisfies the differential equation \(\frac{dy}{dx}\) = m and cuts an intercept 3 on the positive y – axis is:
(a) y = mx + c
(b) y = mx +3
(c) y = mx – 3
(d) y = – mx + 3.
Answer:
(b) y = mx +3

Question 2.
Fill in the blanks:

1. The corresponding differential equation of the equation x² + y² = a² is ……………………………
2. The differential equation of the curve y = e^cx is …………………….. where c is arbitrary constant.
3. The integration factor of the linear differential equation \(\frac{dy}{dx}\) + Py = Q is …………………………….
4. In the linear differential equation \(\frac{dy}{dx}\) + Py = Q. P and Q are …………………………….
5. The form of differential equation (x + y + 1) dy = dx is ……………………………
6. Solution of the differential equation e^-x+y \(\frac{dy}{dx}\) = 1 is ……………………………

Answer:

1. y \(\frac{dy}{dx}\) + x = 0
2. x \(\frac{dy}{dx}\) = y log y
3. e^∫pdx
4. constant
5. linear differential equation
6. e^-y = e^-x + c

Question 3.
Write True/False:

1. Order of differential equation y = x \((\frac { dy }{ dx } )^{ 2 }\) + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) is 2.
2. Degree of differential equation \((\frac { d^{ 3 }y }{ dx^{ 3 } } )^{ 4/5 }\) – 2 \((\frac { dy }{ dx } )\) \((\frac { d^{ 2 }y }{ dx^{ 2 } } )^{ 2 }\) = 0 is 5.
3. The integration factor of linear differential equation x \(\frac{dy}{dx}\) – y = 2x² is e^-x
4. Solution of differential equation dy = sin xdx is y + cos x – c = 0.
5. Solution of differential equation ydx + (x – y³) dy = 0 is xy = \(\frac { y^{ 4 } }{ 4 } \) + c.

Answer:

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

Question 4.
Write the answer is one word/sentence:

1. Write the integration factor of differential equation (1 + y²) + (2xy – cot y) \(\frac{dy}{dx}\) = 0.
2. Find the solution of the differential equation (1 + x²) dy = (1 + y²) dx.
3. Find the sum of order and degree of the differential equation y = x \((\frac { dy }{ dx } )^{ 3 }\) + \(\frac { d^{ 2 }y }{ dx^{ 2 } } \)
4. Find the integration factor of the differential equation x logx \(\frac{dy}{dx}\) + y = 2 log x
5. Find he solution of the differential equation \(\frac{dy}{dx}\) + \(\frac{1}{x}\) = \(\frac { e^{ y } }{ x^{ 2 } } \)

Answer:

1. 1 + y²
2. x – y = c(1 + xy)
3. 3
4. log x
5. 2xe^-y = cx² + 1.

Differential Equations Very Short Answer Type Questions

Question 1.
Find the order and degree of \(\frac{dy}{dx}\) + y = e^-x?
Answer:
1, 1

Question 2.
Find the degree and order of \((\frac { dy }{ dx } )^{ 3 }\) = \(\sqrt { 1+(\frac { dy }{ dx } )^{ 2 } } \)?
Answer:
1, 6

Question 3.
Find the degree and order of \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + \(\sqrt { 1+(\frac { dy }{ dx } )^{ 3 } } \) = 0?
Answer:
2, 2

Question 4.
Find the differential equation corresponding to equation of circle x² + y² = a²?
Answer:
y \(\frac{dy}{dx}\) + x = 0

Question 5.
Find the differential equation of line y = mx + c?
Answer:
\(\frac{dy}{dx}\) = m

Question 6.
Solve the differential equation \(\frac{dy}{dx}\) = 4y?
Answer:
y = c.e^4x

Question 7.
Solve the differential equation x² \(\frac{dy}{dx}\) = 2?
Answer:
y = c – \(\frac{2}{x}\)

Question 8.
Find the solution of differential equation dy = sinxdx?
Answer:
y + cos x = c

Question 9.
Find the solution of differential equation \(\frac{dy}{dx}\) + Px = Q?
Answer:
xe∫^pdy. dy + c

Question 10.
Solve the differential equation (1 – y²) \(\frac{dy}{dx}\) + yx = ay?
Answer:
\(\frac { 1 }{ \sqrt { 1-y^{ 2 } } } \)

Differential Equations Short Answer Type Questions

Question 1.
Solve the differential equation xlog xdy – ydx = 0?
Solution:
x log x dy – ydx = 0 (given)
⇒ x log x = ydx
⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = \(\frac{1}{xlogx}\) dx
⇒ ∫\(\frac{1}{y}\) dy = ∫\(\frac{1}{xlogx}\) dx
⇒ log y = ∫\(\frac{1}{t}\) dt , (let log x = t ⇒ \(\frac{1}{x}\) dx = dt)
⇒ log y = log t + log c
⇒ log y = log log x + log c

Question 2.
Solve the differential equation \(\frac{dy}{dx}\) = e^x-y + x.e^-y?
Solution:
\(\frac{dy}{dx}\) = e^x-y + x.e^-y (given)
⇒ \(\frac{dy}{dx}\) = e^-y (e^x + x)
⇒ e^ydy = (e^x + x) dx
Integrating both sides,
∫e^ydy = ∫(e^x + x) dx
e^y = e^x + \(\frac { x^{ 2 } }{ 2 } \) + c

Question 3.
Prove that solution of y = 4sin 3x is \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + 9y = 0?
Solution:
y = 4 sin 3x (given)
Differentiating with respect to x,
∴\(\frac{dy}{dx}\) = 12 cos 3x
Again differentiating with respect to x,
\(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = – 36 sin 3x = – 9 × 4 sin 3x
⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) = -9y, [from eqn.(1)]
⇒ \(\frac { d^{ 2 }y }{ dx^{ 2 } } \) + 9y = 0. Proved.

Question 4.
Solve the differential equation \(\frac{dy}{dx}\) = sec² x + 2x?
Solution:
\(\frac{dy}{dx}\) = sec x (sec x + tan x) (given)
⇒ dy = (sec² x + sec x tan x ) dx
⇒ ∫dy = ∫sec² dx + ∫sec x tan x dx
∴y = tan x + sec x + c.

Question 5.
Solve the differential equation \(\frac{dy}{dx}\) = sec² x + 3x²?
Solution:
\(\frac{dy}{dx}\) = sec² x + 3x² (given)
⇒ dy = (sec² x + 3x²) dx
⇒ ∫dy = ∫sec² xdx + 3∫x² dx
⇒ y = tan x + \(\frac { 3x^{ 3 } }{ 3 } \) + c
⇒ y = tan x + x³ + c.

Question 6.
Solve the differential equation \(\frac{dy}{dx}\) = sec² x + 2x?
Solution:
Solve as Q.No . 5

Question 7.
Solve the differential equation \(\frac{dy}{dx}\) = (3x² x + 2)?
Solution:
\(\frac{dy}{dx}\) = (3x² + 2) (given)
⇒ dy = (3x² + 2) dx
⇒ ∫dy = ∫(3x² + 2) dx
⇒ y = 3 × \(\frac { x^{ 3 } }{ 3 } \) + 2x = c = x² + 2x + c.

Question 8.
Solve the differential equation x² \(\frac{dy}{dx}\) = 2?
Solution:
x² \(\frac{dy}{dx}\) = 2 (given)
⇒ dy = 2.x^-2 dx
⇒ ∫dy = 2∫x^-2 dx
⇒ y = 2 \((\frac { -1 }{ x } )\) + c

Question 9.
Solve the differential equation \(\frac{dy}{dx}\) = x³ + sin 4x?
Solution:
\(\frac{dy}{dx}\) = x³ + sin 4x (given)
⇒ dy = (x³ + sin 4x) dx
⇒ ∫dy = ∫x³ dx + ∫sin 4x dx
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) + \((\frac { -cos4x }{ 4 } )\) + c
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) – \(\frac { -cos4x }{ 4 } \) + c

Question 10.
Solve the differential equation \(\frac{dy}{dx}\) + 2x = e^3x?
Solution:
\(\frac{dy}{dx}\) + 2x = e^3x (given)
⇒ \(\frac{dy}{dx}\) = e^3x – 2x
⇒ dy = (e^3x – 2x) dx
⇒ ∫dy = ∫^3x dx – 2∫x dx
⇒ y = e^3x. \(\frac{1}{3}\) e^3x – x² + c

Question 11.
Solve the differential equation \(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \) (given)
⇒ \(\frac { 1 }{ cos^{ 2 }y } \) dy = \(\frac { 1 }{ sin^{ 2 }x } \) dx
⇒ sec² ydy = cosec² xdx
⇒ ∫sec² ydy = ∫cosec² xdx
⇒ tan y = – cot x + c

Question 12.
Solve the differential equation (x² + 1) \(\frac{dy}{dx}\) = 1?
Solution:

Question 13.
Solve the differential equation \(\frac{dy}{dx}\) = sin x sin y?
Solution:
\(\frac{dy}{dx}\) = sin x sin y
⇒ cosec y dy = sin x dx
On integrating,
– log_e (cosec y + cot y) = – cos x + c
⇒ cos x – log_e (cosec y + cot y) = c

Question 14.
Solve the differential equation \(\frac{dy}{dx}\) = y sin x?
Solution:
\(\frac{dy}{dx}\) = y sin x
⇒ \(\frac{1}{y}\) \(\frac{dy}{dx}\) = sin x
⇒ ∫\(\frac{1}{y}\) dy = ∫sin x dx
⇒ log y = – cos x + c

Question 15.
Solve the differential equation \(\frac{dy}{dx}\) = x cos x?
Solution:
Given:
\(\frac{dy}{dx}\) = x cos x
⇒ dy = x cos x dx
⇒ ∫dy = ∫x cos x dx
⇒ y = xsin x – ∫1. sin x dx + c
⇒ y = xsin x + cos x + c

Question 16.
Solve the differential equation \(\frac{dy}{dx}\) = x³ + sin 4x?
Solution:
\(\frac{dy}{dx}\) = x³ + sin 4x (given)
⇒ dy = (x³ + sin 4x) dx
⇒ ∫dy = ∫x³ dx + ∫sin 4x dx
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) + \((\frac { -cos4x }{ 4 } )\) + c
⇒ y = \(\frac { x^{ 4 } }{ 4 } \) – \(\frac { cos4x }{ 4 } \) + c

Question 17.
Solve the differential equation \(\frac{dy}{dx}\) + 2x = e^3x?
Solution:

Question 18.
Solve the differential equation \(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \)?
Solution:
\(\frac{dy}{dx}\) = \(\frac { cos^{ 2 }y }{ sin^{ 2 }x } \)
⇒ \(\frac { 1 }{ cos^{ 2 }y } \) dy = \(\frac { 1 }{ sin^{ 2 }x } \) dx
⇒ sec² ydy = cosec² xdx
∴∫sec² ydy = ∫cosec² xdx + c
tan y = – cotx + c

Question 19.
Solve the differential equation \(\frac{dy}{dx}\) = 1 – x + y – xy?
Solution:
\(\frac{dy}{dx}\) = 1 – x + y – xy (given)
⇒ \(\frac{dy}{dx}\) = (1 – x) + y (1 – x)
⇒ \(\frac{dy}{dx}\) = (1 – x) (1 + y)
⇒ \(\frac{dy}{1+y}\) = (1 – x) dx
⇒ ∫\(\frac{dy}{1+y}\) = ∫(1 – x) dx
⇒ log_e (1 + y) = x – \(\frac { x^{ 2 } }{ 2 } \) + c

Question 20.
Solve the differential equation \(\frac{dy}{dx}\) = (1 + x) (1 + y²)?
Solution:
\(\frac{dy}{dx}\) = (1 + x) (1 + y²) (given)
⇒ \(\frac { 1 }{ 1+y^{ 2 } } \) dy = (1 + x) dx
Integrating both sides,
⇒ tan^-1y = x + \(\frac { x^{ 2 } }{ 2 } \) + c.

Question 21.
Solve the differential equation:
\(\frac{dy}{dx}\) = cot² x?
Solution:
\(\frac{dy}{dx}\) = cot² x (given)
⇒ dy = cot² x
⇒ ∫dy = ∫cot² x dx
⇒ y = ∫(cosec² x – 1) dx
⇒ y = – cot x – x + c

Differential Equations Long Answer Type Questions – II

Question 1.
(A) Solve the differential equation \(\frac{dy}{dx}\) + y tan x = sec x?
Solution:
\(\frac{dy}{dx}\) + y tan x = sec x (given)
Comparing the equation with \(\frac{dy}{dx}\) + Py = Q,
P = tan x, Q = sec x
∴L.F. = e^{∫P dx} = e^{∫tanx dx} = e^{log secx}
⇒ L.F. = sec x
Applying formula
y × (1.F.) = ∫Q × (I.F.) dx + c
⇒ ∫sec² x dx + c
⇒ y sec x = tan x + c

(B) Solve the differential equation \(\frac{dy}{dx}\) + y tan x = sin x?
Solution:
Solve as Q.No.1 (A).

Question 2.
Solve the differential equation \(\frac{dy}{dx}\) = \(\frac { \sqrt { 1-y^{ 2 } } }{ \sqrt { 1-x^{ 2 } } } \)?
Solution:

Question 3.
Solve the differential equation 3x² dy = (3xy + y²)dx?
Solution:
3x² dy = (3xy + y²) dx (given)

Putting in eqn. (1),

⇒ – \(\frac{1}{v}\) = \(\frac{1}{3}\) log x + c
∴- \(\frac{1}{y}\) = \(\frac{1}{3}\) log x + c

Question 4.
Solve the differential equation (1 + x²) \(\frac{dy}{dx}\) + 2xy = 4x²?
Solution:

The required solution will be:

Question 5.
Solve the differential equation (1 + x²) \(\frac{dy}{dx}\) + 2xy = cos x?
Solution:
(1 + x²) \(\frac{dy}{dx}\) + 2xy = cos x (given)
⇒ \(\frac{dy}{dx}\) + \(\frac { 2x }{ (1+x^{ 2 }) } \)
Comparing with \(\frac{dy}{dx}\) + Py = Q,

The required solution of eqn. (1) is:

Question 6.
Marginal cost price of making anything is given by equation c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x. Find the total cost price c(x) for making it? [Given c (0) = 100]
Solution:
c'(x) = \(\frac{dc}{dx}\) = 2 + 0.15 x (given)
Integrating both sides
∫c'(x) dx = ∫ (2 + 0.15 x) dx
c(x) = 2x + 0.15 \(\frac { x^{ 2 } }{ 2 } \) + A …………. (1)
If x = 0
c(0) = 2 × 0 + \(\frac { 0.15 }{ 2 } \) × 0² + A
⇒ c(0) = A
∴ A = 100, [∵ c(0) = 100]
Putting in eqn. (1),
c(x) = 2x + 0.075 x² + 100

Question 7.
Solve the differential equation x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0?
Solution:
x\(\sqrt { 1+y^{ 2 } } \) dx + y \(\sqrt { 1+x^{ 2 } } \) dy = 0 (given)
⇒ \(\frac { y }{ \sqrt { 1+y^{ 2 } } } \) dy = – \(\frac { x }{ \sqrt { 1+x^{ 2 } } } \) dx
Integrating both sides,

Question 8.
Solve the differential equation (x + y + 1) \(\frac{dy}{dx}\) = 1?
Solution:
(x + y + 1) \(\frac{dy}{dx}\) = 1 (given)
⇒ \(\frac{dy}{dx}\) = x + y + 1
⇒ \(\frac{dy}{dx}\) – x = y + 1
Comparing the equation with \(\frac{dx}{dy}\) + Px = Q?
P = – 1 and Q = y + 1
∴I.F. = e^∫pdy = e^-∫dy = e^-y
Here the required solution is:
⇒ x.e^∫pdy = ∫e^∫pdy. Qdy
⇒ x.e^-y = ∫e^-y (y + 1) dy + c
⇒ x.e^-y = – (y + 1)e^-y – ∫1(-e^-y) dy + c
⇒ x = -(y + 1) -1 + ce^y
x + y + 2 = ce^y

Question 9.
Solve the differential equation sec²x tany dx + sec² y tan xdy = 0?
Solution:
sec²x tany dx + sec² y tan xdy = 0 (given)

⇒ log y = – log x + log c
⇒ log x + log y = log c
⇒ log xy = log c
⇒ xy = c

Question 10.
Solve the differential equation \(\frac{dy}{dx}\) = y tanx – 2 sin x?
Solution:
\(\frac{dy}{dx}\) – y tanx = – 2 sin x (given)
Comparing with \(\frac{dy}{dx}\) + Py = Q,
P = – tan x, Q = – 2 sin x
L.F. = e^∫pdx = e^-∫tanxdx
= e^log_ecosx = cos x
The required solution is:
y.(I.F.) = ∫Q.I.F. dx + c
⇒ ycos x = -2∫sin x cos x dx + c
⇒ ycosx = -∫sin2xdx + c
⇒ ycosx = \(\frac { cos2x }{ 2 } \) + c.

Question 11.
Solve the differential equation \(\frac{dy}{dx}\) + 2y = 4x?
Solution:
Solution:
\(\frac{dy}{dx}\) + 2y = 4x
Comparing with \(\frac{dy}{dx}\) + Py = Q,
P = 2, Q = 4x
I.F. = e^∫pdx = e^∫2dx = e^2x
The required differential solution is:
y. (I.F) = ∫Q.I.F. dx + c
⇒ ye^2x = ∫4xe^2x dx + c

Question 12.
Solve the differential equation cos² x \(\frac{dy}{dx}\) + y = 2?
Solution:
cos² x \(\frac{dy}{dx}\) + y = 2
⇒ \(\frac{dy}{dx}\) + sec² x. y = 2 sec² x
Comparing with \(\frac{dy}{dx}\) + P y = Q,
P = sec² x, Q = 2 sec² x
I.F. = e^{∫sec2 xdx} 
Required solution is:
y. (I.F) = ∫Q.I.F. dx + c
⇒ y.e^tanx = ∫2 sec² x. e^tanx dx + c
⇒ y.e^tanx = 2.∫e^t dt + c, (Let tan x = t ⇒ sec² xdx = dt)
= 2.e^tanx + c
⇒ e^tanx (y – 2) = c.

Question 13.
Solve the differential equation cos x \(\frac{dy}{dx}\) + y = sin x?
Solution:
cos x \(\frac{dy}{dx}\) + y = sin x
⇒ \(\frac{dy}{dx}\) + sec x.y = tan x
Comparing with \(\frac{dy}{dx}\) + Py = Q,
P = secx, Q = tanx
L.F. = e^∫secxdx
= e^{log_e(sec x + tan x)} dx + c
Required Solution y.I.F. = ∫Q.I.F. dx + c
⇒ y. (sec x + tan x) = ∫tan x.(sec x + tan x) dx + c
= ∫sec x tan x dx + ∫tan² xdx + c
= sec x + ∫(sec² x – 1) dx = sec x + tan x = – x + c.

Question 14.
Solve the differential equation (1 + y²) dx = (tan^-1 y – x) dy? (CBSE 2015)
Solution:
(1 + y²) dx = (tan^-1 y – x) dy (given)

The required solution is:

Question 15.
Solve the differential equation (1 + y²) + ( x – e^{tan-1 y}) \(\frac{dy}{dx}\) = 0? (CBSE 2016)
Solution:
The given equation is:


Comparing with \(\frac{dx}{dy}\) + Px = Q,

The required solution is:

Put tan^-1 y = t,

Question 16.
Solve the differential equation (1 + x²) \(\frac{dy}{dx}\) + 2xy = \(\frac { x }{ 1+x^{ 2 } } \). Where, y = 0 and x = 1? (NCERT)
Solution:

Comparing with \(\frac{dy}{dx}\) + Py = Q,

The required solution is:

y = 0 and x = 1 (given)
0 (1 + 1)² = tan^-1 1 + c
⇒ 0 = \(\frac { \pi }{ 4 } \) + c
⇒ c = – \(\frac { \pi }{ 4 } \)
From eqn. (2),
y. (1 + x²) = tan^-1 x – \(\frac { \pi }{ 4 } \).

Question 17.
Solve the differential equation \(\frac{dy}{dx}\) + y cot x = 4x cosec x. Given: y = 0 and x = \(\frac { \pi }{ 2 } \)? (NCERT; CBSE, 2012)
Solution:
\(\frac{dy}{dx}\) + y cot x = 4x cosec x (given)
Comparing with eqn. (1) \(\frac{dy}{dx}\) + Py = Q,
P = cot x, Q = 4x cosec x
I.F. = e^∫pdx
= e^{∫cot xdx} = e^log(sinx) = sin x
The required solution is:
y.I.F = ∫I.F. × Qdx
⇒ y.sin x = ∫sin x × 4x cosec x dx
⇒ y sin x = 4∫\(\frac { xsinx }{ sinx } \) dx
⇒ y sin x = 4∫xdx
⇒ ysin x = \(\frac { 4x^{ 2 } }{ 2 } \) + c
⇒ y sin x = 2x² + c
When x = \(\frac { \pi }{ 2 } \) and y = 0,
0 (sin \(\frac { \pi }{ 2 } \)) = 2 ( \(\frac { \pi }{ 2 } \) ) ² + c
⇒ 0 = \(\frac { 2\pi ^{ 2 } }{ 4 } \) + c
∴ c = \(\frac { -\pi ^{ 2 } }{ 2 } \)
Putting in eqn. (1),
y. sin x = 2x² – \(\frac { -\pi ^{ 2 } }{ 2 } \).

MP Board Class 12 Maths Important Questions

MP Board Class 12th Biology Important Questions Chapter 13 Organisms and Populations

Organisms and Populations Important Questions

Organisms and Populations Objective Type Questions 

Question 1.
Choose the correct answer:

Question 1.
Example of photophilous plant is:
(a) Sunflower
(b) Abies
(c)Taxus
(d) All of these
Answer:
(a) Sunflower

Question 2.
Example of saprophytic plant is:
(a) Orobanche
(b) Monotropa
(c) Balanophora
(d) Rafflesia
Answer:
(b) Monotropa

Question 3.
Example of total parasitic plant:
(a) Balanophora
(b) Monotropa
(c) Rafflesia
(d) Drosera
Answer:
(c) Rafflesia

Question 4.
Characteristic features of aquatic plants are:
(a) Undeveloped stem with thorns
(b) Leaves with hairs
(c) Soft and flexible stem
(d) None of the above
Answer:
(c) Soft and flexible stem

Question 5.
Sunken stomata are found in:
(a) Xerophytic plants
(b) Aquatic plants
(c) Terrestrial plants
(d) Floating plants
Answer:
(a) Xerophytic plants

Question 6.
Spongy roots are present in :
(a) Vallisnaria
(b) Hydrilla
(c) Trapa
(d) Pistia
Answer:
(b) Hydrilla

Question 7.
Which soil is best for growth of plants : (MP 2016)
(a) Sandy soil
(b) Silty soil
(c) Loamy soil
(d) Clayey soil.
Answer:
(c) Loamy soil

Question 8.
Hydrophytes are characterised by : (MP 2009 Set A)
(a) Spiny less developed stem
(b) Leathery leaves
(c) Delicate and mucilage bearing stem
(d) Sunken stomata.
Answer:
(c) Delicate and mucilage bearing stem

Question 9.
Sunken stomata is found in :
(a) Xerophytes
(b) Hydrophytes
(c) Terrestrial
(d) Floating hydrophytes.
(a) Xerophytes

Question 10.
Spongy roots are found in :
(a) Jussiaea
(b) Eichhomia
(c) Trapa
(d) Pistia.
Answer:
(a) Jussiaea

Question 11.
Which one is the example of mangrove :
(a) Rhizophora
(b) Eichhomia
(c) Avicennia
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 12.
Pneumatophores are the characteristic feature of :
(a) Mangrove plants
(b) Mesophytes
(c) Hydrophytes
(d) Xerophytes
Answer:
(a) Mangrove plants

Question 13.
Vascular bundles are feebly developed in :
(a) Hydrophytes
(b) Xerophytes
(c) Psammophytes
(d) Halophytes.
Answer:
(a) Hydrophytes

Question 14.
Phyllode is a modification of :
(a) Root
(b) Stem
(c) Leaf
(d) None of these
Answer:
(c) Leaf

Question 15.
Phylloclade is a modification of :
(a) Root
(b) Stem
(c) Leaf
(d) Petiole
Answer:
(b) Stem

Question 16.
Rootless plant is :
(a) Wolffia
(b) Eichhomia
(c) Lemna
(d) Jussiaea.
Answer:
(a) Wolffia

Question 17.
Free floating plant is :
(a) Pistia
(b) Eichhomia
(c) Azolla
(d) All of these
Answer:
(d) All of these

Question 18.
Vivipary is found in :
(a) Hydrophytes
(b) Xerophytes
(c) Epiphytes
(d) Mangrove plants.
Answer:
(d) Mangrove plants.

Question 19.
Transpiration Higher among the following plants : (MP 2015)
(a) Mesophytes
(b) Hydrophytes
(c) Xerophytes
(d) Algae cells.
Answer:
(a) Mesophytes

Question 2.
Fill in the blanks:

1. Sandal plant is a …………………… parasite.
2.  …………………… and …………………… are found in lichen’s thallus.
3.  …………………… is the example of rodent.
4. Plants and animals combine togather and make the …………………… factor of the enviroments.
5. Inactivity of animals in summer season is called ……………………
6.  …………………… is prescribed the word ecology.
7. Raffesia is a …………………… plant.
8. Formation of forest ecosystem is the last stage of the ……………………
9. The study of human papulation is called ……………………
10. Respiratony root is found in ……………………  plant
11. Vivipary is the characteristics of …………………… plant
12. Group of animals which are interacted to each other is called ……………………
13. Herbivorous animals which are take the food in the plants is called ……………………
14. Plants they grow low density of light is called ……………………

Answer:

1. Partial parasite
2. Algae and fungi
3. Guinea pig
4. Biotic
5. Summer hybernation
6. Peter
7. Parasitic plant
8. Secession
9. Demography
10. Man – grove
11. Mangrove
12. Community
13. Plant eater
14. Sciophyte.

Question 3.
Match the followings:
I. (MP 2012)

Answer:

1. (d)
2. (a)
3. (b)
4. (e)
5. (c)

II. (MP 2012)

Answer:

1. (c)
2. (b)
3. (a)
4. (e)
5. (d)

Question 4.
Write the answer in one word/sentances:

1. Name any animal which lays eggs in the nest of other animal?
2. Name the plant which has pneumatophores.
3. Name two symbiotic partners of lichen.
4. Write the type of water which is available in plants.
5. Loss of the upper surface of the soil is called.
6. When do we celebrate world environmental day?
7. Name the rootless free floating plant.
8. In which plant, root cap is absent.
9. Give an example of symbiosis.
10. Formation of soil is called.
11. Plants are dependent to other plant for shelter is called.
12. Name an amphibious plant.
13. Name the plant in which stem is modified into leaf like structure and leaves are modified into spines.
14. Give an example of sexual dimorphism.

Answer:

1. Cuckoo
2. Rhizophora
3. Algae and Fungi
4. Capillary water
5. Soil erosion
6. 5 June
7. Wolffia
8. Hydrophytes
9. Pea and Rhizobium bacteria
10. Soil fertility
11. Epiphytes
12. Ranunculus
13. Opuntia
14. Honey bee

Organisms and Populations Very Short Answer Type Questions

Question 1.
Give two examples of each:
Symbiotic, commensal, phytoplankton, zooplankton and rooted floating plant
Answer:

1. Symbiotic – Escherichia, Triconimpha
2. Commensal – Orchid and Trees, Hermit crab
3. Phytoplankton – Nostoc, Anabaena
4. Zooplankton – Paramecium, Euglena
5. Rooted floating plant – Wolffia, Lemna.

Question 2.
Name the speciality of any population.
Answer:
Any population has some speciality, these are :

1. Population density
2. Growth rate
3. Death rate
4. Age distribution
5. Biotic capacity
6. Population growth form
7. Changes of population
8. Population dispersion

Question 3.
Name any two mangrove plants.
Answer:
Rhizophora, Avicennia.

Question 4.
Name the vagetation found in sundarban delta.
Answer:
Mangrove vegetation.

Question 5.
Name the plant in which pneumatophores are found.
Answer:
Rhizophora.

Question 6.
Give the term for scientific study of human population.
Answer:
Demography.

Question 7.
What is sex ratio of India according to 2001 census?
Answer:
933 females per 1000 of njales.

Question 8.
Name any animal which lays egg in the nest of other animal.
Answer:
Koyal.

Question 9.
Give the name of larva of frog, salamander and moths.
Answer:
Tadpole, Axolotal, Catterpillar.

Question 10.
Give the reason for increase in population.
Answer:
Decrease in death rate.

Question 11.
Give example of species polymorphism.
Answer:
Honeybee.

Question 12.
Name the species found in different geographical areas.
Answer: Allopatric

Question 13.
Write the name of any saprophytic angiospermic plant
Answer:
Monotropa.

Question 14.
What is called the relationship between hermit crab and sea – anemone?
Answer:
Protocooperation.

Question 15.
Why insectivores eat insects?
Answer:
To fulfill the need of nitrogen.

Question 16.
Which are main abiotic factor?
Answer:
Temperature, water, light, soil.

Question 17.
Give the name of one insectivores plant
Answer:
Utricularia.

Question 18.
What is ecological competition?
Answer:
Ecological competition is the struggle between two organisms for the same resources within an environment.

Question 19.
A transition area between two biomes.
Answer:
Ecotone.

Question 20.
The number of live births per thousand of population per year.
Answer:
Birthrate.

Organisms and Populations Short Answer Type Question

Question 1.
How is diapauses different from hibernation?
Answer:
Diapauses is a stage of suspended development to cope with unfavorable con-ditions. Many species of zooplankton and insects exhibit diapauses to tide over adverse climatic conditions during their development. On the other hands hibernation or winter sleep is a resting stage where in animals escape winters (cold) by hiding themselves in their shelters. They escape the winter season by entering a state of inactivity by slowing their metabolism. The phenomenon of hibernation is exhibited by bats, squirrels, and other rodents.

Question 2.
If a marine fish is placed in a fresh water aquarium, will the fish be able to survive? Why or why not?
Answer:
If a marine fish is placed in a freshwater aquarium, then its chances of survival will diminish. This is because their bodies are adapted to high salt concentrations of the marine environment. In freshwater conditions, they are unable to regulate the water entering their body (through osmosis). Water enters their body due to the hypotonic environment out¬side. This results in the swelling up of the body, eventually leading to the death of the marine fish.

Question 3.
Give definition of :

1. Population
2. Population density
3. Biotic potential
4. Growth rate
5. Death rate

Answer:

1. Population:
A community of animals, plants or humans among whose members inbreeding occurs.

2. Population density:
Population density is a measurement of population per unit area or unit volume.

3.Biotic potential:
Biotic potential is the ability of a population of living species to increase under ideal environmental conditions sufficient food supply, no predators and a lack of disease. An organisms rate of reproduction and the size of each litter are the primary determining factors for biotic potential.

4. Growth rate:
Human population growth is the increase in the number of individuals in a population. Global human population growth amounts to around 83 million annually.
Growth rate = \(\frac { Annually birth }{ Mid year population}\) x 100

5. Death rate:
The death rate is the number of people per thousand who die in a particular are during a particular period of time.
Death rate = \(\frac { Annually birth }{ Mid year population}\) x 100

Question 4.
What suggestions will you give for awakening people towards population control.
Answer:
Following things can be done for controlling the population:

1. Promotion of education: Misunderstanding can be avoided by promotion of education such as:

• Children are gift of god
• More income from the more children.

2. To introduce for reality of population growth.
3. Uses of family planning techniques.
4. Restriction of more than one marriage.
5. To decrease birth rate.

Question 5.
Write the differences:

1. Species and Population
2. Population growth and Population density,
3. Monospecific and Polyspecific population
4. Competition and Scattering.

Answer:
1. Species and Population:
Population on defined as organisms that belong to the identical species and identical geographical niche or area. The said area should enable these species to interbreed with each other.

2. Population growth and Population density:
In biology, Population growth is the increase in the number of individuals in a population but population density is a measurement of population per unit area, it is a quantity of type number density.

3. Monospecific or Polyspecific population:
Monospecific population is the population of individuals of only one species but poly specific or mixed population is the population of individuals of more than one species and it is generally referred to as a community.

4. Competition and Scattering:
Competition is in general a contest between two or more rivalry between two or more entities, organisms, animals, economic groups or social groups etc. Population scattering is a method which shows equilibrium by interaction of population. In population, scattering seeds away from the parent plant.

Question 6.
Give four reasons of population growth in India.
Answer:
Four reasons of population growth in India are as follows:

1. Increasing growth rate.
2. Decreasing death rate.
3. No importance of education and promotion of education is less.
4. Conservative.

Question 7.
Explain, the formation of new species.
Answer:
Although all life on earth shares various genetic similarities, only certain organisms combine genetic information by sexual reproduction and have offspring that can than successfully reproduce. Scientists call such organisms members of the same biological species.

Question 8.
What do you mean by population equilibrium?
Answer:
Population equilibrium:
When growth pattern of any population is observed then it becomes clear that any population increases at faster rate in the beginning in a new area and reaches to a maximum limit and then remain constant for long time.This phase is called as population equilibrium. At equilibrium birth and death rate of any population become equal.

Question 9.
Write cooperative interactions between the members of a species.
Answer:
The cooperative intraspecific interactions involve help to other members. The cooperative interaction amongst the individuals of a species is necessary for reproduction, perpetuation, parental care of young ones, social organization, territoriality, protection and food, etc.

1. Cooperation for protection and food:
For protection and food, members of a species may form groups. Such organisms which live in group are called as gregarious.

2. Cooperation for reproduction:
It is necessary for reproduction in which adult male and female comes together for mating.

3. Social organization:
Active association for mutual benefits amongst the individuals of same species often bring social organization. Success of these organizations is measured in the terms of the survival or colony. Honeybees, ants, wasps and termites, etc. form well – organized societies showing division of labour and polymorphism. In these social in sects, a large number of individuals of different kinds live together in the colony and work collectively for the benefit of the group.

The insect societies are formed of different castes such as workers, drones (male) and queen which are specialized for the different kind of work. The workers collect and store food, build houses of complicated design and pay special attention to the queen. The drones (male) and queen (female) are mainly concerned with reproduction.

Question 10.
Define phenotypic adaptation. Give one example.
Answer:
Phenotypic adaptation involves changes in the body of an organism in response to genetic mutation or certain environmental changes.These responsive adjustments occur in an organism in order to cope with environmental conditions present in their natural habitats For example, desert plants have thick cuticles and sunken stomata on the surface of their leaves to prevent transpiration. Similarly, elephants have long ears that act as thermoregulators.

Question 11.
Most living organisms cannot surv ive at temperature above 45°C. How are some microbes able to live in habitats with temperatures exceeding 100°C?
Answer:
Archaebacteria (Thermophiles) are ancient forms of bacteria found in hot water springs and deep sea hydrothermal vents. They are able to survive in high temperatures because their bodies have adapted to such environmental conditions. These organisms contain specialized thermo-resistant enzymes, which carry out metabolic functions that do not get destroyed at such high temperatures unlike other enzymes.

Most living – organism can not survive above 45°C because:

• Above 45° C enzymes get denatured.
• Protoplasm precipitates at high temperature.

Question 12.
List the attributes that populations but not individuals possess.
Answer:
A population has the following attributes that an individual does not possess:

1. Birth rates and death rates.
2. Sex ratio.
3. Population density.
4. Age distribution.
5. Population growth.

Question 13.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Answer:
A population grow exponentially if sufficient amounts of food resources are available to the individual. Its exponential growth can be calculated by the following integral form of the exponential growth equation :
N_t = N₀ e^rt
Where,
N_t = Population density after time t.
N_o = Population density at time zero.
r = Intrinsic rate of natural increase.
e = Base of natural logarithms (0.434).
From the above equation, we can calculate the intrinsic rate of increase (r) of a population. Now, as per the question,
Present population density =x
Then, Population density after two years = 2x
t = 3 years
Substituting these values in the formula, we get:
=> 2x = x e^3r
=> 2 = e^3r
Applying log on both sides :
=> log 2 = 3r log e
\(\frac { log 2 }{ 3 loge }\) = r
\(\frac { 0.301 }{ 3 x 0.434 }\) = r
\(\frac { 0.301 }{1.302 }\) = r
0.2311 = r
Hence, the intrinsic rate of increase for the above illustrated population is 0.2311.

Question 14.
Name important defence mechanisms in plants against herbivory.
Answer:
Several plants have evolved various mechanisms both morphological and chemical to protect themselves against herbivory.

1. Morphological defence mechanisms:

• Cactus leaves (Opuntia) are modified into sharp spines (thorns) to deter herbivores from feeding on them.
• Sharp thorns along with leaves are present in Acacia to deter herbivores.
• In some plants, the margins of their leaves are spiny or have sharp edges that prevent herbivores from feeding on them.

2. Chemical defence mechanisms :

• All parts of Calotropis weeds contain toxic cardiac glycosides, which can prove to be fatal in ingested by herbivores.
• Chemical substances such as nicotine, caffeine, quinine and opium are produced in plants as a part of self – defense.

Question 15.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?
Answer:
An orchid growing on the branch of a mango tree is an epiphyte. Epiphytes are plants growing on other plants which however, do not derive nutrition from them. Therefore, the relationship between a mango tree and an orchid is an example of commensalisms, where one species gets benefited while the other remains unaffected. In the above interaction, the orchid is benefited as it gets support while the mango tree remains unaffected.

Question 16.
What is the ecological principle behind the biological control method of managing with pest insects?
Answer:
The ecological principle behind the biological control method of managing with pest insects is to keep their population in check by using their natural predators and parasites.

Question 17.
Distinguish between the following:

1. Hibernation $nd Aestivation
2. Ectotherms and Endotherms.

Answer:
1. Differences between Hibernation and Aestivation :

Hibernation:

• It is called winter sleep.
• Animal rests in warm place.
• It lasts for long winter months.

Aestivation:

• It is called summer sleep.
• Animal rests in cool, moist, shady place.
• It lasts for day time as nights are cooler and animal will come out.

2. Differences between Ectotherms and Endotherms :

Ectotherms:

• These are cold – blooded animals.
• They are unable to regulate their body temperature which changes with temperature of the surrounding environment.
• They hibernate in winters and aestivate in summers.

Endotherms:

• These are warm – blooded animals.
• They are capable of maintaining their body temperature.
• They can regulate their body temperature so, they do not need to show behavioural adaptations like these.

Question 18.
List the various abiotic environmental factors.
Answer:

1. Atmospheric factors : Light, temperature, wind and water.
2. Lithosphere : Rock, soil.
3. Hydrosphere : Pond, river, lake and ocean.
4. Edaphic factors: Soil texture, soil water, soil air, soil microorganism, soil pH, minerals.
5. Topographic factors : Slope, altitude, valley.

Question 19.
Give an example for:

1. An endothermic animal,
2. An ectothermic animal,
3. An organism of benthic zone.

Answer:

1. An endothermic animals : Birds such as crows, sparrows, pigeons, cranes, etc. and mammals such, as bears, cows, rats, rabbits etc. are endothermic animals.
2. An ectothermic animals : Fishes such as sharks, amphibians frogs, and reptiles, tortoise, snakes, and lizards are ectothermic animals.
3. An organism of benthic zone : Decomposing bacteria is an example of an organism found in the benthic zone of a water body.

Question 20.
Define population and community.
Answer:
Population:
Population is a group of individuals of same species, which can reproduce among themselves and occupy a particular area in a given time.

Community:
It is an assemblage of several population in a particular area and time and exhibit interaction and interdependence through trophic relationship.

Question 21.
Define the following terms and give one example for each:

1. Commensalism
2. Parasitism
3. Camouflage
4. Interspecific competition.

Answer:
1. Commensalism:
It is interspecific interaction in which one species is benefited and other the neither harmed nor benefitted. The two individuals may be physically associated. example Sucker fish and shark.

2. Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples Parasitic bacteria:
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

3. Camouflage:
It is a strategy adapted by prey species to escape their predators. Organisms are cryptically coloured so, that they can easily mingle in their surroundings and escape their predators. Many species of frogs and insects camouflage in their surroundings and escape their predators.

4. Interspecific competition:
It is an interaction between individuals of different species where both species get negatively affected. For example, the competition between flamingoes and resident fishes in South American lakes for common food resources i.e., zooplankton.

Question 22.
Explain population dispersal in brief.
Answer:
Population dispersal:
The movement of individuals of a population from a particular place to another is known as population dispersal. Dispersal of members of population may occur for search of food, escape from enemies or for breeding purpose. There are three types of

population dispersal:

1. Immigration – Inward movement of members of other population and settle with local population. It causes increase in population.
2. Emigration – Outward movement of individuals of a population to shift in some other place. It causes decrease in population.
3. Migration – It can be observed in birds. It is periodic departure of organism and returning back to its original place, example Siberian cranes visit India during winters and go back to north during summer.

Question 23.
With the help of suitable diagram describe the logistic population growth curve.
Answer:
Logistic growth:
1. The resources become limited at certain point of time so, no population can grow exponentially.

2. This growth model is more realistic.

3. Every ecosystem or environment or habitat has limited resources to support a particular maximum number of individuals called is carrying capacity (K).

4. When N is plotted in relation to time t, the logistic growth show sigmoid curve and is also called Verhulst-Pearl logistic growth. It is given by the following equation:
\(\frac { dN }{ dt }\) = rN [ \(\frac{K-N}{K}\) ]
Where, N = Population density at time t
R = Intrinsic rate of natural increase
K = Carrying capacity.

a = When resources are not limiting the growth, plot is exponential
b = When resources are limiting the growth. Plot is logistic, k is carrying capacity.

Question 24.
Write characteristic features of aquatic plants.
Answer:
Characteristic features of aquatic plants (Hydrophytes):

1. The cuticle is either absent or very thinly deposited on the epidermis.
2. Usually stomata are absent, if present they are non – functional.
3. Many air chambers or large intercellular spaces are present inside the plant body of hydrophytes.
4. The whole body of the plant helps in the absorption of water and minerals.
5. Root system is feebly developed and usually non – branched or less branched. Some hydrophytes like Wolffia lack foots.
6. The mechanical tissues are less distributed.
7. The vascular tissues are either absent or feebly developed in the case of hydrophytes.

Question 25.
Write short note on population growth form or pattern.
Answer:
Population growth from or pattern: The way of growth of population is known as population growth pattern. Population growth shows two patterns :

1. S – shaped form : In this pattern, the increase in population is very slow in the beginning, then there is a fast growth rate and then becomes stable.

2. J – shaped form : In some organisms the population increases very fast in the beginning because of some environmental factors, the growth is checked abruptly.

Question 26.
Give only five examples of communication of informations in animals.
Answer:
Communication of information within a species:
Communication may be defined as an act which influences the activity of one individual by some behaviour of another. Most of the animal species have some mechanism of exchanging information among their members through sight, sound, touch or chemicals. The evolution of the techniques of communication in the animal kingdom is progressive but complex.

Following are few examples:

1. The honeybee gives the information to the other worker bees of the hive regarding availability of nectar in the vicinity by dancing in a special manner on the surface of the hive. Through the round dance they communicate that the availability of the food is very near to the hive and the waggle dance (by moving abdomen) shows the direction and the distance of the food source from the hive.

2. Croaking of male frog attract female frog during breeding season.

3. The dogs express their various intensions by facial expressions and movement of the tail and by making typical sound.

4. Rabbits inform their group members about any sort of danger by tapping their
tails.

5. Certain chemical compounds called pheromones are secreted by animals to transmit message to other members of the species. Pheromones are detected by smell or taste.

Question 27.
Write distinguish between:

1. Parasitism and Symbiosis,
2. Mutualism and Commensalism,
3. Hydrosere and Xerosere.

Answer:
1. Differences between Parasitism and Symbiosis:

Parasitism:

• In this case, one organism depends on other for their food.
• Only one organism benefitted.
• It produce only disease.
• example Tapeworm and Man.

Symbiosis:

• In symbiosis two organisms lives together in such way that both are benefitted by each other.
• Both organism benefitted.
• It does not produce any disease.
• example Lichens.

2. Differences between Mutualism and Commensalism:

Mutualism:

• It is an interspecific interaction in which both the species are mutually benefited.
• The two individuals may be physically or physiologically associated.
• example Rhizobium and the legume plants.

Commensalism:

• It is interspecific interaction in which one species is benefited and other the neither harmed nor benefitted.
• The two individuals may be physically associated.
• example Sucker fish and shark.

3. Differences between Hydrosere and Xerosere:

Hydrosere:

• Succession in water is called hydrosere.
• Changes are fast.
• It produce aquatic plants.

Xerosere:

• Succession in deserts is called xerosere.
• Changes are slow.
• It produce xerophytic plants.

Question 28.
Write short notes on:

1. Parasitism
2. Species dominance
3. Biotic potential.

Answer:
1. Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples : Parasitic bacteria:
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

2. Species dominance:
Presence of any biotic community or presence of more popu¬lation of any species biosphere shown its dominance and it is called species dominance which has capacity to reduce other species in population. The effect of dominant species is more upon environment or other species.

3. Biotic potential:
Biotic potential is the ability of a population of living species to increase under ideal environmental conditions sufficient food supply, no predators and a lack of disease. An organisms rate of reproduction and the size of each litter are the primary determining factors for biotic potential.

Question 29.
Explain population fluctuation.
Answer:
Population is generally a group of individuals of a particular species occupying a particular area at a specific time.

Population fluctuation:
Any increase or decrease in number of individuals in a population from its equilibrium state is known as population fluctuation. It may occur due to various reasons, such as due to change in climate or due to change in physical environment or due to predators.

Question 30.
Give an example for:

1. Heliophytes
2. Sciophytes
3. Viviparous.

Answer:

1. Heliophytes : Sunflower, Amranthus.
2. Sciophytes : Java moss, Lycopodium, Polytrichus.
3. Viviparous : Rhizophora, Salicomia, Sonneratia.

Organisms and Populations Long Answer Type Questions

Question 1.
Write a short note on :

1. Adaptations of deserts plants and animals
2. Adaptations of plants to water scarcity
3. Behavioural adaptations in animals
4. Importance of light to plants
5. Effect of temperature or water scarcity and the adaptations of animals.

Answer:
1. Adaptations of desert plants and animals :

(i) Adaptations of desert plants:
Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration. In Opuntia, the leaves are entirely modified into spines and photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM (C₄ pathway). It enables the stomata to remain closed during the day to reduce the loss of water through transpiration.

(ii) Adaptations of desert animals:
Animals found in deserts such as desert Kangaroo rats, Lizards, Snakes, etc. are well adapted to their habitat. The Kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and bur¬row themselves in the sand during afternoons to escape the heat of the day. These adaptations occur in desert animals to prevent the loss of water.

2. Adaptations of plants to water scarcity :
Plants found in deserts are well adapted to cope with water scarcity and scorching heat of the desert. Plants have an extensive root system to tap underground water. They bear thick cuticles and sunken stomata on the surface of their leaves to reduce transpiration.

In Opuntia, the leaves are modified into spines and the process of photosynthesis is carried out by green stems. Desert plants have special pathways to synthesize food, called CAM (C₄ pathway). It enables their stomata to remain closed during the day to reduce water loss by transpiration.

3. Behavioural adaptations in animals:
Certain organisms are affected by temperature variations. These organisms undergo adoptions such as hibernation, aestivation, migration, etc. to escape environmental stress to suit their natural habitat. These adaptations in the behaviour of an organism are called behavioural adaptations.

For example, ectothermal animals and certain endosperms exhibit behavioural adaptations. Ectotherms are cold blooded animals such as fish, amphibians, reptiles etc. Their temperature varies with their surroundings. For example, the desert lizard basks in the sun during early hours when the temperature is quite low. However, as the temperature begin? to rise, the lizard burrows itself inside the sand to escape the scorching, sun.

Similar burrowing strategies are exhibited by other desert animals. Certain endotherms (warm blooded animals) such as birds and mammals escape cold and hot weather conditions by hibernating during winters and aestivating during summers. They hide themselves in shelters such as caves, burrows, etc. to protect against tem-perature variations.

4. Importance of light to plants:
Sunlight acts as the ultimate source of energy for plants. Plants are autotrophic organisms, which need right for carrying out the process of photosynthesis. Light also plays an important role in generating photoperiodic responses occurring in plants. Plants respond to changes in intensity of light during various seasons to meet their photoperiodic requirements for flowering. Light also plays an important role in aquatic habitats for vertical distribution of plants in the sea.

5. Effects of temperature or water scarcity and the adaptations of animals:
Temperature is the most important ecological factor. Average temperature on the Earth varies from one place to another. These variations in temperature affect the distribution of animals on the Earth. Animals that can tolerate a wide range of temperature are called eurythermals. Those which can tolerate a narrow range of temperature are called stenothermal animals. Animals also undergo adaptations to suit their natural habitats.

Animals found in polar regions have thick layers of fat below their skin and thick coats of fur to prevent the loss of heat. Some organisms exhibit various behavioural changes to suit their natural habitat. These adaptations present in the behaviour of an organism to escape environmental stresses are called behavioral adaptations. For example, desert lizards are ectotherms. This means that they do not have a temperature regulatory mechanism to escape temperature variations.

These lizards bask in the sun during early hours when the temperature is quite low. As the temperature begins to increase, the lizard burrows itself inside the sand to escape the scorching sun. Similar burrowing strategy is seen in other desert animals. Water scarcity is another factor that forces animals to undergo certain adaptations to suit their natural habitat. Animals found in deserts such as desert kangaroo rats, lizards, snakes, etc. are well adapted to stay in their habitat.

The kangaroo rat found in the deserts of Arizona never drinks water in its life. It has the ability to concentrate its urine to conserve water. Desert lizards and snakes bask in the sun during early morning and burrow in the sand as the temperature rises to escape the heat of the day. Such adaptations can be seen to prevent the loss of water.

Question 2.
List any three important characteristics of a population and explain.
Answer:
The three important characteristic of a population are:

1. Population density : The number of individuals of a species per unit area or a volume is called population density.
P_D = \(\frac { N }{ S }\)
Where, PD = Population density
N = Number of individuals in a region S = Number of unit area in a region.

2. Birth rate : It is expressed as the number of births per 1000 individuals of a population per year.

3. Death rate : It is expressed as the number of deaths per 1000 individuals of a population per year.

Question 3.
Describe the anatomical adaptation of xerophytic plants.
Answer:
The anatomical adaptations of xerophytic plants are:

1. The epidermis of xerophytic plant organs is covered by thick cuticle. It protects plants against heavy transpiration and also provides mechanical strength to some extent.

2. Epidermis may be multiple layered which also reduces the rate of transpiration.
Examples : Banyan, Nerium etc.

3. Except some monocotyledonous plants the stomata are restricted to the lower surface of the leaves. The stomata are generally sunken and surrounded by many hairs.

4. Hypodermis in xerophytes is multilayered and sclerenchymatous which checks evaporation of water.
Example: Pinus needle.

5. Mesophyll tissue is well differentiated into palisade and spongy parenchyma but palisade parenchyma is more in number than spongy parenchyma.
Examples : Pinus needle, Nerium leaf.

6. Vascular bundles are well developed and differentiated into xylem and phloem. Vascular bundles are comparatively more in number. Besides vascular bundles, mechanical tissues are well developed.

7. Leaves of some xerophytes have some rolling devices due to presence of motor cells or bulliform cells in the epidermal zone.
Examples : Poapratensis, Agropyron, Ammophila etc.

8. Stomata in certain desert plants such as Capparis spinosa and Aristida ciliata may sometimes get blocked due to deposition of resinous matter or wax. This adaptation is also known to reduce the loss of water during transpiration.

Question 4.
Describe the morphological adaptations of aquatic plants.
Or
Describe morphological adaptations found in the stem and leaves of xerophytic plants.
Or
Explain in brief, xerophytic plants along with examples.
Answer:
The morphological adaptations of aquatic plants are:

1. Adaptations in roots:

(i) Root system is feebly developed and unbranched. Some floating plants like Wolffia, Utricularia etc. and submerged plants like Ceratophyllum lack roots.

(ii) Root hairs are absent except rooted floating hydrophytes. In most of the hydro¬phytes roots are meaningless as the entire surface of the plant body is in direct contact with water and acts as absorptive surface. This may probably be the reason why roots in hydrophytes are reduced or absent.

(iii) True root caps are absent in some free floating hydrophytes and in lieu of that they develop root pockets or root sheaths which protect their tips from injuries.

(iv) If roots are present they are always of adventitious type e.g., Nymphaea, Nelumbo etc. In Jussiaea modified spongy negative geotrophic floating roots are present which pro¬vide buoyancy to the plants and also do the job of gaseous exchange.

2. Adaptations in stems:

(i) In submerged hydrophytes like Hydrilla, Potamogeton etc. stems are long, slender and flexible.

(ii) In free floating hydrophytes, stems are modified as thick, stout, stoloniferous or as offset and occur horizontal on the water surface. It bears vegetative bud at their apex and helps in vegetative propagation.

(iii) In rooted floating hydrophytes or rooted submerged hydrophytes or in marshy plants, stem may be modified as runner or rhizome.

(iv) The stem may be modified as runner, stolon, tuber to perform vegetative propaga¬tion effectively. Usually stems are of perennial nature.

3. Adaptations in leaves:

(i) Leaves are thin, long, ribbon shaped in the submerged forms.
Example: Vallisneria, Ceratophyllum etc. or they are finely dissected as found in Ranunculus aquatilis.

(ii) In rooted floating plants, like Nymphaea, Nelumbo etc. the petioles of leaves show indefinite power of growth and they keep the laminae of leaves always on the surface of water.

(iii) In free floating plants like Eichornia, Trapa etc. the petioles become charac¬teristically swollen and develop sponginess which keep the plant afloat.

(iv) Occurrence of two or more than two different types of leaves in a plant is called heterophylly. Examples are Limnophila, Heterophylla, Sagittaria, Ranunculus aquatilis etc. They bear leaves of many types viz., submerged leaves, floating leaves, aerial leaves, linear, ribbon shaped leaves or dissected leaves. Floating leaves are entire and lobed. The broad leaves found on the surface of water transpire actively and regu¬late the hydrostatic pressure in the plant body. The submerged leaves act as water absorbing organs. Heterophylly is also associated with quantitative reduction in transpiration.

(v) In the amphibious or marshy plants, the leaves that are exposed to air show typical mesophytic features. They are leathery and more tough than the submerged and floating hydrophytes.

Question 5.
What do you mean by biotic community? Describe characteristic features of any biotic community.
Answer:
Biotic community:
A biotic community is a localized association of several populations of different species living in a given geographic area of habitat. It represents heterogeneous assemblage of different groups of organisms both plants and animals. Biotic community is composed of smaller units of intimately associated members belonging to different species.

The different species of a community share common environment and their relationships are based on direct or indirect functional interactions. The nature or relationship is determined by the requirements of the members of a community.

Characteristics of a community:
Each community has its own characteristics which are not shown by its individual component species.

1. Species diversity:
Each community is made up of much different organisms Plants, animals, microbes, which differ taxonomically from each other. The number of species and population abundance in community also vary greatly.

2. Growth form and Structure:
Each community have a definite growth form. This different growth form determines the structural pattern of a community.

3. Dominance:
In each community, all the species are not equally important. There are relatively only a few of these, which determine the nature of community. These few species exert a major controlling influence on the community. Such species are known as dominants.

4. Succession:
Each community has its own development history. It develops as a result of a directional change in it with time.

5. Trophic structure (Self – sufficiency):
Nutritionally, each community, a group of autotrophic plants as well as heterotrophic animals exists as a self – sufficient, perfectly balanced assemblage of organism.

Question 6.
Write cooperative interactions between the members of a species.
Answer:
The cooperative intraspecific interactions involve help to other members. The co – operative interaction amongst the individuals of a species is necessary for reproduction, perpetuation, parental care of young ones, social organization, territoriality, protection and food, etc.

1. Cooperation for protection and food:
For protection and food, members of a species may form groups. Such organisms which live in group are called as gregarious.

2. Cooperation for reproduction:
It is necessary for reproduction in which adult male and female comes together for mating.

3. Social organization:
Active association for mutual benefits amongst the individuals of same species often bring social organization. Success of these organizations is measured in the terms of the survival or colony. Honey – bees, ants, wasps and termites, etc. form well-organized societies showing division of labour and polymorphism. In these social insects, a large number of individuals of different kinds live together in the colony and work collectively for the benefit of the group.

The insect societies are formed of different castes such as workers, drones (male) and queen which are specialized for the different kind of work. The workers collect and store food, build houses of complicated design and pay special attention to the queen. The drones (male) and queen (female) are mainly concerned with reproduction.

Question 7.
Explain interrelationship between various species of a biotic community along with examples.
Answer:
Odum (1971) distinguished interactions into two broad categories :

1. Positive interactions and
2. Negative interactions.

1. Positive Interactions:
In this type, those type of interactions are considered in which both interacting species are mutually involved to help each other. Here, one interacting species helps the other either one way or on reciprocal terms and may be in the form of nutrition or shelter or substratum or transport. Here interaction may be obligatory or facultative. The positive interactions include three types of interactions. They are :

(i) Commensalism:
This type of interaction occurs in between two organisms of two different species in which one species benefits and the other is neither benefited nor harmed, example Lichens.

(ii) Protocooperation:
The type of interaction where both population are ben¬efited but not obligatory i.e., not essential for the survival of either population is called proto – co – operation. This is also known as non-obligatory mutualism. The relationship between hermit crab and sea anemone is an example of proto – cooperation. The crab uses the gastropod (mollusc) shell as a portable shield and the sea anemone eats the leftover food of the crab, which is protected from its predators by the stinging cells of the sea anemone.

(iii) Mutualism or Symbiosis:
When two different species grow together and are mutually benefited, the plants are known as symbiotic plants and the phenomenon is called symbiosis or mutualism. It is a sort of obligatory association. This type the organisms are dependent upon each other for survival. In this type of association, two types of species are physiologically related. This type of relationship may exist in between two plants or in between one plant and one animal or in between two animals.

2. Negative Interactions:
Those interaction of two different species in which both are harmed or one organism is benefited while other is more or less harmed is referred to as negative interac¬tion. In this type of interaction one population eats the other type of population or one organism does not allow other organisms to grow near it by using the food supply of the other or producing toxic substances. The negative interactions be categorised into following types:

• Competition
• Parasitism
• Predation and
• Antibiosis.

Competition:
When interaction occurs between two species for the use of same resources and when resources are in short supply is referred to as competition. It is a relationship which involves struggle amongst the organisms for food, shelter, water, sunlight and climate.

Parasitism:
Parasite is an organism which is dependent upon other living organism for their food and other requirements. Parasitism is an interspecific interrelation in which the individual of one species lives at the cost of the individual of another species, its host which it harms and eventually kills. In this case host is always larger than the parasite. The parasite that is seen on the surface of host plant is called ectoparasite or external parasite and if a parasite is found inside the plant body is said to be an endoparasite or internal parasite.

Examples : Parasitic bacteria :
Parasitic bacteria cause several diseases in plants and animals. For example, the citrus canker disease of lemon is caused by Xanthomonas citri, black rots of vegetables of Cruciferae group is caused by Xanthomonas campestris. The common diseases caused by bacteria in human beings are diarrhoea, pneumonia, tetanus, leprosy, plague etc.

Predation:
To obtain food by hunting is predation. A predator is an animal or plant that kills and feeds on another organism, its prey. It represents a generalised carnivorous habit. Commonly predators are larger than their prey but this is not always true but predator is in any case always occur as a naturally better equipped hunter than its prey. Most of the predatory organisms are animals but there are some plants (carnivores) also especially fungi which feed upon other animals.

MP Board Class 12th Biology Important Questions