Class 11 – Page 36 – Maharashtra Board Solutions

11th Biology Chapter 14 Exercise Human Nutrition Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Prasanna

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 14 Human Nutrition Textbook Exercise Questions and Answers.

Human Nutrition Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 14 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 14 Exercise Solutions

1. Choose correct option

Question A.
Acinar cells are present in ……………..
a. liver
b. pancreas
c. gastric glands
d. intestinal glands
Answer:
b. pancreas

Question B.
Which type of teeth are maximum in number in human buccal cavity?
a. Incisors
b. Canines
c. Premolars
d. Molars
Answer:
d. Molars

Question C.
Select odd one out on the basis of digestive functions of tongue.
a. Taste
b. Swallowing
c. Talking
d. Mixing of saliva in food
Answer:
c. Talking

Question D.
Complete the analogy:
Ptyalin: Amylase : : Pepsin : …………….. .
a. Lipase
b. Galactose
c. Proenzyme
d. Protease
Answer:
d. Protease

2. Answer the following questions

Question A.
For the school athletic meet, Shriya was advised to consume either Glucon-D or fruit juice but no sugarcane juice. Why it must be so?
Answer:
Sugarcane juice contain disaccharides. Disaccharides take time to digest i.e. breaking into monosaccharides, Glucon — D and fruit juices contain monosaccharide. Therefore, for instant supply of energy during athletic meet Glucon – D or fruit juices are preferred and not sugarcane.

Question B.
Alcoholic people may suffer from liver disorder. Do you agree? Explain your answer.
Answer:

Liver disorder in alcoholic people may occur after years of heavy drinking.
Most of the alcohol in the body is broken down in the liver by an enzyme called alcohol dehydrogenase, which transforms ethanol into a toxic compound called acetaldehyde (CH₃CHO).
ver consumption of alcohol leads to cirrhosis (distorted or scarred liver) and eventually to liver failure.
Therefore, alcoholic people may suffer from liver disorder.

Question C.
Digestive action of pepsin comes to a stop when food reaches small intestine. Justify.
Answer:
Pepsin acts in acidic medium thus it is active in stomach. There is alkaline condition in the small intestine. pH of small intestine is very high for pepsin to work. Therefore, pepsin gets denatured in the small intestine.

Question D.
Small intestine is very long and coiled. Even if we jump and run, why it does not get twisted? What can happen if it gets twisted?
Answer:

Mesentery is a tissue that is located in the abdomen. It attaches the small intestine to the wall of the abdomen and keeps it in place and therefore it does not get twisted while running and jumping.
If small intestine gets twisted, the affected spot may block the food, liquid passing through it. It may sometimes cut off the blood flow if the twist is very severe. If this happens the surrounding tissue may die and can cause serious problems.

3. Write down the explanation

Question A.
Digestive enzymes are secreted at appropriate time in our body. How does it happen?
Answer:

The digestive enzymes and juices are produced in sequential manner and at a proper time.
These secretions are under neurohormonal control.
Sight, smell and even thought of food trigger saliva secretion.
Tenth cranial nerve stimulates secretion of gastric juice in stomach.
Even the hormone gastrin brings about the same effect.

B. Explain the structure of tooth. Explain why human dentition is considered as thecodont, diphydont and heterodont.
Answer:

Structure of tooth:
- A tooth consists of the portion that projects above the gum called crown and the root that is made up of two or three projections which are embedded in gum.
- A short neck connects the crown with the root.
- The crown is covered by the hardest substance of the body called enamel which is made up of calcium phosphate and calcium carbonate.
- Basic shape of tooth is derived from dentin which is a calcified connective tissue.
- The dentin encloses the pulp cavity. It is filled with connective tissue pulp. It contains blood vessels and nerves.
- Pulp cavity has extension in the root of the tooth called root canal.
- The dentin of the root of tooth is covered by cementurn which is a bone like substance that attaches the root to the surrounding socket in the gum.
Human dentition is described as thecodont, diphyodont and heterodont.
It is called the codont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.

Question C.
Explain heterocrine nature of pancreas with the help of histological structure.
Answer:
Pancreas:

Pancreas is a leaf shaped heterocrine gland present in the gap formed by bend of duodenum under the stomach.
Exocrine part of pancreas is made up of acini, the acinar cells secrete alkaline pancreatic juice that contains various digestive enzymes.
Pancreatic juice is collected and carried to duodenum by pancreatic duct.
The common bile duct joins pancreatic duct to form hepato-pancreatic duct. It opens into duodenum.
Opening of hepato-pancreatic duct is guarded by sphincter of Oddi.
Endocrine part of pancreas is made up of islets of Langerhans situated between the acini.
It contains three types of cells a-cells which secrete glucagon, P-cells which secretes insulin and 5 cells secrete somatostatin hormone.
Glucagon and insulin together control the blood-sugar level.
Somatostatin hormone inhibits glucagon and insulin secretion.

4. Write short note on

Question A.
Position and function of salivary glands.
Answer:
Salivary Glands:

There are three pairs of salivary glands which open in buccal cavity.
Parotid glands are present in front of the ear.
The submandibular glands are present below the lower jaw.
The glands present below the tongue are called sublingual.
Salivary glands are made up of two types of cells.
Serous cells secrete a fluid containing digestive enzyme called salivary amylase.
Mucous cells produce mucus that lubricates food and helps swallowing.

Question B.
Jaundice
Answer:

Jaundice is a disorder characterized by yellowness of conjunctiva of eyes and skin and whitish stool.
It is a sign of abnormal bilirubin metabolism and excretion.
Jaundice develops if excessive break down of red blood cells takes place along with increased bilirubin level than the liver can handle or there is obstruction in the flow of bile from liver to duodenum.
Bilirubin produced from breakdown of haemoglobin is either water soluble or fat soluble.
Fat soluble bilirubin is toxic to brain cells.
There is no specific treatment to jaundice.
Supportive care, proper rest are the treatments given to the patient.
[Note: Treatment ofjaundice will depend on the underlying cause of it. For example, hepatitis-induced jaundice would require treatment which includes antiviral or steroid medications ]

Question 5.
Observe the diagram. This is histological structure of stomach. Identify and comment on significance of the layer marked by arrow.

Answer:
The layer marked in the diagram represents glandular epithelium of mucosa.
Significance of the glandular epiihelium of mucosa:
Goblet cells of the epithelial layer of a mucous membrane secrete mucus which lubricates the lumen of the alimentary canal. This helps in movement of food through the gastrointestinal tract.

Question 6.
Find out pH maxima for salivary amylase, trypsin, nucleotidase and pepsin and place on the given pH scale

Answer:
Salivary amylase = 6.8
Trypsin = 8
Nucleotidase = 7.5
Pepsin = 2

Question 7.
Write the name of a protein deficiency disorder and write symptoms of it.
Answer:

Kwashiorkor is a protein deficiency disorder.
This protein deficiency disorder is found generally in children between one to three years of age.
Children suffering from Kwashiorkor are underweight and show stunted growth, poor brain development, loss of appetite, anaemia, protruding belly, slender legs, bulging eye, oedema of lower legs and face, change in skin and hair colour.

Question 8.
Observe the diagram given below label the A, B, C, D, E and write the function of A, C in detail.

Answer:
A- Bile duct, B- Stomach, C- Common hepatic duct, D- Pancreas, E- Gall Riadder

Functions: Bile duct: It carries hile from the gall bladder and empties it into the tipper part of the small intestine. Common hepatic duct: It drains bile from the liver. It helps in transportation of waste from liver and helps in digestion by releasing bile.
[Note: Labels (A) and (O) have been modified for the better understanding of the students]

Practical / Project : Here are the events in the process of digestion. Fill in the blanks and complete the flow chart.

Answer:

11th Biology Digest Chapter 14 Human Nutrition Intext Questions and Answers

Can you recall? (Textbook Page No. 161)

Question 1.
What is nutrition?
Answer:

Nutrition is the sum of the processes by which an organism consumes and utilizes food substances,
WHO defines nutrition as the intake of food, considered in relation to the body’s dietary needs.
The term nutrition includes the process like ingestion, digestion, absorption, assimilation and egestion.

Question 2.
Enlist life processes that provide us energy to perform different activities.
Answer:
The life processes which are essential and provide us energy are nutrition and respiration.

Think about it (Textbook Page No. 161)

Question 1.
Our diet includes all necessary nutrients. Still we need to digest it. Why is it so?
Answer:

Digestion is a very important process of converting complex, noil-diffusible and non-absorbable food substances into simple, diffusible and assimilable substances.
Our diet includes all necessary nutrients, which are in the form of complex substances like carbohydrates, proteins, fats and vitamins.
These complex substances are converted into simple, diffusible and assimilable substances through the process of digestion.
Hence, there is a need for digestion of food.

Human Digestive System (Textbook Page No. 161)

Question 1.
Label the diagram
Answer:

Do you know? (Textbook Page No. 162)

Question 1.
Who controls the deglutition?
Answer:
The process of swallowing is called deglutition. Medulla oblongata controls the deglutition.

Question 2.
Is deglutition voluntary or involuntary?
Answer:

Deglutition consists of three phases: oral phase, pharyngeal phase and oesophagal phase.
The oral phase is voluntary whereas the pharyngeal and oesophagal phases are involuntary.
[Source: Goya!, R. K., & Mashimo, H. (2006,.). Physio!o’ of oral, pharyngeal, and esophageal motility. GI Motility online.]

Use your brain power (Textbook Page No. 165)

Question 1.
Draw a neat labelled diagram of human alimentary canal and associated glands in situ.
Answer:

Question 2.
Write a note on human dentition.
Answer:

Human dentition is described as thecodont, diphyodont and heterodont.
It is called thecodont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.

Question 3.
Muscularis layer in stomach is thicker than that in intestine. Why is it so?
Answer:
Muscularis layer in stomach is thicker than that of intestine because food is churned and gastric juices are mixed in the stomach whereas in intestine only absorption takes place.

Question 4.
Liver is a vital organ. Justify.
Answer:

Kupffer cells of liver destroy toxic substances, dead and worn-out blood cells and microorganisms.
Bile juice secreted by liver emulsifies fats and makes food alkaline.’
Liver stores excess of glucose in the form of glycogen.
Deamination of excess amino acids to ammonia and its further conversion to urea takes place in liver.
Synthesis of vitamins A, D, K and BI2 takes place in liver.
It also produces blood proteins like prothrombin and fibrinogen.
During early development, it acts as haemopoietic organ.
Therefore, liver is a vital organ.

Internet my friend: (Textbook Page No. 171)

Question 1.
Collect the different videos of functioning of digestive system,
Answer:
[Note: Students can scan the adjacent Q.R code to get conceptual clarity with the aid of a relevant video.]

Find out (Textbook Page No. 162)

Question 1.
What will be the dental formula of a three years old child?
Answer:
The dental formula of a three-year-old child will be: I \(\frac{2}{2}\), C \(\frac{1}{1}\), M \(\frac{2}{2}\) = \(\frac{2,1,2}{2,1,2}\)
i. e. 5 × 2 = 10 teeth in each jaw = 20 teeth.
As a child has 20 teeth by the age of three.

Question 2.
What is dental caries and dental plaque? How can one avoid it?
Answer:

Dental caries are tooth decay or cavities caused by acids secreted by bacteria. Dental caries may be yellow or black in color.
Dental plaques also known as tooth plaque is a soft, sticky film which forms on the teeth regularly. It is colourless to pale yellow in colour.
Tooth decay and dental plaque can be prevented by brushing teeth twice a day with a fluoride containing tooth paste.
Rinsing mouth thoroughly with a mouth wash and use of dental floss or interdental cleaners to clean teeth daily can help to avoid dental caries and dental plaque.

Internet my friend (Textbook Page No. 162)

Question 1.
Find out the role of orthodontist and dental technician.
Answer:
a. Orthodontics is a specialization in dental profession. Orthodontist straightens the crooked teeth, locates problem in patients’ teeth and their overall oral development. They might use X-rays, plaster molds or dental appliances like retainers and space maintainers to correct the problems,

b. Dental technicians are the ones which improves patients’ appearance, ability to chew and speech. They make dentures, crowns, bridges and dental braces.

Question 2.
What is a root canal treatment?
Answer:

Root canal treatment is also known as endodontic treatment.
It is a dental treatment of removing infection from inside of a tooth.
Root canal is hollow section of tooth which contains the nerve tissue, blood vessels and other cells, this is also known as pulps.
Crown and root are a part of tooth. Crown is present above the gum while root is embedded in the gum.
e. Pulp which is present inside the root canal nourishes the tooth and provides moisture to the surrounding material.
The nerves present inside the pulp sense hot cold temperatures as pain.
First step of a root canal treatment is removal of dead pulp tissues by making a hole on the surface of tooth.
In second step, the dentist cleans and decontaminates the area and fills the hollow area with adhesive cement in order to seal the canal completely.
The tooth is dead after the therapy and the patient no longer feel any pain but the tooth becomes more fragile than ever.
The last step of root canal is adding a crown or filling. Until the crown or filling is complete, patient is not supposed to chew or bite using that tooth. After the crown or filling patient can use that tooth as before.

Find out (Textbook Page No. 163)

Question 1.
You must have heard about appendicitis. It is inflammation of appendix. Find more information about this disorder.
Answer:

Appendicitis is a condition where there is inflammation of appendix.
Appendix is a vestigial organ. It is a linger shaped pouch that projects from colon on the lower right side of the abdomen.
Appendicitis pain is very severe. It initially starts from the navel and then moves.
It occurs in the people of age group between 10 to 30.
Surgical removal is the standard treatment for appendicitis.
Symptoms: Nausea and vomiting, loss of appetite, low grade fever, constipation, abdominal bloating, severe pain in the right side of the abdomen.
Appendicitis is caused when there is blockage in the lining of the appendix that results in infection. The bacteria multiply rapidly and causes inflammation and it is then filled with pus.
If not treated properly appendix can rupture which can lead to further complications.
[Students can use above answer for reference and find more information about appendicitis.]

Question 2.
What is heartburn? Why do we take antacids to control it?
Answer:
Heart burn is a problem created when stomach contents (acid) are forced back up to oesophagus. It causes a burning pain in lower chest.

Antacids are bases and help to treat heartburn by neutralizing the stomach acid. The key ingredients of antacids are calcium carbonate, magnesium hydroxide, aluminium hydroxide or sodium bicarbonate.

Activity (Textbook Page No. 163)

Make a model of human digestive system in a group.
Answer:
[Students are expected to perform this activity on their own.]

Always Remember (Textbook Page No. 166)

Question 1.
Food remains for a very short time in mouth but action of salivary amylase continues for further IS to 30 minutes till gastric juice mixes with food in the stomach. Why do you think it stops after the food gets mixed with gastric juice?
Answer:

The gastric juices are mixed with food in the stomach.
The pH of the stomach is 1.0-2.0 which is very acidic. Such high level of acidity leads to denaturation of salivary amylase’s protein structure.
On the other hand, pH 6.8 is required for salivary amylase to carry out the activity which is not found in stomach. Thus, activity of salivary amylase is stopped when food is mixed with gastric juice.

Internet my friend (Textbook Page No. 167)

Question 1.
How are bile pigments formed?
Answer:

When old and worn out red blood cells are destroyed by macrophages in liver, the globin portion of hemoglobin is split off and heme is converted to biliverdin.
Most of this biliverdin is converted to bilirubin, which gives bile its major pigmentation.
[Source http://www.biologydiscussion.com/human-physiology/digestive-system/bile-pigments/bile-pigments-origin-and-formation-digestive-juice-human-biology/81803]

Think about it (Textbook Page No. 167)

Question 1.
How can I keep my pancreas healthy? Can a person live without pancreas?
Answer:

Pancreas can be kept healthy by:
- Eating proper balanced and low-fat diet, with plenty of whole grains, fruits and vegetables.
- Regular exercise and maintaining a healthy weight.
- Limiting alcohol consumption and avoid smoking.
- Adequate intake of water.
- Regular checkups.
The pancreas is a gland that secretes digestive enzymes and insulin which is needed for a person to survive.
Without pancreas the person will develop diabetes and will have to take insulin for the rest of the life.
Without pancreas the body’s ability to absorb nutrients also decreases.
Hence, though a person can survive without pancreas he may have to remain dependent on the medicines for survival.

Do it yourself? (Textbook Page No. 167)

Question 1.
You have studied the representation of enzymatic actions in the form of reactions.
Write the reactions of pancreatic enzymes.
Answer:

Do it yourself (Textbook Page No. 168)

Question 1.
Observe the following reactions and explain in words.

Answer:

Maltase acts on maltose to form glucose.
Sucrase acts on sucrose to form glucose and fructose.
Lactase acts on lactose to form glucose and galactose.
Dipeptidase acts on dipeptides to form amino acids.
Emulsified fats are converted into fatty acids and glycerol by lipase.

Use your brain power (Textbook Page No. 168)

Question 1.
Make a flow chart for digestion of carbohydrate.
Answer:

Question 2.
What is a proenzyme? Enlist various proenzymes involved in process of digestion and state their function.
Answer:
Proenzymes are synthesized in cells as an inactive precursor that undergo some modification before becoming catalytically active.
The various proenzymes involved in process of digestion are as follows:

Pepsinogen: Pepsinogen when converted into its active form pepsin acts on proteins to form peptones and proteoses.
Trypsinogen: Trypsinogen when converted to it active form trypsin converts proteins, proteoses and peptones to polypeptides.
Chymotrypsinogen: Chymotrypsinogen when converted to active form chymotrypsin it converts polypeptides to dipeptides.

Question 3.
Differentiate between Chyme and Chyle.
Answer:

No.	Chyme	Chyle
a.	Chyme is a semi-fluid acidic mass of partially digested food.	Chyle is an alkaline slurry which contains various nutrients ready for absorption.
b.	Chyme leaves stomach and enters the small intestine.	Chyle leaves small intestine and enters large intestine.

Question 4.
Digestion of fats take place only after the food reaches small intestine. Give reason.
Answer:
Digestion of fats takes place in small intestine because the presence of fats in small intestine stimulates the release of pancreatic lipase from pancreas and bile from liver. Pancreatic lipases hydrolyze fat molecules into fatty acids and monoglycerides and bile brings about emulsification of fats. Therefore, digestion of fats occur when food reaches small intestine.

Observe and Discuss (Textbook Page No. 169)

Question 1.
Action of digestive juice in your group.
Answer:

Digestive juices	Action
Saliva	Saliva contains salivary amylase which breaks down starch into maltose.
Gastric juice	HC1 breaks converts inactive pepsinogen into its active form pepsin. Pepsin then breakdown proteins into peptones and proteoses.
Pancreatic juice	Pancreatic amylase acts on glycogen and starch and converts those into disaccharides. Enterokinase converts trypsinogen into trypsin (active form). Trypsin converts proteins, proteoses, peptones to polypeptides. Chymotrypsin converts polypeptides to dipeptides. Nucleases digest nucleic acids to pentose sugar.
Intestinal enzymes	Maltase converts maltose to glucose. Sucrase converts sucrose to glucose and fructose. Lactase converts lactose to glucose and galactose. Dipeptidases converts dipeptides to amino acids. Lipase converts emulsified fats into fatty acids and monoglycerides.
Bile juice	It brings about emulsification of fats.

Can you recall? (Textbook Page no. 170)

Question 1.
What is balanced diet?
Answer:
Balanced diet is a diet which contains proper amount of carbohydrates, fats, vitamins, proteins and minerals to maintain a good health.

Question 2.
Explain the terms undernourished, over-nourished and malnourished in details.
Answer:

Undernourished: When supply of nutrients is less than the minimum amount of nutrients or food required for good health is called undernourished.
Over-nourished: The intake of nutrients is excessive. In over-nourished the amount of nutrients exceeds the amount required for normal growth.
Malnourished: Malnourished is a condition where a person’s diet does not contain right amount of nutrients.

Do you know? (Textbook Page No. 170)

Question 1.
What is gross calorific value?
Answer:
The amount of heat liberated by complete combustion of lg food in a bomb calorimeter is termed as gross calorific (gross energy) value.

Question 2.
What is physiological value?
Answer:
The actual energy produced by 1 g food is its physiological value.

Question 3.
Name the following
Energy content of food in animals is expressed in terms of?
Answer:
Heat Energy

Question 4.
Complete the following table representing Gross calorific value and physiological value of food component.

Food Component	Gross calorific value (Kcal/g)	Physiological value (Kcal/g)
Fats	(A)	9.0
(B)	5.65	4.0
Carbohydrates	(C)	(D)

Answer:

Food Component	Gross calorific value (Kcal/g)	Physiological value (Kcal/g)
Fats	9.45	9.0
Proteins	5.65	4.0
Carbohydrates	4.1	4.0

Find out (Textbook Page No. 171)

Question 1.
Find out the status of nialnutrition among children in Maharashtra and efforts taken by the government to overcome the situation. Search for various NGOs working in this field.
Answer:
93,783 children have been diagnosed with severe acute malnutrition and 5.7 lakh with moderate acute malnutrition in Maharashtra.
Steps taken by government to overcome malnutrition:

Promotion of infant and young child feeding practices.
Management of malnutrition at community and facility level by trained service providers.
Treatment of children with severe acute malnutrition at special units called the Nutrition Rehabilitation Centres (NRCs), set up at public health facilities.
A special program to combat micronutrient deficiencies of Vitamin A, Iron and Folic acid.
The initiatives like Mother and Child protection card, village health and nutrition days, are taken by the government for addressing the nutrition concerns in children, pregnant women and lactating mothers.

Various NCOs working in this field:

Akshay Patra
Fight Hunger Foundation,
Feeding India,
No Hungry child
[Source: http://pib.nic.in/newsite/PrintRelease.aspx?relid=l 13725; https://yourstory.com/2016/10/world- food-day-ngosj
[Note: Students can use above answer as reference and find more information from the internet.]

Question 2.
Are jaundice and hepatitis same disorders?
Answer:
Jaundice and Hepatitis are two different disorders.

Jaundice: Jaundice occurs when the rate of bilirubin production exceeds the rate of its elimination. It causes yellowing of skin and eyes.

Hepatitis: It is a disease where there is inflammation of liver. It may be caused because of infection, over alcohol consumption, immune system disorder etc.

Do you know (Textbook Page No. 171)

Question 1.
Alcoholism causes different disorders of liver like steatosis (fatty liver), alcoholic hepatitis, fibrosis and cirrhosis. Collect more information on these disorders and try to increase awareness against alcoholism in society. Collect information about NGOs working against alcoholism.
Answer:
Steatosis (fatty liver): Steatosis is accumulation of fat in the liver. Treatment can help but it cannot be cured. Major risk factors are obesity and Diabetes type II, it is also associated with excessive alcohol consumption. Fatigue, weight loss and abdominal pain are some symptoms. It is a benign condition but in very smaller number of patients it can lead to liver failure. Treatment involves diet and exercise to reduce obesity.

Alcoholic hepatitis: Alcoholic Hepatitis is liver inflammation caused by excessive consumption of alcohol. It occurs in people who drink heavily for many years. Symptoms like yellowing of skin and eye, accumulation of fluid in stomach which leads to increase in stomach size. Treatments like completely stopping of alcohol consumption, hydration and nutrition care are carried out. Administration of steroid drugs reduces liver inflammation.

Fibrosis: There is significant scarring of liver tissue in this condition. Fibrosis itself does not cause any symptoms. Diagnosis includes doctor’s evaluation, blood tests and imaging tests, liver biopsy. Treatments include stopping the consumption of alcohol. There are no such effective drugs for curing of fibrosis.

Cirrhosis: It is a chronic liver damage caused due to various reasons which leads to irreversible scarring of liver and liver failure. Causes of cirrhosis are chronic alcohol abuse and hepatitis. Patients may experience fatigue, weakness and weight loss. In later stages, patients may develop jaundice, abdominal swelling and gastrointestinal bleeding. In advanced stage, a liver transplant is required.

NGOs working against alcoholism:

Muktangan Rehabilitation Centre
Anmol Jeevan Foundation
Sankalp Rehabilitation Trust
Kripa Foundation
Harmony Foundation
Hands for you Rehab Centre

11th Std Biology Questions And Answers:

11th Physics Chapter 12 Exercise Magnetism Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Prasanna

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 12 Magnetism Textbook Exercise Questions and Answers.

Magnetism Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 12 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 12 Exercise Solutions

1. Choose the correct option.

Question 1.
Let r be the distance of a point on the axis of a bar magnet from its center. The magnetic field at r is always proportional to
(A) \(\frac {1}{r^2}\)
(B) \(\frac {1}{r^3}\)
(C) \(\frac {1}{r}\)
(D) Not necessarily \(\frac {1}{r^3}\) at all points
Answer:
(B) \(\frac {1}{r^3}\)

Question 2.
Magnetic meridian is the plane
(A) perpendicular to the magnetic axis of Earth
(B) perpendicular to geographic axis of Earth
(C) passing through the magnetic axis of Earth
(D) passing through the geographic axis
Answer:
(C) passing through the magnetic axis of Earth

Question 3.
The horizontal and vertical component of magnetic field of Earth are same at some place on the surface of Earth. The magnetic dip angle at this place will be
(A) 30°
(B) 45°
(C) 0°
(D) 90°
Answer:
(B) 45°

Question 4.
Inside a bar magnet, the magnetic field lines
(A) are not present
(B) are parallel to the cross sectional area of the magnet
(C) are in the direction from N pole to S pole
(D) are in the direction from S pole to N pole
Answer:
(D) are in the direction from S pole to N pole

Question 5.
A place where the vertical components of Earth’s magnetic field is zero has the angle of dip equal to
(A) 0°
(B) 45°
(C) 60°
(D) 90°
Answer:
(A) 0°

Question 6.
A place where the horizontal component of Earth’s magnetic field is zero lies at
(A) geographic equator
(B) geomagnetic equator
(C) one of the geographic poles
(D) one of the geomagnetic poles
Answer:
(D) one of the geomagnetic poles

Question 7.
A magnetic needle kept nonparallel to the magnetic field in a nonuniform magnetic field experiences
(A) a force but not a torque
(B) a torque but not a force
(C) both a force and a torque
(D) neither force nor a torque
Answer:
(C) both a force and a torque

2. Answer the following questions in brief.

Question 1.
What happens if a bar magnet is cut into two pieces transverse to its length/ along its length?
Answer:
i. When a magnet is cut into two pieces, then each piece behaves like an independent magnet.

ii. When a bar magnet is cut transverse to its length, the two pieces generated will behave as independent magnets of reduced magnetic length. However, the pole strength of all the four poles formed will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,

iii. When the bar magnet is cut along its length, the two pieces generated will behave like an independent magnet with reduced pole strength. However, the magnetic length of both the new magnets will be same as that of the original bar magnet. Thus, the new dipole moment of the smaller magnets will be,

Question 2.
What could be the equation for Gauss’ law of magnetism, if a monopole of pole strength p is enclosed by a surface?
Answer:
i. According to Gauss’ law of electrostatics, the net electric flux through any Gaussian surface is proportional to net charge enclosed in it. The equation is given as,
ø_E = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\)

ii. Similarly, if a monopole of a magnet of pole strength p exists, the Gauss’ law of magnetism in S.I. units will be given as,
ø_E = ∫\(\vec{B}\) . \(\vec{dS}\) = µ₀P

3. Answer the following questions in detail.

Question 1.
Explain the Gauss’ law for magnetic fields.
Answer:
i. Analogous to the Gauss’ law for electric field, the Gauss’ law for magnetism states that, the net magnetic flux (ø_B) through a closed Gaussian surface is zero. ø_B = ∫\(\vec{B}\) . \(\vec{dS}\) = 0

ii. Consider a bar magnet, a current carrying solenoid and an electric dipole. The magnetic field lines of these three are as shown in figures.

iii. The areas (P) and (Q) are the cross – sections of three dimensional closed Gaussian surfaces. The Gaussian surface (P) does not include poles while the Gaussian surface (Q) includes N-pole of bar magnet, solenoid and the positive charge in case of electric dipole.

iv. The number of lines of force entering the surface (P) is equal to the number of lines of force leaving the surface. This can be observed in all the three cases.

v. However, Gaussian surface (Q) of bar magnet, enclose north pole. As, even thin slice of a bar magnet will have both north and south poles associated with it, the number of lines of Force entering surface (Q) are equal to the number of lines of force leaving the surface.

vi. For an electric dipole, the field lines begin from positive charge and end on negative charge. For a closed surface (Q), there is a net outward flux since it does include a net (positive) charge.

vii. Thus, according to the Gauss’ law of electrostatics ø_E = ∫\(\vec{E}\) . \(\vec{dS}\) = \(\frac {q}{ε_0}\), where q is the positive charge enclosed.

viii. The situation is entirely different from magnetic lines of force. Gauss’ law of magnetism can be written as ø_B = ∫\(\vec{B}\) . \(\vec{dS}\) = 0
From this, one can conclude that for electrostatics, an isolated electric charge exists but an isolated magnetic pole does not exist.

Question 2.
What is a geographic meridian? How does the declination vary with latitude? Where is it minimum?
Answer:
A plane perpendicular to the surface of the Earth (vertical plane) and passing through geographic axis is geographic meridian.

i. Angle between the geographic and the magnetic meridian at a place is called magnetic declination (a).
ii. Magnetic declination varies with location and over time. As one moves away from the true north the declination changes depending on the latitude as well as longitude of the place. By convention, declination is positive when magnetic north is east of true north, and negative when it is to the west. The declination is small in India. It is 0° 58′ west at Mumbai and 0° 41′ east at Delhi.

Question 3.
Define the angle of dip. What happens to angle of dip as we move towards magnetic pole from magnetic equator?
Answer:
Angle made by the direction of resultant magnetic field with the horizontal at a place is inclination or angle of dip (ø) at the place.
At the magnetic pole value of ø = 90° and it goes on decreasing when we move towards equator such that at equator value of (ø) = 0°.

4. Solve the following problems.

Question 1.
A magnetic pole of bar magnet with pole strength of 100 Am is 20 cm away from the centre of a bar magnet. Bar magnet has pole strength of 200 Am and has a length 5 cm. If the magnetic pole is on the axis of the bar magnet, find the force on the magnetic pole.
Answer:
Given that, (q_m)₁ = 200 Am
and (2l) = 5 cm = 5 × 10^-2 m
∴ m = 200 × 5 × 10^-2 = 10 Am²
For a bar magnet, magnetic dipole moment is,
m = q_m (21)
For a point on the axis of a bar magnet at distance, r = 20 cm = 0.2 m,
B_a = \(\frac{\mu_{0}}{4 \pi} \times \frac{2 m}{r^{3}}\)
= 10^-7 × \(\frac{2 \times 10}{(0.2)^{3}}\)
= 0.25 × 10^-3
= 2.5 × 10^-4 Wb/m²
The force acting on the pole will be given by,
F = q_m B_a = 100 × 2.5 × 10^-4
= 2.5 × 10^-2 N

Question 2.
A magnet makes an angle of 45° with the horizontal in a plane making an angle of 30° with the magnetic meridian. Find the true value of the dip angle at the place.
Answer:
Let true value of dip be ø. When the magnet is kept 45° aligned with declination 30°, the horizontal component of Earth’s magnetic field.
B’_H = B_H cos 30° Whereas, vertical component remains unchanged.
∴ For apparent dip of 45°,
tan 45° = \(\frac{\mathrm{B}_{\mathrm{V}}^{\prime}}{\mathrm{B}_{\mathrm{H}}^{\prime}}=\frac{\mathrm{B}_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}} \cos 30^{\circ}}=\frac{\mathrm{B}_{\mathrm{v}}}{\mathrm{B}_{\mathrm{H}}} \times \frac{1}{\cos 30^{\circ}}\)
But, real value of dip is,
tan ø = \(\frac {B_V}{B_H}\)
∴ tan 45° = \(\frac {tan ø}{cos 30°}\)
∴ tan ø = tan 45° × cos 30°
= 1 × \(\frac {√3}{2}\)
∴ ø = tan^-1 (0.866)

Question 3.
Two small and similar bar magnets have magnetic dipole moment of 1.0 Am² each. They are kept in a plane in such a way that their axes are perpendicular to each other. A line drawn through the axis of one magnet passes through the centre of other magnet. If the distance between their centres is 2 m, find the magnitude of magnetic field at the midpoint of the line joining their centres.
Answer:
Let P be the midpoint of the line joining the centres of two bar magnets. As shown in figure, P is at the axis of one bar magnet and at the equator of another bar magnet. Thus, the magnetic field on the axis of the first bar magnet at distance of 1 m from the centre will be,

B_a = \(\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}}\)
= 10^-7 × \(\frac {2×1.0}{(1)^3}\)
= 2 × 10^-7 Wb/m²
Magnetic field on the equator of second bar magnet will be,
B_eq = \(\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}\)
= 10^-7 × \(\frac {1.0}{(1)^3}\)
= 1 × 10^-7 Wb/m²
The net magnetic field at P,
B_net = \(\sqrt {B_a^2+B_{eq}^2}\)
= \(\sqrt {(2×10^{-7})^2+(1×10^{-7})^2}\)
= \(\sqrt {(10^{-7})^2×(4+1)}\)
= √5 × 10^-7 Wb/m²

Question 4.
A circular magnet is made with its north pole at the centre, separated from the surrounding circular south pole by an air gap. Draw the magnetic field lines in the gap. Draw a diagram to illustrate the magnetic lines of force between the south poles of two such magnets.
Answer:
i. For a circular magnet:

Question 5.
Two bar magnets are placed on a horizontal surface. Draw magnetic lines around them. Mark the position of any neutral points (points where there is no resultant magnetic field) on your diagram.
Answer:
The magnetic lines of force between two magnets will depend on their relative positions. Considering the magnets to be placed one besides the other as shown in figure, the magnetic lines of force will be as shown.

11th Physics Digest Chapter 12 Magnetism Intext Questions and Answers

Can you recall? (Textbook page no. 221)

Question 1.
What are the magnetic lines of force?
Answer:
The magnetic field around a magnet is shown by lines going from one end of the magnet to the other. These lines are named as magnetic lines of force.

Question 2.
What are the rules concerning the lines of force?
Answer:
i. Magnetic lines of force originate from the north pole and end at the south pole.
ii. The magnetic lines of force of a magnet or a solenoid form closed loops. This is in contrast to the case of an electric dipole, where the electric lines of force originate from the positive charge and end on the negative charge.
iii. The direction of the net magnetic field \(\vec {B}\) at a point is given by the tangent to the magnetic line of force at that point.
iv. The number of lines of force crossing per unit area decides the magnitude of magnetic field \(\vec {B}\)
v. The magnetic lines of force do not intersect. This is because had they intersected, the direction of magnetic field would not be unique at that point.

Question 3.
What is a bar magnet?
Answer:
Bar magnet is a magnet in the shape of bar having two poles of equal and opposite pole strengths separated by certain distance (2l).

Question 4.
If you freely hang a bar magnet horizontally, in which direction will it become stable?
Answer:
A bar magnet suspended freely in air always aligns itself along geographic N-S direction.

Try this (Textbook page no. 221)

You can take a bar magnet and a small compass needle. Place the bar magnet at a fixed position on a paper and place the needle at various positions. Noting the orientation of the needle, the magnetic field direction at various locations can be traced.
Answer:
When a small compass needle is kept at any position near a bar magnet, the needle always aligns itself in the direction parallel to the direction of magnetic lines of force.

Hence, by placing it at different positions, A, B, C, D,… as shown in the figure, the direction of magnetic lines of force can be traced. The direction of magnetic field will be a tangent at that point.

Internet my friend: (Text book page no. 227)

https://www.ngdc.noaa.gov
[Students are expected to visit above mentioned link and collect more information about Geomagnetism.]

11th Std Physics Questions And Answers:

11th Chemistry Chapter 5 Exercise Chemical Bonding Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 5 Chemical Bonding Textbook Exercise Questions and Answers.

Chemical Bonding Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 5 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 5 Exercise Solutions

1. Select and write the most appropriate alternatives from the given choices.

Question A.
Which molecule is linear?
a. SO₃
b. CO₂
c. H₂S
d. Cl₂O
Answer:
b. CO₂

Question B.
When the following bond types are listed in decreasing order of strength (strongest first). Which is the correct order ?
a. covalent > hydrogen > van der waals
b. covalent > vander waal’s > hydrogen
c. hydrogen > covalent > vander waal’s
d. vander waal’s > hydrogen > covalent.
Answer:
a. covalent > hydrogen > van der waals

Question C.
Valence Shell Electron Pair repulsion (VSEPR) theory is used to predict which of the following :
a. Energy levels in an atom
b. the shapes of molecules and ions.
c. the electrone getivities of elements.
d. the type of bonding in compounds.
Answer:
b. the shapes of molecules and ions.

Question D.
Which of the following is true for CO₂?

	C=O bond	CO₂ molecule
A	polar	non-polar
B	non-polar	polar
C	polar	polar
D	non-polar	non-polar

Answer:

	C=O bond	CO₂ molecule
A	polar	non-polar

Question E.
Which O₂ molecule is pargmagnetic. It is explained on the basis of :
a. Hybridisation
b. VBT
c. MOT
d. VSEPR
Answer:
c. MOT

Question F.
The angle between two covalent bonds is minimum in:
a CH₄
b. C₂H₂
c. NH₃
d. H₂O
Answer:
d. H₂O

2. Draw

Question A.
Lewis dot diagrams for the folowing
a. Hydrogen (H₂)
b. Water (H₂O)
c. Carbon dioxide (CO₂)
d. Methane (CH₄)
e. Lithium Fluoride (LiF)
Answer:

[Note: H atom in H₂ and Li atom in LiF attain the configuration of helium (a duplet of electrons).]

Question B.
Diagram for bonding in ethene with sp² Hybridisation.
Answer:

Question C.
Lewis electron dot structures of
a. HF
b. C₂H₆
c. C₂H₄
d. CF₃Cl
e. SO₂
Answer:

Question D.
Draw orbital diagrams of
a. Fluorine molecule
b. Hydrogen fluoride molecule
Answer:
a.

b.

3. Answer the following questions

Question A.
Distinguish between sigma and pi bond.
Answer:

σ (sigma) bond	π (pi) bond
1. It is formed when atomic orbitals overlap along internuclear axis.	1. It is formed when atomic orbitals overlap side-ways (laterally).
2. Electron density is high along the axis of the molecule (i.e., internuclear axis).	2. Electron density is zero along the axis of the molecule (i.e., internuclear axis).
3. In the formation of sigma bond, the extent of overlap is greater, hence, more energy is released.	3. In the formation of pi bond, the extent of overlap is less, hence, less energy is released.
4. It is a strong bond.	4. It is a weak bond.
5. Formation of sigma bonds involves s-s, s-p, p-p overlap and overlap between hybrid orbitals.	5. Formation of pi bonds involves p-p or d-d overlap. The overlap between hybrid orbitals is not involved.

Question B.
Display electron distribution around the oxygen atom in water molecule and state shape of the molecule, also write H-O-H bond angle.
Answer:
Electron distribution around oxygen atom in water molecule:
Shape of water molecule: Angular or V shaped H-O-H bond angle = 104°35′

Question C.
State octet rule. Explain its inadequecies with respect to
a. Incomplete octet
b. Expanded octet
Answer:
Statement: During the formation of chemical bond, atom loses, gains or shares electrons so that its outermost orbit (valence shell) contains eight electrons. Therefore, the atom attains the nearest inert gas electronic configuration.

a. Molecules with incomplete octet: e.g. BF₃, BeCl₂, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl₂ has four electrons while B in BF₃ has six electrons in the valence shell.

b. Molecules with expanded octet: Some molecules like SF₆, PCl₅, H₂SO₄ have more than eight electrons around the central atom.

Question D.
Explain in brief with one example:
a. Ionic bond
b. covalent bond
c. co-ordinate bond
Answer:
a. Formation of calcium chloride (CaCl₂):
i. The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or (2, 8, 8, 2)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
v. Calcium atom changes into Ca²⁺ ion while the two chlorine atoms change into two Cl^– ions. These ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

b. Formation of Cl₂ molecule:
i. The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.
ii. It needs one more electron to complete its valence shell.
iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl₂ molecule is formed.

c. co-ordinate bond:
i. A coordinate bond is a type of covalent bond where both of the electrons that form the bond originate from the same atom
ii. An atom with a lone pair of electrons (non-bonding pair of electrons) is capable of forming a coordinate bond.
iii. For example, reaction of ammonia with boron trifluoride: Before the reaction, nitrogen (N) in ammonia has eight valence electrons, including a lone pair of electrons. Boron (B) in boron trifluoride has only six valence electrons, so it is two electrons short of an octet. The two unpaired electrons form a bond between nitrogen and boron, resulting in complete octets for both atoms. A coordinate bond is represented by an arrow. The direction of the arrow indicates that the electrons are moving from nitrogen to boron. Thus, ammonia forms a coordinate bond with boron trifluoride.

iv. Once formed, a coordinate covalent bond is the same as any other covalent bond.

Question E.
Give reasons for need of Hybridisation.
Answer:
The concept of hybridization was introduced because the valence bond theory failed to explain the following points:
i. Valencies of certain elements:
The maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell. However, valence bond theory failed to explain how beryllium, boron and carbon forms two, three and four covalent bonds respectively.
a. Beryllium: The electronic configuration of beryllium is 1s² 2s². The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl₂.
b. Boron: The electronic configuration of boron is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\). The valency is expected to be 1 but it is 3 as in BF₃.
c. Carbon: The electronic configuration of carbon is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) . The valency is expected to be 2, but observed valency is 4 as in CH₄.

ii. The shapes and geometry of certain molecules:
The valence bond theory cannot explain shapes, geometries and bond angles in certain molecules,
e.g. a. Tetrahedral shape of methane molecule.
b. Bond angles in molecules like NH₃ (107°18′) and H₂O (104°35′).
However, the valency of the above elements and the observe structural properties of the above molecules can be explained by the concept of hybridization. These are the reasons for need of the concept of hybridization.

Question F.
Explain geometry of methane molecule on the basis of Hybridisation.
Answer:
Formation of methane (CH₄) molecule on the basis of sp³ hybridization:
i. Methane molecule (CH₄) has one carbon atom and four hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\);
Electronic configuration of carbon:

iii. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp³hybridization.
iv. One electron from the 2s orbital of carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py and pz mix and recast to form four new sp³ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with H-C-H bond angle of 109°28′. Each hybrid orbital contains one unpaired electron.
v. Each of these sp³ hybrid orbitals with one electron overlap axially with the 1s orbital of hydrogen atom to form one C-H sigma bond. Thus, in CH₄ molecule, there are four C-H bonds formed by the sp³-s overlap.
Diagram:

Question G.
In Ammonia molecule the bond angle is 107°18 and in water molecule it is 104°35′, although in both the central atoms are sp³ hybridized Explain.
Answer:
i. The ammonia molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 107°28′. It is due to the following reasons.

One lone pair and three bond pairs are present in ammonia molecule.
The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.
Due to these repulsions, there is a small decrease in bond angle (~2°) from 109°28′ to 107°18′.

ii. The water molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 104°35′. It is due to the following reasons.

Two lone pairs and two bond pairs are present in water molecule.
The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair.
Due to these repulsions, there is a small decrease in bond angle (~5°) from 109°28′ to 104°35′.

Question H.
Give reasons for:
a. Sigma (σ) bond is stronger than Pi (π) bond.
b. HF is a polar molecule
c. Carbon is a tetravalent in nature.
Answer:
a. i. The strength of the bond depends on the extent of overlap of the orbitals. Greater the overlap, stronger is the bond.
ii. A sigma bond is formed by the coaxial overlap of the atomic orbitals which are oriented along the internuclear axis, hence the extent of overlap is maximum.
iii. A pi bond is formed by the lateral overlap of the atomic orbitals which are oriented perpendicular to the internuclear axis, hence the extent of orbital overlapping in side wise manner is less.
Hence, sigma bond is stronger than pi bond.

b. i. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.
ii. In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial -ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and H-F bond becomes polar. Hence, H-F is a polar molecule.

c. The electronic configuration of carbon is:
1s² 2s² 2p_x¹ 2p_y¹
One electron from ‘2s’ orbital is promoted to the empty ‘2p’ orbital.
Thus, in excited state, carbon has four half-filled orbitals.

Hence, carbon can form 4 bonds and is tetravalent in nature.

Question I.
Which type of hybridization is present in ammonia molecule? Write the geometry and bond angle present in ammonia.
Answer:
The type of hybridization present in ammonia (NH₃) molecule is sp³.
Geometry of ammonia molecule is pyramidal or distorted tetrahedral.
Bond angle in ammonia molecule is 107°18′.

Question J.
Identify the type of orbital overlap present in
a. H₂
b. F₂
c. H-F molecule.
Explain diagramatically.
Answer:
i. s-s σ overlap:
a. The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H₂ molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹. The 1s¹ orbitais of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H₂ molecule.
b. Diagram:

ii. p-p σ overlap:
a. This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F₂ molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\).
During the formation of F₂ molecule, half-filled 2p_z orbital of one F atom overlaps with similar half-filled 2p_z orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
b. Diagram:

iii. s-p σ overlap:
a. In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹ and fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\). During the formation of HF molecule, half-filled Is orbital of hydrogen atom overlaps coaxially with half-filled 2p_z orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
b. Diagram:

Question K.
F-Be-F is a liner molecule but H-O-H is angular. Explain.
Answer:
i. In the BeF₂ molecule, the central beryllium atom undergoes sp hybridization giving rise to two sp hybridized orbitals placed diagonally opposite with an angle of 180°. Thus, F-Be-F is a linear molecule.

ii. In the H₂O molecule, the central oxygen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. There are two lone pairs of electrons in two of the sp³ hybrid orbitals of oxygen. The lone pair-lone pair repulsion distorts the structure. Hence, H-O-H is angular or V-shaped.

Question L.
BF₃ molecule is planar but NH₃ pyramidal. Explain.
Answer:
i. In the BF₃ molecule, the central boron atom undergoes sp² hybridization giving rise to three sp² hybridized orbitals directed towards three comers of an equilateral triangle. Thus, the geometry is trigonal planar.

ii. In the NH₃ molecule, the central nitrogen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. The expected geometry of NH₃ molecule is regular tetrahedral with bond angle 109°28′. There is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen. The lone pair-bond pair repulsion distorts the bond angle. Hence, the structure of NH₃ is distorted and it has pyramidal geometry.

Question M.
In case of bond formation in Acetylene molecule :
a. How many covalend bonds are formed ?
b. State number of sigma and pi bonds formed.
c. Name the type of Hybridisation.
Answer:
a. In acetylene molecule, there are five covalent bonds.
b. In acetylene molecule, there are three sigma bonds and two pi bonds.
c. In acetylene molecule, each carbon atom undergoes sp hybridization.

Question N.
Define :
a. Bond Enthalpy
b. Bond Length
Answer:
a. Bond Enthalpy:
Bond enthalpy is defined as the amount of energy required to break one mole of a bond of one type, present between two atoms in a gaseous state.

b. Bond Length:
Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.

Question O.
Predict the shape and bond angles in the following molecules:
a. CF₄
b. NF₃
c. HCN
d. H₂S
Answer:
a. CF₄: There are four bond pairs on the central atom. Hence, shape of CF₄ is tetrahedral and F-C-F bond angle is 109° 28′.
b. NF₃: There are three bond pairs and one lone pair on the central atom. Hence, shape of NF₃ is trigonal pyramidal and F-N-F bond angle is less than 109° 28′.
c. HCN: There are two bond pairs on the central atom. Hence, shape of HCN is linear and H-C-N bond angle is 180°.
d. H₂S: There are two bond pairs and two lone pairs on the central atom. Hence, shape of H₂S is bent or V-shaped and H-S-H bond angle is slightly less than 109° 28′.

4. Using data from the Table, answer the following :

a. What happens to the bond length when unsaturation increases?
b. Which is the most stable compound?
c. Indicate the relation between bond strength and Bond enthalpy.
d. Comment on overall relation between Bond length, Bond Enthalpy and Bond strength and stability.
Answer:
a. When unsaturation increases, the bond length decreases.
b. The stable compound is ethyne (C₂H₂).
c. Bond strength ∝ Bond enthalpy
Larger the bond enthalpy, stronger is the bond.
d. As bond length decreases, bond enthalpy, bond strength and stability increase.

5. Complete the flow chart

Answer:

6. Complete the following Table

Answer:

7. Answer in one sentence:

Question A.
Indicate the factor on which stalility of ionic compound is measured?
Answer:
The stability of an ionic compound is measured by the amount of energy released during lattice formation.

Question B.
Arrange the following compounds on the basis of lattice energies in decreasing (descending) order: BeF₂, AlCl₃, LiCl, CaCl₂, NaCl.
Answer:
AlCl₃ > BeF₂ > CaCl₂ > LiCl > NaCl

Question C.
Give the total number of electrons around sulphur (S) in SF₆ compound.
Answer:
The total number of electrons around sulphur (S) in SF₆ is 12.

Question D.
Covalant bond is directional in nature. Justify.
Answer:
Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature.

Question E.
What are the interacting forces present during formation of a molecule of a compound ?
Answer:
a. Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
b. Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

Question F.
Give the type of overlap by which pi (π) bond is formed.
Answer:
The type of overlap by which pi (π) bond is formed is p-p lateral overlap.

Question G .
Mention the steps involved in Hybridization.
Answer:
The steps involved in hybridization are:

formation of the excited state and
mixing and recasting of orbitals.

Question H.
Write the formula to calculate bond order of molecule.
Answer:
Bond order of a molecule = \(\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}\)
where, N_b is the number of electrons present in bonding MOs and N_a is the number of electrons present in antibonding MOs.

Question I.
Why is O₂ molecule paramagnetic?
Answer:
The electronic configuration of O₂ molecule is (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p_z)² (π2p_x)² (π2p_y)² (π*2p_x)¹ (π*2p_y)¹
Since the oxygen molecule contains two unpaired electrons, it is paramagnetic.

Question J.
What do you mean by formal charge ? Explain its significance with the help of suitable example.
Answer:
Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.

Structure (I):

Structure (II):

Structure (III):

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.

In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.

11th Chemistry Digest Chapter 5 Chemical Bonding Intext Questions and Answers

(Textbook Page No. 55)

Question 1.
Why are atoms held together in chemical compounds?
Answer:
Atoms are held together in chemical compounds due to chemical bonds.

Question 2.
How are chemical bonds formed between two atoms?
Answer:
There are two ways of formation of chemical bonds:

by loss and gain of electrons
by sharing a pair of electrons between the two atoms.

In either process of formation of chemical bond, each atom attains a stable noble gas electronic configuration.

Question 3.
Which electrons are involved in the formation of chemical bonds?
Answer:
The electrons present in the outermost shell of an atom are involved in the formation of a chemical bond.

Internet my friend (Textbook Page No. 55)

Question 1.
Search more atoms, which complete their octet during chemical combinations.
Answer:
In compounds like KCl, MgCl₂, CaO, NaF, etc, the constituent atoms complete their octet by lose or gain of electrons.
e.g. K → K⁺ + e^–
Cl + e^– → Cl^–
K⁺ + Cl^– → KCl
[Note: Students are expected to search more atoms on their own.]

Use your brainpower. (Textbook Page No. 60)

Question 1.
Which atom in \(\mathrm{NH}_{4}^{+}\) will have formal charge +1?
Answer:
In \(\mathrm{NH}_{4}^{+}\), nitrogen atom (N) will have formal charge of+1.

Use your brainpower. (Textbook Page No. 61)

Question 1.
How many electrons will be around I in the compound IF₇?
Answer:
Lewis structure of IF₇ is:

In IF₇, iodine (I) atom will be surrounded by 14 electrons.

Question 2.
Why is H₂ stable even though it never satisfies the octet rule?
Answer:
The valence shell configuration of hydrogen atom is 1s¹. Two hydrogen atoms approach each other and share their valence electrons. By having two electrons in its valence shell, H atom attains the nearest noble gas configuration of He. H₂ molecule attains stability due to duplet formation. Hence, H₂ is stable even though it never satisfies the octet rule.

(Textbook Page No. 64)

Question 1.
Lowering of energy takes during bond formation. How does this happen?
Answer:
i. When two combining atoms approach each other to form a covalent bond, the following interacting forces come into play.

Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

ii. The balance between attractive and repulsive forces decide whether the bond will be formed or not.
iii. When the magnitude of attractive forces is more than the magnitude of repulsive forces, the energy of the system decreases and a covalent bond is formed.
iv. When the magnitude of repulsive forces becomes more than that of attraction, the total energy of the system increases and a covalent bond is not formed.
Hence, lowering of energy takes during bond formation.

Can you tell? (TextBook Page No. 76)

Question 1.
Which molecules are polar?
H-I, H-O-H, H-Br, Br₂, N₂, I₂, NH₃
Answer:
i. H-I: Polar
ii. H-O-H: Polar
iii. H-Br: Polar
iv. Br₂: Nonpolar
v. N₂: Nonpolar
vi. I₂: Nonpolar
vii. NH₃: Polar

11th Std Chemistry Questions And Answers:

11th Biology Chapter 9 Exercise Morphology of Flowering Plants Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 9 Morphology of Flowering Plants Textbook Exercise Questions and Answers.

Morphology of Flowering Plants Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Biology Chapter 9 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 9 Exercise Solutions

1. Choose correct option

Question (A)
Which one of the following will grow better in moist and shady region?
(a) Opuntia
(b) Orchid
(c) Mangrove
(d) Lotus
Answer:
(b) Orchid

Question (B)
A particular plant had a pair of leaves at each node arranged in one plane. What is the arrangement called?
(a) Alternate phyllotaxy
(b) Decussate phyllotaxy
(c) Superposed phyllotaxy
(d) Whorled phyllotaxy
Answer:
(c) Superposed phyllotaxy

Question (C)
In a particular flower the insertion of floral whorls was in such a manner, so the ovary was below other three whorls, but its stigma was taller than other three whorls. What will you call such flower?
(a) Hypogynous
(b) Perigynous
(c) Inferior ovary
(d) Half superior – half inferior
Answer:
(c) Inferior ovary

Question (D)
Beet and Arum both store food for perennation.
Are the examples for two different types?
(a) Beet is a stem but Arum is a root
(b) Beet is a root but Arum is a stem
(c) Beet is a stem but Arum is a leaf
(d) Beet is a stem but Arum is an inflorescence
Answer:
(b) Beet is a root but Arum is a stem

2. Answer the following questions

Question (A)
Two of the vegetables we consume are nothing but leaf bases. Which are they?
Answer:
Onion, Garlic

Question (B)
Opuntia has spines but Carissa has thorns. What is the difference?
Answer:

In Opuntia, stem is modified into leaf like photosynthetic organ known as phylloclade.
Spines growing on phylloclade of Opuntia are leaves, modified to reduce the loss of water through transpiration.
Thoms in Carissa are modified apical buds. They provide protection against browsing animals.
Thus, spines in Opuntia and thorns in Carissa have different origin and function.

Question (C)
Teacher described Hibiscus as solitary Cyme. What it means?
Answer:
1. In Cymose inflorescence, growth of peduncle is finite and it terminates into flower.
2. In Hibiscus, flower is borne singly at the tip of peduncle. Hence, teacher described Hibiscus as solitary cyme.

3. Write notes on

Question (A)
Fusiform root.
Answer:
Fusiform root:
1. Fusiform root is the modification of tap root for food storage.
2. Fusiform root:
The fusiform root is swollen in the middle and tapering towards both ends forming spindle shaped structure, e.g. Radish (Raphanus sativus)

Question (B)
Racemose inflorescence.
Answer:
Racemose inflorescence

Question (C)
Fasciculated tuberous root.
Answer:
Fasciculated tuberous root:
1. Fasciculated tuberous roots are modification of adventitious roots for storage of food.
2. Fasciculated tuberous roots do not develop any definite shape like modified tap roots.
3. a. A cluster of roots arising from one point which becomes thick and fleshy due to storage of food is known as fasciculated tuberous root.
b. These clusters are seen at the base of the stem, e.g. Dahlia, Asparagus, etc.

Question (D)
Region of cell maturation.
Answer:
Region of maturation/region of differentiation:
a. It is the uppermost major part of the root.
b. The cells of this region are quite impermeable to water due to thick wall.
c. The cells show differentiation and form different types of tissues.
d. This region helps in fixation of plant and conduction of absorbed substances.
e. Development of lateral roots also takes place from this region.

Question (E)
Rhizome.
Answer:
Rhizome:

Rhizome is a modification of underground stem for storage of food.
It is prostrate, dorsiventrally thickened and brownish in colour.
It grows either horizontally or obliquely beneath the soil.
Rhizome shows nodes and intemodes. It bears terminal and axillary buds at nodes.
Terminal bud under favourable conditions produces aerial shoot which degenerates at the end of favourable condition.
Growth of rhizome takes place with lateral buds, such growth is known as sympodial growth, e.g. Ginger (Zingiber officinale), Turmeric {Curcuma domestica), Canna etc.
In plants where rhizomes grow obliquely, terminal bud brings about growth of rhizomes. This is known as monopodial growth, e.g. Nymphea, Nelumbo (Lotus), Pteris (Fern) etc.
Rhizomes perform functions like storage of food, vegetative propagation and perennation.

Question (F)
Stolon.
Answer:
Stolons:
1. The slender lateral branch arising from the base of main axis is known as stolon.
2. In some plants it is above ground (wild strawberry).
3. Primarily stolon shows upward growth in the form of ordinary branch, but when it bends and touches the ground terminal bud grows into new shoot and develops adventitious roots.
e.g. Wild Strawberry, Jasmine, Mentha, etc. [Any one example]

Question (G)
Leaf venation.
Answer:
Leaf venation:

Arrangement of veins and veinlets in leaf lamina is known as venation.
Veins are responsible for conduction of water and minerals as well as food.
The structural framework of the lamina is developed by veins.
There are two types of leaf venation: parallel venation which is found in monocot leaves and reticulate venation which is found in dicot leaves.

Question (H)
Cymose inflorescence.
Answer:
Cymose inflorescence.

Question (I)
Perianth.
Answer:
Perianth (P):
a. Many times, calyx and corolla remain undifferentiated. Such member is known as tepal.
b. The whorl of tepals is known as Perianth.
c. It protects other floral whorls.
d. If all the tepals are free the condition is called as polyphyllous and if they are fused the condition is called as gamophyllous.
e. Sepaloid perianth shows green tepals, while petaloid perianth shows brightly coloured tepals. e.g. Lily, Amaranthus, Celosia, etc.
f. Petaloid tepal helps in pollination and sepaloid tepals can perform photosynthesis.

Question (J)
Write a short note on vexillary aestivation.
Answer:
Vexillary: Corolla is butterfly shaped and consists of five petals. Outermost and largest is known as standard or vexillum, two lateral petals are wings and two smaller fused forming boat shaped structures keel. e.g. Pisum sativum

Question (K)
Write a short note on axile placentation.
Answer:
Axile placentation: Placentation: The mode of arrangement of ovules on the placenta within the ovary is called placentation.
Axile: Ovules are placed on the central axis of a multilocular ovary, e.g. China rose, Cotton, etc.

Question 4.
Identify the following figures and write down the types of leaves arrangement.

Answer:
1. The given figures represent phyllotaxy. It is the arrangement of leaves on the stem and branches in a specific manner.
2. Figure ‘a’ and ‘b’ represents, alternate phyllotaxy. In this type of phyllotaxy, single leaf arises from each node of a stem. e.g. Mango
3. Figure ‘c’ represents opposite decussate phyllotaxy. In this type of phyllotaxy, a pair of leaf arise from each node and the consecutive pair at right angle to the previous one. e.g. Calotropis.

5. Students were on the excursion to a botanical garden. They noted following observation. Will you be able to help them in understanding those conditions?

Question (A)
A wiry outgrowth was seen on a plant arising from in between the leaf and stem.
Answer:
A wiry outgrowth on a plant arising from in between the leaf and stem can be an axillary stem tendril. Stem tendrils:
a. Tendrils are thin, wiry, photosynthetic, leafless coiled structures.
b. They give additional support to developing plant.
c. Tendrils have adhesive glands for fixation.

Question (B)
There was a green plant with flat stem, but no leaves. The entire plant was covered by soft spines.
Answer:
Student must have observed phylloclade, which is a modification of stem.
Phylloclade:
a. Modification of stem into leaf like photosynthetic organ is known as phylloclade.
b. Being stem it possesses nodes and internodes.
c. It is thick, fleshy and succulent, contains mucilage for retaining water e.g. Opuntia, Casuarina (Cylindrical shaped phylloclade) and Muehlenbeckia (ribbon like phylloclade).

Question (C)
Many oblique roots were given out from the lower nodes, apparently for extra support.
Answer:
a. Students must have observed adventitious roots in monocotyledonous plants like maize, sugarcane, wheat, etc.
b. Adventitious roots develop from any part of a plant other than radicle.
c. In such plants, adventitious roots arise from the lower node of a stem and provide extra support to the plant. These roots are also called as stilt roots.

Question (D)
Many plants in the marshy region had upwardly growing roots. They could be better seen during low tide.
Answer:
a. Plants growing in marshy region (halophytes) produce upwardly growing roots called as
pneumatophores or respiratory roots.
b. The main root system of these plants does not get sufficient air for respiration as soil is water logged.
c. Due to this, mineral absorption of plant also gets affected.
d. To overcome this problem underground roots, develop special roots which are negatively geotropic; growing vertically upward.
e. These roots are conical projections present around main trunk of plant.
f. Respiratory roots show presence of lenticels which helps in gaseous exchange.

Question (E)
A plant had leaves with long leaf apex, which was curling around a support.
Answer:
a. Students must have observed leaf tip tendril.
b. In some weak stems, leaf apex modifies into thin, green, wiry, coiled structure called as leaf tendril.
c. Such leaf tendrils, help in climbing by curling around a support.

Question (F)
A plant was found growing on other plant. Teacher said it is not a parasite. It exhibited two types of roots.
Answer:
a. Student must have observed an epiphytic plants like Dendrobium, Vanda growing on other plant.
b. The two types of roots exhibited by this plant must be clinging roots and epiphytic roots.
c. Clinging roots:
1. Clinging roots are tiny roots develop along intemodes, show disc at tips.
2. It exudes sticky substance which enables plant to get attached to the substratum without damaging it.

d. Epiphytic roots:
1. Epiphytic plants like Vanda, Dendrobium grow on branches of trees in dense rain forests and are unable to obtain moisture from soil.
2. Such plants produce epiphytic roots which hang in the air.
3. The roots are provided with a spongy membranous absorbent covering of the velamen tissue.
4. The cells of velamen tissue are hygroscopic and have porous walls, thus they can absorb moisture from air.
5. Epiphytic roots can be silvery white or green and are without root cap.

Question (G)
While having lunch onion slices were served to them. Teacher asked which part of the plant are you eating?
Answer:
a. The edible part of an onion is fleshy leaves.
b. Onion is a bulb, in which stem is highly reduced, discoid and possesses adventitious roots at the base.
c. This stem bears a whorl of fleshy leaves which store food material.
d. The scale leaves or fleshy leaves are arranged in concentric manner over the stem. Some outer scale leaves become thin and dry. Thus, it is also called as tunicated or layered bulb.

Question (H)
Students observed large leaves of coconut and small leaves of Mimosa. Teacher asked it what way they are similar?
Answer:
a. Both large leaves of coconut and small leaves of Mimosa show pinnately compound leaves.
b. In both plants, leaf lamina is divided into number of leaflets.
c. Leaflets are present laterally on a common axis called rachis, which represents the midrib of the leaf.

Question (I)
Teacher showed them Marigold flower and said it is not one flower. What the teacher meant?
Answer:
a. Marigold flower is an inflorescence in which flowers are produced in a definite manner on a peduncle.
b. In Marigold, racemose type of inflorescence can be observed.
c. In this, peduncle condenses to form a flat rounded structure called receptacle.
d. Opening of flower centripetal i.e. younger flowers are towards the centre and open later, while older flowers towards the periphery and open first.

Question (J)
Students cut open a Papaya fruit and found all the seeds attached to the sides. Teacher inquired about the possible placentation of Papaya ovary.
Answer:
a. In Papaya, seeds are attached to the sides of a fruit. Thus, parietal placentation is possible in papaya ovary,
b. In parietal placentation, ovules are placed on the inner wall of unil unilocular ovary of multicarpellary, syncarpus gynoecium.

Question 6.
Match the following.

Answer:
(i-c-1), (ii-e-3), (iii-a-4), (iv-b-5), (v-d-2)
[Note: Another example of palmately compound leaf (Bifoliate) is Balanites roxburghii.]

Question 7.
Observe the following figures and label the different parts.

Answer:

8. Differentiate with diagrammatic representation.

Question (A)
Differentiate with diagrammatic representation: Racemose and cymose inflorescence.
Answer:

Question (B)
Differentiate with diagrammatic representation: Reticulate and parallel venation
Answer:

Question (C)
Differentiate with diagrammatic representation: Taproot and Adventitious roots
Answer:

Practical / Project:

Question 1.
Collect different leaves from nearby region and observe variation in margin, leaf base, apex etc.
[Note: Students can scan the given Q.R code to study the different le
af margin, leaf base and apex.]

Question 2.
Find out and make a note of economically important plant from family Fabaceae, Solanaceae and Liliaceae.
Answer:
1. Economically important plant from family Fabaceae:
Family Fabaceae includes many pulses like gram, arhar, moong, soybean; edible oil seeds like soybean, groundnut; dye (lndigofera); fibres which can be obtained from Sun hemp, Sesbania trifolium which can be used as fodder; some plants are ornamental like lupin, sweet pea; some medicinal plants like muliathi.

2. Economically important plant from family Solanaceae:
Family Solanaceae includes many plants which are good source of food e.g. tomato, brinjal, potato; Spice e.g. chilli; Medicine e.g. belladonna, ashwagandha; Ornamental plants like Petunia.

3. Economically important plant from family Liliaceae:
Family Liliaceae includes many ornamental plants like tulip, Gloriosa, Medicinal plants like Aloe vera. Asparagus and source of colchicine, e.g. Colchicum autumnale.

Question 3.
Collect different leaves from garden and observe their veins and classify it.

11th Biology Digest Chapter 9 Morphology of Flowering Plants Intext Questions and Answers

Use your brainpower. (Textbook Page No. 102)

Why underground stem is different from roots?
Answer:
Underground stems are modified to perform different functions like storage of food, perennation and vegetative propagation. However, they differ from root in having nodes and intemodes.

Use your brainpower. (Textbook Page No. 104)

Why the stem has to perform photosynthesis in xerophytes?
Answer:
1. Xerophytes are the plants which grow in regions with scanty or no rainfall like desert.
2. In Xerophytes, leaves get modified into spines or get reduced in size to check the loss of water due to transpiration.
3. As the leaves are modified into spines, the stem becomes green in colour to do the function of photosynthesis.

Internet My Friend. (Textbook Page No. 106)

Collect information of types of leaf venation.
Answer:
1. Figure ‘R’ shows types of reticulate venation. WTien the veins and veinlets form a network, it is called
reticulate venation.
On the basis of number of mid-veins, reticulate venation is of two types:
a. Pinnate or unicostate: It is with single midrib e.g. Peepal, Mango.
b. Palmate or multicostate: It is with two or more prominent veins. It is further divided into convergent or divergent.
1. Multicostate convergent reticulate: Many prominent veins appear from the base of leaf lamina and converged in a curved manner towards the leaf apex. e.g. Zizyphus
2. Multicostate divergent reticulate: Prominent veins arise from the single point at the base of leaf lamina
and then diverge from one another towards the leaf margin, e.g. Cucurbita

2. Figure ‘P’ shows types of parallel venation. When veins run almost parallel to one another it is called parallel venation. It is of two types:
a. Unicostate: In this, lamina has single prominent mid vein from which many lateral parallel veins arise at regular intervals, e.g. Banana
b. Multicostate: In this, two or more mid veins run parallel to each other. It is further divided into convergent or divergent.

1. Multicostate convergent parallel:
Many prominent veins arise from the leaf base and then converge at leaf apex. e.g. Grasses
2. Multicostate divergent parallel:
Many prominent veins arise from the leaf base and then diverge towards margin, e.g. Borassus flabellifer (Toddy palm)

Observe and Discuss. (Textbook Page No. 112)

Observe and Discuss.
Answer:
1. Figure ‘a’ shows fruit of tomato.

It is a simple fruit as it develops from a single flower with bicarpellary syncarpous gynoecium.
It is a berry, because it has fleshy endocarp and many seeds.

2. Figure ‘b’ shows fruit of Custard apple.

It is an aggregate fruit, because it develops from a single flower with polycarpellary, apocarpous gynoecium.
Here, the ovary of each carpel gives rise to a part of the fruit called fruitlet. Hence, it is called an aggregation of fruitlets.
Custard apple can be further described as Etaerio of berries.

3. Figure ‘c’ shows fruit of pineapple.

It is a composite fruit, because it develops from a complete inflorescence.
Pineapple can be further described as Sorosis, as it develops from catkin type of inflorescence.

4. Figure ‘d’ shows fruit of milkweed.

It is a simple dehiscent dry fruit.
It has many seeds. When pericarp becomes dry and thin, it breaks open by one ventral suture.

Activity. (Textbook Page No. 113)

Study the family Liliaceae, prepare a table of following characteristics.
Answer:

Symmetry of flower	Actinomorphic
Bisexual/ Unisexual	Bisexual
Calyx	Absent
Corolla	Absent
Androecium	Stamens six, arranged in two whorls of 3 each, epiphyllous
Gynoecium	Tricarpellary, syncarpous, trilocular ovary with many ovules
Aestivation	Valvate
a. Calyx	Absent
b. Corolla	Absent
Placentation	Axile
Position of ovary	Superior ovary
Types of fruit	Capsule, rarely berry

11th Std Biology Questions And Answers:

11th Physics Chapter 13 Exercise Electromagnetic Waves and Communication System Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Prasanna

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 13 Electromagnetic Waves and Communication System Textbook Exercise Questions and Answers.

Electromagnetic Waves and Communication System Class 13 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 13 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 13 Exercise Solutions

1. Choose the correct option.

Question 1.
The EM wave emitted by the Sun and responsible for heating the Earth’s atmosphere due to green house effect is
(A) Infra-red radiation
(B) X ray
(C) Microwave
(D) Visible light
Answer:
(A) Infra-red radiation

Question 2.
Earth’s atmosphere is richest in
(A) UV
(B) IR
(C) X-ray
(D) Microwaves
Answer:
(B) IR

Question 3.
How does the frequency of a beam of ultraviolet light change when it travels from air into glass?
(A) depends on the values of p and e
(B) increases
(C) decreases
(D) remains same
Answer:
(D) remains same

Question 4.
The direction of EM wave is given by
(A) \(\bar{E}\) × \(\bar{B}\)
(B) \(\bar{E}\).\(\bar{B}\)
(C) along \(\bar{E}\)
(D) along \(\bar{B}\)
Answer:
(A) \(\bar{E}\) × \(\bar{B}\)

Question 5.
The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(A) h½
(B) h
(C) h^3/2
(D) h²
Answer:
(A) h½

Question 6.
The waves used by artificial satellites for communication purposes are
(A) Microwave
(B) AM radio waves
(C) FM radio waves
(D) X-rays
Answer:
(A) Microwave

Question 7.
If a TV telecast is to cover a radius of 640 km, what should be the height of transmitting antenna?
(A) 32000 m
(B) 53000 m
(C) 42000 m
(D) 55000 m
Answer:
(A) 32000 m

2. Answer briefly.

Question 1.
State two characteristics of an EM wave.
Answer:
i. The electric and magnetic fields, \(\vec{E}\) and \(\vec{B}\) are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.

ii. The cross product (\(\vec{E}\) × \(\vec{B}\)) gives the direction in which the EM wave travels. (\(\vec{E}\) × \(\vec{B}\)) also gives the energy carried by EM wave.

Question 2.
Why are microwaves used in radar?
Answer:
Microwaves are used in radar systems for identifying the location of distant objects like ships, aeroplanes etc.

Question 3.
What are EM waves?
Answer:
Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally at right angles to each other and to the direction of propagation are called EM waves or EM radiation.

Question 4.
How are EM waves produced?
Answer:

According to quantum theory, an electron, while orbiting around the nucleus in a stable orbit does not emit EM radiation even though it undergoes acceleration.
It will emit an EM radiation only when it falls from an orbit of higher energy to one of lower energy.
EM waves (such as X-rays) are produced when fast moving electrons hit a target of high atomic number (such as molybdenum, copper, etc.).
An electric charge at rest has an electric field in the region around it but has no magnetic field.
When the charge moves, it produces both electric and magnetic fields.
If the charge moves with a constant velocity, the magnetic field will not change with time and hence, it cannot produce an EM wave.
But if the charge is accelerated, both the magnetic and electric fields change with space and time and an EM wave is produced.
Thus, an oscillating charge emits an EM wave which has the same frequency as that of the oscillation of the charge.

Question 5.
Can we produce a pure electric or magnetic wave in space? Why?
Answer:
No.
In vacuum, an electric field cannot directly induce another electric field so a “pure” electric field wave cannot exist and same can be said for a “pure” magnetic wave.

Question 6.
Does an ordinary electric lamp emit EM waves?
Answer:
Yes, ordinary electric lamp emits EM waves.

Question 7.
Why light waves travel in vacuum whereas sound wave cannot?
Answer:
Light waves are electromagnetic waves which can travel in vacuum whereas sound waves travel due to the vibration of particles of medium. Without any particles present (like in a vacuum) no vibrations can be produced. Hence, the sound wave cannot travel through the vacuum.

Question 8.
What are ultraviolet rays? Give two uses.
Answer:
Production:

Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth’s atmosphere.

Uses:

Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
Used in burglar alarms and security systems.
Used to distinguish real and fake gems.

Question 9.
What are radio waves? Give its two uses.
Answer:

Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
Thus, by choosing suitable values of the inductance and the capacitance, radio waves of desired frequency can be produced.

Uses:

Radio waves are used for wireless communication purpose.
They are used for radio broadcasting and transmission of TV signals.
Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.

Question 10.
Name the most harmful radiation entering the Earth’s atmosphere from the outer space.
Answer:
Ultraviolet radiation.

Question 11.
Give reasons for the following:
i. Long distance radio broadcast uses short wave bands.
ii. Satellites are used for long distance TV transmission.
Answer:
i. Long distance radio broadcast uses short wave bands because electromagnetic waves only in the frequency range of short wave bands only are reflected by the ionosphere.

ii. a. It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
b. Hence, satellites are helpful in long distance TV transmission.

Question 12.
Name the three basic units of any communication system.
Answer:
Three basic (essential) elements of every communication system are transmitter, communication channel and receiver.

Question 13.
What is a carrier wave?
Answer:
The high frequency waves on which the signals to be transmitted are superimposed are called carrier waves.

Question 14.
Why high frequency carrier waves are used for transmission of audio signals?
Answer:
An audio signal has low frequency (<20 kHz) and low frequency signals cannot be transmitted over large distances. Because of this, a high frequency carrier waves are used for transmission.

Question 15.
What is modulation?
Answer:
The signals in communication system (e.g. music, speech etc.) are low frequency signals and cannot be transmitted over large distances. In order to transmit the signal to large distances, it is superimposed on a high frequency wave (called carrier wave). This process is called modulation.

Question 16.
What is meant by amplitude modulation?
Answer:
When the amplitude of carrier wave is varied in accordance with the modulating signal, the process is called amplitude modulation.

Question 17.
What is meant by noise?
Answer:

A random unwanted signal is called noise.
The source generating the noise may be located inside or outside the system.
Efforts should be made to minimize the noise level in a communication system.

Question 18.
What is meant by bandwidth?
Answer:
The bandwidth of an electronic circuit is the range of frequencies over which it operates efficiently.

Question 19.
What is demodulation?
Answer:
The process of regaining signal from a modulated wave is called demodulation. This is the reverse process of modulation.

Question 20.
What type of modulation is required for television broadcast?
Answer:
Amplitude modulation is required for television broadcast.

Question 21.
How does the effective power radiated by an antenna vary with wavelength?
Answer:

To transmit a signal, an antenna or an aerial is needed.
Power radiated from a linear antenna of length l is, P ∝ (\(\frac {l}{λ}\))²
where, λ is the wavelength of the signal.

Question 22.
Why should broadcasting programs use different frequencies?
Answer:
If broadcasting programs run on same frequency, then the information carried by these waves will get mixed up with each other. Hence, different broadcasting programs should run on different frequencies.

Question 23.
Explain the necessity of a carrier wave in communication.
Answer:

Without a carrier wave, the input signals could be carried by very low frequency electromagnetic waves but it will need quite a bit of amplification in order to transmit those very low frequencies.
The input signals themselves do not have much power and need a fairly large antenna in order to transmit the information.
Hence, it is necessary to impose the input signal on carrier wave as it requires less power in order to transmit the information.

Question 24.
Why does amplitude modulation give noisy reception?
Answer:
i. In amplitude modulation, carrier is varied in accordance with the message signal.

ii. The higher the amplitude, the greater is magnitude of the signal. So even if due to any reason, the magnitude of the signal changes, it will lead to variation in the amplitude of the signal. So its easy for noise to disturb the amplitude modulated signal.

Question 25.
Explain why is modulation needed.
Answer:
Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.

3. Solve the numerical problem.

Question 1.
Calculate the frequency in MHz of a radio wave of wavelength 250 m. Remember that the speed of all EM waves in vacuum is 3.0 × 10⁸ m/s.
Answer:
Given: λ = 250 m, c = 3 × 10⁸ m/s
To find: Frequency (v)
Formula: c = v⁸
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{250}\) = 1.2 × 10⁶ Hz
= 1.2 MHz

Question 2.
Calculate the wavelength in nm of an X-ray wave of frequency 2.0 × 10¹⁸ Hz.
Solution:
Given: c = 3 × 10⁸, v = 2 × 10¹⁸ Hz
To find: Wavelength (λ)
Formula: c = vλ
Calculation. From formula,
λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{2×10^{18}}\) = 1.5 × 10^-10
= 0.15 nm

Question 3.
The speed of light is 3 × 10⁸ m/s. Calculate the frequency of red light of wavelength of 6.5 × 10^-7 m.
Answer:
Given: c = 3 × 10⁸ m/s, λ = 6.5 × 10^-7 m
To find: Frequency (v)
Formula: c = vλ
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{6.5×10^{-7}}\) = 4.6 × 10¹⁴ Hz

Question 4.
Calculate the wavelength of a microwave of frequency 8.0 GHz.
Answer:
Given: v = 8 GHz = 8 × 10⁹ Hz,
c = 3 × 10⁸ m/s
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{8×10^9}\) = 3.75 × 10^-2
= 3.75 cm

Question 5.
In a EM wave the electric field oscillates sinusoidally at a frequency of 2 × 10¹⁰ What is the wavelength of the wave?
Answer:
Given: v = 2 × 10¹⁰ Hz, c = 3 × 10⁸ m
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{2×10^{10}}\) = 1.5 × 10^-2

Question 6.
The amplitude of the magnetic field part of a harmonic EM wave in vacuum is B0 = 5 X 10-7 T. What is the amplitude of the electric field part of the wave?
Answer:
Given: B₀ = 5 × 10^-7 T, c = 3 × 10⁸
To find: Amplitude of electric field (E₀)
Formula: c = \(\frac {E_0}{B_0}\)
Calculation /From formula,
E₀ = c × B₀
= 3 × 10⁸ × 5 × 10^-7
= 150 V/m

Question 7.
A TV tower has a height of 200 m. How much population is covered by TV transmission if the average population density around the tower is 1000/km²? (Radius of the Earth = 6.4 × 10⁶ m)
Answer:
Given: h = 200 m,
Population density (n)
= 1000/km² = 1000 × 10^-6/m² = 10^-3/m²
R = 6.4 ×10⁶ m
To find: Population covered
Formulae: i. A = πd² = π(\(\sqrt{2Rh}\))² = 2πRh
ii. Population covered = nA
Calculation /From formula (i),
A = 2πRh
= 2 × 3.142 × 6.4 × 10⁶ × 200
≈ 8 × 10⁹ m²
From formula (ii),
Population covered = nA
= 10^-3 × 8 × 10⁹
= 8 × 10⁶

Question 8.
Height of a TV tower is 600 m at a given place. Calculate its coverage range if the radius of the Earth is 6400 km. What should be the height to get the double coverage area?
Answer:
Given: h = 600 m, R = 6.4 × 10⁶ m
To find: Range (d)
Height to get the double coverage (h’)
Formula: d = \(\sqrt{2hR}\)
Calculation: From formula,
d = \(\sqrt{2×600×6.4×10^6}\) = 87.6 × 10³ = 87.6 km
Now, for A’ = 2A
π(d’)² = 2 (πd²)
∴ (d’)² = 2d²
From formula,
h’ = \(\frac{(d’)^2}{2R}\)
= \(\frac{2d^2}{2R}\)
= 2 × h ……….. (∵ h = \(\frac{d^2}{2R}\))
= 2 × 600
=1200 m

Question 9.
A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of Earth is 6.4 × 10⁶ m.
Answer:
Given: h_t = 32 m, h_r = 50 m, R = 6.4 × 10⁶ m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation: From formula,
d_t = \(\sqrt{2Rh_t}\) = \(\sqrt{2×6.4×10^6×32}\)
= 20.238 × 10³ m
= 20.238 km
d_r = \(\sqrt{2Rh_t}\)
= \(\sqrt{2×6.4×10^6×50}\)
= 25.298 × 10³ m
= 25.298 km
Now, d = d_t + d_r
= 20.238 + 25.298
= 45.536 km

11th Physics Digest Chapter 13 Electromagnetic Waves and Communication System Intext Questions and Answers

Can you recall? (Textbookpage no. 229)

Question 1.
i. What is a wave?
Answer:
Wave is an oscillatory disturbance which travels through a medium without change in its form.

ii. What is the difference between longitudinal and transverse waves?
Answer:
a. Transverse wave: A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of wave is called transverse wave.
b. Longitudinal wave: A wave in which particles of the medium vibrate in a direction parallel to the direction of propagation of wave is called longitudinal wave.

iii. What are electric and magnetic fields and what are their sources?
Answer:
a. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
b. A magnetic field is produced around a magnet or around a current carrying conductor.

iv. By which mechanism heat is lost by hot bodies?
Answer:
Hot bodies lose the heat in the form of radiation.

Question 2.
What are Lenz’s law, Ampere’s law and Faraday’s law?
Answer:
Lenz’s law:
Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.

Ampere’s law:
Ampere’s law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.

Faraday’s law:
Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.

Internet my friend. (Tpxtboakpage no. 240)

https//www.iiap.res.in/centers/iao
[Students are expected to visit the above mentioned website and collect more information about different EM wave propagations used by astronomical observatories.]

11th Std Physics Questions And Answers:

11th Chemistry Chapter 8 Exercise Elements of Group 1 and 2 Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 8 Elements of Group 1 and 2 Textbook Exercise Questions and Answers.

Elements of Group 1 and 2 Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 8 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 8 Exercise Solutions

1. Explain the following

Question A.
Hydrogen shows similarity with alkali metals as well as halogens.
Answer:

Electronic configuration of hydrogen is 1s¹ which is similar to the outer electronic configuration of alkali metals of group 1 i.e., ns¹.
However, 1s¹ also resembles the outer electronic configuration of group 17 elements i.e., ns² np⁵.
By adding one electron to H, it will attain electronic configuration of the inert gas He which is 1s² and by adding one electron to ns² np⁵ we get ns²np⁶ which is the outer electronic configuration of the remaining inert gases.
Therefore, some chemical properties of hydrogen are similar to those of alkali metals while some resemble halogens.

Hence, hydrogen shows similarity with alkali metals as well as halogens.

Question B.
Standard reduction potential of alkali metals have high negative values.
Answer:

The general outer electronic configuration of alkali metals is ns¹.
They readily lose one valence shell electron to achieve stable noble gas configuration and hence, they are highly electropositive and are good reducing agents.

Hence, standard reduction potentials of alkali metals have high negative values.

Question C.
Alkaline earth metals have low values of electronegativity; which decrease down the group.
Answer:

Electronegativity represents attractive force exerted by the nucleus on shared electrons.
The general outer electronic configuration of alkaline earth metals is ns². They readily lose their two valence shell electrons to achieve stable noble gas configuration. They are electropositive and hence, they have low values of electronegativity.

Question D.
Sodium dissolves in liquid ammonia to form a solution which shows electrical conductivity.
Answer:
i. Sodium dissolves in liquid ammonia giving deep blue coloured solutions which is electrically conducting in nature.
Na + (x + y) NH₃ → [Na(NH₃)_x]+ + [e(NH₃)_y]^–
ii. Due to formation of ions, the solution shows electrical conductivity.

Question E.
BeCl₂ is covalent while MgCl₂ is ionic.
Answer:

Be²⁺ ion has very small ionic size and therefore, it has very high charge density.
Due to this, it has high tendency to distort the electron cloud around the negatively charged chloride ion (Cl^–) which is larger in size.
This results in partial covalent character of the bond in BeCl₂.
Mg²⁺ ion has very less tendency to distort the electron cloud of Cl^– due to the bigger size of Mg²⁺ as compared to Be²⁺.

Hence, BeCl₂ is covalent while MgCl₂ is ionic.

Question F.
Lithium floats an water while sodium floats and catches fire when put in water.
Answer:

When lithium and sodium react with water, hydrogen gas is released. Due to these hydrogen gas bubbles, lithium and sodium floats on water.
eg. 2Na + 2H₂O → 2Na⁺ + 2OH^– + H₂↑
The reactivity of group 1 metals increases with increasing atomic radius and lowering of ionization enthalpy down the group.
Thus, sodium having lower ionization enthalpy, is more reactive than lithium.
Hence, lithium reacts slowly while sodium reacts vigorously with water.
Since the reaction of sodium with water is highly exothermic, it catches fire when put in water.

2. Write balanced chemical equations for the following.

Question A.
CO₂ is passed into concentrated solution of NaCl, which is saturated with NH₃.
Answer:

Question B.
A 50% solution of sulphuric acid is subjected to electrolyte oxidation and the product is hydrolysed.
Answer:

Question C.
Magnesium is heated in air.
Answer:

Question D.
Beryllium oxide is treated separately with aqueous HCl and aqueous NaOH solutions.
Answer:
Beryllium oxide (BeO) is an amphoteric oxide and thus, it reacts with both acid (HCl) as well as base (NaOH) to give the corresponding products.
i. \(\mathrm{BeO}+\underset{(\text { Acid })}{2 \mathrm{HCl}} \longrightarrow \mathrm{BeCl}_{2}+\mathrm{H}_{2} \mathrm{O}\)
ii. \(\mathrm{BeO}+\underset{(\text { Base })}{2 \mathrm{NaOH}} \longrightarrow \mathrm{Na}_{2} \mathrm{BeO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

3. Answer the following questions

Question A.
Describe the diagonal relationship between Li and Mg with the help of two illustrative properties.
Answer:
a. The relative placement of these elements with similar properties in the periodic table is across a diagonal and is called diagonal relationship.
b. Lithium is placed in the group 1 and period 2 of the modem periodic table. It resembles with magnesium which is placed in the group 2 and period 3.

ii. Li and Mg show similarities in many of their properties.
e. g.
a. Reaction with oxygen:
1. Group 1 elements except lithium, react with oxygen present in the air to form oxides (M₂O) as well as peroxides (M₂O₂) and superoxides (MO₂) on further reaction with excess of oxygen.
2. This anomalous behaviour of lithium is due to its resemblance with magnesium as a result of diagonal relationship.
3. As group 2 elements form monoxides i.e., oxides, lithium also form monoxides.

b. Reaction with nitrogen:
1. All the group 1 elements react only with oxygen present in the air to form oxides while group 2 elements react with both nitrogen and oxygen present in the air forming corresponding oxides and nitrides.
2. However, lithium reacts with oxygen as well as nitrogen present in the air due to its resemblance with magnesium.

Question B.
Describe the industrial production of dihydrogen from steam. Also write the chemical reaction involved.
Answer:
Three stages are involved in the industrial production of dihydrogen from steam.
i. Stage 1:
a. Reaction of steam on hydrocarbon or coke (C) at 1270 K temperature in presence of nickel catalyst gives water-gas which is a mixture of carbon monoxide and hydrogen.
1. Reaction of steam with hydrocarbon:

2. Reaction of steam with coke or carbon (C):

b. Sawdust, scrapwood, etc. can also be used in place of carbon.

ii. Stage 2:
Water-gas shift reaction: When carbon monoxide in the water-gas reacts with steam in the presence of iron chromate (FeCrO₄) as catalyst, it gets transformed into carbon dioxide. This is called water-gas shift reaction.

iii. Stage 3: In the last stage, carbon dioxide is removed by scrubbing with sodium arsenite solution.

Question C.
A water sample, which did not give lather with soap, was found to contain Ca(HCO₃)₂ and Mg(HCO₃)₂. Which chemical will make this water give lather with soap? Explain with the help of chemical reactions.
Answer:

Soap does not lather in hard water due to presence of the soluble salts of calcium and magnesium in it. So, the given water sample is hard water.
Hardness of hard water can be removed by removal of these calcium and magnesium salts.
Sodium carbonate is used to make hard water soft as it precipitates out the soluble calcium and magnesium salts in hard water as carbonates. Thus, it will make water give lather with soap.
e.g. Ca(HCO₃)_2(aq) + Na₂CO_3(aq) → CaCO_3(s) + 2NaHCO_3(aq)

Question D.
Name the isotopes of hydrogen. Write their atomic composition schematically and explain which of these is radioactive ?
Answer:
i. Hydrogen has three isotopes i.e., hydrogen \(\left({ }_{1}^{1} \mathrm{H}\right)\), deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) and tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) with mass numbers 1, 2 and 3 respectively.
ii. They all contain one proton and one electron but different number of neutrons in the nucleus.
iii. Atomic composition of isotopes of hydrogen:

iv. Tritium is a radioactive nuclide with half-life period 12.4 years and emits low energy β^– particles.
v. Schematic representation of isotopes of hydrogen is as follows:

4. Name the following

Question A.
Alkali metal with smallest atom.
Answer:
Lithium (Li)

Question B.
The most abundant element in the universe.
Answer:
Hydrogen (H)

Question C.
Radioactive alkali metal.
Answer:
Francium (Fr)

Question D.
Ions having high concentration in cell sap.
Answer:
Potassium ions (K⁺)

Question E.
A compound having hydrogen, aluminium and lithium as its constituent elements.
Answer:
Lithium aluminium hydride (LiAlH₄)

5. Choose the correct option.

Question A.
The unstable isotope of hydrogen is …..
a. H-1
b. H-2
c. H-3
d. H-4
Answer:
c. H-3

Question B.
Identify the odd one.
a. Rb
b. Ra
c. Sr
d. Be
Answer:
a. Rb

Question C.
Which of the following is Lewis acid ?
a. BaCl₂
b. KCl
c. BeCl₂
d. LiCl
Answer:
c. BeCl₂

Question D.
What happens when crystalline Na₂CO₃ is heated ?
a. releases CO₂
b. loses H₂O
c. decomposes into NaHCO₃
d. colour changes.
Answer:
b. loses H₂O

Activity :

1. Collect the information of preparation of dihydrogen and make a chart.
2. Find out the s block elements compounds importance/uses.
Answer:
1.

2. Uses of s-block elements:
Group 1 elements (alkali metals):
a. Lithium: Lithium is widely used in batteries.
b. Sodium:

Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
Sodium is also used as an important reagent in the Wurtz reaction.
It is used in the manufacture of sodium vapour lamp.

c. Potassium:

Potassium has a vital role in biological system.
Potassium chloride (KCl) is used as a fertilizer.
Potassium hydroxide (KOH) is used in the manufacture of soft soaps and also as an excellent absorbent of carbon dioxide.
Potassium superoxide (KO₂) is used as a source of oxygen.

d. Caesium: Caesium is used in devising photoelectric cells.

Group 2 elements (alkaline earth metals):
a. Magnesium: Magnesium hydroxide [Mg(OH)₂] in its suspension form is used as an antacid.
b. Calcium: Compounds of calcium such as limestone and gypsum are used as constituents of cement and mortar.
c. Barium: BaSO₄ being insoluble in H₂O and opaque to X-rays is used as ‘barium meal’ to scan the X-ray of human digestive system.
[Note: Students are expected to collect additional information about preparation of dihydrogen and uses of s-block elements on their own.]

11th Chemistry Digest Chapter 8 Elements of Group 1 and 2 Intext Questions and Answers

Can you recall? (Textbook Page No. 110)

Question 1.
Which is the first element in the periodic table?
Answer:
Hydrogen is the first element in the periodic table.

Question 2.
What are isotopes?
Answer:
Many elements exist naturally as a mixture of two or more types of atoms or nuclides. These individual nuclides are called isotopes of that element. Isotopes of an element have the same atomic number (number of protons) but different atomic mass numbers due to different number of neutrons in their nuclei.

Question 3.
Write the formulae of the compounds of hydrogen formed with sodium and chlorine.
Answer:
Hydrogen combines with sodium to form sodium hydride (NaH) while it reacts with chlorine to form hydrogen chloride (HCl).

Can you tell? (Textbook Page No. 110)

Question 1.
In which group should hydrogen be placed? In group 1 or group 17? Why?
Answer:

Hydrogen contains one valence electron in its valence shell and thus, its valency is one. Therefore, hydrogen resembles alkali metals (group 1 elements) as they also contain one electron in their valence shell (alkali metals tend to lose their valence electron).
However, hydrogen also shows similarity with halogens (group 17 elements) as their valency is also one because halogens tend to accept one electron in their valence shell.
Due to this unique behaviour, it is difficult to assign any definite position to hydrogen in the modem periodic table.

Just think! (Textbook Page No. 112)

Question 1.
\(2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaH}_{(\mathrm{s})}\)
In the above chemical reaction which element does undergo oxidation and which does undergo reduction?
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + H_2(g) → 2Na⁺ + 2H^–
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and each hydrogen atom gains one electron to form H. This can be represented as follows:

iii. Na is oxidised to NaH by loss of electrons while the elemental hydrogen is reduced to NaH by gain of electrons.

Can you recall? (Textbook Page No. 113)

Question i.
What is the name of the family of reactive metals having valency one?
Answer:
The family of reactive metals having valency one is known as alkali metals (group 1).

Question ii.
What is the name of the family of reactive metals having valency two?
Answer:
The family of reactive metals having valency two is known as alkaline earth metals (group 2).

11th Std Chemistry Questions And Answers:

11th Physics Chapter 3 Exercise Motion in a Plane Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Prasanna

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 3 Motion in a Plane Textbook Exercise Questions and Answers.

Motion in a Plane Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Physics Chapter 3 Exercise Solutions Maharashtra Board

Physics Class 11 Chapter 3 Exercise Solutions

1. Choose the correct option.

Question 1.
An object thrown from a moving bus is on example of __________
(A) Uniform circular motion
(B) Rectilinear motion
(C) Projectile motion
(D) Motion in one dimension
Answer:
(C) Projectile motion

Question 2.
For a particle having a uniform circular motion, which of the following is constant ____________.
(A) Speed
(B) Acceleration
(C) Velocity
(D) Displacement
Answer:
(A) Speed

Question 3.
The bob of a conical pendulum undergoes ___________
(A) Rectilinear motion in horizontal plane
(B) Uniform motion in a horizontal circle
(C) Uniform motion in a vertical circle
(D) Rectilinear motion in vertical circle
Answer:
(B) Uniform motion in a horizontal circle

Question 4.
For uniform acceleration in rectilinear motion which of the following is not correct?
(A) Velocity-time graph is linear
(B) Acceleration is the slope of velocity time graph
(C) The area under the velocity-time graph equals displacement
(D) Velocity-time graph is nonlinear
Answer:
(D) Velocity-time graph is nonlinear

Question 5.
If three particles A, B and C are having velocities \(\overrightarrow{\mathrm{v}}_{A}\), \(\overrightarrow{\mathrm{v}}_{B}\) and \(\overrightarrow{\mathrm{v}}_{C}\) which of the following formula gives the relative velocity of A with respect to B
(A) \(\overrightarrow{\mathrm{v}}_{A}\) + \(\overrightarrow{\mathrm{v}}_{B}\)
(B) \(\overrightarrow{\mathrm{v}}_{A}\) – \(\overrightarrow{\mathrm{v}}_{C}\) + \(\overrightarrow{\mathrm{v}}_{B}\)
(C) \(\overrightarrow{\mathrm{v}}_{A}\) – \(\overrightarrow{\mathrm{v}}_{B}\)
(D) \(\overrightarrow{\mathrm{v}}_{C}\) – \(\overrightarrow{\mathrm{v}}_{A}\)
Answer:
(C) \(\overrightarrow{\mathrm{v}}_{A}\) – \(\overrightarrow{\mathrm{v}}_{B}\)

2. Answer the following questions.

Question 1.
Separate the following in groups of scalar and vectors: velocity, speed, displacement, work done, force, power, energy, acceleration, electric charge, angular velocity.
Answer:
Scalars
Speed, work done, power, energy, electric charge.

Vectors
Velocity, displacement, force, acceleration, angular velocity (pseudo vector).

Question 2.
Define average velocity and instantaneous velocity. When are they same?
Answer:
Average velocity:

Average velocity (\(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\)) of an object is the displacement (\(\Delta \overrightarrow{\mathrm{x}}\)) of the object during the time interval (∆t) over which average velocity is being calculated, divided by that time interval.
Average velocity = (\(\frac{\text { Displacement }}{\text { Time interval }}\))
\(\overrightarrow{\mathrm{V}_{\mathrm{av}}}=\frac{\overrightarrow{\mathrm{x}}_{2}-\overrightarrow{\mathrm{x}}_{1}}{\mathrm{t}_{2}-\mathrm{t}_{1}}=\frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}\)
Average velocity is a vector quantity.
Its SI unit is m/s and dimensions are [M⁰L¹T^-1]
For example, if the positions of an object are x +4 m and x = +6 m at times t = O and t = 1 minute respectively, the magnitude of its average velocity during that time is V_av = (6 – 4)1(1 – 0) = 2 m per minute and its direction will be along the positive X-axis.
∴ \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}\) = 2 i m/min
Where, i = unit vector along X-axis.

Instantaneous velocity:

The instantaneous velocity (\(\overrightarrow{\mathrm{V}}\)) is the limiting value of ¡he average velocity of the object over a small time interval (∆t) around t when the value of lime interval goes to zero.
It is the velocity of an object at a given instant of time.
\(\overrightarrow{\mathrm{v}}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\)
where \(\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\) derivative of \(\overrightarrow{\mathrm{x}}\) with respect to t.

In case of uniform rectilinear motion, i.e., when an object is moving with constant velocity along a straight line, the average and instantaneous velocity remain same.

Question 3.
Define free fall.
Answer:
The motion of any object under the influence of gravity alone is called as free fall.

Question 4.
If the motion of an object is described by x = f(t) write formulae for instantaneous velocity and acceleration.
Answer:

Instantaneous velocity of an object is given as,
\(\overrightarrow{\mathrm{v}}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \overrightarrow{\mathrm{x}}}{\Delta \mathrm{t}}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\)
Motion of the object is given as, x = f(t)
The derivative f ‘(f) represents the rate of change of the position f (t) at time t, which is the instantaneous velocity of the object.
∴ \(\vec{v}=\frac{\mathrm{d} \overrightarrow{\mathrm{x}}}{\mathrm{dt}}\) = f'(t)
Acceleration is defined as the rate of change of velocity with respect to time.
The second derivative of the position function f “(t) represents the rate of change of velocity i.e., acceleration.
∴ \(\overrightarrow{\mathrm{a}}=\frac{\Delta \overrightarrow{\mathrm{v}}}{\Delta \mathrm{t}}=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}\) = f”(t)

Question 5.
Derive equations of motion for a particle moving in a plane and show that the motion can be resolved in two independent motions in mutually perpendicular directions.
Answer:

Consider an object moving in an x-y plane. Let the initial velocity of the object be \(\overrightarrow{\mathrm{u}}\) at t = 0 and its velocity at time t be \(\overrightarrow{\mathrm{v}}\).
As the acceleration is constant, the average acceleration and the instantaneous acceleration will be equal.

This is the first equation of motion in vector form.
Let the displacement of the object from time t
= 0 to t be \(\overrightarrow{\mathrm{s}}\)
For constant acceleration, \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}=\frac{\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{u}}}{2}\)

This is the second equation of motion in vector form.
Equations (1) and (2) can be resolved into their x and y components so as to get corresponding scalar equations as follows.
v_x = u_x + a_xt ………….. (3)
v_y = u_y + a_y t …………… (4)
s_x = u_xt + \(\frac{1}{2}\) a_xt² ………….. (5)
s_y = u_yt + \(\frac{1}{2}\) a_yt² ………..(6)
It can be seen that equations (3) and (5) involve only the x components of displacement, velocity and acceleration while equations (4) and (6) involve only the y components of these quantities.
Thus, the motion along the x direction of the object is completely controlled by the x components of velocity and acceleration while that along the y direction is completely controlled by the y components of these quantities.
This shows that the two sets of equations are independent of each other and can be solved independently.

Question 6.
Derive equations of motion graphically for a particle having uniform acceleration, moving along a straight line.
Answer:

Consider an object starting from position x = 0 at time t = 0. Let the velocity at time (t = 0) and t be u and v respectively.
The slope of line PQ gives the acceleration. Thus
∴ a = \(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}-0}=\frac{\mathrm{v}-\mathbf{u}}{\mathrm{t}}\)
∴ v = u + at …………… (1)
This is the first equation of motion.
The area under the curve in velocity-time graph gives the displacement of the object.
∴ s = area of the quadrilateral OPQS = area of rectangle OPRS + area of triangle PQR.
= ut + \(\frac{1}{2}\) (v – u) t
But, from equation (1)
at = v – u
∴ s = ut + \(\frac{1}{2}\) at²
This is the second equation of motion,
The velocity is increasing linearly with time as acceleration is constant. The displacement is given as,

∴ s = (v² – u²) / (2a)
∴ v² – u² = 2as
This is the third equation of motion.

Question 7.
Derive the formula for the range and maximum height achieved by a projectile thrown from the origin with initial velocity \(\vec{u}\) at an angel θ to the horizontal.
Answer:
Expression for range:

Consider a body projected with velocity \(\vec{u}\), at an angle θ of projection from point O in the co-ordinate system of the XY- plane, as shown in figure.
The initial velocity \(\vec{u}\) can be resolved into two rectangular components:

u_x = u cos θ (Horizontal component)
u_y = u sin θ (Vertical component)
The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to
v_y = u_y + a_yt
with a_y = -g and u_y = u sinθ
Thus, the components of velocity of the projectile at time t are given by,
v_x = u_x = u cos θ
v_y = u_x – gt = usin θ – gt
Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
s = (u cosθ) t
s_y = (usinθ)t – \(\frac{1}{2}\) gt²
At the highest point, the time of ascent of the projectile is given as,
t_A = \(\frac{u \sin \theta}{g}\) …………..(2)
The total time in air i.e., time of flight is given as, T = 2t_A = \(\frac{2u \sin \theta}{g}\) …… (3)
The total horizontal distance travelled by the particle in this time T is given as,
R = u_x ∙ T
R = u cos θ ∙ (2t_A)
R = u cos θ ∙ \(\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) ……………[From (3)]
R = \(\frac{u^{2}(2 \sin \theta \cdot \cos \theta)}{g}\)
R = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) ………..[∵ sin 2θ = 2sin∙cosθ]
This is required expression for horizontal range of the projectile.

Expression for maximum height of a projectile:
The maximum height H reached by the projectile is the distance travelled along the vertical (y) direction in time t_A.

Substituting s_y = H and t = t_a in equation (1),
we have,
H = (u sin θ)t_A – \(\frac{1}{2}\) gt_A²

This equation represents maximum height of projectile.

Question 8.
Show that the path of a projectile is a parabola.
Answer:

Consider a body projected with velocity initial velocity \(\vec{u}\) , at an angle θ of projection from point O in the co-ordinate system of the XY-plane. as shown in figure.
The initial velocity \(\vec{u}\) can be resolved into two rectangular components:
u_x = u cos θ (Horizontal component)
u_y = u sin θ (Vertical component)
The horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to,
v_y = u_y + a_y t
with a_y, = -g and u_y = u sinθ
Thus, the components of velocity of the projectile at time t are given by,
v_x = u_x = u cosθ
v_y = u_y – gt = u sinθ – gt
Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
s_x = (u cosθ)t …………..(1)
s_y = (u sinθ)t – \(\frac{1}{2}\) gt² ………………. (2)
As the projectile starts from x = O, we can use
s_x = x and s_y = y.
Substituting s_x = x in equation (1),
x = (u cosθ) t
∴ t = \(\frac{\mathrm{X}}{(\mathrm{u} \cos \theta)}\) …………….. (3)
Substituting, s_y in equation (2),
y = (u sinθ)t – \(\frac{1}{2}\) gt² …………… (4)
Substituting equation (3) in equation (4), we have,
y = u sin θ (\(\frac{\mathrm{X}}{(\mathrm{u} \cos \theta)}\)) – \(\frac{1}{2}\) (\(\frac{\mathrm{X}}{(\mathrm{u} \cos \theta)}\))² g
∴ y = x (tan θ) – (\(\frac{g}{2 u^{2} \cos ^{2} \theta}\))x² ………………. (5)
Equation (5) represents the path of the projectile.
If we put tan θ = A and g/2u²cos²θ = B then equation (5) can be written as y = Ax – Bx² where A and B are constants. This is equation of parabola. Hence, path of projectile is a parabola.

Question 9.
What is a conical pendulum? Show that its time period is given by 2π \(\sqrt{\frac{l \cos \theta}{g}}\), where l is the length of the string, θ is the angle that the string makes with the vertical and g is the acceleration due to gravity.
Answer:
A simple pendulum, Ch i given such a motion that the bob describes a horizontal circle and the string making a constant angle with the vertical describes a cone, is called a conical pendulum.

Consider a bob of mass m tied to one end of a string of length ‘P and other end is fixed to rigid support.
Let the bob be displaced from its mean position and whirled around a horizontal circle of radius ‘r’ with constant angular velocity ω, then the bob performs U.C.M.
During the motion, string is inclined to the vertical at an angle θ as shown in the figure above.
In the displaced position, there are two forces acting on the bob.
- The weight mg acting vertically downwards.
- The tension T acting upward along the string.
The tension (T) acting in the string can be resolved into two components:
- T cosθ acting vertically upwards.
- T sinθ acting horizontally towards centre of the circle.
Since, there is no net force, vertical component T cosθ balances the weight and horizontal component T sinθ provides the necessary centripetal force.
∴ T cos θ = mg ……………. (1)
T sin θ = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = mrω² ……….. (2)
Dividing equation (2) by (1),
tan θ = \(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\) ……….. (3)
Therefore, the angle made by the string with the vertical is θ = tan^-1 (\(\frac{\mathrm{v}^{2}}{\mathrm{rg}}\))
Since we know v = \(\frac{2 \pi \mathrm{r}}{\mathrm{T}}\)

where l is length of the pendulum and h is the vertical distance of the horizontal circle from the fixed point O.

Question 10.
Define angular velocity. Show that the centripetal force on a particle undergoing uniform circular motion is -mω²\(\vec{r}\).
Answer:
Angular velocity of a particle is the rate of change of angular displacement.

Expression for centripetal force on a particle undergoing uniform circular motion:
i) Suppose a particle is performing U.C.M in anticlockwise direction.
The co-ordinate axes are chosen as shown in the figure.
Let,
A = initial position of the particle which lies on positive X-axis
P = instantaneous position after time t
θ = angle made by radius vector
ω = constant angular speed
\(\overrightarrow{\mathrm{r}}\) = instantaneous position vector at time t

ii) From the figure,
\(\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}} \mathrm{x}+\hat{\mathrm{j}} \mathrm{y}\)
where, \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) are unit vectors along X-axis and Y-axis respectively.

Negative sign shows that direction of acceleration is opposite to the direction of position vector. Equation (3) is the centripetal acceleration.
vii) Magnitude of centripetal acceleration is given by a = ω²r

viii) The force providing this acceleration should also be along the same direction, hence centripetal.
∴ \(\overrightarrow{\mathrm{F}}\) = m\(\overrightarrow{\mathrm{a}}\) = -mω²\(\overrightarrow{\mathrm{r}}\)
This is the expression for the centripetal force on a particle undergoing uniform circular motion.

ix) Magnitude of F = mω²r = \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) = mωv

[Note: The definition of angular velocity is not mentioned in this chapter but is in Ch.2 Mathematical Methods.]

3. Solve the following problems.

Question 1.
An aeroplane has a run of 500 m to take off from the runway. It starts from rest and moves with constant acceleration to cover the runway in 30 sec. What is the velocity of the aeroplane at the take off ?
Answer:
Given: Length of runway (s) = 500 m, t = 30 s
To find: Velocity (y)
Formulae. i) s = ut + \(\frac{1}{2}\) at²
ii) v = u + at
Calculation: As the plane was initially at rest, u = 0
From formula (1),
500 = 0 + \(\frac{1}{2}\) × a × (30)²
∴ 500 = 450 a
∴ a = \(\frac{10}{9}\) m/s²
From formula (ii),
v = 0 + \(\frac{10}{9}\) × 30
∴ v = \(\frac{100}{3}\) m/s = (\(\frac{100}{3} \times \frac{18}{5}\)) km/hr
∴ v = 120 km/hr
The velocity of the aeroplane at the take off is 120 km/hr.

Question 2.
A car moving along a straight road with a speed of 120 km/hr, is brought to rest by applying brakes. The car covers a distance of 100 m before it stops. Calculate
(i) the average retardation of the car
(ii) time taken by the car to come to rest.
Answer:
Given: u = 120 kmh^-1 = 120 × \(\frac{5}{18}\) = \(\frac{100}{3}\) ms^-1
s = 100 m, v = 0
To find: i) Average retardation of the car (a)
ii) Time taken by car (t)

Formulae: i) v² – u² = 2as
ii) v = u + at
Calculation: From formula (i),

i) Average retardation of the car is \(\frac{50}{9}\) ms² (in magnitude).
ii) Time taken by the car to come to rest is 6 s.

Question 3.
A car travels at a speed of 50 km/hr for 30 minutes, at 30 km/hr for next 15 minutes and then 70 km/hr for next 45 minutes. What is the average speed of the car?
Answer:
Given: v₁ = 50 km/hr. t₁ = 30 minutes = 0.5 hr,
v₂ = 30 km/hr, t₂ = 15 minutes = 0.25 hr,
v₃ = 70 km/hr, t₃ = 45 minutes 0.75 hr
To find: Average speed of car (v_av)
Formula v_av = \(\frac{\text { total path length }}{\text { total time interval }}\)
Calculation:
Path length,
x₁ = v₁ × t₁ = 50 × 0.5 = 25km
x₂ = v₂ × t₂ = 30 × 0.25 = 7.5 km
x₃ = v₃ × t₃ = 70 × 0.75 = 52.5 km
From formula,
v_av = \(\frac{x_{1}+x_{2}+x_{3}}{t_{1}+t_{2}+t_{3}}\)
∴ v_av = \(\frac{25+7.5+52.5}{0.5+0.25+0.75}=\frac{85}{1.5}\)
∴ v_av = 56.66 km/hr

Question 4.
A velocity-time graph is shown in the adjoining figure.

Determine:

initial speed of the car
maximum speed attained by the car
part of the graph showing zero acceleration
part of the graph showing constant retardation
distance travelled by the car in first 6 sec.

Answer:

Initial speed is at origin i.e. 0 m/s.
Maximum speed attained by car, v_max = speed from A to B = 20 m/s.
The part of the graph which shows zero acceleration is between t = 3 s and t = 6 s i.e., AB. This is because, during AB there is no change in velocity.
The graph shows constant retardation from t = 6 s to t = 8 s i.e., BC.
Distance travelled by car in first 6 s
= Area of OABDO
= A(△OAE) + A(rect. ABDE)
= \(\frac{1}{2}\) × 3 × 20 + 3 × 20
= 30 + 60
∴ Distance travelled by car in first 6 s = 90 m

Question 5.
A man throws a ball to maximum horizontal distance of 80 meters. Calculate the maximum height reached.
Answer:
Given: R = 80m
To find: Maximum height reached (H_max)
Formula: R_max = 4H_max
Calculation: From formula,
∴ H_max = \(\frac{\mathrm{R}_{\max }}{4}=\frac{80}{4}\) = 20 m
The maximum height reached by the ball is 20m.

Question 6.
A particle is projected with speed v₀ at angle θ to the horizontal on an inclined surface making an angle Φ (Φ < θ) to the horizontal. Find the range of the projectile along the inclined surface.
Answer:

i) The equation of trajectory of projectile is given by,
y(tan θ)x – (\(\frac{\mathrm{g}}{2 \mathrm{u}^{2} \cos ^{2} \theta}\))x² …………..(1)

ii) In this case to find R substitute,
y = R sinΦ ………….. (2)
x = R cosΦ ………….. (3)

iii) From equations (1), (2) and (3),
we have,

Question 7.
A metro train runs from station A to B to C. It takes 4 minutes in travelling from station A to station B. The train halts at station B for 20 s. Then it starts from station B and reaches station C in next 3 minutes. At the start, the train accelerates for 10 sec to reach the constant speed of 72 km/hr. The train moving at the constant speed is brought to rest in 10 sec. at next station.
(i) Plot the velocity- time graph for the train travelling from the station A to B to C.
(ii) Calculate the distance between the stations A, B and C.
Answer:

The metro train travels from station A to station B in 4 minutes = 240 s.
The trains halts at station B for 20 s.
The train travels from station B’ to station C in 3 minutes= 180 s.
∴ Total time taken by the metro train in travelling from station A to B to C
= 240 + 20 + 180 = 440 s.
At start, the train accelerates for 10 seconds to reach a constant speed of 72 km/hr = 20 m/s.
The train moving is brought to rest in 10 s at next station.
The velocity-time graph for the train travelling from station A to B to C is as follows:
Distance travelled by the train from station A to station B
= Area of PQRS
= A ( △PQQ’) A (☐QRR’) + A(SRR’)
= (\(\frac{1}{2}\) × 10 × 20 + (220 × 20) + (\(\frac{1}{2}\) 10 × 20)
= 100 + 4400 + 100
= 4600m = 4.6km

Distance travelled by the train from station B’ to station C
= Area of EFGD
= A(△EFF’) + A(☐F’FGG’) + A(△DGG’)
= (\(\frac{1}{2}\) × 10 × 20) × (160 × 20) + (\(\frac{1}{2}\) × 10 × 20)
= 100 + 3200 + 100
= 3400m = 3.4km

Question 8.
A train is moving eastward at 10 m/sec. A waiter is walking eastward at 1.2m/sec; and a fly is flying toward the north across the waiter’s tray at 2 m/s. What is the velocity of the fly relative to Earth.
Answer:
Given

Question 9.
A car moves in a circle at the constant speed of 50 m/s and completes one revolution in 40 s. Determine the magnitude of acceleration of the car.
Answer:
Given: v = 50 m/s, t = 40 s, s = 2πr
To find: acceleration (a)
Formulae: i) v = \(\frac{\mathrm{s}}{\mathrm{t}}\)
ii) a = \(\frac{\mathrm{v}^{2}}{\mathrm{r}}\)
Calculation: From formula (i),
50 = \(\frac{2 \pi \mathrm{r}}{40}\)
∴ r = \(\frac{50 \times 40}{2 \pi}\)
∴ r = \(\frac{1000}{\pi}\) cm
From formula (ii),
a = \(\frac{v^{2}}{r}=\frac{50^{2}}{1000 / \pi}\)
∴ a = \(\frac{5 \pi}{2}\) = 7.85 m/s²
The magnitude of acceleration of the car is 7.85 m/s.

Alternate method:
Given: v = 50 m/s, t = 40 s,
To find: acceleration (a)
Formula: a = rω² = vω
Calculation: From formula,
a = vω
= v(\(\frac{2 \pi}{\mathrm{t}}\))
= 50(\(\frac{2 \times 3.142}{40}\))
= \(\frac{5}{2}\) × 3.142
∴ a = 7.85m/s²

Question 10.
A particle moves in a circle with constant speed of 15 m/s. The radius of the circle is 2 m. Determine the centripetal acceleration of the particle.
Answer:
Given: v = 15 m/s, r = 2m
To find: Centripetal acceleration (a)
Formula: a = \(\frac{\mathrm{v}^{2}}{\mathrm{r}}\)
Calculation: From formula,
a = \(\frac{(15)^{2}}{2}=\frac{225}{2}\)
∴ a = 112.5m/s²
The centripetal acceleration of the particle is 112.5 m/s².

Question 11.
A projectile is thrown at an angle of 30° to the horizontal. What should be the range of initial velocity (u) so that its range will be between 40m and 50 m? Assume g = 10 m s^-2.
Answer:
Given: 40 ≤ R ≤ 50, θ = 300, g = 10 m/s²
To find: Range of initial velocity (u)
Formula: R = \(\frac{\mathrm{u}^{2} \sin (2 \theta)}{\mathrm{g}}\)
Calculation: From formula,
The range of initial velocity,

∴ 21.49m/s ≤ u ≤ 24.03m/s
The range of initial velocity should be between 21.49 m/s ≤ u ≤ 24.03 m/s.

11th Physics Digest Chapter 3 Motion in a Plane Intext Questions and Answers

Can you recall? (Textbook Page No. 30)

Question 1.
What ¡s meant by motion?
Answer:
The change ¡n the position of an object with respect to its surroundings is called motion.

Question 2.
What Is rectilinear motion?
Answer:
Motion in which an object travels along a straight line is called rectilinear motion.

Question 3.
What is the difference between displacement and distance travelled?
Answer:

Displacement is the shortest distance between the initial and final points of movement.
Distance is the actual path followed by a body between the points in which it moves.

Question 4.
What is the difference between uniform and non-uniform motion?
Answer:

A body is said to have uniform motion if it covers equal distances in equal intervals of time.
A body is said to have non-uniform motion if it covers unequal distances in equal intervals of time.

Internet my friend (Textbook Page No. 44)

i. hyperphysics.phy-astr.gsu.eduJhbase/mot.html#motcon
ii. www .college-physics.comlbook/mechanics
[Students are expected to visit the above mentioned webs ires and collect more information.]

11th Std Physics Questions And Answers:

11th Chemistry Chapter 9 Exercise Elements of Group 13, 14 and 15 Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 9 Elements of Group 13, 14 and 15 Textbook Exercise Questions and Answers.

Elements of Group 13, 14 and 15 Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 9 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 9 Exercise Solutions

1. Choose correct option.

Question A.
Which of the following is not an allotrope of carbon ?
a. buckyball
b. diamond
c. graphite
d. emerald
Answer:
d. emerald

Question B.
………… is inorganic graphite.
a. borax
b. diborane
c. boron nitride
d. colemanite
Answer:
c. boron nitride

Question C.
Haber’s process is used for preparation of ………….
a. HNO3
b. NH3
c. NH2CONH2
d. NH4OH
Answer:
b. NH3

Question D.
Thallium shows different oxidation state because ……………
a. of inert pair effect
b. it is inner transition element
c. it is metal
d. of its high electronegativity
Answer:
a. of inert pair effect

Question E.
Which of the following shows most prominent inert pair effect ?
a. C
b. Si
c. Ge
d. Pb
Answer:
d. Pb

2. Identify the group 14 element that best fits each of the following description.

A. Non-metallic element
B. Form the most acidic oxide
C. They prefer +2 oxidation state.
D. Forms strong π bonds.
Answer:
i. Carbon (C)
ii. Carbon
iii. Tin (Sn) and lead (Pb)
iv. Carbon

3. Give reasons.

A. Ga³⁺ salts are better reducing agent while Tl³⁺ salts are better oxidising agent.
B. PbCl₄ is less stable than PbCl₂
Answer:
A. i. Both gallium (Ga) and thallium (Tl) belong to group 13.
ii. Ga is lighter element compared to thallium Tl. Therefore, its +3 oxidation state is stable. Thus, Ga+ loses two electrons and get oxidized to Ga³⁺. Hence, Ga+ salts are better reducing agent.
iii. Thallium is a heavy element. Therefore, due to the inert pair effect, Tl forms stable compounds in +1 oxidation state. Thus, Tl³⁺ salts get easily reduced to Tl¹⁺ by accepting two electrons. Hence, Tl³⁺ salts are better oxidizing agent.
[Note: This question is modified so as to apply the appropriate textual concept.]

B. i. Pb has electronic configuration [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p².
ii. Due to poor shielding of 6s² electrons by inner d and f electrons, it is difficult to remove 6s² electrons (inert pair).
iii. Thus, due to inert pair effect, +2 oxidation state is more stable than +4 oxidation state.
Hence, PbCl₄ is less stable than PbCl₂.

4. Give the formula of a compound in which carbon exhibit an oxidation state of

A. +4
B. +2
C. -4
Answer:
A. CCl₄
B. CO
C. CH₄

5. Explain the trend of the following in group 13 elements :

A. atomic radii
B. ionization enthalpy
C. electron affinity
Answer:
A. Atomic radii:

In group 13, on moving down the group, the atomic radii increases from B to Al.
However, there is an anomaly observed in the atomic radius of gallium due to the presence of 3d electrons. These inner 3d electrons offer poor shielding effect and thus, valence shell electrons of Ga experience greater nuclear attraction. As a result, atomic radius of gallium is less than that of aluminium.
However, the atomic radii again increases from Ga to Tl.
Therefore, the atomic radii of the group 13 elements varies in the following order:
B < Al > Ga < In < Tl

B. Ionization enthalpy:

Ionization enthalpies show irregular trend in the group 13 elements.
As we move down the group, effective nuclear charge decreases due to addition of new shells in the atom of the elements which leads to increased screening effect. Thus, it becomes easier to remove valence shell electrons and hence, ionization enthalpy decreases from B to Al as expected.
However, there is a marginal difference in the ionization enthalpy from Al to Tl.
The ionization enthalpy increases slightly for Ga but decreases from Ga to In.
In case of Ga, there are 10 d-electrons in its inner electronic configuration which shield the nuclear charge less effectively than the s and p-electrons and therefore, the outer electron is held fairly strongly by the nucleus. As a result, the ionization enthalpy increases slightly.
Number of d electrons and extent of screening effect in indium is same as that in gallium. However, the atomic size increases from Ga to In. Due to this, the first ionization enthalpy of In decreases.
The last element Tl has 10 d-electrons and 14 f-electrons in its inner electronic configuration which exert still smaller shielding effect on the outer electrons. Consequently, its first ionization enthalpy increases considerably.

C. Electron affinity:
a. Electron affinity shows irregular trend. It first increases from B to A1 and then decreases. The less electron affinity of boron is due to its smaller size. Adding an electron to the 2p orbital in boron leads to a greater repulsion than adding an electron to the larger 3p orbital of aluminium.

b. From Al to Tl, electron affinity decreases. This is because, nuclear charge increases but simultaneously the number of shells in the atoms also increases. As a result, the effective nuclear charge decreases down the group resulting in increased atomic size and thus, it becomes difficult to add an electron to a larger atom. The electron affinity of Ga and In is same.
Note: Electron affinity of group 13 elements:

6. Answer the following

Question A.
What is hybridization of Al in AlCl₃?
Answer:
Al is sp² hybridized in AlCl₃.

Question B.
Name a molecule having banana bond.
Answer:
Diborane (B₂H₆)

7. Draw the structure of the following

Question A.
Orthophosphoric acid
Answer:

Question B.
Resonance structure of nitric acid
Answer:

8. Find out the difference between

Question A.
Diamond and Graphite
Answer:
Diamond:

It has a three-dimensional network structure.
In diamond, each carbon atom is sp³ hybridized.
Each carbon atom in diamond is linked to four other carbon atoms.
Diamond is poor conductor of electricity due to absence of free electrons.
Diamond is the hardest known natural substance.

Graphite:

It has a two-dimensional hexagonal layered structure.
In graphite, each carbon atom is sp² hybridized.
Each carbon atom in graphite is linked to three other carbon atoms.
Graphite is good conductor of electricity due to presence of free electrons in its structure.
Graphite is soft and slippery.

Question B.
White phosphorus and Red phosphorus
Answer:
White phosphorus:

It consists of discrete tetrahedral P₄ molecules.
It is less stable and more reactive.
It exhibits chemiluminescence.
It is poisonous.

Red phosphorus:

It consists chains of P₄ molecules linked together by covalent bonds.
It is stable and less reactive.
It does not exhibit chemiluminescence.
It is nonpoisonous.

9. What are silicones? Where are they used?
Answer:
i. a. Silicones are organosilicon polymers having R₂SiO (where, R = CH₃ or C₆H₅ group) as a repeating unit held together by

b. Since the empirical formula R₂SiO (where R = CH₃ or C₆H₅ group) is similar to that of ketones (R₂CO), these compounds are named as silicones.

ii. Applications: They are used as

insulating material for electrical appliances.
water proofing of fabrics.
sealant.
high temperature lubricants.
for mixing in paints and enamels to make them resistant to high temperature, sunlight and chemicals.

10. Explain the trend in oxidation state of elements from nitrogen to bismuth.
Answer:

Group 15 elements have five valence electrons (ns² np³). Common oxidation states are -3, +3 and +5. The range of oxidation state is from -3 to +5.
Group 15 elements exhibit positive oxidation states such as +3 and +5. Due to inert pair effect, the stability of +5 oxidation state decreases and +3 oxidation state increases on moving down the group.
Group 15 elements show tendency to donate electron pairs in -3 oxidation state. This tendency is maximum for nitrogen.
The group 15 elements achieve +5 oxidation state only through covalent bonding.
e. g. NH₃, PH₃, ASH₃, SbH₃, and BiH₃ contain 3 covalent bonds. PCl₅ and PF₅ contain 5 covalent bonds.

11. Give the test that is used to detect borate radical is qualitative analysis.
Answer:
i. Borax when heated with ethyl alcohol and concentrated H₂SO₄, produces volatile vapours of triethyl borate, which bum with green edged flame.

ii. The above reaction is Used as a test for the detection and removal of borate radical \(\left(\mathrm{BO}_{3}^{3-}\right)\) in qualitative analysis.

12. Explain structure and bonding of diborane.
Answer:

Electronic configuration of boron is 1s² 2s² 2p¹. Thus, it has only three valence electrons.
In diborane, each boron atom is sp³ hybridized. Three of such hybrid orbitals are half filled while the fourth sp3 hybrid orbital remains vacant.
The two half-filled sp³ hybrid orbitals of each B atom overlap with 1s orbitals of two terminal H atoms and form four B – H covalent bonds. These bonds are also known as two-centred-two-electron (2c-2e) bonds.
When ‘1s’ orbital of each of the remaining two H atoms simultaneously overlap with half-filled hybrid orbital of one B atom and the vacant hybrid orbital of the other B atom, it produces two three-centred-two- electron bonds (3c-2e) or banana bonds.
Hydrogen atoms involved in (3c-2e) bonds are the bridging H atoms i.e., H atoms in two B – H – B bonds.
In diborane, two B atoms and four terminal H atoms lie in one plane, while the two bridging H atoms lie symmetrically above and below this plane.

13. A compound is prepared from the mineral colemanite by boiling it with a solution of sodium carbonate. It is white crystalline solid and used for inorganic qualitative analysis.

a. Name the compound produced.
b. Write the reaction that explains its formation.
Answer:
a. Borax
b. Borax is obtained from its mineral colemanite by boiling it with a solution of sodium carbonate.

14. Ammonia is a good complexing agent. Explain.
Answer:
i. The lone pair of electrons on nitrogen atom facilitates complexation of ammonia with transition metal ions. Thus, ammonia is a good complexing agent as it forms complex by donating its lone pair of electrons.

ii. This reaction is used for the detection of metal ions such as Cu²⁺ and Ag⁺.

15. State true or false. Correct the false statement.

A. The acidic nature of oxides of group 13 increases down the graph.
B. The tendency for catenation is much higher for C than for Si.
Answer:
A. False
The acidic nature of oxides of group 13 decreases down the group. It changes from acidic through amphoteric to basic.
B. True

16. Match the pairs from column A and B.

	Column A		Column B
i.	BCl₃	a.	Angular molecule
ii.	SiO₂	b.	Linear covalent molecule
iii.	CO₂	c.	Tetrahedral molecule
		d.	Planar trigonal molecule

Answer:
i – d,
ii – c,
iii – b

17. Give the reactions supporting basic nature of ammonia.
Answer:
In the following reactions ammonia reacts with acids to form the corresponding ammonium salts which indicates basic nature of ammonia.

18. Shravani was performing inorganic qualitative analysis of a salt. To an aqueous solution of that salt, she added silver nitrate. When a white precipitate was formed. On adding ammonium hydroxide to this, she obtained a clear solution. Comment on her observations and write the chemical reactions involved.
Answer:
i. When silver nitrate (AgNO₃) is added to an aqueous solution of salt sodium chloride (NaCl), a white precipitate of silver chloride (AgCl) is formed.

ii. On adding ammonium hydroxide (NH₄OH) to this, the white precipitate of silver chloride gets dissolved and thus, a clear solution is obtained.

11th Chemistry Digest Chapter 9 Elements of Group 13, 14 and 15 Intext Questions and Answers

Can you recall? (Textbook Page No. 123)

Question 1.
If the valence shell electronic configuration of an element is 3s²3p¹, in which block of the periodic table is it placed?
Answer:
The element having valence shell electronic configuration 3s² 3p¹ must be placed in the p-block of the periodic table as its last electron enters in p-subshell (3p).

Can you recall? (Textbook Page No. 127)

Question 1.
What is common between diamond and graphite?
Answer:
Both diamond and graphite are made up of carbon atoms as they are two allotropes of carbon.

Can you recall? (Textbook Page No. 129)

Question i.
Which element from the following pairs has higher ionization enthalpy?
B and TI, N and Bi
Answer:
Among B and Tl, boron has higher ionization enthalpy while, among N and Bi, nitrogen has higher ionization enthalpy.

Question ii.
Does boron form covalent compound or ionic?
Answer:
Yes, boron forms covalent compound.

Try this. (Textbook Page No. 131)

Question 1.
Find out the structural formulae of various oxyacids of phosphorus.
Answer:

11th Chemistry Chapter 16 Exercise Chemistry in Everyday Life Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 16 Chemistry in Everyday Life Textbook Exercise Questions and Answers.

Chemistry in Everyday Life Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 16 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 16 Exercise Solutions

1. Choose correct option

Question A.
Oxidative Rancidity is …………….. reaction
a. addition
b. subtitution
c. Free radical
d. combination
Answer:
c. Free radical

Question B.
Saponification is carried out by ……………..
a. oxidation
b. alkaline hydrolysis
c. polymarisation
d. Free radical formation
Answer:
b. alkaline hydrolysis

Question C.
Aspirin is chemically named as ……………..
a. Salicylic acid
b. acetyl salicylic acid
c. chloroxylenol
d. thymol
Answer:
b. acetyl salicylic acid

Question D.
Find odd one out from the following
a. dettol
b. chloroxylenol
c. paracetamol
d. trichlorophenol
Answer:
c. paracetamol

Question E.
Arsenic based antibiotic is
a. Azodye
b. prontosil
c. salvarsan
d. sulphapyridine
Answer:
c. salvarsan

Question F.
The chemical used to slow down the browning action of cut fruit is
a. SO₃
b. SO₂
c. H₂SO₄
d. Na₂CO₃
Answer:
b. SO₂

Question G.
The chemical is responsible for the rancid flavour of fats is
a. Butyric acid
b. Glycerol
c. Protein
d. Saturated fat
Answer:
a. Butyric acid

Question H.
Health benefits are obtained by consumption of
a. Saturated fats
b. trans fats
c. monounsaturated fats
d. all of these
Answer:
c. monounsaturated fats

2. Explain the following :

Question A.
Cooking makes food easy to digest.
Answer:

During the cooking process, high polymers of carbohydrates or proteins are hydrolysed to smaller polymeric units.
The uncooked food mixture is a heterogeneous suspension which becomes a colloidal matter on cooking.
As a result, the constituent nutrient molecules present in cooked food are smaller in size and hence, easier to digest, than the uncooked food.

Hence, cooking makes food easy to digest.

Question B.
On cutting some fruits and vegetables turn brown.
Answer:
i. Cutting of fruits and vegetables damage the cells, resulting in release of chemicals.
ii. Depending on the pH of fruits/vegetables, polyphenols are released.
iii. Due to the action of an enzyme, these polyphenols react with oxygen present in the air and get oxidised to form quinones.

iv. Quinones further undergo reactions including polymerization, which results in the formation of brown coloured products called as tannins.
Thus, on cutting, some fruits and vegetables turn brown.

Question C.
Vitmin E is added to packed edible oil.
Answer:

Vitamin E is a very effective natural antioxidant.
The phenolic – OH group present in the structure of vitamin E is responsible for its antioxidant activity.
Also, the long chain of saturated carbon atoms makes it fat soluble.

Therefore, when vitamin E is added to packed edible oil, it prevents the oxidative rancidity of the oil.

Question D.
Browning of cut apple can be prolonged by applying lemon juice.
Answer:

Browning of cut apple is due to the oxidation of polyphenols at a particular pH to quinones, which further undergoes polymerization to form brown coloured tannins.
This browning reaction can be prolonged or slowed down by using reducing agents or by changing the pH.
Applying lemon juice (i.e., citric acid) on the cut apple, lowers the pH at the surface of the apple. This prevents the oxidation reaction. Thus, browning of cut apple can be prolonged by applying lemon juice.

Question E.
A diluted solution (4.8 % w/v) of 2,4,6-trichlorophenol is employed as antiseptic.
Answer:

2,4,6-Trichlorophenol (TCP) is more potent antiseptic than phenol.
It has low corrosive effects as compared to phenol, if used in lower concentrations.

Hence, diluted solution (4.8% w/v) of 2,4,6-trichlorophenol is used as antiseptic.

Question F.
Turmeric powder can be used as antiseptic.
Answer:

Turmeric powder contains an active ingredient called curcumin.
Curcumin has antiseptic properties; thus, it is used for wound healing or applied on bruise.

Hence, turmeric powder can be used as antiseptic.

3. Identify the functional groups in the following molecule :

Answer:

4. Give two differences between the following

Question A.
Disinfectant and antiseptic
Answer:

Disinfectant	Antiseptic
1. Disinfectants are applied on non-living surfaces like floors, instruments, sanitary ware, etc. to kill wide range of microorganisms.	1. Antiseptics are applied on the surface of living tissues in order to sterilise them.
2. Disinfectants cannot be applied on wounds.	2. Antiseptics can be directly applied on wounds.
3. p-chloro-o-benzyl phenol	3. Iodine, boric acid, iodoform, dettol, etc.

Question B.
Soap and synthetic detergent
Answer:

Soap	Synthetic detergent
1. Soaps can be broadly classified into two types, i.e., toilet soaps (prepared using KOH) and laundry soaps (prepared using NaOH).	1. Synthetic detergents are of three types, i.e., anionic, cationic and nonionic detergents.
2. Soaps cannot be used in hard water.	2. Synthetic detergents can be used in soft water as well as in hard water.

Question C.
Saturated and unsaturated fats
Answer:

Saturated fats	Unsaturated fats
1. In saturated fat, long chains of tetrahedral carbon atoms in the fatty acid get closely packed together.	1. In unsaturated fats, the presence of one or more C = C bond in long chains of fatty acids, prevent molecules from packing closely together.
2. In saturated fats, the van der Waals forces between long saturated chains are strong. Hence, their melting points are higher than unsaturated fats.	2. In unsaturated fats, the van der Waals forces between long unsaturated chains are weak. Hence, their melting points are lower than saturated fats.

Question D.
Rice flour and cooked rice
Answer:

Rice flour	Cooked rice
1. Rice flour can be stored for a long period of time. It has a long shelf life.	1. Cooked rice cannot be stored for a longer period of time. It has very short shelf life.
2. Rice flour is uncooked food and hence, it is difficult to digest.	2. Cooked rice is easier to digest.

5. Match the pairs.

A group	B group
A. Paracetamol	a. Antibiotic
B. Chloramphenicol	b. Synthetic detergent
C. BHT	c. Soap
D. Sodium stearate	d. Antioxidant
	e. Analgesic

Answer:
A – e,
B – a,
C – d,
D – c

6. Name two drugs which reduce body pain.
Answer:
Aspirin and paracetamol are the two drugs that reduce body pain.

7. Explain with examples

Question A.
Antiseptics
Answer:
i. Antiseptics are used to sterilise surfaces of living tissue when the risk of infection is very high, such as during surgery or on wounds.
ii. Commonly used antiseptics include inorganics like iodine and boric acid or organics like iodoform and some phenolic compounds.

e.g.

Tincture of iodine (2-3% solution of iodine in alcohol-water mixture) and iodoform serve as powerful antiseptics and is used to apply on wounds.
A dilute aqueous solution of boric acid is a weak antiseptic used for eyes.
Various phenols are used as antiseptics. A dilute aqueous solution of phenol (carbolic acid) is used as antiseptic; however, phenol is found to be corrosive in nature. Many chloro derivatives of phenols are more potent antiseptics than the phenol itself. They can be used with much lower concentrations, which reduce their corrosive effects.
Two of the most common phenol derivatives in use are trichlorophenol (TCP) and chloroxylenol (which is an active ingredient of antiseptic dettol).
Thymol obtained from oil of thyme (a spice plant) has excellent non-toxic antiseptic properties.

Question B.
Disinfectant
Answer:

Disinfectants are non-selective antimicrobials.
They kill a wide range of microorganisms including bacteria.
They are used on non-living surfaces for example, floors, instruments, sanitary ware, etc.
Various phenols can be used as disinfectants.
e.g. p-Chloro-o-benzyl phenol is used as a disinfectant in all-purpose cleaners.

Question C.
Cationic detergents
Answer:
Cationic detergents: These are quaternary ammonium salts having one long chain alkyl group.
e.g. Ethyltrimethylammonium bromide: [CH₃(CH₂)₁₅ – N⁺(CH₃)₃]Br^–

Question D.
Anionic detergents
Answer:
Anionic detergents: These are sodium salts of long chain alkyl sulphonic acids or long chain alkyl substituted benzene sulphonic acids.
e.g. Sodium lauryl sulphate: CH₃(CH₂)₁₀CH₃O\(\mathrm{SO}_{3}^{-}\)Na⁺

Question E.
Non-ionic detergents
Answer:
Nonionic detergents: These are ethers of polyethylene glycol with alkyl phenol or esters of polyethylene glycol with long chain fatty acid.
e.g. a. Nonionic detergent containing ether linkage:

b. Nonionic detergent containing ester linkage: CH₃(CH₂)₁₆ – COO(CH₂CH₂O)_nCH₂CH₂OH

8. Explain : mechnism of cleansing Action of soap with flow chart.
Answer:
The following flow chart shows mechanism of cleansing action of soap:

9. What is meant by broad spectrum antibiotic and narrow spectrum antibiotics?
Answer:
Antibiotics which are effective against wide range of bacteria are known as broad spectrum antibiotics, while antibiotics which are effective against one group of bacteria are known as narrow spectrum antibiotics.

10. Answer in one senetence

Question A.
Name the painkiller obtained from acetylation of salicyclic acid.
Answer:
Aspirin is the pain killer obtained from acetylation of salicylic acid.

Question B.
Name the class of drug often called as painkiller.
Answer:
Analgesics are the class of drug often called as painkiller.

Question C.
Who discovered penicillin?
Answer:
Alexander Fleming discovered penicillin.

Question D.
Draw the structure of chloroxylenol and salvarsan.
Answer:
Structure of chloroxylenol:

Structure of salvarsan:

Question E.
Write molecular formula of Butylated hydroxy toulene.
Answer:
Molecular formula of butylated hydroxytoluene is C₁₅H₂₄O.

Question F.
What is the tincture of iodine ?
Answer:
Tincture of iodine is a 2-3% solution of iodine in alcohol-water mixture.

Question G.
Draw the structure of BHT.
Answer:

Question I.
Write a chemical equation for saponification.
Answer:

Question J.
Write the molecular formula and name of

Answer:
Molecular formula: C₉H₈O₄
Name: Aspirin

11. Answer the following

Question A.
Write two examples of the following.
a. Analgesics
c. Antiseptics
d. Antibiotics
e. Disinfectant
Answer:

No.	Drug type	Examples
i.	Analgesics	Aspirin, paracetamol
ii.	Antiseptics	Dettol, thymol
iii.	Antibiotics	Penicillin, sulphapyridine
iv.	Disinfectant	Phenol, p-Chloro-o-benzyl phenol

Question B.
What do you understand by antioxidant ?
Answer:

An antioxidant is a substance that delays the onset of oxidant or slows down the rate of oxidation of foodstuff.
It is used to extend the shelf life of food.
Antioxidants react with oxygen-containing free radicals and thereby prevent oxidative rancidity.
e.g. Vitamin E is a very effective natural antioxidant.

Activity :

Collect information about different chemical compounds as per their applications in day-to-day life.
Answer:

No.	Chemical compound	Applications
i.	Vinegar(CH₃COOH)	Preservation of food, salad dressing, sauces, etc.
ii.	Magnesium hydroxide [Mg(OH)₂]	Common component of antacids (used to relieve heartburn, acid indigestion and stomach upset.)
iii.	Baking soda (NaHCO₃)	Cooking, antacid, toothpaste, etc.
iv.	Sodium benzoate (C₆H₅COONa)	Used as food preservative

[Note: Students can use the above information as reference and collect additional information on their own.]

11th Chemistry Digest Chapter 16 Chemistry in Everyday Life Intext Questions and Answers

Can you recall? (Textbook Page No. 261)

Question i.
What are the components of balanced diet?
Answer:
Carbohydrates, proteins, lipids (fats and oil), vitamins, minerals and water are the components of balanced diet.

Question ii.
Why is food cooked? What is the difference in the physical states of uncooked and cooked food?
Answer:

Food is cooked in order to make it easy to digest.
Also, the raw or uncooked food may contain harmful microorganisms which may cause illness. Cooking of food at high temperature kills most of these microorganisms.
Raw/uncooked food materials like dried pulses, vegetables, meat, etc. are hard and thus, not easily chewable while cooked food is soft and tender, therefore, easily chewable.

Question iii.
What are the chemicals that we come across in everyday life?
Answer:
Detergents, shampoos, medicines, various food flavours, food colours, etc. are different types of chemicals that we come across in everyday life.

Just think (Textbook Page No. 261)

Question i.
Why is food stored for a long time?
Answer:
Food (like various cereals, pulses, pickles) is stored for a long time to make it available in all seasons.

Question ii.
What methods are used for preservation of food?
Answer:
Various physical and chemical methods are used for preservations of food.

Physical methods like, addition of heat, removal of heat, removal of water, irradiation, etc., are used in order to preserve food.
Chemical methods like, addition of sugar, salt, vinegar, etc. are employed for preservation of food.

Question iii.
What is meant by quality of food?
Answer:
Food quality can be described in terms of parameters such as flavour, smell, texture, colour and microbial spoilage.

Can you recall? (Textbook Page No. 263)

Question i.
How is Vanaspati ghee made?
Answer:
Vanaspati ghee is prepared by hydrogenation of oils. Hydrogen gas is passed through the oils at about 450 K in the presence of nickel catalyst to form solid edible fats like vanaspati ghee.

Question ii.
What are the physical states of peanut oil, butter, animal fat, Vanaspati ghee at room temperature?
Answer:

Example	Physical state
Peanut oil	Liquid
Butter	Semi-solid
Animal fat	Solid/semi-solid
Vanaspati ghee	Solid/semi-solid

Can you tell? (Textbook Page No. 264)

Question 1.
When is an antipyretic drug used?
Answer:
An antipyretic drug is used to reduce fever (that is, it lowers body temperature when a fever is present).

Question 2.
What type of medicine is applied to a bruise?
Answer:
Antiseptic such as tincture of iodine is applied on a bruise in order to prevent the exposed living tissue from getting infected.

Question 3.
What is meant by a broad spectrum antibiotic?
Answer:
Antibiotics which are effective against wide range of bacteria are known as broad spectrum antibiotic.

Question 4.
What is the active principle ingredient of cinnamon bark?
Answer:
Cinnamaldehyde is the principle active ingredient of cinnamon bark.

Can you tell? (Textbook Page No. 268)

Question i.
Can we use the same soap for bathing as well as cleaning utensils or washing clothes? Why?
Answer:
No, we cannot use the same soap for bathing as well as cleaning utensils or washing clothes due to the following reasons:

Chemical composition of each type of soap or cleansing material is different.
Nature, acidity, texture, reactivity towards water (i.e., hard water or soft water), reactivity towards microorganisms, stains are different for each type of soap.
Depending on these qualities, soaps are classified and used accordingly.
e.g. pH of soaps used for bathing purpose is different than that of the soap which is used for cleaning utensils.

Thus, we cannot use the same soap for bathing as well as cleaning utensils or washing clothes.

Question ii.
How will you differentiate between soaps and synthetic detergent using borewell water?
Answer:
Borewell water is hard water. Soaps and synthetic detergents react differently with hard water.

Soap: Soaps are insoluble in hard water. Borewell water (hard water) contains Ca²⁺ and Mg²⁺ ions. Soaps react with these ions to form insoluble magnesium and calcium salts of fatty acids. These salts precipitate out as gummy substance or form scum.
Synthetic detergents: Synthetic detergents can be used in hard water as well. They contain molecules (components) which form soluble calcium and magnesium salts.

Thus, soaps will form scum in borewell water but synthetic detergents will not.

11th Std Chemistry Questions And Answers:

11th Chemistry Chapter 2 Exercise Introduction to Analytical Chemistry Solutions Maharashtra Board

January 7, 2025January 6, 2025 by Bhagya

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 2 Introduction to Analytical Chemistry Textbook Exercise Questions and Answers.

Introduction to Analytical Chemistry Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 2 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 2 Exercise Solutions

1. Choose correct option

Question A.
The branch of chemistry which deals with study of separation, identification, and quantitaive determination of the composition of different substances is called as ………………..
a. Physical chemistry
b. Inorganic chemistry
c. Organic chemistry
d. Analytical chemistry
Answer:
d. Analytical chemistry

Question B.
Which one of the following property of matter is Not quantitative in nature ?
a. Mass
b. Length
c. Colour
d. Volume
Answer:
c. Colour

Question C.
SI unit of mass is ……..
a. kg
b. mol
c. pound
d. m³
Answer:
a. kg

Question D.
The number of significant figures in 1.50 × 10⁴ g is ………..
a. 2
b. 3
c. 4
d. 6
Answer:
b. 3

Question E.
In Avogadro’s constant 6.022 × 10²³ mol^-1, the number of significant figures is ……….
a. 3
b. 4
c. 5
d. 6
Answer:
b. 4

Question F.
By decomposition of 25 g of CaCO₃, the amount of CaO produced will be ……………….
a. 2.8 g
b. 8.4 g
c. 14.0 g
d. 28.0 g
Answer:
c. 14.0 g

Question G.
How many grams of water will be produced by complete combustion of 12g of methane gas
a. 16
b. 27
c. 36
d. 56
Answer:
b. 27

Question H.
Two elements A (At. mass 75) and B (At. mass 16) combine to give a compound having 75.8 % of A. The formula of the compound is
a. AB
b. A₂B
c. AB₂
d. A₂B₃
Answer:
d. A₂B₃

Question I.
The hydrocarbon contains 79.87 % carbon and 20.13 % of hydrogen. What is its empirical formula ?
a. CH
b. CH₂
c. CH₃
d. C₂H₅
Answer:
c. CH₃

Question J.
How many grams of oxygen will be required to react completely with 27 g of Al? (Atomic mass : Al = 27, O = 16)
a. 8
b. 16
c. 24
d. 32
Answer:
c. 24

Question K.
In CuSO₄.5H₂O the percentage of water is ……
(Cu = 63.5, S = 32, O = 16, H = 1)
a. 10 %
b. 36 %
c. 60 %
d. 72 %
Answer:
b. 36 %

Question L.
When two properties of a system are mathematically related to each other, the relation can be deduced by
a. Working out mean deviation
b. Plotting a graph
c. Calculating relative error
d. all the above three
Answer:
b. Plotting a graph

2. Answer the following questions

Question A.
Define : Least count
Answer:
The smallest quantity that can be measured by the measuring equipment is called least count.

Question B.
What do you mean by significant figures? State the rules for deciding significant figures.
Answer:
i. The significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit.
ii. Rules for deciding significant figures:
a. All non-zero digits are significant.
e.g. 127.34 g contains five significant figures which are 1, 2, 7, 3 and 4.
b. All zeros between two non-zero digits are significant, e.g. 120.007 m contains six significant figures.
c. Zeros on the left of the first non-zero digit are not significant. Such a zero indicates the position of the decimal point.
e.g. 0.025 has two significant figures, 0.005 has one significant figure.
d. Zeros at the end of a number are significant if they are on the right side of the decimal point,
e. g. 0.400 g has three significant figures and 400 g has one significant figure.
e. In numbers written is scientific notation, all digits are significant.
e.g. 2.035 × 10² has four significant figures and 3.25 × 10^-5 has three significant figures.

Question C.
Distinguish between accuracy and precision.
Answer:
Accuracy:

Accuracy refers to nearness of the measured value to the true value.
Accuracy represents the correctness of the measurement.
Accuracy is expressed in terms of absolute error and relative error.
Accuracy takes into account the true or accepted value.
Accuracy can be determined by a single measurement.
High accuracy implies smaller error.

Precision:

Precision refers to closeness of multiple readings of the same quantity.
Precision represents the agreement between two or more measured values.
Precision is expressed in terms of absolute deviation and relative deviation.
Precision does not take into account the true or accepted value.
Several measurements are required to determine precision.
High precision implies reproducibility of the readings.

Question D.
Explain the terms percentage composition, empirical formula and molecular formula.
Answer:
Percentage Composition:

The percentage composition of a compound is the percentage by weight of each element present in the compound.
Quantitative determination of the constituent elements by suitable methods provides the percent elemental composition of a compound.
If the percent total is not 100, the difference is considered as percent oxygen.
From the percentage composition, the ratio of the atoms of the constituent elements in the molecule is calculated.

Empirical Formula:
The simplest ratio of atoms of the constituent elements in a molecule is called the empirical formula of that compound.
e.g. The empirical formula of benzene is CH.

Molecular Formula:
1. Molecular formula of a compound is the formula which indicates the actual number of atoms of the constituent elements in a molecule.
e.g. The molecular formula of benzene is C6H6.
2. It can be obtained from the experimentally determined values of percent elemental composition and molar mass of that compound.
3. Molecular formula can be obtained from the empirical formula if the molar mass is known.
Molecular formula = r × Empirical formula

Question E.
What is a limiting reagent ? Explain.
Answer:
Limiting reagent:

The reactant which gets consumed and limits the amount of product formed is called the limiting reagent.
When a chemist carries out a reaction, the reactants are not usually present in exact stoichiometric amounts, that is, in the proportions indicated by the balanced equation.
This is because the goal of a reaction is to produce the maximum quantity of a useful compound from the starting materials. Frequently, a large excess of one reactant is supplied to ensure that the more expensive reactant is completely converted into the desired product.
The reactant which is present in lesser amount gets consumed after some time and subsequently, no further reaction takes place, whatever be the amount left of the other reactant present.

Hence, limiting reagent is the reactant that gets consumed entirely and limits the reaction.

Question F.
What do you mean by SI units ? What is the SI unit of mass ?
Answer:
i. In 1960, the general conference of weights and measures proposed revised metric system, called International system of Units i.e. SI units, abbreviated from its French name.
ii. The SI unit of mass is kilogram (kg).

Question G.
Explain the following terms
(a) Mole fraction
(b) Molarity
(c) Molality
Answer:
(a) Mole fraction: Mole fraction is the ratio of number of moles of a particular component of a solution to the total number of moles of the solution.

If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are n_A and n_B, respectively, then the mole fraction of A and B are given as:

(b) Molarity: Molarity is defined as the number of moles of the solute present in 1 litre of the solution. It is the most widely used unit and is denoted by M.
Molarity is expressed as follows:
Molarity (M) = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in litres }}\)

Molality: Molality is the number of moles of solute present in 1 kg of solvent. It is denoted by m. Molality is expressed as follows:
Molality (m) = \(\frac{\text { Number of moles of solute }}{\text { Mass of solvent in kilograms }}\)

Question H.
Define : Stoichiometry
Answer:
The study of quantitative relations between the amount of reactants and/or products is called stoichiometry.

Question I.
Why there is a need of rounding off figures during calculation ?
Answer:

When performing calculations with measured quantities, the rule is that the accuracy of the final result is limited to the accuracy of the least accurate measurement.
In other words, the final result cannot be more accurate than the least accurate number involved in the calculation.
Sometimes, the final result of a calculation often contains figures that are not significant.
When this occurs, the final result is rounded off.

Question J.
Why does molarity of a solution depend upon temperature ?
Answer:

Molarity is the number of moles of the solute present in 1 litre of the solution. Therefore, molarity depends on the volume of the solution.
Volume of the solution varies with the change in temperature.

Hence, molarity of a solution depends upon temperature.

Question M.
Define Analytical chemistry. Why is accurate measurement crucial in science?
Answer:
The branch of chemistry which deals with the study of separation, identification, qualitative and quantitative determination of the compositions of different substances, is called analytical chemistry.

1. The accuracy of measurement is of great concern in analytical chemistry. This is because faulty equipment, poor data processing, or human error can lead to inaccurate measurements. Also, there can be intrinsic errors in analytical measurement.
2. When measurements are not accurate, this provides incorrect data that can lead to wrong conclusions. For example, if a laboratory experiment requires a specific amount of a chemical, then measuring the wrong amount may result in an unsafe or unexpected outcome.
3. Hence, the numerical data obtained experimentally are treated mathematically to reach some quantitative conclusion.
4. Also, an analytical chemist has to know how to report the quantitative analytical data, indicating the extent of the accuracy of measurement, perform the mathematical operation, and properly express the quantitative error in the result.

3. Solve the following questions

Question A.
How many significant figures are in each of the following quantities ?
a. 45.26 ft
b. 0.109 in
c. 0.00025 kg
d. 2.3659 × 10^-8 cm
e. 52.0 cm³
f. 0.00020 kg
g. 8.50 × 10⁴ mm
h. 300.0 cg
Answer:
a. 4
b. 3
c. 2
d. 5
e. 3
f. 2
g. 3
h. 4

Question B.
Round off each of the following quantities to two significant figures :
a. 25.55 mL
b. 0.00254 m
c. 1.491 × 10⁵ mg
d. 199 g
Answer:
a. 26 mL
b. 0.0025 m
c. 1.5 × 10⁵ mg
d. 2.0 × 10² g

Question C.
Round off each of the following quantities to three significant figures :
a. 1.43 cm³
b. 458 × 10² cm
c. 643 cm²
d. 0.039 m
e. 6.398 × 10^-3 km
f. 0.0179 g
g. 79,000 m
h. 42,150
i. 649.85
j. 23,642,000 mm
k. 0.0041962 kg
Answer:
a. 43 cm³
b. 4.58 × 10⁴ cm
c. 643 cm² (or 6.43 × 10² cm²)
d. 0.0390 m (or 3.90 × 10^-2 m)
e. 6.40 × 10^-3 km
f. 0.0179 g (or 1.79 × 10^-2 m)
g. 7.90 × 10⁴ m
h. 4.22 × 10⁴ (or 42,200)
i. 6.50 × 10²
j. 2.36 × 10⁷ mm
k. 0.00420 kg (or 4.20 × 10^-3 kg)

Question D.
Express the following sum to appropriate number of significant figures :
a. 2.3 × 10³ mL + 4.22 × 10⁴ mL + 9.04 × 10³ mL + 8.71 × 10⁵ mL;
b. 319.5 g – 20460 g – 0.0639 g – 45.642 g – 4.173 g
Answer:
To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added/subtracted.
a. (0.23 × 10⁴ mL) + (4.22 × 10⁴ mL) +(0.904 × 10⁴ mL) + (87.1 × 10⁴ mL)
= (0.23 + 4.22 + 0.904 + 87.1) × 10⁴ mL
= 92.454 × 10⁴ mL
= 9.2454 × 10⁵
= 9.2 × 10⁵ mL
b. 319.5 g – 20460 g – 0.0639 g – 45.642 g – 4.173 g
= – 20190.3789 g
= – 20190 g
Ans: Sum to appropriate number of significant figures = 9.2 × 10⁵ mL
ii. Sum to appropriate number of significant figures = – 20190 g
[Note: In addition and subtraction, the final answer is rounded to the minimum number of decimal point of the number taking part in calculation. If there is no decimal point, then the final answer will have no decimal point.]

4. Solve the following problems

Question A.
Express the following quantities in exponential terms.
a. 0.0003498
b. 235.4678
c. 70000.0
d. 1569.00
Answer:
a. 0.0003498 = 3.498 × 10^-4
b. 235.4678 = 2.354678 × 10²
c. 70000.0 = 7.00000 × 10⁴
d. 1569.00 = 1.56900 × 10³

Question B.
Give the number of significant figures in each of the following
a. 1.230 × 10⁴
b. 0.002030
c. 1.23 × 10⁴
d. 1.89 × 10^-4
Answer:
a. 4
b. 4
c. 3
d. 3

Question C.
Express the quantities in above (B) with or without exponents as the case may be.
Answer:
a. 12300
b. 2.030 × 10^-3
c. 12300
d. 0.000189

Question D.
Find out the molar masses of the following compounds :
a. Copper sulphate crystal (CuSO₄.5H₂O)
b. Sodium carbonate, decahydrate (Na₂CO₃.10H₂O)
c. Mohr’s salt [FeSO₄(NH₄)₂SO₄.6H₂O]
(At. mass : Cu = 63.5; S = 32; O = 16; H = 1; Na = 23; C = 12; Fe = 56; N = 14)
Answer:
a. Molar mass of CuSO₄.5H₂O
= (1 × At. mass Cu) + (1 × At. mass S) + (9 × At. mass O) + (10 × At. mass H)
= (1 × 63.5) + (1 × 32) + (9 × 16) + (10 × 1)
= 63.5 + 32 + 144 + 10
= 249.5 g mol^-1
Molar mass of CuSO₄.5H₂O = 249.5 g mol^-1

b. Molar mass of Na₂CO₃.10H₂O
= (2 × At. mass Na) + (1 × At. mass C) + (13 × At. mass O) + (20 × At. mass H)
= (2 × 23) + (1 × 12) + (13 × 16) + (20 × 1)
= 46 + 12 + 208 + 20
= 286 g mol^-1
Molar mass of Na₂CO₃.10H₂O = 286 g mol^-1

c. Molar mass of [FeSO₄(NH₄)₂SO₄.6H₂O]
= (1 × At. mass Fe) + (2 × At. mass S) + (2 × At. mass N) + (14 × At. mass O) + (20 × At. mass H)
= (1 × 56) + (2 × 32) + (2 × 14) + (14 × 16) + (20 × 1)
= 56 + 64 + 28 + 224 + 20
= 392 g mol^-1
Molar mass of [FeSO₄(NH₄)₂SO₄.6H₂O] = 392 g mol^-1

Question E.
Work out the percentage composition of constituents elements in the following compounds :
a. Lead phosphate [Pb₃(PO₄)₂],
b. Potassium dichromate (K₂Cr₂O₇),
c. Macrocosmic salt – Sodium ammonium hydrogen phosphate, tetrahydrate (NaNH₄HPO₄.4H₂O)
(At. mass : Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14)
Answer:
Given: Atomic mass: Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14
To find: The percentage composition of constituent elements
Formula:

Calculation:
i. Lead phosphate [Pb₃(PO₄)₂]
Molar mass of Pb₃(PO₄)₂ = 3 × (207) + 2 × (31) + 8 × (16) = 621 + 62 + 128 = 811 g mol^-1
Percentage of Pb = \(\frac {621}{811}\) × 100 = 76.57%
Percentage of P = \(\frac {621}{811}\) × 100 = 7.64%
Percentage of O = \(\frac {128}{811}\) × 100 = 15.78%

ii. Potassium dichromate (K₂Cr₂O₇)
Molar mass of K₂Cr₂O₇ = 2 × (39) + 2 × (52) + 7 × (16) = 78 + 104 + 112 = 294 g mol^-1
Percentage of K = \(\frac {78}{294}\) × 100 = 26.53%
Percentage of Cr = \(\frac {104}{294}\) × 100 = 35.37%
Percentage of O = \(\frac {112}{294}\) × 100 = 38.10%

iii. Microcosmic salt – Sodium ammonium hydrogen phosphate, tetrahydrate (NaNH₄HPO₄.4H₂O)
Molar mass of NaNH₄HPO₄.4H₂O = 1 × (23) + 1 × (14) + 1 × (31) + 13 × (1) + 8 × (16)
= 23 + 14 + 31 + 13 + 128 = 209 g mol^-1
Percentage of Na = \(\frac {23}{209}\) × 100 = 11.00%
Percentage of N = \(\frac {14}{209}\) × 100 = 6.70%
Percentage of P = \(\frac {31}{209}\) × 100 = 14.83%
Percentage of H = \(\frac {13}{209}\) × 100 = 6.22%
Percentage of O = \(\frac {128}{209}\) × 100 = 61.24%
Ans: i. Mass percentage of Pb, P and O in lead phosphate [Pb₃(PO₄)₂] are 76.57%, 7.64% and 15.78% respectively.
ii. Mass percentage of K, Cr and O in potassium dichromate (K₂Cr₂O₇) are 26.53%, 35.37% and 38.10% respectively.
iii. Mass percentage of Na, N, P, H and O in NaNH₄HPO₄.4H₂O are 11.00%, 6.70%, 14.83%, 6.22% and 61.24% respectively.

Question F.
Find the percentage composition of constituent green vitriol crystals (FeSO₄.7H₂O). Also find out the mass of iron and the water of crystallisation in 4.54 kg of the crystals. (At. mass : Fe = 56; S = 32; O = 16)
Answer:
Given: i. Atomic mass: Fe = 56; S = 32; O = 16
ii. Mass of crystal = 4.54 kg
To find: i. Mass percentage of Fe, S, H and O
ii. Mass of iron and water of crystallisation in 4.54 kg of crystal
Formula:

i. Molar mass of FeSO₄.7H₂O = 1 × (56) + 1 × (32) + 14 × (1) + 11 × (16)
= 56 + 32 + 14+ 176
= 278 g mol^-1
Percentage of Fe = \(\frac {56}{278}\) × 100 = 20.14%
Percentage of S = \(\frac {32}{278}\) × 100 = 11.51%
Percentage of H = \(\frac {14}{278}\) × 100 = 5.04%
Percentage of O = \(\frac {176}{278}\) × 100 = 63.31%

ii. 278 kg green vitriol = 56 kg iron
∴ 4.54 kg green vitriol = x
∴ x = \(\frac{56 \times 4.54}{278}\)
Mass of 7H₂O in 278 kg green vitriol = 7 × 18 = 126 kg
∴ 4.54 kg green vitriol = y
∴ y = \(\frac{126 \times 4.54}{278}\)
Ans: i. Mass percentage of Fe, S, H and O in FeSO₄.7H₂O are 20.14%, 11.51%, 5.04% and 63.31% respectively.
ii. Mass of iron in 4.54 kg green vitriol = 0.915 kg
Mass of water of crystallisation in 4.54 kg green vitriol = 2.058 kg

Question G.
The red colour of blood is due to a compound called “haemoglobin”. It contains 0.335 % of iron. Four atoms of iron are present in one molecule of haemoglobin. What is its molecular weight ? (At. mass : Fe = 55.84)
Answer:
Given: Iron percentage in haemoglobin = 0.335%
To find: Molecular weight of haemoglobin
Calculation: There are four atoms of iron in a molecule of haemoglobin. Four atoms of iron contribute 0.335% mass to a molecule of haemoglobin.
Mass of one Fe atom = 55.84 u
∴ Mass of 4 Fe atoms = 55.84 × 4 = 223.36 u = 0.335%
Let molecular weight of haemoglobin be x.
Hence,
\(\frac{223.36}{x}\) × 100 = 0.335%
∴ x = \(\frac{223.36}{0.335}\) × 100 = 66674.6 g mol^-1
Ans: Molecular weight of haemoglobin = 66674.6 g mol^-1

Question H.
A substance, on analysis, gave the following percent composition:
Na = 43.4 %, C = 11.3 % and O = 45.3 %. Calculate the empirical formula. (At. mass Na = 23 u, C = 12 u, O = 16 u).
Answer:
Given: Atomic mass of Na = 23 u, C = 12 u, and O = 16 u
Percentage of Na, C and O = 43.4%, 11.3% and 45.3% respectively.
To find: The empirical formula of the compound
Calculation:

Hence, empirical formula is Na₂CO₃.
Ans: Empirical formula of the compound = Na₂CO₃

Question I.
Assuming the atomic weight of a metal M to be 56, find the empirical formula of its oxide containing 70.0% of M.
Answer:
Given: Atomic mass of M = 56
Percentage of M = 70.0%
To find: The empirical formula of the compound
Calculation: % M = 70.0%
Hence, % O = 30.0%, Atomic mass of O = 16 u

Convert the ratio into whole number by multiplying by the suitable coefficient, i.e., 2.
Therefore, the ratio of number of moles of M : O is 2 : 3.
Hence, the empirical formula is M₂O₃.
Ans: Empirical formula of the compound = M₂O₃

Question J.
1.00 g of a hydrated salt contains 0.2014 g of iron, 0.1153 g of sulfur, 0.2301 g of oxygen and 0.4532 g of water of crystallisation. Find the empirical formula. (At. wt. : Fe = 56; S = 32; O = 16)
Answer:
Given: Atomic mass of Fe = 56, S = 32, and O = 16
Mass of iron, sulphur, oxygen and water = 0.2014 g, 0.1153 g, 0.2301 g and 0.4532 respectively.
To find: The empirical formula of the compound
Calculation: Since the mass of crystal is 1 g, the % iron, sulphur, oxygen and water = 20.14%, 11.53%, 23.01% and 4.32 % respectively.

Hence, empirical formula is FeSO₄.7H₂O.
Ans: Empirical formula of the compound = FeSO₄.7H₂O.

Question K.
An organic compound containing oxygen, carbon, hydrogen and nitrogen contains 20 % carbon, 6.7 % hydrogen and 46.67 % nitrogen. Its molecular mass was found to be 60. Find the molecular formula of the compound.
Answer:
Given: Percentage of carbon, hydrogen, nitrogen = 20%, 6.7%, 46.67% respectively.
Molar mass of the compound = 60 g mol^-1
To find: The molecular formula of the compound
Calculation: % carbon + % hydrogen + % nitrogen = 20 + 6.7 + 46.67 = 73.37%
This is less than 100%. Hence, compound contains adequate oxygen so that the total percentage of elements is 100%.
Hence, % of oxygen = 100 – 73.37 = 26.63%

Hence, empirical formula is CH₄N₂O.
Empirical formula mass = 12 + 4 + 28 + 16 = 60 g mol^-1
Hence,
Molar mass = Empirical formula mass
∴ Molecular formula = Empirical formula = CH₄N₂O
Ans: Molecular formula of the compound = CH₄N₂O

Question L.
A compound on analysis gave the following percentage composition by mass : H = 9.09; O = 36.36; C = 54.55. Mol mass of compound is 88. Find its molecular formula.
Answer:
Given: Percentage of H, O, C = 9.09%, 36.36%, 54.55% respectively.
Molar mass of the compound = 88 g mol^-1
To find: The molecular formula of the compound
Calculation:

Hence, empirical formula is C₂H₄O.
Empirical formula mass = 24 + 4 + 16 = 44 g mol^-1
Hence,

Molecular formula = r × empirical formula
Molecular formula = 2 × C₂H₂O = C₄H₈O₂
Ans: Molecular formula of the compound = C₄H₈O₂

Question M.
Carbohydrates are compounds containing only carbon, hydrogen and oxygen. When heated in the absence of air, these compounds decompose to form carbon and water. If 310 g of a carbohydrate leave a residue of 124 g of carbon on heating in absence of air, what is the empirical formula of the carbohydrate ?
Answer:
Given: Mass of carbon residue = 124 g, mass of carbohydrate = 310 g
To find: Empirical formula of the carbohydrate
Calculation: Since the 310 g of compound decomposes to carbon and water and the mass of carbon produced is 124 g, the remaining mass would be of water.
∴ Molar mass of water = 310 – 124 = 186 g

The ratio of number of moles of C : water = C : H₂O = 1 : 1
Hence, empirical formula = CH₂O
Ans: Empirical formula of the carbohydrate = CH₂O

Question N.
Write each of the following in exponential notation :
a. 3,672,199
b. 0.000098
c. 0.00461
d. 198.75
Answer:
a. 3,672,199 = 3.672199 × 10⁶
b. 0.000098 = 9.8 × 10^-5
c. 0.00461 = 4.61 × 10^-3
d. 198.75 = 1.9875 × 10²

Question O.
Write each of the following numbers in ordinary decimal form :
a. 3.49 × 10^-11
b. 3.75 × 10^-1
c. 5.16 × 10⁴
d. 43.71 × 10^-4
e. 0.011 × 10^-3
f. 14.3 × 10^-2
g. 0.00477 × 10⁵
h. 5.00858585
Answer:
a. 3.49 × 10^-11 = 0.0000000000349
b. 3.75 × 10^-1 = 0.375
c. 5.16 × 10⁴ = 51,600
d. 43.71 × 10^-4 = 0.004371
e. 0.011 × 10^-3 = 0.000011
f. 14.3 × 10^-2 = 0.143
g. 0.00477 × 10⁵ = 477
h. 5.00858585 = 5.00858585

Question P.
Perform each of the following calculations. Round off your answers to two digits.

Answer:

Question Q.
Perform each of the following calculations. Round off your answers to three digits.
a. (3.26 × 10⁴) (1.54 × 10⁶)
b. (8.39 × 10⁷) (4.53 × 10⁹)
c. \(\frac{8.94 \times 10^{6}}{4.35 \times 10^{4}}\)
d. \(\frac{\left(9.28 \times 10^{9}\right) \times\left(9.9 \times 10^{-7}\right)}{(511) \times\left(2.98 \times 10^{-6}\right)}\)
Answer:
i. (3.26 × 10⁴) (1.54 × 10⁶) = 5.0204 × 10⁴⁺⁶ = 5.02 × 10¹⁰
ii. (8.39 × 10⁷) (4.53 × 10⁹) = 38.0067 × 10⁷⁺⁹ = 38.0067 × 10¹⁶ = 3.80 x 10¹⁷

Question R.
Perform the following operations :
a. 3.971 × 10⁷ + 1.98 × 10⁴;
b. 1.05 × 10^-4 – 9.7 × 10^-5;
c. 4.11 × 10^-3 + 8.1 × 10^-4;
d. 2.12 × 10⁶ – 3.5 × 10⁵.
Answer:
Solution:
To perform addition/subtraction operation, first the numbers are written in such a way that they have the same exponent. The coefficients are then added/subtracted.
a. 3.971 × 10⁷ + 1.98 × 10⁴ = 3.971 × 10⁷ + 0.00198 × 10⁷ = (3.971 + 0.00198) × 10⁷
= 3.97298 × 10⁷
b. 1.05 × 10^-4 – 9.7 × 10^-5 = 10.5 × 10^-5 – 9.7 × 10^-5 = (10.5 – 9.7) × 10^-5 = 0.80 × 10^-5
= 8.0× 10^-6
c. 4.11 × 10^-3 + 8.1 × 10^-4 = 41.1 × 10^-4 + 8.1 × 10^-4 = (41.1 + 8.1) × 10^-4 = 49.2 × 10^-4
= 4.92 × 10^-3
d. 2.12 × 10⁶ – 3.5 × 10⁵ = 21.2 × 10⁵ – 3.5 × 10⁵ = (21.2 – 3.5) × 10⁵ = 17.7 × 10⁵
= 1.77 × 10⁶

Question S.
A 1.000 mL sample of acetone, a common solvent used as a paint remover, was placed in a small bottle whose mass was known to be 38.0015 g. The following values were obtained when the acetone – filled bottle was weighed : 38.7798 g, 38.7795 g and 38.7801 g. How would you characterise the precision and accuracy of these measurements if the actual mass of the acetone was 0.7791 g ?
Answer:
Precision:

Measurement	Mass of acetone observed (g)
1	38.7798 – 38.0015 = 0.7783
2	38.7795 – 38.0015 = 0.7780
3	38.7801 – 38.0015 = 0.7786

Mean = \(\frac{0.7783+0.7780+0.7786}{3}\) = 0.7783 g

Measurement	Mass of acetone observed (g)	Absolute deviation (g) = \| Observed value – Mean \|
1	0.7783	0
2	0.7780	0.0003
3	0.7786	0.0003

Mean absolute deviation = \(\frac{0+0.0003+0.0003}{3}\) = 0.0002
∴ Mean absolute deviation = ±0.0002 g

ii. Accuracy:
Actual mass of acetone = 0.7791 g
Observed value (average) = 0.7783 g
a. Absolute error = Observed value – True value
= 0.7783 – 0.7791
= – 0.0008 g

Ans: These observed values are close to each other and are also close to the actual mass. Therefore, the results are precise and as well accurate.
i. Relative deviation = 0.0257%
ii. Relative error = 0.1027%
[Note: i. As per the method given in textbook, the calculated value of relative deviation is 0.0257%.
ii. The negative sign in -0.1027% indicates that the experimental result is lower than the true value.]

Question T.
Your laboratory partner was given the task of measuring the length of a box (approx 5 in) as accurately as possible, using a metre stick graduated in milimeters. He supplied you with the following measurements: 12.65 cm, 12.6 cm, 12.65 cm, 12.655 cm, 126.55 mm, 12 cm.
a. State which of the measurements you would accept, giving the reason.
b. Give your reason for rejecting each of the others.
Answer:
a. The metre stick is graduated in millimetres i.e. 1 mm to 1000 mm, and 1 mm = 0.1 cm. Therefore, if length is measured in centimetres, the least count of metre stick is 0.1 cm. The results 12.6 cm has the least count of 0.1 cm and is acceptable result.

b. Since, the least count of metre stick is 0.1 cm or 1mm, the results such as 12.65 cm, 12.655 cm, 126.55 mm cannot be measured using this stick and hence, these results are rejected. The result, 12 cm doesn’t include the least count and is rejected.

Question U.
What weight of calcium oxide will be formed on heating 19.3 g of calcium carbonate ?
(At. wt. : Ca = 40; C = 12; O = 16)
Answer:
Given: Mass of CaCO₃ consumed in reaction = 19.3 g
To find: Mass of CaO formed
Calculation: Calcium carbonate decomposes according to the balanced equation,

So, 100 g of CaCO3 produce 56 g of CaO.

Ans: Mass of CaO formed = 10.81 g

[Calculation using log table:
56 × 0.193
= Antilog₁₀ [log₁₀ (56) + log₁₀ (0.193)]
= Antilog₁₀ [1.7482 + \(\overline{1} .2856\)]
= Antilog₁₀ [1.0338] = 10.81]

Question V.
The hourly energy requirements of an astronaut can be satisfied by the energy released when 34 grams of sucrose are “burnt” in his body. How many grams of oxygen would be needed to be carried in space capsule to meet his requirement for one day ?
Answer:
34 g of sucrose provides energy for an hour.
Hence, for a day, the mass of sucrose needed = 34 × 24 = 816g
The balanced equation is,

Thus, 342 g of sucrose require 384 g of oxygen.
∴ 816 g of sucrose will require = \(\frac{816}{342}\) × 384 = 916 g of O₂
Ans: Astronaut needs to carry 916 g of O₂.

11th Std Chemistry Questions And Answers: