1st PUC

You can Download Chapter 16 Digestion and Absorption Questions and Answers, 1st PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Question Bank Chapter 16 Digestion and Absorption

1st PUC Biology Digestion and Absorption NCERT Text Book Questions and Answers

Question 1.
Choose the correct answer among the following:
(a) Gastric juice contains
(i) Pepsin, lipase and rennin
(ii) Trypsin, lipase and rennin
(iii) Trypsin, pepsin and lipase
(iv) Trypsin, pepsin and renin
Answer:
(i) Pepsin, lipase and rennin

(b) Succus entericus is the name given to
(i) A junction between the ileum and large intestine
(ii) Intestinal juice
(iii) Swelling in the gut
(iv) Appendix
Answer:
(ii) Intestinal juice

Question 2.
Match column I with column II

Column I – Column II

(a) Bilirubin and biliverdin – (i) Parotid
(b) Hydrolysis of starch – (ii) Bile
(c) Digestion of fat – (iii) Lipases
(d) Salivary gland – (iv) Amylases
Answer:
(a) – ii
(b) – iv
(c) – iii
(d) – i

Question 3.
Answer briefly:
(1) Why are villi present in the intestine and not in the stomach?
Answer:
Because villi are supplied with a network of capillaries and a large lymph vessel called a lacteal. The mucosal epithelium has goblet cells which secrete mucus that help in lubrication. Mucosa also forms glands in the stomach (gastric glands) and crypts in between the bases of villi in the intestine (crypts of lieberkuhn).

(2) How does pepsinogen change into its active form?
Answer:
The proenzyme pepsinogen, on exposure to hydrochloric acid, gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach. Pepsin converts proteins into proteases and peptones (peptides).

(3) What are the basic layers of the wall of the alimentary canal?
Answer:
The wall of the alimentary from the esophagus to the rectum possesses four layers namely serosa, muscularity, submucosa, and mucosa. The serosa is the outermost layer and is made up of a thin mesothelium (epithelium of visceral organs) with some connective tissues. Muscularity is formed by smooth muscles usually arranged into an inner circular and an outer longitudinal layer. An oblique muscle layer may be present in some regions. The submucosal layer is formed of loose connective tissue containing nerves, blood, and lymph vessels. In the duodenum, glands are also present in the submucosa. The innermost layer lining the lumen of the alimentary canal is the mucosa.

(4) How does bile help in the digestion of fats?
Answer:
The bile released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol and phospholipids but no enzymes. Bile helps in the emulsification of fats i.e. break down the fats into very small micelles. Bile also activates lipases. Fats are broken down by lipases with the help of bile into di-and monoglycerides.

Question 4.
State the role of pancreatic juice in the digestion of proteins.
Answer:
The pancreatic juice contains inactive enzymes trypsinogen, chymotrypsinogen, procarboxy¬peptidases, amylases, lipases, and nucleases. Trypsinogen is activated by an enzyme enterokinase secreted by intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. These enzymes are concerned with protein, carbohydrate, fats, and nucleic acid digestion.

Question 5.
Describe the process of digestion of protein in the stomach.
Answer:
The proenzyme pepsinogen present in the stomach gets converted to proteolytic enzyme pepsin in presence of Hydrochloric acid. Pepsin converts proteins into proteases and peptones. Rennin is another proteolytic enzyme present in the gastric juice of infants which helps in the digestion of milk proteins.

Question 6.
Give the dental formula of human beings.
Answer:

Question 7.
Bile juice contains no digestive enzymes, yet it is important for digestion. Why?
Answer:
Bile is a watery greenish fluid mixture containing bile pigments, bile salts, cholesterol, and phospholipids. Bile helps in the emulsification of fats i.e. breaking down of the fats into smaller micelles, it also activates lipases. Thus, it is important for digestion.

Question 8.
Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by its source gland?
Answer:
Chymotrypsin is a protein digestive enzyme that breaks down proteins, peptones, and proteoses into dipeptides. The other two proteolytic enzymes are trypsin and carboxyl peptidase.

Question 9.
How are polysaccharides and disaccharides digested?
Answer:
About 30% of starch is hydrolysed by the enzyme salivary amylase into disaccharide maltose in oral cavity.

Carbohydrates in the chyme are hydrolysed by pancreatic amylase into disaccharides. Polysaccharides (starch)

Maltase present in the intestinal juice converts maltose into glucose. Lactase converts lactose into glucose and galactose. Sucrase converts sucrose into glucose and fructose

Question 10.
What would happen if HCl were not secreted in the stomach?
Answer:
Importance of HCl secreted by stomach:
HCl provides the acidic pH (pH 1.8) optimal for pepsins. It converts proenzyme pepsinogen into active enzyme pepsin, the proteolytic enzyme of the stomach.
HCl is also necessary to kill harmful bacteria which may be present in the food.

Question 11.
How does butter In your food get digested and absorbed in the body?
Answer:
Butter is rich in lipids. Digestion of fat starts in the stomach. The gastric lipase hydrolyses a small amount of lipids. Bile helps in the emulsification of fats and activates lipases which break down fats into diglycerides and then to monoglycerides. Monoglycerides and Diglycerides are further broken down to fatty acids and glycerol in presence of a lipase.

Fatty acids and glycerol being insoluble is converted to small droplets called micelles which are reformed to very small protein-coated fat globules called the chylomicrons which are absorbed by the intestine villi.

Question 12.
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Answer:
There is no protein digestion in the oral cavity as there are no proteases present.
Stomach: The proenzyme pepsinogen, on exposure to hydrochloric acid gets converted to active enzyme pepsin. Which converts proteins into proteases and peptones.

Intestine: Pancreatic juice contains inactive enzymes like trypsinogen, chymotrypsinogen, and procarboxy peptidases. Trypsinogen is converted to active trypsin with the help of the enzyme enterokinase. Trypsin, in turn, activates other proteolytic enzymes in the pancreatic juice. Proteins, proteoses, and peptones in the intestine are now converted to dipeptides in presence of trypsin, chymotrypsin, and carboxypeptidase.

The aminopeptidase hydrolyses the peptide bond that attaches the terminal amino acid to the amino end of the peptide. Dipeptidase acts on dipeptides and converts them to amino acids.

Amino acids are the end products of protein digestion.

Question 13.
Explain the terms thecodont and diphyodont.
Answer:
The alimentary canal begins with an anterior opening-the mouth, and it opens out posteriorly through the anus. The mouth leads to the buccal cavity or oral cavity. The oral cavity has a number of teeth and a muscular tongue. Each tooth is embedded in a socket of the jaw bone. This type of attachment is called thecodont. The majority of mammals including humans forms two sets of teeth during their life, a set of temporary milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.

Question 14.
Name different types of teeth and their number in an adult human.
Answer:
Incisors – 8, canine – 4, premolars – 8, molars – 12.

Question 15.
What are the functions of the liver?
Answer:
Functions of Liver:

1. The liver performs several roles in carbohydrate, lipid, and protein metabolism.
2. The liver is responsible for the mainstay of protein metabolism, synthesis as well as degradation.
3. The liver produces and excretes bile (a greenish liquid) required for emulsifying fats.
4. The breakdown of insulin and other hormones
5. The liver breaks down hemoglobin, creating metabolites that are added to bile as a pigment (bilirubin and biliverdin).
6. The liver converts ammonia to urea.

1st PUC Biology Digestion and Absorption Additional Questions and Answers

1st PUC Biology Digestion and Absorption One Mark Questions

Question 1.
What type of medium is required for the activity of trypsin?
Answer:
Alkaline medium.

Question 2.
Which cells secrete HCI?
Answer:
Parietal cells.

Question 3.
What is the other name for Pancreatic amylase?
Answer:
Amylopsin

Question 4.
What is chylomicron?
Answer:
The reconstructed triglycerides combine with phospholipids and cholesterol are released into the lymph in the form of protein-coated water-soluble fat globules or droplets. These are called chylomicron.

Question 5.
Which glands secrete HCI?
Answer:
Gastric glands.

Question 6.
Which digestive Juice is non-enzymatic?
Answer:
BWe.

Question 7.
Which gland secretes Bile?
Answer:
Liver.

Question 8.
Where is caecum located in the alimentary canals?
Answer:
The caecum is located at the junction of the small intestine and colon (large intestine).

Question 9.
What type of muscles present on the wall stomach?
Answer:
Smooth muscles.

Question 10.
Which part of small intestine is the longest?
Answer:
Ileum

Question 11.
Give an example of a vestigial organ.
Answer:
Vermiform appendix. (Oct. 83)

Question 12.
How is the tongue attached to the floor of the buccal cavity?
Answer:
The tongue is attached to the floor of the buccal cavity by the frenulum.

Question 13.
Mention the function of villi. (Oct. 84)
Answer:
The Villi are structures of the intestinal wall that increase the surface of absorption i.e. their chief function is absorption

Question 14.
What is ‘Bile’? (Oct. 85)
Answer:
Bile is the juice secreted by the liver.

Question 15.
What is the mechanical action taking place in the mouth called? (Oct. 90)
Answer:
The mechanical action taking place in the mouth is known as ‘Mastication’.

Question 16.
Name a proteolytic enzyme
Answer:
Pepsin (or trypsin, chymotrypsin, carboxypeptidase Aminopeptidase, tripeptidase, dipeptidase – anyone can be named)

Question 17.
Name the gland which is both exo and endocrine? (April 93)
Answer:
Pancreas.

Question 18.
What is the sphincter of Oddi?
Answer:
The sphincter of Oddi is the muscular structure, that guards the opening of the hepato-pancreatic duct into the duodenum.

Question 19.
Name the Enzyme that hydrolyses lipids.
Answer:
‘Lipases’ hydrolyze lipids. (Oct. 94)

Question 20.
Name the enzyme which does not act on the food materials in the small intestine.
Answer:
The “Enterokinase” enzyme.

Question 21.
What prevents the entry of food into the larynx during swallowing? (April 96)
Answer:
Epiglottis.

Question 22.
Which hormone causes Diabetes mellitus? (April 1996)
Answer:
The deficiency of ‘Insulin’ causes Diabetes mellitus.

Question 23.
Name the pancreatic hormone which acts as a hypoglycemic factor.(April 97, M.Q.P.)
Answer:
Insulin acts as a hypoglycemic factor.

Question 24.
Name the Carbohydrate digesting enzyme present in saliva. (Oct. 97)
Answer:
Salivary amylase or Ptyalin.

Question 25.
What is the basic mechanism of digestion in the intestine? (Oct. 88)
Answer:
‘Hydrolysis’.

Question 26.
The catalytic activity of which enzyme produces Fructose as one of the end products? (Oct. 88)
Answer:
Sucrase acts on the disaccharide sucrose resulting in a molecule of Fructose and a molecule of glucose.

Question 27.
What type of enzyme is pepsin? (Oct. 89)
Answer:
Proteolytic enzyme.

Question 28.
What Is emulsification? (Oct. 98)
Answer:
Emulsification is a process by which fats are converted to small water-soluble muscles by Bile salts making them easily accessible to the action of Fats digesting enzymes called lipases.

Question 29.
Mention the function of microvilli. (April 99)
Answer:
The chief function of the microvilli is absorption (i.e., it increases the surface of absorption).

Question 30.
What Is Chyme? (April 2000, July 2010)
Answer:
The liquified and acidified food present in the stomach is called as chyme.

Question 31.
What are the end products of protein digestion? (April 2001)
Answer:
Amino acids.

Question 32.
What is Peristalsis? (Oct. 2003)
Answer:
The powerful, rhythmic waves of muscular contraction and relaxation in the walls of hollow tubular organs like the digestive tract is called peristalsis.

Question 33.
Name an enzyme present in saliva meant for killing bacteria. (July 2006)
Answer:
Lysozyme.

Question 34.
What Is succus entericus? (April 2007)
Answer:
Succus entericus is the intestinal juice made of water, mucin and 7 digestive enzymes.

Question 35.
Give reason: (July 2008)
In the absence of Enterokinase, protein digestion is incomplete.
Answer:
Enterokinase activates trypsinogen to trypsin which breaks down proteins. Hence the absence of enterokinase results in indigestion of proteins.

Question 36.
Mention two characteristics of mammalian teeth.
Answer:
Heterodont and thecodont.

Question 37.
Why are proteases generally released In – inactive form? (All India 2003)
Answer:
proteases are inactive form, they would hydrolyse the cellular and extracellular proteins of the gut wall in the absence of food.

Question 38.
What are micelles? (All India 1998)
Answer:
The products of fat digestion are incorporated with the help of bile salts and phospholipids, into small spherical, water-soluble droplets called micelles.

Question 39.
Name the passage that leads bile from the liver Into gall bladder. (Delhi 1998)
Answer:
Cystic duct.

Question 40.
Name the different parts of large intestine in humans in their natural sequence. (Delhi 1998C)
Answer:
Caecum, colon and rectum.

Question 41.
What is the number of permanent teeth in an adult human being?
Answer:
32. (Thirty-two).

Question 42.
Name the two sets of teeth a human gets In his life.
Answer:
Milk teeth (deciduous teeth) and permanent teeth.

Question 43.
Name the substance that makes up the chewing surface of teeth.
Answer:
Ename

Question 44.
What are the four different types of teeth present in humans?
Answer:
lncisdrs, canine, premolars, and molars.

Question 45.
How many (I) molar and (II) premolar teeth are there in an adult human?
Answer:

• 12 molars
• 8 premolars.

Question 46.
Where are the taste buds located?
Answer:
Taste buds are located in the papillae on the upper surface of the tongue.

Question 47.
How is the tongue attached to the floor of the buccal cavity?
Answer:
The tongue is attached to the floor of the buccal cavity by the frenulum.

Question 48.
What is the function of epiglottis?
Answer:
Epiglottis prevents the entry of food into the trachea during swallowing.

Question 49.
Name the muscular structure that regulates the movement of food from the esophagus into the stomach.
Answer:
Gastro-Oesophageal sphincter.

Question 50.
Where is the stomach located?
Answer:
Stomach is located in the upper part of the abdominal cavity on the left side, just below the diaphragm.

Question 51.
Name the three regions of the stomach in proper sequence.
Answer:
Cardiac, fundic and pyloric regions.

Question 52.
Which part of the stomach does the oesophagus enter into?
Answer:
Cardiac region.

Question 53.
Which part of the stomach continues into the duodenum?
Answer:
Pyloric region.

Question 54.
Name the three regions of the human small intestine in the proper sequence.
Answer:
Duodenum, jejunum and ileum.

Question 55.
Name the muscular structure that guards the opening of the stomach Into the duodenum.
Answer:
Mr.-Pyloric sphincter.

Question 56.
Name the part of the alimentary canal where symbiotic microbes live.
Answer:
Caecum.

Question 57.
Name the three parts of the colon.
Answer:
Ascending, transverse and descending part.

Question 58.
Name any two structural features of the small intestine, which helps in absorption.
Answer:

• Presence of villi
• Projections called microvilli giving brush border appearance.

Question 59.
What is the name of the major lymph vessel present in the intestinal villi?
Answer:
Lacteal.

Question 60.
Where are the crypts of leiberkuhn located?
Answer:
They are located in between the bases of the villi in the intestine.

Question 61.
Name any two major glands associated with the human alimentary canal.
Answer:
Pancreas, liver, and the salivary glands.

Question 62.
Name the largest gland in our body.
Answer:
Liver.

Question 63.
Name the structural and functional unit of liver.
Answer:
Hepatic lobules.

Question 64.
Name the connective tissue sheath of liver lobules.
Answer:
Glisson’s capsule.

Question 65.
What are the constituents of the common bile duct?
Answer:
Hepatic duct and cystic duct.

Question 66.
How does bile reach the gall bladder?
Answer:
The bile secreted by the hepatic cells passes through the hepatic ducts and enters the gall bladder through the cystic duct.

Question 67.
What is sphincter of Oddi?
Answer:
Sphincter of Oddi is the muscular structure, that guards the opening of the hepato-pancreatic duct into the duodenum.

Question 68.
Name one gland in human body, that secretes both digestive enzymes as well as hormones.
Answer:
Pancreas.

Question 69.
What are the two major functions of buccal cavity?
Answer:
Buccal cavity helps in the mastication of food and facilitates in swallowing.

Question 70.
What is a bolus?
Answer:
When thoroughly masticated food mixes with saliva, the food particles stick together with the help of mucus into what is known as bolus.

Question 71.
Define deglutition.
Answer:
Deglutition is the process of swallowing in which bolus is conveyed into the pharynx and then into the oesophagus.

Question 72.
In humans, starch digestion begins in the buccal cavity, but stops in the stomach. Why?
Answer:
Hydrochloric acid in the gastric juice (acidic pH) inactivates salivary amylase. Hence starch digestion stops in the stomach.

Question 73.
What is the role of the intrinsic factor?
Answer:
lntrinsic factor helps in the absorption of vitamin B₁₂ from the intestine.

Question 74.
Name two enzymes which take part in the digestion of proteins in our body.
Answer:
Pepsin, trypsin, chymotrypsin and carboxypeptidase.

Question 75.
What type of medium (pH) is required for the activity of trypsin?
Answer:
Trypsin requires an alkaline pH.

Question 76.
What is the role of intrinsic factor?
Answer:
Intrinsic factor helps in absorption of vitamin B12from the intestine.

Question 77.
Name two enzymes which take part in the digestion of proteins in our body.
Answer:
Pepsin, Trypsin, Chymotrypsin and carboxypeptidase.

Question 78.
What type of medium (pH) is required for the activity of trypsin?
Answer:
Trypsin requires an alkaline pH.

Question 79.
What provides the alkaline medium for the action of trypsin in small Intestine?
Answer:
Alkaline medium is provided by bile from the liver and bicarbonates from the pancreas and Brunner’s glands.

Question 80.
What are peptones?
Answer:
Peptones are partially hydrolysed proteins.

Question 81.
Which is the food constituent which bile helps to digest and absorb?
Answer:
Fats.

Question 82.
Name the bile pigments?
Answer:
Bilirubin and biliverdin.

Question 83.
Name the end products of digestion of fats.
Answer:
Glycerol, fatty acids, and monoglycerides.

Question 84.
Define absorption.
Answer:
Absorption is the process by which the end products of digestion pass through the intestinal mucosa into the blood or lymph.

Question 85.
Name the form in which fats enter lymph.
Answer:
Chylomicrons.

Question 86.
What are chylomicrons?
Answer:
Chylomicrons are the protein-coated, water-soluble globules of the newly synthesized fats that is combined with phospholipids and liberated into the lymph for circulation.

Question 87.
What is defaecation?
Answer:
The egestion of faeces through the anal opening is called defaecation.

Question 88.
Name the most common disorder of the alimentary system.
Answer:
Inflammation of the intestinal tract.

Question 89.
Which of the body is affected by jaundice?
Answer:
Liver.

Question 90.
What is Constipation?
Answer:
Constipation is the condition where the faeces are retained within the rectum, as the bowel movement occurs irregularly.

Question 91.
What is facilitated transport?
Answer:
Some of the substances like fructose and few amino acids are absorbed by the blood with the help of carrier ions like Na+. This mechanism is called facilitated transport.

1st PUC Biology Digestion and Absorption Two Marks Questions

Question 1.
If the pancreatic duct of a person is blocked, how would it affect the digestion of fat in the duodenum?
Answer:
If the pancreatic duct is blocked, the pancreatic juice cannot reach the duodenum. As a result, the enzymes like trypsin, chymotrypsin, carboxypeptidase, aminopeptidase (help in the digestion of protein); amylase (help in carbohydrate digestion), lipase (help in fat digestion) could not reach the duodenum. Digestion of carbohydrates, proteins, and fats will remain incomplete.

Question 2.
Mention any four proteolytic enzymes Involved in human digestion. (April 90)
Answer:

• Pepsin
• Trypsin
• Chymotrypsin
• Carboxypeptidase.

Question 3.
Name the watery fluid secreted from Brunner’s gland in the duodenum. Mention its any two characteristics. What role does it play inside the duodenum?
Answer:
The watery fluid secreted by Brunner’s gland is called mucoid fluid or mucus.
Characteristics:
(i) It is viscous.
(ii) It is enzyme-free
(iii) It is alkaline in nature.
Functions:
It enables the duodenum to withstand the acidic chyme entering from the stomach.

Question 4.
Write the role of Trypsin and Chymotrypsin in protein digestion. (April 2002)
Answer:
Trypsin is secreted as trypsinogen. It is activated by the enterokinase of the intestinal juice.

Trypsin breaks down proteins into proteases, peptones & polypeptides.

polypeptides.Chymotrypsin is secreted as chymotrypsinogen. It is activated by trypsin. It too converts proteins into protein fragments, Chymotrypsinogen

Question 5.
Mention the three types of salivary glands in man and name the carbohydrate present in the saliva. (Oct. 2002)
Answer:
In man there are 3 pairs of salivary glands, they are are the parotid, Sublingual and Submaxillary glands. Saliva contains the starch splitting enzyme salivary amylase or Ptyalin.

Question 6.
Name the non-digestive enzyme found in intestinal juice. Mention its significance. (Oct. 2004)
Answer:
The non-digestive enzyme in intestinal juice is ‘E-‘nterokinase’ and helps to activate trypsinogen into trypsin which digests proteins.

Question 7.
What are microvilli? State their functions.
Answer:
Microvilli are bristle-like extensions of the free surface of epithelial cells that line the surface of villi. They increase the surface area of epithelium for the absorption of nutrients.

Question 8.
Write any two roles of HCI In human digestion. (July 2010)
Answer:
The major roles played by HCI in human digestion are;

• It activates pepsinogen to pepsin which is required for protein digestion.
• It prevents decay of food in the stomach.

Question 9.
How is our gut lining protected from its own secretion?
Answer:

• Proteases are secreted in an inactive form and pose no threat to the gut lining.
• The mucus and bicarbonates present in the gastric juice play an important role in the protection of the mucosal epithelium from excoriation by the highly concentrated Hydrochloric acid.

Question 10.
What would happen if hydrochloric acid is not secreted in our stomach?
Answer:
If hydrochloric acid is not secreted in the stomach, the following will happen:

• Pepsinogen will not be converted into pepsin.
• An acidic medium needed for the action of proteases will not be created.
• Salivary amylase may continue to function.

Question 11.
Amylase is secreted by two different glands. Name them what is its action on food? (Foreign 2002)
Answer:
Amylase is secreted by the salivary glands into the buccal cavity and also by the pancreas.
Amylase acts on starch and breaks it into two molecules of glucose.

Question 12.
What is the site of fat digestion In humans? Name the enzymes responsible for it and mention its end products.
Answer:
Fat is mainly digested in the small intestine and very little in the stomach.
Lipase is the enzyme responsible for fat digestion. The final products of fat digestion in humans are glycerol, fatty acids and some monoglycerides.

Question 13.
Enumerate the contents of saliva.
Answer:
The saliva secreted into the oral cavity contains electrolytes like Na⁺, K⁺, Cl^–, \({ HCO }_{ 3 }^{ – }\) and enzymes like salivary amylase and lysozyme. Salivary amylase helps in digesting starch and lysozyme acts as an antibacterial agent that prevents infections.

Question 14.
What is rennin? What is its function?
Answer:

• Rennin is the proteolytic enzyme found in the gastric juice of infants.
• It hydrolyses milk proteins and helps in its digestion.

Question 15.
How would it affect the digestion of proteins and carbohydrates, if there is a blockage in the pancreatic duct?
Answer:
If the pancreatic duct is blocked:

• Digestion of proteins will be impaired, as pancreatic juice contains the proteolytic enzymes trypsin, chymotrypsin and carboxypeptidase.
• Digestion of carbohydrates will stop as there will be no enzyme amylase.
• Since the bicarbonates provide an alkaline pH for the action of enzymes in the duodenum, the whole process of digestion in the intestine will be affected.

Question 16.
Give two protein-digesting enzymes of pancreatic Juice with their action.
Answer:
Pancreatic juice contains enzymes like trypsin, chymotrypsin and carboxyptidase which help in digestion of proteins.

• Trypsin acts on the proteins, proteoses and peptones and convert them into small peptides/ dipeptides
• Carboxypeptidase release the last amino acid from the peptide chain and shorten the peptide chain.

Question 17.
What are the basic layers of the wall of the alimentary canal?
Answer:
The wall of the alimentary canal consists of four main concentric layers. Beginning from outside, these layers are

• visceral peritoneum
• muscular layer
• sub-mucosa
• mucosa.

Question 18.
Mention the Important functions of the large intestine.
Answer:

• It helps in the absorption of water, minerals, and certain drugs.
• Secretion of mucus helps in adhering the waste particles together and lubricating it for an easy passage.

Question 19.
Where is the ileocaecal valve present? What are its functions?
Answer:

• The ileo-caecal valve is present at the junction of ileum of small intestine and the caecum of the large intestine.
• It prevents the backflow of the matter from the caecum into the ileum.

Question 20.
What Is diarrhea? What is its consequence In nutrition?
Answer:
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhea. It reduces the absorption of food.

Question 21.
Briefly explain the structure and regions of the stomach.
Answer:
Stomach is a ‘J’ shaped bag-like structure. A muscular sphincter called ‘gastro-oesophageal sphincter regulates the opening of oesophagus into the stomach. The stomach is located in the upper-left portion of the abdominal cavity and has three major parts a cadiac portion into which the oesophagus opens, a fun dic region and a pyloric region which opens into the small intestine.

Question 22.
How Is saliva produced?
Answer:
Saliva is mainly produced by three pairs of salivary glands, the parotids (cheek), the sub-maxillary/sub-mandibular and the sub linguals. These glands are situated just outside the buccal cavity and secrete salivary juice into the buccal cavity.

Question 23.
Where is the pancreas situated? Name the secretions of the pancreas.
Answer:
The pancreas is a compound (both exocrine and endocrine) elongated organ situated between the limbs of the U shaped duodenum. The exocrine portion of the pancreas secretes an alkaline pancreatic juice containing enzymes liketrypsinogen, chymotrypsinogen, procarboxy peptidases, amylases, lipases and nucleases. The endocrine portion secretes hormones, insulin and glucagon.

Question 24.
What constitutes the juice? What does it .contain?
Answer:
The secretions of the brush border cells of the mucosa along with the secretions of the gobletcells constitute the intestinal juice or succus entericus. This juice contains a variety of enzymes like disaccharidases (e.g. maltase) dipeptidases, lipases and nucleosidase.

Question 25.
How are fatty acids and glycerol absorbed into the blood?
Answer:
Fatty acids and glycerol being insoluble, cannot be absorbed into the blood. They are first incorporated into small droplets called micelles which move into the intestinal mucosa. They are re-formed into very small protein-coated fat globules called the chylomicrons which are transported into the lymph vessels in the villi. These lymph vessels ultimately release the absorbed substances into the bloodstream.

Question 26.
Draw a labelled diagram of anatomical regions of the human stomach.
Answer:

1st PUC Biology Digestion and Absorption Three Marks Questions

Question 1.
Draw a neat labelled diagram representing the duct system of liver, gall bladder and pancreas.
Answer

Question 2.
Name any three enzymes secreted by the pancreas Specify the substrate and product of each.
Answer:

• Amylase: It acts on starch and converts it into maltose.
• Trypsin: It acts on proteins, proteases and peptones and converts them into shorter peptides or dipeptides.
• Lipase: It acts on triglycerides and converts them into triglycerides and monoglycerides along with the release of fatty acids.

Question 3.
Draw a neat labelled diagram representing transverse section of gut.
Answer:

Question 4.
What is protein-energy malnutrition? How it is responsible to cause marasmus and kwashiorkor in infants and children?
Answer:
Protein-energy malnutrition (PEM) may affect large sections of the population during drought and famine.
This happened in Ethiopia during the severe drought in the mid-eighties. PEM affects infants and children to produce Marasmus and Kwashiorkor diseases.

• Marasmus is produced by a simultaneous deficiency of proteins and calories. It is found in infants less than a year in age if mother’s milk is replaced too early by other foods which are poor in both proteins and caloric value.
• This often happens if the mother has second pregnancy or childbirth when the older infant is still too young.
• In Marasmus, protein deficiency impairs growth and replacement of tissue proteins; extreme emaciation of the body and thinning of limbs results, the skin becomes dry, thin, and wrinkled.
• Growth rate and bodyweight decline considerably. Even growth and development of the brain and mental abilities are impaired.
• Kwashiorkor is proceeded by protein deficiency unaccompanied by calorie deficiency.
• It resubs from the replacement of mother’s milk by a high calorie-low protein diet in a child more than one year in age. Like marasmus, kwashiorkor shows wasting of muscles, thinning of limbs, failure of growth, and brain development.
• But unlike, marasmus, some fat is still left under the skin; moreover, extensive oedema and swelling of body parts are seen.

Question 5.
Draw a neat labelled diagram of a section of small intestinate mucosa showing villi.
Answer:

Question 6.
Which layer of the stomach contains gastric glands? Name the three types of cells present in gastric glands with their functions.
Answer:
The mucosa of the stomach has gastric glands.
Three major types of cells in gastric glands are:

• Mucus neck cells which secrete mucus
• Peptic or chief cells secrete the proenzyme pepsinogen.
• parietal or oxyntic cells which secrete HCL and intrinsic factors essential for the absorption of vitamin B_12.

Question 7.
How are the nucleic acid fraction of our food digested?
Answer:
Nucleases present in the pancreatic juice convert the nucleic acids to nucleotides.

• Nucleotidases present in the pancreatic juice convert the nucleotides into nucleosides and phosphates.
• Nucleosides present in the intestinal juice convert the nucleosides into nitrogen bases and sugar.

Question 8.
Write a note on hormonal control and co-ordination of digestive parts.
Answer:
The activities of the gastrointestinal tract are under neural and hormonal control for proper co-ordination of different parts. The sight, smell, or the presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also, similarly, stimulated by neural signals. The muscular activities of different parts of the alimentary canal can also be moderated by neural mechanisms, both local and through CNS. Hormonal control of the secretion of digestive juices is carried out by the local hormones produced by the gastric and intestinal mucosa.

Question 9.
Write a note on absorption in different parts of the digestive system
Answer:
Mouth: Certain drugs coming in contact with the mucosa of the mouth and lower side of the tongue are absorbed into the blood capillaries lining them.

Stomach: Absorption of water, simple sugars, and alcohol, etc. takes place in the stomach. Small intestine: Principle organ for the absorption of nutrients. The digestion is completed here and the final products of digestion such as glucose, fructose, fatty acids, glycerol, and amino acids are absorbed through the mucosa into all bloodstream and lymph.

Large Intestine: Absorption of water, some minerals, and drugs take place in the large intestine.

1st PUC Biology Digestion and Absorption Five Marks Questions

Question 1.
Mention the components of human saliva. Give the function. (M.Q.P.)
Answer:
The salivary glands produce a secretion called salivary juice or saliva. It is composed
Nucleosidases of water; mucin; ions like sodium, chlorides, HCO₃ etc; a bactericidal enzyme and a digestive enzyme called ptyalin or salivary amylase.
Functions of Saliva

• Moistens dry food and helps in swallowing
• Keeps mouth and teeth clean.
• Helps in tasting food
• The Salivary amylase helps in digesting the polysaccharides, starch & glycogen to maltose, a disaccharide.
• The bactericidal enzyme kills microbes and checks their load in food.
• Many substances like Iodides, lead etc are excreted through saliva.

Question 2.
Describe the digestion of carbohydrates in the small intestine. (Apr. 83, Oct. 04)
Answer:
The undigested food in terms of polysaccharides, starch and glycogen and disaccharides, sucrose & lactose (got from diet), maltose (produced from digestive action of salivary amylase on polysaccharides or those got directly as a food source) are acted upon by the carbohydrates of the pancreas and intestinal juice in the lumen of the intestine.

(i) The polysaccharides starch (plant source) and glycogen (animal source) are acted upon by the pancreatic amylase (or amylopsin) splitting them into a pool of maltose, a disaccharide.

(ii) The final pool of disaccharides produced by the action of polysaccharides (Pancreatic and salivary amylases) or those obtained directly through diet are acted upon by the di- saccharases Sucrase, Lactase, and Maltase

(disaccharide splitting enzymes) of the intestinal juice in the intestinal lumen as follows, (a) Sucrose (s) are acted upon by sucrase enzyme and split into a molecule of glucose and a molecule of Fructose each.

(b) Maltose (s) are acted upon by Maltase -enzyme and split into two glucose molecules each.

(c) Lactose (s) are acted upon by Lactase enzyme and split into a molecule of glucose and galactose

Finally resulting in an absorbable pool of monosaccharides in the intestinal lumen.

Question 3.
Explain protein digestion In small Intestine? (M.Q.P., Apr. 83, Oct. 99, Mar. 2010, July 2011)
Answer:
The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralised by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotryr.in and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carbxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and amino peptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 4.
Describe the digestive process in the stomach and small intestine of man. (April 84)
Answer:
Digestion in the stomach: The chief organic molecules in food that are digested in the stomach are the proteins (native) and short chain simple lipids. The cells of the mucosal wall produce the gastric juice and it consists of the following enzymes apart from HCI and mucin.

• Pepsin- proteolytic enzyme(endopeptidase)
• Renin – milk curdling enzyme (Calf and infants of humans)
• Gastric lipase – acts on simple lipids. Pepsin acts on native protein molecules of food in the lumen of the stomach converting them into proteoses, peptones, and polypeptides.

In Infants & calf, where the chief source of food is milk, renin is an additional enzyme which helps in the curdling of milk protein casein, and later this curdled casein is digested by the action of pepsin and converted to proteoses, peptones and polypeptides.

proteoses, peptones, and polypeptides. Digestion in Small Intestine: Two types of juices are secreted into the intestinal lumen namely the pancreatic juice and intestinal juice. Both their juices possess enzymes digesting all types of complex organic molecules namely carbohydrates, proteins, lipids & nucleic acids.
The digestion by this enzyme in the small intestine proceeds as follows:

Carbohydrates: Polysaccharides starch (plant source) and glycogen (animal source) are acted upon by pancreatic amylase (amylopsin) in the small intestinal lumen and converted to a pool of maltose, a disaccharide.

The disaccharides, in the intestinal lumen namely maltose (obtained from the digestion of polysaccharides or direct source); sucrose, and Lactose (from diet) are acted upon disaccharide digesting (disaccharidases) enzyme into a pool of monosaccharides, glucose, fructose, and galactose.


Proteins: The proteins reaching the small intestinal lumen are in the form of native proteins, proteoses, peptones, and polypeptides. They are acted upon by the following enzymes.

Thus resulting in a final pool of absorbable amino acids.

Lipids: The Lipids are first emulsified by bile salts in the intestinal lumen. The emulsified lipids are then worked upon by the pancreatic lipase to a pool of glycerol, fatty acids and monoglycerides.

Fattyacids, monoglycerides

Nucleic acids: Nucleases from the pancreas” acts upon the RNA & DNA nucleic acids breaking them into nucleotides and these nucleotides are later acted upon by nucleotidases & nucleosidases of the intestinal juice.


deoxyribose sugar; ribose sugar; nucleic acid bases.

Question 5.
Are mention two enzymes present in pancreatic juice? Explain their role In intestinal digestion. (April 86)
Answer:
The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralized by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotrypsin and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carbxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and aminopeptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 6.
Explain the digestion of food in the small intestine by the enzymes of intestinal glands. (Oct. 87)
Answer:
The secretion produced by the intestinal glands is known as the intestinal juice or succus entries. It is a clear yellow fluid with a pH of around 7.6 and contains water, mucous and some digestive enzymes produced by intestinal cells. The enzymes found in the intestinal juice and their action is as follows on food in the small intestine.

(1) Carbohydrases: Maltase, Lactase, Sucrase these are disaccharides digesting disaccharides Maltose, Lactose, and sucrose in the intestinal lumen as follows at pH 7.1 – 8.2.

(2) Proteases: Aminopeptidase, tripeptidase, and dipeptidase-(Exopeptdases) are proteolytic enzymes digesting the intermediate polypeptides produced by the action of endopeptidases of the stomach & pancreas. Their action is as follows in the small intestinal at pH 7.1 – 8.2


(Enterokinase, an activator enzyme, activates gastric glands to produce gastric juice)

(3) Lipase: A fat-digesting enzyme. It is not as strong as pancreatic lipase but it completes the digestion of fat in the small intestinal lumen.

(4) Nucleases: Act on nucleoside molecules of nucleoproteins, splitting them into corresponding nitrogenous bases and sugar molecules.

Question 7.
Explain the role of the pancreas in the process of digestion. (Oct. 87, 88)
Answer:
The pancreas is a dual gland composed of two portions, an exocrine portion having cells producing enzymes required for digestion of food and an Endocrine portion having groups of cells producing hormones required for metabolism and homeostasis.

As a digestive gland, it plays a very important role in the process of digestion by possessing the most potent or powerful enzymes required for digestion of all types of food molecules namely carbohydrates, proteins, fats, and nucleic acids.

The pancreatic juice secreted by the pancreatic exocrine glands reaches the intestine through the pancreatic duct (duct of wiring) which usually joins the bile duct before opening into the first part of the duodenum. It is a clear, colourless liquid, containing some salts and sodium bicarbonate apart from enzymes.

As soon as food enters the duodenum, a series of neurogenic and hormonal mechanism’s come into play and maintain a steady secretion of these fluids into the duodenum. These secretions mix with the acid chyme released from the stomach’ in the duodenum. The salts and bicarbonates neutralize and alkalize the acid chyme whereas the enzymes work upon the complex food molecules converting them into smaller molecules which are later on acted upon by intestinal juice enzymes converting them into absorbable forms.

The pancreatic digestive action can be summarized as follows :

(1) Carbohydrases (carbohydrate splitting): Pancreatic amylase acts on polysaccharides starch (plant origin) and glycogen (animal origin) converting these into disaccharide molecules, namely maltose.

(2) Proteases (Protein-splitting):

• Trypsin, Chymotrypsin (Endopeptidases) act on native proteins splitting them into proteases, peptones, and polypeptide molecules.
• Carboxypeptidase (exopeptidase) acts on proteins, peptones, and polypeptides splitting them into tripeptides, dipeptides, and amino acids.

(3) Pancreatic lipase: act on fats emulsified by brie salts splitting them into glycerol, fatty acids, and monoglycerides.

(4) Nucleases: Ribonuclease and deoxyribonuclease – splitting nucleic acids (RNA & DNA) into simple nucleotides of respective types. These are later acted upon by nucleotides, splitting them into nucleosides of RNA & DNA and PO

Question 8.
Explain how carbohydrates and protein are digested by intestinal juice. (Oct. 90)
Answer:
The undigested food in terms of polysaccharides, starch and glycogen and disaccharides, sucrose & lactose (got from diet), maltose (produced from digestive action of salivary amylase on polysaccharides or those got directly as a food source) are acted upon by the carbohydrates of the pancreas and intestinal juice in the lumen of the intestine.

(i) The polysaccharides starch (plant source) and glycogen (animal source) are acted upon by the pancreatic amylase (or amylopsin) splitting them into a pool of maltose, a disaccharide.

(ii) The final pool of disaccharides produced by the action of polysaccharides (Pancreatic and salivary amylases) or those obtained directly through diet are acted upon by the di- saccharases Sucrase, Lactase and Maltase

(disaccharide splitting enzymes) of the intestinal juice in the intestinal lumen as follows, (a) Sucrose (s) are acted upon by sucrase enzyme and split into a molecule of glucose and a molecule of Fructose each.

(b) Maltose (s) are acted upon by Maltase -enzyme and split into two glucose molecules each.

(c) Lactose (s) are acted upon by Lactase enzyme and split into a molecule of glucose and galactose

Finally resulting in an absorbable pool of monosaccharides in the intestinal lumen.

The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralised by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotryrin, and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carboxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and amino peptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 9.
Explain the Digestion of protein In the human digestive system. (Oct. 92)
Answer:
Digestion in the stomach: The chief organic molecules in food that are digested in the stomach are the proteins (native) and short-chain simple lipids. The cells of the mucosal wall produce the gastric juice and it consists of the following enzymes apart from HCI and mucin.

• Pepsin- proteolytic enzyme(endopeptidase)
• Renin – milk curdling enzyme (Calf and infants of humans)
• Gastric lipase – acts on simple lipids. Pepsin acts on native protein molecules of food in the lumen of the stomach converting it into proteoses, peptones and polypeptides.

In Infants & calf, where the chief source of food is milk, renin is an additional enzyme which helps in the curdling of milk protein casein and later this curdled casein is digested by the action of pepsin and converted to proteoses, peptones and polypeptides.

proteoses, peptones and polypeptides. Digestion in Small Intestine: Two types of juices are secreted into the intestinal lumen namely the pancreatic juice and intestinal juice. Both their juices possess enzymes digesting all types of complex organic molecules namely carbohydrates, proteins, lipids & nucleic acids.
The digestion by this enzyme in the small intestine proceeds as follows:

Carbohydrates: Polysaccharides starch (plant source) and glycogen (animal source) are acted upon by pancreatic amylase (amylopsin) in the small intestinal lumen and converted to a pool of maltose, a disaccharide.

The disaccharides, in the intestinal lumen namely maltose (obtained from the digestion of polysaccharides or direct source); sucrose and Lactose (from diet) are acted upon disaccharide digesting (disaccharidases) enzyme into a pool of monosaccharides, glucose, fructose and galactose.


Proteins: The proteins reaching the small intestinal lumen are in the form of native proteins, proteoses, peptones and polypeptides. They are acted upon by the following enzymes.

Thus resulting in a final pool of absorbable amino acids.

Lipids: The Lipids are first emulsified by bile salts in the intestinal lumen. The emulsified lipids are then worked upon by the pancreatic lipase to a pool of glycerol, fatty acids, and monoglycerides.

Fatty acids, monoglycerides

Nucleic acids: Nucleases from the pancreas” acts upon the RNA & DNA nucleic acids breaking them into nucleotides and these nucleotides are later acted upon by nucleotidases & nucleosidases of the intestinal juice.


deoxyribose sugar; ribose sugar; nucleic acid bases.

(ii) The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralised by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotryr.in and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carbxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and aminopeptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 10.
What is digestion? Explain the physiology of digestion of carbohydrates and protein In the small intestine. (Oct. 94, March 2011)
Answer:
Digestion is a process by which complex organic food molecules are broken down or hydrolysed into simple (or small) absorbable forms by the action of digestive enzymes.

The undigested food in terms of polysaccharides, starch and glycogen and disaccharides, sucrose & lactose (got from diet), maltose (produced from digestive action of salivary amylase on polysaccharides or those got directly as a food source) are acted upon by the carbohydrates of the pancreas and intestinal juice in the lumen of the intestine.

(i) The polysaccharides starch (plant source) and glycogen (animal source) are acted upon by the pancreatic amylase (or amylopsin) splitting them into a pool of maltose, a disaccharide.

(ii) The final pool of disaccharides produced by the action of polysaccharides (Pancreatic and salivary amylases) or those obtained directly through diet are acted upon by the di- saccharases Sucrase, Lactase and Maltase

(disaccharide splitting enzymes) of the intestinal juice in the intestinal lumen as follows, (a) Sucrose (s) are acted upon by sucrase enzyme and split into a molecule of glucose and a molecule of Fructose each.

(b) Maltose (s) are acted upon by Maltase -enzyme and split into two glucose molecules each.

(c) Lactose (s) are acted upon by the Lactase enzyme and split into a molecule of glucose and galactose

Finally resulting in an absorbable pool of monosaccharides in the intestinal lumen.

The acid chyme released from the stomach into the intestinal lumen may contain digested protein (by pepsin) or native protein (untouched by pepsin if any) apart from carbohydrates and fats. The acid chyme is first neutralised by bicarbonates of pancreatic juice and then acted upon by various enzymes. The proteins in the chyme are acted upon by the proteolytic enzymes of the

• Pancreatic juice – namely trypsin, chymotryr.in and carboxypeptidase.
• ‘Intestinal juice – aminopeptidase, tripeptidase and dipeptidases.

Among these trypsin and chymotrypsin are endopeptidases (attacking internal peptide bonds) acting on native protein and the carboxypeptidases, aminopeptidases, tri and dipeptidases are exopeptidases (attacking superficial peptide bonds at the carbxyl end or amino end respectively) acting on protein fragments produced from whole or native proteins.

The action of the above enzymes on proteins is as follows;

(i) Trypsin and Chymotrypsin act on native proteins splitting or reducing them into shorter protein chains namely proteoses, peptones and polypeptides.

(ii) Carboxypeptidases and amino peptidases act on the protein molecules got from the digestion of native proteins converting them into tripeptides, dipeptides and amino acids.

(iii) The tri and dipeptides resulted from the above reactions are acted upon by tri and dipeptidases and split into amino acids.

Finally, resulting in a pool of amino acids which can be easily absorbed.

Question 11.
Give an account of the digestion of proteins in man. (Oct. 2001)
Answer:
Protein digestion in man:
(a) Protein digestion in the mouth: Proteins do not undergo any change as there is no protease in the mouth.
(b) Protein digestion in the stomach: Gastric juice has two proteases pepsin and rennin.

(i) Pepsin is first secreted as pepsinogen: It is converted to pepsin by HCI. One HCI has converted some pepsinogen, further conversion is autocatalytic. Pepsin is an endopeptidase. It acts on internally situated peptide bonds of proteins. In the end, the proteins are converted to proteoses, peptones and polypeptides.

(ii) Rennin (also called chymosin): It is a milk curdling enzyme that is mostly present in infants. It is secreted as prorennin. Renin acts on the soluble milk protein, casein, converting it into paracasein. In the presence of Ca²⁺ of milk, it is further converted into calcium para caseinate, which coagulates preventing the rapid passage of milk into the duodenum. It is then hydrolyzed by pepsin to proteoses, peptones & polypeptides

As digestion proceeds in the stomach, the food becomes more or less liquified and is referred to as chyme.

(c) Protein digestion in the Intestine: Here it comes under the action of
(i) Pancreatic proteases &
(ii) Intestinal proteases.

(i) Pancreatic proteases: These are Trypsin, chymotrypsin, and carboxypeptidase. The first two are endopeptidases and the third is an exopeptidase. Trypsin is secreted as trypsinogen & it is activated by enterokinase (enteropeptidase) of the intestinal juice.

chymotrypsinogen which is activated by trypsin. It, too, converts proteins into protein fragments.

Carboxypeptidase an endopeptidase, hydrolyzes terminally situated bonds. The products are tripeptides, Depeptides, and amino acids.

(ii) Intestinal proteases: They may be termed peptidases because they act on the protein fragments. Proteoses, peptones, and polypeptides aminopeptidase tripeptides, dipeptides and

The proteins have been completed digested and the products are free amino acids. They are now ready to be absorbed.

Question 12.
Draw a neat labelled diagram of the human alimentary canal. (M.Q.P.)
OR
Draw a neat labelled diagram of digestive
Answer:

Question 13.
Write the functions of intestinal juice.
Answer:

• Intestinal mucus protects the intestinal epithelia.
• Disaccharidases like maltase, lactase, and sucrase bring complete hydrolysis of disaccharides into monosaccharides (see carbohydrate digestion).
• Aminopeptidases hydrolyse peptides of different lengths into amino acids.
• Dipeptidase hydrolyse dipeptides into amino acids, (see protein digestion).
• Nucleotidase and Nucleosidase hydrolyse Nucleotides and convert them into purines, pyrimidine, and inorganic phosphate.

Question 14.
Describe the process of Lipid digestion in small intestine of man. (Oct. 2004)
Answer:
Fats or lipids mainly contain triglycerides i.e. consists of 3 fatty acid molecules attached to a molecule of glycerol. The major fat-digesting enzyme is pancreatic lipase.

The fats are first emulsified i.e. the bile salts combine with fats and break them down into small droplets which form a soapy mixture. The small droplets are acted by lipase which breaks them into fatty acids, glycerol, mono and diglycerides.

The bile salts then combine with the products and form water-soluble complexes which are absorbed.

Question 15.
Briefly explain the disorders of the digestive system.
Answer:
The disorder of the digestive system are:-

• The inflammation of the intestinal tract is the most common ailment due to bacterial or viral infections. The infections are also caused by the parasites of the intestine like tapeworm, roundworm, threadworm, hookworm, pinworm etc.
• Jaundice:- The liver is affected, skin and eyes turn yellow due to the deposit of bile pigments.
• Vomiting: It is the ejection of stomach contents through the mouth. This reflex action is controlled by the vomiting centre in the medulla. A feeling of nausea precedes vomiting.
• Diarrhoea: The abnormal frequency of bowel movement and increased liquidity of the faecet discharge is known as diarrhoea. It reduces the absorption of food.
• Constipation: In constipation, the faeces are retained with the rectum as the bowel movements occur irregularly.
• Indigestion: In this condition, the food is not properly digested leading to a feeling of fullness. The causes of indigestion are inadequate enzyme secretion, anxiety food poisoning, overeating, and spicy food.

Question 16.
Fill in the blanks.

(a) The type of dentition in humans is called ……………..
(b) ……………..has …………….. cells which secrete mucus that help In lubrication.
(c) Each lobule of liver is covered by a thin connective tissue sheath called ……………..
(d) Hepate pancreatic duct is guarded by ……………..
(e) Alcohol is absorbed in ……………..
Answer:
(a) Diphyodont
(b) Mucosal epithelium, goblet cells
(c) Glisson’s capsule
(d) Sphincter of Oddi
(e) Stomach.

Question 17.
Match the following:
(a) Trypslnogen – (i) Goblet cells
(b) Saliva – (ii) Oxyntic cells
(c) HCL- (iii) Pancreas
(d) Pepsinogen – (iv) Parotids
(e) Mucus – (v) Chief cells
Answer:
(a) → (iii)
(b) → (iv)
(c) → (ii)
(d)→ (v)
(e) → (i)

You can Download Chapter 2 Structure of Atom Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 2 Structure of Atom

1st PUC Chemistry Structure of Atom One Mark Questions and Answers

Question 1.
Mention the constituents of atom.
Answer:
Electron, Proton & Neutron.

Question 2.
Who discovered electron.
Answer:
J.J. Thomson

Question 3.
What is the mass of electron?
Answer:
9.108 × 10³¹ kg (0.0005487amu).

Question 4.
Mention the charge of electron in colomb.
Answer:
1.61.6*10¹⁹C.

Question 5.
What is the charge of proton ?
Answer:
Positive charge.

Question 6.
What is the mass of proton ?
Answer:
1.672 x 10²⁷ kg(1.00728amu)

Question 7.
Who discovered proton ?
Answer:
E. Goldstein.

Question 8.
Who invented charge on electron ?
Answer:
Robert Millikan.

Question 9.
Name the experiment by which charge on electron determined.
Answer:
Robert Millikan oil drop experiment.

Question 10.
Who discovered neutron ?
Answer:
James Chadwick.

Question 11.
What is atomic number ?
Answer:
Atomic number *Z’ is the number of protons present in the nucleus of an atom of an element or the number of electrons present in an atom of an element.

Question 12.
What is mass number ?
Answer:
Atomic mass ‘A’. The total number of protons and neutrons present in the nucleus of an atom of an element.

Question 13.
Write the relationship between mass number and atomic number.
Answer:
Number of neutrons = Mass number (A) – Atomic number (Z) = A – Z

Question 14.
How do you represent an atom symbolically with atomic number and mass number?
Answer:
_ZX^A where A = Mass Number, X = atom, Z = Atomic number.

Question 15.
What is the number of proton and neutron in ₉₂X²³⁵ ?
Answer:

Question 16.
Give the number of Protons, Electrons and Neutrons present in the atom – having atomic number 27 and mass number 56.
Answer:
Number of Protons 27, Number of Electrons 27, Number of Neturons 29.

Question 17.
Mention the proton, neutron and electons ₁₇C³⁵.
Answer:
Neutron = 35 – 17 = 18, Electron = Proton = 17 .

Question 18.
Name the species which has no electron.
Answer:
Proton

Question 19.
Name the atom which has no neutron.
Answer:
Hydrogen

Question 20.
What is the ratio between mass of proton and electron ?
Answer:
Mass of proton is 1837 times the mass of electron.

Question 21.
Name the particles which constitute cathode rays.
Answer:
Electrons

Question 22.
Who demonstrated the particle property of an electron ?
Answer:
J. J. Thomson

Question 23.
Who showed the wave property of electrons ?
Answer:
G.P. Thomson

Question 24.
What is the charge on neutrons ?
Answer:
Nil / Zero

Question 25.
Mention the mass of neutron.
Answer:
1.675 × 10²⁷ kg(1.00866 amu)

Question 26.
Name the scientist who proposed the nuclear theory (Solar model of atom) theory and discovered the existence of nucleus of an atom ?
Answer:
Ernest Rutherford.

Question 27.
What is photon ?
Answer:
Electromagnetic radiations are emitted or absorbed or propagated in the form of small pockets of energy called photon.

Question 28.
What is Isotopes ?
Answer:
Atoms of the same element having the same atomic number but different mass numbers are called isotopes.

Question 29.
What is Isobars ?
Answer:
Atoms of different elements having the same mass number but different atomic numbers are called isobars.

Question 30.
What is Isotones ?
Answer:
Atoms of different elements which contain the same number of neutrons, but different mass numbers are known as isotones.

Question 31.
Define the term ‘radiation’
Answer:
Radiation is the emission and transmission of energy through space in the form of waves.

Question 32.
How are velocity, frequency and wavelength of light radiation related ?
Answer:

Question 33.
State Pauli’s exclusion principles.
Answer:
No two electrons in an atom can have the same set of four quantum numbers.

Question 34.
An atom having mass number 40 has 20 neutrons in its nucleus. What is the atomic number of the element ?
Answer:
Mass Number = Number of protons + Number of neutrons
40 = Number of protons + 20
∴Atomic number = Number of protons = 20.

Question 35.
What is aufbau principle ?
Answer:
Aufbau principle is also known as the building up principle. According to this principle, electrons should eneter the subshells in the order of increasing energies.

Question 36.
What is emission spectrum ?
Answer:
The spectrum of radiation emitted by a substance that has absorbed energy is called in emission spectrum.

Question 37.
How is the magnetic moment of paramagnetic species is related to the number of unpaired electrons present in it ?
Answer:
Magnetic moment \(\mu=\sqrt{n(n+2) B M}\) where n is the number of unpaired electrons.

Question 38.
Differentiate between the terms ‘ground state’ and ‘excited state’.
Answer:
Ground state means the least energy state or the most stable state. Excited state means the higher energy state, in which the electrons are in the higher energy level (unstable state).

Question 39.
What is the expression for the energy of a photon ?
Answer:
E = hu where, h = planks constant = 6.626 × 10^-34 Js , γ = frequency or radiation.

Question 40.
Write the unit for frequency of radiation.
Answer:
Cycles per sec (sec^-1) or Hertz (Hz)

Question 41.
Name the experiment which shows that light has particle property.
Answer:
Photoelectric effect.

Question 42.
Name the experiment which shows that light has wave property.
Answer:
Refraction.

Question 43.
How is wave number and wavelength of a wave related?
Answer:

Question 44.
What is the velocity of light?
Answer:
3 × 10⁸ m/sec

Question 45.
Define wavelength.
Answer:
It is the distance between two successive crests (peaks) or trough in a wave.

Question 46.
Define Wave number.
Answer:
It is the number of waves present per meter and is equal to the reciprocal of wavelength. \(\left(\frac{1}{\lambda}\right)=\bar{v}\)

Question 47.
Define frequency of light.
Answer:
The number of waves which pass through a given point in one second is called frequency of light.

Question 48.
What type of waves does light constitute?
Answer:
Electromagnetic waves.

Question 49.
What is orbital (atom orbital) ?
Answer:
Orbital is a three dimensional region in space around the nucleus in which the probability of finding an electron is maximum.

Question 50.
How many electrons can be accommodated in an orbital?
Answer:
Two electrons.

Question 51.
Write the de Broglie’s equation.
Answer:
\(\lambda=\frac{\mathbf{h}}{\mathbf{m} \mathbf{V}}\) where λ – wavelength of wave, h = Plank’s constant, m = mass of the electron, V – velocity of the electron.

Question 52.
Write Rydberg’s formula.
Answer:

Question 53.
Write the Balmer equation.
Answer:

Question 54.
What are the four prominent lines in Balmer series of hydrogen spectrum ?
Answer:
H_α, H_β, H_γ and H_????

Question 55.
What is the value of Rydberg’s constant?
Answer:
10.97 × 10⁶ m^-1

Question 56.
Give the range of wavelengths of visible light.
Answer:
400 nm to 750 nm.

Question 57.
Give the Rydberg equation where R is Rydberg constant?
Answer:

Question 58.
Name the element whose atom contain six protons in the nucleus.
Answer:
Carbon

Question 59.
Name the different series of hydrogen spectrum.
Answer:
(i) Lyman
(ii) Balmer
(iii) Pachen
(iv) Bracket
(v) Pfund

Question 60.
Name two series of hydrogen spectra which fall in the infra – red region.
Answer:
Brackett series and P-fund series.

Question 61.
Match the following:

Answer:
a-iii, b-iv, c-ii, d-i

1st PUC Chemistry Structure of Atom Two Marks Questions and Answers

Question 1.
Write the difference between isotope & isobars.
Answer:

Question 2
Explain Planks quantum theory.
Answer:
Electromagnetic radiations are emitted, absorbed and propagated discontinuously in the form of small packet of energy called photon. A body emits (absorbs) radiations in the integral multiples of ‘photon’.
Energy associated with each photon is E = hv, h = 6.626 × 10 ^-34 Js

Question 3.
Distinguish between emission spectra & atomic spectra.
Answer:
Emission spectra : When an object is strongly heated, it starts to emit light. When the emitted light is subjected to depression, emission spectrum is obtained)
Atomic spectra : When an atom of an element is strongly heated, it starts emitting light energy in different region. This emitted light when dispersed, a large number of closely spaced lines where formed they are called atomic spectra.

Question 4.
What do you mean by electromagnetic spectra ?
Answer:
The complete range of electromagnetic waves in the increasing order of wavelength (decreasing order of frequency) is known as electromagnetic spectra.

Question 5.
Write the electromagnetic spectra in the increasing order of wave length. (Decreasing order of frequency)
Answer:
Increasing order of wave length > Cosmic rays > Gamma rays < X-rays < U.V. rays < Visible rays < IR rays < Microwave rays < Radio waves (i.e, decreasing order of frequency).

Question 6.
Deduce the de-Broglies matter wave equation.
Answer:
Dual nature of electron – de-Broglie’s matter wave equation
According to Einstein’s mass – energy relation, E = mc² … (1)
Where m and c are the mass and velocity of an electron respectively.
According to quantum theory, Energy E = hv = (he) / λ … (2)
Where h is Plank’s constant = 6.626 × 10^-34 Js, u is the frequency of moving electron.
Comparing(1)and(2)me2 = (hc)/λ >me = h/λ > = λ = h /me … (3)
Equation (3) is called de-Broglie’s equation.
For an electron its velocity is given as V, So the de-Broglie’s equation for an electron is given by,
λ = h / mv (mv = p = momentum of electron)
λ = h / p is called de-Broglie’s matter – wave equation

Question 7.
Explain the wave nature of light.
Answer:
Wave nature of light: According to Newton, light is a stream of particles which are also known as corpuscles of light. This could not account for the phenomena of interference and deflection (as Rutherford’s model) but justified reflection and refraction.

Hygen suggested in his wave theory that the fight travels in the form of waves, later James Maxwell proposed that fight and other radiations are transmitted, these waves are associated with oscillating electric and magnetic field.

Question 8.
Distinguish between particle and wave.
Answer:

Question 9.
Explain Werner Heisenberg’s uncertainty principle (qualitative)
Answer:
It is impossible to determine both the momentum (particle nature) and position (wave nature) of a moving sub atomic particle simultaneously with absolute accuracy.
Mathematically ∆x × Ap = h / 4π where ∆x = uncertainty in position :
∆p = uncertainty in momentum ; h = Plank’s constant = 6.626 × 10^-34 Js.

Question 10.
Mention the Merits of Bohr’s theory.
Answer:

1. Explains the formation of hydrogen spectrums
2. It explains the stability of atom
3. Ryd berg constant is calculated by it.

Question 11.
Write any two limitations of Bohr’s theorem.
Answer:

1. Not explained spectra of atoms containing more than one electrons.
2. Not explained fine spectra (when spectroscope of high resolving power is used, it is found that each line in the ordinary specrum spilts into number of components differing in frequency).
3. Not given explanation regarding Stark effect (spiltting of spectral line in electric field) and Zeeman Effect (splitting of spectral fine in magnetic field).

Question 12.
Write the difference between orbit & orbital.
Answer:

Question 13.
Draw the structure of p-orbitals. (Draw the shape of orbital whose Azimuthal quantum no is 1).
Answer:
p orbital has 3 orientations p_x,p_y &p_z

Question 14.
Draw the structure of d-orbital (Orbital whose Azimuthal quantum no = 2).
Answer:
d- orbital has 5 orientations.

Question 15.
What do you mean by electronic configuration ? With the sequence.
Answer:
Electronic Configuration is the distribution of electrons in the various available orbitals of an atom of an element.
The sequenced in which the electron occupy the various oribitals is as follows.
Is, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p….

Question 16.
Explain the electronic Configuration of cation Fe+.
Answer:
Example for electronic configuration of cation.

Question 17.
Explain electronic configuration of anion using N.
Answer:
Example for electronic configuration of anion.

Question 18.
State & Explain Paul’s Exclusion Principle.
Answer:
Paul’s Exculsion principle : No two electrons in the same atom can have all the four quantum numbers same.
Illustration : In Is orbital there are two electrons, the set of quantum number of one electron differ with another in spin quantum number.

Question 19.
State and explain Hunds Rule of maximum multiplicity.
Answer:
Hund’s rule or maximum multiplicity : Electron pairing does not take place until orbitals of same energy are singly occupied.

Question 20.
What are quantum number and Name them ?
Answer:
In order to define state energy and location of electron a set of four numbers are used.
These numbers are called quantum numbers.

• Principal quantum number (n)
• Azimuthal quantum number (1)
• Magnetic quantum number (m)
• Spin quantum number (s).

Question 21.
Write all quantum number values for 3s orbital electrons.
Answer:
S – orbital has two electrons.
3s; n = 3; 1 = 0; m = 0; s = + 1/2 for first electron,
n = 3; 1 = 0; m = 0; s = -1/2 for second electron.

Question 22.
An orbital can contain only two electrons. Why ?
Answer:
An orbital can have only two electrons, provided these two electrons have anti parallel spins. This arrangement is represented, as ↓↑

Question 23.
Write the atomic number at an element with outer configuration,
i) 4s1, ii) 3d3 .
Answer:
(i) Electronic configuration for outer configuration 4s¹ = 1s². 2s²,2p⁶, 3s², 3p⁶, 4s¹.
Atomic number =19
(ii) Exact configuration for outer configuration 3d3.
1s², 2s², 2p⁶, 3s²,3p⁶, 4s², 3d³ = 23 Atomic number = 23

Question 24.
Write the electronic configuration of
1. Cl^– 2. Na⁺ ion
Answer:
Electronic configuration
1) Cl^– = 1s², 2s², 2p⁶, 3s², 3p⁶
2) Na+ = 1s², 2s², 2p⁶

1st PUC Chemistry Structure of Atom Three / Four Marks Questions and Answers

Question 1.
Summarize the Bohr’s Model of an atom,
Answer:
Bohr’s Model of an atom, the postulates are

1. Electrons revolve around the nucleus of an atom in a certain definite path called Orbit or stationary state of shell.
2. The shells are having different energy levels denoted as K, L, M, N
3. As long as the electron remains in an orbit, they neither absorb nor emit energy.
4. The electron can move only in that orbit in which angular momentum is quantized, i.e., the angular momentum of the electron is an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)

Question 2.
Explain the experimental set up and different series of emission spectrum of hydrogen.
Answer:
Hydrogen Spectrum : Hydrogen spectrum is obtained by exciting electrons present in hydrogen gas using a discharge tube under low pressure and high voltage of current. The electrons present in various atoms of Hydrogen absorb energy and jumps to higher energy levels. Later they return back to the lower energy levels by emitting excess of energy in the form of photons. When these electrons are coming to the lower energy state a series of lines were formed called Hydrogen spectrum. Denoted as H_α,H_β,H_γ and H_????H_ε

• Lyman Series : This series is formed when the electron jumps from 2nd, 3rd, 4th, …. Higher energy level to first (K shell) level by emitting excess of energy in UV region.
• Baimer Series ; This series is formed when the electron jumps from 3rd, 4th, 5th,… higher energy level to 2nd (L shell) level by emitting excess of energy in visible region.
• Paschen Series : This series is formed when the electron jumps from 4th, 5th, 6th higher energy level to 3rd (M Shell) level by emitting excess of energy in Infra Red (IR Region).
• Brackett Series : This series is formed when the electron jumps from 5th, 6 th…. higher energy level to 4th (N shell) level by emitting excess of energy in Infra Red (IR) region.
• Pfund Series : This series is formed when the electron jumps from 6th, 7th …. higher energy level to 5th (0 shell) level by emitting excess of energy in Infra Red (IR) region.

Question 3.
What is Wave number, Frequency and Amplitude ? Give its SI Units.
Answer:
Wave Number ( \(\overrightarrow{\mathrm{v}}\) ) : Number of waves per unit length. Units : m^-1
Frequency (v) ; Number of wave peaks that pass a given point in time unit.
Units: Hz
Amplitude (A): Height of the wave above the central line.

1st PUC Chemistry Structure of Atom Four / Five Marks Questions and Answers

Question 1.
Calculate the wave number, wavelength and frequency first line of hydrogen spectrum Or Calculate the maximum wave length of a line in the Lyman series.
Answer:
First line is Lyman Series, where n₁ = 1, n₂ = 2. Maximum wave length corresponds to minimum frequency i.e., n₁ = 1, n₂ = 2.

R = Rydberg constant = 1.097 × 10⁺⁷ m. n₁ = 1 n₂ = 2

Wave length λ = 0.8227 × 107 = 8.227 × 10⁶ m^-1

Question 2.
Calculate the limiting frequency of Balmer series.
Answer:
Limiting frequency for Balmer series possible when n₁ = 2 and n₂ = ∞.

Question 3.
Calculate the wavelength and wave numbers of the first and second lines in the Balmer series of hydrogen spectrum. Given R = 1.096 × 10⁷m^-1
Answer:
For Balmer series n₁ = 2 , for third line n₂ = 3, for fourth line n₂ = 4
i. Wave number \(\overrightarrow{\mathbf{V}}\) of the third line is given by

Question 4.
Calculate the wavelength and wave number of H_γ and H_???? line (R = 10.97 × 106 m^-1)
Answer:
For Balmer series n₁ = 2 , for third line n₂ = 5, for fourth line n₂ = 6
i. Wave number \(\overrightarrow{\mathbf{V}}\) of the third line is given by

= 364 × 10^-9m = 364.9nm
Wavelength of H line = 364.9nm

Question 5.
Calculate the wave number and wavelength of the first spectral line of Lyman series of hydrogen spectrum.
Rydberg constant R = 10.97 × 106 m^-1
Answer:

Question 6.
Calculate the wave number of the spectral line when electron jumps from the seond Bohr orbit to the ground state. R = 1.097 × 10⁷ m^-1.
Answer:

Wavelength of the spectral line = 8.2258 × 10⁶m^-1

Question 7.
In a hydrogen atom, an electron jumps from third orbit to the first orbit. Find out the frequency and wavelength of the spectral line. Given R = 1.097 × 10⁷m^-1

Wavelength of the spectral line = 1025.4 A⁰
Frequency of the spectral line = 2.9257 × 10¹⁵ s^-1

Question 8.
Calculate the wavelength of 2^nd line and limiting line of Balmer series. If wave length of first line of Balmer series is 656 nm.
Answer:

For first line m = 2, m = 3 & λ = 656 nm

For secong line n1 = 2,n2 = 4

∴ Wavelength of the limiting line n₁ = 2, n₂ = ∞

Question 9.
Calculate the wavelength of a wave of frequency 1012 Hz, travelling with the speed of light 3 x 10⁸ ms^-1
Answer:

Question 10.
Calculate the frequency of electromagnetic radiation having the wavelength 3 μ. Calculate the wave number corresponding to it. (1μ = 10^-6 m)
Answer:

Question 11.
Calculate the frequency and energy per quantum of a radiation with a wavelength of 200 nm. (c = 3 × 10⁸ ms^-1 and h = 6.625 × 10^-34 Js)
Answer:

= 1.5 × 10¹⁵ Hz = 1.55 × 10 ¹⁵Hz.
E = hv = 6.625 × 10 ^-34 x 1.5 × 10¹⁵= 9.9375 × 10^-19 J

Question 12.
Calculate the number of photon of light with a wavelength of 6000A that provide I joule of energy.
Answer:

= 0.00331 × 10^-16J = 3.31 × 10^-19 J
Number of photons 3.31 × 10^-19 J make 1 photon
1J = \(\frac{1}{3.31 \times 10^{-19}}\) = 3.021 × 10²⁰ photons

Question 13.
A major line in an atomic emission spectrum occurs at 450 nm. Find the energy decrease, as this photon is emitted.
Answer:

Question 14.
Calculate the wave number, wavelength and frequency of the first line in the Baimer series.
Answer:

For Balmer Series
n₁ = 2 and n₂ = 3 for the first time

Question 15.
The red light of neon signs has a wavelength of 693 nm. Find the energy difference (per mole of atoms) between the two energy levels involved.
Answer:

= 2.8684 × 10¹⁵ J
Energy difference per mole of atoms = 2.8684 × 10¹⁵ × 6.022 × 10²³

Question 16.
Calculate the wavelength of an electron moving with a velocity of 2.5 × 10^-7 ms^-1 h = 6.626 × 10^-34 Js ; mass of an electron = 9.11 × 10^-31 kg.
Answer:

Question 17.
Find the mass of an electrically charged particle moving with a velocity of 3 × 106 ms^-1 and having a de Broglie wavelength of 2Å.
Answer:

Question 18.
Calculate the energy of a photon whose wavelength is 3.864 × 10^-7m.
Answer:
\(\mathrm{E}=\mathrm{h} \vartheta=\frac{\mathrm{h} \mathrm{c}}{\lambda}\)
h = 6.626 × 10^-34
c = 3.0 × 10⁸ ms^-1

λ = 3.864 × 10^-7

Question 19.
Calculate the de Broglie wavelength of (a) an electron of mass 9.11 × 10^-31 kg and moving with a velocity of 1.0 × 106 ms^-1 (b) a bullet of mass 25g moving with a velocity of 100ms^-1
Answer:

You can Download Chapter 15 Waves Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 15 Waves

1st PUC Physics Waves Textbook Questions and Answers

Question 1.
A string of mass 2.50 kg Is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse Jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Mass per unit length of the string,
µ = \(\frac{2.50 \mathrm{kg}}{20 \mathrm{m}}\) = 0.125 kg m^-1
Given tension, T = 200 N.
The speed of wave on the string is given by,
v = \(\sqrt{\frac{T}{\mu}}\) =\(\sqrt{\frac{200}{0.125}}\) = 40 ms^-1
∴ Time taken by the disturbance to reach the other end of the string:
t = \(\frac{\mathrm{s}}{\mathrm{v}}\) = \(\frac{20}{40}\) = 0.5 s

Question 2.
A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When Is the splash heard at the top given that the speed of sound In air Is 340 m s^-1? (g= 9.8 ms^-2)
Answer:
Time taken to hear the splash at the top =Time taken by the stone to reach the pond + Time taken by the sound to travel to the top from the base of tower,
i e., T = T₁ + T₂ ……..(1)
T₁ : We know that s = u t + 1 / 2 a t²
u = 0, s = 300 m, a = g = 9.8 ms^-2
⇒ t = \(\sqrt{\frac{2 \mathrm{s}}{\mathrm{a}}}\) = \(\sqrt{\frac{2 \times 300}{9.8}}\) = 7.82 s
∴ T₁ =7.82 s
T₁ : s = vt ⇒ t = \(\frac{\mathrm{s}}{\mathrm{v}}\) = \(\frac{300}{340}\) = 0.88 s
∴ T₂ = 0.88 s
∴ T₁+ T₂ = 7.82 + 0.88 = 8.70 s

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension In the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 m s^-1?
Answer:
Mass per unit length of the wire,
µ = \(\frac{2.10 \mathrm{kg}}{12.0 \mathrm{m}}\) = 0.175 kg m^-1
The speed of wave on a string is
given by, v=\(\sqrt{\frac{T}{\mu}}\)
Given v = 343 ms^-1
∴ T = µv² =0.175 × 343²
= 20588.575 = 2.06×10⁴ N.

Question 4.
Use the formula v = \(\sqrt{\frac{rP}{\rho}}\) to explain why the speed of sound in air

1. Is Independent of pressure,
2. Increases with temperature,
3. Increases with humidity.

Answer:
Given v = \(\sqrt{\frac{\mathrm{rP}}{\rho}}\) ……….(1)
According ideal gas law, P = \(\frac{\rho \mathrm{RT}}{\mathrm{M}}\) , where
\(\boldsymbol{\rho}\) is the density, T is the temperature, M is the Molecular mass of the gas;
R – universal gas constant.
Substituting for P in (1) we get
v = \(\sqrt{\frac{\mathbf{r} \mathbf{R} \mathbf{T}}{\mathbf{M}}}\)
This shows that v is

1. Independent of pressure
2. Increases with temperature i.e. v ∝ \(\sqrt{T}\)
3. We know that the molecular mass of water (18) is less than that of N₂ (28) and O₂ (32). therefore as humidity increases, the effective molecular mass of air decreases and hence velocity increases.

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x – v t or x + vt, i.e., y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave :
(a) (x – vt )²
(b) log [(x + Vy)/x₀]
(c) 1/(x + vt)
Answer:
The converse is not true. For any function to represent a travelling wave, an obvious requirement is that the function should be finite at all times and finite everywhere.

Function a) and b) do not satisfy this requirement and hence cannot represent a travelling wave. Only function c) satisfies the condition.

Question 6.
A bat emits the ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of

1. The reflected sound,
2.  The transmitted sound? The speed of sound in air is 340 ms^-1 and in water 1486 ms^-1.

Answer:
We know that λ = \(\frac{\mathrm{v}}{\mathrm{f}}\)
where,
λ : wavelength
v : velocity
f: frequency

1.  Reflected sound:
λ = \(\frac{340}{1000}\) = 0.34 m
(∵ v The reflected sound travels in air)

2. Transmitted sound:
λ = \(\frac{1486}{1000}\) = 1.486 m
(∵ v The transmitted sound travels in water)
Note: When sound moves from one medium to other, the frequency does not change.

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1? The operating frequency of the scanner is 4.2 MHz.
Answer:
Given
f = 4.2 × 10⁶ Hz
v = 1.7 × 10³ ms^-1
we know that, λ = \(\frac{\mathrm{v}}{\mathrm{f}}\)
\(=\frac{1.7 \times 10^{3}}{4.2 \times 10^{6}}\) = 4.05 × 10^-4 m
∴ wavelength of sound in tissue = 4.05 × 10^-4 m

Question 8.
A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x+ π/4 ) where x and y are in cm and t in s. The positive direction of x is from left to right.

1. Is this a travelling wave or a stationary wave? If it is traveling, what are the speed and direction of its propagation?
2. What are its amplitude and frequency?
3. What is the initial phase at the origin?
4. What is the least distance between two successive crests in the wave?

Answer:

1. Travelling wave: The wave is travelling from right to left with a speed of \(\frac{36 s^{-1}}{0.018 \mathrm{cm}^{-1}}\) = 2000 cm s^-1 = 20 ms^-1
2. Amplitude: 3 Frequency : \(\frac{36}{2 \pi}\) = 5.73 Hz
3. Initial phase at origin: π/4
4. Distance between two successive crests in the wave = \(\frac{2 \pi}{0.018 \mathrm{cm}^{-1}}\) = 349.06 cm = 3.49 m.

Question 9.
For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0,2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion In traveling wave differ from one point to another: amplitude, frequency or phase?
Answer:

All the graphs have same amplitude of frequency but different initial phase. All graphs are sinusoidal.

Question 10.
For the travelling harmonic wave y(x,t) = 2 cos 2π (10t – 0.0080 x + 0.35) where x and y are In centimetre and t In seconds. Calculate the phase difference between oscillatory motion of 2 points separated by a distance of
a) 4 m
b) 0.5 m
c) λ/2
d) 3λ/2
Answer:
Given
k = 2m × 0.008 cm^-1
We know that, k = \(\frac{2 \pi}{\lambda}\) ⇒ λ = \(\frac{2 \pi}{(0.008 \times 2 \pi)}\) = 125cm
Now, (phase difference) = \(\frac{2 \pi}{\lambda}\) × (path difference)
∴ ∆Φ = \(\frac{2 \pi}{\lambda}\) ∆x

Question 11.
The transverse displacement of a string (clamped at Its both ends) is given by y(x,t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\). cos \((120 \pi \mathrm{t})\) where x and y are in m and t In s The length of the string is 1.5 m and its mass is 3.0 x10^-2 kg.
Answer the following :

1. Does the function represent a travelling wave or a stationary wave?
2. Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
3. Determine the tension in the string.

Answer:
1.  The function represents a stationary wave.

2.

∴ λ = 3 m , f = 60 Hz, v = \(\frac{120 \pi}{2 \pi} \times 3\) = 180 ms^-1
for both waves.
3.  We know that T = µ v² where µ = mass per unit length of the string
\(=\frac{3 \times 10^{-2} \mathrm{kg}}{1.5 \mathrm{m}}\) = 2 × 10^-2 kgm^-1
∴ T = 2 × 10^-2 × 180² = 648 N

Question 12.
(i) For the wave on a string described in Exercise 15.11, do all the points c n the string oscillate with the same
(a) frequency,
(b) phase,
(c) amplitude?
Explain your answers.
(ii) What Is the amplitude of a point 0.375 m away from one end?
Answer:
Given :
y (x, t) = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\) cos (120 πt)

i) Amplitude of wave = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)

Frequency = 60 Hz.
∴ From the above equation of wave, we see that all points on the string oscillate with same phase and frequency except the nodes (endpoints). We can also see that amplitude of the wave changes for different ‘x’.

ii) Amplitude of a point at 0.375 m
i.e., x = 0.375 m
Amplitude = 0.06 sin \(\left(\frac{2 \pi}{3} x\right)\)
= 0.06 sin \(\left(\frac{2 \pi}{3} \times 0.375\right)\)
= 0.042 m

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all:

1. y= 2 cos (3x) sin (10t)
2. y= 2\(\sqrt{x-v t}\)
3. y = 3 sin(5x – 0.5t) + 4cos(5x-0.5t)
4. y = cos x sin t + cos 2x sin 2t

Answer:
1.  y = 2cos (3x) sin (10 t)
Represents a stationary wave

2.  y = 2\(\sqrt{x-v t}\)
Does not represent any wave.

3.  y = 3 sin (5x – 0.5 t) + 4 cos (5x – 0.5t)
= 5 sin \(\left[5 x-0.5 t-\tan ^{-1}(3 / 4)\right]\) Represents a travelling wave,

4.  y = cos x sin t + cos 2x sin 2t
Represents superposition of 2 stationary waves.

Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10^-2 kg and its linear mass density is 4.0 × 10^-2kgm^-1. What is

1. The speed of a transverse wave on the string,
2. The tension in the string?

Answer:
Given :
f = 45 Hz mass of wire
= m =3.5×10^-2 kg
linear mass density
= µ = mass per unit length
= 4 × 10^-2 kg m^-1
∴ length of the wire
\(=\frac{m}{\mu}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}\)
l = 0.875 m

1.  We vibrate in fundamental mode
⇒ λ = 2l

∴ λ =2×0.875 = 1.75 m
∴ Speed of wave
= v =fλ = 45×1.75
= 78.75 ms^-1

2. We know that
T = µV²
∴ Tension, T = 4×10^-2 × 78.75^-2 = 248.06 N

Question 15.
A meter-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Answer:
For a closed organ pipe of length ‘l’, the frequency of n^th mode of vibration is
f = (2n – 1)\(\frac{v}{4 l}\) ………….(1)
Thus the (n + 1)^th mode of vibration of closed pipe is
f = (2n + 1)\(\frac{v}{4 l}\) ………….(2)
Given that the closed pipe vibrates at 340 Hz for tube length = 25.5 cm and 79.3 cm.
Let l₁ = 25.5 cm l₂ = 79.3 cm
Equating (1) and (2) we get

On solving we get n = 1.
On substituting in (1) we have
\(\frac{(2 \times 1-1) v}{4 \times\left(\frac{25.5}{100}\right)}=340\) ⇒ v = 346.8 ms^-1

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
Answer:
Given: Here, length of rod, l = 100 cm = 1 m
Fundamental frequency of longitudinal vibrations,
v = 2.53 kHz = 2.53 x 10³ Hz.
In fundamental mode, wavelength, λ= 21 = 2×1= 2m.
∴ speed of sound in the steel rod.
υ= vλ. = 2.53 x 10³ x 2 = 5.06 X 10³ ms^-1 = 5.06 km s^-1.

Question 17.
A pipe 20 cm long Is closed at one end. Which harmonic mode of the pipe 1s resonantly excited by a 430 Hz source? Will the same source be In resonance with the pipe if both ends are open? (speed of sound In air Is 340 m s^-1).
Answer:
Given :
l = 20 cm = 0.2 m
v = 340 ms^-1
f = 430 Hz
For a pipe closed at one end :
f = \(\frac{(2 n-1) v}{4 l}\) n = 1, 2, 3 ……
⇒ 430 = \(=(2 n-1) \frac{340}{4 \times 0.2}\)
⇒ n = 1
⇒ Resonance occurs only for first/fundamental mode of vibration.
For a pipe open at both ends,
f = \(\frac{\mathrm{nv}}{2 l}\) n = 1, 2, …..
⇒ 430 = \(\frac{\mathrm{n} \times 340}{2 \times 0.2}\)
⇒ n = 0.51
Since n<1, resonance does not occur.

Question 18.
Two sitar strings A and B playing the note ‘Gd’ are slightly out of tune and produce beats of frequency 6 Hz. The tension In the string A Is slightly reduced and the beat frequency Is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what Is the frequency of B?
Answer:
Let f₁ and f₂ be the frequencies of strings A and B respectively.
Given:
f₁ = 324 Hz
Beat frequency = f_b = 6 Hz
∴ f₂= f₁ ± f_b =(324 ±6) Hz
We know that T ∝ v² and v ∝ f
(∵ T=μv² and v=fλ.)
Thus f ∝ \(\sqrt{T}\) …………(1)
Decreasing the tension in string A will decrease the frequency f₁.
Since the beat frequency f_b is reduced to 3 Hz upon decreasing the tension in string A, f₂ = 318 Hz.
Note:
If f₂ = 330 Hz, then the beat frequency would have increased when the tension in string A was reduced.

Question 19.
Explain why (or how):

1. In a sound wave, a displacement node Is a pressure antinode and vice versa,
2. bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,
3. A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
4. Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate In gases, and
5. The shape of a pulse gets distorted during propagation in a dispersive medium.

Answer:

1. In a sound wave, the displacement node is a point where the amplitude of oscillation is zero. But at a displacement node, the pressure changes are maximum. Hence it is also a pressure antinode. Similarly, at displacement antinode, the amplitude of oscillation is maximum whereas pressure changes are minimum. Hence displacement antinode is also a pressure node.
2. Bats emit high-frequency ultrasonic waves. These waves are reflected back from the obstacles on their path and are sensed by bats.
3. Because they emit different harmonies which can be easily differentiated by human ears.
4. This is because solids have both shear and bulk modulus of elasticity whereas gas has only bulk modulus of elasticity.
5. A sound pulse consists of waves with different wavelengths. In a dispersive medium, these waves travel with different velocities which distorts the shape of pulse.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz In still air.
1.  What Is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10 m s^-1,
(b) recedes from the platform with a speed of 10 m s^-1?

2.  What Is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Given, f = 400 Hz v = 340 ms^-1 v_s = 10 ms^-1 : speed of train .
1.
a) train approaches the platform

b) train recedes from the platform

2.  The speed of sound does not change, i.e., it is 340 ms^-1 for both cases.

Question 21.
A train, standing In a station-yard, blows a whistle of frequency 400 Hz In still air. The wind starts blowing in the direction from the yard to the station at a speed of 10 ms^-1. what are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly Identical to the case when the air Is still and the observer runs towards the yard at a speed of 10 m s^-1? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Given f = 400 Hz v = 340 ms^-1 v_m = 1.0 ms^-1 : speed of air
Since there is no relative motion between the source and the observer,
the frequency of sound for the observer = f =400 Hz.
The wind is blowing in the direction from the yard to the station, the effective speed of sound for an observer at the platform = v + v_m = 350 ms^-1
Wavelength of sound : λ= \(\frac{\left(v+v_{m}\right)}{f}\)
= \(\frac{350}{400}\) = 0.875 m
The situation is not identical to the case when the air is still and the observer runs towards the yard at a speed of 10 ms^-1 (= v₀). Because here there is a relative motion between the source and the observer.

Question 22.
A travelling harmonic wave on a string is described by y(x, t) = 7.5 sin (0.005 x +12t + π /4)

1. what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
2. Locate the points of the string which have the same transverse displacements and velocity as the x a 1 cm point at t = 2 s, 5 s, and 11s.

Answer:
Given:
y (x, t) = 7.5 sin (0.005 x + 12t + π /4)
1. At x=1 cm and t=1s
y (1, 1)= 7.5 Sin(0.005 +12 + π /4)
= 7.5 sin (12.005 + π /4)
= 1.67 cm
Velocity of oscillation : v = \(\frac{\mathrm{d}(\mathrm{Y}(\mathrm{x}, \mathrm{t})}{\mathrm{dt}}\)
= \(\frac{\mathrm{d}}{\mathrm{dt}}\) (7.5 sin (0.005 x + 12t + π/4) dt
= 7.5 × 12 cos (0.005 x+ 12t + π/4)
At x = 1 cm and t = 1 s
v = 7.5 × 12 cos(0.005 + 12 + π/4)
= 87.75 cm s^-1
We know that velocity of wave propagation = \(\frac{\mathrm{w}}{\mathrm{k}}\)
Here w = 12 s^-1 and k = 0.005 cm^-1
∴ Velocity of wave propagation
= \(\frac{12 s^{-1}}{0.005 \mathrm{cm}^{-1}}\) = 24 ms^-1

∴ At x = 1 cm and t = 1 s velocity of oscillation is not equal to velocity of wave propagation.

2. In a wave, all the points which are separated by a distance ±λ,±2λ ……..
from x = 1 cm will have same transverse displacements and velocity. For the given
wave , λ= \(\frac{2 \pi}{0.005}\) = ±12.56 cm, +25.12
m….From x = 1 cm will have the same displacements and velocity as at x = 1 cm, t = 2s, 5s and 11 s.

Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
1.  Does the pulse have a definite
(i) frequency,
(ii) wavelength,
(iii) speed of propagation?
2.  If the pulse rate is 1 after every 20 s, (that is the whistle Is blown for a split of seconds after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:
1. In a nondispersive medium, the wave propagates with definite speed but its wavelength and frequency are not definite
2. No, the frequency of the note is not \(\frac { 1 }{ 20 } \) or 0.05 Hz. 0.005 Hz is only the frequency of repetition of the pip of the whistle.

Question 24.
One end of a long string of linear mass density 8.0 × 10^-3 kg m^-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is kg. The pulley end absorbs all the Incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down its transverse displacement y as a function of x and t that describes the wave on the string.
Answer:
Given : Mass per unit length of string = μ = 8×10^-3kgm^-1
f = 256 Hz A = 5 cm : amplitude Tension in the string,
T = 90 kg × 9.8 ms^-2= 882 N
velocity of the wave :
v = \(\sqrt{\frac{T}{\mu}}\)

propagation constant,
k= \(\frac{2 \pi}{\lambda}\) = 4.84 m^-1
Equation of wave y (x, t) = A sin(wt – kx)
∴ y(x, t) = 5 sin(512 π t – 4.84 x)
with x and y in centimetre and t in seconds.

Question 25.
A SONAR system fixed In a submarine operates at a frequency of 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be1450ms^-1.
Answer:
Given: f = 40 kHz v = 1450ms^-1
Speed of enemy submarine = v₀ = 360 km/ hr
=360 × \(\frac{5}{16}\) = 100 m/s
Apparent frequency of sound waves:
(source at rest and observer moving towards the source)

Now this frequency is reflected by the enemy submarine and is observed by the SONAR
∴ \(\mathrm{f}^{\prime}=\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}} \times \mathrm{f}^{\prime}\)
Note: Here the observer (SONAR) is at rest and the source is moving towards the observer at a speed of 360 km h^-1 = 100 ms^-1 (v_s)
\(\therefore \mathrm{f}^{\prime}=\frac{1450}{1450-100} \times 42.76 \mathrm{k}=45.93 \mathrm{kHz}\)

Question 26.
Earthquakes generate sound waves Inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s^-1, and that of the P wave is 8.0 km s^-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 minutes before the first S wave. Assuming the waves travel in a straight line, at what distance does the earthquake occur?
Answer:
Let v_p =8 km s^-1 and v_s = 4 k ms^-1
t = time gap between arrival of s and p waves = 4 min = 240 s
t = t_s – t_p = \(\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{s}}}-\frac{\mathrm{d}}{\mathrm{v}_{\mathrm{p}}}\)
where d :
distance of earthquake centre (kms)
∴ 240 = \(\mathrm{d}\left[\frac{1}{4}-\frac{1}{8}\right]\)
∴ d = \(\frac{240}{0.25-0.125}\) = 1920 km

Question 27.
A bat Is flitting about In a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat Is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Given:
f = 40 kHz
velocity of bat = 0.03 velocity of sound i.e., v_s = 0.03 v
Apparent frequency of sound hitting the wall
\(f^{\prime}=\frac{v}{v-v_{s}} \times f=\frac{v}{v-0.02 v} \times f\)
= 1.03 \(f^{\prime}\) = 1.03 × 40 k
= 41.24 k Hz
This freuency f1 is reflected e ft tne wall and is recieved by the bat moving towards the wall
∴ \(f^{\prime \prime}=\frac{\gamma+v_{s}}{v} \times f^{\prime}=\frac{v+0.03 v}{v} \times f^{\prime}\)
= 1.01 \(f^{\prime}\) = 1.03 × 41.24 k
= 42.47 k Hz

1st PUC Physics Waves one mark Questions and Answers

Question 1.
What is meant by a wave?
Answer:
The disturbance set up in a medium is called a wave.

Question 2.
What property of the medium is essential for the propagation of mechanical wave?
Answer:
Elasticity and Inertia.

Question 3.
Which physical quantity does not change when a wave travels from one medium to another?
Answer:
Frequency

Question 4.
What is a progressive wave?
Answer:
A wave, which travels continuously in a medium in the same direction, is called a progressive wave.

Question 5.
If y = 2sin π (40t – 2x) represents a progressive wave. What is its frequency?
Answer:
20 Hz.
Solution:
Given equation is
y = 2 sin π (40t – 2x)
y =2 sin 40π (t – x/20)
Comparing this with y = a sin ω (t -x/v)
ω = 2 πf = 40π
∴ f= 20Hz

Question 6.
Two waves are represented by the equations Y₁ = a sin(ωt- kx) and Y₂ = a cos (ω t – kx). What is the phase difference between them?
Answer:
π/2 radians.
Y₁ = a sin (ω t – kx).
Y₂= a cos (ω t – kx).
= a sin(ω t – kx + π /2).
∴ Phase difference between Y₁ and Y₂ is π/2 rad.

Question 7.
What is meant by ‘Phase of a Particle’ in a wave?
Answer:
The phase of a particle at any instant represents the state of vibration of the particle at that instant.

Question 8.
What is a mechanical wave?
Answer:
A wave which requires a medium for its propagation is called a mechanical wave.

Question 9.
Give an example of a transverse wave.
Answer:
Light waves.

Question 10.
Give an example of a two-dimensional wave.
Answer:
Water waves.

Question 11.
What is a transverse wave?
Answer:
In transverse waves, particles of the medium are vibrating perpendicular to the direction of wave propagation.

Question 12.
What is the angle between the vibration of the particle of medium and direction of propagation of the wave in a transverse wave?
Answer:
90°

Question 13.
How does velocity of sound vary with pressure?
Answer:
The velocity of sound is independent of pressure.

Question 14.
How does velocity of sound vary with temperature?
Answer:
The velocity of sound in a gas is directly proportional to the square root of the absolute temperature.

Question 15.
How does velocity of sound vary with humidity?
Answer:
The velocity of sound increases with increase in humidity.

Question 16.
Why sound travels faster in moist air than in dry air?
Answer:
The density of moist air is less than that of dry air. As the velocity of sound in a gas is inversely proportional to the square root of its density, the velocity of sound in moist air is greater than that in dry air.

Question 17.
A wave has a velocity of 330 ms^-1 at one atmospheric pressure. What will be its velocity at 4 atmospheric pressure?
Answer:
330 ms^-1
[Reason: Velocity of sound in a gas is independent of pressure].

Question 18.
What are the beats?
Answer:
The rise and fall in the intensity of sound due to the superposition of two sound waves of slightly different frequencies traveling in the same direction is known as beats.

Question 19.
What is beat frequency?
Answer:
The number of beats heard per second is called beat frequency.

Question 20.
Define the beat period.
Answer:
The time interval between two consecutive maxima or minima is called beat period. It is also equal to the reciprocal of beat frequency.

Question 21.
What is the available range of sound frequencies?
Answer:
20 Hz to 20,000 Hz.

Question 22.
By how much does the frequencies of 2 sound sources differ, if they produce 10 beats in 2 seconds.
Answer:
Number of beats per second
\(=\frac{10}{2}=5\)
∴ ∆ f = f₁ ~ f₂ = 5 Hz

Question 23.
Define reverberation.
Answer:
Reverberation is defined as the persistence of audible sound even after the source has ceased to produce the sound.

Question 24.
What is a stationary wave?
Answer:
The wave formed due to the superposition of two identical waves travelling with the same speed in opposite directions is called a stationary wave or standing wave.

Question 25.
What Is fundamental frequency?
Answer:
The vibration of a body with the lowest frequency is called the fundamental frequency.

Question 26.
What are overtones?
Answer:
Frequencies greater than fundamental frequency are called overtones.

Question 27.
What are harmonics?
Answer:
Overtones, which are an integral multiple of the fundamental frequency, are called harmonics.

Question 28.
The length of the vibrating portion of a sonometer wire Is doubled. How does its frequency change?
Answer:
Halved
[Reason: Frequency f ∝ 1//. As / is doubled f is halved].

Question 29.
Give the relation between the fundamental note and overtone in an open pipe.
Answer:
\(f^{\prime}=(n+1) f\)
f – fundamental frequency, n = 1,2, 3….
for 1^st and 2^nd  …. overtones.

Question 30.
What is the distance between a node and the neighbouring antinode of a stationary wave?
Answer:
\(\frac{\lambda}{4}\)

Question 31.
The fundamental frequency produced in a dosed pipe is 500 Hz. What Is the frequency of the first overtone?
Answer:
1500 Hz.
[Solution: Frequency of the first over tone in a closed pipe is \(f^{\prime}\) = 3f = 3 × 500 = 1500 Hz].

Question 32.
Find the distance between node and an adjacent antinode if the wavelength in 4m in a stationary wave.
Answer:
Distance between node and adjacent antinode =\(\frac{\lambda}{4}=\frac{4}{4}\) = 1 m

Question 33.
Explain why is it NOT possible to have interference between the waves produced by 2 sitars?
Answer:
Because the waves produced will not have a constant phase difference.

Question 34.
Which harmonies are present in a closed organ pipe?
Answer:
All the odd harmonies are present.

Question 35.
What will be the resultant amplitude when 2 waves y₁ = a sin ω t are superposed at any point at a particular instant?
Answer:
y = y₁ + y₂ = a sin ω t + a cos ω t
= a (sin ω t + cos ω t)
=\(\sqrt{2} \mathrm{a}(\sin (\omega \mathrm{t}+\pi / 4))\)
∴ Resultant amplitude : \(\sqrt{2}\)a

Question 36.
Write 2 characteristics of a medium which determine the speed of sound waves in the medium.
Answer:
Elasticity and inertia.

Question 37.
State the factors in which the speed of a wave travelling along a stretched Ideal string depends
Answer:
Tension and mass per unit length.

Question 38.
The fundamental frequency of oscillation of a closed pipe is 400 Hs. What will be the fundamental frequency of oscillation of open pipe of the same length?
Answer:
f_e = \(\frac{\mathrm{v}}{4 l}\) = 400 Hz
f_o = \(\frac{\mathrm{v}}{2 l}\) ⇒ f_o = 2 f_e = 800 Hz.

Question 39.
Why is it difficult some times to recognise your friend’s voice on phone?
Answer:
Because of modulation.

Question 40.
Which of the following media can pass longitudinal waves only air, water or Iron?
Answer:
Air

1st PUC Physics Waves two marks Questions and Answers

Question 1.
Explain different types of waves (based on medium)
Answer:
Waves are classified into two types:

1. Mechanical Waves:
Waves, which requires a medium for their propagation are called mechanical waves.
e.g.: Waves on the surface of water, Seismic waves (due to earthquake), Sound waves, Waves on stretched string, waves formed in an air column, shock waves.

2. Non-mechanical Waves:
Waves, which do not require a medium for their propagation are called Non-mechanical Waves.
e.g.: Light waves, heat waves, radio waves, X- rays, ultraviolet rays, Infrared rays, etc.

Question 2.
The equation of a progressive wave is y = 0.2 sin(50t-0.5x). Find the amplitude and the magnitude of the velocity, if ‘x’ and ‘y’ are in metres.
Answer:
Given equation is,
y = 0.2 sin(50t – 0.5x)
= 0.2 sin50(t – x/100)
Comparing with y = a sin ω (t – x/v)
amplitude a = 0.2m,
velocity v = 100m/s

Question 3.
State the principle of superposition. Name the phenomenon produced due to the superposition of waves.
Answer:
When two or more waves superpose, he resultant displacement of particle of the medium is equal to the vector sum of the displacements due to individual waves. Superposition of waves leads to the phenomenon of interference, diffraction, beats, and formation of stationary waves are due to the superposition of waves.

Question 4.
What is a longitudinal wave ? Give an example.
Answer:
If the particles of a medium vibrate along the direction of wave propagation then wave is called longitudinal waves,
e.g: Sound waves in air are longitudinal waves.

Question 5.
The distance between two particles is 0.1m; If the phase difference between these points is π/2 rad calculate the wavelength.
Answer:
∆= 0.1m, Φ = π/2 rad

Question 6.
How does its frequency of a tuning fork change when the prongs are

1. filed
2. waxed.

Answer:

1. When the prongs of a tuning fork are filed its frequency increases.
2. The frequency of a tuning fork decreases when the prongs are waxed.

Question 7.
What is meant by beats? What are its applications ?.
Answer:
The periodic rise and fall (Waxing and waning) in the intensity of sound due to the superposition of two sound waves of slightly different frequencies traveling in the same direction is called beats.
The phenomenon of beats can be used.

1. To determine the unknown frequency of a tuning fork.
2. In tuning musical instruments.

Question 8.
What is the Doppler effect? Give an example.
Answer:
The apparent change in the frequency of sound due to the relative motion between the source and the observer is called the Doppler effect.
E.g.: The apparent frequency of the whistle of a train increases as it. approaches an observer on the platform and decreases when the train passes the observer.

Question 9
What are the uses of the Doppler effect?
Answer:

1. Doppler effect is used in a radar system to detect the speed of automobiles and aeroplanes.
2. It is used to determine the speed submarines (Using SONAR).
3. It is used to determine the speed of stars and planets and other celestial bodies.

Question 10.
When two tuning forks A and X are sounded together produces 6 beats per second. If the frequency of A is 341 Hz. What is the frequency of X?
Answer:
f_A = 341 Hz
f=?
and f_b = 6 beats/s
f_b = f_A ~ f_x
or f = f_A ± f_b = 341 ± 6
= 335 Hz or 347 Hz

Question 11.
What are nodes and antinomies in a stationary wave?
Answer:
The points in a stationary wave where the amplitude of vibration of the particles zero are called nodes The points in a stationary wave where the amplitude of vibration of particles is maximum are called antinodes.

Question 12.
An open pipe and a closed pipe have the same fundamental frequency. Explain how their lengths are related?
Answer:
Fundamental Frequency of an open pipe f₁ = \(\frac { v }{ { 2l }_{ 1 } } \)
Fundamental Frequency of an closed pipe
f₂ = \(\frac{\mathbf{v}}{4 \ell_{2}}\)
Given f₁ = f₂ ∵ 2l₁ = 4l₂
\(\frac{\ell_{1}}{\ell_{2}}=\frac{2}{1}\) ⇒ l₁:l₂ = 2:1

Question 13.
Mention any four characteristics of a stationary wave.
Answer:

1. A stationary wave does not move in any direction.
2. There is no flow of energy.
3. All particles in a loop are in the same phase & they are in opposite phase with respect to the adjacent loop.
4. Amplitude is different for different particles.

Question 14.
The fundamental frequency produced in a closed pipe Is 100Hz. What is the frequencies of first and second overtone?
Answer:
Given, f = 100Hz
For the first overtone f_1,
i.e f₁ = 3f
f₁ = 3 × 100
= 300 Hz
For the second overtone f_2,
i.e f₂ = 5f
= 5 × 100
= 500 Hz.

Question 15.
The equation for the transverse wave on a string is y \(=4 \sin 2 \pi(t / 0.05-x / 50)\) with length expressed Iff centimetre and time In second. Calculate the wave velocity and maximum particle velocity.
Answer:
Given

Particle velocity

Question 16.
Equation of a wave travelling on a string is y = 0.1 sin(3001 – 0.01 x) cm. Here x is in centimetre and t is in seconds. Find

1. wavelength of the wave
2. Time taken by the wave to travel 1 m

Answer:

The wave takes T seconds to travel a distance of λ.
∴ Time taken to travel 1 m is

Question 17.
What is meant by RADAR and SONAR? How are long distances measured using these techniques?
Answer:

1. RADAR – Radio Detecting and Ranging.
2. SONAR – Sound Navigation and Ranging

The waves produced by the devices are sent and are reflected by the bodies and reach them back. If the speed of sound is known and the time taken for to and fro journey, the distance can be estimated.

Question 18.
If the frequency of a tuning fork is 256 Hz and speed of sound in air is 320 ms^-1. Find how far does the sound travel when the fork executes 64 vibrations.
Answer:
We know that
v = f λ f= 256 Hz v = 320 ms^-1
∴ λ = \(\frac{\mathrm{v}}{\mathrm{f}} \) = \(\frac{320}{256} m \)
Also, the distance covered in n vibrations = n λ
∴ Distance covered in 64 vibrations = \(\frac{64 \times 320}{256} \) = 80 m

1st PUC Physics Waves four/five marks Questions and Answers

Question 1.
Distinguish between longitudinal and transverse waves.
Answer:
1. The vibration of particles of the medium is along the direction of wave propagation in longitudinal waves, whereas in transverse waves, the vibration of particles of the medium is perpendicular to the direction of wave propagation.

2. The wave propagates by forming alternate compressions and rarefactions in longitudinal waves, whereas in transverse waves, the wave propagates by forming alternative crests and troughs.

3. Longitudinal waves travel through solids, gases, and liquids, whereas Transverse waves travel through solids and on 1 liquid surfaces.

4. Longitudinal waves cannot be polarised, whereas Transverse waves can be polarised.

5. Distance between two successive compressions or rarefactions is equal to the wavelength of the wave in longitudinal waves, whereas, in transverse waves, the distance between two successive crests or troughs is equal to the wavelength of the wave.

Question 2.
Define the following terms.

1. Wave amplitude
2. Wave period
3. Wave frequency
4. Wavelength
5. Wave velocity

Answer:
1. Wave amplitude (a):
The maximum displacement of a particle of the medium from its mean position is called 1 wave amplitude.

2. Wave period (T):
The time taken by a particle of the medium to complete one vibration or Wave period is the time during which one complete wave is set up in a medium.

3. Wave frequency (f):
The number of vibrations completed by a particle of the medium in one second is called wave frequency or Wave frequency is the number of waves set up in the medium in one second.

4. Wavelength (l):
The distance traveled by the wave in a time equal to its period is called wavelength.

5. Wave velocity (v):
It is the distance traveled by a wave in one second.

Question 3.
What are the characteristics of progressive wave?
Answer:

1. A progressive wave is formed due to continuous vibration of the particles of the medium.
2. The wave travels with a certain velocity.
3. There is a flow of energy in the direction of the wave.
4. No particles in the medium are at rest.
5. The amplitude of all the particles is the same.
6. Phase changes continuously from par¬ticle to particle.

Question 4.
State Newton’s formula for the velocity of sound in a gas. What is the Laplace’s correction? Explain.
Answer:
According to Newton, velocity of sound in any medium is given by v = \(\sqrt{\frac{E}{\rho}}\) where E is the modulus of elasticity and p is the density of the medium.
For gases E = B, bulk modulus
∴ v = \(\sqrt{\frac{\mathrm{B}}{\rho}}\) …………(1)
When sound waves travel through a gas alternate compressions and rarefactions are produced. At the compression region pressure increases and volume decreases and at the rarefaction region pressure decreases and volume increases. Newton assumed that these changes take place under isothermal conditions i.e., at a constant temperature.
Under isothermal condition, B = P, pressure of the gas.
∴ In (1) v= \(\sqrt{\frac{P}{\rho}}\) ………..(2)
This is Newton’s formula for velocity of sound in a gas.
For air at NTP, P = 101.3 kPa and
ρ = 1.293 kgm^-3
Substituting the values of P and ρ in . equation (1) we get v = 280m/s. This is much lower than the experimental value of 332 m/s. Thus Newton’s formula is discarded.

Laplace’s correction:
According to Laplace, in a compressed region temperature increases and in a rarefied region it decreases and these changes take place rapidly. Since air is an insulator, there is no conduction of heat. Thus changes are not isothermal but adiabatic.
Under adiabatic condition, B = γ P, where γ is the ratio of specific heats of the gas.
Substituting in equation (1), v = \(\sqrt{\frac{\gamma P}{\rho}}\)
The above equation is called Newton- Laplace’s equation
Substituting the values of P, ρ and γ in the above equation, gives the velocity of sound in air at NTP to be about 331 m/s. This is in close agreement with the experimental value.

Question 5.
Discuss the variation of velocity of sound with,

1. Pressure
2. Temperature
3. Humidity
4. Wind

Answer:
1. Effect of Pressure:
According to Boyle’s law, for the given mass of a gas, pressure (P) is inversely proportional to its volume (V) at constant temperature
\(P \propto \frac{1}{V}\)
or PV = constant if m is the mass and
ρ is the density of a gas then,
\(\mathrm{P}\left(\frac{\mathrm{m}}{\rho}\right)\) = constant
For given mass of gas \(\frac{P}{\rho}\) = constant.
∴ In the equation v = \(\sqrt{\frac{\gamma P}{\rho}}\)
γ and \(P / \rho\) are constants.
Thus velocity of sound is independent of pressure at constant temperature,

2. Effect of Temperature:
From Charle’s law, for the given mass of a gas, the volume V is directly proportional to its absolute temperature T at constant pressure.

At constant pressure, velocity of sound

Hence velocity of sound in a gas is directly proportional to the square root of the absolute temperature,

3. Effect of Humidity:
The presence of water vapour (humidity or moisture) in the air reduces the density of air.
∴ The density of dry air is greater than the moist air.
As the velocity of sound in a gas is inversely proportional to the square root of the density, the velocity of sound in moist air is greater than that in dry air. Thus, as the humidity increases, the velocity of sound also increases.

4. Effect of wind:
The velocity of sound is greater in the direction of the wind and lesser in the opposite direction. Let v be the velocity of sound waves and v_w the velocity of wind if the wind blows in the direction of sound waves, then the resultant velocity of sound is (v + v_w). If the wind blows against the velocity of sound waves, the resultant velocity of sound is (v- v_w).

Question 6.
Explain the theory of beats.
Answer:
Consider two sound waves of the same amplitude ‘a’ and slightly different frequencies f₁ and f₂ travelling in the same direction. The displacement of a particle in a time t due to the two waves is y₁ = a sin ω₁ t, and y₂ = a sin ω₂ t The resultant displacement of the particle due to the superposition of these two waves is,


Is the amplitude of the resultant wave. The intensity of resultant sound is maximum when A is maximum.
A is maximum when cos 2 π \(\left(\frac{f_{1}-f_{2}}{2}\right) t=\pm 1\)

The interval between successive maxima is \(\frac{1}{t_{1}-t_{2}}\),
The number of times intensity of sound becomes maximum per second is f_B = \(\frac{1}{\mathrm{T}_{\mathrm{B}}}\) = f₁ – f₂
Hence the beat frequency is the differ-ence between the frequencies of the two waves.

Question 7.
Derive a general expression for apparent frequency when the source moves towards the observer and observer moving away from the source.
Answer:

Consider a source S emitting sound of frequency f. Let v be the velocity of sound. Let the source move towards the observer with velocity vs and the observer move away from the source with velocity vQ. In one second the source travels a distance SS’ = v_s and wave travels a distance SP = v. In one second source emits f waves such that these waves will be contained in a length S’ P = v – v_s.
The apparent wavelength of these waves is

These waves approach the observer O with a relative velocity (v – v₀)
∴ Number of waves received by the observer in one second or apparent frequency is,

This is the general expression for apparent frequency.

Question 8.
What is Doppler’s effect? Give the expression for the apparent frequency of the note at different cases.
Answer:
The apparent change in the frequency of sound due to the relative motion between the source and the observes is called the Doppler effect.

Case (i) :
When the source moves towards the observer and observer moves away from the source.

Case (ii) :
When the source and ob¬serves move towards each other.

Case (iii) :
When the’source and ob-serves move away from each other

Case (iv):
Source moving away from the observer and the observer moving towards the source

Case (v):
When the source is in motion and the observer is at rest.
a. when the source moves towards the observer

b. when the source moves away from the observer

Case (vi):
when the source is at rest and the observer is in motion a. when the observer moves towards the source

when the observer moves away from the source

where f – real frequency
\(f^{\prime}\) – apparent frequency
v – velocity of sound
v_o – velocity of observer
v_s – velocity of source.

Question 9.
What are the beats? Define beat frequency. Explain how the frequency of a tuning fork is determined using beats.
Answer:
The periodic rise and fall (Waxing and waning) in the intensity of sound due to the superposition of two sound waves of slightly different frequencies traveling in the same direction is called beats. The number of beats heard per second is called the beat frequency and is equal to the difference in the frequency is of the two sound waves. To determine the unknown frequency of a tuning fork

Step 1:
Consider a tuning fork A of known frequency f and another fork B of unknown frequency \(f^{\prime}\). When A and B are sounded together let m beats are heard/sec,
∴\(f^{\prime}\) = f±m

Step 2:
Let one of the prong of B is loaded with a bit of wax. The two forks are again sounded together and let m be the number of beats heard/sec.
If m’ > m, i.e. betas increases after adding wax, then real frequency of B is \(f^{\prime}\) = f – m
If nr’ < m, i.e. betas decreases or remains same after adding wax, then real frequency of B is \(f^{\prime}\) = f + m

Question 10.
Distinguish between stationary wave and progressive wave.
Answer:

1. A stationary wave is formed by the superposition of two equal progressive waves travelling in opposite directions whereas, a progressive wave is formed due to continuous vibration of the particles of the medium.
2. The wave does not travel in any direction in stationary waves whereas, in progressive waves, the wave travels with a certain velocity.
3. There is no flow of energy in stationary waves whereas, in progressive waves, there is a flow of energy.
4. Particles at the node are at rest in stationary waves whereas, in progressive waves, no particles in the medium are at rest.
5. In stationary waves, amplitude is different for different particles, whereas in progressive waves, amplitude of all the particles is the same.
6. In stationary waves, all particles in a loop are in the same phase and they are in opposite phase with respect to particles in adjacent loops, whereas in progressive waves, phase changes continuously from particle to particle.

Question 11.
Derive the equation for a stationary wave.
Answer:
The equation of two waves having the same amplitude, wavelength, and speed but propagating in opposite directions is

Where a is the amplitude, λ is the wave-length and v is the velocity of the wave. A stationary wave is formed due to the superposition of these two waves. The resultant displacement of a particle is given by,
y = y₁ + y₂

where A = 2a cos\(\frac{2 \pi}{\lambda} x\) represents the amplitude of the resultant wave.

Question 12.
What is a closed pipe? Show that the overtones ira closed pipe are odd harmonics of the fundamental.
Ans: A pipe which is open at are end and closed at the other end is called a closed pipe.

Consider a closed pipe of length l. Let v be the velocity of sound in air. The air column in a closed pipe vibrates in such. a way that always antinodes formed at the open end and node is formed at the closed end. Let f₁, f_2, and f₃ be the frequencies and l₁, l₂ art l₃ be the wavelengths of 1^st, 2^nd and 3^rd modes of vibration respectively.

For 1st mode (fundamental mode) l = \(\frac{\lambda_{1}}{4}\) or  λ₁=4l
but \(\mathrm{f}_{1}=\frac{\mathrm{V}}{\lambda_{1}} \text { or } \quad \mathrm{f}_{1}=\frac{\mathrm{V}}{4l}\) ……….(1)
For i2nd mode (1sl overtone)
\(l=\frac{3 \lambda_{2}}{4} \quad \text { or } \quad \lambda_{2}=\frac{4 l}{3}\)
but \(\mathrm{f}_{2}=\frac{\mathrm{V}}{\lambda_{2}} \text { or } \quad \mathrm{f}_{1}=\frac{3 \mathrm{V}}{4l}=3 \mathrm{f}_{1}\) ………(2)
from (1)
For 3rd mode (2nd overtone)
\(l=\frac{5 \lambda_{3}}{4} \quad \text { or } \quad \lambda_{3}=\frac{4 l}{5}\)
but \(\mathrm{f}_{3}=\frac{\mathrm{V}}{\lambda_{3}} \text { or } \quad \mathrm{f}_{3}=\frac{5 \mathrm{V}}{4l}=5 \mathrm{f}_{3}\) …………(3)
from (1)
From (1), (2) and (3)
f₁: f₂: f₃ = 1:2:3
Therefore in the case of an closed pipe, the frequencies of overtones are odd harmonics of the fundamental.

Question 13.
What is an open pipe Show that overtones In an open pipe are harmonics of the fundamental? OR Discuss the modes of vibration of air in an open pipe. OR Show that all harmonics are present in an open pipe.
Answer:
A pipe which is open at both ends is called open pipe.

Consider an open pipe of length l. Let v be the velocity of sound in air. The air column in an open pipe vibrates in such a way that always antinodes are formed at the open ends. Let f₁ f₂ and f₃ be the frequencies and λ₁, λ₂ and λ₃ be the wavelengths of the 1^st, 2^nd and 3^rd modes of vibration respectively. For first mode (fundamental mode).

From (1), (2) and (3)
f₁: f₂: f₃ = 1:2:3
Therefore in the case of an open pipe, the frequencies of overtones are simple harmonics of the fundamental.

Question 14.
Derive an expression for fundamental frequency In the case of stretched string.
Answer:
In the fundamental mode of vibration of the string, there will be an antinode in between the two nodes a the fixed points.

If l is the length of the string then

Velocity of the-wave along the string is

where T is the tension and m is the mass per unit length (linear density) of the string. Fundamental frequency of vibration of the string is,

Question 15.
Given below are some examples of wave motion. State In each case If the motion is transverse, longitudinal or a combination of both.

1. Motion of a kink In a long coil spring produced by displacing one end of the spring sideways.
2. Wave produced in a cylinder containing water by moving its piston back and forth.
3. Wave produced by a motorboat sailing in water.
4. Ultrasonic waves In air produced by a vibrating crystal.

Answer:

1. Longitudinal wave
2. Transverse wave
3. Combination of both
4. Longitudinal wave

Question 16.
What do you mean by wave motion? Discuss Its four important characteristics.
Answer:
Wave motion is a motion where the energy is transferred without shifting the material particles.
Four characteristics:

1. It is a Simple Harmonic Motion.
2. Energy is transported without material shift.
3. The velocity of waves depends on the medium (only for longitudinal waves).
4. The particles oscillate in SHM.

Question 17.
A simple harmonic wave Is ex-pressed by the equation

y and x are In centimetre and t in seconds. Calculate the following:
i) amplitude
ii) frequency
iii) wavelength
iv) wave velocity
v) phase difference between two particles separated by 17.0 cm
Answer:


= \(\frac{2 \pi}{5}\) rad.

1st PUC Physics Waves numerical problems Questions and Answers

Question 1.
A transverse wave is represented by y = 5 sin (50 πt – πx). Find the wavelength and velocity of the wave.
Solution:
Given equation is,
y = 5 sin (50 πt – πx)
Comparing with the standard transverse wave equation,

∴ Velocity of the wave, u = 50 m/s

Question 2.
A wave travelling at a speed of 200 m/s has a frequency of 1000 Hz. Its amplitude is 2 units. Write down the wave equation.
Solution:
Standard wave equation is y = a sin 2πt \(\left[\frac{t}{T}-\frac{x}{\lambda}\right]\)
Given a = 2 units.
Frequency = 1/T = 1000 Hz
Velocity, u= 200 m/s

Therefore the wave equation is

Question 3.
The distance between two particles on a string is 10 cm. Find the phase difference between these particles if the frequency of the wave is 400 Hz and speed Is 100m/s.
Solution :
If the distance between two points is ∆x, the phase difference is given by
\(\Delta \phi=\frac{2 \pi}{\lambda} \Delta x\)
Here, Wavelength \(\lambda=\frac{v}{f}=\frac{100}{400}=0.25 \mathrm{m}\)
path difference \(\Delta x=10 \mathrm{cm}=0.1 \mathrm{m}\)
∴ Phase difference \(\Delta \phi=\frac{2 \pi}{0.25} \times 0.1\)
= 0.8π radians
=144°.

Question 4.
A sinusoidal wave propagating through air has a frequency of 200Hz. If the wave speed is 300m/s, how far apart are two points in the medium with a phase difference of 45° and 60°?
Solution:
v = 300m/s, f = 200Hz, Φ₁=45°, Φ₂=60°
v = f λ
∴ wavelength \(\lambda=\frac{\mathrm{v}}{\mathrm{f}}=\frac{300}{200}=1.5 \mathrm{m}\)
path difference = \(\frac{\lambda}{2 \pi}\) phase difference

∴ Distance between the points

Question 5.
A wave traveling along a string is described by y(x,t) = 0.00327 Sin(72x-2.72t) in which the numerical constants are in SI units (i.e.,\ 0.00327m, 72.1 rad/m and 2.72 rad/s.) Find the amplitude, wavelength, period and speed of the wave.
Solution:
The equation for a wave travelling along the +ve x direction is
y(x, t) = a sin(kx – ωt)-(1)
The given equation is
y(x, t) = 0.00327sin(72.1 x -2.72t) – (2)
comparing equation (1) and equation (2)
1.  Amplitude a = 0.00327 m

2.  Wavelength:
We have k =72.1 rad m^-1
ω = 2.72 rad s^-1

3.  period, T = \(\frac{2 \pi}{\omega}\)

\(=\frac{2 \pi}{2.72} \quad=2.31 \mathrm{sec}\)

4.  Frequency f = \(\frac{1}{T}\)
\(\frac{1}{2.31}=0.4329 \mathrm{Hz}\)

5.  Speed of the wave: from the equation
v = fλ
= 0.4329×0.0872
= 0.0377ms^-1.

Question 6.
A sinusoidal wave propagating through air has a frequency of 150HZ. If the wave speed is 200 ms^-1 how far apart are two points in the medium which have a phase difference of 45° and 150°?
Solution:
f =150Hz, v = 200ms^-1 φ₁ =45° φ₂=150°
v= fλ
wave length \(\lambda=\frac{v}{f}=\frac{200}{150}=1.333 \mathrm{m}\)
path difference =\(\frac{\lambda}{2 \pi}\) phase difference

∴ Distance between the points is

Question 7.
A wave travelling along a string is represented by the equation,
y = 0.08 Sin (5t – 3x) Where x and y are in metre and t is in second.

1. At t = 0.1 sec, find the displacement at x = 0.2m
2. At x s 0.1m, find the displacement at t = 0.4 sec.

Answer:
1.

(∵sin (- θ) = -sin (θ) and sine function is in radian. It is converted into degree by multiplying by 180/3.14)
y =-7.98x 10^-3m.

2. t = 0.4s and x = 0.1m.
y = 0.08 sin (5t – 3x) = 0.08 sin (5 × 0.4 – 3 × 0.1)
y = 0.08 sin (2 – 0.3) = 0.08 sin (1.7)
y = 0.08 sin \(\left(\frac{1.7 \times 180}{3.14}\right)\) = 0.08 sin (97.45°)
y = 0.08 cos (7.45°) = 0.07932 m
(∵sin (90 + θ) = cos (θ)).

Question 8.
A train moving at a speed of 72kmph towards a situation sounding a whistle of frequency 600 Hz. What are the apparent frequency of the whistle as heard by a man on the platform when the train

1. approaches him?
2. recedes from him? (speed of sound In air Is 340 ms^-1).

Answer:
V_s= 72m/hr = 72x 1000m/3600s,
= 20m/s, V = 340m/s, f = 600Hz

1. Apparent frequency when train approaches the observer
\(f^{\prime}=\left(\frac{V}{V-V_{s}}\right) f=\frac{340 \times 600}{340-20}=637.5 \mathrm{Hz}\)

2. Apparent frequency when train recedes from observer
\(\mathrm{f}^{\prime}=\left(\frac{\mathrm{V}}{\mathrm{V}+\mathrm{V}_{\mathrm{s}}}\right) \mathrm{f}=\frac{340 \times 600}{340+20}=566.7 \mathrm{Hz}\)

Question 9.
Find the ratio of velocity of sound in oxygen and velocity of sound In hydrogen at S.T.P. Given the molecular weight of Oxygen is 32 and that of Hydrogen is 2.
Solution :
Let v₀ and v_H be the velocities of sound in oxygen and hydrogen respectively
Then

Where Po and pH are the densities of oxygen and Hydrogen respectively. X is the ratio of specific heats which is same for hydrogen and oxygen. But density is directly proportional to the molecular weight.

Question 10.
At what temperature will the velocity of sound in air reduces to half of its value at 0° C?
Solution :

Question 11.
Two tuning forks X and Y sounded together produce 10 beats per second. When Y is slightly loaded with wax, the number of beats reduces to 6 per second. If the frequency of X is 480 Hz, find that of Y.
Solution :
Frequency of x = 480 Hz
No. of beats with y = 10
∴ Frequency of y = 480+10 or 480-10
i.e., 490 or 470
On loading y, number of beats = 6
Frequency of y after loading = 486 or 474
∴ The frequency of y before loading can not be 470, because if it were 470 before loading, it must be less than 470 after loading.
∴ Actual frequency of y = 490 Hz.

Question 12.
Two tuning forks A and B gives 4 beats per second. The frequency of A is 510 Hz. When B Is filed 4 beats per second are again produced. Find the frequency of B before and after filing.
Solution:
Frequency of A =510 Hz.
Beats per second = 4
Therefore the frequency of B before filing,
= 510 + 4 or 510 – 4
= 514 or 506
After filing B, 4 beats are produced again the frequency of B after filing
=510 + 4 or 510 – 4
= 514 or 506
The frequency of B before filing can not be 514 Hz because if it is 514 before filing, after filing its frequency must be more than 514.
Therefore the frequency of B before filing must be 506 Hz and after filing it is 514 Hz.

Question 13.
Two sound waves of wavelength 1.34 m and 1.36 m produces 4 beats per second. Calculate the velocity of sound in the medium.
Solution :
Let v be the velocity of sound in the given medium. Then Frequency of the first wave
\(f_{1} \quad=\frac{v}{\lambda_{1}}=\frac{v}{1.34}\)
Frequency of the second wave

Velocity of sound v = \(\frac{4 \times 1.34 \times 1.36}{0.02}\)
= 364.5 ms^-1

Question 14.
A set of 65 tuning forks are arranged In the Increasing order of frequencies such that each gives 3 beats per second with the previous one. Find the frequency of the first tuning fork If the frequency of the last tuning fork Is one Octave above the first one.
Solution :
Let N be the frequency of the first tuning fork. As each tuning fork is giving 3 beats with the preceding one, and they are arranged in the ascending order of frequencies,
Frequency of the second tuning fork = N + 3
Frequency of the third T.F. = N + 6
= N + 3 × 2
Frequency of the fourth T.F = N+9
= N + 3 × 3
Similarly Frequency of the 65th T.F. = N + (65-1 ) × 3
= N + 64 × 3 = N+192 ……………….(1)
But the frequency of the last tuning fork (65th) is one octave above that of the first one.
∴ Frequency of the 65th T.F.= 2N ……………(2)
From (1) and (2) 2N = N + 192
N = 192
∴ Frequency of the first tuning fork = 192 Hz.

Question 15.
A source of ultrasonic wave is emitting ultrasonic waves of frequency 30 kHz. It is placed in a moving car. With what velocity is the car moving If the frequency appears to be 20 kHz to a stationary observer? Velocity of sound in car 340 ms^-1.
Solution:
Here the listener is at rest and the source is moving. As the apparent frequency is lesser than the actual frequency, the source is moving away from the listener.
To find the velocity with the source is moving:
The apparent frequency is given by,
\(\mathrm{f}^{\prime}=\mathrm{f} \frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}} \quad \therefore \frac{\mathrm{f}^{\prime}}{\mathrm{f}}=\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\)
Here, \(f^{\prime}\) is the apparent frequency = 20 kHz
f is the actual frequency = 30 kHz
v is the velocity of sound = 340 ms^-1
On substituting the values, \(\frac{20}{30}=\frac{340}{340+v_{s}}\)
cross multiplying, 2(340 + v_S) = 3 × 340 2 × 340 + 2v_s = 3 × 340
2 . v_s = 3 × 340 – (2 × 340) or 2 . v_s = 340
v_s =\(\frac{340}{2}\) = 170ms^-1
∴ The car is moving with a velocity of 170ms^-1 away from the listener.

Question 16.
An engine moving with a speed of 25 ms^-1 sends out a whistle at a frequency of 280 Hz. Find the apparent frequency of the whistle for a stationary observer

1. when the engine Is approaching him and
2. when it is moving away from him. The velocity of sound is 330 ms^-1.

Solution :

1. When the source is moving towards the observer,
apparent frequency \mathrm{f}^{\prime}=\mathrm{f} \frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}
Here source velocity v_s = 25 ms^-1,
velocity of sound v = 330 ms<sup.-1
frequency f = 280 Hz.
\(\therefore \mathrm{f}^{\prime}=\left(\frac{330}{330-25}\right) \times 280=302.95 \mathrm{Hz}\)

2. When the source is moving away from the observer, apparent frequency is given by,

Question 17.
The apparent frequency of the whistle of an engine changes in the ratio 6:5 as the engine passes a stationary observer. If the velocity of sound is 330ms^-1 what is the velocity of the engine?
Solution:
Let f be frequency of the sound emitted by the engine and vs be its velocity. The apparent frequency f of the whistle as the engine is approaching the observer is, \(\mathrm{f}^{\prime}=f \frac{\mathrm{V}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\)
But v = 330 ms^-1
\(\therefore \mathrm{f}^{\prime}=\mathrm{f} \frac{330}{330-\mathrm{v}_{\mathrm{s}}}\)   ………(1)
Let f” be the apparent frequency of the sound heard when the engine is moving away from the observer. Then
But v = 330 ms^-1

Cross multiplying
6 × (330 – v_s) = 5 × (330 + v_s)
330 × 6 – 6V_s = 5 × 330 + 5 v_s
330 × 6 – 330 × 5 = 5v_s + 6v_s
330 (6 – 5) = 11v_s
330 × 1 = 11 v_s
∴ v_s \(=\frac{330}{11}\) = 30 ms^-1

Question 18.
Two cars approach each other with a common speed of 20ms1. The first car sounds a horn and a passenger in the other car estimates it to be 700 Hz. The speed of sound is 332 ms^-1.

1. Calculate the actual frequency of the horn
2. When the cars move away from each other, what Is the estimated frequency by the same passenger?

Answer:
1. To find true frequency f:

2.

Question 19.
An observer standing by the side of a highway estimates a drop of 20% in the pitch of the horn of a car as it crosses him. If the velocity of sound is 330 m/s, calculate the speed of the car.
Answer:
Given :
v = 330 m/s
Let f₁ and f₂ be the apparent frequencies heard ty the observer, before and after the source crossing respectively.


Solving this we get vs = 36.67 ms^-1

Question 20.
A person standing in front of a mountain at a certain distance beats a drum at regular intervals. The drum¬ming rate is gradually increased, and he finds that the echo is not heard distinctly when the rate becomes 50 per minute. He then moves nearer to the mountain by 100 meters, and finds that the echo is again not heard when the drumming rates become 60 per minute. Calculate:

1. The distance between the mountain and the initial position of the man,
2. Velocity of sound.

Answer:
Let ‘s’ be the distance between the man and the mountain and let ‘v’ be the velocity of sound.
Given :
Drumming rate = 50 per minute.
∴ Interval between successive beats
\(=\frac{60}{50}=1.2 / \mathrm{sec}\)
Time taken by the echo \(=\frac{2 \mathrm{s}}{\mathrm{v}}\)
Given that when the drumming rate is 50 per minute, echo is not heard by the man ⇒ beats overlap in time frame

Similarly when the person moves 100 m
towards the mountain, with drumming rate = 60 per minute

Substituting this in (1) we get.
v = 1000 m/s

Question 21.
An open pipe Is 30 cm long. Find the fundamental frequency of vibration. Which harmonic is excited by a tuning fork of frequency 22 kHz? velocity of sound is 340 ms^-1.
Solution:
The fundamental frequency of an open pipe is

The frequency of the n^th mode of vibration is given by

Thus the 2.2 kHz source will produce 4^th harmonic.

Question 22.
Two open pipes when sounded together produce 10 beats. If the lengths of the pipes are In the ratio of 4:5, calculate their frequencies.
Solution :
Let f₁ and f₂ be the fundamental frequencies of the two pipes and L₁ be their lengths.
Then,

Then,

Question 23.
Two tuning forks A and B gives 6 beats per second. A. is in resonance with a closed pipe of length 15 cm. and B Is In resonance with an open pipe of length 30.5 cm. Calculate the frequencies of A and B.
Solution:
Let f₁ and f2 be the frequencies of the tuning forks A and B respectively.
Then, f₁ – f₂ = 6 ………….(1)
But tuning fork A is in resonance with a closed pipe of length 15 cm
\(\therefore f_{1}=\frac{V}{4 L_{1}}=\frac{v}{4 \times 0.15}=\frac{v}{0.6}\)
Similarly tuning fork B is in resonance with an open pipe of length 30.5 cm

Frequency of the tuning fork A is
\(\mathrm{f}_{1}=\frac{\mathrm{v}}{0.6}=\frac{219.6}{0.6}=336 \mathrm{Hz}\)
Frequency of the tuning fork B is
\(\mathrm{f}_{2}=\frac{v}{0.61}=\frac{219.6}{0.61}=360 \mathrm{Hz}\)

Question 24.
A closed pipe resonates In its fundamental mode of frequency 500 Hz In air. What will be the fundamental frequency if air is replaced by hydrogen at the same temperature? Density of air = 1.20 kg/m³ and density of hydrogen = 0.089 kg/ m³.
Solution:
Let v_a be the velocity of sound in air and v_H be the velocity of sound in hydrogen. Then fundamental frequency of the pipe filled with air is,

If f is the fundamental frequency of the closed pipe, filled with hydrogen then,

Where γ is the ratio of specific heats, P is the pressure and Pa is the density of air.
Similarly, \(v_{H}=\sqrt{\frac{y P}{\rho_{H}}}\)
\(\therefore \quad \frac{v_{\mathrm{a}}}{v_{\mathrm{H}}}=\sqrt{\frac{\rho_{\mathrm{H}}}{\rho_{\mathrm{a}}}}=\sqrt{\frac{0.089}{1.20}}\) =0.2734
On substituting in equation (3),
\(\frac{500}{f}=0.2734\)
\(f=\frac{500}{0.2734}\)
=1836 Hz.

Question 25.
A string vibrates with a frequency of 200 Hz. When its length Is doubled and Its tension is altered it begins to vibrate with a frequency of 300 Hz. What Is the ratio of new tension to the original tension?
Solution:
\(f=\frac{1}{2 L} \sqrt{\frac{T}{m}}\)
Let L be the length of the string and m be the mass per unit length and T₁ be the original tension. Then,
\(200=\frac{1}{2 L} \sqrt{\frac{T_{1}}{m}}\) …………..(1)
In the Second case length = 2L, let T₂ be the tension

Question 26.
Two tuning forks A and B when vibrated together gives 6 beats per second. The tuning fork A is in unison with an air column in a closed pipe 0.15m long vibrating In fundamental mode and tuning fork B is in unison with an air column In an open pipe 0.61 m long vibrating In first overtone. Calculate the frequencies of the tuning forks.
Solution:
Let f and f be the frequencies of tuning forks A and B respectively.

\(\frac{\omega 1}{m} f_{m}-f_{c}=6 \quad \Rightarrow \frac{f_{B}}{f(1)}=6\)
f_B = 360 Hz
∴ f_A = 6+f_B = 366 Hz.

Question 27.
Two tuning forks, when sounded together, produce 5 beats per second. A sonometer wire of length 0.24 m is in unison with one of them. If the length of the wire is increased by 0.01 m, ft is in unison with the other fork. Find the frequencies of the forks.
Solution:
Let f₁ and f₂ be the frequency of the two turning forks.
given f₁ ~ f₂ =5 ……….(1)
l₁ = 0.24m, l₂ = 0.25m.
Frequency of vibration of the wire is given by

Question 28.
A closed organ pipe of length 0.42 m and an open organ pipe, both contain air at 35° C. The frequency of the first overtone of the closed pipe is equal to the fundamental frequency of open pipe. Calculate the length of open pipe and the velocity of sound in air at 0° C. Given that closed pipe is in unison in its fundamental mode with a tuning fork of frequency 210 Hz.
Solution:
Let l₁ be the length of the closed pipe and l₂ be the length of the open pipe. l₁ = 0.42m
Let v_t be the velocity of sound in air at t° c, t = 35° c
Let f be the fundamental frequency of the closed pipe
\(t=v_{1} / 4 / 1\)
Let f be the fundamental frequency of the open pipe
\(f^{\prime}=v_{1} / 2 l_{2}\)
Given that frequency of first overtone of the closed pipe is equal to the fundamental frequency of the open pipe

Also given that closed pipe is in unison with a tuning fork of frequency 210Hz i.e.,
f = 210 Hz
∴ In (1) 210 \(=\frac{v_{t}}{4 \times 0.42}\)
v_t=840×0.42 =352 .8m /s
Velocity of sound in air at 0°C is
\(v_{0}=v_{t} \sqrt{\frac{273}{273+t}}\)
\(=3528 \sqrt{\frac{273}{273+35}}=3322 \mathrm{m} / \mathrm{s}\)

Question 29.
A stretched wire emits a note of fundamental frequency 35 Hz. When the tension Is increased by 0.5 kg. wt., the frequency of the fundamental rises to 40 Hz. Find the initial tension and the length of the wire. (Mass per unit length of the wire = 1.33 × 10^-3 kg/m).
Solution:
Fundamental frequency of vibration in a stretched string is

Question 30.
One end of a horizontal wire is fixed and the other passes over a smooth frictionless pulley and has a heavy body attached to it. The fundamental frequency is 400 Hz. When the body is totally immersed in water, the frequency drops to 350 Hz. Find the density of the body. \(\left[\rho_{\omega}=19 / \mathrm{cm}^{3}\right]\).
Answer:
Given :
f₁= 400 Hz. f₂ = 350 Hz
\(\rho_{\omega}=^{1} 9 / \mathrm{cm}^{3}\)
We know that, for a string under tension T, the frequency of oscillation
\(\mathrm{f}=\frac{\mathrm{n}}{21} \sqrt{\frac{\mathrm{T}}{\mu}}\)
l: length of wire, \(\mu\) : mass per unit length
∴ T₁ = mg   T₂ = mg – B
Now m: \(\mathrm{v} \rho\)   v: volume      ρ: mass density
∴ mg = \(\mathrm{v} \rho\)g     B = \(\mathrm{v} \rho\)_ωg
g = 9.8 m s^-2
Given:

Question 31.
A whistle, emitting a sound of frequency 300 Hz is tied to a string of 2 m length and is rotated with an angular velocity of 15 rad / second in the horizontal phase. Find the range of frequencies heard by the observer stationed at a large distance from the whistle.
Answer:
Given:
radius r = 2 m ;
w = 15 rad s^-1;  v = 330 ms^-1
We know that v_s = r w = 30 m s^-1

The observer will receive maximum frequency when the source is approaching (A) and minimum when its receding (B).

∴ Frequency range = 275 Hz – 330 Hz.

Question 32.
The wavelengths of 2 notes are 7/165^m and 7/167^m. Each note produces 5 beats per second with a 3^rd note of a fixed frequency. Calculate the velocity of sound in air.
Answer:
Given

Question 33.
An open pipe is suddenly closed at one end with the result that the frequency of 3^rd harmonic of closed pipe is found to be higher by 50 Hz than the fundamental frequency of the open pipe. What is the fundamental frequency of open pipe?
Answer:
Let f0 be the fundamental frequency of the pipe of length l. Then,
\(\mathrm{f}_{0}=\frac{\mathrm{v}}{2 l}\) ………..(1)
Let f_e be the 3rd harmonic of closed pipe then,

From (3) and (4) we get f₀ = 100 Hz

Question 34.
Calculate the number of beats heard per second if there are 3 sources of frequencies (n -1), n and (n +1) Hz of equal intensity sounded together.
Answer:
Let us assume each disturbance has an amplitude ‘A’ then the resultant displacement is given by,
y = A sin 2π(n – 1)t + A sin2πnt+ A sin2π(n + 1)t
i.e. y= 2A sin 2πnt cos2πt + A sin2πnt
y= A(1 + 2cos2πt) sin2πnt
∴ Resultant amplitude: A(1 + 2 cos2πt)
Amplitude is maximum when cos2πt = 1
i.e., when 2πt = 2πk k = 0,1,2……….
i.e., when t = 0, 1, 2, 3……….
∴ Time difference between successive maxima = 1 s
Similarly, amplitude is ‘0’ when cos2πt = -1/2
i.e., when cos2πt = 2πk + 2π/3
k = 0, 1, 2………….
i.e. when t = k + 1/3
i.e., when t = 1/3, 4/3, 7/3 …….
Again the time difference between successive minima = 1 s
∴ The frequency of beats is also 1 Hz.
Thus, one beat is heard per second.

Question 35.
Two sound waves originating from the same source, travel along different paths in air and then meet at a point. If the source vibrates at a frequency of 2 kHz and one path is 41.5 cm longer than the other, what will be the nature of interference ? The speed of sound in air is 332 m/s
Answer:
We know that v = fλ
\(\therefore \lambda=\frac{v}{f}\)
Given v = 332 ms^-1
f = 2 k Hz
⇒ \(\lambda=\frac{332}{2 \times 10^{3}}=0.166 \mathrm{m}\)
We know that phase difference (∆Φ)) and path difference (∆x) are related by
\(\Delta \phi=\frac{2 \pi}{\lambda} \quad \Delta x \Rightarrow \Delta \phi=\frac{2 \pi}{0.166} \times 0.415\)
∴ \(\Delta \phi=5 \pi\)
Since phase difference is an odd multiple ot π, the interference is destructive.

Question 36.
A metallic rod of length 2 m is rigidly clamped at its midpoint. Longitudinal stationary waves are set up in the rod in such a way that there are 2 nodes on either side of the midpoint. The amplitude of an antinode is 4 × 10^-6 m. Write an equation of motion of the constituent waves in the rod. (Young’s modulus = 2×10¹¹Nm^-2 and density = 8000 kg m^-3)
Answer:
General equation of a standing wave is y = 2A sink x cos ωt where
\(\mathrm{k}=\frac{2 \pi}{\pi}\) and \(\omega=2 \pi \mathrm{f}\)
Which is obtained by adding 2 identical progressive waves travelling in opposite directions
i.e., y = y₁ + y₂ where
y₁ = A sin (kx – ωt)
y₂ = A sin (kx + ωt)
Let l be the length of the rod: l = 2 m
From the figure below, we see that
\(l=\frac{5 \lambda}{2}\)

We know that velocity of longitudinal wave is given by

∴ Equation of standing wave :
y = 2(4 × 10^-6) sin (2.5 πx) cos (12.5 × 10³π) t
Equation of constituent waves
y₁ = (4 ×10^-6) sin (2.5πx – 12.5 × 10³ πt)
y₂= (4 × 10^-6) sin (2.57πx + 12.5 × 10³ πt)

You can Download Chapter 4 Chemical Bonding and Molecular Structure Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 4 Chemical Bonding and Molecular Structure

1st PUC Chemistry Chemical Bonding and Molecular Structure One Mark Questions and Answers

Question 1.
What is chemical bonding ?
Answer:
Chemical bonding is the attractive force that holds or binds constituents were atoms, ions or molecules together in a substance.

Question 2.
Why do atoms combine ?
Answer:
Atoms have tendency to attain the stable electronic configuration octet structure of the nearest noble gas in the periodic table.

Question 3.
Name the types of chemical bonds.
Answer:
(i) Ionic bond
(ii) Covalent bond
(iii) Co-ordinate bond
(iv) Hydrogen bond
(v) Metallic bond

Question 4.
Noble gas does not form compounds. Why ?
Answer:
Noble gas has stable electronic configuration in the valence shell ns2np6 (octet structure) helium has two electrons only.

Question 5.
What is octet rule ?
Answer:
The tendency of atoms of elements to attain octet structure of the nearest noble gas during formation of a chemical bond is known as octet rule.

Question 6.
What is meant by octet structure ?
Answer:
The atoms having 8 electrons in the outer most orbit or having valence electronic configuration ns2n p6.

Question 7.
Why are rare gases are mono atomic ?
Answer:
Rare gases process octet structure i.e., 8-electrons in the outer most orbit. They have no tendency to react.

Question 8.
Give one molecule that does not obey octet rule.
Answer:
Hydrogen molecule.

Question 9.
Define bond length.
Answer:
It is defined as average distance between the centers of nuclei of the two bonded atoms in a molecule.

Question 10.
Define polarization.
Answer:
The tendency of the anion to get distorted or polarized by the cations is called polarization or polarisability.

Question 11.
Between AICl₃ and AIF₃ which is covalent ?
Answer:
AICl₃.

Question 12.
Name the type of bonds present in ice ?
Answer:
Covalent bonds and intermolecular hydrogen bonds.

Question 13.
What is the type of hybridization exhibited by Boron in Boron trifluoride ?
Answer:
sp_2.

Question 14.
What is H-C-H bond angle in methane ?
Answer:
109° 28′.

Question 15.
How many sigma bonds are present in ethane ?
Answer:

There are 7a bonds in ethane.

Question 16.
What angles are associated with the following orbitals ? sp, sp2 and sp3.
Answer:
sp = 180°
sp² = 120° ‘
sp³ = 109°28′

Question 17.
What is London forces ?
Answer:
It is a weak force of attraction between non-polar molecules like O2 or N2 (i.e, between like atoms).

Question 18.
Mention the type of bond formed between the atoms having same electronegativity.
Answer:
Covalent bond.

Question 19.
Mention the type of bond formed between the atoms having different electronegativity.
Answer:
Ionic bond.

Question 20.
Who received noble price for the work on chemical bonds.
Answer:
Linus pauling. .

Question 21.
What is an ionic bond ?
Answer:
An ionic bond is the electrostatic force of attraction between oppositely charged ions. Or A chemical bond formed by complete transfer of one or more electrons from outermost orbit of electropositive atom to outer most orbit of electronegative atom.

Question 22.
Give an example for ionic crystal ?
Answer:
Sodium chloride crystal, magnesium chloride.

Question 23.
Among HF, HC1, HBr and III which has highest ionic character ?
Answer:
HF

Question 24.
What do you mean by electrovalence ?
Answer:
Number of ions lost (gained) by atoms during the formation of ionic bond.

Question 25.
Give any two factors which favours ionic bond.
Answer:
Low ionization energy for metals and high electron affinity of non-metals.

Question 26.
Define lattice energy.
Answer:
The energy associated when are mole of ionic crystal is formed from its gaseous ions.

Question 27.
What you mean by Born-IIaber cycle ?
Answer:
It is cycle of operation needed for the formation of an ionic crystalline solid from its gaseous ions as element.

Question 28.
What is meant by covalent bond ?
Answer:
A chemical bond formed by the sharing up of one or more electrons between the atoms.

Question 29.
What is covalency ?
Answer:
Number of electrons shared between the atoms during covalent bond formation.

Question 30.
What is the covalency of CH₄ ?
Answer:
4

Question 31.
Who proposed concept of covalent bond ?
Answer:
In 1916, G.N. Lewin

Question 32.
What is the maximum covalency of chlorine ?
Answer:
+7

Question 33.
Who proposed valency bond theory ?
Answer:
In 1927, Heitlen and London.

Question 34.
Draw a diagram to show orbital overlap to form a σ bond.
Answer:

Question 35.
Draw a diagram to show the orbital overlap to form π-bond.
Answer:

Question 36.
Draw a diagram for sigma p-p overlapping.
Answer:

Question 37.
What is a sigma bond ?
Answer:
Sigma bond is a bond formed by the overlapping of atomic orbital along the axis.

Question 38.
What is a pi bond ?
Answer:
A bond formed between two atoms by the sideways overlapping of two half filled p-orbitals is called as a pi bond.

Question 39.
How many sigma and pi bonds are present in an acetylene molecule ?
Answer:
H – C = C – H Four sigma bonds and two pi bonds.

Question 40.
How many pairs of electrons are involved in bond formation in case of nitrogen molecule.
Answer:
Three pairs.

Question 41.
What is hybridization ?
Answer:
It is a process where two or more atomic orbital of comparable energy of valency shell of an atom mix to form the same number of new orbital of same energy.

Question 42.
Mention the overlapping present in
(a) H₂
(b) Cl₂
(c) HCl
Answer:
(a) s – s
(b) p – p
(c) s – p

Question 43.
How many electrons are shared in a double bond between two atoms ?
Answer:
4 (four)

Question 44.
Which type of hybridization is called (a) tetrahedral (b>trigonal ?
Answer:
(a) sp³, b) sp².

Question 45.
Give example for two linear molecule.
Answer:
Ethyne and CO_2.

Question 46.
Who proposed VSEPR theory ?
Answer:
In 1940 by Sidgwick and Powell.

Question 47.
What is the structure of ammonia & water molecule ?
Answer:
Pyramidal and Bent structure respectively.

Question 48.
Mention the band angle in Ammonia & H₂O,
Answer:
107° and 104° respectively.

Question 49.
What is meant by non (homo) polar covalent bond ?
Answer:
In a covalent bond if the pair of electrons shared equally (symmetrically) between the atoms called non-polar covalent bond.

Question 50.
What is meant by hetero (polar) covalent bond ?
Answer:
In a covalent bond if the pair of electrons are shared unequally (asymmetrically) between the atoms are called polar covalent bond.

Question 51.
Give example for polar and non-polar molecule.
Answer:
Polar HCl, H₂O, NH₃, Non-polar H₂, O₂, Cl₂.

Question 52.
Write the relationship between dipole movement and charge.
Answer:
µ = exd (where µ = dipole moment, e = charge d = distance)

Question 53.
Define Hydration energy.
Answer:
The energy released when an ionic solid is dissolved is water resulting is polarization of water molecules to form hydrated ions, called hydration energy.

Question 54.
What is dipole moment?
Answer:
If is the product of the distances of charge (positive or negative) centres and distance between the centre.

Question 55.
Which among the following has zero dipole moment BF₃, µH₃, H₂O ?
Answer:
BF₃

Question 56.
What is coordinate bond ?
Answer:
In a chemical bonding both the electrons for sharing between atoms are contributed (donated) by one atom only.

Question 57.
Give one example for the molecule conducting covalent & coordinate bond.
Answer:

Question 58.
Give one example of a compound containing ionic, covalent and coordinate
Answer:

Question 59.
Which type of bond formed between NH3 and BF3 ?
Answer:
Co-ordinate bond.

Question 60.
Who proposed the concept of co-ordinate bond ?
Answer:
G. A. Perkin in 1921.

Question 61.
What is hydrogen bond ?
Answer:
The attractive force which binds hydrogen atom of one molecule with electronegative atom present is the same or other molecule is called hydrogen bond.

Question 62.
Name two types of hydrogen bonding.
Answer:
Inter molecular and Intra molecular hydrogen bond.

Question 63.
At what temperature water has maximum density ?
Answer:
4°C

Question 64.
How are formic acid molecular are associated ?
Answer:
Formic acid molecules are associated with inter molecular hydrogen bonding

Question 65.
Name the property by which the dissolution of an ionic solid in water is possible.
Answer:
Hydrogen bonding.

Question 66.
What is Vender Waal’s force ?
Answer:
It is a weak inter atomic or inter molecular force of attractions existing in substances without formation of any chemical bond.

1st PUC Chemistry Chemical Bonding and Molecular Structure Give Reasons Questions and Answers

Question 1.
Why “Ionic bond is formed between alkali metals and halogens” ?
Answer:
Because Alkali metals have low first ionization energy while halogens have high electron affinity.

Question 2.
Ionic solids do not conduct electricity.
Answer:
In ionic solids, ions occupy certain fixed positions in the crystal and do not migrate.

Question 3.
Ionic compounds are insoluble in ether.
Answer:
Ionic compounds does not ionize in ether i.e. ether cannot separate the cations and anions of ionic solids.

Question 4.
7π bond is weaker than σ bond.
Answer:
In sigma band the overlapping is maximum where as in pi band the overlapping is minimum-

Question 5.
Dipole moment of CO2 is zero.
Answer:
In the linear molecule O = C = O, the dipole due to oxygen from both sides are equal & opposite. As a result dipole moment is zero.

Question 6.
Water molecule has bent structure even it undergoes sp3 hybridization.
Answer:
Due to the repulsion between the two lone pair of electrons in water molecule, the structure become bent.

Question 7.
Boiling point of water is very high.
Answer:
Due to the presence of intermolecular hydrogen bonding in water.

1st PUC Chemistry Chemical Bonding and Molecular Structure Two Marks Questions and Answers

Give Reason Question:

Question 1.
1. Give any two difference between σ and π bond.
Answer:

Question 2.
Mention the π and π bond present in (1) N₂ (2) O₂ (3) C₂H₄ (4) C₂H₂
Answer:

Question 3.
Write any two conditions for (factors favouring) covalent bond.
Answer:
1) Two atoms have half filled orbitals
2) Shared electrons have opposite spins.
3) Overlapping must be maximum.

Question 4.
Which type of overlapping result in the formation Of (i) sigma bond (ii) pi bond.
Answer:
(i) Sigma bond-overlapping of orbital along the axis.
(ii) pi bond overlapping sideways of two p-orbital.

Question 5.
What is a Bond angle ? Give example.
Answer:
It is defined as the “angle between the oribitals containing bonding electron pairs around the central atom in a molecule or in a complex ion”. It is expressed in degrees. Example : In CO₂ the bond angle is 180°. So CO2 has linear shape.

Question 6.
What is bond energy ? Give example.
Answer:
It is defined as “the amount of energy required to break one mole of bonds of a particular type in the gaseous state.” It is also called bond dissociation enthalpy. Example : C – C bond enthalpy is 348 kJ mol’1.

Question 7.
What is bond order ? Give example.
Answer:
It is the number of covalent bonds holding the atoms in the molecule.
Example : If the bond is formed by the sharing of two electron pairs, then the bond order is 2. O = O or C = C bond in alkenes.

Question 8.
Calculate the bond order of Helium.
Answer:
Helium contains four electrons. Its electronic configuration is H2
Bond order = \(\frac{1}{2}\) (n_b -n_a)
n_b = number of bonding electrons
n_a = number of anti bonding electrons
\(\frac{1}{2}\)(2-2) =0
So, He₂. does not exist.

Question 9.
Explain Van-der-waals force with example.
Answer:
It can be explain with 3 types.
(a) Dipole and Dipole interaction : Dipole – Dipole interaction exists between polar molecules like NH₃, SO₂, HC1… Greater the dipole interaction greater is Vander Waal’s force. Hence such molecules liquefied easily.
(b) Dipole & induced dipole interaction : A permanent dipole molecule can induce polarity in a non-polar molecular when comes near to each other. As a result between polar of non-polar molecule there exists force of attraction.
(c) Induced dipole & Induced dipole interaction. Non polar moleculas like, O₂ and N₂ having induced polarity on each other which is called Landon forces. When these two molecules comes near to each other the force increases.

Question 10.
Explain the term ‘Hydration of Ions”.
Answer:
When ionic solid is dissolved in polar solvent like water, the cations & anions of ionic solid is attracted by oxygen and hydrogen of water respectively. As a result each ions surrounded by many number of water molecule to form hydrated ion. This exothermic process is called Hydration.

* Degree of hydration depends upon the charge density of an ion.
• Degree of hydration decreases as size of ion increases
Example : Li⁺ > Na⁺ > K⁺
• Ionic solid dissolves in water easily, if lattice energy is less than hydration
energy.

Question 11.
Give example for s – p overlapping.
Answer:

Question 12.
Give example for p – p σ overlapping.
Answer:

Question 13.
Why do C show 4 valency ?
Answer:
The ground state electronic configuration of C is 1s²2s22p². In exited state, one electron if 2s jumps to 2p, hence the electronic configuration 1s^{2 2s12p1_x 2p1_y 2p1_z}

Question 14.
Draw the sp³. hybridized C.
Answer:
In sp³ hybridization from 1s²2s¹2p¹_x 2p¹_y 2p¹_z the orbitals 2s²2p¹_x 2p¹_y 2p¹_z hybridization to give sp³ hybridized C as follows.

Question 15.
Draw the sp^{2 hybridized C.
Answer:
In sp2 hybridization, the 2s1,2p1_x,2p1_y orbtials hybridization where as 2p_z is not involves in hybridization.}

Question 16.
Draw the sp hybridized C.
Answer:
In sp hybridized C, the 2s¹ &2p¹_x is hybridizes where as 2p¹_y and 2p¹_z is not.

Question 17.
What do you mean by lone pair of electron ?
Answer:
The pair of electron which involves in hybridization but not involves in covalent bond formation (involves in co-ordinate bond formation).

Question 18.
Mention the condition favouring ionic bond formation.
Answer:

1. Lower Ionisation energy of metals.
2. Higher Electron affinity of non – metal
3. Higher magnitude of lattice energy of bond
4. Higher electromagnativity difference between metal and non-metals.
5. Larger radius of cation of smaller radius of anion.

Question 19.
Mention any two characteristics of ionic compounds.
Answer:

1. They have high Melting point.
2. Many of it are solid in nature.
3. Does not conduct in solid state but good conductors in fused or aqueous state.
4. Soluble in polar solvent & insoluble in non-polar solvent.
5. Ionic bond is non-directional.
6. They do not exhibit Isomerism.

Question 20.
Mention the factors favouring covalent bond.
Answer:
In covalent molecule the combining atoms should have

1. High ionization energy
2. High electron affinity
3. Small or equal electronegative
4. Smaller atomic size
5. Contain unpaired electron with opposition spin
6. Attain octect structure.
7. Smaller intermolecular distance.

Question 21.
Mention the characteristic of covalent molecule.
Answer:

1. They are generally gases or liquid or soft solid
2. Insoluble in polar solvent & soluble in non polar solvent.
3. They do not conduct.
4. They always exists in its molecular state but not in ionic state.
5. They have fixed direction in space.
6. They show isomerism.
7. They do not undergoes electrolysis hence called non-electrolytes.

Question 22.
How do Vander Waal’s force establishes ?
Answer:
Electrons in non-polar molecule oscilates (vibrates). As a result the charges will be v conentrated in a region. Hence non-polar molecule becomes momentarily polarized. This polarized molecule shows induced (artificial) dipole moment which attracts the constituents together – results is Vander Wall’s force.

Question 23.
What do you mean by Co-ordinate Bond ? Explain with example.
Answer:
The bond formed by donation of electron from one of the constituents present in the molecule is called co-ordinate bond.
It is only directional bond & indicated by ” → ”
H₃ → BF₃ [Nitrogen donates pair of electrons to form bond] NH₃BF₃

Question 24.
HC1 form polar bond.
Answer:
In HCl electron is not shared equally, since chlorine is more electronegative than hydrogen and the shared pair shifts more towards chlorine. As a result partially charged H^δ+ – Cl^δ- result polarity.

Question 25.
Ionic solid do not conduct.
Answer:
In it the ions occupy certain fixed lattice positions in the crystals and do not move. Hence they do not conduct.

Question 26.
NH₃ polar but BF₃ non-polar.
Answer:
Ammonia has dipole moment due to pyramidal structure but is BF₃ the dipole moment is zero due to symmetrical planar structure.

Question 27.
HF is liquid where as HCl, HBr…. Other hydrogen halide is liquid.
Answer:
Since F is highly electronegative, name there exists powerful inter molecular hydrogen bonding in HF results in liquid state.

Question 28.
O-Nitrophenol is more volatile than p-Nitrophenol (Or) boling point of p- nitriphenol is greater than O-nitrophenol)
Answer:
In O-nitrophenol there exists intra molecular hydrogen bonding within the molecule where as in p-nitrophenol there exists inter molecular hydrogen bonding between the molecule.

Question 29.
Covalent compounds have law melting point.
Answer:
In covalent compound molecules are held together by weak intermolecular forces.

Question 30.
H₂O is liquid whereas H₂S is gas.
Answer:
In H₂O, hydrogen is combines with highly electronegative element and also forms inter molecular hydrogen bond. But S is less electronegative and in H₂S there is no hydrogen bonding.

Question 31.
C – H bond is polar but CH4 is non polar.
Answer:
Methane has symmetrical tetrahedral structure hence dipole moment is zero.
Therefore it is non polar.

Question 32.
Density of water is maximum at 4°C.
Answer:
Along with covalent bond and hydrogen bond, H₂O molecule exists in best form from tetrahydral from state by leaving vacant space. When heated from 0°C > 4°C the ‘ hydrogen bond brakes and H₂O modecule comes closer. As a result density increases to maximum at 0°C. From 4°C above, due to heat examples takes place by braking the the destroy decreases.

Question 33.
Water is polar in nature.
Answer:
Even though sp³ hybridization is there is H₂O but due to lone pair – lone pair replace
there exists.

Question 34.
Water has high boiling point.
Answer:
Due to intermolecular hydrogen bonding, water has high boiling point.

Question 35.
Ice floats on water or ice is lighter than water.
Answer:
Along with covalent bond & hydrogen bond, ice form of water exists in the tetrahydral shape from bent structure by leaving vacant space. Due to this vacant space is ice if floats in water.

Question 36.
CO₂ and C₂H₂ (acetylene) are non-polar.
Answer:
Because both have linear structure.

Question 37.
σ bond is stronger compare to π bond.
Answer:
In a σ bond the overlapping is along the axis and is maximum extent. Where as π bond overlapping is sidewise and overlapping is minimum.

Question 38.
Ethyl alcohol is an organic compound but still freely soluble in water explain.
Answer:
The solubility of ethyl alcohol in water is due to the presence of intermolecular hydrogen bonding between the two.

Energy is released during the attraction this helps in the dissolution of alcohol in water.

Question 39.
Nitrogen and chlorine have the same electro negativity but only former forms hydrogen bonds. Explain.
Answer:
Both these elements have the same elecro negativity (3.0). But the size of nitrogen . atom (atomic radius = 75pm) is smaller than that of chlorine (atomic radius = 99pm). As a result nitrogen atom can cause greater polarization of N – H bond than the Cl atom in Cl – H bond. Thus hydrogen bonding is present in H – H bonds (e.g., NH₃) but not in a – H bonds (eg. HCl)

1st PUC Chemistry Chemical Bonding and Molecular Structure Three Marks Questions and Answers

Question 1.
Explain the formation of Ionic bond in NaCl.
Answer:
Sodium has electronic configuration 1s² 2s² 2p⁶ 3s¹, when it looses one electron,
attains stability where as chlorine 1s²2s²2p⁶3s²3p⁵, it gain one electron to get stability.
∴ The electron lost by Na is gained by the chlorine forming ionic bond.

Question 2.
Write the Barn Haber cycle for the formation of NaCl.
Answer:

When U = Lattice energy ; ∆H_S – Enthalpy for sublimation, ∆H_i = Enthalpy of ionization, ∆H_d = enthalpy of dissociation, ∆H_Ea = Enthalpy for electron affinity.
The only one step for formation of NaCl from sodium and chlorine is associated with the energy called = ∆H_f = enthalpy of formation.
∆H_f = U + ∆H_S + ∆H_i + ∆H_d + ∆H_Ea
∴ Lattice energy U = ∆H_f= [∆H_S + ∆H_i + ∆_Hd + ∆H_ta ]

Question 3.
Explain the steps involved in Born Haber cycle for the formation of NaCl.
Answer:
Step 1 : Sublimation of metallic sodium of gaseous sodium atom with enthalpy of
sublimation = ∆H_S

Step – 2 : Dissociation of molecule of chlorine to chlorine atoms with enthalpy of dissociation = ∆H_d

Step – 3 : Ionization of gaseous sodium with enthalpy of ionization = ∆H_i

Step – 4 : Addition of electron to gaseous chlorine atom with enthalpy of electron affinity = ∆HEa

Step – 5: Close packing of gaseous sodium ion and chloride ion to form lattice structure of NaCl, with lattice chloride ion to form lattice structure of NaCl, with
Lattice energy = U; NaCl⁺(g) + Cl^–(g) → NaCl∆; ∆H = U

Step – 6 : But sum of all the energies will be equal to the heat of formation of one mole of sodium chloride form its reluctant i.e., ∆Hf
Na(s) + 1/2Cl₂(g) >NaCl(s);∆H = ∆H_f

Question 4.
Define hybridization ? Explain the hybridization in Methane molecule.
Answer:
CH₄ – Methane Molecule.
It is a process where two or more atomic orbital of comparable energy of valency shell of an atom mix to form the same number of new orbital of same energy.

The Molecular formula of Methane is CH4
Electronic configuration of C is ground state – 1s²2s²2p².
Electronic configuration of C in excited state – 1s²2s¹2p³
Valance orbital representation –

The valence orbital contains impaired electron. Hence sp³ hybridized carbon combine with 4 hydrogen atom forms methane molecule.

Question 5.
Explain sp² hybridization in ethane molecule. C₂H₄ – ethene.
Answer:
The molecular formula of ethene is C₂H₄
Electronic configuration of C is ground state – 1s² 2s² 2p²
Electronic configuration of C is excited state – 1s²2s¹2p³
Valence orbital representation

sp² hybridization involves 2 carbon atoms combines with 4 atoms of hydrogen.
In ethane molecule, there is 5 sigma and 1 pi bond. The angle of sp² hybridized ethane is 120°.

Question 6.
Explain sp² hybridization BF₃ molecule.
Answer:
BF₃ molecule
Electronic configuration of B – 1s² 2s² 3P¹
Valance orbital representation

Electronic configuration of F – 1s² 2s² 2P⁵
Valance orbital representation
Only one unpaired electrons is present in fluorine. Hence only 2pz orbital of fluorine involved in bond formation.
Structure – triangular planar.

Question 7.
Explain sp hybridisation with an example.
Answer:
C₂H₂ – ethyne or acetylene
Electronic configuration of C is ground state -1s² 2s² 2P²
Electronic configuration of C excited state – 1s² 2s² 2P³
Valance orbital representation

Sp hybridised, C contains 2 impaired electron and one each in py and pz two sp hybridised C combines with two 1s¹ orbital of H.
Two more hybrid orbitals each carbon overlap with Is atomic orbitals of hydrogen to form sigma bonds. The unhybridised 2pz orbitals of both the carbon atoms now overlap sideways to form carbon-carbon pi-bond.

Question 8.
Explain valence Shell Electron Pair Repulsion (VSEPR) theory.
Ans:
Bonding s The bond formation takes place if there exists impaired electron in the valance shell.

• Shape : The geometry or the shape of a molecule depends on the number of valence shell electrons surroundings the central atom.
• Repulsion : Electron pair tend to repell one another because electron clouds have similar charge.
  Stability s As a result of electron pair repulsion these electron pairs try to stay apart as possible in order to attain minimum energy and maximum stability.
• Repulsive interaction : Lane pair with lane pair electrons are having maximum repulsive interaction, bond pair – bond pair electrons are having minimum repulsive interaction.
• When angle of repulsion decreases angle between the electron pair increases.
  (109°. 28′ less repulsion, 104° more repulsion)

Question 9.
Based on VSEPR theory explain the structure of ammonia.
Answer:
Ammonia (NH₃)
Electronic configuration of N : 1s² 2s² 2p³
Valance orbital representation

1. When 2s² orbital hybridised with 2p_x, 2p_y and 2p_z forms the following sp³ hybridised structure..
2. In 2s² orbital, the electrons are paired up hence they are not involved in bond formation called loan paired of electron (L.P).
3. When sp³ hybridized N combines with 3 hydrogen atom, forms pyramidal structure of ammonia.
Pyramidal structure angle = 107°

Question 10.
Based on VSEPR theory explain the structure of water.
Answer:
Electronic configuration of O: 1s²2s²2p⁴
Valence orbital representation:

1. When 2s² orbital hybridised with 2s_x, 2p_y and 2p^z forms the following sp³ hybridisation structure.
2. In 2s² and 2p_x² , the electrons are paired up hence they are not involved in bond formation called loan paired of electron (L.P.)
3. When sp³ hybridised O combines with 2 hydrogen atom forms Bond structure water.

Question 11.
What are polar and Non Polar covalent bond explain with example.
Answer:
Polar covalent molecules : In diatomic covalent molecule, the shared pair of electrons is displaced towards more electronegative atom present in the molecule is called polar covalent molecule. Example : HC1
Chlorine pulls the shared electron from hydrogen towards itself. Hence it is polar molecule.
H₂O
Oxygen pulls the shared pair of electron from two hydrogen atom surrounding it. Hence it is a polar molecule.
Non Pollar covalent bond : In diatomic covalent molecule, the shared pair of electron is equally distributed between the atoms in the molecule.
Example: H₂, O₂, N₂, O₃ etc.

Question 12.
Define dipole moment ? Comment on structure & dipole moment of CO₂, BF₃
Answer:
It is the product of electric charge and distance between the positive and negative species present in the molecule.

The combining atom should attain octect structure by sharing one or more electron.

Question 13.
Define Hydrogen Bond ? Explain its type with example.
Answer:
Electrostatic force of attraction that exists between hydrogen atom of one molecule and electronegative atom of same or other molecule is called hydrogen bond.
Type of hydrogen bond :

1. Intermolelcular hydrogen bond.
2. Intramolecular hydrogen bond.

1. Intermolecular hydrogen bond : Electrostatic force of attraction that exists between hydrogen atom of one molecule and electronegative atom of another molecule is called intermolecular hydrogen atom.
1. H₂O .

2. Intramolecular Hydrogen bond : Electrostatic force of attraction that exists between hydrogen atom of one molecule electronegative atom of the same molecule. Example : Salicylic acid C₆H₄COOH(OH)

1st PUC Chemistry Chemical Bonding and Molecular Structure Four / Five Marks Questions and Answers

Question 1.
Explain Valence Bond Theory.
Answer:

• The covalent bond is formed by overlapping of those orbital which contains impaired electrons.
• When the overlapping is along axis, the overlapping becomes maximum and strong, called as o sigma bond.
• During overlapping, the electron present in the valance orbit must have opposite spin.
• Covalent molecule have lower energy than combining atoms. Therefore resulting molecule are always stable.
  The extent of overlapping is ss > s-p > p-p.

Question 2.
Salient features of Molecular Orbital Theory (MOT). ‘
Answer:

• The electrons in an atom are found in atomic orbitals, the electrons in a molecule are found in molecular orbitals.
• The molecular orbitals are formed by the combination of atomic orbitals of comparable energies and proper symmetry.
• The BMO has lower energy and hence greater stability than the correspondinb ABMO.
• The molecular orbitals are filled by electrons in accordance with Auibau principle, Pauli’s exclusion principle and the Hund’s rule.

You can Download Chapter 18 Body Fluids and Circulation Questions and Answers, 1st PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Question Bank Chapter 18 Body Fluids and Circulation

1st PUC Biology Body Fluids and Circulation NCERT Text Book Questions and Answers

Question 1.
Name the components of the formed elements in the blood and mention one major function of each of them.
Answer:
Red blood cells (RBC) or Erythrocytes, white blood cells (WBC) or Leucocytes, and platelets are collectively called formed elements and they constitute nearly a 45percent of the blood.
Major Function:

1. RBC: Transport of gases (O₂, and CO₂)
2. WBC: Fight against infection.
3. Platelets: Help in blood clotting.

Question 2.
What is the Importance of plasma proteins?
Answer:
Fibrinogen, globulin, and albumins are the proteins found in plasma. Fibrinogens are needed for clotting blood. Globulins are involved in defence mechanisms of the body and albumins help in osmotic balance.

Question 3.
Match Column I with Column II:
Column I – Column II

(a) Eosinophils – (i) Coagulation
(b) RBC – (ii) Universal Recipient
(c) AB Group – (iii) Resist Infections
(d) Platelets – (iv) Contraction of Heart
(e) Systole – (v) Gas transport
Answer:
(a) – (iii)
(b) – (v)
(c) – (ii)
(d) – (i)
(e) – (iv)

Question 4.
Why do we consider blood as a connective tissue?
Answer:
Blood is considered a connective tissue for two basic reasons.

• Embryologically, it has the same origin as other connective tissue.
• Blood connects the body systems together bringing the needed oxygen, nutrient, hormones and removing the wastes.

Question 5.
What is the difference between lymph and blood?
Answer:

Blood	Lymph
(i) Red in colour (ii) Contains RBC (iii) High protein concentration (iv) Transportation of gases, nutrients, and waste	(i) Colourless liquid (ii) Does not contain RBC (iii) Low protein con­centration (iv) Contains specialized lymphocytes which are responsible for the defense of the body.

Question 6.
What is meant by double circulation? What is its significance?
Answer:
Double circulation: It is the passage of the same blood twice through the heart in order to complete one cycle. One component of the circulation is the passage of deoxygenated blood to lungs for oxygenation. It is called pulmonary circulation. The oxygenated blood comes back to heart for being pumped into various parts of the body for providing oxygen. It is called systemic circulation. The deoxygenated blood comes back to heart for being pumped to lungs again.

Significance:

• Double circulation checks the mixing of oxygenated blood and deoxygenated blood,
• Oxygenated blood carries more oxygen per unit volume. It is, therefore, able to provide more oxygen for metabolism,
• Deoxygenated blood can carry more C02 for removal.

Question 7.
Write the differences between:
(a) Blood and Lymph
(b) Open and Closed system of circulation
(c) Systole and Diastole
(d) P-wave and T-wave
Answer:
(a) Blood is a fluid connective tissue that contains plasma, RBC, WBC, and platelets. Lymph is a tissue fluid formed from the blood. It contains only lymphocytes.

(b) Open circulatory system is present in arthropods and mollusks in which blood pumped by the heart passes through large vessels into open spaces or body cavities called sinuses.

Closed circulatory system is present in annelids and chordates in which blood pumped by the heart is always circulated through a closed network of blood vessels. A closed system can be better regulated than open system.

(c) Systole is the contraction of the heart muscle and diastole is the dilation of the heart muscle. Systole results in increased pressure in heart chambers, whereas diastole results in decreased pressure.

(d) The P – wave represents the electrical excitation (on depolarisation) of the atria, which leads to the contraction of both the arteria. The T – wave represents the return of the ventricles from an excited state to a normal state (repolarisation). The end of T – wave marks the end of systole.

Question 8.
Describe the evolutionary change in the pattern of heart among the vertebrates.
Answer:
All vertebrates possess a muscular chambered heart. Fishes have 2 – chambered heart with an atrium and a ventricle. They have single circulation in which the heart pumps deoxygenated blood, which is oxygenated by gills and supplied to body parts. Amphibians and reptiles have a 3-chambered heart with two atria and a single ventricle.

They possess incomplete double circulation with oxygenated and deoxygenated blood getting mixed up in a single ventricle. Crocodiles, birds and mammals possess a 4 chambered heart with two atria and two ventricles. Two separate circulatory pathways are present in these organisms and hence have double circulation.

Question 9.
Why do we call our heart myogenic?
Answer:
Normal activities of the heart are regulated intrinsically, i.e., auto regulated by specialized muscles (nodal tissue), hence the heart is called myogenic.

Question 10.
The sino-atrial node is called the pacemaker of our heart. Why?
Answer:
In human heart, Sinoatrial node initiates the conduction of the heartbeat. The excitatory wave of the SA node is called cardiac impulse and it determines the rate of heartbeat and sets the pace of the activities of the heart. Hence Sinoatrial node is termed as pacemaker of the heart.

Question 11.
What is the significance of the atrioventricular node and atrioventricular bundle in the functioning of the heart?
Answer:
The atrioventricular node (AVN) is a mass of neuro-muscular tissues and is situated in the wall of the right atrium. The AV node picks up the wave of contraction originated by SAN. Bundle of HIS is a mass of specialised fibres which originates from the AV node. The Bundle of HIS and Purkinje fibres which form an atrioventricular bundle convey impulses of contraction from the AV node to the muscles of the ventricle.

Question 12.
Define a cardiac cycle and the cardiac output.
Answer:
The sequential events that occur from the beginning of one heartbeat to the beginning of the next heartbeat which is cyclically repeated is called the cardiac cycle. The volume of blood pumped by the ventricle per minute is called the cardiac output. It ranges about 5000 ml (5 L) in a healthy individual.

Question 13.
Explain heart sounds.
Answer:
The rhythmic closening and opening of the valves forms the sound of heartbeat. The first sound lub (duration 0.16-0.90 sec) is created by closer of atrioventricular valves. The second sound dub (duration 0.10 sec) is created by the closure of semilunar valves.

Question 14.
Draw a standard ECG and explain the different segments in it.
Answer:

ECG is a graphical representation of the electrical activity of the heart during a cardiac cycle. To obtain a standard ECG (as shown in figure 18.3). a patient is connected to the machine with three electrical leads (one to each wrist and to the left ankle) that continuously monitor the heart activity. For a detailed evaluation of the heart’s function, multiple leads are attached to the chest region. Here, we will talk only about a standard ECG.

(1) Each peak in the; ECG is identified with a letter from P to T that corresponds to a specific electrical activity of the heart.

(2) The P-wave represents the electrical excitation (or depolarization) of the atria, which leads to the contraction of both the atria.

(3) The QRS complex represents the depolarization of the ventricles, which initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole.

(4) The T-wave represents the return of the ventricles from an excited to normal state (repolarisation). The end of the T-wave marks the end of the systole. Obviously, by counting the number of QRS complexes that occur in a given time period, one can determine the heartbeat rate of an individual. Since the ECGs obtained from different individuals have roughly the same shape for a given lead configuration, any deviation from this shape indicates a possible abnormality or disease. Hence it is of great clinical significance.

1st PUC Biology Body Fluids and Circulation Additional Questions and Answers

1st PUC Biology Body Fluids and Circulation One Mark Questions

Question 1.
What do you understand by joint diastole?
Answer:
Joint diastole is a phase in the cardiac cycle during which both atria and ventricles are released simultaneously.

Question 2.
What is fibrinogen?
Answer:
It is a soluble plasma protein which is acted upon by thrombin to form insoluble fibrin clot.

Question 3.
What is Prothrombin?
Answer:
This is an inactive precursor of thrombin formed in the liver.

Question 4.
What is the function of factor VI?
Answer:
It is required for the formation of prothrombin activator by tissue extract.

Question 5.
What are the functions of lymphocytes?
Answer:
Lymphocytes play an important role in cell- mediated immunity

Question 6.
Name the instrument used for measuring blood pressure. (Apr. 1983, Sep. 91, July. 06)
Answer:
Sphygmomanometer.

Question 7.
Expand S.A.N. (April 86)
Answer:
Sino-Atrial Node.

Question 8.
What is pulse? (April 86)
Answer:
The pressure waves created by the heartbeat along the arteries when the left ventricle pumps nearly 70 ml. of blood each systole into the aorta is called as the pulse.

Question 9.
What is blood pressure? (M.Q.P., Oct. 86, 94)
Answer:
Blood pressure is the force with which blood pushes against the wall of the blood vessels (arteries) and is generated by the cardiac output.

Question 10.
What is atherosclerosis?
Answer:
Atherosclerosis is the deposition of lipids (specially cholesterol) on the wall lining of large and medium-sized arteries.

Question 11.
How much is the diastolic pressure of a normal adult human being? (Oct. 87)
Answer:
It is about 80 mm of Hg (mercury)

Question 12.
State the normal blood pressure of man.
Answer:
‘120/80’ (April 89, 91)

Question 13.
What type of heart is found in Man? (April 90, Oct. 93)
Answer:
The heart of man is of the ‘Myogenic type’.

Question 14.
Name the blood vessels from which the heart receives oxygenated blood?
Answer:
Pulmonary Veins. (April 92, 95)

Question 15.
Name the layers of the heart through which heart is covered?
Answer:
Endocardium, myocardium and pericardium.

Question 16.
Mention the meaning and causes of the Myocardial infraction (M.Q.P.)
Answer:
‘Myocardial infarction means death of cardiac muscle in certain locations of the heart due to faulty coronary supply of blood or circulation. It is commonly called ‘Heart Attack’.

It may be caused due to thrombus (or blood clots of different forms) in one of the coronary arteries cutting off blood supply to certain region of the heart (myocardium). Thrombosis may be caused due to hypertension, high cholesterol, levels, smoking & Diabetes mellitus.

Question 17.
Where does the impulse for the heartbeat originate? (Oct. 89)
Answer:
The pacemaker or Sinu-Atrial node (SAN)

Question 18.
Expand SAN and ET. (April 98)
Answer:

• San – Sino Atrial Node (Pace Maker)
• Et – Embryo Transfer.

Question 19.
What is myocardial infarction? (April 00,06)
Answer:
Myocardial infarction/ heart attack is the death of cardiac muscle due to faulty coronary blood supply or circulation.

Question 20.
Name the valve present in between aorta and left ventricle. (Oct. 2000)
Answer:
Bicuspid valve or mitral valve.

Question 21.
What is pericardium? (Oct. 2002)
Answer:
The pericardium is the tough connective tissue sac enclosing the heart and holding it in place.

Question 22.
Name the artery which carries deoxygenated blood. (Oct. 2002)
Answer:
pulmonary trunk.

Question 23.
What are coronary arteries? (Oct. 2003)
Answer:
Coronary arteries are the arteries arising from the aorta and supply blood to the myocardium of the heart.

Question 24.
Give an equation for cardiac output. (M.Q.P., July 2006)
Answer:
Cardiac output is the product of stroke volume and heart rate/minute.
Equation for cardiac output is,
CO = SV x HR i.e = 70 x 72 = 5040ml.
CO is cardiac output, SV = Stroke Volume HR = heart rate (Number of heartbeats/minute).

Question 25.
Write the reason for Cyanosis. (April 2007)
Answer:
Tetralogy of Fallot (interventricular septal defect).

Question 26.
Name the enzyme which converts fibrinogen into fibrin. (March 2008)
Answer:
Thrombin.

Question 27.
Define cardiac output. (July 2008)
Answer:
Cardiac output is the volume of blood ejected out from the ventricles over one minute.

Question 28.
Define stroke volume. (March 2009)
Answer:
Stroke volume is the amount of blood ejected by the ventricle per heartbeat.

Question 29.
Give reason:
Human heart is myogenic. (March 2009)
Answer:
Human heart is myogenic because the heartbeat originates in the heart itself.

Question 30.
What is double circulation? Mention its types. (March 2011)
Answer:
The circulation of blood in which blood passes through the heart twice during one complete circuit is called double circulation. The types are pulmonary and systemic circulation.

Question 31.
Ventricles are thicker than atria. (March 2011)
Answer:
The two ventricles represent the pumping chambers of the heart, hence are thicker than auricles.

Question 32.
Name the type of granulocytes that play an important role in inflammatory reactions.
Answer:
Basophils

Question 33.
Name the type of granulocytes that are significant In allergic reactions and detoxification.
Answer:
Eosinophils.

Question 34.
What transmits the cardiac impulse from the atria to the ventricles? (Delhi 2000 C)
Answer:
Atrio – ventricular bundle.

Question 35.
Which of the four chambers of the human heart has the thickest muscular walls?
Answer:
Left ventricle.

Question 36.
Name one animal in which heart pumps only deoxygenated blood. How many chambers does this heart have? (All India 1998 C)
Answer:
Fish; the heart has two chambers.

Question 37.
What is joint diastole?
Answer:
Joint diastole is the phase in the cardiac cycle during which both atria and ventricles are relaxed simultaneously.

Question 38.
What is the function of the sinoatrial node?
Answer:
The Sinoatrial node generates the action potential and determines the rate of the heart.

Question 39.
What causes the first heart sound? (Foreign 1997)
Answer:
The closure of AV – valves.

Question 40.
Name the two fluids circulated in the body of mammals
Answer:
Blood and lymph

Question 41.
What is blood?
Answer:
Blood is a connective tissue consisting of ‘ a fluid matrix, plasma and formed elements.

Question 42.
Where are RBCs formed from In an adult human?
Answer:
RBCs are formed from the red bone marrow.

Question 43.
Why are erythrocytes red in colour?
Answer:
Erythrocytes contain haemoglobin which gives it a red colour.

Question 44
How many RBCs are present in a mm³ of blood of an adult human?
Answer:
5.0 – 5.5 millions.

Question 45.
Name the leucocytes that are phagocytic.
Answer:
Neutrophils and monocytes.

Question 46.
Name any two substances secreted by basophils.
Answer:
Histamine, serotonin, heparin etc.

Question 47
Name the cells which produce thrombocytes.
Answer:
Mega karyocytes of bone marrow.

Question 48.
Why is blood group ‘O’ called a universal donor?
Answer:
Blood group ‘O’ has no antigen to react with the antibodies of the recipient. So it can be transfused to any person and is called as universal donor.

Question 49.
Why is blood group AB called as universal recipient?
Answer:
Blood group AB does not have any anti-body to react with the antigen of donor, and can accept any blood group. So it is known as universal recipient.

Question 50.
Name the element Involved in blood clotting.
Answer:
Calcium.

Question 51.
Name a reptile that has four-chambered heart.
Answer:
Crocodile.

Question 52.
Name the double-layered membranous covering of the heart.
Answer:
Pericardium.

Question 53.
Where is pericardial fluid present?
Answer:
Pericardial fluid is present between the two layers of the pericardium.

Question 54.
What Is the average number of heartbeats in a man?
Answer:
10 – 75 per minute. (72 is average)

Question 55.
Name the phase of the cardiac cycle in which both atrlo and the ventricles are relaxed simultaneously.
Answer:
Join diastole.

Question 56.
What are Purkinje fibers?
Answer:
The minute fibres throughout the ventricular musculature that arise from the branches of the right and left atrioventricular (AV) bundles on respective sides are called as’ Purkinje fibres.

Question 57.
What is the duration of one cardiac cycle?
Answer:
About 0.8 seconds.

Question 58.
Name two organs affected by high BP.
Answer:
Brain and Kidney.

Question 59.
What is the main symptom of heart failure?
Answer:
Congestion of lungs.

Question 60.
What is the organ heart derived from?
Answer:
Mesoderm.

Question 61.
Why nodal musculature is called auto excitable?
Answer:
The nodal musculature has the ability to generate action potentials without any external stimuli.

Question 62.
What is lymph?
Answer:
Lymph is a colourless liquid present in the lymphatic system containing specialised lymphocytes which are responsible for immune system of the body.

Question 63.
Name two classes with an open circulatory system.
Answer:
Arthropods and molluses.

Question 64.
Who are called Rh-positive and Rh-negative?
Answer:
Individuals having Rh antigen on the surface of their RBCs are called Rh-positive and the one who do not contain these antigens are called Rh-negative.

Question 65.
What separates the atrium and the ventricle of the same side?
Answer:
Atrio – ventricular septum.

1st PUC Biology Body Fluids and Circulation Two Marks Questions

Question 1.
Define Pacemaker. Give an example. (April 93)
Answer:
The pacemaker is a compact mass of muscle fibers which initiates each cardiac cycle, thereby setting or establishing the rate of beating-(pace) of the heart.
Eg: The sinoatrial (Sinuatrial) node situated inferior to the opening of the superior vena cava.

Question 2.
Define Hypotension and Hypertension. (Oct. 93)
Answer:

• Hypotension or Low blood pressure is a clinical condition wherein the SBP (Systolic Blood Pressure) and DBP (Diastolic Blood Pressure) fall below the normal value, i.e below 100 and 50 mmHg respectively.
• Hypertension or High blood pressure is a clinical condition wherein the DBP rises or elevates above the normal value (i.e. more than 90 mm Hg).

Question 3.
Explain pulmonary circulation. (April 97, 2000, 2007)
Answer:
In the pulmonary circulation (or route) the blood leaves the right ventricle of the heart through the pulmonary artery (trunk), carrying deoxygenated blood to the lungs through its left and right branches. After oxygenation of this blood in the lungs it is carried back into the left atrium or auricle through the left and right pulmonary veins.

Question 4.
Why is the human circulation called complete double circulation. (M.Q.P.)
Answer:
In the human heart the chambers and blood vessels (arteries; veins) comprising the circulatory system are arranged and organised to form the pulmonary and systemic circulations to receive, pump, purify (or oxygenate) and repump blood from the body cells in such way that they not only keep the oxygenated and deoxygenated blood separately in the heart at a time but also help to circulate blood through the heart twice during one complete circuit, ensuring continuous supply of purified or oxygenated blood to the body. This type of circulation is known as a complete double circulation.

Question 5.
Explain briefly the systemic circulation. (April 2003)
Answer:
The left atrium is the receiving chamber for the oxygenated blood brought to it by the four pulmonary veins (two from each lung). The blood leaves the left atrium to fill the left ventricle which pumps the blood into the aorta.

The aorta is the main artery of the systemic circulation. From it, blood is contributed to the other arteries, arterioles and capillaries to different organs of the body. Blood is collected from the organs and poured into the venae cavae which bring blood back to the right atrium. This route constitutes the systemic circulation.

Question 6.
What is cardiac output?
Answer:
It is volume of blood ejected out from the heart (ventricles) over one minute. Cardiac output is the product of the stroke volume and the heart rate/ minute.
CO = SV × HR 70 × 72 = 5040 ml.
The cardiac output is influenced by several factors like heart rate, contractility of the heart etc.

Question 7.
What is cardiac cycle? What is heart rate?
Ans:
The alternate systole and diastole of auricles and ventricles constitute the heartbeat and cardiac cycle. The number of heartbeats/minute is called heart rate.

Question 8.
Mention the artery that carries deoxygenated blood and vein that carries oxygenated blood.
Answer:
Pulmonary artery Pulmonary vein.

Question 9.
Give a schematic representation of Pulmonary circulation.
Answer:

Question 10.
Can all the four chambers of the human heart experience systole simultaneously? Justify your answer.
Answer:
No, all chambers cannot experience systole simultaneously.
Arteriole systole occurs first as it is directly connected to SA-node which initiates the cardiac impulse. This impulse is then discharged to atria and transmitted from atrial muscles to the ventricular muscles through atrioventricular node and Av – bundle. The impulse passes slowly across the AV-node and hence ventricular systole begins after the atrial systole is completed.

Question 11.
Name the different types of granulocytes. Give the function of the one which constitutes maximum percentage of the total leucocytes. (Delhi 2004)
Answer:

• Neutrophils, eosinophils and basophils are three types of granulocytes.
• Neutrophils constitute 60 – 65 % of the total leucocytes and are phagocytic cells which destroy foreign organisms entering the body.

Question 12.
Write a note on hepatic portal system.
Answer:
Hepatic portal system is a unique vascular connection that exists between the digestive tract and
the liver.

The hepatic portal vein carries blood from intestine to the liver before it is delivered to the systemic circulation.

Question 13.
Name In a sequence the specialised cardiac muscle fibres responsible for conduction of heartbeat. Also mention their location in the heart. (All India 1999 C)
Answer:
Heart conducting system contains Sino-atrial node, Atrio-ventricular node, AV-bundle and Purkinje fibres.

• SA node is situated in the upper lateral wall of the right atrium.
• AV node is located in the posterior part of the interatrial septum.
• The AV bundle branches into two and enter left and right ventricles.
• Purkinje fibres are located throughout the ventricular wall.

Question 14.
Name the events of one complete heartbeat in the proper sequence.
Answer:
Joint diastole → atrial systole → ventricular systole and atrial diastole → ventricular diastole.

Question 15.
Differentiate between RBCs and WBCs.
Answer:

RBC	WBC
(a)  They do not have nucleus at maturity. (b)  They possess hae­moglobin and hence red. (c) They help in transport of respiratory gases. (d) They are about 5 mil­lion / mm3 of blood.	(a)  They contain a large characteristic nucleus. (b)  They are colourless and contain no pig­ment. (c) They help in defence mechanism. (d) They are about 7000 / mm3 of blood.

Question 16.
Differentiate between open and closed system of circulation
Answer:

Open circulating system	Closed circulating system
(a) Blood flows through open spaces and channels.	(a) Blood flows through heart and closed blood vessels.
(b) Sufficiently high blood pressure cannot be maintained	(b) Sufficiently high blood pressure can be maintained
(c) The volume of blood flowing to different tis­sues cannot be regu­lated.	(c) The volume of blood can be regu­lated.
(d) Blood flows at a slow velocity.	(d) Blood flows at a higher velocity.
(e) Body tissues are in di­rect contact with blood.	(e) Body tissues do not come in con­tact with blood.

Question 17.
Where are bicuspid and tricuspid valves located in humans? What is their function?
Answer:

• The bicuspid valve guards the opening in the atrioventricular septum between the left atrium and left ventricle.
• The tricuspid valve guards the opening in the atrioventricular septum between the right atrium and right ventricle. They allow the flow of blood in one direction.

Question 18.
Define stroke volume. What is its value?
Answer:
The volume of blood pumped by a each ventricle during one cycle is called stroke volume. It is about 70 ml.

Question 19.
Name the four types of blood groups. What is the basis for such grouping?
Answer:
The four types of blood groups are A, B, AB and O. Blood grouping is based on the presence or absence of two surface antigens on the RBCs namely A and B.

1st PUC Biology Body Fluids and Circulation Three Marks Questions

Question 1.
Draw a schematic representing double circulation in humans.
Answer:

Question 2.
Where and from which cells do platelets originate? What Is their life span? How do they act when blood vessels get injured?
Answer:
Platelets originate from the megakaryocytes in the bone marrow. They live for about seven days. They release thromboplastins, which help convert prothrombin of the plasma into thrombin and thus help in clotting of blood.

Question 3.
What kind of circulatory system do the mollusks have? List the characteristics of such a system.
Answer:
Molluscs have an open circulatory system.

• In this type, the blood flows through open spaces (lacunae) and channels (sinuses) and not confined to closed blood vessels.
• Sufficiently high blood pressure cannot be developed in lacunae and sinuses and so blood flows at a very slow velocity.
• Blood directly comes in contact with the body tissues.
• The volume of blood flowing to different tissues and organs cannot be regulated according to the need.

Question 4.
Mention the functions of each of the following :
(a) Basophils
(b) Eosinophils
(c) Monocytes
Answer:
(a)Basophils: They secrete histamine, serotonin, heparin etc. and are involved in inflammatory reactions.
(b) Eosinophils: They resist infections and are also associated with allergic reactions.
(c) Monocytes: They are phagocytic which destray foreign organisms entering the body.

Question 5.
Why Is it necessary to check the Rh-factor of the blood of a pregnant woman?
Answer:
A Rh-negative person, if exposed to Rh +ve blood, will form specific antibodies against the Rh antigens. A special case of Rh incompatibility has been observed between the Rh -ve blood of a pregnant mother with Rh +ve blood of the foetus. Rh antigens of the foetus do not get exposed to the Rh -ve blood of the mother in the first pregnancy as the two bloods are well separated by the placenta.

During the delivery of the first child, there is a possibility of exposure of the maternal blood to small amounts of the Rh +ve blood from the foetus. In such cases, the mother starts preparing antibodies against Rh antigen in her blood. In case of her subsequent pregnancies, the Rh antibodies from the mother (Rh -ve) can leak into the blood of the foetus (Rh +ve) and destroy the foetal RBCs.

This could be fatal to the foetus or could cause severe anaemia and jaundice to the baby. This condition is called as erythroblastosis foetulis. Hence it is necessary to check the Rh factor of the blood of a pregnant woman.

Question 6.
How is cardiac activity regulated?
Answer:
Normal activities of the heart are regulated intrinsically by specialised muscles, and hence the heart is called myogenic. A special neural centre in the medulla oblongata can moderate cardiac function through Autonomic Nervous System. (ANS).

Neural signals through the sympathetic nerves can increase the rate of heartbeat, the strength of ventricular contraction and thereby the cardiac output. Para sympathetic neural signals decrease the rate of heart beat, speed of conduction of action potential and thereby the cardiac output. Adrenal medullary hormones can also increase the cardiac out-put.

Question 7.
What do you mean by myogenic and neurogenic heart?
Answer:

• The myogenic heart is the one that generates its own electrochemical impulse with the help of special muscles called myogenic muscles, e.g., mollusks, chordates.
• The neurogenic heart is the one where the electrochemical impulse for its contraction originates from a nerve ganglion or mass of nerve cells present nearby, e.g., most arthropods, annelids.

1st PUC Biology Body Fluids and Circulation Five Marks Questions

Question 1.
Explain the mechanism of Blood clotting. (Mar. 09)
Answer:
Calcium salts, cell injury which liberates thrombokinase, (thromboplastin) thrombin and fibrin.
Prothrombin + Ca + thrombokinase →
Thrombin.
Thrombin + fibrinogen → Fibrin
Fibrin + blood cells → Clot (Fibrin clot)
Clotting consists 3 phases.

Stage I: Thromboplastin or thrombokinase is liberated from injured tissue or shed blood.
Stage II: The thromboplastin in the presence of ionic calcium acts on prothrombin, a protein present in the blood plasma and coverts it to thrombin.
Stage III: The thrombin in turn acts on fibrinogen, the soluble protein present in plasma and converts it to fibrin. It is insoluble and precipitates as a network of elongated thread-like fibres enmeshing cellular and liquid components in the meshwork and giving it a solid appearance. Subsequently, the fibres shrink and the mesh becomes closely knit. It holds the cellular elements, but exuding out the liquid. The liquid is the serum.

Question 2.
What is conducting system? Explain its role in the functioning of the heart. (Oct. 83, 96, April 90, 94)
Answer:
Blood is pumped by the heart by rhythmic contractions and relaxation (heartbeat) of the auricles and ventricles in a definite order.

The conducting system brings about the rhythmic and automatic working of the heart (i.e. contraction and relaxation of the heart) through the transmission of impulses generated within specialised cardiac muscles and its channels, thereby accomplishing continuous circulation (or pumping) of blood through the heart.

The impulses for initiation of the heartbeat starts in the SAN (Sino-atrial node) or Pacemaker of the heart and is transmitted to the AV node (Atrioventricular node) through 3 special tracts or pathways (composed of specialised cardiac muscle fibres) called internodal pathways. From here the stimulus is carried by a single tract of specialised muscle called the “Bundle of His” in the beginning and diverted to the left and right side of the heart by its branches, (i.e. left and right branches).

These branches later branch profusely into Purkinje fibres. The Purkinje’s fibres transmit the Impulses received from the branches of Bundle of His to all the parts of the Ventricles bringing about the simultaneous contraction of the Ventricle pumping blood from right ventricle into the lungs through the pulmonary artery and pumping blood from left ventricle into the aorta to various parts of the body through its branches. In this way impulses are transmitted through the conducting system ensuring systematic working of heart and blood supply.

Question 3.
Explain the structure of the mammalian heart with a neat labelled diagram.
(Oct. 83, 86, 91,96, 97, April 85, 06, 07)
Answer:
The heart of Mammals is somewhat fist-shaped, broader at the anterior end and triangular, narrow & pointed at the posterior end marked by an apex at its tip.

The heart is made up of cardiac muscle and protected on its outer side with a double membrane, the pericardium, with its pericardial fluid. Internally the heart is 4 chambered with a left

half consisting of the left auricle (receiving chamber) followed by the left ventricle (pumping chamber) and the right half consists of the right auricle (receiving chamber) and the right ventricle (pumping chamber). The left half is separated completely from the right half with the auriculoventricular septum.

The auricles or atrium (the receiving chambers) of the heart are anterior in position followed by the ventricles which posterior in position. The walls of the auricles are thinner than the ventricles and the walls of the ventricles bear special papillary muscles and trabeculae holding the valves present between the auricles and ventricles. The chambers are lined by simple squamous epithelium or endocardium. The auricles and ventricles are separated and guarded by valves allowing the flow of blood only in one direction due to their mode of arrangement.

The right auriculo-ventricular value is a tricuspid (3 flapped) valve opening towards the right ventricle and that between the left auricle and ventricle is the Bicuspid (2 flapped) or mitral valve opening towards the left ventricle.

The left side of the heart receives oxygenated blood through a set of blood vessels and pumps blood through a set of blood vessels towards the body. The right side of the heart receives impure (deoxygenated) blood from the body (cells) and pumps blood to lungs for purification through another set of blood vessels.

They are as follows;
(i) Superior and inferior vena cava:
Bring blood collected (deoxygenated) from various (upper NS and lower) parts of the body into the right
auricle.

(ii) Pulmonary artery and its branches: The pulmonary artery carries deoxygenated blood pumped by the right ventricle to the lungs for purification of blood in terms of respiratory gases (removal CO₂ & addition of O₂) channelized to left & right lobes of the lung with the help of its branches. The pulmonary artery is guarded by semilunar valves.

(iii) The pulmonary veins: Collect the purified blood from the left & right lobes of the lung and carry it into the left auricle.

(iv) Aorta: It is the largest vessel of the heart guarded by semilunar valves.
It transports blood (oxygenated) pumped by the left ventricle (received from the left auricle) into the body through its branches (to the upper and lower parts of the body).

Question 4.
What is double circulation? Describe with reference to a human heart. (April 87, 91, 99, Oct. 95, July 07)
Answer:
The circulation of blood in which blood passes through the heart twice during one complete circuit is called as double circulation.

The heart of man is four-chambered. The left side of the heart is made up of left atrium or auricle (receiving chamber anterior region) and the left ventricle (pumping chamber posterior region). These two chambers are guarded by a bicuspid valve, opening towards the ventricle. The right half of the heart is made up of right atrium or auricle (on the anterior region) and right ventricle (on posterior region) these two chambers are guarded by the tricuspid valves opening towards ventricle. The valves ensure unidirectional flow of blood and prevent its backflow. The right half of the heart receives deoxygenated blood from body

and the left half receives oxygenated blood from the lungs. The two halves are separated by the atrioventricular septum ensuring absolutely no mixing up of the blood between these two halves.

The oxygenated blood leaves the left ventricle through the aorta (the great vessel) and reaches various parts of the body through its branches, arteries and capillaries. After this blood is deoxygenated in cells and takes up CO₂, it is returned back into the right side of the heart by venules and veins forming larger veins, the superior and inferior vena cava into the right auricle.

From here it enters the right ventricle and is pumped into the lungs through the pulmonary artery. (Aorta & pulmonary artery are guarded by semilunar valves). Later the oxygenated blood from lungs is, carried back into the left auricle or atrium by pulmonary- veins. From the left auricle it enters the left ventricle and from here blood is pumped into the aorta for distribution.

Thus blood circulates (without mixing up) continuously in the human heart and passes it twice during one complete circuit keeping the impure or deoxygenated blood separate from the oxygenated or pure blood during the circulation. This is known as double circulation.

Question 5.
Fill in the blanks:

1. ………………. is a plasma protein needed for the coagulation of blood.
2. Granulocyte which is phagocytic ………………..
3. ……………….is the pacemaker of the human heart.
4. Expand ECG: ………………..
5. Sound associated with the closure of the tricuspid and bicuspid valves is ………………..

Answer:

1. Fibrinogen
2. Neutrophils
3. The sinoatrial node (SAN)
4. Electrocardiograph
5. lub.

Question 6.
Draw a neatly labelled structure (section) of a human heart.
Answer:

Question 7.
Match the Following

I	II
(a) Depolarisation of atria (b) Repolarisation (c) Sympathetic nerve (d) Parasympathetic nerve (e) Depolarisation of ventricles.	(i) Increase cardiac out­put (ii) Decrease cardiac output (iii) P – wave (iv) QRS complex (v) T-wave.

Answer:
(a) → (iii)
(b) → (v)
(c) → (i)
(d) →(ii)
(e) → (iv)

Question 8.
Describe step by step what happens during different phases of the cardiac cycle in human beings.
Answer:
Different phases of the cardiac cycle in a human being are as follow:

• Atrial systole: As soon as atria contract due to contraction wave of SA node, the blood is forced into the ventricle through open bicuspid and tricuspid valves.
• Beginning of ventricular systole: As soon the wave of contraction stimulates the ventricle, bicuspid, and tricuspid valves are closed immediately producing lub sound.
• Complete ventricular systole: With the complete contraction of ventricle the blood flows into the pulmonary trunk and aorta opening semilunar valves.
• Beginning of ventricular diastole: It is marked by the closing of semilunar valves producing second heart sound. The ventricles start relaxation.
• Complete ventricular diastole: As the ventricles are relaxed completely, the bicuspid, tricuspid valves open (due to fall in pressure of ventricles) and blood flow in it from atria.

You can Download Chapter 3 Classification of Elements and Periodicity in Properties Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Chemistry Question Bank Chapter 3 Classification of Elements and Periodicity in Properties

1st PUC Chemistry Classification of Elements and Periodicity in Properties One Mark Questions and Answers

Question 1.
Define atomic radius.
Answer:
Atomic radius is the distance from the center of the nucleus to the point where the electron density is effectively zero.

Question 2.
Define Vander Waal’s radius.
Answer:
Vander Waal’s radius is one half of the distance between the nuclei of two non bonded adjacent atoms belonging to two neighboring molecules of an elements in the solid state.

Question 3.
Define ionic radius.
Answer:
Ionic radius is the distance from the nucleus of an ion to the point up to which the nucleus has influence on its electron cloud.
OR
Ionic radius is the distance from the nucleus of an ion to the outer most orbital containing electrons.

Question 4.
What is ionization energy?
Answer:
Ionisation energy is the amount of energy required to remove the most loosely bound electron from an isolated neutral gaseous atom.

Question 5.
Why ionisation potential of inert gases are comparatively higher?
Answer:
Inert gases have completely filled stable electronic configuration. A lot of energy is required to disturb that stable electronic configuration and remove an electron. Hence inert gases have a high ionsation potential.

Question 6.
Among Na⁺, Ca⁺², Al⁺³ which is having smallest size?
Answer:
Al⁺³.

Question 7.
Which has got the smallest size among Fe, Fe⁺² & Fe³⁺?
Answer:
Fe²⁺ is smallest. ,

Question 8.
The electron affinity of Nitrogen is more than that of oxygen, why?
Answer:
Because Nitrogen contains halfly filled two orbitals, which is an extra stable state.

Question 9.
Alkali metals have low ionization energy why?
Answer:
Alkali metals are present in the periodic table after inert gases. By loosing one electron they gets electronic configuration of inert gas.

Question 10.
What are iso electronic ions?
Answer:
Ions having same number of electrons but differ in the atomic number.

Question 11.
How does ionization energy varies along a period and down the group?
Answer:
Ionisation energy increases along the period and decreases down the group.

Question 12.
Define covalent radius.
Answer:
Covalent radius is one half of the distance between nuclei of two covalently bound atoms of the same element in a molecule.

Question 13.
Name the element having highest electron affinity.
Answer:
Chlorine.

Question 14.
Arrange F, Cl, Br and I in the order of increasing electron affinity.
Answer:
I, Br, F, Cl.

Question 15.
What is meant by electro negativity of an atom?
Answer:
It is tendency of an atom in a molecule to attract the shared pair of electron to itself.

Question 16.
Group the following species that are isoelectronic.
Be²⁺, F^–, Fe²⁺, N^3-, He, S^2- , CO³⁺, Ar
Answer:
(Be²⁺,He); (F^–,N^3-); (Fe³⁺,CO³⁺); (S^2-,Ar)

Question 17.
Which one has the larger size : Fe²⁺ or Fe³⁺ ?
Answer:
Fe²⁺

Question 18.
State the modern periodic law.
Answer:
Properties of elements are periodic functions of their atomic numbers.

Question 19.
Name the element which is most electronegative and the element which is least electronegative in the periodic chart.
Answer:
Fluorine is the most electronegative (EN – 4.0) element
Ceasium is the least electronegative (EN = 0.7) element

Question 20.
Write the general outer electronic configurations of the following elements,
a) alkali metals
b) alkaline earth metals
c) halogens
d) nobel gases
Answer:
(a) alkali metals – ns^-1
(b) alkaline earth metals – ns²
(c) halogens – ns²np⁵
(d) nobel gases – ns²np⁶

Question 21.
What is the decreasing order of shielding effect of orbitals s, p, d and f.
Answer:
Decreasing order : s > p > d > f.

Question 22.
Why do alkali metals have lowest ionization energy?
Answer:
They have largest atomic size, therefore, there is less force of attraction between valence electrons, and nucleus.

Question 23.
Which is smallest among Na⁺, Mg²⁺, Al³⁺, and why?
Answer:
Al³⁺ is smallest because it has highest number of protons (13) among Na⁺, Mg²⁺, and Al³⁺ ions, due to which effective nuclear charge is maximum.

Question 24.
Which has largest ionic radius among Ca²⁺, Mg²⁺, Ba²⁺?
Answer:
Ba²⁺.

Question 25.
Define (i) metallic radius, (ii) van der Waal’s radius.
Answer:
(i) Metallic radius is half the distance between centres of nuclei of two atoms of metal held together by metallic bond.
(ii) Van der Waals’ radius is half of the distance between centres of nuclei of two atoms held by weak van der Waal’s forces of attraction.

Question 26.
How does electronegativity vary (i) down the group, (ii) across the period from left to right?
Answer:
(i) Electronegativity goes on decreasing down the group.
(ii) It goes on increasing along the period from left to right.

Question 27.
What is the nature of oxides formed by most of p-block elements?
Answer:
They form mostly acidic oxides. Some of them form amphoteric and neutral oxides also.

Question 28.
Which of the following pairs of elements would you expect to have lower first ionization energy? (i) Cl or F, (ii) Cl or S, (iii) K or Ar, (iv) Kr or Xe.
Answer:
(i) Cl, (ii) S, (iii) K, (iv) Xe.

1st PUC Chemistry Classification of Elements and Periodicity in Properties Two Marks Questions and Answers

Question 1.
Among C, N, B and O which element has the highest ionization potential and which element has the lowest ionization potential. Give reason.
Answer:
Element having highest ionization potential is N. Due to extra stability of half filled 2p orbital. Element having lowest ionization potential is B. Due to shielding effect of completely filled 2s orbital.

Question 2.
Which is the most electronegative element and the most electropositive element in the modern periodic table?
Answer:
Most electronegative element is fluorine. Most electropositive element is Cesium

Question 3.
Which of the following two elements belong to i) the same group ii) same period in the periodic table 4^Be ; 3^Li; 12^Ms, 35^Br
Answer:

Principal quantum number of outermost electron in Be and Li is 2. Hence both Be and Li belong to the same period i.e., 2^nd period. Be and Mg have the same number of outermost electrons, i.e., 2. Hence both Be and Mg belong to the same group in the periodic table.

Question 4.
What is ionization energy? How does it change in a period as well as in a group?
Answer:
The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom. Ionization energy increases along the period and decreases down the group.

Question 5.
What is electro negativity? How does it change in a period as well as in a group?
Answer:
The ability of an atom to attract the shared electron pair (of a covalent bond) in a molecule towards itself is called electro negativity. In a period from left to right the electronegativity increases. Down a group electronegativity value decreases.

Question 6.
What is electron affinity? How does it vary along the period and group?
Answer:
The energy released when an electron is added to the outer most orbit of an isolated
neutral gaseous atom. It increases along the period and decreases down the group.

Question 7.
To which blocks do the elements with following atomic number belong ? 7, 13, 25, 42
Answer:
At No. Electronic configuration Block of the element

Question 8.
How does covalent radii vary in a period as well as in a group in the periodic table? What is the reason?
Answer:
In a period as we move from left to right covalent radius decreases in a period. As we move’ from left to right atomic number or nuclear charge increases. The pull of the electron cloud by the nucleus increases, the electron cloud shrinks and as a result the covalent radius decreases.

In a group, covalent radius increases from top to bottom. In a group as we move from top to bottom one by one new orbitals are added up thereby the size of the atom goes on increasing. The atomic radius also goes on increasing.

Question 9.
What are s, p, d and f block elements?
Answer:
The elements for which the last electron has entered in s- orbital are called s- block elements, p-block elements are those for which the last electron has entered in the p – orbital, d-block elements are those in which the last electron has entered the d- orbital. The elements in which the last electron entered into the f – orbital of their atoms are called f – block elements.

Question 10.
Give four characteristics of s-block elements.
Answer:

1. They are soft metals
2. They are highly electropositive
3. They are good . reducing agents
4. They have low melting and boiling points
5. They impart specific colour to the flame.

Question 11.
Give four defects of Mendeleev’s periodic table.
Answer:

1. Isotopes should be given separate place because they have different atomic mass and character
2. The increasing order of atomic weight is not maintained
3. Some elements in the same group differ in their properties
4. The position of hydrogen is not justified.

Question 12.
Give two reasons, why the number of elements in first period is only 2?
Answer:
It is because 1^st energy level can have only Is orbital which can have two electrons. When n = 1, then 1 = 0.

Question 13.
On the basis of their electronic configurations, explain why alkali metals are highly reactive.
Answer:
Alkali metals have general electronic configuration ns1. They can lose 1 electron to acquire stable electronic configuration. They have large atomic size, therefore, they can lose electron easily and they are most reactive.

Question 14.
Give the order in which the melting points of halides of sodium decrease and why?
Answer:
NaF > NaCl > NaBr > Nal. Greater the difference in electronegativity more will be ionic character, higher will be melting point due to high lattice energy.

Question 15.
Why are group I elements called alkali metals and group 17 are called halogens?
Answer:
Group I elements are called alkali metals because their hydroxides form soluble bases called alkalies and their ashes are alkaline in nature. Group 17 are called halogens because they are salt-producer.

Question 16.
Give four characteristics of d-block elements.
Answer:

1. They show variable oxidation state
2. They form coloured ions
3. They are used as catalyst
4. They form alloys.

Question 17.
Give any two features of Mendeleev’s periodic table.
Answer:

1. It was based on atomic mass
2. It has places for undiscovered elements.

Question 18.
How do the solubilities of alkaline earth metal sulphate and carbonates vary down the group and why?
Answer:
Solubilities of alkali earth metal sulphates and carbonate decrease down the group because lattice energy dominates over hydration energy.

Question 19.
Why is melting point of LiCl lower than NaCl?
Answer:
LiCl has lower melting point than NaCl because it is covalent whereas NaCl is ionic.

Question 20.
Arrange the following in increasing order: (i) BeCO₃, BaCO₃, CaCO₃, MgCO₃ of Thermal stability; (ii) BeCl₂, BaCl₂, SrCl₂, CaCl₂ Ionic character.
Answer:
(i) BeCO₃ < MgCO₃ < CaCO₃ < BaCO₃. (ii) BeCl₂ < CaCl₂ < SrCl₂ < BaCl₂.

Question 21.
Which alkali metal carbonate is thermally unstable and why?
Answer:
Li₂CO₃ is thermally unstable because it is covalent.

Question 22.
Out of O and S which has higher negative electron gain enthalpy and why?
Answer:
S has high electron gain enthalpy because in oxygen there is more inter-electronic
repulsion than sulphur, therefore, more energy is released in case of sulphur on gaining electrons.

Question 23.
Predict which atom in each of the following pairs has the highest first ionization energy, (a) B and C, (b) N and O, (c) F and Ne.
Answer:
(a) C, (b) N, (c) Ne.

Question 24.
Among the elements Li, K, Ca, S and Kr, which one is expected to have the lowest first ionization enthalpy and which one has the highest first ionization enthalpy? ,
Answer:
K has lowest first ionization enthalpy whereas Kr has highest first ionization enthalpy.

Question 25.
Among thp elements of the third period Na to Ar pick out the element:
(i) with highest first ionization enthalpy, (ii) with largest atomic radius,
(iii) that is most reactive non-metal, (iv) that is most reactive metal.
Answer:
(i) Ar, (ii) Na, (iii) Cl, (iv) Na.

Question 26.
Name a species that will be isoelectronic with each of the following atoms or ions: (i) Ne, (ii) Cl^– (iii) Ca²⁺, (iv) Rb.
Answer:
(i) Na⁺, (ii) Ar, (iii) S²⁺, (iv) Y²⁺.

Question 27.
Arrange the following ions in the order of increasing size: Be³⁺, Cl^–, S^2-, Na⁺, Mg⁺, Br^–.
Answer:
Be²⁺ < Mg²⁺ < Na⁺ < Cl^–< S² < Br^–.

Question 28.
Write the characteristics of p-block elements.
Answer:
(i) p-block elements consists of metals, non-metals and metalloids, (ii) p-block consists of solids, liquids and gases.

Question 29.
Describe the characteristic properties of d and f-block elements.
Answer:
(i) All of them are metals, (ii) All of them are solids except Hg. (iii) They show variable oxidation state (valency), (iv) They are good conductor of heat and electricity, (v) Most of them are meltable and ductile, (vi) They form alloys, (vii) They form coloured ions.

Question 30.
Arrange the following elements in increasing order of metallic character: B, Al, Mg. K.
Answer:
B, Al, Mg, K is increasing order of metallic character. .

Question 31.
Arrange the following elements in increasing order of non-metallic character: B, C, Si, N, F.
Answer:
Si, B, C, N, F is increasing order of non-metallic character.

Question 32.
A, B, C, D and E have the following electronic configuration:
A  = 1s² 2s² 2p¹; B = 1s² 2s² 2p⁵ 3s² 3P¹; C = 1s² 2s² 2p⁶ 3s² 3P³; D = 1s² 2s² 2p⁶ 3s² 3P⁵; E = 1s² 2s² 2p⁶ 3s² 3P⁶4s² Which among these belong to the same group in the periodic table?
Answer:
A and B belong to same group of periodic table because they have same number of valence electrons.

Question 33.
Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements: (a) Silicon and oxygen, (b) Aluminium and bromine, (c)-Calcium and iodine, (d) element with atomic number 114 and fluorine, (e) element with atomic number 120 and oxygen.
Answer:
(a) SiO₂ (b) AlBr₃, (c) CaI₂, (d) UnqF₄, .(e) XO where X is element with atomic number 120.

Question 34.
Amongst the elements B, Al, C and Si: (a) Which has the highest first ionization enthalpy? (b) Which has most negative electron gain enthalpy? (c) Which has the largest atomic radius? (d) Which has the most metallic character?
Answer:
(a) Carbon, (b) Carbon, (c) Al. (d) Al.

Question 35.
Which of the elements Na, Mg, Si and P would have greatest difference between first and second ionization enthalpies?
Answer:
Na has greatest difference between first and second ionization enthalpies because Na+ has stable electronic configuration, i.e., 1s² 2s² 2p⁶, therefore, it has very high second ionization energy.

Question 36.
Discuss significance of atomic number as the basis of classification of elements over mass number.
Answer:
Properties of elements depend upon number of valence electrons which depend upon electronic configuration. Atomic number is needed to write electronic configuration of an element. It shows that atomic number is more important to determine chemical properties of elements than atomic mass.

Question 37.
(a) Why do group I metals have lower ionization enthalpy than corresponding group II metals? (b) Why is an anion larger in size than its neutral atoms?
Answer:
(a) Group I elements are larger in size than alkli earth metals (Group II elements), therefore, there is less force of attraction between nucleus and valence electron, that is why their ionization energy is lower, (b) Anions are larger than neutral atom because electrons are more than protons, therefore, effective nuclear charge is less, therefore, distance between centre of nucleus and valence electrons is more.

Question 38.
Define electro negativity. How does it differ from electron affinity?
Answer:
Electronegativity is defined as measure of tendency to attract shared pair of electrons towards itself in a covalently bound molecule. It has arbitrary value whereas electron affinity has absolute value. Electro negativity is property of covalently bound atom whereas electron affinity is property of isolated atom.

1st PUC Chemistry Classification of Elements and Periodicity in Properties Three Mark Questions and Answers

Question 1.
Explain the features that influence/affect the ionization energy.
Answer:

• Effective Nuclear charge:- If effective nuclear charge increases Ionization energy decreases.
• Atomic Size:- If atomic size increases Ionization energy increases.
• Half or completely filled orbitais:- From Half or completely filled orbitais, starting high lE. . . .
• Orbital of same energy: If orbitais having the same principle quantum number n, .
• Shielding (Screening) Effect:- Inner electrons repel the outer valence electrons. They reduce the nuclear force.

Question 2.
Explain classification of elements into different blocks in the periodic table.
Answer:
(a) s-block elements:- An element in which the outermost (differentiating) electron of its atom belongs to s-orbital of valence shell is called s-block elements. Kept in left hand side of the periodic table, group I & II (or IA & lIA).

(b) p-block elements:- An element in which the outermost (differentiating) electron of its atom belongs to p-orbital of valence shell is called p-block elements. Kept in right hand side of the periodic table, group 13 to 18.
Elements in group 18 are called Aerogens or Noble gases.
General electronic configuration of p-block elements is ns² np^1-6.
Both s and p block together is called representative or normal elements.

(c) d .block elements:- The atom of an element in which outermost (differentiating)
electron enters to d- sub shell of pen ultimate (n-1) shell is called d-block elements.
Kept at middle (between s and p block elements) portion of the periodic table, group 3 to 12.
d) f-block elements: An element in which outermost (differentiating) electron of its atom enters to f-sub shell of anti-penultimate (n-2) shell is called f -block elements. Placed separately at the bottom portion of the periodic table.
General electronic configuration of f-block elements is (n – 2) f^1-14 (n-l)d¹ns².

Question 3.
Arrange the following ions in the order of increasing size? Be²⁺, Cl^–, S^2-, Na⁺, Mg²⁺, Br^–.
Answer:
Arranging the given ions into different groups and periods in order of increasing atomic numbers of their respective elements, we have,

Question 4.
Consider the elements N, P, O, S and arrange them in order of (a) increasing first ionization enthalpy, (b) increasing negative electron gain enthalpy, (c) increasing non-metallic character.
Answer:
Arranging all the given elements into different groups and periods in order of their increasing atomic numbers, we have,
(a) Since Δ_iH₁ decreases down a group, therefore, Δ_iH₁ of N and O are higher than those of P and S. Further since N has more stable exactly half-filled electronic configuration in the 2p-subshell, therefore, it is more difficult to remove out an electron from N than from O even though O has higher nuclear charge. Similarly, P has exactly half-filled electronic configuration in the 3p-subshell. Therefore, Δ_iH₁ of P is higher than that of S. Thus, the overall increasing order of first ionization enthalpy of these elements follows the order: S < P < O < N.

(b) Since adding an electron to smaller size 2p-orbital causes greater repulsion than adding an electron to larger 3p-orbital, therefore, electron gain enthalpies of P and S are more negative than those of N and O respectively. Further since S has higher nuclear charge but P has more stable exactly half-filled electronic configuration in the 3p-subshell, therefore, it is easier to add an electron to S than to P. In other words, electron gain enthalpy of S is more negative than that of P. Similarly, N has exactly half-filled electronic configuration in the 2p-subshell but O has higher nuclear charge.

But the addition of an electron to N causes repulsions to such an extent that electron gain enthalpy of N is actually positive while that of O as expected in negative. Combining all the above results, the increasing order of negative electron gain enthalpy of these elements follows the order: N < P < O < S.

(c) Since non-metallic character decreases down a group but increases along a period, therefore, O is the most non-metallic element while P is the least non-metallic element. The actual order of increasing non-metallic character is : P < S < N < O.

Question 5.
The first ionization energy of carbon atom is greater than that of boron whereas the reverse is true for the second ionization energy. Explain.
Answer:
E.C. of C-atom is 1s² 2s² 2p² and E.C. of B-atom is 1s² 2s² 2p¹. The first electron to be removed in both cases is from a 2p-orbital but nuclear charge of C is more than that of – B. Therefore, the Δ_iH₁ of C is greater than that of B. After the removal of first electron, the second electron to be removed from C-atom is from a 2p-orbital whereas that from B-atom is from a 2s orbital. Since a s-orbital is more penetrating and hence is more strongly attracted by the nucleus then a p-orbital, therefore, Δ_iH₂ of B is higher than that of C.

Question 6.
Arrange the following ions in order of their increasing ionic radii: Li⁺, Mg²⁺, K⁺, Al³⁺.
Answer:
(i) The ionic radius of any cation increases as the number of energy shells increases and decreases as the magnitude of the positive charge increases.

(ii) Mg²⁺ (1s² 2s² sp⁶) and Al³⁺ (1s² 2s² 2p⁶) are isoelectronic ions and each one of these has two energy shells. Since the positive charge on Al3+ is higher than that on Mg²⁺, therefore ionic radius of Al³⁺is lower than that of Mg²⁺.

(iii) Since, K+ (1s² 2s² 2p⁶ 3s² 3p⁶) has three shells and Mg²⁺ and Al³⁺ have two shells each, therefore, ionic radius of K+ is the largest followed by Mg²⁺ and then Al³⁺.

(iv) Now Li⁺ (1s²) has one shell and +1 charge but Al3+ (Is2 2s2 2p6) has two shells and +3 charge. Since the increase in the ionic radius of Al³⁺ due to the presence of two shells is more than counter balanced by the decrease in its size due to an increase in charge from +1 in Li+ to +3 in Al³⁺, therefore, the ionic radius of Al³⁺ is lower than that of Li⁺.
Thus, the ionic radii of these four ions increase in the order : Al³⁺ < Li⁺ < Mg²⁺ < K⁺.

Question 7.
Arrange the elements of second period in order of increasing second ionization enthalpies.
Answer:
The electronic configuration of the ions obtained after removal of first electron from the elements of 2^nd period from left to right are: Li⁺ (1s²), Be⁺ (Is² 2s¹), B⁺ (1s² 2s²), O (1s² 2s² 2p¹), N⁺ (1s² 2s² 2p²), O^– (1s² 2s² 2p³), F (1s² 2s² 2p⁴), Ne⁺ (1s² 2s² 2p³).
The following conclusions can be drawn from the above configurations:

(i) Li+ has noble gas, i.e., He gas configuration, therefore, AJH2 of Li is the highest in the second period.

(ii) Since in B+, the electron has to be removed from a more stable fully filled 28- orbital while in Be+, it has to be lost from the less stable half-filled 2s-orbital and furthermore, the loss of an electron from Be+ gives more stable Be2+ ion with noble gas configuration, therefore, AjH2 of Be is lower than that of B.

(iii) Since more energy is required to remove an s-electron than a p-electron of the same energy level, therefore, more energy is required to remove a 2s-electron from B+ (1s² 2s²) than a 2p-electron from C+ (Is² 2s² 2P¹). In other words, Δ_iH₂ of C is lower than that of B.

(iv) As we move from C to N to O, the nuclear charge increases by one unit at a time, therefore, their Δ_iH₂ also increase accordingly. In other words, Δ_iH₂ of O is higher than that of N which, in turn, is higher than that of C.

(v) In case of O⁺ (1s² 2s² 2p³) an electron is to be lost from an exactly half-filled 2p-orbital but in case of F⁺( 1s² 2s² 2p⁴ ) this is not so. However, loss of an electron from F⁺ gives an exactly half-filled 2p-orbital (i.e., F²⁺ ( 1s² 2s² 2p³), therefore, Δ_iH₂ of F should be lower than that of O.

(vi) Like O⁺ (1s² 2s² 2p³) and F+ ( 1s² 2s² 2p⁴), in case of Ne+ (1s² 2s² 2p⁵) also an electron is to be removed from a 2p-orbital. Since Ne has the highest nuclear charge in 2nd period, Δ_iH₂ of Ne is expected to be much higher than that of O or F.
From the above discussion, it follows that Δ_iH₂ of the elements of 2^nd period increase in the order: Be < C < B < N < F < O < Ne < Li.

Question 8.
Classify the elements having atomic numbers as given below into three separate pairs on the basis of similar chemical properties. Give brief electronic explanation: 9, 12, 16, 34, 53, 56.
Answer:
The second period ends at atomic number 10 while the third period ends at atomic number 18. Therefore, 9, 12 and 16 are the first elements in their respective groups. The atomic numbers of the other elements of the same group can be deduced by adding magic numbers of 8, 18, 18 and 32 to elements of 2^nd period and by adding magic numbers of 18,18 and 32 to the elements of 3^rd period. Thus,
9 + 8 + 18 + 18 = 53
12 + 8 + 18 + 18 = 56
16 + 18 = 34
elements with atomic numbers 9 (F) and 53 (I) belong to halogen family (group 17); elements with atomic numbers 12 (Mg) and 56 (Ba) belong to alkaline earth metals (Group 2) while elements with atomic numbers 16 (S) and 34 (Se) belong to oxygen family (Group 16).

Question 9.
Give the name and atomic number of the inert gas atom in which the total number of d-electrons is equal to the difference in numbers of total p and s-electrons.
Answer:
The first inert gas which contains d-electrons is krypton. Its atomic number is 36 and its electronic configuration is: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶.
Total number of d-electrons = 10
Total number of p-electrons = 6 + 6 + 6 = 18.
Total number of s-electrons = 2 + 2 + 2 + 2 = 8.
∴Difference in total number of p- and s-electrons = 18 – 8 = 10.
Thus, the inert gas is krypton.

Question 10.
Give four characteristics of f-block elements. Why are they called inner transition metals?
Answer:
(a) They form coloured ions, (b) They are paramagnetic in nature, (iii) They form basic , oxides and hydroxides, (iv) They get transmitted in air. They are called inner transition metals because inner 4f-orbital is progressively filled.

Question 11.
Arrange the species in each group in order of increasing ionization energy and give reason: (a) K⁺, Cl^–, Ar, (b) Na, Mg, Al, (c) C, N,).
Answer:
(a) Cl^–< Ar < < K⁺ became nuclear change goes on increasing, (b) Na < Al < Mg because Mg has stable electronic configuration and Na has large atomic size due to which it has lowest ionization energy, (c) C < O < N because N has half filled p-orbital which is more stable whereas carbon is larger in size.

Question 12.
What are the factors that affect electron affinity?
Answer:
(i) Atomic size
(ii) Stability of electronic configuration
(iii) Inter-electronic repulsion
(iv) Screening effect.

Question 13.
Explain the terms (i) screening effect, (ii) penetration effect, (iii) metallic character.
Answer:
(i) Screening effect: The inner electrons between valence electron and nucleus shield’s the valence electron from nucleus; it is called shielding effect.
(ii) Penetration effect: Due to shape of the orbital, s-electron penetrates nearer to the nucleus than p, d or ^electrons and more tightly held.
(iii) Metallic character: Lower the ionization energy, more will be tendency to lose electron, higher will be metallic character.

Question 14.
(a) Explain why the second ionization energy of B is significantly higher
than the second ionization energy of C, even though the first ionization energy of B is less than C. ‘
(b) Which has higher 1st ionization energy B or Be and why?
Answer:
(a) B(5) 1s² 2s² 2p¹ C(6) 1s² 2s² 2p²; B after losing one electron, has completely filled s-orbital from which removal of second electron is more difficult than carbon.
(b) Be has higher first ionization energy due to completely filled valence s-orbital.

Question 15.
Give the reasons of the following: (a) Fluorine has less negative electron gain enthalpy than chlorine
(b) Noble gases tend to be less reactive
(c) First ionization enthalpy of Mg is more than that of Na but second ionizationb enthalpy of Mg is less than that of Na.
Answer:
(a) It is due to more inter electronic repulsion in fluorine atom due to smaller atomic size than chlorine.
(b) It is because they have stable electronic configuration, i.e., their octet is complete except He.
(c) First ionization energy of Mg is more than that of Na because it has smaller atomic size but 2nd ionization of Mg is less because after losing one electron sodium acquires nearest noble gas configuration.

Question 16.
(a) Arrange F, Cl, Br and I in increasing order of electron affinity.
(b) Predict the position of the element with atomic number 26 in the periodic table.
(c) Why I.E. of oxygen is less than that of nitrogen?
Answer:
(a) I < Br < F < Cl is increasing order of electron affinity.
(b) It belongs to group 8 of periodic table because its electronic configuration is [Ar] . 4s² 3d⁶.
(c) It is because nitrogen has stable electronic configuration, i.e., half filled p-orbitals. Therefore, its I.E. is more than that of oxygen.

Question 17.
(a) Account for the following: (i) Mg⁺² ion is smaller than O^-2 ion although both have same electronic structure, (ii) Ionisation enthalpy of nitrogen is more than that of oxygen, (b) Write the IUPAC name and the symbol for the element with at. no. 118.
Answer:
(a) (i) It is due to greater nuclear charge and greater effective nuclear charge in Mg²⁺ and O^2-.
(ii) It is due to stability of electronic configuration of nitrogen. .
(b) Uuo (Ununoctium).

Question 18.
(a) Write the general electronic configuration for f-block elements
(b) Which of the following atoms and ions will have the largest and smallest size? Al, Mg, Al⁺³, Mg⁺².
Answer:
(a) (n-2)f^1-14(n-l)d^0-1ns².
(b) Mg will be largest and Al³⁺ will be smallest.
(c) III set because non-metal cannot lose electrons easily therefore, it will have high first and second ionization energies.

Question 19.
(a) Predict the position of the element in periodic table having valency shell electronic configuration of (n-l)d¹ns²; n = 4. (b) Why noble gases have bigger atomic size than halogens? Why electron gains enthalpy of noble gases are positive?
Answer:
(a) (n -l)d¹ns², with n = 4, 3d¹4s² is Scandium belongs to 4^th period and group 3.
(b) It is because we can measure van der Waal’s radii in noble gases which are bigger than covalent radii. It is because electron has to enter the next higher energy level leading to very unstable electronic configuration.

Question 20.
(a) Predict the position of the element in the periodic table satisfying the electronic configuration (n-l)d¹ns² when n = 4. (b) Name the species which is isoelectronic with Cl^–, (c) Why are f-block elements are placed in a separate row at the bottom of periodic table.
Answer:
(a) It belongs to group 3 and fourth period.
(b) Ar is isoelectronic with Cl⁺.
(c) It is because they resemble each other but do not resemble any other group elements.

Question 21.
Some elements are wrongly placed in the decreasing order of the property mentioned. Rectifying the fault, place them in correct order of the property. Also, furnish reason for the correction done
(a) F > O > N > C (second ionization potential
(b) N > Si > C > P (electronegativity of the elements)
(c) Na > Mg > A1 > Si (First ionization potential).
Answer:
(a) O > F > N > C, second ionization potential of oxygen is highest since electron is to be removed from half filled configuration for rest follow the order of size.

(b) N > C > P > Si, nitrogen has smallest size and so has higher tendency to attract shared pair of electrons.

(c) Si > Al > Mg > Na, silicon has highest first ionization potential due to smallest size, Mg is exception i.e., has higher ionization potential than aluminium because of ns2 configuration which is stable.

Question 22.
Which of the second, third or fourth ionization energy values for calcium shows a sudden increase? Why?
Answer:
3rd ionization energy shows a sudden increase because electron is pulled from 3p orbital which is completely filled orbital instead of 4s.

Question 23.
Identify the element out of the choices for which hints are given:
(a) Hint (This metal is extracted from sea water), choices (Mg, Be, Ca, Sr)
(b) Hint (Material used in solar cells contain the metal); choices (Cs, Si, K, Rb).
(c) Hint (most electropositive element amongst alkaline earth metals); choices (Be, Ba, Ca, Mg).
Answer:
(a) Mg because Mg and …. a are present in large amount in sea water.
(b) Si is used in solar cells.
(c) Ba is most electropositive because electrospositive character increases down the group.

Question 24.
Variation of first ionization enthalpies with Z = 1 to 60.

Study the figure given above and answer the following:
(a) Compounds of Xe are known but other noble gases do not form compounds
(b) Giving reason arrange alkali metals in increasing order of first ionization enthalpies
(c) Estimate the first ionization enthalpy values of Mg and Al.
Answer:
(a) Ionisation energy of Xe is minimum, hence electron can be taken out from it stable noble gas configuration allowing it to make some compounds,
(b) Li > Na > K > Rb > Cs. Ionisation energy decreases as size increases. As size increases, outermost elements are away from nuclear pull, so electron can be easily removed.
(c) Mg = 700 kJ/mol. Al = 600 kJ/mol.

Question 25.
Values of two types of radius of sodium are 186 and 102 pm. Which value indicates metallic and which indicates ionic radius of sodium?
Answer:
186 pm represents metallic radius, 102 pm represents ionic radius. Reason: Ionic radius is smaller due to removal of an electron, cation is formed which has smaller radius than its parent atom.

Question 26.
Fill the arrows (I), (II) and (III) in the following diagram choosing appropriately from the options as electronegativity, atomic radius and non metallic character.

Answer:
I. Electronegativity, II. Non-metallic character, III. Atomic radius.

You can Download Chapter 17 Breathing and Exchange of Gases Questions and Answers, 1st PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Question Bank Chapter 17 Breathing and Exchange of Gases

1st PUC Biology Breathing and Exchange of Gases NCERT Text Book Questions and Answers

Question 1.
Define vital capacity. What is its significance?
Answer:
The maximum volume of air a person can breathe in after forced expiration. This includes ERV, TV, and IRV or the maximum volume of air a person can breathe out after a forced inspiration.

It represents the maximum amount of air one can renew in the respiratory system in single respiration. Thus, greater the vital capacity more is the energy available to the body for doing strenuous work. Vital capacity is higher in athletes and mountain dwellers. Young persons would possess more vital capacity as compared to children or older persons.

Question 2.
State the volume of air remaining in the lungs after normal breathing.
Answer:
Functional Residual capacity.

Question 3.
Diffusion of gases occurs in the alveolar region only and not in the other parts of the respiratory system. Why?
Answer:
Only alveolar region is made up of squamous epithelium which is favorable for the diffusion of gases.

Question 4.
What are the major transport mechanisms for CO₂? Explain.
Answer:
CO₂ is carried by haemoglobin as carbamino haemoglobin. This binding is related to the partial pressure of CO₂. When PO₂ is low as in the tissues and PCO₂ is high, more carbon dioxide binding occurs whereas when PCO₂ is low and PO₂ is high as in the alveoli, dissociation of CO₂from carbamino haemoglobin takes place. RBC contains a high concentration of enzyme carbonic anhydrase that converts carbon dioxide to bicarbonates and vice versa.

CO₂ diffuses the blood and forms bicarbonate ions. Thus CO₂ is trapped as bicarbonate at the tissue level and transported to the alveoli and released out as CO₂. A small amount of CO₂ dissolves in the plasma water and forms acid. On reaching the lungs carbonic acid dissociates and releases carbon dioxide. So carbon dioxide is transported as carbamino-hemoglobin, bicarbonates and carbonic acid.

Question 5.
What will be the PO₂ and PCO₂ In the atmospheric air compared to those in the alveolar air?
1. PO₂ lesser, PCO₂ higher
2. PO₂ higher, PCO₂ lesser
3. PO₂ higher, PCO₂ higher
4. PO₂ lesser, PCO₂ lesser
Answer:
(1) Where there is low pO₂ high pCO₂, high H⁺ concentration and higher temperature exist, the conditions are favorable for dissociation of oxygen from the oxyhemoglobin.

(2) Where there is high pO₂, low pCO₂, lesser H⁺ concentration, and lesser temperature, the factors are all favorable for the formation of oxyhemoglobin whereas in the tissues.

(3) When pCO₂, is high and pO₂. is low as in the tissues, more binding of carbon dioxide occurs whereas.

(4) When the pCO₂, is low and pO₂, is high as in the alveoli, dissociation of CO₂ from carbamino-hemoglobin takes place, i.e., CO₂, which is bound to hemoglobin from the tissues is delivered at the alveoli.

Question 6.
Explain the process of inspiration under normal conditions.
Answer:

1. Inspiration occurs when the pressure within the lungs is less than the atmospheric pressure, i.e., there is a negative pressure in the lungs with respect to atmospheric pressure.
2. Inspiration is initiated by the contraction of diaphragm which increases the volume of the thoracic chamber in the anteroposterior axis.
3. The contraction of external intercostal muscles lifts up the ribs and the sternum causing an increase in the volume of a thoracic chamber in the dorsoventral axis.
4. The overall increase in thoracic volume causes a similar increase in pulmonary volume.
5. An increase in pulmonary volume decreases the intrapulmonary pressure, to less than atmospheric pressure which forces the air from outside to move into the lungs i.e., inspiration.

Question 7.
How is respiration regulated?
Answer:
A specialised centre present in the medulla region of the brain called respiratory rhythm centre is primarily responsible for the regulation of respiration. Another centre present in the pons region of the brain called the pneumatic centre can reduce the duration of inspiration and thereby alter the respiratory rate.

A chemosensitive area situated adjacent to the rhythm centre is highly sensitive to CO₂ and hydrogen ions. Increase in these substances can activate this centre, which in turn can signal the rhythm centre to make necessary adjustments in the respiratory process. Receptors associated with aortic arch and carotid artery also can recognise changes in CO₂ and H⁺ concentration and send necessary signals to the rhythm centre for remedial actions.

Question 8.
What is the effect of PCO₂ on oxygen transport?
Answer:
In the alveoli, where there is high PO₂ low pCO₂ lesser FT concentration, and lesser temperature, the factors are all favorable for the formation of oxyhemoglobin whereas, in the tissues, where low pO₂, high pCO₂, high H⁺ concentration, and higher temperature exist, the conditions are favorable for dissociation of oxygen from the oxyhemoglobin. This clearly indicates that O₂ gets bound to hemoglobin in the lung surface and gets dissociated from the tissues. Every 100ml of oxygenated blood can deliver around 5 ml of O₂, to the tissues under normal physiological conditions.

Question 9.
What happens to the respiratory process in a man going up a hill?
Answer:
When a man is going uphill or doing some strenuous exercise then there is more consumption of oxygen. This decreases the partial pressure of oxygen in haemoglobin resulting in more demand for haemoglobin. As a result, there is an increased breathing rate to fill the gap.

Question 10.
What is the site of gaseous exchange in an insect?
Answer:
Insects have a network of tubes (tracheal tubes) to transport atmospheric air within the body so that the cells can directly exchange the gases.

Question 11.
Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Answer:

1. Oxygen dissociation curve: It is a graphic representation of the relationship between partial pressure of oxygen or pO₂ and percentage saturation of haemoglobin with oxygen.
2. The graph is sigmoid as at low p0₂, there is reduced synthesis of oxyhemoglobin. The percentage of oxyhemoglobin rises with higher pO₂ till at about p0₂ is 100mm Hg, the haemoglobin becomes fully saturated with O₂.
3. A further rise in pO₂ cannot increase the value of oxyhemoglobin as the blood is already saturated with it.

Question 12.
Have you heard about hypoxia? Try to gather information about it, and discuss it with your friends.
Answer:
Hypoxia refers to the shortage of oxygen supply to the body. It is of different types:

1. Anemic hypoxia (deficiency of hemoglobin),
2. Cytotoxic hypoxia (impaired utilization as in cyanide poisoning)
3. Stagnant hypoxia. Due to heart failure or reduced pumping activity of the heart.
4. Hypoxic hypoxia. Insufficient oxygen in the air as at high altitude.
5. CO Poisoning. Carbon monoxide binds to hemoglobin irreversibly. Oxygen transport is correspondingly reduced.

Question 13.
Distinguish between
(a) IRV and ERV
(b) Inspiratory capacity and Expiratory capacity.
(c) Vital capacity and Total lung capacity.
Answer:
(a) Inspiratory Reserve volume (IRV) is an additional volume of air a person can inspire by a forcible inspiration. It averages 2500 ml to 3000 ml. Expiratory Reserve volume (ERV) is an additional volume of air that a person can expire by a forcible expiration. This averages 1000 ml to 1100 ml.

(b) Inspiratory capacity (IC) is the total volume of air a person can inspire after a normal expiration. It is the sum of tidal volume and inspiratory reserve volume.
Expiratory capacity (EC) is the total volume of air a person can expire after a normal inspiration. It is the sum of tidal volume and expiratory reserve volume.

(c) Vital capacity is the maximum volume of air a person can breathe in after forced expiration. It is the sum of Tidal volume, expiratory reserve volume and inspiratory reserve volume. It is also the maximum volume of air a person can breathe out after a forced inspiration.

Total lung capacity is the total volume of air accommodated in the lungs at the end of forced inspiration. It is the sum of Residual volume, expiratory reserve volume, tidal volume, and inspiratory reserve volume.

Question 14.
What Is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.
Answer:
The volume of air inspired/breath during normal respiration. It is approximately 500mL.
The number of breaths per minute 12 to 16.
Tidal volume per minute = 500 x 12 to 16 = 6000 – 8000 mL or 6 -8 litres
Tidal volume per hour = 6 to 8 x 60 = 360 – 480 litres.

1st PUC Biology Breathing and Exchange of Gases Additional Questions and Answers

1st PUC Biology Breathing and Exchange of Gases One Mark Questions

Question 1.
What is an expiration?
Answer:
The movement or exit of air from the alveoli of the lungs to the outer atmosphere is called expiration or exhalation.

Question 2.
What is Ventilation? OR What Is breathing?
Answer:
It is a process by which air is exchanged between the atmosphere and the alveoli of lungs or Entry of atmospheric air into the alveoli of lungs and exit of air from the alveoli into the atmosphere is called pulmonary ventilation or breathing.

Question 3.
What is inhalation or inspiration?
Answer:
Entry of air into alveoli of the lungs is called inspiration or inhalation.

Question 4.
Which muscles do you find In diaphragm?
Answer:
Skeletal muscles.

Question 5.
Which process helps in increasing the size of the thorax during respiration? (Oct. 1990, April 1995)
Answer:
The contraction of the external inter-costal muscles of the ribs and the muscles of the diaphragm.

Question 6.
Name the membrane (covering) of the lungs. (April 91, 93, 94)
Answer:
Pleural membrane (Pleuron). [The plural membrane (outer Fibrous & inner serous)]

Question 7.
What art; the functional units of lungs called? (Oct. 92, 99, July 2010)
Answer:
The Alveoli.

Question 8.
Mention the respiratory pigment.
Answer:
Haemoglobin. (April 98)

Question 9.
What is Spirometer? (April 2002)
Answer:
The apparatus used in measuring the amount of air exchanged during breathing & the rate of ventilation is the Spirometer.

Question 10.
Name the last part of the bronchiole tree.
Answer:
Alveoli. (April 2003)

Question 11.
Which is a common passage for both air and food in man ? (July 2006)
Answer:
Trachea

Question 12.
Name the enzyme that acts on carbonic acid in living cells. (Delhi 2006)
Answer:
Carbonic Anhydrase.

Question 13.
Where is carbonic anhydrase found in human body? Give its function.
Answer:
Carbonic anhydrase is found in RBC. It catalyses the formation of carbonic acid from carbon dioxide and water.

Question 14.
What are the two factors that contribute to the dissociation of oxyhemoglobin in the arterial blood to release molecular oxygen in an active tissue? (Delhi 2000)
Answer:
Low PO₂, high PCO₂, high H⁺ concentration and high temperature.

Question 15.
Name the respiratory organs of
(1) butterfly and
(2) frog larva. (All India 1996)
Answer:

1. Trachea
2. Gills.

Question 16.
What is a soundbox?
Answer:
Larynx is a cartilaginous box which helps in sound production and hence called the sound box.

Question 17.
What are alveoli?
Answer:
Alveoli are number of very thin, irregular walled and vascularised bag like structures into which terminal bronchioles end.

Question 18.
How does diaphragm help in inspiration? (All India 998 C)
Answer:
When the diaphragm muscles contract, it moves down towards abdomen, increasing the volume of thoracic cavity, but decrease in air pressure. So air is drawn into the lungs.

Question 19.
What is the maximum number of molecules of oxygen which one molecule of haemoglobin can carry? (All India 1998 C)
Answer:
Four.

Question 20.
What is formed when CO₂ combines with globin part of reduced haemoglobin? Where does it occur? (All India 1998 C)
Answer:
Carbamino haemoglobin.
Its formation occurs in metabolicaily active tissues.

Question 21.
Which part (s) of the brain control (breathing movements? (Foreign 1997)
Answer:
Medulla and pons.

Question 22.
Name two animals where exchange of gases occurs by diffusion across their entire body surface.
Answer:
Hydra, Sponges, flatworms etc.,

Question 23.
What is the function of pleural fluid?
Answer:
Pleural fluid reduces the friction on the lung surface.

Question 24.
State the function of exchange part.
Answer:
Exchange part is the actual site of diffusion of oxygen and carbon dioxide between blood and atmosphere.

Question 25.
What causes the movement of air in and out of lungs?
Answer:
Pressure gradient between the lungs and the atmosphere.

Question 26.
Name the structures that bring about a pressure gradient between lungs and the atmosphere.
Answer:
Diaphragm and intercostal muscles.

Question 27.
Why does exchange of respiratory gases continue to occur in the lungs even after a maximum expiration?
Answer:
Some amount of air, called residual vol-ume, remains in the lungs even after forceful expiration.

Question 28.
Name the primary site of respiratory gas exchange.
Answer:
Alveoli.

Question 29.
Define partial pressure of a gas?
Answer:
The pressure exerted by an individual gas in a mixture of gases is called partial pressure.

Question 30.
Why can more CO₂ diffuse across the respiratory membrane per unit difference in the pressure as compared to oxygen?
Answer:
Tne solubility of CO₂ is 20-25 times higher than that of O₂. So more CO₂ diffuses across the respiratory membrane.

Question 31.
What is oxyhaemoglobln?
Answer:
Oxyhaemoglobin is a complex formed when oxygen combines with the Fe²⁺ part of haemoglobin.

Question 32.
How much of CO₂ is transported by 100 ml of blood
Answer:
About 4 ml.

Question 33.
What is carbamino haemoglobin?
Answer:
Carbamino haemoglobin is the complex formed when carbon dioxide combines with the amine radical of the globin of haemoglobin.

Question 34.
Where is the respiratory rhythm centre located?
Answer:
Medulla.

Question 35.
How does pneumotaxic centre alter the respiratory rate.
Answer:
Pneumotaxic centre can reduce the duration of inspiration and alter the respiratory rate.

Question 36.
Where are the receptors that can sense the changes in CO₂ and H⁺ concentration located?
Answer:
Receptors are located in the aortic arch and carotid artery.

1st PUC Biology Breathing and Exchange of Gases Two Marks Questions

Question 1.
What is inspiration and expiration.
Answer:

• Inspiration: Entry of air into alveoli of the lungs is called inspiration or inhalation.
• Expiration: Movement or exit of air from the alveoli of the lungs to outer atmosphere is called expiration or exhalation.

Question 2.
Mention the functions of trachea.
Answer:

• It is called wind pipe helps for passage of air.
• Cilia move upwards towards the larynx, and this movement keeps out inhaled particles of dust, pollen etc.
• The cartilages which serve to keep the trachea open makes easy passage of air.

Question 3.
Write any two functions of larynx.
Answer:

• Larynx is called voice box. The vibration of vocal cords produce sound.
• It helps in speech.
• It prevents the entry of food into the lungs.
  The epiglottis present in larynx helps to close off the larynx during swallowing.

Question 4.
Mention any four conducting parts of the human respiratory system. (April 83, 92, 99)
Answer:

• Nasal cavities
• Nasopharynx
• Trachea
• Bronchial tree.

Question 5.
Draw a neat labelled diagram of alveolus.
Answer:

Question 6.
Write a note on pleura.
Answer:
The two lungs are covered by a double layered membrane called pleura, with pleural fluid between them. The fluid reduces the friction on the lung surface. The outer pleural membrane is in close contact with the thoracic lining whereas the inner pleural membrane is in contact with the lung surface.

Question 7.
Mention four functions of the conducting part of the human respiratory system.
Answer:

• It transports the air into alveoli
• It clears the air from foreign particles.
• It moistens and humidifies the air.
• It brings the air to body temperature.

Question 8.
Mention the boundaries of the thoracic cavity. What is the use of them?
Answer:
The thoracic cavity is formed dorsally by the vertebral column, ventrally by the sternum, laterally by the ribs and on the lower side by the dome:shaped diaphragm. The anatomical setup of lungs in thorax is such that any change in the volume of the thoracic cavity will be reflected in the lung (pulmonary) cavity, which is essential for breathing as we cannot directly alter the pulmonary volume.

Question 9.
How do partial pressures of respiratory gases determine the diffusion of oxygen from the blood capillaries into the tissues?
Answer:
In the tissues, PCO₂ is high, PO₂ is low and there is high concentration of H⁺ ions and high temperature. Since the PO₂ is lower than that of the blood, oxyhaemoglobin dissociates and releases the oxygen into the tissues.

Question 10.
How are gases transported in human body?
Answer:
Blood is the medium of transport for O₂ and CO₂. About 97% of O₂ is transported by RBCs in the blood. The remaining 3% of O₂ is carried in a dissolved state through the plasma. Nearly 20-25 percent of CO₂ is transported by RBCs whereas 70% of it is carried as bicarbonate. About 7% is carried in a dissolved state through plasma.

Question 11.
Name the factors that affect the binding of oxygen to haemoglobin.
Answer:
The factors are:-

• Partial pressure of oxygen
• Partial pressure of carbon dioxide
• Hydrogen ion (H⁺ ) concentration
• Temperature.

Question 12.
What is carbonic anhydrase? List the three major forms In which the carbon dioxide is transported in the blood?
Answer:
Carbonic anhydrase is an enzyme that is present in RBC which catalyses the formation of carbonic acid from carbon dioxide and water. CO₂ is transported

• In the dissolved form in the plasma
• As bicarbonates in the plasma and RBC
• As carbamino haemoglobin.

Question 13.
Give the values PO₂ and PCO₂ respectively of each of the following
(1) Atmospheric air
(2) Tissues of the body
Answer:

1. Atmospheric air PO₂= 159 mm Hg; PCO₂ = 0.3 mm Hg;
2. Tissues of the body PO₂= 40 mm Hg; PCO₂ = 40 mm Hg;

Question 14.
What is emphysema? What is its major cause?
Answer:
Emphysema is a chromic disorder in which alveolar walls are damaged due to which respiratory surface is decreased. One of the major causes of this is cigarette smoking.

Question 15.
Where is pneumotaxic centre located in humans? What is its significance inbreathing?
Answer:
Pneumotaxic centre is located in the pons region of the brain.
Neural signal from this centre can reduce the duration of inspiration and thereby alter the respiratory rate.

Question 16.
Name the three layers of diffusion membrane.
Answer:

• Squamous epithelium of alveoli
• Endothelium of alveolar capillaries
• Basement substance.

1st PUC Biology Breathing and Exchange of Gases Three Marks Questions

Question 1.
Explain the transport mechanism of oxygen.
Answer:
Oxygen binds with haemoglobin in reversible manner to form oxyhaemoglobin. About 97% of O₂ is transported as oxyhaemoglobin. Each haemoglobin molecule can carry a maximum of four molecules of O₂. Binding of O₂ with haemoglobin depends on partial pressure of oxygen primarily and also on partial pressure of carbon dioxide, hydrogen ion concentration and temperature.

In the alveoli, where there is high pO₂, low pCO₂, lesser H⁺ concentration and lower temperature, the factors are all favourable for the formation of oxyhaemoglobin, whereas in the tissues, where low PO₂, high PCO₂, high H⁺ concentration and higher temperature exist the conditions are favourable for dissociation of oxygen from the oxyhemoglobin. This indicates that O₂ gets bound to haemoglobin in the lung surface and gets dissociated at the tissues.

Question 2.
Describe the role of haemoglobin in the transport of respiratory gases. (Foreign 2001)
Answer:

• Oxygen binds to the Fe²⁺ part of haemoglobin and is transported as oxyhaemoglobin through the RBCs of the blood.
• Each molecule of haemoglobin can transport a maximum of four oxygen molecules.
• CO₂ combines with the amine radial of haemoglobin to form carbamino haemoglobin and about 20 – 25% of CO₂ is transported in this form.

Question 3.
Draw a neat labelled diagram of human respiratory system showing the mechanism of
(a) inspiration
(b) expiration.
Answer:

Question 4.
Explain the process of expiration under normal conditions.
Answer:
Expiration takes place when the intra-pulmonary pressure is higher than the atmospheric pressure. The diaphragm and a specialised set of muscles-external and internal intercostals between the ribs, help in generation of such gradients.

Relaxation of the diaphragm and the intercostal muscles returns the diaphragm and sternum to their normal positions after inspiration which reduces the thoracic volume and thereby the pulmonary volume. This leads to an increase in intra-pulmonary pressure to slightly above the atmospheric pressure causing the expulsion of air from the lungs, i.e. expiration.

Question 5.
Give a diagrammatic representation of exchange of gases at the alveolus and the body tissues with blood and transport of oxygen and carbon dioxide.
Answer:

Question 6.
Draw a labelled diagram of a section of an alveolus with pulmonary capillary.
Answer:

Question 7.
Name and explain few disorders of respiratory system.
Answer:

• Asthma: It is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles.
• Emphysema: It is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased. This is mainly caused by cigarette smoking.
• Occupational Respiratory disorders: In certain industries, involving grinding and stone breaking the dust produced results in inflammation leading to fibrosis and this causing serious lung damage.

1st PUC Biology Breathing and Exchange of Gases Five Marks Questions

Question 1.
Draw a neat labelled diagram of human respiratory system.
Answer:

Question 2.
Give the five steps that is involved in respiration.
Answer:
Respiration involves the following steps:

1. Breathing or pulmonary ventilation by which atmospheric air is drawn in and CO₂ rich alveolar air is released out.
2. Diffusion of gases (O₂ and CO₂ ) across alveolar membrane.
3. Transport of gases by the blood.
4. Diffusion of O₂ and CO₂ between blood and tissues.
5. Utilisation of O₂ by the cells for catabolic reactions and resultant release of CO₂

Question 3.
Define the following:
(a) Inspiratory Reserve volume
(b) Expiratory Reserve volume
(c) Total lung capacity
(d) Residual volume
(e) Functional residual capacity
Answer:
(a) Inspiratory Reserve volume (IRV): Additional volume of air, a person can inspire by a forcible inspiration.

(b) Expiratory Reserve volume (ERV): Additional volume of air, a person can expire by a forcible expiration.

(c) Total lung capacity: Total volume of air accommodated in the lungs at the end of a forced inspiration. This includes RV, ERV, TV and IRV.

(d) Residual Volume (RV): Volume of air remaining in the lungs even after a forcible expiration.

(e) Functional residual capacity (FRC): Volume of air that will remain in the lungs after a normal expiration.
This includes ERV + RV.

Karnataka 1st PUC Geography Question Bank Chapter 12 Cartography

You can Download Chapter 12 Cartography Questions and Answers, Notes, 1st PUC Geography Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

1st PUC Geography Cartography One Mark Questions and Answers

Question 1.
Define the term Cartography.
Answer:
The science and art of making maps, charts, globes and rile models is known as Cartography.

Question 2.
What is a Map?
Answer:
A map is defined as a symbolical and conventional representation of the earth or a portion f it drawn to scale on a flat surface and bounded by the geographical coordinates as viewed from above.

Question 3.
Name any two essential features of a Map?
Answer:
Title, Scale, Direction are essential features of a Map.

Question 4.
What is Scale?
Answer:
A scale is the ratio of the distance between two points on the map and their corresponding distance on the ground.

Question 5.
Mention any two uses of Maps.
Answer:
Maps are essential to a geographer, to present spatial information systematically. They are useful to locate lakes, rivers, vegetation, coastal features and also to understand the distribution of soils, minerals, crops, population, tourist places.

Question 6.
What is map index?
Answer:
The features show on a map is indicated by a guide called map index.

Question 7.
What is Physical Map?
Answer:
These maps show the natural phenomena such as relief, climate, vegetation; soils etc. are known as physical maps.

Question 8.
What is a Cultural map?
Answer:
The maps which are prepared to show the various cultural patterns designed over the earth’s surface are called cultural maps.

Question 9.
What is Astronomical map?
Answer:
These maps show the heavenly bodies such as stars, planets, satellites, nebulae etc.

Question 10.
What is Relief Map?
Answer:
They are also known as orographic maps they show the surface features of a given region.
For example: Mountain, plains, plateaus, valleys, hills etc.

Question 11.
What is soil map?
Answer:
The map which show the distribution of various types of soils in a particular region and help in agricultural planning.

Question 12.
What is Population Map?
Answer:
These maps denote the distribution of human beings over an area, the density, sex ratio, literacy etc of a country.

Question 13.
Identify the latitudes and longitudes for the given places.
Answer:

Question 14.
What is map reading?
Answer:
Map reading means getting the correct visual image of the features shown on a map.

Question 15.
What is Contour?
Answer:
They are the imaginary lines joining all the places which are of the same height above the sea level.

Question 16.
What is Uniform slope?
Answer:
There is no change in the degree of slope such slopes may be steep or gentle are known as uniform slopes.

Question 17.
What is undulating slopes?
Answer:
Such slopes are alternately marked by outward bulges and inward bends. Hence contours are spaced irregularly.

Question 18.
What is location?
Answer:
Maps represent the earth or a part of it. So its is essential to know the location of a place on the globe. The latitudes and longitudes are highly useful in under standing the location.

Question 19.
What are cordial Points?
Answer:
These are point’s represents different directions on north south and east.

1st PUC Geography Cartography Two Marks Questions And Answers

Question 1.
What is Political Map?
Answer:
These are the maps which are prepared to show the political boundaries between different countries or states or between different political units of a country. Capital cities and other important towns, railways and highways are also shown on these maps.

Question 2.
Define the term photogrammetry?
Answer:
The out science and technology of taking reliable measurements on photographs about physical objects and the environment is known as photogrammetry. These measurements are obtained by interpreting photographic images.

Question 3.
What is Weather map?
Answer:
They show the weather conditions at fixed time. Average atmospheric pressure, wind velocity and direction, cloudiness, rainfall, drizzle; know fall, sea conditions and other weather phenomena are shown on these maps. These maps are published daily by the meteorological department.
i.e. Indian Daily Weather Report.

Question 4.
What is an international Map Projection?
Answer:
This is a modified polyconic projection. Following the decision of International Map Committee held in 1909, the projection was introduced for the topographical maps of the whole world on a scale of 1,000,000 in preference to polyconic projections.

Question 5.
What are dot maps?
Answer:
The distribution maps, where the dot method is applied to show the distribution of economic phenomena e.g. population, agricultural crops, industries etc. Dots of uniform size are used where each dot represents a certain number or quantity.

Question 6.
What are cartograms?
Answer:
The diagrammatic representation of a statistical map where purposeful distortion is sought to high! lit the distribution pattern of a particular spatial element. It may be termed as a cartographic cartoon.

1st PUC Geography Cartography Five Mark Questions And Answers

Question 1.
Explain the types of Maps:
Answer:
Maps may be broadly divided into two types.
A. On the basis of Scale Maps are classified into three types.

(a) Large Scale Maps: The Maps drawn on the scale of 1 cm= 1 km or 1:1, 00,000 and 1 inch=1 mile or 1:63,360 eg. Cadastral maps (Village, Town and City maps).

(b) Medium Scale Maps: the Maps drawn on the scale of 1 cnm=1 Km to 1 cm=10 km or 1:1,00,000 and 1:10,00,000 eg. Topographical Maps (Mountains, Plateaus, plains).

(c) Small Scale maps: The Maps drawn on the scale below 1 cm: 15Km or 1:15,00,000 eg. Atlas and Wall Maps. These maps show broad physical and cultural features.

B. On the basis of purpose various types of Maps are prepared.

(i) Topographical maps: To show relief features, forests, land use, river system, roads, railways, pipelines, distribution of rural and urban settlements etc.

(ii) Cadastral maps: The Cadastral maps are drawn to register the ownership of field, farm, building, firm etc.

(iii) Economic Maps: These maps provide information about human economic activities eg. Agriculture, mining, industry, marketing, trade etc.

(iv) Population Maps: These maps show the information about distribution, growth, density, migration, age and sex composition of population. These maps are also drawn to show the distribution of occupational structure, language, social groups of people etc.

(v) Weather Maps: These are useful to analyzed weather condition and distribution of temperature, pressure, humidity, winds, rainfall etc.

Question 2.
Explain the uses of Maps.
Answer:

• Maps are very useful to the government for planning and administrative purposes.
• Maps are essential to a geographer, to present spatial information systematically.
• They are useful to locate lakes, rivers, vegetation, coastal features and also to understand the distribution of soils, minerals, crops, population, tourist places.
• They are very much helpful at the time of war and defence.
• Maps are very important for the army. Military maps are very useful for the overall planning of the strategy of war and for coordinating military action during war.
• Maps are very useful tools for a geographer. Geography cannot be understood and made interesting without maps.
• Maps are also immensely useful to other sciences, like physical and social sciences. For example Geology, climatology, Meteorology etc.
• Maps serve as a permanent record to locate features like rivers, lakes vegetation etc.
• Maps enable us to know details of the landforms and other ground features. Mountains, plateaus, plains, coastal plains etc.
• Maps serve as a permanent record to locate features like rivers, lakes, vegetations etc.
• Maps showing the distribution of objects become very useful to understand the location and distribution of different objects such as minerals, soils, vegetation, agricultural crops, industries, population, etc.
• Maps help to mark political boundary, administrative areas to formulate legislation and legal jurisdiction.

Question 3.
Explain the techniques of measurement of Distance on map?
Answer:
Maps represent the earth’s surface based on a particular ratio called the scale. So the distance between any two points on the globe can be easily any two points on the globe can be easily calculated.

There are several simple methods for distance calculations on the maps.

a. By a strip of paper: With the help of a strip of paper we can take the distance of two point on the map and put it on the scale. With this method one can understand the actual distance between the points.

b. By Thread: When the distance between the two places is Jig-jag a thread can be used effectively for the measurement of distance. After taking the distance with the

c. By divider: Even a divider can be used for the measurement of distance on the map. The distance between any two points on the map may be taken with-the divider and then put it on the sale. So the actual distance can be understood and it is converted to the scale of other map.

Question 4.
Explain the map reading and state its advantages:
Answer:
Map reading refers to getting the correct visual image of the features shown on a map. It plays an important role in the study of maps. The objective of map reading is to provide a clear and an accurate visualization of the features on the ground.
Skills required to an efficient map reader:

1. Map legends: Map reader should have the skill to read the map legend, i.e. the explanations of the symbols and colours on a map, to find out exactly what the symbols represent.
For example: triangular shaped symbols may stand for a forest or an orchard, many symbols may stand for a forest or an orchard. Many symbols have no resemblance to what they represent. Example: triangular sign stands for well. The some symbol may be used to show different features in different maps.

2. Map scale: Map scale is an important element or aspect of a map. So, the map reader should make use of the scale intelligently to find out the extent of an area, the distance between various points etc. Many maps show scale by marking distances on a graphical scale. Each marking represents a certain number of kins or miles. Scale may be expressed by representative fraction (RF) scales are usually found at the bottom of the map.

3. Map index: The features shown on a map are indicated by a guide called map index. It helps to locate places on a map. So, the map reader should use the map index to locate the places. Usually in Atlases, index is given at the end of the Atlas. Each entry in the index is listed with its longitude and latitude.

4. Geographical grids: The network of lines crossing each other at right angles is known as Geographical grid or references. These help map reader to find the location, distance and direction of places. The latitudes and longitudes torrid system is commonly used for the above mentioned purpose. The network of these imaginary lines is known as Gratitude.

1. Draw the diagram to the following:

1. Cycle of Seasons

2. Layers of the Earth’s interior

3. Pressure belts of the globe

4. Orographic rainfall

5. Structure of Atmosphere

1. Draw the outline map of India, mark and name the following.

1. Physical divisions of India

2. Forests of India

3. National park and Wild life in India

4. Rivers and Lakes

7. Biosphere Reserves

1. Nanda Devi Saikhawa
2. Nokrek
3. Manas
4. Dibru
5. Dehang Debang
6. Sunderbans
7. Gulf of Mannar
8. Nilgiri
9. Great Nicobar
10. Similipal
11. Khanghendzonga
12. Panchamarhi
13. Agasthymalai
14. Achanakmari – Amar Kantak

8. Soils of India

Karnataka 1st PUC Geography Question Bank Chapter 11 Natural hazards and disasters

You can Download Chapter 11 Natural hazards and disasters Questions and Answers, Notes, 1st PUC Geography Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

1st PUC Geography Natural hazards and disasters One Mark Questions and Answers

Question 1.
What is Natural Hazard? (T.B Qn)
Answer:
It is a threat of naturally occurring event that will have a negative effect on people or the environment.
Ex: Earthquake, Landslide, Volcanic eruption.

Question 2.
What do you mean by Natural Disaster? (T.B Qn)
Answer:
A natural disaster is a major adverse event resulting from natural processes of the Earth e.g. Earthquakes, floods, drought and famine, cyclones, landslides, coastal erosion.

Question 3.
Mention any two types of disasters. (T.B Qn)
Answer:
Major types of Disaster are Tectonic, Meteorological, and Topographical disasters.

Question 4.
What are floods? (T.B Qn)
Answer:
Floods are temporary inundation of large regions as a result of heavy rainfall, prolonged rain cyclones, storm surge along coast.

Question 5.
Name the most important flood prone area of India. (T.B Qn)
Answer:
The Ganga basin is the most important flood prone area of India.

Question 6.
Why are cyclones are caused in the Bay of Bengal? (T.B Qn)
Answer:
The Bay of Bengal is subject to intense heating, giving rise to humid and unstable air masses that produce cyclones.

Question 7.
What is drought? (T.B Qn)
Answer:
The term drought is applied to an extended period when there is a shortage of water availability due to inadequate precipitation, excessive rate of evaporation and over utilization of water from the reservoirs other storages, including the groundwater.

Question 8.
Which region of India is in the extreme drought prone area? (T.B Qn)
Answer:
The regions are western parts of Rajasthan, Kutch regions of Gujarat and semi-arid and arid regions of Western and North western parts of India.

Question 9.
Why does landslide occur? (T.B Qn)
Answer:
Landslides are occur by severe marine erosion of sea coast, Seismic activity, Heavy rainfall, construction of roads, railway lines, canal construction, mining and quarrying, over grazing deforestation.

Question 10.
Mention the most important avalanche prone area of India. (T.B Qn)
Answer:
The most important avalanche prone area of India are mainly Jammu and Kashmir, Himachal radish, uttarkhand, Sikkim, parts of Arunchal Pradesh etc.

Question 11.
What is an Earthquake?
Answer:
An earthquake is a sudden movement, trembling of the earth’s crust.

Question 12.
Which is the highest earthquake intensity region of India?
Answer:
Himalayan region.

Question 13.
Which is the only one active volcano in India?
Answer:
Barren Volcanic Island in the Andaman Island

Question 14.
When national flood control programme was launched?
Answer:
In 1954.

Question 15.
What is coastal erosion?
Answer:
Coastal erosion means eroding down the coastline by sea waves.

Question 16.
What is mean by avalanche?
Answer:
Avalanches are a hurtling mass of snow, ice and rock debris descending a mountain side.

1st PUC Geography Natural hazards and disasters Two Marks Questions And Answers

Question 1.
Name the two most important seismic zones of India. (T.B Qn)
Answer:
Zone V: This is the most severe seismic zone and is referred as very high damage Risk zone. The areas are Northeastern states, parts of Jammu Kashmir, Uttarkhand, and Bihar and Kutch region.

Zone IV: This zone is second in severity zone. Northern regions of Jammu and Kashmir, Himachal Pradesh, parts of Bihar, UP, Gujarat, West Bengal.

Question 2.
Mention any four factors that cause floods. (T.B Qn)
Answer:
Floods are caused by both natural and man-made factors. They are:
(a) Natural factors

• Continuous rainfall for a long period
• Cyclones
• Obstruction on flow of river water.

(b) Man made factors:

• Deforestation
• Unscientific Agricultural practice
• Urbanization

Question 3.
State two important flood prone areas of the country. (T.B Qn)
Answer:
The Ganga Basin: The badly affected states of the Ganga basin area Uttar Pradesh, Bihar and West Bengal.
The Brahmaputra basin: the Brahmaputra along with its tributaries floods the areas of Assam and North West Bengal regions.

Question 4.
Name any four factors that cause drought and famine. (T.B Qn)
Answer:
The main causes for the occurrence of drought and famine are reduction in annual rainfall, long period scarcity of surface and underground water, scarcity of stored water, excess utilization of freshwater. Overgrazing, deforestation. Improper agricultural practice, mining.

Question 5.
Mention any four consequences of natural hazard and disasters. (T.B Qn)
Answer:
The most important consequences of natural disasters are los of human life and property, animal wealth, destruction of vegetation etc. Natural disasters create fear, anguish and trauma in the human beings leading to various physical, biological and psychological changes. Natural disasters affect population, its distribution and density. It affects on agriculture, cropping pattern, industries, transport and communication, public health, water supply.

Question 6.
What is an earthquake? What are the main causes of an earthquake?
Answer:
An earthquake is a sudden release of energy accumulated in rocks causing the ground to . tremble or shake.
The main causes of earthquake are natural and man – made factors.

1. Tectonic forces
2. Volcanic activity
3. Landslides and Landslips
4. Collapse of underground cave roofs

Question 7.
How can drought prevented in India?
Answer:
We can reduce the intensity and impact of drought through individual and collective actions:

• Community based rainwater harvesting structures should b constructed.
• Watershed programmes should be increased
• Through plantation programmes, forest cover should be increased.
• Encouraging farmers to grow drought-resistant crops.

Question 8.
Mention the important regions of land slides in India.
Answer:
There are three important region.

1. Himalayan zone
2. Western Ghats
3. Southern Plateau.

Question 9.
Mention the different types of drought in India.
Answer:
Drought is a weather hazard, uncertainty of monsoon rainfall, deficient rainfall. So India is more frequently affected by droughts.

India droughts are classified into four types:

1. Meteorological Drought
2. Hydrological Drought
3. Agricultural Drought
4. Ecological Drought

1st PUC Geography Natural hazards and disasters Five Marks Questions And Answers

Question 1.
Explain the major seismic zones of India. (T.B Qn)
Answer:
Zone V: This is the most severe seismic (intensity above 7 in Richter scale) seismic zone and is referred as Very High Damage risk zone. The areas are. Northeastern states, parts of Jammu Kashmir, Uttarkhand, and Bihar and Kutch region.

Zone IV: This zone is second in severity (intensity between 5 and 7 in R.S) to zone VG. This is referred to as High Damage Risk zone. Northern regions of Jammu and Kashmir, Himachal Pradesh, Parts of Bihar, UP, Gujarat, West Bengal lie in this region zone. Northern regions of Jammu and Kashmir, Himachal Pradesh, parts of Bihar, UP, Gujarat, West Bengal.

Zone III: This is termed as Moderate Damage (very strong) Risk zone (intensity between 3 and 5 in R.S). The areas are Gujarat, Madya Pradesh, Rajasthan, Chhattisgarh, Odisha, Maharashtra, Northern Karnataka, Andhra Pradesh, West coastal region etc.

Zone II: This zone is referred to as low Damage (strong) Risk Zone (intensity 2 to 3 R.S). The areas are Rajasthan, Madhya Pradesh, Parts of Karnataka, Andhra Pradesh, Odisha etc.

Zone I: This zone is termed as Very Low Damage (Slight-tremor) Risk Zone. The left out parts of India and Deccan Plateau region.

Question 2.
Briefly explain the distribution of flood prone areas of India. (T.B Qn)
Answer:
a. The Ganga basin: The badly affected states of the Ganga basin are U.P, Bihar and West Bengal. Besides the Ganga River, Sarada, Gandak and Ghagra cause flood in Eastern part of U.P. The Yamuna is famous for flooding Haryana, U.P and Delhi. Bihar experiences massive and dangerous flood every year by the Kosi. Rivers like the Mahanadi, Bhagirathi and Damodar also cause floods.

b. The Brahmaputra basin: The Brahmaputra along with its tributaries floods the areas of Assam and North West Bengal regions.

c. The Central India and Peninsular river basin: In odisha spilling over of river banks by the Mahanadi, Baitarnika and Brahmani causes havoc. Southern and central India experiences floods caused by the Narmada, Godavari, Tapti and Krishna during heavy rainfall. Cyclonic storms in the deltaic regions of the Godavari, Mahanadi and the Krishna flood the coastal regions of Andhra Pradesh.

Question 3.
Explain the major drought prone areas of India. (T.B Qn)
Answer:
On the basis of severity of droughts, India can be divided into three drought prone areas.

a. The Extreme drought prone areas: This is the most important drought prone areas of the country which has been recording continuous drought for many years. The regions are western parts of Rajasthan, Kutch regions of Gujarat and semi-arid regions of Western and North western parts of India.

b. The Severe drought prone areas: This is the second important drought prone areas of the county. The eastern parts of Rajasthan, western parts of Madhya Pradesh, Parts of Maharashtra, interior parts of Andhra Pradesh. North and northeastern parts of Karnataka and Tamil nadu.

c. The Moderate drought prone areas: This region is mainly found in regions of U.P, parts of Gujarat, Maharashtra, Jharkhand, Tamil Nadu and interior parts of Karnataka.

Karnataka 1st PUC Geography Question Bank Chapter 10 Climate, Soil and Forest

You can Download Chapter 10 Climate, Soil and Forest Questions and Answers, Notes, 1st PUC Geography Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

1st PUC Geography Climate, Soil and Forest One Mark Questions and Answers

Question 1.
What type of climate is found in India? (T.B Qn)
Answer:
India has “Tropical Monsoon” type of climate.

Question 2.
Define Monsoon. (T.B Qn)
Answer:
The word ‘Monsoon’ is derived from Arabic word ‘Mousim’ meaning season.

Question 3.
Mention the place which records high range of Temperature. (T.B Qn)
Answer:
In summer the western Rajasthan records more’than 55°C of temperature.

Question 4.
Which is the driest season in India? (T.B Qn)
Answer:
The summer or hot weather Season from March to End of May.

Question 5.
Name the region which receives ‘Monsoon outburst’. (T.B Qn)
Answer:
The.Malabar Coast of Kerala receives ‘Monsoon outburst’.

Question 6.
Which is called ‘Mawsynram of South India’? (T.B Qn)
Answer:
Agumbe of Karnataka is called ‘Mawsynram of South India.’

Question 7.
Why are cyclones formed during North East Monsoon season? (T.B Qn)
Answer:
In this season due to pressure variation between the Bay of Bengal and main land of India variable winds-cyclones and anti -cyclones originate in the Bay of Bengal.

Question 8.
What is mean by ‘Burst of Monsoon’?
Answer:
The sudden violent onset of rainfall during the period of ‘Monsoon’ is called the ‘Burst of Monsoon’.

Question 9.
Name any two local winds which blow in India in the summer season.
Answer:
‘Loo and Kalabaisakhi are the two local winds which blow in India in the summer season.

Question 10.
Which wind is responsible for the rainfall experienced over the greater part of India?
Answer:
A South-west monsoon winds is responsible for the rainfall experienced over the greater part of India.

Question 11.
What is the average annual rainfall of India?
Answer:
The average annual rainfall is 118 cm.

Question 12.
Name the place in Southern India which receives highest rainfall from the summer monsoon.
Answer:
Mahabaleswar receives the highest rainfall in South India from the summer monsoon.

Question 13.
What is Kalabaisaki?
Answer:
The local rainfall of summer season in West-Bengal is called ‘Kalabaisaki’.

Question 14.
In which state the South-west monsoon wind enters first.
Answer:
The south west monsoon enters first to the Malabar Coast of Kerala.

Question 15.
Define Pedology. (T.B Qn)
Answer:
The scientific study of soil is known as ‘pedology’.

Question 16.
Name the soil which covers vast area of the country. (T.B Qn)
Answer:
Alluvial soil

Question 17.
Why Black soil is called Regur soil? (T.B Qn)
Answer:
This soil derived from the weathered basalt rock. This soil holds water form long period and become hard whenever it is dry.

Question 18.
Where do we see Laterite soil? (T.B Qn)
Answer:
Laterite soil found in Western Ghats, parts of Eastern Ghats and North eastern hills of India.

Question 19.
What is Humus?
Answer:
Decomposed organic material found in the soil is called Humus.

Question 20.
State the type of soil that is found in the heavy rainfall regions?
Answer:
The laterite soils re found in the heavy rainfall regions.

Question 21.
Which soil is suitable for cotton cultivation?
Answer:
The black soil is suitable for cotton crop.

Question 22.
Which is highest fertile soil?
Answer:
The mountain soil (Forest soil) is fertile soil

Question 23.
Where is Green Gold? (T.B Qn)
Answer:
The forest and their resources are useful to man in various forms. Therefore, they are called ‘Green Gold’.

Question 24.
Mention the average forest cover of the country. (T.B Qn)
Answer:
The average forest cover of the country is 22.50%.

Question 25.
Which forest has high economic value trees? (T.B Qn)
Answer:
Monsoon forest has high economic value trees.

Question 26.
Where do we find Dehang Debaqg Biosphere reserve? (T.B Qn)
Answer:
Arunachal Pradesh.

1st PUC Geography Climate, Soil and Forest Two Marks Questions And Answers

Question 1.
Why India is called ‘Meteorological Unit’? (T.B Qn)
Answer:
Monsoons are the periodic winds in which there is reversal of wind direction periodically. On account of the variability in climatic conditions, seasonally and regionally, India is called ‘Meteorological Unit’.

Question 2.
Mention any two convectional rainfall of India. (T.B Qn)
Answer:
They are “Mango Showers” in Kerala, “cherry Blossoms” in Karnataka and “Kalabiashaki” in West Bengal and Assam.

Question 3.
Write the significance of Monsoon. (T.B Qn)
Answer:
The climatic conditions of the country are greatly influenced by monsoon winds. The winds blow in a particular direction west monsoon winds blow from south west to north east, while north east, while north east monsoon winds blow from northeast to south west.

Question 4.
Mention the annual average rainfall of different seasons in India.
Answer:
The season-wise distribution of rain fall is

• The south-west monsoon seasons – 75%
• The summer season – 10%
• The winter season – 2%
• The retreating monsoon seasons – 13%

Question 5.
Write a short note on Retreating Monsoon?
Answer:
The season of Retreating Monsoon is the period of Transition. During the period of transition low pressure of the north-west shifts to the Bay of Bengal. It results in the formation of cyclones over the Bay. These cyclones cause havoc on the coasts of Orissa and Andhra Pradesh.

Question 6.
Why does India have a monsoon type of climate?
Answer:
India has monsoon climate because there is a seasonal reversal in the wind system in India. During summer winds blow from sea to land and During winter winds blow from land to sea.

Question 7.
What is meant by the word ‘Monsoon’?
Answer:
The word ‘Monsoon’ is derived form the Arabic word ‘Mausam’ which means season. Hence, the word ‘Monsoon’ implies the seasonal reversal in the wind pattern over the year. It reveals the rhythm of season and changes in direction of winds. There is also ca change in the distribution pattern of rainfall and temperature with the change of seasons. The monsoon winds move six months from sea to land and another six months from land to sea,

Question 8.
Mention major factors affecting the climate of our country.
Answer:

• Location and Relief.
• Latitude
• Altitude
• Pressure and winds.

Question 9.
What are the branches of south-west monsoon winds?
Answer:
The south-west monsoons are divided into two branches. They are:

1. The Arabian Sea branch and
2. Bay of Bengal branch.

Question 10.
What is annual range of temperature? Explain it by giving one example.
Answer:
The difference between the maximum average temperature and minimum average temperature of a place over twelve months is known as annual range of temperature.

Ex: The max. average temperature at Jodhpur is 33.9°C and min. Average temperature is 14.9°C. Hence the annual range of temperature at Jodhpur is 19°C.

Question 11.
Mention the importance of Red soil. (T.B Qn)
Answer:
This soil is formed by the weathered granite rocks. It is red in colour and rich in ferrous content. Red soil covers the second largest area in the country. Largest part of peninsular region is covered with red soil. Tamil Nadu has the largest distribution of this soil in the country. Rice, Ragi, Jowar, Groundnut oil seeds are the main crops cultivated in this soil.

Question 12.
Name any four factors that affect soil erosion. (T.B Qn)
Answer:
High Temperature, Rainfall wind and waves are the natural agents. Deforestation, over grazing, shifting cultivation, unscientific methods of agriculture cause soil erosion.

Question 13.
State four best measures in the conservation of soil. (T.B Qn)
Answer:
Afforestation, control of over grazing, contours ploughing. Terrace farming, Erection of bunds, construction of check dams, crop rotation, and control of shifting Cultivation are the best measures in the soil conservation.

Question 14.
State the characteristics of Laterite soils.
Answer:
The important characteristics of Laterite soils are: Laterite soils are red in colour. They are rich in Iron and Aluminum, but poor in potash, Lime, Nitrogen and Phosphoric Acid. They are less retentive of moisture. They are poor in fertility. But they respond very well to manuring. So, with the help of manuring they can be used for the cultivation of plantation crops, such as Tea, Coffee, Spices, Rubber, etc.

Question 15.
Name the states which have the highest and the lowest forest areas in the country. (T.B Qn)
Answer:
Madhya Pradesh (44.8%) is the highest and Haryana state 2.6% is the lowest forest areas in the country.

Question 16.
Write the salient features of Evergreen forest. (T.B Qn)
Answer:
These forests are found in the regions of heavy rainfall (above 250 cm) and high temperature (above 27° C) Tall umbrella shaped trees with dense assemblage is a prominent feature of this forest. The evergreen forests always look green because, various species of trees are found here and they shed leaves in different seasons.

Question 17.
What is Mangrove forest? Why has it become important in the recent years?
(T.B Qn)
Answer:
Mangroves are trees and shrubs that grow in saline coastal habitats in the tropics and subtropics in India. The trees in these forests are hard, durable and are used in boat making and as fuel, in the recent years mangrove vegetation is being grown in the coastal areas to control effects of tidal waves and coastal erosion.

Question 18.
Mention any four measures of conservation of forest. (T.B Qn)
Answer:
Protection and preservation of forest is known as conservation. The important measures of conservation of forest are:

• Forest fires, pests and diseases should be controlled through the scientific methods.
• Encroachers of forest area should be severely punished.
• Forest education, research and training should be expanded through programmes like vanamahotsava, social forestry, and reforestation.

Industrial and mining activities in the forest regions should be compensated by reforestation.

1st PUC Geography Climate, Soil and Forest Five Marks Questions And Answers

Question 1.
What is Climate? Explain the factors that determine the climate of India. (T.B Qn)
Answer:
The average weather condition of place for a long period like 30-33 years in known in known as climate. India’s climate is said to be “Tropical Monson”.

The main factors are monsoon winds.

(i) Location: The northern part of India lies in sub-tropical and temperate zone and the part lying to the south of the tropic of cancer come under tropical zone. The tropical zone being nearer to the Equator, experiences high temperature throughout the year, with small daily and annual range. Tropic of Caner 23 1/2° N latitude passes through the centre of the country. So India is situated both in the tropical and temperate region.

(ii) Mountain Ranges: The lofty Himalayan Mountains have prevented the cold winds of central Asia, and keep India warm. They are also greatly responsible for the monsoon rains in the country.

(iii) Distribution of Land and Water: India is bounded by the Arabian Sea in the west and Bay of Bengal in the east, Indian Ocean in the south. These adjoining seas have influenced the climate of the country considerably. They influence the rainfall of the coastal region. Even the cyclones which originate from these seas regularly affect the weather condition.

(iv) The relief features of India also affect the temperature, air pressure, direction and speed of wind, the amount and distribution of rainfall. The windward side of Western Ghats and north east received high rainfall from June to September.

(v) Monsoon winds: The climatic conditions of other country are greatly influenced by monsoon winds. The winds blow in a particular direction during one season, but get reversed during the other season.

Question 2.
Explain the South West Monsoon season with the help of map. (T.B Qn)
Answer:
The south-west monsoon winds as starts in June and ends in mid-September. It is also known as advancing monsoon season or rainy season. During this season, India gets more than 75% of its annual rainfall and more than 90% of the country’s area receives downpour. It is the prime season for Kharif crops.

In the middle of June the direct rays of the Sun fall on tropic of caner due to shift in the position of the Sun from Equator towards northern hemisphere. Therefore, there is an increase in temperature from south to north. The temperature in the main land of India and nearby land masses is high compared to water bodies of the Indian Ocean.

a. The Arabian Sea branch: The Arabian Sea branch of the south-west monsoon strikes the western coast of India in Kerala on the 1st June. Arabian sea winds by carrying more moisture blow along the western coast of India and cause heavy rainfall in the western part of Western Ghats due to obstruction. These winds behave like sea breeze and cause continuous rainfall I the wind ward side of the Western Ghats tHl they lose their moisture.

Agumbe of Karnataka receives the highest rainfall during this season. This regions coming under southeast monsoon winds receive good rainfall wherever they get obstruction by hills and plateaus.

b. The Bay of Bengal branch blow from water bodies towards the Indian mainland due to variation in pressure. These winds carry moisture form the Bay of Bengl and blow along eastern coast and finally reach north eastern hills. In its path, whenever this wind receives obstruction, they cause good rainfall. The eastern part of Eastern Ghats and north astern hills receive heavy rainfall. These winds after crossing eastern coast merge the Arabian sea winds.

The Arabian Sea and the Bay of Bengal winds, after merging, blow towards north eastern regions of India. The shape of the Himalayan Mountains and northeastern hills greatly obstruct these winds. Therefore the Meghalaya plateau region, particularly Nokrek areas of Mawsynram and cheerapunji, receive very high rainfall. This place is popularly called Rainiest or wettest place on the Earth.

The southwest monsoon after crossing northeastern region blow towards east. Since the Himalayas obstruct these winds they have to take westerly direction and blow along the foothills of Himalayas. The shift in the direct6 sun rays from Tropic of Cancer towards Equator results in the gradual disappearance of southwest monsoons. Indian economy depends on the Monsoons to a large extent.

Question 3.
Briefly explain the characteristics features of winter and summer. (T.B Qn)
Answer:
The winter season (December to February): It is also called cold weather season. In this season direct rays of the Sun fall on Tropic of Capricorn. The temperature in the country is not uniform from north to south. Regions lying to the north of tropic of cancer record low temperature compared to regions in the south. There is a general decrease in temperature from south to north.

December is the beginning of cold weather season and it extends up to February. The annual average temperature is around 18° C. In the northern parts of the plains temperature falls below 5°C. January is the coldest month in the year. Jammu and Kashmir, Punjab, Haryana, UP and parts of Bihar record very low temperature with snow storms. Though the rainfall is small, in some parts on North India it is beneficial for Rabi crops. Annual rainfall in this season is around 2%.

The Summer Season (March to May): The summer season is also known as hot weather season. It begins in March and continues up to May. During this season there is gradual increase n temperature from south north due to shifting of Sun rays from Tropic of Capricorn towards the Equator. In this period south Indian states- Tamilnadu, Andhra Pradesh, Karnataka, and Kerala record high temperature.

Some part of Andhra Pradesh and Karnataka record more than 40° C of temperature. Sri Ganganagar of Rajasthan has recorded the highest temperature of above 52° C. The average temperature of the country will be around 24° C. In this season some parts of India receive convectional rainfall. During this season the country receives 10% of the annual rainfall.

Question 4.
Give details of North East Monsoon season. (T.B Qn)
Answer:
This season is also called North East Monsoon Season. It starts in the middle of September and extends up to middle of December. On September 23rd the direct rays of the sun falls on Equator. Therefore, there is a change in temperature and pressure in the land and water bodies. In this period the Indian sub-continent. The high pressure formed in the northern part of Bay of Bengal results in movement of wind from northeastern part of India towards southwestern region.

These winds blow along the eastern coast of India and Bay of Bengal. In this season due to pressure variation between the Bay of Bengal and main land of India variable winds cyclones take birth in this season and cause great damage in the eastern coast of India. The coastal areas of Tamilnadu, Andhra Pradesh, Odisha and West Bengal come under the frequent effect of cyclones. Some cyclones recorded in the last few years are Bola, Nargis, Nisha, Laila, jal, Neelam etc.

Question 5.
What is Soil? Explain the major types of soils. (T.B Qn)
Answer:
Soil is the minute or finer rock particles found on the surface of the Earth. It is formed naturally, due to the weathering of rocks, under the influence of climate.

The main types of soil in India are:

1. Alluvial soil: This soil is formed by depositional work of rivers and they are mainly found in the flood plains and deltas. Alluvial soil covers largest geographical are in the country. They are mainly distributed in the river plains of the Ganga, Brahmaputra and the Indus. Uttar Pradesh has the largest area under alluvial soil. It is also found in the deltas of east flowing rivers. Alluvial soils are classified into two types.

• Bhangar: Older alluvium, coarse and pebble like in nature, found at the lower depths of the plain.
• Khadar: New alluvium, finer in nature, found in the low lying flood plains and rich in fertility

2. Black soil: The black soils covered more area in peninsular plateau. This soil is also called ‘Cotton soil’ or “Regur soil”. It is derived from the weathered basalt rocks. This soil holds water from long period and become hard whenever it is dry. It is light-black to dark-black in colour. Maharashtra and Gujarat Madhya Pradesh, Karnataka, Andhra Pradesh and Tamilnadu. Black soils are good for Cotton, Sugarcane, Tobacco, Pulses, Millets, Citrus fruits, etc.

3. Red soil: This soil is formed by the weathered granite rocks. It is red in colour and rich in ferrous content. Red soil covers the second largest area in the country. Largest parts of peninsular region are covered with red soil. TamilNadu has the largest distribution of this soil in the country. Rice, Ragi, Jowar, Groundnut, Tobacco, Millets are the major crops cultivated in this soil.

4. Laterite soil: The hot and humid tropical regions of India are rich in laterite soil. This soil is derived from the fragmentation and disintegration of rocks in the mountain ranges. It is mainly found in the Western Ghats, parts of Eastern Ghats and Northeastern hills of India. Plantation crops like Tea, coffee, Rubber, Cashew nut are cultivated in this soil.

5. Desert soil: This soil is also called arid soil. They are mainly found in the desert and semi-desert regions of Western and North western parts of India. This soil has the least water holding capacity and humus content. Generally it is not suitable for cultivation of crops. This soil is mainly found in Rajasthan, parts of Gujarat and Haryana. With water facility crops like Bajra, Pulses and Guar ar cultivated in this soil.

6. Mountain Soil: The Himalayan mountain valleys and hill slopes are covered with Mountain or Forest soil. It is found in the mountain slopes of Jammu and Kashmir, Himachal Pradesh, Utarkhand regions, Crops like Tea, Almond, saffron are cultivated in this soil.

Question 6.
Explain soil erosion and conservation of soil. (T.B Qn)
Answer:
The removal or wearing away of the top soil by various natural agents and man-made factors is called ‘Soil Erosion’. High temperature, Rainfall wind and waves are the natural agents and deforestation, over grazing, shifting cultivation, improper and unscientific methods of agriculture are human activities cause soil erosion. In the hilly regions rainfall and temperature cause more soil erosion. In coastal area sea waves and in desert winds is the dominant factor s in the soil erosion process.

The prevention of soil erosion as well as the protection and maintenance of the Fertility of the soil. The important measures followed in the Conservation of soil are:

• Afforestation
• Control of overgrazing
• Contour ploughing
• Terrace Farming
• Erection of bunds
• Construction of check dams
• Crop rotation
• Strip Farming
• Mulching
• Literacy and education programmes on soil conservation.

Question 7.
Describe the major types of forest in India.(T.B Qn)
Answer:
The peninsular region of India has the largest forest cover with around 57% of the total forest area.

According to geo-climatic conditions, forests are classified into:

a. Evergreen Forests: These forests are found in the regions of heavy rainfall and high temperature. Tall umbrella shaped trees with dense assemblage is a prominent feature of this forest. The eve4rgree forest always looks green because various species of trees are found here and they shed leaves in different seasons.

The hardwood trees, rose wood, white cedar, toon, gurjan, chaplash, ebony, Mahogany, canes, bamboo, shisham etc. These are found in North-east India, Western Ghats, Andaman and Nicobar islands, parts of Assam and some areas of Himalayan foot hills.

b. The Deciduous forests: The deciduous forest covers a wide range of rainfall regimes. The trees of these forests seasonally shed their leaves. The Indian deciduous forest is found in a range of landscapes from the plains to the hills. These forests provide shelter to most endangered wild life in the country, such as the Tiger, Asian Elephant, Bison, Gaur etc. The deciduous forest are two types

(i) Moist Deciduous forests: The moist deciduous forests are found in wet regions, receiving annual rainfall between 100cm to 200cam and temperature of 25° C to 30° C. The trees of these forests shed their leaves during spring and early summer. They are found on the eastern slopes of the Western Ghats, Chota Nagpur Plateau, the siwaliksetc.

(ii) The Dry Deciduous Forests: The dry deciduous forest are found I the areas where annual rainfall is between 50cm to 150 cm and temperature of 25° C to 30° C. Sal is the most significant tree found in this forest. Varieties of acacia and bamboo are also fund here. These forests are found in areas of central Deccan plateau, South-east of Rajasthan, Punjab, Haryana and parts of Uttar Pradesh and Madhya Pradesh.

(iii) The mountain forests: As the name indicates these forests are confined to the Himalayan region, where the temperature is less compared to other parts of the country. The trees in this forest are cone shape with needle like leaves. The important trees are oak, fir, pin e spruce, silver fir, deodhar, devdar, juniper, picea chestnut etc. They provide softwood for making country boats, packing materials and sport articles.

c. The Desert forests: These forests are found in the areas of very low rainfall. Thorny bushes, shrubs, dry grass, acacia, cacti and babul are the important vegetation found in these forests. The Indian wild date known as ‘Khejurs”, is common in the deserts. They have spine leaves, long roots and thick fleshy stems in which they store water to survive during the long drought. These vegetations are found in Rajasthan, Gujarat, Punjab and Haryana.

d. The Mangrove Forests: These forests occur along the river deltas (Ganga, Mahanadi. Godavari and Krishna) of eastern coast and also concentrated in the coastal areas of Katchch, Kathiawar, and Gulf of Khambar. The mangrove forests in the Ganga delta are called Sunder bans because, they have extensive growth of Sundari trees. The trees in these forests are hard, durable and are used in boat making and as fuel. In the recent years mangrove vegetation is being grown I the coastal areas to control effects of tidal waves and coastal erosion.

Question 8.
Briefly explain the importance of forests. (T.B Qn)
Answer:
Forests are the one of the important natural resources. They provide various benefits to mankind and environment.

The important benefits are:

1. Forests supply fresh air, food and fodder.
2. Forests are the rain bearers, help in causing good rainfall.
3. They control soil erosion and desertification.
4. Forest provides various products like bamboo, timber, resin, lac, gum cane, fuel, wood etc.
5. They provide medicinal trees and plants used in ayurvedic medicines Eg.Neem tree. Basil, Brahmi etc.
6. They provide shelter to various birds and animals.
7. They absorb much of the rainwater and control floods and safeguards against drought.
8. They act as wind breakers and protect the agricultural crops.
9. The forest soils are rich in humus and thereby maintain the fertility of the soil.
10. They provide raw materials to paper, match box, plywood and sports articles industries and they provide pastures for grazing animals.

Question 9.
Explain the important measures of conservation of forest. (T.B Qn)
Answer:
The conservation of forest is concerned with proper utilization of forest, protection from destructive influences, misuses of forests etc.

The important measures of conservation of forest are:

1. Careless felling of tree, over-grazing and shifting cultivation should be avoided. Afforestation should be practiced.
2. Forest fires, pests and diseases should be controlled through the scientific methods.
3. Encroachers of forest area should be severely punished.
4. Forest education, research and training should be expanded through programmes like vanamahotsava, social forestry, and reforestation.
5. Industrial and mining activities in the forest regions should be compensated by reforestation.
6. Development of Green belts in the urban areas.
7. Plantation of trees along the roads, railway lines, river, canal banks, tanks and ponds.
8. Use of fuel wood, wood-charcoal by the tribal people must be prohibited.
9. Government should promote intensive tree planting programmes in urban centers.
10. Massive awareness about the aesthetic of forests should be created through mass media, workshops, live programmes etc.

Question 10.
What are Biosphere reserves? Mention the important biosphere reserves of India. (T.B Qn)
Answer:
A biosphere Reserve is a unique and representative ecosystem of terrestrial and coastal areas .The regions surrounding the biosphere reserves would be utilized for the research and experimentation in developing forest and other products.

The Man and the Biosphere Programme (MAB) of UNESCO was established in 1971 to promote interdisciplinary approaches to management, research and education in ecosystem conservation and sustainable use of natural resources. Eight of the eighteen biosphere reserves are a part of the world network of Biosphere reserves, based on the UNESCO man and the Biosphere Programme list.

The objectives of Biosphere reserves:

• Conservation of biodiversity and ecosystem.
• Association of environment with development.
• International network for research and monitoring.

Sl.No	Name of the Biosphere reserve	State	Estd.Year
1.	Nilgiri Biosphere Reserve	Tamilnadu, Kerala, Karnataka	2000
2.	Gulf of Mannar Biosphere Reserve	Tamil Nadu	2001
3.	Sunder bans Biosphere Reserve	West Bengal	2001
4.	Nanda Devi Biosphere Reserve	Uttarkhand	2004
5.	Nokrek Biosphere Reserve	Meghalaya	2009
6.	Panchmarhi Biosphere Reserve	Madhya Pradesh	2009
7.	Simlipal Biosphere reserve	Odisha	2008
8.	Achanakmar-Amarkantak	Chhattisgarh, Jharkhand	2012