Category: Class 12

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 10 Human Health and Diseases Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Biology Solutions Chapter 10 Human Health and Diseases

1. Multiple Choice Questions

Question 1.
Which of the following is NOT caused by unsterilized needles?
(a) Elephantiasis
(b) AIDS
(e) Malaria
(d) Hepatitis
Answer:
(a) Elephantiasis

Question 2.
Opium derivative is …………………
(a) Codeine
(b) Caffeine
(c) Heroin
(d) Psilocybin
Answer:
(c) Heroin

Question 3.
The stimulant present in tea is …………………
(a) tannin
(b) cocaine
(C) caffeine
(d) crack
Answer:
(c) caffeine

Question 4.
WhIch of the following Is caused by smoking?
(a) Liver cirrhosis
(b) Pulmonary tuberculosis
(c) Emphysema
(d) Malaria
Answer:
(c) Emphysema

Question 5.
An antibody is …………………
(a) molecuic that binds specifically an antigen
(b) WBC which invades bacteria
(c) secretion of mammalian RBC
(d) cellular component of blood
Answer:
(a) molecule that binds specifically an antigen

Question 6.
The antiviral proteins released by a virus-infected cell are called …………………
(a) histamines
(b) interferons
(c) pyrogens
(d) allergens
Answer:
(b) interferons

Question 7.
Both B-cells and T-cells are derived from …………………
(a) lymph nodes
(b) thymus glands
(c) liver
(d) stem cells in bone marrow
Answer:
(b) thymus glands

Question 8.
Which of the following diseases can be contracted by droplet infection?
(a) Malaria
(b) Chicken pox
(c) Pneumonia
(d) Rabies
Answer:
(c) Pneumonia

Question 9.
Confirmatory test used for detecting HIV infection is …………………
(a) ELISA
(b) Western blot
(c) Widal test
(d) Eastern blot
Answer:
(b) Western blot

Question 10.
Elephantiasis is caused by …………………
(a) W. barterofti
(b) P. vivax
(c) Bedbug
(d) Elephant
Answer:
(a) W. bancrofti

Question 11.
Innate immunity is provided by …………………
(a) phagocytes
(b) antibody
(c) T-lymphocytes
(d) B-lymphocytes
Answer:
(c) T-lymphocytes

2. Short Answer Questions

Question 1.
What is the source of cocaine?
Answer:
Source of cocaine is coca plant – Erythroxylum coca.

Question 2.
Name one disease caused by smoking.
Answer:
Emphysema. (Damaged and enlarged lungs causing breathlessness)

Question 3.
Which cells stimulate B-cells to form antibodies ?
Answer:
Helper T-cells stimulate B-cells to form antibodies.

Question 4.
What does the abbreviation AIDS stand for?
Answer:
AIDS stands for Acquired Immuno Deficiency Syndrome.

Question 5.
Name the causative agent of typhoid fever.
Answer:
Salmonella typhi

Question 6.
What is Rh factor?
Answer:
Antigen ‘D’ present on the surface of RBCs is known as Rh factor.

Question 7
What is schizont?
Answer:
Schizont is a ring-like form produced from merozoites inside the erythrocytes of human beings, infected by Plasmodium, which again forms new merozoites.

Question 8.
Name the addicting component found in tobacco.
Answer:
Nicotine

Question 9.
Name the pathogen causing Malaria.
Answer:
Plasmodium vivax

Question 10.
Name the vector of Filariasis.
Answer:
Female Culex mosquito

Question 11.
Name of the causative agent of ringworm.
Answer:
Trichophyton

Question 12.
Health
Answer:
Health is defined as the state of complete physical, mental and social well¬being and not merely the absence of disease or infirmity.

3. Short Answer Questions

Question 1.
What are acquired diseases?
Answer:
Diseases which are developed after the birth of an individual are called acquired diseases. These are of two types, viz. (a) Communicable or infectious diseases and (b) Non- communicable or Non-infectious diseases. Communicable or infectious diseases are transmitted from infected person to another healthy person either directly or indirectly. They are caused due to pathogens like viruses, bacteria, fungi, helminth worms, etc. Non-communicable or Non-infectious diseases cannot be transmitted from infected person to another healthy one either directly or indirectly.

Question 2.
Antigen and antibody.
Answer:

Antigen	Antibody
1. Antigens are foreign proteins which are capable of producing infection.	1. Antibodies are immunoglobulins produced by the body to act against the antigens.
2. The structure of antigens is variable dependent upon the type of pathogen.	2. The structure of antibody is Y-shaped.
3. The antigen is the ‘non-self’ molecule.	3. The antibody is ‘self’ molecule.
4. The antigens have epitope sites which bind with the antibody molecule.	4. The antibodies have paratope sites which bind with the antigen molecule.

Question 3.
Name the infective stage of Plasmodium. Give Symptoms of malaria
Answer:
Sporozoite
I. Symptoms of malaria:

1. Fever accompanied by shivering.
2. Joint pain or arthralgia.
3. Vomiting.
4. Anaemia caused due to rupture of RBCs or haemolysis.
5. Haemoglobinuria.
6. Retinal damage.
7. Convulsions.
8. Cyclical occurrence of sudden coldness followed by rigor and then fever and sweating lasting for four to six hours. This is called a classic symptom of malaria.
9. Splenomegaly or enlarged spleen, severe headache, cerebral ischemia, hepatomegaly, i. e. enlarged liver, hypoglycaemia and haemoglobinuria with renal failure may occur in severe infections.

II. Spread / Transmission of malaria:

1. Malaria parasite is transmitted through the female Anopheles mosquito and hence it is known as mosquito-borne disease. Mosquito acts as a vector.
2. There are four species of Plasmodium, viz., P. vivax, P. falciparum, P. ovale and P. malariae which transmit malaria.

Question 4.
Explain the mode of infection and cause of elephantiasis.
Answer:
Mode of infection, i.e. transmission:

1. The parasite Wuchereria bancrofti is transmitted from a patient to other normal human being by female Culex mosquito.
2. The filarial larvae leave mosquito body and arrive on the human skin where they penetrate the skin and enter inside.
3. They undergo two moultings to become adults. Later they settle in the lymphatic system. They incubate for about 8-16 months.
4. When they settle in lymphatic system, this infection is called lymphatic filariasis.
5. The worms start infecting lymphatic circulation resulting into enlargement of lymph vessels and lymph nodes. The extremities like legs or limbs become swollen which resembles elephant legs. Therefore it is called elephantiasis.
6. This condition is lymphoedema, i.e. accumulation of lymph fluid in tissue causing swelling.

Question 5.
Why is smoking a bad habit?
Answer:

1. Smoking involves inhaling the cigarette smoke which contains nicotine and other toxic substances like N-nitrosodimethlene. There is some amount of carbon monoxide.
   All these substances affect the normal respiratory health.
2. Smoking invites problems like asthma, hypertension, heart disease, stroke, lung damage.
3. The worst impact is that these substances are carcinogenic and hence can cause cancer of larynx, trachea, lung, etc.
4. Smoking not only affects the smokers but also has bad effect on others due to passive smokers.
5. In women, smoking is still hazardous as their ovaries can undergo mutations due to mutagenic chemicals found in smoke.
6. Therefore, smoking is a very bad habit.

Question 6.
What do the abbreviations AMIS and CMIS denote?
Answer:
AMIS is Antibody-mediated immune system or humoral immunity and CMIS is cell- mediated immune system.

Question 7.
What is a carcinogen? Name one chemical carcinogen with its target tissue.
Answer:

1. Carcinogen is the substance or agent that causes cancer.
2. Urinary bladder cancer caused by 2-naphthylamine and 4-aminobiphenyl.

Question 8.
Active immunity and passive immunity.
Answer:

Active immunity	Passive immunity
1. Active immunity is produced in response to entry of pathogens and their antigenic stimuli.	1. Passive immunity is produced due to antibodies that are transferred to the body.
2. Active immunity is the long lasting immunity.	2. Passive immunity is short-lived immunity.
3. In active immunity, the body produces its own antibodies.	3. In passive immunity, antibodies are given to the body from outside.
4. Natural acquired active immunity is obtained due to infections by pathogens.	4. Natural acquired passive immunity is obtained through antibodies of mother transmitted- to baby by placenta or colostrum.
5. Artificial acquired active immunity is obtained through vaccinations. These vaccines contain dead or live but attenuated pathogens.	5. Artificial acquired passive immunity is also obtained through vaccinations, but here the vaccines contain the readymade antibodies which are prepared with the help of other animals such as horses.

4. Short Answer Questions

Question 1.
B-cells and T-cells.
Answer:

B-cells	T-cells.
1. B-cells are type of lymphocyte whose origin is in bone marrow but maturation is in blood.	1. T-cells are type of lymphocytes which originate in bone marrow but maturation occurs in thymus.
2. B-cells Eire type of lymphocytes which are involved in humoral mediated immunity.	2. T-cells are type of lymphocytes which are involved in cell-mediated immunity.
3. 20% of lymphocytes present in the blood are B-cells.	3. 80% of lymphocytes present in the blood are T-cells.
4. Two types of B-cells are Memory cells and Plasma cells.	4. T-cells are of following subtypes : Cytotoxic T-cells, helper T-cells, suppressor T-cells.
5. They are involved in antibody mediated immunity. (AMI)	5. They are involved in cell-mediated immunity (CMI).
6. B-cells produced antibodies with which they fight against pathogens.	6. T-cells do not produce antibodies.
7. B-cells have membrane bound immunoglobulins located on the surface.	7. There is a presence of T cell receptors on the T-cell surface.

Question 2.
What are the symptoms of malaria? How does malaria spread?
Answer:
Symptoms of malaria:

1. Fever accompanied by shivering.
2. Joint pain or arthralgia.
3. Vomiting.
4. Anaemia caused due to rupture of RBCs or haemolysis.
5. Haemoglobinuria.

Question 3.
AIDS.
Answer:
(1) AIDS or the acquired immuno deficiency syndrome, is fatal viral disease caused by a retrovirus (ss RNA) known as the human immuno deficiency virus (HIV) which weakens the body’s immune system. It is called a modern pandemic.

(2) The HIV attacks the immune system which in turn causes many opportunistic infections, neurological disorders and unusual malignancies ultimately leading to death.

(3) AIDS was first noticed in USA in 1981 whereas in India, first confirmed case of AIDS was in April 1986 from Tamil Nadu.

(4) HIV is transmitted through body fluids such as saliva, tears, nervous system tissue, spinal fluid, blood, semen, vaginal fluid and breast milk. However, only blood, semen, vaginal secretions and breast milk generally transmit infection to others.

(5) The transmission of HIV occurs by sexual contact, through blood and blood products and by contaminated syringes, needles, etc. There is also transplacental transmission or through breast milk at the time of nursing.

(6) Accidental needle injury, artificial insemination with infected donated semen and transplantation with infected organs are some of the rare occasions of transmission of HIV.

(7) HIV infection is not spread by casual contact such as hugging, bite of mosquitoes or using other objects touched by a patient.

(8) Acute HIV infection progresses over time to asymptomatic HIV infection and then to early symptomatic HIV infection. Later, it progresses to full blown AIDS when patient shows advanced HIV infection with CD4 T-cell count below 200 cells/mm.

Question 4.
Give the symptoms of cancer.
Answer:
Symptoms of cancer:

1. Presence of lump or tumour.
2. White patches in the mouth.
3. Change in a wart or mole on the skin.
4. Swollen or enlarged lymph nodes.
5. Vertigo, headaches or seizures if cancer affect the brain.
6. Coughing and shortness of breath if lungs are affected due to cancer.

Question 5.
Antigens on blood cells.
Answer:

1. There are about 30 known antigens on the surface of human red blood cells. They decide the type of blood group such as ABO, Rh, Duffy, Kidd, Lewis, P MNS, Bombay.
2. The different blood groups are determined genetically due to presence of a particular antigen.
3. Landsteiner found two antigens or agglutinogens on the surface of human red blood cells which are named as antigen A and antigen B.
4. There is another antigen called Antigen D which decides the Rh status of the blood. If Antigen D is present, the person is said to be RH positive and when it is lacking, the person is Rh negative.
5. These antigens are responsible for types of blood group and the specific transfusions.
6. Antigens present on the RBCs and antibodies present in the serum can cause agglutination reactions if they are non-compatible. Therefore, at the time of transfusion blood groups are checked properly.

Question 6.
Antigen-antibody complex:
Img 1
Answer:

1. Between antigen and antibody there is specificity.
2. Each antibody is specific for a particular antigen.
3. On the antigens there are combining sites which are called antigenic determinants or epitopes.
4. Epitopes react with the corresponding antigen binding sites of antibodies which are called paratopes.
5. The antigen binding sites are located on the variable regions of the antibody. Variable regions have small variations which make each antibody highly specific for a particular antigen.
6. Owing to variable region the antibody can recognize the specific antigen.
7. Antibody thus binds to specific antigen in a lock and key manner, forming an antigen- antibody complex.

Question 7.
What are the various public health measures, which you would suggest as safeguard against infectious diseases?
Answer:
Infectious diseases spread through pathogens, therefore, it is an important duty of each person to decrease the risk of infecting our own self or others. This can be achieved by

1. Washing hands often, especially whenever, we are in contact with food and water. Before and after preparing food, before eating and after using the toilet, hand wash is a must.
2. Vaccinations : Immunization helps us to protect against contracting many diseases. Therefore, timely vaccination should be taken. Especially at the time of epidemic, one must keep distance from infected area or get vaccinated.
3. One must be at home if there are signs and symptoms of an infection. By going out, we may infect other healthy persons.
4. Proper diet and exercise should be followed to improve one’s own immunity.
5. Hygiene should be utmost in the kitchen and dining area. One must take care while eating uncovered and leftover food.
6. Bathroom and toilet should be cleaned daily as there can be a high concentration of bacteria or other infectious agents in these areas.
7. One should have responsible sexual behaviour to avoid sexually transmitted diseases.
8. Personal items such as toothbrush, comb, towel, undergarments or razor blade should never be shared.
9. Travelling should be avoided because we may infect other passengers during travel. Moreover, our illness can be aggravated. Some special immunizations are needed during certain travels, such as anti-cholera vaccine while going to Pandharpur during Ashadhi.

Question 8.
How does the transmission of each of the following diseases take place?
(a) Amoebiasis:
Answer:
Amoebiasis is usually transmitted by the following ways:

1. The faecal-oral route.
2. Through contact with dirty hands or objects.
3. By anal-oral contact.
4. Through contaminated food and water.

(b) Malaria:
Answer:
Symptoms of malaria:

1. Fever accompanied by shivering.
2. Joint pain or arthralgia.
3. Vomiting.
4. Anaemia caused due to rupture of RBCs or haemolysis.
5. Haemoglobinuria.

(c) Ascariasis:
Answer:

1. Unsafe and unhygienic food and drinks contaminated with the eggs of Ascaris are the main mode of transmission.
2. Eggs hatch inside the intestine of the new host.
3. The larvae pass through various organs and settle as adults in the digestive system.

(d) Pneumonia:
Answer:

1. Pneumonia usually spreads by direct person to person contact.
2. It is also spread via droplet infection, i.e. droplets released by infected person.
3. Using clothes and utensils of the patient.

Question 9.
What measures would you take to prevent water-borne diseases?
Answer:

1. To prevent water-borne diseases, use of safe, clean and potable water is a must. Water should be filtered, then boiled and stored in covered container. If possible water purifier systems should be installed at home.
2. One should preferably use bottled water or carry our own water container while travelling.
3. Cleaning of water containers and maintaining personal hygiene near water storage is a must.
4. Megacities offer chlorinated and purified water for citizens. But villages and smaller rural set ups use river water which may be highly contaminated with pathogens. Such water should be purified before consumption to prevent water-borne diseases.

Question 10.
Typhoid.
Answer:
Typhoid is an infective disease caused by Gram-ve bacterium, Salmonella typhi.
(1) It is food and water-borne infection. In the intestinal lumen of infected person this bacteria is found.

(2) The bacterium has “O” – antigen, which is a lipopolysaccharide (LPS), present on surface coat and its flagella has “H” – antigen. Thus it becomes pathogenic.

(3) Signs and Symptoms of typhoid are as follows:
Prolonged and high fever with nausea, fatigue, headache.
Abdominal pain, constipation or diarrhoea. In severe cases rose-coloured rash is seen on skin. Tongue shows white coating and there is cough. Anorexia or loss of appetite is seen. In chronic cases there is breathlessness, irregular heartbeats and haemorrhage.

(4) Poor hygiene habits and poor sanitation and insects like houseflies and cockroaches spread typhoid.

(5) Typhoid is diagnosed by Widal test.

(6) Antibiotics like Chloromycetin can cure typhoid. Preventive vaccines such as oral Ty21a vaccine and injectable typhim vi and typherix against typhoid are also available. Chronic cases need surgical removal of gall bladder.

5. Match the following.

Column I	Column II
(a) AIDS	(i) Antibody production
(b) Lysozyme	(ii) Activation of B-cells
(c) B-cells	(iii) Immunoglobulin
(d) T-helper cells	(iv) Tears
(e) Antibody	(v) Immuno deficiency

Answer:

Column I	Column II
(a) AIDS	(v) Immuno deficiency
(b) Lysozyme	(iv) Tears
(c) B-cells	(i) Antibody production
(d) T-helper cells	(ii) Activation of B-cells
(e) Antibody	(iii) Immunoglobulin

6. Long Answer Questions

Question 1.
Describe the structure of antibody.
Answer:
Img 2

1. Antibodies are highly specific to specific antigens. They are glycoprotein called immunoglobulins (Igs.).
2. They are produced by plasma cells. Plasma cells are in turn formed by B-lymphocytes.
3. About 2000 molecules of antibodies are formed per second by the plasma cells.
4. Antibody is a ‘Y’-shaped molecule. It has four polypeptide chains, two heavy or H-chains and two light or L-chains.
5. Disulfide bonds (-s-s-) hold the polypeptide chains together to form a ‘Y’-shaped structure.
6. The region holding arms and stem of antibody is termed as hinge. Each chain of the antibody has two distinct regions, the variable region and the constant region.
7. Variable regions have a paratope which is an antigen-binding site. This part of antibody recognizes and binds to the specific antigen forming an antigen-antibody complex.
8. Antibodies are called bivalent as they carry two antigen binding sites.

Question 2.
Vaccination.
Answer:

1. Vaccines are prepared from inactivated pathogen, in the form of protein or sugar from pathogen or dead form of pathogen or toxoid from pathogens or attenuated pathogen.
2. These when they are administered to a person to protect against a particular pathogen, it is called vaccination.
3. Vaccination ’teaches’ the immune system to recognize and eliminate pathogenic organism. Because, already in the body the vaccine is injected and body has made antibodies in response to it. Thus, body is prepared before the attack, if at all it is exposed to pathogen.
4. Thus, it is an important form of primary prevention, which reduces the chances of illness by protecting people. It works by exposing the pathogen in a safe form.
5. Vaccinations control spread of diseases like measles, polio, tetanus and whooping cough that once threatened many lives.
6. Vaccination controls the epidemic outbreak of diseases, if all the people Eire pre-vaccinated.
7. Some hazardous diseases like small, pox and polio have been completely eradicated by the vaccination.

Question 3.
What is cancer? Differentiate between benign tumour and malignant tumour. The main five types of cancer
Answer:
I. Cancer : Cancer is a disease caused by uncontrolled cell division due to disturbed cell cycle.

II. Difference between benign tumour and malignant tumour:

Benign tumour	malignant tumour
1. Benign tumour is localized and it does not spread to neighbouring areas.	1. Malignant tumour starts as local but spreads rapidly to neighbouring areas.
2. Benign tumour is enclosed in connective tissue sheath.	2. Malignant tumour is not enclosed in connective tissue sheath.
3. Benign tumour compresses the surrounding normal tissue.	3. Malignant tumour invades and destroys the surrounding tissue.
4. Benign tumours can be removed surgically.	4. Malignant tumours need further treatment after removal.
5. Except for brain tumour, benign tumours are usually not fatal.	5. Malignant tumours are fatal.
6. Benign tumours do not show metastasis.	6. Malignant tumours show metastasis.
7. Benign tumours are well differentiated.	7. Malignant tumours are poorly differentiated.
8. Benign tumours show slow and progressive growth.	8. Malignant tumours show rapid and erratic growth.

III. The main five types of cancer:

Types of Cancer : According to the tissue affected, the cancers are classified into five main types. These are as follows:

1. Carcinoma : Cancer of epithelial tissue covering or lining the body organs is known as carcinoma. E.g. breast cancer, lung cancer, cancer of stomach, skin cancer, etc.
2. Sarcoma : Cancer of connective tissue is called sarcoma. Following are the types of sarcoma osteosarcoma (bone cancer), myosarcoma (muscle cancer),
   chondrosarcoma (cancer of cartilage) and liposarcoma (cancer of adipose tissue).
3. Lymphoma : Cancer of lymphatic tissue is called lymphoma. Lymphatic nodes, spleen and tissues of immune system are affected due to lymphoma.
4. Leukaemia : Leukaemia is blood cancer. In this condition, excessive formation of leucocytes take place in the bone marrow. There are millions of abnormal immature leucocytes which cannot fight infections. Monocytic leukaemia, lymphoblastic leukaemia, etc. are the types of leukaemia.
5. Adenocarcinoma : Cancer of glandular tissues such as thyroid, pituitary, adrenal, etc. is called adenocarcinoma.

Question 4.
Describe the different type of immunity.
Answer: There are two basic types of immunity, viz. innate immunity and acquired immunity.
(A) Innate immunity:

1. Innate immunity is natural, inborn immunity, which helps the body to fight against the invasion of microorganisms.
2. Innate immunity is non-specific because it does not depend on previous exposure to foreign substances.
3. Innate immunity mechanisms consist of various types of barriers such as anatomical barriers, physiological barriers, phagocytic barriers and inflammatory barriers. They prevent entry of foreign agents into the body.

(B) Acquired immunity:

1. The immunity that an individual acquires during his life is called acquired immunity or adaptive immunity or specific immunity. It helps the body to adapt by fighting against specific antigens hence it is called adaptive immunity. Since it is produced specifically against an antigen, it is called specific immunity.
2. Acquired immunity takes long time for its activation.
3. This type of immunity is seen only in vertebrates.
4. Due to acquired immunity, the body is able to defend against any invading foreign agent.

Question 5.
Describe the ill-effects of alcoholism on health.
Answer:

1. Alcohol in any form is toxic for the body. Hence as soon as alcohol is consumed, the liver tries to detoxify it.
2. In low doses it acts as a stimulant but in high dose, it acts on central nervous system, especially the cerebrum and cerebellum. Still higher dose can induce a comatose condition.
3. Alcohol affect the gastrointestinal tract by causing inflammation and damage to gastric 4 mucosa. Ulceration and painful condition arises in alcoholics.
4. Excessive doses of alcohol induce vomiting.
5. The worst effect of alcohol is on liver causing diseases like cirrhosis.
6. Alcohol induces hypertension and cardiac problems.
7. Apart from physical effect, it causes deterioration of mental health and emotional well-being.
8. Alcoholic person cannot think due to numbness in his/her cerebrum.
9. The social health is greatly affected as the alcoholic can cause problems to his family, friends and society in general.

Question 6.
In your view, what motivates the youngsters to take to alcohol or drugs and how can this be avoided?
Answer:
I. Taking drugs or alcohol:

1. Youngsters are at the vulnerable age, where they lack the planning about their future.
2. If they fall into bad company or are facing parental neglect, they get hooked on to alcohol or drugs.
3. Some common causes for addiction among youngsters are insufficient parental supervision and monitoring or excessive pressure and expectations from them. Lack of communication between an adolescent and parents.
4. Poorly defined rules for the family. Continuous family conflicts.
5. Favourable parental attitudes towards alcohol and drug uses. Many a times, at home children are exposed to such habits.
6. Inability to cope up with present and hence switching to the addictions. Risk taking behaviour which is common among youngsters.

II. Methods/measures to avoid drug abuse:

1. There should be complete acceptance for the child, because the adolescent phase is the most crucial phase when the children should be treated with love, care and respect.
2. Many physical, hormonal and psychological transformations are taking place in this phase. Therefore child suffers from stressful situation.
3. Wrong company and bad influence of peer group can trap the child in bad addictive habits. Thus, family should be supportive and communicative to help such children.
4. The sexual thoughts should be sublimed by channelizing energy into healthy pursuits like sports, reading, music, yoga and other extracurricular activities.
5. Ill-effects of drugs or alcohol should be told to youngsters.
6. Education and counselling can control the children from getting hooked on to the addictions.

Question 7.
Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?
Answer:
Friends can influence one to take alcohol and drugs, if a boy or girl is timid and non-communicative with his or her parents and teachers. It also depends on the personality of the indtvidual. In the adolescent age, many fall in trap due to such peer pressure. The confusion in the mind and role of hormones playing on the psyche and thought process makes one unable to understand the hazards of such habits. Also there is curiosity to do these experimentations due to bad influence of media.

If there is complete trust and friendship with sensible parents, then such influence does not work. One should protect himself or herself by a strong denial. Communicating such incidents to an elder in whom a boy or girl can confide, is very important. One should tell his or her friends about the ill-effects of alcohol and drugs. He should be made aware of these aspects that he or she has learnt in this lesson.

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 6 Chemical Kinetics Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Chemistry Solutions Chapter 6 Chemical Kinetics

1. Choose the most correct option.

Question i.
The rate law for the reaction aA + bB → P is rate = k[A] [B]. The rate of reaction doubles if
a. concentrations of A and B are both doubled.
b. [A] is doubled and [B] is kept constant
c. [B] is doubled and [A] is halved
d. [A] is kept constant and [B] is halved.
Answer:
b. [A] is doubled and [B] is kept constant

Question ii.
The order of the reaction for which the units of rate constant are mol dm^-3 s^-1 is
a. 1
b. 3
c. 0
d. 2
Answer:
c. 0

Question iii.
The rate constant for the reaction 2N₂O₅(g) → 2N₂O₄(g) + O₂(g) is 4.98 × 10^-4 s^-1. The order of reaction is
a. 2
b. 1
c. 0
d. 3
Answer:
b. 1

Question iv.
Time required for 90 % completion of a certain first order reaction is t. The time required for 99.9 % completion will be
a. t
b. 2t
c. t/2
d. 3t
Answer:
d. 3t

Question v.
Slope of the graph ln[A]_t versus t for first order reaction is
a. -k
b. k
c. k/2. 303
d. -k/2. 303
Answer:
a. -k

Question vi.
What is the half life of a first order reaction if time required to decrease concentration of reactant from 0.8 M to 0.2 M is 12 h?
a. 12 h
b. 3 h
c. 1.5 h
d. 6 h
Answer:
d. 6 h

Question vii.
The reaction, 3ClO^3Θ ClO₃^Θ + 2 Cl^Θ occurs in two steps,
(i) 2 ClO^– → ClO₂^Θ
(ii) ClO₂^Θ + ClO^Θ → ClO₃^Θ + Cl^Θ

The reaction intermediate is
a. Cl^Θ
b. ClO₂^Θ
c. ClO₃^Θ
d. ClO^Θ
Answer:
b. ClO₂^Θ

Question viii.
The elementary reaction O₂(g) + O(g) → 2O₂(g) is
a. unimolecular and second order
b. bimolecular and first order
c. bimolecular and second order
d. unimolecular and first order
Answer:
c. bimolecular and second order

Question ix.
Rate law for the reaction, 2NO + Cl₂ → 2 NOCl is rate = k[NO₂]²[Cl₂]. Thus k would increase with
a. increase of temperature
b. increase of concentration of NO
c. increase of concentration of Cl₂
d. increase of concentrations of both Cl₂ and NO
Answer:
a. increase of temperature

Question x.
For an endothermic reaction, X ⇌ Y. If E f is activation energy of the forward reaction and Er that for reverse reaction, which of the following is correct?
a. E_f = E_r
b. E_f < E_r
c. E_f > E_r
d. ∆H = E_f – E_r is negative
Answer:
(c) Ef → Er

2. Answer the following in one or two sentences.

Question i.
For the reaction,
N₂(g) + 3 H₂(g) → 2NH₃(g), what is the relationship among \(\frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{dt}}\)\(\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}} \text { and } \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}} ?\)
Answer:
N_2(g) + 3H_2(g) → 2NH_3(g)
From the above reaction, when 1 mole of N₂ reacts, 3 moles of H₂ are consumed and 2 moles of NH₃ are formed.

If the instantaneous rate R of the reaction is represented in terms of rate of the consumption of N₂ then, \(R=-\frac{d\left[\mathrm{~N}_{2}\right]}{d t}\)

Hence the rate of reaction in terms of concentration changes in N₂, H₂ and NH₃ may be represented as,

Question ii.
For the reaction,
CH₃Br(aq) + OH-(aq) → CH₃OH^Θ (aq) +Br^Θ (aq), rate law is rate = k[CH₃Br][OH^Θ]
a. How does reaction rate changes if [OH^Θ] is decreased by a factor of 5?
b. What is change in rate if concentrations of both reactants are doubled?
Solution :
Given :
(a) Rate = R = k [CH₃Br] x [OH^–]
If R₁ and R₂ are initial and final rates of reaction then,

Rate will be increased 4 time.

Question iii.
What is the relationship between coeffients of reactants in a balanced equation for an overall reaction and exponents in rate law. In what case the coeffients are the exponents?
Answer:
Explanation : Consider the following reaction, aA + bB → products

If the rate of the reaction depends on the concentrations of the reactants A and B, then, by rate law,
R α [A]^a [B]^b
∴ R = k [A]^a [B^b
where [A] = concentration of A and
[B] = concentration of B

The proportionality constant k is called the velocity constant, rate constant or specific rate of the reaction.

a and b are the exponents or the powers of the concentrations of the reactants A and B respectively when observed experimentally.

The exponents or powers may not be necessarily a and b but may be different x and y depending on experimental observations. Then the rate R will be,
R = k [A]^x [B]^y
For example, if x = 1 and y = 2, then,
R = k [A] x [B]²

Question iv.
Why all collisions between reactant molecules do not lead to a chemical reaction?
Answer:
(i) Collisions of reactant molecules : The basic re-quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

(ii) Energy requirement (Activation energy) : The colliding molecules must possess a certain mini-mum energy called activation energy required far breaking and making bonds resulting in the reaction. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

(iii) Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activation energy.

This suggests that in addition, the colliding molecules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B-l-C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

Question v.
What is the activation energy of a reaction?
Answer:
Activation energy : The energy required to form activated complex or transition state from the reactant molecules is called activation energy.
OR
The height of energy barrier in the energy profile diagram is called activation energy.

Question vi.
What are the units for rate constants for zero order and second order reactions if time is expressed in seconds and concentration of reactants in mol/L?
Answer:
(a) For a zero order reaction, the rate constant has units, molL^-1s^-1.
(b) For second order reaction,
Rate = k x [Reactant]²

Question vii.
Write Arrhenius equation and explain the terms involved in it.
Answer:
Arrhenius equation is represented as k = A x e^-E_a/RT
where
k = Rate constant at absolute temperature T
E_a = Energy of activation R = Gas constant
A = Frequency factor or pre-exponential factor.

Question viii.
What is the rate determining step?
Answer:
Many chemical reactions take place in a series of elementary steps. Among many steps of the reaction, one of the steps is the slowest step compared to other steps.

Rate determining step : The slowest step in the reaction mechanism which involves many steps is called the rate-determining step.

Question ix.
Write the relationships between rate constant and half life of fist order and zeroth order reactions.
Answer:
(a) For first order reaction, half-life period t_1/2 is, \(t_{1 / 2}=\frac{0.693}{k}\) where k is the rate constant.
(b) For zeroth-order reaction, half half period (t_1/2) is, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where k is the rate constant and [A]₀ is initial concentration of the reactant.

Question x.
How do half lives of the fist order and zero order reactions change with initial concentration of reactants?
Answer:
(A) For the first order reaction, half life, t_1/2 is given by, \(t_{1 / 2}=\frac{0.693}{k}\) where k is rate constant. Hence it is independent of initial concentration of the reactant.

(B) Zero order reaction,
\(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]₀ is initial concentration of the reactant.

Hence, half life period increases with the increase in concentration of the reactant.

3. Answer the following in brief.

Question i.
How instantaneous rate of reaction is determined?
Answer:
(1) The instantaneous rate is expressed as an infinite¬simal change in concentration (- dc) of the reactant with the infinitesimal change in time (dt).
For a reaction, A → B, let an infinitesimal change in A be – dc in time dt, then Rate \(=\frac{d[\mathrm{~A}]}{d t}\).

Hence, it is represented as,
∴ Instantaneous rate \(=-\frac{d[\mathrm{~A}]}{d t}\)

The negative sign indicates a decrease in the concentration of A.

It is obtained by drawing a tangent to the curve obtained by plotting the concentration against the time. Hence, the slope at a given point represents the instantaneous rate of the reaction.

(2) The instantaneous rate can also be expressed as an infinitesimal change (or increase) in the concentration of the product with the infinitesimal change in time (dt).

Let dB be an infinitesimal change in the concentration of product B in time dt, then Rate \(=\frac{d[\mathrm{~B}]}{d t}=\frac{d x}{d t}\).

Hence,
Instantaneous rate \(=\frac{d x}{d t}\)

It is obtained from the slope of the curve obtained by plotting the concentration of the product against time.

The instantaneous rate is more useful in obtaining the rate law integrated equations.

Question ii.
Distinguish between order and molecularity of a reaction.
Answer:

Order	Molecularity
1. It is the sum of the exponents to which the concentration terms in rate law expression are raised.	1. ¡t is the number of molecules (or atoms or ions) of the reaCtants taking part in the elementary reaction.
2. It is experimentally determined and indicates the dependence of the reaction rate on the concentration of particular reactants.	2. It is the oretical property and indicátes the number of molecules of reactant in each step of the reaction.
3. It may have values that are integer, fractional, or zero.	3. It is always an integer.
4. Its value depends upon experimental conditions.	4. Its value does not depend upon experimental conditions.
5. It is the property of elementary and complex reactions.	5. It is the property of elementary reactions only.
6. Rate law expression describes the order of the reaction.	6. Rate law does not describe molecularity.

Question iii.
A reaction takes place in two steps,
1. NO(g) + Cl₂(g) NOCl₂(g)
2. NOCl₂(g) + NO(g) → 2NOCl(g)
a. Write the overall reaction.
b. Identify reaction intermediate.
c. What is the molecularity of each step?
Solution :
Given :
(1) NO_(g) + Cl_2(g) → NOCl_2(g)
(2) NOCl_2(g) + NO_(g) → 2NOCl_(g)

(a) Overall reaction is obtained by adding both the reactions
2NO_(g) + Cl_2(g) → 2NOCl_2(g)
(b) The reaction intermediate is NOCl₂, since it is formed in first step and consumed in the second step.
(c) Since the first step is a slow and rate determin­ing step, the molecularity is two.

Since the second step is a fast step its molecularity is not considered.

Question iv.
Obtain the relationship between the rate constant and half-life of a fist order reaction.
Answer:
Consider the following reaction,

If [A]₀ and [A]t are the concentrations of A at start and after time t, then [A]₀ = a and [A]_t = a – x.

The velocity constant or the specific rate constant k for the first order reaction can be represented as,

where, a is the initial concentration of the reactant A, x is the concentration of the product B after time t, so that (a – x) is the concentration of the reactant A after time t.

Half-life of a reaction : The time required to reduce the concentration of the reactant to half of its initial value is called the half-life period or the half-life of the reaction.

If t_1/2 is the half-life of a reaction, then at t = t_1/2, x = a/2, hence a – x = a – a/2 = a/2

Hence, for a first order reaction, the half-life of the reaction is independent of the initial concentration of the reactant.

Question v.
How will you represent zeroth-order reaction graphically?
Answer:
(1) A graph of concentration against time : In case of a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. The concentration [A]_t of the reactant at a time t is given by
[A]_t = – kt + [A]₀ (y = – mx + c)
where [A]₀ is the initial concentration of the reactant and k is a rate constant.

Hence in case of zero order reaction, when the concentration of the reactant is plotted against time, a straight line with the slope equal to – k is obtained. The concentration of the reactants de-crease with time. The intercept on the concentration axis gives the initial concentration, [A]₀.

(2) A graph of rate of a reaction against the concen-tration of the reactant: Rate of a zero order reaction is independent of the concentration of the reactant.

Rate, R = k [A]⁰ = k

Hence even if the concentration of the reactant decreases, the rate of the reaction remains constant.

Therefore if the rate of a zero order reaction is plotted against concentration, then a straight line with zero slope is obtained indicating, no change in the rate of the reaction with a change in the concentration of the reactants.

(3) A graph of half-life period against concentration : The half-life period of a zero-order reaction is given by, \(t_{1 / 2}=\frac{[\mathrm{A}]_{0}}{2 k}\) where [A]₀ is initial con-centration of the reactant and k is the rate constant. Hence the half-life period is directly proportional to the concentration.

When a graph of t_1/2 is plotted against concentration, a straight line passing through origin is obtained, and the slope gives \(\frac{1}{2 k}\), where k is the rate constant.

Question vi.
What are pseudo-fist order reactions? Give one example and explain why it is pseudo-fist order.
Answer:
Pseudo-first-order reaction : A reaction which has higher-order true rate law but is experimentally found to behave as first order is called pseudo first order reaction.
Explanation : Consider an acid hydrolysis reaction of an ester like methyl acetate.
CH₃COOCH_3(aq) + H₂O₍₁₎ \(\stackrel{\mathrm{H}_{(\mathrm{aq})}^{+}}{\longrightarrow}\) CH₃COOH_(aq) + CH₃OH_(aq)
Since the reaction involves two substances, ester and water, it is a bimolecular reaction and the true rate law should be, Rate = k’ [CH₃COOCH₃] x [H₂O]

Hence the reaction is expected to follow second order kinetics. However experimentally it is found that the reaction follows first order kinetics.

This is because solvent water being in a large excess, its concentration remains constant. Hence, [H₂O] = constant = k”
Rate = k [CH₃COOCH₃] x [H₂O]
= k [CH₃COOCH₃] x k”
= k’ x k” x [CH₃COOCH₃]
If k’ x k” = k, then Rate = k [CH₃COOCH₃],

This indicates that second-order true rate law is forced into first order rate law. Therefore this bimolecular reaction which appears of second order is called pseudo first order reaction.

Question vii.
What are the requirements for the colliding reactant molecules to lead to products?
Answer:
Collisions of reactant molecules : The basic re­quirements of a reaction is that the reacting species (atoms, ions or molecules) must come together and collide for a reaction to occur. Therefore the rate of the reaction shall depend on the rate and frequency of collisions which increase with the amount of reacting species and temperature.

However it is observed that the rate of reaction is very low as compared to the rate of collisions between reacting species in gaseous phase or liquid phase. This suggests that all the collisions are not fruitful leading to a reaction. Hence it is necessary to consider another factor like energy of colliding species along with collision frequency.

Energy requirement (Activation energy) : The colliding molecules must possess a certain mini­mum energy called activation energy required far breaking and making bonds resulting in the reac­tion. This implies that the colliding molecules must have energy equal to or greater than the activation energy. The colliding molecules with less energy do not react.

Orientation of reactant molecules : The concept of activation energy is satisfactory in case of simple molecules or ions but not in case of complex or higher polyatomic molecules. It is observed that the rates of reaction are less as compared to the rates of collisions between activated molecules with activa­tion energy.

This suggests that in addition, the colliding mole­cules must have proper orientations relative to each other during collisions. For example, consider the reaction, A – B + C → A + B – C. For the reaction to occur, C must collide with B while collisions with A will not be fruitful. Since B has to bond with C.

Question viii.
How catalyst increases the rate of reaction? Explain with the help of a potential energy diagram for catalyzed and uncatalyzed reactions.
Answer:
(i) A catalyst is a substance, when added to the reactants, increases the rate of the reaction without being consumed. For example, the decomposition of KClO₃ in the presence of small amount of MnO₂ is very fast but very slow in the absence of MnO₂.

2KClO_3(s) \(\frac{\mathrm{MnO}_{2}}{\Delta}\) 2KCl_(s) + 3O_2(g)

(ii) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.

(iii) The catalyst provides alternative and lower energy path or mechanism for the reaction.

(iv) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

(v) Due to lowering of energy of activation, (E_a) the number of molecules possessing E_a increases, hence the rate of the reaction increases.

(vi) The rate constant = k = A x e^-E_a/RT where A is a frequency factor and hence the rates of the catalysed reaction are higher than those of un-catalysed reactions.

(vii) The catalyst does not change the extent of the reaction but hastens the reaction.

(viii) The catalyst enters the reaction but does not appear in the balanced equation since it is consumed in one step and regenerated in the another.

Question ix.
Explain with the help of the Arrhenius equation, how does the rate of reaction changes with (a) temperature and (b) activation energy.
Answer:
(a) By Arrhenius equation, k = Ax e^-E_a/RT where k is rate constant, A is a frequency factor and Ed is energy of activation at temperature T. As E_a increases, the rate constant and rate of the reaction decreases.

(b) As temperature increases E_a/RT decreases but due to negative sign, k and rate increase with the increase in temperature.

Question x.
Derive the integrated rate law for first order reaction.
Answer:
Consider following gas phase reaction,

Let initial pressure of A be P₀ at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, P_A = P_Q – x; P_B = x and P_c = x

Total pressure will be,

P_T + P₀ – x + x + x = Po + x
∴ x = P_T – P_n

The partial pressures at time t will be,

Question xi.
How will you represent first-order reactions graphically.
Answer:
(1) A graph of rate of a reaction and concentra­tion : The differential rate law for first-order reac­tion, A → Products is represented as, Rate = [/latex]-\frac{d[\mathrm{~A}]}{d t}=k[\mathrm{~A}]\(

∴ Rate = k x [A]_t (y = mx). When the rate of a first order reaction is plotted against concentration, [A]_t, a straight line graph is obtained.

With the increase in the concentration [A]_t, rate R, increases. The slope of the line gives the value of rate constant k.

(2) A graph of concentration against time : When the concentration of the reactant is plotted against time t, a curve is obtained. The concentration [A], of the reactant decreases exponentially with time. The variation in the concentration can be represented as,

where [A]₀ and [A]_t are initial and final concentra­tions the reactant and k is the rate constant. The time required to complete the first order reaction is infinity.

(3) A graph of log₁₀ (a – x) against time t :

When log₁₀(a – x) is plotted against time t, a straight line with negative slope is obtained, from which the velocity constant k can be calculated.

(4) A graph of half-life period and concentration : The half-life period, t_1/2 of a first order reaction is given by, where k is the rate constant.

For the given reaction at a constant temperature, t_1/2 is constant and independent of the concentration of the reactant.

Hence when a graph of t_1/2 is plotted against concentration, a straight line parallel to the concen­tration axis (slope = zero) is obtained.

(5) A graph of log₁₀ [latex]\left(\frac{a}{a-x}\right)\) against time : The rate constant, for a first order reaction is represented as, where [A₀] and [A]_t are the respective initial and final concentrations of the reactant after time t.


When \(\log _{10}\left(\frac{a}{a-x}\right)\) is plotted against time t, a straight line graph passing through the origin is obtained and the slope gives the value of k/2.303. From this slope, the rate constant can be calculated.

Question xii.
Derive the integrated rate law for the first order reaction, A(g) → B(g) + C(g) in terms of pressure.
Answer:
Consider following gas phase reaction,

Let initial pressure of A be P₀ at t = 0. If after time t the pressure of a A decreases by jc then the partial pressures of the substances will be, P_A = P_Q – x; P_B = x and P_c = x

Total pressure will be,

P_T + P₀ – x + x + x = Po + x
∴ x = P_T – P_n

The partial pressures at time t will be,

Question xiii.
What is zeroth-order reaction? Derive its integrated rate law. What are the units of rate constant?
Answer:
Definition : Zero order reaction : A reaction in which the rate of the reaction does not depend on the concentration of any reactant taking part in the reaction is called zero order reaction.

Consider a zero order reaction, A → Products
The rate of the reaction is, Rate \(=\frac{-d[\mathrm{~A}]}{d t}\)

By rate law,
Rate = k x [A]⁰ = k
∴ – d[A] = k x dt

If [A]₀ is the initial concentration of the reactant A at t = 0 and [A]_t is the concentration of A present after time t, then by integrating above equation,

This is the integrated rate law expression for rate constant for zero order reaction.
∴ k x t = [A]₀ – [A]_t
∴ [A]_t = – kt + A₀

For a zero order reaction :
The rate of reaction is R = k [A]⁰ = k
Hence, the velocity constant k has the unit of the rate of the reaction, i.e., mol dm^-3 s^-1.

Question xiv.
How will you determine activation energy: (a) graphically using Arrhenius equation (b) from rate constants at two different temperatures?
Answer:
(a) By Arrhenius equation,
Rate constant = = A x e^-E_a/RT where A is a fre-quency factor.

When log₁₀k is plotted against 1/T a straight line with negative slope is obtained. From the slope of the graph, energy of activation E_a, is obtained as follows :
Slope = \(\frac{E_{\mathrm{a}}}{2.303 R}\)
∴ Ea = 2303R x sloPe

(b) For the given reaction, rate constants k₁ and k₂ are measured at two different temperatures T₁ and T₂ respectively. Then \(\log _{10} \frac{k_{2}}{k_{1}}=\frac{E_{\mathrm{a}}\left(T_{2}-T_{1}\right)}{2.303 R \times T_{1} \times T_{2}}\) where E_a is the energy of activation.

Hence by substituting appropriate values, energy of activation E_a for the reaction is determined.

Question xv.
Explain graphically the effect of temperature on the rate of reaction.
Answer:
(i) It has been observed that the rates of chemical reactions increase with the increase in temperature.
(ii) The kinetic energy of the molecules increases with the increase in temperature. The fraction of molecules possessing minimum energy barrier,
i. e. activation energy E_a increases with increase in temperature.
(iii) Hence the fraction of colliding molecules that possess kinetic energy (E_a) also increases, hence the rate of the reaction increases with increase in temperature.

(iv) The above figure shows that the area that represents the fraction of molecules with kinetic energy exceeding E_a is greater at higher temperature T₂ than at lower temperature T₁. This explains that the rate of the reaction increases at higher temperature.
(v) The shaded area to the right of activation energy E_a represents fraction of collisions of activated molecules having energy E_a or greater.

Question xvi.
Explain graphically the effect of catalyst on the rate of reaction.
Answer:
(i) The phenomenon of catalysed reaction is called catalysis and depends on nature of the catalyst. In heterogeneous catalysis, the reactant molecules are adsorbed on the solid catalyst surface while in case of homogeneous catalysis, the catalyst reacts with one of the reactants, forms intermediate and decomposes reforming original catalyst and the products.
(ii) The catalyst provides alternative and lower energy path or mechanism for the reaction.
(iii) In the presence of the catalyst, the activation energy of the reaction is lowered. The height of activation energy barrier is less than that in the uncatalysed reaction.

Question xvii.
For the reaction 2A + B → products, find the rate law from the following data.

[A]/M	[A]/M	rate/M s-1
0.3	0.05	0.15
0.6	0.05	0.30
0.6	0.2	1.20

Solution :
Given : 2A + B → Products
Rates : R₁ = 0.15 Ms^-1 R₂ = 0.3 Ms^-1
[A]₁ = 0.3 M [A]₂ = 0.6 M
[B]₁ = 0.05 M [B]₂ = 0.05 M
(i) If order of the reaction in A is x and in B is y then, by rate law,

∴ y = 1. Hence the reaction has order one in B.
The order of overall reaction = n = n_A + n_B = 1 + 1 = 2

Answer:
(i) Rate law : Rate = fc [A] x [B]
Rate constant = k = 10M^-1s^-1
Order of the reaction = 2

4. Solve

Question i.
In a first order reaction, the concentration of reactant decreases from 20 mmol dm^-3 to 8 mmol dm^-3 in 38 minutes. What is the half life of reaction? (28.7 min)
Solution :
Given: [A]₀ =20 mmol dm^-3;
[A]_t=8 mmol dm^-3; t=38 mm;

Answer:
Half life period = 28.74 min

Question ii.
The half life of a first order reaction is 1.7 hours. How long will it take for 20% of the reactant to react? (32.9 min)
Solution :
Given : t_1/2 = 1.7 hr; [A]₀ = 100;
[A]_t = 100 – 20 = 80; t =?
\(t_{1 / 2}=\frac{0.693}{k}\)

Answer:
Time required = t = 32.86 min

Question iii.
The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at 25 0C is 3.7 × 10^-5 s^-1. What is the rate constant at 30⁰C? (R = 8.314 J/K mol) (7.4 × 10^-5)
Solution :

Answer:
k₂ = 7.382 x 10^-5 s^-1

Question iv.
What is the energy of activation of a reaction whose rate constant doubles when the temperature changes from 303 K to 313 K? (54.66 kJ/mol)
Solution :
Given : k₂ = 2k_t, T₁ = 303 K; T₂ = 313 K; E_a = ?

Answer:
Energy of activation = E_a = 54.66 kJ

Question v.
The rate constant of a reaction at 5000C is 1.6 × 10³ M^-1 s^-1. What is the frequency factor of the reaction if its activation energy is 56 kJ/mol. (9.72 × 106 M^-1 s^-1)
Solution :

Answer:
Frequency factor = A = 9.727 x 10⁶ M^-1s^-1

Question vi.
Show that time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.
Solution :
Given : For 99.9 % completion, if [A]₀ = 100,

If t₁ and t₂ are the times required for 99.9 % and 90 % completion of reaction respectively, then

Answer:
Time required for 99.9 % completion of a first order reaction is three time the time required for 90 % completion of the reaction.

Question vii.
A first order reaction takes 40 minutes for 30% decomposition. Calculate its half life. (77.66 min)
Solution :

Answer:
Half life period = 77.70 min.

Question viii.
The rate constant for the first order reaction is given by log₁₀ k = 14.34 – 1.25 × 10⁴ T. Calculate activation energy of the reaction. (239.3 kJ/mol)
Solution :
Given : log₁₀ k = 14.34 – \(\frac{1.25 \times 10^{4}}{T}\) ……………………. (1)
From Arrhenius equation we can write,
\(\log _{10} k=\log _{10} A-\frac{E_{\mathrm{a}}}{2.303 R \times T}\) ……………………. (2)
By comparing equations (1) and (2),
\(\frac{E_{\mathrm{a}}}{2.303 \times R}\) = 1.25 x 10⁴
∴ Ea = 1.25 x 10⁴ x 2.303 x R
= 1.25 x 10⁴ x 2.303 x 8.314
= 23.93 x 10⁴ = 239.3 kJ mol^-1

[Note : Frequency factor A may also be calculated as follows : log₁₀ A = 14.34
∴ A = Antilog 14.34 = 2.188 x 10⁴
Answer:
Energy of activation = E_a = 239.3 kJ mol^-1.

Question ix.
What fraction of molecules in a gas at 300 K collide with an energy equal to activation energy of 50 kJ/mol? (2 × 10^-9)
Solution :
Given : T = 300 K; E_a = 50 kJ mol^-1
= 50 x 10³ mol^-1
The fraction of molecules undergoing fruitful collisions is

Answer:
Fraction of molecules undergoing collision = 2 x 10^-9

Activity :
1. If you wish to determine the reaction order and rate constant for the reaction, 2AB₂ → A₂ + 2B₂.
a) What data would you collect?
b) How would you use these data to determine whether the reaction is zeroth or first order?

2. The activation energy for two reactions are E_a and E’_a with E_a > E’_a. If the temperature of reacting system increases from T₁ to T₂, predict which of the following is correct?

k values are rate constants at lower temperatures and k values at higher temperatures.

12th Chemistry Digest Chapter 6 Chemical Kinetics Intext Questions and Answers

(Textbook Page No 121)

Question 1.
Write the expressions for rates of reaction for :
2N₂O_5(g) → 4NO_2(g) + O_2(g)?
Answer:
For the given reaction, Rate of reaction =
\(=R=-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}\)
\(\begin{aligned}
&=+\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t} \\
&=+\frac{d\left[\mathrm{O}_{2}\right]}{d t}
\end{aligned}\)

Problem 6.1: (Textbook Page No 121)

Question 1.
For the reaction,
\(\mathbf{3 I}_{(a q)}^{-}+\mathbf{S}_{2} \mathbf{O}_{8(a q)}^{2-} \longrightarrow \mathbf{I}_{3(\text { (aq) }}^{-}+2 \mathbf{S O}_{4(\mathrm{aq})}^{2-}\)
Calculate (a) the rate of formation of I₃^–,
(b) the rates of consumption of 1 and S₂O and (c) the overall rate of reaction if the rate of formation of \(\mathrm{SO}_{4}^{2-}\) is 0.O22 moles dm^-3 sec^-1.
Answer:


∴ (a) Rate of formation of \(\mathrm{I}_{3}^{-}\) = 0.011 mol dm^-3 s^-1
(b) Rate of consumption of I^– = 0.033 mol dm^-3 s^-1
(c) Rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) = 0.011 mol dm^-3 s^-1
(d) Overall rate of reaction = Rate of consumption of reactant = Rate of formation of product

Try this….. (Textbook Page No 122)

Question 1.
For the reaction :
NO_2(g) + CO_(g) → NO_(g) + CO_2(g), the rate of reaction is experimentally found to be proportional to the square of the concentration of NO2 and independent that of CO. Write the rate law.
Answer:
Since the rate of the reaction is proportional to [NO₂]² and [CO]⁰, the rate law is R = k[NO₂]² x [CO]⁰
∴ R = k[NO₂]².

Try this….. (Textbook Page No 124)

Question 1.
The reaction,
CHCl_3(g) + Cl_2(g) → CCl_4(g) + HCl_(g) is first order in CHCl₃ and 1/2 order in Cl₂. Write the rate law and overall order of reaction.
Answer:
Since the reaction is first order in CHCl₃ and 1/2 order in Cl₂, the rate law for the reaction will be, Rate = k[CHCl₃] X [Cl₂]^1/2
The overall order (n) of the reaction will be, n = l + = \(\frac{1}{2}=\frac{3}{2}\)

Use your brain power! (Textbook Page No 124)

Question 1.
The rate of the reaction 2A + B → 2C + D is 6 x 10^-4 mol dm^-3 s^-1 when [A] =[B] = O.3 mol dm^-3 If the reaction is of first order in A and zeroth order in B, what is the rate constant?
Answer:
For the reaction,
2A + B → 2C + D,

(Problem 6.7) (Textbook Page No 126)

Question 1.
A reaction occurs in the following steps :
(i) NO_2(g) + F_2(g) → NO₂F_(g) + F_(g) (slow)
(ii) F_(g) + NO_2(g) → NO₂F_(g) (fast)
(a) Write the equation of overall reaction.
(b) Write down rate law.
(c) Identify the reaction intermediate.
Solution :
(a) The addition of two steps gives the overall reaction as
2NO_2(g) + F_2(g) → 2NO₂ F_(g)
(b) Step (i) is slow. The rate law of the reaction is predicted from its stoichiometry. Thus, rate = k [NO₂] [F₂]
(c) F is produced in step (i) and consumed in step (ii) hence F is the reaction intermediate.

Try this….. (Textbook Page No 126)

Question 1.
A complex reaction takes place in two steps :
(i) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(ii) NO_3(g) + O_(g) → NO_2(g) + O_2(g)
The predicted rate law is rate = k [NO] [O₃]. Identify the rate-determining step. Write the overall reaction. Which is the reaction inter-mediate? Why?
Answer:
(i) NO_(g) + O_3(g) → NO_3(g) + O_(g)
(ii) NO_3(g) + O_(g) → NO_2(g) + O_2(g)
(a) The first step is slow and rate determining step since the rate depends on concentrations of NO_(g) and O_3(g). (Given : Rate = k [NO] x [O])
(b) The overall reaction is the combination of two steps.
NO_(g) ⁺ O_3(g) → NO_2(g) + O_2(g)
(c) NO_3(g) and O_(g) are reaction intermediates. They are formed in first step (i) and removed in the second step (ii).

Try this….. (Textbook Page No 129)

Question 1.
The half-life of a first-order reaction is 0.5 min. Calculate (a) time needed for the reactant to reduce to 20% and (b) the amount decomposed in 55 s.
Answer:

Try this….. (Textbook Page No 123)

Question 1.
For the reaction 2A + 2B → 2C + D, if concentration of A is doubled at constant [B] the rate increases by a factor of 4. If the concentration of B is doubled with [A] being constant the rate is doubled. Write the rate law of the reaction.
Answer:
Rate = R₁ = k[A]^x [B]^y
When concentration of A = [2A] and

Hence order with respect to A is 2 and with respect to B is 1. By rate law,
Rate = A: [A]² [B]

Question 2.
The rate law for the reaction A + B → C is found to be rate = k [A]² x [B]. The rate constant of the reaction at 25 °C is 6.25 M^-2 S^-1. What is the rate of reaction when [A] = 1.0 mol dm^-3 and [B] = 0.2 mol dm^-3?
Answer:
Rate = k x [A]² x [B]
= 6.25 x 1² x 0.2
Rate = 1.25 x 10² mol dm^-3 s^-1

Balbharti Maharashtra State Board 12th Physics Textbook Solutions Chapter 14 Dual Nature of Radiation and Matter Textbook Exercise Questions and Answers.

Maharashtra State Board 12th Physics Solutions Chapter 14 Dual Nature of Radiation and Matter

1. Choose the correct answer.

i) A photocell is used to automatically switch on the street lights in the evening when the sunlight is low in intensity. Thus it has to work with visible light. The material of the cathode of the photocell is
(A) zinc
(B) aluminum
(C) nickel
(D) potassium
Answer:
(D) potassium

ii) Polychromatic (containing many different frequencies) radiation is used in an experiment on the photoelectric effect. The stopping potential
(A) will depend on the average wavelength
(B) will depend on the longest wavelength
(C) will depend on the shortest wavelength
(D) does not depend on the wavelength
Answer:
(C) will depend on the shortest wavelength

iii) An electron, a proton, an α-particle and a hydrogen atom are moving with the same kinetic energy. The associated de Broglie wavelength will be longest for
(A) electron
(B) proton
(C) α-particle
(D) hydrogen atom
Answer:
(A) electron

iv) If N_Red and N_Blue are the number of photons emitted by the respective sources of equal power and equal dimensions in unit time, then
(A) N_Red < N_Blue
(B) N_Red = N_Blue
(C) N_Red > N_Blue
(D) N_Red ≈ N_Blue
Answer:
(C) N_Red > N_Blue

v) The equation E = pc is valid
(A) for all sub-atomic particles
(B) is valid for an electron but not for a photon
(C) is valid for a photon but not for an electron
(D) is valid for both an electron and a photon
Answer:
(C) is valid for a photon but not for an electron

2. Answer in brief.

i) What is photoelectric effect?
Answer:
The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of appropriate frequency is incident on it is known as photoelectric effect.

ii) Can microwaves be used in the experiment on photoelectric effect?
Answer:
No

iii) Is it always possible to see photoelectric effect with red light?
Answer:
No

iv) Using the values of work function given in Table 14.1, tell which metal will require the highest frequency of incident radiation to generate photocurrent.

Answer:
Gold.
[ Note : W₀ = hv₀, where h is Planck’s constant. The larger the work function (W₀), the higher is the threshold frequency (v₀). ]

v) What do you understand by the term wave-particle duality? Where does it apply?
Answer:
Depending upon experimental conditions or structure of matter, electromagnetic radiation and material particles exhibit wave nature or particle nature. This is known as wave-particle duality.

It applies to all phenomena. The wave nature and particle nature are liked by the de Broglie relation λ = h/p, where λ is the wavelength of matter waves, also called de Broglie waves / Schrodinger waves, p is the magnitude of the momentum of a particle or quantum of radiation and h is the universal constant called Planck’s constant.

[Note : It is the smallness of h (= 6.63 × 10^-34 J∙s) that is very significant in wave-particle duality.]

Question 3.
Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.
Answer:
We have V₀e = \(\frac{h c}{\lambda}\) – Φ, where V₀ is the stopping potential, e is the magnitude of the charge on the electron, h is Planck’s constant, c is the speed of light in free space, λ is the wavelength of the electromagnetic radiation incident on a metal surface and Φ is the work function for the metal, h, c and e are constants. Φ is constant for a particular metal.
Hence, it follows that as \(\frac{1}{\lambda}\) increases, V₀ increases.
The plot of V₀ verses \(\frac{1}{\lambda}\) is linear. This is because the energy associated with a quantum of radiation (photon) is directly proportional to the frequency of radiation and hence inversely proportional to the wavelength of radiation.

Question 4.
It is observed in an experiment on photoelectric effect that an increase in the intensity of the incident radiation does not change the maximum kinetic energy of the electrons. Where does the extra energy of the incident radiation go? Is it lost? State your answer with explanatory reasoning.
Answer:
When electromagnetic radiation with frequency greater than the threshold frequency is incident on a metal surface, there is emission of electrons. It is observed that not every incident photon is effective in liberating an electron. In fact, the number of electrons emitted per second is far less than the number of photons incident per second. The photons that are not effective in liberation of electrons are reflected (or scattered) or absorbed resulting in rise in the temperature of the metal surface. The maximum kinetic energy of a photoelectron depends on the frequency of the incident radiation and the threshold frequency for the metal. It has nothing to do with the intensity of the incident radiation. The increase in intensity results in increase in the number of electrons emitted per second.

Question 5.
Explain what do you understand by the de Broglie wavelength of an electron. Will an electron at rest have an associated de Broglie wavelength? Justify your answer.
Answer:
Under certain conditions an electron exhibits wave nature. Waves associated with a moving electron are called matter waves or de Broglie waves or- Schrodinger waves. The de Broglie wavelength of these matter waves is given by X = h/p, where h is Planck’s constant and p is the magnitude of the momentum of the electron.

If an electron is at rest, its momentum would be zero, and hence the corresponding de Broglie wavelength would be infinite indicating absence of a matter wave. However, according to quantum mechanics/wave mechanics, this is not possible.

Question 6.
State the importance of Davisson and Germer experiment.
Answer:
The Davisson and Germer experiment directly indicated the wave nature of material particles and quantitatively verified the de Broglie hypothesis for the existence of matter waves.

[Note : The aim of the experiment was not to verify wave like properties of electrons. The realisation came only later, an example of serendipity.]

[Note : Like X-rays, electrons exhibit wave nature under suitable conditions. When the wavelength of matter waves associated with moving electrons is comparable to the inter-atomic spacing in a crystal, electrons show diffraction effects. In 1927, Sir George Thomson (1892 – 1975), British physicist, with his student Alex Reid, observed electron diffraction with a metal foil. It is found that neutrons, atoms, molecules, Œ-particles, etc. show wave nature under suitable conditions.]

Question 7.
What will be the energy of each photon in monochromatic light of frequency 5 × 10¹⁴ Hz?
Answer:
Data: y = 5 × 10¹⁴ Hz, h = 6.63 × 10^-34 Js,
1eV=1.6 × 10^-19 J
The energy of each photon,
E = hv = (6.63 × 10^-34 J.s)(5 × 10¹⁴ Hz)
= 3.315 × 10^-19 J
= \(\frac{3.315 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 2.072 eV

Question 8.
Observations from an experiment on photoelectric effect for the stopping potential by varying the incident frequency were plotted. The slope of the linear curve was found to be approximately 4.1 × 10^-15 V s. Given that the charge of an electron is 1.6 × 10^-19 C, find the value of the Planck’s constant h.
Answer:
Data : Slope=4.1 × 10^-15 V∙s, e = 1.6 ×10^-19 C
V₀e = hv – hv₀
∴ V₀ =\(\left(\frac{h}{e}\right) v-\left(\frac{h v_{0}}{e}\right)\)
∴ Slope = \(\frac{h}{e}\) ∴ Planck’s constant,
h = (slope) (e)=(4.1 × 10^-15 V∙s)(1.6 × 10^-19 C)
= 6.56 × 10 34J. (as 1 V = \(\frac{1 \mathrm{~J}}{1 \mathrm{C}}\))

Question 9.
The threshold wavelength of tungsten is 2.76 × 10^-5 cm. (a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 × 10^-5 cm.(b) What will be the maximum kinetic energy of electrons ejected in each of the following cases
(i) if ultraviolet radiation of wavelength λ = 1.80 × 10^-5 cm and
(ii) radiation of frequency 4 × 10¹⁵ Hz is made incident on the tungsten surface.
Answer:
Data: λ₀ = 2.76 × 10^-5 cm = 2.76 × 10^-7 m,
λ =1.80 × 10^-5 cm = 1.80 × 10^-7 m,
v = 4 × 10¹⁵ Hz, h = 6.63 × 10^-34 J∙s,c = 3 × 10⁸ m/s
(a) For λ > λ₀, v < v₀ (threshold frequency).
∴ hv < hv₀. Hence, no photoelectrons are emitted.

(b) Maximum kinetic energy of electrons ejected
= hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)\)
=(6.63 × 10^-34)(3 × 10⁸)\(\left(\frac{10^{7}}{1.8}-\frac{10^{7}}{2.76}\right)\)J
= (6.63 × 10^-19)(0.5555 – 0.3623)
= (6.63)(0.1932 × 10^-19)J = 1.281 × 10^-19 J
= \(\frac{1.281 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 0.8006 eV

(c) Maximum kinetic energy of electrons ejected
= hv – \(\frac{h c}{\lambda_{0}}\)
=(6.63 × 10^-34(4 × 10¹⁵) – \(\frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{2.76 \times 10^{-7}}\)
= 26.52 × 10^-19 – 7.207 × 10^-19
= 19.313 × 10^-19 J
= \(\frac{19.313 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\) = 12.07eV

Question 10.
Photocurrent recorded in the micro ammeter in an experimental set-up of photoelectric effect vanishes when the retarding potential is more than 0.8 V if the wavelength of incident radiation is 4950 Å. If the source of incident radiation is changed, the stopping potential turns out to be 1.2 V. Find the work function of the cathode material and the wavelength of the second source.
Answer:
Data: V₀ = 0.8 V, λ = 4950 Å = 4.950 × 10^-7 m,
V₀‘ = 1.2V, h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s.
(i) V₀e = hv – Φ = \(\frac{h c}{\lambda}\) – Φ
∴ The work function of the cathode material,

Question 11.
Radiation of wavelength 4500 Å is incident on a metal having work function 2.0 eV. Due to the presence of a magnetic field B, the most energetic photoelectrons emitted in a direction perpendicular to the field move along a circular path of radius 20 cm. What is the value of the magnetic field B?
Answer:
Data: λ = 4500Å = 4.5 × 10^-7 m,
Φ = 2.0eV = 2 × 1.6 × 10^-19 J = 3.2 × 10^-19 J,
h = 6.63 × 10^-34 J∙s, c = 3 × 10⁸ m/s,
r = 20 cm = 0.2 m, e= 1.6 × 10^-19 C,
m = 9.1 × 10^-31kg

This is the value of the magnetic field.

Question 12.
Given the following data for incident wavelength and the stopping potential obtained from an experiment on photoelectric effect, estimate the value of Planck’s constant and the work function of the cathode material. What is the threshold frequency and corresponding wavelength? What is the most likely metal used for emitter?

Answer:
Data: λ = 2536Å = 2.536 × 10^-7 m,
λ’ = 3650Å = 3.650 ×10^-7 m, V₀ = 1.95V, V₀‘ = 0.5V,
c = 3 × 10⁸ mIs, e = 1.6 × 10^-19 C

(i) V₀e = \(\frac{h c}{\lambda}\) – Φ and V₀‘e =\(\frac{h c}{\lambda^{\prime}}\) – Φ
∴ (V₀ – V₀‘)e = hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda^{\prime}}\right)\)
∴ (1.95 – 0.5(1.6 × 10^-19)
= h (3 × 10⁸\(\left(\frac{10^{7}}{2.536}-\frac{10^{7}}{3.650}\right)\)
∴ 2.32 × 10^-19 = h(3 × 10¹⁵)(0.3943 – 0.2740)
∴ h = \(\frac{2.32 \times 10^{-34}}{0.3609}\) = 6.428 × 10^-34 J∙s
This is the value of Planck’s constant.

(ii) Φ = \(\frac{h c}{\lambda}\) – V₀e

This is the work function of the cathode material.

(iii) Φ = hv₀
∴ The threshold frequency, v₀ = \(\frac{\phi}{h}\)
= \(\frac{4.484 \times 10^{-19} \mathrm{~J}}{6.428 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}\) = 6.976 × 10¹⁴ Hz

(iv) v₀ = \(\frac{c}{\lambda_{\mathrm{o}}}\) ∴ The threshold frequency, λ₀ = \(\frac{c}{v_{\mathrm{o}}}\)
= \(\frac{3 \times 10^{8}}{6.976 \times 10^{14}}\) = 4.300 × 10^-7 m = 4300 Å

(v) The most likely metal used for emitter : calcium

Question 13.
Calculate the wavelength associated with an electron, its momentum and speed
(a) when it is accelerated through a potential of 54 V
Answer:
Data : V = 54 V, m = 9.1 × 10^-31 kg, e
e = 1.6 × 10^-19 C, h = 6.63 × 10^-34 J.s, KE = 150 eV
(a) We assume that the electron is initially at rest.
∴ Ve = \(\frac{1}{2}\)mv²
∴ v = \(\sqrt{\frac{2 V e}{m}}=\sqrt{\frac{2(54)\left(1.6 \times 10^{-19}\right)}{9.1 \times 10^{-31}}}\)
= \(\sqrt{19 \times 10^{12}}\) = 4.359 × 10⁶ m/5
This is the speed of the electron.
p = mv = (9.1× 10^-31)(4.359 × 10⁶)
= 3.967 × 10^-24 kg∙m/s
This is the momentum of the electron. The wavelength associated with the electron,
λ = \(\frac{h}{p}=\frac{6.63 \times 10^{-34}}{3.967 \times 10^{-24}}\) = 1.671 × 10^-10 m
= 1.671 Å = 0.1671 nm

(b) when it is moving with kinetic energy of 150 eV.
Answer:
As KE ∝ \(\sqrt{V}\), we get
\(\frac{v^{\prime}}{v}=\sqrt{\frac{150}{54}}\) = 1.666
∴ v’ = 1.666v = (1.666)(4.356 × 10⁶)
= 7.262 × 10⁶ m/s
This is the speed of the electron.
p’ = mv’’=(9.1 × 10^-31)(7.262 × 10⁶)
= 6.608 × 10^-24 kg∙m/s
This is the momentum of the electron. The
wavelength associated with the electron,
λ = \(\frac{h}{p^{\prime}}=\frac{6.63 \times 10^{-34}}{6.608 \times 10^{-24}} \) = 1.003 × 10^-10 m
= 1.003 Å = 0.1003 nm

Question 14.
The de Broglie wavelengths associated with an electron and a proton are same. What will be the ratio of
(i) their momenta
(ii) their kinetic energies?
Answer:
Data : λ (electron) = λ (proton)
m (proton) = 1836 m (electron)
(i) λ = \(\frac{h}{p}\) As λ (electron) = λ (proton),
\(\frac{p(\text { electron })}{p \text { (proton) }}\) = 1, where p denotes the magnitude of momentum.

(ii) Assuming v «c,
KE = \(\frac{1}{2}\)mv² = \(\frac{1}{2} \frac{m^{2} v^{2}}{m}=\frac{p^{2}}{2 m}\)
∴ \(\frac{\mathrm{KE} \text { (electron) }}{\mathrm{KE} \text { (proton) }}=\frac{m \text { (proton) }}{m \text { (electron) }}\) = 1836 as p is the same for the electron and the proton.

Question 15.
Two particles have the same de Broglie wavelength and one is moving four times as fast as the other. If the slower particle
is an α-particle, what are the possibilities for the other particle?
Answer:
Data : λ₁ = λ₂, v₁ = 4v₂
λ = \(\frac{h}{p}=\frac{h}{m v}\) ∴ λ₁ = \(\frac{h}{m_{1} v_{1}}\), λ₂ = \(\frac{h}{m_{2} v_{2}}\)
∴ m₁ = m₂ \(\frac{v_{2}}{v_{1}}\) = m₂\(\left(\frac{1}{4}\right)=\frac{m_{2}}{4}\)
As particle 2 is the a-particle, particle 1 (having the mass \(\frac{1}{4}\) times that of the a-particle) may be a proton or neutron.

Question 16.
What is the speed of a proton having de Broglie wavelength of 0.08 Å?
Answer:
Data : λ = 0.08 Å = 8 × 10^-12m, h = 6.63 × 10^-34 J∙s, m = 1.672 × 10^-27 kg
λ = \(\frac{h}{m v}\) ∴ v = \(\frac{h}{\lambda m}=\frac{6.63 \times 10^{-34}}{\left(8 \times 10^{-12}\right)\left(1.672 \times 10^{-27}\right)}\)
∴ v = 4.957 × 10⁴ m/s
This is the speed of the proton.

Question 17.
In nuclear reactors, neutrons travel with energies of 5 × 10^-21 J. Find their speed and wavelength.
Answer:
Data : KE = 5 × 10^-21 J, m = 1.675 × 10^-27 kg, h = 6.63 × 10^-34 J∙s
KE = \(\frac{1}{2}\) mv² = 5 × 10^-21 J
∴ v = \(\sqrt{\frac{2 \mathrm{KE}}{m}}=\sqrt{\frac{(2)\left(5 \times 10^{-21}\right)}{1.675 \times 10^{-27}}}\)
= 2.443 × 10³ m/s
This is the speed of the neutrons. The de Broglie wavelength associated with the neutron,
λ = \(\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{\left(1.675 \times 10^{-27}\right)\left(2.443 \times 10^{3}\right)}\)
= 1.620 × 10^-10 m = 1.620 Å

Question 18.
Find the ratio of the de Broglie wavelengths of an electron and a proton when both are moving with the (a) same speed, (b) same energy and (c) same momentum? State which of the two will have the longer wavelength in each case?
Answer:
Data: m_p = 1836 m_e

12th Physics Digest Chapter 14 Dual Nature of Radiation and Matter Intext Questions and Answers

Remember This (Textbook Page No. 316)

Question 1.
Is solar cell a photocell?
Answer:
Yes

Remember This (Textbook Page No. 317)

Question 1.
Can you estimate the de Broglie wavelength of the Earth?
Answer:
Taking the mass of the Earth as (about) 6 × 10²⁴ kg, and the linear speed of the earth around the Sun as (about) 3 × 10⁴ m/s, we have, the de Brogue wave length of the Earth as
λ = \(\frac{h}{p}=\frac{h}{M v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{\left(6 \times 10^{24} \mathrm{~kg}\right)\left(3 \times 10^{4} \mathrm{~m} / \mathrm{s}\right)}\)
= 3.683 × 10^-63 m (extremely small)

Question 2.
The expression p = E/c defines the momentum of a photon. Can this expression be used for momentum of an electron or proton?
Answer:
No

Remember This (Textbook Page No. 319)

Diffraction results described above can be produced in the laboratory using an electron diffraction tube as shown in figure. It has a filament which on heating produces electrons. This filament acts as a cathode. Electrons are accelerated to quite high speeds by creating large potential difference between the cathode and a positive electrode. On its way, the beam of electrons comes across a thin sheet of

graphite. The electrons are diffracted by the atomic layers in the graphite and form diffraction rings on the phosphor screen. By changing the voltage between the cathode and anode, the energy, and therefore the speed, of the electrons can be changed. This will change the wavelength of the electrons and a change will be seen in the diffraction pattern. By increasing the voltage, the radius of the diffraction rings will decrease. Try to explain why?
Answer:
When the accelerating voltage is increased, the kinetic energy and hence the momentum of the electron increases. This decreases the de Brogue wavelength of the electron. Hence, the radius of the diffraction ring decreases.

Remember This (Textbook Page No. 320)

Question 1.
On what scale or under which circumstances are the wave nature of matter apparent?
Answer:
When the de Brogue wavelength of a particle such as an electron, atom, or molecule is comparable to the interatomic spacing in a crystal, the wave nature of matter is revealed in diffraction/interference.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 1.
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of (i) 2 successes (ii) at least 3 successes (iii) at most 2 successes.
Solution:
X: Getting an odd no.
p: Probability of getting an odd no.
A die is thrown 4 times
∴ n = 4
∵ p = \(\frac{3}{6}=\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∵ X ~ B(3, \(\frac{1}{2}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(Two Successes)

(ii) P(Atleast 3 Successes)

(iii) P(Atmost 2 Successes)

Question 2.
A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
n: No. of times die is thrown = 3
X: No. of doublets
p: Probability of getting doublets
Getting a doublet means, same no. is obtained on 2 throws of a die
There are 36 outcomes
No. of doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Question 3.
There are 10% defective items in a large bulk of items. What is the probability that a sample of 4 items will include not more than one defective item?
Solution:
n: No of sample items = 4
X: No of defective items
p: Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
X ~ B(4, 0.1)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} \mathrm{q}^{n-x}\)
P(Not include more than 1 defective)
P(X ≤ 1) = p(0) + p(1)
= ⁴C₀ (0.1)⁰ (0.9)⁴ + ⁴C₁ (0.1)¹ (0.9)^4-1
= 1 × 1 × (0.9)⁴ + 4 × 0.1 × (0.9)³
= (0.9)³ [0.9 + 0.4]
= (0.9)³ × 1.3
= 0.977

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that (i) all the five cards are spades, (ii) only 3 cards are spades, (iii) none is a spade.
Solution:
X: No. of spade cards
Number of cards drawn
∴ n = 5
p: Probability of getting spade card

(i) P(All five cards are spades)

(ii) P(Only 3 cards are spades)

(iii) P(None is a spade)

Question 5.
The probability that a bulb produced by a factory will use fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of (i) X = 0, (ii) X ≤ 1, (iii) X > 1, (iv) X ≥ 1.
Solution:
X : No. of bulbs fuse after 200 days of use
p : Probability of getting fuse bulbs
No. of bulbs in a sample
∴ n = 5
∴ p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
∵ X ~ B(5, 0.2)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(X = 0) = ⁵C₀ (0.2)⁰ (0.8)^5-0
= 1 × 1 × (0.8)⁵
= (0.8)⁵

(ii) P(X ≤ 1) = p(0) + p(1)
= ⁵C₀ (0.2)⁰ (0.8)^5-0 + ⁵C₁ (0.2)¹ (0.8)^5-1
= 1 × 1 × (0.8)⁵ + 5 × 0.2 × (0.8)⁴
= (0.8)⁴ [0.8 + 1]
= 1.8 × (0.8)⁴

(iii) P(X > 1) = 1 – [p(0) + p(1)]
= 1 – 1.8 × (0.8)⁴

(iv) P(X ≥ 1) = 1 – p(0)
= 1 – (0.8)⁵

Question 6.
10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
Solution:
X : No. of balls drawn marked with the digit 0
n : No. of balls drawn
∴ n = 4
p : Probability of balls marked with 0.
∴ p = \(\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
p(x) = \({ }^{n} C_{x} p^{x} q^{n-x}\)
P(None of the ball is marked with digit 0)

Question 7.
In a multiple-choice test with three possible answers for each of the five questions, what is the probability of a candidate getting four or more correct answers by random choice?
Solution:
n: No. of Questions
∴ n = 5
X: No. of correct answers by guessing
p: Probability of getting correct answers

Question 8.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
X : No. of sixes in 6 throws
n : No. of times dice thrown
∴ n = 6
p : Probability of getting six
∴ p = \(\frac{1}{6}\)
∴ q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
∵ X ~ B(6, \(\frac{1}{6}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
P(At most 2 sixes)
P(X ≤ 2) = p(0) + p(1) + p(2)

Question 9.
Given that X ~ B(n, p),
(i) if n = 10 and p = 0.4, find E(X) and Var(X).
(ii) if p = 0.6 and E(X) = 6, find n and Var(X).
(iii) if n = 25, E(X) = 10, find p and Var(X).
(iv) if n = 10, E(X) = 8, find Var(X).
Solution:
∵ X ~ B (n, p), E(X) = np, V(X) = npq, q = 1 – p
(i) E(X) = np = 10 × 0.4 = 4
∵ q = 1 – p = 1 – 0.4 = 0.6
V(X) = npq = 10 × 0.4 × 0.6 = 2.4

(ii) ∵ p = 0.6
∴ q = 1 – p = 1 – 0.6 = 0.4
E(X) = np
∴ 6 = n × 0.6
∴ n = 10
∴ V(X) = npq = 10 × 0.6 × 0.4 = 2.4

(iii) E(X) = np
∴ 10 = 25 × p
∴ p = 0.4
∴ q = 1, p = 1 – 0.4 = 0.6
∴ S.D.(X) = √V(X)
= \(\sqrt{n p q}\)
= \(\sqrt{25 \times 0.4 \times 0.6}\)
= √6
= 2.4494

(iv) ∵ E(X) = np
∴ 8 = 10p
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
∵ V(X) = npq = 10 × 0.8 × 0.2 = 1.6

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Ex 3.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Ex 3.2

Question 1.
For bivariate data.
\(\bar{x}\) = 53, \(\bar{x}\) = 28, b_yx = -1.2, b_xy = -0.3
Find,
(i) Correlation coefficient between X and Y.
(ii) Estimate Y for X = 50
(iii) Estimate X for Y = 25
Solution:
(i) r² = b_yx . b_xy
r² = (-1.2)(-0.3)
r² = 0.36
r = ±0.6
Since, b_yx and bxy are negative, r = -0.6

(ii) Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
Y – 28 = -1.2(50 – 53)
Y – 28 = -1.2(-3)
Y – 28 = 3.6
Y = 31.6

(iii) Regression equation of X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 53) = -0.3(25 – 28)
X – 53 = -0.3(-3)
X – 53 = 0.9
X = 53.9

Question 2.
From the data of 20 pairs of observation on X and Y, following result are obtained \(\bar{x}\) = 199, \(\bar{y}\) = 94, \(\sum\left(x_{i}-\bar{x}\right)^{2}\) = 1200, \(\sum\left(y_{i}-\bar{y}\right)^{2}\) = 300
\(\sum\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)\) = -250
Find
(i) The line of regression of Y on X.
(ii) The line of regression of X on Y.
(iii) Correlation coefficient between X on Y.
Solution:

Question 3.
From the data of 7 pairs of observations on X and Y following results are obtained.
Σ(x_i – 70 ) = -35, Σ(y_i – 60) = -7, Σ(x_i – 70)² = 2989, Σ(y_i – 60)² = 476, Σ(x_i – 70) (y_i – 60) = 1064 [Given √0.7884 = 0.8879]
Obtain
(i) The line of regression of Y on X.
(ii) The line of regression of X on Y.
(iii) The correlation coefficient between X and Y.
Solution:


(i) Line of regression Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 59) = 0.36(x – 65)
(Y – 59) = 0.36x – 23.4
Y = 0.36x + 35.6

(ii) Line of regression X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 65) = 2.19(y – 59)
(X – 65) = 2.19y – 129.21
X = 2.19y – 64.21

(iii) r² = b_yx . b_xy
r² = (0.36) (2.19)
r² = 0.7884
r = ±√0.7884 = ±0.8879
Since b_yx and b_xy are positive.
∴ r = 0.8879

Question 4.
You are given the following information about advertising expenditure and sales.

Correlation coefficient between X and Y is 0.8
(i) Obtain two regression equations.
(ii) What is the likely sales when the advertising budget is ₹ 15 lakh?
(iii) What should be the advertising budget if the company wants to attain sales target of ₹ 120 lakh?
Solution:
Given, \(\bar{x}\) = 10, \(\bar{y}\) = 90, σ_x = 3, σ_y = 12, r = 0.8
byx = \(\frac{r \sigma_{y}}{\sigma_{x}}=0.8 \times \frac{12}{3}\) = 3.2
bxy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.8 \times \frac{3}{12}\) = 0.2
(i) Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 90) = 3.2(x – 10)
Y – 90 = 3.2x – 32
Y = 3.2x + 58
Regression equation of X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 10) = 0.2(y – 90)
X – 10 = 0.2y + 18
X = 0.2y – 8

(ii) When x = 15,
Y = 3.2(15) + 58
= 48 + 58
= 106 lakh

(iii) When y = 120
X = 0.2(120) – 8
= 24 – 8
= 16 lakh

Question 5.
Bring out inconsistency if any, in the following:
(i) b_yx + b_xy = 1.30 and r = 0.75
(ii) b_yx = b_xy = 1.50 and r = -0.9
(iii) b_yx = 1.9 and b_xy = -0.25
(iv) b_yx = 2.6 and b_xy = \(\frac{1}{2.6}\)
Solution:
(i) Given, b_yx + b_xy = 1.30 and r = 0.75
\(\frac{b_{y x}+b_{x y}}{2}=\frac{1.30}{2}\) = 0.65
But for regression coefficients b_yx and b_xy
\(\left|\frac{b_{y x}+b_{x y}}{2}\right| \geq r\)
Here, 0.65 < r = 0.75
∴ The data is inconsistent
(ii) The signs of b_yx, b_xy and r must be same (all three positive or all three negative)
∴ The data is inconsistent.

(iii) The signs of b_yx and b_xy should be same (either both positive or both negative)
∴ The data is consistent.

(iv) b_yx . b_xy = 2.6 × \(\frac{1}{2.6}\) = 1
∴ 0 ≤ r² ≤ 1
∴ The data is consistent.

Question 6.
Two sample from bivariate populations have 15 observation each. The sample means of X and Y are 25 and 18 respectively. The corresponding sum of square of deviations from respective means are 136 and 150. The sum of product of deviations from respective means is 123. Obtain the equation of line of regression of X on Y.
Solution:
Given, n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, Σ(x – \(\bar{x}\)) = 136, Σ(y – \(\bar{y}\)) = 150, Σ(x – \(\bar{x}\)) (y – \(\bar{y}\)) = 123
Regression equation of X on Y is (X – \(\bar{x}\)) = bxy (Y – \(\bar{y}\))
(X – 25) = 0.82(y – 18)
(X – 25) = 082y – 14.76
X = 0.82y + 10.24

Question 7.
For a certain bivariate data

And r = 0.5 estimate y when x = 10 and estimate x when y = 16
Solution:
Given, \(\bar{x}\) = 25, \(\bar{y}\) = 20, σ_x = 4, σ_y = 3, r = 0.5
b_yx = \(\frac{r \sigma_{y}}{\sigma_{y}}=0.5 \times \frac{3}{4}\) = 0.375
Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 20) = 0.375(x – 25)
Y – 20 = 0.375x – 9.375
Y = 0.375x + 10.625
When, x = 10
Y = 0.375(10) + 10.625
= 3.75 + 10.625
= 14.375
b_xy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.5 \times \frac{4}{3}\) = 0.67
Regression equation of X on Y is
(X – \(\bar{x}\)) = b_yx (Y – \(\bar{y}\))
(X – 25) = 0.67(y – 20)
(X – 25) = 0.67y – 13.4
X = 0.67y + 11.6
When, Y = 16
x = 0.67(16) + 11.6
= 10.72 + 11.6
= 22.32

Question 8.
Given the following information about the production and demand of a commodity obtain the two regression lines:

Coefficient of correlation between X and Y is 0.6. Also estimate the problem when demand is 100.
Solution:
Given \(\bar{x}\) = 85, \(\bar{y}\) = 90, σ_x = 5, σ_y = 6 and r = 0.6
bxy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.6 \times \frac{5}{6}\) = 0.5
byx = \(\frac{r \sigma_{y}}{\sigma_{x}}=0.6 \times \frac{6}{5}\) = 0.72
Regression equation of X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 85) = 0.5(y – 90)
(X – 85 ) = 0.5y – 45
X = 0.5y + 40
When y = 100,
x = 0.5 (100) + 40
= 50 + 40
= 90
Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 90) = 0.72(x – 85)
(Y – 90) = 0.72x – 61.2
Y = 0.72x + 28.8

Question 9.
Given the following data, obtain linear regression estimate of X for Y = 10
Solution:
\(\bar{x}\) = 7.6, \(\bar{y}\) = 14.8, σ_x = 3.2, σ_y = 16 and r = 0.7
b_xy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.7 \times \frac{3.2}{16}\) = 0.14
Regression equation of X on Y is
(X – \(\bar{y}\)) = b_xy (Y – \(\bar{y}\))
(X – 7.6) = 0.14(y – 14.8)
X – 7.6 = 0.14y – 2.072
X = 0.14y + 5.528
When y = 10
x = 0.14(10) + 5.528
= 1.4 + 5.528
= 6.928

Question 10.
An inquiry of 50 families to study the relationship between expenditure on accommodation (₹ x) and expenditure on food and entertainment (₹ y) gave the following result:
Σx = 8500, Σy = 9600, σ_x = 60, σ_y = 20, r = 0.6
Estimate the expenditure on food and entertainment when expenditure on accommodation is ₹ 200
Solution:
n = 50 (given)

Regression equation of Y on X is
Y – \(\bar{y}\) = b_yx (X – \(\bar{x}\))
(Y – 192) = 0.2(200 – 170)
Y – 192 = 0.2(30)
Y = 192 + 6
Y = 198

Question 11.
The following data about the sales and advertisement expenditure of a firms is given below (in ₹ crores)

Also correlation coefficient between X and Y is 0.9
(i) Estimate the likely sales for a proposed advertisement expenditure of ₹ 10 crores.
(ii) What should be the advertisement expenditure if the firm proposes a sales target ₹ 60 crores
Let the sales be X and advertisement expenditure be Y
Solution:
Given, \(\bar{x}\) = 40, \(\bar{y}\) = 6, σ_x = 10, σ_y = 1.5, r = 0.9
b_yx = \(\frac{r \sigma_{y}}{\sigma_{x}}=0.9 \times \frac{1.5}{10}\) = 0.135
b_xy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.9 \times \frac{10}{1.5}\) = 6
(i) Regression equation of X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 40) = 6(y – 6)
X – 40 = 6y – 36
X = 6y + 4
When y = 10
x = 6 (10) + 4
= 60 + 4
= 64 crores

(ii) Regression equation Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 6) = 0.135 (x – 40)
Y – 6 = 0.135x – 5.4
Y = 0.135x + 0.6
When x = 60
Y = 0.135 (60) + 0.6
= 8.1 + 0.6
= 8.7 crores

Question 12.
For certain bivariate data the following information are available

Correlation coefficient between x and y is 0.6, estimate x when y = 15 and estimate y when x = 10.
Solution:
Given, \(\bar{x}\) = 13, \(\bar{y}\) = 17, σ_x = 3, σ_y = 2, r = 0.6
b_yx = \(\frac{r \sigma_{y}}{\sigma_{x}}=0.6 \times \frac{2}{3}\) = 0.4
b_xy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.6 \times \frac{3}{2}\) = 0.9
Regression equation of Y on X
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
Y – 17 = 0.4(x – 13)
Y = 0.4x + 11.8
When x = 10
Y = 0.4(10) + 11.8
= 4 + 11.8
= 15.8
Regression equation of X on Y
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 13) = 0.9(y – 17)
X – 13 = 0.9y – 15.3
X = 0.9y – 2.3
When y = 15
X = 0.9(15) – 2.3
= 13.5 – 2.3
= 11.2

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Ex 3.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Ex 3.1

Question 1.
The HRD manager of the company wants to find a measure which he can use to fix the monthly income of persons applying for the job in the production department. As an experimental project. He collected data of 7 persons from that department referring to years of service and their monthly incomes.

(i) Find the regression equation of income on years of service.
(ii) What initial start would you recommend for a person applying for the job after having served in a similar capacity in another company for 13 years?
Solution:


(i) Regression equation of Y on X is (Y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(Y – 8) = 0.75(x – \(\bar{x}\))
Y = 0.75x + 2
(ii) When x = 13
Y = 0.75(13) + 2 = 11.75
Recommended income for the person is ₹ 11750.

Question 2.
Calculate the regression equations of X on Y and Y on X from the following date:

Solution:


Regression equation of X on Y is (X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 14) = 1(Y – 8)
X – 14 = Y – 8
X = Y + 6
Regression equation Y on X is (Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 8) = 0.87(X – 14)
Y – 8 = 0.87X – 12.18
Y = 0.87X – 4.18

Question 3.
For a certain bivariate data on 5 pairs of observations given
Σx = 20, Σy = 20, Σx² = 90, Σy² = 90, Σxy = 76
Calculate (i) cov(x, y), (ii) b_yx and b_xy, (iii) r
Solution:


Sine byx and bxy are negative, r = -0.4

Question 4.
From the following data estimate y when x = 125

Solution:
Let u = x – 122, v = y – 14

Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 13.5) = -0.21(x – 121.5)
Y – 13.5 = -0.21x + 25.52
Y = -0.21x + 39.02
When x = 125
Y = -0.21(125) + 39.02
= -26.25 + 39.02
= 12.77

Question 5.
The following table gives the aptitude test scores and productivity indices of 10 works selected at workers selected randomly.

Obtain the two regression equation and estimate
(i) The productivity index of a worker whose test score is 95.
(ii) The test score when productivity index is 75.
Solution:


Regression equation of Y on X,
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{y}\))
(Y – 65) = 1.16 (x – 65)
Y – 65 = 1.16x – 75.4
Y = 1.16x – 10.4
(i) When x = 95
Y = 1.16(95) – 10.4
= 110.2 – 10.4
= 99.8
Regression equation of X on Y,
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 65) = 0.59(y – 65)
(X – 65) = 0.59y – 38.35
X = 0.59y + 26.65
(ii) When y = 75
x = 0.59(75) + 26.65
= 44.25 + 26.65
= 70.9

Question 6.
Compute the appropriate regression equation for the following data.

Solution:
Since x is the independent variable, and y is the dependent variable,
we need to find regression equation of y on x


Regression equation of y on x is (y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 10) = -13.4(x – 6)
y – 10 = -1.34x + 8.04
y = -1.34x + 18.04

Question 7.
The following are the marks obtained by the students in Economic (X) and Mathematics (Y)

Find the regression equation of Y and X.
Solution:
Let u = x – 61, v = y – 80

Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 80.4) = 0.3(x – 61)
Y – 80.4 = 0.3x – 18.3
Y = 0.3x + 62.1

Question 8.
For the following bivariate data obtain the equation of two regressions lines:

Solution:


Regression equation of Y on X
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 9) = 2(x – 3)
Y – 9 = 2x – 6
Y = 2x + 3
Regression equation of X on Y
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 3) = 0.5(y – 9)
(X – 3) = 0.5y – 4.5
X = 0.5y – 1.5

Question 9.
Find the following data obtain the equation of two regression lines:

Solution:


Regression of Y on X,
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 8) = 0.65(x – 6)
Y – 8 = -0.65x + 3.9
Y = -0.65x + 11.9
Regression of X on Y
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 6) = -1.3(y – 8)
(X – 6) = -1.3y + 10.4
X = -1.3y + 16.4

Question 10.
For the following data, find the regression line of Y on X

Hence find the most likely value of y when x = 4
Solution:

(Y – 3) = 2(x – 2)
Y – 3 = 2x – 4
Y = 2x – 1
When x = 4
Y = 2(4) – 1
= 8 – 1
= 7

Question 11.
Find the following data, find the regression equation of Y on X, and estimate Y when X = 10.

Solution:

Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 5) = (0.63)(x – 3.5)
Y – 5 = 0.63x – 2.2
Y = 0.63x + 2.8
When x = 10
Y = 0.63(10) + 2.8
= 6.3 + 2.8
= 9.1

Question 12.
The following sample gave the number of hours of study (X) per day for an examination and marks (Y) obtained by 12 students.

Obtain the line of regression of marks on hours of study.
Solution:
Let u = x – 5, v = y – 70

∴ Equation of marks on hours of study is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 70.83) = 6.6(x – 4.92)
Y – 70.83 = 6.6x – 32.47
∴ Y = 6.6x + 38.36

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 2 Insurance and Annuity Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Insurance and Annuity Miscellaneous Exercise 2

(I) Choose the correct alternative.

Question 1.
“A contract that pledges payment of an agreed-upon amount to the person (or his/her nominee) on the happening of an event covered against” is technically known as
(a) Death coverage
(b) Saving for future
(c) Life insurance
(d) Provident fund
Answer:
(c) Life insurance

Question 2.
Insurance companies collect a fixed amount from their customers at a fixed interval of time. This amount is called
(a) EMI
(b) Installment
(c) Contribution
(d) Premium
Answer:
(d) Premium

Question 3.
Following are different types of insurance.
I. Life insurance
II. Health insurance
III. Liability insurance
(a) Only I
(b) Only II
(c) Only III
(d) All the three
Answer:
(d) All the three

Question 4.
By taking insurance, an individual
(a) Reduces the risk of an accident
(b) Reduces the cost of an accident
(c) Transfers the risk to someone else
(d) Converts the possibility of large loss to the certainty of a small one
Answer:
Converts the possibility of large loss to the certainty of a small one

Question 5.
You get payments of ₹ 8,000 at the beginning of each year for five years ta 6%, what is the value of this annuity?
(a) ₹ 34,720
(b) ₹ 39,320
(c) ₹ 35,720
(d) ₹ 40,000
Answer:
(c) ₹ 35,720

Question 6.
In an ordinary annuity, payments or receipts occur at
(a) Beginning of each period
(b) End of each period
(c) Mid of each period
(d) Quarterly basis
Answer:
(b) End of each period

Question 7.
The amount of money today which is equal to a series of payments in the future is called
(a) Normal value of the annuity
(b) Sinking value of the annuity
(c) Present value of the annuity
(d) Future value of the annuity
Answer:
(c) Present value of the annuity

Question 8.
Rental payment for an apartment is an example of
(a) Annuity due
(b) Perpetuity
(c) Ordinary annuity
(d) Installment
Answer:
(b) Perpetuity

Question 9.
_________ is a series of constant cash flows over a limited period of time.
(a) Perpetuity
(b) Annuity
(c) Present value
(d) Future value
Answer:
(b) Annuity

Question 10.
A retirement annuity is particularly attractive to someone who has
(a) A severe illness
(b) Risk of low longevity
(c) Large family
(d) Chance of high longevity
Answer:
(d) Chance of high longevity

(II) Fill in the blanks.

Question 1.
An installment of money paid for insurance is called _________
Answer:
premium

Question 2.
General insurance covers all risks except _________
Answer:
life

Question 3.
The value of insured property is called _________
Answer:
property value

Question 4.
The proportion of property value to insured is called _________
Answer:
policy value

Question 5.
The person who receive annuity is called _________
Answer:
Annuitant

Question 6.
The payment of each single annuity is called _________
Answer:
installment

Question 7.
The intervening time between payment of two successive installments is called as _________
Answer:
payment period

Question 8.
An annuity where payments continue forever is called _________
Answer:
perpetuity

Question 9.
If payments of an annuity fall due at the beginning of every period, the series is called _________
Answer:
annuity due

Question 10.
If payments of an annuity fall due at the end of every period, the series is called annuity _________
Answer:
immediate

(III) State whether each of the following is True or False.

Question 1.
General insurance covers life, fire, and theft.
Answer:
False

Question 2.
The amount of claim cannot exceed the amount of loss.
Answer:
True

Question 3.
Accident insurance has a period of five years.
Answer:
False

Question 4.
Premium is the amount paid to the insurance company every month.
Answer:
True

Question 5.
Payment of every annuity is called an installment.
Answer:
False

Question 6.
Annuity certainly begins on a fixed date and ends when an event happens.
Answer:
True

Question 7.
Annuity contingent begins and ends on certain fixed dates.
Answer:
False

Question 8.
The present value of an annuity is the sum of the present value of all installments.
Answer:
True

Question 9.
The future value of an annuity is the accumulated value of all installments.
Answer:
False

Question 10.
The sinking fund is set aside at the beginning of a business.
Answer:
True

(IV) Solve the following problems.

Question 1.
A house valued at ₹ 8,00,000 is insured at 75% of its value. If the rate of premium is 0.80%. Find the premium paid by the owner of the house. If the agent’s commission is 9% of the premium, find the agent’s commission.
Solution:
Property value = ₹ 8,00,000
Policy value = 75% × 8,00,000 = ₹ 6,00,000
∵ Rate of Premium = 0.80%
∴ Amount of Premium = 0.80% × 6,00,000 = ₹ 4,800
∵ Rate of commission = 9%
∴ Agent commission = 9% × 4800 = ₹ 432

Question 2.
A shopkeeper insures his shop and godown are valued at ₹ 5,00,000 and ₹ 10,00,000 respectively for 80% of their values. If the rate of premium is 8%, find the total annual premium.
Solution:
Property value of shop = ₹ 5,00,000
∴ Policy value = 80% × 5,00,000 = ₹ 4,00,000
∵ Rate of Premium = 8%
∴ Amount of premium = 8% × 4,00,000 = ₹ 32,000
∵ Property value of Godown = ₹ 10,00,000
∴ Policy value = 80% × 10,00,000 = ₹ 8,00,000
∵ Rate of Premium = 8%
∴ Amount of Premium = 8% × 8,00,000 = ₹ 64,000
∴ Total annual Premium = 64,000 + 32,000 = ₹ 96,000

Question 3.
A factory building is insured for \(\left(\frac{5}{6}\right)^{\text {th }}\) of its value at a rate of premium of 2.50%. If the agent is paid a commission of ₹ 2,812.50, which is 7.5% of the premium, find the value of the building.
Solution:
Let the Property value be ₹ x
∴ Policy value = ₹ \(\frac{5 x}{6}\)
∵ Rate of premium = 2.50%
∴ Amount of premium = \(\frac{5 x}{6}\) × 2.50% = ₹ \(\frac{x}{48}\)
∵ Rate of Agent commission = 7.5%
∴ Agent commission = 7.5% × \(\frac{x}{48}\)
∴ 2812.50 = \(\frac{x}{640}\)
∴ 2812.50 × 640 = x
∴ x = ₹ 18,00,000
∴ Value of the building is ₹ 18,00,000.

Question 4.
A merchant takes a fire insurance policy to cover 80% of the value of his stock. Stock worth ₹ 80,000 was completely destroyed in a fire. While the rest of the stock was reduced to 20% of its value. If the proportional compensation under the policy was ₹ 67,200, find the value of the stock.
Solution:
Let the Property value be ₹ x
∴ Policy value 80% × x = ₹ \(\frac{4 x}{5}\)
∵ Complete loss = ₹ 80,000
∴ Partial loss = 20% × (x – 8,00,000) = \(\frac{x-80,000}{5}\)
∴ Total loss = 80,000 + \(\frac{x-80,000}{5}\) = \(\frac{x}{5}\) + 64,000
∵ Claim = ₹ 67,200

∴ x = ₹ 1,00,000
∴ The value of the stock is ₹ 1,00,000.

Question 5.
A 35-year old person takes a policy for ₹ 1,00,000 for a period of 20 years. The rate of premium is ₹ 76 and the average rate of bonus is ₹ 7 per thousand p.a. If he dies after paying 10 annual premiums, what amount will his nominee receive?
Solution:
Policy value = ₹ 1,00,000
Period of Policy = 20 years
∵ Rate of premium = ₹ 76 per thousand
∴ Amount of premium = \(\frac{76}{1,000}\) × 1,00,000 = ₹ 7,600
∴ Total Premium = 7,600 × 10 = ₹ 76,000
∴ Rate of Bonus = ₹ 7 per thousand p.a
∴ Total Bonus = \(\frac{7}{1,000}\) × 1,00,000 = ₹ 7,000
∴ Amount received by Nominee = Policy value + Bonus earned
= 1,00,000 + 7,000
= ₹ 1,07,000

Question 6.
15,000 articles costing ₹ 200 per dozen were insured against fire for ₹ 1,00,000. If 20% of the articles were burnt completely and 2,400 other articles were damaged to the extent of 80% of their value, find the amount that can be claimed under the policy.
Solution:
Total Articles = 15,000
∴ Property value = \(\frac{15,000}{12}\) × 200 = 2,50,000
∵ Policy value = ₹ 1,00,000
∴ Complete loss = 20% × 2,50,000 = ₹ 50,000
∴ Partial loss = 80% × \(\frac{2,400}{12}\) × 200 = ₹ 3,20,000
∴ Total loss = 32,000 + 50,000 = ₹ 82,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{1,00,000}{2,50,000}\) × 82,000
= ₹ 32,800

Question 7.
For what amount should a cargo worth ₹ 25,350 be insured so that in the event of a total loss, its value, as well as the cost of insurance, may be recovered when the rate of premium is 2.5%.
Solution:
Let the policy value be ₹ 100 which includes the cost of insurance and premium
∴ Property value = 100 – 2.50 = ₹ 97.50
If the value of the cargo is ₹ 97.50, then the policy value is ₹ 100.
If the value of the cargo is ₹ 25,350, then
Policy value = \(\frac{100 \times 25,350}{97.50}\) = ₹ 26,000

Question 8.
A cargo of grain is insured at \(\left(\frac{3}{4}\right)\)% to cover 70% of its value. ₹1,008 is the amount of premium paid. If the grain is worth ₹ 12 per kg, how many kg of the grain did the cargo contain?
Solution:
Let the Property value be ₹ x
∴ policy value = 70% × x = ₹ \(\frac{7 x}{10}\)
∵ Rate of premium = \(\frac{3}{4}\)%
∴ Amount of premium = Rate × Policy value

∴ x = ₹ 1,92,000
∵ Rate of Jowar = ₹ 12/kg
∴ Quantity of Jowar = \(\frac{1,92,000}{12}\) = 16,000 kgs

Question 9.
4,000 bedsheets worth ₹ 6,40,000 were insured for \(\left(\frac{3}{7}\right)^{t h}\) of their value. Some of
the bedsheets were damaged in the rainy season and were reduced to 40% of their value. If the amount recovered against damage was ₹ 32,000. Find the number of damaged bedsheets.
Solution:
∵ Property value = ₹ 6,40,000
∴ Policy value = 6,40,000 × \(\frac{3}{7}\) = ₹ \(\frac{19,20,000}{7}\)
∴ Cost of one Bedsheet = \(\frac{6,40,000}{4,000}\) = ₹ 160
Let ‘x’ bedsheets be damaged.
∴ Cost of x bedsheets = ₹ 160x

∴ 875 Bedsheets damaged.

Question 10.
A property valued at ₹ 7,00,000 is insured to the extent of ₹ 5,60,000 at \(\left(\frac{5}{8}\right)\)% less 20%. Calculate the saving made in the premium. Find the amount of loss that the owner must bear, including premium, if the property is damaged to the extent of 40% of its value.
Solution:
∵ Property value = ₹ 7,00,000
∵ Policy value = ₹ 5,60,000
∵ Rate of premium = \(\frac{5}{8}\)%
∴ Amount of premium = \(\frac{5}{8}\)% × 5,60,000 = ₹ 3,500
New rate of premium = \(\frac{5}{8}\)% less 20%
= \(\frac{5}{8}\) – [20% x \(\frac{5}{8}\)]
= \(\frac{5}{8}\) – \(\frac{1}{8}\)
= \(\frac{1}{2}\)%
∴ Amount of premium = \(\frac{1}{2}\)% × 5,60,000 = ₹ 2,800
∴ Saving made in premium = 3,500 – 2,800 = ₹ 700
∴ Loss = 7,00,000 × 40% = 2,80,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{5,60,000}{7,00,000}\) × 2,80,000
= ₹ 2,24,000
∴ Loss bear by owner = loss – claim + premium
= 2,80,000 – 2,24,000 + 2,800
= ₹ 58,800

Question 11.
Stocks in a shop and godown worth ₹ 75,000 and ₹ 1,30,000 respectively were insured through an agent who receive 15% of the premium as commission. If the shop was insured for 80% and godown for 60% of the value, find the amount of agent’s commission when the premium was 0.80% less 20%. If the entire stock in the shop and 20% stock in the godown is destroyed by fire, find the amount that can be claimed under the policy.
Solution:
∵ Rate of premium = 0.80% less 20%
= 0.80 – 20% × 0.80
= 0.80 – 0.16
= 0.64%
For Shop
∵ Property value = ₹ 75,000
∴ Policy value = 80% × 75,000 = ₹ 60,000
∴ Premium = 0.64% × 60,000 = ₹ 384
∵ Loss = ₹ 75,000
∵ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{60,000}{75,000}\) × 75,000
= ₹ 60,000
For Godown
∵ Property value = ₹ 1,30,000
∴ Policy value = 60% × 1,30,000 = ₹ 78,000
∴ Premium = 0.64% × 78,000 = ₹ 499.2
Loss = 20% × 1,30,000 = ₹ 26,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{78,000}{1,30,000}\) × 26,000
= ₹ 15,600
Total claim = 16,600 + 60,000 = ₹ 75,600
∵ Rate of commission = 15%
∴ Agent commission = 15% × [384 + 499.2]
= 15% × 883.2
= ₹ 132.48

Question 12.
A person holding a life policy of ₹ 1,20,000 for a term of 25 years wants to discontinue after paying a premium for 8 years at the rate of ₹ 58 per thousand p.a. Find the amount of paid-up value he will receive on the policy. Find the amount he will receive if the surrender value granted is 35% of the premium paid, excluding the first year’s premium.
Solution:
Policy value = ₹ 1,20,000
∵ Rate of premium = ₹ 58 per thousand p.a.
∴ Premium for 8 years = \(\frac{8 \times 58}{1000}\) × 1,20,000 = ₹ 55,680
∴ Amount of 1st premium = \(\frac{55,680}{8}\) = ₹ 6,960
∵ Paid-up value of policy = \(\frac{\text { No of Premium paid }}{\text { Terms of policy }}\) × Policy value
= \(\frac{8}{25}\) × 1,20,000
= ₹ 38,400
∵ Surrender value = 35% × [Total premium – 1st year premium]
= 35% × [55,680 – 6,960]
= 35% × 48,720
= ₹ 17,052

Question 13.
A godown valued at ₹ 80,000 contained stock worth ₹ 4,80,000. Both were insured against fire. Godown for ₹ 50,000 and stock for 80% of its value. A part of stock worth ₹ 60,000 was completely destroyed and the rest was reduced to 60% of its value. The amount of damage to the godown is ₹ 40,000. Find the amount that can be claimed under the policy.
Solution:
For Godown
∵ Property value = ₹ 80,000
∵ Policy value = ₹ 50,000
∵ Loss = ₹ 40,000
∵ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{50,000}{80,000}\) × 40,000
= ₹ 25,000
For stock
∵ Property value = ₹ 4,80,000
∵ Policy value = 80% × 4,80,000 = ₹ 3,84,000
∵ Complete loss = ₹ 60,000
∴ Partial loss = (100 – 60)% × [4,80,000 – 60,000]
= 40% × 4,20,000
= ₹ 1,68,000
∴ Total loss = 1,68,000 + 60000 = ₹ 2,28,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{3,84,000}{4,80,000}\) × 2,28,000
= ₹ 1,82,400
∴ Total claim = 25,000 + 1,82,400 = ₹ 2,07,400

Question 14.
Find the amount of an ordinary annuity if a payment of ₹ 500 is made at the end of every quarter for 5 years at the rate of 12% per annum compounded quarterly. [Given: (1.03)²⁰ = 1.8061]
Solution:
∵ C = ₹ 500
∵ r = 12% p.a. compounded quarterly,
∴ r = \(\frac{12}{4}\) = 3%
∵ n = 5 years
But, payment is made quarterly
∴ n = 5 × 4 = 20

Question 15.
Find the amount a company should set aside at the end of every year if it wants to buy a machine expected to cost ₹ 1,00,000 at the end of 4 years and interest rate is 5% p.a. compounded annually.
Solution:
∵ A = ₹ 1,00,000
∵ r = 5% p.a.
∴ i = \(\frac{r}{100}=\frac{5}{100}\) = 0.05
∵ n = 4 years
∵ A = \(\frac{C}{i}\left[(1+\mathrm{i})^{n}-1\right]\)
∴ 1,00,000 = \(\frac{C}{0.05}\)[(1 + 0.05)⁴ – 1]
∴ 1,00,000 × 0.05 = C [(1.05)⁴ – 1]
∴ 5,000 = C(1.2155 – 1)
∴ 5,000 = C × 0.2155
∴ \(\frac{5,000}{0.2155}\) = C
∴ C = ₹ 23,201.86

Question 16.
Find the least number of years for which an annuity of ₹ 3,000 per annum must run in order that its amount exceeds ₹ 60,000 at 10%compounded annually. [Given: (1.1)¹¹ = 2,8531, (1.1)¹² = 3.1384]
Solution:
∵ A = ₹ 60,000
∵ C = ₹ 3,000
∵ r = 10% p.a.
∴ i = \(\frac{r}{100}=\frac{10}{100}\) = 0.1
∵ A = \(\frac{C}{i}\left[(1+i)^{n}-1\right]\)
∴ 60,000 = \(\frac{3,000}{0.1}\left[(1+0.1)^{n}-1\right]\)
∴ 60,000 = 30,000[(1.1)ⁿ – 1]
∴ \(\frac{60,000}{30,000}\) + 1 = (1.1)ⁿ
∴ 2 + 1 = (1.1)ⁿ
∴ 3 = (1.1)ⁿ
Taking log
∴ log 3 = log (1.1)ⁿ
∴ log 3 = n log(1.1)
∴ \(\frac{\log 3}{\log 1.1}\) = n
∴ n = \(\frac{0.4771}{0.0414}\) = 11.52 ~ 12 years

Question 17.
Find the rate of interest compounded annually if an ordinary annuity of ₹ 20,000 per year amounts to ₹ 41,000 in 2 years.
Solution:
∵ C = ₹ 20,000
∵ A = ₹ 41,000
∵ n = 2 years

∴ r = 5% p.a.

Question 18.
A person purchases a television by paying ₹ 20,000 in cash and promising to pay ₹ 1,000 at the end of every month for the next 2 years. If money is worth 12% p.a., converted monthly. Find the cash price of the television. [Given: (1.01)^-24 = 0.7880]
Solution:
Down payment = ₹ 20,000
∵ n = 2 years
But, EMI Payable monthly
∴ n = 2 × 12 = 24
∵ r = 12% p.a. compounded monthly

∴ P = 1,00,00 × 0.2120
∴ P = ₹ 21,200
Cash price = Present value + Down payment
= 21,200 + 20,000
= ₹ 41,200

Question 19.
Find the present value of an annuity immediate of ₹ 20,000 per annum for 3 years at 10% p.a. compounded annually. [Given: (1.1)^-3 = 0.7513]
Solution:
∵ C = ₹ 20,000
∵ n = 3 years
∵ r = 10% p.a.

∴ P = 2,00,000 [1 – 0.7513]
∴ P = 2,00,000 [0.2487]
∴ P = ₹ 49,740

Question 20.
A man borrowed some money and paid it back in 3 equal installments of ₹ 2,160 each. What amount did he borrow if the rate of interest was 20% per annum compounded annually? Also, find the total interest charged. [Given: (1.2)^-3 = 0.5788]
Solution:
∵ C = ₹ 2,160
∵ n = 3
∵ r = 20% p.a.

∴ P = ₹ 6,251.04
∴ Total amount paid = 2,160 × 3 = ₹ 6,480
∴ Interest = 6,480 – 6,251.04 = ₹ 228.96

Question 21.
A company decides to set aside a certain amount at the end of every year to create a sinking fund that should amount to ₹ 9,28,200 in 4 years at 10% p.a. Find the amount to be set aside every year. [Given: (1.1)⁴ = 1.4641]
Solution:
∵ A = ₹ 9,28,200
∵ n = 4 years
∵ r = 10% p.a.

∴ 9,28,200 × 0.1 = C[1.4641 – 1]
∴ 92,820 = C × 0.4641
∴ \(\frac{92,820}{0.4641}\) = C
∴ C = ₹ 2,00,000

Question 22.
Find the future value after 2 years if an amount of ₹ 12,000 is invested at the end of every half-year at 12% p.a. compounded half-yearly. [Given: (1.06)⁴ = 1.2625]
Solution:
∵ n = 2 years
Payable half yearly, n = 2 × 2 = 4
∵ C = ₹ 12,000
∵ r = 12% p.a. Compounded half yearly


∴ A = 1,00,000 [1.2625 – 1]
∴ A = 1,00,000 × 0.2625
∴ A = ₹ 26,250

Question 23.
After how many years would an annuity due of ₹ 3,000 p.a. accumulated ₹ 19,324.80 at 20% p.a. compounded annually? [Given: (1.2)⁴ = 2.0736]
Solution:
∵ C = ₹ 3,000
∵ A = ₹ 9,324.80
∵ r = 20% p.a.

∴ 19,324.80 = 15,000 × 1.2[(1.2)ⁿ – 1]
∴ 19,324.80 = 18,000[(1.2)ⁿ – 1]
∴ \(\frac{19,324.80}{18,000}\) + 1 = (1.2)ⁿ
∴ 1.0736 + 1 = (1.2)ⁿ
∴ 2.0736 = (1.2)ⁿ
∴ (1.2)⁴ = (1.2)ⁿ
∴ n = 4 years

Question 24.
Some machinery is expected to cost 25% more over its present cost of ₹ 6,96,000 after 20 yeas. The scrap value of the machinery will realize ₹ 1,50,000. What amount should be set aside at the end of every year at 5% p.a. compound interest for 20 years to replace the machinery? [Given: (1.05)²⁰ = 2655]
Solution:
Present cost = ₹ 6,96,000
Expected cost = 25% × 6,96,000 + 6,96,000
= 1,74,000 + 6,96,000
= ₹ 8,70,000
∴ Scrap value = ₹ 1,50,000
∴ Sinking fund = 8,70,000 – 1,50,000 = ₹ 7,20,000
∴ A = ₹ 7,20,000, n = 20 years, r = 5% p.a.

∴ 7,20,000 × 0.05 = C[(1.05)²⁰ – 1]
∴ 36,000 = C[2.655 – 1]
∴ 36,000 = C × 1.655
∴ \(\frac{36,000}{1.655}\) = C
∴ C = ₹ 21,752.27

Balbharti Yuvakbharati English 12th Digest Chapter 2.4 Have you Earned Your Tomorrow Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 12 English Yuvakbharati Solutions Chapter 2.4 Have you Earned Your Tomorrow

12th English Digest Chapter 2.4 Have you Earned Your Tomorrow Textbook Questions and Answers

Question 1.
Complete the following web

Answer:

Question 2.
Discuss with your paitner about the different idioms/proverbs related to word ‘tomorrow’.
Answer:
(a) Never put off until tomorrow what you can do today.
(b) Tomorrow’s battle is won during today’s practice. (Japanese Proverb)
(c) Today must not borrow from tomorrow. [German Proverb]
(d) Yesterday, today and tomorrow – these are the three days of man.. [Chinese Proverb]

Question 3.
When you make your future plans you think of:
Answer:

1. Career
2. Higher studies
3. Retirement-plans/Financial security
4. Family life
5. Goal in life to be accomplished

Question 4.
‘Plan your tomorrow’ by completing the table given below.
Answer:

Examination	College Function	Function at your home
Complete studying portion	Preparing the list of duties	Cleaning the house
Clarify doubts	Delegate jobs	Arrangements for sending invites, preparation and service of food
Revision	Confirm date/ time with resource people	Seating arrangements and other conveniences for guests

(A1)

Question 1.
Discuss with your friend how she/he spent the whole day that was beneficial for others.
Points:
(a) visited retirement home/time spent with elderly residents
(b) spent time conversing about their children/ grandchildren/ looking at photos,
(c) taught some of them how to use the internet to communicate/learned some traditional recipes
(d) promised to visit often regularly

(A2)

Question (i)
……..’was it well or sorely spent’? Explain the meaning and give illustrations.
Answer:
The poet asks the reader again and again if he/she spent each day well or wasted it -‘sorely spent’. We all are busy with our own lives, acting for our own benefit. The poet inspires us to be mindful and must be of use to the world around us. The poet prompts us to speak kindly and unselfishly help, at least one fellow human everyday. The message is implied throughout the poem.

He asks if you have made one person happy, some stranger who had lost all hope, to find some hope again. So he will speak well of you. Is someone grateful to you at the end of (each) the day?

Question (ii)
‘As you close your eyes in slumber do you think that God would say,
You have earned one more tomorrow by the work you did today?’
Elaborate the idea expressed in these lines.
Answer:
The poet indirectly means that each day we exist we must make our living useful. We are not sure if we shall wake in the morning. We pray we do. But for God to grant us one more day -tomorrow – we have to justify our existence today. Did we put today to use? Did we help at least one fellow-human? Did we extend the help without expecting any returns?

Did at least one person feel grateful for your act of help? If nothing, we can make the effort to speak a kind word to a stranger we may pass by in our daily hurry. So one has live mindfully, unselfishly and look for ways to be of help to someone in need. That is the minimum expected by God. Or to have lived as a human would be futile.

Question (iii)
The poet suggests that one should do good to others. Complete the table by giving examples of doing good to following people.
Answer:

Family members	Friends	Neighbours
Help with household chores/run errands	Be ready to help in unexpected situations.	Be mindful not to intrude or cause disturbance.
Take care if someone is sick.	Share resources whenever possible.	Be cooperative when we all have to work together for the common good.

(A3)

Question (i)
Pick out the describing words from the poem and add a noun of your own.
Answer:

(Toiling) time	(Toiling) time
(Happier) anybody	(Kindly) word
(Cheerful) greeting	(Churlish) howdy
(Grateful) someone	(Rejoicing) heart
(Fading) hopes	(Slipping) days

Question (ii)
Match the words given in column A with their meaning in column B:

A	B
1. Cheerful	(a) With the feeling of disappointment
2. Selfish	(b) lack of satisfaction
3. Sorely	(c) happy
4. Discontent	(d) concerned with one’s own pleasure

Answer:

1. Cheerful – happy
2. Selfish – concerned with one’s own pleasure
3. Sorely – with the feeling of disappointment
4. Discontent – lack of satisfaction

Question (iii)
There are a few examples of homonyms in the poem. For example ‘spoke’. List homonyms from the poem and give their meanings.
Answer:
Passed:

1. (of a candidate) be successful in (an examination, test, or course).
2. went past/left behind Deed:
3. an action that is performed intentionally or consciously.
4. a legal document that is signed and delivered, especially one regarding the ownership of property or legal rights.

Waste:
1. use or expend carelessly, extravagantly, or to no purpose.
(of a person or a part of the body) become progressively weaker and more emaciated.
2. (of a material, substance, or by-product) eliminated or discarded as no longer useful or required after the completion of a process.

Question (iv)
Find out expressions/phrases which denote, ‘going away’ from each stanza.
Answer:

1. Stanza 1: “is almost over”
2. Stanza 1: “passed his way”
3. Stanza 1: “is almost over”
4. Stanza 2: “vanish in the throng”
5. Stanza 2: “rushed along”
6. Stanza 3: “were fading now”

(A4)

Question (i)
The poet has used different poetic devices like Alliteration and Interrrogation in the poem. Identify them and pick out the lines.
Answer:

Poetic Device	Lines
1. Alliteration	‘Were you selfish pure and simple as you rushed along the way’ ‘As you close your eyes in slumber do you think that God would say’, (The sounds ‘s’ & ‘sh’ are repeated in both lines.)
2. Interrogation	The first, the second and the fourth lines of stanzas 1, 2 and 4 are all questions – Interrogation. The second and fourth lines in the stanza 3 are questions.

Question (ii)
The rhyme scheme of the first stanza is ‘aabb’. Find the rhyme scheme of other stanzas.
Answer:
The rhyme scheme of all the stanzas is ‘aabb’.

(A5)

Question (i)
Write the appreciation of this poem based on the points given below :

• About the poem/ poet and the title
• The theme
• Poetic style
• The language/poetic devices used in the poem
• Special features
• Message, Values, Morals in the poem
• Your opinion about the poem

Answer:
The poet, Edgar Guest’s “Have you Earned your Tomorrow”, is a thought provoking composition. The title itself pushes our mind to wonder if today we have done something useful.

It Urges the reader to be thoughtful in everyday life about the people around them. The poet puts forward questions. Each question forces us to ask ourselves if we are kind, unselfish, patient and thoughtful. In our everyday rush, to live our life only for our own benefit, we forget to consider the people nearby who may be less fortunate.

There is interrogation in eleven lines of the sixteen-line poem. The language is simple. There is alliteration and rhyme. The poem has four stanzas of four lines each. The first stanza has four lines, each having fourteen syllables. The second, third and fourth stanzas also with four lines, have fifteen syllables each.

The clear message of the poem is one’s life is meaningful only if it is useful for humanity at large. The poet says one’s conscience has to know that your existence is justified. Or one cannot feel he has the right to ask for one more day of life. It is an uplifting poem. We can take the message and begin implementing it in our life immediately and every day.

Question (ii)
Prepare a mind map on ‘How to plan a goal for tomorrow’ or ‘My future goal’. Take the help of points given in ‘Writing Skills Section’ for preparing a mind map.
Answer:

Question (iii)
Write a set of 8 to 10 interview questions to be asked to a social worker. Take the help of the following points:

• Childhood
• Education
• Service
• Difficulties
• Future plans
• Achievements
• Message

Answer:

1. Good evening Rima ma’am. I would like to know a bit about your life. Could we begin with a walk down memory lane to your childhood?
2. What was your hobby/past-time in your childhood?
3. Which is the best memory during your school/high school/college years? Which phase did you enjoy the most?
4. You have moved to different cities due to your father’s job. Which is the city/town which you loved living in the most?
5. How did you get into social service? Who was your role model or inspiration?
6. What were the challenges and difficulties that caused any setback in your life?
7. What plans do you have for the future? Would you mind sharing a little of those with your fans?
8. There are many achievements you have seen. Which is the most important according to you?
9. What is the message you want to give to those in this noble field? What would be your tip especially for youngsters?

Question (iv)
Compose 4-6 lines on your own on ‘Good deeds’.
Answer:
‘Good deeds’
The tree gives shade and fruits it does not eat
The river flows cool and sweet of water it doesn’t drink.
When a stranger sad or in need you may meet
Be sure you lift him up, not let him into despair sink.

(A6)

Question (i)
Find out different career opportunities in the field of social work.

Question (ii)
Collect information of the NGOs working for the underprivileged section of the society.

Yuvakbharati English 12th Digest Chapter 2.4 Have you Earned Your Tomorrow Additional Important Questions and Answers

Read the poem and complete the activities given below:

Personal Response:

Question 1.
Describe the various ways you use to greet your elder.
Answer:
Whenever we meet our elders we greet them with great respect and love. Through the length and breadth of our country touching the feet of elders is the tradition. We also fold our palms in the very Indian greeting of ‘Namaste’. This comes from the word ‘Namaskaar’. In south India touching people is not a normal custom. Younger people prostrate full length before elders such as parents, uncles-aunts, gurus and even older siblings. In north India the younger bend before the elders and ladies cover their head with the shawl or sari edge. Age is a very significant factor. The greeting is always a gesture of respect and the elders respond affectionately by showering blessings.

Poetic Devices:

Question 1.
Identify an example of synecdoche from the poem.
Answer:
‘Is a single heart rejoicing over what you did or said;’
The word ‘heart’, 3rd line of the 3rd stanza is an example of synecdoche.
The word heart – a part – refers to a whole or the person who is rejoicing.

Balbharti Yuvakbharati English 12th Digest Chapter 2.3 The Inchcape Rock Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 12 English Yuvakbharati Solutions Chapter 2.3 The Inchcape Rock

12th English Digest Chapter 2.3 The Inchcape Rock Textbook Questions and Answers

Question 1.
Prepare a word register related to marine life:
Answer:
sailors; ship; tides; winds; seabed; anchor; captain; submarine; international-waters; port; harbour; shipyard; patrol; trawler; sail; port; starboard; deep-sea.

Question 2.
The functions of a lighthouse are:

Answer:

Question 3.
Discuss in pairs the various famous rocks in the world and mention the places where they are.
Answer:

Famous Rock	Place
Balancing Rock (Krishna’s butter-ball) 250 tons – balanced on a slope attempts to move it for safety remains v unsuccessful The Trimurti Cave-dedicated to trinity Brahma, Vishnu and Shiva Protected by ASI and UNESCO	Mahabalipuram
Ayer’s Rock Called Uluru by Australian Aboriginal has carvings- paintings. Composed of sandstone The rock changes colour according to position of Sun; most striking at sunset, coloured a fiery orange-red	Central Australia
Giant’s Causeway – Most of the columns hexagonal, – some four/ five/ seven/ eight sided made up of some 40,000 interlocking basalt columns one of the great natural wonders – World Heritage Site	Northern Ireland
Sigiriya rock plateau, formed from magma of an extinct volcano, 200 metres high; UNESCO Heritage Site Ancient hydraulic system – canals, locks, lakes, dams, bridges, fountains, surface/underground water pumps. In rainy season, water begins to circulate in Sigiriya. Fountains built in Fifth century – oldest in the world.	Sri Lanka

Question 4.
Narrate in the class a story about someone who destroyed or spoilt someone else’s good work.
(Points: A bright Student-Punctual, cheerful, intelligent-Helpful to classmates, explains and lends notes-Tutored junior class students- Jealous group tears up notes before exams-Is able to study with the friends whom he/she helped-The jealous group is outwitted)

Question 5.
Discuss the following expressions in pairs/groups. Take the help of your teacher.
(a) As you sow so shall you reap.
(b) Crime gets its own punishment
(c) What goes around comes around
(d) Tit for tat
(e) Evil digs a pit for others but falls into the same.
Answer:
All the above are idioms and proverbs. They all convey the same meaning. They all mean that when a person acts with a certain intention, the results will be the same as the action. If the intentions are good the person will benefit from rewards. If the intentions are evil he will be punished.

(A1)

Question 1.
Narrate in groups the scene described in the beginning of the poem.
Points:
A clear calm day at sea
The sea was quiet – the ship is still
The wind is not blowing – the sails unmoving
The waves do not move the Bell
All these point fin first 3 stanzas] to a quiet sea and
calm weather one morning in spring.

(A2)

Question (i)
Complete the following statement:
The Abbot of Aberbrothok placed a bell on the Inchcape Rock because
Answer:
The Abbot of Aberbrothok placed a bell on the Inchcape Rock because there were dangerous rocks near the coast which would wreck ships.

Question (ii)
Given below are the events that give the theme of the poem in a jumbled form. Arrange in a proper sequence as per their occurrence.
(a) The waves were so small that they did not move enough to ring the bell at the Inchcape Rock.
(b) The Abbot of Aberbrothok had placed the bell on a buoy on the rock.
(c) There was a thick haze spread over the atmosphere.
(d) Ralph bent over from the boat.
(e) Sir Ralph cursed himself in despair and in his frustration tore his hair.
Answer:
(b) The Abbot of Aberbrothok had placed the bell on a buoy on the rock.
(a) The waves were so small that they did not move enough to ring the bell at the Inchcape Rock.
(d) Ralph bent over from the boat.
(c) There was a thick haze spread over the atmosphere.
(e) Sir Ralph cursed himself in despair and in his frustration tore his hair.

Question (iii)
Describe the qualities of the Abbot of Aberbrothok in your own words.

Answer:

Question (iv)
‘Jealousy’ is the most incurable defect, Justify.
Answer:
When someone is in a better position of money or success or fame, there are people who feel that they should destroy that. This is jealousy. We can see people who have more, and we can also work hard to reach that position. But when someone wants to destroy that person who has reached the better position that is wickedness. The jealous person is not willing to work for that state. They will not accept a lesser place also. So a jealous mind-set slowly becomes completely evil.

Question (v)
‘But the Rover’s mirth was wickedness’. Explain this line in your own words with the help of the extract.
Answer:
The season of spring made everyone feel happy and light-hearted. The Rover was whistling and singing. But this joyful mood made him reckless. He wanted to trouble the Abbot. The Abbot had put a Bell there as a warning about the Inchcape Rock. Ralph rashly decided to undo his good work. The Rover was jealous of the Abbot who was blessed by grateful sailors. He wanted to trouble the Abbot of Aberbrothok.

(A3)

Question 1.
Some words in the poem are related to different parts of a ship or a mariner’s life. Given below are the meanings of those terms. Identify the word.
Answer:
(a) Helps in steering the ship-wheel
(b) The lowest part of the ship – keel
(c) Floating object that shows direction- buoy
(d) Another name for a ship-vessel
(e) Sinking – gurgling

(A4)

Question 1.
Select the appropriate figure of speech from the box given below and complete the table.
Answer:

Example	FOS	Explanation
No stir in the air, no stir in the sea	Repetition	Emphasizes the quiet stillness
On a buoy in the storm it floated and swung	Alliteration	The sound of the vowel ‘o’ is repeated
The ship was as still as she could be	Personification	The ship is spoken of as ‘she’ as if a human being

(A5)

Question (i)
Write the appreciation of this poem based on the points given below :

• About the poem/poet and the title
• The theme
• Poetic style
• The language/poetic devices used in the poem
• Special features
• Message, Values, Morals in the poem
• Your opinion about the poem

Answer:
The Poem “The Inchcape Rock’ is about a real stretch of treacherous rocks near the Scottish coast. I Robert Southey wrote prose and other poems too. But this poem is well-liked. The title gives the clue that the rock is a part of an interesting story.

The theme is about an Abbot and a pirate. The Abbot is concerned for his fellow humans and helps to save sailors. He put the Inchcape Bell on a buoy to warn ships day and night of the terrible Inchcape Rock, during storms. [According to records, warning bell was placed.]

But the Rover in a fit of madness, on a spring day, cut the bell just to trouble the Abbot. Many months later, when the pirate was sailing towards Scotland, the weather was different. As the frightened sailors were caught in the dark stormy sea the pirate realised he had not troubled the Abbot but brought ruin for himself and his sailors.

The poem is a ballad. The story is told in stanzas of four lines, with aabb rhyme. The story is told in easy language. The poet uses many Old English words like ‘blest’, ‘Quoth’, and ‘canst’. The poet begins with spring, a metaphor for the pleasant mood, with a calm sea, still air and the ship in quiet waters. Repetition emphasizes the gladness in the heart.

The mood changes from mischief to wickedness. When the mist blocks the sun, metaphor makes the story gloomy, suspenseful. The nightfall is the metaphor for the dark situation for the ship, its sailors. They finally meet a violent end. There is alliteration which adds to the beauty of the poem.

The poem is a didactic one with a clear message – “When we try to trouble others, trouble first comes to the doer.” The story has a moral and is useful even in these times.

Question (ii)
Compose 4 to 6 lines on ‘Sea’ :
Answer:
Sea
I meet the sky far away, brothers of the same colour.
I mirror his white woolly sheep and birds.
I pull and push; deep down or sometimes upwards
In my cool-world, small and big creatures, softly slither.

(A6)

Question (i)
Expand the ideas on your own on the following topics:
(a) Pride goes before a fall.
(b) Time and tide wait for none.
(c) Man proposes, God disposes.
(d) Look before you leap.

Question (a)
Pride goes before a fall.
Answer:
There is a saying in Sanskrit that translates as “Knowledge brings humility.” The opposite would be that only an ignorant person would be proud or arrogant. A person becomes overconfident about himself or what he has. He starts thinking lowly of others. Only a harsh experience makes him see his stupidity.

There is a story about the God of riches who was drunk on his wealth. He invited all the other gods to a grand feast so that his wealth would be seen by them. He also invited Shiva and Parvati. They gently told him they would not be able to come, They said their son Ganesha would come instead. The host welcomed his guests.

Ganesha also arrived. The guests seated in a dazzling hall ate their fill of the lavish food. They praised the food, the hospitality and took leave impressed by the grandeur of everything there. But Ganesha was still being served. The host was stunned to see the servants running frantically to serve at the little boy’s speed of eating. The cooks were preparing more food. The puzzled King saw to it that Ganesha was served what he wanted.

Then word came from the kitchen that supplies were needed. Soon the supplies in adjacent villages were empty. Ganesha in anger chased the King till he ran to Shiva’s abode. Ganesha complained he was not fed. The King realized his foolishness trying jto impress the Lord and Mother with his riches. He went humbled, not able to feed one child.

Hence how much ever one possesses one must not think lowly of others. The right kind of knowledge makes a person more and more humble. Like the tree full of fruits bends lower and lower.

Question (ii)
The poem begins with:
‘Without either sign or sound of their shock, The waves flowed over the Inchcape Rock.’
It ends with:
On the basis of these lines explain the change in mood of the poem.
Answer:
At the beginning of the poem the season is spring, the weather is mild and the sea-waters are calm. The waves pour softly over the Inchcape Bell. The Heavy Bell on a buoy would ring due to strong waves only in stormy weather.

When the Rover cut the Bell it was spring season. The mood was happy, light-hearted. He was up to mischief in a rash, jolly mood on a lovely spring day. He wanted to only trouble the Abbot.

After undoing the Abbot’s good work the Rover went away on his criminal voyages. But when he was returning the sea was stormy. Wild winds threw the ship off course. The mood is of confusion and fear because a thick fog covered them from the sun. The mood is of suspense, the sailors are lost.

By nightfall they did not still know where they were. They are really and metaphorically in the dark. They could hear the waves crashing yet they did not know which land was near. There is fear. There was no wild wind but the rough sea was pulling their ship along. They desperately wanted some clue to help them to know their location. The ship shattered onto the rocks as the Rover yelled and cursed. The dramatic end is violent and filled with despair.

(A7)

Question (i)
Read the tree diagrams and information given on pages 109-110 of the textbook and find out more information about opportunities in ‘on and off the shore’ the Indian Navy.

Question (ii)
Required qualifications and various fields/opportunities for women to join in the Navy.

Question (iii)
Colleges that provide education in oceanography-
National Institute of Oceanography, Goa
National Institute of Oceanography, Mumbai
MBA (Logistic Shipping Management), IIKM Business School, Calicut, Kerala
Indira Gandhi College of Distance Education IGCDE, Tamil Nadu

Yuvakbharati English 12th Digest Chapter 2.3 The Inchcape Rock Additional Important Questions and Answers

Read the extract and complete the activities given below:

Global Understanding:

Question 1.
Give reasons for the sailors’ appreciation of the Abbot.
Answer:
There were some dangerous rocks near the Scottish coast. The Abbot of Aberbrothok had placed a buoy and fixed a bell on it, near those rocks. If the sea was rough sailors could spot the buoy. Even in the darkness the rough seas made the bell ring. So by day or night the Abbot’s bell saved the sailors and their ships from the rocks, and they blessed him.

Question 2.
Complete the following:
‘Wheel’d round’ here implies
Answer:
Wheel’d around here implies a flock of birds flying round in circles, which looks like a wheel.

Question 3.
Describe the state of mind of Ralph.
Answer:
Ralph the Rover also felt the effects of the season of spring. He felt very cheerful; he whistled and sang as he walked about on the deck. He was in an extremely happy state of mind but his joy was evil in intentions.

Question 4.
Complete the following:
Answer:
The pirate asked his men to row him over to the Inchcape Bell. He then bent over and cut the Bell from the buoy. He did so that the sailors of the next ship would no longer bless the Abbot for placing the warning Bell.

Question 5.
Choose the words and phrases that could describe Sir Ralph the Rover.
(a) Criminal
(b) Jealous
(c) Arrogant
(d) Vicious
(e) Spiteful
Answer:
All of the above

Question 6.
Choose the correct option:
On spotting the bell, Rover cut the bell from the buoy. This was an act of:
(a) Hatred
(b) Anger
(c) Jealousy
(d) Frustration
Answer:
(c) Jealousy

Question 7.
‘O Christ! It is the Inchcape Rock’ – Give reasons for Ralph’s Exclaimation.
Answer:
The Rover’s ship had struck the terrible rocks feared by sailors. Some time ago he himself had cut off the Bell put there by the blessed Abbot. Now his own ship had hit the Inchcape Rock and was going to sink with all his riches. He too was sure about to die.

Question 8.
Complete the following statements:
Answer:
1. The result of the thick haze that covered the sky was that the sailors had no way of knowing in which direction they were sailing.
2. The Rover in frustration pulled his hair and cursed himself because he himself had cut the Bell which would have rang and the sound would have helped them to save themselves from those killer-rocks.

Inference/Interpretation/Analysis:

Question 1.
The pirate is given the title ‘Sir’ though he was a feared criminal. He is called a ‘rover’. Give reasons for the same.
Answer:
Though he was a feared criminal Ralph was the captain of his ship. The crew may have addressed him ‘Sir’ which explains it attached to his name. A rover is a person, animal, or thing that roves, or wanders, Ralph the pirate roamed around on the seas looking for ships to attack and loot. Maybe that is why he was called Ralph the Rover.

Question 2.
The poet gives hints to the reader in the second stanza of the extract. Find the significant line from the extract and give reason for your answer.
Answer:
The second stanza of this extract tells about what the pirate did after removing the Bell. He roamed the seas and carried on his evil activities, killing and looting.

The last line is the hint of what is to happen later. ‘He steers his course for Scotland’s shore.’ The rover set the course ‘for Scotland’s shores’. This is significant because the treacherous Inchcape Rock was on the Scottish shores. So we get an idea that something may happen there.

Question 3.
Read the following lines and say what the situation was:
‘For me thinks we should be near the shore’. ‘Now where we are I cannot tell,’
Answer:
The sailors could hear the waves crashing on the shore. But they had been blown about by wild winds all day and so did not where they had reached. They did not know which land or shore was near. The situation was that danger was near.

Question 4.
Explain the danger implied in the two lines:
‘They hear no sound, the swell is strong; Though the wind hath fallen they drift along,’
Answer:
There was no sound except the breakers crashing on a nearby shore. There was no wind. But the sea was rough and the strong sea pulled the helpless ship along. The sailors were confused and could not make out which was the safe direction.

Personal Response:

Question 1.
Write an account of something which you did out of concern for others.
Answer:
In our colony there is a young couple living with twin toddlers and elderly parents. The young man is a doctor and his working hours are sometimes unpredictable.

My family is aware of this and we help in small ways. I help the elderly lady to take a walk on the street and my brother helps the gentleman. I also help the young mother to mind the small children if she has to go out shopping. I sometimes run errands for them too.

Question 2.
Give your opinion on the following line and explain its significance.
‘Quoth Sir Ralph, ‘The next who comes to the Rock
Won’t bless the Abbot of Aberbrothok.’
Answer:
The Pirate says these words after he cut the Bell placed by the Abbot. The Abbot had placed it for saving others. This act had brought fame for the Abbot and also the blessings of the many sailors that were saved. But the pirate was jealous of the fame. He cut the Bell thinking to harm the Abbot. When someone is concerned about others they are not looking for fame. But a selfish person is blinded by jealousy. They behave foolishly and cause trouble only for themselves.

Question 3.
‘Now where we are I cannot tell,
But I wish I could hear the Inchcape Bell’ From these lines describe the thoughts of
(a) ……… the sailors in the Rover’s ship.
(b) ……. the Rover’s.
Answer:
(a) The sailors must have been terrified. They may have been feeling angry with their Captain for his senseless act of cutting off the Inchcape Bell. It would be useful now to save them.
(b) The captain of the pirates must have been going mad with fear of the possible crash and sure death of everyone on board. He did not know where his ship was located. He was wondering if they were going to Crash on the Inchcape Rock. He had ensured his own destruction and death by cutting the Bell.

Poetic Devices:

Question 1.
Pick out the examples of imagery from the extract, state what kind it is and explain.
Answer:
Example of Visual imagery from the extract:
1. ‘The ship was as still as she could be’.
2. ‘Her keel was steady in the ocean’. Both the lines depict how the ship was on the sea, almost unmoving.
3. ‘The waves flow’d over the Inchcape Rock;
So little they rose, so little they fell,
They did not move the Inchcape Bell.’
These lines in the second stanza describe the very mild sea and the small waves.

Examples of Sound imagery from the extract:

1. And over the waves its warning rung. The line describes the loud warning sounded by the Inchcape Bell in a storm,
2. ‘And there was joyance in their sound.’ These lines show an air of joy in that scene. Even the birds seemed to be flying round and round – like a wheel – ‘wheel’d round’, ‘joyance in their sound’.

Question 2.
Pick out the examples of imagery from the extract, state what kind it is and explain.
Answer:
The lines with Visual imagery:
1. The Sun in heaven was shining gay,
The Sun shone bright and made the morning cheerful.
2. The sea birds screamed as they wheel’d round,
The birds seemed to be flying round and round in joy, like a wheel.

Examples of Sound imagery from the extract:
1. And over the waves its warning rung. The line describes the loud warning sounded by the Inchcape Bell in a storm.
2. ‘The sea-birds scream’d as they wheel’d round
And there was joyance in their sound.’
These lines show an air of joy in that scene. Even the birds seemed to be happy as their calls seemed like they were screaming in joy, ‘joyance in their sound’.

Question 3.
‘Gurgling sound’. Find the figure of speech.
Answer:
This is onomatopoeia. The pronunciation of the word resembles the meaning – the sound of an object sinking and bubbles rising and bursting.

Question 4.
Pick out an example of imagery from the extract.
Answer:
‘So thick a haze o’erspreads the sky,
They cannot see the Sun on high.’
The reader is able to imagine the fog so dense that the sun is blocked out. This is visual imagery.

Poetic Creativity:

Question 1.
Compose 2-4 lines using “A Song in the Air” as the theme. You could begin with…
‘The leaves rustle gently….’
Answer:
‘The leaves rustle gently and flowers nod. The droplets gather into a bigger drop The birds shake their plumes, bright-eyed. A song is in the air, the new day, a pretty bride.’

Question 2.
Compose 2-4 lines with one of the following as the theme : anger/ hatred/jealousy
Answer:
The Enemy Inside

I don’t know where he hides everyday He flashes in my eyes, in some words I say
To elders, family, friends. I am surprised
By my own words, my actions, only later I cried.

Question 3.
Compose 2-4 lines with one of the following as the theme:
Answer:
I Regret

I think of the harsh remark
The careless action I threw
I vow not to repeat anymore
As I begin this day new.

Writing Skills:

Question (a)
Time and tide wait for none Points:
There is a time for doing each thing

• Postponing action is laziness
• If the time for the action is lost the opportunity is lost for ever.
• Only regret remains.

Question (b)
Man proposes, God disposes

Points:

• It is in one’s power to plan a way of doing things
• It is a smart thing to prepare in advance
• In spite of planning we may not be able to carry on with the plan due to circumstances
• We must accept the unexpected circumstances and yet go ahead by some other method
• We must be flexible and find an alternative way
• It is smart to always have a Plan B ready

Question (c)
Look before you leap Points:

• Before we act one must think of the results
• only a fool will act without thinking of the future consequences of the present action
• if we think the results are going to harm someone, one must not do that
• It is also a good thing to take the advice of experienced or elders when making important decisions.

Balbharti Yuvakbharati English 12th Digest Chapter 2.2 Indian Weavers Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 12 English Yuvakbharati Solutions Chapter 2.2 Indian Weavers

12th English Digest Chapter 2.2 Indian Weavers Textbook Questions and Answers

Question 1.
Artisans are also called craftsmen. They are creators of diverse goods and use their hands to create unique, functional and also decorative items using traditional techniques. Now complete the web given below:

Answer:

Question 2.
Discuss with your partner the seasons/ occasions when we need:
Answer:
(a) Woollen clothes – in winter
(b) Casual clothes – when at home
(c) Rich silk clothes – festivals, weddings
(d) Colourful, comfortable clothes – travel, socializing

Question 3.
Let’s play a game. The teacher will ask the students some questions. Students will understand that there are some exceptions to the general rules. Let’s start:
Answer:
(a) One who weaves is a – weaver
(b) One who plays a game is a – sportsman
(c) One who sings is a – singer
(d) One who dances is a – dancer
(e) One who teaches is a – teacher
(f) One who cooks is a – cook/chef

Question 4.
We have often seen the picture of Gandhiji spinning on his charkha. Discuss the reasons behind this.
(a) To give rural people an opportunity to earn their livelihood.
(b) To give the message to wear hand-spun clothes.
(c) To convey the message of simple living.
(d) To be self-reliant.

Question 5.
Name some tools used by the weavers.
(a) Loom
(b) Shuttle
(c) Bobbin
(d) Scissors

Question 6.
Name some types of yarns used by the weavers.
(a) Linen
(b) Cotton
(c) Jute
(d) Flax
(e) Silk

(A1)

Question 1.
Discuss with your partner the following vocations:
(a) Weaving:
[Points:
India has rich heritage of weaving.

• Handloom weaving the second most widespread occupation next to agriculture.
• Each region of India has its typical yarn, style and fabric.
• India is tropical – people like to wear cotton
• Silks from different regions of India are prized by women worldwide.]

(b) Tailoring:
[Points:

• A tailor can make someone a star with style and fitting.
• Can have own machines, and stitch clothes according to client’s needs.
• Could be under a designer creating high fashion garments.
• Could be under a fashion-house label, working at computerized machines.
• hey create the trending fashions world- wide.]

(c) Knitting:
[Points:

• Knitting is weaving with two needles with woollen yarn.
• Many women are skilled.
• Women in cold regions do knitting to make warm clothes for the family.
• Knitting is done on machines too.]

(d) Embroidering:
[Points:

• Embroidery needs skilful fingers and patience.
• From hankies to wall-hangings are laboriously made with silk threads and tiny needles.
• Many other beautiful things like tablecloths, dress-collars, sari-edges, bedcovers and cushion sleeves.
• Machine embroidery also is done.]

(A2)

Question (i)
Discuss the various products made by the weavers in the poem.
Answer:
The poet asks the weavers what they are weaving at daybreak. It is a brightly coloured cloth and she asks the reason. The weavers reply that the robes, in the gay colour of the wild kingfisher, are for a newborn.

The poet next asks the weavers at dusk, for what they are weaving that bright cloth. The purple and green shaded fabric is for a queen’s wedding veil, reply the weavers. The poet asks what they are making in the chill night in the moonlight. The weavers are weaving a cloth as white as a feather, as a cloud. They say they are weaving a shroud for a dead man’s funeral.

Question (ii)
The words in the three stanzas of the poem mention different times of a day. Complete the table.
Answer:

Time of the day	Words/ phrases	Weaver’s work
Early morning	break of day	Weavers weave robes for the newborn child
Late in the evening	fall of night	Weavers weave a wedding veil for a queen
Night	Moonlight chill	Weavers weave a shroud for a dead man’s funeral

Question (iii)
The poem reveals three phases of life. Fill in the blanks with feelings and colours appropriate to the phase of life.
Answer:

	Newborn/ Childhood	Youth/ Adulthood	Old age/ Death
Colour	Blue	Peacock colours of purple-green	White as a feather or cloud
Feeling	Hopes and expectations	Expectations, responsibilities, romance, energy	Frailness, peace, wisdom

Question (iv)
Complete the sentence:
The weavers weave in the chill moonlight ……………….. .
Answer:
The weavers weave in the chill moonlight a cloth as white as a feather, as a cloud, a shroud, for a dead man’s funeral.

Question (v)
Pick out two words used to describe the weavers in the last stanza. Also state their importance.
Answer:
‘Solemn’ and ‘still’ are the two words used in the last stanza to describe the weavers. The words describe the occasion for which the craftsmen are working. A shroud is being woven for a funeral. The workers’ mood is also serious and unsmiling because of the occasion.

Question (vi)
Express your views about the present conditions of weavers.
Answer:
Weaving has existed for thousands of years in India. It was second only to agriculture. The weaves and fabrics from various regions were known around the world. From the Bengal muslin, to Kashmiri, Banarasi and Kancheevaram silks the cloth from India was renowned for the quality and fineness, the designs and richness.

Industrialization then brought problems for them. Power-looms are faster and manufacture large quantities in short time, Fashion-houses buy fabric from the weavers, put their labels and sell off the cloth at a very high price compared to the price paid to the craftsmen.

The craftsmen remain unknown, their craft under-valued and their life is in poverty. Ancient skills are lost and some take loans to somehow struggle. When debts are too much they commit suicide. Master weavers send away sons to cities to take up jobs.

They do not want the sons to struggle. Parents will not give girls in marriage to weavers. They lack facilities like lighting and water supply. Though the government has given subsidies for weavers most of it is lost to bribes and the weavers receive negligible sums.

Question (vii)
Describe in your own words the steps or measures that can be taken to solve the problems of the weavers.
Answer:
Weavers are unorganized. Buyers offer very low prices and if one weaver refuses, they go to another weaver. So the prices are at lowest. The government allocates crores for Handloom Promotion Council, but the weavers get nothing. Weavers’ organization can help the situation.

The weavers are forced to sell their creations to designers at low rates. No one helps to update weavers of the latest fashions trends. Also their creations are not commercially advertised. Handloom industry is eco-friendly in every way. It should be promoted by the Government, designers and supported by the public.

Subsidies for buying yarns and dyes should be given to the craftsmen. Clean water and proper lighting facilities are needed for the craftsmen. Some designers and activists for the weaver communities have begun working for the upliftment of the community and marketing the product.

The buyer can buy directly from the weavers and cut out corrupt middlemen. Exhibitions for selling wares to the public will bring the craftsmen and buyer closer. We can hope that soon the craftsmen of handloom will regain the lost glory which they deserve.

Question (viii)
Express your own views and opinions from the weavers point of view and complete the following table:
Answer:

Stanza	Activity	Views/Opinion
First stanza	Robes for a new- born child	The weavers feel happy because they are enthusiastic to weave the bright blue robes for the newborn.
Second stanza	A purple- green veil for the wedding of a queen	The weavers are joyous to dress the royalty on the happiest and most important day of her life
Third stanza	Shroud for the funeral of a dead man	The weavers are solemn and quiet as the cycle has closed for a person and it is true for everyone.

(A3)

Question (i)
Pick out the rhyming words from the poem.
Answer:
1st stanza: day – gaywild – child
2nd stanza: night – bright; green – queen
3rd stanza: still – chill; cloud – shroud

Question (ii)
Give antonyms and synonyms of the following and make sentences –
Answer:

Word	Antonym/ Synonym	Sentence
New	Antonym: old Synonym: fresh	Everyone must grow old. We begin every morning with fresh life.
Bright	Antonym: dull Synonym: shining	One feels dull without a shower after leaving bed. Children look at shining things with bright and curious eyes.
Dead	Antonym: alive Synonym: lifeless	The last time we met, he brought the room alive with laughter and cheer. The children were malnourished and their eyes were lifeless.
Still	Antonym: lively Synonym: unmoving	The crowd was lively and cheering the players. The lake was covered with oil and lay dirty and unmoving.
Wild	Antonym: civilized Synonym: untamed	After many years of instruction they finally changed him into a civilized city-dweller. Though we think of elephants as gentle giants, untamed ones are dangerous.
Fall	Antonym: rise Synonym: drop	The rise of the new star of Indian tennis is stunning. When prices rise suddenly, everyone hopes it will drop soon.
Child	Antonym: adult Synonym: young one	An adult has a fully developed immune system till about 60 years of age. Young ones are innocent till they start imitating grown-ups.

Question (iii)
Make a word register for clothes/attire/ dress:
Answer:
garments, togs, outfit, wardrobe, raiment, apparel, get-up, gear.

(A4)

Question (i)
Complete the following table:
Answer:

Figures of speech	Line
Simile	3rd – Blue as the wing of halcyon wild. 7th – like the plumes of peacock, purple and green 11th – white as a feather and white as a cloud
Imagery	1st – break of day, gay, blue (beginning of life) 4th – newborn child 5th – fall of night, green-purple royal (marriage and family) 8th – marriage-veil 9th – solemn and still, (Moonlight in the night, chillness of death, gloom, end) Colourless, cold white.
Metaphor	1st – break of day – day is born, a newborn child (day) for 1st stage of life 5th – fall of night – twilight, the romantic half-light, beginning of wedded bliss, for 2nd stage of life 8th – marriage-veils 10th – chill indicating death, moon indicating night and darkness. For the last stage of life.
Alliteration	Each line of the poem has the ‘w’ in the words ‘we’, ‘weave’, ‘why’ ‘weaver’, wild’, ‘wing’, ‘what’, ‘white’. The words ‘we’, ‘weaver’, ‘weave’, ‘why’ occur in each stanza multiple times adding to the musical quality of the poem.

Question (ii)
The rhyme scheme in the first stanza is ‘aabb’. Find rhyme schemes in the second and third stanzas:
Answer:
The rhyme scheme in the second stanza is – aabb.
The rhyme scheme for the third stanza is – aabb.

(A5)

Question (i)
The poet has asked a question at the beginning of every stanza. Explain the effect it creates on the reader.
Answer:
Each stanza of the poem begins with a question to the weaver. The first is at dawn as the poet asks why they are working so early. The second question asks why they make the bright coloured garment at dusk. The third stanza begins asking them why they are working so late, in the cold darkness.

In the question the reader knows about the time of day, the colour of the cloth. The reply gives information about the purpose of the new cloth and why that colour is chosen. The poem is like a conversation between the poet and craftsmen. It conveys the metaphor using time of day and stage of life, the colour and cloth suited for that stage. The mood of the weavers matches the time and purpose of their work. The poem thus flows easily.

Question (ii)
Write an appreciation of the poem with the help of following points:

• About the poem/poet and the title
• The theme
• Poetic style
• The language/poetic devices used in the poem
• Special features
• Message, values, morals in the poem
• Your opinion about the poem

Answer:
The poet Sarojini Naidu’s poem ‘Indian Weavers’ tells about the work of India’s famous handloom craftsmen. The three stanzas mark the three stages of life itself. The weavers reply to questions about why they are weaving that particular piece of cloth, of a certain colour at that time of day.

The theme is cycle of life. The weavers use colours associated with birth, marriage and death through weaving cloths for a newborn, a queen- bride and a dead man. Three stanzas of four line each in the form of questions & answers. The conversational tone gives a flow like life, one stage moving into the next.

The poem is a metaphor for the cycle of life: new life-dawn, marriage-dusk, and death-night. Simile compares the woven garments to objects in colours apt for that stage of life. The sound of ‘w’ occurs a total of 20 times, at least once in all lines except one. This alliteration gives a musical quality.

The poem is dedicated to the talented weavers and the fabrics of India which were world famous. It shows the hard work of craftsmen and how we use their products in every occasion of our life. This poem is a beautiful way of the poet to salute the weavers of India. I find that the weavers are not only skillful but also talented. They combine colours and create patterns that are eye-catching. They know which colours are apt for occasions. The poet brings out their talent as well their hard-working nature.

Question (iii)
Compose four lines on ‘Importance of clothes’.
Answer:
The attire indicates the man he is
His coat brings him confidence and protects too
The colour indicates her mood and occasion The weaver makes the christening, wedding and celebration.

Question (iv)
Write an appeal to use handloom products in our daily life.
Answer:
An Appeal

THREADS -The Handloom Research And Development Society, Maharashtra is proud to present to you the wonderful creations of THREADS for the first-time.
The weaves come directly from the craftsmen to you, the customer.
The range of products includes Pure silk Silk-cotton Cotton Linen Jute
Come and see, appreciate the skills and support the craftsmen who practice our ancient art.
There are wedding saris and Punjabi suits in pure silk, casual wear in all the materials, bed linen, towels, door mats, carpets, hand towels, handkerchiefs, scarves to suit every budget.
Come support our artisans and appreciate the fine quality of our finest Indian weaves!!
Come. Support us. You will Love our Indian fabrics. We need your support.

Contact: Incharge Mrs. Das.
99123 xxxx Email: MrsDas@xxx.

Question (v)
Visit a handloom factory near your locality and write a report of it.
Answer:
Weavers Of Dreams
– by Team A, SYJC -Com. 25/02/2020. The SYJC students of the S.K.Nayak Junior College went on an educational tour to Moinbad village to take a tour of the handloom factory there.

The students viewed many of the processes involved in the weaving of a saree. The students were taken on a guided tour by a local who is a family member of one of the weaver – households of the village. In the first stage the bundles of white yarn were strung on iron frames. The frames held by two men were dipped into hot iron vats of boiling dye.

The yarn is dipped several times, thoroughly drenched for the colour to coat the yarn evenly. After several minutes the men transfer each bundle on to short thick wooden rods. They twist the sticks to wring the bundle dry. A third man takes the hot bundle on another stick and lays the bundles to dry.

The next stage the students saw was coloured yarn stretched on frames several metres long. This was outside in the open. The guide told us it was for a sari. It was a blue yarn in the middle with purple yarn on both sides lengthwise, for borders. Each yarn is stretched, the number of threads for each colour counted.

There were long buildings which are the workshops. Here are frames, looms and the weavers seated before each loom. Each weaver was working on a saree. The guide told us that one saree averagely takes four to seven days. If the design is complicated and in different colours it takes longer.

The finished products are folded and packaged for the market. The tour ended with the students speaking to the weavers at the looms and craftsmen dyeing the yarns. The craftsmen spoke to the visitors giving interesting details about the popular colour combinations. They told us about auspicious colours for special occasions. The artisans also spoke about a different place where traditional nine-yard long are woven. Those are mostly for weddings. The students took photographs of the various processes and the artisans happily posed for them.

Question (vi)
A handicraft exhibition is being organized in your college. You are given the task to compere the inaugural function. Write the script for compering.
Answer:
1. Introduction: A very Good evening and warm welcome to all. This is for a very special kind of occasion we have gathered here this morning. We want to bring the spotlight on that section of our nation’s citizenry who are carrying thousands of years of our heritage on their shoulders! No in their skilful fingers!! Yes our very talented craftsmen. We have our craftsmen with their various talents to showcase their beautiful creations.

We have the weavers of the rich Paithani silks, the Warli painters, the makers of the famous Kolhapur Leather – Footwear, the intricate Bidri brassware, the Dhurrie Weavers, Banjara Embroidery…. all these from Maharashtra. We have men and women with magic in their hands from other states too. Craftsmen have come from distant Meghalaya and Nagaland, from neighbouring Gujarat. We have the makers of the amazing Kashmiri embroidery to the wooden toymakers from Andhra. And so any more.

2. Welcome speech: Our Respected Principal, Sri. Harsh Nayak, our beloved teachers, staff and all my friends join me in welcoming the Honourable Chief Guest, the textile Minister, Shrimati Mandakini Gadge, to this exciting and colourful programme. A very warm welcome to you Madam! It is a great pleasure and an honour to have you here. And a warm welcome to all the parents and all guests.

3. Inaugural Ceremony – Lighting of the Lamp: I request our Chief Guest Shrimati Mandakini Gadge and our Principal to kindly come to the dais. We request you to light the ceremonial lamp, in the traditional Indian way, to declare the exhibition open.

4. Prayer song: And now kindly take your seats for a short programme before we go around viewing the exhibition. Music is such a positive influence. So starting on a note of gratitude we have the prayer song by the stars of our music-club, Nandini, Sonia, Bhaskar, Jay and Kavita. Manjeet is accompanying them on the violin.

5. Welcome Programme: Thank you for that lovely song to begin the programme my friends. Now for a unique performance! We have a Fashion-show!! It is unique because the students of our college and our participating craftsmen-guests worked together for the last few days. This is a first! A big round of applause for our models on the catwalk!

6. Concluding Remarks: Thank you for your encouragement! Wasn’t that wonderful? You may have noticed for yourself, but let me clarify. The stylish saris, salwar suits, elegant kurtas, those shawls, the stunning neck-pieces, the baskets on the ladies’ arms, the wooden screens in the backdrop, the lampshades on the stage, the carpet on the catwalk… and all the decor you can see on the stage are all creations of the masters who are displaying their creations in the exhibition! You now have an idea of what is in store for dressing stylishly, or doing up your home elegantly. That was only starters! The main course is still waiting for you. And there is dessert as well…!!

1. I now request our Chief Guest Shrimati Gadge to address the audience.
2. Thank you, for those words of praise and appreciation of the craftsmen Madam. They richly deserve them.
3. This is the first time a college is hosting such an exhibition. We welcome you all to go around and view the stunning collection of handicraft products. There are master workers who will show you how some of their handicraft is created. They will happily demonstrate their ancient skills. You can watch how the lovely designs we wear are made at the loom.

There are demos to show the yarn being dyed and informative presentations showing the stages in the process. The beadwork artists can help guests to select stones and make them into elegant neck-pieces. There are stalls selling numerous stunning products. Apparel, Decor pieces for your homes, gifts for the festival season.

I invite our Honourable Chief Guest to formally declare the exhibition open and take a leisurely round of the displays.
4. Thank you all for making this festival of crafts a resounding success.

(A6)

Question 1.
Go to your college library and collect and read the poems written by Sarojini Naidu.

Question 2.
Find various career opportunities in Small Scale Industries like Handloom, Art and Craft, Block Printing, etc.

Question 3.
Find out information about the Mahavastra of Maharashtra – Paithani

Yuvakbharati English 12th Digest Chapter 2.2 Indian Weavers Additional Important Questions and Answers

Read the poem and complete the activities given below:

Global Understanding:

Question 1.
The weavers reply to the poet’s questions in each stanza. What is the common factor? What do you understand from that?
1st stanza: ‘Blue as the wing of a halcyon wild’
2nd stanza: ‘Like the plumes of a peacock, purple and green,’
3rd stanza: ‘White as a feather ..’
Answer:
We find reference to a bird in all the three comparisons. In all of them the weavers refer to the colour of the birds’ feathers. We could take it not only refers to the brightness but also to the lightness of the cloth. The fabric is as soft and light as feathers for the tender newborn, or for the transparent veil of the queen. The pure white shroud for the dead is soft too.