Month: October 2024

JAC Board Class 9th Social Science Notes  Geography Chapter 2 Socialism in Europe and the Russian Revolution

→ The Age of Social Change

• The French Revolution opened up the possibility of creating a dramatic change in the way in which society was structured.
• The social and political changes that took place all over the European continent can be traced to the French Revolution.
• But not everyone in Europe wanted a complete transformation of the society. Some were conservatives, while others were ‘liberals’, and ‘radicals’.
• Through the revolution in Russia, socialism became one of the most significant and powerful ideas to shape the society in the twentieth century.

→ Liberals, Radical sand Conservatives

• Liberals:
  • These leaders (group of people) wanted a nation which tolerated all religions,
  • Opposed the uncontrolled power of dynastic rulers,
  • Wanted to safeguard the rights of individuals against governments,
  • They argued for a representative, elected parliamentary government, subject to laws -interpreted by a well-trained judiciary that was independent of rulers and officials.
• Radicals:
  • They wanted a nation in which government was based on the majority of a country’s population,
  • They supported women suffragette movements,
  • They opposed the privileges of landowners and wealthy factory owners,
  • They disliked the concentration of property in the hands of a few people.
• Conservatives: They accepted that some changes were inevitable but believed that the past had to be respected and changes had to be brought through a gradual process.

→  Industrial Society and Social Change

• It was the beginning of the industrial revolution. Men, women and children were pushed into the factories for low wages.
• Liberals and radicals who were the factory owners felt that efforts must be encouraged so that benefits of industrialisation may be passed on to the workers.

→ The Coming of Socialism in Europe

• Some socialists believed in the ideas of cooperatives, while others demanded that governments encourage cooperatives.
• These cooperatives were to be the associations of people who produced goods together and divide the profits according to the work done by members. Karl Marx and Friedrich Engels believed that a socialist society would free the workers from capitalism. This would be a communist society.

→ Support for Socialism

• Workers in Germany and England began forming associations to fight for better living conditions.
• They set up funds for members in distress. They wanted reduction of working hours and right to vote.
• By 1905, socialists and trade unionists formed ‘Labour Party’ in Britain and ‘Socialist Party’ in France.
• However, till 1914, socialists did not succeed in forming a government in Europe.

→ The Russian Revolution

• The government in Russia was taken over by the socialists through the October Revolution of 1917.
• The fall of monarchy in February 1917 in Russia and the events of the October Revolution are normally called the Russian Revolution.→ The Russian Tsars had built a vast empire which included part of Poland, Finland, Latvia, Lithuania, Estonia, Ukraine and Belarus.

→ The Russian Empire in 1914

• The major religion of Russia was the orthodox Christianity which grew out the Greek Orthodox Church.
• The empire also included Catholics, Protestants, Muslims and Buddhists.
• At the beginning of the twentieth century, 85 per cent of the Russian population were agriculturists.

→ Economy and Society

• Russia was a major exporter of foodgrains. Russian cultivators produced for the market and for their own needs.
• In Russia, prominent industrial areas were St. Petersburg and Moscow.
• In 1890s, many factories started due to the expansion of the railway network and there was an increase in foreign investment in industries.
• Most industries were private property of industrialists.
• Government supervised large factories to ensure minimum wages and limited hours of work.
• Workers were divided as social groups or by skin. Workers were sometimes united to participate in strikes.
• In the countryside, peasants cultivated most of the land but the nobility, the crown and the orthodox church owned large properties.
• Russian peasants wanted the land of the nobles to be given to them. They refused to pay rent and even murdered the landlords.

→ Socialism in Russia

• All political parties were illegal in Russia before 1914.
• The Russian Social Democratic Workers Party was founded in 1898 by socialists who respected Marx’s ideas.
• Socialists were active in the countryside through the late nineteenth century. They formed the Socialist Revolutionary Party in 1900.
• This party struggled for peasants’ rights and demanded that the property belonging to nobles be transferred to peasants.
• The Social Democratic Party was far divided into two wings, the Bolsheviks and the Mensheviks.
• Vladimir Lenin led the Bolshevik group.

→ A Turbulent Time: The 1905 Revolution

• In the beginning of the 20th century, Russia was still an autocracy and the Tsar was not subject to parliament.
• The year 1904 was bad for Russian workers because of increase in prices of essential goods. Moreover, the real wages declined by 20 percent.
• In 1905, on one Sunday, the procession of workers led by Father Gapon reached the Winter Palace. It was attacked by the police and the Cossacks.
• Over 100 workers were killed and about 300 wounded. This incident, known as Bloody Sunday, started a series of events that came to known as the 1905 Revolution.
• Strike took place all over the country. During the 1905 Revolution, the Tsar allowed the creation of elected consultative parliament or Duma.
• The Tsar dismissed the first Duma within 75 days and the re-elected second Duma within 3 months.
• The Tsar did not want any questioning of his supreme authority.

→ The First World War and the Russian Empire

• In 1914, the First World War started between two European alliances—Germany, Austria and Turkey (the central powers) and France, Britain and Russia (later, Italy and Romania also).
• Russian armies lost badly in Germany and Austria between 1914 and 1916. There were over 7 million Russian casualties by 1917.
• The war had a severe impact on industry. Large supplies of grain were sent to feed the vast army. Food scarcity became common, which sometimes led to riots in bread shops.

→  The February Revolution in Petrograd

• In the winter of 1917, conditions in the capital, Petrograd, were grim.
• In February 1917, a lockout of a factory in Petrograd led to many strikes and demonstrations.
• The Duma was suspended, which ultimately lead to the resign of the Tsar.
• Soviet leaders and Duma leaders formed a provisional movement to-run the country.

→ After February

• In April 1917, the Bolshevik leader Lenin returned from exile and made three demands known as the ‘April Theses’.
• Throughout the summer, the workers’ movement spread.
• Peasants seized land between July and September 1917.

→ The Revolution of October 1917

• The conflict between the provisional government and the Bolsheviks grew.
• On 16 October, 1917, Lenin persuaded the petrograd Soviet and the Bolshevik party to agree to a socialist seizure of power.
• The uprising began on 24th October, 1917.
• There was he ivy fighting between pro-government troops and the Bolsheviks. By December 1917 the Bolsheviks controlled the Moscow-Petrograd area.

→ What changed after October

• The Bolsheviks nationalised banks and industries in November, 1917.
• They declared land as social property and allowed peasants to seize the land of nobles.
• The Bolshevik party was renamed as the Russian Communist Party (Bolshevik).
• In November 1917, the Bolsheviks conducted the elections to the Constituent Assembly, but they failed to gain majority support.
• In January 1918, the Assembly rejected Bolshevik measures and Lenin dismissed the Assembly.
• The Bolsheviks became the only party to participate in the elections to the all Russian Congress of Soviets which was the parliament of the country.

→ The Civil War

• When the Bolsheviks ordered land redistribution, the Russian army began to break up. Soldiers, mostly peasants, wished to go home for the redistribution and deserted.
• The pro Tsarists (the whites) and the socialists (the greens) fought a civil war with the Bolsheviks (the reds) during 1918 and 1919.
• Most non-Russian nationalists were given political autonomy in the USSR created by the Bolsheviks in 1922.

→ Making a Socialist Society

• During the civil war, the Bolsheviks kept industries and banks nationalised.
• They permitted peasants to cultivate the seized land to show collective work.
• A process of centralised planning was introduced and officials made five year plans for the improvement of the economy.
• Rapid construction brought poor working conditions for workers.
• An extended schooling system was developed and arrangements were made for factory workers and peasants to enter universities.
• Creches were established in factories for the children of women workers.

→  Stalinism and Collectivisation

• The period of the early planned economy was linked to the disasters of the collectivisation of agriculture.
• By 1927-1928, the towns in Soviet Russia were facing an acute problem of grain supplies.
• Stalin, who headed the party after the death of Lenin, introduced firm emergency measures.
• He introduced collective farming to reduce the shortage of grains in the country. However, bad harvests during 1930-1933 led to famines with over 4 million people dying.
• Those who criticised Stalin’s policies were charged with conspiracy against socialism.
• By 1939, over 2 million people were imprisoned or sent to labour camps.

→ The Global Influence of Russian Revolution and the USSR

• The impact of the Russian Revolution was felt globally with communist parties being formed in many countries.
• By the time of the outbreak of the Second World War, the USSR had given socialism a global face and world stature. The USSR became a great power.
• Its industries and agriculture had developed and the poor were being fed, But it had denied the essential freedom to its citizens and adopted repressive policies for its developmental projects.
• By the end of the 20th century, the international reputation of the USSR as a socialist country had declined, though it was recognised that socialist ideals still enjoyed respect among its people.
• In each country, the ideas of socialism were re-thought in a variety of different ways.
• Many Indian writers like Jawaharlal Nehru and Rabindranath Tagore were impressed by the Russian Revolution and its ideals.

→ Important Dates and Related Events

• 1850-1880: Debates over socialism in Russia.
• 1898: Formation of the Russian Social Democratic Workers Party.
• 1905: The Bloody Sunday incident and the Revolution of 1905. Formation of Labour Party in Britain and Socialist Party in France.
• 1914: The First World War started.
• 1917: Abdication of the Tsar on 2nd March; Bolshevik uprising in Petrograd on 24th October.
• 1918-20: The Civil War in Russia.
• 1919: Formation of Comintern.
• 1929 : Beginning of Collectivisation in farming.

→ Conservatives: Group of people who believed that the past had to be respected and change had to be brought through a slow process.

→ Liberals: Group of people who opposed the uncontrolled power of dynastic rulers.

→ Radicals: Those who favoured the formation of a government based on the majority of a country’s population.

→ Democrats: Group of people who believed in universal adult Franchise.

→ Suffragette movement: A movement to provide women the right to vote.

→ Jadidists: Muslim reformers within the Russian empire.

→ Real Wage: Reflects the quantities of goods which the wages will actually buy.

→ Duma: Russian Parliament which was created in 1905 for the first time.

→ April Theses: A set of three demands made by Vladimir Lenin in April 1917  War be brought to a close, land be transferred to the peasants and banks be nationalised.

→ Autonomy: The right of a person, an organization, a region etc. to govern his/her/ its own affairs.

→ Nomadism: Lifestyle of those who do not live at one place but move from one place to another to earn their living.

→ Kulaks: The name for well-to-do peasants.

→ Deported: Forcibly removed from one’s own country.

→ Exiled: Forced to live away from one’s own country.

→ Capitalism: It is an economic system under which the means of production are in the hands of private individual or individuals.

→ Socialism: It is an economic system under which the means of production are controlled by the society.

→ Bolsheviks: The majority group of Russian Social Democratic Workers Party
formed in 1898. This group led by Lenin was based on the ideology of Karl Marx and Friedrich Engels. They believed that party should be disciplined and should control the number and quality of its members.

→ Mensheviks: The minority group of Russian Social Democratic Workers Party formed in 1898. They thought that the party should be open to all.

→ Russian Steam Roller: The imperial Russian army came to be known as the Russian steam roller. It was the largest armed force in the world. When this army shifted its loyalty and began supporting the revolutionaries, Tsarist power collapsed.

→ Tsar: The title of the Emperor of Russia.

→ Giuseppe Mazzini: An Italian nationalist politician, journalist and activist who worked to set up a nation where all citizens would have equal rights. He was the spearhead of the Italian revolutionary movement.

→ Robert Owen: A leading English manufacturer, sought to build a co-operative community called New Harmony in Indiana (U.S.A.).

→ Karl Marx: Gave birth to the idea of socialism.

→ Friedrich Engels: Worked along with Karl Marx to give shape to the idea of Socialism.

→ Vladimir Lenin: Led the Bolshevik group to a successful revolution in Russia.

→ Father Gapon: Led the procession of workers that marched to the Winter Palace in 1904.

→ Tsarina Alexandra: Wife of Tsar Nicholas II.

→ Rasputin: Tsarina’s close confidant who had a strong influence on her.

→ Leon Trotskii: Headed the Military Revolutionary Committee appointed by the Soviet.

→ Kerenskii: Prime Minister of Russia in October, 1917.

→ Stalin: Headed the ruling party in Russia after the death of Lenin.

JAC Class 9 Social Science Notes

JAC Board Class 9th Social Science Notes History Chapter 1 The French Revolution

• The French Revolution started on 14th July, 1789 with the storming of the fortress prison- the Bastille, hated by everybody, because it stood for the despotic power of the King. The fortress was demolished.
• The beginning of chain of events started by the middle-class affected and shook the lower class and led to the execution of king in France followed by a revolt against monarchy.

→ Introduction

• The French revolution is a landmark in the history of European continent and the world.
• This revolution ended the monarchical system in France.
• The slogan of French Revolution “liberty, equality and fraternity” became important ideas of the new era.

→ French Society During the Late Eighteenth Century

• In 1774, Louis XVI of Bourbon family became the king of France.
• Due to long years of war and maintenance of an extravagant court of the palace of Versailles, he got an empty treasury.
• King Louis also helped thirteen American colonies to gain their indendence from Britain.
• This war added more than 1 billion livres (unit of currency in France) to the already
  existing debt of more than 2 billion livres. .
• To meet the expenses like maintaining army, court, running government offices or universities etc., the Franch government was forced to increase the taxes.
• French society in the 18th century was divided into three estates and only members of the third estate paid taxes.
• The three estates of France at that time were First Estate (the clergy), Second estate
  (the nobility), Third Estate (businessmen, merchants, court officials, lawyers, peasants, artisans, ladies, labour, servants etc.).

→ The Struggle to Survive

• During 1715-1789, the population of France increased rapidly which led to rapid increase in demand for foodgrains.
• Insufficient production increased the price of bread. But the wages of the workers did not keep pace with the rise in prices.
• Situation became worse when bad weather conditions reduced the harvest. This condition created subsistence crisis.

→ Emergence of Middle Class

• In the 18th century, a new social group emerged which was known as the middle class. They had become rich by expansion of overseas trade and manufacturing goods.
• This group believed that no group of society should be privileged by birth.
• Philosophers like John Locke, Jean Jacques Rousseau and Montesquieu considered that the middle class was solely responsible for the revolution.
• They spread the ideas of freedom, equal laws and opportunities for all.

→ The Outbreak of the Revolution

• On 5th May, 1789 Louis XVI called for an assembly of the Estates General to pass the proposals for new taxes.
• First, second and third estates sent their representatives.
• Peasants, artisans and women were denied entry to the assembly but they sent their demands and grievances through their representatives.
• The members of the third estate demanded that voting should be conducted by taking assembly as a whole. But King Louis XVI rejected this proposal and members of the third estate walked out of the assembly in protest.
• On 20th June, 1789, the representatives of the third estate assembled in the hall of an indoor tennis court in Versailles.
• They declared themselves a National Assembly.
• They also swore to draft a new constitution for France that would limit the powers of the monarch.
• The representatives of the third estate were led by Mirabeau and Abbe Sieyes.
• Due to severe winter, harvest was severely affected in France, which led to the increase in prices of essential commodities.
• After spending many hours in long queues at the bakery, crowds of angry women raided the shops.
• On 14th July, 1789, an agitated craud stormed and destroyeed the Bastille.
• Seeing the power of the revolt, King Louis XVI recognised the proposal of National Assembly that his powers would be checked by a constitution.
• On the night of 4th August, 1789. France passed the law for abolishing the feudal system of obligations and taxes.

→ France Becomes a Constitutional Monarchy

• The National Assembly completed the draft of the constitution in 1791. Its main objective was to limit the powers of the monarch.
• Under the new constitution, the laws were to be made by the National Assembly.
• The constitution began with a declaration of the right of man and citizen.

→  France Abolishes Moranchy and Becomes a Republic

• The National Assembly declared war against Prussia and Austria in April, 1792.
• While men were busy fighting at the front, women had to earn a living and look after their families.
• A large segment of the population was convinced to carry the revolution further, as the constitution of 1791 gave political rights only to the richer section.
• On 10th August, 1792, the Jacobins attacked the palace of the Tuileries with a large number of Parisians.
• They killed the King’s guards and held the king himself as hostage for several hours.
• New election were held and all the men above 21 years of age were allowed to vote.
• The newly elected assembly, the Convention, abolished monarchy on 21st September, 1792 and declared France as a‘Republic’.
• King Louis XVI was sentenced to death by a court on the charge of treason.

→ The Reign of Terror

• The period from 1793 to 1794 is reffered to as the ‘Reign of Terror’ in France.
• Robespierre followed a policy of severe control and punishment.
• Due to his harsh policies, his supporters left him in the end. He was finally convicted by a court and guillotined in July 1794.

→ A Directory Rules France

• After the fall of the Jacobin government, the wealthier middle classes seized power.
• The new constitution was introduced which denied the right to vote to non-propertied society.
• It provided for two elected legislative councils. These councils appointed a Directory, an executive made up of five members.
• The political in stability of the Directory paved the way for the rise of a military dictator- Napoleon Bonaparte.

→  Did Women have a Revolution?

• Women played a very significant role in the French Revolution.
• Women in France were disappointed with the constitution of 1791, as it reduced them to passive citizens who had no
  political rights.
• They demanded political rights, viz., right to vote to be elected to the Assembly and to hold political office.
• Finally in 1946, French women won the right to vote.

→ The Abolition of Slavery

• The most important social reform made by the Jacobin government was to abolish slavery in French colonies.
• After long debates, the National Convention passed a law in 1794. It declared slavery illegal and freed all the slaves in French overseas colonies.
• After two years, Napoleon reintroduced slavery.
• Finally, slavery was abolished in the French colonies in 1848.

→ The Revolution arid Everyday Life

• After 1789, many changes took place in the lives of man, woman, and children in France.
• One important law that come into effect after the storming of the Bastille in the sum¬mer of 1789 was the abolition of censorship.
• With the abolition of censorship and the Declaration of the Right of man and citizen, freedom of speech and expression became a natural right of people.

→ Conclusion

• In 1804, Nepoleon Bonaparte crowned himself as the Emperor of France.
• He conquered the neighbouring European countries, dispossed dynasties and created kingdoms where he placed members of his family.
• He introduced many laws such as the protection of private property and a uniform system of weights and measures provided by the decimal system.
• Napoleon was defeated at the Battle of Waterloo in 1815.
• The ideas of liberty and democratic rights were the most important legacy of the French Revolution.
• Tipu Sultan and Raja Ram Mohan Roy are the two examples of Indian individuals who were inspired by the ideas of the French Revolution.

→ Important Dates and Related Events

• 1774: Louis XVI of the Bourbon family ascended the throne of France and faced empty treasury and growing discontent within the society of the Old Regime.
• 1789: Convocation of Estates General, Third Estate formed National Assembly, the Bastille was demolished, peasants revolted in the countryside, Assembly passed a decree abolishing the feudal system of obligations and taxes.
• 1719: Constitution framed to limit the powers of the king and to guarantee basic rights to all human beings.
• 1792- 93: France became a republic nation. Overthrow of the Jdcobin republic, a Directory which ruled France.
• 1804: Napoleon became the emperor of France, annexed a large part of Europe.
• 1815: Napoleon was defeated at Waterloo.
• 1848: Abolition of slavery in French colonies.
• 1945: Women in France won the right to vote.

→ Revolution: As a historial process, revolution refers to a movement, often violent, to overthrow an old regime and effect complete change in the fundamental institution of society.

→ Livre: Unit of currency in France, discontinued in 1794.

→ Feudal System: This system existed in the Middle Ages in Europe In this system, people received land and protection from a Lord for which they worked and fought.

→ Clergy: Group of persons invested with special functions in the Church.

→ Tithe: A tax levied by the Church, comprising one-tenth of the agricultural produce.

→ Taille: Tax to be paid directly to the state by the members of the Third Instate.

→ Subsistence Crisis: An extreme situation where the basic means of livelihood are endangered.

→ Anonymous: One whose name remains unknown.

→ Manor: An estate consisting of the Lord’s lands and his mansion.

→ Chateau: Castle or stately residence belonging to a king or a nobleman.

→ Marseillaise: The National Anthem of France.

→ Convent: Building belonging to a community devoted to a religious life.

→ Sans-culottes: Jacobins came to be known as the Sans-culottes, literally meaning ‘those without knee breeches’. Sans-culottes men wore in addition the red cap that symbolised liberty.

→ Convention: The newly elected assembly of France in 1792 was called the convention.

→ Republic: A form of government in which the people elect the government including the head of the government.

→ Aristocracy: The highest class in some societies.

→ Guillotine: A machine or a device consisting of two poles and a blade with which a person is beheaded. It was named after Dr. Guillotin who invented it.

→ Directory: It was an executive made up of five members.

→ Treason: Betrayal of one’s country or government.

→ Negroes: A term used for the indigenous people of Africa, south of the Sahara. It is a derogatory term, not in common use any longer.

→ Emancipation: The act of freeing.

→ Militia: An organization that operates like an army but whose members are not professional soldiers.

→ Bastille: A fort in the eastern part of Paris, used as a state prison.

→ Despot: Someone, such as a ruler, who uses power in a cruel and unfair way.

→ Sous: Subordinate (a French prefix).

→ Souvenir: A thing that is kept as a reminder of a person, place or event.

→ Jacobin: A political club of people which was formed to discuss the government policies and plan their own forms of action.

→ Louis XVI: He ascended the throne of France in 1774 and ruled over France during the French Revolution.

→ Rousseau: He was a great philosopher and writer of the book ‘The Social Contract’.

→ Montesquieu: A great philosopher. He wrote the book ‘The Spirit of the Laws’.

→ Abbe Sieyes: He was a priest who led National Assembly which was created in 1789. He wrote an influential pamphlet named “What is the Third Estate”.

→ Mirabeau: He also led the National Assembly. He was born in a noble family. He brought out a journal and delivered powerful speeches.

→ Maximilien Robespierre: Political leader of Jacobins who ruled over France from 1793 to 1794.

JAC Class 9 Social Science Notes

JAC Board Class 9th Social Science Important Questions Economics Chapter 4 Food Security in India

I. Objective Type Questions

1. What is meant by food security?
(a) Availability of food to poor people
(b) Availability of food to rich people
(c) Availability of food to all sections of people
(d) None of these
Answer:
(c) Availability of food to all sections of people

2. Which two food crops have been intensively cultivated in the Green Revolution in India,
(a) Wheat and maize
(b) Wheat and rice
(c) Wheat and mustard
(d) None of these
Answer:
(b) Wheat and rice

3. Buffer stock is created to:
(a) Maintain minimum support price
(b) Distribute food grains in the deficit areas
(c) Run the food-for-work programme
(d) All of the above
Answer:
(b) Distribute food grains in the deficit areas

4. Integrated child development service was introduced in:
(a) 1976
(b) 1978
(c) 1985
(d) 1975
Answer:
(d) 1975

5. In which state of India is the famous cooperative AMUL located:
(a) Rajasthan
(b) Gujarat
(c) Haryana
(d) Tamil Nadu
Answer:
(b) Gujarat

II. Very Short Answer Type Questions

Question 1.
What do you mean by Food Security?
Answer:
Food security means availability, accessibility, and affordability of food to all people at all times.

Question 2.
Who are more vulnerable to food insecurity?
Answer:
The poor are more vulnerable to food insecurity.

Question 3.
What are the three dimensions of food security?
Answer:
The three dimensions of food security are:

1. Availability of food,
2. Accessibility of food,
3. Affordability of food.

Question 4.
When did the great famine of Bengal occur in India?
Answer:
1943.

Question 5.
Who were affected the most by the famine of Bengal:
Answer:

1. The agricultural labourers,
2. Fishermen,
3. Transport workers,
4. Casual labourers.

Question 6.
Name any two states of India that have faced acute food scarcity?
Answer:

1. Bihar
2. West Bengal.

Question 7.
Write two dimensions of hunger.
Answer:

1. Chronic hunger,
2. Seasonal hunger.

Question 8.
Who suffers from chronic hunger?
Answer:
Poor people suffer from chronic hunger.

Question 9.
Which type of hunger is prevalent in rural areas?
Answer:
Seasonal hunger is prevalent in rural areas.

Question 10.
What are the two components of food security System of India?
Answer:
The two components of food security system of India are:

1. Buffer Stock,
2.  Public Distribution System.

Question 11.
What is Buffer stock?
Answer:
Buffer stock is the stock of food grains (wheat and rice) procured by the government through the food corporation of India.

Question 12.
What is PDS?
Answer:
Public Distribution System refers to a system through which the food procured by the FCI is distributed among the poor through government-regulated ration shops by using ration cards.

Question 13.
FCI stands for what?
Answer:
FCI stands for Food Corporation of India.

Question 14.
Why are food-for-work schemes security implemented?
Answer:
To ensure food and nutritional to people living below poverty level.

Question 15.
What is the need of fair price shop?
Answer:
To ensure distribution of food grains to even the poorest of poor at lower than market price. ’

Question 16.
Why does the government give subsidy?
Answer:
To make sure that price of essential commodities should be within the reach of the poorest people.

Question 17.
Which social group is the target of the Annapurna Scheme?
Answer:
The Annapurna Scheme is for indigent senior citizens who are not financially sup-ported by anybody.

Question 18.
Name any two food cooperatives in India.
Answer:

1. Amul,
2. Mother Dairy.

III. Short Answer Type Questions

Question 1.
What are the three dimensions of food security?
Or
Explain the dimensions of food security?
Answer:
Food security has the following dimensions:
1. Availability of food:
It means food production within the country, food imports and the previous years stock stocked in government granaries.

2. Accessibility:
It means food is within reach of every person.

3. Affordability:
It means an individual has enough money to buy sufficient, safe and nutritious food to meet his dietary need.

Question 2.
What was declared in World Food Summit, 1995?
Answer:
The World Food Summit, 1995 declared, “Food security at the individual, household, regional, national and global levels exists when all people, at all times, have physical and economic access to sufficient, safe and nutritious food to meet their dietary needs and food preferences for an active and healthy life.”

Question 3.
How is food security affected during a calamity?
Answer:
Following are the ways in which food security gets affected dring a natural calamity.

1. Due to natural calamity, say drought, total product ion of food grains decreases.
2. Low food production leads to shortage of food in the affected areas.
3. Due to shortage of food, prices go up. At such high prices, many people cannot afford to buy food.
4. If such a calamity happens in a very widespread area or is stretched over a longer time period, it may cause a situation of starvation.
5. A massive starvation might take the turn of famine.

Question 4.
What is a famine? How are people affected by a famine?
Answer:
A famine is characterised by widespread deaths due to starvation and epidemics caused by forced use of contaminated water or decaying food and loss of body re-sistance due to weakening from starvation. This occurs due to a severe shortage of . food resulting from crop failure or other calamity, which increases the price of food, making it unaffordable to the weaker sections of the population.

Question 5.
Why is food security needed in India?
Answer:
The poorest section of the society might be food insecure most of the times while persons above the poverty line might also be food insecure when the country faces a national disaster like earthquake, drought, flood, tsunami, famine, etc. Even today, there are places like Kalahandi and Kashipur in Odisha where famine-like conditions have been existing for many years and where some starvation deaths have also been reported.

Starvation deaths have also been reported in Baran district of Rajasthan, Palamau district of Jharkhand and many other remote areas during the recent years.
Therefore, food security is needed in India to ensure food at all times.

Question 6.
Explain the impact of green revolution on food security in India.
Answer:
Following is the impact of green revolution on food security.

1. It made India self sufficient in food production.
2. After green revolution no food scarcity was felt even during adverse weather conditions.
3. With the increase in per hectare crop production as well as extension of agri-cultural activities in disadvantaged areas, it ensured regular food supply.

Question 7.
What is the need of maintaining buffer stock?
Answer:
Maintenance of Buffer stock is required for the following reasons:

1. It ensures continuous supply of commodities to fair-price shops.
2. It is the stock of country which can be used during any calamity.
3. It helps farmers to sell their surplus crop at a reasonable price.

Question 8.
Write a brief note on the national food security Act, 2013.
Answer:
The national food security Act, 2013 provides for food and nutritional security life at affordable prices and enables people to live their life with dignity. Under this act 75 percent of the rural population and 50 percent of the urban population have been categorised as eligible households for food security aid.

Question 9.
What are the three kinds of ration cards?
Answer:
The three kinds of ration cards are as follows:
1. Antyodaya Cards:
Antyodaya cards for the poorest of the poor.

2. BPL Cards:
Below Poverty Line (BPL) cards for those who are below poverty line.

3. APL Cards:
Above Poverty Line (APL) cards for all others.

Question 10.
Explain any two types of PDS.
Answer:
Two types of PDS are as follows:
1. Revamped Public Distribution System (RPDS):
It was introduced in 1992 in 1,700 blocks in the country. The launch of RPDS was to provide the benefits of PDS to remote and backward areas.

2. Targetted Public Distribution System (TPDS): TPDS was introduced in June 1997 to adopt the principle of targetting the “poor in all areas”. It was for the first time that a differential price policy was adopted for poors and non-poor.

Question 11.
List any three characteristics of public distribution system.
Answer:
Following are the three characteristics of public distribution system.

1. It procures grain from FCl and distributes it among the poorer sections of society.
2. There are almost 5.5 lakh shops under this system reaching the farthest comer of country.
3. It distributes food grains through ration card.

Question 12.
How do fair-price shops help food distribution in India?
Answer:
There are more than 5.5 lakh fair-price shops to help food distribution in India. They provide food stuff like wheat, rice, sugar, and kerosene oil on a monthly basis to ration card holders at lower than the market rates. These rates depend on the category of family, whether they are APL, BPL or Antyodaya families. This enables the poorer sections of society to have food security.

Question 13.
what is MSP? How does it help in food security?
Answer:
The Food Corporation of India (FCI) purchases wheat and rice for the government from the farmers of surplus states at pre-announced prices. This price is called minimum support price (MSP). It helps in ensuring food security in several ways:

1. For farmers it gives surety to their crop price and encourages them to grow certain crops.
2. For government, it ensures that they have enough crop to buy for their buffer stock and public welfare programmes.
3. For public, this ensures that prices of a commodity will be stable and within their reach.

IV. Long Answer Type Questions

Question 1.
Describe the measures adopted by the government of India to achieve self-sufficiency in food grains;
Answer:
Following measures were adopted by government of India to achieve self-sufficiency in food grains.
1. India adopted a new strategy in agriculture, which resulted in green revolution especially in the production of wheat and rice.

2. The government of India has made the provision of buffer stock meant to dis-tribute food grains in the deficit areas and among the poorer strata of society at a price lower than the market price. This also helps to resolve the problem of shortage of food during adverse weather conditions or during the periods of calamity.

3. The government has made provision of Public Distribution System (PDS). This is meant for distribution of food grains trough government regulated ration shops (Fair price shops) among the poorer sections of the society.

4. The government has also initiated other food intervention programmes like Integrated Child Development Services (ICDS), Food-for-Work programme (FFW), Mid-Day meals, Antyodaya Anna Yojana (AAY) etc.

Question 2.
Explain two special schemes launched by the government in 2000.
Answer:
In 2000, two special schemes were launched viz. Antyodaya Anna Yojana and the Annapurna Scheme with special target groups of ‘poorest of the poor’ and ‘indigent senior citizens’, respectively. The functioning of these two schemes was linked with the existing network of the PDS.

1. Antyodaya Anna Yojana (AAY):
This scheme was launched in December 2000. Under the scheme one crore of the poorest among the BPL families covered under the Targetted Public Distribution System were identified. Poor families were identified by the respective state rural development departments through a Below Poverty Line (BPL) survey.

Twenty-five kilograms of food grains were made available to each eligible family at a highly subsidised rate of ₹ 2 per kg for wheat and ₹ 3 per kg for rice. This quantity has been enhanced from 25 to 35 kgs with effect from April 2002. The scheme has been further expanded twice to additional 50 lakh BPL families in June 2003 and in August 2004. With this increase, 2 crore families have been covered under the AAY.

2. Annapurna Scheme:
Annapurna Scheme (APS) was introduced in 2000 as Public Distribution Scheme to support indigent senior citizens with 10 kilograms of food grains free of cost.

JAC Class 9 Social Science Important Questions

JAC Board Class 9th Social Science Important Questions Economics Chapter 3 Poverty as a Challenge

I. Objective Type Questions

1. In India, approximately how many people live in poverty?
(a) 30 crore
(b) 27 crore
(c) 20.2 core
(d) 88.3 crore.
Answer:
(b) 27 crore

2. In rural areas, average calories required person per day are
(a) 2100 calories
(b) 3000 calories
(c) 2400 calories
(d) 1700 calories.
Answer:
(c) 2400 calories

3. The average number of calories required per person per day in urban areas of India are
(a) 3000 calories
(b) 2400 calories
(c) 2900 calories
(d) 2100 calories.
Answer:
(d) 2100 calories.

4. Which of the following states of India has the highest poverty ratio?
(a) Odisha
(b) Bihar
(c) Uttar Pradesh
(d) Assam
Answer:
(b) Bihar

5. When was the Rural Employment Generation programme (REGP) launched?
(a) In 1992
(b) In 2005
(c) In 1993
(d) In 1995
Answer:
(d) In 1995

II. Very Short Answer Type Questions

Question 1.
What is Poverty?
Answer:
Poverty refers to a situation in which a person is not able to get the minimum basic necessities of life, e.g. food, clothing, shelter, etc. for his or her sustenance.

Question 2.
What is one of the biggest challenges of independant India?
Answer:
One of the biggest challenges of independent India is to bring millions of its people out of abject poverty.

Question 3.
What did Mahatma Gandhi always emphasize?
Answer:
Mahatma Gandhi always emphasized that India would be truly independent only when the poorest of its people become free of human suffering.

Question 4.
Name some social indicators through which poverty is seen.
Answer:

1. Illiteracy,
2. Lack of general resistance due to malnutrition,
3. Lack of access to healthcare,
4. Lack of job opportunities,
5. Lack of access to safe drinking water, sanitation, etc.

Question 5.
Define vulnerability to poverty?
Answer:
Vulnerability to poverty is a measure which describes the greater probability of certain communities or individuals becoming or remaining poor in the coming years.

Question 6.
What is poverty line?
Answer:
Poverty line is an imaginary line used by any country to determine its poverty level.

Question 7.
When is a person considered poor?
Answer:
A person is considered poor if his or her income or consumption level falls below a given “minimum level” necessary to fulfil the basic needs.

Question 8.
What is considered when determining the poverty line in India?
Answer:
While determining the poverty line in India a minimum level of food requirement, clothing, footwear, fuel and light, educational and medical requirement, etc. are determined for subsistence.

Question 9.
Why is the calorie requirement of people higher in rural areas as compared to urban areas?
Answer:
The calorie requirement of people in rural areas in higher because the rural people are engaged in more physical labour than people in urban areas.

Question 10.
What is the accepted average calorie requirement in India?
Answer:
The accepted average calorie requirement in India is 2400 calories per person per day in rural areas and 2100 calories per person per day in urban areas.

Question 11.
Which agency conducts the periodical sample surveys for estimating the poverty line in India?
Or
Which organization in India carries the periodical survey for the estimation of poverty?
Answer:
Poverty line is estimated periodically by the National Sample Survey Organisation (NSSO) by conducting sample surveys, generally after every 5 years.

Question 12.
What is the full form of NSSO?
Answer:
The full form of NSSO is National Sample Survey Organisation.

Question 13.
Which are the most Vulnerable groups of poverty?
Answer:
Scheduled castes, scheduled tribes, rural agricultural labour households and urban casual labour households are vulnerable to poverty.

Question 14.
What was the percentage of people living below the poverty line in 2011-12 in India.
Answer:
In 2011-12, 22 per cent of the population was living below the poverty line in India.

Question 15.
Name the two poorest states in India.
Answer:

1. Bihar
2. Odisha.

Question 16.
Name five states having the highest percentage of people below poverty line.
Answer:

1. Bihar,
2. Odisha,
3. Assam,
4. Madhya Pradesh,
5. Uttar Pradesh.

Question 17.
Name five states having the least percentage of people below poverty line.
Answer:

1. Kerala,
2. Himachal Pradesh,
3. Punjab,
4. Andhra Pradesh,
5. Haryana.

Question 18.
How have states like Punjab and Haryana been successful in reducing poverty?
Answer:
States like Punjab and Haryama have been successful in reducing poverty with high agricultural growth rates.

Question 19.
By what method has the state of West Bergal reduced poverty?
Answer:
Proper implementation of land reforms have helped to reduce poverty in West Bengal.

Question 20.
Why has Kerala succeeded in reducing poverty?
Answer:
By focusing more on human resource development.

Question 21.
Which standard is used by the World Bank for the estimation of poverty line?
Answer:
The World Bank uses a uniform standard for poverty line, which is the minimum income of the equivalent of $ 1.90 per person per day.

Question 22.
Why is World Bank important in estimating poverty line?
Answer:
Different countries have different poverty lines as per their existing level of devel¬opment. The World Bank compares countries by presenting a uniform standard for poverty line which is acceptable to all countries.

Question 23.
List some countries having the highest percentage of people below poverty line.
Answer:

1. Nigeria,
2. Bangladesh,
3. India,
4. Pakistan,
5. China.

Question 24.
Name any two causes of poverty in India.
Answer:

1. Low level of economic development during British rule,
2. Inefficient administration.

Question 25.
The current anti-poverty strategy of the government is based broadly on two planks Mention these.
Or
The present anti-poverty strategy of the Government of India is broadly based on what factors?
Answer:
The present anti-poverty strategy of the Government of India is broadly based on two planks :

1. Promotion of economic growth
2. targetted anti-poverty programmes.

Question 26.
Name any two anti-poverty programs.
Answer:

1. Prime Minister Rozgar Yojana (PMRY)
2. Swarna Jayanti Gram Swarozgar Yojana (SGSY).

Question 27.
Write the full form of MNREGA.
Answer:
Full form of MNREGA is Mahatma Gandhi National Rural Employment Guarantee Act.

Question 28.
Which act guarantees minimum 100 days employment per person per year in rural areas?
Answer:
Mahatma Gandhi National Rural Employment Guarantee Act.

Question 29.
When was MNREGA passed?
Answer:
MNREGA was passed in September 2005.

Question 30.
Which scheme has been started to create self-employment opportunities for educated unemployed youth in rural areas?
Answer:
Prime Minister Rozgar Yojana (PMRY).

III. Short Answer Type Questions

Question 1.
How is poverty seen by social scientists?
Answer:
Poverty as seen by social scientists in the following way:

1. Poverty relates to the level of income and consumption.
2. Apart from this, poverty is looked at through other social indicators like illiteracy level, lack of general resistance due to malnutrition, lack of access to healthcare, lack of job opportunities, lack of access to safe drinking weather, sanitation, etc.

Question 2.
What is social exclusion? Explain with an example.
Answer:
Social exclusion is a process through which individuals or groups are excluded from facilities, benefits and opportunities that others enjoy.
Example: The caste system in India in which people belonging to certain castes are” excluded from equal opportunities.

Question 3.
Does poverty line vary with time and place?
Answer:
Yes, poverty line may vary with time and place. A person is considered poor, if his or her income or consumption level falls below a given minimum level that is necessary to satisfy basic needs. What is necessary to satisfy basic needs is different at different times and in different countries. Therefore, poverty line may vary with time and place.

Example: A person not having a car in the USA may be considered poor but in India owning a car is still considered a luxury.

Question 4.
What are the calories and rupees fixed for rural and urban areas for measuring the poverty line?
Answer:
The poverty line fixed for the rural and urban areas in India in the year 2012 was ₹ 186 and ₹ 100 per person per day for urban and rural areas, respectively. It is higher in urban areas because of higher prices of many essential commodities in urban centres. The accepted average calories requirement in India is 2400 calories in rural areas and 2100 in urban areas. It is high in the rural areas because of more physical work done by the rural people.

Question 5.
Briefly explain the principle measures taken in Punjab, Haryana, Kerala and West Bengal to reduce poverty.
Answer:
1. Punjab and Haryana:
These states have succeeded in reducing poverty due to high agricultural growth.

2. Kerala:
It has succeeded in reducing poverty by investing on human resource development. Kerala has the highest literacy rate in the country.

3. West Bengal:
In West Bengal land reform measures have helped in reducing poverty.

Question 6.
How are sociocultural and economic factors responsible for poverty in India?
Answer:
1. Sociocultural causes of poverty: In order to fulfil social obligations and observe religious ceremonies, most people in India including the very poor spend a lot of money. This takes them back to poverty.

2. Economic Causes of Poverty: Small farmers need money to buy agricultural inputs like seeds, fertilizer, pesticides etc. Since poor people hardly have any savings, they borrow money and when there is crop failure they become indebted. This takes pusher them into poverty.

Question 7.
Which circumstances compelled the government to start targetted anti-poverty programmes?
Answer:
Following circumstances compelled the government to start targetted anti-poverty programmes:

1. Since the poor are not able of take advantage of the opportunities created for them, the growth in the agriculture sector is much below expectations.
2. In agriculture sector, a large number of poor people are unemployed for most part of the year, therefore, the need for targetted anti-poverty programmes was felt.

Question 8.
Give the main feature of Rural Employment Generation Programme.
Answer:
Rural Employment Generation Programme (REGP) was launched in 1995. The main features of REGP are:

1. Creation of self-employment opportunities in rural areas and towns.
2. Creation of 25 lakh near Jobs under the tenth five-year plan.

Question 9.
Why were the poverty alleviation programmes not successful in most parts of India?
Answer:
The poverty alleviation programmes were not successful in most parts of India for these reasons:

1. Lack of proper implementation and right targetting.
2. There has been a lot of overlapping of schemes.
3. Every year a huge number is added to the population pool of the country. This makes schemes ineffective.
4. Despite good intentions, the benefits of these schemes do not fully reach to the deserving poor.

IV. Long Answer Type Questions

Question 1.
Analyse poverty on the basis of social exclusion and vulnerability.
Answer:
The analysis of poverty is based on social exclusion and vulnerability as follows:
1. Social exclusion:
Social exclusion means living in a poor surroundings with poor people, excluded from enjoying social equality of better off people in better surrounding. In India, Caste system is based on social exclusion. People belonging to certain castes were prevented from enjoying equal facilities, benefits and opportunities. This caused more poverty than that caused by lower income.

2. Vulnerability:
Vulnerability to poverty is a measure, which describes the greater probability of certain communities becoming or remaining poor in the coming years, e.g. members of a backward caste or individuals like widows, physically handicapped persons and so on. Vulnerability is determined by various options available to different communities in terms of assets, education, job, healthcare, etc. and analyses their ability to face various risks like natural disasters.

Question 2.
Explain the main features of the global poverty scenario.
Answer:
Following are the main features of global poverty scenario:

1. The priducortion of people in developing countries living in extreme economic poverty defined by the World Bank as living on less than $ 1.90 per day has come down from from 36 per cent in 1990 to 10 per cent in 2015.
2. Poverty J.eclined substantially in China and South-East Asian countries as a result ox rapid economic growth and massive investment in human resource development.
3. In the countries of South Asia: India, Pakistan, Sri Lanka, Nepal, Bangladesh, Bhutan, the decline has also been rapid-34 per cent in 2005 to 16.2 per cent in 2013.
4. In sub-Saharan Africa, poverty in fact declined from 51 per cent in 2005 to 41 per cent in 2005.
5. In Latin America, the ratio of poverty has also declined from 10 per cent in 2005 to 4 per cent in 2015.
6. Poverty has also resurfaced in some of the former socialist countries like Russia, where officially it was non-existent earlier.

Question 3.
“There is a strong link between economic growth and poverty reduction” Explain the statement.
Answer:
It is clear that there is a strong link between economic growth and poverty reduction. The higher growth rates have helped significantly in the reduction of poverty. In the 1980s, India’s economic growth was one of the fastest in the world. The growth rate rose from the average of about 3.5% a year in the 1970s to about 6 percent during the 1980s and 1990s.

Economic growth widens opportunities and provides the resources needed to invest in human development. This also encourages people to send their children, including the girl child, to schools in the hope of getting better economic returns from investing in education. Thus, it can be concluded that there is a strong link between economic growth and poverty reduction.

Question 4.
Describe any four poverty alleviation programmes currently being implemented in India.
Or
Describe any four anti-poverty programmes.
Answer:
Following are the four major poverty alleviation programmes implemented by the
Government of India.
1. Mahatma Gandhi National Rural Employment Guarantee Act (MNREGA)

1. This act was passed in September 2005.
2. This act provides 100 days assured employment every year to every rural household.
3. It also aims at sustainable development by addressing the cause of drought, deforestation and soil erosion.
4. One-third of the proposed jobs have been reserved for women.
5. The scheme provided employment to 220 crore mandays of employment to 4.78 crore households.
6. The range of wage rate for different states and union territories lies in between ₹ 281 per day to ₹ 168 per day.

2. Prime Minister Rozgar Yojana (PMRY) .

1. This programme was launched in 1993.
2. It is aimed at providing self-employment opportunities to educated unemployed youth in the rural and small towns.
3. Under this programme, scheduled banks provide loans at a lower interest rate to start small businesses and set up industries.

3. Rural Employment Generation Programme (REGP)

1. This programme was launched in 1995.
2. The aim of the programme is to provide self-employment opportunities to educated unemployed youth in the rural areas and small towns.
3. A target for creating 25 Lakh new Jobs has been set for this programme under the Tenth five-year plan.

4. Pradhan Mantri Gramodaya Yojana (PMGY)

1. This programme was launched in 2000.
2. Under this programme, the Central Government provides additional assistance to the State Government for improving basic services in the village.
3. The major basic services covered under this programme are primary health, primary education, rural shelter, rural drinking water and rural electrification.

Question 5.
Review the future challenges in the context of poverty in India.
Answer:
Despite many plans to reduce poverty in India, the main reason for its partial success is lack of proper implementation and proper targetting. The benefits of these schemes could not be passed on to the poor, even after a good deal of sincere efforts have been made. Poverty has come down due to the implementation of many schemes, but the biggest problem of completely eradicating poverty is still standing before us.

Wide disparities in poverty are visible between rural and urban areas and among . different states. Certain social and economic groups are more vulnerable to poverty. Further, poverty should include not only the factor of adequate amount of food, but also other factors like education, healthcare, shelter, job security, gender equality, dignity and so on. These give as the true concept of human poverty.Poverty reduction is expected to be more effective in next 10-15 years.

In addition to anti-poverty measures, government should focus on the following to reduce poverty:

1. Higher economic growth
2. Universal free elementary education,
3. Decrease in population
4. Empowerment of women and weaker sections.

The official definition of poverty, however, captures only a limited part of what poverty really means to people. It is about a “minimum subsistence level of living rather than a “reasonable” level of living. Many scholars advocate that we must broaden the concept into human poverty. A large number of people many have been able to feed themselves.

But do they have education? or shelter? or healthcare? or job security? or self confidence? Are they free from caste and gender discrimination? Is the practice of child labour still common? World wide experience shows that with development, the definition of poverty is always a moving target.

Hopefully, we will be able to provide the minimum “necessary” in terms of only income to all people by the end of the next decade. But the target will move on for many of the bigger challenges that still remain: Providing healthcare, education and job security for all, and achieving gender equality, and dignity for the poor.

JAC Class 9 Social Science Important Questions

Jharkhand Board JAC Class 10 Sanskrit Solutions Shemushi Chapter 3 व्यायामः सर्वदा पथ्यः Textbook Exercise Questions and Answers.

JAC Board Class 10th Sanskrit Solutions Shemushi Chapter 3 व्यायामः सर्वदा पथ्यः

JAC Class 10th Sanskrit व्यायामः सर्वदा पथ्यः Textbook Questions and Answers

प्रश्न 1.
एकपदेन उत्तरं लिखत- (एक शब्द में उत्तर लिखिए-)
(क) परमम् आरोग्यं कस्मात् उपजायते ? (परमारोग्य किससे होता है?)
(ख) कस्य मासं स्थिरीभवति? (किसका मांस स्थिर होता है?)
(ग) सदा कः पथ्य? (सदा क्या अनुकूल है?)
(घ) कै पुभिः सर्वेषु ऋतुषु व्यायामः कर्त्तव्यः? (किन लोगों को सभी ऋतुओं में व्यायाम करना चाहिए?)
(ङ) व्यायामस्विन्नगात्रस्य समीपं के न उपसर्पन्ति? (व्यायाम से उत्पन्न पसीना वाले शरीर के समीप कौन नहीं आते?)
उत्तरम्म् :
(क) व्यायामात् (व्यायाम से)।
(ख) व्यायामाभिरतस्य (व्यायाम में लगे हुए का)
(ग) व्यायामः (कसरत)।
(घ) पुम्भिरात्महितैषिभिः (अपना हित चाहने वालों को)।
(ङ) व्याधयः (रोग)।

प्रश्न 2.
अधोलिखितानां प्रश्नानाम् उत्तराणि संस्कृतभाषया लिखत –
(निम्नलिखित प्रश्नों के उत्तर संस्कृत भाषा में लिखिए)
(क) कीदृशं कर्म व्यायामसंज्ञितं कथ्यते ?
(कैसा कर्म व्यायाम नाम से पुकारा जाता है ?)
उत्तरम् :
शरीरायासजननं कर्म व्यायामसंज्ञितं कथ्यते।
(शरीर को थकाने वाला कर्म व्यायाम कहलाता है।)

(ख) व्यायामात् किं किमुपजायते ?
(व्यायाम से क्या पैदा होता है ?)
उत्तरम् :
शरीरोपचयः, कान्तिः, गात्राणां सुविभक्तता, दीप्ताग्नित्वम्, अनालस्यं, स्थिरत्वं, लाघवं, मृजा, श्रमक्लमपिपासा, उष्ण-शीतादीनां सहिष्णुता परमं च आरोग्यम् व्यायामात् उपजायते।
(शरीर की वृद्धि, चमक, सुडौलता, दीप्ति, पाचनशक्ति की वृद्धि, आलस्यहीनता, स्थिरता, फुर्ती, स्वच्छता, परिश्रम, थकान, प्यास, गर्मी-सर्दी आदि को सहन करने की क्षमता और महान स्वस्थता व्यायाम से उत्पन्न होती है। )

(ग) जरा कस्य सकाशं सहसा न समधिरोहति ?
(बुढ़ापा अचानक किस पर हावी नहीं होता ?)
उत्तरम् :
व्यायामाभिरतस्य सकाशं जरा सहसा न समधिरोहति।
(व्यायाम में संलग्न के पास बुढ़ापा अचानक नहीं आता।)

(घ) कस्य विरुद्धमपि भोजनं परिपच्यते?
(किसका विरुद्ध भोजन भी पचता है?)
उत्तरम् :
व्यायामम् कुर्वतः नित्यं विरुद्धमपि भोजनं परिपच्यते।
(व्यायाम करते हुए का विरुद्ध भोजन भी पचता है।)

(ङ) कियता बलेन व्यायामः कर्त्तव्यः ?
(कितने बल से व्यायाम करना चाहिए ?)
उत्तरम् :
अर्धनबलेन व्यायामः कर्त्तव्यः।
(आधे बल से व्यायाम करना चाहिए।)

(च) अर्धबलस्य लक्षणं किम् ?
(आधे बल का लक्षण क्या है ?)
उत्तरम् :
यदा व्यायामं कुर्वतः हृदिस्थानस्थितो वायुः वक्त्रं प्रपद्यते, तद् बलार्धस्य लक्षणम्।
(जब व्यायाम करने वाले के हृदय में स्थित वायु मुख की ओर प्रवृत्त हो जाती है वह आधे बल का लक्षण है।)

प्रश्न 3.
उदाहरणमनुसृत्य कोष्ठकगतेषु पदेषु तृतीयाविभक्तिं प्रयुज्य रिक्तस्थानानि पूरयत –
(उदाहरण के अनुसार कोष्ठक में दिए शब्दों में तृतीया विभक्ति लगाकर प्रयोग कीजिए)
यथा – व्यायामः ……….. हीनमपि सुदर्शनं करोति। (गुण)
व्यायामः गुणैः हीनमपि सुदर्शनं करोति।

(क) ……….. व्यायामः कर्त्तव्यः। (बलस्यार्ध)
उत्तरम् :
बलस्यार्धेन व्यायामः कर्त्तव्यः।
(आधे बल से व्यायाम करना चाहिए।)

(ख) ……….. सदृशं किञ्चित् स्थौल्यापकर्षणं नास्ति। (व्यायाम)।
उत्तरम् :
व्यायामेन सदृशं किञ्चित् स्थौल्यापकर्षणं नास्ति।
(व्यायाम के समान कुछ भी मोटापा दूर करने वाला नहीं है।)

(ग) ……….. विना जीवनं नास्ति। (विद्या)
उत्तरम् :
विद्यया विना जीवनं नास्ति।
(विद्या के बिना जीवन नहीं है।)

(घ) सः ……….. खञ्जः अस्ति। (चरण)
उत्तरम् :
सः चरणेन खञ्जः अस्ति।
(वह पैर से लँगडा है।)

(ङ) सूपकारः ……….. भोजनं जिघ्रति। (नासिका)
उत्तरम् :
सूपकारः नासिकया भोजनं जिघ्रति।
(रसोइया नाक से भोजन सूंघता है।)

प्रश्न 4.
(अ) स्थूलपदमाधृत्य प्रश्ननिर्माणं कुरुत (मोटे शब्दों के आधार पर प्रश्नों का निर्माण कीजिए)
(क) शरीरस्य आयासजननं कर्म व्यायामः इति कथ्यते। (शरीर को थकाने वाला कर्म व्यायाम कहलाता है।)
(ख) अरयः व्यायामिनं न अर्दयन्ति। (शत्रु व्यायामी को नहीं कुचलते।)
(ग) आत्महितैषिभिः सर्वदा व्यायामः कर्त्तव्यः। (आत्महितैषियों द्वारा हमेशा व्यायाम करना चाहिए।)
(घ) व्यायाम कुर्वतः विरुद्धं भोजनम् अपि परिपच्यते। (व्यायाम करने वाले को विपरीत भोजन भी पच जाता है।)
(ङ) गात्राणां सुविभक्तता व्यायामेन संभवति। (अंगों का अच्छा विभाजन व्यायाम से संभव होता है।)
उत्तरम् :
(क) कस्य आयासजननं कर्म व्यायाम इति कथ्यते ? (किसको थकाने वाला कर्म व्यायाम कहलाता है ?)
(ख) के व्यायामिनं न अर्दयन्ति ? (कौन व्यायामी को नहीं कुचलते ?)
(ग) कैः सर्वदा व्यायामः कर्त्तव्यः ? (किन्हें हमेशा व्यायाम करना चाहिए ?)
(घ) व्यायाम कुर्वतः कीदृशं भोजनम् अपि परिपच्यते ? (व्यायाम करने वाले को कैसा भोजन भी पच जाता है ?)
(ङ) केषां सुविभक्तता व्यायामेन सम्भवति ? (किनकी सुविभक्तता व्यायाम से सम्भव होती है ?)

(आ) षष्ठश्लोकस्य भावमाश्रित्य रिक्तस्थानानि पूरयत – (छठे श्लोक के भाव पर आश्रित रिक्तस्थानों की पूर्ति कीजिए)
यथा : ………. समीपे उरगाः न ………. एवमेव व्यायामिनः जनस्य समीपं ……………………. न गच्छन्ति। व्यायामः वयोरूपगुणहीनम् अपि जनम् ………. करोति।
उत्तरम् :
यथा – वैनतेयसमीपे उरगाः न उपसर्पन्ति एवमेव व्यायामिनः जनस्य समीपे व्याधयो न गच्छन्ति। व्यायामः वयोरूपगुणहीनम् अपि जनम् सुदर्शनम् करोति। (जैसे गरुड़ के समीप साँप नहीं आते वैसे ही व्यायामी व्यक्ति के समीप रोग नहीं आते हैं। व्यायाम उम्र, रूप और गुणहीन को भी सुन्दर कर (बना) देता है।)

प्रश्न 5.
(अ) ‘व्यायामस्य लाभाः’ इति विषयम् अधिकृत्य पञ्चवाक्येषु ‘संस्कृतभाषया’ एकम् अनुच्छेदं लिखत।
(‘व्यायाम के लाभ’ इस विषय पर पाँच वाक्यों में संस्कृत भाषा में एक अनुच्छेद लिखिए।)
उत्तरम् :
व्यायामात् शरीरोपचयः, कान्तिः, अङ्गानां सुविभाजनं, दीप्तिः, जठराग्निवृद्धिः, आलस्यहीनता, स्थिरता, स्फूर्तिः, स्वच्छता, पिपासा, उष्ण-शीतादीनां सहिष्णुता आरोग्यं च उपजायते। व्यायामः मनुष्यस्य पीनत्वं दूरं करोति, जरागमनं अवरोधयति। व्यायामिनः व्याधयो नोपसर्पन्ति। तस्य गरिष्ठभोजनमपि पचति। व्यायामेन मनुष्यः गुणहीनः अपि सुदर्शनः भवति। (व्यायाम से शरीर की वृद्धि, चमक, अंग-विभाजन, भूख-वृद्धि, आलस्यहीनता, स्थिरता, फुर्ती, स्वच्छता, प्यास, सर्दी-गर्मी को सहन करने की सामर्थ्य और नीरोगता होती है। व्यायाम मनुष्य के मोटापे को दूर करता है, बुढ़ापे को आने से रोकता है। व्यायाम करने वाले के पास रोग नहीं आते। उसको गरिष्ठ भोजन भी पचता है। व्यायाम से मनुष्य गुणहीन होते हुए भी सुन्दर होता है।)

(आ) यथा निर्देशमुत्तरत (निर्देशानुसार उत्तर दीजिए)
(क) ‘तत्कृत्वा तु सुखं देहम्’ अत्र विशेषण पदं किम्?
(‘उसे करके सुखी शरीर होता है’ यहाँ विशेषण पद क्या है?)
(ख) ‘व्याधयोः नोपसर्पन्ति वैनतेयमिवोरगाः।’ अस्मिन् वाक्ये क्रियापदं किम्?’
(‘रोग ऐसे समीप नहीं आते जैसे गरुड़ के पास सर्प’ इस वाक्य में क्रियापद क्या है?)
(ग) ‘पुम्भिरात्महितैषिभिः’ अत्र ‘पुरुषैः’ इत्यर्थे किं पदं प्रयुक्तम्?
(‘हित चाहने वाले पुरुषों द्वारा’ यहाँ ‘पुरुषैः’ अर्थ में कौन-सा पद प्रयोग हुआ है?)
(घ) ‘दीप्ताग्नित्वमनालस्यं स्थिरत्वं लाघं मृजा’ इति वाक्यात् ‘गौरवम्’ इति पदस्य विपरीतार्थकं पदं चित्वा लिखत।’
(जठराग्निदीपन, अनालस्य, स्थिरता, हलकापन, सफाई इस वाक्य से गौरवम् पद का विलोम चुनकर लिखिए)
(ङ) ‘न चास्ति सदृशं तेन किञ्चित् स्थौल्यापकर्षणम्’ अस्मिन् वाक्ये ‘तेन’ इति सर्वनाम पदं कस्मै प्रयुक्तम्?
(‘उसके समान मुटापे को दूर करने वाला और कुछ नहीं’ इस वाक्य में ‘तेन’ सर्वनाम पद किसके लिए प्रयोग हुआ है?)
उत्तराणि :
(क) सुखम् (देहं का विशेषण)
(ख) उपसर्पन्ति (पास आती है)
(ग) पुम्भिः (पुमान् शब्द का तृतीया विभक्ति बहुवचन)
(घ) लाघवम् (हल्कापन)
(ङ) व्यायामेन (व्यायाम के समान)।

प्रश्न 6.
(अ) निम्नलिखितानाम् अव्ययानां रिक्तस्थानेषु प्रयोगं कुरुत – (निम्नलिखित अव्ययों को रिक्तस्थानों में प्रयोग कीजिए)
सहसा, अपि, सदृशं, सर्वदा, यदा, सदा, अन्यथा।
(क) …………. व्यायामः कर्त्तव्यः।
(ख) ………….. मनुष्यः सम्यक्पे ण व्यायामं करोति तदा सः ………. स्वस्थः तिष्ठति।
(ग) व्यायामेन असुन्दराः ………. सुन्दराः भवन्ति।
(घ) व्यायामिनः जनस्य सकाशं वार्धक्यं ………. नायाति।
(ङ) व्यायामेन ………. किञ्चित् स्थौल्यापकर्षणं नास्ति।
(च) व्यायाम समीक्ष्य एव कर्तव्यम् ………. व्याधयः आयान्ति।
उत्तरम् :
(क) सदा
(ख) यदा, सर्वदा
(ग) अपि
(घ) सहसा
(ङ) सदृशम्
(च) अन्यथा।

(आ) उदाहरणमनुसृत्य वाच्यपरिवर्तनं कुरुत (उदाहरण के अनुसार वाच्य परिवर्तन कीजिए)

प्रश्न 7.
अधोलिखितेषु तद्धितपदेषु प्रकृति/प्रत्ययं च पृथक् कृत्वा लिखत –
(निम्नलिखित तद्धित पदों में प्रकृति/प्रत्यय अलग-अलग करके लिखिए) –
उत्तरम् :

योग्यताविस्तार यह पाठ आयुर्वेद के प्रसिद्ध ग्रन्थ ‘सुश्रुतसंहिता’ के चिकित्सा स्थान में वर्णित 24वें अध्याय से संकलित है। इसमें आचार्य सुश्रुत ने व्यायाम की परिभाषा बताते हुए उससे होने वाले लाभों की चर्चा की है। शरीर में सुगठन, कान्ति, स्फूर्ति, सहिष्णुता, नीरोगता आदि व्यायाम के प्रमुख लाभ है।

(क) सुश्रुत: आयुर्वेदस्य, सुश्रुतसंहिताश इत्याख्यस्य ग्रन्थस्य रचयिता। अस्मिन् ग्रन्थे शल्यचिकित्सायाः प्राधान्यमस्ति। सुश्रुतः शल्याशास्त्रज्ञय दिवोदासस्य शिष्यः आसीत्। दिवोदासः सुश्रुतं वाराणस्याम् आयुर्वेदम् अपाठयत्। सुश्रुतः दिवोदासस्य उपदेशान् स्वग्रन्थेऽलिखत्।

(ख) उपलब्धासु आयुर्वेदीय-संहितासु ‘सुश्रुतसंहिता’ सर्वश्रेष्ठः शल्यचिकित्साप्रधानो ग्रन्थः। अस्मिन् ग्रन्थे 120 अध्यायेषु क्रमेण सुत्रस्थाने मौलिकसिद्धान्तानां शल्यकर्मोपयोगि-यन्त्रादीनां, निदानस्थाने प्रमुखाणां रोगाणां, शरीरस्थाने शरीरशास्त्रस्य चिकित्सास्थाने, शल्यचिकित्साया: कल्पस्थाने च विषाणां प्रकरणानि वर्णितानि। अस्य उत्तरमतंत्रे 66 अध्यायाः सन्ति।

(ग) वैनतेयमिवोरगा: – कश्यप ऋषि की दो पत्नियाँ थीं-कट्ठ और विनता। विनता का पुत्र गरुड़ और कट्ठ का पुत्र सर्प थे। विनता का पुत्र होने के कारण गरुड़ को वैनतेय कहा जाता है। (विनतायाः अयम् वैनतेयः ढक (एय) प्रत्यये कृते)। गरुड़ सर्प से अधिक ताकतवर होता है, भयवश साँप गरुड़ के पास जाने का साहस नहीं करता। यहाँ व्यायाम करने वाले मनुष्य की तुलना गरुड़ से तथा व्याधियों की तुलना साँप से की गई है। जिस प्रकार गरुड़ के समक्ष साँप नहीं जाते। उसी प्रकार व्यायाम करने वाले व्यक्ति के पास रोग नहीं फटकते।

भाषिकविस्तार: गुणवाचक शब्दों से भाव अर्थ में ष्यज् अर्थात् य प्रत्यय लगाकर भाववाची पदों का निर्माण किया जाता है। शब्द के प्रथम स्वर में वृद्धि होती है और अन्तिम अ का लोप होता है।

(क) शूरस्य भावः शौर्यम् – शूर + ष्यज्
(ख) सुन्दरस्य भावः सौन्दर्यम् – सुन्दर + ष्यज्
(ग) सुखंस्य भावः सौख्यम् – सुख + ष्य
(घ) विदुषः भावः वैदुष्यम् – विद्वस् + ष्यज्
(ड) मधुरस्य भावः माधुर्यम् – मधुर + ष्यज्
(च) स्थूलस्य भावः स्थौल्यम् – स्थूल + ष्यज्
(छ) अरोगस्य भावः आरोग्यम् – अरोग + ष्यज्
(ज) सहितस्य भावः साहित्यम् – सहित + ष्यज्

थाल्-प्रत्यय – ‘प्रकार’ अर्थ में थाल प्रत्यय का प्रयोग होता है।
जैसे – तेन प्रकारेण – तथा
येन प्रकारेण – यथा
अन्येन प्रकारेण – अन्यथा
सर्वप्रकारेण – सर्वथा
उभयकारेण – उभयथा

भावविस्तार:

(क) शरीरमाद्यं खलु धर्मसाधनम्।
(ख) लाघवं कर्मसामर्थ्य स्थैर्य क्लेशसहिष्णुता।
दोषक्षयोऽग्निवृद्धिश्च व्यायामदुपजायते।

(ग) यथा शरीरस्य रक्षायै उचितं भोजनम्, उचितश्च व्यवहारः आवश्यकोऽस्ति तथैव शरीरस्य स्वास्थाय व्यायामः अपि आवश्यक।

(घ) युक्ताहारविहारस्य युक्तचेष्टस्य कर्मसु।
युक्तस्वप्नावबोधस्य योगो भवति दुःखहा।

(ड) पक्षिण: आकाशे उड्डीयन्तं तेषाम् उड़यनमेव तेषां व्यायामः। पशवोऽपि इतस्तत: पलायन्ते पलायनमेव तेषां व्यायामः।
शैशवे शिशुः स्वहस्तपादौ चालयति, अयमेव तस्य व्यामामः।
वि + आ + यम् धातोः घञ् प्रत्ययात् निष्पन्नः ‘व्यायाम’ शब्दः विस्तारस्य विकासस्य च वाचकः। यतो हि व्यायामेन अङ्गानां विकासः भवति। अतः सुखपूर्वकं जीवनं यापयितुं मनुष्यैः नित्यं व्यायामः करणीयः।

JAC Class 10th Sanskrit व्यायामः सर्वदा पथ्यः Important Questions and Answers

शब्दार्थ चयनम् –

अधोलिखित वाक्येषु रेखांकित पदानां प्रसङ्गानुकूलम् उचितार्थ चित्वा लिखत –

प्रश्न 1.
शरीरायासजननं कर्म व्यायामसंज्ञितम्।
(अ) कार्यम्
(ब) व्यायाम
(स) संज्ञितम्
(द) सुखम्
उत्तर :
(अ) कार्यम्

प्रश्न 2.
शरीरोपचयः कान्तिर्गात्राणां सुविभक्तता –
(अ) सुविभक्तता
(ब) दीप्ताग्नित्वम्
(स) अङ्गगानाम्
(द) लाघवं
उत्तर :
(स) अङ्गगानाम्

प्रश्न 3.
श्रमक्लमपिपासोष्ण-शीतादीनां सहिष्णुता –
(अ) सहत्वं
(ब) श्रमक्लमः
(स) सहिष्णुता
(द) उपजायते
उत्तर :
(अ) सहत्वं

प्रश्न 4.
न चास्ति सदृशं तेन किञ्चित्स्थौल्यापकर्षणम् –
(अ) सदृशं
(ब) तुल्यं
(स) व्यायामिनम्
(द) मर्त्यम्
उत्तर :
(ब) तुल्यं

प्रश्न 5.
व्याधयो नोपसर्पन्ति वैनतेयमिवोरगा: –
(अ) च पद्भ्याम्
(ब) वैनतेयम्
(स) कुर्यात्
(द) समीपं न आयान्ति
उत्तर :
(द) समीपं न आयान्ति

प्रश्न 6.
व्यायामं कुर्वतो नित्यं विरुद्धमपि भोजनम् –
(अ) अशनम्
(ब) कुर्वतः
(स) विरुद्धम्
(द) परिपच्यते
उत्तर :
(अ) अशनम्

प्रश्न 7.
व्यायामो हि सदा पथ्यो बलिनां स्निग्धभोजिनाम् –
(अ) बलिनाम्
(ब) बलवतां
(स) व्यायामो हि
(द) स्मृतः
उत्तर :
(ब) बलवतां

प्रश्न 8.
बलस्यार्धन कर्त्तव्यो व्यायामो हन्त्यतोऽन्यथा –
(अ) ऋतुषु
(ब) नाशयति
(स) पुम्भिः
(द) अतोऽन्यथाः
उत्तर :
(ब) नाशयति

प्रश्न 9.
व्यायामं कुर्वतो जन्तोस्तबलार्धस्य लक्षणम् –
(अ) प्रपद्यते
(ब) वक्त्रम्
(स) संकेतम्
(द) लक्षणम्
उत्तर :
(स) संकेतम्

प्रश्न 10.
समीक्ष्य कुर्याद् व्यायाममन्यथा रोगमाप्नुयात्
(अ) च देश
(ब) अशनानि
(स) परीक्ष्य
(घ) कुर्यात्
उत्तर :
(स) परीक्ष्य

संस्कृतमाध्यमेन प्रश्नोत्तराणि –

एकपदेन उत्तरत (एक शब्द में उत्तर दीजिए)

प्रश्न 1.
देहं कथं विमृद्नीयात्?
(देह की कैसे मालिश करनी चाहिए ?)
उत्तरम् :
सुखम् (सुखपूर्वक)।

प्रश्न 2.
कीदृशं कर्म व्यायामसंज्ञितम् ?
(कैसा कर्म व्यायाम संज्ञक है ?)
उत्तरम् :
शरीरायासजननम् (शरीर को थकाने वाला)।

प्रश्न 3.
व्यायामेन गात्राणां का स्थितिः भवति ?
(व्यायाम से अंगों की क्या स्थिति होती है ?)
उत्तरम् :
सुविभक्तता (सम्यक् विभाजन)।

प्रश्न 4.
अग्नि: केन प्रदीप्तः भवति ?
(जठराग्नि किससे प्रदीप्त होती है ?)
उत्तरम् :
व्यायामेन (व्यायाम से)।

प्रश्न 5.
लाघवं केन भवति ?
(हलकापन किससे होता है ?)
उत्तरम् :
व्यायामेन (व्यायाम से)।

प्रश्न 6.
शीतोष्णसहिष्णुता कस्मात् उपजायते?
(सर्दी-गर्मी की सहन-क्षमता किससे होती है ?)
उत्तरम् :
व्यायामेन (व्यायाम से)।

प्रश्न 7.
व्यायामात् कीदृशम् आरोग्यम् उपजायते ?
(व्यायाम से कैसा आरोग्य पैदा होता है ?)
उत्तरम् :
परमम् (महान्)।

प्रश्न 8.
व्यायामेन शरीरे का परिणतिः भवति ?
(व्यायाम से शरीर में क्या परिवर्तन होता है ?)
उत्तरम् :
स्थौल्यापकर्षणम्
(मोटापा कम करना)।

प्रश्न 9.
व्यायामिनं केन अर्दयन्ति ?
(व्यायाम करने वाले को कौन नहीं पीड़ित करते ?)
उत्तरम् :
अरयः (शत्रु)।

प्रश्न 10.
उरगाः कस्मात् बिभ्यति ?
(साँप किससे डरते हैं ?)
उत्तरम् :
गरुडात् (गरुड़ से)।

प्रश्न 11.
बलिनां सदा पथ्यः कः ?
(बलवानों के लिए सदा पथ्य क्या है ?)
उत्तरम् :
व्यायामः।

प्रश्न 12.
पूर्णबलेन कृतः व्यायामः किं करोति?
(पूर्ण बल से किया व्यायाम क्या करता है ?)
उत्तरम् :
हन्ति (नष्ट कर देता है)।

प्रश्न 13.
हृदिस्थानं स्थितः वायुः कदा वक्त्रं प्रपद्यते ?
(हृदय में स्थित वायु कब मुँह तक पहुँचती है ?)
उत्तरम् :
बलार्द्धप्रयोगेन
(आधे बल के प्रयोग में)।

प्रश्न 14.
शरीरायासजननं किं कुर्यात् ?
(शरीर को थकाने के लिए क्या करना चाहिए?).
उत्तरम् :
व्यायामः (व्यायाम)।

प्रश्न 15.
देहं समन्ततः किं कुर्यात् ?
(शरीर का चारों ओर से क्या करना चाहिए ?)
उत्तरम् :
विमर्दनम् (मालिश)।

प्रश्न 16.
व्यायामेन कस्योपचयः भवति ?
(व्यायाम से किसकी वृद्धि होती है?)
उत्तरम् :
शरीरस्य (शरीर की)

प्रश्न 17.
व्यायामेन केषां सुविभक्तता भवति ?
(व्यायाम से किनका सुविभाजन होता है ?)
उत्तरम् :
गात्राणाम् (अंगों का)।

प्रश्न 18.
आरोग्य केनोपजायते ?
(आरोग्य किससे पैदा होता है ?)
उत्तरम् :
व्यायामेन (व्यायाम से)।

प्रश्न 19.
शीतादीनां व्यायामात् का उपजायते ?
(शीत आदि का व्यायाम से क्या पैदा होता है ?)
उत्तरम् :
सहिष्णुता (सहनशीलता)।

प्रश्न 20.
स्थौल्यापकर्षणं केन भवति ?
(मोटापा किससे दूर होता है ?)
उत्तरम् :
व्यायामेन (व्यायाम से)।

प्रश्न 21.
कं मनुष्यम् अरयः न अर्दयन्ति ?
(किस मनुष्य को शत्रु पीड़ित नहीं करते ?)
उत्तरम् :
व्यायामिनम्
(व्यायाम करने वाले को)।

प्रश्न 22.
जरा कं न समधिरोहति ?
(बुढ़ापा किसे नहीं घेरता ?)
उत्तरम् :
व्यायामिनम् (व्यायाम करने वाले को)।

प्रश्न 23.
व्यायामरतस्य किं स्थिरी भवति ?
(व्यायामरत का क्या स्थिर होता है ?)
उत्तरम् :
मासम्।

प्रश्न 24.
उगाः कं नोपसर्पन्ति ?
(साँप किसके पास नहीं जाते ?)
उत्तरम् :
वैनतेयस्य (गरुड़ के)।

प्रश्न 25.
व्यायामः वयोरूपगुणैीनमपि किं करोति ?
(व्यायाम आयु, रूप और गुणहीन को भी क्या करता है?)
उत्तरम् :
सुदर्शनम् (सुन्दर)।

प्रश्न 26.
किं कुर्वतः नित्यं विरुद्धमपि भोजनं परिपच्यते ?
(क्या करते हुए का विरुद्ध भोजन भी पचता है।)
उत्तरम् :
व्यायामम् (व्यायाम को)।

प्रश्न 27.
विरुद्धम् अपि भोजनं कस्य पचति ?
(विरुद्ध भोजन भी किसका पच जाता है ?)
उत्तरम् :
व्यायामिनः
(व्यायाम करने वाले का)।

प्रश्न 28.
कीदृशां बलिनां व्यायामः सदा पथ्यः ?
(कैसे बलवानों का व्यायाम सदा पथ्य है ?)
उत्तरम् :
स्निग्धभोजिनाम् (चिकना खाने वाले का)।

प्रश्न 29.
शीते वसन्ते च बलिनां व्यायामः कीदृशः स्मृतः ?।
(सर्दी और वसन्त में बलवानों का व्यायाम कैसा कहा गया है ?)
उत्तरम् :
पथ्यतमः (अधिक हितकर)।

प्रश्न 30.
कीदृशैः पुरुषैः व्यायामः करणीयः?
(कैसे मनुष्यों को व्यायाम करना चाहिए ?)
उत्तरम् :
आत्महितैषिभिः (अपना हित चाहने वालों को)।

प्रश्न 31.
कियत् बलेन व्यायामः करणीयः ?
(कितनी शक्ति से व्यायाम करना चाहिए ?)
उत्तरम् :
बलस्यार्द्धन (आधे बल से)।

प्रश्न 32.
वायुः कुत्र स्थितः भवति ?
(वायु कहाँ स्थित होती है ?)
उत्तरम् :
हृदिस्थाने (हृदय में)।

प्रश्न 33.
श्लोके कस्य लक्षणं कृतम् ?
(श्लोक में किसका लक्षण किया है ?)
उत्तरम् :
बलार्धस्य (आधे बल का)।

प्रश्न 34.
यः असमीक्ष्य व्यायामं करोति सः किं प्राप्नोति ?
(जो बिना सोचे-समझे व्यायाम करता है वह क्या प्राप्त करता
उत्तरम् :
रोगम् (बीमारी)।

प्रश्न 35.
अशनानि समीक्ष्य किं कुर्यात् ?
(भोजन की परख करके क्या करना चाहिए ?)
उत्तरम् :
व्यायामम्।

पूर्णवाक्येन उत्तरत –
(पूरे वाक्य में उत्तर दीजिए)

प्रश्न 36.
कीदृशं कर्म व्यायाम इति कथ्यते ?
(कैसा कर्म व्यायाम कहलाता है ?)
उत्तरम् :
शरीरायासजननं कर्म व्यायाम इति कथ्यते।
(शरीर थकाने वाला कर्म व्यायाम कहलाता है।)

प्रश्न 37.
व्यायामात् किम् उपजायते ?
(व्यायाम से क्या पैदा होता है?)
उत्तरम् :
व्यायामात् परमारोग्यमुपजायते।
(व्यायाम से महान् आरोग्य पैदा होता है।)

प्रश्न 38.
व्याधयः कमिव नोपसर्पन्ति व्यायामिनम् ?
(रोग किसकी तरह व्यायामी के पास नहीं आते ?)
उत्तरम् :
व्याघयः उरगाः इव नोपसर्पन्ति व्यायामिनम्।
(साँप की तरह रोग व्यायामी के पास नहीं आते)।

प्रश्न 39.
व्यायामः कं सुदर्शनं करोति ?
(व्यायाम किसको सुन्दर बनाता है ?)
उत्तरम् :
व्यायामः वयौरूपगुणै नमपि सुदर्शनं करोति।।
(व्यायाम आयु, रूप और गुणों से हीन व्यक्ति को भी सुन्दर बनाता है।)

प्रश्न 40.
केषां व्यायामः पथ्यः ?
(कैसों का व्यायाम पथ्य है ?)
उत्तरम् :
व्यायामः सदा बलिनां स्निग्धभोजिनां च पथ्यः।
(व्यायाम शक्तिशालियों और स्निग्ध भोजन करने वालों के लिए पथ्य है।)

प्रश्न 41.
व्यायामः कदा कं हन्ति ?
(व्यायाम कब और किसे मारता है ?)
उत्तरम् :
यः मनुष्यः पूर्णबलेन व्यायामं करोति तदा व्यायामः तं हन्ति।
(जो मनुष्य पूर्ण बल से जब व्यायाम करता है तब व्यायाम उसे नष्ट करता है।)

प्रश्न 42.
कस्य जन्तोः हृदि स्थितः वायुः वक्त्रं प्रवर्तते ?
(किस प्राणी के हृदय में स्थित वायु मुँह की ओर बढ़ती है ?)
उत्तरम् :
व्यायाम कुर्वतो जन्तोः हृदिस्थित: वायुः वक्त्रं प्रवर्तते।
(व्यायाम करते हुए प्राणी के हृदय में स्थित वायु मुख की ओर बढ़ती है।)

प्रश्न 43.
कति वस्तूनि समीक्ष्य व्यायाम कुर्यात् ?
(कितनी वस्तुओं को परखकर व्यायाम करना चाहिए ?)
उत्तरम् :
षड् वस्तूनि समीक्ष्य व्यायाम कुर्यात्।
(छः वस्तुएँ को परखकर व्यायाम करना चाहिए।)

प्रश्न 44.
पुरुषः रोगं कथं प्राप्नोति ? (मनुष्य रोग क्यों प्राप्त करता है ?)
उत्तरम् :
यः पुरुषः वयः आदि असमीक्ष्य व्यायामं करोति सः रोगम् आप्नोति।
(जो मनुष्य उम्र आदि का विचार किए बिना व्यायाम करता है वह रोग प्राप्त करता है।)

प्रश्न 45.
व्यायामं कृत्वा किं कुर्यात् ?
(व्यायाम करके क्या करना चाहिए ?)
उत्तरम् :
व्यायामं कृत्वा समन्ततः देहं सुखं विमृदूनीयात्।
(व्यायाम करके सभी ओर से शरीर की सुखपूर्वक मालिश करनी चाहिए।)

प्रश्न 46.
लाघवं केनोपजायते ?
(फुर्ती किससे होती है ?)
उत्तरम् :
व्यायामेन लाघवं जायते।
(व्यायाम से फुर्ती पैदा होती है।)

प्रश्न 47.
व्यायामेन केषां सहिष्णुतोपजायते ?
(व्यायाम से किसकी सहने की क्षमता होती है ?)
उत्तरम् :
व्यायामेन श्रमक्लमपिपासोष्ण-शीतादीनां सहिष्णुतोपजायते।
(व्यायाम से श्रम, थकावट, प्यास, गर्मी और सर्दी आदि की सहन क्षमता पैदा होती है।)

प्रश्न 48.
व्यायामेन सदृशं किञ्चित् अन्यत् कस्मात् न भवति ?
(व्यायाम के समान और कुछ क्यों नहीं होता ?)
उत्तरम् :
यत: व्यायामेन स्थौल्यापकर्षणं भवति।
(क्योंकि व्यायाम से मोटापा कम होता है।)

प्रश्न 49.
व्यायामिनं जरा कथं न समधिरोहति ?
(व्यायामी को बुढ़ापा कैसे नहीं घेरता ?)
उत्तरम् :
व्यायामिनं जरा सहसा आक्रम्य न समधिगच्छति।।
(व्यायामी को बुढ़ापा अचानक आक्रमण करके नहीं घेरता है।)

प्रश्न 50.
व्याधयः कस्य समीपं नोपसर्पन्ति ?
(व्याधियाँ किसके समीप नहीं आती ?)
उत्तरम् :
व्याधयः व्यायामस्विन्नगात्रस्य पद्भ्यामुवर्तितस्य च नोपसर्पन्ति।
(व्याधियाँ व्यायाम के कारण पसीने से भीगे और पैरों से मर्दित के पास नहीं आती।)

प्रश्न 51.
व्यायाम कुर्वतः भोजनं कथं परिपच्यते ?
(व्यायाम करते हुए का भोजन कैसे पचता है ?)
उत्तरम् :
व्यायाम कुर्वतः भोजनं निर्दोषं परिपच्यते।
(व्यायाम करने वाले का भोजन बिना दोष के पच जाता है।)

प्रश्न 52.
बलिनां व्यायामः कदा पथ्यतमः स्मृतः ?
(बलियों का व्यायाम कब हितकर अथवा लाभकारी कहा गया
उत्तरम् :
बलिनां व्यायामः शीते वसन्ते च पथ्यतमः स्मृतः।
(बलियों का व्यायाम सर्दी और वसन्त में पथ्यतम होता है।)

प्रश्न 53.
आत्महितैषिभिः व्यायामः कदा करणीयः ?
(अपना हित चाहने वालों को व्यायाम कब करना चाहिए ?)
उत्तरम् :
आत्महितैषिभिः पुरुषैः सर्वेषु ऋतुषु प्रतिदिनं व्यायामः करणीयः।
(अपना हित चाहने वाले पुरुषों को सभी ऋतुओं में प्रतिदिन व्यायाम करना चाहिए।)

प्रश्न 54.
कानि समीक्ष्य व्यायाम कुर्यात् ? (क्या परखकर व्यायाम करना चाहिए ?)
उत्तरम् :
वयोबलशरीराणि देशकाल-अशनानि च समीक्ष्य व्यायाम कुर्यात्।।
(आयु-बल-शरीर-देश-काल और भोजन की समीक्षा करके व्यायाम करना चाहिए।)

अन्वय-लेखनम –

अधोलिखितश्लोकस्यान्वयमाश्रित्य रिक्तस्थानानि मञ्जूषातः समुचितपदानि चित्वा पूरयत।
(नीचे लिखे श्लोक के अन्वय के आधार पर रिक्तस्थानों की उचित पद चुनकर कीजिए।)

1. शरीरायासजननं …………………………… विमृद्नीयात् समन्ततः।।
मञ्जूषा – समन्ततः, संज्ञितम्, कर्म, देहं।
शरीर-आयास-जननं (i) …………. व्यायाम-(i) ……………., तत् कृत्वा तु (iii) …………… सुखं (iv) ………….. विमृद्नीयात्।
उत्तरम् :
(i) कर्म (ii) संज्ञितम् (iii) देहं (iv) समन्ततः।

2. शरीरोपचयः कान्तिर्गात्राणां ………………………. स्थिरत्वं लाघवं मृजा।।
मञ्जूषा – अनालस्यं, कान्तिः, शरीरोपचयः, लाघवं।

सुविभक्तता (i) ……………… गात्राणां (ii) ……………… दीप्ताग्नित्वम् (iii) ………………. स्थिरत्वं (iv) …………….. मृजा।
उत्तरम् :
(i) शरीरोपचयः (ii) कान्तिः (iii) अनालस्यं (iv) लाघवं।

3. श्रमक्लमपिपासोष्ण-शीतादीनां ……………………. व्यायामादुपजायते।।
मञ्जूषा – अपि, शीतादीनां, क्लम, परमं।

श्रम (i) …………………, पिपासा, उष्ण, (ii) ………………. सहिष्णुता (iii) ……………… च आरोग्यम् (iv) ……………… व्यायामादुपजायते।
उत्तरम् :
(i) क्लम (ii) शीतादीनां (iii) परमं (iv) अपि।

4. न चास्ति सदृशं तेन …………………………….. मर्त्यमर्दयन्त्यरयो बलात्।।
मञ्जूषा – मर्त्यम्, सदृशं, अर्दयन्ति, अपकर्षम्।

तेन च (i) ……………….. किञ्चित् स्थौल्य-(ii) …………… नास्ति व्यायामिनं (iii) ……………… अरयः बलात् न (iv) ……………… च।
उत्तरम् :
(i) सदृशं (ii) अपकर्षम् (iii) मर्त्यम् (iv) अर्दयन्ति।

5. न चैनं सहसाक्रम्य ………………… व्यायामाभिरतस्य च।।
मञ्जूषा – सहसा, स्थिरी, च, समधिरोहति।

च एनं (i) ………………. आक्रम्य जरा न (ii) ……………….. व्यायामाभिरतस्य (iii) ……………… मांसं (iv) ………….. भवति।
उत्तरम् :
(i) सहसा (ii) समधिरोहति (iii) च (iv) स्थिरी।

6. व्यायामस्विन्नगात्रस्य ……………………… कुर्यात्सुदर्शनम्।।
मञ्जूषा – सुदर्शनं, इव, पद्भ्यामुवर्तितस्य, नोपसर्पन्ति।

व्यायामस्विन्नगात्रस्य च (i) ………………. वैनतेयम् उरगाः (ii) ……………….. व्याधयो (iii) …………………। वयोरूपगुणहीनमपि (iv) ………………… कुर्यात्।
उत्तरम् :
(i) पद्भ्यामुवर्तितस्य (ii) इव (iii) नोपसर्पन्ति (iv) सुदर्शन।

7. व्यायाम कुर्वता …………………………… निर्दोषं परिपच्यते।।
मञ्जूषा – व्यायाम, निर्दोष, अविदग्धम्, विरुद्धम्।

नित्यं (i) ……………….. कुर्वतो भोजनं (ii) ……………… अपि विदग्धम् (iii) ……………….. वा (iii)……. परिपच्यते।
उत्तरम् :
(i) व्यायाम (ii) विरुद्धम् (iii) अविदग्धम् (iv) निर्दोषं।

8. व्यायामो हि सदा ……………………………………………. पथ्यतमः स्मृतः।।
मञ्जूषा – पथ्यतमः पथ्यः, बलिनां, शीते।

स्निग्धभोजिनाम् (i)……. व्यायामो हि सदा (ii)……. सः च तेषां (iii)……. वसन्ते च (iv)……. स्मृतः।
उत्तरम् :
(i) बलिनां (ii) पथ्यः (iii) शीते (iv) पथ्यतमः

9. सर्वेष्वृतुष्वहरहः ………………………………………………. हन्त्यतोऽन्यथा।।
मञ्जूषा-कर्त्तव्यः, बलस्यार्थेन, अहरहः, अन्यथा।

सर्वेषु ऋतुषु (i)…… पुम्भिः आत्महितैषिभिः, (ii)….. व्यायामः (i)….. अत: (iii)……, (व्यायाम:) हन्ति।
उत्तरम् :
(i) अहरहः (ii) बलस्यार्धेन (iii) कर्त्तव्यः (iv) अन्यथा।

10. हृदिस्थानास्थितो वायुर्यदा ……………………………………… लक्षणम्।।
मञ्जूषा – व्यायामकुर्वतः, प्रपद्यते, बलार्धस्य, हृदिस्थानास्थितो। यदा (i)……. वायुः वक्त्रं (ii)……. तद् (iii)……. जन्तोः (iii)……. लक्षणम्।
उत्तरम् :
(i) हृदिस्थानास्थितो (ii) प्रपद्यते (iii) व्यायामकुर्वतः (iv) बलार्धस्य।

11. वयोबलशरीराणि ……………………………. रोगमाप्नुयात्।।
मञ्जूषा – शरीराणि, अन्यथा, व्यायाम, अशनानि।

वय:-बल-(i)……. च देश-काल-(ii)……. समीक्ष्य (iii)……. कुर्यात् (iii)…….रोगमाप्नुयात्।
उत्तरम् :
(i) शरीराणि (ii) अशनानि (iii) व्यायाम (iv) अन्यथा।

प्रश्ननिर्माणम् –

अधोलिखित वाक्येषु स्थूलपदमाधृत्य प्रश्ननिर्माणं कुरुत –

1. शरीरायासजननं कर्म व्यायामसंज्ञितम्।
(शरीर के श्रम से उत्पन्न कर्म व्यायाम संज्ञक है।)
2. व्यायामं कृत्वा सुखेन देहं समन्ततः विमृद्नीयात्।।
(व्यायाम करके सुखपूर्वक देह का सभी ओर से मर्दन करना चाहिए।)
3. व्यायामिनं मर्त्यम् अरयः बलात् न अर्दयन्ति।
(व्यायामी मनुष्य को शत्रु बलपूर्वक नहीं कुचल डालते हैं।)
4. व्यायामिनं सहसाक्रम्य जरा न समधिरोहति।
(व्यायामी व्यक्ति को अचानक आक्रमण करके बुढ़ापा आरूढ़ नहीं होता।)
5. व्यायामरतस्य मांसं स्थिरीभवति।
(व्यायामरत (मनुष्य) का मांस स्थिर रहता है।)
6. व्यायामस्विन्नगात्रस्य व्याधयो नोपसर्पन्ति।
(व्यायाम के कारण पसीने से लथपथ शरीर वालों के व्याधियाँ पास नहीं आती।)
7. वैनतेयम् उरगाः नोपसर्पन्ति ?
(गरुड़ के पास सर्प नहीं आते हैं।)
8. व्यायाम कुर्वतः विरुद्धमपि भोजनं निर्दोषं परिपच्यते।
(व्यायाम करते हुए व्यक्ति को विरुद्ध भोजन भी निर्दोष पचता है।)
9. व्यायामो हि सदा पथ्यो बलिनां स्निग्धभोजिनाम्।
(बलवान् और स्नेहयुक्त भोजन करने वालों के लिए व्यायाम सदा पथ्य होता है।)
10. आरोग्यम् चापि व्यायामाद् उजायते।
(आरोग्य आदि व्यायाम से प्राप्त होते है।)
उत्तराणि :
1. कीदृशं कर्म व्यायामसंज्ञितम् ?
2. किं कृत्वा सुखं देहं समन्ततः विमृद्नीयात् ?
3. कं मर्त्यम् अरयः बलात् न अर्दयन्ति ?
4. व्यायामिनं सहसाक्रम्य का न समधिरोहति ?
5. कस्य मांसं स्थिरी भवति ?
6. कस्य व्याधयः नोपसर्पन्ति ?
7. कम् उरगाः नोपसर्पन्ति ?
8. कस्य विरुद्धमपि भोजनं निर्दोषं परिपच्यते ?
9. व्यायामो हि सदा कस्य पथ्यः ?
10. आरोग्यम् चापि कस्मात् उपजायते?

भावार्थ-लेखनम् –

अधोलिखित पद्यांशानां संस्कृते भावार्थं लिखत –

1. शरीरायासजननं कर्म ………………………….. विमृद्नीयात् समन्ततः ॥

भावार्थ – वपुसः उद्यमेन या श्रान्ति भवति तत् कर्म व्यायाम इति नाम्ना ज्ञायते। तत्कर्म व्यायाम वा विधाय तु मानवस्य शरीरं सुखी भवति। व्यायामं कृत्वा सर्वतः शरीरस्य अभ्यजनं कुर्यात्।

2. शरीरोपचयः ……………………. लाघवं मृजा ॥

भावार्थ – अनेन व्यायामेन मानवशरीरस्य सुविभक्तता शारीरिक सौष्ठवम् वा, अङ्गानां विकास: जठराग्ने: प्रवर्धने, आलस्य प्रमाद हीनता वा स्थैर्यम् शरीरे स्फूर्तिः स्वच्छीकरणं च भवति।

3. श्रमक्लमपिपासोष्ण-शीतादीनां …………………….. व्यायामादुपजायते ॥

भावार्थ – श्रमजनित शैथिल्यं, जलपातुमिच्छा, तापः, शैत्यादीनां सो, क्षमता सामर्थ्य वा अत्यधिकं नैरोग्यं अपि व्यायामात् सम्भवति।

4. न चास्ति सदृशं तेन ………………………… मर्त्यमर्दयन्त्यरयो बलात् ॥

भावार्थ – अमुना (व्यायामेन) तुल्यं किमपि नास्ति यः स्थूलतां पीवरतां वा दूरीकर्तुं शक्नोति। अर्थात् व्यायामात् पीनत्वं दूरीकरणं तनुता सम्भवति। श्रमरतं मानवं शत्रवोऽपि बलपूर्वकं न पीडयन्ति।

5. न चैनं सहसाक्रम्य …………………….. व्यायामाभिरतस्य च ॥

भावार्थ – च इमं व्यायामिनं अकस्मादेव वार्धक्यमपि आरूढं न भवति। परिश्रमे संलग्नस्य पिशितमपि शान्तं जायते।

6. व्यायामस्विन्नगात्रस्य ……………………………………. कुर्यात्सुदर्शनम् ॥

भावार्थ – परिश्रमजनित श्रम जलेन क्लिन्न शरीरस्य पादाभ्यां च उन्नमितस्य समीपे व्याधय रोगाः वा न आयान्ति। यथा गरुडस्य समीपे सर्पाः न आयान्ति। सौन्दर्यरहितं गुणविशेष रहितमपि व्यायामः सुदर्शनीयं करोति।

7. व्यायाम कुर्वतो नित्यं …………………………………. निर्दोषं परिपच्यते ॥

भावार्थ – यः मनुष्य सदैव प्रतिदिनं उद्यमं करोति तस्य प्रतिकूलम् अपि, सुपक्वम् अपक्वम् वापि भोजनं दोषं विनैव पचति।

8. व्यायामो हि सदा पथ्यो …………………….. तेषां पथ्यतमः स्मृतः ॥

भावार्थ-तैलघृतादि स्नेहयुक्त अशनानि भुञ्जानानां शक्तिशालिनां परिश्रमः सर्वदा सर्वाधिक हितकरः कथ्यते।

शेमुषी-द्वितीयो भागः

9. सर्वेष्वृतुष्वहरहः ………………………………………. व्यायामो हन्त्यतोऽन्यथा ॥

संस्कृत व्याख्या: – अखिलेषु कालेषु प्रतिदिनं स्वकीयं हिताभिलाषुकैः पुरुषैः आत्मनः शरीरस्य शक्तेः अर्धांशेन एव परिश्रमः करणीयः। अस्मात् अधिक: व्यायामः मानवं नाशयति। पूर्ण सामर्थ्येन विहितः व्यायामः व्याधिमामन्ययति।

10. हृदिस्थानास्थितो …………………………………………. लक्षणम् ॥

भावार्थ – यस्मिन् काले हृदयस्थले विद्यमानः वातः मुखं प्रवर्तते तत् व्यायामिनः प्राणिनः अर्ध बलस्य सङ्केत भवति।

11. वयोबलशरीराणि ……………………………… रोगमाप्नुयात् ॥

भावार्थ – आयुः बलं (शक्ति) गात्राणि, स्थानं समयं भोजनं च सम्यक् परीक्ष्य एव परिश्रमं (व्यायाम) कुर्वीत् नोचेत् रुग्णताम् आप्नोति।

अधोलिखितसूक्तीनां भावबोधं हिन्दी-अंग्रेजी-संस्कृतभाषया वा लिखत।
(निम्नलिखित सूक्तियों का भावार्थ हिन्दी, अंग्रेजी अथवा संस्कृत भाषा में लिखिए।)

(i) शरीरायासजननं कर्म व्यायामसंज्ञितम्।
भावार्थ – येन कार्येण श्रान्तिः श्रमं वा भवति तत् कर्म व्यायाम इति नाम्ना ज्ञायते।
(जिस कार्य से थकावट अथवा श्रम होता है वह काम व्यायाम के नाम से जाना जाता है।)

(ii) आरोग्यं चापि परमं व्यायामादुपजायते।
भावार्थ – व्यायामः तु नीरोगतां स्वस्थतां वा अपि करोति।
(व्यायाम तो रोगरहित अथवा स्वस्थ भी करता है।)

(iii) न चास्ति सदृशं तेन किञ्चित्स्थौल्यापकर्षणम्।
भावार्थ – अतः व्यायामेन तुल्यः न अन्यत् किञ्चित् कार्यं यत् शरीरस्य स्थूलतां गुरुतां वा अपकर्षेत्।
(अतः व्यायाम के समान अन्य कोई कार्य नहीं जो शरीर का मोटापा दूर करे।)

(iv) व्यायामो हि सदा पथ्यः।
भावार्थ – व्यायामः श्रमः वा सदैव मानवस्य हितकारकः भवति।
(व्यायाम अथवा श्रम सदैव मानव का हितकारक होता है।)

(v) बलस्यार्धन कर्तव्यो व्यायामः।
भावार्थ – मानवेन स्वस्य अर्द्धसामर्थ्येन व्यायामः करणीयः न तु पूर्ण बलेन।
(मनुष्य को अपनी आधी क्षमता से ही व्यायाम करना चाहिए न कि पूरे बल से।)

व्यायामः सर्वदा पथ्यः Summary and Translation in Hindi

पाठ-परिचय – संस्कृत में आयुर्विज्ञान विषय पर प्रभूत साहित्य की रचना हुई है। इसी क्रम में तीसरी शताब्दी में आचार्य सुश्रुत ने ‘सुश्रुतसंहिता’ की रचना की। ‘सुश्रुतसंहिता’ में शल्य-चिकित्सा की प्रधानता है। इस ग्रन्थ में रोगों की चिकित्सा के साथ-साथ उनके पथ्यों का भी उल्लेख किया गया है। ‘सुश्रुतसंहिता’ छः खण्डों में विभक्त है, जिन्हें ‘स्थान’ कहा गया है।

प्रस्तुत पाठ आयुर्वेद के इसी प्रसिद्ध ग्रन्थ के चिकित्सास्थान के 24वें अध्याय से संकलित है। इसमें आचार्य सुश्रुत ने व्यायाम की परिभाषा बताते हुए उससे होने वाले लाभों की चर्चा की है। शरीर में सुगठन, कान्ति, स्फूर्ति, सहिष्णुता, नीरोगता आदि व्यायाम के प्रमुख लाभ हैं।

मूलपाठः, अन्वयः, शब्दार्थाः, सप्रसंग हिन्दी-अनुवादः

1. शरीरायासजननं कर्म व्यायामसंज्ञितम्।
तत्कृत्वा तु सुखं देहं विमृद्नीयात् समन्ततः ।।1।।

अन्वयः – शरीर-आयास-जननं कर्म व्यायाम-संज्ञितम्, तत् कृत्वा तु देहं सुखं समन्ततः विमृद्नीयात्।

शब्दार्थाः – शरीर = गात्रे (शरीर में), आयास = श्रमम् (थकावट), जननम् = उत्पादकम् (पैदा करने वाला), कर्म = कार्यम् (कार्य), व्यायाम = (कसरत), संज्ञितम् = नामधेयम् (नामवाला है), तत् = तं व्यायामम् (उस व्यायाम को), कृत्वा = सम्पाद्य, विधाय (करके), तु = तदा (तब), देहम् = गात्रम् (शरीर को), सखम् = सुखपूर्वकम, सहजभावेन (सुखपूर्वक), समन्ततः = सर्वतः (पूरी तरह से, सभी ओर से), विमृदुनीयात् = मर्दयेत् (मालिश करनी चाहिए।

सन्दर्भ-प्रसङ्गश्च – यह श्लोक हमारी शेमुषी पाठ्यपुस्तक के ‘व्यायामः सर्वदा पथ्यः’ पाठ से उद्धृत है। यह पाठ आचार्य सुश्रुत रचित ‘सुश्रुत-संहिता’ से सङ्कलित है। इस श्लोक में आचार्य व्यायाम की परिभाषा प्रस्तुत करते हैं। वे कहते हैं –

हिन्दी-अनुवादः – शरीर में थकावट पैदा करने वाला कार्य व्यायाम नाम वाला है अर्थात् व्यायाम कहलाता है। उस व्यायाम को करके तो (तब) सुखपूर्वक (सहज रूप से) पूरी तरह (सभी ओर से शरीर के अंगों की) मालिश करनी चाहिए।

2. शरीरोपचयः कान्तिर्गात्राणां सुविभक्तता।
दीप्ताग्नित्वमनालस्यं स्थिरत्वं लाघवं मृजा ।।2।।

अन्वयः – सुविभक्तता शरीरोपचयः गात्राणां कान्तिः दीप्ताग्नित्वम् अनालस्यं स्थिरत्वं लाघवं मृजा।

शब्दार्थाः – (व्यायामेन) सुविभक्तता = शारीरिकं सौष्ठवम् (शारीरिक सौन्दर्य), (सम्यक विभक्तस्य) शरीरोपचयः = गात्रस्य अभिवृद्धिः, अंगानां विकासः (शरीर की वृद्धि), गात्राणाम् = अङ्गानाम् (अंगों की), कान्तिः = आभा (चमक), दीप्ताग्नित्वम् = जठराग्नेः प्रवर्धनम् (जठराग्नि का प्रदीप्त होना अर्थात् भूख लगना), अनालस्यम् = आलस्यहीनता (आलस्यहीनता), स्थिरत्वम् = स्थिरता (स्थिरता), लाघवं = स्फूर्तिः (फुर्ती), मृजा = स्वच्छीकरणम् (स्वच्छता), (भवन्ति = होती हैं)।।

सन्दर्भ – प्रसङ्गश्च – यह श्लोक हमारी ‘शेमुषी’ पाठ्य पुस्तक के ‘व्यायामः सर्वदा पथ्यः’ पाठ से लिया गया है। यह पाठ आचार्य सुश्रुत द्वारा रचे गये ग्रन्थ ‘सुश्रुत संहिता’ से संकलित है। इस श्लोक में आचार्य व्यायाम के लाभों को बताकर उसका महत्व प्रदर्शित करते हैं।

हिन्दी-अनुवादः – (व्यायाम से) शारीरिक सौन्दर्य, शरीर की वृद्धि, अंगों की चमक, जठराग्नि का प्रदीप्त होना अर्थात् भूख लगना, आलस्यहीनता, स्थिरता, फुर्ती एवं स्वच्छता (होती हैं)।

3. श्रमक्लमपिपासोष्ण-शीतादीनां सहिष्णुता।।
आरोग्यं चापि परमं व्यायामादुपजायते ।।3।।

अन्वयः – श्रम क्लम, पिपासा, उष्ण, शीतादीनां सहिष्णुता परमं च आरोग्यम् अपि व्यायामादुपजायते।

शब्दार्थाः – श्रमक्लमः = श्रमजनितं शैथिल्यम् (थकान), पिपासा = पातुम् इच्छा (प्यास), उष्णः = तापः (गर्मी), शीतादीनाम् = शैत्यादीनाम् (ठण्ड आदि की), सहिष्णुता = सहत्वं/सोढुं क्षमता (सहन करने की शक्ति/सामर्थ्य), च = (और), परमम् = अत्यधिकम् (महान्), आरोग्यम् अपि = स्वस्थता/नीरोगिता अपि (रोगहीनता भी), व्यायामाद् = श्रमाद् (कसरत करने से), उपजायते = सम्भवति (होता है)।

सन्दर्भ-प्रसङ्गश्च – यह श्लोक हमारी पाठ्य पुस्तक शेमुषी के ‘व्यायाम सर्वदा पथ्यः’ पाठ से उद्धृत है। यह पाठ आचार्य सुश्रुत रचित ‘सुश्रुत संहिता’ ग्रन्थ से संकलित है। इस श्लोक में आचार्य व्यायाम से होने वाले गुणों का प्रतिपादन करते हुए कहते हैं।

हिन्दी-अनुवादः – थकान, प्यास, गर्मी, ठण्ड आदि को सहन करने की शक्ति और महान् स्वस्थता (नीरोगता) अर्थात् रोग-हीनता व्यायाम से होती है।

4. न चास्ति सदृशं तेन किञ्चित्स्थौल्यापकर्षणम्।
न च व्यायामिनं मर्त्यमर्दयन्त्यरयो बलात् ।।4।।

अन्वयः – तेन च सदशं किञ्चित स्थौल्य-अपकर्षम नास्ति व्यायामिनं मर्त्यम अरयः बलात न अर्दयन्ति च।

शब्दार्थाः – तेन च = अमुना च (और इसके), सदृशं = समानं, तुल्यम् (समान), किञ्चित् = (कुछ), स्थौल्यापकर्षणम् = अतिमांसलत्वम्, पीनताम् दूरीकरणम् (मोटापा दूर करना, कम करना), नास्ति = न वर्तते (नहीं है), व्यायामिनम् = श्रमरतं, व्यायामनिरतम् (व्यायाम करने वाले व्यक्ति को), मर्त्यम् = मनुष्यम् (मानव को), अरयः = शत्रवः (शत्रुगण), बलात् = बलपूर्वकम् (जबरदस्ती), न अर्दयन्ति = अर्दनं न कुर्वन्ति, न पीडयन्ति (नहीं कुचल डालते हैं)।

सन्दर्भ-प्रसङ्गश्च – प्रस्तुत श्लोक हमारी संस्कृत की पाठ्यपुस्तक ‘शेमुषी’ (द्वितीयो भागः) के ‘व्यायामः सर्वदा पथ्यः’ शीर्षक पाठ से उद्धृत किया गया है। मूलतः यह पाठ आचार्य सुश्रुत द्वारा विरचित आयुर्वेद के सुप्रसिद्ध ग्रन्थ ‘सुश्रुतसंहिता’ के चिकित्सास्थान में वर्णन किये गये 24वें अध्याय से संकलित किया गया है। इस श्लोक में व्यायाम करने से मोटापा दूर किये जाने का लाभ वर्णित करते हुए कहा गया है कि व्यायाम करने वाले को शत्रु भी पराजित नहीं कर पाते हैं।

हिन्दी-अनुवादः – और उस व्यायाम के समान मोटापा दूर (कम) करने वाला कोई (साधन) नहीं है। व्यायाम करने वाले व्यक्ति को शत्रुगण जबर्दस्ती नहीं कुचल सकते हैं। (या पीड़ा नहीं देते।)

5. न चैनं सहसाक्रम्य जरा समधिरोहति।
स्थिरीभवति मांसं च व्यायामाभिरतस्य च।। 5।।

अन्वयः – च एनं सहसा आक्रम्य जरा न समधिरोहति व्यायामाभिरतस्य च मांसं स्थिरी भवति।

शब्दार्थाः – च एनम् = च इमं व्यायामिनम् (और इस व्यायाम करने वाले को), सहसा = अकस्मात् (अचानक), जरा = वृद्धत्वं, वार्धक्यम् (बुढ़ापा), न समधिरोहति = न आरूढं भवति (सवार नहीं होता है, नहीं दबोचता), व्यायामाभिरतस्य च = व्यायामे/परिश्रमे संलग्नस्य च (और व्यायाम/परिश्रम में संलग्न/लगे हुए व्यक्ति का), मांसं = आमिषम्, पिशितम् (मांस), स्थिरी = शान्तः (स्थिर), भवति = जायते (होता है)।

सन्दर्भ-प्रसङ्गश्च – यह श्लोक हमारी शेमुषी पाठ्य पुस्तक के ‘व्यायाम सर्वदा पथ्यः’ पाठ से लिया गया है। यह पाठ आचार्य सुश्रुत-विरचित ‘सुश्रुत संहिता’ आयुर्वेद ग्रन्थ से सङ्कलित है। इस श्लोक में आचार्य सुश्रुत कहते हैं कि व्यायाम करने वाले व्यक्ति का मांस स्थिर हो जाता है तथा उसे अकस्मात् बुढ़ापा भी नहीं घेरता है।

हिन्दी-अनुवादः- और इस व्यायाम (परिश्रम) करने वाले को बुढ़ापा अचानक नहीं दबोचता (आरूढ़ होता) है। व्यायाम (परिश्रम) में संलग्न (लगे हुए) व्यक्ति का मांस स्थिर (शान्त) रहता है।

6. व्यायामस्विन्नगात्रस्य पद्भ्यामुवर्तितस्य च।
व्याधयो नोपसर्पन्ति वैनतेयमिवोरगाः
वयोरूपगुणैीनमपि कुर्यात्सुदर्शनम्।।6।।

अन्वयः – व्यायामस्विन्नगात्रस्य च पद्भ्यामुवर्तितस्य वैनतेयम् उरगाः इव व्याधयो नोपसर्पन्ति। वयोरूपगुणहीनमपि सुदर्शनं कुर्यात्।

शब्दार्थाः – व्यायाम = व्यायामेन, परिश्रमेण (व्यायाम या परिश्रम से), स्विन्नगात्रस्य = श्रमजलेन, स्वेदेन सिक्तस्य शरीरस्य (पसीने से लथपथ शरीर वाले का), च पद्भ्याम् = च पादाभ्याम् (और पैरों से), उद्वर्तितस्य = उन्नमितस्य (ऊपर उठने वाले, पैरों से रौंधने वाले, व्यायाम करने वाले के), वैनतेयम् = गरुड़ (गरुड़ के), उरगाः इव = सर्पाः इव (साँपों के समान) व्याधयः = रोगाः (बीमारियाँ), नोपसर्पन्ति = समीपं न आयान्ति (पास नहीं आती हैं), वयो = आयुसं (आयु), रूप = सौन्दर्य (सुन्दर रूप), गुण = गुण विशेष (गुण विशेष), हीनम् अपि = रहितमपि (रहित को भी). सुदर्शनम् = शोभनीयं (सुन्दर दिखाई देने वाला), कुर्यात् = करोति (बना देता है)।

सन्दर्भ-प्रसङ्गश्च – प्रस्तुत श्लोक हमारी संस्कृत की पाठ्यपुस्तक ‘शेमुषी’ के ‘व्यायामः सर्वदा पथ्यः’ शीर्षक पाठ से उद्धृत किया गया है। मूलतः यह पाठ आचार्य सुश्रुत द्वारा विरचित आयुर्वेद के सुप्रसिद्ध ग्रन्थ ‘सुश्रुतसंहिता’ के चिकित्सास्थान में वर्णन किये गये 24वें अध्याय से संकलित किया गया है। इस श्लोक में व्यायाम के महत्त्व को प्रदर्शित करते हुए कहा गया है कि व्यायाम करने वाला कभी रोगी नहीं बनता है।

हिन्दी-अनुवादः- व्यायाम या परिश्रम के कारण पसीने से लथपथ शरीर वाले तथा पैरों से रौंधे हुए शरीर वाले व्यक्ति के पास व्याधियाँ गरुड़ के पास साँपों की तरह नहीं आती हैं। (व्यायाम) आयु, रूप और गुणरहित व्यक्ति को भी दर्शनीय बना देता है।

7 व्यायाम कुर्वतो नित्यं विरुद्धमपि भोजनम्।
विदग्धमविदग्धं वा निर्दोषं परिपच्यते ।।7।।

अन्वयः – नित्यं व्यायामं कुर्वतो भोजनं विरुद्धम् अपि विदग्धम् अविदग्धम् वा निर्दोषं परिपच्यते।।

शब्दार्थाः – नित्यम् = सदैव, प्रतिदिनम् (हमेशा, रोजाना), व्यायामम् = श्रमम् (व्यायाम), कुर्वतः = कुर्वाणस्य (करने वाले के), भोजनम् = खाद्यम्, अशनम् (भोजन के), विरुद्धम् = प्रतिकूलम् (विपरीत होने पर भी), वा = अथवा (भोजनम् = खाना), विदग्धम् = सुपक्वम् (भली-भाँति पका हुआ), अविदग्धम् = अपक्वम् (न पका हुआ), निर्दोषं = दोषं विना (बिना किसी दोष के), परिपच्यते = जीर्यते (आसानी से पच जाता है)।

सन्दर्भ-प्रसङ्गश्च – यह श्लोक हमारी शेमुषी पाठ्य पुस्तक के व्यायामः सर्वदा पथ्यः पाठ से लिया गया है। यह पाठ आचार्य सुश्रुत-रचित सुश्रुत-संहिता आयुर्वेद ग्रन्थ से सङ्कलित है। इस श्लोक में आचार्य प्रतिपादित करते हैं कि व्यायाम से सब प्रकार का भोजन अच्छी तरह पचता है।

हिन्दी-अनुवादः – रोजाना व्यायाम करने वाले मनुष्य को विपरीत होते हुए भी भली-भाँति पके हुए तथा कम पके हुए भोजन आसानी से बिना परेशानी के पच जाते हैं।

8 व्यायामो हि सदा पथ्यो बलिनां स्निग्धभोजिनाम्।
स च शीते वसन्ते च तेषां पथ्यतमः स्मृतः।।8।।

अन्वयः – स्निग्धभोजिनाम् बलिनां व्यायामो हि सदा पथ्यः सः च तेषां शीते वसन्ते च पथ्यतमः स्मृतः।

शब्दार्थाः – स्निग्धभोजिनाम् = स्नेहयुक्त अशनानि खादताम् (चिकना भोजन खाने वालों), बलिनाम् = बलवताम्, शक्तिशालिनाम् (बलवान् व्यक्तियों के लिए), व्यायामो हि = व्यायामः, परिश्रमः (व्यायाम), सदा = सर्वदा (हमेशा), पथ्यतमः = सर्वाधिकः हितकरः (सबसे अधिक हितकर), स्मृतः = कथ्यते (कहा जाता है)। . सन्दर्भ-प्रसङ्गश्च-यह श्लोक हमारी शेमुषी पाठ्यपुस्तक के व्यायामः सर्वदा पथ्यः पाठ से लिया गया है। यह पाठ आचार्य सुश्रुत-रचित सुश्रुत-संहिता आयुर्वेद ग्रन्थ से सङ्कलित है। इस श्लोक में आचार्य व्यायाम का उचित काल बताते हैं। शीतकाल और बसन्त काल में व्यायाम करना सबसे अधिक लाभदायक होता है।

हिन्दी-अनुवादः – चिकना भोजन खाने वालों और बलवान् व्यक्तियों का व्यायाम शीत और वसंत में सदैव सबसे . अधिक हितकर कहलाता है।

9. सर्वेष्वृतुष्वहरहः पुम्भिरात्महितैषिभिः।
बलस्यार्धेन कर्त्तव्यो व्यायामो हन्त्यतोऽन्यथा।।9।।

अन्वयः – सर्वेषु ऋतुषु अहरहः पुम्भिः आत्महितैषिभिः, बलस्यार्धेन व्यायामः कर्त्तव्यः अतः अन्यथा, (व्यायामः) हन्ति।

शब्दार्थाः – सर्वेषु = अखिलेषु (सभी), ऋतुषु = कालेषु (ऋतुओं में), अहरहः = प्रतिदिनम् (हर रोज), पुम्भिः = पुरुषैः (मनुष्यों द्वारा, मनुष्यों को), आत्महितैषिभिः = स्वकीयं हितं आत्मनः हितं अभिलाषुकैः (अपना हित चाहने वालों द्वारा, या अपना हित चाहने वालों को), (आत्मनः शरीरस्य = अपने शरीर के), बलस्यार्धेन = अर्द्ध बलेन, शक्तेः अर्धांशेन (आधे बल से ही), व्यायामः कर्तव्यः = परिश्रमः करणीयः (व्यायाम करना चाहिए), अतोऽन्यथा = अस्मात् अधिकतरः व्यायामः (इससे अधिक व्यायाम मनुष्य को), हन्ति = मारयति, नाशयति (मार देता है)।

सन्दर्भ-प्रसङ्गश्च – यह श्लोक हमारी शेमुषी पाठ्य पुस्तक के ‘व्यायामः सर्वदा पथ्यः’ पाठ से लिया गया है। यह पाठ आचार्य सुश्रुत-रचित सुश्रुत-संहिता आयुर्वेद ग्रन्थ से सङ्कलित है। इस श्लोक में आचार्य व्यायाम की मात्रा का वर्णन करते। कहते हैं कि अपना भला चाहने वालों द्वारा व्यायाम आधे बल से ही करना चाहिए अन्यथा हानिकारक होता है।

हिन्दी-अनुवादः – सभी ऋतुओं में हर रोज अपने हित के अभिलाषी (चाहने वाले) मनुष्यों को अपने शरीर के आधे बल से व्यायाम करना चाहिए। इससे अधिक व्यायाम मनुष्य को मार देता है।

10. हृदिस्थानास्थितो वायुर्यदा वक्त्रं प्रपद्यते।
व्यायाम कुर्वतो जन्तोस्तबलार्धस्य लक्षणम् ।।10।।

अन्वयः – यदा हृदिस्थानास्थितो वायुः वक्त्रं प्रपद्यते तद् व्यायामकुर्वतः जन्तोः बलार्धस्य लक्षणम्।

शब्दार्थाः – यदा = यस्मिन् काले (जिस समय), हृदिस्थानास्थितः = हृदयस्थले विद्यमानः (हृदय स्थल में विद्यमान), वायुः = वातः (हवा, वात), वकाम् = मुखम् (मुँह पर), प्रपद्यते = प्रवर्तते (प्रवर्तित होती है), तद् = असौ (वह), व्यायामकुर्वतः = व्यायामिनः (व्यायाम करने वाले का), जन्तोः = प्राणिनः (प्राणी के), बलार्धस्य = अर्धबलस्य, शक्तेः अर्धांशस्य (आधे सामर्थ्य, बल), लक्षणम् = सङ्केतम्, लक्षणस्य प्रतीतिः (लक्षण की प्रतीति है)।

सन्दर्भ-प्रसङ्गश्च – यह श्लोक हमारी शेमुषी पाठ्य पुस्तक के ‘व्यायामः सर्वदा पथ्यः’ पाठ से लिया गया है। यह पाठ आचार्य सुश्रुत-रचित सुश्रुत-संहिता आयुर्वेद ग्रन्थ से सङ्कलित है। इस श्लोक में आचार्य अर्धबल का लक्षण कहते हैं।

हिन्दी-अनुवादः – जिस समय (जब) हृदयस्थल में विद्यमान वात (हवा) मुख की ओर प्रवृत्त होती है तब व्यायाम करने वाले प्राणी के आधे बल (सामर्थ्य) का लक्षण (संकेत) प्रतीत होता है।

11. वयोबलशरीराणि देशकालाशनानि च।
समीक्ष्य कुर्याद् व्यायाममन्यथा रोगमाप्नुयात् ।।11।।

अन्वयः – वयः-बल-शरीराणि च देश-काल-अशनानि समीक्ष्य व्यायाम कुर्यात् अन्यथा रोगमाप्नुयात्।

शब्दार्थाः – वयः = आयुः, अवस्थाम् (उम्र), बल = शक्ति (ताकत), शरीराणि = गात्राणि (शरीर), च देश = स्थानम् (स्थान), काल = समय (समय), अशनानि = आहाराः, भोजनानि (भोजन को), समीक्ष्य = परीक्ष्य (परीक्षण करके/परख करके), व्यायाम = परिश्रमं (कसरत), कुर्यात् = कुर्वीत (करना चाहिए), अन्यथा = नोचेत् (नहीं तो, अन्यथा), रोगम् = रुग्णताम् (रोग), आप्नुयात् = लभते (प्राप्त करता है)।

सन्दर्भ-प्रसङ्गश्च-यह श्लोक हमारी शेमुषी पाठ्य पुस्तक के व्यायामः सर्वदा पथ्यः पाठ से लिया गया है। मूलतः यह पाठ आचार्य सुश्रुत-रचित ‘सुश्रुत-संहिता’ आयुर्वेद ग्रन्थ से संकलित है। इसमें आचार्य कहते हैं कि मनुष्य को व्यायाम अपनी आयु, बल, शारीरिक क्षमता, स्थान व उचित समय देख कर ही करना चाहिए।

हिन्दी-अनुवादः- (मनुष्य) आयु, बल, शरीर, स्थान, समय और भोजन का परीक्षण करके व्यायाम करे अन्यथा (नहीं तो) रुग्णता को प्राप्त होता है अर्थात् बीमार हो जाता है।

Jharkhand Board JAC Class 10 Maths Important Questions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Important Questions and Answers.

JAC Board Class 10th Maths Important Questions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन

लघुत्तरात्मक / निबन्धात्मक प्रश्न :

प्रश्न 1.
एक अर्धगोलाकार टैंक पानी से भरा है, जिसका पानी एक पाइप द्वारा \(\frac{25}{7}\) लीटर प्रति सेकण्ड की दर से खाली किया जा रहा है। ज्ञात कीजिए कि इस टैंक को आधा खाली करने में कितना समय लगेगा, यदि टैंक के आधार का व्यास 3 मी. है।
हल:
दिया है,
अर्धगोलाकार टैंक का व्यास = 3 मी
∴ अर्धगोलाकार टैंक की त्रिज्या (r) = \(\frac{3}{2}\) मी
अर्धगोलाकार टैंक का आयतन = \(\frac{2}{3}\) πr³

∵ \(\frac{25}{7}\) लीटर पानी को खाली करने में लगा समय
= 1 सेकण्ड
∴ \(\frac{99000}{28}\) लीटर पानी को खाली करने में लगा समय
= \(\frac{7}{25} \times \frac{99000}{28}\)
= 990 सेकण्ड
= \(\frac{990}{60}\) मिनट
= 16.5 मिनट
अतः आधे टैंक को खाली करने में लगा समय = 16.5 मिनट।

प्रश्न 2.
10 सेमी भुजा वाले एक घनाकार ब्लॉक के ऊपर एक अर्धगोला रखा हुआ है। अर्धगोले का अधिकतम व्यास क्या हो सकता है ? इस प्रकार बने ठोस के संपूर्ण पृष्ठीय क्षेत्र को पेंट करवाने का ₹ 5 प्रति 100 वर्ग सेमी की दर से व्यय ज्ञात कीजिए। [π = 3.14 लीजिए]
हल:
अर्द्धगोले का अधिकतम व्यास:
= घन के एक किनारे की ल. (a) = 10 सेमी

∴ अर्द्धगोले की त्रिज्या, r = \(\frac{10}{2}\) = 5 सेमी
अब ठोस का सम्पूर्ण पृष्ठीय क्षे. = घन का सम्पूर्ण पृष्ठीय क्षेत्रफल + अर्द्धगोलीय का वक्र पृष्ठीय क्षेत्रफल – अर्द्धगोले के आधार का क्षेत्रफल
= 6a² + 2πr² – πr²
= 6 × 10² + 2 × 3.14 × (5)² – 3.14 × (5)² सेमी²
= 600 + 157 – 78.5 cm²
= 678.5 cm²
पेंट करवाने का खर्चा = ₹ 5 प्रति 100 सेमी²
∴ ठोस को पेंट करवाने का खर्चा
= 678.5 × \(\frac{5}{100}\)
= ₹ 33.90 (लगभग)
अतः पेंट करवाने का व्यय = ₹ 33.90 (लगभग)।

प्रश्न 3.
एक ठोस धातु के बेलन के दोनों किनारों से उसी व्यास के अर्द्धगोले के रूप में धातु निकाली गई। बेलन की ऊँचाई 10 सेमी तथा इसके आधार की त्रिज्या 4.2 सेमी है। शेष बेलन को पिघलाकर 1.4 सेमी मोटी बेलनाकार तार बनाई गई तार की लम्बाई ज्ञात कीजिए।
हल:
ठोस धातु के बेलन की ऊँचाई = 10 सेमी
ठोस धातु के बेलन की त्रिज्या = 4.2 सेमी
∴ सभी अर्द्धगोले की त्रिज्या = ठोस धातु के बेलन की त्रिज्या; r = 4.2 सेमी
अब शेष बचे हुए बेलन का आयतन = बेलन का आयतन – 2 × अर्द्धगोले का आयतन
= πr²h – 2 × \(\frac{2}{3}\)πr³
= πr²\(\left(h-\frac{4}{3} r\right)\) = π × (4.2)² \(\left(10-\frac{4}{3} \times 4.2\right)\)
= π(4.2)² × 4.4 सेमी³
बेलनाकार तार की मोटाई = 1.4 सेमी
बेलनाकार तार की त्रिज्या = \(\frac{2}{3}\) = 7 सेमी
माना तार की लम्बाई = H सेमी
प्रश्नानुसार-
बेलनाकार तार का आयतन = शेष बचे हुए बेलन का आयतन
⇒ π × .7 × .7 × H = π × (4.2)².4.4
⇒ H = \(\frac{4.2 \times 4.2 \times 4.4}{.7 \times .7}\)
= 158.4
अतः तार की लम्बाई = 138.4 सेमी

प्रश्न 4.
पानी से भरी हुई अर्द्धगोलाकार टंकी को एक पाइप द्वारा 5 लीटर प्रति सेकण्ड की दर से खाली किया जाता है। यदि टंकी का व्यास 3.5 मीटर है तो कितने समय में आधी खाली हो जायेगी।
हल:
दिया है,
अर्द्धगोलाकार टैंकी का व्यास = 13.5 मीटर
∴ त्रिज्या r = \(\frac{3.5}{2}\) मीटर
अर्द्धगोलाकार टंकी की आयतन

…….5 लीटर पानी को खाली करने में लगा समय = 1 सेकण्ड
∴ \(\frac{67375}{21}\) लीटर पानी को खाली करने में लगा समय
= \(\left(\frac{1}{5} \times \frac{67375}{21}\right)\)
\(\left(\frac{1}{5} \times \frac{67375}{21}\right)\) से. = 641.66 से. = 10.69 मिनट
अतः आधी टंकी को खाली करने में लगभग = 10.69 मिनट।

प्रश्न 5.
आकृति में, PQRS एक वर्गाकार लॉन है जिसकी भुजा PQ = 42 मीटर है। दो वृत्ताकार फूलों की क्यारियाँ भुजा PS तथा QR पर हैं जिनका केन्द्र इस वर्ग के विकणों का प्रतिच्छेन बिन्दु O है। दोनों फूलों की क्यारियों (छायांकित भाग) का कुल क्षेत्रफल ज्ञात कीजिए।

हल:
वर्गाकार लॉन का क्षेत्रफल PQRS = 42 मी × 42 मी
माना OP = OS = x मी
इसलिए x² + x² = (42)²
⇒ 2x² = 42 × 42
x² = 21 × 42 …..(1)
अब भाग POS का क्षेत्रफल = \(\frac{90}{360}\) πx² = \(\frac{1}{4}\) πx²
= \(\frac{1}{4} \times \frac{22}{7}\) × 21 × 42 मी² …..(2)
पुन: ΔPOS का क्षे. = \(\frac{1}{4}\) × वर्गाकार लॉन PQRS का क्षे.
= \(\frac{1}{4}\) × 42 × 42 मी² …..(3)
∠POQ = 90°
∴ फूलों की क्यारियों PSP का क्षे. = खण्ड POS भाग का क्षे. – ΔPOS का क्षे.
= \(\frac{1}{4} \times \frac{22}{7}\) × 21 × 42 – \(\frac{1}{4}\) × 42 × 42
= 33 × 21 – 441 = 693 – 441 = 252
दोनों फूलों की क्यारियों का कुल = 2 × 252 = 504 मी²
अतः दोनों फूलों की क्यारियाँ का कुल क्षे. 504 मी² है।

प्रश्न 6.
3.5 सेमी व्यास तथा 3 सेमी ऊँचे 504 शंकुओं को पिघलाकर एक धात्विक गोला बनाया गया। गोले का व्यास ज्ञात कीजिए। अतः इसका पृष्ठीय क्षेत्रफल ज्ञात कीजिए। [π = \(\frac{22}{7}\) लीजिए]
हल:
शंकु का व्यास = 3.5 सेमी
(r) = \(\frac{3.5}{2}\) सेमी
ऊँचाई (h) = 3 सेमी
अब प्रत्येक शंकु का आयतन = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times\left(\frac{3.5}{2}\right)^2 \times 3\) = 9.625 सेमी³
∴ 504 शंकुओं का आयतन = 504 × 9.625 = 4851 सेमी³
माना धात्विक गोले की त्रिज्या = R
प्रश्नानुसार,
गोले का आयतन = 504 × (शंकु का आयतन)
\(\frac{4}{3}\) πR³ = 4851
R³ = \(\frac{4851 \times 3}{4 \times 3.14}\) = 1157.625
R = 10.5 सेमी
गोले का व्यास = 2R = 2 × 10.5 = 21 सेमी
∴ गोले का पृष्ठीय क्षेत्रफल = 4πR² = 4 × \(\frac{22}{7}\) × (10.5)²
= 1386 सेमी²

प्रश्न 7.
दो घनों, जिनमें से प्रत्येक का आयतन 27 सेमी³ है, तो संलग्न फलकों को मिलाकर एक ठोस बनाया जाता है। प्राप्त घनाभ का पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
हल:

माना
धन की भुजा = a
घन का आयतन = 27 सेमी³
घन का आयतन = 27 भुजा³
27 सेमी³ = a³
a³ = 27
a = 3 सेमी
∴ घनाभ का सम्पूर्ण पृष्ठीय क्षे.
= 2 (ल. × चौ. + चौ. x ऊँ. + ऊँ. x ल.)
= 2 ( 3 × 3 + 3 × 6 + 6 × 3)
= 3(9 + 18 + 18)
= 3(45) = 135 सेमी²

प्रश्न 8.
एक ठोस अर्धगोले का सम्पूर्ण पृष्ठीय क्षेत्रफल 462 वर्ग सेमी है। इसकी त्रिज्या ज्ञात कीजिए।
हल:
माना अर्धगोले की त्रिज्या r सेमी है।
प्रश्नानुसार,
अर्धगोले का सम्पूर्ण पृष्ठीय क्षेत्रफल = 462 वर्ग सेमी
⇒ 3πR² = 462
⇒ 3 × \(\frac{22}{7}\) × r² = 462
⇒ r² = \(\frac{462 \times 7}{3 \times 22}\) = 49
⇒ r = \(\sqrt{49}\) = 7 सेमी
अतः अर्धगोले की त्रिज्या = 7 सेमी

प्रश्न 9.
एक चाँदी के घनाभ, जिसकी विमाएँ 8 सेमी × 9 सेमी × 11 सेमी है, को पिघलाकर समान त्रिज्या के सात गोले बनाए गए है। एक चाँदी के गोले की त्रिज्या ज्ञात कीजिए।
हल:
माना गोले की त्रिज्या = r सेमी
घनाभ का आयतन = लम्बाई × चौड़ाई × ऊँचाई
= 8 × 9 × 11 घन सेमी
∵ घनाभ को पिघलाकर सात गोले बनाए गए हैं,
∴ घनाभ का आयतन = 7 गोलों का आयतन
⇒ 8 × 9 × 11 = 7 × \(\frac{4}{3}\) πR³
⇒ 8 × 9 × 11 = \(7 \times \frac{4}{3} \times \frac{22}{7} \times r^3\)
⇒ r³ = \(\frac{8 \times 9 \times 11 \times 3}{4 \times 22}\) = 27
⇒ r³ = (3)³
⇒ r = 3 सेमी
अतः एक गोले की त्रिज्या = 3 सेमी

प्रश्न 10.
कोई बर्तन एक खोखले अर्धगोले के आकार का है। अर्धगोले का व्यास 14 सेमी है। इस बर्तन का आन्तरिक पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
हल:
दिया है, अर्धगोले का व्यास = 14 सेमी
∴ त्रिज्या = 7 सेमी
∵ गोला खोखला है,
∴ अर्धगोले का आन्तरिक पृष्ठीय क्षेत्रफल = 2πr²
= 2 × \(\frac{22}{7}\) × (7)² वर्ग सेमी
= 308 वर्ग सेमी

प्रश्न 11.
7 मी व्यास वाला एक कुआँ खोदा जाता है और खोदने से निकली हुई मिट्टी को समान रूप से फैलाकर 22 मी × 14 मी × 2.5 मी वाला एक चबूतरा बनाया गया है। कुएं की गहराई ज्ञात कीजिए।
हल:
कुआँ बेलनाकार होता है।
माना कुएँ की गहराई h मी है।
∴ कुएँ से निकली मिट्टी का आयतन = बेलनाकार कुएँ का आयतन
= πr²h
= \(\frac{22}{7}\) × (7)² × घन मी
घनाभाकार चबूतरे का आयतन = 22 × 14 × 2.5 घन मी
प्रश्नानुसार,
\(\frac{22}{7}\) × (7)² × h = 22 × 14 × 2.5
⇒ h = \(\frac{22 \times 14 \times 2.5}{22 \times 7}\)
⇒ h = 17.5 मी
अतः कुएँ की गहराई = 17.5 मी

प्रश्न 12.
एक ठोस बेलन के आकार का है जिसके दोनों सिरे अर्धगोलाकार है। ठोस की कुल लम्बाई 20 सेमी है तथा बेलन का व्यास 7 सेमी है। ठोस का कुल आयतन ज्ञात कीजिए। (π = \(\frac{22}{7}\) प्रयोग कीजिए)
हल:
माना ABCD एक ठोस बेलन है जिसके दोनों सिरों पर दो एक-समान अर्धगोले हैं।

दिया है, पूरे ठोस की लम्बाई = 20 सेमी हैं, 2r = 7
∴ AB की लम्बाई = 20 – 2r
= 20 – 2 × \(\frac{7}{2}\)
h = 13 सेमी
अब, बेलन ABCD का आयतन = πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 13\)
= 500.5 सेमी³
दोनों अर्धगोलों का आयतन = 2 × एक अर्धगोले का आयतन
= 2 × \(\frac{2}{3}\)πr³
= \(2 \times \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\)
= 179.67 सेमी³
ठोस का कुल आयतन = बेलन ABCD का आयतन + दोनों अर्धगोलों का आयतन
= 179.67 + 500.5
= 680.17 सेमी³

प्रश्न 13.
एक ही धातु के दो गोलों का भार 1 किलोग्राम तथा 7 किलोग्राम है। छोटे गोले की त्रिज्या 3 सेमी है। दोनों गोलों को पिघलाकर एक बड़ा गोला बनाया गया। नए गोले का व्यास ज्ञात कीजिए।
हल:
माना छोटे गोले की त्रिज्या r₁ तथा बड़े गोले की त्रिज्या r₂ है और माना इनसे मिलकर बनने वाले नए गोले की त्रिज्या R है।
छोटे गोले का आयतन = \(\frac{4}{3}\)πr₁³ [∵ r₁ = 3 ( दिया है)]
= \(\frac{4}{3}\)π(3)³ = 36π घन सेमी
अब,
छोटे गोले के पदार्थ का घनत्व = गोले का द्रव्यमान / गोले का आयतन
छोटे गोले के पदार्थ का आयतन = \(\frac{1}{36 \pi}\)
दोनों गोले एक ही पदार्थ से बने हैं, अतः इनके पदार्थ का घनत्व भी समान होगा।

अतः छोटे गोले का आयतन + बड़े गोले का आयतन = नए गोले का आयतन
⇒ \(\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^3=\frac{4}{3} \pi R^3\)
⇒ r₁³ + r₂³ = R³
⇒ 27 + 189 = R³
⇒ R³ = 216
R = 6 ⇒ D = 12 सेमी
अतः नए गोले का व्यास 12 सेमी है।

प्रश्न 14.
त्रिज्या 6 सेमी और ऊँचाई 15 सेमी वाले एक लंबवृत्तीय बेलन के आकार का बर्तन आइसक्रीम से पूरा भरा हुआ है। इस आइसक्रीम को 10 बच्चों में बाँटने के लिए बराबर-बराबर शंकुओं में भरा जाना है, जिनका ऊपरी सिरा अर्धगोले के आकार का है। यदि शंक्वाकार भाग की ऊँचाई इसके आधार की त्रिज्या का 4 गुना है, तो आइसक्रीम शंकु की त्रिज्या ज्ञात कीजिए।
हल:
माना R तथा H क्रमशः लंबवृत्तीय बेलन की त्रिज्या तथा ऊँचाई है।
दिया है, R = 6 सेमी तथा H = 15 सेमी
लंब वृत्तीय बेलन में आइसक्रीम का आयतन
= πR²H
= π × 36 × 15
= 540π सेमी³
अब माना कि शंकु के आधार की त्रिज्या r सेमी हैं. तब इसकी ऊँचाई (h) = 4r

शंकु में आइसक्रीम का आयतन = शंकु का आयतन + अर्द्धगोले का आयतन
= \(\frac{1}{3}\)πr²h + \(\frac{2}{3}\)πr³
= \(\frac{1}{3}\)πr²(h + 2r)
= \(\frac{1}{3}\)πr²(4r + 2r) (∵ h = 4r)
= \(\frac{1}{3}\)πr² × 6r
= 2πr³
अब, 10 × शंकु में आइसक्रीम का आयतन = बेलन में आइसक्रीम का आयतन
⇒ 10 × 2πr³ = 540π
⇒ r³ = 27
⇒ r = 3 सेमी
अतः आइसक्रीम शंकु की त्रिज्या 3 सेमी है।

प्रश्न 15.
6 मी चौड़ी और 1.5 मी गहरी एक नहर में पानी 10 किमी/घंटा की चाल से बह रहा है। 30 मिनट में, यह नहर कितने क्षेत्रफल की सिंचाई कर पाएगी, जबकि सिंचाई के लिए 8 सेमी गहरे पानी की आवश्यकता होती है।
हल:
दिया है, नहर की चौड़ाई = 6 मी तथा गहराई 1.5 मी है व नहर में बहने वाले पानी की चाल 10 किमी/घंटा है, तो नहर में बहने वाला पानी घण्टे में 10 किमी या 10000 मी दूरी तय करेगा।
तथा माना सिंचाई हेतु क्षेत्रफल x मी है।
तब 30 मिनट में पानी का आयतन = 6 × 1.5 × 10000 × \(\frac{30}{60}\)
= 45000 मी³
अब, सिंचाई क्षेत्र का आयतन = नहर के पानी का आयतन
x × \(\frac{8}{100}\) = 45000
x = \(\frac{45000 \times 100}{8}\)
x = 562500 मी² या 56.25 हेक्टेयर

प्रश्न 16.
शंकु के छिन्नक के आकार की ऊपर से खुली एक बाल्डी का आयतन 12308.8 घन सेमी है। इसके ऊपरी तथा निचले वृत्तीय सिरों की त्रिज्याएँ क्रमशः 20 सेमी तथा 12 सेमी हैं। बाल्टी की ऊंचाई तथा इसके बनाने में लगी धातु की चादर का क्षेत्रफल ज्ञात कीजिए। [π = 1.732 लीजिए]
हल:
दिया है, r₁ = 20 सेमी, r₂ = 12 सेमी तथा छिन्नक का आयतन 12308.8 सेमी³ है।

अब,
छिन्नक का आयतन = \(\frac{1}{3}\) × π × h (r₁² + r₂² + r₁r₂)
12308.8 = \(\frac{1}{3}\) × 3.14 × (20² + 12² + 20 × 12)
= \(\frac{1}{3}\) × 3.14 × h(400 + 144 + 240)
12308.8 = \(\frac{1}{3}\) × 3.14(784) × h
h = \(\frac{12308.8 \times 3}{3.14 \times 784}\)
h = 15 सेमी
धातु का क्षेत्रफल = πl(r₁ + r₃) + πr₂²
यहाँ l = \(\sqrt{\left(r_1-r_2\right)^2+h^2}\)
= \(\sqrt{(20-12)^2+15^2}\)
= \(\sqrt{289}\)
= 17 सेमी
धातु का क्षेत्रफल = 3.14 × 17 (20 + 12) + 3.14 × 12²
= 1708.16 + 452.16
= 2160.32 सेमी²

प्रश्न 17.
लकड़ी के एक ठोस बेलन के प्रत्येक सिरे पर एक अर्थ गोला खोद कर निकालते हुए, एक वस्तु बनाई गई, जैसा कि दी गई आकृति में दर्शाया गया है। यदि बेलन की ऊँचाई 10 सेमी है और आधार की त्रिज्या 3.5 सेमी है, तो इस वस्तु का सम्पूर्ण पृष्ठीय क्षेत्रफल ज्ञात कीजिए।

हल:
वस्तु का सम्पूर्ण पृष्ठीय क्षेत्रफल = बेलन का वक्रपृष्ठीय क्षेत्रफल + 2 × अर्द्धगोले का वक्रपृष्ठीय क्षेत्रफल

= 2πrh + 2 × 2πr²
= 2πr[h + 2r]
= 2 × \(\frac{22}{7} \times \frac{35}{10}\)[10 + 2 × 3.5]
= 22[17]
= 374 सेमी²

प्रश्न 18.
चावल की एक ढेरी शंकु के आकार की है जिसके आधार का व्यास 24 मी तथा ऊँचाई 3.5 मी है। चावलों का आयतन ज्ञात कीजिए इस ढेरी को पूरा-पूरा ढकने के लिए कितने कैनवस की आवश्यकता हैं?
हल:
शंक्वाकार ढेरी का व्यास = 24 मी
त्रिज्या, r = 12 मी
ऊँचाई, h = 35 मी
तिर्यक ऊँचाई, l = \(\sqrt{h^2+r^2}\)
= \(\sqrt{(3.5)^2+(12)^2}\)
= \(\sqrt{12.25+144}\)
= \(\sqrt{156.25}\)
= 12.5 मी
चावलों का आयतन = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3} \times \frac{22}{7} \times(12)^2 \times(3.5)\)
= \(\frac{11}{3}\) × 144
= \(\frac{1584}{3}\)
= 528 मी³
ढेरी को ढकने के लिए कैनवास की आवश्यकता = शंकु का वक्रपृष्ठीय क्षेत्रफल
= πrl
= \(\frac{22}{7}\) × 12 × 12.5
= \(\frac{3300}{7}\)
= 471 \(\frac{3}{7}\) मी²

प्रश्न 19.
शंकु के छिन्नक के आकार की एक बाल्टी के निचले तथा ऊपरी किनारों के व्यास क्रमशः 10 सेमी तथा 30 सेमी हैं। यदि बाल्टी की ऊँचाई 24 सेमी है, तो ज्ञात कीजिए :
(i) बाल्टी को बनाने में लगने वाली धातु की शीट का क्षेत्रफल।
(ii) बाल्टी बनाने में सामान्य प्लास्टिक को क्यों नहीं लगाना चाहिए? [π = 3.14 लीजिए]
हल:
बाल्टी के ऊपरी सिरे का व्यास = 30 सेमी
बाल्टी के ऊपरी सिरे की त्रिज्या R = 15 सेमी
बाल्टी के निचले सिरे का व्यास = 10 सेमी
बाल्टी के निचले सिरे की त्रिज्या, r = 5 सेमी
बाल्टी की ऊँचाई, h = 24 सेमी
तिर्यक ऊँचाई, l = \(\sqrt{h^2+(R-r)^2}\)
= \(\sqrt{(24)^2+(15-5)^2}\)
= \(\sqrt{576+100}\)
= \(\sqrt{676}\)
= 26 सेमी

(i) बाल्टी बनाने में लगी धातु की शीट का क्षेत्रफल = छिन्नक का वक्रपृष्ठीय क्षेत्रफल + आधार का क्षेत्रफल
= πl(R + r) + πr²
= π[l(R + r) + r²]
= 3.14[26(15 + 5) + 5²]
= 3.14[520 + 25]
= 3.44(545)
= 1711.3 सेमी²
(ii) बाल्टी बनाने में सामान्य प्लास्टिक इसलिए नहीं लगानीं चाहिये, क्योंकि उसकी शक्ति तथा गलनांक बहुत कम होता है और प्लास्टिक वातावरण के लिए हानिकारक है।

प्रश्न 20.
5.4 मी चौड़ी और 1.8 मी गहरी एक नहर में पानी 25 किमी / घण्टा की गति से बह रहा है। इससे 40 मिनट में कितने प्रतिशत क्षेत्रफल की सिंचाई हो सकती है, यदि सिंचाई के लिए 10 सेमी गहरे पानी की आवश्यकता है।
हल:
नहर की चौड़ाई = 5.4 मी
नहर की गहराई = 1.8 मी
नहर में 1 घण्टे में बहे पानी की लम्बाई = 25 किमी = 25000 मी
नहर में 1 घण्टे में बहे पानी का आयतन
= l × b × h
= 5.4 × 1.8 × 25000
= 243000 मी³
40 मिनट में पानी का आयतन = 243000 × \(\frac{40}{60}\)
= 162000 मी³
10 सेमी ऊँचाई के सींचे जाने वाले खेत का का क्षेत्रफल

= 162000 मी²
= 162 हेक्टेयर

प्रश्न 21.
एक शंकु के छिन्नक की तिर्यक ऊँचाई 4 सेमी है तथा इसके वृत्तीय सिरों की परिमाप 18 सेमी और 6 सेमी है। इस छिन्नक का व्रक पृष्ठीय क्षेत्रफल ज्ञात कीजिए।
हल:
छिन्नक को तिर्यक ऊँचाई = 4 सेमी
ऊपरी हिस्से का परिमाप = 18 सेमी
⇒ 2πR = 18 सेमी
⇒ R = \(\frac{9}{4}\) सेमी
निचले हिस्से का परिमाप = 6 सेमी
2πr = 6 ⇒ r = \(\frac{3}{8}\) सेमी
छिन्नक का वक्रपृष्ठ = πl[R + r]
= π × 4 × \(\left[\frac{9}{\pi}+\frac{3}{\pi}\right]\)
= π × 4 × \(\frac{12}{\pi}\) = 48 सेमी²

प्रश्न 22.
एक ठोस लोहे के घनाभ की विमाएँ 4.4 मी × 2.6 मी × 2.0 मी है। इसे पिघलाकर 30 सेमी आन्तरिक त्रिज्या और 5 सेमी मोटाई का एक खोखला बेलनाकार पाइप बनाया गया है। पाइप की लंबाई ज्ञात कीजिए।
हल:
पाइप की आन्तरिक त्रिज्या, r = 30 सेमी
पाइप की मोटाई = 5 सेमी
पाइप की बाहरी त्रिज्या = 30 + 5
R = 35 सेमी
माना, पाइप की लम्बाई = h सेमी
खोखले पाइप की आयतन = घनाभ की आयतन
πrh[R² – r²] = l × b × h
\(\frac{22}{7}\) × h[35² – 30²] = 4.4 × 2.6 × 1 × 100 × 100 × 100
\(\frac{22}{7}\) × h × 65 × 5 = 44 × 26 × 1 × 100 × 100
h = \(\frac{44 \times 26 \times 100 \times 100 \times 7}{22 \times 65 \times 5}\)
= 11200 सेमी
⇒ h = 112 मीटर

प्रश्न 23.
किसी वर्षा जल संग्रहण तन्त्र में, 22 मी × 20 मी की छत से वर्षा जल बहकर 2 मी आधार के व्यास तथा 3.5 मी ऊँचाई के एक बेलनाकार टैंक में आता है। यदि टैंक भर गया हो, तो ज्ञात कीजिए कि सेमी में कितनी वर्षा हुई। जल संरक्षण पर अपने विचार व्यक्त कीजिए।
हल:
एकत्रित पानी का आयतन = बेलनाकार टैंक का आयतन
L × B × H = πr²h
20 × 20× H = \(\frac{22}{7}\) × 1 × 1 × 3.5
22 × 20 × H = 11
H = \(\frac{11}{22 \times 20}=\frac{1}{40}\) मीटर
= \(\frac{1}{40} \times 100\)
= \(\frac{5}{2}\) = 2.5 सेमी
अतः 2.5 सेमी वर्षा हुई।
जल संरक्षण अति आवश्यक है, वह सूखे कि स्थिति के समय में कारगर सिद्ध होता है।

प्रश्न 24.
5 सेमी आंतरिक त्रिज्या तथा 24 सेमी ऊँचाई के एक शंक्वाकार बर्तन का \(\frac{3}{4}\) भाग पानी से भरा है। इस पानी को 10 सेमी आंतरिक त्रिज्या के बेलनाकार बर्तन में खाली किया जाता है। बेलनाकार बर्तन में पानी की ऊँचाई ज्ञात कीजिए।
हल:
प्रश्नानुसार,
शंक्वाकार बर्तन में \(\frac{3}{4}\) भाग का आयतन = बेलनाकार बर्तन का आयतन

अतः बेलनाकार बर्तन में पानी की ऊँचाई 1.5 सेमी है।

वस्तुनिष्ठ प्रश्न :

रिक्त स्थानों की पूर्ति कीजिए :

प्रश्न (क).
1. ऐसी वस्तुओं को जो स्थान घेरती हैं ………….. आकृतियाँ लाती हैं।
2. ठोस आकृति की सीमा बनाने वाली समतल आकृतियों का क्षेत्रफल ……………. क्षेत्रफल कहलाता है।
3. कोई वस्तु जितना स्थान घेरती है, उसे उस वस्तु का ……………… कहते है।
4. एक …………… के आयताकार समतल फलक होते हैं।
5. किसी आयत के उसकी एक स्थिर भुजा के परितः घुमाने से प्राप्त आकृति लम्बवृत्तीय ………….. कहलाती हैं।
उत्तर:
1. ठोस,
2. पृष्ठीय,
3. आयतन,
4. घनाभ,
5. बेलन।

निम्न में सत्य / असत्य बताइए :

प्रश्न (ख).
1. बेलन का आयतन 2πrh वर्ग इकाई होता है।
2. खोखले बेलन में लगे पदार्थ का आयतन दोनों बेलनों के आयतनों के योग के बराबर होता है।
3. गोले का आयतन 4πr² घन इकाई होता है।
4. अर्द्धगोले का सम्पूर्ण पृष्ठ = 3πr² वर्ग इकाई ।
5. शंकु का आयतन = πrl घन इकाई
उत्तर:
1. असत्य,
2. असत्य,
3. असत्य,
4. सत्य,
5. असत्य।

(ग) बहुविकल्पीय प्रश्न :

प्रश्न 1.
12π घन सेमी आयतन वाले गोले की त्रिज्या (सेमी में) है :
(A) 3
(B) 3\(\sqrt{3}\)
(C) 3^2/3
(D) 3^1/2
हल:
गोले का आयतन = \(\frac{4}{3}\)πr³
12π = \(\frac{4}{3}\)πr³
r³ = \(\frac{12 \pi \times 3}{4 \pi}\) = 9
r³ = 3²
⇒ r = 3^2/3
अत: सही विकल्प (C) है।

प्रश्न 2.
एक ठोस अर्धगोले का कुल पृष्ठीय क्षेत्रफल हैं:
(A) 3πr²
(B) 2πr²
(C) 4πr²
(D) \(\frac{2}{3}\)πr³
हल:
कुल पृष्ठीय क्षेत्रफल = 2πr² + πr² = 3πr²
अत: सही विकल्प (A) है।

प्रश्न 3.
एक 22 सेमी. आंतरिक किनारे वाले खोखले घन को 0.5 सेमी व्यास वाले गोलाकार कंचो से भरा जाता है तथा यह कल्पना की जाती है कि घन का भाग भरा नहीं जा सकता है। तब घन में समावेशित होने वाले कंचों की संख्या है :
(A) 142296
(B) 142396
(C) 142496
(D) 142596
हल:
दिया है,
घन की आन्तरिक भुजा = 22 सेमी.
तथा गोलाकार कंचे का व्यास = 0.5 सेमी.
∴ गोलाकार कंचे की त्रिज्या (r) = \(\frac{0.5}{2}\) सेमी.
= \(\left(\frac{5}{20}\right)\) सेमी.
∴ 1 गोलाकार कंचे का आयतन = \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7} \times\left(\frac{5}{20}\right)^3\)
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{5 \times 5 \times 5}{20 \times 20 \times 20}\)
माना घन में समावेशित होने वाले कंचों की संख्या x है।
तब x कंचों का आयतन = \(x \times \frac{4}{3} \times \frac{22}{7} \times \frac{5 \times 5 \times 5}{20 \times 20 \times 20}\)
= \(\frac{x \times 55}{21 \times 40}\) घन सेमी. ……….(1)
घन का आन्तरिक आयतन = 22 × 22 × 22 घन सेमी.
……. खोखले घन को कंचों से भरा जाता है, तो पन का \(\frac{1}{8}\) भाग भरा नहीं जा सकता है।
∴ खोखले घन का कंचों से भरा गया भाग = 1 – \(\frac{1}{8}\) = \(\frac{7}{8}\)
खोखले घन का कंचों से भरे भाग का आयतन = \(\frac{7}{8}\) × 22 × 22 × 22 घन सेमी. …..(2)
समीकरण (1) तथा (2) से
\(\frac{x \times 55}{21 \times 40}\) = \(\frac{7}{8}\) × 22 × 22 × 22
⇒ x = \(\frac{7 \times 22 \times 22 \times 22 \times 21 \times 40}{55 \times 8}\)
= 14 × 22 × 22 × 21 = 142296
अत: सही विकल्प (A) है।

प्रश्न 4.
यदि दो शंकुओं की त्रिज्याओं में अनुपात 3 : 1 और ऊँचाइयों में अनुपात 1 : 3 है, तो उनके आयतनों में अनुपात होगा :
(A) 2 : 1
(B) 3 : 1
(C) 1 : 3
(D) 1 : 2
हल:
माना दो शंकुओं की त्रिज्याएँ क्रमशः r₁ और r₂ ऊँचाइयाँ h₁ और h₂ हैं।
प्रश्नानुसार, \(\frac{r_1}{r_2}=\frac{3}{1}\) और \(\frac{h_1}{h_2}=\frac{1}{3}\)
दोनों शंकुओं के आयतनों का अनुपात

दोनों शंकुओं के आयतनों का अनुपात = 3 : 1
अत: सही विकल्प (B) है।

प्रश्न 5.
14 सेमी भुजा के एक घन से एक बड़े से बड़ा शंकु काटा जाता है। शंकु का आयतन है:
(A) 766.18 घन सेमी
(B) 817.54 घन सेमी
(C) 1232 घन सेमी
(D) 718.66 घन सेमी।

हल:
14 सेमी भुजा के घन से बड़े से बड़ा शंकु काटा जाता है।
अतः शंकु की ऊँचाई (h) = 14 सेमी
शंकु के आधार का व्यास = 14 सेमी
अत: शंकु के आधार की त्रिज्या = व्यास / 2 = \(\frac{14}{2}\)
= 7 सेमी
शंकु का आयतन = \(\frac{1}{3}\)πr²

अत: सही विकल्प (D) है।

प्रश्न 6.
दो गोलों के आयतनों का अनुपात 64 : 27 है। उनके पृष्ठीय क्षेत्रफलों का अनुपात है :
(A) 3 : 4
(B) 4 : 3
(C) 9 : 16
(D) 16 : 9
हल दिया है,
माना कि दो गोलों की त्रिज्याएँ क्रमशः r₁ तथा r₂ है। पहले गोले का आयतन : दूसरे गोले का आयतन = 64 : 27

अत: सही विकल्प (B) है।

प्रश्न 7.
क्रमशः आन्तरिक और बाहरी व्यास 4 सेमी और 8 सेमी वाले धातु के गोलाकार खोल को पिघलाकर आधार व्यास 8 सेमी. के एक शंकु के आकार में ढाला जाता है। इस शंकु की ऊँचाई है:
(A) 12 सेमी
(B) 14 सेमी
(C) 15 सेमी
(D) 18 सेमी
हल:
दिया है,
गोलाकार खोल की आन्तरिक व्यास = 4 सेमी
∴ गोलाकार खोल की आन्तरिक त्रिज्या
(r₁) = \(\frac{4}{2}\) = 2 सेमी
गोलाकार खोल की बाहरी व्यास = 8 सेमी
∴ गोलाकार खोल की बाहरी त्रिज्या
(r₂) = \(\frac{8}{2}\) = 4 सेमी
∴ शंकु का व्यास (r) = \(\frac{8}{2}\) = 4 सेमी
माना शंकु की ऊँचाई h सेमी है।
……. गोलाकार खोल को पिघलाकार शंकु के आकार में ढाला जाता है।
∴ गोलाकार खोल का आयतन = शंकु का आयतन
⇒ \(\frac{4}{3}\)π[r₂³ – r₁³] = \(\frac{1}{3}\)πr²h
⇒ 4[4³ – 2³] = 4² × h
⇒ 4 × [64 – 8] = 16 × h
⇒ 4 × 56 = 16 × h
⇒ h = \(\frac{4 \times 56}{16}\) = 14 सेमी
अत: सही विकल्प (B) है।

प्रश्न 8.
आधार व्यास 2 सेमी. और ऊँचाई 16 सेमी. वाले धातु के एक ठोस बेलन को पिघलाकर समान माप के बारह ठोस गोले बनाए जाते हैं। प्रत्येक गोले का व्यास है:
(a) 4 सेमी
(b) 3 सेमी
(c) 2 सेमी
(d) 6 सेमी
हल:
दिया है,
धातु के ठोस बेलन की ऊंचाई (h) = 16 सेमी
और धातु के ठोस बेलन का व्यास = 2 सेमी
∴ धातु के ठोस बेलन की त्रिज्या (r₁) = \(\frac{2}{2}\) = 1 सेमी
माना कि पिघलाकर समान माप के प्रत्येक ठोस गोले की त्रिज्या r₂ सेमी है।
….. ठोस बेलन को पिघलाकर समान माप के 12 ठोस गोले बनाए जाते हैं।
∴ ठोस बेलन का आयतन = 12 ठोस गोलों का आयतन
⇒ πr₁²h = 12 × \(\frac{4}{3}\)πr₂³
⇒ r₁²h = 16r₂³
⇒ 1² × 16 = 16r₂³
⇒ \(\frac{16}{16}\) = r₂³
⇒ r₂³ = 1 सेमी.
⇒ r₂ = \(\sqrt[3]{1}\) = 1 सेमी
∴ ठोस गोले का व्यास = 2 × 1 = 2 सेमी.
अंत: सही विकल्प (C) है।

प्रश्न 9.
यदि 11 सेमी × 3.5 सेमी × 2.4 सेमी मोम के एक घनाभ से 2.8 सेमी व्यास की एक मोमबत्ती बनाई जाती है। मोमबत्ती की लम्बाई होगी:
(A) 14 सेमी
(B) 15 सेमी
(C) 25 सेमी
(D) 12 सेमी
हल:
दिया है,
घनाभ की विमाएँ = 11 सेमी. × 3.5 सेमी × 2.4 सेमी.
∴ घनाभ का आयतन = 11 × 3.5 × 2.4
= 92.4 घन सेमी
माना मोमबत्ती की ऊँचाई h सेमी है।
मोमबत्ती का व्यास = 2.8 सेमी
मोमबत्ती की त्रिज्या (r) = \(\frac{2.8}{2}\) = 1.4 सेमी
प्रश्नानुसार,
मोमबत्ती का आयतन = घनाभ का आयतन
⇒ πr²h = 92.4
⇒ \(\frac{22}{7}\) × (1.4)² × h = 92.4
⇒ \(\frac{22}{7}\) × 1.4 × 14 × h = 92.4
⇒ h = \(\frac{92.4 \times 7}{22 \times 1.4 \times 1.4}\)
⇒ h = \(\frac{4.2}{0.2 \times 1.4}\) = 15 सेमी
मोमबत्ती की ऊँचाई = 15 सेमी
अत: सही विकल्प (B) है।

प्रश्न 10.
एक जलाशय लम्बवृत्तीय शंकु के छिन्नक के आकार में है। इसका ऊपरी सिरा 8 मीटर तथा पेंदी वाला सिरा 4 मीटर चौड़ा है। यदि यह 6 मीटर गहरा हो, तो इसकी क्षमता है:
(A) 176 मीटर³
(B) 196 मीटर³
(C) 200 मीटर³
(D) 110 मीटर³
हल:
दिया है,
छिन्नक के ऊपरी सिरे का व्यास = 8 मीटर
छिन्नक के ऊपरी सिरे की त्रिज्या (r₁) = \(\frac{8}{2}\) = 4 मीटर
छिन्नक के निचले सिरे का व्यास = 4 मीटर
छिन्नक के निचले सिरे की त्रिज्या (r₂) = \(\frac{4}{2}\) = 2 मीटर
छिन्नक की ऊँचाई (h) = 6 मीटर
छिन्नक का आयतन = \(\frac{1}{3}\)π × (r₁² + r₂² + r₁r₂)h
= \(\frac{1}{3} \times \frac{22}{7}\)(4² + 2² + 4 × 2) × 6
= \(\frac{22}{7}\) × (16 + 4 + 8) × 2
= \(\frac{22 \times 56}{7}\) = 22 × 8 = 176 घन मीटर
अत: सही विकल्प (A) है।

प्रश्न 11.
एक तेल की कुप्पी शंकु के छिन्नक को बेलन से जोड़ने पर बनी है। कुप्पी की ऊँचाई 12 सेमी है। बेलनाकार भाग की त्रिज्या 4 सेमी है और कुप्पी के. ऊपरी सिरे की त्रिज्या 9 सेमी है, तो शंकु के छिन्नक की तिर्यक ऊंचाई होगी:
(A) 13 सेमी
(B) 23 सेमी
(C) 26 सेमी
(D) 13.5 सेमी
हल:
कुप्पी के ऊपरी सिरे की त्रिज्या (r₁) = 9 सेमी
कुप्पी के निचले सिरे की त्रिज्या (r₂) = 4 सेमी
कुप्पी की ऊँचाई (h) = 12 सेमी
शंकु के छिनक की तिर्यक ऊँचाई (l) = \(\sqrt{h^2+\left(r_1-r_2\right)^2}\)
= \(\sqrt{(12)^2+(9-4)^2}\)
= \(\sqrt{(12)^2+(5)^2}\)
= \(\sqrt{144+25}\)
= \(\sqrt{169}\)
= 13 सेमी
अतः सही विकल्प (A) है।

प्रश्न 12.
9 सेमी त्रिज्या के धातु के गोले को पिघलाकर 3 सेमी त्रिज्या और 6 सेमी ऊँचाई के शंकु बनाये जा सकने वाले शंकुओं की संख्या है:
(A) 54
(B) 45
(C) 55
(D) 44
हल:
दिया है,
धातु के गोले की त्रिज्या (R) = 9 सेमी
शंकु की त्रिज्या (r) = 3 सेमी
तथा शंकु की ऊँचाई (h) = 6 सेमी
9 सेमी त्रिज्या वाले गोले का आयतन
\(\frac{4}{3}\)πR² = \(\frac{4}{3}\) × π × (9)²
= \(\frac{4}{3}\) × π × 9 × 9 × 9
= 4 × π × 3 × 81
= 972π घन सेमी
अब 3 सेमी त्रिज्या और 6 सेमी ऊँचाई वाले शंकु का आयतन
\(\frac{1}{3}\)πr²h = \(\frac{1}{3}\) × π × (3)² × 6
= \(\frac{1}{3}\) × π × 9 × 6
= π × 3 × 6 = 18π घन सेमी
गोले का आयतन शंकुओं की संख्या

अत: सही विकल्प (A) है।

प्रश्न 13.
एक ठोस लम्ब वृत्तीय शंकु को उसकी ऊँचाई के बीचों बीच से होकर जाते, शंकु के आधार के समान्तर एक तल द्वारा दो भागों में काटा गया है। इस प्रकार छोटे शंकु के आयतन का पूरे शंकु के आयतन से अनुपात है:
(A) 1 : 2
(B) 1 : 4
(C) 1 : 6
(D) 1 : 8
हल:
माना OAB एक शंकु है जिसकी ऊँचाई h तथा त्रिज्या R है। शंकु को बीचों बीच से आधार के समान्तर काटा गया है। अतः छोठे शंकु की ऊँचाई (h’) = \(\frac{h}{3}\)
माना कि छोटे शंकु की त्रिज्या r है।

LD || MB
⇒ ∠OLD = ∠OMB (संगत कोण)
ΔOLD तथा ΔOMB में,
∠OLD = ∠OMB
∠LOD = ∠MOB (उभयनिष्ठ कोण)
ΔOLD ~ ΔOMB (AA समरूपता गुणधर्म से)
⇒ \(\frac{O L}{O M}=\frac{L D}{M B}\)
[समरूप त्रिभुजों की भुजाएँ समानुपाती होती हैं]
⇒ \(\frac{\frac{h}{2}}{h}=\frac{r}{R}\)
⇒ \(\frac{h}{2 h}=\frac{r}{R}\)
⇒ \(\frac{1}{2}=\frac{r}{R}\)
⇒ R = 2r
छोटे शंकु का आयतन : पूरे शंकु का आयतन
= πr²h’ : πR²h
= \(\frac{\pi \times r^2 \times \frac{h}{2}}{\pi \times(2 r)^2 \times h}\)
= \(\frac{r^2 \times h}{2 \times 4 r^2 \times h}=\frac{1}{8}\) = 1 : 8
अतः विकल्प (D) सही है।

प्रश्न 14.
तीनों घनों की कोर क्रमशः 3 सेमी, 4 सेमी और सेमी हैं। इनसे बनने वाले एक घन की भुजा है:
(A) 6 सेमी
(B) 5 सेमी
(C) 7 सेमी
(D) 4 सेमी
हल:
पहले घन का आयतन = (3)³ = 27 घन सेमी
दूसरे घन का आयतन = (4)³ = 64 घन सेमी
तीसरे घन का आयतन = (5)³ = 125 घन सेमी
इन तीनों घनों का कुल आयतन = 27 + 64 + 125
= 216 घन सेमी
∵ तीनों घनों से एक नया घन बनता है।
नये घन का आयतन = तीनों घनों का कुल आयतन
(भुजा)³ = 216
भुजा = \(\sqrt[3]{216}\)
= \(\sqrt[3]{6 \times 6 \times 6}\) = 6 सेमी
अतः सही विकल्प (A) है।

प्रश्न 15.
एक घनाभ की माप 18 सेमी × 12 सेमी × 9 सेमी है। इस घनाभ को पिघलाकार 3 सेमी भुजा वाले कितने घन बनाये जा सकते हैं ?
(A) 60
(B) 55
(C) 69
(D) 72
हल:
दिया है,
घनाभ की माप = 18 सेमी × 12 सेमी × 9 सेमी
दिये गये घनाभ का आयतन = 18 × 12 × 9
= 18 × 108
= 1944 घन सेमी
3 सेमी भुजा वाले घन का आयतन = (भुजा)³
= (3)³ = 27 घन सेमी
घनाभ को पिघलाकर बनाये गये घनों की संख्या = घनाभ का आयतन / 1 घन का आयतन
= \(\frac{1944}{27}\) = 72
अतः सही विकल्प (D) है।

Jharkhand Board JAC Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.3

Question 1.
Find the sum of following APs:
1. 2, 7, 12, ……, to 10 terms.
2. -37, -33, -29, ……, to 12 terms.
3. 0.6, 1.7, 2.8, ……, to 100 terms.
4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots\) to 11 terms.
Solution:
1. For the given AP 2, 7, 12, ……., a = 2.
d = 7 – 2 = 5, n = 10 and S_n is to be found.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₁₀ = \(\frac{10}{2}\)[4 + (10 – 1) 5]
= 5(49) = 245
Thus, the sum of first 10 terms of the given AP is 245.

2. For the given AP -37, -33, -29, a = -37, d = (-33) – (-37) = 4, n = 12 and S_n is to be found.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₁₂ = \(\frac{12}{2}\)[-74 + (12 – 1)4]
= 6(-30) = – 180
Thus, the sum of first 12 terms of the given AP is -180.

3. For the given AP 0.6, 1.7, 2.8,…… a = 0.6, d = 1.7 – 0.6 = 1.1, n = 100 and S_n is to be found.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀₀ = \(\frac{100}{2}\)[1.2 + (100 – 1) (1.1)]
= 50[1.2 + 108.9]
= 50 × 110.1 = 5505
Thus, the sum of first 100 terms of the given AP is 5505.

4.

Thus, the sum of first 11 terms of the given AP is \(\frac{33}{20}\).

Question 2.
Find the sums given below:
1. 7 + 10\(\frac{1}{2}\) + 14 + … + 84
2. 34 + 32 + 30 + … + 10
3. (-5) + (-8) + (-11) + … + (-230)
Solution:
1. 7 + 10\(\frac{1}{2}\) + 14 + …… + 84
Here, a = 7; d = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\); last term l = 84.
Let the last term be nth term.
a_n = a + (n – 1)d
∴ 84 = 7 + (n – 1) (3\(\frac{1}{2}\))
∴ 77 = \(\frac{7}{2}\)(n – 1)
∴ (n – 1) = 22
∴ n = 23
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{23}{2}\)(7 + 84)
= \(\frac{23}{2}\) × 91 = 1046\(\frac{1}{2}\)
Thus, the required sum is 1046\(\frac{1}{2}\).

2. 34 + 32 + 30 + … + 10
Here, a = 34; d = 32 – 34 = (-2); last term l = 10.
Let the last term be nth term.
a_n = a + (n – 1)d
∴ 10 = 34 + (n – 1)(-2)
∴ -24 = -2(n – 1)
∴ (n – 1) = 12
∴ n = 13
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{13}{2}\)(34 + 10)
= 13 × 22 = 286
Thus, the required sum is 286.

3. (-5) + (-8) + (-11) + … + (-230)
Here, a = (-5); d = (-8) – (-5) = (-3):
last term l = (-230).
Let the last term be nth term.
a_n = a + (n – 1)d
∴ -230 = -5 + (n – 1)(-3)
∴ -225 = -3 (n – 1)
∴ n – 1 = 75
∴ n = 76
Now,
S_n = \(\frac{n}{2}\)(a + l), where l is the last term.
= \(\frac{76}{2}\)[(-5) + (-230)]
= 38(-235) = -8930
Thus, the required sum is -8930.

Question 3.
In an AP:
1. Given a = 5, d = 3, a_n = 50, find n and S_n.
2. Given a = 7, a₁₃ = 35, find d and S₁₃.
3. Given a₁₂ = 37, d = 3, find a and S₁₂.
4. Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
5. Given d = 5, S₉ = 75, find a and a₉.
6. Given a = 2, d = 8, S_n = 90, find n and a_n.
7. Given a = 8, a_n = 62, S_n = 210, find n and d.
8. Given a_n = 4, d = 2, S_n = -14, find n and a.
9. Given a = 3, n = 8, S_n = 192, find d.
10. Given l = 28, S_n = 144, and there are total 9 terms. Find a.
Solution:
1. a = 5, d = 3, a_n = 50, n = ? S_n = ?
a_n = a + (n – 1)d
∴ 50 = 5 + (n – 1)3
∴ 45 = 3(n – 1)
∴ 15 = n – 1
∴ n = 16
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₆ = \(\frac{16}{2}\)(5 + 50)
∴ S₁₆ = 8 × 55
∴ S₁₆ = 440

2. a = 7, a₁₃ = 35, d = ?, S₁₃ = ?
a_n = a + (n – 1)d
a₁₃ = a + (13 – 1) d
∴ 35 = 7 + 12d
∴ 28 = 12d
∴ d = \(\frac{28}{12}\)
∴ d = \(\frac{7}{3}\)
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₃ = \(\frac{13}{2}\)(17 + 35)
∴ S₁₃ = 13 × 21
∴ S₁₃ = 273

3. a₁₂ = 37, d = 3, a = ?, S₁₂ = ?
a_n = a + (n – 1)d
∴ a₁₂ = a + 11d
∴ 37 = a + 11 (3)
∴ a = 4
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₂ = \(\frac{12}{2}\)(4 + 37)
∴ S₁₂ = 246

4. a₃ = 15, S₁₀ = 125, d = ?, a₁₀ = ?
a_n = a + (n – 1)d
∴ a₃ = a + 2d
∴ a + 2d = 15 …….(1)
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀ = \(\frac{10}{2}\)[2a + 9d]
∴ S₁₂₅ = 5(2a + 9d)
∴ 2a + 9d = 25 …….(2)
Solving equations (1) and (2), we get
d = -1 and a = 17.
a_n = a + (n – 1)d
∴ a₁₀ = a + 9d
∴ a₁₀ = 17 + 9(-1)
∴ a₁₀ = 8

5. d = 5, S₉ = 75, a = ?, a₉ = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₉ = \(\frac{9}{2}\)[2a + (9 – 1) d]
∴ 75 = \(\frac{9}{2}\)[2a + 8(5)]
∴ 75 = 9(a + 20)
∴ \(\frac{25}{3}\) = a + 20
∴ a = \(\frac{25}{3}\) – 20
∴ a = –\(\frac{35}{3}\)
a_n = a + (n – 1) d
∴ a₉ = a + 8d
∴ a₉ = (-\(\frac{35}{3}\)) + 8(5)
∴ a₉ = –\(\frac{35}{3}\) + 40
∴ a₉ = \(\frac{85}{3}\)

6. a = 2, d = 8, S_n = 90, n = ?, a_n = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 90 = \(\frac{n}{2}\)[4 + (n – 1)8]
∴ 90 = \(\frac{n}{2}\)[8n – 4]
∴ 90 = n (4n – 2)
∴ 4n² – 2n – 90 = 0
∴ 2n² – n – 45 = 0
∴ 2n² – 10n + 9n – 45 = 0
∴ 2n (n – 5) + 9 (n – 5) = 0
∴ (n – 5) (2n + 9) = 0
∴ n – 5 = 0 or 2n + 9 = 0
∴ n = 5 or n = –\(\frac{9}{2}\)
Since n is a positive integer, n ≠ –\(\frac{9}{2}\)
∴ n = 5
a_n = a + (n – 1)d
∴ a₅ = a + 4d
∴ a₅ = 2 + 4(8)
∴ a₅ = 34

7. a = 8, a_n = 62, S_n = 210, n = ? d = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ S_n = \(\frac{n}{2}\)(a + a_n)
∴ 210 = \(\frac{n}{2}\)(8 + 62)
∴ 420 = n (70)
∴ n = 6
a_n = a + (n – 1)d
∴ a₆ = a + 5d
∴ 62 = 8 + 5d
∴ 54 = 5d
∴ d = \(\frac{54}{5}\)

8. a_n = 4, d = 2, S_n = -14, n = ?, a = ?
a_n = a + (n – 1)d
∴ 4 = a + (n – 1) (2)
∴ 4 = a + 2n – 2
∴ a = 6 – 2n
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ -14 = \(\frac{n}{2}\)[2 (6 – 2n) + (n – 1) (2)] (by (1))
∴ -14 = \(\frac{n}{2}\)[12 – 4n + 2n – 2]
∴ -14 = \(\frac{n}{2}\)[-2n + 10]
∴ -14 = n(-n + 5)
∴ -14 = -n² + 5n
∴ n² – 5n – 14 = 0
∴ (n – 7)(n + 2) = 0
∴ n – 7 = 0 or n + 2 = 0
∴ n = 7 or n = -2
Since n is a positive integer, n ≠ -2.
∴ n = 7
By (1), a = 6 – 2n
∴ a = 6 – 2(7)
∴ a = -8

9. a = 3, n = 8, S_n = 192, d = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 192 = \(\frac{8}{2}\)[6 + (8 – 1) d]
∴ 192 = 4[6 + 7d]
∴ 48 = 6 + 7d
∴ 42 = 7d
∴ d = 6

10. l = 28, S_n = 144, n = 9, a = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ 144 = \(\frac{9}{2}\)(a + 28)
∴ 32 = (a + 28)
∴ a = 4

Question 4.
How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636?
Solution:
Here, a = 9; d = 17 – 9 = 8; S_n = 636, n = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 636 = \(\frac{n}{2}\)[18 + (n – 1)8]
∴ 636 = \(\frac{n}{2}\)[10 + 8n]
∴ 636 = n[4n + 5]
∴ 4n² + 5n – 636 = 0
Here, a = 4; b = 5; c = -636
b² – 4ac = (5)² – 4(4)(-636)
= 25 + 10176
= 10201
∴ \(\sqrt{b^2-4 a c}=\sqrt{10201}=101\)
Then, n = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ n = \(\frac{-5 \pm 101}{8}\)
∴ n = \(\frac{96}{2}\) or n = \(\frac{-106}{8}\)
∴ n = 12 or n = \(-\frac{53}{4}\)
As n denotes the numbers of terms, it is a positive integer.
∴ n = \(-\frac{53}{4}\) is not possible.
∴ n = 12
Thus, 12 terms of the AP 9, 17, 25,… must be taken to give a sum of 636.

Question 5.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
Here, a = 5; l = 45; S_n = 400; n = ?; d = ?
S_n = \(\frac{n}{2}\)(a + l)
∴ 400 = \(\frac{n}{2}\)(5 + 45)
∴ 800 = n (50)
∴ n = 16
l = a_n = a + (n – 1)d
∴ a₁₆ = a + 15d
∴ 45 = 5 + 15d
∴ 40 = 15d
∴ d = \(\frac{40}{15}\)
∴ d = \(\frac{8}{3}\)
Thus, the number of terms is 16 and the common difference is \(\frac{8}{3}\).

Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
Here, a = 17; l = a_n = 350; d = 9; n = ?; S_n = ?
a_n = a + (n – 1)d
∴ 350 = 17 + (n – 1)9
∴ 333 = 9 (n – 1)
∴ n – 1 = 37 ∴ n = 38
Again, S_n = \(\frac{n}{2}\)(a + l)
∴ S₃₈ = \(\frac{38}{2}\)(17 + 350)
∴ S₃₈ = 19 × 367
∴ S₃₈ = 6973
Thus, there are 38 terms and their sum is 6973.

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Here, a₂₂ = 149; d = 7; S₂₂ = ?
a_n = a + (n – 1) d
∴ a₂₂ = a + 21d
∴ 149 = a + 21 × 7
∴ a = 149 – 147
∴ a = 2
Now, S_n = \(\frac{n}{2}\)(a + l)
∴ S₂₂ = \(\frac{22}{2}\)(2 + 149)
∴ S₂₂ = 11 × 151
∴ S₂₂ = 1661
Thus, the sum of first 22 terms of the given AP is 1661.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Here, a₂ = 14; a₃ = 18; S₅₁ = ?
a_n = a + (n – 1)d
∴ a₂ = a + d = 14 ……..(1)
∴ a₃ = a + 2d = 18 ……..(2)
Solving equations (1) and (2), we get
d = 4 and a = 10.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₅₁ = \(\frac{51}{2}\)[20 + 50 × 4]
∴ S₅₁ = 51 × 110
∴ S₅₁ = 5610
Thus, the sum of first 51 terms of the given AP is 5610.

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Here, S₇ = 49; S₁₇ = 289: S_n = ?
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₇ = \(\frac{7}{2}\)[2a + 6d]
∴ 49 = 7(a + 3d)
∴ a + 3d = 7 ……..(1)
Again S₁₇ = \(\frac{17}{2}\)[2a + 16d]
∴ 289 = 17(a + 8d)
∴ a + 8d = 17 ……..(2)
Solving equations (1) and (2), we get d = 2 and a = 1.
∴ S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S_n = \(\frac{n}{2}\)[2 + (n – 1)2]
∴ S_n = \(\frac{n}{2}\)[2 + 2n – 2]
∴ S_n = \(\frac{n}{2}\)(2n)
∴ S_n = n²
Thus, the sum of first n terms of the given AP is n².

Question 10.
Show that a₁, a₂, ……., a_n, ……. form an AP where a is defined as below:
1. a_n = 3 + 4n
2. a_n = 9 – 5n
Also find the sum of the first 15 terms in each case.
Solution:
1. a_n = 3 + 4n
a₁ = 3 + 4(1) = 7,
a₂ = 3 + 4(2) = 11,
a₃ = 3 + 4(3) = 15,
a₄ = 3 + 4(4) = 19 and so on.
Here, a₄ – a₃ = a₃ – a₂ = a₂ – a₁ = 4.
Hence, a_k+1 – a_k remains the everywhere.
Hence, a₁, a₂, a₃, ….. defined as a_n = 3 + 4n form an AP in which a = 7 and d = 4.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₁₅ = \(\frac{15}{2}\)[14 + 14 × 4]
∴ S₁₅ = 15 × 35
∴ S₁₅ = 525
The sum of first 15 terms of the given AP is 525.

2. a_n = 9 – 5n
a₁ = 9 – 5(1) = 4,
a₂ = 9 – 5(2) = -1,
a₃ = 9 – 5(3) = -6,
a₄ = 9 – 5(4) = -11 and so on.
Here, a₄ – a₃ = a₃ – a₂ = a₂ – a₁ = -5.
Hence, a_k+1 – a_k remains the same everywhere.
Hence, a₁, a₂, a₃, ….. defined as a_n = 9 – 5n form an AP in which a = 4 and d = -5.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₅ = \(\frac{15}{2}\)[8 + 14 (-5)]
∴ S₁₅ = \(\frac{15}{2}\)(-62)
∴ S₁₅ = 15 × (-31)
∴ S₁₅ = -465
The sum of first 15 terms of the given AP is -465.

Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
For the given AP
S_n = 4n – n²
∴ S₁ = 4(1) – (1)² = 4 – 1 = 3,
S₂ = 4 (2) – (2)² = 8 – 4 = 4,
S₃ = 4(3) – (3)² = 12 – 9 = 3,
S₉ = 4 (9) – (9)² = 36 – 81 = -45,
S₁₀ = 4 (10) – (10)² = 40 – 100 = -60
Now, the first term = a – a₁ = S₁ = 3
The sum of first two terms S₂ = 4
The second term a₂ = S₂ – S₁ = 4 – 3 = 1
The third term a₃ = S₃ – S₂ = 3 – 4 = -1
The tenth term a₁₀ = S₁₀ – S₉
= -60 – (-45) = -15
Now, S_n = 4n – n²
∴ S_n-1 = 4(n – 1) – (n – 1)²
= 4n – 4 – n² + 2n – 1
= -n² + 6n – 5
Now nth term a_n = S_n – S_n-1
∴ a_n = (4n – n²) – (-n² + 6n – 5)
∴ a_n = 4n – n² + n² – 6n + 5
∴ a_n = -2n + 5

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The first 40 positive integers divisible by 6 form the AP 6, 12, 18, ……, 240.
Here, a = 6; d = 12 – 6 = 6; n = 40 and l = 240.
S_n = \(\frac{n}{2}\)(a + l)
∴ S₄₀ = \(\frac{40}{2}\)(6 + 240)
∴ S₄₀ = 20 × 246
∴ S₄₀ = 4920
Thus, the required sum is 4920.

Alternate method:
Required sum
= 6 + 12 + 18 + … + 240
= 6(1 + 2 + 3 + … + 40)
= 6 × \(\frac{40 \times 41}{2}\) (1 + 2 + 3 + …. + n = \(\frac{n(n+1)}{2}\))
= 4920

Question 13.
Find the sum of the first 15 multiples of 8.
Solution:
The first 15 multiples of 8 form the AP 8, 16, 24, ….., 120.
Here, a = 8, d = 16 – 8 = 8, n = 15 and l = 120.
S_n = \(\frac{n}{2}\)(a + l)
∴ S₁₅ = \(\frac{15}{2}\)(8 + 120)
∴ S₁₅ = 15 × 64
∴ S₁₅ = 960
Thus, the required sum is 960.

Alternate method:
Required sum
= 8 + 16 + 24 + … + 120
= 8(1 + 2 + 3 + … + 15)
= 8 × \(\frac{15 \times 16}{2}\) (1 + 2 + 3 + … + n = \(\frac{n(n+1)}{2}\))
= 960.

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
The odd numbers between 0 and 50 form the AP 1, 3, 5, ….., 49.
Here, a = 1, d = 3 – 1 = -2, l = 49.
Let the last term be the nth term.
a_n = a + (n – 1) d
49 = 1 + (n – 1)²
2(n – 1) = 48
n – 1 = 24
n = 25
Now, S_n = \(\frac{n}{2}\)(a + l)
∴ S₂₅ = \(\frac{25}{2}\)(1 + 49)
∴ S₂₅ = 25 × 25
∴ S₂₅ = 625
Thus, the required sum is 625.

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc.. the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Solution:
The sums (in rupees) of penalty for delay of completion form the AP 200, 250, 300, …..
Here, a = 200; d = 250 – 200 = 50; n = 30 as the contractor has delayed the work by 30 days. The total penalty amount (in rupees) to be paid by the contractor will be given by S₃₀.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S₃₀ = \(\frac{30}{2}\)[400 + (30 – 1)50]
∴ S₃₀ = 15 × 1850
∴ S₃₀ = 27750
Thus, the contractor has to pay a penalty of ₹ 27,750.

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If cach prize is 20 less than its preceding prize, find the value of each of the prizes.
Solution:
₹ 700 is to be distributed as seven prizes such that each prize is ₹ 20 less than its preceding prize. Let the highest prize, i.e., the first prize be ₹ a. Then, the second prize will be of ₹ a – 20, the third prize will be of ₹ a – 40 and so on up to seven prizes. Hence, the amount (in rupees) of these prizes form a finite AP with seven terms as a, a – 20, a – 40, a – 60, a – 80, a – 100 and a – 120.
Here, the first term = a; d = (a – 20) – a = -20;
n = 7 and the sum of all the terms = S₇ = 700.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 700 = \(\frac{7}{2}\)[2a + (7 – 1) (-20)]
∴ 200 = 2a + 6(-20)
∴ 200 = 2a – 120
∴ 2a = 320
∴ a = 160
Then, a – 20 = 140; a – 40 = 120; a – 60 = 100; a – 80 = 80; a – 100 = 60 and a – 120 = 40.
Thus, the values (in rupees) of those seven prizes are 160, 140, 120, 100, 80, 60 and 40.

Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class. will plant, will be the same as the class, in which they are studying, e.g.. a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
The number of trees that the three sections of Class I will plant = 1 + 1 + 1 = 3.
The number of trees that the three sections of Class II will plant = 2 + 2 + 2 = 6.
This system will continue till Class XII.
The number of trees that the three sections of Class XII will plant = 12 + 12 + 12 = 36.
Thus, the number of trees that will be planted will form a finite AP with 12 terms as 3, 6, 9, ….., 36.
Here, a = 3, d = 6 – 3 = 3, n = 12 and S₁₂ will give the total number of trees that will be planted.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₂ = \(\frac{12}{2}\)[6 + (12 – 1)³]
∴ S₁₂ = 6 × 39
∴ S₁₂ = 234
Thus, 234 trees will be planted by the students.

Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, as shown in the given figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = \(\frac{22}{7}\))

Hint: Length of successive semicircles is l₁, l₂, l₃, l₄, …with centres at A, B, A, B, ….., respectively.]
Solution:
We know that the length of a semicircle = πr, where r is the radius.
Length of 1st semicircle with centre A and radius 0.5 cm = l₁ = π × 0.5 cm.
Length of 2nd semicircle with centre B and radius 1 cm = l₂ = π × 1 cm.
Length of 3rd semicircle with centre A and radius 1.5 cm = l₃ = π × 1.5 cm.
This system continues till 13 semicircles are drawn.
Then, the 13th semicircle will be drawn with centre A and radius 6.5 cm. Length of 13th semicircle with centre A and radius 6.5 cm = l₁₃ = π × 6.5 cm.
Now, the total length of the spiral
= l₁ + l₂ + l₃ + … + l₁₃
= (π × 0.5) + (π × 1) + (π × 1.5) + … + (π × 6.5)
= π(0.5 + 1 + 1.5 + … + 6.5)
The sum inside the brackets is the sum of all the 13 terms of the finite AP 0.5, 1, 1.5, ….., 6.5.
For this AP a = 0.5; d = 1 – 0.5 = 0.5 and n = 13.
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
∴ S_n = [1 + (13 – 1) (0.5)]
∴ S_n = \(\frac{13}{2}\) × 7
Hence, the total length of the spiral
= π\(\left(\frac{13}{2} \times 7\right)\)
= \(\frac{22}{7} \times \frac{13}{2} \times 7\)
= 143 cm
Thus, the total length of the spiral made up of thirteen consecutive semicircles is 143 cm.

Question 19.
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see the given figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:
The number of logs stacked in the first row from the bottom = 20.
The number of logs stacked in the second row from the bottom = 19.
The number of logs stacked in the third row from the bottom = 18.
This system continues till all the 200 logs are stacked.
Thus, the number of logs stacked in the rows form the finite AP 20, 19, 18,….. up ton-terms and the sum of those n terms is 200. Here, a = 20 and d = 19 – 20 = -1.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ 200 = \(\frac{n}{2}\)[40 + (n – 1) (-1)]
∴ 400 = n(40 – n + 1)
∴ 400 = n(41 – n)
∴ 400 = 41n – n²
∴ n² – 41n + 400 = 0
∴ n² – 16n – 25n + 400 = 0
∴ n (n – 16) – 25 (n – 16) = 0
∴ (n – 16) (n – 25) = 0
n – 16 = 0 or n – 25 = 0
n = 16 or n = 25
Here, both the answers are admissible. Hence, we verify by the value of 16th term and 25th term.
a_n = a + (n – 1) d
∴ a₁₆ = 20 + 15(-1) = 5
∴ a₂₅ = 20 + 24(-1) = -4
Thus, for the 25th row, the number of logs in the row becomes negative. This is inadmissible.
Hence, n ≠ 25.
∴ n = 16 and a₁₆ = 5.
Thus, the 200 logs are placed in 16 rows and in the top row there are 5 logs.

Question 20.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see the given figure).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)].
Solution:
The distance (in metres) to be covered to pick up first potato = 2 × 5 = 10.
The distance (in metres) to be covered to pick up second potato = 2 × (5 + 3) = 16.
The distance (in metres) to be covered to pick up third potato = 2 × (5 + 3 + 3) = 22.
Thus, the distances to be covered to pick up 10 potatoes form the finite AP 10, 16, 22, …. up to 10 terms.
Here, a = 10, d = 16 – 10 = 6, n = 10 and Sn will give the total distance (in metres) that the competitor has to run.
S_n = \(\frac{n}{2}\)[2a + (n – 1) d]
∴ S₁₀ = \(\frac{10}{2}\)[20 + (10 – 1)6]
∴ S₁₀ = 5 × 74
∴ S₁₀ = 370
Thus, the total distance that the competitor has to run is 370 metres.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
The class interval having the maximum frequencies is 35 – 45.
f₁ = 23, l = 35, h = 10, f₀ = 21, f₂ = 14


Maximum number of patients admitted in the hospital are of the age 36.8 years. The average age of the patient admitted to the hospital is 35.37 years.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical

Determine the modal lifetimes of the components.
Class interval having the maximum frequency is 60 – 80.
f₁ = 61, f₀ = 52, f₂ = 38, l = 60, h = 20.

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (in Rs.)	No. of families
1000 – 1500	24
1500 – 2000	40
2000 – 2500	33
2500 – 3000	28
3000 – 3500	30
3500 – 4000	22
4000 – 4500	16
4500 – 5000	7

Solution:

Step deviation method: \(\bar{x}\) = a + \(\left[\frac{\sum f_i u_i}{\sum f_i}\right]\)h
= 3250 + \(\left[\frac{-235}{200}\right]\) × 500
= 3250 – \(\left[-\frac{1175}{2}\right]\)
= 3250 – 587.5
= 2662.5.
Mean expenditure Rs. 2662.50.
l = 1500, f₁ = 40, f₂ = 33, f₀ = 24, h = 500

Modal monthly expenditure = 1847.83.

Question 4.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher	No. of states/U.T.
15 – 20	3
20 – 25	8
25 – 30	9
30 – 35	10
35 – 40	3
40 – 45	0
45 – 50	0
50 – 55	2

Solution:

l = lower limit of the CI = 30, f₁ = 10, f₀ = 9, f₂ = 3, h = 5
Mode = \(l+\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right] \times \mathrm{h}\)
= \(=30+\left[\frac{10-9}{20-9-3}\right] \times 5\)
= \(30+\left[\frac{1}{8} \times 5\right]=30+\frac{5}{8}\)
= 30 + 0.625 = 30.625.
Most states/UT’s have a student-teacher ratio of 30.6. On an average this ratio is 29.2.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored	Number of batsmen
3000 – 4000	4
4000 – 5000	18
5000 – 6000	9
6000 – 7000	7
7000 – 8000	6
8000 – 9000	3
9000 – 10000	1
10000 – 11000	1

Find the mode of the data.
Solution:
l = 4000, f₁ = 18, f₀ = 4, f₂ = 9, h = 1000

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.

Solution:
Class interval having the maximum frequency is 40 – 50. f₁ = 20, f₀ = 12, f₂ = 11, l = 40, h = 10.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1

Unless stated otherwise, use π = \(\frac{22}{7}\)

Question 1.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution :
Circumference of circle 1 = 2πr
= 2 × π × 19 cm = 2π × 19
Circumference of circle 2 = 2 × π × 9 cm = 2π × 9
Sum of the circumferences = (2π × 19) + (2π × 9)
= 2π(19 + 9)
= 2π 28
= 56 π.
Circumference of circle 3 = 56π.
2πr = 56π
r = \(\frac{56π}{2π}\)
= 28 cm.

Alternative Method:
r₁ = 19 cm, r₂ = 9 cm, R = ?
2πr₁ + 2πr₂ = 2πR (given)
2π(г₁ + r₂) = 2πR
(19 + 9) = R
∴ R = 28 cm.

Question 2.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution :
Area of circle 1 = π × 8² cm²
Area of circle 2 = π × r² = π × 6²
Sum of the areas = π(8² + 6²)
= π(64 + 36)
= 100π
Area of circle 3 = 100 π
πr² = 100 π
r² = \(\frac{100π}{π}\) = 100
∴ r = 10 cm.

Alternative Method:
πr₁² + πr₂² = πR²
π(r₁² + r₂²) = πR²
8² + 6² = R²
64 + 36 = R²
\(\sqrt{100}\) = R
∴ R = 10 cm.

Question 3.
The figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
Solution :

KG = GE = EC = CA = AO = OB = 10.5 cm
Area of the gold band = πr²
= \(\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}\)
r = \(\frac{AB}{2 }=\frac{21}{2}\) = 10.5
= \(\frac{33 \times 21}{2}=\frac{693}{2}\)
= 346.5 cm².

Area of the red band d = 21 + 10.5 + 10.5 = 42 cm.
= (\(\frac{22}{7} \times \frac{42}{2} \times \frac{42}{2}\)) – 346.5(area of the gold circle)
= (66 × 21) – 346.5
= 1386 – 346.5 = 1039.5 cm².

Area of the blue band d = 6 × 10.5 = 63
= (\(\frac{22}{7} \times \frac{63}{2} \times \frac{63}{2}\)) – area of the red circle
= \(\frac{99 \times 63}{2}\) – (66 × 21)
= \(\frac{6237}{2}\) – 1386
= 3118.5 – 1386.0 = 1732.5 cm²

Area of the black band = d = GH = 8 × 10.5 = 84
= (\(\frac{22}{7} \times \frac{84}{2} \times \frac{84}{2}\)) – area of the blue circle
= \(\frac{22}{7} \times \frac{84}{2} \times \frac{84}{2}\) – (\(\frac{22}{7} \times \frac{63}{2} \times \frac{63}{2}\))
= (132 × 42) – \(\frac{99 \times 63}{2}\)
= 5544.0 – 3118.5 = 2425.5 cm²

Area of the white band = d = KL = 10 × 10.5 = 105
= (\(\frac{22}{7} \times \frac{105}{2} \times \frac{105}{2}\)) – (132 × 42) (area of black circle)
= \(\frac{165 \times 105}{2}\) – (132 × 42)
= \(\frac{17325}{2}\) – 5544
= 8662.5 – 5544.0 = 3118.5 cm².

Question 4.
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Solution :
Distance travelled in one revolution = Circumference of the wheel
= 2πr d = 80
= 2 × \(\frac{22}{7} \times \frac{80}{2}=\frac{22 \times 80}{7}\) cm
Distance travelled by the car in 10 min — ?
Speed of the car = 66 km/hr.
Speed of the car = \(\frac{66}{60}\) × 10 = 11 km/min.
= 11 × 1000 mts./min.

Question 5.
Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) π units
(C) 4 units
(D) 7 units.
Solution :
2πr = πr²
2r = r²
Solution :
(A). r = 2 units.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.3

Question 1.
Evaluate:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°.
Solution :
\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) sin θ = cos (90° – θ)
sin 18° = cos (90° – 18°) = cos 72°
∴ \(\frac{sin 18}{cos 72}\) = \(\frac{cos 72°}{cos 72°}\) = 1

(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\) tan θ = cot (90 – θ)
tan 26 = cot (90 – 26) = cot 64°.
∴ \(\frac{tan 26°}{cot 64°}\) = \(\frac{cot 64}{cot 64}\) = 1

(iii) cos 48° – sin 42° = cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0.

(iv) cosec 31° – sec 59° . cosec θ = sec (90 – θ)
cosec 31 = sec (90 – 31) = sec 59°.
cosec 31° – sec 59° = sec 59° – sec 59° = 0.

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1.
(ii) cos 38° cos 52° – sin 38° sin 52° = 0.
Solution :
(i) tan 48° tan 23° tan 42° tan 67° = 1.
LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° . tan 23° tan (90° – 48°) . tan (90° – 23°)
= tan 48° . tan 23° . cot 48° . cot 23° [∵ tan (90° – θ) = cot θ]
= tan 48° . tan 23° . \(\frac{1}{tan 48°}\) . \(\frac{1}{tan 23°}\) [∵ cot θ = \(\frac{1}{tan θ}\)]
= 1 = RHS.

(ii) cos 38° cos 52° – sin 38° sin 52° = 0.
LHS = cos 38° . cos 52° – sin 38° sin 52°
= cos (90° – 52°) cos (90° – 38°) – sin 38°. sin 52°
= sin 52° . sin 38° – sin 38° . sin 52° [∵ cos (90° – θ) = sin θ]
= 0 = RHS.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution :
tan θ = cot (90° – θ)
tan 2A = cot (90° – 2A)
tan 2A = cot (A – 18°) ∵ cot (90° – 2A) = cot (A – 18°)
90° – 2A = A – 18
– 2A – A = – 18 – 90
– 3A = – 108
A = \(\frac{-108}{-3}\) = 36°.

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution :
tan A = cot B (90° – A)
tan A = cot B
∴ cot (90° – A) = cot B
∴ 90° – A = B
90° = B + A
A + B = 90°.

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution :
sec 4A = cosec (A – 20°)
cosec (90° – 4A) = cosec (A – 20°)
90° – 4A = A – 20°
– 4A – A = – 110°
– 5A = – 110°
A = 22°.

Question 6.
If A, B and C are interior angles of a triangle ABC, then show that
Solution :
sin (\(\frac{B+C}{2}\)) = cos \(\frac{A}{2}\)
sin θ = cos (90° – θ)

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution :
sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°. (Since cos (90° – θ) = sin θ and sin (90° – θ) = cos θ)

Jharkhand Board JAC Class 10 Maths Solutions Chapter 10 Circles Ex 10.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 10 Circles Exercise 10.2

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution :

OQ² = TO² – TQ²
= 25² – 24²
= (25 + 24) (25 – 24)
= 49 × 1 = 49
OQ = \(\sqrt{49}\) = 7 cm.

Question 2.
In the figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°
Solution :

POQT is a cyclic quadrilateral because \(\hat{\mathbf{P}}\) + \(\hat{\mathbf{O}}\) = 180°.
∴ \(\hat{\mathbf{O}}\) + \(\hat{\mathbf{T}}\) = 180°
latex]\hat{\mathbf{O}}[/latex] = 110°.
latex]\hat{\mathbf{T}}[/latex] = 180° – 110° = 70°.

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to

A) 50°
B) 60°
C) 70°
D) 80°
Solution :
In ΔPAO, P\(\hat{A}\)O = 90° A\(\hat{P}\)O = 40°
∴ A\(\hat{O}\)P = 180° – 90° – 40° = 50°.

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution :

Data: O is the centre of the circle.
AB is a diameter. CD and EF are tangents drawn at A and B.
To prove: CD || EF.
Proof: O\(\hat{A}\)C = 90°
O\(\hat{B}\)E = 90°
B\(\hat{A}\)C + A\(\hat{B}\)E = 180°
These are co-interior angles.
Radius is ⊥ to the tangent at the point of contact

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution :

Let us consider a circle with centre O.
Let AB be a tangent which touches the circle at P.
We have to prove that the line perpendicular to AB at P passes through the centre O.
We shall prove this by contradiction method.
Let us assume that the perpendicular to AB at P does not pass through centre O.
Let it pass through another point O’. Join OP and O’P.
As the perpendicular to AB at P passes through O’, therefore,
∠O’PB = 90° ……..(1)
O is the centre of the circle and P is the point of contact. We know that the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.
∴ ∠OPB = 90° ……..(2)
Comparing equations (1) and (2), we obtain
∠O’PB = ∠OPB ….(3)
From the figure, it can be observed that,
∠O’PB < ∠OPB ………..(4)
Therefore, ∠O’PB = ∠OPB is not possible. It is only possible when the line O’P coincides with OP. Therefore, the perpendicular to AB at P passes through the centre O.

Question 6.
The length of a tangent from a point A at a distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution :
A\(\hat{B}\)O = 90° (The radius is perpendicular to the tangent at the point of contact)
BO² + AB² = AO²
BO² = AO² – AB²
= 5² – 4²
= 25 – 16 = 9
BO = \(\sqrt{9}\) = 3 cm.
Hence, radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution :

O is the centre of the concentric circles C₁ and C₂.
AB is a chord of the bigger circle C₁.
AB is a tangent to circle C₂.
OC = 3 cm, OB = 5 cm.
To find AB.
Proof: In ΔOCB,
O\(\hat{C}\)B = 90° (The tangent AB to C₂ is ⊥ to the radius CO drawn from the point of contact C)
In ΔOCB, O\(\hat{C}\)B = 90°
OC² + CB² = OB²
3² + CB² = 5²
CB² = 5² – 3²
= 25 – 9 = 16
CB = 4 cm.
In circle C₁ AB is a chord. OC ⊥ AB.
∴ AC = CB = 4 cm. The ⊥ drawn to a chord from the centre bisects the chord.
AB = AC + CB
= 4 + 4 = 8 cm.

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Solution :

A, B, C and D are external points. Tangents drawn to the circle from these points are equal.
AP = AS ….(1) (Tangents from A)
BP = BQ ….(2) (Tangents from B)
CR = CQ ….(3) (Tangents from C)
DR = DS ….(4) (Tangents from D)
Adding the results
AP + BP + CR + DR = AS + BQ+CQ + DS
AB + CD = AS + DS + BQ +CQ
∴ AB + CD = AD + BC.

Question 9.
In the figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.
Solution :

Given: XY || X’Y’.
XPA, X’QY’ and AB are the tangents.
To prove: ∠AOB = 90°
Construction: Join OC.
Proof: In ΔOAP and ΔOAC
AP = AC (Theorem 4.2)
OA = OA (Common)
OP = OC (Radii of circle)
AOAP AOAC (S.S.S. congruency)
Since XY || X’Y’
∠PAB + ∠QBA = 180° (Co-interior angles)
\(\frac{1}{2}\)∠PAB + \(\frac{1}{2}\)∠QBA = 90°
∴ ∠OAB + ∠OBA = 90°
In ΔAOB, ∠AOB = 180° – (∠OAB + ∠OBA)
= 180° – 90° = 90°
∴ ∠AOB = 90°.
The line joining the external point to the centre, bisects the angle between the tangents.

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution :

Data: O is the centre of the circle.
AB and AC are tangents drawn from A.
To prove: B\(\hat{\mathbf{A}}\)C + B\(\hat{\mathbf{O}}\)C = 180°.
Proof: AB is a tangent. BO is the radius drawn from the point of contact B.
∴ O\(\hat{\mathbf{B}}\)A = 90°
AC is a tangent. CO is the radius drawn from the point of contact C.
∴ O\(\hat{\mathbf{C}}\)A = 90°
∴ O\(\hat{\mathbf{B}}\)A + O\(\hat{\mathbf{C}}\)A = 90°.
∴ B\(\hat{\mathbf{O}}\)C + B\(\hat{\mathbf{A}}\)C = 180° [Sum of angles of a quadrilateral is 360°]

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution :

Data: ABCD is a circumscribed parallelogram.
AB = CD, AD = BC
To prove: ABCD is a rhombus.
AB = BC – CD – DA
Proof: ABCD is a circumscribed quadrilateral.
∴ AB + CD = AD + BC (Proved in problem 8)
AB = DC, AD = BC (Opposite sides of the parallelogram)
AB + AB = AD + AD
2AB = 2AD
AB = AD
AB = CD, AD = BC
AB = BC = CD = DA. [Since adjacent sides of a parallelogram are equal]
∴ ABCD is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Solution :

In the adjoining figure, the circle touches AB at F, BC at D, CA at E.
AE = AF = x. (Tangents to the circle from A)
Tangent BF = Tangent BD = 8 cm
Tangent CE = Tangent CD = 6 cm
AB = AF + FB = (x + 8) cm
AC = AE + EC = (x + 6) cm
CB = CD + DB = 6 + 8 = 14 cm.
2S(perimeter of AABC) = AB + BC + CA
= (x + 8) + (6 + 8) + (x + 6).
= 2x + 28
S = x + 14

Area of ΔABC = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 14 × 4 = 28 cm².
Area of ΔOAB = \(\frac{1}{2}\) × 4(x + 8) = 2(x + 8)
Area of ΔOAC = \(\frac{1}{2}\) × 4(x + 6) = 2(x + 6)
Area of ΔOBC = \(\frac{1}{2}\) × 4 × 14 = 28 cm.
Total area = Area of ΔOBC + Area of ΔOAB + Area of ΔOAC
= 28 + 2x + 12 + 2x + 16
∴ 4x + 56 = 4(x + 14)
∴ 4\(\sqrt{3 x^2+42 x}\) = 4(x + 14)
\(\sqrt{3 x^2+42 x}\) = x + 14
Squaring both sides 3x² + 42x = (x + 14)² = x² + 28x + 196
= 2x² + 145 – 196 = 0 x² + 7x – 98 = 0
∴ (x + 14) (x – 7) = 0 x = – 14 or x = 7
AC = x + 8
= 7 + 8
= 15 cm.

AC = x + 6
= 7 + 6
= 13 cm.

BC = 6 + 8
= 14 cm.

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution :

Construction: Join OP, OQ, OR and OS.
Proof: The two tangents drawn from an external point to a circle subtend equal angles at the centre.
∴ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8.
Now, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360° (Sum of angles subtended at a point)
2(∠2 + ∠3 + ∠6 + ∠7) = 360° and
2(∠1 + ∠8 + ∠4 + ∠5) = 360°
⇒ (∠2 + ∠3) + (∠6 + ∠7) = 180°
(∠1 + ∠8) + (∠4 + ∠5) = 180°
∴ ∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180°.